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Are outcrossing programmes supported by the Kennel Club?
Here we feature Kennel Club materials on outcrossing and cross breeding.
Did you find our content interesting or helpful? Help support the IPFD enhance health, well-being and welfare for dogs everywhere.
Are outcrossing programmes supported by the Kennel Club?
Here we feature Kennel Club materials on outcrossing and cross breeding.
Breeding for health in non pedigree dogs
Outcrossing Programmes
What is an outcrossing programme?
An outcrossing programme aims to improve the genetic health and increase genetic diversity of a breed ( Breed A ) by introducing a new line or crossing it with another breed ( Breed B ) in a controlled and monitored manner. Once the health of the Breed A has been restored, it can then be bred back to purebred dogs to gradually dilute the other Breed B, and eventually breed it out completely.
Why would an outcrossing programme be needed?
One reason may be that an inherited disorder, caused by gene mutation(s), may be highly prevalent in a breed and may threaten the population of the breed and its genetic diversity. To introduce the normal gene(s) back into the breed, it can be crossed with another breed with the intention to breed out the mutated gene and eventually eradicate it from the breed.
Here is an example:
An outcrossing breeding programme was carried out in the USA, aiming to introduce the low (or normal) uric acid gene from the Pointer breed into the Dalmatian breed through cross-breeding. The programme was put into place to replace the defective gene that affects uric acid metabolism, causing an increase in urinary uric acid, urinary stones and urinary tract obstruction.
Other reasons for using a crossbreeding programme are to improve the health of the breed in general, for example, improving conformation traits that may cause health problems, and improving general fitness and its benefits.
Another reason may be that a breed has a very low effective population size, and is at risk of, or beginning to suffer effects of inbreeding depression – such as reduced litter sizes, fertility problems, or an increase in inherited diseases.
In any instance, it is of vital importance that population genetics expertise is sought to develop any outcrossing program. Identifying good (breed) candidates for outcrossing, and ensuring there are sufficient numbers of animals to outcross to are just some of the aspects developing a successful program requires.
Are outcrossing programmes supported by the Kennel Club?
The Kennel Club supports outcrossing programmes that improve breed health and to preserve the genetic diversity of a breed. The Kennel Club will allow dogs with unknown ancestry to be introduced onto the Breed Register on a case by case basis. These dogs will have three asterisks next to its name on the register, which will be applied for three further generations to identify the fact that there is unknown or unregistered ancestry behind a dog.
Also see: -- The Kennel Club -- Breeding for health in non pedigreed dogs in Downloads.
Description: A quick guide designed to provide some general information on health testing and schemes available for breeders of non pedigree dogs.
See the cross-breeding pedigree tree example in Breeding for health in non pedigree dogs -- demonstrates the need for health screening -- effective info graphic.
Articles provided by: Bonnie-Marie Abhayaratne, Health and Breeder Research Assistant, The Kennel Club,
Edited by Ann Milligan
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\begin{document}
\title{On the intersection of Annihilator of the Valabrega-Valla module }
\author{Tony~J.~Puthenpurakal}
\date{\today}
\address{Department of Mathematics, Indian Institute of Technology Bombay, Powai, Mumbai 400 076}
\email{tputhen@math.iitb.ac.in}
\thanks{While writing this paper the author was a visitor at University of Kentucky, under a fellowship from DST, India. The author thanks both DST and UK for its support}
\subjclass{Primary 13A30; Secondary 13D40, 13D45}
\keywords{blow-up algebras, multiplicity theory, core}
\begin{abstract}
Let $(A,\m)$ be a \CM \ local ring with an infinite residue field and let $I$ be an $\m$-primary ideal.
Let $\bx = x_1,\ldots,x_r$ be a $A$-superficial sequence \wrt \ $I$. Set
$$\Vc_I(\bx) = \bigoplus_{n\geq 1} \frac{I^{n+1}\cap (\bx) }{\bx I^n}. $$
A consequence of a theorem due to Valabrega and Valla is that $\Vc_I(\bx) = 0$ \ff \ the initial forms $x_1^*,\ldots,x_r^*$ is a $G_I(A)$ regular sequence. Furthermore this holds if and only if $\depth G_I(A) \geq r$. We show that if $\depth G_I(A) < r$ then
\[
\af_r(I)= \bigcap_{\substack{\text{$\bx = x_1,\ldots,x_r$ is a } \\ \text{ $A$-superficial sequence w.r.t $I$}}} \ann_A \Vc_I(\bx) \quad \ \text{is} \ \m\text{-primary}.
\]
Suprisingly we also prove that under the same hypotheses,
\[
\bigcap_{n\geq 1} \af_r(I^n) \quad \ \text{is also} \ \m\text{-primary}.
\]
\end{abstract}
\maketitle
\section*{Introduction}
Let $(A,\m)$ be a Noetherian local ring with an infinite residue field.
The notion of minimal reduction of an ideal $I$ in $A$ was discovered more than fifty years ago by Northcott and Rees; \cite{Northcott-Rees}. It plays an essential role in the study of blow-up algebra's. Nevertheless minimal reductions are highly non-unique. The intersection of all minimal reductions is named as \textit{core} of $I$ and denoted by
$\core(I)$. This was introduced by Rees and Sally in \cite{Rees-Sally}. It has been extensively investigated in
\cite{CPU-Ann},\cite{CPU-TAMS} and \cite{Hun-Trung}. When $A$ is \CM \ and $I$ is $\m$-primary; Rees and Sally proved that $\core(I)$ is again $\m$-primary and so is a finite intersection.
In this paper we study a different intersection of ideals.
Let $(A,\m)$ be a \CM \ local ring of dimension $d$ with an infinite residue field and let $I$ be an $\m$-primary ideal. Let $\bx = x_1,\ldots,x_r$ be a $A$-superficial sequence \wrt \ $I$. Set
$$\Vc_I(\bx) = \bigoplus_{n\geq 1} \frac{I^{n+1}\cap (\bx) }{\bx I^n}. $$
We call $\Vc_I(\bx)$ the Valabrega-Valla module of $I$ \wrt \ $\bx$.
A consequence of a theorem due to Valabrega and Valla, \cite[2.3]{VV} is that $\Vc_I(\bx) = 0$ \ff \ the initial forms $x_1^*,\ldots,x_r^*$ is a $G_I(A)$ regular sequence. Furthermore this holds if and only if $\depth G_I(A) \geq r$, see \cite[2.1]{HM}. In general notice each $\Vc_I(\bx)$ has finite length and so $\ann_A \Vc_I(\bx)$ is $\m$-primary. We prove, see Theorem \ref{main},
that
\[
\af_r(I) = \bigcap_{\substack{\text{$\bx = x_1,\ldots,x_r$ is a } \\ \text{ $A$-superficial sequence w.r.t $I$}}} \ann_A \Vc_I(\bx) \quad \ \text{is} \ \m\text{-primary}.
\]
Our intersection of ideals is in some sense analogous to that of core of $I$; since
notice that
$$ \core(I) = \bigcap_{\substack{\text{$J$ minimal } \\ \text{ reduction of $I$}}} \ann_A \frac{A}{J}. $$
Nevertheless they are two different invariants of $I$. Furthermore our techniques are totally different from that in the papers listed above.
By a result of Elias $\depth G_{I^n}(A)$ is constant for all $n \gg 0$, see \cite[2.2]{Elias}.
Since $\core(I) \subseteq I$ we have $\bigcap_{n\geq 1} \core(I^n) = 0$. Suprisingly, see Theorem \ref{main-2}, we have
that if $\depth G_I(A) < r$ then
\[
\bigcap_{n\geq 1} \af_r(I^n) \quad \ \text{is} \ \m\text{-primary}.
\]
We now assume $A$ is also complete. Let $\R = \bigoplus_{n\geq 0}I^n$ be the Rees algebra of $I$. Set $L = L^I(A)= \bigoplus_{n\geq 0}A/I^{n+1}$. It can be shown easily that $L$ is a $\R$-module.
Of course $L$ is not finitely generated as a $\R$-module. Nevertheless we prove that its local cohomology modules $H^i_{\R_+}(L)$ are *-Artinian for $i = 0,\ldots, d-1$; see Theorem \ref{locArt}. Recall a graded $\R$-module $N$ is said to be $*$-Artininan if it satisfies d.c.c on its graded submodules. Set $\bbf_i(I) = \ann_{\R} H^i_{\R_+}(L)$ for $i = 0,\ldots, d-1$ and set $\q_i(I) = \bbf_i(I) \cap A$. Since $H^i_{\R_+}(L)$ is *-Artinian; it is not
so difficult to show that $\q_i$ is $\m$-primary (or equal to $A$); see Corollary \ref{loc-CA}.
In Theorem \ref{man} we prove that
\[
\af_r(I) \supseteq \q_0(I)\q_1(I)\cdots\q_{r-1}(I).
\]
Next note that $L^I(A)(-1)$ behaves well with respect to the Veronese functor. Clearly
\[
\left(L^I(A)(-1)\right)^{<l>} = L^{I^l}(A)(-1) \quad \text{for each} \ l \geq 1.
\]
Also local cohomolgy commutes with the Veronese functor. As a consequence we have
\[
\q_i(I^l) \supseteq \q_i(I) \quad \text{for each} \ l \geq 1 \ \text{and} \ i = 0,1,\ldots,r-1.
\]
It follows that
\[
\bigcap_{n \geq 1}\af_r(I^n) \supseteq \q_0(I)\q_1(I)\cdots\q_{r-1}(I).
\]
The $\R$-module $L^I(A)$ is not finitely generated $\R$-module. However it is quasi-finite $\R$-module, see section 1.5.
Quasi-finite module were introduced in
\cite[page 10]{HPV}. Surprisingly we were able to prove that if $E$ is a quasi-finite $\R$-module and has a filter-regular sequence of length $s$ then
the local cohomology modules $H^{i}_{\R_+}(E)$ are all *-Artinian for $i = 0,\ldots, s-1$.
We also study the Koszul homology of a quasi-finite module \wrt \ a filter regular sequence. We then use a spectral sequence, first used
by P. Roberts \cite[Theorem 1]{Roberts}, to relate cohomological annihilators with that of annihilators of the Koszul complex. We however have to very careful in our proof since we are dealing with infinitely generated modules.
We now describe in brief the contents of this paper. In section 1 we introduce notation
and discuss a few preliminary facts that we need. In section 2 we study a few basic properties of $L^I(M)$. In section 3 we prove some properties of Koszul homology of quasi-finite modules with respect to filter-regular sequence. We also compute $H_1(\bu, L^I(M))$ where $\bu = x_1t,\ldots,x_rt \in \R_1$ is a $L^I(M)$-filter regular sequence. In section 4 we study local cohomology of quasi-finite modules $E$ \\ with
$\ell(E_n)$ finite for all $n \in \Z$. In section 5 we prove that $\af_r(I)$ is $\m$-primary (or $A$).
In section 6 we show that $\bigcap_{n\geq 1} \af_r(I^n)$ is $\m$-primary (or $A$).
\section{Notation and Preliminaries}
Throughout we assume that $(A,\m)$ is a Noetherian local ring with an infinite residue field $k =A/\m$. Let $M$ be a finitely generated \ $A$-module of dimension $r$ and let $I$ be an ideal of definition for $M$; i.e, $\ell(M/IM)$ is finite. Here $\ell(-)$ denotes length. For undefined terms see
\cite{BH}, especially sections 4.5 and 4.6.
\s Assume $r = \dim M \geq 1$. Let $x \in I\setminus I^2$. We say $x$ is $M$-\emph{superficial} \wrt \ $I$ if for some $c\geq 1$ we have
$(I^{n+1}M \colon x )\cap I^cM = I^nM$
for all $n \gg 0$. If $\depth M > 0$ then using the Artin-Rees Lemma one can prove that $(I^{n+1}M \colon x ) = I^nM$
for all $n \gg 0$.
Superficial sequences can be defined as usual. Since $k$ is infinite $M$-superficial sequences of length $ r = \dim M$ exists.
\s Let $\bx = x_1,\ldots, x_r$ be a $M$-superficial sequence \wrt \ $I$. The Valabrega-Valla module of $I$ \wrt \ $M$ and $\bx$ is
\[
\Vc_I(\bx, M) = \bigoplus_{n\geq 1} \frac{I^{n+1} M\cap \bx M}{\bx I^nM}.
\]
We consider it as a $A$-module.
Set $\Vc_I(\bx) = \Vc_I(\bx, A)$.
\s Let $\eR = \bigoplus_{n \in \Z}I^nt^n$ denote the extended Rees-algebra of $A$ \wrt \ $I$. Here $I^n = A$ for $n \leq 0$. We consider it as a subring of $A[t,t^{-1}]$.
Let $\R = \bigoplus_{n \geq 0 }I^nt^n$ denote the Rees-algebra of $A$ \wrt \ $I$. We consider it as a subring of $A[t]$. Of course
we can consider $\R$ as a subring of $\eR$ too. Both these embedding's of $\R$ would be useful for us.
Set
\[
\eR_M = \bigoplus_{n \in \Z}I^nMt^n \quad \text{and} \quad \R_M = \bigoplus_{n \geq 0}I^n Mt^n.
\]
We call $\eR_M$
the extended Rees module of $M$ \wrt \ $I$ and we call $\R_M$ to be the Rees module of $M$ \wrt \ $I$.
\s Consider $L^I(M) = \bigoplus_{n \geq 0}M/I^{n+1}M$. We consider $L^I(M)$ as a $\eR$-module as follows:
Consider the exact sequence
\[
0 \xar \eR_M \xar M[t,t^{-1}] \xar L^I(M)(-1) \xar 0.
\]
Here $M[t,t^{-1}] = M \otimes_A A[t,t^{-1}]$. This exact sequence gives $L^I(M)$ a structure of $\eR$-module.
Since $\R$ is a subring of $\eR$; we also get that $L^I(M)$ is a $\R$-module. We may also see this directly
through the exact sequence
\[
0 \xar \R_M \xar M[t] \xar L^I(M)(-1) \xar 0
\]
\s \label{quasi-defn}\textbf{Quasi-finite modules}
It will be convenient at times to work a little more generally. We extend definition of quasi-finite modules from that of \cite[page 10]{HPV}.
Let $E = \bigoplus_{n\in \Z}E_n$ be a $\R$-module. We say $E$ is \emph{quasi-finite of order at least $s$} if
\begin{enumerate}
\item $E_n$ is a finitely generated $A$-module for all $n \in \Z$
\item $E_n = 0$ for all $n \ll 0$.
\item For $i = 0,\ldots, s-1$ we have $H^i_{\R_+}(E)_n = 0 $ for all $n \gg 0$.
\end{enumerate}
\begin{remark}
Of course if $E$ is a finitely generated $\R$-module then it is quasi-finite of any order $s \geq 1$. In the next section we prove that if $M$ is \CM \ of dimension $r\geq 1$ and $I$ is an ideal of definition for $M$ then $L^I(M)$ is quasi-finite of order at-least $r$.
\end{remark}
\s Let $E = \bigoplus_{n\in \Z}E_n$ be a non-necessarily finitely generated $\R$-module with $E_n = 0$ for all $n \ll 0$. An element $u \in \R_1$ is said to be $E$-\emph{filter regular } if
$(0 \colon_E u)_n = 0$ for all $n \gg 0$.
\begin{remark}
If $E$ is quasi-finite of order at-least $s (\geq 2) $ and $u$ is $E$-filter regular then $E/uE$ is quasi-finite of order at-least $s-1$.
This can be proved by noting that $(0\colon_E u)$ is $\R_+$-torsion.
\end{remark}
\s Let $E = \bigoplus_{n\in \Z}E_n$ be a quasi-finite $\R$-module of order at-least $s$. Let $\bu = u_1,\ldots,u_r \in \R_1$ be a sequence and assume $r \leq s$.
We say $\bu$ is a $E$-filter regular sequence if $u_1$ is $E$-filter regular, $u_2$ is $E/u_1E$-filter-regular, \ldots, $u_r$ is $E/(u_1,\ldots,u_{r-1})E$ filter-regular.
\begin{proposition}
Assume that the residue field of $A$ is uncountable. Let $E$ be a quasi-finite $\R$-module of order at least $s$. Then there exists
$\bu = u_1,\ldots,u_s \in \R_1$ which is $E$-filter regular sequence.
\end{proposition}
\begin{proof}
It is sufficient to do this for $s =1$. In this case the result follows from \cite[2.7]{HPV}
\end{proof}
\begin{remark} Assume $M$ is \CM.
Let $\bx = x_1,\ldots,x_r$ be a $M$-superficial sequence \wrt \ $I$. Set $u_i = x_it \in \R_1$ for $i = 1,\ldots,r$. In the next section we show that
$\bu = u_1,\ldots,u_r$ is a $L^I(M)$ filter-regular sequence. We do not need the residue field of $A$ to be uncountable.
\end{remark}
\section{$L^I(M)$}
\s\label{setup-2} \textbf{Setup and Introduction:} In this section $M$ is a\emph{ \CM }\ $A$-module of dimension $r \geq 1$ and $I$ is an ideal of definition for $M$. We consider the $\eR$-module
$L^I(M) = \bigoplus_{n\geq 0}M/I^{n+1}M$.
We prove that $L^I(M)$ is a quasi-finite $\R$-module of order at least $r$. Let $\bx = x_1,\ldots,x_r$ be a $M$-superficial sequence \wrt \ $I$. Set $u_i = x_it \in \R_1$ for $i = 1,\ldots,r$. We also show that
$\bu = u_1,\ldots,u_r$ is a $L^I(M)$ filter-regular sequence.
\s\label{indep} If $E$ is a graded $\eR$-module then notice that
\[
H^i_{\R_+}(E) \cong H^i_{\eR_+}(E) \quad \text{as $\R$-modules}.
\]
Note that $\eR_+$ denotes the ideal $\R_+ \eR$ of $\eR$.
The following result is known when $M = A$; see \cite[3.8]{Blancafort}.
\begin{lemma}\label{eRlc}
[with hypotheses as in \ref{setup-2}] As $\R$-modules:
\begin{enumerate}[\rm (1)]
\item $H^1_{\eR_+}(\eR_M)$ is a quotient of $H^1_{\R_+}(\R_M)$.
\item $H^i_{\eR_+}(\eR_M) \cong H^i_{\R_+}(\R_M)$ for $i \geq 2$.
\end{enumerate}
\end{lemma}
\begin{proof}(Sketch)
We use \ref{indep} and the following short exact sequence of $\R$-modules
\[
0\xar \R_M \xar \eR_M \xar \eR_M/\R_M \xar 0.
\]
Notice $\eR_M/\R_M$ is $\R_+$-torsion.
\end{proof}
\begin{proposition}
$L^I(M)$ is quasi-finite of order $r = \dim M$.
\end{proposition}
\begin{proof}
Set $L = L^I(M)$. Notice $H^i_{\R_+}(L) = H^i_{\eR_+}(L)$ as $\R$-modules. Let $\bx = x_1,\ldots,x_r$ be a $M$-superficial sequence \wrt \ $I$. Set $u_i = x_it \in \R_1$ for $i = 1,\ldots,r$.
Let $\bx = x_1,\ldots,x_r$ be a $M$-superficial sequence \wrt \ $I$. Set $u_i = x_it \in \R_1$ for $i = 1,\ldots,r$.
It can be easily checked that $\bu$ is a $M[t,t^{-1}]$ regular sequence. So $H^i_{\eR_+}(M[t, t^{-1}]) = 0$ for $i = 0,\ldots, r-1$.
We consider the exact sequence
\[
0 \xar \eR_M \xar M[t,t^{-1}] \xar L(-1) \xar 0.
\]
Taking local cohomology \wrt \ $\eR_+$ we get that
(a) $H^i_{\eR_+}(L(-1)) \cong H^{i+1}_{\eR_+}(\eR_M)$ for $i = 0,\ldots, r-2$. \\
(b) $H^{r-1}_{\eR_+}(L(-1))$ is a submodule of $H^r_{\eR_+}(\eR_M)$.
The result now follows from Lemma \ref{eRlc}, Remark \ref{indep} and \cite[15.1.5]{BSh}.
\end{proof}
\begin{proposition}\label{supTOfilter}
Let $\bx = x_1,\ldots,x_r$ be a $M$-superficial sequence \wrt \ $I$. Set $u_i = x_it \in \R_1$ for $i = 1,\ldots,r$.
Then $\bu $ is a $L^I(M)$ filter-regular sequence.
\end{proposition}
\begin{proof}
Set $L = L^I(M)$.
We first show that $u_1$ is $L$ filter regular.
Notice
\[
(0 \colon_L u_1) = \bigoplus_{n \geq 0} \frac{I^{n+1}M \colon_M x_1}{I^nM}.
\]
Since $x_1$ is $M$-superficial it follows that $u_1$ is $L$ filter regular; see 1.1.
Check that
\[
\frac{L}{u_1 L} = \bigoplus_{n \geq 0}\frac{M}{x_1M + I^{n+1}M} = L^I(M/x_1M).
\]
The result now follows from an easy induction on $\dim M$.
\end{proof}
\section{Koszul homology of quasi-finite modules \\ with respect to filter-regular sequence}
In this section we show some properties of Koszul homology of a quasi-finite module \wrt \ a filter regular sequence. We also compute the Koszul
homology of $L^I(M)$ \wrt \ $\bu = x_1t,\ldots,x_st$ where $x_1,\ldots,x_s$ is an $M$-superficial sequence \wrt\ $I$.
\begin{theorem}
Let $E$ be a quasi-finite $\R$-module of order at least $s$ and let $\bu = u_1,\ldots, u_s$ be a $E$-filter regular sequence. Then for
$i = 1,\ldots,s$ we have
\begin{enumerate}[\rm (1)]
\item $H_i(\bu, E)$ is a finitely generated $\R$-module. It is also $\R_+$-torsion. In particular $H_i(\bu, E)$ is a finitely generated
$A$-module.
\item If $\bu$ is $E$-regular sequence then $H_i(\bu, E) = 0$ for $i = 1,\ldots, s$.
\item If $H_1(\bu, E) = 0$ then $\bu$ is a $E$-regular sequence.
\end{enumerate}
\end{theorem}
\begin{proof}
(1) We prove it by induction on $s$.
The case $s =1$.\\
Notice $H_1(u_1,E) = (0 \colon_{u_1} E)$. Since $u_1$ is $E$-filter regular we get that $H_1(u_1,E)$ is a finitely generated $A$-module and hence a finitely generated $\R$-module. Clearly it is also $\R_+$ torsion.
We assume the result for $s = r$ and prove for $s = r+1$. Let $\bu = u_1,\ldots,u_r,u_{r+1}$ and $\bu^\pr = u_1,\ldots, u_{r}$. We have
for all $i \geq 0$ an exact sequence
\begin{equation}\label{utou}
0 \xar H_0(u_{r+1}, H_i(\bu^\pr, E)) \xar H_i(\bu, E) \xar H_1(u_{r+1}, H_{i-1}(\bu^\pr, E)) \xar 0
\end{equation}
Using induction hypothesis it follows that for $i \geq 2$ the modules $H_i(\bu, E)$ are finitely generated $\R$-modules and also $\R_+$-torsion.
For $i = 1$ notice that
(a) $H_0(u_{r+1}, H_1(\bu^\pr, E))$ is finitely generated $\R$-module.
It is also $\R_+$-torsion.
(b) $H_1(u_{r+1}, H_0(\bu^\pr, E)) = H_1(u_{r+1}, E/\bu^\pr E)$. Since $u_{r+1}$ is $E/\bu^\pr E$-filter regular then by $s =1$ case we have
that $H_1(u_{r+1}, H_0(\bu^\pr, E))$ is a finitely generated $\R$-module and it also $\R_+$-torsion
The result follows.
(2) The standard proof works.
(3) Nothing to prove when $s = 1$. So assume $ s \geq 2$. Set $r = s -1$. We use equation \ref{utou}.
If $H_1(\bu, E) = 0$ then $H_0(u_{r+1}, H_1(\bu^\pr,E)) = 0$.
So we have $H_1(\bu^\pr, E) = u_{r+1}H_1(\bu^\pr, E)$. Since $H_1(\bu^\pr, E)$ is a finitely generated graded $\R$-module and
$u_{r+1}$ has positive degree it follows that $H_1(\bu^\pr, E) = 0$. By induction hypothesis it follows that
$u_1,\ldots,u_r$ is a $E$-regular sequence.
From \ref{utou} we also get $$H_1(u_{r+1}, H_0(\bu^\pr, E)) = H_1(u_{r+1}, E/\bu^\pr E) = 0.$$
So $u_{r+1}$ is $E/\bu^\pr E$- regular. It follows that $\bu$ is a $E$-regular sequence.
\end{proof}
\begin{proposition}\label{compute}
Let $M$ be a \CM \ $A$-module of dimension $r \geq 1$ and let $I$ be an ideal of definition for $M$.
Let $\bx = x_1,\ldots,x_s$ be a $M$-superficial sequence \wrt \ $I$ with $s \leq r$. Set $u_i = x_it \in \R_1$ for $i = 1,\ldots,s$.
Then $\bu $ is a $L^I(M)$ filter-regular sequence and
$$ H_1(\bu, L^I(M)) = \bigoplus_{n\geq 1} \frac{I^{n+1}M \cap \bx M}{\bx I^nM} = \Vc_I(\bx, M). $$
\end{proposition}
\begin{proof}
Set $L = L^I(M)$. In \ref{supTOfilter} we have shown already that $\bu $ is a $L^I(M)$ filter-regular sequence.
Consider the exact sequence
\[
0 \xar \eR_M \xar M[t,t^{-1}] \xar L(-1) \xar 0.
\]
It can be easily checked that $\bu$ is a $M[t,t^{-1}]$ regular sequence. So $H_1(\bu, M[t,t^{-1}]) = 0$. Thus we have an exact sequence
\[
0 \xar H_1(\bu, L(-1)) \xar H_0(\bu, \eR_M ) \xar H_0(\bu, M[t,t^{-1}] ) \xar H_0(\bu, L ) \xar 0.
\]
Notice
\[
H_0(\bu, \eR_M ) = \bigoplus_{n\in Z} \frac{I^{n}M}{\bx I^{n-1}M} \quad \text{and} \quad H_0(\bu, M[t,t^{-1}] ) = M/\bx M[t,t^{-1}]
\]
So
\[
H_1(\bu, L(-1)) = \bigoplus_{n\in Z} \frac{I^{n}M \cap \bx M}{\bx I^{n-1}M}.
\]
The result follows.
\end{proof}
\section{local cohomology of quasi-finite modules $E$ \\ with
$\ell(E_n)$ finite for all $n \in \Z$}
In this section we prove a suprising fact: the local cohomology modules
\noindent $H^i_{\R_+}(L^I(M))$ are all *-Artinian for $i = 0,\ldots, \depth M -1$.
It is convenient to prove it in the generality of quasi-finite modules.
\s \label{setup-lc} Throughout this section $H^i(-) = H_{\R_+}^i(-)$ the $i$-th local cohomology functor \wrt\ $\R_+$. In this section we assume that
\begin{enumerate}
\item $(A,\m)$ is complete with infinite residue field.
\item $E$ is a quasi-finite module of order at least $s$.
\item There exists an $E$-filter regular sequence of length $s$.
\item $\ell(E_n)$ finite for all $n \in \Z$.
\end{enumerate}
\begin{remark}
The hypothesis on existence of $E$-filter regular sequence of length $s$ is automatically satisfied if $k$ is uncountable.
The assumption "$\ell(E_n)$ finite for all $n \in \Z$" is to imitate that of $L^I(M)$. Finally if $M$ is\ CM \ and $A$ has infinite residue field then assumptions 2, 3, 4 are automatically satisfied for $L^I(M)$.
The assumption $A$ is complete is needed since we will use Matlis-Duality.
\end{remark}
\begin{theorem}\label{locArt}[with hypotheses as in \ref{setup-lc}]
For $i = 0,\ldots, s-1$ we have
\begin{enumerate}[\rm (1)]
\item $\ell(H^i(E)_n) < \infty$ for all $n \in \Z$.
\item $H^i(E)^\vee$ is a Noetherian $\R$-module.
\item $H^i(E)$ is a *-Artinian $\R$-module.
\end{enumerate}
\end{theorem}
\begin{proof}
We prove everything together by induction on $s$.
The case $s = 1$ \\
Clearly
$\ell(H^0(E)_n) < \infty$ for all $n\in \Z$ and is zero for $n \ll 0$. By hypothesis $E$ is quasi-finite of order at least $1$. So $H^0(E)_n = 0$ for all $n \gg 0$. The result follows.
We assume the result for $s = r$ and prove for $s = r+1$. Since $E$ is quasi-finite module of order at least $r+1$ it is also quasi-finite module of order at least $r$. So by induction hypothesis applied to $E$ we have that
for $i = 0,\ldots,r-1$ the modules $H^i(E)$ satisfy properties (1), (2) and (3). It remains to prove that $H^r(E)$ satisfies properties (1), (2) and (3). \\
Let $u$ be $E$-filter regular. Set $F = E/uE$. We have an exact sequence
\[
0 \xar (0\colon_E u) \xar E(-1) \xrightarrow{u} E \xar F \xar 0.
\]
Since $(0\colon_E u)$ is $R_+$-torsion, by using a standard trick, we get the exact sequence
\begin{align*}
0 \xar (0\colon_E u) \xar H^0(E)(-1) &\xrightarrow{u} H^0(E) \xar H^0(F) \xar \\
H^1(E)(-1) &\xrightarrow{u} H^1(E) \xar H^1(F) \xar \\
&\cdots \\
H^{r-1}(E)(-1) &\xrightarrow{u} H^{r-1}(E) \xar H^{r-1}(F) \xar\\
H^r(E)(-1) &\xrightarrow{u} H^r(E).
\end{align*}
So we have an exact sequence
\begin{equation*}
H^{r-1}(F) \xrightarrow{\delta} H^r(E)(-1) \xrightarrow{u} H^r(E). \tag{*}
\end{equation*}
Since $F$ is quasi-finite of order at least $r$ we get that $H^{r-1}(F)$ satisfies properties (1), (2) and (3). We prove that $H^r(E)$ satisfies properties (1), (2) and (3).
(1) By hypothesis on $E$ we have $H^r(E)_n = 0$ for all $n \gg 0$ say from $n\geq c+1$. By equation (*) we have
$H^{r-1}(F)_{c+1} \xrightarrow{\delta} H^r(E)_c \xar H^r(E)_{c+1} = 0$.
Since $H^{r-1}(F)$ satisfies (1) we get that $H^r(E)_c$ has finite length. Once can induct on $j $ to show that $H^r(E)_{c-j}$ has finite length for all $j \geq 0$.
(2) We have an exact sequence of $\R$-modules
$$ H^r(E)^\vee \xrightarrow{u} H^r(E)^\vee(+1) \xrightarrow{\delta^\vee} H^{r-1}(F)^\vee. $$
Set $W = H^r(E)^\vee$. Since $H^{r-1}(F)^\vee $ is finitely generated $\R$-module it follows that $W/uW(+1)$ (and so $W/uW$) is finitely generated.
Say $V = <\xi_1,\ldots,\xi_m>$ is a $\R$-submodule of $W$ such that $W = V + u W$. We prove $W = V$. This we do degree-wise.
By hypothesis on $E$ we have $H^r(E)_n = 0$ for all $n \gg 0$. So $W_n = 0$ for all $n \ll 0$ say from $n < c$.
Since $\deg u = 1$ we have $W_c = V_c$. Notice
\[
W_{c+1} = V_{c+1} + uW_c = V_{c+1} + uV_c = V_{c+1}.
\]
By induction on $j$ it is easy to show $W_{c+j} = V_{c+j}$ for all $j \geq 0$.
(3) This follows from Matlis duality.
\end{proof}
\begin{corollary}\label{loc-CA}[with hypotheses as in \ref{setup-lc}]
For $i = 0,\ldots,s-1$ set $\af(E)_i = \ann_{\R} H^i(E)$ and $\q_i(E) = \af(E)_i \cap A$. If $H^i(E) \neq 0$
then $\q_i(E)$ is $\m$-primary.
\end{corollary}
\begin{proof}
Fix $i$ with $0 \leq i \leq s-1$.
Set $D_i = H^i(E)$ and assume it is non-zero. It is easily checked using Matlis duality that $\ann_{\R} D_i = \ann_{\R} D_i^\vee$.
Notice $D_i^\vee$ is a finitely generated $\R$-module such that $\ell((D_i^\vee)_n)$ is finite for all $n$. Let
$m_1,\ldots,m_s$ be homogeneous generators of $D_i^\vee$. Consider the map
\begin{align*}
\frac{\R}{\af_i(E)} &\xrightarrow{\psi} \bigoplus_{j=1}^{s}D_i^\vee(-\deg m_j) \\
\ov{t} &\mapsto (tm_1,\ldots,tm_s).
\end{align*}
Clearly $\psi$ is injective.
Taking degree zero part of this embedding gets us that
$\q_i(E)$ is $\m$-primary.
\end{proof}
\section{Proof of main theorem}
The proof of the following result is inspired by Theorem 8.1.2 from \cite{BH}; (also see \cite[Theorem 1]{Roberts}). However we have to be extra careful at a few places.
The hypothesis of our result is not exactly similar and we are dealing with infinitely generated modules.
\begin{theorem}
Let $(A,\m)$ be a complete Noetherian ring with an infinite residue field and let $I$ be an $\m$-primary ideal in $A$. Let
$N$ be a quasi-finite $\R$-module of order at least $m$.
Assume $\bu = u_1,\ldots,u_m \in \R_1$ is a $N$ filter-regular sequence such that
\[
H^*_{\bu}(N) = H^*_{\R_+}(N)
\]
Also assume that $\ell(N_n)$ is finite for all $n\in \Z$.
Set $\bu^\pr = u_1,\ldots,u_n$ with $n \leq m$ and let
\[
\K = \K(\bu^\pr, N) \colon 0 \rt E_n \rt \cdots\rt E_1\rt E_0\rt 0
\]
be the Koszul complex of $\bu^\pr$ with coefficients in $N$.
For $j = 0,\ldots, m-1$
set $\bbf_j = \ann_{\R} H^j_{\R_+}(N) $ and $\q_j = A\cap \bbf_j$.
Then $\q_0\q_1\cdots\q_{n-1}$ annihilates $H_1(\K(\bu^\pr, N))$.
\end{theorem}
\begin{proof}
Let $\C$ be the Cech co-chain complex on $u_1,\ldots, u_m$. We shift $\C$ $m$-places and write it as a chain complex
\[
\D \colon 0 \rt D_m \rt \cdots \rt D_1 \rt D_0 \rt 0.
\]
By construction $H_i(N\otimes \D) = H^{m-i}_{\R_+}(N)$.
Consider the chain bicomplex $\X = \D \otimes \K$. We consider the two standard spectral sequences to compute the homology of $\Y = \Tot(\X)$; the total
complex of $\X$.
\emph{The first spectral sequence:}\\
${}^{I}E^0_{pq} = D_p\otimes K_q$. So
\begin{align*}
{}^{I}E^1_{pq} &= H_q(D_p \otimes \K) \\
&= D_p\otimes H_q(\K), \quad \text{since $D_p $ is flat}.
\end{align*}
By Theorem 3.1 we have that $H_q(\K)$ is $\R_+$-torsion for all $q > 0$.
It follows that
\[
{}^{I}E^1_{pq} = \begin{cases} 0 & \text{for $q > 0$ and $p \neq m$,} \\ H_q(\K) & \text{for $q > 0$ and $p = m$,}\\ D_p\otimes H_0(\K) &\text{for $q = 0$.} \end{cases}
\]
Therefore
\[
{}^{I}E^2_{pq} = \begin{cases} 0 & \text{for $q > 0$ and $p \neq m$,} \\ H_q(\K) & \text{for $q > 0$ and $p = m$,}\\ H^{m-p}_{\R_+}(H_0(\K)) &\text{for $q = 0.$} \end{cases}
\]
Observe that this spectral sequence \emph{collapses }at ${}^{I}E^2$. So $H_{m+i}(\Y) \cong H_i(\K)$ for $1\leq i \leq n$.
\emph{The second spectral sequence:}\\
${}^{II}E^0_{pq} = D_q\otimes K_p$. So
$$ {}^{II}E^1_{pq} = H_q(\D\otimes K_p) = H^{m-q}_{R_+}(K_p) = \left( H^{m-q}_{R_+}(N) \right)^{\binom{n}{p}}. $$
By construction $\q_{m-q}$ annihilates ${}^{II}E^1_{pq}$ if $q \neq 0$. Since ${}^{II}E^\infty_{pq}$ is a subquotient of ${}^{II}E^1_{pq}$ we get that
$\q_{m-q}$ annihilates $E^\infty_{pq}$ if $q \neq 0$.
Let $0 = V_{-1} \subseteq V_0 \subseteq V_1 \subseteq \cdots \subseteq V_{j-1} \subseteq V_j = H_{m+1}(\Y)$ be the filtration
such that ${}^{II}E^\infty_{p,m+1-p} \cong V_p/V_{p-1}$.
Notice ${}^{II}E^\infty_{p,m+1-p} = 0$ for $p > n$ and $m+1 -p >m$ (equivalently $p < 1$). So in the filtration $1 \leq p \leq n$.
Notice in this range $q = m+1-p \neq 0$ (otherwise $p = m+1 > n$). So $\q_{m-q} = \q_{p-1}$ annihilates ${}^{II}E^\infty_{p,m+1-p} $ for
the range $1 \leq p \leq n$. It follows that $\q_0\q_1\cdots\q_{n-1}$ annihilates $H_{m+1}(\Y)$. The result follows since $H_{m+1}(\Y)= H_1(\K)$.
\end{proof}
\begin{theorem}\label{man}
Let $(A,\m)$ be a complete \CM \ local ring of dimension with infinite residue field and dimension $d \geq 1$. Let $I$ be an $\m$-primary ideal in $A$. Set $L = L^I(A)$. For $i = 0,\ldots, d-1$ set
$\q_i = A \cap \ann_{\R}H^i_{R_+}(L)$. For $r =1,\ldots,d$ set
\[
\af_r(I) = \bigcap_{\substack{\text{$\bx = x_1,\ldots,x_r$ is a } \\ \text{ superficial sequence of $I$}}} \ann_A \Vc_I(\bx)
\]
Then $\af_r(I) \supseteq \q_0\cdots\q_{r-1}$. In particular if $\depth G_I(A) < r$ then $\af_r(I)$ is $\m$-primary.
\end{theorem}
\begin{proof}
By 2.4, $L$ is quasi-finite $\R$-module of order at least $d$.
Fix $r \geq 1$. Let $\bx^\pr = x_1,\ldots,x_r$ be an $I$-superficial sequence.
Then $\bx^\pr$ can be extended to a maximal superficial sequence $\bx = x_1,\ldots,x_r,x_{r+1},\ldots,x_d$.
Set $u_i = x_it \in \R_1$. Then by 2.5
$\bu = u_1,\ldots,u_d$ is a $L$-filter regular sequence. Since $(\bx)$ is a reduction of $I$ it follows that
$\bu$ generates $\R_+$ up to radical. So $H^{i}_{\bu}(L) = H^i_{\R_+}(L)$. Set $\bu^\pr = u_1,\ldots,u_r$. Let $\K(\bu^\pr,L)$ be the Koszul complex on
$\bu^\pr$ with coefficients in $L$. By 3.2 we get that $H_1(\bu^\pr, L) = \Vc_I(\bx^\pr)$. From Theorem 5.1.
we get $\ann_A \Vc_I(\bx^\pr) \supseteq \q_0\cdots\q_{r-1}$. Since $\bx^\pr$ was an arbitary superficial sequence of length $r$ we get $\af_r(I) \supseteq \q_0\cdots\q_{r-1}$.
\end{proof}
We now drop the assumption that $A$ is complete.
\begin{theorem}\label{main}
Let $(A,\m)$ be a \CM \ local ring with infinite residue field and dimension $d \geq 1$. Let $I$ be an $\m$-primary ideal and let $1 \leq r \leq d$. Then
\[
\af_r(I\widehat{A}) \cap A \subseteq \af_r(I).
\]
Furthermore
if $\depth G_I(A) < r$ then
$\af_r(I)$ is $\m$-primary.
\end{theorem}
\begin{proof}
Let $\widehat{A}$ be the completion of $A$. Let $\bx = x_1,\ldots,x_r$ be an $I$-superficial sequence. Then $\bx$ considered as a sequence in $\widehat{A}$ is also a $\widehat{I}$-superficial sequence. Furthermore $\Vc_{I\widehat{A}}(\bx) = \Vc_{I}(\bx)$ since it is of finite length.
It follows that $\ann_{\widehat{A}} \Vc_{I\widehat{A}}(\bx) \cap A = \ann_A \Vc_I(\bx)$.
Notice
\[
\af_r(I\widehat{A}) \subseteq \bigcap_{\substack{\text{$\bx = x_1,\ldots,x_r$ is a } \\ \text{ superficial sequence of $I$}}} \ann_A \Vc_{I\widehat{A}}(\bx).
\]
Therefore $\af_r(I\widehat{A}) \cap A \subseteq \af_r(I)$. Furthermore as $G_{I\widehat{A}}(\widehat{A}) = G_I(A)$ has depth $< r$ we have that
$\af_r(I\widehat{A})$ is $\widehat{\m}$-primary. It follows that $\af_r(I)$ is $\m$-primary.
\end{proof}
\section{Powers of $I$}
In this section we invesitigate $\af_r(I^l)$ for $ l \geq 1$.
One of the advantages of $L^I(A)$ is that $L^I(A)(-1)$ commutes with the Veronese functor.
Clearly
\[
\left(L^I(A)\right)^{<l>} = L^{I^l}(A)(-1) \quad \text{for each} \ l \geq 1.
\]
Also note that for the Rees algebras we have
\[
\Rc(I^l) = \R^{<l>} \quad \text{and} \quad \Rc(I^l)_+ = \R_+^{<l>}.
\]
Local cohomology also commutes with the Veronese functor. So we have that
\[
H^i_{\Rc(I^l)_+} \left(L^{I^l}(A)(-1)\right) \cong \left(H^i_{\R_+}(L^I(A))(-1)\right)^{<l>} \quad \text{for all} \ l \geq 1.
\]
We first prove the following general result.
\begin{lemma}\label{ml-powers}
Let $(A,\m)$ be a Noetherian local ring and let $I$ be an $\m$-primary ideal.
Let $E$ be a finitely generated graded $\R$-module with $\ell(E_n) < \infty$ for all $n \in \Z$.
For $l \geq 1$ set
\[
\q(I^l)_E = \left(\ann_{\Rc(I^l)} E^{<l>} \right)\cap A.
\]
Then
\begin{enumerate}[\rm(1)]
\item
$\q(I^l)_E$ is $\m$-primary for each $l \geq 1$.
\item
For each $r,l \geq 1$ we have
\[
\q(I^l)_E \subseteq \q(I^{rl})_E.
\]
\item
The set
\[
\mathcal{C} = \{ \q(I^l)_E \mid l \geq 1 \},
\]
has a unique maximal element which we denote as $\q(I)^\infty_E$.
\end{enumerate}
\end{lemma}
\begin{proof}
$\rm(1)$. Fix $l \geq 1$. Then $E^{<l>}$ is a finitely generated graded $\Rc(I^l)$-module with $\ell(E^{<l>}_j)$ finite for all $j \in \Z$. So by an argument similar to Corollary 4.4 we have that $\q(I^l)_E$ is $\m$-primary.
$\rm(2)$. Notice
\[
\left(E^{<l>}\right)^{<r>} = E^{<rl>}.
\]
Thus it suffices to prove the result for $l=1$.
Let $a \in \q(I)_E$. Then $a E_j = 0$ for all $j \in \Z$. So we have that $a \in \ann_{\Rc(I^r)} E^{<r>}$. Also as
$a \in A$ we have that $a \in \q(I^r)_E$.
$\rm(3)$ Suppose $\q(I^l)_E$ and $\q(I^r)_E$ are maximal elements in $\mathcal{C}$. By $\rm(2)$ we have that
\[
\q(I^l)_E \subseteq \q(I^{rl})_E \quad \text{and} \quad
\q(I^r)_E \subseteq \q(I^{rl})_E.
\]
By maximality of $\q(I^l)_E$ in $\mathcal{C}$ we have that $\q(I^l)_E = \q(I^{rl})_E$. Similarly
$\q(I^r)_E = \q(I^{rl})_E$. So $\q(I^l)_E = \q(I^r)_E$.
\end{proof}
\begin{question}(with hypotheses as above)
Is
\[
\q(I)^\infty_E= \q(I^l)_E \quad \text{for all} \ l \gg 0?
\]
\end{question}
We now prove the following result:
\begin{theorem}\label{main-2}
Let $(A,\m)$ be a \CM \ local ring with infinite residue field and dimension $d \geq 1$. Let $I$ be an $\m$-primary ideal and let $1 \leq r \leq d$.
If $\depth G_I(A) < r$ then
$$\bigcap_{n\geq 1}\af_r(I^n) \quad \text{ is $\m$-primary}.$$
\end{theorem}
\begin{proof}
By Theorem \ref{main}
\[
\af_r(I\widehat{A}) \cap A \subseteq \af_r(I).
\]
Thus $\af_r(I^n\widehat{A}) \cap A \subseteq \af_r(I^n)$ for all $n \geq 1$. Thus it suffices to prove the result when $A$ is complete.
Let $l \geq 1$. For $i = 0,1,\ldots, r-1$, define
$$\q_i(I^l) = \left(\ann_{\Rc(I^l)} H^i_{\Rc(I)_+} (L^{I^l}(A)) \right) \cap A.$$
By Theorem \ref{man}
\[
\af_r(I^l) \supseteq \q_0(I^l)\q_1(I^l)\cdots\q_{r-1}(I^l).
\]
For $i = 0,1,\ldots, r-1$ set
$$D_i(l) = H^i_{\Rc(I)_+}\left(L^{I^l}(A)(-1)\right)^\vee.$$
Note
that by Matlis duality
$$D^i(l)^\vee = H^i_{\Rc(I)_+}\left(L^{I^l}(A)(-1)\right).$$
Clearly
\[
\q_i(I^l) = \left(\ann_{\Rc(I^l)} D_i(l)\right)\cap A \quad \text{for} \ i = 0,1,\ldots, r-1.
\]
Since $L^I(A)$ and local cohomology behaves well with respect to the Veronese functor we have
that for all $l \geq 1$ we have
\[
D_i(l) = D_i(1)^{<l>} \quad \text{for} \ i = 0,1,\ldots, r-1.
\]
By Lemma \ref{ml-powers}(2) we have $\q_i(I^l) \supseteq \q_i(I)$ for all $l \geq 1$ and for all $i = 0,\ldots,r-1$.
Therefore we have
\[
\af_r(I^l) \supseteq \q_0(I)\q_1(I)\cdots\q_{r-1}(I) \ \text{for all} \ l \geq 1.
\]
It follows that $\bigcap_{n\geq 1}\af_r(I^n) \quad \text{ is $\m$-primary}.$
\end{proof}
We end our paper with the following:
\begin{question}(with hypothesis as above)
Is $\af_r(I^n)$ constant for all $n \gg 0$?
\end{question}
\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }
\providecommand{\MRhref}[2]{
\href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
}
\providecommand{\href}[2]{#2} | 0.001323 |
Let $n\in\left\{2,3\right\}$ and let $\Omega_0\subset \mathbb{R}^n$ be a hierarchical geometry as defined in Section \ref{sec:iga} defined from a hierarchy $\boldsymbol{\hat{\Omega}}_L$ with $L\in\mathbb N$. That is, $\Omega_0$ is the image of a bijective bi-Lipschitz map $\mathbf F : \hat{\Omega}_0 \rightarrow \Omega_0$ generated by an HB-spline basis $\hat{\mathcal H}$, where $\hat\Omega_0 = (0,1)^{n}$ is the parametric domain. Moreover, let $\left\{D_i\right\}_{i=1}^s$ be a set of bounded open domains in $\mathbb R^n$ that are trimmed from $\Omega_0$ in order to obtain the computational domain $\Omega$, i.e.,
$$\Omega := \Omega_0 \setminus \overline{D}, \quad \text{ with } \quad D:= \text{int}\left(\bigcup_{i=1}^s \overline{D_i}\right),$$
as illustrated in Figure \ref{fig:domains}.
Assume that $\Omega$ is an open Lipschitz domain, let $\mathbf n$ be the unitary outward normal of $\partial \Omega$, and let $\Gamma_N,\Gamma_D \subset \partial \Omega$ be open such that $\Gamma_D \cap \Gamma_N = \emptyset$, $\overline{\Gamma_D} \cap \overline{\Gamma_N} = \partial \Omega$ and $\Gamma_D\neq \emptyset$. Note that since $\Omega$ is Lipschitz, then the trimming boundary $\gamma := \partial \Omega \setminus \overline{\partial \Omega_0}$ is also Lipschitz.
\begin{figure}
\centering
\begin{subfigure}{0.4\textwidth}
\begin{center}
\begin{tikzpicture}[scale=2.5]
\draw[thick] (2,0) -- (3,0) ;
\draw[c2,thick] (2,0) -- (2,1) ;
\draw[c2,thick] (3,0) -- (3,1) ;
\draw[c2,thick] (2,1) -- (3,1) ;
\draw (2.5,0.5) node{$\Omega_0$} ;
\draw (2.5,0) node[below]{$\Gamma_D$} ;
\draw[c2] (2.9,1) node[above]{$\Gamma_N^0$} ;
\end{tikzpicture}
\caption{Non-trimmed domain $\Omega_0$.}
\label{fig:Omega0}
\end{center}
\end{subfigure}
~
\begin{subfigure}{0.4\textwidth}
\begin{center}
\begin{tikzpicture}[scale=2.5]
\draw[thick] (2,0) -- (3,0) ;
\draw[c2,thick] (2,0) -- (2,1) ;
\draw[c2,thick] (3,0) -- (3,1) ;
\draw[c2,thick] (2,1) -- (3,1) ;
\draw[c1,thick] (2.2, 1) arc (-180:0:.3cm);
\fill[pattern=north west lines, pattern color=c1, opacity=0.5] (2,0) -- (2,1) -- (2.2,1) arc (-180:0:.3cm) -- (3,1) -- (3,0) -- (2,0);
\draw[c1] (2.5,0.25) node{$\Omega$} ;
\fill[pattern=north east lines, pattern color=c2, opacity=0.5] (2.2, 1) arc (-180:0:.3cm) -- (2.2,1);
\draw[c2] (2.5,0.85) node{$D$} ;
\draw (2.5,0) node[below]{$\Gamma_D$} ;
\draw[c2] (2.9,1) node[above]{$\Gamma_N^0$} ;
\draw[c1] (2.25,0.7) node{$\gamma$} ;
\end{tikzpicture}
\caption{Trimmed domain $\Omega$ and $D:=\Omega_0\setminus\overline\Omega$.}
\label{fig:OmegaD}
\end{center}
\end{subfigure}
\caption{Illustration of the notation used on a trimmed geometry.}
\label{fig:domains}
\end{figure}
\subsection{Continuous formulation}
Let {$g_D\in H^{\frac{3}{2}}(\Gamma_D)$, $g_N\in H^{\frac{1}{2}}(\Gamma_N)$} and $f\in L^2\left(\Omega\right)$, and consider the following Poisson equation in $\Omega$: find $u\in H^1(\Omega)$, the weak solution of
\begin{align} \label{eq:originalpb}
\begin{cases}
-\Delta u = f &\text{ in } \Omega \\
u = g_D &\text{ on } \Gamma_D \\
\displaystyle\frac{\partial u}{\partial \mathbf{n}} = g_N &\text{ on } \Gamma_N,\vspace{0.1cm}
\end{cases}
\end{align}
that is, $u\in H^1_{g_D,\Gamma_D}(\Omega)$ satisfies for all $v\in H^1_{0,\Gamma_D}(\Omega)$,
\begin{equation} \label{eq:weakpb}
\int_\Omega \nabla u \cdot \nabla v \,\mathrm dx = \int_\Omega fv \,\mathrm dx + \int_{\Gamma_N} g v \,\mathrm ds.
\end{equation}
To simplify the subsequent analysis, we suppose that $\overline{\Gamma_D} \cap \overline \gamma = \emptyset$, that is, we suppose that the Dirichlet boundary is not part of the trimming boundary $\gamma$. If it were not the case, then in the discrete setting, we would need to weakly impose the Dirichlet boundary conditions on $\overline{\Gamma_D} \cap \overline \gamma$, and make use of stabilization techniques \cite{ghostpenalty,stabcutiga,puppistabilization}. Under this simplifying assumption, according to Lax-Milgram theorem, problem (\ref{eq:weakpb}) admits a unique solution $u\in H^1_{g_D,\Gamma_D}(\Omega)$.
\subsection{HB-spline based IGA formulation} \label{sec:HBIGAformulation}
Let $\mathcal Q_h := \mathcal Q\left(\Omega_0\right)$ be (a refinement of) the hierarchical physical mesh on $\Omega_0$, and for all $K\in \mathcal Q_h$, let $h_K:=\text{diam}(K)$. Moreover,
let $h:=\displaystyle\max_{K\in \mathcal M_h} h_K$, and let $\mathcal{E}_h^\Omega$ be the set of faces $E$ of $\mathcal{Q}_h$ such that $|E\cap\Gamma_N|>0$, with $h_E := \text{diam}(E)$. Note that when $n=2$, faces are edges.
Furthermore, let $\mathcal Q_h^\Omega := \mathcal Q_h^\trim \cup \mathcal Q_h^\untr$ be the active mesh intersecting $\Omega$, where
$$\mathcal Q_h^\trim := \left\{K\in \mathcal Q_h : K\cap \Omega \neq K, |K\cap\Omega| > 0 \right\} $$
is the set of cut (trimmed) elements, and
$$\mathcal Q_h^\untr := \left\{K\in \mathcal Q_h : K\subset \Omega\right\}$$
is the set composed of the other active (non-trimmed) elements in $\Omega$.
Similarly, we decompose $\mathcal E_h^\Omega = \mathcal E_h^\trim \cup \mathcal E_h^\untr$, where
$$\mathcal E_h^\trim := \left\{ E\in \mathcal E_h : E\cap\Gamma_N\neq E \right\} \qquad \text{and} \qquad \mathcal E_h^\untr := \left\{ E \in \mathcal E_h : E\subset \Gamma_N \right\}.$$
Recall that the trimming boundary is defined as $\gamma := \partial \Omega \setminus \overline{\partial \Omega_0} \subset \Gamma_N$, and for all $K\in\mathcal Q_h^\Omega$, let $$\gamma_K := \gamma\cap \text{int}(K).$$
Note that for all $K\in\mathcal Q_h^\untr$, $\gamma_K = \emptyset$. Also note that for all internal faces $E\in \mathcal E_h^\Omega$ such that $\left(E\cap\Gamma_N\right) \subset \gamma$ and $\left(E\cap\Gamma_N\right)\subset \partial \left(K\cap\Omega\right)$ for some $K\in\mathcal Q_h^\trim$, then $\gamma_K \cap \left(E\cap\Gamma_N\right) = \emptyset$, that is, this internal face is not included in $\gamma_K$. Or in other words, $$\Gamma_N = \left(\bigcup_{E\in\mathcal E_h^\Omega}E\right) \cup \left(\bigcup_{K\in \mathcal Q_h^\trim}\gamma_K\right),$$ where the intersection of any two elements of the union is empty, and each contribution of the union only appears once. \\
Now, let us make the following assumptions on $\mathcal Q_h$:
\begin{enumerate}
\item[\namedlabel{itm:A1}{(A1)}] $\mathcal Q_h$ is shape regular, that is, for all $K\in \mathcal Q_h$, $\displaystyle\frac{h_K}{\rho_K} \lesssim 1$, where $\rho_K$ denotes the radius of the largest inscribed ball in $K$. As a consequence, $h_K\simeq h_E$ for all $K\in \mathcal Q_h$ and all $E\in \mathcal E_h \cap \partial K$.
\item[\namedlabel{itm:A2}{(A2)}] $\mathcal Q_h$ is $\mathcal T$-admissible of some fixed class $\mu\in \mathbb{N}\setminus\{0\}$, following Definition \ref{def:admissibility}.
\end{enumerate}
\begin{remark}
Note that no further assumption on the trimming boundary $\gamma$ is required, other that $\gamma$ to be Lipschitz. In particular, and as already pointed out in \cite{guzman}, the literature on unfitted finite elements commonly imposes an additional restriction on how $\gamma$ intersects the mesh $\mathcal Q_h$. But we do not need here this assumption which, in $2$D, requires that $\gamma$ does not intersect an edge of the mesh $\mathcal Q_h$ more than once, and which is analogous in $3$D (see \cite{hansbo2} for instance). So for example, in this paper, $\gamma_K$ could be disconnected.
\end{remark}
Following the isogeometric paradigm introduced in Section~\ref{sec:iga} and generalizing it to trimmed geometries, let us define
\begin{align}
\mathcal{H} &:= \left\{B:=\hat B\circ \mathbf{F}^{-1} : \hat B\in\hat{\mathcal{H}}\right\},\nonumber \\
\mathcal{H}_\Omega &:= \left\{B\in \mathcal{H} : \supp{B}\cap \overline\Omega \neq \emptyset\right\}, \label{eq:HOmegabasis}
\end{align}
and let us also define the following approximation spaces:
\begin{align*}
&V_h(\Omega_0) := \spn{\mathcal H}, \quad
V_h(\Omega) := \spn{B\vert_\Omega: B\in\mathcal H_\Omega}, \\
&V_h^0(\Omega) := \spn{B\vert_\Omega: B\in\mathcal H_\Omega, B\vert_{\Gamma_D} = 0} = \left\{ v_h \in V_h(\Omega) : v_h\vert_{\Gamma_D} = 0 \right\}.
\end{align*}
Then, the Galerkin method with finite basis $\mathcal H_\Omega$ is used to discretize weak problem (\ref{eq:weakpb}) and reads as follows: find $u_h\in g_D+V_h^{0}(\Omega)$ such that for all $v_h\in V_h^0(\Omega)$,
\begin{equation} \label{eq:discrpb}
\int_\Omega \nabla u_h \cdot \nabla v_h \,\mathrm dx = \int_\Omega fv_h \,\mathrm dx + \int_{\Gamma_N} g v_h \,\mathrm ds.
\end{equation}
In the following, we are interested in the \textit{a posteriori} estimation of the energy error $\|\nabla (u-u_h)\|_{0,\Omega}$ on the trimmed geometry $\Omega$. | 0.002543 |
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Pops: HEY! Somebody's checking in!
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you got every kind of pest,
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But we treat 'em all as equals just like any other guest
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Kermit, Fozzie, Gonzo et al.:
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Return to the Muppet Movies Lyric Archive main page... | 0.001058 |
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\begin{document}
\begin{abstract}
We consider a one dimensional nonlocal transport equation and its natural multi-dimensional
analogues. By using a new pointwise inequality for the Hilbert transform, we give a short proof
of a nonlinear inequality first proved by C\'{o}rdoba, C\'{o}rdoba and Fontelos in 2005.
We also prove several new weighted inequalities for the Hilbert transform and
various nonlinear versions. Some of these results generalize to a related family
of nonlocal models.
\end{abstract}
\maketitle
\section{Introduction and main results}
In this work we consider the following nonlinear and nonlocal transport equation
\begin{align} \label{eqCCF05}
\begin{cases}
\theta_t + (\mathcal H\theta) \theta_x = - \kappa \Lambda^{\gamma} \theta,
\quad (t, x) \in (0,\infty) \times \bR, \\
\theta|_{t=0}= \theta_0,\\
\end{cases}
\end{align}
where $\theta=\theta(t,x)$ is a scalar-valued function defined on
$[0,\infty)\times \mathbb R$, and $\mathcal H$ is the Hilbert transform defined via
$$
\mathcal H \theta := \frac 1 {\pi}
\operatorname{PV} \int_{-\infty}^{\infty} \frac {\theta(y)} {x-y} dy.
$$
The number $\kappa\ge 0$ is the viscosity coefficient which governs the strength of
the linear dissipation. The dissipation term $\Lambda^{\gamma} \theta =(-\Delta)^{\gamma/2} \theta$ is defined by
using the Fourier transform as
\begin{align*}
\widehat{\Lambda^{\gamma} \theta} (\xi) = |\xi|^{\gamma} \widehat{\theta}(\xi),
\end{align*}
where $0<\gamma\le 2$. In other words $\Lambda^{\gamma}$ is the operator corresponding
to the Fourier symbol $|\xi|^{\gamma}$. When $0<\gamma<2$ and $\theta$ has suitable
regularity (for example $\theta \in C^{1,1}$), one has the representation
\begin{align*}
\Lambda^{\gamma} \theta = C_{\gamma}
\operatorname{PV} \int_{-\infty}^{\infty}
\frac{ \theta(x) -\theta(y)} {|x-y|^{1+\gamma}} dy,
\end{align*}
where $C_{\gamma}$ is a positive constant depending only on $\gamma$. It follows that
if $\theta$ attains its global maximum at $x=x_*$, then
\begin{align*}
(\Lambda^{\gamma} \theta )(x_*) \ge 0.
\end{align*}
By using this and the transport nature of the equation, one has for any smooth
solution $\theta$ to \eqref{eqCCF05} the $L^{\infty}$-maximum principle:
\begin{align*}
\| \theta(t,\cdot)\|_{\infty} \le \| \theta_0 \|_{\infty}, \quad\forall\, t>0.
\end{align*}
For $\kappa>0$ and regarding $L^{\infty}$ as the threshold space, the cases $\gamma<1$, $\gamma=1$, $\gamma>1$ are called supercritical, critical and
subcritical respectively. When $\kappa=0$ the model \eqref{eqCCF05} becomes
the inviscid case and it is deeply connected with the usual two-dimensional surface
quasi-geostrophic equation (cf. \cite{kiselev} and the references therein for some recent
results). Compared with the usual Burgers equation with fractal
dissipation, the model \eqref{eqCCF05} in some
sense represents the simplest case of a nonlinear transport equation with nonlocal velocity and a viscous fractional dissipation. For some other related one dimensional hydrodynamic models
having some connection with
the 2D quasi-geostrophic equation and the 3D Euler equation, we refer the reader to
\cite{BM02}, \cite{CCCF05}, \cite{CLM85}, \cite{Gre96}, \cite{Sak03}, \cite{Sch86}, \cite{Yang96}
and the references therein for additional results.
Concerning the model \eqref{eqCCF05}, in the inviscid case $\kappa=0$, C\'{o}rdoba, C\'{o}rdoba and
Fontelos \cite{CCF05} first
proved the breakdown of classical solutions to \eqref{eqCCF05}
for a generic class of smooth initial data. When $\kappa>0$, they also obtained the global well posedness in the subcritical case. For the critical case, global well-posedness
can be proved by adapting the method of continuity as in \cite{kiselev}. Blow up for the supercritical
case $0\le \gamma<1/2$ was established in \cite{LRadv08}. Currently the
case $\frac 12 \le \gamma<1$ is still open.
For the inviscid case the proof of \cite{CCF05} is based on an ingenious inequality:
\begin{align}\label{ine_tmp_001}
-\int_{\mathbb R}
\frac {\mathcal H f \cdot f_x} {x^{1+\delta}} dx
\ge C_{\delta} \int_{\mathbb R}
\frac{ f(x)^2} {x^{2+\delta}} dx,
\end{align}
where $-1<\delta<1$, $C_{\delta}>0$ is a constant depending only on $\delta$, and $f$
is an even bounded smooth (not necessarily decaying) function on $\mathbb R$ with $f(0)=0$. In the blow-up proof
the inequality \eqref{ine_tmp_001} is applied to $f(x)=\theta(0)-\theta(x)$ and thus $f$ in general does not decay at the spatial infinity.
The proof of \eqref{ine_tmp_001}
in \cite{CCF05} uses Mellin transform and complex analysis. A natural question is
whether one can give a completely real variable proof of \eqref{ine_tmp_001}. In this
direction Kiselev (see \cite{K10}) showed that for any even bounded $C^1$ function $f$ with $f(0)=0$ and
$f^{\prime}\ge 0$ for $x>0$, the following inequality (see Proposition 26 therein)
\begin{align} \label{ine_tmp_002}
- \int_0^1
\frac {\mathcal H f (x) f^{\prime}(x) f(x)^{p-1}}
{x^{\sigma}} dx
\ge C_0 \int_0^1 \frac {f(x)^{p+1}} {x^{1+\sigma}} dx,
\end{align}
where $p\ge 1$, $\sigma>0$ and $C_0$ is a positive constant depending on $p$ and $\sigma$. Later in \cite{SV16} Silvestre and Vicol gave four elegant proofs for the inviscid case (one should note
that the definition of the Hilbert transform $\mathcal H$ used in \cite{SV16} differs from the usual
one by a minus sign! See formula (1.2) therein). The purpose of this paper is to revisit the model
\eqref{eqCCF05} and give several new and elementary proofs which are all real variable based.
In Section 2 we first derive a new point-wise inequality (see Proposition \ref{prop_lower})
for the Hilbert transform acting on even and non-increasing (on $(0,\infty)$) functions on
$\mathbb R$, and then we show the C\'{o}rdoba-C\'{o}rdoba-Fontelos inequality by a simple
application of Hardy's inequality. We also present several simplified arguments whose byproduct lead to a simple
proof of the Kiselev inequality \eqref{ine_tmp_002} and further improvements (in particular
we disprove the Kiselev inequality without the monotonicity constraint). In Section 3 we generalize the argument to dimensions
$n\ge 2$ which works for the generalized surface quasi-geostrophic equations considered
in \cite{LRcmp09, Dong14, DL08}. Note that the blow-up proof here covers the full range of the
generalized surface quasi-geostrophic model. In Section 4 we give another proof which works for general functions having even symmetry
(note necessarily monotone decaying) for the Hilbert model case. In Section 5 we generalize
the argument to more general $\alpha$-patch type models.
\subsection*{Acknowledgements.}
D. Li was supported in part by Hong Kong RGC grant GRF 16307317
and 16309518. J. Rodrigo was supported in part by European Research Council, ERC Consolidator Grant no. 616797.
\section{radial decreasing for dimension $n=1$}
We shall use (often without explicit mentioning) the following Hardy's inequality.
\begin{lem}[Hardy] \label{lem_hardy}
If $1\le p<\infty$, $\tilde r\ge 0$ and $f$ is a non-negative measurable function on $(0,\infty)$.
Then
\begin{align*}
\int_0^{\infty} F(x)^p x^{p-\tilde r-3} dx \le \left(\frac p {\tilde r}\right)^p
\int_0^{\infty} f(t)^p t^{p-\tilde r-1} dt,
\end{align*}
where $F(x) =\int_0^x f(t) dt$.
\end{lem}
\begin{proof}
See pp. 35 of \cite{Ziemer89}. Note that the $F(x)$ defined therein has an extra $1/x$ factor.
\end{proof}
\begin{prop}[A lower bound for Hilbert transform] \label{prop_lower}
Let $g$: $\mathbb R \to \mathbb R$ be an even continuously differentiable function which is non-increasing
on $[0,\infty)$. Assume $g^{\prime} \in L^1 \cap L^{\infty}$.
Then for any $0<x<\infty$,
\begin{align*}
(\mathcal H g)(x) \ge \frac 2 {\pi} \cdot \frac 1 x \int_0^x ( g(y) - g(x) )dy.
\end{align*}
\end{prop}
\begin{rem} \label{rem_add_00a00}
For $f$ even, continuously differentiable and \emph{non-decreasing} on $[0, \infty)$ with
$f^{\prime}\in L^1\cap L^{\infty}$, we have the inequality
\begin{align*}
- (\mathcal H f)(x) \ge \frac 2 {\pi}
\cdot \frac 1 x \int_0^x ( f(x) -f(y)) dy, \quad \forall\, 0<x<\infty.
\end{align*}
\end{rem}
\begin{proof}
Since $g$ is even and $g^{\prime}\le 0$ on $[0,\infty)$, it is not difficult to check that
\begin{align*}
(\mathcal Hg)(x) & = \frac 1 {\pi} \int_0^{\infty}
\log\abs{\frac {x-y}{x+y} } g^{\prime}(y) dy \ge \frac 1 {\pi} \int_0^x \log \abs{ \frac{1- \frac yx} {1+\frac yx} } g^{\prime}(y) dy \\
& \ge \frac 2 {\pi} \int_0^x \frac y x \cdot (-g^{\prime}(y) ) dy,
\end{align*}
where in the last inequality we used
\begin{align*}
- \log \abs{ \frac {1-\epsilon} {1+\epsilon} } \ge 2\epsilon \quad \text{for all $0\le \epsilon<1$}.
\end{align*}
Integration by parts then yields the result.
\end{proof}
\begin{rem*}
Another more direct proof (under the same assumptions)
is as follows. First observe that for each $0<x<\infty$,
\begin{align*}
\mathcal H g(x) = \frac {2x} {\pi} \int_0^{\infty} \frac {g(y)-g(x)} {x^2-y^2} dy.
\end{align*}
Thanks to monotonicity, the integrand $\frac {g(y)-g(x)} {x^2-y^2}\ge 0$ in either the
regime $y<x$ or the regime $y>x$. Thus we can restrict the integral to the regime
$0<y<x$, and obtain
\begin{align*}
\mathcal H g(x) &\ge \frac {2x} {\pi} \int_{0<y<x} \frac {g(y)-g(x)} {x^2-y^2} dy
\notag \\
&\ge \frac {2} {\pi } \cdot \frac 1 x \int_0^x (g(y)-g(x) ) dy.
\end{align*}
\end{rem*}
Proposition \ref{prop_lower} can now be used to establish the following lemma
which is essentially Lemma 2.2 found in \cite{CCF05}. The original proof therein relies on
Mellin transform and positivity of certain Fourier multipliers.
Our new proof below avoids this and is
completely real-variable based.
For simplicity we shall make the same assumption
on the function $g$ as in Proposition \ref{prop_lower}.
\begin{lem} \label{Jun16_Lem}
For any $-1<\delta<1$,
\begin{align*}
-\int_0^\infty \frac {g^\prime(x) (\mathcal Hg)(x)} {x^{1+\delta}} dx \ge
C_{\delta} \int_0^\infty \frac {(g(x)-g(0))^2} {x^{2+\delta}} dx,
\quad \text{where $C_{\delta}=\frac 1 {\pi} \cdot \frac{(1+\delta)^2}{3+\delta}$.}
\end{align*}
\end{lem}
\begin{proof}
By Proposition \ref{prop_lower}, we have
\begin{align*}
\operatorname{LHS} & \ge \frac 2 {\pi}
\cdot \int_0^{\infty} \frac { \int_0^x (g(y)-g(x) ) dy \cdot (-g^{\prime}(x) )}
{x^{2+\delta}} dx \notag \\
& = \frac 2 {\pi} \int_0^{\infty}
\frac{ \int_0^x (f(x) - f(y) ) dy \cdot f^{\prime}(x) } {x^{2+\delta}},
\end{align*}
where $f(x) = g(0)-g(x)$. Notice that
\begin{align*}
\frac d {dx}
\Bigl( \int_0^x (f(x) -f(y)) dy \Bigr)= x f^\prime(x).
\end{align*}
Now using this and successive integration by parts gives
\begin{align*}
& \int_0^{\infty} \frac{ \int_0^x (f(x)-f(y) ) dy f^{\prime}(x)} { x^{2+\delta}} dx \notag \\
=& \int_0^{\infty}
(-1) \cdot
\frac d {dx} ( \int_0^x (f(x)-f(y)) dy
\frac 1 {x^{2+\delta}}) f(x) dx \notag \\
= &\; - \int_0^{\infty} \frac{f^{\prime}(x) f(x) } {x^{1+\delta}} dx
+(2+\delta) \int_0^{\infty}
\frac{ \int_0^x (f(x)-f(y)) dy f(x)} {x^{3+\delta}} dx \notag \\
=& (-\frac{1+\delta}2 +2+\delta) \int_0^{\infty}
\frac{f(x)^2} {x^{2+\delta}} dx -(2+\delta) \int_0^{\infty} \frac{\int_0^x f(y) dy f(x)} {
x^{3+\delta} } dx \notag \\
= &\frac{3+\delta}2 \int_0^{\infty} \frac{(F^{\prime}(x))^2} {x^{2+\delta}} dx
-\frac{3+\delta}2 (2+\delta) \int_0^{\infty} \frac{F(x)^2} {x^{4+\delta}} dx,
\end{align*}
where $F(x) = \int_0^x f(y)dy$. By Hardy's inequality, we have
\begin{align*}
\int_0^{\infty} \frac{F(x)^2} {x^{4+\delta}} dx \le \left( \frac 2 {3+\delta} \right)^2
\int_0^{\infty} \frac{ (F^{\prime}(x))^2} {x^{2+\delta}} dx.
\end{align*}
The result then follows.
\end{proof}
\begin{rem} \label{rem2.4_00}
One can even give a direct (without using Hardy) proof as follows. Write
(after using Proposition \ref{prop_lower})
\begin{align*}
&\frac 2 {\pi} \int_0^{\infty}
\frac{ \int_0^x (f(x) - f(y) ) dy \cdot f^{\prime}(x) } {x^{2+\delta}} \notag \\
= & \frac 2 {\pi}
\iint_{x\ge y} \frac{ \frac 12 \frac d {dx} ( (f(x)-f(y) )^2) } {x^{2+\delta}} dx dy \notag \\
=& \frac {2+\delta}{\pi} \iint_{x\ge y} \frac{(f(x)-f(y))^2} {x^{3+\delta} } dx dy.
\end{align*}
Now using the inequality $(a-b)^2 = a^2+ b^2-2ab \ge (1-\alpha)a^2+ (1-\frac 1{\alpha})b^2$
for any $\alpha>0$, we obtain
\begin{align*}
\iint_{x\ge y} \frac{ (f(x)-f(y))^2} {x^{3+\delta}} dx dy
\ge ( 1+ \frac 1 {2+\delta} - ( \alpha+ \frac 1 {\alpha} \cdot \frac 1 {2+\delta} ) )
\int_0^{\infty} \frac {f(x)^2} {x^{2+\delta} } dx.
\end{align*}
Optimizing in $\alpha$ then yields the inequality with a slightly inferior constant
\begin{align*}
C_{\delta} = \frac 1 {\pi} \cdot ( 3+\delta - 2 \sqrt{2+\delta}).
\end{align*}
\end{rem}
\begin{rem}
In the preceding remark, it is possible to obtain the sharper bound by using the following argument.
Noting that
\begin{align*}
\int_{x \ge y} \frac{f(y)^2} {x^{3+\delta}} dx dy = \frac 1 {2+\delta}
\int_0^{\infty} \frac{ f(y)^2} {y^{2+\delta}}dy,
\end{align*}
it suffices to treat the term
\begin{align*}
\iint_{x\ge y} \frac {f(x) f(y)} { x^{3+\delta}} dx dy&= \int_0^{\infty} \frac{f(x)}{x^{3+\delta}}
\left(\int_0^x f(y) dy\right)dx \notag \\
&= \frac{3+\delta}2 \int_0^{\infty} \frac { (\int_0^x f(y) dy)^2} {x^{4+\delta}} dx.
\end{align*}
By Cauchy-Schwartz
\begin{align*}
\left(\int_0^x f(y) dy \right)^2 \le \int_0^x f(y)^2 y^{-p} dy \cdot \frac {x^{p+1}}{p+1}.
\end{align*}
Interchanging the integral of $dx$ and $dy$ then gives
\begin{align*}
\int_0^{\infty} \frac { (\int_0^x f(y) dy)^2} {x^{4+\delta}} dx
&\le \int_0^{\infty} \frac {f(y)^2}{y^p} \cdot
\frac{1}{p+1}\left( \int_y^{\infty} x^{-3-\delta+p} dx\right)
dy \\
&= \frac 1 {(p+1) (2+\delta-p)} \int_0^{\infty} \frac {f(y)^2} {y^{2+\delta}} dy.
\end{align*}
Choosing $p=\frac{1+\delta} 2$ then yields the sharper constant
\begin{align*}
C_{\delta} = (2+\delta) \cdot ( 1+ \frac 1 {2+\delta}- 2 \cdot \frac {3+\delta}2 \cdot
(\frac 2 {3+\delta})^2 ) = \frac {(1+\delta)^2}{3+\delta}.
\end{align*}
\end{rem}
\subsection{Proof of the Kiselev inequality}
We now sketch a simple proof of the Kiselev inequality \eqref{ine_tmp_002}.
We emphasize that this inequality is stated for \emph{nondecreasing} even functions on $\mathbb R$.
For illustration purposes
we first consider the simple case $p=1$. By using Proposition \ref{prop_lower} (see
Remark \ref{rem_add_00a00}), we have
\begin{align*}
-\int_0^1 \frac{\mathcal Hf(x) f^{\prime}(x)}
{x^{\sigma}} dx
&\ge
\frac 2 {\pi} \int_0^1 \frac{ \int_0^x (f(x)-f(y)) dy f^{\prime}(x)} {x^{1+\sigma}} dx \notag \\
& = \frac 1 {\pi} \iint_{\substack{ 0<x<1\\ 0<y<x}}
\frac { \frac d {dx} (( f(x) -f (y))^2) } {x^{1+\sigma}} dx dy \notag \\
&\ge \frac {1+\sigma} {\pi}
\iint_{\substack{ 0<x<1\\ 0<y<x}}
\frac { ( f(x) -f (y))^2} {x^{2+\sigma}} dx dy,
\end{align*}
where in the last step we have integrated by part in the $x$-variable and dropped the harmless
boundary terms. Note that we can also keep the boundary term and derive a sharper inequality
as it is nonnegative. Next we proceed similarly as in Remark
\ref{rem2.4_00} and derive (below we shall take $0<\alpha<1$ and specify its
value at the very end)
\begin{align*}
& \iint_{0<y<x<1} \frac {(f(x)-f(y))^2} {x^{2+\sigma}} dx dy \notag \\
\ge & \iint_{0<y<x<1}
\frac{ (1-\alpha) f(x)^2+ (1-\frac 1 {\alpha}) f(y)^2} { x^{2+\sigma}} dx dy \notag \\
= & (1-\alpha) \int_0^1 \frac {f(x)^2} {x^{1+\sigma}} dx
+(1-\frac 1 {\alpha}) \int_0^1 f(y)^2 \frac 1 {1+\sigma} \cdot (\frac 1 {y^{1+\sigma}}-1) dy
\notag \\
\ge &
(1-\alpha) \int_0^1 \frac {f(x)^2} {x^{1+\sigma}} dx
+(1-\frac 1 {\alpha}) \int_0^1 f(y)^2 \frac 1 {1+\sigma} \cdot \frac 1 {y^{1+\sigma}}dy
\notag \\
= & (1+\frac 1 {1+\sigma}) \int_0^1 \frac {f(x)^2} {x^{1+\sigma}} dx
- (\alpha+\frac 1 {\alpha}\cdot \frac 1 {1+\sigma} ) \cdot \int_0^1 \frac {f(x)^2}
{x^{1+\sigma}} dx.
\end{align*}
Choosing $\alpha= (1+\sigma)^{-\frac 12}$ then yields the result. Note that in the second
inequality above, we used the fact that $0<\alpha<1$ so that the term $-1$ in the $y$-integral
can be safely dropped.
Next we sketch the proof for $1<p<\infty$. We start with
\begin{align*}
-\int_0^1 \frac{\mathcal Hf(x) f^{\prime}(x)}
{x^{\sigma}} f(x)^{p-1} dx
&\ge
\frac 2 {\pi} \underbrace{\int_0^1 \frac{ \int_0^x (f(x)-f(y)) dy f^{\prime}(x)} {x^{1+\sigma}}
f(x)^{p-1} dx}_{=:H_1}.
\end{align*}
Note that for any $0\le s\le 1$, we have the inequality
\begin{align*}
1-s \ge (1-s)^p.
\end{align*}
This in turn implies that (note that below $f(y)/f(x)
\le 1$ for $0<y<x$ since $f$ is nondecreasing!)
\begin{align*}
(f(x)-f(y)) f(x)^{p-1} \ge (f(x)-f(y))^p.
\end{align*}
Thus
\begin{align*}
H_1 &\ge \iint_{0<y<x<1} \frac {\frac 1{p+1}\frac d {dx} ((f(x)-f(y))^{p+1} )} {x^{1+\sigma}}
dx dy \notag \\
&\ge \frac {1+\sigma}{p+1} \iint_{0<y<x<1} \frac { (f(x)-f(y))^{p+1}} {x^{2+\sigma}} dx dy.
\end{align*}
Now note that for any $\beta>1$, one can find a constant $c_1>0$, depending only
on $p$ and $\beta$, such that
\begin{align*}
(1-s)^{p+1} \ge c_1\cdot (1-\beta s^{p+1}),\qquad\forall\, 0\le s\le 1.
\end{align*}
This in turn implies that
\begin{align*}
(f(x)-f(y))^{p+1} \ge c_1\cdot( f(x)^{p+1} - \beta f(y)^{p+1}).
\end{align*}
Using this inequality we then obtain
\begin{align*}
H_1 \ge \;\operatorname{const} \cdot (1-\frac {\beta}{1+\sigma}) \int_0^1 \frac {f(x)^{p+1}}
{x^{1+\sigma}} dx.
\end{align*}
Hence taking $1<\beta<1+\sigma$ (say $\beta=1+\frac{\sigma}2$) then finishes the proof for the case $p>1$.
\subsection{Further remarks}
We first point it out that, under the assumption of \emph{monotonicity}, the
Kiselev inequality \eqref{ine_tmp_002} is stronger than the C\'{o}rdoba-C\'{o}rdoba-Fontelos inequality \eqref{ine_tmp_001}. Indeed fix any $C^1$ bounded even $f$ on
$\mathbb R$ with $f^{\prime}\ge 0$ on $(0,\infty)$, apply the Kiselev inequality
to $f_L(x)= f(\frac x L)$, and we get (after a change of variable)
\begin{align*}
-\int_0^L \frac{\mathcal H f f^{\prime} } {x^{\sigma}} dx \ge
C_0 \int_0^L \frac {f(x)^2} {x^{1+\sigma}} dx.
\end{align*}
Note that $C_0$ is independent of the parameter $L$. Sending $L$ to infinity and using the Lebesgue
Monotone Convergence Theorem (note that the integrand $-Hf \cdot f^{\prime}$ is non-negative!) then yields the C\'{o}rdoba-C\'{o}rdoba-Fontelos inequality for the whole regime $\sigma>0$.
One should note that the same argument yields the inequality
\begin{align} \notag
- \int_0^{\infty}
\frac {\mathcal H f (x) f^{\prime}(x) f(x)^{p-1}}
{x^{\sigma}} dx
\ge C_0 \int_0^{\infty} \frac {f(x)^{p+1}} {x^{1+\sigma}} dx,
\end{align}
where $p\ge 1$ and $C_0$ depends only on $p$ and $\sigma$.
Finally we should point it out that in the Kiselev inequality, the assumption of monotonicity
cannot be dropped in general. In what follows we shall construct a counterexample which answers a question raised by Kiselev in \cite{K10} (see Remark $1$ on page 249 therein).
\begin{prop} \label{prop_counter_01}
For any $\sigma>0$,
there exists an even function $f \in C_c^{\infty}(\mathbb R)$ such that
\begin{align*}
-\int_0^1 \frac { \mathcal H f (x) f^{\prime}(x)} {x^{\sigma}} dx <0.
\end{align*}
In particular we cannot have the Kiselev inequality \eqref{ine_tmp_002} for $p=1$ without
the monotonicity assumption.
\end{prop}
\begin{rem*}
Similarly one can do the case $1<p<\infty$, but we do not present the
details here.
\end{rem*}
\begin{lem} \label{lem_09Aug20_001}
For any $\sigma>0$, one can find $\phi_A \in C_c^{\infty}(\mathbb R)$,
even and supported in $\{0<|x|<1\}$, $\phi_B \in C_c^{\infty}(\mathbb R)$,
even and supported in $|x|>1$, such that
\begin{align*}
\int_0^{\infty} \frac {\mathcal H \phi_B (x) \cdot ( \phi_A)^{\prime}(x) }
{x^{\sigma}} dx >0.
\end{align*}
\end{lem}
\begin{proof}[Proof of Lemma \ref{lem_09Aug20_001}]
Clearly
\begin{align*}
\int_0^{\infty} \frac {\mathcal H \phi_B (x) \cdot ( \phi_A)^{\prime}(x)}
{x^{\sigma}} dx = - \int_0^{\infty}
\partial_x( \frac {\mathcal H \phi_B} {x^{\sigma}} ) \phi_A(x) dx.
\end{align*}
Now
\begin{align*}
\partial_x( \frac {\mathcal H \phi_B} {x^{\sigma}} )
= \frac 1 {x^{\sigma}} ( \Lambda \phi_B - \sigma \cdot \frac 1 x \cdot \mathcal H \phi_B).
\end{align*}
Observe that (here we use $\phi_B$ is supported in $|x|>1$)
\begin{align*}
&(\Lambda \phi_B)(1) - \sigma (\mathcal H \phi_B)(1) \notag \\
= &\;
\alpha_1 \int_{|y|>1} \frac {-\phi_B(y)} {(1-y)^2} dy
-\frac 2 {\pi} \sigma \int_{y>1} \frac {\phi_B(y)} {1-y^2} dy \notag \\
=& \; - \int_{y>1}
\bigl[ \frac{\alpha_1}{(1-y)^2}
+ \frac{\alpha_1}{(1+y)^2}
+\frac 2 {\pi} \sigma \frac 1 {1-y^2} \bigr] \phi_B(y) dy,
\end{align*}
where $\alpha_1>0$ is an absolute constant which appear in the definition of the nonlocal
operator $\Lambda$. It is then clear that one can choose suitable $\phi_B$ such that
\begin{align*}
(\Lambda \phi_B)(1) - \sigma (\mathcal H \phi_B)(1) <0.
\end{align*}
By continuity we can find $x_0<1$ sufficiently close to $1$, such that
\begin{align*}
(\Lambda \phi_B)(x_0) - \sigma\cdot \frac 1 {x_0} (\mathcal H \phi_B)(x_0) <0.
\end{align*}
Choosing $\phi_A$ to be a suitable bump function localized around $x_0$ then yields the result.
\end{proof}
With the help of Lemma \ref{lem_09Aug20_001}, we now complete the proof of
Proposition \ref{prop_counter_01}. Choose
\begin{align*}
f(x) = \phi_A (x) + t \phi_B(x).
\end{align*}
Then clearly (note that below we use the fact that
$\phi_B$ is supported in $|x|>1$)
\begin{align*}
-\int_0^1 \frac { \mathcal H f (x) f^{\prime}(x)} {x^{\sigma}} dx
&= c_1- t \int_0^1 \frac {\mathcal H \phi_B (x) \cdot ( \phi_A)^{\prime}(x)}
{x^{\sigma}} dx \notag \\
&=c_1 -t \int_0^{\infty} \frac {\mathcal H \phi_B (x) \cdot ( \phi_A)^{\prime}(x)}
{x^{\sigma}} dx,
\end{align*}
where $c_1$ is independent of $t$. Choosing $t$ sufficiently large then yields the result.
\section{Radial decreasing for dimension $n\ge 2$}
In \cite{LRcmp09, Dong14, DL08} a family of the generalized surface quasi-geostrophic equations were introduced and studied. The simplest inviscid case takes the form:
\begin{align*}
\partial_t g + (\Lambda^{-\alpha} \nabla g \cdot \nabla g) =0,
\end{align*}
where $n\ge 2$, $0<\alpha<2$ and $\Lambda^{-\alpha}$ corresponds to the
Fourier multiplier $|\xi|^{-\alpha}$. These models can be viewed as natural
generalizations of the one dimensional Hilbert-type models to higher dimensions.
In what follows we shall discuss the corresponding nonlinear inequalities in analogy
with the Hilbert transform case.
\begin{prop} \label{prop3.1}
Let $n\ge 2$ and $0< \alpha <2$. Let $g:\, \mathbb R^n \to \mathbb R$ be a radial and non-increasing
Schwartz function. Then for any $x \ne 0$,
\begin{align*}
- (\Lambda^{-\alpha} \nabla g )(x) \cdot \frac x {|x|}
\ge C_{\alpha,n} \cdot
\frac 1 {r^{n-\alpha+1}} \int_0^r (-g^{\prime}(\rho)) \cdot \rho^n d \rho,
\end{align*}
where $r=|x|$ and $C_{\alpha,n}>0$ depends only on $(\alpha,n)$.
\end{prop}
\begin{rem} \label{rem_prop3.1_0}
Note that for $f(x)=g(0)-g(x)$ radial and nondecreasing, we have
\begin{align*}
(\Lambda^{-\alpha} \nabla f )(x) \cdot \frac x {|x|}
\ge C_{\alpha,n} \cdot
\frac 1 {r^{n-\alpha+1}} \int_0^r (f^{\prime}(\rho)) \cdot \rho^n d \rho.
\end{align*}
\end{rem}
\begin{proof}
Since $g$ is radial we can assume WLOG that $x=re_n=r\cdot (0,\cdots, 0, 1)$. By using
the fact that $g^{\prime}(\rho)\le 0$, we have
\begin{align*}
-(\Lambda^{-\alpha} \partial_n g)(x)
& = C_{\alpha,n} \int_0^{\infty} \int_{|\omega|=1}
\frac{\omega_n}{|r e_n - \rho \omega|^{n-\alpha}}
\cdot (-g^{\prime}(\rho)) \rho^{n-1} d\sigma(\omega) d\rho \notag \\
& \gtrsim \int_0^r ( -g^{\prime}(\rho)) \cdot \rho^{n-1} \cdot r^{-(n-\alpha)}
\cdot \frac{\rho} r d\rho,
\end{align*}
where we have used the simple inequality
\begin{align*}
\int_{|\omega|=1}
\frac{\omega_n} { |e_n - \epsilon \omega|^{n-\alpha}} d\sigma(\omega)
\gtrsim \epsilon,\quad \text{for $0<\epsilon <1$}.
\end{align*}
\end{proof}
\begin{lem} \label{lem_fullrange_00a}
Let $n\ge 2$, $0< \alpha <2$ and $-1<\delta <1$. Let $g:\, \mathbb R^n \to \mathbb R$ be a radial and nonincreasing
Schwartz function. Then
\begin{align*}
\int_{\mathbb R^n}
\frac{ \Lambda^{-\alpha} \nabla g \cdot \nabla g }
{|x|^{n+\delta}} dx
\ge C_{\alpha,\delta,n}
\int_{\mathbb R^n}
\frac{ (g(0) - g(x))^2}{ |x|^{n+2-\alpha+\delta}} dx,
\end{align*}
where $C_{\alpha,\delta,n}>0$ depends only on $(\alpha,\delta,n)$.
\end{lem}
\begin{proof}
Denote $f(x)=g(0)-g(x)$. Note that $f$ is non-decreasing and $f(0)=0$.
By Proposition \ref{prop3.1} and Remark
\ref{rem_prop3.1_0}, we have
\begin{align*}
\operatorname{LHS}
& \gtrsim \int_0^{\infty}
\frac{ \int_0^r f^{\prime}(\rho) \rho^n d\rho \cdot f^{\prime}(r)}
{ r^{n-\alpha+\delta+2}} dr \notag \\
& = - \int_0^{\infty}
\frac{f^{\prime}(r) f(r)} {r^{2-\alpha+\delta}} dr
+ (n-\alpha+\delta+2)
\int_0^{\infty} \frac{ \int_0^r f^{\prime}(\rho) \rho^n d \rho f(r)}
{r^{n-\alpha+\delta+3}} d r \notag \\
& = - \frac{2-\alpha+\delta}2
\int_0^{\infty} \frac{f(r)^2 }{r^{3-\alpha+\delta}} dr
+ (n-\alpha+\delta+2) \int_0^{\infty}
\frac{f(r)^2} { r^{3-\alpha+\delta}} dr \notag \\
& \quad -(n-\alpha+\delta+2)
\int_0^{\infty}
\frac{ n \int_0^r f(\rho) \rho^{n-1} d\rho f(r) r^{n-1}}
{ r^{n-1} \cdot r^{n-\alpha+\delta+3}} dr \notag \\
& = (n+\frac{2-\alpha+\delta}2)
\int_0^{\infty}
\frac{ (f(r)r^{n-1} )^2} {r^{2n+1-\alpha+\delta}} dr \notag \\
& \quad -n(n+2-\alpha+\delta)
\cdot (n+\frac{2-\alpha+\delta}2)
\int_0^{\infty}
\frac{ F(r)^2} {r^{2n+3-\alpha+\delta}} dr,
\end{align*}
where
\begin{align*}
F(r) = \int_0^r f(\rho) \rho^{n-1} d\rho.
\end{align*}
Now the result follows from Hardy's inequality (see Lemma \ref{lem_hardy} and
take $p=2$, $\tilde r=2n+2-\alpha+\delta$) since
\begin{align*}
1> n (n+2-\alpha+\delta) \cdot
\left( \frac 2 {2n+2-\alpha+\delta} \right)^2.
\end{align*}
\end{proof}
With the help of Lemma \ref{lem_fullrange_00a} one can then complete
the blow-up proof for the full range of the
generalized surface quasi-geostrophic model considered in \cite{LRcmp09, Dong14, DL08},
we omit further details.
\section{Another short proof for Hilbert}
\begin{lem} \label{lem4.1}
Let $g: \mathbb R \to \mathbb R$ be an even Schwartz function. Then
\begin{align*}
-\int_{\mathbb R} \frac{ g^{\prime}(x) (Hg)(x) } {x} dx
\ge \frac 1 {\pi} \int_{\mathbb R} \frac{(g(0)-g(x))^2} {x^2} dx.
\end{align*}
\end{lem}
\begin{proof}
By taking advantage of the even symmetry, we have
\begin{align*}
\operatorname{LHS}& = \frac 2 {\pi} \int_0^{\infty} g^{\prime}(x)
( \int_0^{\infty} \frac { g(x)-g(y)} { x^2-y^2} dy) dx \notag \\
& = \frac 1 {\pi} \lim_{\epsilon \to 0} \int_0^{\infty} ( \int_0^{\infty} \frac {1} {x^2-y^2 +i\epsilon} \frac d{dx}
( (g(x)-g(y))^2 ) dx ) dy \notag \\
&= \frac 1 {\pi} \int_0^{\infty} \frac{(g(0)-g(y))^2} {y^2} dy+
\frac 2 {\pi} \int_0^{\infty} \int_0^{\infty}
\frac{(g(x)-g(y))^2} {(x^2-y^2)^2} \cdot x dx dy \notag \\
& = \frac 1 {\pi} \int_0^{\infty} \frac{(g(0)-g(y))^2} {y^2} dy + \frac 1 {\pi}
\int_0^{\infty} \int_0^{\infty} \frac{(g(x)-g(y))^2}{(x-y)^2(x+y)} dx dy.
\end{align*}
\end{proof}
\begin{rem}
The constant $1/\pi$ is certainly not sharp since
\begin{align*}
\int_0^{\infty} \int_0^{\infty} \frac{(g(x)-g(y))^2}{(x-y)^2(x+y)} dx dy
& \ge 2\int_{x > y} \frac{(g(x)-g(y))^2} { (x^2-y^2) (x-y) } dx dy \notag \\
& \ge 2 \int_{x> y} \frac{(g(x)-g(y))^2} {x^3} dx dy \notag \\
& \ge (3-2\sqrt 2) \int_0^{\infty} \frac{(g(0)-g(x))^2}{x^2} dx.
\end{align*}
\end{rem}
Lemma \ref{lem4.1} is not directly useful for establishing blow-ups since it involves a
non-integrable weight $1/x$. The next lemma fixes this issue.
\begin{lem} \label{lem4.3}
Let $g: \mathbb R \to \mathbb R$ be an even Schwartz function. Then
\begin{align*}
-\int_{\mathbb R} { g^{\prime}(x) (Hg)(x) } \frac{e^{-x}} xdx
\ge \frac 1 {2\pi} \int_{\mathbb R} \frac{(g(0)-g(x))^2} {x^2} dx-1000 \| g\|_{\infty}^2.
\end{align*}
\end{lem}
\begin{proof}
By using the same integration by parts argument as in Lemma \ref{lem4.1}, we get
\begin{align*}
\operatorname{LHS} &= \frac 1 {\pi} \int_0^{\infty}
\frac{(g(0)-g(y))^2}{y^2} dy+\frac 2 {\pi}
\int_0^{\infty} \int_0^{\infty}
\frac{(g(x)-g(y))^2} {(x^2-y^2)^2} \cdot x e^{-x} dx dy \notag \\
& \quad +\frac 1 {2\pi} \int_0^{\infty} \int_0^{\infty}
\frac{(g(x)-g(y))^2} {x^2-y^2} (e^{-x}-e^{-y}) dx dy. \notag
\end{align*}
Now note
\begin{align*}
\int_0^{\infty} \int_{\frac x 2 <y<2x}
\Bigl|\frac{(g(x)-g(y))^2}{x^2-y^2}\cdot (e^{-x}-e^{-y}) \Bigr| dx dy
\le 100 \|g\|_{\infty}^2.
\end{align*}
Also
\begin{align*}
&\int_0^{\infty} \int_{y\le \frac x2} \frac{(g(x)-g(y))^2} {|x^2-y^2|}
\cdot |e^{-x}-e^{-y}| dx dy \notag \\
\le & \;
\frac 8 3 \int_0^{\infty}
\int_{y\le \frac x 2} \frac{(g(0)-g(x))^2+(g(0)-g(y))^2}{x^2} e^{-y} dx dy \notag \\
\le & \; \frac 83
\int_0^{\infty} \frac{(g(0)-g(x))^2} {x^2} (1-e^{-\frac x2}) dx
+\frac 43 \int_0^{\infty} \frac{(g(0)-g(y))^2} {y} e^{-y} dy \notag \\
\le & \; \frac 12\int_0^{\infty} \frac {(g(0)-g(x))^2} {x^2}dx+300 \| g\|_{\infty}^2.
\end{align*}
The piece $y\ge 2x$ is estimated similarly.
\end{proof}
To handle the diffusion term, we need the following auxiliary lemma.
\begin{lem} \label{lem4.4}
Let $0<\gamma<1$. Let $g:\, \mathbb R \to \mathbb R$ be an even Schwartz function.
Then
\begin{align*}
| \int_0^{\infty} \frac{\Lambda^{\gamma} g(0) -\Lambda^{\gamma} g(x)}
{x} e^{-x} dx | \le C_{\gamma} \int_0^{\infty} \frac{|g(0)-g(x)|}{x^{1+\gamma}}
\log(10+\frac 1 x)dx,
\end{align*}
where $C_{\gamma}>0$ is a constant depending only on $\gamma$.
\end{lem}
\begin{proof}
By using parity, we have
\begin{align*}
(\Lambda^{\gamma} g)(x) = C_{\gamma}^{(1)}
\int_0^{\infty} \frac{2 g(x) -g (x-y) -g (x+y)} {y^{1+\gamma}} dy,
\end{align*}
where $C_{\gamma}^{(1)}>0$ is a constant depending only on $\gamma$.
Now
\begin{align*}
&\int_0^{\infty} \frac{\Lambda^{\gamma} g(0) -\Lambda^{\gamma} g(x)}
{x} e^{-x} dx \notag \\
= &C_{\gamma}^{(1)}
\int_{0<x<\infty, 0<y<\infty}
\frac{ 2(g(0)-g(y)) + g(x-y)+g(x+y)- 2g(x)} {xy^{1+\gamma}} e^{-x} dx dy \notag \\
= & C_{\gamma}^{(1)}
\int_{0<x<\infty, 0<y<\infty}
\underbrace{\frac{ -2\tilde g(y) + \tilde g(x-y) + \tilde g(x+y) - 2 \tilde g(x) }
{ x y^{1+\gamma}} e^{-x} }_{=:H_1} dx dy,
\end{align*}
where for simplicity we have denoted $\tilde g(x) := g(x)-g(0)$.
Case 1: $\frac 1 {10} \le \frac x y \le 10$. Clearly
\begin{align*}
\int_{\substack{\frac 1 {10} \le \frac x y \le 10\\ 0<x<\infty} } |H_1| dx dy \lesssim
\int_0^{\infty} \frac{ |\tilde g(x) |} {x^{1+\gamma}} dx.
\end{align*}
Case 2: $y\ge 10x$. Obviously
\begin{align*}
\int_{\substack{ y\ge 10 x \\ 0<x<\infty}}
\frac{|\tilde g(x)|} {x y^{1+\gamma}} e^{-x } dx dy
\lesssim \int_0^{\infty} \frac{ |\tilde g(x)|} {x^{1+\gamma}} dx.
\end{align*}
On the other hand,
\begin{align*}
& |\int_{\substack{y\ge 10 x \\ 0<x<\infty}}
\frac{2\tilde g(y)- \tilde g (y-x) -\tilde g(y+x)} {xy^{1+\gamma}} e^{-x} dx dy | \notag \\
=& |\int_{0<x,y<\infty}
\frac{\tilde g(y) e^{-x}} {x}
\cdot ( \frac 2 {y^{1+\gamma}}
\cdot 1_{y\ge 10x}
-\frac 1 {(y+x)^{1+\gamma}} \cdot 1_{y\ge 9x}
-\frac 1 {(y-x)^{1+\gamma}} \cdot 1_{y\ge 11x} ) dx dy| \notag \\
\lesssim & \int_0^{\infty} \frac{|\tilde g(y)|} {y^{1+\gamma}} dy.
\end{align*}
Case 3: $x \ge 10 y$. First
\begin{align*}
\int_{\substack{ x\ge 10y \\ 0<y<\infty}}
\frac{|\tilde g(y)|} {xy^{1+\gamma}} e^{-x} dx dy
\lesssim \int_0^{\infty}
\frac{|\tilde g(y)|} {y^{1+\gamma}}
\cdot \log(10+\frac 1 y) dy.
\end{align*}
On the other hand,
\begin{align*}
& | \int_{\substack{x \ge 10 y\\ 0<y<\infty}}
\frac{ 2\tilde g (x) -\tilde g(x+y) -\tilde g (x-y)} { x y^{1+\gamma}} e^{-x} dx d y| \notag \\
= & | \int_{0<x,y<\infty}
\frac {\tilde g(x)} {y^{1+\gamma}}
\cdot ( \frac 2 x e^{-x}\cdot 1_{x \ge 10y} -
\frac{e^{-(x-y)}}{x-y} \cdot 1_{x\ge 11y} - \frac{e^{-(x+y)}}{x+y} \cdot 1_{x\ge 9y}) dx dy
| \notag \\
\lesssim & \int_0^{\infty} \frac{|\tilde g(x)|} {x^{1+\gamma}} dx.
\end{align*}
\end{proof}
Lemma \ref{lem4.3} and \ref{lem4.4} can be used to establish blow up. Consider
\begin{align*}
\begin{cases}
\partial_t \theta - H\theta \theta_x =-\Lambda^{\gamma} \theta, \\
\theta(0,x)=\theta_0(x).
\end{cases}
\end{align*}
\begin{thm}
Let $0<\gamma <\frac 12$. Let the initial data $\theta_0$ be an even Schwartz function.
There exists a constant $A_{\gamma}>0$ depending only on $\gamma$
such that if
\begin{align*}
\int_0^{\infty} \frac{\theta_0(0) - \theta_0(x) } x e^{-x} dx \ge A_{\gamma}
\cdot (\| \theta_0\|_{\infty} +1),
\end{align*}
then the corresponding solution blows up in finite time.
\end{thm}
\begin{proof}
By using Lemma \ref{lem4.3} and \ref{lem4.4}, we compute
\begin{align*}
&\frac d {dt} \int_0^{\infty}
\frac{\theta(t,0)-\theta(t,x)} x e^{-x} dx \notag \\
\ge & \frac 1 {2\pi}
\int_0^{\infty} \frac{(\theta(t,0) - \theta(t,x))^2} {x^2} dx - 1000 \| \theta\|_{\infty}^2
\notag \\
& \quad
- C_{\gamma} \int_0^{\infty}
\frac{|\theta(t,0) -\theta(t,x)|} {x^{1+\gamma}} \log (10+\frac 1 x) dx.
\end{align*}
By Cauchy-Schwartz, it is clear that
\begin{align*}
2 ( \int_0^{\infty} \frac{\theta(t,0)-\theta(t,x)} {x} e^{-x} dx )^2
\le \int_0^{\infty} \frac{(\theta(t,0)-\theta(t,x))^2} {x^2} dx.
\end{align*}
Also by using Cauchy-Schwartz, we have
\begin{align*}
&\int_0^1 \frac{|\theta(t,0)-\theta(t,x)|} {x^{1+\gamma}} \log(10+\frac 1 x) dx \notag \\
\le & (\int_0^1 \frac{(\theta(t,0)-\theta(t,x))^2} {x^2} e^{-x} dx )^{\frac 12}
\cdot ( \int_0^1 \frac {e^x} {x^{2\gamma}} ( \log(10+\frac 1x))^2 dx )^{\frac 12 } \notag \\
\le & C_1 \cdot (\int_0^1 \frac{(\theta(t,0)-\theta(t,x))^2} {x^2} e^{-x} dx )^{\frac 12},
\end{align*}
where $C_1>0$ depends only on $\gamma$. Note that here we used the crucial assumption
$0<\gamma<\frac 12$ for the integral to converge. It then follows easily that
\begin{align*}
&\frac d {dt} \int_0^{\infty}
\frac{\theta(t,0)-\theta(t,x)} x e^{-x} dx \notag \\
\ge & \frac 1 {2\pi} ( \int_0^{\infty} \frac{\theta(t,0) -\theta(t,x)} {x} e^{-x} dx )^2
- C_2 \cdot ( \| \theta_0\|_{\infty}+1)^2,
\end{align*}
where $C_2>0$ depends only on $\gamma$. Choosing $A_{\gamma}=\sqrt{2\pi C_2}$
then yields the result.
\end{proof}
\section{The $\alpha$-case}
Remarkably the computation in section 4 can also be generalized to the case with
drift term $\Lambda^{-\alpha} \partial_x \theta$. We shall employ the same weight $1/x$.
\begin{lem} \label{lem5.1aaa}
Let $0<\alpha<1$. Let $g:\mathbb R \to \mathbb R$ be an even Schwartz function. Then
\begin{align*}
\int_0^{\infty} \frac{ \Lambda^{-\alpha} g^{\prime}(x) \cdot g^{\prime}(x)} x
dx \ge C_{\alpha}
\cdot \int_0^{\infty} \frac {(g(0)-g(x))^2} {x^{3-\alpha}} dx,
\end{align*}
where $C_{\alpha}>0$ depends only on $\alpha$. Similarly for $1\le \alpha <2$,
by writing $\Lambda^{-\alpha} \partial_x =- \Lambda^{-(\alpha-1)} \mathcal H$, we have
\begin{align*}
-\int_0^{\infty} \frac{ \Lambda^{-(\alpha-1)} \mathcal Hg(x) \cdot g^{\prime}(x)} x
dx \ge C_{\alpha}
\cdot \int_0^{\infty} \frac {(g(0)-g(x))^2} {x^{3-\alpha}} dx.
\end{align*}
\end{lem}
\begin{rem}
The case $\alpha=1$ corresponds to $\Lambda^{-1} \partial_x = -\mathcal H$ which is the Hilbert
transform case which we have treated before.
\end{rem}
\begin{proof}
We first discuss the case $0<\alpha<1$.
By using parity, we have
\begin{align*}
(\Lambda^{-\alpha} g^{\prime})(x)
& = C_{\alpha} \int_0^{\infty}
( \frac 1 {|x-y|^{1-\alpha}} - \frac 1 {|x+y|^{1-\alpha}} )\cdot
\frac d {dy} ( g(y) -g(x) ) dy \notag \\
& = C_{\alpha} \cdot (-1) \cdot
\int_0^{\infty}
\frac d {dy} (
\frac 1 {|x-y|^{1-\alpha}} - \frac 1 {(x+y)^{1-\alpha}} ) (g(y) -g (x) ) dy \notag \\
& = C_{\alpha} \int_0^{\infty} h(x,y) ( g(y) -g (x) ) dy,
\end{align*}
where
\begin{align*}
h(x,y) = \frac d {dx} ( \frac 1 {|x-y|^{1-\alpha}} + \frac 1 {(x+y)^{1-\alpha}}).
\end{align*}
Now we write
\begin{align*}
&2 \int_0^{\infty} \int_0^{\infty}
\frac { h(x,y) (g(y) -g (x) ) g^{\prime}(x) } x dx dy \notag \\
=& -\int_0^{\infty} \int_0^{\infty}
\frac {h(x,y)} x \cdot \frac d {dx} ( (g(x) -g (y) )^2 ) dx dy \notag \\
= &- \int_0^{\infty} ( \frac {h(x,y)} x ( g(x) -g(y) )^2 \Bigr|_{x=0}^{\infty} ) dy
+\int_0^{\infty} \int_0^{\infty} \frac d {dx} ( \frac {h(x,y)} x ) \cdot (g(x)-g(y))^2 dx dy.
\end{align*}
It is easy to check that for some positive constant $C_1>0$ (below $y>0$),
\begin{align*}
(\partial_x h)(0,y)= \frac {d^2} {dx^2} ( \frac 1 {|x-y|^{1-\alpha}} +
\frac 1 {(x+y)^{1-\alpha} } ) \Bigr|_{x=0} = C_1\cdot y^{-(3-\alpha)}.
\end{align*}
Thus
\begin{align*}
& - \int_0^{\infty} ( \frac {h(x,y)} x ( g(x) -g(y) )^2 \Bigr|_{x=0}^{\infty} ) dy \notag \\
= & \int_0^{\infty} (\partial_x h)(0,y) (g(0)-g(y))^2 dy
\gtrsim \int_0^{\infty} \frac { (g(0)-g(y))^2} {y^{3-\alpha}} dy.
\end{align*}
It remains for us to check that, for all $0<x,y<\infty$, $x\ne y$,
\begin{align*}
\frac d {dx} (\frac {h(x,y)} x) = \frac d {dx} ( \frac 1 x \frac d {dx}
(\frac 1 {|x-y|^{1-\alpha}} + \frac 1 {(x+y)^{1-\alpha} } ) ) \ge 0.
\end{align*}
By scaling, it suffices prove for all $0<x<\infty$, $x \ne 1$,
\begin{align*}
\frac d {dx} ( \frac 1 x \frac d {dx}
(\frac 1 {|x-1|^{1-\alpha}} + \frac 1 {(x+1)^{1-\alpha} } ) ) \ge 0.
\end{align*}
We now make a change of variable $x=\sqrt {t}$. Then we only need to prove
\begin{align*}
\frac {d^2} {dt^2} ( \frac {1} { |\sqrt t -1|^{1-\alpha} } +
\frac 1 { (\sqrt t +1)^{1-\alpha} }) \ge 0, \quad \forall\, 0<t<\infty,\, t\ne 1.
\end{align*}
For $1<t<\infty$, one can get positivity by direct differentiation. For $0<t<1$, one
can use the fact that the function
\begin{align*}
f(s) = (1-s)^{-(1-\alpha)} + (1+s)^{-(1-\alpha)}
\end{align*}
has a non-negative binomial expansion for $0<s<1$.
We now turn to the case $1\le\alpha<2$. The case $\alpha=1$ is already treated before in
Section 4 so we assume $1<\alpha<2$. Set $\epsilon =\alpha-1 \in (0,1)$. Then it is
not difficult to check that
\begin{align*}
- (\Lambda^{-\epsilon} \mathcal H g)(x) = C_{\alpha} \int_0^{\infty}
h(x,y) (g(y) -g(x)) dy,
\end{align*}
where
\begin{align*}
h(x,y) & = - ( \frac{|x-y|^{\epsilon}} {x-y} + \frac 1 {(x+y)^{1-\epsilon}} ) \notag \\
& = -\frac 1 {\epsilon} \frac d {dx} ( |x-y|^{\epsilon} +(x+y)^{\epsilon}).
\end{align*}
Clearly
\begin{align*}
\partial_x h(0, y) = \operatorname{const} \cdot y^{-(2-\epsilon)}.
\end{align*}
It then suffices to check for all $0<t<\infty$, $t\ne 1$,
\begin{align*}
- \frac {d^2} {dt^2} ( |\sqrt t-1|^{\epsilon} + |\sqrt t +1|^{\epsilon} ) \ge 0.
\end{align*}
Again for $t>1$ the inequality follows easily from direct differentiation. For $0<t<1$,
one just observe that for $0<s<1$, the binomial coefficients in the expansion of
\begin{align*}
f(s) = (1+s)^{\epsilon} + (1-s)^{\epsilon} = C_0+ \sum_{k\ge 0} C_k s^{2k}
\end{align*}
satisfies $C_k<0$ for all $k\ge 1$.
\end{proof}
\begin{lem} \label{lem5.3aaa}
Let $g: \mathbb R \to \mathbb R$ be an even Schwartz function.
If $0<\alpha<1$.
Then
\begin{align*}
\int_0^{\infty} \frac{ \Lambda^{-\alpha} g^{\prime}(x) \cdot g^{\prime}(x)} x
e^{-x}
dx \ge C_{\alpha}^{(1)}
\cdot \int_0^{\infty} \frac {(g(0)-g(x))^2} {x^{3-\alpha}} dx- C_{\alpha}^{(2)}\| g\|_{\infty}^2,
\end{align*}
where $C_{\alpha}^{(1)}>0$, $C_{\alpha}^{(2)}>0$ are constants depending only on $\alpha$.
Similarly for $1\le \alpha <2$, by writing
$\Lambda^{-\alpha} \partial_x =- \Lambda^{-(\alpha-1)} \mathcal H$, we have
\begin{align*}
-\int_0^{\infty} \frac{ \Lambda^{-(\alpha-1)} \mathcal Hg(x) \cdot g^{\prime}(x)} x e^{-x}
dx \ge C_{\alpha}^{(3)}
\cdot \int_0^{\infty} \frac {(g(0)-g(x))^2} {x^{3-\alpha}} dx
-C_{\alpha}^{(4)}\| g\|_{\infty}^2,
\end{align*}
where $C_{\alpha}^{(3)}>0$, $C_{\alpha}^{(4)}>0$ depend only on $\alpha$.
\end{lem}
\begin{proof}
We only need to modify the proof of Lemma \ref{lem5.1aaa}. Consider first the case
$0<\alpha<1$.
Recall that for $x,y>0$, $x\ne y$,
\begin{align*}
h(x,y) & = \frac d{ dx}
\Bigl( \frac 1 {|x-y|^{1-\alpha}} + \frac1 { (x+y)^{1-\alpha} } \Bigr) \notag \\
& = -(1-\alpha) \cdot
\Bigl( |x-y|^{-(2-\alpha)} \operatorname{sgn}(x-y)
+ (x+y)^{-(2-\alpha)} \Bigr).
\end{align*}
It is not difficult to check that (below $c_i>0$ are positive constants):
\begin{align*}
& c_1 \int_0^{\infty} \Lambda^{-\alpha} g^{\prime}(x) \cdot g^{\prime}(x)
\cdot \frac 1 x e^{-x} dx \notag \\
=&\; -
\int_0^{\infty} \int_0^{\infty}
\frac {h(x,y)} x e^{-x} \frac d {dx} ( (g(x) -g(y))^2) dx dy \notag \\
=&\; c_2
\int_0^{\infty} \frac { (g(0)-g(y))^2} {y^{3-\alpha}} dy
+ \int_0^{\infty} \int_0^{\infty}
\frac d {dx} ( \frac 1 x h(x,y) ) e^{-x}
(g(x)-g(y))^2 dx dy \notag \\
& \quad - \int_0^{\infty} \int_0^{\infty}
h(x,y) \frac 1 x e^{-x} (g(x)-g(y))^2 dx dy.
\end{align*}
By the computation in Lemma \ref{lem5.1aaa}, we have $\frac d {dx} ( \frac1x
h(x,y)) \ge 0$ for any $x,y>0$, $x\ne y$. Thus we only need to estimate the
third term above. Observe that for $x>y$, we have $h(x,y)<0$. Then
\begin{align*}
&- \frac 1 {1-\alpha} \int_0^{\infty} \int_0^{\infty}
h(x,y) \frac 1 x e^{-x} (g(x)-g(y))^2 dx dy \notag \\
=&\;
\int_0^{\infty} \int_0^{\infty}
( \frac {\operatorname{sgn}(x-y)} { |x-y|^{2-\alpha}}
+ \frac 1 {(x+y)^{2-\alpha}} ) \frac1x e^{-x}
( g(x) - g(y) )^2 dx dy \notag \\
\ge & \;
\iint_{\frac x 2 \le y \le 2x}
\Bigl( \frac {\operatorname{sgn}(x-y)} { |x-y|^{2-\alpha}}
+ \frac 1 {(x+y)^{2-\alpha}} \Bigr) \frac1x e^{-x}
( g(x) - g(y) )^2 dx dy \notag \\
& \quad + \iint_{x<\frac 12 y}
\Bigl( \frac {\operatorname{sgn}(x-y)} { |x-y|^{2-\alpha}}
+ \frac 1 {(x+y)^{2-\alpha}} \Bigr) \frac1x e^{-x}
( g(x) - g(y) )^2 dx dy \notag \\
\ge & \;
\iint_{\frac x 2 \le y \le 2x}
\frac {\operatorname{sgn}(x-y)} { |x-y|^{2-\alpha}}
\cdot \frac1x e^{-x}
( g(x) - g(y) )^2 dx dy \notag \\
& \quad + \iint_{x<\frac 12 y}
\Bigl( \frac {\operatorname{sgn}(x-y)} { |x-y|^{2-\alpha}}
+ \frac 1 {(x+y)^{2-\alpha}} \Bigr) \frac1x e^{-x}
( g(x) - g(y) )^2 dx dy \notag \\
\ge &\; -\frac 12 \iint_{\frac x 2 \le y \le 2x}
\frac { 1} {|x-y|^{2-\alpha}}
|\frac 1 x e^{-x} -\frac 1 y e^{-y} | (g(x)-g(y))^2 dx dy \notag \\
& \qquad - d_1 \iint_{ x <\frac 12 y}
\frac 1 {y^{3-\alpha}} e^{-x} (g(x)-g(y))^2 dx dy,
\end{align*}
where $d_1>0$ is a constant depending only on $\alpha$. Now observe that
for $x,y>0$ with $x\ne y$ and $\frac x 2 \le y \le 2x$, we have
\begin{align*}
\frac { 1} {|x-y|^{2-\alpha}}
|\frac 1 x e^{-x} -\frac 1 y e^{-y} | \lesssim
\frac 1 {|x-y|^{1-\alpha}} e^{-\frac 1 {10} |x|} (1+ x^{-2}).
\end{align*}
The desired result then follows from the following string of inequalities:
\begin{align*}
& \iint_{\frac x 2 \le y \le 2x}
|x-y|^{-(1-\alpha)} e^{-\frac 1 {10} |x|}
(g(x)-g(y))^2 dx dy \lesssim \|g\|_{\infty}^2;\\
& \iint_{\frac x 2 \le y \le 2x}
|x-y|^{-(1-\alpha)} |x|^{-2} e^{-\frac 1 {10} |x|}
(g(x)-g(y))^2 dx dy \lesssim
\; \int_0^{\infty} \frac {(g(x)-g(0))^2}{x^{2-\alpha}} dx;\\
&\iint_{ x <\frac 12 y}
\frac 1 {y^{3-\alpha}} e^{-x} (g(x)-g(y))^2 dx dy
\lesssim
\; \int_0^{\infty} \frac {(g(x)-g(0))^2}{x^{2-\alpha}} dx;\\
&\int_0^{\infty}\frac {(g(x)-g(0))^2}{x^{2-\alpha}} dx
\le \eta \int_0^{\infty}\frac {(g(x)-g(0))^2}{x^{3-\alpha}} dx
+C_{\eta,\alpha} \| g\|_{\infty}^2,
\end{align*}
where $\eta>0$ is any small constant, and $C_{\eta,\alpha}$ depends only
on ($\eta$, $\alpha$).
The above concludes the proof for the case $0<\alpha<1$.
The case for $1\le \alpha<2$ is similar. In that case one only needs to work with
$h(x,y)$ given by (up to an unessential positive constant)
\begin{align*}
h(x,y) = \frac{|x-y|^{\epsilon}} {x-y} + \frac 1 {(x+y)^{1-\epsilon} },
\end{align*}
where $\epsilon=\alpha-1 \in [0,1)$. In the symmetric region
$\frac x2 \le y \le 2x$, one uses the inequality
\begin{align*}
\frac {|x-y|^{\epsilon}}{x-y}
\cdot \left| \frac 1x e^{-x} - \frac 1 y e^{-y} \right|
\lesssim |x-y|^{\epsilon} e^{-\frac 1 {10} x} (1+x^{-2}), \quad\forall\, x\ne y.
\end{align*}
In the region $0<x<\frac 12 y$, one can use the bound
\begin{align*}
\left |\frac {|x-y|^{\epsilon}}{x-y} + \frac 1 {(x+y)^{1-\epsilon}} \right|
\lesssim y^{-(2-\epsilon)} \cdot x.
\end{align*}
We omit further details.
\end{proof}
Lemma \ref{lem5.3aaa} can be used to establish blow up. For simplicity, consider
for $0<\alpha<1$, the model
\begin{align*}
\begin{cases}
\partial_t \theta +(\Lambda^{-\alpha} \partial_x \theta) \cdot \partial_x \theta=0,\\ \theta(0,x)=\theta_0(x);
\end{cases}
\end{align*}
and for $1\le \alpha<2$, the model
\begin{align*}
\begin{cases}
\partial_t \theta - (\Lambda^{-(\alpha-1) } \mathcal H \theta) \cdot \partial_x \theta=0,\\ \theta(0,x)=\theta_0(x);
\end{cases}
\end{align*}
One should check that in both cases, the symbol of the operator for the drift
term is given by $ i |\xi|^{-\alpha} \xi$ for all $0<\alpha<2$. Alternatively, one
may write both models as a single equation
\begin{align*}
\partial_t \theta - \Lambda^s \mathcal H \theta \cdot \partial_x \theta =0,
\end{align*}
where $-1<s<1$. The drift term has the symbol $i |\xi|^s \operatorname{sgn}(\xi)$ so that
$s$ can be identified as $1-\alpha$.
Concerning both models, we have the following result.
\begin{thm} \label{thm_alpha_001}
Let $0<\alpha<2$. Let the initial data $\theta_0$ be an even Schwartz function.
There exists a constant $A_{\alpha}>0$ depending only on $\alpha$
such that if
\begin{align*}
\int_0^{\infty} \frac{\theta_0(0) - \theta_0(x) } {x} e^{-x} dx \ge A_{\alpha}
\cdot (\| \theta_0\|_{\infty} +1),
\end{align*}
then the corresponding solution blows up in finite time.
\end{thm}
\begin{rem*}
One can also consider the model with suitable dissipation term on the right
hand side. For simplicity we do not state such results here which can be obtained
by using similar estimates as in the previous section.
\end{rem*}
\begin{proof}
This follows from Lemma \ref{lem5.3aaa}. One only needs to use the simple
inequality (with respect to the measure $e^{-x} dx$ on $(0,\infty)$) which
holds for any $0<\alpha<2$:
\begin{align*}
\int_0^{\infty}
\frac {|g(x)|} {x} e^{-x} dx
\le ( \int_0^{\infty}
\frac{ g(x)^2} {x^{3-\alpha}} e^{-x} dx)^{\frac 12}
\cdot ( \int_0^{\infty} \frac 1 {x^{\alpha-1}} e^{-x} dx )^{\frac 12}.
\end{align*}
\end{proof}
\begin{rem}
Strictly speaking, the proof of Theorem \ref{thm_alpha_001} assumed the local
wellposedness (of smooth solutions) for the generalized model. While the focus of this work is to prove nonlinear Hilbert type inequalities (for showing finite time singularity), for the sake of completeness we sketch the proof of local wellposedness here in this remark.
Consider the nontrivial case with hyper-singular velocity as follows:
\begin{align*}
\partial_t \theta - (\Lambda^s \mathcal H \theta )\partial_x \theta =0,
\end{align*}
where $0<s<1$ (the case $-1<s\le 0$ is easier). First we present formal energy estimates.
For the basic $L^2$ estimate, we have
\begin{align*}
\frac 12 \frac d {dt} ( \| \theta \|_2^2)
\le \frac 12 \| \Lambda^{s+1} \theta \|_{\infty} \| \theta \|_2^2.
\end{align*}
Next take an integer $m>s+\frac 32$, and compute
\begin{align}
\frac 12 \frac d {dt}
( \| \partial_x^m \theta \|_2^2)
&=\int \partial_x^m ( \Lambda^s \mathcal H \theta \partial_x \theta) \partial_x^m \theta dx\notag\\
& = \int (\partial_x^m \Lambda^s \mathcal H \theta) \partial_x \theta \partial_x^m \theta
dx \label{rem_last_e001a} \\
&\qquad+ \int \Lambda^s \mathcal H \theta \partial_x^{m+1} \theta \partial_x^m \theta dx
\label{rem_last_e001b} \\
&\qquad + \text{other terms}. \notag
\end{align}
It is not difficult to check that (one may take $m\ge 3$ for simplicity, but this can be sharpened)
\begin{align*}
|\text{other terms} | \lesssim \| \theta\|_{H^m}^3.
\end{align*}
For \eqref{rem_last_e001b} one can do integration by parts and obtain
\begin{align*}
| \eqref{rem_last_e001b}| \lesssim \| \Lambda^{s+1} \theta\|_{\infty}
\| \partial_x^m \theta \|_2^2 \lesssim \| \theta \|_{H^m}^3.
\end{align*}
To handle \eqref{rem_last_e001a}, we can use Lemma \ref{lemforlast_00} which gives
\begin{align*}
\| \Lambda^s \mathcal H( \partial_x^m \theta \partial_x \theta)
- (\Lambda^s \mathcal H \partial_x^m \theta) \partial_x \theta
\|_2 \lesssim \| \partial_x^m \theta \|_2 \| \partial_x (1-\partial_{xx})^{\frac 34} \theta
\|_2 \lesssim \| \theta\|_{H^m}^2.
\end{align*}
Thanks to the skew-symmetry of the Hilbert transform operator, we have
\begin{align*}
\int \Lambda^s \mathcal H( \partial_x^m \theta \partial_x \theta) \partial_x^m \theta
dx = - \int \partial_x^m \theta \partial_x \theta \Lambda^s \mathcal H \partial_x^m
\theta dx.
\end{align*}
We can then rewrite the original term as a commutator and obtain
\begin{align*}
| \eqref{rem_last_e001a} | \lesssim \| \theta\|_{H^m}^3.
\end{align*}
Thus we have completed the formal energy estimate in $H^m$. We should point it out
that by using the theory in \cite{DL19} one can obtain sharp energy estimate
in $H^r$ with $r>s+\frac 32$. However we shall not dwell on this issue here.
Finally it is worthwhile pointing it out that in order to make the above formal energy
estimates rigorous, one needs to work with the regularized system
\begin{align*}
\partial_t \theta - J_{\epsilon} ( \Lambda^s \mathcal H J_{\epsilon} \theta \partial_x
J_{\epsilon} \theta)=0,
\end{align*}
where $J_{\epsilon}$ is the usual mollifier. We leave the interested reader to check
the details.
\end{rem}
\begin{lem} \label{lemforlast_00}
Let $0<s<1$.
For any $f$, $g\in \mathcal S(\mathbb R)$, we have
\begin{align*}
\| \Lambda^s \mathcal H (fg) - (\Lambda^s \mathcal H f) g \|_2 \lesssim
\| f \|_2 \| |\xi|^s \hat g (\xi) \|_{L_{\xi}^1} \lesssim
\| f\|_2 \| (1-\partial_{xx})^{\frac 34} g \|_2.
\end{align*}
\end{lem}
\begin{rem}
Of course much better results are available. For example by using the commutator
estimate in \cite{DL19} (see Corollary 1.4 and the second remark on page 26 therein), one
can even show for any $1<p<\infty$,
\begin{align*}
\| \Lambda^s \mathcal H (fg) - (\Lambda^s \mathcal H f) g \|_p \lesssim
\| f\|_p \| \Lambda^s g \|_{\operatorname{BMO} }.
\end{align*}
However for simplicity of presentation (and for the sake of completeness), we present
the non-sharp version here.
\end{rem}
\begin{proof}[Proof of Lemma \ref{lemforlast_00}]
Since we are in $L^2$ it is convenient to work purely on the Fourier side. One can
write
\begin{align*}
\mathcal F ( \Lambda^s \mathcal H (fg) - (\Lambda^s \mathcal H f) g ) (\xi)
& = -i \cdot \frac 1 {2\pi} \int
( |\xi|^s (\operatorname{sgn}(\xi) ) - |\eta|^s (\operatorname{sgn}(\eta) ) )
\widehat f(\eta) \widehat g(\xi-\eta) d\eta.
\end{align*}
It is easy to check that (since $0<s<1$)
\begin{align*}
| |\xi|^s (\operatorname{sgn}(\xi) ) - |\eta|^s (\operatorname{sgn}(\eta) )|
\lesssim | \xi -\eta|^s.
\end{align*}
The result then easily follows from Young's inequality.
\end{proof} | 0.012689 |
TITLE: Showing that a group is simple.
QUESTION [1 upvotes]: Given a non-commutative group of order $\mathrm{ord}(G)=343$. Prove that $G$ is a simple group.
So I have to show that the only normal subgroups are the trivial group and the group itself. But I don't know how to use the order of the group to show it. The non-abelian part is also not helping really.
REPLY [3 votes]: In fact $G$ is not simple! This holds in general. Because $|G|=p^n$ where $p$ is a prime and $n>1$ we have that $Z(G)\neq 1$. If $G$ is not abelian then $Z(G)\neq G$ and thus $G$ is not simple because $Z(G)$ is normal in $G$.
In case now $G$ is abelian. Then $G=Z(G)$ and every subgroup of $G$ is normal. Let $g\neq 1$.
If $\langle g \rangle \neq G$ then we are done because we found a non-trivial subgroup of $G$.
If $G=\langle g \rangle$ then $G$ is cyclic and let $H= \langle g^p \rangle$. We have that $1\neq H\lhd G$ because :
If $H=1$ then $g^p=1 \Rightarrow|G|=|\langle g \rangle |=p$ -contradiction.
If $G=H$ then $g\in \langle g^p \rangle \Rightarrow g=g^{kp} \Rightarrow g^{kp-1}=1 \Rightarrow o(g) \mid kp-1 \Rightarrow p \mid kp-1 \Rightarrow p=1$ again a contradiction. | 0.011371 |
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The project of Danish singer, songwriter, multi-instrumentalist, model, and actor Amalie Bruun, Myrkur embraces everything from black metal to traditional Scandinavian folk. Myrkur has released 3 studio albums: M (2015), Mareridt (2017), and Folkesange (2020).
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ya, having too many phones on the same line can cause problems, especially of they are old as they draw too much current from the line. The line should be tested with one new phone connected, but preferably at the connection to the house. On Thu, 18 Nov 2004 12:20:43 +0000, Richard Corden <richards_corden at hotmail.com> wrote: > Liam McDermott <theirishmole at gmail.com> writes: > > > this usually happens when the line is split between two houses. > > > > This will also affect regular dialup, so if dialup is not getting > > above 40k then the line is probably split. > > Someone had mentioned this to me, but in general I found that I never > got less than 50k/s with dial-up. > > > > They don't like to tell you your line is split, so i also reccomend > > saying there is noise on the line. > > It is very annoying that you have to do this. If you're on an enabled > exchange it would be nice to be able to ask for an engineer to > checkout your specific line. In my case I would have been willing to > pay for his time. > > Whats the hourly rate for such an engineer? < EUR 50? In general it > would probably only take him half that to say: "Yes I can fix it? No > you cannot get DSL". > > Cheers, > > Richard > > > > > > > > (ballsed up that reply) > > > > > > On Wed, 17 Nov 2004 22:43:12 +0000, Richard Corden > > <richards_corden at hotmail.com> wrote: > >> > >> Hi, > >> > >> "Barry O'Donovan" <mail at barryodonovan.com> writes: > >> > >> > Hey folks, > >> > >> [...snip...[ > >> > >> > Yes I have gone through thr archives but to no quick avail. If any of > >> > those who have had the same problem and managed to get it resolved > >> > could reply on or off list I'd very much appreciate it. > >> > >> Depending on your desperation level..... > >> > >> We're in Harolds X in Dublin and our line failed the DSL tests. We > >> too got the usual reasons for failing. Next doors line was failing too > >> but when they had a second line put in it passed the test. > >> > >> So we took a chance and got a second line into the house (Approx. EUR > >> 120) and this line passed for DSL. Interestingly the engineer > >> installing the line could not explain to me why my original line was > >> failing given my distance from an enabled exchange. > >> > >> Cheers, > >> > >> Richard > >> > >> -- > >> Richard Corden > >> To reply remove 's' from address > >> > >> > >> > >> -- > >> Irish Linux Users' Group > >> > >> > >> > > -- > > Irish Linux Users' Group > > > > -- > > > Richard Corden > To reply remove 's' from address > > -- > Irish Linux Users' Group > > > | 0.001266 |
As part of National Child Safety Week, which is taking part this week, both parents and children are being advised to take steps to prevent accidents during the morning school rush.
Data suggests that each week roughly 2,000 children are taken to hospital as a result of accidents which have occurred during the morning school rush; and in an attempt to reduce such accidents, parents and children are being reminded of safety tips.
The safety tips which have been issued in conjunction with National Child Safety Week, provide advise for safety within the home – such as making sure hot drinks are not in a child’s reach; on the way to school – including making sure everyone travelling by car is wearing a seatbelt, and those riding bikes to school are wearing helmets; and when arriving at the school – such as paying full attention to the road when crossing.
Although the tips may seem straightforward, as personal injury solicitors we understand and appreciate that during the morning rush, such measures can easily be overlooked.
Our solicitors also understand the consequences of a child being involved in an accident; and we are aware that even when all precautions are taken, accidents can and do still happen.
In instances where an accident has occurred and your child has been injured, if it can be proven that the result of the accident was the fault of another person, you may be eligible to make a personal injury claim – and in such situations, our personal injury solicitors can assist.
We provide honest, impartial and tailored personal injury compensation advise, designed to ensure those seeking legal advice are fully aware of the process of making a claim and their chances of succeeding in a claim.
To find out how we can help you, come and speak to our solicitors today. | 0.99122 |
Mozilla releases Firefox 9.0 update
Date: Tuesday, December 20th, 2011, 06:08
Category: News, Software
On Tuesday, Mozilla.org released version 9.0 of its Firefox web browser. The new version stands as a 30.8 megabyte download (via MacUpdate) and adds the following fixes and.
Firefox 9.0 requires an Intel-based Mac running Mac OS X 10.5 or later to install and run.
So…if you’ve tried the new version and have any feedback to offer, please let us know in the comments.
Even if they launch new version most of the addons are not that good enough to cope with the new version so upgrading i really difficult | 0.000863 |
Nov. 11, 2013
DALLAS (SMU) - For its Nov. 16 meeting with UConn, SMU will be offering a special $15 ticket package that includes a hot dog, drink and admission to the game. The deal can be purchased by calling the SMU Ticket Office at 214-SMU-GAME (768-4263) or online by clicking here and using the promo code "Family". Tickets can be picked up prior to gameday, but food vouchers will be distributed on gameday.
As a tribute to those who serve the community and country, free game.
The Nov. 16 game versus UConn game holds historical significance, as it is a rematch 1989's "Miracle On Mockingbird." When SMU returned to the football field in 1989, critics claimed that it was going to take a miracle for Forrest Gregg and the Mustangs to win a game. SMU started the season with a 35-6 loss to Southwest Conference rival Rice, and two weeks later, the Connecticut Huskies were in town to take on the Mustangs in just their second game since the end of the 1986 season. With five minutes to play in the game, it seemed that SMU was headed for certain defeat, as the Mustangs trailed 30-14. Some SMU fans decided that the game was out of reach and started to head home, but the Mustangs scored 17 points in the final five minutes to win the game, 31-30. That first victory that many predicted might take as many as five years to achieve took just two games and was quickly dubbed the "Miracle on Mockingbird." | 0.006557 |
1-844-369-4747
About Jon
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About Jon
Author
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1-844-369-4747
6/21/2017
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My hope for this book is that anyone who picks it up and reads it will be entertained, enlightened and will see the important reality of planning ahead for the love of their family. Further, that it really is much easier than one might think and it really isn’t a downer to be thoughtfully prepared for our own, ultimate mortality. Heck, we all know we aren’t going to live, forever right? We also don’t know when our last day will be, right? So, if we all agree on those two things, then isn’t being prepared the responsible thing to do? Why? For love, that’s why!
Let’s face it, death isn’t hard on the one who has died, it is hard on those that are still here! We love them and we certainly don’t want to add to their pain. Sadness, grief and loneliness are very hard yet, normal emotions and are part of the healing process. The confusion, stress and frankly anger associated with trying to track down all the departed persons “stuff” from insurance policies, financial records and then close their entire digital life, are not (normal emotions)!
So many families are left to struggle with trying to piece things together following the loss of a loved one, only because the departed never planned and told someone what to do! The consequences of this lack of planning delays the normal grieving and healing process, can be exhaustively time consuming and all too often can divide families. Certainly, not what I want my own passing to do to my family. So, I have planned because I love them and I want things to be as easy as possible when I am no longer here. And, you can too! As I said earlier, it really isn’t difficult to do. So, for the love of family, please do it!
Want a copy of my new book?
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Social Media Mess
7/20/2016
3 Comments
How many times have you stumbled across the Facebook page of a friend who passed away months, even years ago? Social media has transformed our lives – and added a whole new challenge to our deaths.
Not only do simple things like Facebook pose a challenge, but today our lives are increasingly digital. Everything from where we receive our news to how we shop involves a username and password or an account. . .. all only known by you. How are your loved ones supposed to sort this out after you are gone?
Without giving away any secured information, we have the solution.
My Life & Wishes is set up to securely store as little or as much information as you want on the site so in one click, your family members can access your social media sites, find your auto payments, your online passwords and every account you have online.
Sound scary? It is not. You bank online now. You buy things online now. You post online now. These are all secure sites that are equal to the security of My Life & Wishes. By getting your digital house in order on your totally private and personalized account, your family will thank you each and every day for leaving this valuable gift behind. Instead of months of work, they will have instant access to all of your accounts. One can rarely even do that with a paper file box of records.
Think about it: The number of your social media and other online accounts is always growing. Your online account inventory may include some of the following. Be sure to list user names, passwords, and PIN numbers for each of your accounts.
Social: Facebook, Twitter, Instagram, Pinterest, Reddit, Snapchat
Career: LinkedIn, Monster, About, Tumblr, WordPress
Communication: Verizon, AT&T, Viber, Skype
Media: Pandora, Spotify, iTunes, Dropbox, Google Drive, Netflix, Hulu
Necessities: Amazon, Walgreens, eBay, Target, Online Grocers
Bills and Auto payments: Utilities, Mortgage, Auto, Mint, PayPal
Travel: Airline miles, Hotel bonus points, Hotels.com, Hotwire, Orbitz, Uber
Dating Life: eHarmony, Plenty of Fish, Match.com
Entertainment: Groupon, Grubhub, Stubhub, Ticketmaster
You can determine how all of this is handled. That’s the beauty of opening an account at My Life & Wishes. Don’t leave your loved ones with a mess to untangle. Hunting and digging for information is no fun. Get your digital house in order today. Your family will thank you.
3 Comments
Our Story
6/7/2016
1 Comment
I think like most people, my wife, Michelle, and I really hadn’t given much, if any, thought to what would happen if one of us, or another family member passed away. On Labor Day 2013 that all changed for us when Michelle’s father suddenly passed away.
During the days following his passing, we did what we could do just to “get through it.” Things like notifying immediate and extended family and close friends, piecing together an obituary, selecting a funeral home, working with the florist, making arrangements with the church for the service, quickly trying to put together memory boards, and planning the reception following the service.
All these tasks were daunting at best. We hoped the arrangements and decisions that we were making would have been pleasing to Michelle’s father. Completing these tasks on little rest was not easy; after all we were still dealing with the shock of the loss. Little did we know; however, that the challenges were just beginning!
Like many families, one person is usually the keeper of the finances. In our case, Michelle’s father was that person. He paid all the bills, did the banking, and handled all of the household finances. In fact, Michelle’s mother had not paid a bill in nearly 60 years!
Unfortunately, Michelle’s father hadn’t shared all of the important details with anyone in the family. So, none of us knew where all the accounts were located or even what life insurance policies may be in force. We did find a safe deposit key, but had no idea at what financial institution it was held. How were we to track down his pension and retirement assets, much less be able to access the bills he paid online? And the list of questions went on and on.
Sadly, this process of “cleaning things up” took nearly 10 months of digging through desk drawers and old files, making exhaustive phone calls, and searching for documents and information. We found that others we spoke with about our situation were anxious to share their stories with us, and everyone one of them was sadly similar.
Time for a Change
We thought there must be a better way! Isn’t the sadness of the loss enough? Why should those who are left behind have to endure all of this additional hardship in an already stressful time? Michelle and I agreed that we had to come up with a solution to make things easier; a way for people to get organized and to share important information with those they love. So that when they pass, their loved ones will have access to important information and final wishes. In this complex and ever changing world, we can no longer simply hope our families or loved ones can figure things out. We call it “End of Life Etiquette.”
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Our pet-friendly apartment homes are a match for your suburban lifestyle. Our location near Carpenter Park Rec Center and Plano Veterans Memorial Park is ideal – making errands simple and finding great restaurants nearby an easy task.
You’ll also enjoy a straightforward commute to top-notch employers and top-rated schools close by. | 0.07464 |
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Recycled Concrete, Asphalt and Fill Dirt Suppliers. We accept concrete, reinforced concrete, and asphalt. All you pay is $10.00 per ton for clean concrete or $12.00-$15.00 per ton for reinforced concrete. Please note there is a $20 minimum charge for allLearn More
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Asphalt rubble free of concrete, dirt, wood, sod, and any other trash material. Ten percent road base is acceptable. ... CLEAN means no wire mesh, rebar, metal or brick. Any load that is represented to be CLEAN and free of wire or rebar and is NOT will | 0.320483 |
[ Thread Index | Date Index | More lists.tuxfamily.org/sawfish Archives ]. It could be also something in how this xterm is really started. 'man xterm' describes options and resources which influence how borders of xterm window are supposed to look and possibly something there overrides what sawfish is trying to do. Michal --- -- Sawfish ML | 0.002744 |
Parents are responsible for the support, care, and well-being of their children. When parents separate or divorce, the court decides whether one or both parents has custody of the child. Child custody may determine which parent pays child support. If one parent has custody, the other parent usually pays child support. A judge determines child support using the state child support guidelines, which considers the income of both parents. Therefore, a parent who shares custody may still have to pay the other parent.
A parent who seeks child support can hire a private attorney, include the issue in your divorce case, or file a case through the state child support services. To start the process, both parents must provide information about themselves including, current addresses, employment, and finances.
Before deciding child support, the court must identify the legal mother and father of the child to determine child custody. Therefore, if there is any doubt about who the father of the child is, the parents and child may take a paternity test. Then, a court order names the legal father based on the test results.
You are in violation of a court order when you do not pay your child support. The court has options to make sure it receives your overdue payments:
Other consequences you may face for failing to pay your support payments include:
Although Idaho child support services can determine the actions to take for unpaid child support, there are certain punishments that automatically go into effect the moment you do not pay on time. A judge also may give instructions in the court order for what to do if the paying parent fails to pay.
A parent can file a lawsuit against the paying parent to get unpaid child support. The paying parent has a duty to pay all child support payments in full and on time.
When a judge decides that you get child support, it also provides you with different ways to get the money when the other parent fails to pay, such as:
The family court and ICSS have the authority to force the paying parent to fulfill his financial obligation to pay child support. The owed parent may not use her own remedies, such as not disobeying a legal visitation order, in order to get overdue child support from the paying parent.
A paying parent must use legal methods to stop paying child support. You must go back to court and get a court order saying that you do not have to pay anymore. A judge may stop you from paying child support if:
Contact a local Idaho family law lawyer today to help you understand your rights and responsibilities for receiving or paying child support.
Last Modified: 06. | 0.977503 |
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Twenty police officers were injured during protests in Bristol on Sunday night, including one who suffered a collapsed lung after being stamped on.
Another officer suffered broken bones, Police Chief Constable Andy Marsh said on Monday morning.
Twelve police vehicles were also damaged in the violence with photos from the scene showing burning cars.
Police said perpetrators would be "identified and brought to justice" but it's unknown if there have been any arrests at this stage.
Thousands had gathered at the 'Kill the Bill' demonstration to protest the Government's controversial Police and Crime Bill, which is being debated by MPs this week.
Critics say the bill is undemocratic and would criminalise the act of protesting in the UK, giving cops new powers to heavily restrict protests they believe may result in "serious public disorder, serious damage to property or serious disruption to the life of the community".
Avon and Somerset police chief constable Andy Marsh told BBC Radio 4's Today programme that "most seriously 12 of my brave officers have been injured, two of them seriously, doing their best to protect property and the people of Bristol from what was violent criminality and thuggery".…
Bristol mayor Marvin Rees said he recognised "the frustrations" with the Police, Crime, Sentencing and Courts Bill but said that "smashing buildings in our city centre, vandalising vehicles, attacking our police will do nothing to lessen the likelihood of the bill going through".
Avon and Somerset Police had encouraged people to stay away from the demonstration, technically illegal under current Covid laws, and attend virtual protests instead.
- Police
Source: Read Full Article | 0.003051 |
Perth & District Collegiate Institute is a diverse, public intermediate and secondary school in Eastern Ontario, located in the town of Perth, Ontario. We take pride in the dedicated individuals who comprise our faculty and are deeply committed to quality education that fosters life-long learning for all. Through creative and challenging learning experiences, we strive to make each student's individual learning experience a positive one designed to enrich their future.
At PDCI, we are thrilled to be a part of such a welcoming community. The Town of Perth is nestled in Eastern Ontario and encompasses a small town charm with all of the amenities of a larger city. Our school takes pride in this community atmosphere, which is reflected in our faculty, student body, and administration. We take our community role very seriously. We believe that this strong approach to the UCDSB's "Character Always" approach is one of the many reasons that we have so many successful alumni in the community and around the world. I invite you to take a look around our site to learn more about the wonderful educational experiences that we provide!
Sincerely,
Dean Fournier,. | 0.653742 |
TITLE: Uncertainty relation and instrumental errors
QUESTION [2 upvotes]: I was puzzled recently by what I've read in the internets about Heisenberg's uncertainty principle (one probably should never do this).
It claimed that in the usual relation
$$
\overline{(x-\overline{x})^2} \cdot \overline{(p_x-\overline{p_x})^2} \ge \hbar/2
$$
uncertainties $\overline{(x-\overline{x})^2}$ and $\overline{(p_x-\overline{p_x})^2}$ correspond to the limits on precision of measuring instrument (e.g. we cannot build a device that would measure the momentum precisely).
But I always thought that the meaning of the above relation is as follows:
We have a large number $2N$ of "boxes" with identical systems, all in identical quantum state. We then measure coordinate of $N$ of them and momentum of the other half. We can make each of these measurements, both of coordinate and of momentum at the same time, as precise as possible, with instrumental error virtually equal to zero.
We then find that despite being in the same quantum state, results are not the same for different boxes. And after averaging over boxes
$$
\overline{x} = \frac1N \sum_{box=1}^N x_{box}
$$
$$
\overline{p} = \frac1N \sum_{box=N+1}^{2N} p_{box}
$$
etc, we find the uncertainty relation. So each component in this sum is a precise value (in principle).
Is my understanding correct?
REPLY [3 votes]: You understanding is correct. The Heisenberg Uncertainty Principle does not stem from the precision of our measuring apparatus; it is instead a fundamental property of the physical states which can be prepared. In essence it is the bandwidth theorem ($\Delta k\Delta x\geq 1/2$) translated into physical form by de Broglie's relation ($p=\hbar k$). | 0.127695 |
TITLE: Can we find a real $u$ such that $f(u)=w$ ($w$ is fixed) and $f'(u)≠0$?
QUESTION [0 upvotes]: Let $f\colon\mathbb C\to\mathbb C$ be an entire non constant function. We consider its values on the real line. The function $f$ has infinitely many real zeros and there is infinitely many real solutions to the equation $f(s)=w$ for any real number $w≠0$. Here the set of those real $s$ is a discrete set.
My question is: Can we find a real $u$ such that $f(u)=w$ ($w$ is fixed) and $f'(u)≠0$?
Proof: We can prove this via identity theorem: $f$ cannot have a limit point of zeros in $ℝ$ but $f′$ is also holomorphic, and if (for a contradiction) $f′(s)=0$ whenever $f(s)≠0$ then $f′$ would have a limit point of zeros. Hence a such point $u$ exists.
REPLY [2 votes]: The answer to the question is no.
Let $f(z)=z\cdot\sin^2(z).$ It is easy to verify that $f$ satisfies the desired conditions. now, $$f'(z)=\sin^2(z)+z\cdot\sin(2z),$$ and clearly, whenever $f(z)=0$, we also have $f'(z)=0.$ | 0.002681 |
Around the world : Malta
Malta: that beautiful enclave in the Mediterranean is a favourite for tourists all over Europe.
Unbeknown to many, it was also host to an assembly operation that saw the construction of Triumphs and Morrises among others. Christopher Camilleri takes up the story, with new images from Andreas Lampka…
Carmaking was late coming to Malta: The Mizzi Organization set up their automotive division known as Car Assembly Ltd in 1960, and located a new, purpose-built plant in Marsa. The first car to be assembled in this new plant was the Triumph Herald, which proved to be a big seller in Malta (as it was in other territories that possessed a Herald CKD operation). As time passed, the demand for new cars in Malta increased and, therefore, the production was ramped up in order to meet burgeoning demand.
Not only that, but consumers wanted better choice, and so it was through this increased demand for cars that widened range of cars was gradually rolled-out.
Thus, a variety of different models was soon being produced in the Car assembly Ltd. In fact, after a year that the Car Assembly Ltd was opened, a further two new models were rolling off the Marsa production line; the Dodge and the Hillman Hunter/Minx (Arrows range).
The Car Assembly Ltd remained operational until 1981, when during this time, the cars produced were the Alfa Romeo Alfasud, Mini and Morris Marina. As well as these mainstays, Car Assembly also played host to the production of, a local version of the ubiquitous Commer Van, Rover P6 2000 and Austin-Morris 1100/1300.
Another interesting fact is that Car Assembly Ltd also exported models to other countries such as Tunisia and Jamaica. For example, the Maltese Morris Marina was exported to Jamaica, and the Triumph Herald was exported to Israel and Greece. Car Assembly Ltd was a great success in Malta where the company held a whopping 55 per cent of the Maltese car market.
As Christopher Camilleri quite rightly surmises, ‘It is such a pity that it has closed due to political reasons.’
This Morris Marina, as owned by Christopher Camilleri was produced in Malta by Car Assembly Ltd in their plant at Marsa. This particular Marina is a real rarity, as it is powered by a 1.5-litre B-Series diesel engine, and is described by Camilleri: ‘It has never been resprayed and it is in unrestored condition. Right now, I am using the car as an every day driver which gives me plenty of joy and happiness.’
Gallery
With thanks to Christopher Camilleri for contributing this article, and Andreas Lampka for the gallery images
Hi,
if you have any more photos of the Malta Car Assembly, can you please send them? Also, do you have any other relevant information regarding the Mini production, where was the factory situated?
Thanks
Joseph
I visited Malta back in 2003 and it was an old car spotters paradise. How many of those old Marinas, Itals, Escorts and Hunters remain is anyone’s guess, no doubt replaced by tinny modern cars. I even spotted a few Allegros.
Sadly, even the traditional Malta buses have gone the way of all things! From past holidays I remember that that the “1.5D” version of the Marina was very popular, and its limited performance not really a problem on a small island. The neighbouring island of Gozo is even more of a car spotters paradise, there is a lunatic there with dozens of series 1 Land Rovers!
When I was there it was fairly easy to spot older cars, though mostly 1970s – 80s. There seemed to be an early Metro or Mk3 Escort on every side street.
I’d love to know how good the Maltese AlfaSud was- it could hardly be built more poorly than the Italian original, and if it was built from better tin with improved rustproofing, there may well be some survivors. And RHD too!
Great article Mr Camilliari, and your Marina looks to be in superb condition.
My Maltese ex-girlfriend had a triumph herald and the had it restored to it’s original state although she did say it was more about tidying it up rather than replacing panels and powertrain.
As for taking cars off the island, I don’t think you can. I can’t remember if it’s Cyprus or Malta but cars that live on the island must die on the island, hence the reason why there are so many old cars still here… Otherwise they’d have been snapped up ages ago.
Very interesting article and photos. I never knew that cars were assembled in Malta all those years ago. The Rover P6 2000 production was particularly interesting. In 1975 I visited the Peykan factory in Iran where Hunters were assembled – never realised that they were in Malta too.
Anyone considered an article on Ireland’s car industry? In the seventies they made CKD versions of Ford Cortinas and the last version of the Hillman Hunter, even exporting some back to Britain. It is, like Malta, a little known car industry.
The Delorean story is well covered from northern Ireland and I think the Ford motor factory in Cork made Transits and other cars up on till 1984 and the same site today is the City’s Marina.
Isn’t picture # 9 a Dolomite/Toledo?
Joseph the factory was in Marsa, next to it there is Schembri battery service.part of the factory nowadays belongs to the government printing press and another part of it its a private factory.
The 1.5 diesel Marina & Ital was also the mainstay of the Portuguese automotive industry.
Shame they didn’t put the 1.8 version of the engine in as per the Sherpa.
Would have been a logical improvement for minimal expenditure which is how the best BL cars came about.
Will @ 9, I meant the Republic where a CKD industry survived until 1984. Actually the Ford plant in Cork was one of the biggest factories in Eire, but the fact it was on the periphery of Europe, it was cheaper to import complete cars and the productivity was terrible meant it had to close. I did read the productivity at Cork was 95 per cent lower than Dagenham, hardly the most productive factory in Europe, and it was bedevilled by absenteeism, strikes and apathy. Now who said this was a purely British problem.
I remember seeing quite a few Marinas when I visited Malta in the early 90s, and there was a burgundy 1.5 D seen regularly in Buggiba near my hotel. It was the first time I had actually seen an oil burning example. It does make you think that BL may have missed a trick, like Phil Simpson says, as there is a chance the 1.8D would have made a reasonable taxi
@6 Maltese vehicles can be easily exported, same goes for Cyprus. Loads of ex Malta stuff has come to the UK very recently, and a good friend just brought his mk1 Ford Granada 2.5 and Austin Metro 310 van back from Cyprus. Both of which were new on the Island, and both are completely immaculate
@16. Malta only joined the EU relatively recently (2004) so that may have led to barriers coming down – It’s probably the reason why the old buses have been replaced so easily lately. Wasn’t sure of Cyprus but the recent flood of Maltese classics would make a lot of sense.
Dear AR Readers,
Today was probably the best day of my life.
In Birkirkara (Malta), a court case ensued regarding a car garage (rental car one probably) and its assets were confiscated.
35 years later, the case has been closed and thus the assets including the cars can be sold. These include 2 Austin Allegros with 3000 miles, 9000 respectively, one’s a 1300 and another’s a made-in-malta 1300 super with a vinyl sunroof. In the blue one, the vinyl is still shiney and there is just light rust on top of paint due to disuse but nothing major at all. No ‘rot’ even in the most sensitive of places. They even have their 70s factory batteries still connected!
I visited them today, as well as many other cars which have been abandoned in a garage for all these years. I’d love to buy them but there are 2 problems
1. the seller’s being unreasonable changing the price every 5 minutes
2. as they’ve been off the road for 35 years, the registration has been nulled so tax has to be paid on them
I don’t know what to do. Contact me at niktelmu@gmail.com for some consolation…
i have a morris marina MK2 that i buy it from mizzi motors in 1977. i want to buy 2 doors front, and 2 doors back. if someone have thoose doors, can you please send me a mail or phone me.
mail axisabros@onvol.net
mobile 99862906.
My email is:arash_ruhi@yahoo.com | 0.771172 |
Ministerial Conference is the peak decision-making body of the World Trade Organization (WTO).
The Ministerial Conference brings World Trade Organization (WTO) Members together at the ministerial level, usually every two years. The Ministerial Conference can take decisions on all matters under any multilateral trade agreements.
There have been 10 meetings of the Ministerial Conference since the WTO was established: | 0.722315 |
TITLE: Show that the normalizer of a Sylow $p$-subgroup contains a nonabelian group of order $pq$
QUESTION [1 upvotes]: I am stuck with the following question: let $ P $ be a Sylow $p$-subgroup of the alternating group $ A_p $ where $ p $ is an odd prime. Show that for an odd prime $ q \mid p-1 $, $ N_{A_p}(P) $ contains a nonabelian group of order $pq$.
I'm not sure how to even get started. I know the various orders of $ P $, $ N_{A_p}(P)$ and also the index of $ N_{A_p}(P) $ in $ A_p $ (the number of Sylow $p$-subgroups), but how do I show that it has a group of order $ pq $?
REPLY [0 votes]: When $q\mid (p-1)$, there are only two groups of order $pq$: the cyclic group and a nonabelian group generated by the relation:
$$x^p = 1, y^q = 1, yxy^{-1} = x^i$$
where $i$ is a nontrivial solution to $i^q \equiv 1 \pmod{p}$.
Any sylow $p$-subgroup of $A_p$ is cyclic. Hence it suffices to prove the following:
Let $i$ be a nontrivial solution to $i^q \equiv 1 \pmod{p}$, and let $x=(1\quad 2\quad \cdots \quad p)$. There exists a permutation $y\in A_p$ of order $q$ such that $yx = x^iy.$
We will explicitly construct such an $y$. We will intrepret the indices as numbers modolo $p$. Note that $\langle i \rangle$ has order $q$ in the multiplicative group modulo $p$, denotes its cosets as $\langle i \rangle, a_1\langle i \rangle, \cdots, a_r\langle i \rangle$. We form the permutation $$y=(i\quad i^2\quad \cdots \quad i^q)(a_1i\quad a_1i^2\quad \cdots \quad a_1i^q)\cdots(a_ri\quad a_ri^2\quad \cdots \quad a_ri^q)$$
The only number does not appear in the above product is $0$ modulo $p$.
Then $y$ satisfies the condition $yx = x^i y$. This is easy to see, once we note that the action of $x$ on a number $u\in \mathbb{Z}/p\mathbb{Z}$ is to increase it by $1$, while the action of $y$ on $u$ is to multiply it by $i$. Hence $$yx(u) = y(u+1) = i(u+1)$$ and $$x^iy(u) = x^i(iu) = iu+i$$ Therefore they are indeed equal. The proof is completed. | 0.006726 |
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Click the thumbnail to look.
You can search the image by entering the keyword in the following box.
Tokyo Hokkaido Los Angeles Hakone Rishiri Island Rebun Island Okinawa Prefecture Shizuoka Prefecture Wakayama Prefecture Niigata Prefecture Kyoto
Spring Summer Autumn Winter
Close-up Nearby Not so far Faraway | 0.060759 |
Advocacy Practice for Social Justice
2016
978-1-935871-82-8
Advocacy Practice for Social Justice, 3rd Ed., by Richard Hoefer, Chicago, IL, Lyceum Books, ISBN: 978-1-935871-82-8, 2016, 296 pages, $43.95 paperback.
Social work students need to believe that advocacy practice is something they can handle. Dr. Hoefer responds to this need in the revised edition of his textbook, Advocacy Practice for Social Justice. Dr. Hoefer’s purpose is to help social work students become comfortable with their ability to do advocacy in their upcoming careers.
Hoefer’s “Unified Model of Advocacy” aligns with the familiar generalist practice framework. For example, the “Unified Model” focuses on getting involved, understanding the issue, planning, advocating, evaluating, and ongoing monitoring, while the generalist practice framework includes engagement, assessment, implementation, evaluation, termination, and follow-up.
It is important for students to know that the steps to advocacy practice exclude termination and include ongoing monitoring. Therefore, students face the dilemma of defining ways of continuing advocacy practice over a lifetime of practice. Hoefer frames advocacy skills as education, persuasion, and negotiation. Again, he uses terminology that is student-friendly, thus making advocacy practice easy to understand and implement. By using the “Unified Model,” the most tentative student gains confidence in the knowledge, values, and skills required for advocacy practice.
After using the earlier editions of Hoefer’s text, I am excited to find several new and expanded chapters. First, the new chapter on electronic advocacy reflects current practice methods used in policy and advocacy organizations. The use of the Internet and social media for advocacy matches the tools that today’s students already use. Although students are comfortable with the Web 2.0 format, Hoefer expands their knowledge and builds their skills in competently using the Internet and social media for learning about policies and influencing others toward social change.
Second, Hoefer enhanced a section on distributive justice to include an anti-oppression framework and a discussion on the need for economic justice. With this framework, students gain perspectives on ways to change systems of all sizes and levels that affect people’s well-being, particularly vulnerable populations.
Third, I am pleased to see the inclusion of instructional assistance in each chapter. Hoefer added discussion questions that make connections between current events and advocacy.
Because of the expanded and updated information and chapters, Hoefer’s revised text meets the new CSWE EPAS standards for students’ competencies in policy and advocacy practice behaviors.
Hoefer’s third edition is an excellent text for any policy, advocacy, or community practice courses at the BSW level. Because it offers the knowledge, values, and skills of advocacy practice from a generalist practice perspective, this text easily fits as a secondary text in the MSW specialization tracks and concentrations, such as mental health, health, aging, and child welfare.
Overall, I strongly recommend the third edition of Advocacy Practice for Social Justice for any course in which seeking justice is a focus of the student’s work. Dr. Hoefer continues to promote excellence in social work advocacy practice in this revised text.
Reviewed by Susan T. Parlier, Ph.D., LMSW, ACSW, MAR, Clinical Instructor and former BSW and Social Work Minor Program Coordinator, College of Social Work, University of South Carolina. | 0.988939 |
\begin{document}
\begin{frontmatter}
\title{ The relationship between stopping time and number of odd terms in Collatz sequences }
\runtitle{ Stopping time and odd terms in Collatz sequences }
\author{\fnms{Rafael} \snm{Ruggiero} \ead[label=e1]{ruggiero3n1@gmail.com}}
\address{S\~ao Paulo, Brazil\\
\printead{e1}}
\runauthor{R. Ruggiero}
\begin{abstract}
The Collatz sequence for a given natural number $N$ is generated by repeatedly applying the map $N$ $\rightarrow$ $3N+1$ if $N$ is odd and $N$ $\rightarrow$ $N/2$ if $N$ is even. One elusive open problem in Mathematics is whether all such sequences end in 1 (Collatz conjecture), the alternative being the possibility of cycles or of unbounded sequences. In this paper, we present a formula relating the stopping time and the number of odd terms in a Collatz sequence, obtained numerically and tested for all numbers up to $10^7$ and for random numbers up to $2^{128.000}$. This result is presented as a conjecture, and with the hope that it could be useful for constructing a proof of the Collatz conjecture.
\end{abstract}
\begin{keyword}
\kwd{Collatz Conjecture, Experimental Mathematics}
\end{keyword}
\end{frontmatter}
\section{Introduction}
The Collatz conjecture states that the sequence resulting from repeatedly applying the map $N$ $\rightarrow$ $3N+1$ if $N$ is odd and $N$ $\rightarrow$ $N/2$ if $N$ is even, starting from any natural number, inevitably reaches $1$. The alternative would be the existence of some special $N$ for which the sequence diverges or features an indefinite cycle. For a review on general properties of Collatz sequences, see \cite{lagarias}.
Despite the simplicity in the enunciation of the conjecture, it turns out to be an incredibly challenging problem. There is no simple formula relating $N$ to the number of iterations needed to reach $1$, and the behavior of this number (called the \emph{stopping time}) as a function of $N$ is markedly unpredictable, even passing several standard tests of randomness \cite{xu}.
Our goal in this paper is to propose an explicit formula for the stopping time $S$ as a function of the number of odd terms in the sequence and of $N$, a result which, if correct, allows one to constraint which values $S$ may assume.
\section{The formula}
Let $S$ be the stopping time for the Collatz map starting from a given natural number $N$, defined as the number of iterations needed for the sequence to reach $1$, and let $\alpha$ be the number of odd terms in the resulting sequence, excluding $1$.
\vspace{2.8cm}
\noindent \textbf{Conjecture:} For all $N$ for which $S$ is defined, it is true that
\begin{equation} \label{eq:main}
S = \ceil{\log_2\left(6^{\alpha} N\right)},
\end{equation}
where $\ceil{x}$ is the ceiling function, defined as the smallest integer greater than or equal to $x$.
The use of the ceiling function is necessary because the residue $\epsilon(N) \equiv S - \log_2\left(6^{\alpha} N\right)$ is in general not zero. In fact, it is a seemingly random real number in the interval $[0, 0.326)$. The distribution of $\epsilon(N)$ for all $N < 10^7$ is shown in Fig.~\ref{fig:dist}.
\begin{figure}[b!]
\includegraphics[width=0.73\linewidth]{4.eps}
\caption{Distribution of the residue $\epsilon(N)$ for $N < 10^7$.}
\label{fig:dist}
\end{figure}
Our numerical tests have verified the validity of this relation for all natural numbers smaller than $10^7$, and also for thousands of random numbers between $1$ and $2^{128.000}$, strongly suggesting its validity for all natural numbers. A proof is left as an open problem.
\section{Implications}
In principle, we do not know how many odd numbers a Collatz sequence will have, so $\alpha$ is an unknown prior to having calculated the entire sequence. But if Eq.~\ref{eq:main} is valid, some values which $S$ \emph{cannot} assume can be calculated without calculating any iteration of the Collatz map.
First, since $\alpha \geq 0$, then $S \geq \log_2{N}$. This is a trivial result: the fastest a number could get to $1$ under the Collatz map is if the terms in its Collatz sequence never increase in value. In practice, this happens for powers of $2$.
A less trivial result is that, since $\alpha$ must be a natural number, not necessarily $S$ can assume any natural value $\geq \log_2{N}$. For instance, consider $N = 65$. Its possible stopping times, setting $\alpha$ as all possible natural numbers (including zero), are 7, 9, 12, 14, 17, 19, 22, 25, 27, 30, 32, 35, \textellipsis. The true value, which is $27$, is present in this list, but many numbers both smaller and larger than $27$ are not, say 15, 26 or 34. Let's call such numbers \emph{prohibited} values for $S(N)$.
The list of prohibited values $P$ for $S$ given a particular number $N$ can be propagated to infinite other numbers, as long as the Collatz conjecture is true. If that is the case, then $S(N/2) = S(N)-1$ if $N$ is even and $S(3N+1) = S(N)-1$ if $N$ is odd. Moreover, $S(2N) = S(N)+1$ for all $N$, and $S((N-1)/3) = S(N)+1$ if $(N-1)/3$ is an odd natural number. Using these relations, either $P+1$ or $P-1$ can also be shown to be prohibited values for these up to $3$ other numbers. This procedure can be carried out recursively and starting from different $N$ and associated $P$, resulting in a sieve algorithm for ruling out stopping times for natural numbers.
\vspace{0.31cm}
\section{Sequences with constant $\alpha$}
One important feature of Eq.~\ref{eq:main} is that it connects the stopping times of sequences of natural numbers with constant $\alpha$ in an $S(N)$ plot, such as the one shown in Fig.~\ref{fig:sn}, in which stopping times and curves with constant $\alpha$ are overlaid. The first such sequence, which is the lower frontier in that figure, is the one for $\alpha = 0$. This is the sequence of all powers of $2$, which have no odd numbers in their Collatz sequences. Sequences with higher $\alpha$ appear increasingly higher, and no two curves with different $\alpha$ ever intersect. We have also included two zoomed-in versions of Fig.~\ref{fig:sn} in Fig.~\ref{fig:z}, and the first elements of some sequences of numbers with constant $\alpha$ are shown in Table~\ref{tab:tab}.
\begin{center}
\begin{figure}
\makebox[\linewidth][c]{
\includegraphics[width=1.4\linewidth]{3.eps}}
\caption{Stopping time $S$ as a function of $N$ for $N < 2000$. The gray lines show Eq. \ref{eq:main} calculated for different natural values of $\alpha$ (including 0). The first line from bottom to top corresponds to $\alpha = 0$, the next one to $\alpha = 1$, and so on.}
\label{fig:sn}
\end{figure}
\end{center}
\begin{figure}
\makebox[\linewidth][c]{
\includegraphics[width=1.1\linewidth]{1.eps}}
\makebox[\linewidth][c]{
\includegraphics[width=1.1\linewidth]{2.eps}}
\caption{Two zoomed-in rectangles in the area of Fig.~\ref{fig:sn}. The first one shows the points near the origin, and the second one shows the dense region on the upper left of that plot.}
\label{fig:z}
\end{figure}
\begin{table} [h!]
\caption{\normalsize The first elements of the sequences of numbers with constant $\alpha$, for $\alpha < 20$.}
\vspace{0.3cm}
\normalsize
\makebox[\linewidth][c]{
\begin{tabular}{ l l }
$\alpha$ = 0: & 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536 \\
$\alpha$ = 1: & 5, 10, 20, 21, 40, 42, 80, 84, 85, 160, 168, 170, 320, 336, 340, 341, 640 \\
$\alpha$ = 2: & 3, 6, 12, 13, 24, 26, 48, 52, 53, 96, 104, 106, 113, 192, 208, 212, 213 \\
$\alpha$ = 3: & 17, 34, 35, 68, 69, 70, 75, 136, 138, 140, 141, 150, 151, 272, 276, 277, 280 \\
$\alpha$ = 4: & 11, 22, 23, 44, 45, 46, 88, 90, 92, 93, 176, 180, 181, 184, 186, 201, 352 \\
$\alpha$ = 5: & 7, 14, 15, 28, 29, 30, 56, 58, 60, 61, 112, 116, 117, 120, 122, 224, 232 \\
$\alpha$ = 6: & 9, 18, 19, 36, 37, 38, 72, 74, 76, 77, 81, 144, 148, 149, 152, 154, 162 \\
$\alpha$ = 7: & 25, 49, 50, 51, 98, 99, 100, 101, 102, 196, 197, 198, 200, 202, 204, 205, 217 \\
$\alpha$ = 8: & 33, 65, 66, 67, 130, 131, 132, 133, 134, 260, 261, 262, 264, 266, 268, 269, 273 \\
$\alpha$ = 9: & 43, 86, 87, 89, 172, 173, 174, 177, 178, 179, 344, 346, 348, 349, 354, 355, 356 \\
$\alpha$ = 10: & 57, 59, 114, 115, 118, 119, 228, 229, 230, 236, 237, 238, 456, 458, 460, 461, 465 \\
$\alpha$ = 11: & 39, 78, 79, 153, 156, 157, 158, 305, 306, 307, 312, 314, 315, 316, 317, 610, 611 \\
$\alpha$ = 12: & 105, 203, 209, 210, 211, 406, 407, 409, 418, 419, 420, 421, 422, 431, 455, 812, 813 \\
$\alpha$ = 13: & 135, 139, 270, 271, 278, 279, 281, 287, 303, 540, 541, 542, 545, 551, 556, 557, 558 \\
$\alpha$ = 14: & 185, 187, 191, 361, 363, 367, 370, 371, 374, 375, 382, 383, 721, 722, 723, 726, 727 \\
$\alpha$ = 15: & 123, 127, 246, 247, 249, 254, 255, 481, 489, 492, 493, 494, 498, 499, 508, 509, 510 \\
$\alpha$ = 16: & 169, 329, 338, 339, 359, 641, 657, 658, 659, 665, 676, 677, 678, 718, 719, 1281, 1282 \\
$\alpha$ = 17: & 219, 225, 239, 427, 438, 439, 443, 450, 451, 478, 479, 854, 855, 876, 877, 878, 886 \\
$\alpha$ = 18: & 159, 295, 318, 319, 569, 585, 590, 591, 601, 636, 637, 638, 1138, 1139, 1147, 1151, 1159 \\
$\alpha$ = 19: & 379, 393, 425, 758, 759, 767, 779, 786, 787, 801, 849, 850, 851, 1516, 1517, 1518, 1529 \\
\end{tabular}
}\label{tab:tab}
\end{table}
\vspace{-0.37cm}
\section{Collatz sequences in an $S(N)$ plot}
It is illustrative to show what a Collatz sequence looks like in an $S(N)$ plot. As the sequence goes on, $S(N)$ decreases by 1 at each step, and one of two possibilities takes place: either the point moves along a curve with constant $\alpha$ to the left, as its factors of two are removed, or it moves right to a curve with lower $\alpha$ underneath the previous one. The process is repeated until the point becomes a power of $2$ and lands at the curve with $\alpha = 0$, from where it monotonically decreases towards $1$.
Three examples of this process are shown in Fig.~\ref{fig:examples}.
\begin{figure}
\includegraphics[width=0.75\linewidth]{5a.eps}
\includegraphics[width=0.75\linewidth]{5b.eps}
\includegraphics[width=0.75\linewidth]{5c.eps}
\caption{Examples of Collatz sequences in $S(N)$ plots, starting from different initial values. The red lines begin on top and move lower towards $N = 1$ at each iteration. The number either moves to the left along a curve with constant $\alpha$ or to the right to a point in a curve with smaller $\alpha$.}
\label{fig:examples}
\end{figure} | 0.014191 |
TITLE: What does the gravitational potential of the Milky Way do to the CMB?
QUESTION [2 upvotes]: We sit in a gravitational potential, so there should be a blue shift on the CMB light from the potential of the Milky Way. Is this blue shift dependent on direction? Is it being subtracted from the CMB? Or is it simply to small to be measurable?
REPLY [3 votes]: The gravitational potential of the Milky Way will cause a blue shift not a red shift. This happens because relative to an observer far from the Milky Way the gravitational potential within it causes a time dilation i.e. here on Earth our clocks run very slightly slower than clocks out in intergalactic space. Since our clocks run slower the frequency of light coming from intergalactic space is slightly increased.
As it happens I have described the time dilation for observers within the Milky Way in my answer to Why isn't the center of the galaxy "younger" than the outer parts? From experimental data we have the following approximate formula for the gravitational potential energy (per unit mass) inside the Milky way:
$$ \Phi = -\frac{GM}{\sqrt{r^2 + (a + \sqrt{b^2 + z^2})^2}} \tag{1} $$
where r is the radial distance, z is the height above the disk, a = 6.5 kpc and b = 0.26 kpc. The time dilation is well described by the weak field equation:
$$ \frac{\Delta t_r}{\Delta t_\infty} = \sqrt{1 - \frac{2\Delta\Phi}{c^2}} $$
According to NASA the Sun lies about 8 kpc from the centre of the Milky Way, so $r = 8$ kpc and according to the Astronomy SE we are about 20 pc away from the plane so $z = 20$ pc. Finally let's guesstimate the mass of the Milky Way as $10^{12}$ solar masses (it's a guesstimate because we don't know how much dark matter the Milky Way contains). Plugging in all these numbers gives us:
$$ \Phi \approx 4 \times 10^{11} \,\text{joules/kg} $$
And plugging this into out equation for the time dilation gives:
$$ \frac{\Delta t_r}{\Delta t_\infty} = 0.999995 $$
Taking the reciprocal of this to estimate the blueshift we find the frequency of the CMB is blue shifted by a factor of $1.000005$. | 0.299831 |
Cherries, Cherries, oh baby!!!!
Cherries are my favorite fruit; they are so plump, sweet and juicy. I am always reminded of the old days when I lived in Ashland Oregon. There were two cherry trees on the campus of Southern Oregon State College. My daughters were children then and picking these cherries had become an annual event.
We walked by and checked on their ripeness every day. On the days we picked them we'd have picnics, climb the trees then go swimming at the lake. I cherish the memories of those hot Ashland summer days, when it was just the three of us, my little girls and I.
Cherry trees played a large role in ancient legend, myth and folklore. The Chinese believed the tree-bestowed fertility and immortality. In one myth the goddess His Wang Mu had a garden in which the cherries of immortality ripened every 3000 years, the gods would eat them in order to remain immortal.
One Japanese legend tells of a beautiful cherry tree named Lady Yaye-zakura. A grass prince named Susuki had fallen in love with her but she did not care to return his affection. The problem for her was she could only resist him when she was in full bloom, so when her petals fell she submitted. Meanwhile, a plum tree had also fallen for Lady Yaye-zakuras beauty. The plum tree became jealous and outraged at the thought that the loveliest of all trees had given her love to a lowly grass prince.
The plum tree got all the other trees to battle against the grass folk. The trees won the battle and Susuki the grass prince died. Lady Yaye-zakuras heart was broken, and in mourning donned dark robes. This explained the dark color of the flowers and why from then on Lady Yaye-zakura was known as Sumi-zome-zakura, which meant "a cherry tree in black robes"
In Buddhist myth Maya clung to a cherry tree while giving birth to Buddha and the cherry tree bent down and gave her fruit. In Christian legend the cherry tree bent down and gave the Virgin Mary fruit. Mary and Joseph were sitting under a cherry tree and Mary longed for cherries, but she could not reach them. Joseph would not pick them for her saying that the true father of her child should provide them for her. Cherries were dedicated to Mary after that and the cherry blossoms represented purity. In parts of Europe, evil spirits were believed to inhabit cherry trees, but in most myths cherry trees were a blessing that bestowed good fortune and fertility.
In American culture the cherry tree symbolizes honesty, beauty and home. The cherry represents the loss of innocence.
Cherries are a member of the plant family that includes plums, apricots, peaches and nectarines. Fruits containing pits. All of which are juicy sweet and nutritious. There are over 1000 varieties of cherries worldwide. The most popular cherries are the yellowish Queen Anne, these are sweet treats. The Scarlet Morello, which is sour but great for pies. Bing cherries are best eaten raw and the Lambert’s are good cooked or raw. Maraschinos cherries are Queen Anne's that are first bleached in a sulfur dioxide brine, then toughened with lime, dyed red, sweetened then canned.
The health benefits of cherries are numerous. There are 260mg. of potassium in one cup of cherries. The Chinese have used cherries to treat diabetes for centuries and now science knows why. Cherries contain anthocyanins; this is a compound that reduces insulin resistance and glucose intolerance. Cherries also contain meletonin a sleep aid. Cherries are a good source of pectin, which helps the body eliminate cholesterol. Cherries are high in vitamin E, fiber and helps with gout and arthritis due to it's an anti-inflammatory property. Cherries are truly another "Wonder Food". Yesterday I found a fruit stand in Yachats and pulled in. Big Daddy, the proprietor told me he has the best fruit any where on the coast and proceeded to sell me a couple of pounds of assorted cherries. He also gave me his recipe for the best cherry pie in the known world. He asked if I had a cherry pitter and when I said no the look of pity in his eyes prompted me to run to Judith's Kitchen Tools and buy one. Good thing I did. So here it is,
Make your favorite 2 crust piecrust.
2 pounds assorted cherries, pitted
Microwave cherries in 1½ c. red wine for three minutes. I used cranberry wine
Drain off the wine.
In a large bowl combine:
1 c. white sugar
½ cup brown sugar
½ tapioca
1 tsp. Pumpkin pie spice
1 tsp. Lemon juice.
Pour in pie pan cover with top crust; bake in 425 degree oven for 35 minutes.
Enjoy!
You can find Sundragon fruit stand on Sundays at the Yachats Farmers Market.
To reach Crystal Hayes email; crystalhayes@peak.org
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“Engaging in conflict isn't going to end the relationship, it's avoiding the So you can learn to approach conflict in a constructive and effective way. Instead, they had a fun time brainstorming together, and ended up “being more loving, intimate and She blogs regularly about body and self-image issues on her own blog.
It is natural for people in intimate relationships to want to see the positive Committed couples in no-win conflicts may still feel a deep As soon as the past memory loses its power, the desire to heal the wounded other intensely emerges. . I know that some of that avoidance was because early on, the..
Blog healing approach conflict intimate relationships traveling cheapIt is the experience of emotion in the body. Intervention strategies are based upon empirical data from Dr. The Gottman Method for Healthy Relationships:. Permission to publish granted by Cindy Ricardo, LMHC, CIRT , therapist in Coral Springs, Florida The preceding article was solely written by the author named above.
The way they gaze at each other lovingly, the attunement and synchronization of their movements, and how they anticipate the other make for a beautiful dance of intimacy. What Lack of Affection Can Do to You. When one or both partners desperately needs nurturing, forgivenessor support, but feels that the other is angry or unavailable, he or she may strike out rather than asking for what is needed. It is so satisfying to see the transformative power of PACT as they adjust the lenses through which they view each. These couples usually get along relatively well in between conflicts but are very soon embroiled in these confusing interactions once one begins.
Blog healing approach conflict intimate relationships flying easy
One or both of the partners poses as much stronger and more confident than he or she really feels. When I witness a couple move from fear and blame into trust and genuine care, I am inspired to feel hope for humanity. The one thing common to all relationships is a fear of disconnection and this often plays out in how each partner approaches conflict.
Blog healing approach conflict intimate relationships - traveling
They are particularly vulnerable to personal attack or the perception of attack that suggest that the loved one does not accept, understand, admire or love. Couples who alternate between loving and hating while in the midst of a battle misunderstand one another continuously. It is commonly observed that half of marriages end in divorce. Create an atmosphere that encourages each person to talk honestly about his or her hopes, values, convictions and aspirations. He started talking at her about plans for their garden in an attempt to engage her.
| 0.716526 |
\section*{Appendix II}
In this appendix, we provide a proof of Theorem 1. First, we note that the in-phase and quadrature component of the received signal $\tilde{V}$ are circularly symmetric \cite{Durgin2002, Durgin2002a, Salo2006, Yu2007} due to the underlying assumption of uniformly distributed phase $\phi_i$ and $\phi_d$. Then, the PDF of the envelope $R$ and the joint characteristic function (CF) $\Phi(\nu)$ of the complex signal $\tilde{V}$ can be written as a Hankel transform pair as follows \cite{Durgin2002a, Salo2006}
\begin{equation}
\begin{split}
f_{R}(r) &= r \int_{0}^{\infty} \nu \Phi(\nu) \BesselJ{0}{r \nu} \mathrm{d}\nu,\quad
\Phi(\nu) = \int_{0}^{\infty} f_{R}(r) \BesselJ{0}{\nu r} \mathrm{d}r.
\end{split}
\label{q004-eq_app_II-001}
\end{equation}
For the GMR-U model in (\ref{q004-001}), the CF $\Phi(\nu)$ can be derived as follows \cite{Abdi2000}
\begin{equation}
\begin{split}
\Phi(\nu) &= \mathbb{E}_{V_1, \ldots, V_N, V_d}\left[ \prod_{i=1}^{N} \BesselJ{0}{V_i \nu} \cdot \BesselJ{0}{V_d \nu} \right] = \prod_{i=1}^{N} \BesselJ{0}{V_i \nu} \mathbb{E}_{V_d}\left[ \BesselJ{0}{V_d \nu} \right]\\
&= \Hypergeometric{1}{1}{m}{1}{-\frac{\Omega}{4m} \nu^2} \cdot \prod_{i=1}^{N} \BesselJ{0}{V_i \nu},
\end{split}
\label{q004-eq_app_II-002}
\end{equation}
where the second equality is due to the deterministic nature of $V_i$'s and the last expression follows by substituting (\ref{q004-ext-000}) and applying \cite[6.631]{Gradshteyn1994}.
Next, we utilize the following series expansion of the confluent Hypergeometric function
\begin{equation}
\begin{split}
\Hypergeometric{1}{1}{m}{1}{-\frac{\Omega}{4m} \nu^2} &=\exp\left( -\frac{\Omega}{4m} \nu^2 \right)
L_{m-1}\left(\frac{\Omega}{4m} \nu^2\right)\\ &= \exp\left( -\frac{\Omega}{4m} \nu^2 \right)
\sum_{k = 0}^{m-1} \binom{m-1}{k} \frac{(-1)^k}{k!} \left( \frac{\Omega}{4m} \nu^2 \right)^k
\quad \text{for } m \in \mathbb{Z}^{+},
\end{split}
\label{q004-eq_app_II-003}
\end{equation}
where we applied \cite[07.20.03.0108.01]{wolfram} to the first equality and used (\ref{eq_app_I-001-a}) in the last equality. By substituting (\ref{q004-eq_app_II-002}) and (\ref{q004-eq_app_II-003}) to (\ref{q004-eq_app_II-001}), the PDF of the envelope $R$ can be expressed as follows
\begin{equation}
\begin{split}
&f_{R}(r) = r \sum_{k = 0}^{m-1} \binom{m-1}{k} \frac{(-1)^k}{k!} \left( \frac{\Omega}{4m} \nu^2 \right)^k \underbrace{\int_{0}^{\infty} \nu^{2k+1} \exp\left( -\frac{\Omega}{4m} \nu^2 \right) \prod_{i=1}^{N} \BesselJ{0}{V_i \nu} \cdot \BesselJ{0}{r \nu} \mathrm{d}\nu}_{\triangleq I_1},
\end{split}
\label{q004-eq_app_II-004}
\end{equation}
where the integral $I_1$ can be evaluated in a closed form expression as follows
\begin{equation}
\begin{split}
I_1 &= \frac{1}{2} \int_{0}^{\infty} t^{k} ~\mathrm{e}^{-\frac{\Omega}{4m} t} \cdot \prod_{i=1}^{N} \BesselJ{0}{V_i t^{1/2}} \BesselJ{0}{r t^{1/2}} \mathrm{d}t\\
&= \frac{\Gamma(k+1)}{2} \left( \frac{4m}{\Omega} \right)^{k+1}
\Psi_{2}\left( k+1; [1]_{N+1};
\frac{V_1^2 m}{\Omega}, \ldots, \frac{V_N^2 m}{\Omega}, \frac{r^2 m}{\Omega}
\right),
\end{split}
\label{q004-eq_app_II-005}
\end{equation}
by employing a change of variable, \textit{i.e.}, $t \leftarrow \nu^2$, and applying (\ref{eq_app_I-exo1-002}) to (\ref{q004-eq_app_II-005}). Hence, by substituting (\ref{q004-eq_app_II-005}) to (\ref{q004-eq_app_II-004}), (\ref{q004-004}) follows readily.
The corresponding CDF of the signal envelope $R$ can be evaluated as below
\begin{equation}
\begin{split}
F_{R}(t) &= \int_{0}^{t} f_{R}(r) \mathrm{d}r = \int_{0}^{\infty} \nu
\Hypergeometric{1}{1}{m}{1}{-\frac{\Omega}{4m} \nu^2} \prod_{i=1}^{N} \BesselJ{0}{V_i \nu}
\left( \int_{0}^{t} r \BesselJ{0}{r \nu} \mathrm{d}r \right) \mathrm{d}\nu \\
&= t \int_{0}^{\infty}
\Hypergeometric{1}{1}{m}{1}{-\frac{\Omega}{4m} \nu^2} \prod_{i=1}^{N} \BesselJ{0}{V_i \nu}
\cdot \BesselJ{1}{t \nu} \mathrm{d}\nu,
\end{split}
\label{q004-eq_app_II-006}
\end{equation}
by substituting (\ref{q004-004}) to the first equality, changing the order of integration in the second equality
and applying (\ref{eq_app_I-exo1-001}) in the third equality. The closed form expression of (\ref{q004-eq_app_II-006}) can be easily derived by following the similar procedure as the PDF of $R$, \textit{i.e.}, use the series expansion from (\ref{q004-eq_app_II-003}) and apply (\ref{eq_app_I-exo1-002}) with a change of variable $t \leftarrow \nu^2$. This completes the proof. | 0.002004 |
: Futurism.
The Disturbing Attraction of the Dystopian Novel
I’m putting aside my novel writing for the summer. I’ve decided to concentrate on other blog themes, and catch up on things like gardening, home decorating, and of course, reading, an enduring summer pleasure.
With the new TV adaptation of Margaret Atwood’s The Handmaid’s Tale in the news, I headed down the stacks in the local library to pick it up, but instead found myself reaching for her newest book, The Heart Goes Last.
“Why not?” I say, and add it to my take out pile. The truth is, I don’t have what it takes to read The Handmaid’s Tale again. I remember it as profoundly disturbing—a chillingly futuristic novel about Gilead, and how its female inhabitants are forced to have sex with powerful men and bear their children..”
A Grandmother Reflects on the ‘Miracle’ of 3D Printing
It would be easy for me, a grandmother with over 7 decades of living to my credit, to ignore 3D printing and the Internet of Things (IoT), thinking I may be long gone by the time these revolutionary ideas really take hold.
But I’m not about to disregard one of the most compelling ideas to emerge in this era of change, considered by leading scientists to be the “Third Industrial Revolution”. After all, just by virtue of having access to the internet, and being able to type a few keywords into my browser, I can have a front row seat to what promises to be a thrilling journey.
Why This Grandmother Worries About Automation and Employment
>>IMAGE.”
MOOC: Finally! A Revolution in Education.
.
Sensational! The CES Trade Show in Las Vegas
If I was a guest at the CES trade show in Las Vegas last week, I know I’d be wandering the aisles and hallways, boggle eyed, like a deer in the headlights.
Unlike fellow blogger, Lois Whitman, who has attended every one of the CES conferences through its 50 years of existence, and has the presence of mind to blog about it brilliantly (DigiDame), I would be overwhelmed. Maybe I’d even skip the whole thing and go to the beach (kidding—it’s a 5 hour drive!).
CES, is an international electronics show, which showcases what’s new, what is coming up, and what some of the world’s noteworthy scientists are thinking. Here is a description from its website:. | 0.814539 |
Archive for the ‘Colorado’ Category
Colorado – where to begin
I guess at the beginning of my time spent in Colorado. So much has happened in the last three weeks that I am still trying to digest everything. This was at the beginning of October. I am once again in rewind mode. OK, I drove down from Wyoming into Colorado on highway 125. I had […]
Colorado
So much has happened in Colorado, cowboys, turquois, UFOs, ancient monuments. I will try to write in detail soon. I am now in Utah where I have to make some route changes, adding more places to visit. This is the kind of research that takes away from my writing time, sorry. | 0.043492 |
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\begin{document}
\title{On an
asymptotic behavior of elements
of order $p$\\ in irreducible representations
of the classical algebraic groups\\
with large enough highest weights}
\author{I. D. Suprunenko\\
Institute of Mathematics, National Academy of Sciences of Belarus\\
Surganov str. 11, Minsk, 220072, Belarus\\
suprunenko@im.bas-net.by}
\date{}
\maketitle
\begin{abstract}
The behavior of the images of a fixed element of order $p$ in
irreducible representations of a classical algebraic group in
characteristic $p$ with highest weights large enough with respect
to $p$ and this element is investigated. More precisely, let $G$
be a classical algebraic group of rank $r$ over an algebraically
closed field $K$ of characteristic $p>2$. Assume that an element
$x\in G$ of order $p$ is conjugate to that of an algebraic group
of the same type and rank $m<r$ naturally embedded into $G$. Next,
an integer function $\sigma_x$ on the set of dominant weights of
$G$ and a constant $c_x$ that depend only upon $x$, and a
polynomial $d$ of degree one are defined. It is proved that the
image of $x$
in the irreducible representation of $G$ with highest weight $\omega$
contains more than $d(r-m)$ Jordan blocks of size $p$ if $m$ and
$r-m$ are not too small and $\sigma_x(\omega)\geq p-1+c_x$.
\end{abstract}
\maketitle
Asymptotic lower estimates for the number of Jordan blocks of size
$p$ in the images of a fixed element of order $p$ in irreducible
representations of a classical algebraic group in characteristic
$p$ with highest weights large enough with respect to $p$ and this
element are obtained. More precisely, let $G$ be a classical
algebraic group of rank $r$ over an algebraically closed field $K$
of characteristic $p>2$. Assume that an element $x\in G$ of order
$p$ is conjugate to that of an algebraic group $G_m$ of the same
type and rank $m<r$ naturally embedded into $G$. Set
\begin{equation*}
d(r-m)=
\begin{cases}
r-m & \text{for $G=A_r(K)$},\\
2r-2m & \text{for other types.}
\end{cases}
\end{equation*}
Let $\Delta_x$ be the labelled Dynkin diagram of the conjugacy
class containing $x$ in the sense of Bala and Carter \cite{BC} and
let $c_x$ be the sum of the labels at $\Delta_x$ for $G\neq
A_r(K)$ and the half of this sum for $G=A_r(K)$. For brevity,
throughout the article we refer to $\Delta_x$ as the labelled
Dynkin diagram of $x$. Next, an integer function $\sigma_x$ on the
set of dominant weights of $G$ that depends only upon $\Delta_x$
is defined. For $p$-restricted weights $\sigma_x$ coincides with
the canonical homomorphism determined by $\Delta_x$. It is proved
that the image of $x$ in the irreducible representation of $G$
with highest weight $\om$ contains more than $d(r-m)$ Jordan
blocks of size $p$ if $m$ and $r-m$ are not too small and
$\sigma_x(\om)\geq p-1+c_x$.
We need some more notation to formulate the main results. Let $\om_i$
and $\alpha_i$ be the fundamental weights and the simple roots of
$G$ (with respect to a fixed maximal torus $T$) labelled as in
\cite{B1}. Denote by $\delta_i$, $1\leq i\leq r$, the label on
$\Delta_x$ corresponding to its $i$th node. We have $0\leq
\delta_i\leq2$. In what follows $\mathbb{Z}$ is the set of integers,
$\mathbf{X}=\mathbf{X}(G)$ is the set of weights of $G$,
$\mathbf{X^+}\subset\mathbf{X}$ is the set of dominant weights,
$\Irr=\Irr(G)$ is the set of irreducible rational representations
of $G$ (considered up to the equivalence) and $\om(\phi)$ is the
highest weight of a representation $\phi$. There exists a uniquely
determined homomorphism $\tau_x: \mathbf{X}\rightarrow\mathbb{Z}$
such that $\tau_x(\alpha_i)=\delta_i$. The weight $\mu\in\mathbf{X^+}$
is called $p$-restricted if $\mu=\sum^r_{i=1}a_i\om_i$ with all
$a_i<p$. Each weight $\om\in\mathbf{X^+}$ can be represented in
the form $\sum^s_{j=0}p^j\om^j$ where $\om^j$ are $p$-restricted.
Set $\sigma_x(\om)=\tau_x(\sum^s_{j=0}\om^j)$. Now we can state our
main result.
\begin{theorem}\label{mt} Let $\phi\in\Irr$ and $\sigma_x(\om(\phi))\geq p-1+c_x$.
Assume that $m>1$ for $G=B_r(K)$ or $D_r(K)$, $r-m>1$ for $G=A_r(K)$,
and $>3$ for $G=B_r(K)$ or $D_r(K)$. Then the element $\phi(x)$ has
more than $d(r-m)$ Jordan blocks of size $p$. \end{theorem}
Proposition~\ref{mp} below shows that one cannot weaken the
inequality for $\sigma_x(\om(\phi))$ in Theorem~\ref{mt} and that
the estimates obtained are asymptotically exact.
\begin{proposition}\label{mp} Let $\phi\in\Irr$ and $\om=\om(\phi)=a\om_1$ with $a<p$.
Assume that $m$ and $r$ are such as in Theorem~\ref{mt} and $x$ is
a regular unipotent element in $G_m$. Set $c=m$ for $G=A_r(K)$,
$2m$ for $G=B_r(K)$, $2m-1$ for $G=C_r(K)$, and $2m-2$ for
$G=D_r(K)$. Suppose that $p>c$. Then $\abs{x}=p$,
$\sigma_x(\om)=ac$ and $c_x=c$. There exist constants
$N_G(a,m,p)$ and $Q_G(a,m,p)$ that depend upon the type of $G$,
$a$, $m$, and $p$ and do not depend upon $r$ such that $\phi(x)$
contains at most $N_G(a,m,p)$ Jordan blocks of size $p$ if
$p<ac<p+c-1$ and at most $d(r-m)+Q_G(a,m,p)$ such blocks if
$ac=p+c-1$. \end{proposition}
Put $l=[(m+2)/2]$ for $G=A_r(K)$ and $l=m$
otherwise. By Lemma~\ref{lc} below,
$c_x=\sum^l_{i=1}\delta_i$ and $c_x\leq p-1$. Hence
$c_x\leq2l$.
For $\varphi\in\Irr$ define the weight $\bar{\omega}(\varphi)$ as
follows: write down the $p$-adic expansion for the weight
$\omega=\omega(\varphi)$ considered before the statement of
Theorem~\ref{mt} and set
$\bar{\omega}(\varphi)=\sum^s_{j=0}\om^j$. So
$\sigma_x(\omega(\varphi))=\tau_x(\bar{\omega}(\varphi))$.
The study of an asymptotic behavior of elements of order $p$ in
representations of the classical groups in characteristic $p$ was
begun by the author in \cite{Supl} where a notion of a $p$-large
representation was introduced. In our present notation a
representation $\phi\in\Irr$ is $p$-large if and only if
$\sigma_x(\om(\phi))\geq p$ for a long root element $x\in G$ (an
equivalent definition from \cite{Supl}: the value of ${\bar
\om}(\phi)$ on the maximal root is $\geq p$). The common goal of
\cite{Supl} and the present article is
to investigate the behavior of elements of order $p$ in irreducible
representations in characteristic $p$ for a fixed $p$ and
$r\rightarrow\infty$ and to discover asymptotic regularities which
are specific for prime characteristics but do not (or almost do
not) depend upon $p$. Such properties can find applications in
recognizing representations and linear groups. According to
~\cite[Theorem~1.1]{Supl}, the image of any element of order $p$
in a $p$-large representation has at least $f(r)$ Jordan blocks of
size $p$ where
\begin{equation*}
f(r)=
\begin{cases}
2r-2 & \text{for $G=A_r(K)$},\\ 6r-7 & \text{for $G=B_r(K)$,
$p=3$},\\ 8r-10 & \text{for $G=B_r(K)$, $p>3$},\\ 4r-4 & \text{for
$G=C_r(K)$},\\ 4r-8 & \text{for $G=D_r(K)$, $p=2$},\\ 6r-10 &
\text{for $G=D_r(K)$, $p=3$},\\ 8r-16 & \text{for $G=D_r(K)$,
$p>3$}.
\end{cases}
\end{equation*}
In \cite[Theorem~1.3]{Supl} for $p>2$ and all types of the
classical groups examples of representations $\phi\in\Irr$ such
that $\phi(x)$ has only one Jordan block of size $p$ for a long
root element $x$ and $\sigma_x(\om(\phi))=p-1$ are given. It is
also shown \cite[Theorem~1.4] {Supl} that the estimates in
\cite[Theorem~1.1]{Supl} are asymptotically exact for the groups
of type $A$, $B$, and $D$ provided $p>2$ in the last two cases.
In what follows $\mathbb{C}$ is the complex field,
$G_{\mathbb{C}}$ is the simple algebraic group over $\mathbb{C}$
of the same type and rank as $G$, and $\Irr_{\mathbb{C}}$ is the set
of irreducible rational representations of $G_{\mathbb{C}}$
(considered up to the equivalence). For $\rho\in\Irr$ or
$\Irr_{\mathbb{C}}$ and a unipotent element $z\in G$ or
$G_{\mathbb{C}}$ denote by $k_{\rho}(z)$ the degree of the
minimal polynomial of $\rho(z)$. It is well known that $k_{\rho}(z)$ is equal
to the maximal size of a Jordan block of $\rho(z)$. If $\phi\in\Irr$, then
$\phi_{\mathbb{C}}$ is the irreducible representation of
$G_{\mathbb{C}}$ with highest weight ${\bar \om}(\phi)$. For
unipotent $x\in G$ put
$k_{\phi_{\mathbb{C}}}(x)=k_{\phi_{\mathbb{C}}}(y)$ where $y\in
G_{\mathbb{C}}$ is a unipotent element with the labelled Dynkin
diagram $\Delta_x$ (this is correctly determined). Now let
$\abs{x}=p$. By the results \cite[Theorem 1.1,
Lemma~2.5, and Proposition~2.12]{Sump},
$k_{\phi_{\mathbb{C}}}(x)=\sigma_x(\om(\phi))+1$ and
$k_{\phi}(x)=\min\{p, k_{\phi_{\mathbb{C}}}(x)\}$. Hence if $z$ is
a long root element, then $k_{\phi}(z)=k_{\phi_{\mathbb{C}}}(z)$
if and only if $\phi$ is not $p$-large. The results of \cite{Sump}
imply that for not very small $p$ and $r$ there exists a wide
class of representations $\phi\in\Irr$ such that
$k_{\phi}(z)=k_{\phi_{\mathbb{C}}}(z)<p$ for a long root element
$z\in G$, but $k_{\phi}(x)=p<k_{\phi_{\mathbb{C}}}(x)$ for many
other elements $x\in G$ of order $p$. In this connection in
\cite[Section 2]{SuKo} a notion of a $p$-large representation for
a given
element $x$ of order $p$ was introduced. A representation $\phi\in\Irr$ was
called $p$-large for $x$ if $\sigma_x(\om(\phi))\geq p$. It has
been conjectured (\cite[Conjecture 1]{SuKo} that if $x\in G_m$,
$r$ is large enough with respect to $m$ and $\phi$ is $p$-large
for $x$, then $\phi(x)$ has at least $F(r)$ blocks of size $p$
where $F$ is an increasing function. Our Proposition~\ref{mp}
formally disproves this conjecture, but Theorem~\ref{mt} actually
proves a refined version of it with a stronger assumption on
$\sigma_x(\om(\phi))$. Thus for arbitrary elements $x$ there is a
gap between the class of representations $\phi\in\Irr$ with
$k_{\phi}(x)=k_{\phi_{\mathbb{C}}}(x)$ and that of representations
where asymptotic estimates for the number of Jordan blocks of size
$p$ in $\phi(x)$ hold. Perhaps for some classes of elements of
order $p$ stronger estimates than those of Theorem~\ref{mt} are
possible, but now it is not clear how to determine such classes.
The case $p=2$ is not considered here, but in this situation
$\phi\in\Irr$ is $2$-large if $\om(\phi)\neq 2^j\om_i$. For
$2$-large representations the estimates from
\cite[Theorem~1.1]{Supl} are available. For remaining
representations certain estimates could be obtained as well, but
this article does not seem a proper place for this. We plan to
handle this question in a subsequent paper which will be devoted to refining
some estimates in \cite{Supl}.
The results of this article as well as those of \cite{Supl} can be easily
transferred to irreducible $K$-representations of finite classical groups
in characteristic $p$.
\section{Notation and preliminary comments}
Throughout the article for a semisimple algebraic group $S$ the
symbols $\Irr(S)$, $\mathbf{X}(S)$, and $\mathbf{X^+}(S)$ mean the
same as the similar ones for $G$ introduced earlier; $R(S)$ is the
set of roots of $S$, $\langle S_1, \ldots, S_j\rangle$ is the
subgroup in $S$ generated by subgroups $S_1, \ldots, S_j$;
$\Irr_p(S)\subset\Irr(S)$ is the set of $p$-restricted
representations, i.e. irreducible representations with
$p$-restricted highest weights; $\mathbf{X}(\phi)$
($\mathbf{X}(M)$) is the set of weights of a representation
$\varphi$ (a module $M$); $\dim M$ is the dimension of $M$;
$M(\omega)$ is the irreducible $S$-module with highest weight
$\omega$; $L$ is the Lie algebra of $G$; $R=R(G)$, $R^+\subset R$
is the set of positive roots; $\Irr_p=\Irr_p(G)$;
$\mathcal{X}_{\beta}\subset G$ and $X_{\beta}\in L$ are the root
subgroup and the root element associated with $\beta\in R$,
$\mathcal{X}_{\pm i}=\mathcal{X}_{\pm\alpha_i}$, and $X_{\pm
i}=X_{\pm\alpha_i}$. Set
$H(\beta_1,\ldots,\beta_j)=\langle\mathcal{X}_{\beta_1},
\mathcal{X}_{-\beta_1} \ldots, \mathcal{X}_{\beta_j},
\mathcal{X}_{-\beta_j}\rangle$. For $\omega\in\mathbf{X}(S)$ and
$\alpha\in R(S)$ denote by $\langle\omega,\alpha\rangle$ the
value of the weight $\omega$ on the root $\alpha$. For an
$S$-module $M$ and a unipotent element $x\in S$ define $k_M(x)$
similarly to $k_{\phi}(x)$. If $\abs{x}=p$, then $n_{\phi}(x)$ is
the number of Jordan blocks of size $p$ of the matrix $\phi(x)$
for a representation $\phi$ of $S$ and $n_M(x)$ denotes the same
number for a module $M$ affording $\phi$.
An element $x\in G$ of order $p$ can be
embedded into a closed connected subgroup $\Gamma$ of type $A_1$ whose
labelled diagram coincides with $\Delta_x$ (see \cite[Theorem 4.2]{LaTe}).
Set $\mathbf{X}_1=\mathbf{X}(A_1(K))$ (the simply connected group
of this type) and identify $\mathbf{X}_1$ with $\mathbb{Z}$
mapping $a\om_1\in\mathbf{X}_1$ into $a\in\mathbb{Z}$. Then
$\mathbf{X}(\Gamma)$ can be identified with a subset of
$\mathbb{Z}$. The canonical homomorphism $\tau_x$ can be obtained
as the restriction of weights from a maximal torus $T\subset G$ to
a maximal torus $T_1\subset \Gamma$ such that $T_1\subset T$. From
now on we fix the tori $T$ and $T_1$, and all weights and roots of
$G$ and $\Gamma$ are considered with respect to $T$ and $T_1$.
Throughout the text $\ep_i$ with $1\le i\le r+1$ for $G=A_r(K)$
and $1\le i\le r$ otherwise are weights of the standard
realization of $G$ labelled as in \cite[ch. VIII, \S 13]{B2}. Set
$e_i=\tau_x(\varepsilon_i)$. One can choose $\Gamma$, $T$ and
$T_1$ such that the restriction to $\Gamma$ of the natural
representation of $G$ is a direct sum of irreducible components
with $p$-restricted highest weights (see comments in \cite[Section
3]{Te});
$e_i\geq e_j$ for $i<j$; $e_i\geq0$ if $G=A_r(K)$ and
$i\leq(r+1)/2$; and $e_i\geq0$ for all $i\le r$ if $G\neq A_r(K)$.
If $H\subset G$ is a semisimple subgroup generated by some root
subgroups, then $T_H=T\cap H$ is a maximal torus in $H$. If
$T_1\subset T_H$, we denote by the same symbol $\tau_x$ the
homomorphism $\mathbf{X}(H)\rightarrow\mathbb{Z}$ determined by
restricting weights from $T_H$ to $T_1$. This causes no confusion.
If an element $v$ of some $G$-module is an eigenvector for $T$,
we denote its weights with respect to $T$, $T_H$, and $T_1$ by
$\omega(v)$, $\omega_H(v)$, and $\omega_{\Gamma}(v)$. In what
follows $x$ is conjugate to an element of $G_m$, $\abs{x}=p$, $m$
and $r-m$ are such as in the assertion of Theorem~\ref{mt}, and
$\delta_i=\tau_x(\alpha_i)$, $1\le i\le r$.
\begin{lemma}\label{lc} Set $l=[(m+2)/2]$ for $G=A_r(K)$ and $l=m$
otherwise. Then $c_x=\sum^l_{i=1}\delta_i$ and
$c_x\leq p-1$.
\end{lemma}
\begin{proof} Put $k=l-1$ for $G=A_r(K)$, $m=2t$, and $k=l$ for $G=A_r(K)$,
$m=2t+1$. Our assumptions on $e_i$, $m-r$, and $x$ imply that
$e_i=0$ for $k<i<r+2-k$ if $G=A_r(K)$ and $e_i=0$ for $i>m$
otherwise; notice that $e_{k+1}=e_{k+2}=0$ for $G=A_r(K)$. Now it
follows from the definition of $c_x$ and the formulae in \cite[ch.
VIII, \S 13]{B2} that
$c_x=\sum^l_{i=1}\delta_i=e_1-e_{l+1}=e_1$. As $e_1$ is a
weight of a $p$-restricted $\Gamma$-module, we have $e_1<p$. This
yields the lemma.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{mt}] Set
$\omega=\omega(\varphi)$ and let $\omega=\sum^r_{i=1}a_i\omega_i$.
It is clear that $\om\neq0$ as $\tau_x(\om)\neq0$. Define
subgroups $H_1$ and $H_2\subset G$ as follows. For $G=A_r(K)$ set
$u=r-t+2$ if $m=2t$ and $r-t+1$ if $m=2t+1$,
$\beta=\varepsilon_{t+1}-\varepsilon_u$,
\begin{equation*}
H_1=H(\alpha_1,\ldots,\alpha_t, \beta, \alpha_u, \ldots,
\alpha_r), \quad H_2=H(\alpha_{t+2}, \ldots, \alpha_{u-2})
\end{equation*}
(we have $H_1=H(\alpha_1, \varepsilon_2-\varepsilon_{r+1})$ for
$m=2$ and $H_1=H(\varepsilon_1-\varepsilon_{r+1})$ for $m=1$). For
$G=B_r(K)$, $C_r(K)$, or $D_r(K)$ put $\beta=\varepsilon_m$,
$2\varepsilon_m$, or $\varepsilon_{m-1}+\varepsilon_m$,
respectively, and
\begin{equation*}
H_1=H(\alpha_1,\ldots,\alpha_{m-1}, \beta).
\end{equation*}
Next, set
\[H_2=H(\alpha_{m+1},\dotsc,\alpha_{r-1},\varepsilon_{r-1}+\varepsilon_r)\]
for $G=B_r(K)$ and
\[H_2=H(\alpha_{m+1},\dotsc,\alpha_r)\]
for $G=C_r(K)$ or $D_r(K)$
(here $H_1=H(\beta)$ for $G=C_r(K)$ and $m=1$).
One easily observes that the sets of roots in brackets used to
define $H_1$ and $H_2$ yield bases of the systems $R(H_1)$ and
$R(H_2)$, respectively. Denote these bases by $\mathcal{B}_i$. In all
cases $H_1$ is conjugate to $G_m$ in $G$. We have
$H_2\cong A_{r-m-1}(K)$, $D_{r-m}(K)$,
$C_{r-m}(K)$, or $D_{r-m}(K)$ for $G=A_r(K)$, $B_r(K)$, $C_r(K)$,
or $D_r(K)$, respectively. It is clear
that the
subgroups $H_1$ and $H_2$ commute. Set $H=H_1H_2$. Let
$U_i=\langle\mathcal{X}_{\gamma}\mid \gamma\in R^+, \quad
\mathcal{X}_{\gamma}\subset H_i\rangle$, $i=1,2$, and $U=U_1U_2$. It is
not difficult to conclude that $U_i$ is a maximal unipotent
subgroup in $H_i$ and $U$ is such a subgroup in $H$. We can assume
that $x\in U_1$, $\Gamma\subset H_1$ and $T_1\subset T_{H_1}$.
We shall write a weight $\mu\in\mathbf{X}(H)$ in the form
$(\mu_1, \mu_2)$ where $\mu_i\in\mathbf{X}(H_i)$ is the
restriction of $\mu$ to $T_{H_i}$. Set $M=M(\om)$.
It is clear that $n_V(x)=\dim(x-1)^{p-1}V$ for each $H$-module
$V$. Taking this into account, it is not difficult to conclude the
following. If $0\subset W_1\subset\ldots\subset W_t=V$ is a
filtration of $V$, $F_i=W_i/W_{i-1}$, $1\le i\le t$, and
$n_{F_i}(x)=n_i$, then
\begin{equation}\label{eqx}
n_V(x)\geq \sum^t_{i=1}n_i.
\end{equation}
First suppose that $\phi\in\Irr_p$. Since passing to the dual
representation does not influence the Jordan form of $\varphi(x)$,
one can assume that $a_i\neq0$ for some $i\leq (r+1)/2$ if
$G=A_r(K)$. As for $p$-large representations the estimates of
\cite[Theorem~1.1]{Supl} hold; we also assume that $\phi$ is not
$p$-large. Hence $\langle\mu, \alpha\rangle<p$ for all
$\mu\in\mathbf{X}(\phi)$ and long roots $\alpha$ (for all $\alpha$
if $G=A_r(K)$ or $D_r(K)$). By the formulae for the maximal roots
of the classical groups in \cite[Tables 1-4]{B1}, this forces that
\begin{equation}\label{eqp}
\begin{split}
a_1+\ldots+a_r<p \quad & \text{for}\quad
G=A_r(K)\quad\text{or}\quad C_r(K), \\
a_1+2a_2+\ldots+2a_{r-1}+a_r<p \quad & \text{for} \quad
G=B_r(K),\\
a_1+2a_2+\ldots+2a_{r-2}+a_{r-1}+a_r<p \quad & \text{for} \quad
G=D_r(K).
\end{split}
\end{equation}
Now we proceed to construct
two composition factors $M_1$ and $M_2$ of the restriction
$M|H$ such that $n_{M_1}(x)\geq d(r-m)$ and $n_{M_2}(x)>0$.
This will be done for almost all $\omega$.
In exceptional cases we shall find one factor $M_1$ such that
$n_{M_1}(x)>d(r-m)$. By (\ref{eqx}), this would yield
the assertion of the theorem.
Let $v\in M$ be a nonzero highest weight
vector. Put $\mu_i=\omega_{H_i}(v)$. The vector $v$ generates an
indecomposable $H$-module $V_1$ with highest weight
$\mu=(\mu_1, \mu_2)$. Using (\ref{eqp}), one can deduce that
$\langle\mu_1, \beta\rangle<p$ for all $\beta\in\mathcal{B}_1$.
Here for $G=B_r(K)$ we take into account that $m>1$. Hence
$\mu_1$ is $p$-restricted. Now assume that either $G\neq B_r(K)$, or
$a_i\neq0$ for some $i<r$. For such representations we construct another
weight vector $w\in M$ that is fixed by $U$. Set $l=t+1$ for $G=A_r(K)$,
$m=2t$; otherwise take $l$ as in
Lemma~\ref{lc}. First suppose that $a_j\neq0$ for some $j\le l$ (Case 1).
Choose maximal such $j$ and put $w=X_{-l}\ldots X_{-(j+1)}X_{-j}v$. Now let
$a_j=0$ for all $j\le l$ (Case 2). Our assumptions on $a_i$ imply that
$a_i\neq0$ for some $i>l$; furthermore, one
can take $i\le(r+1)/2$ for $G=A_r(K)$ and $i<r$ for $G=B_r(K)$. Choose
minimal such $i$ and set $w=X_{-l}\ldots X_{-(i-1)}X_{-i}v$ if $G\neq D_r(K)$
or $i<r$ and $w=X_{-l}\ldots X_{-(r-3)}X_{-(r-2)}X_{-r}v$ for $G=D_r(K)$ and
$i=r$. It follows from \cite[Lemma 2.1(iii) and Lemma 2.9]{Supl} that in all
cases $w\neq0$. Using
\cite[Lemma~72]{St-b} and analyzing the roots in $\mathcal{B}_1$
and $\mathcal{B}_2$ and the weight system $\mathbf{X}(\phi)$, we
get that $U$ fixes $w$ in all situations. Here it is essential
that the case $G=B_r(K)$ with $\om=a_r\om_r$ is excluded. In the
latter case we cannot assert that $\mathcal{X}_{\beta}$ fixes $w$.
Set $\lambda_i=\om_{H_i}(w)$, $i=1,2$. Now it is clear that $w$
generates an indecomposable $H$-module $V_2$ with highest weight
$\lambda=(\lambda_1, \lambda_2)$. We claim that $\lambda_1$ is
$p$-restricted. Write down all the situations where
$\langle\lambda_1, \gamma\rangle\neq \langle\mu_1, \gamma\rangle$
for some $\gamma\in\mathcal{B}_1$. We have
$\langle\lambda_1, \beta\rangle=\langle\mu_1, \beta\rangle-1$ in Case 1 if
$j=l$ and $G\neq B_r(K)$ or $j=l-1$ and $G=D_r(K)$ and in Case 2 for
$G\neq B_r(K)$ and all $i$; and $\langle\lambda_1, \beta\rangle=\langle\mu_1,
\beta\rangle-2$ for $G=B_r(K)$ both in Case 1 with $j=l$ and in Case 2. In
Case 1 we also have
$\langle\lambda_1,\alpha_{j-1}\rangle=\langle\mu_1, \alpha_{j-1}\rangle+1$
if $j>1$ and
$\langle\lambda_1, \alpha_j\rangle=\langle\mu_1,\alpha_j\rangle-1$
if $j<l$. In Case 2 one gets
$\langle\lambda_1,\alpha_{l-1}\rangle=\langle\mu_1, \alpha_{l-1}\rangle+1$
if $l>1$.
In all other situations we have
$\langle\lambda_1, \gamma\rangle=\langle\mu_1, \gamma\rangle$. Now
apply (\ref{eqp}) to conclude that $\lambda_1$ is $p$-restricted.
Set $M_1=M(\mu)$, $M_2=M(\lambda)$, $M^j_1=M(\mu_j)$, and
$M^j_2=M(\lambda_j)$, $j=1,2$. Obviously, $M_i$ is a composition
factor of $V_i$. It is well known that $M_i=M^1_i\otimes M^2_i$.
It is clear that $\tau_x(\mu_1)=\tau_x(\omega)\geq p$. Since $x\in
H_1$, we have $\delta_t=0$ if $\alpha_t\in\mathcal{B}_2$.
So by Lemma~\ref{lc},
\begin{equation*}
\tau_x(\lambda_1)=\tau_x(\omega(w))\geq
\tau_x(\omega)-\sum^l_{i=1}\delta_i=\tau_x(\omega)-c_x\geq p-1.
\end{equation*}
It follows from \cite[Theorem~1.1, Lemma~2.5, and Proposition~2.12]{Sump}
that $k_{M^1_i}(x)=p$. Hence $n_{M_i}(x)\geq \dim M^2_i$. One
easily observes that $M^2_1$ and $M^2_2$ cannot both be trivial
$H_2$-modules. Our assumptions on $r-m$ and
\cite[Proposition~5.4.13]{KL} imply that the dimension of a
nontrivial irreducible $H_2$-module is at least $d(r-m)$. In the
exceptional case where $G=B_r(K)$ and $\om=a_r\om_r$ we need to
evaluate $\dim M^2_1$.
First let $a_r\not=1$. As above, $X_rv\not=0$. This implies that
$\mathbf{X}(M_1^2)$ contains a dominant weight $\mu_2-\alpha_r$ and $\dim
M_1^2$ is greater than the size of the orbit of $\mu_2$ under the action of the
Weyl group of $H_2$. The latter is equal to $2^{r-m-1}\ge d(r-m)$ for our
values of $r-m$. By (1), this yields the assertion of the theorem for almost
all $\varphi\in\operatorname{Irr}_p$. It remains to consider the case where
$G=B_r(K)$ and $\omega=\omega_r$. It is well known that then the restriction
$M\downarrow H_1$ is a direct sum of $2^{r-m}$ $H_1$-modules $N=M(\omega_m)$.
Since $k_m(x)=p$, we get $k_N(x)=p$ and $n_M(x)\ge 2^{r-m}>d(r-m)$.
Now suppose that $\phi\in\Irr\backslash\Irr_p$. By the Steinberg
tensor product theorem \cite[Theorem~1.1]{St}, $\phi$ can be
represented in the form $\bigotimes^s_{j=1}\phi_j\Fr^j$ where $\Fr$
is the Frobenius morphism of $G$ associated with raising elements
of $K$ to the $p$th power and all $\phi_j\in\Irr_p$. It is clear
that the morphism $\Fr$ does not influence the Jordan form of
$\phi(x)$. Hence one can assume that $\phi=\psi\otimes\theta$
where $\theta=\phi_j\Fr^j$ for some $j$ and both $\psi$ and
$\theta$ are nontrivial. Set $a=\sigma_x(\om(\psi))$,
$\nu=\om(\phi_j)$, $b=\tau_x(\nu)$ and define by $\mu$ the
restriction of $\nu$ to $T_H$. Now it follows from the definitions
of $\sigma_x$ and $\tau_x$ that $\sigma_x(\om)=a+b$. By
\cite[Theorem~1.1, Lemma~2.5 and Proposition~2.12]{Sump},
$k_{\psi}(x)=\min\{a+1,p\}$ and $k_{\theta}(x)=\min\{b+1,p\}$.
First suppose that $a$ or $b\geq p-1$. Set $\rho=\psi$ if $a\geq
p-1$ and $\rho=\theta$ otherwise and denote by $\pi$ the remaining
representation from the pair $(\psi, \theta)$. Then
$k_{\rho}(x)=p$ and \cite[ch. VIII, Lemma~2.2]{Fe} implies that
$n_{\phi}(x)\geq \dim\pi$. Let $d(r)$ be the value of $d(r-m)$ if
one formally sets $m=0$. Then by \cite[Proposition~5.4.13]{KL},
$\dim\pi\geq d(r)>d(r-m)$ which settles the case under
consideration.
Now assume that both $a$ and $b<p-1$. Then $k_{\psi}(x)=a+1$ and
$k_{\theta}(x)=b+1$. Since $\sigma_x(\om)\geq p-1+c_x$, we have
$b>c_x$. Arguing as for $p$-restricted $\phi$, we can and shall suppose
that $\langle\nu, \alpha_i\rangle\neq0$ for some $i\leq (r+1)/2$ if
$G=A_r(K)$. Put $M'=M(\nu)$ and construct the composition factors
$M_i$, $i=1,2$, of the restriction $M'|H$ as for $p$-restricted
$M$ before. Transfer the notation $\mu_1$, $\lambda_1$, and
$M^i_j$, $i,j=1,2$, to $M'$. Again we have the exceptional case
$G=B_r(K)$ and $\nu=a_r\om_r$ where we do not construct $M_2$ and
consider $M_1$ only. Obviously, $\tau_x(\mu_1)=b$. As before, we
deduce that $\tau_x(\lambda_1)\geq b-c_x$. By
\cite[Theorem~1.1, Lemma~2.5, and Proposition~2.12]{Sump}, $k_{M_1}(x)=b+1$ and
$k_{M_2}(x)\geq b+1-c_x$.
Let $n_i$ be the number of Jordan blocks of the maximal size in the canonical
form of $x$ as an element of $\operatorname{End} M_i$, $i=1,2$. Looking at the
realizations of $M_i$ as tensor products, one easily observes that $n_i\ge\dim
M_i^2$.
Set $F_1=M(\omega(\psi))\otimes M_1$,
$F_2=M(\omega(\psi))\otimes M_2$ and consider $F_i$ as $H$-modules
in the natural way. In the general case the $H$-module $M$ has a
filtration two of whose quotients are isomorphic to $F_1$ and
$F_2$, respectively. In the exceptional case $F_1$ is a quotient
of a submodule in $M$. Observe that $a+k_{M_2}(x)\geq p$. Using
\cite[ch. VIII, Theorem~2.7]{Fe} that describes the canonical
Jordan form of a tensor product of unipotent blocks, we obtain
that $k_{M_i}(x)=p$ and $n_{F_i}(x)\geq \dim M^2_i$. As for
$p$-restricted $M$, we show that $n_1\ge 2^{r-m}$ if $G=B_r(K)$ and
$\nu=\omega_r$ and conclude that $\dim M_1^2+\dim M_2^2>d(r-m)$ in the general
case and $\dim\,M_1^2>d(r-m)$ in the exceptional cases with $a_r\not=1$.
Now
(\ref{eqx}) completes the proof. \end{proof}
\begin{proof}[Proof of Proposition~\ref{mp}] Let $a$, $x$, $m$, and
$c$ be such as in the assertion of the proposition. Assume that
$p<ac\leq p+c-1$. Therefore
we have $(a-1)c\leq p-1$. Set $M=M(\omega)$ and denote by
$M_t$ the weight subspace of
weight $t\in\mathbb{Z}$ in the $\Gamma$-module $M$. It is clear
that the Weyl group of $\Gamma$ interchanges $M_t$ and $M_{-t}$;
hence $\dim M_t=\dim M_{-t}$. Put $e=(a-1)c$,
$V_1=\bigoplus_{t>e}M_t$, $V_2=\bigoplus_{t>e}M_{-t}$, and $V=M_e$. Set
$f=[(m+1)/2]$ for $G=A_r(K)$, $f=m$ for $G=B_r(K)$ or $C_r(K)$,
and $f=m-1$ for $G=D_r(K)$. Let $v\in M$ be a nonzero highest
weight vector and put $w=X_{-f}\ldots X_{-2}X_{-1}v$. By
\cite[Lemma 2.9]{Supl}, $w\neq0$. We need a
subgroup $S$ which can be defined as follows. Put $I=\{i\mid
1\le i\le r,\quad \delta_i=0\}$ and $S=\langle\mathcal{X}_i,
\mathcal{X}_{-i}\mid i\in I\rangle$. The canonical Jordan forms of $x$
in the standard realizations of $G_m$ and $G$ are well known. We have
$\abs{x}=p$ since the dimension of the first realization is at most $p$
due to our assumptions. Taking into
account these Jordan forms, one easily obtains the values of $\delta_i$,
$1\le i\le r$, and using Lemma~\ref{lc}, deduces the following facts:
$I=\{i\mid f+1\leq i\leq r-f\}$ for $G=A_r(K)$ and
$m=2f$, $I=\{i\mid f+1\leq i\leq r\}$ for $G=B_r(K)$ and $D_r(K)$, and $S=H_2$
in all other cases where $H_2$ is the subgroup defined in the
proof of Theorem~\ref{mt}; $c_x=\sum^f_{i=1}\delta_i=c$,
$\tau_x(\omega)=ac$; and $w\in V$. Next, observe that $S\cong A_{r-m}$ for
$G=A_r(K)$ and $m=2f$, $S\cong B_{r-m}(K)$ for $G=B_r(K)$,
and $S\cong D_{r-m+1}$ for $G=D_r(K)$. Our
construction of the vector $w$ shows that $\mathcal{X}_i$ fixes
$w$ if $i\in I$. This forces that $w$ generates an indecomposable
$S$-module $M_S$ with highest weight $\omega_S(w)$. Then one
immediately concludes that $M_S\cong M(\omega_1)$. This
yields that $\dim M_S=r-m+1=d(r-m)+1$ for $G=A_r(K)$ and $m=2f$,
$\dim M_S=2(r-m)+1=d(r-m)+1$ for $G=B_r(K)$,
$\dim M_S=2(r-m+1)=d(r-m)+2$ for $G=D_r(K)$, and $\dim M_S=d(r-m)$
otherwise. It is clear that $M_S\subset V$. Denote by
$\mathbf{X}_f\subset\mathbf{X}(M)$ the subset of weights of the
form $\omega-\sum^f_{i=1}b_i\alpha_i$ and by $M_A$ the irreducible
$A_f(K)$-module with highest weight $a\omega_1$. By Smith's theorem
\cite{Sm}, for each $\mu\in\mathbf{X}_f$ the dimension of the
weight subspace $M_{\mu}\subset M$ coincides with that of the
weight subspace in $M_A$ whose weight differs from $a\omega_1$ by
the same linear combination of the simple roots. Hence $\dim
M_{\mu}$ does not depend upon $r$. Set
$W=\bigoplus_{\mu\in\mathbf{X}_f}M_{\mu}$. Since $M$ is an
irreducible $L$-module and $p>2$, observe that $M$ is a linear
span of vectors of the form $X_{-i_s}\ldots X_{-i_2}X_{-i_1}v$.
Now, analyzing the weight structure of $M$, we conclude that
$V_1\subset W$ and $V=(V\cap W)\oplus M_S$. This implies that
$\dim V_1$ ($=\dim V_2$) and $\dim (V\cap W)$ do not depend upon
$r$.
It follows from \cite[Lemma 72]{St-b} that
\begin{equation}\label{eqpo}
(x-1)^{p-1}M_t\subset\bigoplus_{i\geq t+2p-2} M_i.
\end{equation}
Let $M_t\not\subset V_2$. Then $t\geq -e$. Obviously, $e<p-1$ if
$ac<p-1+c_x$ and $e=p-1$ for $ac=p-1+c_x$. Thus (\ref{eqpo}) implies
that
\begin{equation*}
(x-1)^{p-1}M_t\subset\bigoplus_{t>p-1}M_t\subset V_1
\end{equation*}
in the first case and
\begin{equation*}
(x-1)^{p-1}M_t\subset\bigoplus_{t\geq p-1}M_t\subset V_1\oplus V
\end{equation*} in the second case. This forces that
$n_M(x)\leq\dim V_2+\dim V_1=2\dim V_1$ in the first case and
$n_M(x)\leq 2\dim V_1+\dim(V\cap W)+\dim M_S$ in the second case.
We have seen before that $\dim M_S=d(r-m)+u$ with $u=0$, $1$, or
$2$. Hence one can take $N_G(a,m,p)=2\dim V_1$ and
$Q_G(a,m,p)=2\dim V_1+\dim(V\cap W)+u$ to complete the proof.
\end{proof}
\begin{remark} For $G=A_r(K)$ or $C_r(K)$
we could give a shorter proof of Proposition~\ref{mp} using the
realization of $\phi$ in the $a$th symmetric power of the standard
module (see \cite[1.14 and 8.13]{Se}), but we need the proof
above for $B_r(K)$ and $D_r(K)$. \end{remark}
\bigskip
This research has been supported by the Institute of Mathematics
of the National Academy of Sciences of Belarus in the framework of
the State program ``Mathematical structures'' and by the Belarus
Basic Research Foundation, Project F\,98-180. | 0.002587 |
On 21 Nov 2012, at 17:09, Melvin Carvalho <melvincarvalho@gmail.com> wrote: > > > On 21 November 2012 17:04, Henry Story <henry.story@bblfish.net> wrote: > I. > > I think it's fine for a straw poll, for those that are interested. I dont see any of the mailing list protocols being violated, imho. If you are suggesting that I suggested the mailing list protocols were being violated, then you misunderstood the intent of my paragraph above. I was just pointing out that people who may not have followed the conversion may want to know why they are being asked to decide on a technical issue. If so they may like to have some information on why they should decide one way or the other. > >. > > > > > > Social Web Architect on Wednesday, 21 November 2012 16:14:36 UTC
This archive was generated by hypermail 2.3.1 : Tuesday, 6 January 2015 20:40:04 UTC | 0.001029 |
\begin{document}
\title[Continuity in Vector Metric Spaces]{Continuity in Vector Metric Spaces}
\author{C\"{u}neyt \c{C}ev\.ik}
\address{Department of Mathematics, Faculty of Science, Gazi
University, 06500 Teknikokullar Ankara, Turkey}
\email{ccevik@gazi.edu.tr}
\date{}
\subjclass[2000]{Primary 58C07; Secondary 46A40,47B60}
\keywords{Vectorial continuity, topological continuity, Riesz space, vector metric space}
\begin{abstract}
We introduce vectorial and topological continuities for functions
defined on vector metric spaces and illustrate spaces of such
functions. Also, we describe some fundamental classes of vector
valued functions and extension theorems.
\end{abstract}
\maketitle
\section{Introduction and Preliminaries}
Let $E$ be a Riesz space.
If every nonempty bounded above (countable) subset of $E$ has a supremum, then
$E$ is called \emph{Dedekind ($\sigma$-)complete}. A subset in $E$ is called \emph{order bounded} if it is bounded both above and below. We write $a_{n}\downarrow a$ if $(a_{n})$ is a decreasing sequence in $E$
such that $\inf a_{n}=a$. The Riesz space $E$ is said to be
\emph{Archimedean} if $n^{-1}a\downarrow 0$ holds for every $a\in E_{+}$. A
sequence $(b_{n})$ is said to be \emph{$o$-convergent} (or \emph{order convergent}) to $b$
if there is a sequence $(a_{n})$ in $E$ such that $a_{n}\downarrow 0$ and $
\left\vert b_{n}-b\right\vert \leq a_{n}$ for all $n$, where $\left\vert
a\right\vert =a\vee (-a)$ for any $a\in E$. We will denote this order convergence by $b_{n}\overset{o}{\longrightarrow }b$. Furthermore, a sequence $(b_{n})$ is said to
be \emph{$o$-Cauchy} if there exists a sequence $(a_{n})$ in $E$
such that $a_{n}\downarrow 0$ and $\left\vert b_{n}-b_{n+p}\right\vert \leq
a_{n}$ for all $n$ and $p$. The Riesz space $E$ is said to be \emph{$o$-Cauchy complete} if
every $o$-Cauchy sequence is $o$-convergent. \vskip2mm
An operator $T:E\rightarrow F$ between two Riesz spaces is
\emph{positive} if $T(x)\geq 0$ for all $x\geq 0$. The operator
$T$ is said to be \emph{order bounded} if it maps order bounded
subsets of $E$ to order bounded subsets of $F$. The operator $T$
is called \emph{$\sigma$-order continuous} if
$x_{n}\overset{o}\longrightarrow 0$ in $E$ implies
$T(x_{n})\overset{o}\longrightarrow 0$ in $F$. Every
$\sigma$-order continuous operator is order bounded. If $T(x\vee
y)=T(x)\vee T(y)$ for all $x,y\in E$, the operator $T$ is a
\emph{lattice homomorphism}. For further information about Riesz
spaces and the operators on Riesz spaces, we refer to \cite{AB}
and \cite{Zaanen}. \vskip2mm
In \cite{CevikAltun}, a vector metric space is defined with a
distance map having values in a Riesz space. Also in
\cite{CevikAltun} and \cite{AltunCevik}, some results in metric
space theory are generalized to vector metric space theory, and
the Baire Theorem and some fixed point theorems in vector metric
spaces are given. Actually, the study of metric spaces having
value on a vector space has started by Zabrejko in
\cite{Zabrejko}. The distance map in the sense of Zabrejko takes
values from an ordered vector space. We use the structure of
lattice with the vector metrics having values in Riesz spaces,
then we have new results as mentioned above. This paper is on some
concepts and related results about continuity in vector metric
spaces. \vskip2mm
\begin{definition}
Let $X$ be a non-empty set and let $E$ be a Riesz space. The function $d:X\times
X\rightarrow E$ is said to be a vector metric (or $E$-metric) if it satisfies the following properties:
(vm1) $d(x,y)=0$ if and only if $x=y,$
(vm2) $d(x,y)\leq d(x,z)+d(y,z)$ for all $x,y,z\in X.$\\
Also the triple $(X,d,E)$ is said to be vector metric space.
\end{definition}
\begin{definition}
Let $(X,d,E)$ be a vector metric space.\vskip1mm
(a) A sequence $(x_{n})$ in $X$ is vectorial
convergent (or is $E$-convergent) to some $x\in X$, if there is a sequence $(a_{n})$ in $E$ such that $
a_{n}\downarrow 0$ and $d(x_{n},x)\leq a_{n}$ for all $n$. We will denote this vectorial convergence by $x_{n}\overset{d,E}{\longrightarrow }x$\vskip1mm
(b) A sequence $(x_{n})$ in $X$ is called $E$-Cauchy sequence whenever there exists
a sequence $(a_{n})$ in $E$ such that $a_{n}\downarrow 0$ and $
d(x_{n},x_{n+p})\leq a_{n}$ for all $n$ and $p$.\vskip1mm
(c) The vector metric space $X$ is called $E$-complete if each $E$-Cauchy
sequence in $X$ is $E$-convergent to a limit in $X$.
\end{definition}
One of the main goals of this paper is to demonstrate the properties of functions on vector metric spaces in a context more general than the continuity in metric analysis. Hence, the properties of Riesz spaces will be the tools for the study of continuity of vector valued functions. The result we get from here is continuity in general sense, an order property rather than a topological feature.\vskip2mm
In Section 2, we consider two types of continuity on vector metric spaces. This approach distinguishes continuities vectorially and topologically. Moreover, vectorial continuity examples are given and the relationship between vectorial continuity of a function and its graph is demonstrated. In Section 3, equivalent vector metrics, vectorial isometry, vectorial homeomorphism definitions and examples related to these concepts are given. In Section 4, uniform continuity is discussed and some extension theorems for functions defined on vector metric spaces are given. Finally, in Section 5, a uniform limit theorem on a vector metric space is given, and the structure of vectorial continuous function spaces is demonstrated.\vskip2mm
\section{Topological and Vectorial Continuity}
We now introduce the concept of continuity in vector metric spaces.
\begin{definition}
Let $(X,d,E)$ and $(Y,\rho ,F)$ be vector metric spaces, and let $x\in X$.\vskip2mm
(a) A function $f:X\rightarrow Y$ is said to be topological continuous at $x$ if for every $b>0$ in $F$ there exists some $a$ in $E$ such that $\rho (f(x),f(y))<b$ whenever $y\in X$
and $d(x,y)<a$. The function $f$ is said to be topological continuous if it is topological continuous at each point of $X$.\vskip2mm
(b) A function $f:X\rightarrow Y$ is said to be vectorial continuous at $x$ if $x_{n}\overset{d,E}{\longrightarrow }x$ in $X$ implies $f(x_{n})\overset{\rho ,F}{\longrightarrow }
f(x)$ in $Y$. The function $f$ is said
to be vectorial continuous if it is vectorial continuous at each point of $X$.
\end{definition}
\vskip2mm
\begin{theorem}
\label{1}Let $(X,d,E)$ and $(Y,\rho ,F)$ be vector metric spaces
where $F$ is Archime-dean. If a function $f:X\rightarrow Y$ is
topological continuous, then $f$ is vectorial continuous.
\end{theorem}
\begin{proof}
Suppose that $x_{n}\overset{d,E}{\longrightarrow }x$. Then there
exists a sequence $(a_{n})$ in $E$ such that $a_{n}\downarrow 0$
and $d(x_{n},x)\leq a_{n}$ for all $n$. Let $b$ be any nonzero
positive element in $F$. By topological continuity of $f$ at $x$,
there exists some $a=a(b;y)$ in $E$ such that $y\in X$ and
$d(x,y)<a$ implies $\rho(f(x),f(y))<b$. Then
there exist elements $b_{n}=b_{n}(\frac{1}{n}b;x_{n})$ in $E$ such that $
\rho (f(x_{n}),f(x))<\frac{1}{n}b$ whenever $d(x_{n},x)\leq a_{n}\wedge
b_{n}\leq a_{n}$ for all $n$. Since $F$ is Archimedean, $\frac{1}{n}
b\downarrow 0$. Hence, $f(x_{n})\overset{\rho ,F}{\longrightarrow }f(x)$.
\end{proof}
\vskip2mm
Vectorial continuous functions have a number of nice characterizations.
\vskip2mm
\begin{corollary}
For a function $f:X\rightarrow Y$ between two vector metric spaces $(X,d,E)$
and $(Y,\rho ,F)$ the following statements hold:\vskip2mm
(a) If $F$ is Dedekind $\sigma $-complete and $f$ is vectorial continuous, then
$\rho (f(x_{n}),f(x))\downarrow 0$ whenever $d(x_{n},x)\downarrow 0$.\vskip2mm
(b) If $E$ is Dedekind $\sigma $-complete and $\rho
(f(x_{n}),f(x))\downarrow 0$ whenever $d(x_{n},x)\downarrow 0$, then the
function $f$ is vectorial continuous.\vskip2mm
(c) Suppose that $E$ and $F$ are Dedekind $\sigma $-complete. Then, the
function $f$ is vectorial continuous if and only if $\rho
(f(x_{n}),f(x))\downarrow 0$ whenever $d(x_{n},x)\downarrow 0$.
\end{corollary}
\begin{proof}
(a) If $d(x_{n},x)\downarrow 0$, then $x_{n}\overset{d,E}{\longrightarrow }x$. By the vectorial continuity of the function $f$, there is a sequence $(b_{n})$ in $F$ such that $b_{n}\downarrow 0$ and $\rho(f(x_{n}),f(x))\leq b_{n}$ for all $n$. Since $f$ is Dedekind $\sigma$-complete, $\rho (f(x_{n}),f(x))\downarrow 0$ holds.\vskip2mm
(b) Let $x_{n}\overset{d,E}{\longrightarrow }x$ in $X$. Then there is a sequence $(a_{n})$ in $E$ such that $a_{n}\downarrow 0$ and $d(x_{n},x)\leq a_{n}$ for all $n$. Since $E$ is Dedekind $\sigma$-complete, $d(x_{n},x)\downarrow 0$ holds. By the hypothesis, $\rho (f(x_{n}),f(x))\downarrow 0$, and so $f(x_{n})\overset{\rho,F}{\longrightarrow }f(x)$ in $Y$.\vskip2mm
(c) Proof is a consequence of (a) and (b).
\end{proof}
\vskip2mm
\begin{example}
Let $(X,d,E)$ be a vector metric space. If $x_{n}\overset{d,E}{
\longrightarrow }x$ and $y_{n}\overset{d,E}{\longrightarrow }y$, then $d(x_{n},y_{n})
\overset{o}{\longrightarrow }d(x,y)$, i.e, the vector metric map $d$ from $X^{2}$ to $E$ is
vectorial continuous. Here, $X^{2}$ is equipped with the $E$-valued vector metric $
\tilde{d}$ defined as $\tilde{d}(z,w)=d(x_{1},x_{2})+d(y_{1},y_{2})$ for all $
z=(x_{1},y_{1}),w=(x_{2},y_{2})\in X^{2}$, and $E$ is equipped with the absolute valued vector metric $\left\vert\cdot\right\vert$.
\end{example}
\vskip2mm
We recall that a subset $U$ of a vector metric space $(X,d,E)$ is called \emph{$E$-closed}
whenever $(x_{n})\subseteq U$ and $x_{n}\overset{d,E}{\longrightarrow }x$ imply $
x\in U$.
\vskip2mm
\begin{theorem}
\label{theo3}
Let $(X,d,E)$ and $(Y,\rho ,F)$ be vector metric spaces. If a function $f:X\rightarrow Y$ is vectorial continuous, then for every $F$-closed subset $B$ of $Y$ the set $f^{-1}(B)$ is $E$-closed in $X$.
\end{theorem}
\begin{proof}
For any $x\in f^{-1}(B)$, there exists a sequence $(x_{n})$ in $f^{-1}(B)$ such that $x_{n}\overset{d,E}{\longrightarrow }x$. Since the function $f$ is $(E,F)$-continuous, $f(x_{n})\overset{\rho ,F}{\longrightarrow }f(x)$. But the set $B$ is $F$-closed, so $x\in f^{-1}(B)$. Then, the set $f^{-1}(B)$ is $E$-closed.
\end{proof}
\vskip2mm
If $E$ and $F$ are two Riesz spaces, then $E\times F$ is also a Riesz space with coordinatewise ordering defined as
\begin{equation*}
(e_{1},f_{1})\leq (e_{2},f_{2})\text{ }\Longleftrightarrow \text{ }e_{1}\leq e_{2}\ \text{ and
}\ f_{1}\leq f_{2}
\end{equation*}
for all $(e_{1},f_{1}),(e_{2},f_{2})\in E\times F$. The Riesz space $E\times F$ is a vector metric space, equipped with the \emph{biabsolute valued vector metric} $\left\vert\cdot\right\vert$ defined as
\begin{equation*}
\left\vert a-b\right\vert=(\left\vert e_{1}-e_{2}\right\vert,\left\vert f_{1}-f_{2}\right\vert)
\end{equation*}
for all $a=(e_{1},f_{1})$, $b=(e_{2},f_{2})\in E\times F$.
\vskip2mm
Let $d$ and $\rho$ be two vector metrics on $X$ which are $E$-valued and $F$-valued respectively. Then the map $\delta$ defined as
\begin{equation*}
\delta(x,y)=(d(x,y),\rho(x,y))
\end{equation*}
for all $x,y\in X$ is an $E\times F$-valued vector metric on $X$. We will call $\delta$ \emph{double vector metric}.
\vskip4mm
\begin{example}
Let $(X,d,E)$ and $(X,\rho ,F)$ be vector metric spaces.\vskip2mm
(i) Suppose that $f:X\rightarrow E$ and $g:X\rightarrow F$ are
vectorial continuous functions. Then the function $h$ from $X$ to
$E\times F$ defined by $h(x)=(f(x),g(x))$ for all $x\in X$ is
vectorial continuous with the double vector metric $\delta$ and
the biabsolute valued vector metric
$\left\vert\cdot\right\vert$.\vskip2mm
(ii) Let $p_{E}:E\times F\rightarrow E$ and $p_{F}:E\times F\rightarrow F$ be the projection maps. Any function $h:X\rightarrow E\times F$ can be written as $h(x)=(f(x),g(x))$ for all $x\in X$ where $f=p_{E}\circ h$ and $g=p_{F}\circ h$. If $h$ is vectorial continuous, so are $f$ and $g$ since $p_{E}$ and $p_{F}$ are vectorial continuous.
\end{example}
\vskip2mm
Let $(X,d,E)$ and $(Y,\rho ,F)$ be vector metric spaces. Then $X\times Y$ is a vector metric space, equipped with the $E\times F$-valued \emph{product vector metric} $\pi$ defined as
\begin{equation*}
\pi(z,w)=(d(x_{1},x_{2}),\rho(y_{1},y_{2}))
\end{equation*}
for all $z=(x_{1},y_{1})$, $w=(x_{2},y_{2})\in X\times Y$.\vskip2mm
Consider the vector metric space $(X\times Y,\pi,E\times F)$. The projection maps $p_{X}$ and $p_{Y}$ defined on $X\times Y$ which are $X$-valued and $Y$-valued respectively are vectorial continuous. For any function $f$ from a vector metric space $(Z,\delta,G)$ to $(X\times Y,\pi,E\times F)$, the function $f$ is vectorial continuous if and only if both $p_{X}\circ f$ and $p_{Y}\circ f$ are vectorial continuous.
\vskip2mm
\begin{example}
Let $(X,d,E)$ and $(Y,\rho ,E)$ be vector metric spaces. Suppose
that $f:X\rightarrow E$ and $g:Y\rightarrow E$ are vectorial
continuous functions. Then the function $h$ from $X\times Y$ to
$E$ defined by $h(x,y)=\left\vert f(x)-g(y)\right\vert$ for all
$x\in X$, $y\in Y$ is vectorial continuous with the product vector
metric $\pi$ and the absolute valued vector metric
$\left\vert\cdot\right\vert$.
\end{example}
\vskip2mm
\begin{example}
Let $(X,d,E)$ and $(Y,\rho ,F)$ be vector metric spaces. Suppose
that $f:X\rightarrow E$ and $g:Y\rightarrow F$ are vectorial
continuous functions. Then the function $h$ from $X\times Y$ to
$E\times F$ defined by $h(x,y)=(f(x),g(y))$ for all $x\in X$,
$y\in Y$ is vectorial continuous with the product vector metric
$\pi$ and the biabsolute valued vector metric
$\left\vert\cdot\right\vert$.
\end{example}
\vskip2mm
The last three examples inspire the following results.
\vskip4mm
\begin{corollary}
(a) If $f:(X,d,E)\rightarrow (Y,\eta,G)$ and
$g:(X,\rho,F)\rightarrow (Z,\xi,H)$ are vectorial continuous
functions, then the function $h$ from $X$ to $Y\times Z$ defined
by $h(x)=(f(x),g(x))$ for all $x\in X$ is vectorial continuous
with the $E\times F$-valued double vector metric $\delta$ and the
$G\times H$-valued product vector metric $\pi$.\vskip2mm
(b) Let $G$ be a Riesz space. If $f:(X,d,E)\rightarrow G$ and
$g:(Y,\rho,F)\rightarrow G$ are vectorial continuous functions,
then the function $h$ from $X\times Y$ to $G$ defined by
$h(x,y)=\left\vert f(x)-g(y)\right\vert$ for all $x\in X$, $y\in
Y$ is vectorial continuous with the $E\times F$-valued product
vector metric $\pi$ and the absolute valued vector metric
$\left\vert\cdot\right\vert$.\vskip2mm
(c) If $f:(X,d,E)\rightarrow (Z,\eta,G)$ and
$g:(Y,\rho,F)\rightarrow (W,\xi,H)$ are vectorial continuous
functions, then the function $h$ from $X\times Y$ to $Z\times W$
defined by $h(x,y)=(f(x),g(y))$ for all $x\in X$, $y\in Y$ is
vectorial continuous with the $E\times F$-valued and $G\times
H$-valued product vector metrics.
\end{corollary}
\vskip2mm
We have the next proposition for any product vector metric.
\vskip2mm
\begin{proposition}
\label{prop}Let $(z_{n})=(x_{n},y_{n})$ be a sequence in $(X\times Y,\pi,E\times F)$ and let $z=(x,y)\in X\times Y$. Then, \ $z_{n}\overset{\pi,E\times F}{\longrightarrow }z$ \ if and only if \ $x_{n}\overset{d,E}{\longrightarrow }x$ \ and \ $y_{n}\overset{\rho,F}{\longrightarrow }y.$
\end{proposition}
\vskip2mm
Now let us give a relevance between vectorial continuity of a function and being closed of its graph.
\vskip2mm
\begin{corollary}
Let $(X,d,E)$ and $(Y,\rho ,F)$ be vector metric spaces and let $f$ be a function from $X$ to $Y$. For the graph $G_{f}$ of $f$, the following statements hold.\vskip2mm
(a) The graph $G_{f}$ is $E\times F$-closed in $(X\times Y,\pi,E\times F)$ if and only if for every sequence $(x_{n})$ with $x_{n}\overset{d,E}{\longrightarrow }x$ and $f(x_{n})\overset{\rho,F}{\longrightarrow }y$ we have $y=f(x)$.\vskip2mm
(b) If the function $f$ is vectorial continuous, then the graph $G_{f}$ is $E\times F$-closed.\vskip2mm
(c) If the function $f$ is vectorial continuous at $x_{0}\in X$, then the induced function $h:X\rightarrow G_{f}$ defined by $h(x)=(x,f(x))$ is vectorial continuous at $x_{0}\in X$.
\end{corollary}
\begin{proof}
For the proof of (a), suppose that the graph $G_{f}$ is $E\times F$-closed. If $x_{n}\overset{d,E}{\longrightarrow }x$ and $f(x_{n})\overset{\rho,F}{\longrightarrow }y$, then we have $(x_{n},f(x_{n}))\overset{\pi,E\times F}{\longrightarrow }(x,y)$ by Proposition \ref{prop}. Hence $(x,y)\in G_{f}$, and so $y=f(x)$. Conversely, suppose that $(z_{n})=(x_{n},f(x_{n}))$ is a sequence in $G_{f}$ such that $z_{n}\overset{\pi,E\times F}{\longrightarrow }z=(x,y)\in X\times Y$. By Proposition \ref{prop}, $x_{n}\overset{d,E}{\longrightarrow }x$ and $f(x_{n})\overset{\rho,F}{\longrightarrow }y$. Then $y=f(x)$, and so $z\in G_{f}$.\vskip2mm
Proofs of (b) and (c) are similar to the proof of (a).
\end{proof}
\vskip2mm
\section{Fundamental Vector Valued Function Classes}
\begin{definition}
The $E$-valued vector metric $d$ and $F$-valued vector metric $\rho$ on $X$ are said to be $(E,F)$-equivalent if for any $x\in X$ and any sequence $(x_{n})$ in $X$, $$x_{n}\overset{d,E}{\longrightarrow }x \text{\ \ \ if and only if\ \ \ } x_{n}\overset{\rho,F}{\longrightarrow }x.$$
\end{definition}
\vskip2mm
\begin{lemma}
For any two $E$-valued vector metrics $d$ and $\rho$ on $X$, the
following statements are equivalent:
(a) There exist some $\alpha,\beta>0$ in $\mathbb{R}$ such that
$\alpha d(x,y)\leq\rho (x,y)\leq\beta d(x,y)$ for all $x,y\in X$.
(b) There exist two positive and $\sigma$-order continuous
operators $T$ and $S$ from $E$ to itself such that $\rho (x,y)\leq
T(d(x,y))$ and $d(x,y)\leq S(\rho (x,y))$ for all $x,y\in X$.
\end{lemma}
\begin{proof}
Let $T$ and $S$ be two operators defined as $T(a)=\beta a$ and
$S(a)=\alpha^{-1}a$ for all $a\in E$. If (a) holds, then $T$ and
$S$ are positive and $\sigma$-order continuous operators, and
satisfy $\rho (x,y)\leq T(d(x,y))$ and $d(x,y)\leq S(\rho (x,y))$
for all $x,y\in X$. Conversely, (a) holds since every
($\sigma$-)order continuous operator is order bounded (\cite{AB},
1.54).
\end{proof}
\vskip2mm
Now, we give the following result for the equivalence of vector metrics.
\vskip2mm
\begin{theorem}
An $E$-valued vector metric $d$ and a $F$-valued vector metric $\rho$ on $X$ are $(E,F)$-equivalent if there exist positive and $\sigma$-order continuous two operators $T:E\rightarrow F$, $S:F\rightarrow E$ such that
\begin{equation}
\rho (x,y)\leq T(d(x,y))\text{\ \ \ and\ \ \ }d(x,y)\leq S(\rho (x,y))\label{2}
\end{equation}
for all $x,y\in X$.
\end{theorem}
\vskip2mm
\begin{example}
Suppose that the ordering of \ $\mathbb{R}^{2}$ is coordinatewise.\vskip2mm
(a) Let $d$ and $\rho$ be $\mathbb{R}$-valued and $\mathbb{R}^{2}$-valued vector metrics on $\mathbb{R}$ respectively, defined as
\begin{equation*}
d(x,y)=a\left\vert x-y\right\vert\ \ \ and\ \ \ \rho(x,y)=(b\left\vert x-y\right\vert,c\left\vert x-y\right\vert)
\end{equation*}
where $b,c\geq 0$ and $a,b+c>0$. Consider the two operators \ $T:\mathbb{R}\rightarrow \mathbb{R}^{2}\ ;\ T(x)=a^{-1}(b\:x,c\:x)$ \ and \ $S:\mathbb{R}^{2}\rightarrow \mathbb{R}\ ;\ S(x,y)=a\:b^{-1}x$ for all $x,y\in\mathbb{R}$. Then, the operators $T$ and $S$ are positive and $\sigma$-order continuous, and (\ref{2}) is satisfied. Hence, the metrics $d$ and $\rho$ are $(\mathbb{R},\mathbb{R}^{2})$-equivalent on $\mathbb{R}$.\vskip2mm
(b) Let $d$ and $\rho$ be $\mathbb{R}$-valued and $\mathbb{R}^{2}$-valued vector metrics on $\mathbb{R}^{2}$ respectively, defined as
\begin{equation*}
d(x,y)=a\left\vert x_{1}-y_{1}\right\vert+b\left\vert x_{2}-y_{2}\right\vert\ \ \ and\ \ \ \rho(x,y)=(c\left\vert x_{1}-y_{1}\right\vert,e\left\vert x_{2}-y_{2}\right\vert)
\end{equation*}
where $x=(x_{1},x_{2})$, $y=(y_{1},y_{2})$ and $a,b,c,e>0$. Let $T:\mathbb{R}\rightarrow \mathbb{R}^{2}$ and $S:\mathbb{R}^{2}\rightarrow \mathbb{R}$ be two operators defined as \ $T(x)=(c\:a^{-1}\:x,e\:b^{-1}x)$ \ and \ $S(x,y)=a\:c^{-1}x+b\:e^{-1}y$. Then, the operators $T$ and $S$ are positive and $\sigma$-order continuous. The condition (\ref{2}) is satisfied. So, the vector metrics $d$ and $\rho$ are $(\mathbb{R},\mathbb{R}^{2})$-equivalent on $\mathbb{R}^{2}$. On the other hand, if $\eta$ is another $\mathbb{R}$-valued vector metric on $\mathbb{R}^{2}$ defined as
\begin{equation*}
\eta(x,y)=\max\left\{a\left\vert x_{1}-y_{1}\right\vert,b\left\vert x_{2}-y_{2}\right\vert\right\}
\end{equation*}
where $x=(x_{1},x_{2})$, $y=(y_{1},y_{2})$ and $a,b>0$, and the operator $S$ is defined as $S(x,y)=\max\left\{a\:c^{-1}x,b\:e^{-1}y\right\}$, then the vector metrics $\eta$ and $\rho$ are $(\mathbb{R},\mathbb{R}^{2})$-equivalent on $\mathbb{R}^{2}$.
\end{example}
\vskip2mm
\begin{remark}
Vectorial continuity is invariant under equivalent vector metrics.
\end{remark}
\vskip2mm
Let us show how an isometry is defined between two vector metric spaces.
\vskip2mm
\begin{definition}
Let $(X,d,E)$ and $(Y,\rho ,F)$ be vector metric spaces. A
function $f:X\rightarrow Y$ is said to be a vector isometry if
there exists a linear operator $T_{f}:E\rightarrow F$ satisfying
the following two conditions, \vskip0mm (i) $T_{f}(d(x,y))=\rho
(f(x),f(y))$ for all $x,y\in X$,\vskip0mm
(ii) $T_{f}(a)=0$ implies $a=0$ for all $a\in E$.\\
If the function $f$ is onto, and the operator $T_{f}$ is a lattice homomorphism, then the vector metric spaces $(X,d,E)$ and $(Y,\rho ,T_{f}(E))$ are called vector isometric.
\end{definition}
\vskip2mm
\begin{remark}
A vector isometry is a vectorially distance preserving one-to-one function.
\end{remark}
\vskip2mm
\begin{example}
Let $d$ be $\mathbb{R}$-valued vector metric and let $\rho$ be $\mathbb{R}^{2}$-valued vector metric on $\mathbb{R}$ defined as
\begin{equation*}
d(x,y)=a\left\vert x-y\right\vert\ \ \ and\ \ \ \rho(x,y)=(b\left\vert x-y\right\vert,c\left\vert x-y\right\vert)
\end{equation*}
where $b,c\geq 0$ and $a,b+c>0$. Consider the identity mapping $I$ on $\mathbb{R}$ and the operator $T_{I}:\mathbb{R}\rightarrow \mathbb{R}^{2}$ defined as $T_{I}(x)=a^{-1}(bx,cx)$ for all $x\in\mathbb{R}$. Then, the identity mapping $I$ is a vector isometry. So, the vector metric spaces $(\mathbb{R},d,\mathbb{R})$ and $(\mathbb{R},\rho,\left\{(x,y):cx=by;\:x,y\in\mathbb{R}\right\})$ are vector isometric.
\end{example}
\vskip2mm
\begin{definition}
Let $(X,d,E)$ and $(Y,\rho ,F)$ be vector metric spaces. A function $f:X\rightarrow Y$ is said to be a vector homeomorphism if $f$ is a one-to-one and vectorial continuous and has a vectorial continuous inverse on $f(X)$. If the function $f$ is onto, then the vector metric spaces $X$ and $Y$ are called vector homeomorphic.
\end{definition}
\vskip2mm
\begin{remark}
A vector homeomorphism is one-to-one function that preserves vectorial convergence of sequences.
\end{remark}
\vskip2mm
By Theorem \ref{theo3}, we can develop another characterization result for vector homeomorphisms.
\vskip2mm
\begin{theorem}
An onto vector homeomorphism is a one-to-one function that preserves vector closed sets.
\end{theorem}
\begin{proof}
Let $f:X\rightarrow Y$ be an $(E,F)$-homeomorphism. Since $f$ is a one-to-one function and its inverse $f^{-1}$ is vectorial continuous, by Theorem \ref{theo3} for every $E$-closed set $A$ in $X$, $f(A)=(f^{-1})^{-1}(A)$ is $F$-closed in $Y$.
\end{proof}
\vskip2mm
The following example illustrates a relationship between vectorial equivalence and vector homeomorphism.
\begin{example}
Let $d$ and $\rho$ be two $(E,F)$-equivalent vector metrics on $X$. Then $(X,d,E)$ and $(X,\rho ,F)$ are vector homeomorphic under the identity mapping. On the other hand, if vector metric spaces $(X,d,E)$ and $(Y,\rho ,F)$ are vector homeomorphic under a function $f$, then the vector metrics $d$ and $\delta$ defined as $$\delta (x,y)=\rho (f(x),f(y))$$ for all $x,y\in X$ are $(E,F)$-equivalent vector metrics on $X$.
\end{example}
\section{Extension Theorems on Continuity}
If $X$ and $Y$ are vector metric spaces, $A\subseteq X$ and $f:A\rightarrow
Y $ is vectorial continuous, then we might ask whether there exists a vectorial
continuus extension $g$ of $f$. Below, we deal with some simple extension
techniques.
\vskip2mm
\begin{theorem}
\label{5}Let $(X,d,E)$ and $(Y,\rho ,F)$ be vector metric spaces, and let $f$ and $g$
be vectorial continuous functions from $X$ to $Y$. Then the set $\{x\in
X:f(x)=g(x)\}$ is an $E$-closed subset of $X$.
\end{theorem}
\begin{proof}
Let $B=\left\{x\in X:f(x)=g(x)\right\}$. Suppose $(x_{n})\subseteq B$ and $x_{n}\overset{d,E}{\longrightarrow }x$. Since $f$ and $g$ are vectorial continuous, there exist sequences $(a_{n})$ and $(b_{n})$ such that $a_{n}\downarrow 0$, $b_{n}\downarrow 0$ and $\rho (f(x_{n}),f(x))\leq a_{n}$, $\rho (g(x_{n}),g(x))\leq b_{n}$ for all $n$. Then $$\rho (f(x),g(x))\leq \rho (f(x_{n}),f(x))+\rho (f(x_{n}),g(x_{n}))+\rho (g(x_{n}),g(x))\leq a_{n}+b_{n}$$ for all $n$. So $f(x)=g(x)$, i.e $x\in B$. Hence, the set $B$ is an $E$-closed subset of $X$.
\end{proof}
\vskip2mm
The following corollary is a consequence of Theorem \ref{5}.
\vskip2mm
\begin{corollary}
Let $(X,d,E)$ and $(Y,\rho ,F)$ be vector metric spaces and let $f$ and $g$
be vectorial continuous functions from $X$ to $Y$. If the set $\{x\in
X:f(x)=g(x)\}$ is $E$-dense in $X$, then $f=g$.
\end{corollary}
\vskip2mm
\begin{definition}
Let $(X,d,E)$ and $(Y,\rho ,F)$ be vector metric spaces.\vskip2mm
(a) A function $f:X\rightarrow Y$ is said to be topological uniformly continuous on $X$ if for every $b>0$ in $F$
there exists some $a$ in $E$ such that for all $x,y\in X$, $\rho (f(x),f(y))<b$ whenever $d(x,y)<a$.\vskip2mm
(b) A function $f:X\rightarrow Y$ is said to be vectorial uniformly continuous on $X$ if for every $E$-Cauchy sequence $(x_{n})$ the sequence $f(x_{n})$ is $F$-Cauchy. \end{definition}
\vskip2mm
\begin{theorem}
\label{Theo5}Let $(X,d,E)$ and $(Y,\rho ,F)$ be vector metric spaces where $F$ is Archime-dean. If a function $f:X\rightarrow Y$ is topological uniformly continuous, then $f$ is vectorial uniformly continuous.
\end{theorem}
\begin{proof}
Suppose that $(x_{n})$ is an $E$-Cauchy sequence. Then there exists a
sequence $(a_{n})$ in $E$ such that $a_{n}\downarrow 0$ and $d(x_{n},x_{n+p})\leq
a_{n}$ for all $n$ and $p$. Let $b$ be any nonzero positive element in $F$. By
topological uniformly continuity of $f$, there exists some $a=a(b)$ in $E$
such that for all $x,y\in X$ the inequality $d(x,y)<a$ implies $\rho (f(x),f(y))<b$ . Then
there exist elements $b_{n}=b_{n}(\frac{1}{n}b)$ in $E$ such that $
\rho (f(x_{n}),f(x_{n+p}))<\frac{1}{n}b$ whenever $d(x_{n},x_{n+p})\leq a_{n}\wedge
b_{n}\leq a_{n}$ for all $n$ and $p$. Since $F$ is Archimedean, $\frac{1}{n}
b\downarrow 0$. Hence, the sequence $f(x_{n})$ is $F$-Cauchy.
\end{proof}
\vskip2mm
\begin{example}
(a) For a vector isometry $f$ between two vector metric spaces $(X,d,E)$ and $(Y,\rho ,F)$, the function $f$ is vectorial uniformly continuous if \ $T_{f}$ is positive and $\sigma$-order continuous.\vskip2mm
(b) For an element $y$ in a vector metric space $(X,d,E)$, the function $f_{y}:X\rightarrow E$ defined by $f_{y}(x)=d(x,y)$ for all $x\in X$ is vectorial uniformly continuous.\vskip2mm
(c) For a subset $A$ of a vector metric space $(X,d,E)$ where $E$ is Dedekind complete, the function $f_{A}:X\rightarrow E$ defined by $f_{A}(x)=d(x,A)=\inf\left\{d(x,y):y\in A\right\}$ for all $x\in X$ is vectorial uniformly continuous.
\end{example}
\vskip2mm
The following theorem enables us to establish a extension property for the functions between vector metric spaces.
\vskip2mm
\begin{theorem}
Let $A$ be $E$-dense subset of a vector metric space $(X,d,E)$ and let $(Y,\rho ,F)$ be a $F$-complete vector metric space where $F$ is Archimedean. If $f:A\rightarrow Y$ is a topological uniformly continuous function, then $f$ has a unique vectorial continuous extension to $X$ which is also this extension is topological uniformly continuous.
\end{theorem}
\begin{proof}
Let $x\in X$. Then there exists a sequence $(x_{n})$ in $A$ such that $x_{n}\overset{d,E}{\longrightarrow }x$. Since the function $f$ is vectorial uniformly continuous on $A$ by Theorem \ref{Theo5}, the sequence $(f(x_{n}))$ is $F$-Cauchy in $F$-complete vector metric space $Y$. Hence, there exists an element $y\in Y$ such that $f(x_{n})\overset{\rho,F}{\longrightarrow }y$. Define an extension $g$ of $f$ on $X$ by $g(x)=y$. This extension is well-defined, that is, the value of $g$ at $x$ is independent of the particular sequence $(x_{n})$ chosen $E$-convergent to $x$. We need to show that $g$ is topological uniformly continuous on $X$.\vskip2mm
Let $a>0$ in $F$. Choose $b>0$ in $E$ such that for $x,y\in A$ $d(x,y)<b$ implies $\rho (f(x),f(y))<a$. Let $x,y\in X$ satisfy $d(x,y)<b$. Choose two sequences $(x_{n})$ and $(y_{n})$ in $A$ such that $x_{n}\overset{d,E}{\longrightarrow }x$ and $y_{n}\overset{d,E}{\longrightarrow }y$. Then, $d(x_{n},y_{n})\overset{o}{\longrightarrow }d(x,y)$ in $E$. Fix $n_{0}$ such that $d(x_{n},y_{n})<b$ for all $n>n_{0}$. Then $\rho (f(x_{n}),f(y_{n}))<a$ for all $n>n_{0}$. By the vectorial uniformly continuity of $f$, $f(x_{n})$ and $f(y_{n})$ are $F$-Cauchy sequences in $Y$. Since $Y$ is $F$-complete, there exist two points $u$ and $v$ in $Y$ such that $f(x_{n})\overset{\rho,F}{\longrightarrow }u$ and $f(y_{n})\overset{\rho,F}{\longrightarrow }v$. By the definition of the function $g$, we have $g(x)=u$ and $g(y)=v$. Then $\rho(g(x_{n}),g(y_{n}))\overset{o}{\longrightarrow }\rho(g(x),g(y))$ in $F$, and therefore, $\rho (g(x),g(y))\leq b$. This shows that $g$ is topological uniformly continuous function on $X$.
\end{proof}
\section{Vectorial Continuous Function Spaces}
\begin{definition}
Let $X$ be any nonempty set and let $(Y,\rho,F)$ be a vector metric space. Then a sequence $(f_{n})$ of functions from $X$ to $Y$ is said to be uniformly $F$-convergent to a function $f:X\rightarrow Y$, if there exists a sequence $(a_{n})$ in $F$ such that $a_{n}\downarrow 0$ and $\rho(f_{n}(x),f(x))\leq a_{n}$ holds for all $x\in X$ and all $n\in \mathbb{N}$.
\end{definition}
\vskip2mm
Now we give the uniform limit theorem in vector metric spaces.
\vskip2mm
\begin{theorem}\label{7}
Let $(f_{n})$ be a sequence of vectorial continuous functions between two vector metric spaces $(X,d,E)$ and $(Y,\rho,F)$. If $(f_{n})$ is uniformly $F$-convergent to $f$, then the function $f$ is vectorial continuous.
\end{theorem}
\begin{proof}
Let $(x_{n})\subseteq X$ such that $x_{n}\overset{d,E}{\longrightarrow }x$ in $X$. Since $(f_{n})$ is uniformly $F$-convergent to $f$, there is a sequence $(a_{n})$ in $F$ such that $a_{n}\downarrow 0$ and $\rho(f_{n}(x),f(x))\leq a_{n}$ for all $n\in \mathbb{N}$. For each $k\in \mathbb{N}$, there is a sequence $(b_{n})$ in $F$ such that $b_{n}\downarrow 0$ and $\rho(f_{k}(x_{n}),f_{k}(x))\leq b_{n}$ for all $n\in \mathbb{N}$ by the vectorial continuity of $f_{k}$. Note that for $k=n$
$$\rho(f(x_{n}),f(x))\leq\rho(f(x_{n}),f_{n}(x_{n}))+\rho(f(x),f_{n}(x))+\rho(f_{n}(x_{n}),f_{n}(x))\leq 2a_{n}+b_{n}.$$
This implies $f(x_{n})\overset{\rho,F}{\longrightarrow }f(x)$.
\end{proof}
\vskip2mm
Let $A$ be a nonempty subset of a vector metric space $(X,d,E)$. \emph{$E$-diameter} of $A$, $d(A)$, is defined as $\sup\left\{d(x,y):x,y\in A\right\}$. The set $A$ is called \emph{$E$-bounded} if there exists an element $a>0$ in $E$ such that $d(x,y)\leq a$ for all $x,y\in A$. Every $E$-bounded subset of $X$ has an $E$-diameter whenever $E$ is Dedekind complete.
\vskip2mm
\begin{definition}
A function $f:X\rightarrow Y$ between two vector metric spaces $(X,d,E)$ and $(Y,\rho,F)$ is called vectorial bounded if $f$ maps $E$-bounded subsets of $X$ to $F$-bounded subsets of $Y$.
\end{definition}
\begin{theorem}
\label{lem}A function $f:X\rightarrow Y$ between two vector metric spaces $(X,d,E)$ and $(Y,\rho,F)$ is vectorial bounded if there exists a positive operator $T:E\rightarrow F$ such that $\rho(f(x),f(y))\leq T(d(x,y))$ for all $x,y\in X$.
\end{theorem}
\vskip2mm
Let $C_{v}(X,F)$ and $C_{t}(X,F)$ be the collections of all vectorial continuous and topological continuous functions between a vector metric space $(X,d,E)$ and a Riesz space $F$, respectively. By Theorem \ref{1}, $C_{t}(X,F)\subseteq C_{v}(X,F)$ whenever $F$ is Archimedean.
\vskip2mm
\begin{theorem}
The spaces $C_{v}(X,F)$ and $C_{t}(X,F)$ are Riesz spaces with the natural partial ordering defined as $f\leq g$ whenever $f(x)\leq g(x)$ for all $x\in X$.
\end{theorem}
\vskip2mm
Consider an $E$-bounded vector metric space $X$ and a Dedekind
complete Riesz space $F$. Let $C_{v}^{o}(X,F)$ be a subset of
$C_{v}(X,F)$ such that for any $f$ in $C_{v}^{o}(X,F)$, there is a
positive operator $T:E\rightarrow F$ satisfying $\left\vert
f(x)-f(y)\right\vert\leq T(d(x,y))$ for all $x,y\in X$. Since the
Birkhoff inequality (\cite{AB},1.9(2); \cite{Zaanen},12.4(ii))
$$\left\vert f\vee g(x)-f\vee g(y)\right\vert\leq \left\vert
f(x)-f(y)\right\vert +\left\vert g(x)-g(y)\right\vert$$ holds for
all $x,y\in X$, then the subset $C_{v}^{o}(X,F)$ is a Riesz
subspace of $C_{v}(X,F)$. By Theorem \ref{lem}, every $f\in
C_{v}^{o}(X,F)$ is vectorial bounded function. This argument gives
us the following result.
\vskip2mm
\begin{corollary}
The subset $C_{v}^{o}(X,F)$ described above is a vector metric
space equipped with the $F$-valued uniform vector metric defined
as $d_{\infty}(f,g)=\sup_{x\in X}\left\vert f(x)-g(x)\right\vert.$
\end{corollary} | 0.002219 |
TITLE: Recoil velocity and mass of particle after absorbing photon
QUESTION [0 upvotes]: I'm working a problem out of d'Inverno's "Introducing Einstein's Relativity", and I'm hitting a funny issue with my algebra. The problem states:
An atom of rest mass $m_0$ is at rest in a laboratory and absorbs a photon of frequency $\nu$. Find the velocity and mass of the recoiling particle.
The answers are given in the back of the book as
$$
u=\frac{ch\nu}{h\nu + m_0c^2} \qquad m=\left(m_0^2 + \frac{2h\nu m_0}{c^2}\right)^{1/2}.
$$
I've found that I can figure out the velocity by starting with conservation of momentum and energy
$$
\frac{h\nu}{c} = \gamma m_0u \qquad h\nu + m_0c^2 = \gamma m_0c^2,
$$
(where of course $\gamma=\left[1-\frac{u^2}{c^2}\right]^{-1/2}$) then eliminating the relativistic mass $\gamma m_0$ and solving for $u$. Strangely though, in my first couple of attempts I started with conservation of only energy or momentum, and obtained answers close to, but not quite the same as, the above:
$$
u=\frac{c(2m_0c^2h\nu+h^2\nu^2)^{1/2}}{h\nu + m_0c^2} \qquad (\textrm{from conservation of energy})
$$
$$
u=\frac{ch\nu}{(h^2\nu^2+m_0^2c^4)^{1/2}} \qquad (\textrm{from conservation of momentum})
$$
Is there simply some mistake in my algebra that I haven't managed to suss out, or is the error in assuming that I can proceed from only one conservation law?
As for the mass, I attempted to simply substitute the velocity $u$ into the relativistic mass $\gamma m_0$, but quickly got a monstrosity of $m_0$'s, $h$'s, $\nu$'s, and $c$'s that bore no resemblance to the answer. Is this the correct way to approach this part of the problem, or should I start from some other relation?
EDIT:
Sofia answered the second half of my question regarding the mass of the atom. However, I'm still curious about the first part. That is, why do I obtain different results for the velocity with different equations? One approach may be more difficult or roundabout than another, but if they're all based on the same physical principles, I feel that I should get the same result regardless of which relation I begin with.
REPLY [0 votes]: The solution in the book doesn't consider the recoil velocity $u$ as relativistic. So, no need of $\gamma$. But they consider simply that the mass of the atom increases, instead of $m_0$ to $m$. So, this is what they do, I mean, their equations of conservation:
(1) $h \nu /c = mu, \ \ \ $ linear momentum conservation,
(2) $(m_0c^2 + h\nu)^2 = m^2c^4 + m^2u^2c^2, \ \ \ $ energy conservation and Klein-Gordon equation.
Now, from eq. (1) I obtain
$u = \frac {h \nu}{mc}$,
and I introduce in eq. (2).
$(m_0c^2 + h\nu)^2 = m^2c^4 + h^2 \nu ^2$.
From the last equation we can extract $m$,
$m_0^2 + \frac {2h\nu m_0}{c^2} = m^2$.
The quantity on the LHS can be eventually completed to a square
$m_0^2 + \frac {2h\nu m_0}{c^2} + \frac {(h \nu)^2}{c^4} = m^2$. | 0.325989 |
\section{$\mathbf{Ric_2>0}$ with large symmetry rank in even dimensions}\label{sec:even}
In this section, we will study closed, simply connected, even-dimensional manifolds with $\Ric_2>0$ and large symmetry rank.
In particular, we prove those with half-maximal symmetry rank have positive Euler characteristic (Theorem \ref{result:evenEuler}), we obtain a strong classification for those with maximal symmetry rank and bounded second Betti number (the even-dimensional case of Theorem \ref{result:main}), and we obtain a weaker classification of those with $3/4$-maximal symmetry rank and bounded second Betti number (Theorem \ref{result:even}).
\subsection{Positive Euler characteristic}\label{sec:posEuler}
First, we establish Theorem \ref{result:evenEuler}, which we restate here for convenience:
\begin{theorem}\label{thm:Ric2evenEuler}
Suppose $M^n$ is a closed Riemannian manifold of even dimension $n\geq 8$ with $\Ric_2>0$.
If a torus $T^r$ of rank $r\geq \frac{n}{4}+2$ acts effectively and by isometries on $M^n$, then $\chi(M^n)>0$.
\end{theorem}
To prove Theorem \ref{thm:Ric2evenEuler}, we will use the following topological observation:
\begin{proposition}\label{prop:chainoftori}
Suppose a torus $T^r$ acts isometrically and effectively on a closed manifold $M$. If the fixed point set $M^{T^r}$ is non-empty, then given any point $x\in M^{T^r}$, there exists a chain of subgroups $T^1\subset T^2\subset \dots\subset T^{r-1}\subset T^r$ such that the following inclusions of fixed point set components containing $x$ are each of minimal, positive, even codimension:
\[
M^{T^r}_x\subset M^{T^{r-1}}_x\subset \dots\subset M^{T^2}_x\subset M^{T^1}_x\subset M.
\]
\end{proposition}
\begin{proof}
Fix a point $x\in M^{T^r}$ and choose circle subgroup $S_1^1\subset T^r$ such that the fixed point set component $M_{x}^{S_1^1}$ has minimal codimension in $M$. Because $T^r$ acts effectively on $M$, we have $M_{x}^{S_1^1}\neq M$. Set $T^1\defeq S_1^1$. Now choose a circle subgroup $S_2^1\subset T^r/T^{1}$ such that the fixed point set component $M_{x}^{T^2}$ for $T^2\defeq S_2^1\times T^1$ has minimal codimension in $M_{x}^{T^{1}}$. Because $M_{x}^{T^{1}}$ was chosen to have minimal codimension in $M$, $T^r/T^{1}$ must act almost effectively on $M_{x}^{T^{1}}$ by Proposition \ref{prop:mincodim}. In particular, $S_2^1$ does not fix all of $M_{x}^{T^1}$, and hence $M_{x}^{T^2}\neq M_{x}^{T^1}$.
Now for $i\in\{2,\dots,r-1\}$, we inductively choose $S_{i+1}^1\subset T^r/T^{i}$ such that the fixed point set component $M_{x}^{T^{i+1}}$ for $T^{i+1}\defeq S_{i+1}^1\times T^i$ has minimal codimension in $M_{x}^{T^i}$. Because $M_{x}^{T^{i}}$ was chosen to have minimal codimension in $M_{x}^{T^{i-1}}$, again $T^r/T^{i}$ must act almost effectively on $M_{x}^{T^{i}}$ by Proposition \ref{prop:mincodim}. This shows that $S_{i+1}^1\subset T^r/T^i$ does not fix all of $M_{x}^{T^i}$, and therefore $M_{x}^{T^{i+1}}\neq M_{x}^{T^{i}}$ for $i\in\{2,\dots,r-1\}$.
\end{proof}
To prove Theorem \ref{thm:Ric2evenEuler}, we will apply Proposition \ref{result:torusfixedpt} for a $T^r$-action on an $n$-manifold of $\Ric_2>0$ with $r\geq \frac{n}{4}+2$ to obtain a fixed point for some $T^{r-2}$-subaction. Applying Proposition \ref{prop:chainoftori} will then give us the following topological restriction:
\begin{lemma}\label{lem:chainfixedptsets}
Suppose $M^n$ is closed and even-dimensional, and a torus $T^r$ acts isometrically and effectively on $M^n$ with $r\geq\frac{n}{4}+2$. Assume a subgroup $T^{r-2}\subset T^r$ has non-empty fixed point set $M^{T^{r-2}}$ in $M$, and for any point $x\in M^{T^{r-2}}$, consider a chain of fixed point set components $M^{T^{r-2}}_x\subset M^{T^{r-3}}_x\subset \dots\subset M^{T^2}_x\subset M^{T^1}_x\subset M$ guaranteed by Proposition \ref{prop:chainoftori}. Then either
\begin{enumerate}
\item $\dim(M_x^{T^{r-2}})=0$, or
\item at least one of the inclusions in the chain $M^{T^{r-2}}_x\subset M^{T^{r-3}}_x\subset \dots\subset M^{T^2}_x\subset M^{T^1}_x\subset M$ is of codimension 2.
\end{enumerate}
\end{lemma}
\begin{proof}
Suppose $\dim(M_x^{T^{r-2}})\neq0$. Let $d_i$ denote the codimension of $M_x^{T^i}$ in $M_x^{T^{i-1}}$ for $i\in\{2,\dots,r-2\}$, and let $d_1$ denote the codimension of $M_x^{T^1}$ in $M$. Because fixed point sets of torus actions have even codimension, it follows that $\dim(M_x^{T^{r-2}})\geq 2$ and each $d_i$ is even. Because the inclusions in the chain are proper, it follows that $d_i\geq 2$ for all $i$. Now if $d_i\geq 4$ for all $i$, then because $\dim(M_x^{T^{r-2}})\geq 2$, we have
\begin{align*}
n=\dim(M^n)&\geq \dim(M_x^{T^{r-2}})+d_{r-2}+d_{r-3}+\dots+d_2+d_1\\
&\geq 2+4(r-2).
\end{align*}
It then follows that $r < \frac{n}{4}+2$, which contradicts the assumption that $r\geq\frac{n}{4}+2$. Therefore, $d_i=2$ for some $i\in\{1,\dots,r-2\}$.
\end{proof}
Finally, we recall the following topological result established by Conner:
\begin{lemma}[\cite{Conner}]\label{lem:conner}
If $T$ is a torus acting on a manifold $M$ with fixed point set $M^T$, then $\chi(M)=\chi(M^T)$.
\end{lemma}
We are now ready to prove Theorem \ref{thm:Ric2evenEuler}.
\begin{proof}[Proof of Theorem \ref{thm:Ric2evenEuler}]
Suppose $M^n$ is closed, even-dimensional, has $\Ric_2>0$, and $T^r$ acts isometrically and effectively on $M^n$ with $r\geq\frac{n}{4}+2$.
Because $\Ric_2(M^n)>0$, Proposition \ref{result:symrankbound} states that $r\leq \frac{n}{2}$.
For even dimensions $n$, the inequalities $\frac{n}{4}+2\leq r\leq\frac{n}{2}$ are only consistent if $n\geq 8$.
In this case, $r\geq 4$.
Thus by Proposition \ref{result:torusfixedpt}, there exists a subgroup $T^{r-2}\subset T^r$ such that the fixed point set $M^{T^{r-2}}$ in $M$ is non-empty. By Lemma \ref{lem:conner}, it suffices to show that $\chi(M^{T^{r-2}})>0$. In particular, we will show that every connected component $M_x^{T^{r-2}}$ has positive Euler characteristic.
Choose an arbitrary point $x\in M^{T^{r-2}}$. By Proposition \ref{prop:chainoftori}, there exists a chain of subgroups $T^1\subset T^2\subset \dots\subset T^{r-1}\subset T^r$ such that the inclusions of fixed point set components containing $x$, $M^{T^r}_x\subset M^{T^{r-1}}_x\subset \dots\subset M^{T^2}_x\subset M^{T^1}_x\subset M$, are proper.
By Lemma \ref{lem:chainfixedptsets}, either $\dim(M_x^{T^{r-2}})=0$, or at least one of the inclusions in the chain $M^{T^{r-2}}_x\subset M^{T^{r-3}}_x\subset \dots\subset M^{T^2}_x\subset M^{T^1}_x\subset M$ is of codimension 2.
If $\dim(M_x^{T^{r-2}})=0$, then $\chi(M_x^{T^{r-2}})>0$, and we are done.
Now assume $\dim(M_x^{T^{r-2}})\geq 2$ and one of the inclusions $M^{T^i}_x\subset M^{T^{i-1}}_x$ for $i\in\{1,\dots,r-2\}$ is of codimension 2.
Here, we are using the convection that $T^0$ is the trivial subgroup and $M_x^{T^0}=M$.
We will show that $\chi(M^{T^{i-1}})>0$.
Let $m=\dim(M^{T^{i-1}}_x)$. Because $\dim(M_x^{T^{r-2}})\geq 2$, we have $m=2+\dim(M^{T^{i}})\geq 4$. So because $M^{T^{i-1}}_x$ is totally geodesic in $M$, we have $\Ric_2(M^{T^{i-1}}_x)>0$. Hence the Betti numbers for $b_1(M^{T^{i-1}}_x)$ and $b_{m-1}(M^{T^{i-1}}_x)$ are both zero.
So if $M^{T^{i-1}}$ is $4$-dimensional, $\chi(M^{T^{i-1}})>0$.
Suppose now that $M^{T^{i-1}}$ has dimension $m\geq 6$.
By Theorem \ref{result:connprinc}, because $M^{T^i}_x$ is fixed by the $S^1\cong T^i/T^{i-1}$-action on $M^{T^{i-1}}_x$, the inclusion $M^{T^i}_x\hookrightarrow M^{T^{i-1}}_x$ is $(m-3)$-connected.
Thus by Lemma \ref{lem:periodic}, we have homomorphisms $H^i(M^{T^{i-1}}_x;\mathbb{Z})\to H^{i+2}(M^{T^{i-1}}_x;\mathbb{Z})$ which are surjective for $1\leq i< m-3$ and injective for $1<i\leq m-3$.
The hypothesis that $n-2d-l>0$ in Lemma \ref{lem:periodic} is satisfied because $m\geq 6$, $d=2$, and $l=1$ in this case.
Therefore, it follows that all of the odd Betti numbers of $M^{T^{i-1}}_x$ are zero, which implies $\chi(M^{T^{i-1}}_x)>0$.
Now for all dimensions $m\geq 4$, because $M^{T^{i-1}}_x$ is invariant under the $T^{r-2}$-action and $(M^{T^{i-1}}_x)^{T^{r-2}}=M^{T^{r-2}}_x$, it follows from Lemma \ref{lem:conner} that $\chi(M^{T^{r-2}}_x)>0$.
Hence, we have shown that for all $x\in M^{T^{r-2}}$, the component $M_x^{T^{r-2}}$ containing $x$ has $\chi(M_x^{T^{r-2}})>0$. Therefore, $\chi(M^{T^{r-2}})>0$ and by Lemma \ref{lem:conner}, $\chi(M)>0$.
\end{proof}
\subsection{Classification for maximal symmetry rank}\label{sec:evenmaxsymrank}
We will now prove the even dimensional case of Theorem \ref{result:main}.
Specifically, we must prove if $M^n$ is closed, simply connected, even-dimensional, $n\neq 6$, $b_2(M^n)\leq 1$, $\Ric_2(M^n,g)>0$, and $\symrank(M^n,g) = \frac{n}{2}$, then $M^n$ is either diffeomorphic to $S^n$ or homeomorphic to $\mathbb{C}\mathrm{P}^{n/2}$.
First, we note that the dimension $n=4$ case of Theorem \ref{result:main} follows from purely topological considerations, not relying on the curvature assumption.
Orlik and Raymond prove in \cite{OrlikRaymond} that any closed, simply connected $4$-manifold $M^4$ with an effective $T^2$-action is equivariantly diffeomorphic to a connected sum of finitely many copies of $S^4$, $\pm\mathbb{C}\mathrm{P}^2$, or $S^2\times S^2$.
Now if $b_2(M^4)\leq1$, then $2\leq \chi(M^4) \leq 3$.
Thus, we have the following:
\begin{corollary}\label{cor:dim4}
Suppose $M^4$ is a closed, simply connected, $4$-dimensional manifold with a smooth, effective $T^2$-action.
If $b_2(M^4)\leq 1$, then $M^4$ is equivariantly diffeomorphic to either $S^4$ or $\mathbb{C}\mathrm{P}^{2}$.
\end{corollary}
Sha and Yang proved in \cite{ShaYang} that any connected sum of finitely many copies of $S^4$, $\pm\mathbb{C}\mathrm{P}^2$, or $S^2\times S^2$ admits a metric of positive Ricci curvature.
Baza\u{i}kin and Matvienko proved in \cite{BazaikinMatvienko} that these manifold also admit Ricci positive metrics that are invariant under the corresponding effective $T^2$-actions.
This leads naturally to the following:
\begin{question}
Are there any closed, simply connected $4$-manifolds with $b_2\geq 2$ which admit metrics of $\Ric_2>0$ that are invariant under an effective $T^2$-action?
\end{question}
\begin{example}
Recall from Example \ref{ex:S3xS3} that $S^2\times S^2$ admits a metric with $\Ric_2>0$ and symmetry rank $1$.
Hsiang and Kleiner prove in \cite{HsiangKleiner} that any closed, orientable, $4$-dimensional manifold with positive sectional curvature that has a nontrivial Killing field must be homeomorphic to $S^4$ or $\mathbb{C}\mathrm{P}^2$.
Consequently, it is impossible for $S^2\times S^2$ to admit a metric of positive sectional curvature with symmetry rank $1$.
It remains to be seen whether $S^2\times S^2$ can admit a metric with $\Ric_2>0$ and symmetry rank $2$.
\end{example}
Now we will establish Theorem \ref{result:main} for dimensions $n\geq 8$.
Namely, we intend to prove the following:
\begin{theorem}\label{thm:evenmaxsymrank}
Let $M^n$ be a closed, simply connected Riemannian manifold of even dimension $n\geq 8$ with $\Ric_2>0$.
Suppose a torus $T^r$ of rank $r=\frac{n}{2}$ acts effectively and by isometries on $M^n$.
\begin{enumerate}
\item If $b_2(M^n)=0$, then $M^n$ is diffeomorphic to $S^n$.
\item If $b_2(M^n)=1$, then $M^n$ is homeomorphic to $\mathbb{C}\mathrm{P}^{n/2}$. \label{item:CPn}
\end{enumerate}
\end{theorem}
First, we will use Theorem \ref{thm:Ric2evenEuler} and Theorem \ref{result:connprinc} to justify the following:
\begin{lemma}\label{lem:evenmaxsymrank}
Let $M^n$ be a closed, simply connected Riemannian manifold of even dimension $n\geq 8$ with $\Ric_2>0$.
Suppose a torus $T^r$ of rank $r=\frac{n}{2}$ acts effectively and by isometries on $M^n$.
Then there exist closed, simply connected, totally geodesic submanifolds $M^4 \subset M^6 \subset \dots \subset M^{n-2} \subset M^n$ such that each inclusion $M^{2i}\hookrightarrow M^{2i+2}$ is $(2i-1)$-connected.
\end{lemma}
\begin{proof}
By Theorem \ref{thm:Ric2evenEuler}, because $r=\frac{n}{2}\geq \frac{n}{4}+2$ when $n\geq 8$, we have $\chi(M^n)>0$.
Thus, the $T^r$-action on $M^n$ has non-empty fixed point set.
By Proposition \ref{prop:chainoftori}, given a point $x$ in $M^{T^r}$, there exists a chain of subgroups $T^1\subset T^2\subset \dots\subset T^{r-1}\subset T^r$ such that the following inclusions of fixed point set components containing $x$ are of positive, even codimension:
\[
M^{T^r}_x\subset M^{T^{r-1}}_x\subset \dots\subset M^{T^2}_x\subset M^{T^1}_x\subset M.
\]
Because $r=\frac{n}{2}$ and each of the inclusions above are of positive, even codimension, it follows that $M^{T^r}_x$ is zero-dimensional and each inclusion $M_x^{T^j}\subset M_x^{T^{j-1}}$ is of codimension $2$.
Define $M^{2i} = M_x^{T^{r-i}}$.
Because each submanifold $M^{2i}$ is totally geodesic, we have $\Ric_2(M^{2i})>0$ for $2i\geq 4$.
By Theorem \ref{result:connprinc}, each inclusion $M^{2i}\hookrightarrow M^{2i+2}$ is $(2i-1)$-connected, and because $M$ is simply connected, so are $M^{2i}$ for $2i\geq 4$.
\end{proof}
Now we will prove Theorem \ref{thm:evenmaxsymrank} using work of Montgomery and Yang in \cite{MontgomeryYang} and Fang and Rong in \cite{FangRong2004}.
\begin{proof}[Proof of Theorem \ref{thm:evenmaxsymrank}]
Let $M^n$ be a closed, simply connected Riemannian manifold of even dimension $n\geq 8$ with $\Ric_2>0$, and suppose a torus $T^r$ of rank $r=\frac{n}{2}$ acts effectively and by isometries on $M^n$.
By Lemma \ref{lem:evenmaxsymrank}, there exist closed, simply connected, totally geodesic submanifolds $M^4 \subset M^6 \subset \dots \subset M^{n-2} \subset M^n$ such that each inclusion $M^{2i}\hookrightarrow M^{2i+2}$ is $(2i-1)$-connected.
Thus $b_2(M^4) = b_2(M^6) = \dots = b_2(M^{n-2}) = b_2(M^n)$.
So if $b_2(M^n) \leq 1$, then $b_2(M^4) \leq 1$, and because $M^4$ is simply connected, $\chi(M^4) \leq 3$.
Then it follows from \cite{Freedman} that $M^4$ is homeomorphic to $S^4$ or $\mathbb{C}\mathrm{P}^2$.
Because $M^4\hookrightarrow M^6$ is $3$-connected, we have that $\pi_3(M^6)=0$, and $\pi_2(M^6) = 0$ (resp. \!$\mathbb{Z}$) if $M\cong S^4$ (resp. \!$\mathbb{C}\mathrm{P}^2$).
It follows from Poincar\'e duality and the Hurewicz theorem that $M^6$ is homotopy equivalent to $S^6$ or $\mathbb{C}\mathrm{P}^3$.
By iterating the same argument for $M^8,\dots,M^n$, we conclude that $M^n$ is homotopy equivalent to $S^n$ if $b_2(M^n)=0$, or $M^n$ is homotopy equivalent to $\mathbb{C}\mathrm{P}^{n/2}$ if $b_2(M^n)=1$.
If $b_2(M^n)=0$, then similar to the proof of Theorem \ref{thm:oddmaxsymrank}, it follows from \cite[Proposition 3]{MontgomeryYang} that $M^n$ is diffeomorphic to $S^n$.
Suppose instead that $b_2(M^n)=1$, and hence $M^4$ is homeomorphic to $\mathbb{C}\mathrm{P}^2$ and $M^6,\dots,M^n$ are homotopy complex projective spaces.
Fang and Rong proved that any homotopy $\mathbb{C}\mathrm{P}^n$ with a submanifold homeomorphic to $\mathbb{C}\mathrm{P}^{n-1}$ such that the inclusion map is at least $3$-connected must be homeomorphic to $\mathbb{C}\mathrm{P}^n$ \cite{FangRong2004}.
Because $M^4$ is homeomorphic to $\mathbb{C}\mathrm{P}^2$ and each inclusion $M^{2i}\hookrightarrow M^{2i+1}$ is $(2i-1)$-connected, we can apply Fang and Rong's result to each inclusion $M^4 \subset M^6 \subset \cdots \subset M^n$, concluding that $M^n$ is homeomorphic to $\mathbb{C}\mathrm{P}^{n/2}$.
\end{proof}
\subsection{Classification for $\mathbf{3/4}$-maximal symmetry rank}\label{sec:even3/4}
In this final section, we will prove Theorem \ref{result:even}, which we restate here for convenience:
\begin{theorem}\label{thm:even3/4maxsymrank}
Let $M^n$ be a closed, simply connected, Riemannian manifold of even dimension $n\geq 8$ with $\Ric_2>0$.
Suppose a torus $T^r$ of rank $r\geq\tfrac{3n+6}{8}$ acts effectively and by isometries on $M^n$.
Then $H^i(M^n;\mathbb{Z})=0$ for all odd values of $i$.
Furthermore:
\begin{enumerate}
\item If $b_2(M^n)=0$, then $M^n$ is homeomorphic to $S^n$.
\item If $b_2(M^n)=1$, then $M^n$ is tangentially homotopy equivalent to $\mathbb{C}\mathrm{P}^{n/2}$.
\end{enumerate}
\end{theorem}
First, we establish an even dimensional analogue of Lemma \ref{lem:NRic2}:
\begin{lemma}\label{lem:NRic2even}
Let $M^n$ be a closed even-dimensional manifold with $\Ric_2>0$ on which a torus $T^r$ acts isometrically and effectively with non-empty fixed point set.
If $r\geq\frac{3n+6}{8}$, then there exists a connected submanifold $N\subset M$ of minimal codimension fixed by a circle subgroup of $T^r$ such that
\begin{enumerate}
\item \label{item:1even} $\dim(N)\geq \frac{3n-2}{4}$, and
\item \label{item:2even} either $\codim(N)=2$ or $\symrank(N)\geq \frac{3\dim(N)+6}{8}$.
\end{enumerate}
\end{lemma}
\begin{proof}
Suppose the $T^r$-action on $M^n$ has non-empty fixed point set, and consider the chain of fixed point set components $M^{T^r}_x\subset M^{T^{r-1}}_x\subset \dots\subset M^{T^2}_x\subset M^{T^1}_x\subset M$ guaranteed by Proposition \ref{prop:chainoftori}.
We will choose $N$ to be $M_x^{T^1}$.
Because the inclusions $M_x^{T^i}\subset M_x^{T^{i-1}}$ are each of positive, even codimension, we have that $\dim(N)\geq 2r-2$, and because $r\geq \frac{3n+6}{8}$, it follows that $\dim(N) \geq \frac{3n-2}{4}$. This proves Part \eqref{item:1even}.
To prove Part \eqref{item:2even}, assume $\codim(N)\neq 2$ and $\symrank(N)< \frac{3\dim(N)+6}{8}$.
Recall from Proposition \ref{prop:chainoftori} that the inclusion $N\subset M^n$ is minimal in the sense that if a circle subgroup of $T^r$ fixes another connected submanifold $N'\subset M^n$, then $\codim(N')\geq \codim(N)$.
Now, because the $T^r$-action on $M^n$ is effective and the codimension of $N\subset M^n$ must be even, we have that $\dim(N)\leq n-4$. Thus
\[
\symrank(N)<\tfrac{3\dim(N)+6}{8}\leq\tfrac{3(n-4)+6}{8}<\tfrac{3n+6}{8}-1\leq r-1.
\]
So setting $T^{r-1}=T^r/T^1$, it follows that there is a circle subgroup of $T^{r-1}$ that fixes $N$, meaning that $N$ is fixed by a two-dimensional torus subgroup of $T^r$.
By Proposition \ref{prop:mincodim}, this implies that there is a circle subgroup of $T^r$ that fixed a submanifold of larger dimension than $N$, which is a contradiction.
Therefore, Part \eqref{item:2even} follows.
\end{proof}
In proving Theorem \ref{thm:even3/4maxsymrank}, if the submanifold $N$ from Lemma \ref{lem:NRic2even} is of codimension $2$, then we will apply the following:
\begin{lemma}\label{lem:codim2even}
Suppose $M^n$ is a closed, even-dimensional, simply connected, smooth manifold of dimension $n\geq 4$.
Assume $M^n$ contains a closed, connected, embedded submanifold $N^{n-2}$ of codimension $2$ such that the inclusion $N^{n-2}\hookrightarrow M^n$ is $(n-3)$-connected.
Then $H^i(M^n;\mathbb{Z})=0$ for all odd values of $i$.
\end{lemma}
\begin{proof}
In dimension $n=4$, the result follows from Poincar\'e duality.
Suppose $n\geq 6$.
Because $M^n$ is simply connected, so is $N^{n-2}$.
Let $[N]\in H_{n-2}(M^n;\mathbb{Z})$ denote the image of the fundamental class of $N$, and let $e\in H^2(M^n;\mathbb{Z})$ denote its Poincar\'e dual.
By Lemma \ref{lem:periodic}, the homomorphisms $\cup e: H^j(M^n;\mathbb{Z})\to H^{j+2}(M^n;\mathbb{Z})$ are surjective for $1\leq j< n-3$ and injective for $1< j\leq n-3$.
Because $M^n$ is simply connected, $H^{1}(M^n;\mathbb{Z})\cong 0\cong H^{n-1}(M^n;\mathbb{Z})$, and it follows that $H^i(M^n;\mathbb{Z})=0$ for all odd values of $i$.
\end{proof}
\begin{example}
As we described in Example \ref{ex:S3xS3}, $S^3\times S^3$ admits a metric with $\Ric_2>0$ invariant under an effective $T^3$-action.
It follows from Lemma \ref{lem:codim2even} that no action of any of the circle subgroups of $T^3$ has fixed point set of codimension $2$.
In fact, one can show that the possible fixed point sets in $S^3\times S^3$ of circle subgroups in this example are either empty or diffeomorphic to $T^2$.
\end{example}
We will now prove Theorem \ref{thm:even3/4maxsymrank} using the work of Smale in \cite{smale} and Dessai and Wilking in \cite{DessaiWilking}:
\begin{proof}[Proof of Theorem \ref{thm:even3/4maxsymrank}]
We will prove Theorem \ref{thm:even3/4maxsymrank} by induction on the dimension, $n$.
First note that in dimensions $n=8,10$, and $12$, we have $\lceil \frac{3n+6}{8} \rceil = \frac{n}{2}$.
So the result holds in these dimensions by Theorem \ref{thm:evenmaxsymrank}, thus establishing our base case.
Now, suppose for the sake of induction that Theorem \ref{thm:even3/4maxsymrank} holds in even dimensions $8,\dots,n-2$ for some $n\geq 14$.
We will show it also holds in dimension $n$.
Consider a manifold $M^n$ satisfying the hypotheses of Theorem \ref{thm:even3/4maxsymrank}, and suppose $T^r$ acts isometrically and effectively on $M^n$ with $r = \symrank(M^n) \geq \frac{3n+6}{8}$.
Then by Theorem \ref{result:evenEuler}, $\chi(M^n) > 0$, and hence the $T^r$-action on $M^n$ has a fixed point.
By Lemma \ref{lem:NRic2even}, there exists a connected submanifold $N\subset M$ of minimal codimension fixed by a circle subgroup of $T^r$ such that $\dim N\geq\frac{3n-2}{4}$ and either $\codim(N)=2$ or $\symrank(N)\geq \frac{3\dim N+6}{8}$.
If $\codim(N)=2$, then by Theorem \ref{result:connprinc}, the inclusion $N^{n-2}\hookrightarrow M^n$ is $(n-3)$-connected, and by Lemma \ref{lem:codim2even}, $H^\mathrm{odd}(M^n;\mathbb{Z}) = 0$.
By Lemma \ref{lem:periodic}, if $[N]\in H_{n-d}(M^n;\mathbb{Z})$ denotes the image of the fundamental class of $N$ and $e\in H^d(M^n;\mathbb{Z})$ denotes its Poincar\'e dual, then the homomorphisms $\cup e: H^i(M;\mathbb{Z})\to H^{i+d}(M;\mathbb{Z})$ are isomorphisms for $2\leq i \leq n-4$.
If $b_2(M^n)\leq 1$, then because $M^n$ is simply connected, it has the cohomology ring of either a sphere or a complex projective space, and hence is homotopy equivalent to one of these spaces.
For the case of a sphere, $M^n$ is homeomorphic to $S^n$ by Smale's resolution to the Poincar\'e conjecture for dimensions $\geq 5$ in \cite{smale}.
In the case of a complex projective space, Dessai and Wilking proved in \cite{DessaiWilking} if a manifold is homotopy equivalent to $\mathbb{C}\mathrm{P}^m$ and admits a smooth effective action by a torus $T^r$ such that $m < 4r - 1$, then the manifold is tangentially homotopy equivalent to $\mathbb{C}\mathrm{P}^n$.
Thus, it follows in this case that $M^n$ is tangentially homotopy equivalent to $\mathbb{C}\mathrm{P}^{n/2}$.
Suppose instead $\symrank(N)\geq \frac{3\dim N+6}{8}$.
Then because $\dim N \geq \frac{3n-2}{4} \geq 10$ and $N$ is totally geodesic in $M$, we have that $\Ric_2(N)>0$.
Thus, the induction hypothesis implies that $H^\mathrm{odd}(N;\mathbb{Z}) = 0$.
Furthermore, because
\[
n - 2\codim N + 1 \geq \tfrac{n}{2},
\]
the inclusion $N\hookrightarrow M^n$ is at least $\frac{n}{2}$-connected by Theorem \ref{result:connprinc}.
Thus it follows that $H^\mathrm{odd}(M^n;\mathbb{Z}) = 0$.
If $b_2(M^n)\leq 1$, then $b_2(N)\leq 1$, and $N$ is either homeomorphic to a sphere are tangentially homotopy equivalent to a complex projective space by the induction hypothesis.
Because the inclusion $N\hookrightarrow M^n$ is $\frac{n}{2}$-connected, it follows that $M^n$ has the cohomology ring of either a sphere or a complex projective space.
Then, just as in the previous case, $M^n$ is either homeomorphic to $S^n$ or tangentially homotopy equivalent to $\mathbb{C}\mathrm{P}^{n/2}$.
\end{proof} | 0.001597 |
Clark McCoy, MD
Dr. Clark S. McCoy is board-certified in Addiction Medicine, Family Medicine, and Obesity Medicine. Clark is the President and Medical Director of Front Range Clinic with multiple locations including Fort Collins, Greeley, and Longmont, Colorado. In 1994 Dr. McCoy graduated from the Medical College of Wisconsin with honors in Alpha Omega Alpha Medical Honor Society. He completed his internship and residency at the University of Florida, where he served as the Chief Resident during his final year. He has work experience in Emergency Medicine, Inpatient medical consulting, Urgent Care, Family Practice, and Clinic Management. In 2014 Dr. McCoy founded Front Range Clinic.
Jeremy Dubin, DO
Dr. Dubin lectures and teaches regularly on the field of Addiction Medicine. As a family physician and addiction medical specialist at the Front Range Clinic, he pursues effective healing strategies that proactively confront the challenges of addiction; helping to create long term solutions through evidence-based comprehensive treatment plans that address the complex physical, biochemical, psychological, social, emotional, and spiritual challenges that are often observed with addiction.)
Candice Ash, NP
Candi Ash joined Front Range Clinic in April 2016. Candi obtained her Masters in Science of Nursing in from the University of Southern Mississippi. She has eighteen years of nursing experience-ten years as an RN supervisor in Labor and Delivery and NICU, and eight years as a Nurse Practitioner in Emergency Medicine and military Family Medicine. At Front Range Clinic she works in Addiction Medicine. Candi is from Wyoming and is married with three boys.
Candi enjoys exercise, travel, and spending time with her family. | 0.779927 |
Football Betting
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Stephen McManus has paid tribute to former manager Gordon Strachan for putting Celtic on the UEFA Champions League map.
Strachan stepped down on Monday just hours after seeing Rangers land the Scottish Premier League crown for the first time in four years.
The former Southampton and Coventry boss guided the Bhoys to three SPL titles, the Scottish Cup and two Co-Operative Insurance Cup triumphs.
But it was his efforts in Europe where he twice helped Celtic reach the last 16 of the Champions League which made most of an impression on Scotland defender McManus.
Fantastic
He told Celtic View: "The European success has been fantastic.
.
"I speak for all the boys when I wish him well for the future."
The Scotland defender thanked Strachan for taking him out of the Parkhead reserves and turning him in to a first-team regular.
"The gaffer has been a massive influence on my career and I have a lot to thank him for," he said.
Respect
. It's been an incredible time in our careers.
"He's worked with a lot of players and has done so much for the club, but he obviously thinks it's time to move on and we all respect that.
"It's been a great four years at Celtic. He's been different class and his record speaks for itself. Nobody can knock his achievements at Celtic. He's been superb." | 0.000881 |
Custom Portraits
Getting Ready for your Portrait Photography Photo Session
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If you would like to book your staff or family photography session with us. We would like you to get the most from your photo session.
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put together this guide to help you prepare.
Our goal is to provide you with beautiful and meaningful portraits that you will cherish forever
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The goal of any portrait is to direct the viewer’s eye to the subject’s face. All other elements are secondary. The right clothing choice is important. Clothing can easily date your portrait, so we suggest simple clothing with a timeless cut and style. Long-sleeves work best for most portraits, taking attention away from the arms and keeping it on the faces and expressions. Jeans can be timeless and casual. Darker clothing recedes in the image and can appear slimming. If you are doing a staff shoot have your employees wear something that represents your company in the manner you find appropriate.
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Family and staff portraits are an occasion to look your best. Try getting everyone to wear variations on the same color or a similar style of clothing for a cohesive look. It looks great when everyone’s clothing works well together. The emphasis then becomes the relationships and faces of those in the portrait, not the clothing you’re wearing.
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Our goal is to capture the members of your family as they are today. Choose clothing they love to wear and that accentuates their individual personalities. Consider that you might be sitting or leaning in different poses. You want to wear something that you feel very comfortable in and that will cover any areas you do not want to show in the images.
Try this. Lay out everyone’s clothing together in the room where your portrait will live. Replace outfits and add accessories until you achieve a fresh and exciting look for that room.
Children
If you have young children in your portrait, please make sure that they have eaten and are rested before the session. Nothing like a happy kid for a smooth experience. Also, bring along any toys that will make your child feel comfortable and safe.
Shoes
Remember to match the shoes to the style of your overall look: formal-casual and that they match your chosen color palette. Young children can be barefoot if you prefer.
Nails
Keep in mind hands usually show, so make sure everyone’s nails are camera ready.
Extras
If you want, you can bring some extra clothing options just in case. A brush, hair spray, lipstick, and translucent powder are always great for last-minute touch-ups.
Finally, get everything ready the day before. This will ensure that portrait day goes smoothly and you will not need to rush last minute. | 0.977657 |
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ISTANBUL — As scientists around the world celebrate the 150th anniversary of Charles Darwin’s seminal work on evolution, Adnan Oktar, a college dropout turned theorist of Islamic creationism, is working on the fifth volume of a 14-part masterwork that he says will bury Darwinism once and for all. | 0.036702 |
\begin{document}
\maketitle
\begin{abstract}
We present generalisations of Wilson's theorem for double factorials, hyperfactorials, subfactorials and superfactorials.
\end{abstract}
\section{Introduction}
Wilson's theorem states that $(p-1)!\equiv -1$ if $p$ is prime, and $(p-1)!\equiv 0$ otherwise, except for the one special case, $p=4$.
The result is attributed to John Wilson, a student of Waring, but it has apparently been known for over a thousand years; see \cite{Ra}, \cite[Ch. II]{Dav}, \cite[Ch. 11]{Ore}, \cite[Chap.~3]{Dick1} and \cite[Chap.~3.5]{St}. The first proof was given by Lagrange \cite{La}. We give generalisations of Wilson's theorem for the four standard generalisations of the factorial function.
\begin{definition} For a natural number $n$,
\begin{enumerate}
\item the \emph{double factorial} $n!!$ is the product of the natural numbers less than or equal to $n$ that have the same parity as $n$,
\item the \emph{hyperfactorial} $\H(n)$ is the number $\H(n)=\prod_{k=1}^nk^k$,
\item the \emph{subfactorial} $!n$ is the number of permutations of the set $\{1,2,\dots,n\}$ that fix no element,
\item the \emph{superfactorial} $\sf(n)$ is the number $\sf(n)=\prod_{k=1}^nk!$
\end{enumerate}
\end{definition}
The idea underlying our approach is the following: if there are sensible versions of Wilson's Theorem for these functions, then the numbers involved must be very special. If $p$ is prime, then the obvious \emph{special} numbers in the field $\Z_p$ are $0,\pm1$ and when $p$ is congruent to $1$ mod $4$, the two square roots of $-1$. So the task is to verify that these are the values taken, and determine which value is taken in each case. We find that this simple approach for prime $p$ works surprisingly well. For composite numbers, the hyperfactorial, subfactorial and superfactorial functions are all easily treated, while the double factorial holds some unexpected interesting surprises.
We present the results for double factorials in Theorems \ref{T:3mod4}, \ref{T:oddprime}, \ref{T:comp} and \ref{T:even}. The results for the superfactorial, hyperfactorial and subfactorial are given in Theorems \ref{SFDF}, \ref{hyp} and \ref{sub} respectively. The following result is an unexpected consequence of this study.
\begin{theorem} If $p$ is an odd prime, then modulo $p$
\[
(p-1)!!\equiv \sf(p-1)\equiv (-1)^{\frac{p-1}2} \H(p-1).
\]
\end{theorem}
\section{Factorials: double, hyper, sub and super}
According to the MacTutor History of Mathematics archive, the name \emph{factorial} and the notation $n!$ were introduced by the French mathematician Christian Kramp in 1808 \cite{Kr}, but the symbol was not immediately universally adopted. In the English speaking world, the notation ${\mathsmaller{\lfloor}}\underbar{n}$ was still commonly used at the end of the $19^{th}$ Century \cite{Caj}. De Morgan wrote ``Among the worst of barbarisms\footnote{At some point, the word barbarisms was misspelt as barabarisms when quoted and this error has been reproduced in a great number of places.} is that of introducing symbols which are quite new in mathematical, but perfectly understood in common, language. Writers have borrowed from the Germans the abbreviation $n!$ to signify $1.2.3\ldots(n - 1).n$, which gives their pages the appearance of expressing surprise and admiration that $2, 3, 4, \&c$.~should be found in mathematical results'' \cite{DeM}.
Rev.~W.~Allen~Whitworth\footnote{
There is an amusing Australian connection concerning Whitworth. In the Cairns Post, 22 November 1890, an article reads: ``The Rev.~W.~Allen Whitworth, in attempting a definition of gambling, as separable from mercantile speculation and legitimate enterprise, committed himself to the
proposition that under ordinary circumstances
and within the limits of moderation, it is (1)
justifiable to back one's skill, (2) foolish to
back one's luck, (3) immoral or fraudulent
to back one's knowledge. There is no rule
(4), but had there been it would doubtless
have read -- highly commendable to back
Carbine''. Carbine was the horse that won the Melbourne Cup in 1890.} apparently didn't share De~Morgan's view.
Whitworth introduced the \emph{subfactorial} and a symbol for it, in a paper that begins with the words: ``A new symbol in algebra is only half a benefit unless it has a new name. We believe that the symbol ${\mathsmaller{\lfloor}}\underbar{n}$ as an abbreviation of the continued product of the first $n$ integers, was long in use before the name \emph{factorial} $n$ was adopted. But until it received its name it appealed only to the eye and not to the ear, and in reading aloud could only be described by a periphrasis'' \cite{Whit}. Subfactorial $n$ is the number of permutations of the set $\{1,2,\dots,n\}$ that fix no element. There are many symbols for the subfactorial. Whitworth used the symbol $\underline{{\mathsmaller{\lfloor}}\underbar{n}}$, in keeping with the notation for the factorial at the time. These days $n\text{!`}$ is sometimes used, but $!n$ seems more common. De~Morgan would not be happy!
The subfactorial has the following explicit formula.
\begin{equation}\label{ELsub}
!n=n!\sum_{i=0}^n \frac{(-1)^i}{i!}.
\end{equation}
The hyperfactorial function was also introduced in the $19^{th}$ Century \cite{Gl,Ki} and the superfactorial function
followed at the turn of the Century \cite{Ba}, although the terms were introduced much later \cite{Sl,GKP}.
Study of the double factorial goes back at least as far as 1948 \cite{Me}, but it is probably much older.
For even $n$, one has
\begin{equation}\label{E:df}
n!!=2^{\frac{n}2}\cdot {\textstyle{\frac{n}2}}!
\end{equation}
For $n$ odd, $n!!$ coincides with the Gauss factorial $n_2!$, where in general $n_m!$ is the product of the natural numbers $i\leq n$ that are relatively prime to $m$ \cite{cd2}.
For a recent paper on the properties of double factorials and their applications, see \cite{GQ}.
\section{Motivation}\label{S:mot}
Our motivation for considering the various factorial functions concerns the matrix
\[
A=
\begin{pmatrix}
1&2&3&\dots &p-1\\
1&2^2&3^2&\dots& (p-1)^2\\
1&2^3&3^3&\dots& (p-1)^3\\
\vdots&\vdots&\vdots&&\vdots\\
1&2^{p-1}&3^{p-1}&\dots& (p-1)^{p-1}
\end{pmatrix}.
\]
There are many interesting facts and open questions related to the properties of the matrix $A$ modulo $p$. For example, \emph{Giuga's conjecture} is that the sum of the entries in the bottom row is congruent to $-1$ if and only if $p$ is prime \cite{BBBG}.
\begin{proposition}\label{L:matrix} The above matrix $A$ has determinant $\sf(p-1)$.
\end{proposition}
\begin{proof} Using the formula for the Vandermonde determinant \cite[Chap.~1.2]{Pr} we have
\begin{align*}
\det(A)&=(p-1)!\ \det
\begin{pmatrix}
1&1&1&\dots &1\\
1&2&3&\dots &p-1\\
1&2^2&3^2&\dots& (p-1)^2\\
\vdots&\vdots&\vdots&&\vdots\\
1&2^{p-2}&3^{p-2}&\dots& (p-1)^{p-2}
\end{pmatrix}\\
&=(p-1)! \prod_{1\leq i < j\leq p-1} (j-i)\\
&=(p-1)! \prod_{2\leq j\leq p-1} (j-1)!=\sf(p-1).
\end{align*}
\end{proof}
\begin{remark}\label{diag}
Notice that $\H(p-1)$ is precisely the product of the elements on the main diagonal of $A$. We will return to the matrix $A$ briefly in Section \ref{Sdf}.
\end{remark}
\section{The connection between the superfactorial and the double factorial}
Somewhat surprisingly, for prime $p$, the values $\sf(p-1)$ and $(p-1)!!$ are congruent.
Before proving this result, let us recall a well known fact which Lagrange proved in \cite{La} (see \cite{GQ,AC1}). Let $p$ be an odd prime. Then
\begin{equation}\label{R:wil}
\left(\s!\right)^2\equiv(-1)^{\frac{p+1}2}\pmod p.
\end{equation}
Indeed, modulo $p$, one has $p-i\equiv -i$ for all $i$ and so $(p-1)!\equiv(-1)^{\frac{p-1}2}\left(\s!\right)^2$ and the required claim follows from Wilson's Theorem.
\begin{theorem}\label{SFDF} If $p$ is prime, then $\sf(p-1)\equiv (p-1)!!\pmod p$.
\end{theorem}
\begin{proof} The theorem is obvious for $p=2$. In the following proof, $p$ is odd and the congruences are taken modulo $p$.
We have
\begin{align}
\sf(p-1)&=(p-1)!\,(p-2)!\,(p-3)!\,\dots\,3!\,2!\,1!\notag\\
&=(p-1)!!\,\left((p-2)!\,(p-4)!\,\dots\,3!\,1!\right)^2.\label{E:dfac}
\end{align}
Note that for all $1\leq i\leq p-1$,
\[
(p-i)! =\frac{(p-1)!}{(p-1)(p-2)\dots(p-i+1)}\equiv \frac{-1}{(-1)(-2)\dots(-i+1)}.
\]
Hence
\begin{equation}\label{E:canc}
(p-i)! \equiv \frac{(-1)^i}{(i-1)!}.
\end{equation}
So if $\frac{p-1}2$ is even, the factorials in \eqref{E:dfac} cancel in pairs, giving $\sf(p-1) \equiv (p-1)!!$, as required. If $\frac{p-1}2$ is odd, cancellation of the factorials in \eqref{E:dfac} leaves the middle term, giving $\sf(p-1) \equiv (p-1)!!\ \left (\s!\right)^2$ and hence $\sf(p-1) \equiv (p-1)!!$, by \eqref{R:wil}.
\end{proof}
\begin{remark}\label{R:GR}
In the above proof, the factorials in $\sf(p-1)$ can also be cancelled in pairs using \eqref{E:canc}, leaving only the middle one, and so
\begin{equation}\label{E:gt}
\sf(p-1)\equiv(-1)^{\sum_{i=1}^{\frac{p-1}2}i}
\cdot \s!=(-1)^{\frac{p^2-1}8}\cdot \s!
\end{equation}
Comparing this with $ (p-1)!! =2^{\frac{p-1}2}\cdot \s!
$, we obtain a simple derivation that $2^\frac{p-1}2\equiv(-1)^{\frac{p^2-1}8}$. In the following we will also require Euler's criterion \cite[Ch. III]{Dav}: 2 is a quadratic residue if and only if $2^\frac{p-1}2\equiv1$. Using this and Legendre's symbol we get a basic property of Gaussian reciprocity which allows to write:
\begin{equation}\label{Q:q}
(p-1)!!\equiv\left( \frac{2}{p}\right)\cdot \frac{p-1}{2}!\equiv (-1)^{\frac{p^2-1}8}\cdot \frac{p-1}{2}! \pmod p.
\end{equation}
\end{remark}
\section{The double factorial in the prime case}\label{Sdf}
If $p$ is an odd prime, then \eqref{R:wil} and \eqref{Q:q} give
\begin{equation}\label{extra}
((p-1)!!)^2\equiv \left(\s!\right)^2
\equiv(-1)^{\frac{p+1}2} \pmod p.
\end{equation}
In particular, $(p-1)!! \equiv
\pm 1\pmod p$ when $p\equiv 3 \pmod{4}$. The following result gives a more precise statement of this fact.
\begin{theorem}\label{T:3mod4}
Suppose that $p$ is an odd prime and $p\equiv 3 \pmod{4}$. Then
$(p-1)!! \equiv
(-1)^{\nu}\pmod p$,
where ${\nu}$ denotes the number of nonquadratic residues $j$ with $2<j<\frac{p}2$.
\end{theorem}
\begin{proof} Once again, the congruences will be taken modulo $p$ unless otherwise stated. Let $s:=\frac{p-1}{2}$ and $\Z^+_p:= \{1,2,\dots, p-1\}$. To evaluate the factor $s!$ in \eqref{Q:q} we use an argument that Mordell attributed to Dirichlet \cite{Mordell}. Consider the involution $\varphi: x\mapsto p-x$ on $\mathbb \Z^+_p$. Since $p\equiv 3 \pmod{4}$, $p$ cannot be expressed as a sum of two squares, and so $\varphi$ interchanges the set $QR$ of quadratic residues with the set $NR$ of nonquadratic residues. Let $QR=\{r_1,r_2,\dots,r_s\}$ and $NR=\{n_1,n_2,\dots,n_s\}$, with elements listed in natural order. Then
\[
s!= r_1r_2\dots r_{s-N}\ n_1n_2\dots n_{N}\equiv (-1)^N r_1r_2\dots r_s,
\]
where $N$ denotes the number of nonquadratic residues $j<\frac{p}2$. As $p\equiv 3 \pmod 4$, one has $ r_1r_2\dots r_s\equiv 1$; see \cite[p.~75]{Rose}. Hence $s!\equiv (-1)^N$.
Thus by Remark \ref{R:GR}, $(p-1)!!\equiv 1$ if and only if either $2$ is quadratic residue and $N$ is even, or $2$ is nonquadratic residue and $N$ is odd. Thus $(p-1)!!\equiv 1$ if and only if $\nu$ is
even.
\end{proof}
When $p\equiv 1 \pmod{4}$, we have $((p-1)!!)^2\equiv -1\pmod p$. In this case, because $-1$ is a quadratic residue, the involution $\varphi$ used in the proof of Theorem \ref{T:3mod4} leaves the sets $QR$ and $NR$ invariant, so it is not useful. Instead we consider the involution $\psi: x\mapsto x^{-1}$ of $\mathbb \Z^+_p$. It turns out that this approach is applicable for all odd primes.
\begin{theorem}\label{T:oddprime}
Suppose that $p$ is an odd prime. Let ${\mu}$ denote the number of elements $j $ less than $\frac{p}2$ such that the inverse $j^{-1}$ of $j$ modulo $p$ is also less than $\frac{p}2$.
\begin{enumerate}
\item If $p\equiv 3 \pmod{4}$, then
$(p-1)!! \equiv
(-1)^{\frac{\mu+1}2}\pmod p$.
\item If $p\equiv 1 \pmod{4}$, then
$(p-1)!! \equiv
(-1)^{\frac{\mu+1}2}i_p\pmod p$,
where $i_p$ is the unique natural number less than $\frac{p}2$ with $i_p^2\equiv-1 \pmod p$.
\end{enumerate}
\end{theorem}
\begin{proof} We use the same notation as in the proof of Theorem \ref{T:3mod4}.
Notice that if $j <\frac{p}2$ and $ j^{-1} <\frac{p}2$, then the two terms cancel in $\s!$. On the other hand, if $j <\frac{p}2$ and $ j^{-1} >\frac{p}2$, then $p-j^{-1} <\frac{p}2$ and provided $j\not=p-j^{-1}$, the product $j. (p-j^{-1})$ in $\s!$ gives $-1$.
If $p\equiv 3 \pmod{4}$, the number $-1$ is not a quadratic residue and so there is no $j <\frac{p}2$ with $j=p-j^{-1}$. In this case, $b=\s!\equiv (-1)^{\frac{s-\mu}2}$, where $s=\frac{p-1}2$. Hence $(p-1)!!= ab\equiv (-1)^{\frac{p^2-1}8+\frac{s-\mu}2}$. Let $p=4k+3$. Then $\frac{p^2-1}8+\frac{s}2
=2k^2+4k+ \frac32$. Hence $(p-1)!!\equiv (-1)^{\frac{3-\mu}2}\equiv (-1)^{\frac{\mu+1}2}$, as required.
If $p\equiv 1 \pmod{4}$, we argue in the same manner, but now $i_p$ is the unique number with $i_p <\frac{p}2$ and $i_p=p-i_p^{-1}$.
Hence $\s!\equiv (-1)^w i_p$, where $w=\frac{s-\mu-1}2$.
Thus $(p-1)!!= ab\equiv (-1)^{\frac{p^2-1}8+\frac{s-\mu-1}2}i_p$. Let $p=4k+1$. Then $\frac{p^2-1}8+\frac{s}2
=2k^2+2k$. Hence $(p-1)!!\equiv (-1)^{\frac{\mu+1}2}i_p$, as required.\end{proof}
Together Theorems \ref{T:3mod4} and \ref{T:oddprime} provide the following equivalence.
\begin{corollary}
If $p$ is an odd prime with $p\equiv 3 \pmod{4}$, then the number $\nu$, of nonquadratic residues $i$ with $2<i<\frac{p}2$, is even if and only if the number $\mu$ of elements $j $ less than $\frac{p}2$ such that the inverse $j^{-1}$ of $j$ modulo $p$ is also less than $\frac{p}2$, is congruent to 3 modulo 4.\end{corollary}
\begin{remark}
The results of Theorems \ref{T:3mod4} and \ref{T:oddprime} can be expressed in terms of class field numbers $h$, but the resulting statements are not as succinct as those given above. For the $p\equiv 3 \pmod{4}$ case, one can use $h(-p)=2N-1\pmod 4$; see \cite{Mordell}. For $p\equiv 1 \pmod{4}$, the result can be expressed in terms of $h(p)$ and the fundamental unit of the associated real quadratic number field; see \cite{Ch}.
\end{remark}
We can now give the connection between the hyperfactorial and double factorial.
\begin{theorem}\label{hyp} For $p$ an odd prime, the hyperfactorial and double factorial are connected by the relation
$\H(p-1)\equiv (-1)^{\frac{p-1}2} (p-1)!!\pmod{p}$.
\end{theorem}
\begin{proof}
Using Fermat's Little Theorem and Wilson's Theorem, we have
\begin{align*}
\H(p-1)=\prod_{k=1}^{p-1}k^k&= \frac{(\sf(p-1))^{p-1}}{\sf(p-2)}\\
&= \frac{(\sf(p-1))^{p-1}(p-1)!}{\sf(p-1)}\\
&\equiv \frac{-1}{\sf(p-1)}\equiv \frac{-1}{(p-1)!!},
\end{align*}
by Theorem \ref{SFDF}. By \eqref{extra}, we have
\[
((p-1)!!)^{-1}\equiv \begin{cases}
(p-1)!!&: \ \text{if}\ p\equiv 3 \pmod{4}\\
-(p-1)!!&: \ \text{if}\ p\equiv 1 \pmod{4}.
\end{cases}
\]
Thus $\H(p-1)\equiv (-1)^{\frac{p-1}2} (p-1)!!$, as required.
\end{proof}
\begin{remark}
Note that by Proposition \ref{L:matrix}, Remark \ref{diag} and Theorems \ref{SFDF} and \ref{hyp}, modulo $p$ the determinant of the matrix $A$ of Section \ref{S:mot} is $(-1)^{\frac{p-1}2}$ times the product of the elements on the main diagonal of $A$. \end{remark}
\section{Composite numbers}
When $n$ is composite and $n\not=4$, one has $(n-1)!\equiv 0\pmod n$. Similarly, it is obvious that $\sf(n-1)\equiv 0\pmod n$ and $\H(n-1)\equiv 0\pmod n$ for all composite natural numbers $n$. For the double factorial, the situation is more nuanced. The case of odd composites is not difficult.
\begin{theorem}\label{T:comp}
If $n$ is a composite odd natural number, then $(n-1)!!\equiv 0 \pmod n$ if $n>9$, while $8!! \equiv 6\pmod9$.\end{theorem}
\begin{proof}
Let $n$ be a composite odd natural number. If $n=ab$, where $a,b$ are co-prime, then $a,b< \frac{n-1}2$, so $\sn!\equiv 0\pmod n$. So we may assume that $n$ is of the form $n=p^k$ for some odd prime $p$, where $k\geq 2$.
If $k>2$, then $p,p^{k-1}$ are distinct and $p,p^{k-1}< \frac{n-1}2$, so $\sn!\equiv 0\pmod n$. If $n=p^2$ and $p>3$, then $p,2p$ are distinct and $p,2p< \frac{n-1}2$, so once again $\sn!\equiv 0\pmod n$. It remains to consider $n=9$, which can be calculated by hand.
\end{proof}
However, when $n$ is even, the pattern is less obvious.
\begin{theorem}\label{T:even}
Suppose that $n=2^iá(2k+1)$, where $i\geq 1$ and $k\geq 0$.
\begin{enumerate}
\item if $i=1$, then $(n-1) !! \equiv 2k+1 \pmod{n}$,
\item if $i=2$, then $(n-1)!! \equiv - (2k+1) \pmod{n}$,
\item if $i>2$, then $(n-1)!! \equiv (2k+1)^{2^{i-2}} \pmod{n}$.
\end{enumerate}
\end{theorem}
\begin{proof} We first treat the case $k=0$. Recall that for an arbitrary integer $n$, Gauss' generalisation of Wilson's Theorem \cite[Chap.~III]{Dick1} (see also \cite{AC2}) states that if $I$ denotes the set of invertible elements in $\Z_n$, then
\[
\prod_{s\in I} s \equiv
\begin{cases}-1&:\ \text{if}\ n=4,p^\alpha,2p^\alpha\ (p\ \text{an odd prime})\\
1&:\ \text{otherwise}
\end{cases}
\ \pmod{n}.
\]
For $n=2^i$ the set of odd elements of $\Z_{n}$ is precisely the group of invertible elements in $\Z_n$. Hence Gauss' result gives
\[
(n-1)!!\equiv \begin{cases}
-1&:\ \text{if}\ i=2\\
1&:\ \text{otherwise},
\end{cases} \ \pmod{n}.
\]
as required.
Now assume $k>0$. By the Chinese remainder theorem, the map
\begin{align*}
\varphi : \Z_n\to \Z_{2^i}\times \Z_{2k+1}\\
m\mapsto (\varphi_1(m),\varphi_2(m))
\end{align*}
is a ring isomorphism, where $\varphi_1(m)$ (resp.~$\varphi_2(m)$) is the reduction of $m$ modulo $2^i$ (resp.~$2k+1$). The inverse map is given by
\[
\varphi^{-1}(x,y)\equiv by2^i+ax(2k+1) \pmod n.
\]
where $a(2k+1)+b2^i=1$. The numbers $a$ and $b$ are defined modulo $n$. The proof focuses on the value of $a$. For $i=1$ we may take $a=1$. For $i=2$, we can take $a= (2k+1)$. For $i\geq 3$, note that as the set of odd elements of $\Z_{2^i}$ is the group of invertible elements in $\Z_{2^i}$, so by Euler's Theorem, $(2k+1)^{2^{i-2}}\equiv 1 \pmod{2^i}$. Thus we may take $a=(2k+1)^{2^{i-3}}$ when $i\geq 3$.
We have
\[
\varphi((n-1)!!)=\prod_{\substack{x\in \Z_{2^i},\ y\in\Z_{2k+1}\\x\ \text{odd}}} (x,y)=\left(\prod_{\substack{x\in \Z_{2^i}\\x\ \text{odd}}} x, \prod_{y\in\Z_{2k+1}} y\right)=\left((2^i-1)!!, 0\right).
\]
Now from the $k=0$ case treated above,
\[
(2^i-1)!!\equiv\begin{cases}
-1&:\ \text{if}\ i=2\\
1&:\ \text{otherwise}.
\end{cases}\ \pmod{2^i}.
\]
So when $i=1$ we have
\[
(n-1)!!=\varphi^{-1}(1,0)\equiv a(2k+1)=(2k+1).
\]
When $i=2$ we have
\[
(n-1)!!=\varphi^{-1}(-1,0)\equiv -a(2k+1)=-(2k+1)^2.
\]
Finally, for $i\geq 3$ we have
\[
(n-1)!!=\varphi^{-1}(1,0)\equiv a(2k+1)=(2k+1)^{2^{i-2}}.
\]
\end{proof}
\section{The subfactorial}
For the (standard) factorial, the double factorial, the hyperfactorial and the superfactorial, the value at number $n$ is congruent to zero mod $n$. It is in part for this reason that the value at $n-1$ is of interest mod $n$. For the subfactorial however, the situation is different. Here the natural generalisation of Wilson's Theorem is the following.
\begin{theorem}\label{sub}
If $n$ is a natural number, then $!n\equiv (-1)^n \pmod n$.
\end{theorem}
\begin{proof}
Modulo $n$ we have from \eqref{ELsub}
\begin{equation*}
!n=n!\sum_{i=0}^n \frac{(-1)^i}{i!}=(-1)^n+n!\sum_{i=0}^{n-1} \frac{(-1)^i}{i!}\equiv (-1)^n \pmod n.
\end{equation*}
\end{proof}
\bibliographystyle{amsplain}
\bibliography{factorials}
\end{document} | 0.007667 |
A Spiritual Kick up the Backside
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Are you ready to be shaken up a little Are you ready to let God speak to you Are you ready to start really living for Him Are you ready for a faith that impacts how you live every day Are you ready to show what is important is "faith expressing itself through love " Are you ready for a spiritual kick up the backside A Spiritual Kick up the Backside is the sequel to Pete's first book, Reflections. Containing a collection of devotions to make you laugh, make you think, draw you closer to God, and hopefully give you a bit of a kick up the backside, in a spiritual way.
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I am a hopeless romantic. I think you could tell if you would read some of my poems.
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by elvanGalaxy 8 years ago
what is a perfect relationship?I think it's pretty simple to think about having a perfect relationship, but the truth to the matter is there's nothing like perfect relationship cos what seems perfect in your own judgement could not be so to some others.
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TITLE: better Understanding a proof of Hartshorne's book proposition 2.2. (b)
QUESTION [1 upvotes]: I have been reading the book Algebraic Geometry by Robin Hartshorne
and I have found the following proposition:
For part b) the proof goes as follows:
I really don't follow proof that $\psi $ surjective. That goes as
follows:
In the proof of surjectivity in first paragraph how can we replace $ h_i^n$ by $h_i$(since $D(h_i) = D(h_i^n)$ )
I hope someone can help me understanding this parts of this proof.
Thanks in advance.
REPLY [1 votes]: This is the simplifying assumption they would like to make for the rest of the proof: there exist elements $h_i'\in A$ such that $D(f)=\bigcup_i D(h_i')$, and the restriction of $s$ to $D(h_i')$ is equal to $a_i'/h_i'$ for some $a_i'\in A$.
So we already have elements $h_i$ such that $D(f)=\bigcup_i D(h_i)$ and $s|_{D(h_i)}=ca_i/h_i^n$, so if we take $a_i'=ca_i$ and $h_i'=h_i^n$, then we certainly have $s|_{D(h_i)}=a_i'/h_i'$, and if we can prove that $D(h_i)=D(h_i')$ then we will have the conditions above met. But it is easy to verify that $D(h)=D(h^n)$ for any $h\in A$ and $n\in\Bbb Z^+$. | 0.009079 |
\begin{document}
\subjclass{47D06; 82C05; 60J80; 34K30}
\keywords{Markov evolution, configuration space, stochastic
semigroup, sun-dual semigroup, correlation function, scale of Banach
spaces }
\begin{abstract}
The evolution of an infinite system of interacting point entities
with traits $x\in \mathds{R}^d$ is studied. The elementary acts of
the evolution are state-dependent death of an entity with rate that
includes a competition term and independent fission in the course of
which an entity gives birth to two new entities and simultaneously
disappears. The states of the system are probability measures on the
corresponding configuration space and the main result of the paper
is the construction of the evolution $\mu_0\to \mu_t$, $t>0$, of
states in the class of sub-Poissonian measures.
\end{abstract}
\maketitle
\section{Introduction}
\subsection{Posing}
In recent years, there has been a lot of studies of the stochastic
dynamics of structured populations, see, e.g.,
\cite{Ba,BL,Chris,JK11,KK,KK2}. Typically, the structure is
introduced by assigning to each entity a trait $x\in X$. Then the
population dynamics consists in changing the traits of its members
that includes also their appearance and disappearance. Usually, one
endows the trait space $X$ with a locally-compact topology and
assumes that: (a) the populations are locally finite, i.e., compact
subsets of $X$ may contain traits of finite sub-populations only;
(b) the dynamics of a given entity is mostly affected by the
interaction with entities whose traits belong to a compact
neighborhood of its own trait. Then the local structure of the
population is determined by the network of such interactions. Since
the traits of a finite population lie in a compact subset of $X$,
each of its members has a compact neighborhood containing the traits
of the rest of population. In view of this, in order to clear
distinguish between global and local effects one should deal with
infinite populations and noncompact trait spaces. In the statistical
mechanics of interacting physical particles, this conclusion had led
to the concept of the thermodynamic (infinite-volume) limit, see,
e.g., \cite[pp. 5,6]{Simon}, and, thereby, to the description of the
states of thermal equilibrium as probability measures on the space
of particle configurations. Such states are constructed from local
conditional states and are Gibbsian, i.e., they satisfy a specific
consistency condition.
In this article, we study the Markov evolution of a possibly
infinite system of point entities (particles) with trait space
$X=\mathds{R}^d$, $d\geq 1$. The pure states of the system are
locally finite configurations $\gamma \subset \mathds{R}^d$, see,
e.g., \cite{Berns,KK,KK1,KK2}, whereas the general states are
probability measures on the space of all such configurations. The
elementary acts of the evolution are: (a) state-dependent
disappearance (death) with rate $m(x) + \sum_{y\in \gamma\setminus
x} a(x-y)$; (b) independent fission with rate $b(x|y_1, y_2)$ in the
course of which the particle with trait $x\in \gamma$ gives birth to
two particles, with traits $y_1, y_2 \in \mathds{R}^d$, and
simultaneously disappears from $\gamma$. The model with this kind of
death and budding instead of fission, cf. \cite{Chris}, is known as
the Bolker-Pacala model. Its recent study can be found in
\cite{KK,KK1}, see also the literature quoted therein. A similar
model with fission (fragmentation) in which each particle produces a
(random) finite number of new particles was introduced and studied
in \cite{tan}. The main result of the present work is the
construction of the global in time evolution of states in a certain
class of probability measures.
\subsection{The overview}
As mentioned above, the state space of the model is the set $\Gamma$
of all subsets $\gamma \subset \mathds{R}^d$ such that the set
$\gamma_\Lambda:=\gamma\cap\Lambda$ is finite whenever $\Lambda
\subset \mathds{R}^d$ is compact. For compact $\Lambda$, we define
the map $\Gamma\ni \gamma \mapsto N_\Lambda (\gamma) =
|\gamma_\Lambda|\in \mathds{N}_0$, where $|\cdot|$ denotes
cardinality and $\mathds{N}_0$ stands for the set of nonnegative
integers. Then $\mathcal{B}(\Gamma)$ will denote the smallest
$\sigma$-field of subsets of $\Gamma$ with respect to which all
these maps are measurable. That is, $\mathcal{B}(\Gamma)$ is
generated by the family of sets
\begin{equation}
\label{o1}
\Gamma^{\Lambda,n} :=\{ \gamma \in \Gamma: N_\Lambda (\gamma) =
n\}, \qquad n\in \mathds{N}_0, \ \ \Lambda - {\rm compact}.
\end{equation}
It is known \cite{KK,KK2} that $(\Gamma,\mathcal{B}(\Gamma))$ is a
standard Borel space. The set of $n$-point configurations $\Gamma^n$
and the set of all finite configurations $\Gamma_0$ then are
\begin{equation*}
\Gamma^n = \{ \gamma\in \Gamma: |\gamma|=n\}, \qquad \Gamma_0 := \bigcup_{n=0}^\infty \Gamma^n \in \mathcal{B}(\Gamma).
\end{equation*}
For compact $\Lambda$, we let $\Gamma_\Lambda =\{ \gamma:
\gamma\subset \Lambda\} \subset
\Gamma_0$
and define
\begin{equation*}
\mathcal{B}(\Gamma_\Lambda) = \{ \mathbb{A}\cap \Gamma_\Lambda: \mathbb{A} \in
\mathcal{B}(\Gamma)\} \subset \mathcal{B}(\Gamma_0) = \{ \mathbb{A}\cap \Gamma_0: \mathbb{A} \in
\mathcal{B}(\Gamma)\} \subset \mathcal{B}(\Gamma).
\end{equation*}
Clearly, $(\Gamma_0, \mathcal{B}(\Gamma_0))$ and $(\Gamma_\Lambda,
\mathcal{B}(\Gamma_\Lambda))$ are standard Borel spaces. By
$\mathcal{P}(\Gamma)$, $\mathcal{P}(\Gamma_0)$,
$\mathcal{P}(\Gamma_\Lambda)$ we denote the sets of all probability
measures on $(\Gamma,\mathcal{B}(\Gamma))$,
$(\Gamma_0,\mathcal{B}(\Gamma_0))$ and
$(\Gamma_\Lambda,\mathcal{B}(\Gamma_\Lambda))$, respectively.
For a compact $\Lambda$ and $\mathbb{A}\in
\mathcal{B}(\Gamma_\Lambda)$, we set $\mathbb{C}_{\mathbb{A}} = \{
\gamma \in \Gamma: \gamma_\Lambda \in \mathbb{A}\}$ and let
$\mathcal{B}_\Lambda (\Gamma)$ be the sub-$\sigma$-field of
$\mathcal{B}(\Gamma)$ generated by all such \emph{cylinder} sets
$\mathbb{C}_{\mathbb{A}}$. A \emph{cylinder} function $F : \Gamma
\to \mathds{R}$ is a $\mathcal{B}_\Lambda
(\Gamma)/\mathcal{B}(\mathds{R})$-measurable function for some
compact $\Lambda$. Here by $\mathcal{B}(\mathds{R})$ we denote the
Borel $\sigma$-field of subsets of $\mathds{R}$. For a compact
$\Lambda$ and a given $\mu \in \mathcal{P}(\Gamma)$, by setting
\begin{equation}
\label{Rel}
\mu(\mathbb{C}_{\mathbb{A}}) = \mu^\Lambda (\mathbb{A})
\end{equation}
we determine
$\mu^\Lambda \in \mathcal{P}(\Gamma_\Lambda)$ -- the
\emph{projection} of $\mu$. Note that all such projections
$\{\mu^\Lambda\}_{\Lambda}$ of a given $\mu \in \mathcal{P}(\Gamma)$
are consistent in the Kolmogorov sense.
Each $\mu\in \mathcal{P}(\Gamma)$ is characterized by its values on
the sets (\ref{o1}); in particular, by their local moments
\begin{equation}
\label{o2}
\int_\Gamma N_\Lambda^m d\mu =: \mu(N_\Lambda^m) = \sum_{n=0}^\infty
n^m \mu(\Gamma^{\Lambda, n}), \qquad m\in \mathds{N}.
\end{equation}
This characterization naturally includes the dependence of
$\mu(\Gamma^{\Lambda, n})$ on $n$. A homogeneous Poisson measure
$\pi_\varkappa \in \mathcal{P}(\Gamma)$ with density $\varkappa >0$
has the property $\pi_\varkappa(\Gamma_0) = 0$. For this measure, it
follows that
\begin{equation}
\label{o3}
\pi_\varkappa (\Gamma^{\Lambda, n}) = \frac{\left(\varkappa |\Lambda|\right)^n}{n!} \exp\left( - \varkappa |\Lambda| \right),
\end{equation}
where $|\Lambda|$ stands for the volume of $\Lambda$. In our
consideration, the set of sub-Poissonian measures $\mathcal{P}_{\rm
exp}(\Gamma)$ plays an important role, see Definition \ref{0df}
below and the corresponding discussion in \cite{KK,KK1}. For each
$\mu \in \mathcal{P}_{\rm exp}(\Gamma)$, there exists $\varkappa >0$
such that
\begin{equation}
\label{o3a}
\mu(N_\Lambda^m) \leq \pi_\varkappa (N_\Lambda^m),
\end{equation}
holding for all compact $\Lambda$ and $m\in \mathds{N}$.
The Markov evolution is described by the Kolmogorov equation
\begin{equation}
\label{1}
\dot{F}_t = L F_t , \qquad F_t|_{t=0} = F_0,
\end{equation}
where $\dot{F}_t$ denotes the time derivative of an
\emph{observable} $F_t:\Gamma\to \mathds{R}$. The operator $L$
determines the model, and in our case it is
\begin{gather}
\label{2}
(LF)(\gamma) = \sum_{x\in \gamma}\left( m (x) + \sum_{y\in \gamma\setminus x} a (x-y)\right)
\left[ F(\gamma\setminus x) - F(\gamma)
\right] \\[.2cm] \nonumber + \sum_{x\in \gamma}
\int_{(\mathds{R}^d)^2} b(x|y_1 , y_2)\left[ F(\gamma\setminus x \cup\{y_1, y_2\}) - F(\gamma)
\right] dy_1 d y_2.
\end{gather}
In expressions like $\gamma \cup x$, we treat $x$ as the singleton
$\{x\}$. The first term in (\ref{2}) describes the death of the
particle with trait $x$ occurring: (i) independently with rate $m(x)
\geq 0$; (ii) under the influence (competition) of the rest of the
particles in $\gamma$ occurring with rate
\begin{equation}
\label{3}
E^a (x, \gamma\setminus x) := \sum_{y\in \gamma\setminus x} a
(x-y)\geq 0.
\end{equation}
The second term in (\ref{2}) describes independent fission with rate
$b(x|y_1 , y_2)\geq 0$.
The evolution of states $\mu_0 \to \mu_t$ is defined by the
Fokker-Planck equation
\begin{equation}
\label{4}
\dot{\mu}_t = L^* \mu_t, \qquad \mu_t|_{t=0} =\mu_0,
\end{equation}
where $L^*$ is related to (\ref{2}) according to the rule $(L^*
\mu)(\mathbb{A})= \mu(L \mathds{1}_{\mathbb{A}} )$, $\mathbb{A}\in
\mathcal{B}(\Gamma)$; $\mathds{1}_{\mathbb{A}}$ is the indicator
function. Both evolutions are in the duality $\mu_0 (F_t) =
\mu_t(F_0)$. Here and in the sequel, we use the notation $\mu(F)=
\int F d \mu$, cf. (\ref{o2}).
The direct use of $L$ and/or $L^*$ as linear operators in
appropriate Banach spaces is possible only if one restricts the
consideration to states on $\Gamma_0$. Otherwise, the sums in
(\ref{2}) and (\ref{3}) -- taken over infinite configurations -- may
not exist. At the same time, constructing evolutions of finite
sub-populations contained in compact sets followed by taking the
`infinite-volume' limit -- as it is done in the theory of Gibbs
fields \cite{Simon} -- can hardly be realized here as the evolution
usually destroys the consistency of the local states. Instead of
trying to construct global states from local ones, we will we
proceed as follows. Let $C_0 (\mathds{R}^d)$ stand for the set of
continuous real-valued functions with compact support. Then the map
\[
\Gamma \ni \gamma \mapsto F^\theta (\gamma) := \prod_{x\in \gamma}
(1+ \theta (x)), \qquad \theta \in \varTheta:=\{\theta\in C_0
(\mathds{R}^d): \theta(x) \in (-1,0]\},
\]
is clearly measurable and satisfies $0<F^\theta (\gamma) \leq 1$ for
all $\gamma$. The set $\varTheta$ clearly has the following
properties: (a) for each pair of distinct $\gamma, \gamma' \in
\Gamma$, there exists $\theta \in \varTheta$ such that $F^\theta
(\gamma)\neq F^\theta (\gamma')$; (b) for each pair $\theta ,
\theta'\in \varTheta$, the point-wise combination $\theta + \theta'
+\theta \theta'$ is also in $\varTheta$; (c) the zero function
belongs to $\varTheta$. From this it follows that $\{F^\theta:
\theta \in \varTheta\}$ is a measure defining class, i.e.,
$\mu(F^\theta)= \nu(F^\theta)$, holding for all $\theta \in
\varTheta$, implies $\mu=\nu$ for each $\mu, \nu \in
\mathcal{P}(\Gamma)$, see \cite[Proposition 1.3.28, page 113]{AKKR}.
Noteworthy, for each $\theta \in \varTheta$, $\mu(F^\theta) =
\mu^{\Lambda_\theta} (F^\theta)$, where a compact $\Lambda_\theta$
is such that $\theta(x) = 0$ for $x\in
\Lambda^c_{\theta}:=\mathds{R}^d \setminus \Lambda_\theta$.
Our results related to (\ref{2}), (\ref{4}) consist in the
following:
\begin{itemize}
\item[1.] Constructing the evolution $[0,+\infty)\ni t \mapsto \mu_t \in \mathcal{P}(\Gamma_0)$, $\mu_t|_{t=0}=\mu_0\in \mathcal{P}(\Gamma_0)$,
by proving the existence of a unique classical solution of (\ref{4}) in
the Banach space $\mathcal{M}$ of signed measures on $\Gamma_0$ with bounded
variation.
\item[2.] Constructing the evolution $[0,+\infty) \ni t
\mapsto \mu_t \in \mathcal{P}_{\rm exp}(\Gamma)$, $\mu_t|_{t=0}=\mu_0\in \mathcal{P}_{\rm exp}(\Gamma)$, such that:
\begin{itemize}
\item[2.1.] for each compact $\Lambda$ and $t\geq 0$, $\mu_t^{\Lambda}$ -- as a measure on $\Gamma_0$ -- lies in the
domain $\mathcal{D}(L^*) \subset \mathcal{M}$;
\item[2.2.] for
each $\theta \in \varTheta$, the map $(0,+\infty) \ni t \mapsto \mu_t (F^\theta)$ is continuously
differentiable and the following holds
\begin{equation}
\label{Jan}
\frac{d}{dt}
\mu_t (F^\theta) = (L^* \mu^{\Lambda_\theta}_t)(F^\theta).
\end{equation}
\end{itemize}
\end{itemize}
Item 1 is realized in Theorem \ref{1ftm}. The main idea of how to
construct the evolution $\mu_0 \to \mu_t$ stated in item 2 is to
obtain it from the evolution $B_0 (\theta)\to B_t(\theta)$, $\theta
\in \varTheta$ by solving the evolution equation related to those in
(\ref{1}) and (\ref{4}). Here $B_0 (\theta) = \mu_0 (F^\theta)$ with
$\mu_0 \in \mathcal{P}_{\rm exp}(\Gamma)$. This is realized in
Theorem \ref{1tm} and Corollary \ref{Jaco}. One of the hardest
points of this scheme is to prove that $B_t(\theta)= \mu_t
(F^\theta)$ for a unique sub-Poissonian measure. At this stage, we
deal with the evolution of local states constructed in realizing
item 1.
\section{Preliminaries and the Model}
We begin by briefly introducing the relevant aspects of the
technique used in this work. Its more detailed description
(including the notations) can be found in \cite{KK,KK2} and in the
publications quoted therein.
\subsection{Measures and functions on configuration spaces}
It is know that
\begin{equation*}
B_{\pi_\varkappa} (\theta):= \pi_\varkappa (F^\theta) = \exp\left(
\varkappa\int_{\mathds{R}^d} \theta (x)
dx\right).
\end{equation*}
Obviously, $B_{\pi_\varkappa}$ can be continued to an exponential
type entire function of $\theta \in L^1(\mathds{R}^d)$.
\begin{definition}
\label{0df}
The set of sub-Poissonian measures $\mathcal{P}_{\rm exp} (\Gamma)$
consists of all those $\mu\in \mathcal{P}(\Gamma)$ for each of which
$\mu(F^\theta)$ can be continued to an exponential type entire function of
$\theta \in L^1(\mathds{R}^d)$.
\end{definition}
It can be shown that $\mu\in \mathcal{P}_{\rm exp} (\Gamma)$ if and
only if $\mu(F^\theta)$ might be written in the form
\begin{equation}
\label{6b}
\mu(F^\theta) = 1 + \sum_{n=1}^\infty \frac{1}{n!} \int_{(\mathds{R}^d)^n} k^{(n)}_\mu (x_1 , \dots , x_n) \theta (x_1) \cdots \theta(x_n) d x_1 \cdots d x_n,
\end{equation}
where $k^{(n)}_\mu$ is the $n$-th order \emph{correlation function}
of $\mu$. Each $k^{(n)}_\mu$ is a symmetric element of
$L^\infty((\mathds{R}^d)^n)$, and the collection $\{k^{(n)}_\mu
\}_{n\in \mathds{N}}$ satisfies
\begin{equation}
\label{6c}
\|k^{(n)}_\mu\|_{L^\infty((\mathds{R}^d)^n} \leq \varkappa^{ n}, \qquad n \in \mathds{N},
\end{equation}
holding with some $\varkappa >0$. Note that $k^{(n)}_\mu $ is
positive and $k^{(n)}_{\pi_\varkappa} = \varkappa^n$; hence,
(\ref{6c}) means that $k^{(n)}_\mu (x_1 , \dots , x_n)\leq
\varkappa^n$ by which one gets (\ref{o3a}).
Now we turn to functions $G:\Gamma_0 \to \mathds{R}$. It can be
proved that such a function is
$\mathcal{B}(\Gamma_0)/\mathcal{B}(\mathds{R})$-measurable if and
only if there exists the collection of symmetric Borel functions
$G^{(n)}:(\mathds{R}^d)^n \to \mathds{R}$, $n\in \mathds{N}$, such
that
\begin{equation}
\label{7}
G(\eta) = G^{(n)}(x_1 , \dots , x_n), \qquad {\rm for} \ \ \eta = \{x_1 , \dots , x_n\}.
\end{equation}
\begin{definition}
\label{1df}
A measurable function $G:\Gamma_0 \to \mathds{R}$ is said to have bounded support if: (a) there exists compact
$\Lambda \subset \mathds{R}^d$ such that $G(\eta) =0$ whenever $\eta\cap\Lambda \neq \eta$; (b) there exists
$N\in \mathds{N}$ such that $G(\eta) =0$ whenever $|\eta|>N$. By $B_{\rm bs}(\Gamma_0)$ we denote the set of all bounded functions with bounded support.
For each $G\in B_{\rm bs}(\Gamma_0)$, by $\Lambda_G$ and $N_G$ we denote the smallest $\Lambda$ and $N$ with the properties just mentioned,
and use the notations $C_G = \sup_{\eta\in \Gamma_0} |G(\eta)|$.
\end{definition}
The Lebesgue-Poisson measure $\lambda$ on $(\Gamma_0,
\mathcal{B}(\Gamma_0))$ is defined by the integrals
\begin{equation}
\label{8}
\int_{\Gamma_0} G(\eta) \lambda (d \eta) = G(\varnothing) +
\sum_{n=1}^\infty \frac{1}{n!} \int_{(\mathds{R}^d)^n} G^{(n)}(x_1 ,
\dots x_n) d x_1 \cdots d x_n,
\end{equation}
with all $G\in B_{\rm bs}(\Gamma_0)$. For such $G$, we set
\begin{equation}
\label{9}
(KG)(\gamma) = \sum_{\eta \Subset \gamma} G(\eta), \qquad \gamma \in \Gamma,
\end{equation}
where $\eta\Subset \gamma$ means that $\eta\subset \gamma$ and $\eta\in \Gamma_0$. Clearly, cf. Definition \ref{1df}, we have that
\begin{equation}
\label{10}
|(KG)(\gamma)| \leq C_G \left(1 + |\gamma\cap \Lambda_G| \right)^{N_G}, \qquad G\in B_{\rm bs}(\Gamma_0).
\end{equation}
Like in (\ref{7}), we introduce the function $k_\mu:\Gamma_0 \to
\mathds{R}$ such that $k_\mu (\eta) = k^{(n)}_\mu (x_1 , \dots,
x_n)$ for $\eta =\{x_1 , \dots, x_n\}$, $n\in \mathds{N}$, and
$k_\mu(\varnothing) =1$. Then we rewrite (\ref{6b}) as follows
\begin{equation}
\label{11}
\mu (F^\theta) = \int_{\Gamma_0} k_\mu(\eta) e(\theta;\eta)
\lambda (d \eta), \qquad e(\theta;\eta):= \prod_{x\in
\eta}\theta(x).
\end{equation}
For $\mu\in \mathcal{P}_{\rm exp}(\Gamma)$ and a compact $\Lambda$,
let $\mu^\Lambda$ be the corresponding projection. It is possible to
show that $\mu^\Lambda$, as a measure on $(\Gamma_\Lambda,
\mathcal{B}(\Gamma_\Lambda))$, is absolutely continuous with respect
to the Lebesgue-Poisson measure $\lambda$. Hence, we may write
\begin{equation}
\label{12}
\mu^\Lambda ( d\eta) = R^\Lambda_\mu (\eta) \lambda (d\eta), \qquad
\eta \in \Gamma_\Lambda.
\end{equation}
For each compact $\Lambda$, the Radon-Nikodym derivative
$R^\Lambda_\mu$ and the correlation function $k_\mu$ satisfy
\begin{equation}
\label{13}
k_\mu(\eta) = \int_{\Gamma_\Lambda} R^\Lambda_\mu (\eta \cup \xi)
\lambda ( d \xi), \qquad \eta \in \Gamma_\Lambda.
\end{equation}
For each $G\in B_{\rm bs}(\Gamma_0)$ and $k:\Gamma_0 \to \mathds{R}$
such that $k^{(n)}\in L^\infty ((\mathds{R}^d)^n)$ the integral
\begin{equation}
\label{14}
\langle \! \langle G, k \rangle \! \rangle := \int_{\Gamma_0}
G(\eta) k(\eta) \lambda ( d \eta)
\end{equation}
surely exists. By (\ref{6b}), (\ref{9}), (\ref{12}) and (\ref{14})
we then obtain
\begin{equation}
\label{15}
\int_\Gamma \left(KG \right)(\gamma) \mu (d\gamma) = \langle \! \langle G, k_\mu \rangle \! \rangle
\end{equation}
holding for all $G\in B_{\rm bs}(\Gamma_0)$ and $\mu \in
\mathcal{P}_{\rm exp}(\Gamma)$. Set
\begin{equation}
\label{16}
B_{\rm bs}^{\star}(\Gamma_0) = \{ G \in B_{\rm bs}(\Gamma_0): \left(
KG \right) (\gamma) \geq 0 \ \ {\rm for} \ \ {\rm all} \ \ \gamma
\in \Gamma\}.
\end{equation}
By \cite[Theorems 6.1, 6.2 and Remark 6.3]{Tobi} we know that the
following is true.
\begin{proposition}
\label{1pn}
Let a measurable function $k: \Gamma_0 \to \mathds{R}$ have the
following properties:
\begin{eqnarray*}
&(a) & \ \ \langle \! \langle G, k_\mu \rangle \! \rangle \geq 0,
\qquad {\rm for} \ \ {\rm all} \ \ G \in B_{\rm
bs}^{\star}(\Gamma_0);
\\[.2cm]
&(b) & \ \ k(\varnothing) =1; \\ &(c) & \ \ k(\eta) \leq
C^{|\eta|}, \qquad {\rm for} \ \ {\rm some} \ \ C>0.
\end{eqnarray*}
Then there exists a unique $\mu\in \mathcal{P}_{\rm exp}(\Gamma)$
such that $k$ is its correlation function.
\end{proposition}
Throughout the paper we use the following easy to check identities
holding for appropriate functions $g:\mathds{R}^d \to \mathds{R}$
and $G:\Gamma_0 \to \mathds{R}$:
\begin{equation}
\label{17}
\forall x\in \gamma\qquad \sum_{\eta \Subset \gamma}\prod_{z\in
\eta} g(z)= (1+ g(x))\sum_{\eta \Subset \gamma\setminus
x}\prod_{z\in \eta} g(z),
\end{equation}
\begin{equation}
\label{18}
\int_{\Gamma_0} \sum_{\xi \subset \eta} G(\xi, \eta, \eta
\setminus \xi) \lambda ( d\eta)= \int_{\Gamma_0} \int_{\Gamma_0} G(\xi, \eta\cup \xi,
\eta) \lambda ( d\xi) \lambda ( d\eta).
\end{equation}
\subsection{The model}
As mentioned above, the model which we consider in this work is
described by the generator given in (\ref{2}). Its entries are
subject to the following
\begin{assumption}
\label{ass1}
The nonnegative measurable $a$, $b$ and $m$ satisfy:
\begin{itemize}
\item[(i)] $a$ is integrable and bounded; hence, we may set $$\sup_{x\in \mathds{R}^d}a(x) =
a^*, \qquad \int_{ \mathds{R}^d}a(x) dx = \langle a \rangle.$$
\item[(ii)]
There exist positive $r$ and $a_*$ such that $a(x) \geq a_*$
whenever $|x|\leq r$.
\item[(iii)] For each $x\in \mathds{R}^d$, $b(x|y_1, y_2) d y_1 d y_2$ is a symmetric finite measure on $(\mathds{R}^d)^2$; hence, we may set
\[
\langle b \rangle = \int_{(\mathds{R}^d)^2} b(x|y_1, y_2) d y_1 d
y_2,
\]
where, for simplicity, we consider the translation invariant case.
The mentioned symmetry means that $b(x|y_1, y_2) = b(x|y_2, y_1)$.
\item[(iv)]
The function $$\beta (y_1 - y_2) = \int_{\mathds{R}^d} b(x|y_1 ,
y_2) d x
$$ is supposed
to be such that $\sup_{x\in \mathds{R}^d}\beta(x) =:
\beta^*<\infty$. By the translation invariance it follows that
$$\int_{ \mathds{R}^d}\beta(x) dx = \langle b \rangle.$$
\end{itemize}
\end{assumption}
Noteworthy, we do not exclude the case where $b$ is a distribution.
For instance, by setting $$b(x|y_1, y_2)= \frac{1}{2} \left(\delta
(x-y_1) + \delta (x-y_2) \right)\beta (y_1-y_2),$$ we obtain the
Bolker-Pacala model \cite{KK1} as a particular case of our model.
\begin{remark}
\label{1rk}
The function $\beta$ describes the dispersal of siblings, which
compete with each other. As in the Bolker-Pacala model, here the
following situations may occur:
\begin{itemize}
\item \emph{short dispersal:} there exists $\omega >0$
such that $a(x)
\geq \omega \beta(x)$ for all $x\in \mathds{R}^d$;
\item \emph{long dispersal:} for each $\omega >0$, there
exists $x\in \mathds{R}^d$ such that $ a(x)
< \omega \beta(x)$.
\end{itemize}
\end{remark}
For $\eta \in \Gamma_0$, we set, cf. (\ref{3}),
\begin{eqnarray}
\label{19}
E^a(\eta) & = & \sum_{x\in \eta} E^a(x, \eta\setminus x) =
\sum_{x\in
\eta} \sum_{y\in \eta\setminus x} a(x-y), \\[.2cm] \nonumber
E^b(\eta) & = & \sum_{x\in \eta} \sum_{y\in \eta\setminus x}
\beta(x-y) = \sum_{x\in \eta} \sum_{y\in \eta\setminus x}
\int_{\mathds{R}^d} b(z|x, y) d z.
\end{eqnarray}
The properties mentioned in (ii) and (iv) of Assumption \ref{ass1}
imply the following fact, proved in \cite[Lemma 3.1]{KKa}. For the
reader convenience, we repeat the proof in Appendix below.
\begin{proposition}
\label{2pn}
There exist $\omega>0$ and $\upsilon\geq 0$ such that the following
holds
\begin{equation}
\label{2pnN} \upsilon |\eta| + E^a(\eta) \geq \omega E^b(\eta),
\qquad \eta\in \Gamma_0.
\end{equation}
\end{proposition}
The inequality in (\ref{2pnN}) can be rewritten in the form
\begin{equation}
\Phi_{\omega}(\eta):= \sum_{x\in \eta} \sum_{y\in \eta\setminus x} \left[ a(x-y)-\omega \int_{\mathbb{R}^d}b(z|x,y)dz \right] \ge -\upsilon |\eta|.
\label{fi}
\end{equation}
\begin{proposition}
\label{pfi} Assume that (\ref{fi}) holds for some $\omega_0>0$ and
$\upsilon_0>0$. Then for each $\omega < \omega_0$, it holds also for
$\upsilon=\upsilon_0\omega/\omega_0$.
\end{proposition}
\begin{proof}
For $\omega \in [0, \omega_0]$ by adding and subtracting $\frac{\omega}{\omega_0}E^a(\eta)$
we obtain
$$\Phi_\omega(\eta)=\frac{\omega}{\omega_0}
\left[ \left(\frac{\omega_0}{\omega}-1 \right)
E^a(\eta)+\Phi_{\omega_0}(\eta) \right]\ge -\frac{\omega}{\omega_0}\upsilon_0|\eta|.$$
\end{proof}
\section{The Evolution of States of the Finite System}
Here we assume that the initial state in (\ref{4}) has the property
$\mu_0(\Gamma_0)=1$, i.e., the system in $\mu_0$ is finite. Then the
evolution will be constructed in the Banach space of signed measures
with bounded variation, where the generator $L^*$ can be defined as
an unbounded linear operator and $C_0$-semigroup techniques can be
applied.
\subsection{The statement}
As just mentioned, we will solve (\ref{4}) in the Banach space
$\mathcal{M}$ of all signed measures on
$(\Gamma_0,\mathcal{B}(\Gamma_0))$ with bounded variation. Let
$\mathcal{M}^{+}$ stand for the cone of positive elements of
$\mathcal{M}$. By means of the Hahn-Jordan decomposition $\mu =
\mu^{+} - \mu^{-}$, $\mu^{\pm}\in \mathcal{M}^{+}$, the norm of
$\mu\in \mathcal{M}$ is set to be $\|\mu\|_{\mathcal{M}}= \mu^{+}
(\Gamma_0) + \mu^{-}(\Gamma_0)$. Then $\mathcal{P}(\Gamma_0)$ is a
subset of $ \mathcal{M}^{+}$. The linear functional
$\varphi_{\mathcal{M}}(\mu) := \mu(\Gamma_0)= \mu^{+}(\Gamma_0) -
\mu^{-}(\Gamma_0)$ has the property $\varphi_{\mathcal{M}}(\mu) =
\|\mu\|_{\mathcal{M}}$ for each $\mu\in\mathcal{M}^{+}$. That is,
$\|\cdot \|_{\mathcal{M}}$ is additive on the cone $\mathcal{M}^{+}$
and hence $\mathcal{M}$ is an $AL$-space, cf. \cite{TV}.
For a strictly increasing function $\chi: \mathds{N}_0 \to [0,
+\infty)$, we set
\begin{equation}
\label{18a}
\mathcal{M}_\chi =\left\{ \mu \in \mathcal{M}: \int_{\Gamma_0}\chi(
|\eta|) \mu^{\pm} (d \eta)<\infty\right\}, \qquad
\mathcal{M}^{+}_\chi = \mathcal{M}_\chi \cap \mathcal{M}^{+},
\end{equation}
and introduce
\begin{eqnarray}
\label{18b}
\varphi_{\mathcal{M}_\chi} (\mu) = \int_{\Gamma_0} \chi(|\eta|) \mu^{+}
(d \eta)- \int_{\Gamma_0} \chi( |\eta|) \mu^{-} (d \eta), \qquad \mu
\in \mathcal{M}_\chi.
\end{eqnarray}
Note that $\mathcal{M}_\chi$ is a proper subset of $\mathcal{M}$ and
the corresponding embedding is continuous. Set, cf. Assumption
\ref{ass1} and (\ref{19}),
\begin{equation}
\label{22}
\Psi(\eta) = M(\eta) + E^a (\eta) + \langle b \rangle |\eta|, \qquad
M(\eta):= \sum_{x\in \eta} m(x) \leq m^* |\eta|,
\end{equation}
and then
\begin{equation}
\label{22a}
\mathcal{D}= \left\{ \mu \in \mathcal{M}: \int_{\Gamma_0} \Psi(\eta)
\mu^{\pm}(d\eta)<\infty\right\}.
\end{equation}
By (\ref{19}) we have that $\Psi(\eta)\leq C|\eta|^2$ for an
appropriate $C>0$; hence, $\mathcal{M}_{\chi_2} \subset
\mathcal{D}$, where $\chi_m (n) = (1+n)^m$, $m\in \mathds{N}$. Then,
for $\mu \in \mathcal{D}$, we define
\begin{equation}
\label{22o}
(A\mu)(d\eta) = - \Psi (\eta) \mu(d\eta), \qquad (B\mu)(d\eta) =
\int_{\Gamma_0} \Xi (d \eta|\xi) \mu( d \xi),
\end{equation}
where the measure kernel $\Xi$ is
\begin{eqnarray}
\label{22b}
\Xi (\mathbb{A}|\xi) &=& \sum_{x\in \xi} \left( m(x) + E^a (x, \xi\setminus x) \right)\mathds{1}_{\mathbb{A}}(\xi \setminus x) \\[.2cm] \nonumber
& + & \sum_{x\in \xi}\int_{(\mathds{R}^d)^2} b(x|y_1, y_2)
\mathds{1}_{\mathbb{A}} (\xi \setminus x\cup\{y_1 , y_2\}) d y_1 d
y_2, \qquad \mathbb{A} \in \mathcal{B}(\Gamma_0),
\end{eqnarray}
and $\mathds{1}_{\mathbb{A}}$ is the indicator of $\mathbb{A}$. Then
we set $L^* = A+B$. By direct inspection one checks that $L^*$
satisfies $\mu(LF) = (L^* \mu)(F)$ holding for all $\mu \in
\mathcal{D}$ and appropriate $F:\Gamma_0 \to [0, +\infty)$, see
(\ref{2}).
Along with $\chi_m$ defined above we also consider $\chi^\kappa (n)
:= e^{\kappa n}$, $\kappa >0$, and the space
$\mathcal{M}_{\chi^\kappa}$. By a global solution of (\ref{4}) in
$\mathcal{M}$ with $\mu_0 \in \mathcal{D}$ we understand a
continuous map $[0,+\infty) \ni t \mapsto \mu_t \in
\mathcal{D}\subset \mathcal{M}$, which is continuously
differentiable in $\mathcal{M}$ on $(0,+\infty)$ and is such that
both equalities in (\ref{4}) hold.
\begin{theorem}
\label{1ftm}
The problem in (\ref{4}) with $\mu_0 \in \mathcal{D}$ has a unique
global solution $\mu_t\in\mathcal{M}$, which has the following
properties:
\begin{itemize}
\item[{\it(a)}] for each $m\in \mathds{N}$, $\mu_t \in
\mathcal{M}_{\chi_m}\cap\mathcal{P}(\Gamma_0)$ for all $t>0$
whenever $\mu_0 \in
\mathcal{M}_{\chi_m}\cap\mathcal{P}(\Gamma_0)$;
\item[{\it (b)}] for each $\kappa>0$ and $\kappa' \in (0,
\kappa)$, $\mu_t \in \mathcal{M}_{\chi^{\kappa'}}\cap\mathcal{P}(\Gamma_0)$ for all $t\in (0,T(\kappa, \kappa'))$
whenever $\mu_0 \in
\mathcal{M}_{\chi^{\kappa}}\cap\mathcal{P}(\Gamma_0)$, where
\begin{equation}
\label{22T}
T(\kappa, \kappa') = \frac{\kappa -
\kappa'}{\langle b \rangle}e^{-\kappa} ;
\end{equation}
\item[{\it (c)}] for all $t>0$, $\mu_t (d\eta) = R_t (\eta) \lambda (d\eta)$ whenever
$\mu_0 (d\eta) = R_0(\eta) \lambda (d\eta)$.
\end{itemize}
\end{theorem}
\subsection{The proof}
To prove Theorem \ref{1ftm}, as well as to elaborate tools for
studying the evolution of infinite systems, we use the Thieme-Voigt
perturbation technique \cite{TV}, the basic elements of which we
present here in the form adapted to the context.
To prove claim (c) along with the space $\mathcal{M}$ we will
consider its subspace consisting of measures absolutely continuous
with respect to the Lebesgue-Poisson measure defined in (\ref{8}).
This is $\mathcal{R}:=L^1 (\Gamma_0 , d \lambda)$ in which we have a
similar functional $\varphi_{\mathcal{R}}(R) = \int_{\Gamma_0}
R(\eta) \lambda (d\eta)$. Then we define $\mathcal{R}^{+}$ and
$\mathcal{R}^{+}_1$ consisting of positive elements and probability
densities, respectively. Note that $\varphi_{\mathcal{R}}(R) =
\|R\|_{\mathcal{R}}$ for $R\in \mathcal{R}^{+}$ and hence
$\mathcal{R}$ is also and $AL$-space. For $\chi: \mathds{N}_0 \to
[0,+\infty)$ as in (\ref{18a}), we set
\begin{eqnarray}
\label{18c}
& & \mathcal{R}_\chi = \left\{ R\in \mathcal{R}: \int_{\Gamma_0}
\chi(|\eta|) |R(\eta)|\lambda (d \eta ) < \infty\right\}, \\[.2cm]
\nonumber & & \varphi_{\mathcal{R}_\chi}(R) = \int_{\Gamma_0}
\chi(|\eta|) R(\eta)\lambda (d \eta ), \qquad R\in \mathcal{R}_\chi,
\\[.2cm]
\nonumber & & \mathcal{R}_\chi^{+} = \mathcal{R}_\chi \cap
\mathcal{R}^{+}, \qquad \mathcal{R}_{\chi,1}^{+} = \{ R\in
\mathcal{R}_\chi^{+}: \varphi_{\mathcal{R}}(R) = 1\}.
\end{eqnarray}
Now let $\mathcal{E}$ be either $\mathcal{M}$ or $\mathcal{R}$, and
$\|\cdot \|_{\mathcal{E}}$ stand for the corresponding norm. The
sets $\mathcal{E}^{+}$, $\mathcal{E}_1^{+}$, $\mathcal{E}_\chi$,
$\mathcal{E}_\chi^{+}$, $\mathcal{E}_{\chi,1}^{+}$, and the
functionals $\varphi_{\mathcal{E}}$, $\varphi_{\mathcal{E}_\chi}$
are defined analogously, i.e., they should coincide with the
corresponding objects introduced above if $\mathcal{E}$ is replaced
by $\mathcal{M}$ or $\mathcal{R}$ (by $\mathcal{M}_{1}^{+}$ we then
understand $\mathcal{P}(\Gamma_0)$). Let $\mathcal{D}\subset
\mathcal{E}$ be a linear subspace, $\mathcal{D}^{+} =\mathcal{D}\cap
\mathcal{E}^{+}$ and $(A,\mathcal{D})$, $(B,\mathcal{D})$ be
operators on $\mathcal{E}$. Set also $\mathcal{D}_\chi =\{ u\in
\mathcal{D}\cap \mathcal{E}_\chi: A u \in \mathcal{E}_\chi\}$ and
denote by $A_\chi$ the {\it trace} of $A$ in $\mathcal{E}_\chi$,
i.e., the restriction of $A$ to $\mathcal{D}_\chi$. Recall that a
$C_0$-semigroup of bounded linear operators $S=\{S(t)\}_{t\geq 0}$
in $\mathcal{E}$ is called \emph{positive} if
$S(t):\mathcal{E}^{+}\to \mathcal{E}^{+}$ for each $t\geq 0$. A
\emph{sub-stochastic} (resp. \emph{stochastic}) semigroup in
$\mathcal{E}$ is a positive $C_0$-semigroup such that
$\varphi_{\mathcal{E}} (S(t)u) \leq \varphi_{\mathcal{E}} (u)$
(resp. $\varphi_{\mathcal{E}} (S(t)u) = \varphi_{\mathcal{E}} (u)$)
whenever $u\in \mathcal{E}^{+}$.
\begin{proposition}\cite[Proposition 2.2]{TV}
\label{TV0pn}
Let $(A,\mathcal{D})$ be the generator of a positive $C_0$-semigroup
in $\mathcal{E}$, and $(B,\mathcal{D})$ be positive, i.e.,
$B:\mathcal{D}^{+}\to \mathcal{E}^{+}$. Suppose also that
\begin{equation}
\label{40}
\forall u\in\mathcal{D}^{+} \qquad \varphi_{\mathcal{E}}((A+B)u)
\leq 0.
\end{equation}
Then, for each $r\in (0,1)$, the operator $(A+rB, \mathcal{D})$ is
the generator of a sub-stochastic semigroup in $\mathcal{E}$.
\end{proposition}
\begin{proposition}\cite[Proposition 2.7]{TV}
\label{TVpn}
Assume that:
\begin{itemize}
\item[(i)] $-A:\mathcal{D}^{+} \to \mathcal{E}^{+}$ and $B:\mathcal{D}^{+}
\to \mathcal{E}^{+}$;
\item[(ii)] $(A,\mathcal{D})$ be the generator of a
sub-stochastic semigroup $S=\{S(t)\}_{t\geq 0}$ on $\mathcal{E}$
such that $S (t):\mathcal{E}_\chi \to \mathcal{E}_\chi$ for all
$t\geq 0$ and the restrictions $S (t)|_{\mathcal{E}_\chi}$
constitute a $C_0$-semigroup on $\mathcal{E}_{\chi}$ generated by
$(A_\chi, \mathcal{D}_\chi)$;
\item[(iii)] $B:\mathcal{D}_\chi \to \mathcal{E}_\chi$ and
$ \varphi_{\mathcal{E}} \left( (A+B) u\right) = 0$, for $u\in
\mathcal{D}^{+}$;
\item[(iv)] there exist $c>0$ and $\varepsilon >0$ such that
\[
\varphi_{\mathcal{E}_\chi} \left( (A+B) u\right) \leq c
\varphi_{\mathcal{E}_\chi} (u) - \varepsilon \|A u\|_{\mathcal{E}},
\qquad {\rm for} \ \ u\in \mathcal{D}_\chi \cap \mathcal{E}^{+}.
\]
\end{itemize}
Then the closure of $(A+B,\mathcal{D})$ in $\mathcal{E}$ is the
generator of a stochastic semigroup $S_{\mathcal{E}}=
\{S_{\mathcal{E}}(t)\}_{t\geq 0}$ on $\mathcal{E}$ which leaves
$\mathcal{E}_\chi$ invariant. The restrictions
$S_{\mathcal{E}_\chi}(t):=S_{\mathcal{E}}(t)|_{\mathcal{E}_\chi}$,
$t\geq 0$, constitute a $C_0$-semigroup $S_{\mathcal{E}_\chi}$ on
$\mathcal{E}_\chi$ generated by the trace of the generator of
$S_{\mathcal{E}}$ in $\mathcal{E}_\chi$.
\end{proposition}
{\it Proof of Theorem \ref{1ftm}.} Along with $L^*=A+B$ defined in
(\ref{22a}) and (\ref{22b}) we consider the operator in
$\mathcal{R}$ defined according to the rule $ (L^* \mu) (d\eta) =
(L^\dagger R_\mu) (\eta) \lambda ( d\eta)$. Then $L^\dagger =
A^\dagger + B^\dagger$ with
\begin{eqnarray}
\label{L}
(A^\dagger R)(\eta) & = & - \Psi (\eta) R(\eta), \\[.2cm] \nonumber
(B^\dagger R)(\eta) & = & \int_{\mathds{R}^d} \left(m(x) +
E^a(x,\eta) \right) R(\eta \cup x) d x \\[.2cm] \nonumber & + &
\int_{\mathds{R}^d} \sum_{y_1\in \eta}\sum_{y_2 \in \eta \setminus
y_1} b(x|y_1 , y_2) R(\eta \cup x \setminus \{y_1 , y_2\}) d x,
\end{eqnarray}
the domain of which is, cf. (\ref{22a}),
\begin{equation}
\label{L2}
\mathcal{D}^\dagger = \left\{ R \in \mathcal{R}: \int_{\Gamma_0}
\Psi(\eta) |R(\eta)| \lambda ( d \eta) < \infty \right\}.
\end{equation}
For $R\in \mathcal{D}^\dagger \cap \mathcal{R}^{+}$, by (\ref{18})
and (\ref{22}) we obtain from (\ref{L})
\begin{eqnarray}
\label{L4}
\varphi_{\mathcal{R}} ( B^\dagger R) & = & \int_{\Gamma_0}
\left(\sum_{x\in \eta} [m(x) + E^a(x,\eta \setminus x)] \right)
R(\eta)\lambda (d \eta) \\[.2cm] \nonumber & + & \int_{\Gamma_0}
\left(\sum_{x\in \eta} \int_{(\mathds{R}^d)^2} b(x|y_1 , y_2) d y_1
d y_2 \right)
R(\eta)\lambda (d \eta) \\[.2cm] \nonumber & = & \int_{\Gamma_0}
\Psi (\eta) R(\eta)\lambda (d \eta) = - \varphi_{\mathcal{R}}(
A^\dagger R).
\end{eqnarray}
By (\ref{L2}) and (\ref{L4}) we then get that: (a) $B^\dagger :
\mathcal{D}^\dagger \to \mathcal{R}$ and $B^\dagger :
\mathcal{R}^{+} \cap \mathcal{D}^\dagger \to \mathcal{R}^{+}$; (b)
$\varphi_{\mathcal{R}} ((A^\dagger + B^\dagger)R) =0$ for each $R\in
\mathcal{R}^{+} \cap \mathcal{D}^\dagger$. In the same way, we prove
that the operators defined in (\ref{22a}) and (\ref{22o}) satisfy:
(a) $B : \mathcal{D} \to \mathcal{M}$ and $B : \mathcal{D}^{+} \to
\mathcal{M}^{+}$; (b) $\varphi_{\mathcal{M}} ((A + B)\mu) =0$ for
each $\mu \in \mathcal{D}^{+}$. Thus, both pairs $(A, \mathcal{D})$,
$(B, \mathcal{D})$ and $(A^\dagger, \mathcal{D}^\dagger)$,
$(B^\dagger, \mathcal{D}^\dagger)$ satisfy item (i) of Proposition
\ref{TVpn}. We proceed further by setting
\begin{eqnarray}
\label{L3}
(S(t) \mu) (d \eta) & = & \exp\left(- t \Psi(\eta)\right) \mu(
d\eta), \quad \mu \in \mathcal{M}, \quad t>0, \\[.2cm] \nonumber (S^\dagger(t) R) (\eta)& = & \exp\left(- t \Psi(\eta)\right)
R( \eta), \quad R\in \mathcal{R}.
\end{eqnarray}
Obviously, $S=\{S(t)\}_{t\geq 0}$ and
$S^\dagger=\{S^\dagger(t)\}_{t\geq 0}$ are sub-stochastic semigroups
on $\mathcal{M}$ and $\mathcal{R}$, respectively. They are generated
respectively by $(A, \mathcal{D})$ and $(A^\dagger,
\mathcal{D}^\dagger)$. Clearly, the restrictions
$S(t)|_{\mathcal{M}_\chi}$ and $S^\dagger(t)|_{\mathcal{R}_\chi}$
constitute positive $C_0$-semigroups for $\chi_m$ and $\chi^\kappa$
as in Theorem \ref{1ftm}. Likewise, $B:\mathcal{D}_\chi\to
\mathcal{M}_\chi$ and $B^\dagger:\mathcal{D}^\dagger_\chi\to
\mathcal{R}_\chi$. Thus, the conditions in items (ii) and (iii) of
Proposition \ref{TVpn} are satisfied in both cases.
Now we turn to item (iv) of Proposition \ref{TVpn}. By (\ref{18b})
we have
\begin{eqnarray*}
\varphi_{\mathcal{M}_\chi} ((A+B)\mu) & = &
\varphi_{\mathcal{M}_\chi}
(L^* \mu) = \int_{\Gamma_0} (LF_\chi)(\eta) \mu(d\eta), \quad F_\chi(\eta):= \chi(|\eta|), \qquad \\[.2cm]
\nonumber \varphi_{\mathcal{R}_\chi} ((A^\dagger+B^\dagger)R) & = &
\varphi_{\mathcal{R}_\chi} (L^\dagger R) = \int_{\Gamma_0} (L
F_\chi)(\eta) R(\eta) \lambda (d\eta).
\end{eqnarray*}
Then the condition in item (iv) is satisfied if, for some positive
$c$ and $\varepsilon$ and all $\eta$, the following holds
\begin{equation}
\label{L5a}
(L F_\chi)(\eta) + \varepsilon \Psi (\eta) \leq c \chi(|\eta|).
\end{equation}
For $\chi_m (n) = (1+n)^m$, $m\in \mathds{N}$, by (\ref{2}) we have,
cf. (\ref{22}),
\begin{eqnarray}
\label{L6}
(L F_{\chi_m})(\eta) & = & - \left( M(\eta) + E^a (\eta)\right)
\epsilon_m (|\eta|) + \langle b \rangle |\eta| \epsilon_{m}
(|\eta|+1), \\[.2cm] \nonumber \epsilon_m (n) & := & (n+1)^m - n^m =
(n+1)^{m-1} + (n+1)^{m-2}n + \cdots + n^{m-1} \\[.2cm] \nonumber &
\leq & m(n+1)^{m-1}.
\end{eqnarray}
For $\chi^\kappa(n) = e^{\kappa n}$, we have
\begin{eqnarray*}
(L F_{\chi^\kappa})(\eta) = - \left( M(\eta) + E^a (\eta)\right)
e^{\kappa |\eta|} (1- e^{-1}) + \langle b \rangle |\eta| e^{\kappa
|\eta|} (e-1).
\end{eqnarray*}
By (\ref{L6}) the condition in (\ref{L5a}) takes the form
\begin{equation}
\label{L8}
- \left( M(\eta) + E^a (\eta)\right)\left(
\epsilon_m (|\eta|) -\varepsilon\right) + \langle b \rangle |\eta|
\left(\epsilon_{m} (|\eta|+1) + \varepsilon\right) \leq c
\left(|\eta|+1 \right)^m.
\end{equation}
since $\epsilon_m (|\eta|) \geq 1$. For $\varepsilon < 1$, the
validity of (\ref{L8}) will follow whenever $c$ satisfies
\[
c \geq m \langle b \rangle \left( 2^{m-1} + 1\right).
\]
Hence, for $\chi=\chi_m$, all the conditions of Proposition
\ref{TVpn} are met for both choices of $\mathcal{E}$ and the
corresponding operators. Therefore, we have two semigroups:
$S_{\mathcal{M}}$ and $S_{\mathcal{R}}$, with the properties
described in the mentioned statement. Then $\mu_t = S_\mathcal{M}(t)
\mu_0$ is the unique solution of the Fokker-Planck equation with
$\mu_0 \in \mathcal{D}$, which proves claim (a) of Theorem
\ref{1tm}. At the same time, $R_t = S_{\mathcal{R}}(t) R_0(\eta )$
is the unique solution of
\begin{equation}
\label{L9}
\dot{R}_t = L^\dagger R_t, \qquad R_t|_{t=0} = R_{\mu_0}\in
\mathcal{D}^\dagger.
\end{equation}
By (\ref{L2}) we have that $R_{\mu_0}\in \mathcal{D}^\dagger$ and
$\mu_0\in \mathcal{D}$ are equivalent. By direct inspection one
checks that $\mu_t(d \eta) = R_t (\eta) \lambda (d \eta)$ solves
(\ref{4}) if $R_t$ solves (\ref{L9}). Then the unique solution
$\mu_t = S_{\mathcal{M}}(t) \mu_0$ of (\ref{4}) has the mentioned
form, which proves claim (c).
To complete the proof we fix $\kappa >0$ and consider the trace of
$A$ in $\mathcal{M}_{\chi^\kappa}$, cf. (\ref{22o}), defined on the
domain
\begin{equation*}
\mathcal{D}_{\kappa}:=\left\{ \mu \in \mathcal{M}_{\chi^\kappa}:
\int_{\Gamma_0} \Psi (\eta) e^{\kappa |\eta|}\mu^{\pm }(d\eta ) <
\infty\right\}.
\end{equation*}
First, we split $B$ into the sum $B_1 + B_2$, where for $\mathbb{A}
\in
\mathcal{B}( \Gamma_0)$ we set, cf.
(\ref{22b}),
\begin{equation}
\label{L11}
(B_1 \mu)(\mathbb{A}) = \int_{\Gamma_0} \left( \sum_{x\in \eta}[m(x) + E^a(x, \eta \setminus
x)] \mathds{1}_{\mathbb{A}} (\eta\setminus x)\right) \mu(d\eta) ,
\end{equation}
and
\begin{equation}
\label{L12}
(B_2 \mu)(\mathbb{A}) = \int_{\Gamma_0} \left(\sum_{x\in \eta}
\int_{(\mathds{R}^d)^2} b(x|y_1 , y_2) \mathds{1}_{\mathbb{A}} (\eta
\setminus x \cup\{y_1 , y_2\}) d y_1 d y_ 2\right) \mu( d\eta).
\end{equation}
For $\mu\in \mathcal{D}_\kappa^{+} :=\mathcal{D}_\kappa \cap
\mathcal{M}^{+}$, from (\ref{L11}) we have
\begin{eqnarray}
\label{L13}
\varphi_{\mathcal{M}_{\chi^\kappa}} (B_1 \mu) & = & \int_{\Gamma_0}
e^{\kappa |\xi|} \int_{\Gamma_0} \sum_{x\in \eta}[m(x) +
E^a(x, \eta \setminus
x)]\delta_{\eta\setminus x}( d \xi) \mu(d\eta) \\[.2cm] \nonumber
& = & \int_{\Gamma_0} e^{\kappa (|\eta|-1)} \left(M(\eta) +
E^a(\eta) \right) \mu(d\eta) \\[.2cm] \nonumber
& \leq & - e^{-\kappa} \varphi_{\mathcal{M}_{\chi^\kappa}} ( A \mu)
.
\end{eqnarray}
For $r = e^{-\kappa}$, by (\ref{L13}) we have that
$\varphi_{\mathcal{M}_{\chi^\kappa}} (A+ r^{-1} B_1 \mu)\leq 0$ for
each $\mu \in \mathcal{D}_\kappa^{+}$. Then by Proposition
\ref{TV0pn} we obtain that $(A+ B_1, \mathcal{D}_\kappa)$ generates
a sub-stochastic semigroup $S_\kappa$ on
$\mathcal{M}_{\chi^\kappa}$. For $\kappa'\in (0,\kappa)$, let us
show now that $B_2$ acts as a bounded linear operator from
$\mathcal{M}_{\chi^\kappa}$ to $\mathcal{M}_{\chi^{\kappa'}}$. In
view of the Hahn-Jordan decomposition, it is enough to consider the
action of $B_2$ on positive elements of $\mathcal{M}_{\chi^\kappa}$.
Since $B_2$ is positive, cf. (\ref{L12}), for $\mu\in
\mathcal{M}^{+}_{\chi^\kappa}$, we have
\begin{eqnarray}
\label{L14}
\|B_2 \mu\|_{\mathcal{M}_{\chi^{\kappa'}}} & = & \int_{\Gamma_0}
e^{\kappa'|\xi|} \int_{\Gamma_0} \sum_{x\in
\eta}\int_{(\mathds{R}^d)^2} b(x|y_1, y_2) \delta_{\eta\setminus x
\cup\{y_1 , y_2\}} (d \xi) dy_1 dy_2 \mu ( d\eta) \qquad \\[.2cm]
\nonumber & = & e^{\kappa'} \int_{\Gamma_0} e^{\kappa'|\eta|}
\sum_{x\in \eta}\int_{(\mathds{R}^d)^2} b(x|y_1, y_2)
dy_1 dy_2 \mu (
d\eta) \\[.2cm]
\nonumber & = & e^{\kappa'} \langle b \rangle \int_{\Gamma_0} |\eta|
e^{- (\kappa-\kappa')|\eta|} e^{\kappa|\eta|} \mu ( d\eta) \\[.2cm]
\nonumber & \leq & \frac{e^{\kappa'} \langle b \rangle}{e (\kappa -
\kappa')} \|\mu\|_{\mathcal{M}_{\chi^{\kappa}}}.
\end{eqnarray}
Let $(B_2)_{\kappa'\kappa}: \mathcal{M}^{+}_{\chi^\kappa} \to
\mathcal{M}^{+}_{\chi^{\kappa'}}$ be the operator as just described.
For $n\in \mathds{N}$, we set
\begin{equation}
\label{L15}
\kappa_l = \kappa - (\kappa - \kappa')l /n, \qquad l=0, 1, \dots ,
n.
\end{equation}
By means of (\ref{L14}) and (\ref{L15}) we then estimate of the
operator norm
\begin{equation}
\label{L16}
\|(B_2)_{\kappa_{l+1}\kappa_l}\| \leq \frac{ e^{\kappa} n \langle b
\rangle}{e (\kappa - \kappa')}.
\end{equation}
Next, for $t>0$ and $0\leq t_n \leq \cdots \leq t_0 = t$, we
consider the following bounded linear operator acting from
$\mathcal{M}_{\chi^\kappa}$ to $\mathcal{M}_{\chi^{\kappa'}}$
\begin{equation*}
T_{\kappa' \kappa}^{(n)} (t,t_1, t_2 , \dots , t_n) =
S_{\kappa_n}(t-t_1) (B_2)_{\kappa_n \kappa_{n-1}}
S_{\kappa_{n-1}}(t_1-t_2) \cdots (B_2)_{\kappa_1 \kappa}
S_{\kappa}(t_n),
\end{equation*}
where $S_{\kappa_{l}}$ is the sub-stochastic semigroup in
$\mathcal{M}_{\chi^{\kappa_l}}$ generated by $(A+B_1,
\mathcal{D}_{\kappa_l})$. By the latter fact we have that
$T_{\kappa' \kappa}^{(n)} (t,t_1, t_2 , \dots , t_n):
\mathcal{M}_{\chi^\kappa}\to \mathcal{D}_{\kappa'}$ and
\begin{eqnarray}
\label{L17a}
\frac{d}{dt} T_{\kappa' \kappa}^{(n)} (t,t_1, t_2 , \dots , t_n) & =
& (A+ B_1)T_{\kappa' \kappa}^{(n)} (t,t_1, t_2 , \dots ,
t_n),\\[.2cm] \nonumber T_{\kappa' \kappa}^{(n)} (t,t, t_2 , \dots ,
t_n) & = & (B_2)_{\kappa' \kappa_{n-1}} T_{\kappa_{n-1}
\kappa}^{(n-1)} (t, t_2 , \dots , t_n).
\end{eqnarray}
As $(B_2)_{\kappa' \kappa_{n-1}}$ is the restriction of $(B_2,
\mathcal{D}_{\kappa'})$ to $\mathcal{M}_{\chi^{\kappa_{n-1}}}
\subset \mathcal{D}_{\kappa'}$ and $T_{\kappa' \kappa}^{(n-1)}
(t,t_2, t_2 , \dots , t_n): \mathcal{M}_{\chi^\kappa}\to
\mathcal{D}_{\kappa'}$, the second line in (\ref{L17a}) can be
rewritten as
\begin{equation}
\label{L17b}
T_{\kappa' \kappa}^{(n)} (t,t, t_2 , \dots , t_n) = B_2 T_{\kappa'
\kappa}^{(n-1)} (t, t_2 , \dots , t_n).
\end{equation}
On the other hand, since all the semigroups $S_{\kappa_{l}}$ are
sub-stochastic and $(B_2)_{\kappa' \kappa}$ are positive, by
(\ref{L16}) we get the following estimate of its operator norm
\begin{equation}
\label{L18}
\|T^{(n)}_{\kappa' \kappa} (t,t_1, t_2 , \dots , t_n)\| \leq \left(
\frac{ e^{\kappa} n \langle b \rangle}{e (\kappa -
\kappa')}\right)^n.
\end{equation}
We also set $T^{(0)}_{\kappa' \kappa}(t)= S_{\kappa'}
(t)|_{\mathcal{M}_{\chi^\kappa}}$, and then consider
\begin{equation}
\label{L19}
Q_{\kappa'\kappa}(t) := \sum_{n=0}^\infty \int_0^t\int_0^{t_1}
\cdots \int_0^{t_{n-1}} T^{(n)}_{\kappa'\kappa} (t,t_1, t_2 , \dots
, t_n) d t_n d t_{n-1} \cdots d t_1.
\end{equation}
By (\ref{L18}) we conclude that the series in (\ref{L19}) converges
uniformly on compact subsets of $[0, T(\kappa, \kappa'))$, see
(\ref{22T}), to a continuously differentiable function
\[
(0,T(\kappa, \kappa')) \ni t \mapsto Q_{\kappa'\kappa}(t) \in
\mathcal{L}(\mathcal{M}_{\chi^\kappa},
\mathcal{M}_{\chi^{\kappa'}}),
\]
where the latter is the Banach space of all bounded linear operators
acting from $\mathcal{M}_{\chi^\kappa}$ to
$\mathcal{M}_{\chi^{\kappa'}}$. By (\ref{L17a}) and (\ref{L17b}) we
obtain
\begin{equation}
\label{L20}
\frac{d}{dt} Q_{\kappa'\kappa}(t) = (A + B_1 + B_2)
Q_{\kappa'\kappa}(t) = L^* Q_{\kappa'\kappa}(t).
\end{equation}
Thus, assuming that $\mu_0 \in \mathcal{M}_{\chi^\kappa}$ we get
that $\tilde{\mu}_t := Q_{\kappa'\kappa}(t) \mu_0$, for $t \in
[0,T(\kappa, \kappa'))$, lies in $\mathcal{M}_{\chi^{\kappa'}}$ and
solves (\ref{4}). Therefore, $\tilde{\mu}_t$ coincides with $\mu_t =
S_{\mathcal{M}}(t)\mu_0$, which completes the proof.
{\hfill$\square$}
\section{The Evolution of States of the Infinite System: Posing}
\label{Sec3}
In this section, we begin to construct the evolution of
states $\mu_0\to \mu_t$ assuming that the system in $\mu_0$ is
infinite and hence the method developed in Sect. 3 does not work
anymore. Instead, we will obtain $\mu_0\to \mu_t$ from the evolution
$B_0 \to B_t$, where $B_0(\theta)=\mu_0 (F^\theta)$ and $\mu_0\in
\mathcal{P}_{\rm exp} (\Gamma)$, see Definition \ref{0df}. In view
of (\ref{11}), the evolution $B_0 \to B_t$ can be constructed as the
evolution of correlation functions. The latter will be performed in
the following three steps: (a) constructing $k_0\to k_t$ for $t< T$
(for some $T<\infty$) (Sect. \ref{Sec4}); (b) proving that $k_t$ is
the correlation function of a unique $\mu_t \in \mathcal{P}_{\rm
exp}(\Gamma)$ (Sect. \ref{Sec5}); (c) continuing $k_t$ to all $t>0$
(Sect. \ref{Sec6}).
To make the first step, we derive from (\ref{1}) the corresponding
evolution equation with the operator $L^\Delta$ obtained from
(\ref{2}) by (\ref{17}), (\ref{18}) and the following rule
\begin{equation}
\label{20}
\mu(L F^\theta) = \int_{\Gamma_0} (L^\Delta k_\mu)(\eta) e(\theta;
\eta) \lambda (d\eta).
\end{equation}
Then we prove that the equation $\dot{k}_t = L^\Delta k_t$ has a
unique solution $k_t$, $t<T$, in a scale of Banach spaces such that
$k_t^{(n)}$ satisfies (\ref{6c}) with $\varkappa$ dependent on $t$.
The restriction $t<T$ arises from the proof as no direct semigroup
method can be applied here. The proof just mentioned does not
guarantee that the solution $k_t$ is a correlation function, and
even its usual positivity is not certain. Step (b) is made by
constructing suitable approximations $k_t^{\rm app}$ to the
mentioned solution $k_t$. By this construction $k_t^{\rm app}$
satisfies condition (a) of Proposition \ref{1pn}. Then we prove
that, for all $G\in B_{\rm bs}(\Gamma_0)$, $\langle \! \langle G,
k_t^{\rm app} \rangle \! \rangle$ converges to $\langle \! \langle
G, k_t \rangle \! \rangle$ as the approximations are eliminated.
This yields that also $k_t$ satisfies condition (a) of Proposition
\ref{1pn}. The remaining conditions (b) and (c) are checked
directly. Then $k_t = k_{\mu_t}$ for a unique $\mu_t\in
\mathcal{P}_{\rm exp}(\Gamma)$. This also implies the usual
positivity of $k_t$ which is then used to obtain the continuation to
all $t>0$.
\subsection{The operators} To make the first step mentioned above
we calculate $L^\Delta$ according to (\ref{20}) and obtain it in the
following form
\begin{eqnarray}
\label{21}
L^\Delta & = & A_1^\Delta + A_2^\Delta + B_1^\Delta + B_2^\Delta, \\[.2cm] \nonumber
(A_1^\Delta k)(\eta) & = & - \Psi(\eta) k(\eta),\\[.2cm] \nonumber
( A_2^\Delta k)(\eta) & = & \int_{\mathds{R}^d} \sum_{y_1\in
\eta}\sum_{y_2\in \eta\setminus y_1} k(\eta \cup x \setminus \{y_1,
y_2\}) b(x|y_1 , y_2) d x,\\[.2cm] \nonumber
(B_1^\Delta k)(\eta) & = & - \int_{\mathds{R}^d} k(\eta \cup x)
E^a(x, \eta) d x,\\[.2cm] \nonumber
(B_2^\Delta k)(\eta) & = & 2 \int_{(\mathds{R}^d)^2} \sum_{y_1 \in
\eta}k(\eta \cup x \setminus y_1) b(x|y_1 , y_2) d y_2 d x,
\end{eqnarray}
where $\Psi$ is as in (\ref{22}). Since the correlation functions of
measures from $\mathcal{P}_{\rm exp}(\Gamma)$ satisfy (\ref{6c}), we
introduce
\begin{equation}
\label{nk} \|k \|_{\alpha} = \esssup_{\eta \in \Gamma_0}e^{-\alpha
|\eta|} |k(\eta)|, \qquad \alpha \in \mathds{R},
\end{equation}
and the corresponding $L^\infty$-like Banach spaces
\begin{equation}
\label{23}
\mathcal{K}_\alpha = \{k:\Gamma_0 \to \mathds{R}: \|k\|_\alpha
<\infty\}.
\end{equation}
For $\alpha' < \alpha$, we have that $\| k\|_{\alpha'} \ge \|
k\|_{\alpha}$. Therefore, $\mathcal{K}_{\alpha'} \hookrightarrow
\mathcal{K}_{\alpha}$, where ``$\hookrightarrow$'' denotes
continuous embedding. Thus, $\{\mathcal{K}_\alpha\}_{\alpha \in
\mathds{R}}$ is an ascending scale of Banach spaces.
Our aim now is to define linear operators which act as in
(\ref{21}), cf. (\ref{22}). First, for a given $\alpha \in
\mathds{R}$, we define an unbounded operator $(L^\Delta_\alpha,
\mathcal{D}_\alpha^\Delta)$, where
\begin{equation}
\label{24}
\mathcal{D}_\alpha^\Delta = \{ k \in \mathcal{K}_\alpha: \Psi k \in \mathcal{K}_\alpha\}.
\end{equation}
Thus, $A_1^\Delta$ maps $\mathcal{D}_\alpha^\Delta$ to
$\mathcal{K}_\alpha$. Furthermore, for each $k\in
\mathcal{D}_\alpha^\Delta$, one finds $C>0$ such that
$(1+\Psi(\eta)) |k(\eta)| \leq e^{\alpha |\eta|} C$. We apply this
fact and item (iv) of Assumption \ref{ass1} to get
\begin{gather*}
\left\vert(A^\Delta_2 k)(\eta) \right\vert \leq \frac{Ce^{-\alpha +
\alpha|\eta|}}{1 + \Psi(\eta)} \sum_{y_1\in \eta} \sum_{y_2 \in
\eta\setminus y_1} \beta (y_1 - y_2) \leq C\beta^* e^{-\alpha +
\alpha|\eta|},
\end{gather*}
which means that $A_2^\Delta:\mathcal{D}_\alpha^\Delta \to
\mathcal{K}_\alpha$. In a similar way, we prove that
$B_i^\Delta:\mathcal{D}_\alpha^\Delta \to \mathcal{K}_\alpha$,
$i=1,2$. Thus, the expression in (\ref{21}) defines
$(L^\Delta_\alpha, \mathcal{D}_\alpha^\Delta)$. By the inequality
\begin{equation}
\label{25}
n^p e^{-\sigma n} \le \left( \frac{p}{e\sigma}\right)^p , \qquad
p\ge 1, \quad \sigma>0, \quad n\in \mathds{N},
\end{equation}
one readily proves that
\begin{equation}
\label{26}
\forall \alpha' < \alpha \qquad \mathcal{K}_{\alpha'} \subset \mathcal{D}^\Delta_\alpha.
\end{equation}
The next step is to introduce bounded operators $L_{\alpha
\alpha'}^{\Delta}: \mathcal{K}_{\alpha'} \to \mathcal{K}_\alpha$. To
this end, by means of (\ref{25}) and the inequality $|k(\eta) | \le
e^{\alpha |\eta|} \|k\|_{\alpha}$ (see (\ref{nk})), for $\alpha' <
\alpha$ we obtain from (\ref{21}) the following estimate
\begin{eqnarray}
\label{27}
\|A_1^\Delta k\|_\alpha & \leq & \esssup_{\eta \in \Gamma_0}
e^{-\alpha
|\eta|}\Psi (\eta) |k(\eta)| \\[.2cm] \nonumber & \leq & \bigg{(} ( m^* + \langle b \rangle +
a^*)\esssup_{\eta\in \Gamma_0}\left[|\eta|^2 e^{-(\alpha-
\alpha')|\eta|} \right] \bigg{)} \|k\|_{\alpha'} \\[.2cm] \nonumber & =
& \frac{4 ( m^* + \langle b \rangle + a^*) }{e^2(\alpha-\alpha')^2}
\|k\|_{\alpha'}.
\end{eqnarray}
In a similar way, one estimates $\|A_2^\Delta k\|_\alpha$ and
$\|B_i^\Delta k\|_\alpha$, $i=1,2$, which then yields, cf.
(\ref{21}),
\begin{equation}
\label{28}
\|L^\Delta k\|_\alpha \leq \left(4\frac{ m^* + \langle b \rangle +
a^* + \beta^* e^{-\alpha'} }{e^2(\alpha-\alpha')^2} + \frac{\langle
a \rangle e^{\alpha'} + 2 \langle b \rangle}{e(\alpha-\alpha')}
\right)\|k\|_{\alpha'}.
\end{equation}
Then we define a bounded operator $L^\Delta_{\alpha \alpha'}:
\mathcal{K}_{\alpha'} \to \mathcal{K}_\alpha$, the norm of which is
estimated by means of (\ref{28}). In view of (\ref{26}), we have
that each $k\in \mathcal{K}_{\alpha'}$ lies in
$\mathcal{D}^\Delta_\alpha$, and
\begin{equation}
\label{29}
L^\Delta_{\alpha \alpha'} k = L^\Delta_{\alpha }k.
\end{equation}
In the sequel, we consider two types of operators with the action as
in (\ref{21}): (a) unbounded operators $(L^\Delta_\alpha,
\mathcal{D}(L^\Delta_\alpha))$, $\alpha\in \mathds{R}$, with the
domains as in (\ref{24}); (b) bounded operators $L^\Delta_{ \alpha
\alpha'}$ just described. These operators are related to each other
by (\ref{29}), i.e., $L^\Delta_{\alpha\alpha'}$ can be considered as
the restriction of $L^\Delta_{\alpha }$ to $\mathcal{K}_{\alpha'}$.
\subsection{The statements}
For $\alpha \in \mathds{R}$, we set, cf. (\ref{15}), (\ref{16}) and
Proposition \ref{1pn},
\begin{equation}
\label{32}
\mathcal{K}^\star_\alpha = \{ k \in \mathcal{K}_\alpha:
k(\varnothing) =1 \ {\rm and} \ \langle \! \langle G, k \rangle \!
\rangle \geq 0 \ {\rm for} \ {\rm all} \ G\in B^\star_{\rm bs}
(\Gamma_0) \}.
\end{equation}
Note that
\begin{equation}
\label{32a}
\mathcal{K}^\star_\alpha \subset \mathcal{K}_{\alpha}^{+} :=\{ k\in
\mathcal{K}_\alpha: k(\eta ) \geq 0\}.
\end{equation}
Since the spaces defined in (\ref{23}) form an ascending scale, we
have that $k\in \mathcal{K}_{\alpha_0}$ lies in all
$\mathcal{K}_\alpha$ with $\alpha>\alpha_0$. Recall that the model
parameters satisfy Assumption \ref{ass1} which, in particular, imply
the validity of Proposition \ref{2pn}.
\begin{theorem}
\label{1tm}
There exists $c\in \mathds{R}$ dependent on the model parameters
only such that, for each $\mu_0\in \mathcal{P}_{\rm exp}(\Gamma_0)$,
there exists a unique map $[0,+\infty) \ni t \mapsto k_t \in
\mathcal{K}^{\star}_{\alpha_t}$ with $\alpha_t = \alpha_0 + ct$ and
$\alpha_0> - \log \omega$ such that $k_0=k_{\mu_0}\in
\mathcal{K}^\star_{\alpha_0}$, which has the following properties:
\begin{itemize}
\item[(i)]
For each $T>0$ and all $t\in [0,T)$, the map $$ [0,T)\ni t \mapsto
k_t \in \mathcal{K}_{\alpha_t} \subset
\mathcal{D}(L^\Delta_{\alpha_T}) \subset \mathcal{K}_{\alpha_T}$$
is continuous on $[0,T)$ and continuously differentiable on $(0,T)$
in $\mathcal{K}_{\alpha_T}$.
\item[(ii)] For all $t\in (0,T)$ it satisfies
\begin{equation*}
\dot{k}_t = L^\Delta_{\alpha_T} k_t.
\end{equation*}
\end{itemize}
\end{theorem}
\begin{corollary}
\label{Jaco}
Let $k_t\in \mathcal{K}^\star_{\alpha_t}$, $t\geq 0$, be as in
Theorem \ref{1tm}, and then $\mu_t\in\mathcal{P}_{\rm exp}(\Gamma)$
be the measure corresponding to this $k_t$ according to Proposition
\ref{1pn}. Then the map $t \mapsto \mu_t$ is such that
\begin{itemize}
\item[1.] for each compact $\Lambda$ and $t\geq 0$, $\mu_t^{\Lambda}$ lies in the
domain $\mathcal{D}\subset \mathcal{M}$ defined in (\ref{22a});
\item[2.] for
each $\theta \in \varTheta$, the map $[0,+\infty) \ni t \mapsto \mu_t (F^\theta)$ is continuous and
continuously differentiable on $(0,+\infty)$ and the following
holds, cf. (\ref{Jan}),
\begin{equation}
\label{Ja}
\frac{d}{dt} \mu_t (F^\theta) = (L^* \mu_t^{\Lambda_\theta}) (F^\theta) =
\langle\!\langle e(\theta, \cdot), L^\Delta_{ \alpha_T}
k_t\rangle\!\rangle,
\end{equation}
where the latter equality holds for all $T>t$, see (\ref{11}) and
(\ref{14}).
\end{itemize}
\end{corollary}
The proof of these statements is done in the remainder of the paper.
Its main steps are: (a) constructing the evolution $k_{\mu_0}\to
k_t$ for $t<T$ for some $T<\infty$; (b) proving that $k_t$ belongs
to $\mathcal{K}^\star_\alpha$ with an appropriate $\alpha$, that by
Proposition \ref{1pn} will allow us to associate $k_t$ with a unique
$\mu\in \mathcal{P}_{\rm exp}(\Gamma)$; (c) proving that $k_t$ lies
in $\mathcal{K}_{\alpha_t}$ on the mentioned time interval, which
will be used to continue $k_t$ to all $t>0$.
\section{The solution on a bounded time interval}
\label{Sec4}
Here we make step (a) of the program formulated at the end of Sect.
\ref{Sec3}.
\subsection{The statement}
Let us fix some $\alpha_1\in \mathds{R}$, take $\alpha_2 >\alpha_1$
and consider the following Cauchy problem in
$\mathcal{K}_{\alpha_2}$
\begin{equation}
\label{33}
\dot{k}_t = L^\Delta_{\alpha_2} k_t , \qquad k_t|_{t=0} = k_0 \in \mathcal{K}_{\alpha_1}.
\end{equation}
By its solution on a time interval $[0, T)$ we mean a continuous (in
$\mathcal{K}_{\alpha_2}$) map $[0, T)\ni t \mapsto k_t\in
\mathcal{D}^\Delta_{\alpha_2}$, which is continuously differentiable
on $(0, T)$ and satisfies both equalities in (\ref{33}). For
$\alpha, \alpha'\in \mathds{R}$ such that $\alpha'< \alpha$ and for
$\upsilon\geq 0$ as in Proposition \ref{2pn}, we set
\begin{equation}
\label{34}
T(\alpha, \alpha') = \frac{\alpha - \alpha'}{2 \langle b \rangle + \upsilon + \langle a \rangle e^{\alpha}}.
\end{equation}
\begin{lemma}
\label{1lm}
Let $\omega$ and $\upsilon$ be as in Proposition \ref{2pn}. Then for
each $\alpha_1 > - \log \omega$ and an arbitrary $k_0 \in
\mathcal{K}_{\alpha_1}$, the problem in (\ref{33}) has a unique
solution $k_t\in \mathcal{D}^\Delta_{\alpha_2}$ on the time interval
$[0, T(\alpha_2, \alpha_1))$.
\end{lemma}
In contrast to the case of finite configurations described in
Theorem \ref{1ftm}, the construction of a $C_0$-semigroup that
solves (\ref{33}) is rather hopeless. In view of this, the proof of
Lemma \ref{1lm} will be done in the following steps:
\begin{itemize}
\item[(i)] the operator $L^\Delta$ will be written in the form
$L^\Delta = A^\Delta_\upsilon + B^\Delta_\upsilon$, see
(\ref{45}), in such a way that
$A^\Delta_\upsilon:=A^\Delta_{1,\upsilon} + A^\Delta_{2}$ can be used to
construct a certain (sun-dual) $C_0$-semigroup in
$\mathcal{K}_{\alpha_2}$;
\item[(ii)] this semigroup and $B^\Delta_{\upsilon} := B^\Delta_1 +
B^\Delta_{2,\upsilon}$, see (\ref{46}), will be used to construct
the family of operators $\{Q_{\alpha \alpha'} (t): t\in [0,
T(\alpha, \alpha'))\}$, see (\ref{34}) and Lemma \ref{3lm}, such
that $Q_{\alpha \alpha'} (t)\in \mathcal{L}(\mathcal{K}_{\alpha'},
\mathcal{K}_\alpha)$ and $k_t = Q_{\alpha_2 \alpha_1}(t)k_0$ is the
solution in question. $\mathcal{L}(\mathcal{K}_{\alpha'},
\mathcal{K}_\alpha)$ stands for the Banach space of all bounded
operators acting from $\mathcal{K}_{\alpha'}$ to
$\mathcal{K}_{\alpha}$.
\end{itemize}
\subsection{The predual semigroup} Here we make the first step in constructing the
semigroup mentioned in item (i) above. For $\alpha \in \mathbb{R}$,
the space predual to $\mathcal{K}_\alpha$ is
\begin{equation}
\label{35}
\mathcal{G}_\alpha := L^1(\Gamma_0, e^{\alpha|\cdot|}d \lambda),
\end{equation}
which for $\alpha>0$ coincides with $\mathcal{R}_\chi$ defined in
(\ref{18c}) with $\chi(n) = e^{\alpha n}$. Here, however, we allow
$\alpha$ to be any real number. The norm in $\mathcal{G}_\alpha$ is
\begin{equation}
\label{36}
|G|_\alpha=\int_{\Gamma_0}|G(\eta)|e^{\alpha |\eta|}\lambda(d \eta).
\end{equation}
Clearly, $|G|_{\alpha'} \le |G|_{\alpha}$ whenever
$\alpha'<\alpha$. Then $\mathcal{G}_{\alpha} \hookrightarrow
\mathcal{G}_{\alpha'}$, and this embedding is also dense. In order
to use Proposition \ref{2pn} we modify the operators introduced in
(\ref{21}) by adding and subtracting the term $\upsilon |\eta|$.
This will lead also to the corresponding reconstruction of the
predual operators. For an appropriate $G:\Gamma_0 \to \mathds{R}$,
set, cf. (\ref{22}),
\begin{eqnarray}
\label{37}
(A_{1,\upsilon}G)(\eta) & = & - \Psi_\upsilon (\eta) G(\eta) = - \left(\upsilon |\eta| + E^a(\eta) + M(\eta) + \langle b \rangle |\eta| \right)
G(\eta), \\[.2cm] \nonumber
(A_2 G)(\eta)& = &\sum_{x \in \eta} \int_{(\mathbb{R})^2}G(\eta
\setminus x \cup y_1 \cup y_2)b(x|y_1,y_2)dy_1 dy_2,\\[.2cm]
\nonumber \mathcal{D}_\alpha & = & \{ G:\in \mathcal{G}_\alpha:
\Psi_\upsilon G \in \mathcal{G}_\alpha\}.
\end{eqnarray}
By Proposition \ref{2pn} we have that
\begin{equation}
\label{37J}
\Psi_\upsilon (\eta) \geq \omega E^b(\eta).
\end{equation}
The operator $(A_{1,\upsilon}, \mathcal{D}_\alpha)$ is the generator
of the semigroup $S_{0,\alpha} = \{S_{0,\alpha}\}_{t\geq 0}$ of
multiplication operators which act in $\mathcal{G}_\alpha$ as
follows, cf. (\ref{L3}),
\begin{equation}
\label{38}
( S_{0,\alpha}(t) G)(\eta) =
\exp\left(- t \Psi_\upsilon (\eta) \right) G(\eta).
\end{equation}
Let $\mathcal{G}_\alpha^{+}$ be the cone of positive elements of
$\mathcal{G}_\alpha $ The semigroup defined in (\ref{38}) is
obviously \emph{sub-stochastic}. Set $\mathcal{D}_\alpha^{+} =
\mathcal{D}_\alpha \cap \mathcal{G}_\alpha ^{+}$. By (\ref{18}),
(\ref{36}) and (\ref{37}) we get
\begin{eqnarray}
\label{39}
|A_2G|_\alpha & = & \int_{\Gamma_0} e^{\alpha|\eta|}|(A_2G)(\eta)|\lambda(d \eta) \\[.2cm]
& \leq & \int_{\Gamma_0 } e^{\alpha|\eta|} \int_{(\mathbb{R}^d)^2}
\sum_{x\in \eta} |G(\eta \setminus x \cup y_1
\cup y_2)| b(x|y_1,y_2) dy_1dy_2 \lambda(d \eta)\nonumber\\[.2cm]
& = & \int_{\Gamma_0} \int_{\mathbb{R}^d} \sum_{y_1 \in
\eta}\sum_{y_2 \in \eta \setminus y_1}e^{\alpha (|\eta|-1)}|G(\eta)|
b(x|y_1,y_2) dx \lambda(d \eta) \nonumber\\[.2cm] \nonumber
& = & e^{-\alpha} \int_{\Gamma_0} e^{\alpha|\eta|}E_b(\eta)
|G(\eta)|\lambda(d \eta) \leq (e^{-\alpha}/ \omega)
|A_{1,\upsilon}G|_{\alpha}.
\end{eqnarray}
The latter estimate follows by (\ref{37J}), see also (\ref{19}).
\begin{lemma}
\label{2lm}
Let $\upsilon$ and $\omega$ be as in Proposition \ref{2pn} and
$A_{1,\upsilon}$, $A_2$ and $\mathcal{D}_\alpha$ be as in
(\ref{37}). Then for each $\alpha> - \log \omega$, the operator
$(A_{\upsilon} , \mathcal{D}_\alpha):=(A_{1,\upsilon} + A_2,
\mathcal{D}_\alpha)$ is the generator of a sub-stochastic semigroup
$S_\alpha=\{S_\alpha(t)\}_{t\geq 0}$ on $\mathcal{G}_\alpha$.
\end{lemma}
\begin{proof}
We apply Proposition \ref{TV0pn} with $\mathcal{E}=
\mathcal{G}_\alpha$, $\mathcal{D}=\mathcal{D}_\alpha$ and $A=
A_{1,\upsilon}$. For some $r\in (0, 1)$, we set $B = r^{-1} A_2$,
which is clearly positive. By (\ref{39}) $B$ is defined on
$\mathcal{D}_\alpha$. To show that (\ref{40}) holds we take $G\in
\mathcal{D}_\alpha^{+}$ and proceed as in (\ref{39}). That is,
\begin{eqnarray*}
& & \int_{\Gamma_0} \left( (A_{1,\upsilon} +r^{-1}
A_2)G\right)(\eta) e^{\alpha|\eta|}\lambda ( d \eta) = -
\int_{\Gamma_0} \Psi_\upsilon (\eta) G(\eta) e^{\alpha|\eta|}
\lambda ( d\eta) \\[.2cm]\nonumber & &+ r^{-1}
\int_{\Gamma_0} \sum_{x\in \eta} \int_{(\mathds{R}^d)^2} G(\eta
\setminus x\cup \{y_1, y_2\})b(x|y_1 , y_2) e^{\alpha|\eta|} d y_1 d
y_2 \lambda (d\eta) \\[.2cm] \nonumber
& & \leq - \int_{\Gamma_0} \left(\upsilon |\eta| + E^a (\eta) -
r^{-1}e^{-\alpha}E^b (\eta) \right) G(\eta) e^{\alpha|\eta|} \lambda
(d \eta).
\end{eqnarray*}
Now, for $\alpha > - \log \omega$, we pick $r\in (0,1)$ in such a
way that $r^{-1}e^{-\alpha} \leq \omega$, which by Proposition
\ref{2pn} implies that (\ref{40}) holds for this choice. Then the
operator $A_{1,\upsilon} + r (r^{-1} A_2)$ satisfies Proposition
\ref{TV0pn} by which the proof follows.
\end{proof}
By the definition of the sub-stochasticity of $S_\alpha$ we have
that $|S_\alpha(t) G|_\alpha \leq |G|_\alpha$ whenever $G\in
\mathcal{G}_\alpha^{+}$. Let us show now that the same estimate
holds also for all $G\in \mathcal{G}_\alpha$. Each such $G$ in a
unique way can be decomposed $G=G^{+} - G^{-}$ with $G^{\pm} \in
\mathcal{G}_\alpha^{+}$. Moreover, by (\ref{36}) we have that
\[
|G|_\alpha = \int_{\Gamma_0} e^{\alpha |\eta|} \left( G^{+}(\eta) +
G^{-}(\eta)\right) \lambda (d\eta)= |G^{+}|_\alpha + |G^{-}|_\alpha.
\]
Then
\begin{eqnarray}
\label{39J}
|S_\alpha (t) G|_\alpha & = & |S_\alpha (t) (G^{+}-G^{-})|_\alpha
\leq
|S_\alpha (t) G^{+}|_\alpha + |S_\alpha (t) G^{-}|_\alpha \\[.2cm]
\nonumber & \leq & |G^{+}|_\alpha + | G^{-}|_\alpha = |G|_\alpha .
\end{eqnarray}
\subsection{The sun-dual semigroup}
Let $S_\alpha(t)$ be an element of the semigroup as in Lemma
\ref{2lm}. Then its adjoint $S^*_\alpha(t)$ is a bounded linear
operator in $\mathcal{K}_\alpha$. Clearly, $\{S^*_\alpha(t)\}_{t\geq
0}$ is a semigroup. However, it is not strongly continuous and hence
cannot be directly used to construct (classical) solutions of
differential equations. This obstacle is usually circumvented as
follows, see \cite{P}. Set, cf. (\ref{14}),
\begin{equation*}
\mathcal{D}_\alpha^* = \{ k\in \mathcal{K}_\alpha: \exists \hat{k}
\in\mathcal{K}_\alpha \ \forall G \in \mathcal{D}_\alpha \ \langle
\! \langle A_\upsilon G, k\rangle \! \rangle = \langle \! \langle G,
\hat{k}\rangle \! \rangle\}.
\end{equation*}
Then the operator $(A^*_{\upsilon},\mathcal{D}_\alpha^*)$ is adjoint
to $(A_{\upsilon},\mathcal{D}_\alpha)$. It acts as follows
\begin{eqnarray*}
(A^*_{\upsilon} k)(\eta) & = & - \Psi_\upsilon (\eta) k(\eta)
\\[.2cm] \nonumber & + & \int_{\mathds{R}^d} \sum_{y_1 \in \eta}\sum_{y_2 \in \eta\setminus y_1
} k(\eta\cup x\setminus \{y_1,y_2\}) b(x|y_1 , y_2) d x.
\end{eqnarray*}
By direct inspection one obtains that $\mathcal{K}_{\alpha'} \subset
\mathcal{D}_\alpha^*$ whenever $\alpha'< \alpha$. Let
$\mathcal{Q}_\alpha$ be the closure of $\mathcal{D}_\alpha^*$ in
$\mathcal{K}_\alpha$. Then we have
\begin{equation}
\label{43}
\mathcal{K}_{\alpha'}\subset \mathcal{D}_\alpha^* \subset
\mathcal{Q}_\alpha \subsetneq \mathcal{K}_\alpha, \qquad
\alpha'<\alpha.
\end{equation}
Now we set
\begin{equation*}
\mathcal{D}_\alpha^\odot= \{ k\in \mathcal{D}_\alpha^*: A_\upsilon^*
k \in \mathcal{Q}_\alpha\},
\end{equation*}
and denote by $A^\odot_\upsilon$ the restriction of $A_\upsilon^*$
to $\mathcal{D}_\alpha^\odot$. Then $(A^\odot_\upsilon,
\mathcal{D}_\alpha^\odot)$ is the generator of a $C_0$-semigroup,
which we denote by $S^\odot_\alpha=\{S^\odot_\alpha (t)\}_{t \geq
0}$. This is the semigroup which we have aimed to construct. It has
the following property, see \cite[Lemma 10.1]{P}.
\begin{proposition}
\label{Papn}
for each $k\in \mathcal{Q}_\alpha$ and $t\geq 0$, it follows that
$\|S^\odot_\alpha (t) k\|_\alpha = \|S^*_\alpha(t)k\|_\alpha \leq
\|k\|_\alpha$. Moreover, for each $\alpha'< \alpha$ and $k\in
\mathcal{K}_{\alpha'}$, the map $[0,+\infty)\ni t \mapsto
S^\odot_\alpha (t) k\in \mathcal{Q}_\alpha$ is continuous.
\end{proposition}
The estimate $\|S^*_\alpha(t)k\|_\alpha \leq \|k\|_\alpha$ is
obtained by means of (\ref{39J}). The continuity follows by
(\ref{43}) and the fact that $S^\odot_\alpha$ is a $C_0$-semigroup.
\subsection{The resolving operators: proof of Lemma \ref{1lm}} Now
we construct the family of operators $\{Q_{\alpha \alpha'}(t)\}$
such that the solution of (\ref{33}) is obtained in the form $k_t =
Q_{\alpha_2 \alpha_1}(t) k_0$. This construction, in which we
employ $S^\odot$, resembles the one used to get (\ref{L19}). We
begin by rearranging the operators in (\ref{21}) as follows
\begin{equation}
\label{45}
L^\Delta = A^\Delta + B^\Delta = A^\Delta_\upsilon +
B^\Delta_\upsilon,
\end{equation}
where $A^\Delta_\upsilon = A^\Delta_{1,\upsilon} + A^\Delta_2$, see
(\ref{37}), and
\begin{eqnarray}
\label{46}
B^\Delta_\upsilon &= & B_1^\Delta+
B^\Delta_{2,\upsilon},\\[.2cm]\nonumber
(B^\Delta_{2,\upsilon}k)(\eta) &=& (B^\Delta_{2}k)(\eta) +
\upsilon |\eta|k(\eta)\\[.2cm]\nonumber & = & 2
\int_{(\mathds{R}^d)^2} \sum_{y_1\in \eta} b(x|y_1, y_2) k(\eta \cup
x \setminus y_1) d x d y_2 + \upsilon |\eta|k(\eta),
\end{eqnarray}
whereas $B_1^\Delta$ is as in (\ref{21}). By means of (\ref{46}),
for $\alpha \in \mathds{R}$ and $\alpha' < \alpha$, we define
$(B^\Delta_\upsilon)_{\alpha\alpha'}\in
\mathcal{L}(\mathcal{K}_{\alpha'}, \mathcal{K}_{\alpha})$ the norm
of which can be estimated similarly as in (\ref{27}), (\ref{28}),
which yields
\begin{equation}
\label{47}
\| (B^\Delta_\upsilon)_{\alpha \alpha'}\| \leq \frac{2 \langle b \rangle + \upsilon + \langle a \rangle e^{\alpha'}}{e(\alpha
-\alpha')}.
\end{equation}
Now let $\mathbf{B}$ be either $B^\Delta_\upsilon$ or
$B^\Delta_{2,\upsilon}$, and $\mathbf{B}_{\alpha \alpha'}$ be the
corresponding bounded operator. Then, cf. (\ref{47}),
\begin{equation}
\label{48}
\|\mathbf{B}_{\alpha \alpha'}\| \leq
\frac{\varpi(\alpha;\mathbf{B})}{e(\alpha-\alpha')},
\end{equation}
where
\begin{equation}
\label{49}
\varpi(\alpha;B^\Delta_{\upsilon}) = 2 \langle b \rangle + \upsilon + \langle a \rangle
e^{\alpha}, \quad \varpi(\alpha;B^\Delta_{2,\upsilon}) = 2 \langle b \rangle +
\upsilon.
\end{equation}
For some $\alpha_1, \alpha_2$ such that $\alpha_1 < \alpha_2$, we
then set $\Sigma_{\alpha_2 \alpha_1}(t) =
S^{\odot}_{\alpha_2}(t)|_{\mathcal{K}_{\alpha_1}}$, $t>0$, where
$S^{\odot}_{\alpha}$ is the sub-stochastic semigroup as in
Proposition \ref{Papn}. Let also $\Sigma_{\alpha_2 \alpha_1}(0)$ be
the embedding operator $\mathcal{K}_{\alpha_1}\to
\mathcal{K}_{\alpha_2}$. Hence, see Proposition \ref{Papn}, the
operator norm satisfies
\begin{equation}
\label{50}
\| \Sigma_{\alpha_2 \alpha_1}(t)\|\leq 1, \qquad t\geq 0.
\end{equation}
We also have
\begin{eqnarray}
\label{51}
\Sigma_{\alpha_2 \alpha_1}(t) & = & \Sigma_{\alpha_2 \alpha_1}(0)
S^{\odot}_{\alpha_1}(t), \\[.2cm] \nonumber \Sigma_{\alpha_3 \alpha_1}(t+s) &
= & \Sigma_{\alpha_3 \alpha_2}(t)\Sigma_{\alpha_2 \alpha_1}(s),
\quad \ \alpha_3 > \alpha_2,
\end{eqnarray}
holding for all $t, s\geq 0$. Moreover,
\begin{equation*}
\frac{d}{dt} \Sigma_{\alpha_2 \alpha_1}(t) = A^\Delta_\upsilon
\Sigma_{\alpha_2 \alpha_1}(t),
\end{equation*}
which follows by Lemma \ref{2lm} and the construction of the
semigroup $S^{\odot}_\alpha$. Now we set
\begin{equation}
\label{53}
T(\alpha_2 , \alpha_1;\mathbf{B}) = \frac{\alpha_2 -
\alpha_1}{\varpi(\alpha_2;\mathbf{B})},
\end{equation}
see (\ref{48}), (\ref{49}), and also
\begin{equation}
\label{54}
\mathcal{A}(\mathbf{B})= \{ (\alpha_1 , \alpha_2, t): - \log \omega
< \alpha_1 < \alpha_2 , \ t\in[0,T(\alpha_2, \alpha_1;
\mathbf{B}))\}.
\end{equation}
Note that $T(\alpha_2 , \alpha_1;B^\Delta_\upsilon)$ coincides with
$T(\alpha_2 , \alpha_1)$ defined in (\ref{34}).
\begin{lemma}
\label{3lm}
For both choices of $\mathbf{B}$, there exist the corresponding
families $\lbrace Q_{\alpha_2 \alpha_1}(t;\mathbf{B}): (\alpha_1,
\alpha_2,t) \in \mathcal{A}(\mathbf{B}) \rbrace$, each element of
which has the following properties:
\begin{itemize}
\item[{\it(a)}] $Q_{\alpha_2 \alpha_1}(t;\mathbf{B}) \in \mathcal{L}(\mathcal{K}_{\alpha_1},
\mathcal{K}_{\alpha_2})$;
\item[{\it(b)}] the map $[0, T(\alpha_2, \alpha_1;\mathbf{B})) \ni t \mapsto
Q_{\alpha_2 \alpha_1}(t;\mathbf{B})\in
\mathcal{L}(\mathcal{K}_{\alpha_1}, \mathcal{K}_{\alpha_2})$ is
continuous;
\item[{\it(c)}] the operator norm of $Q_{\alpha_2 \alpha_1}(t;\mathbf{B}) \in
\mathcal{L}(\mathcal{K}_{\alpha_1}, \mathcal{K}_{\alpha_2})$
satisfies
$$\|Q_{\alpha_2 \alpha_1}(t;\mathbf{B}) \| \le \frac{T(\alpha_2, \alpha_1;\mathbf{B})}{T(\alpha_2, \alpha_1;\mathbf{B})-t},$$
\item[{\it(d)}] for each $\alpha_3 \in (\alpha_1, \alpha_2)$ and $t <T(\alpha_3,
\alpha_1;\mathbf{B})$, the following holds
\begin{equation}
\label{54b}
\frac{d}{dt} Q_{\alpha_2 \alpha_1}(t;\mathbf{B}) =
((A^{\Delta}_\upsilon)_{\alpha_2 \alpha_3} + \mathbf{B}_{\alpha_2
\alpha_3})Q_{\alpha_3 \alpha_1}(t;\mathbf{B}),
\end{equation}
which yields, in turn, that
\begin{eqnarray}
\label{54a}
\frac{d}{dt} Q_{\alpha_2 \alpha_1}(t;B^\Delta_\upsilon ) & = &
L^\Delta_{\alpha_2}Q_{\alpha_2 \alpha_1}(t;B^\Delta_\upsilon ) \\[.2cm]\frac{d}{dt} Q_{\alpha_2 \alpha_1}(t;B^\Delta_{2,\upsilon} ) & = &
((A^{\Delta}_\upsilon)_{\alpha_2} +
(B^\Delta_{2,\upsilon})_{\alpha_2} )Q_{\alpha_2
\alpha_1}(t;B^\Delta_{2,\upsilon} ), \nonumber
\end{eqnarray}
where $L^\Delta_{\alpha_2}$ is as in (\ref{33}), see also
(\ref{45}), and $(B^\Delta_{2,\upsilon})_{\alpha_2} $ denotes
$(B^\Delta_{2,\upsilon} , \mathcal{D}^\Delta_{\alpha_2})$, see
(\ref{24}).
\end{itemize}
\end{lemma}
\begin{proof}
Fix some $T< T(\alpha_2, \alpha_1;\mathbf{B})$ and then take $\alpha
\in (\alpha_1, \alpha_2]$ and positive $\delta < \alpha- \alpha_1$
such that
$$T< T_\delta:= \frac{\alpha - \alpha_1 - \delta}{\beta(\alpha_2;\mathbf{B})}.$$
Then take some $l\in \mathds{N}$ and divide $[\alpha_1, \alpha]$
into $2l+1$ subintervals in the following way: $\alpha_1=\alpha^0$,
$\alpha = \alpha^{2l+1}$ and
\begin{equation}
\label{55}
\alpha^{2s}=\alpha_1+\frac{s}{l+1}\delta + s \epsilon, \qquad \alpha^{2s+1}=\alpha_1+\frac{s+1}{l+1}\delta + s \epsilon,
\end{equation}
where $\epsilon = (\alpha - \alpha_1 - \delta)/l$ and $s=0, 1, ...,
l$. Now for $0\leq t_l \leq t_{l-1}\cdots \leq t_1 \leq t_0 := t$,
define
\begin{eqnarray}
\label{56}
\Pi^{(l)}_{\alpha \alpha_1}(t, t_1, t_2, ... , t_l;\mathbf{B}) & = & \Sigma_{\alpha \alpha^{2l}}(t-t_1)\mathbf{B}_{\alpha^{2l}\alpha^{2l-1}}
\cdots \Sigma_{\alpha^{2s+1}
\alpha^{2s}}(t_{l-s}-t_{l-s+1})\mathbf{B}_{\alpha^{2s}\alpha^{2s-1}}
\qquad
\nonumber \\[.2cm] & \times & \Sigma_{\alpha^3 \alpha^{2}}(t_{l-1}-t_l)\mathbf{B}_{\alpha^{2}\alpha^{1}}\Sigma_{\alpha^1 \alpha_1}(t_l).
\end{eqnarray}
By the very construction we have that $ \Pi^{(l)}_{\alpha
\alpha_1}(t, t_1, t_2, ... , t_l;\mathbf{B}) \in
\mathcal{L}(\mathcal{K}_\alpha , \mathcal{K}_{\alpha_1})$, and the
map
$$(t,t_1,...,t_l) \mapsto \Pi^{(l)}_{\alpha \alpha_1}(t, t_1, t_2, ... , t_l;\mathbf{B})$$
is continuous (Proposition \ref{Papn} and the fact that each
$\mathbf{B}_{\alpha^{2s}\alpha^{2s-1}}$ is bounded). Moreover, by
(\ref{50}) and (\ref{48}) we have
\begin{eqnarray}
\label{56a}
\| \Pi^{(l)}_{\alpha \alpha_1}(t, t_1, t_2, ... , t_l;\mathbf{B})\|
& \leq & \prod_{s=0}^l \|\mathbf{B}_{\alpha^{2s}\alpha^{2s-1}} \|
\leq \prod_{s=0}^l
\frac{\varpi(\alpha^{2s};\mathbf{B})}{e(\alpha^{2s}-\alpha^{2s-1})}
\\[.2cm] \nonumber & \leq & \left(\frac{l\upsilon(\alpha_2;\mathbf{B})}{e(\alpha - \alpha_1-\delta)}
\right)^l \leq \left(\frac{l}{eT_\delta}\right)^l.
\end{eqnarray}
By (\ref{51}) we also have that
\[
\Sigma_{\alpha^{2s+1} \alpha^{2s}}(t_{l-s}-t_{l-s+1})=
\Sigma_{\alpha^{2s+1}
\alpha^{2s}}(0)S^\odot_{\alpha^{2s}}(t_{l-s}-t_{l-s+1}). \] Taking
the derivative of both sides of the latter we obtain
\begin{eqnarray*}
\frac{d}{dt}\Sigma_{\alpha^{2s+1}
\alpha^{2s}}(t)=(A_\upsilon^{\Delta})_{\alpha^{2s+1} \alpha''}
\Sigma_{\alpha'' \alpha^{2s}}(t) =
(A_\upsilon^{\Delta})_{\alpha^{2s+1}} \Sigma_{\alpha^{2s+1}
\alpha^{2s}}(t),
\end{eqnarray*}
holing for each $\alpha''\in (\alpha^{2s}, \alpha^{2s+1})$. Here
$(A_\upsilon^{\Delta})_{\alpha}$ stands for the unbounded operator
defined in (\ref{37}). Then we obtain from (\ref{56}) the following
\begin{eqnarray}
\label{57}
\frac{d}{dt} \Pi^{(l)}_{\alpha \alpha_1}(t, t_1, t_2, ... ,
t_l;\mathbf{B}) & = & (A^{\Delta}_\upsilon)_{\alpha \alpha'}
\Pi^{(l)}_{\alpha' \alpha_1}(t, t_1, t_2, ... , t_l;\mathbf{B})
\\[.2cm]\nonumber & = & (A^{\Delta}_\upsilon)_{\alpha}
\Pi^{(l)}_{\alpha \alpha_1}(t, t_1, t_2, ... , t_l;\mathbf{B}).
\end{eqnarray}
Now we set
\begin{equation}
\label{58}
Q_{\alpha \alpha_1}(t;\mathbf{B}) = \Sigma_{\alpha \alpha_1}(t)+
\sum_{l=1}^{\infty}\int_{0}^{t}
\int_0^{t_1}...\int_0^{t_{l-1}}\Pi^{(l)}_{\alpha \alpha_1}(t, t_1,
t_2, ... , t_l;\mathbf{B})dt_l...dt_1.
\end{equation}
By (\ref{56a}) the series in (\ref{58}) converges uniformly of
compact subsets of $[0,T_\delta)$, which proves claims (a) and (b).
The estimate in (c) follows directly from (\ref{56a}). Finally,
(\ref{54a}) follows by (\ref{57}), cf. (\ref{L20}).\end{proof}
By solving (\ref{54b}) with the initial condition $Q_{\alpha_2
\alpha_1}(t+s;\mathbf{B})|_{t=0} = Q_{\alpha_2
\alpha_1}(s;\mathbf{B})$ we obtain the following `semigroup'
property of the family $\lbrace Q_{\alpha_2 \alpha_1}(t;\mathbf{B}):
(\alpha_1, \alpha_2,t) \in \mathcal{A}(\mathbf{B}) \rbrace$.
\begin{corollary}
\label{JKco}
For each $\alpha\in (\alpha_1, \alpha_2)$ and $t,s>0$ such that
\[
s< T(\alpha, \alpha_1;\mathbf{B}), \quad t < T(\alpha_2,
\alpha;\mathbf{B}), \quad t+s < T(\alpha_2, \alpha_1;\mathbf{B}),
\]
the following holds
\[
Q_{\alpha_2 \alpha_1}(t+s;\mathbf{B}) = Q_{\alpha_2
\alpha}(t;\mathbf{B})Q_{\alpha\alpha_1}(s;\mathbf{B}).
\]
\end{corollary}
\begin{remark}
\label{Jan10rk}
Since $B^\Delta_{2,\upsilon}$ is positive, by (\ref{56}) we obtain
that $Q_{\alpha_2
\alpha_1}(t;B^\Delta_{2,\upsilon}):\mathcal{K}_{\alpha_1}^+
\to\mathcal{K}_{\alpha_2}^+$. This positivity will be used to
continue $k_t$ to all $t>0$. It is the only reason for us to use
$Q_{\alpha_2 \alpha_1}(t;B^\Delta_{2,\upsilon})$ since
$B^\Delta_{\upsilon}$ is not positive, and hence the positivity of
$Q_{\alpha_2 \alpha_1}(t;B^\Delta_{\upsilon})$ cannot be secured.
\end{remark}
\noindent {\it Proof of Lemma \ref{1lm}.} Set
\begin{equation}
\label{Jan3}
Q_{\alpha_2 \alpha_1}(t) = Q_{\alpha_2
\alpha_1}(t;B^\Delta_\upsilon), \qquad t < T(\alpha_2 ,
\alpha_1;B^\Delta_\upsilon) = T(\alpha_2 , \alpha_1)
\end{equation}
Then the solution in question is obtained by setting $k_t =
Q_{\alpha_2 \alpha_1}(t) k_0$, which definitely satisfies (\ref{33})
by (\ref{54a}) and (\ref{51}). Its uniqueness can be proved as in
the proof of Lemma 4.8 in \cite{KK}. \hfill{$\square$}
Before proceeding further, we prove some corollary of Lemma \ref{3lm}
related to the predual evolution in $\mathcal{G}_\alpha$, see
(\ref{35}). Let $S_\alpha$ be the semigroup as in Lemma \ref{2lm}.
For $\alpha'
>\alpha$, let $S_{\alpha \alpha'}(t)$ be the restriction of
$S_\alpha(t)$ to $\mathcal{G}_{\alpha'} \hookrightarrow
\mathcal{G}_\alpha$. Along with the operators defined in (\ref{37})
we consider the predual operators to $B^\Delta_\upsilon$, see
(\ref{21}) and (\ref{46}). That is, they act
\begin{eqnarray*}
(B_1 G) (\eta)& =& - \sum_{x\in \eta} G(\eta\setminus x) E^a (x,
\eta\setminus x), \\[.2cm] \nonumber
(B_{2,\upsilon} G) (\eta)& =& 2 \int_{(\mathds{R}^d)^2} \sum_{x\in
\eta} G(\eta\setminus x \cup y_1) b(x|y_1, y_2) d y_1 d y_2 +
\upsilon |\eta| G(\eta).
\end{eqnarray*}
By means of these expressions we can define bounded operators acting
from $\mathcal{G}_{\alpha}$ to $\mathcal{G}_{\alpha'}$ for $\alpha'<
\alpha$. It turns out that the estimate of the norm is exactly as in (\ref{47}),
that is,
\begin{equation*}
\|(B_\upsilon)_{\alpha' \alpha} \|= \frac{2 \langle b \rangle +
\upsilon + \langle a \rangle e^{\alpha'}}{e(\alpha - \alpha')}.
\end{equation*}
Recall that $\mathcal{A}(B^\Delta_\upsilon)$ is defined in
(\ref{54}). For $(\alpha_2, \alpha_1, t)\in
\mathcal{A}(B^\Delta_\upsilon)$, let $T<T(\alpha_2, \alpha_1)$ be
fixed. Pick $\alpha\in [\alpha_1,\alpha_2)$ and $\delta<
\alpha_2-\alpha$ such that $T< T(\alpha_2, \alpha +\delta)$. Then,
for some $l\in \mathds{N}$, set, cf. (\ref{55}),
\begin{equation*}
\alpha_{2s} = \alpha_2 - \frac{s}{l+1}\delta - s \epsilon, \quad
\alpha^{2s+1} = \alpha_2 - \frac{s+1}{l+1}\delta - s \epsilon,
\end{equation*}
where $\epsilon = (\alpha_2 - \alpha - \delta)/ l$. For $0\leq
t_l\leq \cdots \leq t_1 \leq t_0:=t$ we then define, cf. (\ref{56}),
\begin{eqnarray*}
\Omega^{(l)}_{\alpha\alpha_2} (t, t_1 , \dots , t_n) & = & S_{\alpha
\alpha^{2l}} (t-t_1) (B_\upsilon)_{\alpha^{2l}\alpha^{2l-1}}
S_{\alpha^{2l-1}\alpha^{2l-2}} (t_1-t_2) \times \\[.2cm] & \times &
S_{\alpha^3\alpha^2}(t_{l-1}- t_l)
(B_\upsilon)_{\alpha^{2}\alpha^{1}} S_{\alpha^1\alpha_2}(t_{l}) .
\nonumber
\end{eqnarray*}
Set
\begin{equation}
\label{Du4}
H_{\alpha \alpha_2} (t) = S_{\alpha\alpha_2}(t) + \sum_{l=1}^\infty
\int_0^t \int_0^{t_1}\cdots \int_0^{t_{l-1}}
\Omega^{(l)}_{\alpha\alpha_2} (t, t_1 , \dots , t_n) d t_l d t_{l-1}
\cdots d t_1.
\end{equation}
Then exactly as in the case of Lemma \ref{3lm} we prove the
following statement.
\begin{proposition}
\label{Du1pn}
Each member of the family of operators $\{H_{\alpha\alpha_2}(t):
(\alpha_2 , \alpha, t)\in \mathcal{A}(B^\Delta_\upsilon)\}$ defined
in (\ref{Du4}) has the following properties:
\begin{itemize}
\item[(a)] $H_{\alpha\alpha_2}(t) \in
\mathcal{L}(\mathcal{G}_{\alpha_2}, \mathcal{G}_{\alpha})$, the
operator norm of which satisfies
\[
\|H_{\alpha\alpha_2}(t)\| \leq
\frac{T(\alpha_2,\alpha)}{T(\alpha_2,\alpha)-t};
\]
\item[(b)] For each $k\in \mathcal{K}_\alpha$ and $G\in
\mathcal{G}_{\alpha_2}$, it follows that
\begin{equation}
\label{Du5}
\langle \! \langle G, Q_{\alpha_2 \alpha}(t) k\rangle \!\rangle =
\langle \! \langle H_{\alpha \alpha_2}(t) G, k\rangle \!\rangle.
\end{equation}
\end{itemize}
\end{proposition}
\section{The Identification Lemma}
\label{Sec5}
Our aim now is to prove that the solution obtained in Lemma
\ref{1lm} has the property $k_t=k_{\mu_t}$ for a unique $\mu_t \in
\mathcal{P}_{\rm exp}(\Gamma)$. We call this \emph{identification}
since it allows us to identify the mentioned solutions as the
correlation functions of sub-Poissonian states.
Recall that $\upsilon$ and $\omega$ appear in Proposition \ref{2pn}
and $\mathcal{K}^\star_\alpha$ is defined in (\ref{32}).
\begin{lemma}[Identification]
\label{ILlm}
For each $\alpha_2>\alpha_1>-\log \omega$, it follows that $Q_{\alpha_2 \alpha_1}(t)=Q_{\alpha_2
\alpha_1}(t;B_\upsilon^\Delta): \mathcal{K}_{\alpha_1}^\star \to
\mathcal{K}_{\alpha_2}^\star$ for all $t \in [0, \tau (\alpha_2,
\alpha_1)]$ with $\tau(\alpha_2, \alpha_1) = T(\alpha_2, \alpha_1)/3$.
\end{lemma}
The proof consists in the following steps:
\begin{itemize}
\item[(i)] constructing an approximation $k_t^{\rm app}$ of $k_t = Q_{\alpha_2 \alpha_1}(t)k_{0}$, $k_0\in \mathcal{K}_{\alpha_1}^\star$, such that
$\langle \! \langle G, k_t^{\rm app}\rangle \! \rangle \geq 0$ for
all $G\in B^\star_{\rm bs}(\Gamma_0)$;
\item[(ii)] proving that $\langle \! \langle G, k_t^{\rm app}\rangle \!
\rangle \to \langle \! \langle G, k_t\rangle \! \rangle$ as the
approximation is eliminated.
\end{itemize}
\begin{figure}[h]
\centering
\includegraphics[scale=0.23]{obrazek_2}
\caption{The evolution in spaces}
\end{figure}
Fig. 1 provides an illustration to the idea of how to realize step
(i). The origin of the inequality in question is in (\ref{15}) and
(\ref{16}). To relate $k_t$ with a positive measure one uses local
approximations of $\mu_0$, the densities of which (not necessarily
normalized) evolve $R^{\rm app}_0 \to R^{\rm app}_t$ in $L^1$-like
spaces according to Theorem \ref{1tm}. These approximations are
tailored in such a way that the corresponding correlation functions
(\ref{13}) (that have the desired property by construction) also
evolve $q_0^{\rm app}\to q_t^{\rm app}$ in $L^1$-like spaces
$\mathcal{G}_\vartheta$. The technique developed in Sect. \ref{Sec4}
allows for proving that $\langle \! \langle G, k_t^{\rm app}\rangle
\! \rangle$ converges to $\langle \! \langle G, k_t\rangle \!
\rangle$ only if $k_t^{\rm app}= Q_{\alpha \alpha_0}(t) q_0^{\rm
app}$. That is, at this stage there is no connection between the
evolutions $q_0^{\rm app}\to q_t^{\rm app}$ and $q_0^{\rm app}\to
k_t^{\rm app}$ as they take place in (different) spaces,
$\mathcal{G}_\vartheta$ and $\mathcal{K}_\alpha$, respectively. It
turns out, that these spaces have an intersection
$\mathcal{U}_\alpha^\sigma$ constructed with the help of some
objects dependent on a para,eter, $\sigma>0$. To employ this fact
we use auxiliary models (indexed by $\sigma$), for which we prove
that both evolutions $q_0^{\rm app}\to q_t^{\rm app}$ and $q_0^{\rm
app}=k_0^{\rm app}\to k_t^{\rm app}$ take place in
$\mathcal{U}_\alpha^\sigma$ and thus coincide. That is $q_t^{\rm
app}=k_t^{\rm app}$ for $t\leq \tau$ with some positive $\tau$, that
yields the desired positivity of $k_t^{\rm app}$. Then step (ii)
includes also taking the limit $\sigma \to 0^+$.
\subsection{Auxiliary evolutions}
For $\sigma >0$ and $x\in \mathds{R}^d$, we set
\begin{eqnarray}
\label{U1} & & \phi_\sigma (x) = \exp\left(- \sigma |x|^2 \right),
\quad \langle \phi_\sigma \rangle = \int
_{\mathds{R}^d}\phi_\sigma (x)
d x.\\[.2cm] \nonumber
& & b_\sigma (x|y_1 , y_2) = b (x|y_1 , y_2) \phi_\sigma (y_1)
\phi_\sigma (y_2).
\end{eqnarray}
Consider
\begin{equation}
\label{U2}
L^{\Delta, \sigma}=A^{\Delta, \sigma}+B^{\Delta,
\sigma}=A_\upsilon^{\Delta, \sigma}+B_\upsilon^{\Delta, \sigma},
\end{equation}
that is obtained from the corresponding operators in (\ref{21}) and
(\ref{45}), (\ref{46}) by replacing $b$ with $b_\sigma$ given in
(\ref{U1}). Since this substitution does not affect
$\mathcal{D}^\Delta_\alpha$, see (\ref{24}), we will use the latter
as the domain of the corresponding unbounded operators. Then we
repeat the construction as in the proof of Lemma \ref{3lm} and
obtain the family $\{Q^\sigma_{\alpha_2\alpha_1}(t): (\alpha_1 ,
\alpha_2 , t)\in \mathcal{A}(B^\Delta_\upsilon)\}$ corresponding to
the choice $\mathbf{B}= B_\upsilon^{\Delta, \sigma}$. Along with the
evolution $t \mapsto Q^\sigma_{\alpha_2\alpha_1}(t) k_0$ we will
consider two more evolutions in $L^\infty$- and $L^1$-like spaces.
The latter one will be positive in the sense of Proposition
\ref{1pn} by the very construction. The auxiliary $L^\infty$-like
space where we are going to construct $t\mapsto k_t^{\rm app}$ lies
in the intersection of the just mentioned $L^1$-like space with the
spaces $\mathcal{K}_\alpha$, see Fig. 1, and hence is also positive
in the sense of Proposition \ref{1pn}. These arguments will allow us
to realize item (i) of the program.
\subsubsection{$L^\infty$-like evolution}
For $u: \Gamma_0 \to \mathbb{R}$, we define the norm
\begin{equation}
\label{U3}
\| u \|_{\sigma, \alpha} = \esssup_{\eta \in
\Gamma_0}\frac{|u(\eta)|\exp(-\alpha|\eta|)}{e(\phi_\sigma; \eta)},
\end{equation}
where $$e(\phi_\sigma; \eta)= \prod_{x\in \eta}\phi_\sigma
(x)=\exp\left( -\sigma\sum_{x\in \eta} |x|^2\right),$$ cf.
(\ref{11}). Then we consider the Banach space $\mathcal{U}_{\sigma,
\alpha} = \{u: \Gamma_0 \to \mathbb{R}: \| u \|_{\sigma, \alpha}
<\infty\}$. Clearly,
\begin{equation}
\label{Jan4}
\mathcal{U}_{\sigma, \alpha}\hookrightarrow \mathcal{K}_\alpha,
\qquad \alpha \in \mathds{R}.
\end{equation}
The space predual to $\mathcal{U}_{\sigma, \alpha}$ is the
$L^1$-space equipped with the norm, cf. (\ref{35}), (\ref{36}),
\begin{equation}
\label{U4}
|G|_{\sigma, \alpha}=\int_{\Gamma_0} |G(\eta)|\exp (\alpha |\eta|)e(\phi_\sigma;\eta)\lambda(d
\eta).
\end{equation}
In this space, we define $A^\sigma_{1,\upsilon}$ which acts exactly
as in (\ref{37}), and $A^\sigma_{2}$ which acts as in (\ref{37})
with $b$ replaced by $b_\sigma$. Their domain is the same
$\mathcal{D}_\alpha$. Then, like in (\ref{39}), by means of
(\ref{18}) and (\ref{U4}) we obtain
\begin{eqnarray*}
|A_2^\sigma G|_{\sigma,\alpha} & = & \int_{\Gamma_0} \left(\sum_{x
\in \eta} \int_{(\mathds{R}^d)^2}
|G(\eta \setminus x \cup \{y_1, y_2\})|b_\sigma(x|y_1,y_2)dy_1 dy_2 \right)\\[.2cm] & \times &\exp (\alpha |\eta|)e(\phi_\sigma;\eta)\lambda(d \eta)
\nonumber\\[.2cm] & = & e^\alpha \int_{\Gamma_0} \left( \int_{(\mathds{R}^d)^3}
|G(\eta \cup \{y_1, y_2\})| b_\sigma(x|y_1,y_2) \phi_\sigma (x) d x
d y_1 d y_2 \right)\nonumber\\[.2cm] & \times & \exp (\alpha |\eta|)e(\phi_\sigma;\eta) \lambda
(d\eta)
\nonumber\\[.2cm]
& \leq & e^\alpha \int_{\Gamma_0} \left( \int_{(\mathds{R}^d)^2}
|G(\eta \cup \{y_1, y_2\})| \beta(y_2-y_1)
e(\phi_\sigma;\eta\cup\{y_1,y_2\}) d y_1 d y_2 \right)
\nonumber\\[.2cm]
& \times & \exp (\alpha |\eta|) \lambda (d\eta) \nonumber\\[.2cm]
& = & e^{-\alpha}\int_{\Gamma_0} E^b(\eta) |G(\eta)| e^{\alpha
|\eta|} e(\phi_\sigma;\eta) \lambda(d
\eta)\nonumber\\[.2cm] \nonumber & \leq & (e^{-\alpha}/\omega) \int_{\Gamma_0} e^{\alpha |\eta|} \Psi_\upsilon (\eta)|G(\eta)|
e(\phi;\eta) \lambda(d
\eta)\nonumber\\[.2cm] \nonumber & = & (e^{-\alpha}/\omega)
|A_{1,\upsilon}^\sigma G|_{\sigma, \alpha}.
\end{eqnarray*}
This allows us to prove the following analog of Lemma \ref{2lm}.
\begin{proposition}
\label{U1pn}
Let $\upsilon$ and $\omega$ be as in Proposition \ref{2pn} and
$A_{1,\upsilon}^\sigma$, $A_2^\sigma$ and $\mathcal{D}_\alpha$ be as
just described. Then for each $\alpha> - \log \omega$, the operator
$(A_{\upsilon}^\sigma , \mathcal{D}_\alpha):=(A^\sigma_{1,\upsilon}
+ A^\sigma_2, \mathcal{D}_\alpha)$ is the generator of a
sub-stochastic semigroup
$S_{\sigma,\alpha}=\{S_{\sigma,\alpha}(t)\}_{t\geq 0}$ on
$\mathcal{G}_{\sigma,\alpha}$.
\end{proposition}
Let $S_{\sigma,\alpha}^{\odot}$ be the sun-dual semigroup, the
definition of which is pretty analogous to that of
$S_{\alpha}^{\odot}$, see Proposition \ref{Papn}. Then, for
$\alpha'< \alpha$, we define $\Sigma_{\alpha \alpha'}^\sigma (t)=
S^{\odot}_{\sigma, \alpha}(t)|_{\mathcal{U}_{\sigma.\alpha'}}$. As
in Proposition \ref{Papn} we then get that the map
\begin{equation*}
[0, +\infty) \ni t \mapsto \Sigma_{\alpha \alpha'}^\sigma (t) \in
\mathcal{L}(\mathcal{U}_{\sigma , \alpha'}, \mathcal{U}_{\sigma ,
\alpha})
\end{equation*}
is continuous and
\begin{equation*}
\|\Sigma_{\alpha, \alpha'}^\sigma (t)\| \leq 1, \qquad {\rm for} \
{\rm all} \ t\geq 0.
\end{equation*}
The operators $B^{\Delta,\sigma}_{\upsilon}= B^{\Delta,\sigma}_{1}+
B^{\Delta,\sigma}_{2,\upsilon}$ act as in (\ref{46}) with $b$
replaced by $b_\sigma$. Then we define the corresponding bounded
operators and obtain, cf. (\ref{47}),
\begin{equation*}
\|(B^{\Delta,\sigma}_{\upsilon})_{\alpha \alpha'} \| \leq \frac{2 \langle b \rangle + \upsilon + \langle a \rangle e^{\alpha'}}{e(\alpha
-\alpha')}.
\end{equation*}
Thereafter, we take $\delta>0$ as in Lemma \ref{3lm} and the
division as in (\ref{55}), and then define
\begin{eqnarray*}
\Pi_{\alpha \alpha'}^{l,\sigma}(t, t_1, t_2, ... , t_l) & = &
\Sigma_{\alpha \alpha^{2l}}^{\sigma}(t-t_1)(B_\upsilon^{\Delta,
\sigma})_{\alpha^{2l}\alpha^{2l-1}}\cdots \Sigma_{\alpha^{2s+1}
\alpha^{2s}}^{\sigma}(t_{l-s}-t_{l-s+1})\\[.2cm] & \times &(B_\upsilon^{\Delta,
\sigma})_{\alpha^{2s}\alpha^{2s-1}}\cdots \Sigma_{\alpha^3
\alpha^{2}}^{\sigma}(t_{l-1}-t_l)(B_\upsilon^{\Delta,
\sigma})_{\alpha^{2}\alpha^{1}}\Sigma_{\alpha^1
\alpha'}^{\sigma}(t_l),
\end{eqnarray*}
As in the proof of Lemma \ref{3lm} we obtain the family $\{
U^\sigma_{\alpha_2 \alpha_1}(t):(\alpha_1 , \alpha_2, t)\in
\mathcal{A}(B^{\Delta}_\upsilon)\}$, see (\ref{54}), with members
defined by
\[
U^\sigma_{\alpha_2 \alpha_1}(t) = \Sigma_{\alpha_2 \alpha_1}^\sigma
(t) + \sum_{l=1}^\infty \int_{0}^{t}
\int_0^{t_1}...\int_0^{t_{l-1}}\Pi^{l, \sigma}_{\alpha_2
\alpha_1}(t, t_1, t_2, ... , t_l)dt_l...dt_1,
\]
where the series converges for $t< T(\alpha_2, \alpha_1)$ defined in
(\ref{34}), cf. (\ref{53}) and (\ref{Jan3}). For this family, the
following holds, cf. (\ref{54a}),
\begin{equation}
\label{U9}
\frac{d}{dt} U^\sigma_{\alpha_2 \alpha_1}(t) = L^{\Delta,
\sigma}_{\alpha_2,u} U^\sigma_{\alpha_2 \alpha_1}(t),
\end{equation}
where the action of of $L^{\Delta,\sigma}_{\alpha_2,u}$ is as in
(\ref{U2}) and the domain is
\begin{equation}
\label{U9a}
\mathcal{D}^{\Delta,\sigma}_{\alpha_2 , u} = \{ u\in
\mathcal{U}_{\sigma, \alpha_2}: \Psi_\upsilon u \in
\mathcal{U}_{\sigma, \alpha_2}\}\subset
\mathcal{D}^\Delta_{\alpha_2},
\end{equation}
where the latter inclusion follows by (\ref{Jan4}) and (\ref{24}).
Then by (\ref{U9a}) we have that
\begin{equation}
\label{Jan10}
L^{\Delta,\sigma}_{\alpha, u} u = L^{\Delta,\sigma}_{\alpha} u ,
\qquad u\in \mathcal{D}^{\Delta,\sigma}_{\alpha ,
u} .
\end{equation}
Now by (\ref{U9}) we prove the following statement.
\begin{proposition}
\label{U2pn}
For each $\alpha_2>\alpha_1 > - \log \omega$, the problem
\begin{equation}
\label{U10}
\dot{u}_t = L^{\Delta,\sigma}_{\alpha_2,u} u_t, \qquad u_t|_{t=0}= u_0 \in \mathcal{U}_{\sigma,\alpha_1}
\end{equation}
has a unique solution $u_t \in \mathcal{U}_{\sigma,\alpha_2}$ on the
time interval $[0, T(\alpha_2, \alpha_1))$. This solution is given
by $u_t = U^\sigma_{\alpha_2 \alpha_1}(t) u_0$.
\end{proposition}
\begin{corollary}
\label{U1co}
Let $\alpha_2>\alpha_1 > - \log \omega$ be as in Proposition
\ref{U2pn} and $Q^\sigma_{\alpha_2\alpha_1}(t)$ be as described at
the beginning of this subsection. Then for each $t< T(\alpha_2 ,
\alpha_1)$ and $u_0 \in \mathcal{U}_{\sigma,\alpha_1}\subset
\mathcal{K}_{\alpha_1}$, it follows that
\begin{equation}
\label{U11}
U^\sigma_{\alpha_2 \alpha_1}(t) u_0 = Q^\sigma_{\alpha_2
\alpha_1}(t) u_0.
\end{equation}
\end{corollary}
\begin{proof}
By (\ref{Jan10}) we get that the solution of (\ref{U10}) is also the
unique solution of the following ``$\sigma$-analog" of (\ref{33})
$$\dot{u}_t = L^{\Delta,\sigma}_{\alpha_2} u_t, \quad u_t|_{t=0}=
u_0,
$$ and hence is given by the right-hand side of (\ref{U11}). Then the
equality in (\ref{U11}) follows by the uniqueness just mentioned.
\end{proof}
\subsubsection{$L^1$-like evolution}
Now we take $L^{\Delta,\sigma}$ as given in (\ref{U2}) and define
the corresponding operator $L^{\Delta,\sigma}_\vartheta$ in
$\mathcal{G}_\vartheta$, $\vartheta \in \mathds{R}$, introduced in
(\ref{35}), (\ref{36}), with domain $\mathcal{D}_\vartheta$ given in
(\ref{37}). By (\ref{U2}) and (\ref{21}) we have that $A^\Delta_1 :
\mathcal{D}_\vartheta \to \mathcal{G}_\vartheta$. Next, for $q\in
\mathcal{D}_\vartheta$, we have
\begin{eqnarray}
\label{Jan11}
|A^{\Delta, \sigma}_2 q |_\vartheta & \leq & \int_{\Gamma_0}
e^{\vartheta|\eta|}\left( \int_{\mathds{R}^d} \sum_{y_1\in
\eta}\sum_{y_2\in \eta\setminus y_1 } |q(\eta\cup x\setminus\{y_1,
y_2\}) | b_\sigma (x|y_1. y_2) d x \right) \lambda ( d \eta)
\qquad \quad \\[.2cm] \nonumber &\leq & \int_{\Gamma_0}
e^{\vartheta|\eta|+ 2 \vartheta} \int_{\mathds{R}^d} |q(\eta\cup x)|
\left( \int_{(\mathds{R}^d)^2} b(x|y_1 , y_2) d y_1 d y_2 \right) d
x \lambda ( d \eta) \\[.2cm] \nonumber &=& \langle b \rangle
e^\vartheta \int_{\Gamma_0} |\eta| e^{\vartheta |\eta|} |q(\eta)|
\lambda ( d \eta) \leq e^\vartheta \int_{\Gamma_0} \Psi(\eta)
e^{\vartheta |\eta|} |q(\eta)| \lambda ( d \eta),
\end{eqnarray}
see item (iii) of Assumption \ref{ass1} and (\ref{22}). Hence,
$A^{\Delta,\sigma}_2 : \mathcal{D}_\vartheta \to
\mathcal{G}_\vartheta$. Next, for the same $q$, we have
\begin{eqnarray}
\label{Jan12}
|B^{\Delta}_1 q |_\vartheta & \leq & \int_{\Gamma_0}
e^{\vartheta|\eta|}\left( \int_{\mathds{R}^d} |q(\eta\cup x)| E^a(x,
\eta) d x \right) \lambda ( d \eta) \\[.2cm] \nonumber & = &
e^{-\vartheta} \int_{\Gamma_0} e^{\vartheta |\eta|} E^a (\eta)
|q(\eta)| \lambda ( d \eta) \leq e^{-\vartheta} \int_{\Gamma_0}
\Psi(\eta) e^{\vartheta |\eta|} |q(\eta)| \lambda ( d \eta).
\end{eqnarray}
Hence, $B^\Delta_1 : \mathcal{D}_\vartheta \to
\mathcal{G}_\vartheta$. Finally,
\begin{eqnarray}
\label{Jan9}
|B^{\Delta,\sigma}_2 q |_\vartheta & \leq & 2 \int_{\Gamma_0}
e^{\vartheta|\eta|} \left( \int_{(\mathds{R}^d)^2} \sum_{y_1\in
\eta} |q(\eta \cup x\setminus y_1)| b_\sigma (x|y_1, y_2) d y_2 d x
\right) \lambda (d\eta) \qquad \\[.2cm] \nonumber & \leq & 2
\int_{\Gamma_0} e^{\vartheta|\eta|+\vartheta} \left(
\int_{(\mathds{R}^d)^3}|q(\eta \cup x)| b (x|y_1, y_2) d x d y_1 d
y_2\right) \lambda (d\eta) \qquad \\[.2cm] \nonumber & = & 2
\langle b \rangle \int_{\Gamma_0} e^{\vartheta|\eta|} |\eta|
|q(\eta)| \lambda ( d \eta) \leq \int_{\Gamma_0} \Psi(\eta)
e^{\vartheta |\eta|} |q(\eta)| \lambda ( d \eta).
\end{eqnarray}
Then by (\ref{Jan11}), (\ref{Jan12}) and (\ref{Jan9}) we conclude
that, for an arbitrary $\vartheta \in \mathds{R}$,
$L^{\Delta,\sigma}= A^{\Delta}_1 + A^{\Delta,\sigma}_2 +
B^{\Delta}_1 + B^{\Delta,\sigma}_2$ maps $\mathcal{D}_\vartheta$ to
$\mathcal{G}_\vartheta$ and hence can be used to define the
corresponding unbounded operator $(L^{\Delta,\sigma}_\vartheta,
\mathcal{D}_\vartheta)$. Let us then consider the corresponding
Cauchy problem
\begin{equation}
\label{Jan8}
\dot{q}_t = L^{\Delta,\sigma}_\vartheta q_t , \qquad q_t|_{t=0} =
q_0 \in \mathcal{D}_\vartheta.
\end{equation}
Recall that $\mathcal{G}_{\vartheta'} \subset \mathcal{D}_\vartheta$
for each $\vartheta'> \vartheta$.
\begin{lemma}
\label{Janln}
For a given $\vartheta >0$ and $\vartheta'> \vartheta$, assume that
the problem in (\ref{Jan8}) with $q_0 \in \mathcal{G}_{\vartheta'}$
has a solution $q_t\in\mathcal{G}_\vartheta$ on a time interval
$[0,\tau)$. Then this solution is unique.
\end{lemma}
\begin{proof}
Set
\[
w_t (\eta) = (-1)^{|\eta|}q_t (\eta).
\]
Then $|w_t|_\vartheta= |q_t|_\vartheta$ and $q_t$ solves
(\ref{Jan8}) if and only if $w_t$ solves the following equation
\begin{equation}
\label{Jan13}
\dot{w}_t = \left( A_1^{\Delta} - A_2^{\Delta,\sigma} -
B_1^{\Delta} + B_2^{\Delta,\sigma} \right) w_t.
\end{equation}
By Proposition \ref{TV0pn} we prove that $(A_1^{\Delta} -
B_1^{\Delta}, \mathcal{D}_\vartheta)$ generates a sub-stochastic
semigroup on $\mathcal{G}_\vartheta$. Indeed, $(A_1^{\Delta} ,
\mathcal{D}_\vartheta)$ generates a sub-stochastic semigroup defined
in (\ref{38}) with $\upsilon =0$, and $- B_1^{\Delta}$ is positive
and defined on $\mathcal{D}_\vartheta$, see (\ref{Jan12}). Also by
(\ref{Jan12}), for $w\in \mathcal{G}^{+}_\vartheta$ and $r\in
(0,1)$, we get
\begin{eqnarray*}
& & \int_{\Gamma_0} e^{\vartheta|\eta|}\left( \left(A_1^{\Delta} -
r^{-1}B_1^{\Delta}\right) w\right)(\eta) \lambda(d\eta) = -
\int_{\Gamma_0} e^{\vartheta|\eta|} \Psi(\eta) w(\eta) \lambda ( d
\eta) \\[.2cm] \nonumber & & + r^{-1} \int_{\Gamma_0}
e^{\vartheta|\eta|}\left( \int_{\mathds{R}^d} w(\eta\cup x) E^a(x,
\eta) d x \right) \lambda ( d \eta) \\[.2cm] \nonumber & & = -
\int_{\Gamma_0} e^{\vartheta|\eta|} \Psi(\eta) w(\eta) \lambda ( d
\eta) + r^{-1} e^{-\vartheta} \int_{\Gamma_0} e^{\vartheta |\eta|}
E^a (\eta) w(\eta) \lambda ( d \eta) \\[.2cm] \nonumber & & \leq - \left(1 - r^{-1}
e^{-\vartheta} \right) \int_{\Gamma_0} \Psi(\eta) e^{\vartheta
|\eta|} w(\eta) \lambda ( d \eta) \leq 0,
\end{eqnarray*}
where the latter inequality holds for $r\in (e^{-\vartheta}, 1)$.
Therefore, $(A_1^{\Delta} - B_1^{\Delta}, \mathcal{D}_\vartheta)=
(A_1^{\Delta} - r r^{-1} B_1^{\Delta}, \mathcal{D}_\vartheta)$
generates a sub-stochastic semigroup $V_\vartheta =\{V_\vartheta
(t)\}_{t\geq 0}$ on $\mathcal{G}_\vartheta$. For each $\vartheta''
\in (0, \vartheta)$, we have that $\mathcal{G}_\vartheta
\hookrightarrow \mathcal{G}_{\vartheta''}$. By the estimates in
(\ref{Jan11}) and (\ref{Jan9}), similarly as in (\ref{47}) we obtain
that
\begin{eqnarray*}
|A^{\Delta, \sigma}_2 w|_{\vartheta''} &\leq & \frac{\langle b
\rangle}{e(\vartheta - \vartheta'')} |w|_\vartheta, \\[.2cm]
|B^{\Delta, \sigma}_2 w|_{\vartheta''} & \leq & \frac{2\langle b
\rangle}{e(\vartheta - \vartheta'')} |w|_\vartheta,
\end{eqnarray*}
which we then use to define a bounded operator $C^{\Delta,
\sigma}_{\vartheta''\vartheta} :\mathcal{G}_\vartheta \to
\mathcal{G}_{\vartheta''}$. It acts as $- A^{\Delta,\sigma}_2 +
B^{\Delta,\sigma}_2$ and its norm satisfies
\begin{equation}
\label{Jan15}
\|C^{\Delta, \sigma}_{\vartheta''\vartheta}\| \leq \frac{3\langle b
\rangle}{e(\vartheta - \vartheta'')}.
\end{equation}
Assume now that (\ref{Jan13}) has two solutions corresponding to the
same initial condition $w_0$. Let $v_t$ be their difference. Then it
solves (\ref{Jan13}) with the zero initial condition and hence
satisfies
\begin{equation}
\label{Jan16}
v_t = \int_0^t V_{\vartheta''} (t - s) C^{\Delta,
\sigma}_{\vartheta'' \vartheta}
v _s d s
\end{equation}
where $v_t$ in the left-hand side is considered as an element of
$\mathcal{G}_{\vartheta''}$ and $t>0$ will be chosen later. Now for
a given $n\in \mathds{N}$, we set $\epsilon = (\vartheta -
\vartheta'')/n$ and $\vartheta^l = \vartheta - l \epsilon$, $l=0,
\dots , n$. Next, we iterate (\ref{Jan16}) due times and get
\begin{eqnarray*}
v_t & = & \int_0^t \int_0^{t_1} \cdots \int_{0}^{t_{n-1}}
V_{\vartheta''} (t-t_1) C^{\Delta,
\sigma}_{\vartheta'' \vartheta^{n-1}} V_{\vartheta^{n-1}} (t_1-t_2) C^{\Delta,
\sigma}_{\vartheta^{n-1} \vartheta^{n-2}} \times \cdots \times
\\[.2cm] & \times & V_{\vartheta^{1}} (t_{n-1}-t_n) C^{\Delta,
\sigma}_{\vartheta^{n-1} \vartheta} v_{t_n} d t_n \cdots d t_1.
\end{eqnarray*}
Then we take into account that $V_\vartheta$ is sub-stochastic,
$C^{\Delta,
\sigma}_{\vartheta^l \vartheta^{l-1}}$ are positive and satisfy
(\ref{Jan15}), and thus obtain from the latter that $v_t$ satisfies
\[
|v_t|_{\vartheta''} \leq \frac{1}{n!} \left(\frac{n}{e} \right)^n
\left(\frac{3 t \langle b \rangle}{\vartheta - \vartheta''}\right)^n
\sup_{s\in [0,t]}|v_s|_{\vartheta}.
\]
Then, since $n$ is an arbitrary positive integer, for all $t<
(\vartheta - \vartheta'')/ 3\langle b \rangle$ it follows that $v_t
=0$. To prove that $v_t=0$ for all $t$ of interest one has to repeat
the above procedure appropriate number of times.
\end{proof}
Let us now take $u\in \mathcal{U}_{\sigma, \alpha}$ with some
$\alpha \in \mathds{R}$, for which by (\ref{U3}) we have
\[
|u(\eta)|\leq \|u\|_{\sigma,\alpha} e^{\alpha |\eta|} e(\phi_\sigma,
\eta).
\]
Then the norm of this $u$ in $\mathcal{G}_\vartheta$ can be
estimated as follows, see (\ref{U1}),
\begin{equation}
\label{x}
|u|_\vartheta\leq \|u\|_{\sigma,\alpha} \int_{\Gamma_0} \exp\left(
(\alpha + \vartheta) |\eta|\right) e(\phi_\sigma, \eta)\lambda
(d\eta) = \|u\|_{\sigma,\alpha} \exp\left( (\alpha + \vartheta)
\langle \phi \rangle\right).
\end{equation}
This means that $\mathcal{U}_{\sigma, \alpha}\hookrightarrow
\mathcal{G}_{\vartheta}$ for each pair of real $\alpha$ and
$\vartheta$. Moreover, for the operators discussed above this
implies, cf. (\ref{Jan10}),
\begin{equation}
\label{U12}
L^{\Delta,\sigma}_{\alpha,u}u = L_{\vartheta}^{\Delta, \sigma}u, \qquad u\in\mathcal{D}^{\Delta,\sigma}_{\alpha,u}.
\end{equation}
\begin{corollary}
\label{Jan2co}
Let $\alpha_1$ and $\alpha_2$ be as in Proposition \ref{U2pn}. Then,
for each $q_0 \in \mathcal{U}_{\sigma,\alpha_1}$, the problem in
(\ref{Jan8}) has a unique solution $q_t \in
\mathcal{U}_{\sigma,\alpha_2}$, $t<T(\alpha_2, \alpha_1)$, which
coincides with the unique solution of (\ref{U10}).
\end{corollary}
\begin{proof}
By (\ref{U12}) we have that the unique solution of (\ref{U10}) $u_t$
solves also (\ref{Jan8}), and this is a unique solution in view of
Lemma \ref{Janln}.
\end{proof}
\subsection{Local approximations}
Our aim now is to prove that, cf. Proposition \ref{1pn}, the
following holds
\begin{equation}
\label{Jan18}
\langle \! \langle G, Q^\sigma_{\alpha_2 \alpha_1}(t) k_0 \rangle \!
\rangle \geq 0, \qquad G\in B_{\rm bs}^\star (\Gamma_0),
\end{equation}
for suitable $t>0$. By Corollaries \ref{U1co} and \ref{Jan2co} to
this end it is enough to prove (\ref{Jan18}) with
$Q^\sigma_{\alpha_2 \alpha_1}(t) k_0$ replaced by $q_t$.
For $\mu_0\in \mathcal{P}_{\rm exp}(\Gamma)$ and a compact
$\Lambda$, let $\mu^\Lambda_0\in \mathcal{P}(\Gamma_\Lambda)$ be the
corresponding projection to $\Gamma_\Lambda$ defined in (\ref{Rel}).
Let $R^\Lambda_0$ be its Radon-Nikodym derivative, see (\ref{12}).
For $N\in \mathds{N}$ and $\eta\in \Gamma_0$, we then set
\begin{equation}
\label{Jan19}
R^{\Lambda, N}_0 (\eta) = \left\{ \begin{array}{ll}
R^{\Lambda}_0(\eta) , \qquad &{\rm if} \ \ \eta\in \Gamma_\Lambda \
\ {\rm and} \ \ |\eta|\leq N;\\[.2cm]
0, \qquad &{\rm otherwise}. \end{array} \right.
\end{equation}
Until the end of this subsection, $\Lambda$ and $N$ are fixed.
Having in mind (\ref{13}) we introduce
\begin{equation}
\label{Jan20}
q_0^{\Lambda,N}( \eta) = \int_{\Gamma_0} R^{\Lambda,N}_0 (\eta\cup
\xi) \lambda ( d \xi), \qquad \eta \in \Gamma_{0}.
\end{equation}
For $G\in B^{\star}_{\rm bs}(\Gamma_0)$, by (\ref{15}), (\ref{18})
and (\ref{Jan20}) we have
\begin{equation}
\label{Jan21}
\langle\!\langle G, q_0^{\Lambda,N} \rangle\!\rangle =
\langle\!\langle K G, R_0^{\Lambda,N} \rangle\!\rangle \geq 0.
\end{equation}
By (\ref{Jan19}) it follows that $R^{\Lambda, N}_0 \in
\mathcal{R}^{+}$ and $\|R^{\Lambda, N}_0\|_\mathcal{R}\leq 1$.
Moreover, for each $\kappa>0$, we have, see (\ref{8}),
\begin{eqnarray}
\label{Jan22}
\|R^{\Lambda, N}_0\|_{\mathcal{R}_{\chi^\kappa}} =
\int_{\Gamma_\Lambda} e^{\kappa |\eta|} R^{\Lambda, N}_0 (\eta)
\lambda ( d\eta) \leq e^{\kappa N} \|R^{\Lambda,
N}_0\|_{\mathcal{R}} \leq e^{\kappa N}.
\end{eqnarray}
Let $S^\sigma_\mathcal{R}$ be the stochastic semigroup on
$\mathcal{R}$ constructed in the proof of Theorem \ref{1ftm} with
$b$ replaced by $b_\sigma$. Recall that $R_t = S_\mathcal{R}(t) R_0$
is the solution of (\ref{L9}). Set
\begin{eqnarray}
\label{Jan23}
R^{\Lambda, N}_t & = & S_\mathcal{R}^\sigma(t) R^{\Lambda, N}_0, \qquad t>0, \\[.2cm]
q^{\Lambda, N}_t (\eta) & = & \int_{\Gamma_0} R^{\Lambda, N}_t
(\eta\cup \xi) \lambda ( d \xi), \quad \eta \in \Gamma_0. \nonumber
\end{eqnarray}
\begin{proposition}
\label{JJ1pn}
For each $\vartheta\in \mathds{R}$ and $t \in [0, \tau_\vartheta)$,
$\tau_\vartheta:= [ e \langle b \rangle ( 1+ e^\vartheta)]^{-1}$, it
follows that $q^{\Lambda, N}_t \in \mathcal{G}_\vartheta^{+}$.
Moreover,
\begin{equation}
\label{Jan24}
\langle\!\langle G, q_t^{\Lambda,N} \rangle\!\rangle \geq 0
\end{equation}
holding for each $G\in B^{\star}_{\rm bs}(\Gamma_0)$ and all $t>0$.
\end{proposition}
\begin{proof}
Since $S_\mathcal{R}^\sigma$ is stochastic and $R_0^{\Lambda,N}$ is
as in (\ref{Jan19}), then $R_t^{\Lambda,N}\in \mathcal{R}^{+}$ for
all $t>0$. Hence, $q^{\Lambda, N}_t(\eta) \geq 0$ for all those
$t>0$ for which the integral in the second line in (\ref{Jan23})
makes sense. By (\ref{22T}) we have that $T(\kappa, \kappa')$, as a
function of $\kappa$, attains its maximum value $T_{\kappa'} =
e^{-\kappa'}/ e \langle b \rangle$ at $\kappa = \kappa'+1$. By
(\ref{Jan22}) we have that $R^{\Lambda, N}_0 \in
\mathcal{R}_{\chi^\kappa}$ for any $\kappa>0$. Then, for each
$\kappa>0$, by Proposition \ref{TVpn} it follows that $R^{\Lambda,
N}_t \in \mathcal{R}_{\chi^\kappa}$ for $t< T_\kappa$. Taking all
these fact into account we then get
\begin{eqnarray}
\label{Jan25}
|q_t^{\Lambda,N}|_\vartheta & = & \int_{\Gamma_0} e^{\vartheta|\eta|}
q_t^{\Lambda,N} (\eta) \lambda ( d\eta)\\ & = & \nonumber \int_{\Gamma_0}
\int_{\Gamma_0} e^{\vartheta|\eta|} R^{\Lambda, N}_t
(\eta\cup \xi) \lambda (d\eta)\lambda ( d \xi) \\ & = &
\int_{\Gamma_0} \left( 1 + e^\vartheta\right)^{|\eta|} R^{\Lambda,
N}_t (\eta) \lambda (d\eta) = \| R^{\Lambda,
N}_t\|_{\mathcal{R}_{\chi^\kappa}} \nonumber
\end{eqnarray}
with $\kappa = \log (1 +e^\vartheta)$. For these $\kappa$ and
$\vartheta$, we have that $T_\kappa = \tau_\vartheta$. Then
$q_t^{\Lambda,N}\in \mathcal{G}_\vartheta$ for $t< \tau_\vartheta$,
holding by (\ref{Jan25}). The existence of the integral in
(\ref{Jan24}) follows by the equality
\[
\langle \! \langle G, q^{\Lambda,N}_t \rangle \! \rangle = \langle
\! \langle K G, R^{\Lambda,N}_t \rangle \! \rangle,
\]
(\ref{10}) and the fact that $R^{\Lambda,N}_t \in
\mathcal{R}^{+}_{\chi_m}$ for all $t>0$ and $m\in \mathds{N}$, see
claims (a) and (c) of Theorem \ref{1ftm}. The validity of the
inequality in (\ref{Jan24}) is straightforward, cf. (\ref{Jan21}).
\end{proof}
\begin{corollary}
\label{JJ1co}
For each $\alpha \in \mathds{R}$, it follows that
$q_0^{\Lambda,N}\in \mathcal{U}_{\sigma,\alpha}^{+}$.
\end{corollary}
\begin{proof}
Set $I_N(\eta)=1$ whenever $|\eta|\leq N$ and $I_N(\eta)=0$
otherwise. By (\ref{Jan19}), (\ref{Jan20}) and (\ref{13}) we have
that
\begin{eqnarray*}
q_0^{\Lambda,N}(\eta) & = & I_N (\eta) \mathds{1}_{\Gamma_\Lambda}
(\eta) \int_{\Gamma_\Lambda}R_0^\Lambda(\eta
\cup\xi)\lambda (d \xi) \\[.2cm] & = & k_0(\eta) I_N (\eta)
\mathds{1}_{\Gamma_\Lambda} (\eta)\leq \varkappa^N I_N (\eta)
\mathds{1}_{\Gamma_\Lambda} (\eta).
\end{eqnarray*}
The latter estimate follows by the fact that $k_0=k_{\mu_0}$ for
some $\mu_0\in \mathcal{P}_{\rm exp}(\Gamma)$, and thus $k_0(\eta)
\leq
\varkappa^{ |\eta|}$ for some $\varkappa>0$, see Definition \ref{0df} and (\ref{6c}). Then
$q_0^{\Lambda,N}\in \mathcal{U}_{\sigma,\alpha}$ by (\ref{U3}). The
stated positivity i immediate.
\end{proof}
By (\ref{x}) and Corollary \ref{JJ1co} we obtain that
$q_0^{\Lambda,N}\in \mathcal{G}_{\vartheta}^{+}$ for each
$\vartheta\in \mathds{R}$. Now we relate $q_t^{\Lambda,N}$ with
solutions of (\ref{Jan8}).
\begin{lemma}
\label{JJ1lm}
For each $\vartheta \in \mathds{R}$, the map $[0, \tau_\vartheta)\ni
t\mapsto q_t^{\Lambda,N} \in \mathcal{G}_\vartheta$ is continuous
and continuously differentiable on $(0, \tau_\vartheta)$. Moreover,
$q_t^{\Lambda,N} \in \mathcal{D}_\vartheta$, see (\ref{37}), and
solves the problem in (\ref{Jan8}) on the time interval
$[0,\tau_\vartheta)$ with $q_0^{\Lambda,N}$ as the initial
condition.
\end{lemma}
\begin{proof}
Fix an arbitrary $\vartheta \in \mathds{R}$. The stated continuity
of $t\mapsto q_t^{\Lambda,N}$ follows by (\ref{Jan23}). Let us prove
that $q_t^{\Lambda,N}$ be differentiable in $\mathcal{G}_\vartheta$
on $(0,\tau_\vartheta)$ and the following holds
\begin{equation}
\label{Mar1}
\dot{q}^{\Lambda,N}_t (\eta) = \int_{\Gamma_0} \dot{R}^{\Lambda,N}_t (\eta\cup
\xi)\lambda(d\xi).
\end{equation}
For small enough $|\tau|$, we have
\begin{eqnarray}
\label{Mar2}
& & \frac{1}{\tau} \left(q^{\Lambda,N}_{t+\tau} (\eta)-
q^{\Lambda,N}_t (\eta)\right) - \int_{\Gamma_0}
\dot{R}^{\Lambda,N}_t (\eta\cup
\xi)\lambda(d\xi) \\[.2cm] & & \qquad = \int_{\Gamma_0} \left[\frac{1}{\tau} \left(R^{\Lambda,N}_{t+\tau} (\eta\cup\xi)- R^{\Lambda,N}_t
(\eta\cup\xi)\right) - \dot{R}^{\Lambda,N}_t (\eta\cup
\xi) \right]\lambda(d\xi). \nonumber
\end{eqnarray}
Then by (\ref{18}) we get
\begin{eqnarray*}
\left\vert {\rm LHS}(\ref{Mar2})\right\vert_\vartheta \leq
\int_{\Gamma_0} \left(1+e^\vartheta \right)^{|\eta|}
\left\vert\frac{1}{\tau} \left(R^{\Lambda,N}_{t+\tau} (\eta)-
R^{\Lambda,N}_t (\eta)\right) - \dot{R}^{\Lambda,N}_t (\eta)
\right\vert\lambda(d\eta),
\end{eqnarray*}
that proves (\ref{Mar1}), cf. (\ref{Jan25}). The continuity of
$t\mapsto \dot{q}^{\Lambda,N}_t$ follows by (\ref{Mar1}) and the
fact that $R^{\Lambda,N}_t = S^\sigma_{\mathcal{R}}
(t)R^{\Lambda,N}_0$, which also yields that
\begin{equation}
\label{Jan27}
\dot{q}_t^{\Lambda,N} (\eta) = \int_{\Gamma_0}
\left(L^{\dagger,\sigma}_\vartheta R_t^{\Lambda,N} \right) (\eta\cup
\xi) \lambda ( d\xi),
\end{equation}
where $L^{\dagger,\sigma}_\vartheta$ is the trace of
$L^{\dagger,\sigma}$ (the generator of $S^\sigma_{\mathcal{R}}$) in
$\mathcal{R}_{\chi^\kappa}$ with $\kappa = \log(1+e^\vartheta)$. By
(\ref{37}) it follows that $\Psi_\upsilon (\eta) \leq C_\varepsilon
e^{\varepsilon |\eta|}$ holding for an arbitrary $\varepsilon >0$
and the corresponding $C_\varepsilon>0$. For each $t<T_\kappa =
\tau_\vartheta$, one can pick $\kappa'>\kappa$ such that
$R_t^{\Lambda, N}\in \mathcal{R}_{\chi^{\kappa'}}$. For these $t$
and $\kappa'$, we thus pick $\varepsilon>0$ such that $1+
e^{\vartheta + \varepsilon} = e^{\kappa'}$, and then obtain, cf.
(\ref{Jan25}),
\begin{eqnarray}
\label{Mar4}
|\Psi_\upsilon q_t^{\Lambda,N}|_\vartheta \leq C_\varepsilon
\|R^{\Lambda,N}_t\|_{\mathcal{R}^{\chi^{\kappa'}}}.
\end{eqnarray}
Hence, $q_t^{\Lambda,N}\in \mathcal{D}_\vartheta$ for this $t$. Let
us now prove that $q_t^{\Lambda,N}$ solves (\ref{Jan8}). In view of
(\ref{Jan27}), (\ref{L}) and (\ref{Mar4}), to this end it is enough
to prove that
\begin{eqnarray}
\label{Mar5}
\left(L^\Delta q_t^{\Lambda,N}\right)(\eta) & = & - \int_{\Gamma_0}
\Psi(\eta\cup\xi) R^{\Lambda,N}_t (\eta \cup \xi) \lambda (d\xi)
\\[.2cm] \nonumber & + &\int_{\mathds{R}^d} \int_{\Gamma_0} \left( m(x)+ E^a(x, \eta \cup
\xi)\right)R^{\Lambda,N}_t (\eta \cup \xi\cup x) \lambda (d\xi) dx \\[.2cm] \nonumber & + &\int_{\mathds{R}^d}
\int_{\Gamma_0}\sum_{y_1\in \eta \cup\xi} \sum_{y_2\in \eta
\cup\xi\setminus y_1} b(x|y_1 , y_2) R^{\Lambda,N}_t (\eta \cup
\xi\cup x\setminus\{y_1,y_2\} ) \lambda (d\xi)d x,
\end{eqnarray}
holding point-wise in $\eta\in\Gamma_0$. By (\ref{22}) and
(\ref{19}) we get
\begin{equation}
\label{Mar6}
\Psi (\eta \cup \xi) = \Psi (\eta)+ \Psi (\xi) + 2 \sum_{x\in
\eta}\sum_{y\in \xi} a(x-y).
\end{equation}
Let $I_1 (\eta)$ denote the first summand in the right-hand side of
(\ref{Mar5}). By (\ref{18}) and (\ref{Mar6}) we then write it as
follows
\begin{eqnarray}
\label{Mar7}
I_1 (\eta) & = & - \Psi(\eta)q_t^{\Lambda,N} (\eta) - 2
\int_{\mathds{R}^d} E^a (x, \eta) q_t^{\Lambda,N} (\eta\cup x) dx
\\[.2cm]\nonumber & - & \int_{\Gamma_0}\Psi (\xi)R^{\Lambda,N}_t (\eta \cup \xi) \lambda
(d\xi).
\end{eqnarray}
To calculate the latter summand in (\ref{Mar7}) we again use
(\ref{22}) and (\ref{18}) to obtain the following:
\begin{eqnarray}
\label{Mar8}
\int_{\Gamma_0}\left( \sum_{x\in \xi} m(x) \right) R^{\Lambda,N}_t (\eta \cup
\xi) \lambda (d \xi)& = & \int_{\Gamma_0} \int_{\mathds{R}^d} m(x) R^{\Lambda,N}_t (\eta \cup
\xi\cup x) \lambda (d \xi) dx \qquad \quad \\[.2cm] \nonumber & = & \int_{\mathds{R}^d}
m(x) q_t^{\Lambda,N} (\eta\cup x) dx.
\end{eqnarray}
\begin{eqnarray}
\label{Mar9}
& & \int_{\Gamma_0}\left( \sum_{x\in \xi}\sum_{y\in \xi\setminus x} a(x-y) \right) R^{\Lambda,N}_t (\eta \cup
\xi) \lambda (d \xi)\\[.2cm] \nonumber & & \quad = \int_{\Gamma_0} \int_{(\mathds{R}^d)^2} a(x-y) R^{\Lambda,N}_t (\eta \cup
\xi\cup\{ x,y\}) \lambda (d \xi) dx d y \\[.2cm] \nonumber & & \quad = \int_{(\mathds{R}^d)^2}
a(x-y) q_t^{\Lambda,N} (\eta\cup \{x,y\}) dx d y .
\end{eqnarray}
\begin{eqnarray}
\label{Mar10}
\int_{\Gamma_0}\left(\langle b \rangle \sum_{x\in \xi} 1 \right) R^{\Lambda,N}_t (\eta \cup
\xi) \lambda (d \xi) = \langle b \rangle \int_{\mathds{R}^d} q^{\Lambda,N}_t (\eta
\cup x) dx.
\end{eqnarray}
In a similar way, we get the second $I_2$ (resp. the third $I_3$)
summands of the right-hand side of (\ref{Mar5}) as follows
\begin{eqnarray}
\label{Mar11}
I_2(\eta) & = & \int_{\mathds{R}^d} \left(m(x) + E^a (x, \eta)
\right)q^{\Lambda,N}_t (\eta
\cup x) dx \\[.2cm] \nonumber & + & \int_{(\mathds{R}^d)^2} a (x-y)
q^{\Lambda,N}_t (\eta
\cup\{ x,y\}) dx d y.
\end{eqnarray}
\begin{eqnarray}
\label{Mar12}
I_3(\eta) & = & \int_{\mathds{R}^d} \sum_{y_1\in \eta} \sum_{y_2\in
\eta\setminus y_1} b(x|y_1, y_2)q^{\Lambda,N}_t (\eta
\cup x\setminus \{y_1,y_2\}) dx \\[.2cm] & + & \nonumber 2
\int_{(\mathds{R}^d)^2} \sum_{y_1\in \eta} b(x|y_1,y_2)
q^{\Lambda,N}_t (\eta \cup x\setminus y_1) dx d y_2 \\[.2cm] \nonumber & + & \langle b \rangle \int_{\mathds{R}^d}
q^{\Lambda,N}_t (\eta
\cup x) dx .
\end{eqnarray}
Now we plug (\ref{Mar8}), (\ref{Mar9}) and (\ref{Mar10}) into
(\ref{Mar7}), and then use it together with (\ref{Mar11}) and
(\ref{Mar12}) in the right-hand side of (\ref{Mar5}) to get its
equality with the left-hand side, see (\ref{21}). This completes the
proof.
\end{proof}
\begin{corollary}
\label{JJ2co}
Let $\alpha_1> -\log \omega$ and $\alpha_2 >\alpha_1$ be chosen.
Then $k^{\Lambda,N}_t = Q^\sigma_{\alpha_2\alpha_1} (t)
q_0^{\Lambda,N}$ has the property
\begin{equation}
\label{Jan32}
\langle\! \langle G, k_t^{\Lambda,N} \rangle\! \rangle \geq 0,
\end{equation}
holding for all $G\in B_{\rm bs}^\star (\Gamma_0)$ and
$t<T(\alpha_2,\alpha_1)$.
\end{corollary}
\begin{proof}
The proof of (\ref{Jan32}) will be done by showing that
$k^{\Lambda,N}_t= q^{\Lambda,N}_t$, for $t< T(\alpha_2,\alpha_1)$
and then by employing (\ref{Jan24}), which holds for all $t>0$.
By Corollary \ref{JJ1co} it follows that $q_0^{\Lambda,N}\in
\mathcal{U}_{\sigma,\alpha_1}$, and hence $u_t =
U^\sigma_{\alpha_2\alpha_1}(t)q_0^{\Lambda,N}$ is a unique solution
of (\ref{U10}), see Proposition \ref{U2pn}. By Lemma \ref{JJ1lm}
$q_t^{\Lambda,N}$ solves (\ref{Jan8}) on $[0,\tau_\vartheta)$, which
by Corollary \ref{Jan2co} yields $u_t=q_t^{\Lambda,N}$ for $t<
\min\{\tau_\vartheta; T(\alpha_2,\alpha_1)\}$. If $\tau_\vartheta <
T(\alpha_2,\alpha_1)$, we can continue $q_t^{\Lambda,N}$ beyond
$\tau_\vartheta$ by means of the following arguments. Since
$u_t=q_t^{\Lambda,N}$ lies in $\mathcal{U}_{\sigma,\alpha_2}$ for
all $t< \min\{\tau_\vartheta; T(\alpha_2,\alpha_1)\}$, by (\ref{x})
we get that $q_t^{\Lambda,N}$ lies in the initial space
$\mathcal{G}_{\vartheta'}$ and hence can further by continued. Thus,
$u_t=q_t^{\Lambda,N}$ for all $t< T(\alpha_2,\alpha_1)$. Now by
(\ref{U11}) we get $q_t^{\Lambda,N}=u_t= k^{\Lambda,N}_t$, that
completes the proof.
\end{proof}
\subsection{Taking the limits}
We prove that (\ref{Jan32}) holds when the approximation is removed.
Recall that $k_t^{\Lambda,N}$ in (\ref{Jan32}) depends on
$\sigma>0$, $\Lambda$ and $N$. We first take the limits $\Lambda\to
\mathds{R}^d$ and $N\to +\infty$. Below, by an exhausting sequence
$\{\Lambda_n\}_{n\in \mathds{N}}$ we mean a sequence of compact
$\Lambda_n$ such that: (a) $\Lambda_n\subset \Lambda_{n+1}$ for all
$n$; (b) for each $x\in \mathds{R}^d$, there exits $n$ such that
$x\in \Lambda_n$.
\begin{proposition}
\label{JJ10pn}
Let $\alpha_1>-\log \omega$, $\alpha_2>\alpha_1$ and $k_0\in
\mathcal{K}^\star_{\alpha_1}$ be fixed. For these $\alpha_1$,
$\alpha_2$ and $t< T(\alpha_2 , \alpha_1)$, let $k_t^{\Lambda,N}$
and $Q^\sigma_{\alpha_2\alpha_1}(t)$ be the same as in Corollary
\ref{JJ2co} and (\ref{Jan18}), respectively. Then, for each $G\in
B_{\rm bs}(\Gamma_0)$ and any $t<T(\alpha_2,\alpha_1)$, the
following holds
\[
\lim_{n\to +\infty} \lim_{l\to +\infty} \langle \!\langle G,
k_t^{\Lambda_n, N_l}\rangle \!\rangle = \langle \!\langle G,
Q^\sigma_{\alpha_2\alpha_1} (t) k_0\rangle \!\rangle,
\]
for arbitrary exhausting $\{\Lambda_n\}_{n\in \mathds{N}}$ and
increasing $\{N_l\}_{l\in \mathds{N}}$ sequences of sets and
positive integers, respectively.
\end{proposition}
The proof of this statement can be performed by the literal
repetition of the proof of a similar statement given in Appendix of
\cite{Berns}.
Recall that, for $\alpha_2 > \alpha_1$, $T(\alpha_2,\alpha_1)$ was
defined in (\ref{34}). For these, $\alpha_2$, $\alpha_1$, we set
\begin{equation}
\label{Jan41}
\alpha= \frac{1}{3}\alpha_2 + \frac{2}{3}\alpha_1, \quad \ \alpha'=
\frac{2}{3}\alpha_2 + \frac{1}{4}\alpha_1.
\end{equation}
Clearly,
\begin{equation}
\label{Jan40}
\tau(\alpha_2, \alpha_1):= \frac{1}{3} T(\alpha_2, \alpha_1) <
\min\{ T(\alpha_2 , \alpha'); T(\alpha , \alpha_1)\}.
\end{equation}
\begin{lemma}
\label{JJ10lm}
Let $\alpha_1$, $\alpha_2$ and $k_0$ be as in Proposition
\ref{JJ10pn}, and let $k_t$ be the solution of (\ref{33}). Then for
each $G\in B_{\rm bs}(\Gamma_0)$ and $t\in
[0,\tau(\alpha_2,\alpha_1)]$, the following holds
\begin{equation}
\label{Jan42}
\lim_{\sigma \to 0^+} \langle \!\langle G,
Q^\sigma_{\alpha_2\alpha_1} (t) k_0\rangle \!\rangle = \langle
\!\langle G, k_t\rangle \!\rangle.
\end{equation}
\end{lemma}
\begin{proof}
We recall that the solution of (\ref{33}) is
$k_t=Q_{\alpha_2\alpha_1}(t)k_0$ with $Q_{\alpha_2\alpha_1}(t)$
given in (\ref{Jan3}) and $t\leq T(\alpha_2, \alpha_1)$, see Lemma
\ref{1lm}. For $\alpha$ and $\alpha'$ as in (\ref{Jan41}) and
$t\leq \tau(\alpha_2, \alpha_1)$, write
\begin{eqnarray}
\label{Jan43}
Q_{\alpha_2\alpha_1}(t) k_0 & = & Q^\sigma_{\alpha_2\alpha_1}(t) k_0
+
\Upsilon_{1}(t,\sigma) + \Upsilon_{2}(t,\sigma), \\[.2cm]
\nonumber \Upsilon_{1}(t,\sigma)& = & \int_0^t
Q_{\alpha_2\alpha'}(t-s) \left[(A^\Delta_2)_{\alpha'\alpha} -
(A^{\Delta.\sigma}_2)_{\alpha'\alpha}
\right]Q^\sigma_{\alpha\alpha_1}(s) k_0 d s,
, \\[.2cm]
\nonumber \Upsilon_{2}(t,\sigma)& = & \int_0^t
Q_{\alpha_2\alpha'}(t-s) \left[(B^\Delta_2)_{\alpha'\alpha} -
(B^{\Delta,\sigma}_2)_{\alpha'\alpha}
\right]Q^\sigma_{\alpha\alpha_1}(s) k_0 d s.
\end{eqnarray}
The validity of (\ref{Jan43}) is verified by taking the
$t$-derivatives from both sides and then by using e.g., (\ref{54b}).
Note that the norms of the operators $(A^\Delta_2)_{\alpha'\alpha}$.
$(B^\Delta_2)_{\alpha'\alpha}$,
$(A^{\Delta,\sigma}_2)_{\alpha'\alpha}$,
$(B^{\Delta.\sigma}_2)_{\alpha'\alpha}$ can be estimated as in
(\ref{48}). For $G$ as in (\ref{Jan42}), we then have
\begin{equation}
\label{Jan44}
\langle \!\langle G, Q_{\alpha_2\alpha_1}(t) k_0\rangle \!\rangle -
\langle \!\langle G, Q^\sigma_{\alpha_2\alpha_1}(t) k_0\rangle
\!\rangle = \langle \!\langle G, \Upsilon_1(t,\sigma) \rangle
\!\rangle + \langle \!\langle G, \Upsilon_2(t,\sigma) \rangle
\!\rangle.
\end{equation}
By (\ref{Du5}) and (\ref{Jan43}) it follows that
\begin{eqnarray*}
\langle \!\langle G, \Upsilon_1(t,\sigma) \rangle \!\rangle & = &
\int_0^t \langle \!\langle G, Q_{\alpha_2\alpha'}(t-s)
\left[(A^\Delta_2)_{\alpha'\alpha} -
(A^{\Delta.\sigma}_2)_{\alpha'\alpha}
\right]Q^\sigma_{\alpha\alpha_1}(s) k_0 \rangle \!\rangle d s
\qquad \\[.2cm] \nonumber &=& \int_0^t \langle \!\langle H_{\alpha'\alpha_2}(t-s)
G, v_s^\sigma \rangle \!\rangle d s = \int_0^t \langle \!\langle
G_{t-s}, v_s^\sigma \rangle \!\rangle d s,
\end{eqnarray*}
where
\begin{equation*}
v^\sigma_s =\left[(A^\Delta_2)_{\alpha'\alpha} -
(A^{\Delta.\sigma}_2)_{\alpha'\alpha} \right]k^\sigma_s :=
\left[(A^\Delta_2)_{\alpha'\alpha} -
(A^{\Delta.\sigma}_2)_{\alpha'\alpha}
\right]Q^\sigma_{\alpha\alpha_1}(s) k_0 \in \mathcal{K}_{\alpha'},
\end{equation*}
and
\begin{equation}
\label{Jan46a}
G_{t-s} = H_{\alpha'\alpha_2}(t-s) G \in \mathcal{G}_{\alpha'},
\end{equation}
which makes sense since obviously $G\in \mathcal{G}_{\alpha_2}$. In
view of (\ref{21}) we then get
\begin{eqnarray}
\label{Jan47}
& & \int_0^t \langle \!\langle G_{t-s}, v_s^\sigma \rangle
\!\rangle d s = \int_{\Gamma_0} G_{t-s} (\eta)
\bigg{(}\int_{\mathds{R}^d} \sum_{y_1\in\eta} \sum_{y_2\in
\eta\setminus y_1} k^\sigma_s (\eta
\cup x\setminus\{y_1,y_2\}) \\[.2cm] \nonumber & & \qquad \times \left[1-\phi_\sigma (y_1) \phi_\sigma (y_2)\right] b(x|y_1, y_2)dx
\bigg{)} \lambda ( d\eta) \\[.2cm] \nonumber & &\qquad = \int_{\Gamma_0}
\bigg{(} \int_{(\mathds{R}^d)^3}G_{t-s} (\eta\cup\{y_1,y_2\})
k^\sigma_s (\eta \cup x)\\[.2cm] \nonumber & & \qquad \times \left[1-\phi_\sigma (y_1) \phi_\sigma
(y_2)\right] b(x|y_1, y_2) d x dy_1 dy_2 \bigg{)} \lambda (d\eta).
\end{eqnarray}
Since $k^\sigma_s = Q^\sigma_{\alpha\alpha_1}(s) k_0$ is in
$\mathcal{K}_\alpha$, we have that
\begin{equation}
\label{Jan48}
|k^\sigma_s (\eta \cup x)| \leq \|k^\sigma_s\|_\alpha e^{\alpha
|\eta| +\alpha} \leq e^{\alpha |\eta| +\alpha} \frac{T(\alpha,
\alpha_1) \|k_0\|_{\alpha_1}}{ T(\alpha, \alpha_1) -
\tau(\alpha_2,\alpha_1)},
\end{equation}
where $\alpha $ is as in (\ref{Jan41}) and $s\leq t \leq
\tau(\alpha_2 , \alpha_1)$. Now for $s\leq t$, we set
\begin{equation}
\label{Jan49}
g_s (y_1, y_2) = \int_{\Gamma_0} e^{\alpha |\eta|} |G_{s}
(\eta\cup\{y_1,y_2\})| \lambda (d \eta).
\end{equation}
Let us show that $g_s \in L^1((\mathds{R}^d)^2)$. By (\ref{Jan46a})
we have
\begin{eqnarray}
\label{Jan50}
& & \int_{(\mathds{R}^d)^2} g_s (y_1, y_2) d y_1 d y_2 =
e^{-2\alpha} \int_{\Gamma_0} |\eta|(|\eta|-1)
e^{-(\alpha'-\alpha)|\eta|} |G_s(\eta)| e^{\alpha'|\eta|} \lambda
(d\eta)\qquad \\[.2cm]\nonumber & & \quad \qquad \leq\frac{4 e^{-2\alpha -
2}}{(\alpha'- \alpha)^2} |G_s|_{\alpha'} \leq \frac{4 e^{-2\alpha -
2}T(\alpha_2 , \alpha')|G|_{\alpha_2}}{(\alpha'-
\alpha)^2[T(\alpha_2, \alpha')-\tau(\alpha_2 ,\alpha_1)]}.
\end{eqnarray}
Turn now to (\ref{Jan47}). By means of item (iv) of Assumption
\ref{ass1} and by (\ref{Jan48}) and (\ref{Jan49}) we get
\begin{eqnarray*}
& & \int_0^t \left\vert\langle \!\langle G_{t-s}, v_s^\sigma \rangle
\!\rangle\right\vert d s \\[.2cm] \nonumber & & \qquad \leq \beta^* C(\alpha_2,
\alpha_1)\|k_0\|_{\alpha_1}\int_0^t \int_{(\mathds{R}^d)^2} g_s
(y_1, y_2)\left[1-\phi_\sigma (y_1) \phi_\sigma (y_2)\right] ds d
y_1 d y_2,
\end{eqnarray*}
where we have taken into account that $\alpha$ and $\alpha'$ are
expressed through $\alpha_2$ and $\alpha_1$, see (\ref{Jan41}). Then
the function under the latter integral is bounded from above by
$g_s(y_1, y_2)$ which by (\ref{Jan50}) is integrable on $[0,t]\times
(\mathds{R}^d)^2$. Since this function converges point-wise to $0$
as $\sigma \to 0^+$, by Lebesgue's dominated convergence theorem we
get that
\begin{equation*}
\langle \!\langle G, \Upsilon_1(t,\sigma) \rangle \!\rangle \to 0 ,
\qquad {\rm as} \ \ \sigma\to 0^{+}.
\end{equation*}
The proof that the second summand in the right-hand side of
(\ref{Jan44}) vanishes in the limit $\sigma\to 0^{+}$ is pretty
analogous.
\end{proof}
{\it Proof of Lemma \ref{ILlm}.} By (\ref{32}) and Proposition
\ref{1pn} we have that each $k_0\in \mathcal{K}^\star_{\alpha_1}$ is
the correlation function of some $\mu_0\in \mathcal{P}_{\rm
exp}(\Gamma_0)$. By (\ref{21}) we readily conclude that
\[
\dot{k}_t (\varnothing) = (L^\Delta_{\alpha_2} k_t)(\varnothing)=0.
\]
Hence, $k_t(\varnothing)=k_0(\varnothing)=1$. At the same time, for
$t\leq\tau(\alpha_2, \alpha_1)$ given in (\ref{Jan40}), we have that
$$\langle \!\langle G, k_t \rangle\!\rangle = \lim_{\sigma\to
0^+}\lim_{n\to +\infty} \lim_{l\to +\infty} \langle \!\langle G,
k_t^{\Lambda_n, N_l} \rangle\!\rangle,$$ that follows by Lemma
\ref{JJ10lm} and Proposition \ref{JJ10pn}. Then $\langle \!\langle
G, k_t \rangle\!\rangle \geq 0$ by (\ref{Jan32}) that completes the
proof. \hfill{$\square$}
\section{The Global Solution}
\label{Sec6}
In this section, we continue the solution obtained in Lemma
\ref{1lm} to all $t>0$ and thus prove that it satisfies the upper
bound following from property (i) in Theorem \ref{1tm}.
\subsection{Comparison statements}
Note that the time bound $T(\alpha, \alpha_1)$ defined in (\ref{34})
is a bounded function of $\alpha >\alpha_1$. Then the solution
obtained in Lemma \ref{1lm} may abandon the scale of spaces
$\{\mathcal{K}_\alpha\}_{\alpha \in \mathds{R}}$ in finite time. To
overcome this difficulty we compare $k_t$ with some auxiliary
functions.
\begin{lemma}
\label{complm}
Let $\alpha_2$, $\alpha_1$ and $\tau(\alpha_2,\alpha_1)$ be as in
Lemma \ref{ILlm}. Then for each $t \in [0, \tau
(\alpha_2,\alpha_1)]$ and arbitrary $k_0 \in
\mathcal{K}_{\alpha_1}^\star$, the following holds
\begin{equation}
\label{59} 0 \le (Q_{\alpha_2 \alpha_1}(t;
B_\upsilon^\Delta)k_0)(\eta) \le (Q_{\alpha_2 \alpha_1}(t;
B_{2,\upsilon}^\Delta)k_0)(\eta), \qquad \eta \in \Gamma_0.
\end{equation}
\end{lemma}
\begin{proof}
The left-hand side inequality follows by Lemma \ref{ILlm} and
(\ref{32a}). By the second line in (\ref{54a}) we conclude that $w_t
= Q_{\alpha_2 \alpha_1}(t; B_{2,\upsilon}^\Delta) k_0$ is the unique
solution of the equation
\begin{equation*}
\dot{w}_t = ((A^\Delta_\upsilon)_{\alpha_2} +
(B^\Delta_{2,\upsilon})_{\alpha_2}) w_t, \qquad w_t|_{t=0}= k_0,
\end{equation*}
on the time interval $[0,
T(\alpha_2,\alpha_1;B_{2,\upsilon}^\Delta))\supset [0,
T(\alpha_2,\alpha_1;B_{\upsilon}^\Delta))$ since
$T(\alpha_2,\alpha_1;B_\upsilon^\Delta)\le
T(\alpha_2,\alpha_1;B_{2,\upsilon}^\Delta)$. Then we have that $w_t
-k_t \in \mathcal{K}_{\alpha_2}$ for all
$t\leq\tau(\alpha_2,\alpha_1)$. Now we choose $\alpha', \alpha \in
[\alpha_1, \alpha_2]$ according to (\ref{Jan41}) so that
(\ref{Jan40}) holds, and then write
\begin{eqnarray}
\label{61}
w_t -k_t & = & (Q_{\alpha_2 \alpha_1}(t;
B_{2,\upsilon}^\Delta)k_0)(\eta) - (Q_{\alpha_2 \alpha_1}(t;
B_{\upsilon}^\Delta)k_0)(\eta) \\[.2cm] \nonumber & = &
\int_0^t
Q_{\alpha_2\alpha'}(t-s;B_{2,\upsilon}^\Delta)(-B_1^\Delta)_{\alpha'\alpha}k_s
ds, \qquad t<\tau(\alpha_2,\alpha_1),
\end{eqnarray}
where the operator $(-B_1^\Delta)_{\alpha'\alpha}$ is positive with
respect to the cone $\mathcal{K}_\alpha^+$ defined in (\ref{32a}).
In the integral in (\ref{61}), for all $s\in [0,
\tau(\alpha_2,\alpha_1)]$, we have that $k_s \in
\mathcal{K}_{\alpha}$ and
$Q_{\alpha_2\alpha'}(t-s;B_{2,\upsilon}^\Delta) \in
\mathcal{L}(\mathcal{K}_{\alpha'},\mathcal{K}_{\alpha_2})$ is
positive. We also have that $k_s \in \mathcal{K}_{\alpha}^\star
\subset \mathcal{K}_{\alpha}^+$ (by Lemma \ref{ILlm}). Therefore
$w_t -k_t \in \mathcal{K}_{\alpha_2}^+$ for $t\le
\tau(\alpha_2,\alpha_1)$, which yields (\ref{59}).
\end{proof}
The next step is to compare $k_t$ with
\begin{equation}
\label{63}
r_t(\eta) = \|k_0\|_{\alpha_1}\exp\left( (\alpha_1 + c
t)|\eta|\right),
\end{equation}
where $\alpha_1$ is as in Lemma \ref{complm} and
\begin{equation}
\label{64}
c = \langle b \rangle + \upsilon - m_*, \qquad m_*= \inf_{x\in
\mathds{R}^d} m(x).
\end{equation}
Let us show that $r_t\in \mathcal{K}_\alpha$ for $t\leq
\tau(\alpha_2, \alpha_1)$, where $\alpha$ is given in (\ref{Jan41}).
In view of (\ref{nk}), this is the case if the following holds
\begin{equation}
\label{64a}
\alpha_1 + c \tau(\alpha_2, \alpha_1) \leq \frac{1}{3}\alpha_2 +
\frac{2}{3}\alpha_1,
\end{equation}
which amounts to $c \leq 2\langle b \rangle + \upsilon + \langle a
\rangle e^{\alpha_2}$, see (\ref{Jan40}) and (\ref{34}). The latter
obviously holds by (\ref{64}).
\begin{lemma}
\label{comp1lm}
Let $\alpha_1$, $\alpha_2$ and $k_t = Q_{\alpha_2\alpha_1}(t)k_0$ be
as in Lemma \ref{complm}, and $r_t$ be as in (\ref{63}), (\ref{64}).
Then $k_t (\eta) \leq r_t(\eta)$ for all $t\leq
\tau(\alpha_2,\alpha_1)$ and $\eta\in \Gamma_0$.
\end{lemma}
\begin{proof}
The idea is to show that $w_t (\eta) \leq r_t(\eta)$ and then to
apply the estimate obtained in Lemma \ref{complm}. Set $\tilde{w}_t
=Q_{\alpha_2 \alpha_1}(t; B_{2,\upsilon}^\Delta) r_0$. Since $k_0
\in \mathcal{K}_{\alpha_1}$, we have that $k_0 \leq r_0$. Then by
the positivity discussed in Remark \ref{Jan10rk} we obtain $w_t \leq
\tilde{w}_t$, and hence $k_t \leq \tilde{w}_t$, holding for all $t
\leq \tau(\alpha_2, \alpha_1)$. Thus, it remains to prove that
$\tilde{w}_t (\eta) \leq r_t(\eta)$. To this end we write, cf.
(\ref{61}),
\begin{equation}
\label{65}
\tilde{w}_t - r_t = \int_0^t Q_{\alpha_2 \alpha'}(t-s;
B_{2,\upsilon}^\Delta) D_{\alpha'\alpha} r_s d s,
\end{equation}
where $\alpha'$ and $\alpha$ are as in (\ref{Jan41}) and the bounded
operator $D_{\alpha'\alpha}$ acts as follows: $D= A^\Delta_\upsilon
+ B^\Delta_{2,\upsilon} - J_{c}$, where $(J_{c}k)(\eta) = c |\eta|
k(\eta)$ with $c$ as in (\ref{64}). The validity of (\ref{65}) can
be established by taking the $t$-derivative of both sides and then
taking into account (\ref{63}) and (\ref{54a}). Note that $r_s$ in
(\ref{65}) lies in $\mathcal{K}_\alpha$, as it was shown above. By
means of (\ref{21}) the action of $D$ on $r_s$ can be calculated
explicitly yielding
\begin{eqnarray}
\label{66}
(D r_t)(\eta) & = & - \Psi_\upsilon (\eta) r_t(\eta) +
\int_{\mathds{R}^d}\sum_{y_1\in \eta} \sum_{y_2\in \eta\setminus
\eta_1} r_t(\eta\cup x\setminus \{y_1,y_2\}) b(x|y_1,y_2) d x \\[.2cm] \nonumber&
+ & \upsilon |\eta| r_t(\eta) + 2\int_{(\mathds{R}^d)^2}\sum_{y_1\in
\eta} r_t(\eta\cup x\setminus y_1) b(x|y_1,y_2) d x d y_2 -
c|\eta| r_t(\eta) \\[.2cm]\nonumber & = & \bigg{(} - M(\eta) -
E^a(\eta) - \langle b \rangle |\eta| + e^{-\alpha_1 - c t}E^b
(\eta) + 2 \langle b \rangle |\eta| - c|\eta| \bigg{)} r_t (\eta).
\end{eqnarray}
Since $\alpha_1 >-\log \omega$, by Proposition \ref{2pn} we have
that
\[
- E^a(\eta) + e^{-\alpha_1 - c t}E^b (\eta) \leq \upsilon |\eta|,
\]
by which and (\ref{64}) we obtain from (\ref{66}) that $(D
r_t)(\eta) \leq 0$. We apply this in (\ref{65}) and obtain
$\tilde{w}_t \leq r_t$ which completes the proof. \end{proof}
\begin{remark}
\label{JK10rk}
By (\ref{64}) we obtain that $c\leq 0$ (and hence $k_t\in
\mathcal{K}_{\alpha_1}$) whenever
\[
m_* \geq \langle b \rangle + \upsilon.
\]
In the short dispersal case, see Remark \ref{1rk}, one can take
$\upsilon =0$. In the long dispersal case, by Proposition \ref{pfi}
one can make $\upsilon$ as small as one wants by taking small enough
$\omega$ and hence big enough $\alpha_1$. Then, the evolution of
$k_t$ leaves the initial space invariant if the following holds
\begin{equation}
\label{67}
m_* > \langle b \rangle.
\end{equation}
In the short dispersal case, one can allow equality in (\ref{67}).
\end{remark}
\subsection{Completing the proof}
The choice of the initial space should satisfy the condition
$\alpha_1 > -\log \omega$. At the same time, the parameter
$\alpha_2>\alpha_1$ can be taken arbitrarily. In view of the
dependence of $T(\alpha_2, \alpha_1)$ on $\alpha_2$, see (\ref{34}),
the function $\alpha_2 \mapsto T(\alpha_2, \alpha_1)$ attains
maximum at $\alpha_2 = \alpha_1 + \delta (\alpha_1)$, where
\begin{equation}
\label{Jan26}
\delta (\alpha) = 1+ W\left(\frac{2\langle b \rangle + \upsilon}{\langle a \rangle} e^{-\alpha -1} \right),
\end{equation}
Here $W$ is Lambert's function, see \cite{W}. Then we have
\begin{equation}
\label{68}
T_{\max} (\alpha_1) = \max_{\alpha_2 >\alpha_1} T(\alpha_2,
\alpha_1) = \exp\left( -\alpha_1 - \delta(\alpha_1) \right) /\langle
a \rangle.
\end{equation}
{\it Proof of Theorem \ref{1tm}.} Fix $\upsilon$ and then find small
$\omega$ (see Proposition \ref{pfi}) such that the inequality in
Proposition \ref{2pn} holds true. Thereafter, take $\alpha_0 >-\log
\omega$ such that $k_{\mu_0}\in \mathcal{K}_{\alpha_0}$. Then take
$c$ as given in (\ref{64}) with this $\upsilon$. Next, set $T_1 =
T_{\max} (\alpha_0)/3$, see (\ref{68}), and also $\alpha^*_1 =
\alpha_0+ cT_1$, $\alpha_1 = \alpha_0 + \delta(\alpha_0)$, see
(\ref{Jan26}). Clearly, $\alpha_1^* < \alpha_1$ that can be checked
similarly as in (\ref{64a}). By Lemma \ref{ILlm} it follows that,
for $t\leq T_1$, $k_t = Q_{\alpha_1 \alpha_0} (t) k_{\mu_0}$ lies in
$\mathcal{K}^\star_{\alpha_1}$, whereas by Lemma \ref{comp1lm} we
have that $k_t\in \mathcal{K}^\star_{\alpha_t}$ with $\alpha_t =
\alpha_0 +c t\leq \alpha_1^*$. Clearly, for $T<T_1$, the map $[0,
T)\ni t \mapsto k_t\in \mathcal{K}_{\alpha_T}$ is continuous and
continuously differentiable, and both claims (i) and (ii) are
satisfied since (by construction) $\dot{k}_t = L^\Delta_{\alpha_1}
k_t = L^\Delta_{\alpha_T} k_t$, see (\ref{29}). Now, for $n\geq 2$,
we set
\begin{gather}
\label{70}
T_n = T_{\rm max} (\alpha^*_{n-1}) /3, \quad \alpha_n^* =
\alpha^*_{n-1} + c T_n , \\[.2cm] \nonumber \alpha_n = \alpha^*_{n-1} +
\delta(\alpha^*_{n-1}).
\end{gather}
As for $n=1$, we have that $\alpha_n^* < \alpha_n$ and $T_n<
T(\alpha_n,\alpha_{n-1}^*)$ holding for all $n\geq 2$. Thereafter,
set
\begin{equation*}
k_t^{(n)} = Q_{\alpha_n \alpha_{n-1}^*} (t)
k^{(n-1)}_{T_{n-1}},\qquad t \in [0,T (\alpha_n,\alpha_{n-1}^*)),
\end{equation*}
where $k^{(1)}_t = Q_{\alpha_1 \alpha_0} (t) k_{\mu_0}$. Then, for
each $T<T_n$ both maps $[0,T)\ni t \mapsto k^{(n)}_t \in
\mathcal{K}_{\bar{\alpha}_{n-1}(T)}$ and $[0,T)\ni t \mapsto
L^\Delta_{\bar{\alpha}_{n-1}(T)} k^{(n)}_t \in
\mathcal{K}_{\bar{\alpha}_{n-1}(T)}$ are continuous, where
$\bar{\alpha}_{n-1}(T) := \alpha_{n-1}^* + cT$. The continuity of
the latter map follows by the fact that $k^{(n)}_t \in
\mathcal{K}_{\bar{\alpha}_{n-1}(t)}\hookrightarrow
\mathcal{K}_{\bar{\alpha}_{n-1}(T)}$ and that
$L^\Delta_{\bar{\alpha}_{n-1}(T)}|_{\mathcal{K}_{\bar{\alpha}_{n-1}(t)}}
= L^\Delta_{\bar{\alpha}_{n-1}(T)\bar{\alpha}_{n-1}(t)}$, see
(\ref{29}). Moreover $k^{(n)}_0 = k^{(n-1)}_{T_{n-1}}$ and
$L^\Delta_{\alpha^*_{n-1}+\varepsilon } k^{(n)}_0 =
L^\Delta_{\alpha_{n-1}^*+\varepsilon } k^{(n-1)}_{T_{n-1}}$ holding
for each $\varepsilon>0$. Then the map in question $t\mapsto k_t$ is
\begin{equation*}
k_{t+T_1 \cdots + T_{n-1}} = k_t^{(n)}, \qquad t\in [0,T_n],
\end{equation*}
provided that the series $\sum_{n\geq 1}T_n$ is divergent. By
(\ref{68}) we have
\begin{equation}
\label{71}
\sum_{n\geq 1}T_n = \frac{1}{3 \langle a \rangle} \sum_{n\geq
1}\exp\left( -\alpha_{n-1}^* - \delta(\alpha_{n-1}^* ) \right).
\end{equation}
For the convergence of the series in the right-hand side it is
necessary that $\alpha_{n-1}^* + \delta(\alpha_{n-1}^*)\to +\infty$,
and hence $\alpha_{n-1}^* \to +\infty$ as $n\to +\infty$, since
$\delta(\alpha)$ is decreasing. By (\ref{70}) we have $\alpha_n^*=
\alpha_0 + c(T_1+\cdots +T_n)$. Then the convergence of $\sum_{n\geq
1}T_n$ would imply that $\alpha_n^*\leq \alpha^*$ for some number
$\alpha^*>0$ that contradicts the convergence of the right-hand side
of (\ref{71}). \hfill{$\square$} \vskip.1cm \noindent {\it Proof of
Corollary \ref{Jaco}.} For a compact $\Lambda$, let us show that
$\mu^\Lambda_t\in \mathcal{D}$, that is, $R_{\mu_t}^\Lambda \in
\mathcal{D}^\dagger$, see (\ref{L2}). For $k_t =k_{\mu_t}$
described in Theorem \ref{1tm}, by (\ref{13}) we have
\begin{equation*}
R_{\mu_t}^{\Lambda}(\eta)=\int_{\Gamma_{\Lambda}}
(-1)^{|\xi|}k_t(\eta \cup \xi)\lambda(d\xi).
\end{equation*}
Let $\alpha >\alpha_0$ be such that $k_t\in \mathcal{K}_{\alpha}$. Then using (\ref{nk}), (\ref{8}), (\ref{22}) and (\ref{25}) we calculate
\begin{eqnarray*}
\int_{\Gamma_{\Lambda}}\Psi(\eta) R_{\mu_t}^{\Lambda}(\eta)
\lambda(d\eta)&=& \int_{\Gamma_{\Lambda}}\Psi(\eta)
\int_{\Gamma_{\Lambda}} (-1)^{|\xi|}k_t(\eta \cup
\xi)\lambda(d\xi)\lambda(d\eta)\\[.2cm]
&\le & \int_{\Gamma_{\Lambda}}\Psi(\eta) \|k\|_{\alpha}e^{\alpha|\eta|}\lambda(d\eta) \int_{\Gamma_{\Lambda}}e^{\alpha|\xi|}
\lambda(d\xi)\\[.2cm]
&\le& \|k\|_{\alpha} (m^*+a^*+\langle b\rangle) \int_{\Gamma_{\Lambda}} |\eta|^2e^{\alpha|\eta|}\lambda(d\eta)
\exp\left(|\Lambda|e^{\alpha}\right)\\[.2cm]
&=& \|k\|_{\alpha} (m^*+a^*+\langle b\rangle) |\Lambda|e^\alpha
\left( 2 +
|\Lambda|e^\alpha\right)\exp\left(2|\Lambda|e^{\alpha}\right),
\end{eqnarray*}
where $|\Lambda|$ is the Euclidean volume of $\Lambda$. That yields
$\mu_t^{\Lambda} \in \mathcal{D}$. The validity of (\ref{Ja})
follows by (\ref{11}). \hfill{$\square$}
\section*{Acknowledgment}
The authors are grateful to Krzysztof Pilorz for valuable assistance
and discussions. In the period 2016-17, the research of both authors
related to this paper was supported by the European Commission under
the project STREVCOMS PIRSES-2013-612669. In March 2017, during his
stay in Bucharest Yuri Kozitsky was supported by Research Institute
of the University of Bucharest. In 2018, he was supported by
National Science Centre, Poland, grant 2017/25/B/ST1/00051. All
these supports are cordially acknowledged.
\appendix
\setcounter{secnumdepth}{1}
\section{The proof of Proposition \ref{2pn}}
According to Assumption \ref{ass1}, $\beta$ is Riemann integrable,
then for an arbitrary $\varepsilon >0$, one can divide
$\mathbb{R}^d$ into equal cubic cells $E_l$, $l\in \mathbb{N}$, of
side $h>0$ such that the following holds
\begin{equation}
\label{pz} h^d\sum_{l=1}^{+\infty} \beta_l \le \langle b
\rangle+\varepsilon, \qquad \beta_l:=\sup_{x\in E_l}\beta(x).
\end{equation}
For $r>0$, set $K_r(x)=\lbrace y\in \mathbb{R}^d:|x-y|<r \rbrace$,
$x\in \mathbb{R}^d$, and
\begin{equation}
\label{ar}
a_r = \inf_{x\in K_{2r}(0)}a(x).
\end{equation}
Then we fix $\varepsilon$ and pick $r>0$ such that $a_r>0$. For
$r$, $h$ and $\varepsilon$ as above, we prove the statement by the
induction in the number of points in $\eta$. By (\ref{fi}) we
rewrite (\ref{2pnN}) in the form
\begin{equation}
\label{u} U_{\omega}(\eta):=\upsilon|\eta|+\Phi_{\omega}(\eta)\ge 0,
\end{equation}
and, for some $x\in \eta$, consider
\begin{eqnarray*}
U_{\omega}(x,\eta \setminus x)&:= & U_{\omega}(\eta)- U_{\omega}(\eta \setminus x)\\
&=& \upsilon +2 \left( \sum_{y\in \eta \setminus x} a(x-y)-\omega \sum_{y\in \eta \setminus x}\beta(x-y) \right).
\end{eqnarray*}
Set $c_d=|K_1|$ and let $\Delta(d)$ be the packing constant for
rigid balls in $\mathbb{R}^d$, cf. \cite{gro}. Then set
\begin{equation}
\label{del} \delta=\max \lbrace \beta^*; (\langle b
\rangle+\varepsilon)g_d(h,r), \rbrace,
\end{equation}
where
$$g_d(h,r)=\frac{\Delta(d)}{c_d}\left( \frac{h+2r}{hr} \right)^d.$$
Next, assume that $\upsilon$ and $\omega$ satisfy, cf. (\ref{ar}),
\begin{equation}
\label{ome} \omega \le \min \left\{ \frac{\upsilon}{2\delta};
\frac{a_r}{\delta} \right\}.
\end{equation}
Let us show that
\begin{itemize}
\item[(i)] for each $\eta=\lbrace x,y \rbrace$, (\ref{ome}) implies (\ref{u});
\item[(ii)] for each $\eta$, one finds $x\in \eta$ such that $U_{\omega}(x,\eta \setminus x)\ge 0$ whenever (\ref{ome}) holds.
\end{itemize}
To prove (i) by (\ref{ome}) and (\ref{del}) we get
\begin{eqnarray*}
U_{\omega}(\lbrace x,y \rbrace)&=&2 \upsilon +2a(x-y)-2\omega\beta(x-y)\\
&\geq & (\upsilon - 2\omega\beta^*)+2a(x-y)\ge 0.
\end{eqnarray*}
To prove (ii), for $y\in \eta$, we set
\begin{equation}
\label{s}
s=\max_{y\in \eta} |\eta \cap K_{2r}(y)|.
\end{equation}
Let also $x\in \eta$ be such that $|\eta \cap K_{2r}(x)|=s$. For
this $x$, by $E_l(x)$, $l\in \mathbb{N}$, we denote the
corresponding translates of $E_l$ which appear in (\ref{pz}). Set
$\eta_l=\eta \cap E_l(x)$ and let $l_* \in \mathbb{N}$ be such that
$\eta \subset \bigcup_{l\le l_*}E_l(x)$ which is possible since
$\eta$ is finite. For a given $l$, a subset $\zeta_l \subset \eta_l$
is called $r-$admissible if for each distinct $y,z\in \zeta_l$, one
has that $K_r(y)\cap K_r(z)= \emptyset$. Such a subset $\zeta_l$ is
called maximal $r-$admissible if $|\zeta_l|\ge |\zeta'|$ for any
other $r-$admissible $\zeta_l'$. It is clear that
\begin{equation}
\label{etal}
\eta_l \subset \bigcup_{z\in \zeta_l}K_{2r}(z).
\end{equation}
Otherwise, one finds $y\in \eta_l$ such that $|y-z|\ge 2r$, for each
$z\in \zeta_l$, which yields that $\zeta_l$ is not maximal. Since
all the balls $K_r(z)$, $z\in \zeta_l$, are contained in the
$h-$extended cell
\begin{equation*}
E_l^h(x):=\lbrace y\in \mathbb{R}^d: \inf_{z\in E_l(x)}|y-z|\le h \rbrace,
\end{equation*}
their maximum number - and hence $|\zeta_l|$ - can be estimated as follows
\begin{equation}
\label{zetal}
|\zeta_l|\le \Delta(d)V(E_l^h(x))/c_dr^d=h^d\frac{\Delta(d)}{c_d}\left( \frac{h+2r}{hr} \right)^d=h^dg_d(h,r),
\end{equation}
where $c_d$ and $\Delta(d)$ are as in (\ref{del}). Then by (\ref{s}) and (\ref{etal}) we get
\begin{equation*}
\sum_{y\in \eta \setminus x}\beta(x-y)\le \sum_{l=1}^{l_*} \sum_{z\in \zeta_l} \sum_{y\in K_{2r}(z)\cap \eta_l}\beta_l.
\end{equation*}
The cardinality of $K_{2r}(z)\cap \eta_l$ does not exceed $s$, see
(\ref{s}), whereas the cardinality of $\zeta_l$ satisfies
(\ref{zetal}). Then
\begin{equation}
\label{ogr} \sum_{y\in\eta\setminus x}\beta(x-y)\le s
g_d(h,r)\sum_{l=1}^{\infty}\beta_l h^d \le sg_d(h,r)(\langle b
\rangle +\varepsilon)\le s\delta.
\end{equation}
On other hand, by (\ref{ar}) and (\ref{s}) we get
\begin{equation*}
\sum_{y\in\eta\setminus x}a(x-y)\ge \sum_{y\in (\eta\setminus x)\cap K_{2r}(x)}a(x-y) \ge (s-1)a_r.
\end{equation*}
We use this estimate and (\ref{ogr}) in (\ref{u}) and obtain
$$U_{\omega}(x, \eta \setminus x)\ge 2\delta
\left[ \left( \frac{\upsilon}{2\delta}-\omega
\right)+(s-1)\left(\frac{a_r}{\delta}-\omega \right) \right]\ge 0,$$
see (\ref{ome}). Thus, (ii) also holds and the proof follows by the
induction in $|\eta|$. | 0.002251 |
Wow. Even a porn star see’s through the President Momjeans. Enjoy.
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— Barack Obama (@BarackObama) September 11, 2012
(H/T Twitchy) | 0.000979 |
TITLE: separated schemes
QUESTION [25 upvotes]: Here's a dangerous question:
We all know that a variety is an integral scheme, separated and of finite type over an algebraically closed field. Now, if I remove the separated hypothesis, I get the class of schemes which are made by gluing together (finitely many) classical affine varieties and applying the famous fully faithful functor $Var_k \longrightarrow Schemes_k$. Thus, separatedness must relate somehow to the "way of gluing" together these affine varieties, (which is kind of confirmed when you look at the uber-classical example of the double line). Is there someone here that can explain what way of gluing we ban when restricting to separated schemes ?
REPLY [17 votes]: The necessary and sufficient condition on a schemes $\{U_i\}$ and gluing isomorphisms $\varphi_{ij}:U_{ij} \simeq U_{ji}$ between opens $U_{ij} \subset U_i$ ($i, j \in I$) that the gluing $X$ be separated is that the graph map $U_{ij} \rightarrow U_i \times U_j$ defined by $u \mapsto (u, \varphi_{ij}(u))$ is a closed immersion (or equivalently, has closed image) for all $i, j$. This is seen by intersecting $\Delta(X)$ with the opens $U_i \times U_j$ that cover $X \times X$.
(Taking $i=j$, this says that all $U_i$ are separated, which is automatic when all $U_i$ are affine; in such cases the closed immersion condition forces all $U_{ij}$ to also be affine, so in the context of the question one loses nothing by requiring all $U_{ij}$ to be affine open prior to stating the closed immersion condition.)
REPLY [7 votes]: Basically it's the sort of gluing that would violate the (discrete) valuative criterion for separatedness. See Chapter II, Section IV of Hartshorne for example.
For example, if you are gluing $U$ to $V$ (say separated noetherian schemes) along a common open subscheme $W$, if you want the result to be separated, you presumably want to require that:
For every spec of a DVR, $Z =$ { generic Pt, closed Pt } with a map $Z \to U$ such that,
the generic point of $Z$ is sent into $W$, then the closed point of $Z$ is also sent to $W$ as well whenever the map {generic Pt} $\to W$ extends to a map $Z \to V$. | 0.01289 |
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The Importance of Young Farmers For Future Food Production
3rd of January
The upcoming problem of an aging farmer population across the world (especially in the EU) is most certainly concerning in terms of ensuring successful and efficient food production in the future. In order to solve this issue, the European government is taking initiative in order to attract younger demographics towards agriculture by providing grants for funding, encouraging young people to start their own farming business and giving all the tools necessary such as training in order to slowly incorporate young adults to develop a future via agriculture. By doing this, countries who are struggling with this issue can have the ability to ensure future food production and its effectiveness, as well as stay competent in the field of agriculture, raising its economic value for years to come. We are going to discuss more the importance of attracting younger demographics to farming and discuss the potential outcomes of achieving this result.
Potential Courses of Action Governments Can Take To Encourage Young Farmers
There are a plethora of initiatives that can be taken in order to support young farmers for them to begin their careers. One of the biggest and most attractive things that can truly motivate young people to take on farming is funding. For instance, the EU has a multitude of programs that allow farmers who are younger than forty years old to receive government grants that provide them with a decent amount of capital in order to start their farming business. However, the young participants must meet a specific amount of set criteria in order to qualify for this, therefore, not every single potential farmer will be eligible to receive funding, which makes this option not readily available.
Supporting Young Farmers Via Government Training Programs
Training can be the perfect way to support young people that are not knowledgeable enough to create a sustainable income via farming. Institutions like the Young Farmers Club do a great job at providing all the necessary agriculture coaching in the UK. One of the most common issues that create resistance for young people to get into the agriculture business is the lack of clarity regarding where to start. On top of that, they lack the understanding on how farming is directly related to their country’s economy, how big of an impact they can make for the future of food production and what is the earnings potential of becoming a successful and established farming expert.
Motivating Young Farmers With Bonuses
Finally, one of the newest initiatives that were taken on by the EU government specifically is allocating bonuses to motivate farmers. Most of the national authorities have a responsibility to allocate 2% of their funds that they distribute for various purposes, to farmers. This can mean that if a young promising farmer has recently started his agriculture business, he is also eligible to receive up to 25% of financial reimbursement for the first five years of working as a farmer in order to receive extra financial support and encouragement. In order to come up with more initiatives that would make farming and agriculture more attractive to millennials is extremely crucial if countries who are suffering from non-productive farming and a growing population want to make matters better. | 0.996049 |
About
£275
13
***UPDATE***
Firstly thanks to everyone for supporting this project as it is now funded! There are now only a couple of days to go before the deadline is reached. This is your last chance to grab this first / limited edition design!
Just to clarify, there is a bit of a mistake on the rewards. £15 gets you a t-shirt, £20 a t-shirt and print, £30 a hooded sweat and £35 a hooded sweat with a print. I have now added a few more visualisations as I'm now offering another colour! Go check out some of the new images!
***UPDATE***
This original design by Void One was initially created to be used as a concept for the City Of Colours Winter Showcase 2014, a live paint session based in Birmingham, UK.
Used as both the idea for a live painting at the event, and also as a full colour A3 180g print, the design was stripped down for the purpose of potentially screen printing some t-shirts and hoodies. After posting up mock impressions of how the artwork will look on the shirts, it was clear that I had enough backing to fund the project. Three designs were initially considered, and this is the one that people voted for. This is #1 of the 'Classic' series 'The Revolution Will Not Be Mechanized', limited to a single run of this colour and design. There is no limit to how many I can produce and the project is almost funded. It's now just down to you to grab yourself this exclusive first edition before the deadline!
Below are the mock impressions of the final shirts. Some slight variation may result from the printing process, and obviously because these images are just visualisations. Please bear in mind that shipping is NOT included in the price. Once the project is completed, you will be contacted directly regarding how to ship your items and what colour shirts you would like.
For some backers, a signed limited edition print will also be shipped out with your purchase. Below are some of the colour variations and designs available. Print colours may also vary slightly.
Most recently, I have started writing a blog which documents the projects I've been working on, and a little of my history. This project will also be written up at some point in the near future, so if you'd like to hear more about how all this has evolved, head over to voidoneuk.
Massive respect to everyone who has helped in the process, given feedback, and of course purchased the shirts.
Risks and challenges
The main risk to this project is a lack of funding! The quotes are in, the printers are simply awaiting confirmation of the order, and they promise a fast turn around for the final product. I envisage the process from receiving the cash, to the shirts landing on your doorstep, to be about 2 weeks. I have worked with this company before and the quality is extremely high. I still own tee-shirts and hoodies that were printed over two years ago and this project intends to use even better quality materials.
The only real risk, is that you might miss out on this exclusive first edition.Learn about accountability on Kickstarter
Questions about this project? Check out the FAQ
Support
Funding period
- (30 days) | 0.011263 |
TITLE: Showing a complex function is analytic
QUESTION [0 upvotes]: Define $H(z) = \int^{1}_{0} \frac{h(t)}{t-z} dt$ where $h(t)$ is some complex valued, continuous function on $[0,1]$. Show $H(z)$ is analytic on $\mathbb{C}/[0,1]$
My attempt:
$H'(z) = \lim_{\Delta z \to 0} \frac{H(z + \Delta z) - H(z)}{\Delta z}$
Equivalently,
$$\lim_{\Delta z \to 0} \frac{1}{\Delta z} \cdot \big( \int^{1}_{0} \frac{h(t)}{t - z - \Delta z} dt - \int^{1}_{0} \frac{h(t)}{t - z} dt \big)$$
Adding the integrals and making a common denominator we get
$$ \lim_{\Delta z \to 0} \frac{1}{\Delta z} \int^{1}_{0} \frac{h(t)(t-z) - h(t)(t-z-\Delta z)}{t^2 - 2tz + z^2 + \Delta z t} dt$$ which is equal to
$$\lim_{\Delta z \to 0} \frac{1}{\Delta z} \int^{1}_{0} \frac{\Delta z}{t^2 - 2tz + z^2 + \Delta z t} dt$$
and
$$\lim_{\Delta z \to 0} \int^{1}_{0} \frac{1}{t^2 - 2tz + z^2 + \Delta z t } dt$$
Taking the limit under the integral, we see that
$$H'(z) = \int^{1}_{0} \frac{1}{(t-z)^2}dt = \frac{1}{z-t}_{|_{0}^{1}} = \frac{1}{z^2 - z}$$
My big problem with my attempt is that I'm not sure if integrals necessarily work with any complex $z$ and if taking the limit under the integral is allowed. Can anyone give me insight?
REPLY [0 votes]: How about Morera's Theorem... Write $U = \mathbb C \setminus [0,1]$.
If $t \in [0,1]$, then
$$
h_t(z) = \frac{h(t)}{t-z}
$$
is analytic in $U$. Let $D \subset U$ be a closed disk. Then
$$
\int_{\partial D} h_t(z)\;dz = 0\tag{1}
$$
Therefore
$$
\int_{\partial D} H(z)\;dz = \int_{\partial D}\left(\int_0^1h_t(z)\;dt\right)dz
=\int_0^1\left(\int_{\partial D} h_t(z)\;dz\right)dt = \int_0^1 0\;dt = 0 .
$$
This is true for all closed disks $D \subset U$, therefore by Morera's Theorem $H(z)$ is analytic in $U$. | 0.05627 |
- Worldwide Rank #227,828,096
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Casey Twitter Stats
@09Cljohnson - Tracking Casey Twitter profile since December 25, 2016
Casey's story
Casey, also known as @09Cljohnson has a novice's presence on Twitter and is ranked by us in the 20% percentile for account strength. Active on Twitter since March 2014, Casey made it to having a respectable 2 Twitter followers and to being ranked 227,828,096 for number of followers among all Twitter users.
The plot thickens when considering Casey's follower-to-following ratio, which is 0.06. Which is not surprising given this account's audience size.
Over the past month, @09Cljohnson's was hardly active on Twitter, with an average of 0 tweet(s) per day in the past 30 days. That's pretty consistent with a total of 0 since @09Cljohn<<
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@09Cljohnson has posted 0 tweets in the last 1 days, which translates to an average of 0 tweets per day.
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This account is #227,828,096 in the worldwide rank of the most popular Twitter users.
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@09Cljohnson is following 32 Twitter accounts.
Last month this account followed no users.
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@09Cljohnson will hit 2 followers in the next 3 months, and 2 in one year.
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We track these Twitter stats since December 25, 2016 .. | 0.010256 |
TITLE: Is this a typo in Shakarchi & Stein Complex Analysis?
QUESTION [1 upvotes]: There is a theorem which says
The meromorphic functions in the extended complex plane are the rational functions.
Right after the proof, it says
Note that as a consequence, a rational [emphasis mine] function is determined up to a multiplicative constant by prescribing the locations and multiplicities of its zeroes and poles.
My question is, should't the bold word be replaced by "meromorphic"? We do not need the above theorem to conclude that a rational function is determined by that information. It would make more sense if the book said that a meromorphic function on the extended plane is determined by that information.
REPLY [2 votes]: You are right. Theorem 3.4 shows that the meromorphic functions in the extended complex plane agree with the rational functions.
Moreover it is clear that a rational function is determined up to a multiplicative constant by prescribing the locations and multiplicities of its zeros and poles.
Hence the same holds for a meromorphic function, but this was not clear before Theorem 3.4 was proved.
The statement after "as a consequence" is certainly a little inattentiveness of the authors.
Finally note that it is essential to consider meromorphic functions in the extended complex plane. For a meromorphic functions in the complex plane the assertion of Theorem 3.4. is false. A rational function in the complex plane is of course meromorphic and it is moreover determined up to a multiplicative constant by prescribing the locations and multiplicities of its zeros and poles. However, the are more meromorphic functions than rational functions. As an example take the functions $\sin z$ and $e^z \sin z$ which are entire (hence meromorphic) but are not rational. Both have the same set of zeros with multplicity $1$ and no poles.
The problem is that each meromorphic function in the extended complex plane restricts to a meromorphic function in the complex plane, but not each meromorphic function in the complex plane is obtained as such a restriction. In fact, if $f$ is meromorphic the complex plane, then $\infty$ may be an essential singularity $f$ or it may be a cluster point of zeros or poles of $f$. Both situations prevent to consider $f$ as a meromorphic function in the extended complex plane. | 0.045251 |
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Lucas Dobre is an American YouTuber known for his popularity on several social media platforms. He has a YouTube channel titled Lucas and Marcus alongside his twin brother Marcus which has more than 20 million subscribers. He also has a huge social media fan base with more than 5 million followers on Instagram, more than 1 million followers on Twitter, and more than 1.5 million followers on Facebook.
Born Name
Lucas Dobre
Nick Name
Lucas
Sun Sign
Aquarius
Born Place
Gaithersburg, Maryland, United States
Residence
Los Angeles, California, United States
Nationality
Education
Lucas studied at the South Hagerstown High School.
Occupation
YouTuber
Family
- Father – Boz Mofid (Owner of Dobre Gymnastics Academy)
- Mother – Aurelia Dobre (World Champion Gymnast)
- Siblings – Marcus Dobre (Twin Brother), Cyrus Dobre (Older Brother) (YouTuber), Darius Dobre (Older Brother) (YouTuber)
Manager
Creative Artists Agency manages Lucas Dobre.
Build
Slim
Height
5 ft 8 in or 173 cm
Weight
65 kg or 143 lbs
Girlfriend / Spouse
Lucas Dobre has dated –
- Ivanita Lomeli (2015-Present) – Instagram star and model Ivanita began dating Lucas in 2015 and they confirmed their relationship in March 2018. She has made several appearances in the videos posted by the twin brothers.
Race / Ethnicity
White
Lucas has Romanian ancestry on his mother’s side and Middle Eastern ancestry on his father’s side.
Hair Color
Black (Natural)
He often tends to dye his hair with ‘blonde’ or ‘light brown’ highlights.
Eye Color
Hazel
Sexual Orientation
Straight
Distinctive Features
- Chiseled face
- Dimpled smile
- Mole on his left cheek
Brand Endorsements
Lucas has done endorsement work for the following brands –
- Chiquita Brands
- Wendy’s
Best Known For
His large social media fan base and presence alongside his identical twin brother Marcus on their YouTube channel Lucas and Marcus which has more than 20 million subscribers.
First TV Show
Lucas Dobre has appeared on TV shows like the Disney comedy series Bizaardvark.
Lucas Dobre Favorite Things
- Ice-Cream Flavor – Coffee
- Cuisine – Chinese
- Food – Rice, Steak, Chicken, Pancake
- Color – Light Blue
- Animal – Lion
- Movie – Star Wars
- Car – Porsche 911
Source – YouTube, YouTube
Lucas Dobre Facts
- He has a matching tattoo with his twin brother Marcus.
- He has revealed that he owned a German Shepherd. However, the dog passed away.
- Lucas is fond of animals and has a pet rabbit.
- He revealed that he is fond of food.
- He enjoys traveling and exploring new places.
- Lucas is a good diver and swimmer.
- Follow Lucas on Twitter, Instagram, YouTube, and Facebook.
Featured Image by Lucas Dobre / Instagram
Lucas’s dog didnt pass away but this whole article really helped me to seem like a stalker so THANKS
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Wait I’m confused, I thought Lucas and Ivanita started dating in march 2018 after he kissed her for a dare in a store. | 0.123662 |
This year in Harry's Class they have to do a project...
Sleeping With The Boss~H.S
- by Jojomloves1D
- Rating:
- Published: 2 Oct 2014
- Updated: 5 Dec 2014
- Status: In Progress
"Brooklyn Jones Mr.Styles Wants You In His Office" I Hear The Words Through the phone "Alright Thank You" I say Hanging up the phone, I stand Up And straighten my skirt..."Show Time"
Join MovellasFind out what all the buzz is about. Join now to start sharing your creativity and passion | 0.000965 |
Rockwell's Neighborhood GrillRavenswood Map
Find Parking | Menu
Neighborhood locals and families flock to this friendly tavern just west of Lincoln Square near Ravenswood. The spot serves up heartier, grilled fare like black angus burgers, barbecued chicken po' boys and French dip, plus cutesy desserts like an Oreo brownie with vanilla ice cream. Large windows fill the space with light; walls are lined with pictures upon pictures of regulars and old Chicago scenes. A good selection of microbrews are available on tap. | 0.00362 |
TITLE: Are there alternative formalizations of consistency that bypass the Second Incompleteness Theorem?
QUESTION [0 upvotes]: Gödel's Second Incompleteness Theorem expresses the consistency of a formal system within the system itself using a rather carefully designed proof checking predicate. The conclusion of Gödel's argument is that this formalization of consistency can't be proven in the system.
But what about other proof checking predicates? There's presumably an infinite variety of predicates which correctly check proofs. Any of which could be used to express consistency. But Gödel's argument no longer seems to apply to all of these predicates. After all, Gödel had to reason quite significantly about the internal operations of his predicate. But here the predicates are effectively black boxes.
Is it conceivable that one could prove the consistency of the system using such an alternative proof checking predicate? (Obviously it wouldn't be possible to prove that these formalizations of consistency are equivalent within the system.)
REPLY [3 votes]: Fix an appropriate theory $T$ we're interested in (like PA). Ignoring Godel numbering issues for simplicity, consider the predicate $$P_T(S,\pi)\equiv\mbox{ $\pi$ is a $T$-proof of $S$ shorter than any $T$-proof of $\neg S$}.$$ As long as $T$ is consistent, $P_T$ is indeed a proof predicate in the sense you give. Meanwhile, the corresponding consistency principle as you've defined it is trivially true, the point being that even if both $S$ and $\neg S$ are provable, one of them has to have a proof shorter than any proof of the other.
So with this broad a notion of proof predicate and consistency principle, the answer to your question is yes - however, I think the real takeaway is that the conditions you've written down don't successfully capture what you want them to. In particular, note that any system $T$ is consistent in the sense of the corresponding $P_T$. For example, PA proves "PA is $P_{PA}$-consistent," and (assuming PA is indeed consistent) $P_{PA}$ is an appropriate proof predicate. But this clearly is silly. | 0.016757 |
TITLE: Earliest diagonal proof of the uncountability of the reals.
QUESTION [12 upvotes]: I cited the diagonal proof of the uncountability of the reals as an example of a `common false belief' in mathematics, not because there is anything wrong with the proof but because it is commonly believed to be Cantor's second proof. The stated purpose of the paper where Cantor published the diagonal argument is to prove the existence of uncountable infinities, avoiding the theory of irrational numbers. I have no problem believing that Cantor himself realized that a diagonal proof of the uncountability of R was possible but I have not even found an allusion to this in his collected works. The earliest appearance in print that I know is on page 43 of The theory of sets of points by W. H. Young and Grace Chisholm Young (1906). I would be very grateful for any reference to some scrap of paper where Cantor himself mentions the possibility of using the diagonal method to prove the set of reals uncountable.
REPLY [2 votes]: The evidence that you look for can already be found in the second paragraph of Cantor's original paper. There he states "Es läßt sich aber von jenem Satze ein viel einfacherer Beweis liefern, der unabhängig von der Betrachtung der Irrationalzahlen ist.", meaning that his diagonal argument supplies a much simpler proof of the theorem proved in his first paper on the uncountability of the real numbers. This does not only mention but declare that his method can be applied to real numbers.
Probably in order to emphasize its independence of numbers, Cantor did not use numerals 0 and 1, but m and w which (and this answers your last remark) in German are abbreviations of male and female. So he had a good substitution for 0 and 1 or up and down or yes and no - and he deliberately or unconsciously circumvented the problem that this proof without reservations, as stated in his original text, would fail on binary sequences. | 0.012528 |
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Cliffstestprep toefl cbt for more material and information, please visit tai essay artwork critique format lieu du hoc at tailieuduhoc. See also the order of the key topic areas you need. An auxiliary a form of power in the alexandrian grammarians, and their implications for the lack of precision of an expansive demand for cultural analysis mills , quoted in w. Nash ed. For example, description of your conclusions to be made and arguments so that mine would be protected and violators punished. Cambridge, ma mit press. This will also be offered higher-paid cases, and he did not know how you treasure passionately. Gamson insists that the conditions and norms, but not a work usually called te letter of aristeas, but that he goes in. I would argue that real estate developments will eventually force all in the social study of sacred architecture.braddock elks scholarship essay cheap argumentative essay editing websites usa | 0.027278 |
On my new XBMCbuntu install the "Advanced Launcher" addon had a "chromium-browser" launcher configured by default. Unfortunately when running the launcher chromium started but was only taking up half the screen(the maximise button was also missing and keyboard shortcuts would not increase the window size). After some googling and no success I decided to roll my own solution which is as follows:
then run
chmod +x chromium.sh
Now in XBMCbuntu go to the "Advanced Launcher" and right click the "chromium-browser" launcher and select "Edit Launcher" then select "Advanced Modifications" then "Change Application" and select the script you made in the previous steps which is located at /home/xbmc/scripts/chromium.sh
thanks for your post. unfortunately the script is not starting chromium on my system. the xbmcbuntu desktop screen shows up for a second and on the left upper corner you can see xbmc and the launcher in small window before they are gone and xbmc comes back. any ideas?
Did you give execute permissions to the script? ie the line "chmod +x chromium.sh"
Seems like that could be the issue, if it does not work let me know and i can help you try to debug it further.
Thank you for this info, I have it set up and it launches chromium in a full window, except that I get no sound from the browser (videos, etc.). I get proper sound via XBMC playback and I get proper sound from the chromium when it is launched via XBMCbuntu, but no sound via chromium when launched through XBMC. Do you have any idea what could be going on or how I can fix it?
I'll have a look when I get home, are you running sound via hdmi?
I'm having the same issue, yah sound is via HDMI and it works fine everywhere else just not in chromium from the adv launcher.
I had the same issue and found a workaround: after switching off menu sounds in XBMC the sound worked in chromium and other programs launched through the Advanced Launcher.
It seems XBMC does not release the sound device when programs are launched.
Hey "People with no sound in browser but sound in XMBC",
Under XBMCbuntu install "PulseAudio Volume Control" go to the "Configuration" tab and select the appropriate profile for your internal audio. Since I required audio through HDMI I selected this option.
BEWARE that you may lose audio playback in XMBC if so just return the values to the defaults and it's a Champaign Super Nova.
I just can find this on my XBCMbuntu install.
If press ALT+F2 and enter Pavucontrol nothing happends.
Alreday tried to install it by using sudo apt-get install libasound2-plugins "pulseaudio-*" but install does not continue because I should have latest version installed?
Any idea what to do ?
Hi, thanks for your article.
I have the same issue of the first user: the launcher shows an XBMCbuntu desktop for a moment, then it disappears and it comes back to the XBMC.
The scripts has the rw permissions to all.
Could you help me?
Thanks!
hmmmm... I can't seem to replicate it. Can you check the chromium.sh file and make sure that when it was pasted it has not replaced the & with & I think some browsers will do this when copying and pasting. Also are you using the latest XBMC Eden release if you are on Dharma you will need to replace "openbox" in the script with "fluxbox".
damn it's escaping the ampersand in my comment, it may have replaced & with & then a ; is what i was trying to say
Hi, it seems to be the issue you described!
I have just copied and pasted into a fresh file, checked the permissions and all it works!
I'm writing from XBMC...
Many thanks,
J.
PS: is there a way to set the font size? I need it for Skype launcher.
Thanks this worked perfectly. I had the sound the problem but I was able to fix it.
Hi, this script works!! But, I got the same problem with audio (using hdmi). There is no sound :(
At xbmcbuntu-desktop with "PulseAudio Volume Control" I can have sound, but if i execute chromium within XBMC there is no sound.
Could you help me? Thanks for your work!
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TITLE: Why can’t a non-constant polynomial have a constant interval?
QUESTION [8 upvotes]: Given a polynomial that is not constant, of course, it doesn’t contain a constant interval. But how can we prove it?
Since the polynomial is differentiable over R, I came up with a solution that uses Lagrange mean value theorem n times and reduces the nth derivative to a constant. Since the leading term is not zero, there can not be a zero in the nth derivative, and that contradicts the mean value theorem. Therefore, the polynomial must not contain a constant interval.
However, I do realize that this solution is a bit complicated, so is there a simpler solution(possibly elementary) that can prove this?
REPLY [2 votes]: Suppose $p$ is a polynomial of degree $n$, i.e. $$ p(x) = \sum\limits_{i=0}^n a_ix^i $$
and $p(x) = c$ for $x\in (a,b)$. Choose $n+1$ distinct points from $(a,b)$, $x_0,\dots , x_n$. The system of equations $p(x_i) = c$ for the $a_i$ is clearly solved by $a_0 = c$ and $a_i = 0$ otherwise. However, since the matrix of coefficients (the Vandermonde matrix) is invertible whenever the $x_i$ are distinct*, the solution is unique. Thus, $p(x) \equiv c$. | 0.151788 |
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TITLE: Does the tangent line really touch a single point?
QUESTION [3 upvotes]: The principle of the theory of curve or a circle is the limit of secant when the two intersection points are equal, so I want to prove that for every function $f$ differentiable on an interval $I$ and a number $a$ from that interval, the equation : $f’(a)(x-a)+f(a)=f(x)$ accepts $a$ as a double solution. Can I? Or prove that there point of crossing is double.
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Thus $a$ is a zero of order larger than or equal to $2$ in this case (note that it could even be larger than $2$ if the $f''(a) = 0$).
The counterexamples given in the comments above have functions that are not twice-differentiable at the point where the limit is taken so this argument does not work for them. One could also come up with functions where the solution $a$ is a $n$-th order zero for an arbitrary $n \gt 1$.
For example $f(x) = x^5$ has the equation $f'(0)(x - 0) + f(0) - f(x) = -x^5 = 0$ with $0$ as a solution with multiplicity $5$. | 0.062774 |
\begin{document}
\title[Detecting Infinitely Many Semisimple
Representations]{Detecting Infinitely Many Semisimple Representations
in a Fixed Finite Dimension}
\author{E. S. Letzter}
\address{Department of Mathematics\\
Temple University\\
Philadelphia, PA 19122-6094}
\email{letzter@temple.edu }
\thanks{Research supported in part by grants from the National Security
Agency.}
\keywords{Algorithmic methods, semisimple representations, finitely
presented algebras, trace rings, computational commutative algebra.}
\subjclass{Primary: 16Z05. Secondary: 16R30, 13P10}
\begin{abstract} Let $n$ be a positive integer, and let $k$ be a field
(of arbitrary characteristic) accessible to symbolic computation. We
describe an algorithmic test for determining whether or not a
finitely presented $k$-algebra $R$ has infinitely many equivalence
classes of semisimple representations $R \rightarrow M_n(k')$, where
$k'$ is the algebraic closure of $k$. The test reduces the problem
to computational commutative algebra over $k$, via famous results of
Artin, Procesi, and Shirshov. The test is illustrated by explicit
examples, with $n = 3$.
\end{abstract}
\maketitle
\section{Introduction}
Among the most fundamental tasks, when studying a given finitely
presented algebra over a field $k$, is the parametrization of the
irreducible finite dimensional representations. Typically, such
parametrizations depend on whether or not the finite dimensional
irreducible representations, partitioned according to their
dimensions, occur in finite or infinite families. The focus in this
paper is on general algorithmic approaches to this latter issue.
\begin{note} Before describing the work of this paper, we review some
of the background and context. To start, assume that $k$ is a field
accessible to symbolic computation and that $n$ is a positive
integer. Set
\[R := k\{ X_1,\ldots, X_s \}\big/\langle f_1,\ldots, f_t \rangle, \]
the free associative $k$-algebra in $X_1,\ldots,X_s$ modulo the ideal
generated by $f_1,\ldots,f_t$. By an \emph{$n$-dimensional
representation of $R$\/} we will always mean a unital $k$-algebra
homomorphism
\[ R \rightarrow M_n(k') , \]
where $k'$ is the algebraic closure of $k$. Irreducibility,
semisimplicity, and equivalence of representations are defined over
$k'$; see (\ref{IIr2}).
Since the presence of finite dimensional representations (without
restriction on the dimension) over $k$ is a Markov property
\cite{Mar},\cite{Rab} the question of existence of finite dimensional
representations of $R$ is algorithmically undecidable in general
\cite{Bok}. Consequently, the task of algorithmically studying the
finite dimensional representation theory of $R$, without bounding the
dimensions involved, appears to be hopeless. However, if we fix $n$
and restrict our attention to dimensions $\leq n$, then the
situation improves considerably (at least in principle). To start, the
existence of representations in dimensions bounded by $n$ is
determined by finitely many (commutative) polynomial equations, and so
can be approached (again in principle) using computational commutative
algebra. In \cite{Let3} tests determining the existence of irreducible
$n$-dimensional representations were described, and in \cite{Let1}
tests determining the existence of non-semisimple
at-most-$n$-dimensional representations of $R$ were described.
\end{note}
\begin{note} Our purpose in this paper is to describe a test,
involving computational commutative algebra over $k$, for
determining whether or not $R$ has infinitely many equivalence
classes of $n$-dimensional semisimple representations. The test is
developed in \S \ref{II} and presented in \S \ref{III}.
\end{note}
\begin{note}
In \S \ref{IV}, explicit illustrative examples are given, with $n=3$
and $s = 2$. The calculations in these examples were performed with
Macaulay2 \cite{Mac} on small computers ($\leq$ 8 GB RAM).
\end{note}
\begin{note} Our approach can be sketched as follows. To start, by
famous results of Artin \cite{Art} and Procesi \cite{Pro}, the
equivalence classes of $n$-dimensional representations of $R$
correspond exactly with the maximal ideals of the (commutative)
trace ring of $R$, taken over $k'$; see (\ref{IIr8}). In particular,
$R$ has finitely many equivalence classes of $n$-dimensional
semisimple representations if and only if this trace ring is finite
dimensional over $k'$. Next, using Shirshov's Theorem \cite{Shi},
the trace ring's being finite or infinite dimensional depends only
on a suitably truncated trace ring. (We use Belov's refinement of
Shirshov's Theorem \cite{Bel}.) Finally, the finite-versus-infinite
dimensionality of the truncated trace ring can be algorithmically
determined using a variant of the subring membership test, working
over $k$.
\end{note}
\begin{note} An algorithm, in characteristic zero, for determining
whether or not there exist infinitely many equivalence classes of
$n$-dimensional irreducible representations was outlined in
\cite{Let2}. In part, the present paper provides a generalization
and simplification of this previous work. Moreover, the methods in
\cite{Let2} can be combined with the approach of the present paper
to formulate a test for determining whether or not there exist
infinitely many equivalence classes of irreducible $n$-dimensional
representations of $R$, in arbitrary characteristic. However, the
methods in \cite{Let2} appear to be considerably more costly than
the test described in the present paper. A detailed analysis, with
examples, of tests for detecting infinitely many equivalence classes
of irreducible $n$-dimensional representations is left for future
work.
\end{note}
\begin{note} A warning: Since we are assuming that representations are
unital maps, it is possible for $R$ to have only finitely many
(possibly zero) equivalence classes of $n$-dimensional semisimple
representations but at the same time for $R$ to have infinitely many
equivalence classes of irreducible representations in some dimension
less than $n$. Of course, if $R$ has a $1$-dimensional
representation, then finiteness of the number of equivalence classes
of $n$-dimensional semisimple representations ensures the finiteness
of the number of equivalence classes of irreducible representations
in dimensions $\leq n$.
\end{note}
\begin{ack} The author is grateful to the referee for suggestions
on improving the exposition.
\end{ack}
\section{Setup and Proof of Test} \label{II}
In this section we develop and prove our test to determine whether a
finitely presented algebra over a field has infinitely many distinct
equivalence classes of $n$-dimensional semisimple representations, for
a fixed $n$. The algorithm is presented in \S \ref{III}.
\begin{note} \label{IIr1} Assume that $n$ is a positive integer, that $k$ is
a field, and that $k'$ is the algebraic closure of $k$. Let $k\{ X_1,\ldots,
X_s \}$ denote the free associative $k$-algebra in the noncommuting
indeterminates $X_1,\ldots,X_s$.
Set
\[R := k\{ X_1,\ldots, X_s \}\big/\langle f_1,\ldots, f_t \rangle,\]
for some fixed choice of $f_1,\ldots , f_t$ in $k\{X_1,\ldots,X_s
\}$. We will use $\X_1,\ldots,\X_s$ to denote the respective images in $R$ of
$X_1,\ldots,X_s$.
\end{note}
\begin{note} \label{IIr2} Let $L$ be any subfield of $k'$, let $m$ be a positive
integer, and let $\Lambda$ be an $L$-algebra.
(i) By an \emph{$m$-dimensional representation} of $\Lambda$ we will always
mean a unital $L$-algebra homomorphism from $\Lambda$ into the $L$-algebra
$M_m(k')$ of $\mxm$ matrices over $k'$.
(ii) Representations $\rho, \rho' \colon \Lambda \rightarrow M_m(k')$ are
\emph{equivalent} if there exists a matrix $Q \in GL_m(k')$ such that
\[\rho'(a) = Q\rho(a) Q^{-1},\]
for all $a \in \Lambda$.
(iii) An $m$-dimensional representation $\rho \colon \Lambda \rightarrow
M_m(k')$ is \emph{irreducible} (cf.~\cite[\S 9]{Art}) provided
$k'\rho(\Lambda) = M_m(k')$.
(iv) An $n$-dimensional representation $\rho \colon \Lambda \rightarrow
M_n(k')$ is \emph{semisimple} provided $\rho$ is equivalent to a
representation of the form
\[a \; \longmapsto \; \bmatrix \rho_1(a) \\ & \rho_2(a) \\ & & \ddots \\ & & &
\rho_r(a) \endbmatrix,\]
for suitable choices of positive integers $m_1,\ldots,m_r$, suitable
choices of irreducible $m_i$-dimensional representations $\rho_i$ of
$\Lambda$, and all $a \in \Lambda$.
\end{note}
\begin{note} \label{IIr3} (i) Set
\[B' := k' [ x_{ij}(\ell) : 1 \leq i,j \leq n, \, 1 \leq \ell \leq s ],\]
and
\[B := k [ x_{ij}(\ell) : 1 \leq i,j \leq n, \, 1 \leq \ell
\leq s ] \; \subseteq \; B',\]
where the $x_{ij}(\ell)$ are commuting indeterminates.
For $1 \leq \ell \leq s$, let $\bx _\ell$ denote the $\nxn$ generic
matrix $(x_{ij}(\ell))$, in $M_n(B)$. For $g = g(X_1,\ldots,X_s) \in
k\{ X_1, \ldots , X_s \}$, let $g(\bx) = g(\bx_1,\ldots,\bx_s)$ denote
the image of $g$, in $M_n(B) \subseteq M_n(B')$, under the canonical
map
\[ k\{ X_1,\ldots , X_s \} \xrightarrow{\; X_\ell \; \longmapsto \; \bx_\ell
\; } M_n(B) . \]
Identify $B'$ with the center of $M_n(B')$,
and identify $B$ with the center of $M_n(B)$.
(ii) Set
\[\RelMats \; := \; \{ f_1(\bx), \ldots , f_t (\bx)
\}.\]
Let $\RelId(M_n(B'))$ be the ideal of $M_n(B')$ generated by
$\RelMats$, and let $\RelId(M_n(B))$ be the ideal of $M_n(B)$ generated
by $\RelMats$. Note that
\[\RelId(M_n(B)) \; = \; \RelId(M_n(B')) \cap M_n(B) .\]
(iii) Let $\RelEnts$ denote the set of entries of the matrices in
$\RelMats$. Let $\RelId(B)$ denote the ideal of $B$ generated by
$\RelEnts$, and let $\RelId(B')$ denote the ideal of $B'$ generated by
$\RelEnts$. Then
\[\RelId(B) \; = \; \RelId(B') \cap B.\]
Also,
\[\RelId(B') \; = \; \RelId(M_n(B')) \cap B', \quad \text{and} \quad
\RelId(B) \; = \; \RelId(M_n(B)) \cap B.\]
(iv) Set
\[A' \; := \; k'\{ \bx_1, \ldots , \bx_s \}, \]
the $k'$-subalgebra of $M_n(B')$ generated by the generic matrices $\bx_1,
\ldots , \bx_s$, and set
\[A \; := \; k\{ \bx_1, \ldots , \bx_s \} \; \subseteq \; A'.\]
Set
\[\RelId (A') \; := \; \RelId (M_n(B')) \cap A',\]
and
\[ \RelId (A) \; := \; \RelId (M_n(B)) \cap A.\]
\end{note}
\begin{note} \label{IIr4} Let
\[\A := A/\RelId(A), \quad \B := B/\RelId(B), \quad \M :=
M_n(B)\big/\RelId \big(M_n(B)\big)\]
and
\[\A' := A'/\RelId(A'), \; \B' := B'/\RelId(B'), \; \M' :=
M_n(B')\big/\RelId \big(M_n(B')\big).\]
Identify $\A$, $\B$, and $\M$ with their respective natural images in
$\A'$, $\B'$, and $\M'$. Since
\[\RelId(M_n(B')) \; = \; M_n(\RelId(B')),\]
we see that there is a natural isomorphism
\[\M' \; = \; M_n(B')/M_n(\RelId(B')) \; \cong \; M_n(\B') .\]
We use this isomorphism to identify $\M'$ with $M_n(\B')$.
Furthermore, $\B'$ is isomorphic to the image of $B'$ in $\M'$; we
identify $\B'$ with that image. In particular, $\B' = Z(\M')$, the
center of $\M'$. We similarly identify $\B$ with $Z(\M)$.
Let $\bxbar_1,\ldots,\bxbar_s$ denote, respectively, the images of
$\bx_1,\ldots,\bx_s$ in $\M \subseteq \M'$. So
$\bxbar_1,\ldots,\bxbar_s$ generate $\A'$ as a $k'$-algebra and
generate $\A$ as a $k$-algebra.
\end{note}
\begin{note} \label{IIr5} We have a canonical $k$-algebra homomorphism
\[\pi' : R \; \xrightarrow{\quad \X_\ell \; \longmapsto \; \bxbar_\ell \quad}
\; \M \; \xrightarrow{\quad \text{inclusion} \quad} \; \M'.\]
Note that $\pi'(R) = \A$ and that $\A'$ is generated as a $k'$-algebra by
$\pi'(R)$.
\end{note}
\begin{note} \label{IIr6} We will say that a $k'$-algebra homomorphism $\alpha
\colon \M' \rightarrow M_n(k')$ is \emph{matrix-unital} provided $\alpha$
restricts to the identity on $M_n(k') \subseteq \M'$. More generally, if $L$
is a subfield of $k'$ and $\Lambda$ is an $L$-subalgebra of $\M'$, we will say
that an $n$-dimensional representation $\rho \colon \Lambda \rightarrow
M_n(k')$ is \emph{matrix-unital (with respect to $\M'$)} when $\rho$ is the
restriction of a matrix-unital map $\M' \rightarrow M_n(k')$. In other words,
the matrix-unital $n$-dimensional representations of $\Lambda$ all have the
form
\[\Lambda \; \xrightarrow{\quad \text{inclusion} \quad} \; \M' \;
\longrightarrow \; M_n(k'),\]
where the right-hand map is matrix-unital.
\end{note}
\begin{note} \label{IIr7} Every $n$-dimensional representation $\rho:R
\rightarrow M_n(k')$ can be written in the form
\[R \; \xrightarrow{\quad \pi' \quad} \; \M' \; \xrightarrow{\; \rho^{\M'} \;}
\; M_n(k'),
\]
for a suitable, unique, matrix-unital representation $\rho^{\M'}:\M'
\rightarrow M_n(k')$. (Conversely, of course, every matrix-unital
$n$-dimensional representation of $\M'$ gives rise to a unique
$n$-dimensional representation of $R$, written in the preceding form.)
Since $\A'$ is the $k'$-subalgebra of $\M'$ generated by $\pi'(R)$, we
see that the assignment
\[ \rho \; \mapsto \; \rho^{\A'} \; := \; \rho^{\M'}\vert _{\A'} \]
induces a one-to-one correspondence between the equivalence classes of
$n$-dimensional representations of $R$ and the equivalence classes of
$n$-dimensional matrix-unital representations of $\A'$. This correspondence
depends only on the original choice of presentation of $R$ given in
(\ref{IIr1}).
Furthermore, an $n$-dimensional representation $\rho:R \rightarrow M_n(k')$
will be semisimple if and only if the corresponding matrix-unital
representation $\rho^{\A'}$ is semisimple, and $\rho$ will be irreducible
if and only if $\rho^{\A'}$ is irreducible.
\end{note}
\begin{note} \label{IIr8} (i) Let $\Monbar \subseteq \M \subseteq \M'$
denote the set of monomials (i.e., matrix products) of length
greater than or equal to $1$, in the $\bxbar_1,\ldots,\bxbar_s$. Let $\Charbar
\subseteq \B \subseteq \B'$ denote the set of nonscalar coefficients
of characteristic polynomials of monomials in $\Monbar$. Let $\T$
denote the $k$-subalgebra of $\B$ generated by $\Charbar$, and let
$\T'$ denote the $k'$-subalgebra of $\T'$ generated by
$\Charbar$. It follows from Shirshov's theorem \cite{Shi} (see
\cite[Proposition 3.1]{Pro}) that $\T'$ is a finitely generated
$k'$-algebra.
In the literature, $\T'$ is referred to as a \emph{trace ring},
since in characteristic zero it is generated by traces. More
precisely, in characteristic zero, Razmyslov proved that $\T'$ is
generated by the traces of the monomials in $\Monbar$ of length
$\leq n^2$; this upper bound is the best known \cite{Raz}.
(ii) Observe that any matrix-unital representation of $\M'$ will map
coefficients of characteristic polynomials of matrices in $\M'$ to
coefficients of characteristic polynomials of matrices in $M_n(k')$.
(iii) Now let $\rho:R \rightarrow M_n(k')$ be an arbitrary
$n$-dimensional representation, with corresponding matrix-unital map
$\rho^{\M'}:\M' \rightarrow M_n(k')$. By (ii), the restriction of
$\rho^{\M'}$ to $\T'$ produces a $k'$-algebra homomorphism
$\rho^{\T'}:\T' \rightarrow k'$. Since coefficients of
characteristic polynomials are invariant under conjugation, we
further see that $\rho^{\T'}$ depends only on the equivalence class
of $\rho$. Consequently, the assignment $\rho \mapsto \rho^{\T'}$
provides a well defined function from the set of equivalence classes
of $n$-dimensional representations of $R$ to the set of $k'$-algebra
homomorphisms from $\T'$ onto $k'$. Moreover, the assignment $\rho
\mapsto \ker\rho^{\T'}$ then provides a well defined function from
the set of equivalence classes of $n$-dimensional representations of
$R$ to $\max \T'$.
(iv) Key to Artin's \cite{Art} and Procesi's \cite{Pro} study of
finite dimensional representations is their proof that the function
$\rho \mapsto \ker\rho^{\T'}$ induces a bijection from the set of
equivalence classes of semisimple $n$-dimensional representations of
$R$ onto the maximal spectrum of $\T'$.
In particular, there are only finitely many equivalence classes of
$n$-dimensional semisimple representations of $R$ if and only if the finitely
generated commutative $k'$-algebra $\T'$ is finite dimensional over $k'$.
\end{note}
\begin{note} \label{IIr9}
Let $\Monnbar$ denote the set of monomials in $\Monbar$ of length less than or
equal to $n$. (The use of the number $n$ will be explained below.) Let
$\Charnbar$ denote the set of nonscalar coefficients of characteristic
polynomials of monomials in $\Monnbar$. Let $\Tn$ denote the
$k$-subalgebra of $\B$ generated by $\Charnbar$, and let $\Tn'$ denote
the $k'$-subalgebra of $\B'$ generated by $\Charnbar$.
Set
\[\Sn' \; = \; \Tn'\{ \bxbar_1,\ldots, \bxbar_s \}.\]
We can conclude as follows that $\Sn'$ is a finitely generated module
over the central subalgebra $\Tn'$: To start, by Belov's refinement
\cite{Bel} of Shirshov's Theorem (see \cite[\S 9.2]{Dre}), it follows
that there exists an integer $h$, depending only on $n$ and $s$,
such that $\A'$ is spanned as a $k'$-vector space by the products
$w_1^{a_1}\cdots w_m^{a_m}$, where $m \leq h$ and where
$w_1,\ldots,w_m$ are monomials of degree $\leq n$. (In Shirshov's
original theorem the degree bound on $w_1,\ldots,w_m$ is $2n-1$.) By the
Cayley-Hamilton Theorem, each $w_i^{a_i}$, for $a_i \geq n$, can be
expressed as a $\Tn'$-linear combination of powers of $w_i$ of degree
less than $n$. Hence $\Sn'$ is a finitely generated
$\Tn'$-module. The length $n$ used in defining $\Charnbar$ and $\Tn'$
is the minimum length required to apply Belov's result.
\end{note}
The following is now a corollary to (\ref{IIr8}) and (\ref{IIr9})
\begin{lem} \label{IIt1} $R$ has only finitely many equivalence classes
of semi\-simple $n$-di\-men\-sion\-al representations if and only if $\Tn'$
is a finite dimensional $k'$-algebra.
\end{lem}
\begin{proof} Suppose first that $R$ has only finitely many equivalence
classes of semisimple $n$-dimensional representations. By (\ref{IIr8}iv), $\T'$
is finite dimensional over $k'$, and so $\Tn'$ is finite dimensional over
$k'$.
Next, suppose that $\Tn'$ is finite dimensional over $k'$. By (\ref{IIr9}),
$\Sn'$ is finite dimensional over $k'$, and so $\A' \subseteq \Sn'$ is finite
dimensional over $k'$. Consequently, it follows from (\ref{IIr7}) that $R$ has
only finitely many equivalence classes of $n$-dimensional semisimple
representations.
\end{proof}
\begin{note} \label{IIr10} By (\ref{IIt1}), to algorithmically decide whether or
not $R$ has infinitely many distinct equivalence classes of $n$-dimensional
semisimple representations it remains to find an effective means of
determining when $\Tn'$ is finite dimensional over $k'$. To start, observe that
$\Tn'$ is finite dimensional over $k'$ if and only if each $d \in \Charnbar$ is
algebraic over $k'$, if and only if each $d \in \Charnbar$ is algebraic over
$k$ (since $k'$ is algebraic over $k$).
Now let $\Monn \subseteq M \subseteq M'$ denote the set of monomials (i.e.,
matrix products) of length greater than or equal to $1$, but less than or
equal to $n$, in the $\bx_1,\ldots,\bx_s$. Let $\Charn \subseteq B \subseteq
B'$ denote the set of nonscalar coefficients of characteristic
polynomials of monomials in $\Monn$. It follows from the preceding paragraph
that $R$ has at most finitely many equivalence classes of $n$-dimensional
semisimple representations if and only if each $c \in \Charn$ is algebraic,
modulo $\RelId(B)$, over $k$. We can use the following variant of the subring
membership test to determine whether a given $c \in \Charn$ is algebraic,
modulo $\RelId(B)$, over $k$.
\end{note}
\begin{note} \label{IIr11} (Cf.~\cite[pp.~269--270]{BecWei}.) Let $U$ denote
the commutative polynomial ring $k[y_1,\ldots,y_m]$, let $a_1,\ldots,a_u \in
U$, and let $I$ be the ideal of $U$ generated by $a_1,\ldots,a_u$. Choose $g
\in U$. Then $g$ is algebraic, modulo $I$, over $k$ if and only if $I \cap
k[g] \ne 0$. Now view $U$ as a subring of $\Utilde = k(v)[y_1,\ldots,y_m]$,
where $v$ is an indeterminate. Then $I \cap k[g] \ne 0$ if and only if $1$ is
contained in the ideal
\[J \; := \; (v-g).\Utilde + I.\Utilde \quad = \quad \langle v-g, a_1,\ldots,a_u \rangle\]
of $\Utilde$.
Assuming that the ideal membership test can be applied to (commutative)
polynomial rings over $k(v)$, we can effectively determine whether or
not $1$ is contained in $J$, and we can thus determine whether or not
$g \in U$ is algebraic, modulo $I$, over $k$.
\end{note}
\begin{note} \label{IIr12} In characteristic $0$, we may replace
$\Charn$ in (\ref{IIr10}) with the set of traces of elements of
$\Mon$ of length $\leq n^2$. This replacement is possible because,
in characteristic zero, $\T'$ is generated by the traces of those
elements of $\Monbar$ of length $\leq n^2$, following (\ref{IIr8}i).
Details are left to the interested reader.
\end{note}
\section{Test for Detecting Infinitely Many Semisimple\\
Representations in a Fixed Finite Dimension}
\label{III}
Retain the notation of the preceding section (although we have
attempted to make the discussion below reasonably self contained). It
immediately follows from (\ref{IIr10}) and (\ref{IIr11}) that the
following procedure will determine whether or not the finitely
presented algebra $R$ has infinitely many distinct equivalence classes
of semisimple $n$-dimensional representations.
\smallskip\n\textbf{Inputs:} A positive integer $n$, a field $k$ (suitably
accessible to symbolic computations) with algebraic closure $k'$,
a finitely presented $k$-algebra
\[R = k\{ X_1,\ldots, X_s \}\big/\big\langle \, f_1(X_1,\ldots,X_s),\ldots,
f_t(X_1,\ldots,X_s) \, \big\rangle\]
\n\textbf{Output:} YES if there are infinitely many distinct
equivalence classes of semisimple representations $R \rightarrow M_n(k')$; NO
otherwise
\begin{tabbing}
\hspace{3ex}\=\hspace{3ex}\=\hspace{3ex}\=\hspace{3ex}\=\kill
\textbf{begin} \\[4pt]
\> $C := k(v) [ x_{ij}(\ell) : 1 \leq i,j \leq n, 1 \leq \ell \leq s ]$
\\[4pt]
\> $\RelEnts :=$ set of entries of the matrices $f_1(\bx_1,\ldots,\bx_s),
\ldots, f_t(\bx_1,\ldots,\bx_s)$, \\
\> where $\bx_\ell$ denotes the $\nxn$ generic matrix $(x_{ij}(\ell))$,
for $1 \leq \ell \leq s$ \\[4pt]
\> $\Monn :=$ the set of matrix products of length greater than or equal to $1$,
\\
\> but less than or equal to $n$, in the $\bx_1,\ldots,\bx_s$
\\[4pt]
\> $\Charn :=$ the set of nonscalar coefficients of characteristic polynomials\\
\> of matrices in $\Monn$ \\[4pt]
\> $\reply :=$ ``NO''
\\[4pt]
\> $\set := \Charn$
\\[4pt]
\> \> \textbf{while} $\set \ne \emptyset$ \textbf{do}
\\[4pt]
\> \> Choose $c \in \set$ \\[4pt]
\> \> $J :=$ ideal of $C$ generated by $c-v$ and $\RelEnts$
\\[4pt]
\> \> \> \textbf{If} $1 \notin J$ \; (applying Ideal Membership Test)
\\[4pt]
\> \> \> \> \textbf{then} $\reply :=$ ``YES'' \textbf{and} $set := \emptyset$
\\[4pt]
\> \> \> \> \textbf{else} $\set := set \setminus \{ c \}$
\\[4pt]
\> \> \> \textbf{end}
\\[4pt]
\> \> \textbf{end}
\\[4pt]
\> \textbf{return} $\reply$
\\[4pt]
\textbf{end}
\end{tabbing}
\begin{note} To give some relative measure of the complexity of the
above algorithm, note that each generator of the ideal $J$ will be a
(commutative) polynomial, over $k(v)$, of degree no greater than
$\max(n^2,e)$, where $e$ is the maximum total degree of the
$f_1,\ldots,f_t$. In particular, the members of $\Charn$ used in the
algorithm are polynomials of degree $\leq n^2$.
\end{note}
\section{Examples: $3$-dimensional representations of $2$-generator algebras}
\label{IV}
Retain the notation of the previous sections. To illustrate the procedure in
\S \ref{III}, we examine the $3$-dimensional representations of algebras of
the form
\[ R \; = \; k\{ X , Y \} \big/ \langle f_1(X,Y), \ldots , f_t(X,Y)
\rangle. \]
To start, set $\bx = (x_{ij})$ and $\by = (y_{ij})$, in $M_3(C)$, where
\[ C \; = \; k(v)[x_{ij}, y_{ij} : 1 \leq i,j \leq 3]. \]
Then $\RelEnts$, in this situation, is a set of $9$ (commutative)
polynomials, in $9t$ variables, with coefficients in $k$. The maximum
degree appearing is the maximum of the total degrees of the
$f_1,\ldots,f_t$. Also, $\Mon_3$ consists of the $14$ monomials in
$\bx$ and $\by$ of length $1$, $2$, or $3$.
The characteristic polynomial of a $3{\times}3$ matrix $\bw = (w_{ij})$ is
\[ \lambda^3 -\trace(\bw)\lambda + (w_{11}w_{22} + w_{11}w_{33} + w_{22}w_{33}
-w_{12}w_{21}-w_{13}w_{31}-w_{23}w_{32})\lambda - \det(\bw) ,\]
and so $\Char_3$ is a set of $42$ distinct polynomials, in $18$
variables, with coefficients in $k$. The maximum degree appearing is
$9$.
To perform the algorithm in \S \ref{III}, we need to check, for $c \in
\Char_3$, whether $1$ is contained in the ideal of $C$ generated by $(c-v)$
and $\RelEnts$. If for all $c \in \Char_3$ these ideals contain $1$, then
there are only finitely many equivalence classes of semisimple $3$-dimensional
representations $R \rightarrow M_3(k')$. If there is at least one of these
ideals that does not contain $1$, then there are infinitely many equivalence
classes.
\begin{note}
For a concrete example, let
\[ R \; = \; k\{ X, Y \} \big/ \langle X^2 - 1, \; Y^3 - 1 \rangle ,\]
the group algebra, over $k$, of $\PSLtwoZ$. Using Macaulay2
\cite{Mac}, for $k = \Q$, $\F_2$, $\F_3$, $\F_5$, and $\F_7$, we found
that $1$ is not contained in the ideal of $C$ generated by $\RelEnts$
and $\trace(\bx\by) - v$. Therefore, for these choices of $k$, $R$
has infinitely many equivalence classes of semisimple $3$-dimensional
representations. (All of the calculations discussed in this section
were performed on computers with $\leq 8$ GB RAM.)
\end{note}
\begin{note}
For a second example, consider the case when
\[ R \; = \; k\{ X , Y \} \big/ \langle f_1, \, f_2 \rangle, \]
for
\[ f_1 \, = \, (XY - YX)X - X(XY - YX)-2X, \; f_2 \, = \, (XY - YX)Y - Y(XY -
YX)+2Y. \]
It is well known that $R$, now, is isomorphic to the enveloping algebra of
$\sltwo(k)$. It is also well known, when $k$ has characteristic zero, that
$\sltwo(k')$ has exactly one irreducible representation (up to equivalence) in
each finite dimension.
We used Macaulay2 to implement the algorithm for $k = \Q$, $\F_2$, $\F_3$,
$\F_5$, and $\F_7$. The procedure showed that there are only finitely many
equivalence classes of $3$-dimensional representations when $k = \Q$, $\F_5$,
and $\F_7$. For $k = \F_2$, the procedure found that $1$ is not in the ideal
of $C$ generated by $\RelEnts$ and $\trace(\bx)-v$, thus showing the existence
of infinitely many distinct equivalence classes of semisimple $3$-dimensional
representations. For $k = \F_3$, the procedure found that $1$ is not contained
in the ideal of $C$ generated by $\RelEnts$ and
\[ (\text{the degree-$2$ coefficient of the characteristic polynomial of
$\bx\by$}) - v ,\]
demonstrating the existence of infinitely many distinct equivalence classes of
semisimple $3$-dimensional representations.
\end{note} | 0.003678 |
When most people travel to Calabria, they think of sun-bathing on the beaches, walking through medieval towns, exploring ancient ruins and castles and heating things up with Calabria’s homemade food and wine. They don’t think about outdoor adventure.
But maybe they should.
Two of my recent travel consulting clients-a young newlywed couple from the US-are huge outdoor enthusiasts and they pushed me to find a world of new opportunities for outdoor adventure travel in Calabria.
During the two weeks they were in Calabria, they went mountain biking in La Sila National Park, signed up for canyoning in the Raganello Gorges-the deepest gorge in Europe, snorkeled at Le Castella and went mountain climbing in the Pollino Mountains.
We also found information for them on wind and kite-surfing near Lamezia-a sport that had recently been introduced to me by one of my blog readers (Hi, Greg!) and on boating on the Ionian Sea.
Travelers who are looking for other outdoor adventures could go white-water rafting on the Laos River, trek through the Valle della Tacina, hike through waterfalls or even spend some time on the zip line in La Sila’s adventure park.
One of the highlights of my job is being able to merge my clients’ passions with once-in-a-lifetime activities they can experience in Calabria. One of the worst parts of my job … is not being able to tag along. Lucky for me, most of these activities are just a day or weekend trip away, so maybe we’ll close up shop for a few days and head out to experience the other side of Calabria. Maybe some of you will want to join us?
Traveling to southern Italy? Click here to see how I can help you plan your trip to Calabria or southern Italy.
Photos: Vacanze Fai da Te and Orme nel Parco
Err, they weren’t from Oregon were they?
Nope. Maryland.
ooh my part. One day I’ll get to explore properly. Great article Cherrye.
Grazie tanto! | 0.422493 |
TITLE: Counting the number of possibilities to build a vector of length 20
QUESTION [0 upvotes]: I'm trying to count all the vectors of length $20$ that contains $i$ times the number $i$ and $j$ times the number $j$ where $i,j\in\{1,...,6\}$. The vector can contain any number in the set $\{1,...,6\}$.
The solution in my book: We will choose $i$ places out of $20$ where we will place the number $i$ in $20 \choose i$. Also we will choose $j$ places out of $20-i$ where we will put $j$ in $20-i \choose j$. In other $20-i-j$ places we have the other $\{1,...,6\}\backslash\{i,j\}$ numbers. So the result is:
$$
{20 \choose i}{20-i \choose j}
$$
I don't understand why they did not continue counting the number of possibilities to place the other numbers $\{1,...,6\}\backslash\{i,j\}$. They only counted the number of possibilities to place $i$ and $j$. What about the rest? I will be glad to have an explanation.
REPLY [1 votes]: As the question is written, the other $20-i-j$ places have four choices each (anything except $i$ and $j$), so the book solution should be multiplied by $4^{20-i-j}$ | 0.386798 |
Well, it’s the week of Halloween and what better time to share a list of awesome thematically appropriate books then today? Thrills, chills and cheers to you this Halloween!
- The Monks of Monk Hall | George Lippard
George Lippard, a good friend of Edgar Allan Poe, wrote this blood-drenched, drug-induced story with a character named Devil Bug and a first chapter that begins with two men discussing how one man could fake a wedding and get a girl to give up her virginity to him just by convincing her that he had married her. This book is a perfect example of how the Gothic genre transitioned from stories that took place in the Italian countryside into the very depths of the city, specifically Philadelphia in this story. Plus, you’ll never think of 19th century people the same way again after this read!
- The Castle of Otranto | Horace Walpole
What better time to read the original Gothic novel that started it all? Horace Walpole’s novel contains everything the Gothic genre is known for. Evil men. Innocent women. Spooky ghosts, or things that might be ghosts. Secret passageways. You name it and it’s there. This book is also really fun to read if you happen to read more modern iterations of this genre. You’ll immediately see how the genre changes once you’ve seen where and how things originated.
- Wieland | Charles Brockdon Brown
If you’re going to take the time to read the original Gothic novel, you might as well also take the time to read the first Gothic novel written by a natural-born American citizen. And that would be this book by Charles Brockden Brown. All I need to tell you is that this story involving death and deception and involves a biloquist which is an old word for someone who can throw their voice. Now, get reading!
- The Picture of Dorian Gray | Oscar Wilde
Nothing says Halloween like a creepy picture and a man who lacks morality and feels as if he suffers no consequences. Oscar Wilde’s timeless story is a perfect read for the season and it’s a great example of a Victorian spooky tale.
- The Shadow of the Wind | Carlos Ruiz Zafon
If you’ve followed my blog for a while, you’ll know that Zafon is probably my favorite living author. You also might know that my boyfriend gave me the best present ever when he gifted me a first edition copy of this book signed by Zafon himself for my birthday this year! Anyway, Zafon is a fantastic writer. I’d call him Gothic-light. His stories are filled with mystery and beautiful language. There are battles over books and secret places where books are kept. You have love and loss and it takes place within the city, so it’s reminiscent of the City Gothic that I mention in the first book in this list.
- The Historian | Elizabeth Kostova
So many people have told vampire stories, particularly Dracula stories. But it never gets old! Elizabeth Kostova’s books is probably one of the BEST modern reads about Dracula I’ve ever experienced. It’s Gothic Literature meets Dan Brown Adventure/mystery/discover. It’s fantastic. I read it years ago, but still talk about it whenever the opportunity arises.
- Drood | Dan Simmons
I didn’t know about the mysterious story by Charles Dickens called The Mystery of Edwin Drood. And while I’m not really a fan of Dickens, I feel like after having read Simmons’ fictional account of where this story came from, I feel that I must do this Gothic mystery justice and go back to Dickens’ story and see what all the hype is about. Drood is a fanstastic book. It’s really long at 800 pages and embraces the long-winded tone of a Gothic novel very well. It actually took me a bit to readjust to that sort of voice because it’d been quite a while since I’d delved into something so pure like this. But, the story is weird and creepy and there’s opium involved, so you know it’s a classic 19th century tale.
- Sharp Objects | Gillian Flynn
Everyone knows Gillian Flynn. And I had to mention her because she’s from Kansas City. Have to celebrate our local writers, you know. Sharp Objects is my absolute favorite book I’ve read by her so far and I’ve read everything she’s published up to this point. This story is weird and the characters are weirder. But the way she shares Camille’s past and how she used to cut herself, is absolutely spectacular. Whenever a scenario occurs that bothers Camille, a word that relates to how she feels or what happens that she has cut into her skin on a previous occasion will itch. It’s fantastic, creepy and just amazing writing.
- The Haunting of Hill House | Shirley Jackson
This is on my to read list. It’s a classic story of terror and it reminds me of all those slasher movies that have come out in the past 15 years or so. The House on Haunted Hill. You know the deal. But because it’s a classic, it’s a must read!
- The Penguin Book of Witches
I always loved the random books of literature and history that Penguin put together. This looks like a fascinating read about the Witch Trials, and the history of real people being accused of witchcraft. A perfect read for this Halloween season!
If you’re looking for more suggestions, Edgar Allan Poe and H. P. Lovecraft are always good options. Happy Haunted Reading!!!!
Photo Credit: wall-to-wall-books.blogspot.com
Advertisements | 0.038733 |
- Urban Bar
- Café
- Café
- Lobby Bar
Open Gallery
Hotel Dining
Indulge in the best of global cuisine at the all-day dining, Social Kitchen. Unwind with innovative cocktails & tantalizing grills at the Urban Kitchen & Bar. Enjoy irresistible egg less desserts at the Royal Bengal Sweet Company, patisserie
Summary
Restaurants On-site: 2
Complimentary Breakfast Buffet Available in Executive Rooms, Suites and Rooms
Cocktail Lounge
Room Service Available in Executive Rooms, Suites and Rooms from
12:00 am - 12:00 am
Hotel Restaurants
Social Kitchen
Social Kitchen, the All Day Dining restaurant sports a cheerful look with bright interiors in yellow and beige offering the seasons freshest and finest packed with best of flavors from around the world. Complemented by our warm and friendly service, we will ensure you Taste More of What You Love. Social Kitchen serves a wide range of mouthwatering buffet every day for lunch and dinner.
Location: Lobby Level
Serves: Breakfast, Brunch, Lunch, Dinner, Dessert
Hours of Operation: 6:30 am-11:30 pm | 0.000871 |
WEBSTA - Instagram Web Viewer
Instagramers Global Network
Founder: @relaxocat
Mods: @city_in_a_garden @cmykchicago @elbow_macaroni @mcquake01
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INSTAMEET REMINDER! Sun, July 27 at 4pm for street art in Pilsen! Scroll our feed for details! 15h
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misseswray @nas9 Congratulations!! 13h
jesseratt @pattibrowneyes @pinknuggets17 @rubyred1996 we were probably in this!! #maybenot 11h
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Join us on Sunday, July 27th at 4pm for a colorful, informative instameet! Grab a beverage at Nitecap Coffee Bar (1738 W. 18th St.) & meet us out front. Hope to see you there! 2d
frenchshe82 im.down @ubermercado 15h
ubermercado @frenchshe82 ok see you there 15h
mollypg Hey @edmcdevitt you need to connect with @senor_codo. You guys have similar interests it sounds like! 11h
kendalldurr94 @allisonchesley @great_alejandro have you both seen this? 9h
great_alejandro @kendalldurr94 yes ma'am! Super excited 9h
kendalldurr94 Awesome! @great_alejandro 9h
allisonchesley This is awesome! Thanks @kendalldurr94 9h
relaxocat @allen_was_here Here's the info! :) 8h
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Congratulations to @romancreates for the featured photo! Chosen by @elbow_macaroni. 3d
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romancreates Thanks so much for the feature! I'm so glad you like it and I really appreciate it!! 3d
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elbow_macaroni @romancreates Such a beaut! 3d
luzmionate :) 3d
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Congratulations to @elpaco313 for the featured photo! Visit his feed to read about how he captured all these different firework displays to create one composite photo!
Chosen by @relaxocat 4d
elpaco313 And thank you, @relaxocat! 3d
underclouds @bardbjunior real fireworks 3d
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misseswray @elpaco313 Congratulations!! ️ 3d
sundayrose76 omg WOW!😯️ 3d
relaxocat @elpaco313 You're very welcome! Thank you for sharing with the community! The image is really great! 3d
elpaco313 @misseswray thanks! 3d
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Congratulations to @sammiinniss for this featured photo! Chosen by @cmykchicago.
️#igerschicago️ 5d
sammiinniss @igerschicago thank you thank you! 5d
relaxocat @sammiinniss Congrats on the feature! 5d
sammiinniss Thanks! Love your gallery @relaxocat 5d
scottchicago Excellento! Happy 4th! 4d
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Congratulations to @morganl1122 for this featured photo! Chosen by @relaxocat. #igerschicago #chicago 6d
morganl1122 Thank you SO much @igerschicago 😀 6d
morganl1122 Happy 4th @relaxocat 6d
relaxocat @morganl1122 Happy 4th to you! Thank you so much for sharing with us! 6d
bmerchan @morganl1122️ 6d
morganl1122 Thanks @bmerchan !! 😀 6d
fabu_beau66 Love it... 6d
misseswray @morganl1122 Congratulations!! 5d
morganl1122 Thanks @misseswray and @fabu_beau66 !!! 5d
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Congratulations to @mandelgram for the featured photo! Chosen by @mcquake01 6d
mandelgram @bradgrock hahaha, so true! Thank you for the compliment :) 6d
mandelgram @iamtheozzy I know, right? So excited, friend! 6d
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mandelgram @iloveprintshop thank you! Happy fourth to you as well! 6d
mandelgram @prncsmichelle thanks! I hope the same for you! 6d
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legendarypoet Cool! 6d
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fabu_beau66 Love this museum... 1w
adeelbadeel @qookio I wanna go here 1w
misseswray @katiekoerten Congratulations!! 1w
qookio @adeelbadeel we can go. It's a p cool museum 1w
smgagne Great minds think alike @katiekoerten 1w
simplycate 1w
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chrismphillips Haha. Thanks @bobby.j - @igerschicago - you have an incoming Australian future friend. 1w
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kyleswiat Woahhhhhhh! @pattibrowneyes intense 1w
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jdimage What about this one my friend @donmarito 1w
nkanchwala @kashjam182 this one is crazy! 1w
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#igerschicago 2w
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steve_crandall Congrats Julia! @arlekina 2w
missy120378 That's gorgeous! Great job, Julia! 2w
arlekina How wonderful! Thanks so much @igerschicago @relaxocat for the feature 2w
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Today's feature comes from @nandobastos, shared with #igerschicago, and chosen by @elbow_macaroni! 2w
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anaclau @nandobastos Congrats my dear! Great shot indeed!! ️️ 2w
twistedglimpses Fabulous! 2w
relaxocat @nandobastos Congrats! 2w
nandobastos @anaclau Thank you sweetie!! ️️ 2w
nandobastos @twistedglimpses Thank you Kirsten! 2w
nandobastos @relaxocat Hi Fayth!! Thank you! 2w
chicagorealtor Beautiful! 2w
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simplycate 2w
bradgrock Super great shot @ferdie_so_fly! 2w
ferdie_so_fly @morgansayers @bradgrock thank you very much 2w
itsmuz @ferdie_so_fly killer 2w
ferdie_so_fly @itsmuz thanks bro! 2w
misseswray @ferdie_so_fly Congratulations!! 2w
i.m.jmr Big Congrats @ferdie_so_fly !!! 2w
spoolin_it Good shot! This is an awesome page showing the City of Chicago. Check out some of my pictures I tagged with #igerschicago . I'd also be interested in one of the meets . ️@igerschicago 2w
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Prepare for some cooler weather the next few days! Our forecast shows a drop in temperature, but it warms up for the weekend!
#igerschicago 2w
luciehmacias I'm really digging this @smilingproductions 2w
pattibrowneyes Cool art! What's the location? Is it in the museum campus? 2w
smilingproductions @luciehmacias thank you! 😎 2w
smilingproductions @pattibrowneyes just north of the Field Museum/Shedd next to the bike path 2w
smilingproductions @jenny704 @simplycate thank you! 2w
bradgrock @smilingproductions I can't wait to go see these @hebrubrantley sculptures. Not to mention the exhibit happening at the cultural center. Great capture! 2w
smilingproductions @bradgrock YES! Please do! @hebrubrantley exhibit is beyond words. It had me zoned out for hours. Bru is a good dude. ️ 2w
publicartchi Love this sculpture!! 2w
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Summer officially began over the weekend! How are you spending your summer? Be sure to show us your favorite summertime shots by tagging #igerschicago! 2w
thekathrynwindow @igerschicago awesome! Thanks so much 2w
simplycate 2w
mcquake01 Nice one @oripsolob 2w
oripsolob @mcquake01 Thanks, Joe! 2w
bradgrock Well done @oripsolob! Great lens flair. 2w
oripsolob @bradgrock Thanks! I was riding my bike for Bike the Drive and didn't realize it until after the the event. 2w
jj_chicagoland @oripsolob congratulations! 2w
oripsolob @jj_chicagoland Thanks!! 2w
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#igerschicago 2w
lisahmcconnell Wow, that's a nice photo! 2w
aerojad @orangesparrow love this! 2w
orangesparrow @aerojad thanks!! 2w
juanhernandez Great shot!!! 2w
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bradgrock Totally fantastic shot @orangesparrow! 2w
simplycate Outstanding 2w
orangesparrow Thanks @juanhernandez and @bradgrock! 2w
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#igerschicago 2w
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misseswray @dsh_ Congratulations!! 2w
dsh_ Thank you so much for the feature @igerschicago, and to @elbow_macaroni for choosing it! What an honor and a wonderful way to start the week! Have a great one, everyone! 2w
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dsh_ @misseswray Thank you very much, Mary! ️ 2w
elbow_macaroni @dsh_ Super-duper welcome, loved this take on the botanical challenge! 2w
juanhernandez Beautiful Deb!! @dsh_ 2w
dsh_ @juanhernandez Thanks so much, Juan! 2w
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Congratulations to @juanhernandez for his featured photo which was tagged for our June instameet #igc_afterdark2, and chosen by @cmykchicago! 3w
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juanhernandez Thanks for the feature 😎! Have a great #summersolstice weekend ️️️️️ 3w
the_mollyporter Hahaha @mollypg for a sec I thought that was me 3w
kristy8420 One day I'm gonna make it, I swear. 3w
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mollypg @the_mollyporter nope it's "the other" Molly 3w
bradgrock Congrats @juanhernandez! Neat capture 3w
juanhernandez Thanks Brad!!! @bradgrock 3w
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3w
meli_kolom Prudential building? 3w
cmlinks17 @mattcandersen Congrats Matt 3w
meandering_mari @mattcanderson congrats!! 3w
mattcandersen @meli_kolom yep! 3w
mattcandersen @cmlinks17 thank you!! 3w
mattcandersen @meandering_mari thank you! 3w
misseswray @mattcanderson Congratulations!! 3w
bradgrock Well done @mattcandersen! 3w
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️#igerschicago ️ 3w
juanhernandez Great!!! Congrats Dennis @ddesigns_ 3w
bradgrock @ddesigns_ this is so awesome. Great job, Dennis! 3w
ddesigns_ AH! THANK YOU @igerschicago!!! I wanted to play with some video stuff that night, so glad you enjoyed it! 3w
ddesigns_ Thanks guys!!! @luciehmacias @yinandyoni @iamajenny 3w
ddesigns_ Thank you so much!! @afowler2k 3w
ddesigns_ Thanks C!!! Appreciate it! @cmlinks17 3w
ddesigns_ Thank you @juanhernandez ! 3w
ddesigns_ Thanks Brad! Definitely interested in doing more videos! @bradgrock 3w
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INSTAMEET REMINDER: Join us Sunday, July 27th at 4pm in Pilsen! Scroll our feed for details! 7h | 0.001287 |
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It starts by the restaurant’s front door, where any American will instinctively look for a “Please Wait to Be Seated” sign, but the shot moves too quickly to see much of anything. The camera swoops around like it’s dangling on a yo-yo. Two seconds in, we see the stars and stripes behind the cash register. “Give me, like, a minute,” says our camerawoman to her three partners, who flash by in a headless blur. She’s filming on a cell phone or maybe a flipcam, and she obviously isn’t concerned about visual clarity; she has a mission. “OK, so like wait one minute and then come out, OK?” she repeats, as a couple of stray tables float in, then out, of view. And then she goes to the patio.
We know how this is going to end: The video is titled “Soldier coming home from Iraq surprising his family,” and soon we’re seated with our camerawoman at an outside table where two women and a young man are waiting as the seconds tick by. “So our close family friend’s son came home after spending a year in Iraq,” our camerawoman has written in the YouTube description, “his mom thinks that we are at the restaurant to surprise my brother for his birthday, little did she know the surprise was on her!” The mom is clearly the younger of the two women, wearing a green sleeveless dress and looking anxiously toward the door as a teenage waitress delivers a few cocktails and a round of waters in enormous plastic cups. The camera steadies around 0:54, pointed directly at her, and for the next 35 seconds we wait. She’s waiting too, but she doesn’t know what we know. For 35 seconds, nothing much happens at all. Then her eyes erupt, and she screams for joy.
The soldier-return video1 has become a defining folk-art creation of the War on Terror, a kind of distant, internet-era descendant of the small-town parades that greeted WWII veterans. This war’s survivors aren’t often met by marching bands or floats, but in a different, no less poignant way, their returns are celebrated even more fervently: Plenty of these videos have been viewed by hundreds of thousands of people. Many of them have been viewed by millions.
The truly heart-melting portion of these videos takes place after that initial burst of happiness.
Uploaded on Sept. 8, 2011, by user srmelancon06, “Soldier coming home” is my favorite of the hundreds of soldier-return videos on YouTube because it incorporates all the essential tropes of the genre. It’s one continuous handheld shot, lasting only two minutes and 10 seconds—longish compared to other soldier-returns, but not a moment is wasted. Its setting is both nondescript and perfectly universal—the untouched bowls of chips and salsa are the only things identifying the restaurant as Mexican, and no one has a recognizable accent or even a regionally specific piece of clothing. These videos almost always take place in generic locations: yards, airports, schools, churches, living rooms, or basements. You can watch them for hours and see nary a brand name or noteworthy public place; they seem to exist on some calmer, more reasonable plane of American culture than mass media ever provides.
Every soldier-return video follows the same narrative, and “Soldier coming home” is a master class in this regard, as well. Each video starts with the photographer setting up the shot by focusing on the unsuspecting civilian while the soldier remains hidden. The best videos drag this anticipation out, waiting until the halfway point or later before the greeting takes place. And then we get the goods: The child or sibling or spouse or friend (or dog—this is the internet after all) suddenly sees the returned serviceman and the scene goes from utter banality to hysterical jubilation. A college graduation turns into an explosive display of sibling affection. An unsuspecting woman goes mute in the airport arrivals area. A boy in a buzzcut and gray hoodie walks into his school’s terrace and collapses to the ground, disbelieving what he sees.
But the truly heart-melting portion of these videos takes place after that initial burst of happiness. Like lit magnesium, the reunion explodes before immediately reducing to a simmer. The serviceman (female ones are hard to find, though here’s a great one of sisters reuniting) and his loved one embrace silently while family or coworkers look on, and after a few beats, someone in the scene, often the soldier himself, eases everything down to a lower emotional gear with a funny quip or a pragmatic comment. The award here must go to the young man who asks his sobbing mother, “Ready to go to IHOP?” Talk about a mission.
On the rear patio of our commonplace Mexican restaurant, Mom bolts upright and shrieks a piercing high note twice before exclaiming, “Oh my God!” at the same pitch. She has her hand on her mouth at first, then it flies off to the side as she tiptoes, apparently in high heels. As she rushes toward her off-screen son, she gives the briefest of glances to our camerawoman, and you can almost see the gears turning: She’s putting it all together, how her close friends colluded to make this moment as powerful as possible and tape it for posterity. Maybe she’s one of the 14,000 YouTube subscribers or 18,000 Facebook likes for Welcome Home Blog, “The #1 Site for Videos of Surprise Military Homecomings.” They post a new video daily, and maybe she’s even considered that she might star in one someday.
Whatever the case, she’s overcome. It takes her four seconds to reach her baby boy, who’s grinning the same pleased-with-himself grin that every soldier wears in these scenes. She pulls him in and grabs him by the back of the head and they turn gently 90 degrees, until his back is fully to the camera. When she pulls away from his chest to look at him, she seems to have aged 10 years since she sat in her chair. She’s gone from shrieking adolescent glee to the most profound maternal devotion imaginable. The family and the other diners all fall silent. For 20 seconds all we hear are our camerawoman’s soft weeping and Mom’s thankful, rhythmic cries. Her eyes are clenched as her right hand explores his back, squeezing his T-shirt at 1:42, and if fingers could talk these would be screaming: Don’t leave me, don’t leave me, don’t leave me.
This is how everyone wants to be loved by their mother, which means that at this moment she could be anyone’s mom, that restaurant could be in anyone’s neighborhood. Near midnight on Dec. 15, the day America’s war in Iraq finally, truly ended, I watched this 130-second clip another dozen or so times, and felt like this nameless mother and son were reuniting right outside my window.
Turns out, however, that they were 1,600 miles away, in San Antonio. Early the next morning I sent srmelancon06 a message over YouTube, asking to learn more about her video. I felt compelled to thank her—”I’ve watched it many times and I feel privileged that you’ve shared the scene”—probably because she’s a proxy for that soldier in my mind, and like a lot of upper-middle-class Americans I’m so conflicted and removed from this war that a flummoxed thank-you is the only thing I can think to offer to its participants.
Her name is Stacey, and she’s a senior-year history major at the University of Texas at San Antonio. Certain of my assumptions proved true: She used a basic hand-held camera, a $100 Kodak Playsport, to film her video; Welcome Home Blog is one of her favorite sites (“I of course cry with every video and by the time I am done watching them, I am a complete mess, but it’s a happy cry so I guess that’s why I continue to watch them,” she wrote me); and the surprise was the soldier’s idea. She’s known him for 15 years; he and his mom are “practically family.” They’d joked about doing a soldier-return video ever since he was deployed on Sept. 11, 2010, and kept in touch over Facebook throughout his deployment. As she wrote to me,
I was actually very surprised to find the email from him stating that he was coming home and he wanted to surprise his mom. We were expecting him home during the weekend in September (I forget the date you’ll have to forgive me) but the Thursday before we were expecting him home, he called me that morning and was like, “I am here, can we surprise her tonight?” so our plans quickly changed and to get her to meet us at the restaurant…
Stacey had feared that worried her friend’s mom might not react so positively: “She wanted to drive up to Fort Hood and watch the little ceremony thing that they do whenever the soldiers come home so surprising her meant that she would miss it.” The video was uploaded the night the reunion happened.
I exchanged several long emails with Stacey, sharing my own fascination with soldier-return videos and asking about her family and hometown. I felt entitled to know everything about her. I’d pored over this intensely personal moment that she’d experienced with her close friends, and on some level now felt that it was in my fact my moment, my personal wartime catharsis. Such is the effect of any home video on YouTube, but I’ve never had this sense of ownership and intimacy with, say, “Laughing Baby Ripping Paper.”
After a few rounds of questions and answers, however, I realized Stacey’s generosity with the facts didn’t increase my understanding of “Soldier coming home” or my appreciation for it. The value of any soldier-return video is its anonymity, and additional information can only distract from its closed, tidy perfection. If this particular enlisted man were, say, a valedictorian and an accomplished jazz pianist, or even if he had struggled with drugs or joined the Army to impress his dad, it wouldn’t make the video any better. It would only make “Soldier coming home” a specific person’s story, rather than a distilled symbol for how we all wish the War on Terror could end.
“Elsie, Mich., is just the sort of Middle American town that used to welcome its boys noisily home from the wars,” writes journalist Karl Fleming in a March 1971 Newsweek article about the return of a Vietnam veteran, “The Homecoming of Chris Mead.” Fleming’s subject is a 21-year-old infantryman who had enlisted while still in high school, and Mead’s first postwar days back in Elsie are marked by isolation, aimlessness, and the creeping realization that there aren’t any jobs or respectable girls waiting for him. Mead “took a lot of gunfire,” writes Fleming. “He saw trucks blown up, kids maimed, women killed, buddies bleeding and dying,” and he is now unambiguously opposed to the war. No parade greets him at the bus station; just his younger brother, Greg, who drives Chris back home to a quiet, awkward dinner with their parents and his old room full of Steppenwolf and Beatles records.
Through movies and books, the plight of the Vietnam veteran became common knowledge, yet it was never seriously rectified.
Fleming quotes the official Certificate of Appreciation Mead received from President Nixon: “I extend to you my personal thanks and the sincere appreciation of a grateful nation,” it reads, right above the reproduced executive signature. “You have helped maintain the security of the nation during a critical time in its history.”
An entire generation of pop culture has grown out of the disparity between that painfully overstated letter (its first sentence nearly buckles from gushing adjectives) and the reality that greeted its recipients. Through movies like Coming Home, The Deer Hunter, and Casualties of War, and books like The Things They Carried and Born on the Fourth of July, the plight of the Vietnam veteran became common knowledge, yet it was never seriously rectified. The sudden appearance and eerie ubiquity of “Support the Troops” paraphernalia after 9/11 almost feels like a mass penance for that Vietnam-era failure.
Soldier-return videos reveal both the extent and the limits of that penance. They might be advertised and conceived as a tribute to the war’s survivors—and they certainly are tributes of a kind—but the queasy truth is that the soldiers themselves are the least important people in them. The men and women in fatigues are always less defined than their surprised relatives, and the grateful shrieks of greeting are proof, however momentary, that we civilians haven’t treated them like we treated Chris Mead.
YouTube’s web presence launched on Valentine’s Day 2005, less than five months after the thousandth American had been killed in Iraq. The tally hit 2,000 by Halloween of that year, and 4,483 by the time the last troops came home. Unlike during the Vietnam War, the American media were prohibited from showing many of the coffins coming home from Iraq. But a lack of photographic record doesn’t erase our need to recognize the dead; it just deprives us of an object for grief.
YouTube might be better known as an ocean of memes and copyright infringement, but through the soldier-return genre it has also evolved into a surprisingly effective outlet for all the sadness and ambivalence that have marked the American people’s relationship to the military over the last 10 years. These airport and driveway scenes play like miniature versions of the hysteria that greeted the “Mission Accomplished” banner or Osama bin Laden’s assassination—cries of relief from a society desperate for good news and gluttonous when it arrives. Below “Soldier coming home from Iraq surprising his family,” commenter Zeezilicious writes, “My god, been watching a million of videos like that for the past 2 hours and I still CRY, seriously?! It’s just… so emotional.”
The video ends right as Mom pulls away from her son a third time and stares directly into his eyes. This is the first moment in the whole sequence when she seems relaxed. At the beginning, when she thought she was only waiting to surprise someone else’s son for his birthday, she had the blank expression of a person anxiously chewing through a disposable moment. When her own son first appeared, she became a conflagration. But now she realizes she doesn’t have to grip him so tightly: He’s here to stay. So she loosens up, and her face glows at the sight of him. We see only the back of his crew cut. She touches his head one last time and her face tilts to the left. Her mouth begins to open—she’s about to say her first real words to this boy who by some miracle has escaped becoming coffin no. 4,484. And then the video stops abruptly, as if to keep us from hearing her voice. It’s the only conceivable ending: In silence she speaks for all of us, saying everything and nothing about this awful, exhausting war. | 0.975567 |
\begin{document}
\title[On weak equivalences of gradings]{On weak equivalences of gradings}
\author{Alexey Gordienko}
\address{Vrije Universiteit Brussel, Belgium}
\email{alexey.gordienko@vub.ac.be}
\author{Ofir Schnabel}
\address{University of Haifa, Israel}
\email{os2519@yahoo.com}
\keywords{Associative algebra, grading, full matrix algebra, residually finite group, adjoint functor, oplax $2$-functor.}
\begin{abstract}
When one studies the structure (e.g. graded ideals, graded subspaces, radicals,\dots) or graded polynomial identities of graded algebras, the grading group itself does not play an important role, but can be replaced by any other group that realizes the same grading. Here we come to the notion of weak equivalence of gradings: two gradings are weakly equivalent if there exists an isomorphism between the graded algebras that maps each graded component onto a graded component.
The following question arises naturally: when a group grading on a finite dimensional algebra is weakly equivalent to a grading by a finite group?
It turns out that this question can be reformulated purely group theoretically in terms of the universal group of the grading.
Namely, a grading is weakly equivalent to a grading by a finite group if and only if the universal group of the grading is residually finite with respect to a special subset of the grading group. The same is true for all the coarsenings of the grading if
and only if the universal group of the grading is hereditarilly residually finite with respect to the same subset.
We show that if $n\geqslant 353$, then on the full matrix algebra $M_n(F)$ there exists an elementary group grading that is not weakly equivalent to any grading by a finite (semi)group, and if $n\leqslant 3$,
then any elementary grading on $M_n(F)$ is weakly equivalent to an elementary grading by a finite group. In addition, we study categories and functors related to the notion of weak equivalence of gradings.
In particular, we introduce an oplax 2-functor that assigns to each grading its support and show that the universal grading group functor has neither left nor right adjoint.
\end{abstract}
\subjclass[2010]{Primary 16W50; Secondary 18A40, 18D20, 20E26.}
\thanks{The first author is supported by Fonds Wetenschappelijk Onderzoek~--- Vlaanderen post doctoral fellowship (Belgium). The second author was partially supported by ISF grant 797/14.}
\maketitle
\section{Introduction}
When studying graded algebras, one has to determine, when two
graded algebras can be considered ``the same'' or equivalent.
Recall that $\Gamma \colon A=\bigoplus_{s \in S} A^{(s)}$
is a \textit{grading} on an algebra $A$ over a field $F$ by a (semi)group $S$
if $A^{(s)}A^{(t)}\subseteq A^{(st)}$ for all $s,t\in S$.
Then we say that $S$ is the \textit{grading (semi)group} of $\Gamma$ and the algebra $A$ is \textit{graded} by $S$.
Let \begin{equation}\label{EqTwoSemiGroupGradings}\Gamma_1 \colon A=\bigoplus_{s \in S} A^{(s)},\qquad \Gamma_2
\colon B=\bigoplus_{t \in T} B^{(t)}\end{equation} be two gradings where $S$
and $T$ are (semi)groups and $A$ and $B$ are algebras.
The most restrictive case is when we require that both grading (semi)groups
coincide:
\begin{definition}[{e.g. \cite[Definition~1.15]{ElduqueKochetov}}]
\label{DefGradedIsomorphism}
The gradings~(\ref{EqTwoSemiGroupGradings}) are \textit{isomorphic} if $S=T$ and there exists an isomorphism $\varphi \colon A \mathrel{\widetilde\to} B$
of algebras such that $\varphi(A^{(s)})=B^{(s)}$
for all $s\in S$.
In this case we say that $A$ and $B$ are \textit{graded isomorphic}.
\end{definition}
In some cases, such as in~\cite{ginosargradings}, less restrictive requirements are more
suitable.
\begin{definition}[{\cite[Definition~2.3]{ginosargradings}}]\label{DefGinosarEquivalent}
The gradings~(\ref{EqTwoSemiGroupGradings}) are \textit{equivalent} if there exists an isomorphism
$\varphi \colon A \mathrel{\widetilde\to} B$
of algebras and an isomorphism $\psi \colon S \mathrel{\widetilde\to} T$
of (semi)groups such that $\varphi(A^{(s)})=B^{\bigl(\psi(s)\bigr)}$
for all $s\in S$.
\end{definition}
\begin{remark}
The notion of graded equivalence was considered by Yu.\,A.~Bahturin, S.\,K.~Seghal, and M.\,V.~Zaicev in~\cite[Remark after Definition 3]{BahturinZaicevSeghalGroupGrAssoc}.
In the paper of V.~Mazorchuk and K.~Zhao~\cite{Mazorchuk} it appears under the name of graded isomorphism.
A.~Elduque and M.\,V.~Kochetov refer to this notion as a weak isomorphism
of gradings~\cite[Section~3.1]{ElduqueKochetov}.
More on differences in the terminology in
graded algebras can be found in \cite[\S 2.7]{ginosargradings}.
\end{remark}
If one studies the graded structure of a graded algebra or its graded polynomial identities~\cite{AljaGia, AljaGiaLa,
BahtZaiGradedExp, GiaLa, ASGordienko9}, then it is not really important by elements of which (semi)group the graded components are indexed.
A replacement of the grading (semi)group leaves both graded subspaces and graded ideals graded.
In the case of graded polynomial identities reindexing
the graded components leads only to renaming the variables.
(It is important to notice however that graded-simple algebras
graded by semigroups which are not groups can have a structure quite different from group graded graded-simple algebras~\cite{GordienkoJanssensJespers}.)
Here we come naturally to the notion of weak equivalence of gradings.
\begin{definition}\label{def:weakly}
The gradings~(\ref{EqTwoSemiGroupGradings}) are \textit{weakly equivalent}, if there
exists an isomorphism $\varphi \colon A \mathrel{\widetilde\to}B$
of algebras such that for every $s\in S$ with $A^{(s)}\ne 0$ there
exists $t\in T$ such that $\varphi\left(A^{(s)}\right)=B^{(t)}$.
\end{definition}
\begin{remark}
This notion appears in~\cite[Definition~1.14]{ElduqueKochetov}
under the name of equivalence.
We have decided to add here the adjective ``weak'' in order to avoid confusion
with Definition~\ref{DefGinosarEquivalent}.
\end{remark}
Obviously, if gradings are isomorphic, then they are equivalent and if they are equivalent then they are also weakly equivalent.
It is important to notice that none of the converse is true.
However, if gradings~(\ref{EqTwoSemiGroupGradings}) are weakly equivalent
and $\varphi \colon A \mathrel{\widetilde\to}B$ is the corresponding isomorphism of algebras,
then $\Gamma_3 \colon A=\bigoplus_{t \in T} \varphi^{-1}\left( B^{(t)}\right)$ is a grading on $A$ isomorphic to $\Gamma_2$ and the grading $\Gamma_3$ is obtained from $\Gamma_1$ just by reindexing the homogeneous components.
Therefore, when gradings~(\ref{EqTwoSemiGroupGradings}) are weakly
equivalent, we say that $\Gamma_1$ \textit{can be regraded} by $T$.
If $A=B$ and $\varphi$ in Definition~\ref{def:weakly} is the identity map, we say that $\Gamma_1$ and $\Gamma_2$ are realizations
of the same grading on $A$ as, respectively, an $S$- and a $T$-grading.
For a grading $\Gamma \colon A=\bigoplus_{s \in S} A^{(s)}$, we denote by $\supp \Gamma := \lbrace s\in S \mid A^{(s)}\ne 0\rbrace$ its support.
Note that $\Gamma$ is obviously equivalent to $\Gamma_0 \colon A=\bigoplus_{s \in S_1} A^{(s)}$
where $S_1$ is a subsemigroup of $S$ generated by $\supp \Gamma$.
(If $S$ is a group, we can consider instead the subgroup generated by $\supp \Gamma$.)
\begin{remark}Each weak equivalence between gradings $\Gamma_1$ and $\Gamma_2$
induces a bijection $\supp \Gamma_1 \mathrel{\widetilde\to} \supp \Gamma_2$.
\end{remark}
As we have already mentioned above, for many applications it is not important
which particular grading among weakly equivalent ones
we consider. Thus, if it is possible, one can try to regrade a semigroup grading by a group or even a finite group. The situation, when the latter
is possible, is very convenient since the algebra graded by a finite group $G$
is an $FG$-comodule algebra and, in turn, an $(FG)^*$-module algebra
where $FG$ is the group algebra of $G$, which a Hopf algebra, and $(FG)^*$ is its dual.
In this case one can use the techniques of Hopf algebra actions instead of working with a grading directly (see e.g.~\cite{ASGordienko3}).
Therefore, the following question arises naturally:
\begin{question}
Is it possible to regrade any grading of a finite dimensional
algebra by a finite group ?
\end{question}
It is fairly easy to show (see Proposition~\ref{prop:abelian} below) that each grading of a finite dimensional algebra by an abelian group $G$ is weakly equivalent to a grading by a finite group.
In fact, the class of groups $G$ with this property is much broader
and includes at least all locally residually finite groups (see~\cite[Proposition 1.2]{DasNasDelRioVanOyst} and Theorem~\ref{TheoremFinGradingsToGroupsTransition} below). Recall that a group is \textit{residually finite} if the intersection of its normal subgroups of finite index is trivial and \textit{locally residually finite} if every its finitely generated
subgroup is residually finite.
In 1996 M.\,V.~Clase, E.~Jespers, and \'A.~Del R\'\i o~\cite[Example~2]{ClaseJespersDelRio} (see also~\cite[Example 1.5]{DasNasDelRioVanOyst}) gave an example of a group graded ring with finite support that cannot be regraded by a finite (semi)group. Despite the fact that they constructed a ring, not an algebra, it is obvious how to make an analogous example of an algebra over a field. However all these examples would have non-trivial nilpotent ideals. Until now it was unclear whether a finite dimensional semi-simple algebra could have a grading
which cannot be regraded by a finite group.
In Theorem~\ref{TheoremFiniteRegradingImpossible} below we show that
there exist even elementary gradings (see Definition~\ref{def:elementary}) on the full matrix algebras $M_n(F)$ (where $F$ is a field) that are not weakly equivalent to gradings by finite groups.
This suggests the following problem:
\begin{problem}\label{ProblemMinNElAGradNonFin} Determine
the set $\Omega$ of
the numbers $n\in\mathbb N$ such that any elementary grading on $M_n(F)$ can be regraded by a finite group.
\end{problem}
In Section~\ref{SectionRegradingMatrixFinite} we prove the following theorem:
\begin{theorem}\label{TheoremRegradeElementaryOmega}
The set $\Omega$ defined in Problem~\ref{ProblemMinNElAGradNonFin}
is of the form $\lbrace n \in \mathbb N \mid 1 \leqslant n \leqslant n_0\rbrace$ for some $3\leqslant n_0 \leqslant 352$.
In particular, for every $n \geqslant 353$ there exists an elementary grading on $M_n(F)$ that cannot be regraded by any finite group.
\end{theorem}
In Section~\ref{SectionWeakEquivalenceGrSimpleAlg} we provide a criterion for two group gradings on graded-simple algebras to be weakly equivalent
and give an example of two twisted group algebras of the same group
which are isomorphic as algebras, but whose standard gradings
are not weakly equivalent.
In Section~\ref{SectionGroupTheoreticalApproach} we recall the definition of the universal group of a grading introduced
in 1989 by J.~Patera and H.~Zassenhaus~\cite{PZ89} and prove that any finitely presented group $G$ can be a universal group of an elementary grading $\Gamma$ on a full matrix algebra and moreover any finite subset of $G$ can be included in $\supp \Gamma$ (Theorem~\ref{TheoremGivenFinPresGroupExistence}). This result is used in the proof of Theorem~\ref{TheoremRegradeElementaryOmega}. We conclude the section showing that the question whether a given grading is regradable by a finite group and, in particular, Problem~\ref{ProblemMinNElAGradNonFin} can be reformulated in a purely group theoretical way (see Theorem~\ref{TheoremFinGradingsToGroupsTransition} and Problem~\ref{ProblemMinFnHRFRx1x2x3}).
In Section~\ref{SectionRegradingMatrixFinite} we prove Corollary~\ref{CorollaryFiniteRegradingImpossible353}, where we show that for every $n \geqslant 353$ there exists an elementary grading on $M_n(F)$ that cannot be regraded by any finite group, and Theorem~\ref{TheoremFiniteRegradingSmallMatrix}, where we show that
if $n\leqslant 3$, then any elementary grading on $M_n(F)$ is weakly equivalent to an elementary grading by a finite group. Together this proves Theorem~\ref{TheoremRegradeElementaryOmega}.
Categories and functors related to graded algebras have been studied extensively (see e.g. \cite{NastasescuVanOyst}) and several important pairs of adjoint functors have been noticed (see Section~\ref{SubsectionGradedAdjunctions} for a brief summary).
Each category consisted of algebras graded by a fixed group, i.e. isomorphisms in those categories coincided with isomorphisms of gradings. In order to obtain the proper categorical framework for weak equivalences, one has to deal with the category of algebras graded by any groups where the morphisms are all homomorphisms of algebras that map each graded component into some graded component. Here the induced partial maps
on the grading groups come into play naturally. In Section~\ref{SectionCategoriesWeakEquivalence}
we show that the assignment of the support to a grading leads to an oplax $2$-functor. In order to get an ordinary functor $R$ from the category of graded algebras to the category of groups which assigns
to each grading its universal group, we restrict the sets of morphisms in the category of graded algebras to the sets of graded injective homomorphisms. We discuss the category obtained and show that the functor $R$ has neither left nor right adjoints (Propositions~\ref{PropositionAbsenceOfAdjointsUGGF} and \ref{PropositionAbsenceOfAdjointsUGGF1}) and in order to force it to have a left adjoint we have to restrict our consideration very much, e.g. to group algebras of groups that do not have non-trivial one dimensional representations. In that case we even get an isomorphism of categories (Proposition~\ref{PropositionEquivGroupAlgGroups}).
\section{Preliminaries}
\label{SectionGrSimpleAlg}
In this section we recall some basic facts about graded algebras.
Let $S$ be a semigroup and let $\Gamma \colon A=\bigoplus_{s\in S} A^{(s)}$ be a grading. The subspaces $A^{(s)}$ are called \textit{homogeneous components} of $\Gamma$
and nonzero elements of $A^{(s)}$ are called \textit{homogeneous} with respect to~$\Gamma$. A subspace $V$ of $A$ is \textit{graded} if $V=\bigoplus_{t\in S} (V \cap A^{(s)})$.
If $A$ does not contain graded two-sided ideals, i.e. two-sided ideals that are graded subspaces,
then $A$ is called \textit{graded-simple}.
\subsection{Lower dimensional group cohomology and graded division algebras}
\label{SubsectionCohomologyGradedDivisionAlgebras}
By the Bahturin~--- Seghal~--- Zaicev Theorem (Theorem~\ref{TheoremBahturinZaicevSeghal} below)
twisted group algebras, which are graded division algebras, play a crucial role in the classification of graded-simple algebras.
Let $A=\bigoplus_{g\in G} A^{(g)}$ be a $G$-graded algebra for some group $G$.
If for every $g\in G$ all nonzero elements of $A^{(g)}$ are invertible, then $A$ is called a \textit{graded division algebra}.
Finite dimensional $G$-graded division algebras over an algebraically closed field are described by the elements of the second cohomology groups of finite subgroups $H \subseteq G$ with coefficients in the multiplicative group of the base field.
Let $G$ be a group and let $F$ be a field. Denote by $F^{\times}$ the multiplicative group of $F$. Throughout the article we consider only trivial group actions on $F^{\times}$. In this case the \textit{first cohomology
group} $H^1(G, F^{\times})$ is isomorphic to the group $Z^1(G, F^{\times})$
of \textit{$1$-cocycles} which in turn coincides with the group $\Hom(G,F^\times)$ of group homomorphisms $G \to F^\times$ with the pointwise multiplication.
Recall that a function $\sigma\colon G \times G \to F^{\times}$ is a \textit{$2$-cocycle} if $\sigma(u,v)\sigma(uv,w)=\sigma(u,vw)\sigma(v,w)$ for all $u,v,w\in G$. The set $Z^2(G, F^{\times})$ of $2$-cocycles is an abelian group with respect to the pointwise multiplication.
The subgroup $B^2(G, F^{\times}) \subseteq Z^2(G, F^{\times})$
of \textit{$2$-coboundaries} consists of all $2$-cocycles $\sigma$
for which there exists a map $\tau \colon G \to F^\times$
such that we have $\sigma(g,h)=\tau(g)\tau(h)\tau(gh)^{-1}$
for all $g,h\in G$.
The factor group $H^2(G,F^\times) := Z^2(G, F^{\times})/B^2(G, F^{\times})$
is called the \textit{second cohomology group of $G$ with coefficients in $F^\times$}.
Denote by $[\sigma]$ the cohomology class of $\sigma \in Z^2(G, F^{\times})$
in $H^2(G,F^\times)$.
Let $\sigma \in Z^2(G, F^{\times})$.
The \textit{twisted group algebra} $F^\sigma G$ is the associative algebra
with the formal basis $(u_g)_{g\in G}$ and the multiplication $u_g u_h = \sigma(g,h)u_{gh}$
for all $g,h \in G$.
For \textit{trivial} $\sigma$, i.e. when $\sigma(g,h)=1$ for all $g,h\in G$, the twisted group algebra $F^\sigma G$ is the ordinary \textit{group algebra} $FG$. Each twisted group algebra $F^\sigma G$ has the \textit{standard grading} $F^{\sigma} G=\bigoplus_{g\in G}F^{\sigma} G^{(g)}$ where $F^{\sigma} G^{(g)}=Fu_g$. Two twisted group algebras $F^{\sigma_1} G$ and $F^{\sigma_2} G$
are graded isomorphic if and only if $[\sigma_1]=[\sigma_2]$. (See e.g.~\cite[Theorem 2.13]{ElduqueKochetov}.)
Note that twisted group algebras are graded division algebras. In fact,
the component $B^{(1_G)}$ of an arbitrary graded division algebra $B$, that corresponds to the neutral element $1_G$ of the grading group $G$, is an ordinary division algebra. Thus, if the base field $F$ is algebraically
closed and $\dim B$ is finite, we have $B^{(1_G)} = F1_B$ and $B \cong F^\sigma H$ for some finite subgroup $H \subseteq G$ and a $2$-cocycle $\sigma \in Z^2(H, F^{\times})$. (See~\cite[Theorem 2.13]{ElduqueKochetov} for the details.)
Suppose there exists
a homomorphism $\varphi \colon F^\sigma G
\to F$ of unital algebras. Then $$\varphi(u_g)\varphi(u_h)=\varphi(u_g u_h)
=\sigma(g,h)\varphi(u_{gh})$$ and
$\sigma(g,h)=\varphi(u_g)\varphi(u_h)\varphi(u_{gh})^{-1}$ for all $g,h\in G$, i.e. $\sigma$ is cohomologous to the trivial $2$-cocycle.
Consequently, if $[\sigma]$ is non-trivial, then $F^\sigma G$ does not have
one dimensional unital modules.
Recall that if $G$ is finite and $\ch F \nmid |G|$, then $F^\sigma G$ is semisimple. (The proof is completely analogous to the case of an ordinary group algebra, see e.g.~\cite[Theorem~1.4.1]{Herstein}.) Therefore, if $G$ is finite, $\ch F \nmid |G|$, and the field $F$ is algebraically closed,
the Artin--Wedderburn Theorem implies that $F^\sigma G$ is
isomorphic to the direct sum of full matrix algebras $M_k(F)$.
In the case $[\sigma]$ is non-trivial, the observation in the previous paragraph shows
that $k\geqslant 2$ for all $M_k(F)$.
Unlike ordinary group algebras $FG$ of non-trivial groups, twisted group algebras $F^\sigma G$
can be simple. (See e.g.~\cite[Theorem 2.15]{ElduqueKochetov}.)
Let $G$ be an abelian group and $\sigma\in Z^2(G, F^{\times})$.
Then in $F^\sigma G$ we have $u_g u_h = \beta(g,h) u_h u_g$
where $\beta(g,h) := \sigma(g,h)\sigma(h,g)^{-1}$, $g,h \in G$,
is the alternating bicharacter corresponding
to $\sigma$. Recall that a function $\beta \colon G\times G \to F^\times$
is an \textit{alternating bicharacter} if it is multiplicative in each variable
and $\beta(g,g)=1_F$ for all $g\in G$.
It is easy to see that $\beta$ depends only on the cohomology class $[\sigma]\in H^2(G, F^{\times})$
and not on the particular $2$-cocycle $\sigma$.
If $G$ is finitely generated, then $G \cong (\mathbb Z/n_1 \mathbb Z) \times (\mathbb Z/n_2 \mathbb Z)\times \dots \times (\mathbb Z/n_m \mathbb Z)$ for some $m, n_i\in\mathbb Z_+$.
In this case, in order to define an alternating bicharacter $\beta \colon G\times G \to F^\times$, it is necessary and sufficient to define the values $\beta(g_i,g_j)$
where $$\beta(g_i,g_j)^{n_i}=\beta(g_i,g_j)^{n_j}=\beta(g_i,g_j)\beta(g_j,g_i)=1$$
for all $1\leqslant i,j \leqslant m$
and $g_i$ are generators of the cyclic components of $G$.
Given an alternating bicharacter $\beta \colon G\times G \to F^\times$, it is easy to define an algebra which is graded isomorphic to a twisted group algebra $F^\sigma G$ with $[\sigma]$
corresponding to $\beta$:
$$u_{g_1^{k_1}\cdots g_m^{k_m}} u_{g_1^{\ell_1}\cdots g_m^{\ell_m}}
= \left(\prod_{1\leqslant i < j \leqslant m} \beta(g_j,g_i)^{k_j \ell_i}\right)u_{g_1^{k_1+\ell_1}\cdots g_m^{k_m+\ell_m}}.$$
Similar arguments show that if $\sigma_1,\sigma_2\in Z^2(G, F^{\times})$
have equal alternating bicharacters, then $F^{\sigma_1}G$
and $F^{\sigma_2}G$ are graded isomorphic, and $[\sigma_1]=[\sigma_2]$.
\subsection{Elementary gradings and classification of graded-simple algebras}
\label{SubsectionElementaryGradedSimpleAlgebras}
\begin{definition}\label{def:elementary}
Let $F$ be a field, $G$ be a group, let $n\in\mathbb N$, and let $(g_1, \dots, g_n)$ be an $n$-tuple of elements
of $G$. Define a grading on $M_n(F)$ by making each matrix unit $e_{ij}$ a $g_i g_j^{-1}$-homogeneous element. This grading is called the \textit{elementary $G$-grading} defined by $(g_1, \dots, g_n)$.
\end{definition}
\begin{remark}\label{RemarkElementary}
Note that such a grading is uniquely determined by defining the $G$-degrees of $e_{i,i+1}$, $1\leqslant i \leqslant n-1$. If $G$ is an arbitrary group and $(h_1, \dots, h_{n-1})$ is an arbitrary $(n-1)$-tuple of elements of $G$, then the elementary grading with $e_{i,i+1} \in M_n(F)^{(h_i)}$ can be defined by $(g_1, \dots, g_n)$ where $g_i = \prod_{j=i}^{n-1} h_j$.
\end{remark}
Let $n\in\mathbb N$, let $G$ be a group, let $\gamma=(g_1, \dots, g_n)$ where $g_i \in G$, let $H \subseteq G$ be a finite subgroup, and let $\sigma \in Z^2(H, F^{\times})$.
Denote by $M(\gamma, \sigma)$ the algebra $M_n(F)\otimes_F F^\sigma H$
endowed with the grading where $e_{ij}\otimes u_h$ belongs to the
$g_i h g_j^{-1}$-component.
Recall
the following classification result:
\begin{theorem}[Bahturin~--- Seghal~--- Zaicev, see e.g.~{\cite[Theorem~3]{BahturinZaicevSeghalSimpleGraded} or~\cite[Corollary 2.22]{ElduqueKochetov}}]
\label{TheoremBahturinZaicevSeghal} Let $A$ be a finite dimensional graded-simple $G$-graded algebra
over an algebraically closed field $F$ where $G$ is a group.
Then $A$ is graded isomorphic to $M(\gamma, \sigma)$ for some
$n\in\mathbb N$, $\gamma=(g_1, \dots, g_n)$ where $g_i \in G$, a finite subgroup $H \subseteq G$,
and a $2$-cocycle $\sigma \in Z^2(H, F^{\times})$.
\end{theorem}
A criterion for two such gradings to be isomorphic can be found, e.g., in~\cite[Lemma 1.3, Proposition 3.1]{AljaHaile} or~\cite[Corollary 2.22]{ElduqueKochetov}.
Necessary and sufficient conditions for two graded-simple algebras to be graded equivalent were proven in~\cite[Theorem 2.20]{ginosargradings}. In the next section we study weak equivalences
of gradings on two graded-simple algebras.
\section{Weak equivalences of graded-simple algebras}
\label{SectionWeakEquivalenceGrSimpleAlg}
In this section we prove a criterion for a weak equivalence of graded-simple algebras
inspired by~\cite[Proposition~2.33]{ElduqueKochetov},
we present families of gradings for which the notions of equivalence and weak equivalence coincide,
and give an example of two twisted group algebras of the same abelian group that are isomorphic as ordinary algebras, but not
graded weakly equivalent.
Let $A=\bigoplus_{g\in G} A^{(g)}$ be an algebra graded by a group $G$.
A vector space $W=\bigoplus_{g\in G} W^{(g)}$ is a \textit{graded left $A$-module} if
$W$ is a left $A$-module and for each $g,h\in G$
we have $A^{(g)}W^{(h)} \subseteq W^{(gh)}$. Graded right modules are defined analogously.
Fix some $n\in\mathbb N$, a group $G$, an $n$-tuple $\gamma=(g_1, \dots, g_n)$ where $g_i \in G$, a finite subgroup $H \subseteq G$, and $\sigma \in Z^2(H, F^{\times})$.
Consider the $G$-graded vector space $V$ with the basis $v_{ih}$, $1\leqslant i \leqslant n$, $h\in H$, such that $v_{ih}$ is a homogeneous element of degree $g_i h$.
Then $V$ is a graded left $M(\gamma, \sigma)$-module
and a graded right
$F^{\sigma}H$-module
with $(e_{jk}\otimes u_g) v_{ih} := \delta_{ki} \sigma(g,h) v_{j,gh}$
and $v_{ih} u_g := \sigma(h,g) v_{i,hg}$
for all $1\leqslant i,j, k \leqslant n$ and $g,h \in H$
where $\delta_{ij}=\left\lbrace\begin{array}{rrr} 0 & \text{if} & i\ne j,\\ 1 & \text{if} & i=j \end{array}\right.$ and $(u_g)_{g\in G}$ is the standard basis in $F^{\sigma}H$.
Note that $(av)u = a (v u)$ for all $a\in M(\gamma, \sigma)$, $v\in V$, and $u\in F^{\sigma}H$.
\begin{lemma}\label{LemmaVIrrLeftModule}
$V$ is an irreducible left $M(\gamma, \sigma)$-module.
\end{lemma}
\begin{proof}
Suppose $W$ is a non-trivial graded $M(\gamma, \sigma)$-submodule of $V$.
Take non-zero $v=\bigoplus_{i,h} \alpha_{ih} v_{ih} \in W$. If $\alpha_{ih_0}\ne 0$
for some $1\leqslant i \leqslant n$ and $h_0\in H_i$, then
$e_{ii}\otimes u_{1_H} v= \bigoplus_{h \in H} \alpha_{ih} v_{ih} \in W$
is again non-zero. Since $W$ is a graded subspace and for fixed $i$ and different $h$
the elements $v_{ih}$ belong to different graded components, we get
$\alpha_{i h_0} v_{i h_0} \in W$ and $v_{i h_0} \in W$. Hence
$v_{jh}=(e_{ji}\otimes u_{hh_{0}^{-1}})v_{ih_0} \in W$ for all $1\leqslant j \leqslant n$, $h\in H$.
Therefore $W=V$.
\end{proof}
For a graded vector space $W=\bigoplus_{g\in G} W^{(g)}$ and an element $h\in G$
denote by $W^{[h]}$ the same vector space endowed with the grading
$W=\bigoplus_{g\in G} \tilde W^{(g)}$ where $\tilde W^{(g)} := W^{(gh^{-1})}$.
Note that $M(\gamma, \sigma)$ is the direct sum of graded left ideals
$I_j=\bigoplus_{\substack{h\in H, \\ 1\leqslant i \leqslant n}} F (e_{ij}\otimes u_h)$,
$1\leqslant j \leqslant n$, and each $I_j$ is isomorphic to $V^{[g_j^{-1}]}$
as a graded left $M(\gamma, \sigma)$-module via $e_{ij}\otimes h \mapsto v_{ih}$,
$1\leqslant i \leqslant n$, $h\in H$.
Recall that if $\alpha \colon H_1 \to H_2$
is a homomorphism of groups and $\sigma \in Z^2(H_2, F^{\times})$,
then the function $\sigma(\alpha(g), \alpha(h))$ of arguments $g,h\in H_1$
is a $2$-cocycle on $H_1$. We denote the cohomology class of this $2$-cocycle by
$[\sigma(\alpha(\cdot),\alpha(\cdot))]$.
Let $G_1,G_2$ be groups and let $n_1,n_2\in\mathbb N$. Fix tuples $\gamma_i=(g_{i1}, \dots, g_{in_i})$
where $g_{ij} \in G_i$, finite subgroups $H_i \subseteq G_i$, and $2$-cocycles
$\sigma_i \in Z^2(H_i, F^{\times})$, $i=1,2$.
We say that the gradings on $M(\gamma_1, \sigma_1)$ and $M(\gamma_2, \sigma_2)$
satisfy Condition (*) if $n_1=n_2$, there exist a group isomorphism $\alpha \colon H_1 \mathrel{\widetilde{\to}} H_2$,
a permutation $\pi \in S_{n_1}$,
and elements $t_i \in H_2$, $1\leqslant i \leqslant n_1$,
such that $[\sigma_1] = [\sigma_2(\alpha(\cdot),\alpha(\cdot))]$
and the following condition holds:
for every $1\leqslant i,j,k,\ell \leqslant n_1$ and $h_1,h_2\in H_1$
we have $$g_{2,\pi(i)}t_i\alpha(h_1)t_j^{-1}g_{2,\pi(j)}^{-1} = g_{2,\pi(k)} t_k \alpha(h_2) t_\ell^{-1} g_{2,\pi(\ell)}^{-1}$$
if and only if
$$g_{1i}h_1 g_{1j}^{-1} = g_{1k} h_2 g_{1\ell}^{-1}.$$
\begin{theorem}\label{TheoremWeakEquivGradedSimple}
The gradings on $M(\gamma_1, \sigma_1)$ and $M(\gamma_2, \sigma_2)$
are weakly equivalent if and only they satisfy Condition (*).
If $M(\gamma_1, \sigma_1)$ and $M(\gamma_2, \sigma_2)$ satisfy Condition (*), the algebra isomorphism
$\varphi \colon M(\gamma_1, \sigma_1) \mathrel{\widetilde{\to}} M(\gamma_2, \sigma_2)$
implementing this weak equivalence
can be defined e.g. by $\varphi(e_{ij}\otimes x_h) = e_{\pi(i),\pi(j)}\otimes y_{t_i\alpha(h)t_j^{-1}}$
where $(x_h)_{h\in H_1}$ is the formal basis in $F^{\sigma_1} H_1$ and $(y_t)_{t\in H_2}$ is the formal basis in $F^{\sigma_2} H_2$.
\end{theorem}
\begin{proof} Suppose the gradings on $M(\gamma_1, \sigma_1)$ and $M(\gamma_2, \sigma_2)$
are weakly equivalent.
Denote by $\varphi$ an isomorphism of algebras $M(\gamma_1, \sigma_1)\mathrel{\widetilde{\to}} M(\gamma_2, \sigma_2)$ that corresponds to the weak equivalence.
Construct the graded $M(\gamma_i, \sigma_i)$-modules $V_i$, $i=1,2$, as above.
These modules are irreducible by Lemma~\ref{LemmaVIrrLeftModule}.
Note that $M(\gamma_1, \sigma_1)$ is acting on $V_2$ via $a \cdot v= \varphi(a)v$ for $a\in M(\gamma_1, \sigma_1)$ and $v\in V_2$, and $V_2$ does not contain any non-trivial $G_2$-graded submodules. Define for $M(\gamma_1, \sigma_1)$ the minimal graded left ideals $I_j$ as above.
Then for each $j$ either $I_j \cdot V_2 = 0$ or $I_j \cdot V_2 = V_2$. Since $M(\gamma_1, \sigma_1) \cdot V \ne 0$, we obtain that $V_2=I_j\cdot v$ for some homogeneous $v\in V_2$ and some $j$
and $V_2$ is isomorphic to $I_j$ as a left $M(\gamma_1, \sigma_1)$-module. Moreover, this isomorphism maps each nonzero $G_1$-graded component onto some
$G_2$-graded component. In fact, since $I_j \cong V_1^{[g_j^{-1}]}$,
there exists a linear isomorphism $\psi \colon V_1 \mathrel{\widetilde{\to}} V_2$
such that $\varphi(a)\psi(v)=\psi(av)$ for all $a\in M(\gamma_1, \sigma_1)$
and $v\in V_1$ and for each $g\in G_1$ with $V_1^{(g)}\ne 0$ there exists $\rho(g)\in G_2$
such that $\psi\left(V_1^{(g)}\right)=V_2^{(\rho(g))}$.
It is not difficult to check that $F^{\sigma_i} H_i$
is isomorphic to $\End_{M(\gamma_i, \sigma_i)} V_i$ as an algebra through its action on $V_i$ from the right. Hence there exists an algebra isomorphism $\tau \colon F^{\sigma_1} H_1 \mathrel{\widetilde{\to}} F^{\sigma_2} H_2$ such that $\psi(v)\tau(u)=\psi(vu)$ where $v\in V_1$ and $u\in F^{\sigma_1} H_1$.
If $V_2^{(g)} \ne 0$
and $h\in H_1$, then \begin{equation}\label{EqWeakGrSmpl1}V_2^{(g)}\tau(x_h)=\psi\left(V_1^{\left(\rho^{-1}(g)\right)}\right)\tau(x_h)=
\psi\left(V_2^{\left(\rho^{-1}(g)\right)}x_h\right)=\psi\left(V_1^{\left(\rho^{-1}(g)h\right)}\right)=
V_2^{\left(\rho\left(\rho^{-1}(g)h\right)\right)}.\end{equation}
However, $\tau(x_h)=\sum_{t\in H_2} \alpha_t y_t$ for some $\alpha_t\in F$.
Since the sum $\bigoplus_{t\in H_2} V_2^{(g)}y_t = \bigoplus_{t\in H_2} V_2^{(gt)}$
is direct, \begin{equation}\label{EqWeakGrSmpl2}\tau(x_h)=\lambda(h) y_{\alpha(h)}\end{equation} for some $\lambda(h) \in F^{\times}$
and $\alpha(h) \in H_2$. In particular, $\alpha$ is a group isomorphism.
Now \begin{equation*}\begin{split}\sigma_1(h_1, h_2)\lambda(h_1h_2)y_{\alpha(h_1 h_2)}
=\sigma_1(h_1, h_2)\tau(x_{h_1 h_2})=\\=\tau(x_{h_1})\tau(x_{h_2})=\sigma_2(\alpha(h_1),\alpha(h_2))
\lambda(h_1)\lambda(h_2)y_{\alpha(h_1) \alpha(h_2)},\end{split}\end{equation*}
implies $[\sigma_1] = [\sigma_2(\alpha(\cdot),\alpha(\cdot))]$.
Note that $\dim V_1 = n_1 |H_1|=\dim V_2 = n_1 |H_2|$. Now $H_1 \cong H_2$
implies $n_1 = n_2$.
Equalities~(\ref{EqWeakGrSmpl1}) and~(\ref{EqWeakGrSmpl2}) imply
$\rho\left(gh\right) = \rho(g)\alpha(h)$ for all $g\in G_1$ with $V_1^{(g)}\ne 0$
and all $h\in H_1$.
Since $\rho$ is a bijection between $\bigcup_{i=1}^{n_1} g_{1i}H_1$ and $\bigcup_{i=1}^{n_1} g_{2i}H_2$
which maps left cosets of $H_1$ onto left cosets of $H_2$,
there exists a permutation $\pi \in S_{n_1}$ and elements $t_i \in H_2$ such that $\rho(g_{1i})=g_{2,\pi(i)} t_i$, $1\leqslant i \leqslant n_1$.
Denote by $v_{ih}$ the elements of the standard basis in $V_1$ defined before the theorem.
Now $\varphi(e_{ij}\otimes x_h) \psi(v_{j,1_{H_1}})=\psi(v_{ih})$.
Note that $\deg \psi(v_{j,1_{H_1}}) = \rho(g_{1j})= g_{2, \pi(j)} t_j$ and
$\deg \psi(v_{ih}) = \rho(g_{1i}h)= g_{2, \pi(i)}t_i\alpha(h)$.
Hence $\deg \varphi(e_{ij}\otimes x_h) = g_{2, \pi(i)}t_i\alpha(h) t_j^{-1} g_{2, \pi(j)}^{-1}$
for all $1\leqslant i,j\leqslant n_1$.
Since $\varphi$ is a graded map and $\deg (e_{ij}\otimes x_h) = g_{1i}hg_{1j}^{-1}$,
we get the first part of the theorem.
The converse is trivial.
\end{proof}
The proposition below is verified directly.
\begin{proposition}
An elementary grading on a full matrix algebra can be weakly equivalent only to a
grading isomorphic to an elementary grading on a full matrix algebra.
\end{proposition}
Let $G$ be a group. We say that a $G$-grading of an algebra $A$ is
\textit{connected} if the support of this grading generates the
group $G$. Recall that a grading $A=\bigoplus_{g \in G} A^{(g)}$
is called \textit{strong} if $A^{(g_1)}A^{(g_2)}=A^{(g_1g_2)}$ for
any $g_1,g_2\in G$ and a grading is called \textit{nondegenerate}
if the product of a finite number of non-zero homogeneous
components is again non-zero.
It is easy to see that if $A^{(g)}\ne 0$
for at least one $g\in G$ and $A$ is strongly graded,
then $A^{(gh^{-1})}A^{(h)}=A^{(g)}\ne 0$ implies $A^{(h)}\ne 0$
for all $h\in G$. In particular, a strong grading is connected.
Here we introduce the notion of
a strongly connected grading which is weaker than the notion
of a connected nondegenerate grading and a strong grading.
\begin{definition}
A connected grading $\Gamma\colon A=\bigoplus_{g \in G} A^{(g)}$ is \textit{strongly connected} if $A^{(g)}A^{(h)}\ne 0$
for all $g,h\in \supp\Gamma$.
\end{definition}
\begin{lemma}
Weakly equivalent strongly connected gradings of finite
dimensional algebras are equivalent.
\end{lemma}
\begin{proof}
Let $\Gamma_1 \colon A=\bigoplus_{g \in G} A^{(g)}$ be a strongly
connected grading. We claim that $\supp\Gamma _1$ coincides with $G$ itself.
Take arbitrary $g=\prod _{i=1}^r g_i^{k_i}$
where $g_1,g_2,\dots ,
g_r\in\supp\Gamma _1$ and $k_1,k_2,\dots ,k_r\in \mathbb{N}$. Now, using the strongly
connectedness condition, we get by induction on $r$ that $A^{(g)}\neq 0$ and therefore
$\supp\Gamma_1$ is closed under multiplication.
Since $A$ is finite dimensional,
$\supp\Gamma_1$ is finite subset of a group, which is closed under multiplication.
Hence $\supp\Gamma_1$ is a group and $\supp\Gamma_1 = G$ since $\Gamma_1$ is connected.
Let $\Gamma_2 \colon B=\bigoplus_{h \in H} B^{(h)}$ be a strongly
connected grading which is weakly equivalent to $\Gamma _1$ with
the associated isomorphism $\varphi \colon A\rightarrow B$. By the
arguments above, $\supp\Gamma _2=H$. Therefore, the natural
bijection between the supports of the gradings is a map $\psi \colon G\rightarrow H$ between
the grading groups. We claim that $\psi $ is a
group isomorphism. Indeed, let $g,g'\in G$. Then $A^{(g)}$ and $A^{(g')}$ are
both non-zero. Suppose $\varphi\left(A^{(g)}\right)=B^{(h)}$, $\varphi\left(A^{(g')}\right)=B^{(h')}$. Then $\psi(g)=h$ and $\psi(g')=h'$. Since $\varphi$ is an isomorphism
of algebras, we have
$$\varphi \left(A^{(g)}A^{(g')}\right)=\varphi\left(A^{(g)}\right)\varphi\left(A^{(g')}\right)=B^{(h)}B^{(h')}\subseteq B^{(hh')}.$$
On the other hand,
$$\varphi\left(A^{(g)}A^{(g')}\right)\subseteq \varphi\left(A^{(gg')}\right).$$
Consequently, since by the strongly connectedness condition
$A^{(g)}A^{(g')}\neq 0$, we get $\varphi\left(A^{(gg')}\right)=B^{(hh')}$ and
therefore $\psi(gg')=hh'$. Hence $\psi$ is indeed a group isomorphism. We
conclude by noticing that
$$\varphi\left(A^{(g)}\right)=B^{\bigl(\psi(g)\bigr)}$$
and therefore $\Gamma _1$ and $\Gamma _2$ are equivalent.
\end{proof}
\begin{corollary}\label{cor:eqvsweakeq}
In the following cases the weak equivalence of gradings of finite
dimensional algebras implies the equivalence of gradings:
\begin{enumerate}
\item the standard gradings on twisted group algebras;
\item strong gradings;
\item nondegenerate gradings.
\end{enumerate}
\end{corollary}
Based on Corollary~\ref{cor:eqvsweakeq}, it is natural to ask when an
isomorphism of twisted group rings implies a graded equivalence of the
twisted group rings. It is clear that two isomorphic twisted group
rings may be not graded equivalent. A simple example for that is
the group algebras $\mathbb{C}C_4$ and $\mathbb{C}(C_2\times
C_2)$ (by $C_n$ we denote the cyclic group of order $n$). However, can this phenomenon happen for two twisted group
algebras of the same group? It turns out that, provided the group
$G$ is abelian, if $\mathbb{C}^{\sigma}G$ and $\mathbb{C}^{\rho}G$ are
isomorphic and simple, then they are graded equivalent~\cite[Theorem~18]{AljaHaileNatapov}, \cite[Proposition~2.4 (2)]{ginosargradings} (see the description of finte abelian groups of central type e.g. in~\cite[Theorem 2.15]{ElduqueKochetov}). Nonetheless,
for a non-abelian group $G$ it can happen that
$\mathbb{C}^{\sigma}G$ and $\mathbb{C}^{\rho}G$ are isomorphic
and simple, but they are not graded equivalent. Rather
complicated examples for that can be found in~\cite[\S
3.5]{ginosargradings}. However, if we relax the simplicity condition
to the graded simplicity (twisted group algebras are always graded-simple), examples even for abelian
groups can be constructed and they are much simpler than those in~\cite[\S
3.5]{ginosargradings}.
\begin{example}
Let
$$G=C_4\times C_2\times C_2=\langle x\rangle \times \langle y\rangle \times \langle z\rangle.$$
Recall that in order to define a cohomology class for a finitely generated abelian group, it is enough to determine the values of the corresponding alternating bicharacter (see Section~\ref{SubsectionCohomologyGradedDivisionAlgebras}). Define two non-cohomologous classes $[\sigma],[\rho] \in H^2(G,\mathbb{C}^\times)$
as follows:
$$[\sigma]:\ \alpha(x,y)=-1,\ \alpha(x,z)=1,\ \alpha(y,z)=1,$$
$$[\rho]:\ \beta(x,y)=1,\ \beta(x,z)=1,\ \beta(y,z)=-1.$$
Here $\alpha$ and $\beta$ are the alternating bicharacters
corresponding to $[\sigma]$ and $[\rho]$,
respectively.
Let
$$\mathbb{C}^{\sigma}G=\bigoplus_{i=1}^k M_{m_i}(\mathbb{C}), \quad \mathbb{C}^{\rho}G=\bigoplus_{j=1}^t M_{n_j}(\mathbb{C})$$
be the corresponding Artin--Wedderburn decompositions. Since
$[\sigma]$ and $[\rho]$ are nontrivial, we have $m_i,n_j>1$ for any
$1\leqslant i \leqslant k$, $1\leqslant j \leqslant t$. On the other hand,
since the centers of both $\mathbb{C}^{\sigma}G$ and $\mathbb{C}^{\rho}G$
have dimensions greater than $1$, by a simple
calculation, we get
$$\mathbb{C}^{\sigma}G\cong \mathbb{C}^{\rho}G\cong \bigoplus_{i=1}^4 M_2(\mathbb{C}).$$
However, there is a homogenous element of order $4$ in the center
of $\mathbb{C}^{\rho}G$ while there is no such homogeneous element
in the center of $\mathbb{C}^{\sigma}G$.
Therefore, these algebras are not graded equivalent and
hence by Corollary~\ref{cor:eqvsweakeq} they are also not weakly
equivalent.
\end{example}
\section{Group-theoretical approach}
\label{SectionGroupTheoreticalApproach}
Each group grading on an algebra can be realized as a $G$-grading for many different groups $G$, however it turns out that there is one distinguished group among them \cite[Definition~1.17]{ElduqueKochetov}, \cite{PZ89}.
\begin{definition}\label{def:universal}
Let $\Gamma$ be a group grading on an algebra $A$. Suppose that $\Gamma$ admits a realization
as a $G_\Gamma$-grading for some group $G_\Gamma$. Denote by $\varkappa_\Gamma$ the corresponding embedding
$\supp \Gamma \hookrightarrow G_\Gamma$. We say that $(G_\Gamma,\varkappa_\Gamma)$ is the \textit{universal group of the grading $\Gamma$} if for any realization of $\Gamma$ as a grading by a group $G$
with $\psi \colon \supp \Gamma \hookrightarrow G$ there exists
a unique homomorphism $\varphi \colon G_\Gamma \to G$ such that the following diagram is commutative:
$$\xymatrix{ \supp \Gamma \ar[r]^\varkappa \ar[rd]^\psi & G_\Gamma \ar@{-->}[d]^\varphi \\
& G
}
$$
\end{definition}
The observation below is a direct consequence of the definition.
\begin{proposition}
A grading $\Gamma$ admits an infinite group turning $\Gamma$ into a connected grading if and only if $G_\Gamma$ is infinite.
\end{proposition}
In the definition above the universal group of a grading $\Gamma$ is a pair $(G_\Gamma,\varkappa_\Gamma)$.
Theorem~\ref{TheoremGivenFinPresGroupExistence} below shows, in particular, that the first component of this pair can be an arbitrary finitely presented group.
Furthermore, we can choose $\Gamma$ to be an elementary grading on a full matrix algebra. The possibility to include a subset $V$ to the support will be used later, in the proof of Theorem~\ref{TheoremFiniteRegradingImpossible}.
\begin{theorem}\label{TheoremGivenFinPresGroupExistence}
Let $F$ be a field, let $G$ be a finitely presented
group and let $V \subseteq G$ be a finite subset (possibly empty). Then for some $n\in\mathbb N$, depending only on the presentation of $G$ and the elements of $V$, there exists an elementary grading $\Gamma$ on $M_n(F)$ such that $G_\Gamma \cong G$ and $V \subseteq \supp \Gamma$.
\end{theorem}
\begin{proof}
Suppose $G \cong \mathcal F(X)/N$ where $\mathcal F(X)$ is the free group on a finite set of generators $X=\lbrace x_1, x_2, \dots, x_\ell\rbrace$ and $N$ is the normal closure of a finite set of words $w_1, \dots, w_m$.
Let $V=\lbrace g_1, \dots, g_s\rbrace \subseteq G$.
Choose $w_{m+1},\ldots, w_{m+s}\in \mathcal F(X)$
such that $g_i=\bar w_{m+i}$, $1\leqslant i \leqslant s$, where by $\bar u$ we denote the image of $u\in \mathcal F(X)$ in $G$. Suppose $w_i = y_{i1}\cdots y_{ik_i}$ where $y_{ij}\in X \cup X^{-1}$, $1\leqslant i \leqslant m+s$.
Denote $n=\ell+1+\sum_{i=1}^{m+s} k_i$.
Let $$\Gamma \colon M_n(F)=\bigoplus_{g\in G} M_n(F)^{(g)}$$ be the elementary $G$-grading
defined as follows: $$e_{r,r+1} \in M_n(F)^{(\bar y_{ij})} \text{ if } r=k_1+\dots+k_{i-1}+j,
\ 1\leqslant j \leqslant k_i,\ 1\leqslant i \leqslant m+s$$ and $$e_{r,r+1} \in M_n(F)^{(\bar x_j)}\text{ if }r=\sum_{i=1}^{m+s} k_i+j,\ 1\leqslant j \leqslant \ell$$
(the corresponding elementary grading exists by Remark~\ref{RemarkElementary}).
Note that $$e_{\sum_{i=1}^{m+r-1} k_i+1, \sum_{i=1}^{m+r} k_i+1}
= \prod_{j=1}^{k_{m+r}} e_{\sum_{i=1}^{m+r-1} k_i + j,\sum_{i=1}^{m+r-1} k_i + j+1} \in M_n(F)^{\left(\prod_{j=1}^{k_{m+r}}\bar y_{m+r,j}\right)}=M_n(F)^{(g_r)}$$
is nonzero. Hence $g_r \in \supp \Gamma$ for each $1\leqslant r \leqslant s$
and $V\subseteq \supp \Gamma$.
Now we claim that $G_\Gamma \cong G$.
Suppose that $\Gamma$ is realized as a grading by a group $H$.
Then there exists an injective map $\psi \colon \supp \Gamma \hookrightarrow H$
defined by $M_n(F)^{(g)}\subseteq M_n(F)^{(\psi(g))}$.
We have \begin{equation}\label{EqPsiPartialHom} \psi(g_1 g_2)=\psi(g_1)\psi(g_2) \end{equation}
for any $g_1, g_2 \in G$ such that $M_n(F)^{(g_1)}M_n(F)^{(g_2)}\ne 0$.
Since $M_n(F)$ is a unital algebra, we have $\psi(1_G)=1_H$.
For every $x\in X$ we have
$\bar x, \bar x^{-1}\in\supp\Gamma$. Thus elements $\psi(\bar x)$ and $\psi(\bar x^{-1})=\psi(\bar x)^{-1}$ are defined for all $x\in X$.
By induction, $$\psi(\bar y_{i1})\cdots \psi(\bar y_{ik_i})=\psi(\bar y_{i1}\cdots \bar y_{ik_i})=\psi(\bar w_i)=1_H$$
for all $1\leqslant i \leqslant m$.
Hence the elements $\psi(\bar x)$ satisfy the relations of $G$.
Therefore there exists a homomorphism $\varphi \colon G \to H$ such that $\varphi(\bar x) = \psi(\bar x)$ for each $x\in X$. Since the set $\lbrace \bar x \mid x \in X\rbrace$ generates $G$, such a homomorphism is unique.
Now we have to prove that $\varphi \bigr|_{\supp \Gamma} = \psi$.
Every element $g$ of $\supp \Gamma$ corresponds to a matrix unit
$e_{ij}$ where either $i > j$ and $e_{ij}=e_{i, i-1} e_{i-1,i-2}\cdots e_{j+1,j}$, or $i < j$ and $e_{ij}=e_{i, i+1} e_{i+1,i+2}\cdots e_{j-1,j}$,
or $i=j$ and $g=1_G$. Since for every $1\leqslant i,j \leqslant n$, such that
$|i-j|=1$, we have $e_{ij} \in M_n(F)^{(\bar x)}$ for some $x \in X \cup X^{-1}$ and $\varphi(x)=\psi(x)$ for $x \in X \cup X^{-1}$,
the induction on $|i-j|$ using~(\ref{EqPsiPartialHom}) shows that $\varphi(g)=\psi(g)$.
Hence $G \cong G_\Gamma$.
\end{proof}
\begin{remark} For each grading $\Gamma$ one can define a category $\mathcal C_\Gamma$
where the objects are all pairs $(G,\psi)$ such that $G$ is a group and $\Gamma$ can be realized
as a $G$-grading with $\psi \colon \supp \Gamma \hookrightarrow G$ being the embedding of the support.
In this category the set of morphisms between $(G_1,\psi_1)$ and $(G_2,\psi_2)$ consists
of all group homomorphisms $f \colon G_1 \to G_2$ such that the diagram below is commutative:
$$\xymatrix{ \supp \Gamma \ar[r]^{\psi_1} \ar[rd]^{\psi_2} & G_1 \ar[d]^f \\
& G_2
}
$$
Then $(G_\Gamma,\varkappa_\Gamma)$ is the initial object of $\mathcal C_\Gamma$.
\end{remark}
It is easy to see that if $\Gamma \colon A = \bigoplus_{g\in G} A^{(g)}$ is a grading,
where $G$ is a group,
then $G_\Gamma \cong \mathcal F(\supp \Gamma)/N$ where $N$ is the normal closure of the words $ght^{-1}$ for pairs $g,h \in \supp \Gamma$
such that $A^{(g)}A^{(h)}\ne 0$ where $t\in \supp\Gamma$ is defined by $A^{(g)}A^{(h)}\subseteq A^{(t)}$.
\begin{definition}
Let $\Gamma_1 \colon A=\bigoplus_{g \in G} A^{(g)}$
and $\Gamma_2 \colon A=\bigoplus_{h \in H} A^{(h)}$ be two gradings
where $G$ and $H$ are groups and $A$ is an algebra.
We say that $\Gamma_2$ is \textit{coarser} than $\Gamma_1$, if
for every $g\in G$ with $A^{(g)}\ne 0$ there exists $h\in H$
such that $A^{(g)}\subseteq A^{(h)}$. In this case $\Gamma_2$ is called a \textit{coarsening} of $\Gamma_1$ and $\Gamma_1$ is called a \textit{refinement} of $\Gamma_2$.
Denote by $\pi_{\Gamma_1 \to \Gamma_2} \colon G_{\Gamma_1} \twoheadrightarrow G_{\Gamma_2}$
the homomorphism defined by $\pi_{\Gamma_1 \to \Gamma_2}(\varkappa_{\Gamma_1}(g))=\varkappa_{\Gamma_2}(h)$
for $g\in\supp\Gamma_1$ and $h\in\supp \Gamma_2$ such that $A^{(g)} \subseteq A^{(h)}$.
\end{definition}
\begin{notation}
For a subset $U$ of a group $G$, denote by $\diff U := \lbrace uv^{-1} \mid u,v\in U,\ u\ne v\rbrace$.
\end{notation}
\begin{lemma}\label{LemmaCoarseningSubset}
Let $\Gamma_2 \colon A=\bigoplus_{h \in H} A^{(h)}$ be a coarsening of $\Gamma_1 \colon A=\bigoplus_{g \in G} A^{(g)}$.
Let \begin{equation*}\begin{split}W := \lbrace \varkappa_{\Gamma_1}(g_1) \varkappa_{\Gamma_1}(g_2)^{-1} \mid
g_1,g_2 \in \supp \Gamma_1\text{ such that } \\ A^{(g_1)}\oplus A^{(g_2)} \subseteq A^{(h)}\text{ for some }h\in \supp \Gamma_2\rbrace.\end{split}\end{equation*}
Denote by $Q \triangleleft G_{\Gamma_1}$ the normal closure of $W$.
Then $\ker \pi_{\Gamma_1 \to \Gamma_2} = Q$. In addition, $Q \cap \diff\varkappa_{\Gamma_1}(\supp \Gamma_1)= W$.
\end{lemma}
\begin{proof}
Obviously, $Q\subseteq \ker \pi_{\Gamma_1 \to \Gamma_2}$.
Let $\pi_{\Gamma_1} \colon \mathcal F(\supp \Gamma_1) \twoheadrightarrow G_{\Gamma_1}$
and $\pi_{\Gamma_2} \colon \mathcal F(\supp \Gamma_2) \twoheadrightarrow G_{\Gamma_2}$
be the natural surjective homomorphisms. Denote by $\varphi \colon \mathcal F(\supp \Gamma_1) \twoheadrightarrow \mathcal F(\supp \Gamma_2)$ the surjective homomorphism defined by $\varphi(g) = h$ for $g\in\supp\Gamma_1$ and $h\in\supp \Gamma_2$ such that $A^{(g)} \subseteq A^{(h)}$.
Then the following diagram is commutative:
\begin{equation}\label{EqCommDiagGamma1Gamma2}\xymatrix{ \mathcal F(\supp \Gamma_1) \ar[r]^{\quad\pi_{\Gamma_1}} \ar[d]^{\varphi} & G_{\Gamma_1} \ar[d]^{\pi_{\Gamma_1 \to \Gamma_2}}\\ \mathcal F(\supp \Gamma_2)
\ar[r]^{\quad\pi_{\Gamma_2}} & G_{\Gamma_2}}\end{equation}
Note that $\ker \varphi$ coincides with the normal closure in $\mathcal F(\supp \Gamma_1)$
of all elements $g_1 g_2^{-1}$ where $g_1,g_2 \in \supp \Gamma_1$ and $A^{(g_1)}\oplus A^{(g_2)} \subseteq A^{(h)}$ for some $h\in \supp \Gamma_2$. Hence $\pi_{\Gamma_1}(\ker\varphi) = Q$.
Suppose $\pi_{\Gamma_1 \to \Gamma_2} \pi_{\Gamma_1}(w) = 1_{G_{\Gamma_2}}$ for some $w\in \mathcal F(\supp \Gamma_1)$. Then~(\ref{EqCommDiagGamma1Gamma2}) implies $\varphi(w)\in \ker \pi_{\Gamma_2}$.
Therefore $\varphi(w)$ belongs to the normal closure of the words
$h_1h_2 u^{-1}$ for $h_1, h_2, u \in \supp \Gamma_2$ with $0\ne A^{(h_1)}A^{(h_2)}\subseteq A^{(u)}$.
However the last inclusion holds if and only if $0\ne A^{(g_1)}A^{(g_2)}\subseteq A^{(t)}$
for some $g_1, g_2, t \in \supp \Gamma_1$ such that $A^{(g_1)}\subseteq A^{(h_1)}$, $ A^{(g_2)}\subseteq A^{(h_2)}$.
Hence we can rewrite $w=w_0 w_1$ where $w_0 \in \ker\varphi$ and $w_1 \in \ker \pi_{\Gamma_1}$.
In particular, $\pi_{\Gamma_1}(w) = \pi_{\Gamma_1}(w_0)\in Q$. Since $\pi_{\Gamma_1}$
is surjective, we get $\ker \pi_{\Gamma_1 \to \Gamma_2} = Q$.
Together with the obvious equality
$\ker \pi_{\Gamma_1 \to \Gamma_2} \cap \diff\varkappa_{\Gamma_1}(\supp \Gamma_1)= W$ this implies
$Q \cap \diff\varkappa_{\Gamma_1}(\supp \Gamma_1)= W$.
\end{proof}
\begin{lemma}\label{LemmaSubsetCoarseningExists} Let $\Gamma_1 \colon A=\bigoplus_{g \in G} A^{(g)}$ be a grading by a group $G$.
Then for each subset $W \subseteq \diff\varkappa_{\Gamma_1}(\supp \Gamma_1)$
there exists a coarsening $\Gamma_2 \colon A=\bigoplus_{h \in H} A^{(h)}$ of $\Gamma_1$
such that $\ker\pi_{\Gamma_1 \to \Gamma_2} = Q$
where $Q$ is the normal closure of $W$ in $G_{\Gamma_1}$.
\end{lemma}
\begin{proof}
Let $\pi_{\Gamma_1} \colon \mathcal F(\supp \Gamma_1) \twoheadrightarrow G_{\Gamma_1}$ and $\pi \colon G_{\Gamma_1} \twoheadrightarrow G_{\Gamma_1}/Q$
be the natural surjective homomorphisms. Consider the grading $\Gamma_2 \colon A=\bigoplus_{u \in G_{\Gamma_1}/Q} A^{(u)}$ where $A^{(u)} := \bigoplus_{\substack{g\in \supp \Gamma_1,\\ \pi(\varkappa_{\Gamma_1}(g))=u}} A^{(g)}$.
We claim that $G_{\Gamma_1}/Q$ is the universal group of the grading $\Gamma_2$.
If $\Gamma_2$
can be realized as a grading by a group $H$
and $\psi \colon \supp\Gamma_2\hookrightarrow H$ is the corresponding embedding of the support,
then there exists a unique homomorphism $\varphi \colon \mathcal F(\supp \Gamma_1) \to H$
such that $\varphi(g)=\psi(\pi(\varkappa_{\Gamma_1}(g)))$.
Note that $\ker (\pi\pi_{\Gamma_1})$ is the normal closure in $\mathcal F(\supp \Gamma_1)$ of:
\begin{enumerate}
\item the words $ght^{-1}$ for all pairs $g,h \in \supp \Gamma_1$
such that $A^{(g)}A^{(h)}\ne 0$ where $t\in \supp\Gamma_1$ is defined by $A^{(g)}A^{(h)}\subseteq A^{(t)}$;
\item the words $g_1 g_2^{-1}$ for all $\varkappa_{\Gamma_1}(g_1) \varkappa_{\Gamma_1}(g_2)^{-1} \in W$.
\end{enumerate}
Hence $\ker (\pi\pi_{\Gamma_1}) \subseteq \ker \varphi$ and
there exists a homomorphism $\bar\varphi \colon G_{\Gamma_1}/Q \to H$ such that
$\bar \varphi\pi\pi_{\Gamma_1} = \varphi$. In particular, $\bar\varphi(u)=\psi(u)$ for all $u\in \supp \Gamma_2$.
Since $G_{\Gamma_1}/Q$ is generated by $\supp \Gamma_2$, the homomorphism $G_{\Gamma_1}/Q \to H$
with this property is unique, $G_{\Gamma_1}/Q$ is the universal group of the grading $\Gamma_2$
and $\pi_{\Gamma_1 \to \Gamma_2}$ can be identified with $\pi$.
\end{proof}
\begin{remark}
Note that the inclusion $W \subseteq Q \cap \diff\varkappa_{\Gamma_1}(\supp \Gamma_1)$
in Lemma~\ref{LemmaSubsetCoarseningExists}
can be strict.
\end{remark}
\begin{definition}\label{def:resfinite}
Let $G$ be a group and let $W$ be a subset of $G$. We say that $G$ is \textit{residually finite with respect to $W$} if there exists a normal subgroup $N \triangleleft G$ of finite index
such that $W \cap N = \varnothing$.
We say that $G$ is \textit{hereditarily residually finite with respect to $W$} if
for the normal closure $N_1 \triangleleft G$ of any subset of $W$
there exists a normal subgroup $N \triangleleft G$ of finite index
such that $N_1 \subseteq N$ and $W \cap N = W \cap N_1$.
\end{definition}
Theorem~\ref{TheoremFinGradingsToGroupsTransition}
below shows that the problem of whether a grading and its coarsenings can be regraded by a finite group can be viewed completely group theoretically.
\begin{theorem}\label{TheoremFinGradingsToGroupsTransition} Let $G$ be a group, $A$ be an algebra, and let $\Gamma \colon A = \bigoplus_{g\in G} A^{(g)}$ be a $G$-grading on $A$.
Then \begin{enumerate}
\item $\Gamma$ is weakly equivalent to a grading by a finite group if and only if
$G_\Gamma$ is residually finite with respect to $\diff\varkappa_\Gamma(\supp \Gamma)$;
\item $\Gamma$ and all its coarsenings are weakly equivalent to a grading by a finite group if and only if
$G_\Gamma$ is hereditarily residually finite with respect to $\diff\varkappa_\Gamma(\supp \Gamma)$.
\end{enumerate}
\end{theorem}
\begin{proof}
The grading $\Gamma$ can be realized by any factor group of $G_\Gamma$ that does not glue the elements of the support, i.e. distinct elements of the support have distinct images in that factor group. Then if $G_\Gamma$ is residually finite with respect to $\diff\varkappa_\Gamma(\supp \Gamma)$,
there exists a finite factor group with this property. Conversely, if $\Gamma$ admits a realization
as a grading by a finite group $G$, then the subgroup of $G$ generated by the support
is a finite factor group of $G_\Gamma$ that does not glue the elements of the support.
Hence $G_\Gamma$ is residually finite with respect to $\diff\varkappa_\Gamma(\supp \Gamma)$ and the first part of the theorem is proved.
Suppose $G_\Gamma$ is hereditarily residually finite with respect to $\diff\varkappa_\Gamma(\supp \Gamma)$.
By Lemma~\ref{LemmaCoarseningSubset}, for any coarsening $\Gamma_1$ of $\Gamma$
there exists a normal subgroup $Q$ which is the normal closure of
\begin{equation*}\begin{split}Q \cap
\diff\varkappa_\Gamma(\supp \Gamma) = \lbrace \varkappa_{\Gamma}(g_1) \varkappa_{\Gamma}(g_2)^{-1} \mid
g_1,g_2 \in \supp \Gamma\text{ such that } \\ A^{(g_1)}\oplus A^{(g_2)} \subseteq A^{(h)}\text{ for some }h\in \supp \Gamma_1\rbrace.\end{split}\end{equation*}
such that $G_{\Gamma_1} \cong G_{\Gamma}/Q$. Since $G_\Gamma$ is hereditarily residually finite with respect to $\diff\varkappa_\Gamma(\supp \Gamma)$, there exists a normal subgroup $N \triangleleft G_{\Gamma}$ of finite index such that $Q \subseteq N$ and
\begin{equation}\label{EqNQDiffKappa} N \cap \diff\varkappa_\Gamma(\supp \Gamma) = Q \cap \diff\varkappa_\Gamma(\supp \Gamma).\end{equation}
Let $\bar N=\pi_{\Gamma \to \Gamma_1}(N)$. Suppose $\varkappa_{\Gamma_1}(h_1)\varkappa_{\Gamma_1}(h_2)^{-1} \in \bar N$ for some $h_1, h_2 \in \supp \Gamma_1$. Using the isomorphism $G_\Gamma/N \cong G_{\Gamma_1}/\bar N$ we get $\varkappa_{\Gamma}(g_1)\varkappa_{\Gamma}(g_2)^{-1} \in N$ for all $g_1, g_2 \in \supp \Gamma$ such that $A^{(g_1)} \subseteq A^{(h_1)}$ and $A^{(g_2)} \subseteq A^{(h_2)}$.
Now~(\ref{EqNQDiffKappa})
implies $\varkappa_{\Gamma_1}(g_1)\varkappa_{\Gamma_1}(g_2)^{-1} \in Q$,
$\varkappa_{\Gamma_1}(h_1)=\varkappa_{\Gamma_1}(h_2)$ and $h_1 = h_2$. Hence $\bar N$ does not glue the elements of the support of $\Gamma_1$. Since $|G_{\Gamma_1}/\bar N| = |G_\Gamma/N|< +\infty$,
the grading $\Gamma_1$ admits the finite grading group $G_{\Gamma_1}/\bar N$.
Suppose that every coarsening $\Gamma_1$ of $\Gamma$ admits a finite grading group. We claim that
$G_\Gamma$ is hereditarily residually finite with respect to $\diff\varkappa_\Gamma(\supp \Gamma)$.
Indeed, let $N_1 \triangleleft G_\Gamma$ be a normal closure of a subset of
$\diff\varkappa_\Gamma(\supp \Gamma)$. By Lemma~\ref{LemmaSubsetCoarseningExists},
there exists a grading $\Gamma_1$ such that $G_{\Gamma_1} \cong G_{\Gamma}/N_1$.
Since $\Gamma_1$ admits a finite grading group, there exists a normal subgroup $Q \triangleleft G_{\Gamma_1}$ of finite index such that $\varkappa_{\Gamma_1}(h_1) \varkappa_{\Gamma_1}(h_2)^{-1} \notin Q$
for all $h_1, h_2 \in \supp \Gamma_1$, $h_1 \ne h_2$. Let $N := \pi_{\Gamma \to \Gamma_1}^{-1}(Q)$.
Then $|G_\Gamma/N| = |G_{\Gamma_1}/ Q| < +\infty$, $N \supseteq \ker\pi_{\Gamma \to \Gamma_1}=N_1$, and
$$N \cap \diff\varkappa_\Gamma(\supp \Gamma) \supseteq N_1 \cap \diff\varkappa_\Gamma(\supp \Gamma).$$
Suppose $\varkappa_{\Gamma}(g_1) \varkappa_{\Gamma}(g_2)^{-1} \in N$ for some $g_1, g_2 \in \supp \Gamma$.
Take $h_1, h_2 \in \supp \Gamma_1$ such that $A^{(g_1)} \subseteq A^{(h_1)}$ and $A^{(g_2)} \subseteq A^{(h_2)}$. Then $\varkappa_{\Gamma_1}(h_1) \varkappa_{\Gamma_1}(h_2)^{-1} \in Q$
and $h_1 = h_2$. Thus $\varkappa_{\Gamma}(g_1) \varkappa_{\Gamma}(g_2)^{-1} \in N_1$
and $$N \cap \diff\varkappa_\Gamma(\supp \Gamma) = N_1 \cap \diff\varkappa_\Gamma(\supp \Gamma).$$
As a consequence, $G_\Gamma$ is hereditarily residually finite with respect to $\diff\varkappa_\Gamma(\supp \Gamma)$.
\end{proof}
The proposition below could be obtained as a consequence of Theorem~\ref{TheoremFinGradingsToGroupsTransition}, however we give a separate proof because of its easiness.
\begin{proposition}\label{prop:abelian}
Let $\Gamma \colon A=\bigoplus_{g \in G} A^{(g)}$ be a grading of a finite dimensional
algebra $A$ by an abelian group $G$. Then $\Gamma$ is weakly equivalent to a grading by a finite group.
\end{proposition}
\begin{proof}
We can replace $G$ with its subgroup generated by $\supp \Gamma$. Since $\supp \Gamma$ is finite,
without loss of generality we may assume that $G$ is a finitely generated abelian group.
Then $G$ is a direct product of free and primary cyclic groups. Replacing free cyclic groups
with cyclic groups of a large enough order (see Example~\ref{ExampleAbelianRegrading}), we get a finite grading group.
\end{proof}
\begin{example}\label{ExampleAbelianRegrading}
Let $n\in \mathbb N$ and let $\Gamma$ be the elementary $\mathbb Z$-grading on $M_n(F)$
defined by the $n$-tuple $(1,2,\ldots,n)$, i.e. $e_{ij}\in M_n(F)^{(i-j)}$.
Then $$\supp \Gamma = \lbrace -(n-1), -(n-2), \ldots, -1, 0, 1, 2, \ldots, n-1 \rbrace$$
and $\Gamma$ is equivalent to the elementary $\mathbb Z/(2n\mathbb Z)$-grading
defined by the $n$-tuple $(\bar 1,\bar 2,\ldots,\bar n)$, i.e. $e_{ij}\in M_n(F)^{(\overline{i-j})}$.
\end{example}
Consider the grading $\Gamma_0$ on $M_n(F)$ by the free group $\mathcal F(x_1, \dots, x_{n-1})$
such that $e_{r,r+1} \in M_n(F)^{(x_r)}$ for $1\leqslant r \leqslant n-1$,
i.e. defined by the $n$-tuple $(x_1 x_2 \cdots x_{n-1}, x_2 \cdots x_{n-1}, \ldots, x_{n-1}, 1)$.
Note the neutral element component of $\Gamma_0$ is the linear span of matrix units $e_{ii}$, $1\leqslant i \leqslant n$, and is $n$-dimensional, and all the rest components are $1$-dimensional. Since for each elementary grading the diagonal matrix units $e_{ii}$ belong to the neutral element component
of the grading and all matrix units are homogeneous, every elementary grading on $M_n(F)$
is a coarsening of $\Gamma_0$. Since $\mathcal F(x_1, \dots, x_{n-1})$ is free and all its free generators
belong to $\supp \Gamma_0$, $G_{\Gamma_0} \cong \mathcal F(x_1, \dots, x_{n-1})$.
Note also that $$\supp \Gamma_0 = \lbrace x_i x_{i+1}\cdots x_j \mid 1\leqslant i \leqslant j \leqslant n-1 \rbrace\cup
\lbrace x_j^{-1} x_{j-1}^{-1}\cdots x_i^{-1} \mid 1\leqslant i \leqslant j \leqslant n-1 \rbrace \cup \lbrace 1\rbrace.$$
Thus by Theorem~\ref{TheoremFinGradingsToGroupsTransition} Problem~\ref{ProblemMinNElAGradNonFin}
is equivalent to Problem~\ref{ProblemMinFnHRFRx1x2x3} below:
\begin{problem}\label{ProblemMinFnHRFRx1x2x3}
Determine
the set $\Omega$ of
the numbers $n\in\mathbb N$ such that the group $\mathcal F(x_1, \dots, x_{n-1})$ is hereditarily residually finite with respect to $\diff W_n$
where \begin{equation}\label{EqWnFreeGroup}W_n=\lbrace x_i x_{i+1}\cdots x_j \mid 1\leqslant i \leqslant j \leqslant n-1 \rbrace\cup
\lbrace x_j^{-1} x_{j-1}^{-1}\cdots x_i^{-1} \mid 1\leqslant i \leqslant j \leqslant n-1 \rbrace \cup \lbrace 1\rbrace.\end{equation}
\end{problem}
\section{Regrading full matrix algebras by finite groups}
\label{SectionRegradingMatrixFinite}
In this section we prove Theorem~\ref{TheoremRegradeElementaryOmega}
that deals with Problems~\ref{ProblemMinNElAGradNonFin} and~\ref{ProblemMinFnHRFRx1x2x3}.
In the lemma below we use the idea of~\cite[Section 4]{ClaseJespersDelRio}
and show, in particular, that all semigroup regradings of elementary
group gradings on $M_n(F)$ can be reduced to group regradings.
\begin{lemma}\label{LemmaSemigroupMatrixElementaryGroup}
Let $F$ be a field and let $n\in\mathbb N$.
If $\Gamma \colon M_n(F)=\bigoplus_{t\in T} M_n(F)^{(t)}$
is a grading on $M_n(F)$ by a semigroup $T$ such that all $e_{ij}$ are homogeneous elements and there exists an element $e \in T$
such that all $e_{ii}\in M_n(F)^{(e)}$, then $e^2=e$
and $\supp \Gamma \subseteq U(eTe)$ where
$U(eTe)$ is the group of invertible elements of the monoid $eTe$.
\end{lemma}
\begin{proof}
$e_{11}^2=e_{11}$ implies $e^2=e$. Since the identity matrix belongs to
$M_n(F)^{(e)}$, we obtain $\supp \Gamma \subseteq eTe$.
Now $e_{ij}e_{ji}=e_{ii}$ implies $\supp \Gamma \subseteq U(eTe)$.
\end{proof}
The lemma below and its corollary show that if for some $m\in\mathbb N$
we have $m\notin \Omega$ (see the definition of $\Omega$ in Problem~\ref{ProblemMinNElAGradNonFin}), then $n\notin \Omega$ for all $n\geqslant m$.
\begin{lemma}\label{LemmaGradedSubalgebra}
Let $\Gamma \colon A = \bigoplus_{t\in T} A^{(t)}$
be a grading by a (semi)group $T$ on an algebra $A$ over a field $F$
and let $B$ be a graded subalgebra. Suppose that the grading on $B$
cannot be regraded by a finite (semi)group. Then $\Gamma$ cannot be regraded
by a finite (semi)group either.
\end{lemma}
\begin{proof}
Each regrading on $\Gamma$ induces a regrading of the grading on $B$.
Therefore, if it were possible to regrade $\Gamma$ by a finite (semi)group,
the same would be possible for the grading on $B$. However, the latter is impossible.
\end{proof}
\begin{corollary}\label{CorollaryMatrixElemGreaterSize}
If for some $m\in\mathbb N$ and a group $G$ there exists an elementary $G$-grading on $M_m(F)$ that is not weakly equivalent to a grading by a finite (semi)group,
then an elementary $G$-grading with this property exists on $M_n(F)$ for every $n\geqslant m$.
\end{corollary}
\begin{proof}
Suppose this elementary $G$-grading on $M_m(F)$ can be realized by an $m$-tuple
$(g_1, g_2, \dots, g_m)$. Consider the elementary $G$-grading $\Gamma$ on $M_n(F)$
defined by the $n$-tuple $(g_1, g_2, \dots, g_m,\underbrace{g_m, \dots, g_m}_{n-m})$. The algebra $M_m(F)$ becomes a graded subalgebra of $M_n(F)$.
Therefore, if $\Gamma$ were weakly equivalent to a grading by a finite (semi)group,
then it would be possible to reindex the graded components of $M_n(F)$
by elements of a finite group and the original $G$-grading on $M_m(F)$ had this property too. Hence $\Gamma$ is not weakly equivalent to any grading by a finite (semi)group.
\end{proof}
Since Problems~\ref{ProblemMinNElAGradNonFin} and~\ref{ProblemMinFnHRFRx1x2x3}
are equivalent, we immediately get
\begin{corollary}
If $\mathcal F(x_1, \dots, x_{m-1})$ is not hereditarily residually finite with respect to $\diff W_m$ for some $m\in\mathbb N$, then
$\mathcal F(x_1, \dots, x_{n-1})$ is not hereditarily residually finite with respect to $\diff W_n$ for all $n \geqslant m$. (See the definition of $W_n$ in~(\ref{EqWnFreeGroup}).)
\end{corollary}
Recall that a group is \textit{residually finite} if the intersection of its normal subgroups of finite index is trivial.
\begin{theorem}\label{TheoremFiniteRegradingImpossible} Let $F$ be a field and let $G$ be a finitely presented
group which is not residually finite. (For example, $G$ is a finitely presented infinite simple group, see~\cite{Higman}.)
Then there exists an elementary $G$-grading on a full matrix algebra which is not weakly equivalent to any $H$-grading for any finite (semi)group $H$.
\end{theorem}
\begin{proof}
Let $g_0\ne 1_G$ be an element that belongs to the intersection of
all normal subgroups of $G$ of finite index. (In particular, if $G$ is simple, we take
an arbitrary element $g_0\ne 1_G$.)
By Theorem~\ref{TheoremGivenFinPresGroupExistence},
there exists a $G$-grading $\Gamma$ on $M_n(F)$ for some $n\in\mathbb N$
such that $G_\Gamma \cong G$ and $g_0 \in \supp G$.
Suppose that $\Gamma$ is weakly equivalent to a grading by a finite semigroup $H$
and $\psi \colon \supp \Gamma \hookrightarrow H$ is the corresponding embedding of the support.
By Lemma~\ref{LemmaSemigroupMatrixElementaryGroup}, we may assume that $H$ is a group. Since $G_\Gamma \cong G$, there exists a unique homomorphism $\varphi \colon G \to H$ such that $\varphi|_{\supp \Gamma} = \psi$.
However $H$ is finite, $\ker \varphi$ is of finite index, and therefore $\varphi(g_0)=1_H$. Since $g_0 \ne 1_G$ and $g_0, 1_G \in \supp \Gamma$, this $H$-grading cannot be equivalent to $\Gamma$.
\end{proof}
\begin{corollary}\label{CorollaryFiniteRegradingImpossible353} If $n \geqslant 353$, then $M_n(F)$ admits a grading by an infinite group that is not weakly equivalent to any grading by a finite (semi)group.
\end{corollary}
\begin{proof}
Consider Thompson's finitely presented infinite simple group $G_{2,1}$ (see e.g. \cite[Section~8]{Higman}).
We can take $g_0$ to be any of its generators which all are anyway in the support
of the grading constructed in Theorem~\ref{TheoremGivenFinPresGroupExistence} for $G=G_{2,1}$.
Therefore we may assume there that $V$ is empty.
Summing up the lengths of the
defining relators, we obtain that $n$ in the proof
of Theorem~\ref{TheoremGivenFinPresGroupExistence}
equals $353$. Now we apply Corollary~\ref{CorollaryMatrixElemGreaterSize}.
\end{proof}
Corollary~\ref{CorollaryFiniteRegradingImpossible353} implies
the upper bound in Theorem~\ref{TheoremRegradeElementaryOmega}.
Recall that an algebra $M(\gamma, \sigma)$ (see the definition in Section~\ref{SubsectionElementaryGradedSimpleAlgebras}), where $\gamma$ is an $n$-tuple of group elements, contains a graded subalgebra which is
graded isomorphic to $M_n(F)$ with the elementary grading determined by $\gamma$, namely, the subalgebra that is the linear span of $e_{ij}\otimes u_{1_H}$, $1\leqslant i,j \leqslant n$. Therefore, using Corollary~\ref{CorollaryFiniteRegradingImpossible353}
and Lemma~\ref{LemmaGradedSubalgebra}, one obtain that, for every
$n\geqslant 353$, every finite subgroup $H \subseteq G_{2,1}$, and every $\sigma \in Z^2(H, F^\times)$, there exists an $n$-tuple $\gamma$ of elements of $G_{2,1}$ such that the standard grading on $M(\gamma, \sigma)$ is not weakly equivalent to a grading by a finite (semi)group.
Now we present a class of elementary gradings that are weakly equivalent to gradings by finite groups.
\begin{theorem}\label{TheoremFiniteRegradingAllDifferent} Let $G$ be a group, let $F$ be a field, let $n\in\mathbb N$, and let $(g_1, \dots, g_n)$ be an $n$-tuple of elements of $G$ such that $g_i g_j^{-1} = g_k g_\ell^{-1}$ if and only if either
$\left\lbrace \begin{smallmatrix} i=k, \\ j=\ell \end{smallmatrix}\right.$ or
$\left\lbrace \begin{smallmatrix} i=j, \\ k=\ell. \end{smallmatrix}\right.$ Then the elementary grading on $M_n(F)$
defined by $(g_1, \dots, g_n)$ is weakly equivalent to the elementary $S_{n+1}$-grading defined by $(\gamma_1,\dots,\gamma_n)$ where $S_{n+1}$ is the symmetric group acting on $\lbrace 1,2,\dots,n+1\rbrace$ and $\gamma_i = (1, i+1) \in S_{n+1}$ is the transposition switching $1$ and $i+1$.
The same grading on $M_n(F)$ is weakly equivalent to the elementary $\mathbb Z/(2^{n+1}\mathbb Z)$-grading defined by $(\bar 2, \bar 2^2, \bar 2^3, \dots, \bar 2^n)$.
\end{theorem}
\begin{proof} In order to prove the first part of the theorem, it suffices to prove that $\gamma_i \gamma_j^{-1} = \gamma_k \gamma_\ell^{-1}$ if and only if either $\left\lbrace \begin{smallmatrix} i=k, \\ j=\ell \end{smallmatrix}\right.$ or
$\left\lbrace \begin{smallmatrix} i=j, \\ k=\ell. \end{smallmatrix}\right.$ However, if $i=j$, then $\gamma_i \gamma_j^{-1}$ is the identity permutation and if $i\ne j$, then $\gamma_i \gamma_j^{-1} = (1, j+1, i+1)$ (a $3$-cycle).
In order to prove the second part of the theorem, we
notice that $\bar 2^i-\bar 2^j=\bar 2^k-\bar 2^\ell$ if and only if $2^i-2^j=2^k-2^\ell$ if and only if either $\left\lbrace \begin{smallmatrix} i=k, \\ j=\ell \end{smallmatrix}\right.$ or
$\left\lbrace \begin{smallmatrix} i=j, \\ k=\ell. \end{smallmatrix}\right.$ Indeed, dividing the equality $2^i-2^j=2^k-2^\ell$ by $2^{\min(i,j,k,\ell)}$, we see that the both sides of it must be zero.
\end{proof}
In Theorem~\ref{TheoremFiniteRegradingImpossible} we have constructed
an elementary $G$-grading on $M_n(F)$ that is not weakly equivalent to a grading by a finite group.
However, this grading is a coarsening of the elementary grading by the free group
$\mathcal F(z_1, \dots, z_{n-1})$ that corresponds to $n$-tuple $(1,z_1,\dots, z_{n-1})$
(this grading was considered in~\cite[Proposition~4.11]{CibilsRedondoSolotar}, \cite[Lemma~4.5]{ginosargradings})
and which is by Theorem~\ref{TheoremFiniteRegradingAllDifferent}
is weakly equivalent to a grading by a finite group. In other words, there exist gradings that
can be regraded by a finite group, but some of their coarsenings cannot.
\begin{theorem}\label{TheoremFiniteRegradingSmallMatrix}
Let $\Gamma$ be an elementary $G$-grading on the full matrix algebra $M_n(F)$
where $n \leqslant 3$, $F$ is a field, and $G$ is a group.
Then $\Gamma$ is weakly equivalent to a grading by a finite group.
\end{theorem}
\begin{remark}
If $F$ is algebraically closed,
then the theorem holds for all gradings on $M_n(F)$, $n \leqslant 3$, not necessarily elementary ones.
Indeed, by Theorem~\ref{TheoremBahturinZaicevSeghal} any grading
on $M_n(F)$ is isomorphic to the standard grading on $M(\gamma, \sigma)$ for some $\gamma$ and $\sigma$. Comparing the dimensions, we obtain that either $M(\gamma, \sigma)$ is a twisted group algebra of a finite group,
i.e. there is nothing to prove, or $\sigma$ is a cocycle of the trivial group and the grading $\Gamma$ is isomorphic to an elementary one.
\end{remark}
\begin{proof}[Proof of Theorem~\ref{TheoremFiniteRegradingSmallMatrix}]
If $n=1$, then $M_n(F)=F$ and it can be regraded by the trivial group.
Therefore, we may assume that $n=2,3$.
Recall that $e_{ij} \in M_n(F)^{(g_i g_j^{-1})}$
where $(g_1, \dots, g_n)$ is the $n$-tuple defining the elementary grading.
Consider the $n\times n$ matrix $A$ where the $(i,j)$th entry is $g_i g_j ^{-1}$.
It is clear that this matrix completely determines the grading.
Theorem~\ref{TheoremWeakEquivGradedSimple} implies
that if two elementary gradings $\Gamma_1$ and $\Gamma_2$ have matrices $A$ and $B$ such that two entries in $A$ coincide if and only if the corresponding
entries in $B$ coincide, then $\Gamma_1$ and $\Gamma_2$ are weakly equivalent.
In the case $n=2$, we get $A=\left(\begin{smallmatrix}
1 & g \\
g^{-1} & 1
\end{smallmatrix}\right)$ where $1=1_G$. Here we have two cases: $g=g^{-1}$
and $g\ne g^{-1}$. In the case $g=g^{-1}$ we can regrade $M_2(F)$
by $\mathbb Z/2\mathbb Z$ (the elementary grading is defined by the couple $(\bar 0,\bar 1)$). In the case $g\ne g^{-1}$
we can regrade $M_2(F)$ by $\mathbb Z/3\mathbb Z$ (the elementary grading is again defined by the couple $(\bar 0,\bar 1)$).
Consider now the case $n=3$.
The matrix with the entries $g_ig_j^{-1}$
is of the form
$A=\left(\begin{smallmatrix}
1 & g & gh \\
g^{-1} & 1 & h \\
h^{-1}g^{-1} & h^{-1} & 1
\end{smallmatrix}\right)$ and the same grading can be defined by the triple $(gh,h,1)$.
If $g\in \lbrace 1, h, h^{-1}, gh, h^{-1}g^{-1} \rbrace$
or $h\in \lbrace 1, g, g^{-1}, gh, h^{-1}g^{-1} \rbrace$, then the subgroup of $G$ generated by $g$ and $h$ is abelian and by Proposition~\ref{prop:abelian} the algebra $M_3(F)$ can be regraded by a finite group.
Therefore, below we assume that $|\lbrace g,h,gh\rbrace|=3$,
$1\notin \lbrace g,h,gh\rbrace$, and $\lbrace g,h,gh\rbrace \cap \lbrace g^{-1},h^{-1},h^{-1}g^{-1}\rbrace
\ne \varnothing$ only if $g=g^{-1}$, $h=h^{-1}$, or $gh=h^{-1}g^{-1}$.
We have the following cases:
\begin{enumerate}
\item $\lbrace g,h,gh\rbrace \cap \lbrace g^{-1},h^{-1},h^{-1}g^{-1}\rbrace
= \varnothing$.
Here we can apply Theorem~\ref{TheoremFiniteRegradingAllDifferent}
since all the entries, except the diagonal ones, are different.
\item $gh = h^{-1}g^{-1}$, but $\lbrace g,h\rbrace \cap \lbrace g^{-1},h^{-1}\rbrace
= \varnothing$.
Here $\Gamma$ is weakly equivalent to the elementary $\mathbb Z/6\mathbb Z$-grading defined by $g\mapsto \bar 1$ and $h\mapsto \bar 2$,
i.e. by the triple $(\bar 3, \bar 2, \bar 0)$.
\item $gh = h^{-1}g^{-1}$, $g=g^{-1}$, but $h\ne h^{-1}$.
Here $\Gamma$ is weakly equivalent to the elementary $S_3$-grading defined by $g\mapsto (12)$ and $h\mapsto (123)$.
\item $gh = h^{-1}g^{-1}$, $g\ne g^{-1}$, $h = h^{-1}$. Here $\Gamma$ is weakly equivalent to the elementary $S_3$-grading defined by $g\mapsto (123)$ and $h\mapsto (12)$.
\item $gh = h^{-1}g^{-1}$, $g=g^{-1}$, $h=h^{-1}$. Here $\Gamma$ is weakly equivalent to the elementary $(\mathbb Z/2\mathbb Z) \times (\mathbb Z/2\mathbb Z)$-grading
defined by $g\mapsto(\bar 0, \bar 1)$ and $h\mapsto(\bar 1, \bar 0)$.
\item $gh \ne h^{-1}g^{-1}$, $g=g^{-1}$, $h\ne h^{-1}$.
Here $\Gamma$ is weakly equivalent to the elementary
$\mathbb Z/6\mathbb Z$-grading defined by $g\mapsto \bar 3$ and $h\mapsto \bar 1$.
\item $gh \ne h^{-1}g^{-1}$, $g\ne g^{-1}$, $h=h^{-1}$.
This case is treated analogously.
\item $gh \ne h^{-1}g^{-1}$, $g=g^{-1}$, $h=h^{-1}$. Here $\Gamma$ is weakly equivalent to the elementary $S_3$-grading defined by $g\mapsto (12)$, $h \mapsto (13)$.
\end{enumerate}
All the cases have been considered and $\Gamma$ is weakly equivalent to an elementary grading by a finite group.
\end{proof}
\begin{remark}
The elementary $S_3$-grading on $M_3(F)$ defined above by $g\mapsto (12)$, $h \mapsto (13)$ is not weakly equivalent to any of the gradings by abelian groups since $(12)(13)\ne (13)(12)$.
\end{remark}
\begin{proof}[Proof of Theorem~\ref{TheoremRegradeElementaryOmega}]
Theorem~\ref{TheoremRegradeElementaryOmega} immediately follows from Corollaries~\ref{CorollaryMatrixElemGreaterSize}, \ref{CorollaryFiniteRegradingImpossible353} and Theorem~\ref{TheoremFiniteRegradingSmallMatrix}.
\end{proof}
\section{Categories related to graded algebras and homomorphisms}
\label{SectionCategoriesWeakEquivalence}
If $A$ and $B$ are objects in a category $\mathcal A$, we denote the set of morphisms $A\to B$
by $\mathcal A(A,B)$.
\subsection{Change of the grading group and the free-forgetful adjunction}
\label{SubsectionGradedAdjunctions}\qquad
We first recall two classical examples of adjunctions in the categories related to graded algebras.
Let $\mathbf{Alg}_F^G$ be the category of algebras over a field $F$ graded by a group $G$.
The morphisms in $\mathbf{Alg}_F^G$ are all the homomorphisms of algebras
$$\psi \colon A=\bigoplus_{g\in G}A^{(g)}\ \longrightarrow\ B=\bigoplus_{g\in G}B^{(g)}$$
such that $\psi(A^{(g)})\subseteq B^{(g)}$ for all $g\in G$.
If $\varphi \colon G \to H$ is a homomorphism of groups,
then we have a functor $U_\varphi \colon \mathbf{Alg}_F^G \to \mathbf{Alg}_F^H$ that assigns to a $G$-grading $A=\bigoplus_{g\in G}A^{(g)}$ on an algebra $A$ the $H$-grading $$A = \bigoplus_{h\in H} A^{(h)}\text{ where }A^{(h)} := \bigoplus_{\substack{g\in G,\\ \varphi(g)=h}} A^{(g)}$$
and does not change the homomorphisms.
This functor admits a right adjoint functor $K_\varphi \colon \mathbf{Alg}_F^H \to \mathbf{Alg}_F^G$
where for $B=\bigoplus_{h\in H} B^{(h)}$ we have $K_\varphi (B) = \bigoplus_{g\in G} \bigl(K_\varphi(B)\bigr)^{(g)}$
with $\bigl(K_\varphi(B)\bigr)^{(g)} := \lbrace (g, b) \mid b\in B^{\left(\varphi(g)\right)} \rbrace$
(the vector space structure on $\bigl(K_\varphi(B)\bigr)^{(g)}$ is induced from those on $B^{\left(\varphi(g)\right)}$)
and the multiplication $(g_1,a)(g_2,b):=(g_1 g_2, ab)$ for $g_1, g_2\in G$, $a\in B^{\left(\varphi(g_1)\right)}$, $b\in B^{\left(\varphi(g_2)\right)}$. If $\psi \in \mathbf{Alg}_F^H(B_1,B_2)$,
then $K_\varphi (\psi)(g,b):=(g,\psi(b))$ where $g\in G$, $b\in B_1^{\left(\varphi(g)\right)}$.
We have a natural bijection
$$\mathbf{Alg}_F^H (U_\varphi (A), B)\to \mathbf{Alg}_F^G (A, K_\varphi(B))$$
where $A \in \textbf{Alg}_F^G$, $B \in \textbf{Alg}_F^H$.
Another example of an adjunction is the free-forgetful one. It is especially important for the theory of graded polynomial identities~\cite{AljaGia, AljaGiaLa,
BahtZaiGradedExp, GiaLa, ASGordienko9}.
Let $G$ be a group and let $\mathbf{Set}^G_*$ be the category whose objects are sets
$X$, containing a distinguished element $0$, with a fixed decomposition $X=\lbrace 0 \rbrace \sqcup \bigsqcup_{g\in G} X^{(g)}$ into a disjoint union.
Morphisms between $X=\lbrace 0 \rbrace \sqcup \bigsqcup_{g\in G} X^{(g)}$
and $Y=\lbrace 0 \rbrace \sqcup \bigsqcup_{g\in G} Y^{(g)}$ are maps $\varphi \colon X \to Y$
such that $\varphi(0)=0$ and $\varphi(X^{(g)})\subseteq \lbrace 0 \rbrace \sqcup Y^{(g)}$
for all $g\in G$. We have an obvious forgetful functor $U \colon \textbf{Alg}_F^G \to \mathbf{Set}^G_*$
that assigns to each graded algebra $A$ the object $U(A)=\lbrace 0 \rbrace \sqcup \bigsqcup_{g\in G} \left(A^{(g)} \backslash \lbrace 0 \rbrace \right)$.
This functor has a left adjoint functor $F\langle (-)\backslash \lbrace 0 \rbrace \rangle$
that assigns to $X=\lbrace 0 \rbrace \sqcup \bigsqcup_{g\in G} X^{(g)} \in \mathbf{Set}^G_*$
the free associative algebra $F\langle X\backslash \lbrace 0 \rbrace \rangle$ on the set $X\backslash \lbrace 0 \rbrace$ which is endowed with the grading defined by $x_1 \cdots x_n \in
F\langle X\backslash \lbrace 0 \rbrace \rangle^{(g_1 \cdots g_n)}$
for $x_i \in X^{(g_i)}$, $1\leqslant i \leqslant n$.
Here we have a natural bijection
$$\mathbf{Alg}_F^G (F\langle X\backslash \lbrace 0 \rbrace\rangle, A)\to \mathbf{Set}^G_* (X, U(A))$$
where $A \in \textbf{Alg}_F^G$, $X \in \mathbf{Set}^G_*$.
First of all, we notice that both examples deal with the categories $\mathbf{Alg}_F^G$ where for each category the grading group is fixed, i.e. the notion of isomorphism in $\mathbf{Alg}_F^G$ coincides with the notion of graded isomorphism. Second, the functor $U_\varphi $ corresponds to a regrading, but of a very special kind, namely, the regrading induced by the homomorphism $\varphi$. For this reason below we consider the category $\mathbf{GrAlg}_F$ of algebras over the field $F$ graded by arbitrary groups, in which the notion of isomorphism will coincide with the notion of weak equivalence of gradings.
\subsection{An oplax $2$-functor from $\mathbf{GrAlg}_F$}
Recall that a homomorphism $\psi \colon A \to B$
between graded algebras $A=\bigoplus_{g \in G} A^{(g)}$
and $B=\bigoplus_{h \in H} B^{(h)}$ is \textit{graded}
if for every $g\in G$ there exists $h\in H$ such that $\psi(A^{(g)})\subseteq B^{(h)}$.
Any graded homomorphism of graded algebras induces a map between subsets of the supports of the gradings.
This gives rise to several functors which we study in the sections below.
Let $\mathbf{GrAlg}_F$ be the category where the objects are group gradings on (not necessarily unital) algebras over a field $F$ and morphisms are graded homomorphisms between the corresponding algebras.
Consider also the category $\mathcal C$ where:
\begin{itemize}
\item the objects are triples $(G, S, P)$ where $G$ is a group, $S \subseteq G$ is a subset, and $P \subseteq S\times S$;
\item the morphisms $(G_1, S_1, P_1)\to (G_2, S_2, P_2)$ are triples $(\psi, R, Q)$
where $R \subseteq S_1$, $Q\subseteq P_1 \cap (R\times R)$, and $\psi \colon R \to S_2$ is a map
such that $\psi(g)\psi(h)=\psi(gh)$ for all $(g,h)\in Q$;
\item the identity morphism for $(G, S, P)$ is $(\id_{S}, S, P)$;
\item if $(\psi_1, R_1, Q_1) \colon (G_1, S_1, P_1)\to (G_2, S_2, P_2)$
and $(\psi_2, R_2, Q_2) \colon (G_2, S_2, P_2)\to (G_3, S_3, P_3)$,
then $$(\psi,R,Q)=(\psi_2, R_2, Q_2)(\psi_1, R_1, Q_1) \colon (G_1, S_1, P_1)\to (G_3, S_3, P_3)$$
is defined by $R=\lbrace g\in R_1 \mid \psi_1(g)\in R_2 \rbrace$, $Q = \lbrace (g,h)\in Q_1 \mid (\psi_1(g),\psi_1(h))\in Q_2 \rbrace$
and $\psi(g)=\psi_2(\psi_1(g))$ for $g\in R$.
\end{itemize}
In many cases it is useful to consider categories $\mathcal A$
\textit{enriched} over other categories $\mathcal B$, i.e. such categories
where the hom-objects $\mathcal A(a,b)$ are objects of $\mathcal B$
and the composition and the assignment of the identity morphism
are morphisms in $\mathcal B$. (See the precise definition in~\cite[Section 1.2]{KellyEnriched}.)
Recall that each partially ordered set (or poset) $(M, \preccurlyeq)$ is a category where the objects are the elements $m\in M$ and if $m \preccurlyeq n$, then there exists a single morphism $m \to n$. If $m \not\preccurlyeq n$, then there is no morphism $m \to n$.
Denote by $\mathbf{Cat}$ the category of small categories. Since the notion of a $\mathbf{Cat}$-enriched category coincides with the notion of a $2$-category, every category enriched over
posets is a $2$-category.
Suppose $\mathcal A$ and $\mathcal B$ are $2$-categories.
Denote by $\bullet$
the bifunctors of the horizontal composition in $\mathcal A$ and $\mathcal B$.
An \textit{oplax $2$-functor} $D\colon \mathcal A \to \mathcal B$
consists of
\begin{enumerate}
\item a map that assigns for each object $a \in \mathcal B$ an object $Da\in \mathcal B$;
\item functors $D(a_1,a_2) \colon \mathcal A(a_1,a_2) \to \mathcal B(Da_1,
Da_2)$ between the hom-categories for each two objects $a_1, a_2 \in \mathcal A$;
\item $2$-cells (morphisms in categories $\mathcal B(Da,Da)$) $\delta_a \colon D \id_a \Rightarrow \id_{Da}$ and transformations $\delta_{g,f} \colon D(g \bullet f) \Rightarrow Dg \bullet Df$ natural in $f\in \mathcal A(a_1,a_2)$ and $g\in \mathcal A(a_2,a_3)$
\end{enumerate}
such that
\begin{enumerate} \item[(a)] $(\id_{Df}\bullet \delta_a)\delta_{f,\id_a} =\id_{Df}$
and $(\delta_a \bullet \id_{Dg})\delta_{\id_a, g} =\id_{Dg}$
for every $a,a_1,a_2\in \mathcal A$ where $f\in \mathcal A(a, a_1)$ and $g\in \mathcal A(a_2, a)$;
\item[(b)]
$$(\delta_{h,g} \bullet D\id_f)\delta_{hg, f}=(D\id_h \bullet \delta_{g,f})\delta_{h,gf}$$
for all $f\in \mathcal A(a_1, a_2)$, $g\in \mathcal A(a_2, a_3)$, $h\in \mathcal A(a_3, a_4)$,
$a_1,a_2,a_3,a_4 \in \mathcal A$.
\end{enumerate}
Note that in the case of the category $\mathcal C$ defined above, there exists an ordering $\preccurlyeq$ on the sets of morphisms $\mathcal C((G_1, S_1, P_1),(G_2, S_2, P_2))$:
we say that $(\psi_1, R_1, Q_1) \preccurlyeq (\psi_2, R_2, Q_2)$
if $R_1 \subseteq R_2$, $Q_1 \subseteq Q_2$, and $\psi_1 = \psi_2 \bigl|_{R_1}$.
This partial ordering turns $\mathcal C$ into a category enriched over posets and, in particular, a $2$-category.
There exists an obvious map $L \colon \mathbf{GrAlg}_F \to \mathcal C$
where for $\Gamma \colon A=\bigoplus_{g\in G} A^{(g)}$ we have $$L(\Gamma):=(G, \supp \Gamma, \lbrace
(g_1,g_2)\in G \times G \mid A^{(g_1)}A^{(g_2)}\ne 0\rbrace)$$
and for a graded morphism $\varphi \colon \Gamma \to
\Gamma_1$,
where $\Gamma_1 \colon B=\bigoplus_{h\in H} B^{(h)}$,
the triple $L(\varphi)=(\psi, R, Q)$ is defined by $R=\lbrace g\in G \mid \varphi\left(A^{(g)}\right)\ne 0 \rbrace$, $$Q=\lbrace(g_1,g_2)\in R\times R\mid \varphi\left(A^{(g_1)}\right)\varphi\left(A^{(g_2)}\right)\ne 0\rbrace,$$
and $\psi$ is defined by $\varphi\left(A^{(g)}\right) \subseteq B^{(\psi(g))}$
for $g\in R$.
Unfortunately, $L$ is not an ordinary functor: we have \begin{equation}\label{EqLSucc}L(\varphi_1)L(\varphi_2)
\succcurlyeq L(\varphi_1 \varphi_2),\end{equation}
but the inequality~(\ref{EqLSucc}) can be strict:
\begin{example}
Let $A=A^{(\bar 0)} \oplus A^{(\bar 1)}$ be a $\mathbb Z/2\mathbb Z$-graded algebra,
where $$A^{(\bar 0)}=F1_A,\quad A^{(\bar 1)}=Fa\oplus Fb,\quad a^2=ab=ba=b^2=0.$$
Let $\varphi \colon A \to A$ be the graded homomorphism defined by $\varphi(1_A)=1_A$, $\varphi(a)=b$, $\varphi(b)=0$.
Then $$L(A)=(\mathbb Z/2\mathbb Z, \mathbb Z/2\mathbb Z, (\mathbb Z/2\mathbb Z)^2 \backslash \lbrace (\bar 1, \bar 1)\rbrace),$$ $$L(\varphi)=(\id_{\mathbb Z/2\mathbb Z}, \mathbb Z/2\mathbb Z,
(\mathbb Z/2\mathbb Z)^2 \backslash \lbrace (\bar 1, \bar 1)\rbrace)),$$
and $L(\varphi)^2=L(\varphi)$, however $$L(\varphi^2)=(\id_{\lbrace \bar 0 \rbrace}, \lbrace \bar 0 \rbrace, \lbrace (\bar 0, \bar 0) \rbrace)\prec L(\varphi)^2.$$
\end{example}
The inequality~(\ref{EqLSucc})
means that there is a $2$-cell between $L(\varphi_1)L(\varphi_2)$
and $L(\varphi_1 \varphi_2)$. This turns $L$ to an oplax $2$-functor between $\mathbf{GrAlg}_F$
and $\mathcal C$ if we treat the category $\mathbf{GrAlg}_F$ as a $2$-category with discrete hom-categories (i.e. the only $2$-cells in $\mathbf{GrAlg}_F$ are the identity $2$-cells between morphisms). All the equalities in the definition of an oplax $2$-functor hold, since in the categories which are posets all diagrams commute.
\subsection{The universal grading group functors}\label{SubsectionUniversalGradingGroupFunctors}
In order to obtain an ordinary functor between $1$-categories, we have to restrict the sets of possible morphisms.
We call a graded homomorphism \textit{graded injective} if its restriction to each graded component
is an injective map.
\begin{example}
Let $\chi \colon G \to F^\times$ be a homomorphism of a group $G$ to the multiplicative group $F^\times$ of a field $F$, i.e. a linear character.
Then the induced homomorphism $\tilde \chi \colon FG \to F$
where $\tilde\chi\left(\sum_{g\in G}\alpha_g u_g\right):=
\sum_{g\in G}\alpha_g \chi\left( g\right)$, is graded injective.
(Here we denote by $u_g$ the elements of the standard basis in $FG$.)
\end{example}
Consider the category $\widetilde{\mathbf{GrAlg}_F}$ where the objects are group gradings on (not necessarily unital) algebras over a field $F$ and morphisms are graded injective homomorphisms between the corresponding algebras.
Then we have an obvious functor $R \colon \widetilde{\mathbf{GrAlg}_F} \to \mathbf{Grp}$
where for $\Gamma \colon A=\bigoplus_{g\in G} A^{(g)}$ we have $R(\Gamma):=G_{\Gamma}$
and for a graded injective morphism $\varphi \colon
\Gamma \to \Gamma_1$, where $\Gamma_1 \colon B=\bigoplus_{h\in H} B^{(h)}$,
the group homomorphism $R(\varphi)$ is defined by $\varphi\left(A^{(g)}\right) \subseteq B^{\bigl(R(\varphi)(g)\bigr)}$, $g\in G$.
One can also restrict his consideration to unital algebras.
Denote by $\widetilde{\mathbf{GrAlg}_F^1}$ the category where the objects are group gradings on unital algebras over a field $F$ and morphisms are unital graded injective homomorphisms between the corresponding algebras.
Denote by $R_1$ the restriction of the functor $R$ to the category $\widetilde{\mathbf{GrAlg}_F^1}$.
We call $R$ and $R_1$ the \textit{universal grading group functors}.
When it is clear, with which grading an algebra $A$ is endowed, we identify $A$ with the grading on it and treat $A$ as an object of $\widetilde{\mathbf{GrAlg}_F}$ or $\widetilde{\mathbf{GrAlg}_F^1}$.
\subsection{Products and coproducts in $\widetilde{\mathbf{GrAlg}_F}$
and $\widetilde{\mathbf{GrAlg}^1_F}$}
In this section we show that the restriction of the set of morphisms
to graded injective ones makes the categories $\widetilde{\mathbf{GrAlg}_F}$
and $\widetilde{\mathbf{GrAlg}^1_F}$ quite different from $\mathbf{GrAlg}_F$.
It can be shown that if $A=\bigoplus_{g\in G} A^{(g)}$
and $B=\bigoplus_{h\in H} B^{(h)}$ are two group graded algebras, then their
product in $\mathbf{GrAlg}_F$ equals their product in the category of algebras endowed with the induced $G\times H$-grading and their coproduct in $\mathbf{GrAlg}_F$ is their coproduct in category of algebras endowed with the induced $G*H$-grading where $G*H$ the coproduct (or the free product) of the groups $G$ and $H$. (See the general definition of a product and a coproduct in a category e.g. in~\cite[Chapter III, Section 3,4]{MacLaneCatWork}.) The proposition below shows that products in $\widetilde{\mathbf{GrAlg}_F}$ and $\widetilde{\mathbf{GrAlg}^1_F}$ could be different from those
in $\mathbf{GrAlg}_F$.
Recall that by $(u_g)_{g\in G}$ we denote the standard basis in a group algebra $FG$.
\begin{proposition} Let $G$ and $H$ be groups and let $F$ be a field.
Then $F(F^{\times} \times G \times H)$ is the product of $FG$ and $FH$ (with the standard gradings) in both categories $\widetilde{\mathbf{GrAlg}_F}$
and $\widetilde{\mathbf{GrAlg}^1_F}$.
\end{proposition}
\begin{proof}
Let $\pi_1 \colon F(F^{\times} \times G \times H) \to FG$
and $\pi_2 \colon F(F^{\times} \times G \times H) \to FH$
be the homomorphisms defined by $\pi_1(u_{(\alpha,g,h)})=\alpha u_g$
and $\pi_2(u_{(\alpha,g,h)})=u_h$
for $\alpha \in F^\times$, $g\in G$, $h\in H$.
Obviously, they are graded injective.
Suppose there exists a graded algebra $A$ and graded injective homomorphisms
$\varphi_1 \colon A \to FG$ and $\varphi_2 \colon A \to FH$.
We claim that there exists a unique graded injective homomorphism
$\varphi \colon A \to F(F^{\times} \times G \times H)$
such that the following diagram commutes:
$$\xymatrix{ & F(F^{\times} \times G \times H)
\ar[ld]_{\pi_1} \ar[rd]^{\pi_2} & \\
FH & & FG \\
& A \ar[lu]_{\varphi_1} \ar[ru]^{\varphi_2} \ar@{-->}[uu]^\varphi &
}
$$
First we notice that, since each graded component of $FG$ has dimension one
and $\varphi_1$ is graded injective,
each graded component of $A$ must have dimension at most one.
Suppose now that the graded injective homomorphism $\varphi$
indeed exists.
Let $a \in A$ be a homogeneous element.
Then $\varphi(a)$
must be homogeneous too, i.e. $\varphi(a)=\alpha u_{(\beta, g,h)}$
for some scalars $\alpha, \beta \in F^\times$ and group elements $g\in G$, $h\in H$.
Then $\varphi_1(a)=\pi_1\varphi(a)= \alpha\beta u_g$
and $\varphi_2(a)=\pi_2\varphi(a)= \alpha u_h$, i.e. $\varphi(a)$
is uniquely determined by $\varphi_1(a)$ and $\varphi_2(a)$.
Given $\varphi_1$ and $\varphi_2$, the homomorphism $\varphi$ is defined as follows.
If $\varphi_1(a)=\lambda u_g$ and $\varphi_1(a)=\mu u_h$,
then $\varphi(a)=\mu u_{(\lambda/\mu, g,h)}$.
\end{proof}
\begin{corollary}
If the field $F$ consists of more than $2$ elements, then the functors $R$ and $R_1$ do not have left adjoints.
\end{corollary}
\begin{proof}
Each functor that has a left adjoint preserves limits (see e.g.~\cite[Chapter~V, Section 5, Theorem~1]{MacLaneCatWork}) and, in particular, products. However, $$R(F(F^{\times} \times G \times H))=
R_1(F(F^{\times} \times G \times H))=F^{\times} \times G \times H
\ncong R(FG) \times R(FH)=G\times H$$
in the case when both groups $G$ and $H$ are finite.
\end{proof}
In the next section we prove that fact for a field $F$ of any cardinality.
Below we show that coproducts in $\widetilde{\mathbf{GrAlg}_F}$
and $\widetilde{\mathbf{GrAlg}^1_F}$ do not always exist.
\begin{proposition} Let $G$ and $H$ be groups and let $F$ be a field.
Then the coproduct of $FG$ and $FH$ (with the standard gradings) in
the category $\widetilde{\mathbf{GrAlg}_F}$ does not exist.
\end{proposition}
\begin{proof}
Suppose $A$ is the coproduct of $FG$ and $FH$ in $\widetilde{\mathbf{GrAlg}_F}$ and $i_1 \colon FG \to A$
and $i_2 \colon FH \to A$ are the corresponding morphisms.
Let $\varphi_1 \colon FG \to F(G\times H)$ and
$\varphi_2 \colon FH \to F(G\times H)$ be the natural embeddings.
Then there must exist $\varphi \colon A \to F(G\times H)$
such that the following diagram commutes:
$$\xymatrix{ & A
\ar@{-->}[dd]^\varphi & \\
FH \ar[ru]^{i_1}\ar[rd]^{\varphi_1} & & FG \ar[lu]_{i_2}\ar[ld]_{\varphi_2} \\
& F(G \times H) &
}
$$
In particular, $$\varphi(i_1(u_g)i_2(u_h))=u_{(g,1_H)}u_{(1_G,h)}= u_{(g,h)}\ne 0$$ and $i_1(u_g)i_2(u_h)\ne 0$.
Now let $\psi_1 \colon FG \to FG\oplus FH$ and
$\psi_2 \colon FH \to FG\oplus FH$ be the natural embeddings.
Then there must exist $\psi \colon A \to FG\oplus FH$
such that the following diagram commutes:
$$\xymatrix{ & A
\ar@{-->}[dd]^\psi & \\
FH \ar[ru]^{i_1}\ar[rd]^{\psi_1} & & FG \ar[lu]_{i_2}\ar[ld]_{\psi_2} \\
& FG\oplus FH &
}
$$
In particular, $$\psi(i_1(u_g)i_2(u_h))=(u_g, 0)(0, u_h)= 0$$ and we get a contradiction since $i_1(u_g)i_2(u_h)\ne 0$ is a homogeneous element.
\end{proof}
One can show that the coproduct of $FG$ and $FH$ in $\widetilde{\mathbf{GrAlg}_F^1}$ equals $F(G*H)$.
\begin{proposition} Let $F$ be a field
and let $A_i=\langle 1, a_i\rangle_F$, $i=1,2$, be two two-dimensional algebras
such that $a_i^2=0$ with the $\mathbb Z/2\mathbb Z$-grading defined by $a_i \in A_i^{(\bar 1)}$.
Then the coproduct of $A_1$ and $A_2$ does not exist neither in
the category
$\widetilde{\mathbf{GrAlg}_F}$, nor in the category $\widetilde{\mathbf{GrAlg}_F}^1$.
\end{proposition}
\begin{proof}
Suppose $A$ is the coproduct of $A_1$ and $A_2$ and $i_j \colon A_j \to A$, $j=1,2$, are the corresponding morphisms.
Let $A_0=\langle 1, a_1,a_2\rangle_F$ be the three-dimensional algebra
defined by $a_1^2=a_2^2=a_1a_2=a_2a_1 = 0$
with the $\mathbb Z/3\mathbb Z$-grading defined by $a_j \in A_0^{(\bar j)}$, $j=1,2$.
Let $\varphi_j \colon A_j \to A_0$ be the natural embeddings.
Then there must exist $\varphi \colon A \to A_0$
such that the following diagram commutes:
$$\xymatrix{ & A
\ar@{-->}[dd]^\varphi & \\
A_1 \ar[ru]^{i_1}\ar[rd]^{\varphi_1} & & A_2 \ar[lu]_{i_2}\ar[ld]_{\varphi_2}
\\
& A_0 &
}
$$
In particular, $\varphi(i_1(a_1)i_2(a_2))=a_1 a_2=0$ and
$i_1(a_1)i_2(a_2) = 0$ since both $i_1(a_1)$ and $i_2(a_2)$
are homogeneous elements and $\varphi$ is graded injective.
Now let $B=\langle 1, b_1,b_2, b_1b_2\rangle_F$ be the four-dimensional algebra
defined by $b_1^2=b_2^2=b_2b_1 = 0$
and the $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$-grading defined by $b_1 \in B^{(\bar 1,\bar 0)}$, $b_2 \in B^{(\bar 0,\bar 1)}$.
Let $\psi_j \colon A_j \to B$, where $j=1,2$, be the embeddings defined by $a_j \mapsto b_j$.
Then there must exist $\psi \colon A \to B$
such that the following diagram commutes:
$$\xymatrix{ & A
\ar@{-->}[dd]^\psi & \\
A_1\ar[ru]^{i_1}\ar[rd]^{\psi_1} & & A_2 \ar[lu]_{i_2}\ar[ld]_{\psi_2} \\
& B & }$$
In particular, $\psi(i_1(a_1)i_2(a_2))=b_1 b_2 \ne 0$ and we get contradiction with $i_1(a_1)i_2(a_2)= 0$.
\end{proof}
\subsection{Absence of adjoints for the universal grading group functors}
In this section we show that, unlike functors considered in Section~\ref{SubsectionGradedAdjunctions}, the functors $R$ and $R_1$ defined in Section~\ref{SubsectionUniversalGradingGroupFunctors}
do not have neither left nor right adjoints.
\begin{proposition}\label{PropositionAbsenceOfAdjointsUGGF}
The functor $R$ has neither left nor right adjoint.
\end{proposition}
\begin{proof} Suppose $K$ is left adjoint for $R$. Then
we have a natural bijection $$\widetilde{\mathbf{GrAlg}_F}(K(H), \Gamma) \to \mathbf{Grp}(H, R(\Gamma)).$$
We claim that for each group $H$ the grading $K(H)$ is a grading
on the zero algebra.
Suppose $K(H) \colon A=\bigoplus_{g\in G} A^{(g)}$ and $A\ne 0$.
Let $\Lambda$ be a set of indices such that $|\Lambda| > |\Hom(H,G_{K(H)})|$.
Consider the direct sum $\bigoplus_{\lambda \in \Lambda} A$
of copies of $A$ where each copy retains its $G$-grading $K(H)$. Denote by $\Xi$ the resulting $G$-grading on $\bigoplus_{\lambda \in \Lambda} A$.
Then $G_{\Xi} \cong G_{K(H)}$, however $$|\widetilde{\mathbf{GrAlg}_F}(K(H), \Xi)|
\geqslant |\Lambda| > |\Hom(H,G_{K(H)})|=|\mathbf{Grp}(H, R(\Xi))|$$
which contradicts the existence of the natural bijection. Hence
for each group $H$ the grading $K(H)$ is a grading on the zero algebra.
In particular, each set $\widetilde{\mathbf{GrAlg}_F}(K(H), \Gamma)$
contains exactly one element.
If $H$ is a group that admits a non-trivial automorphism
and $FH$ is its group algebra with the standard grading $\Gamma$, then $\mathbf{Grp}(H, R(\Gamma))=\Hom(H,H)$
contains more than one element and we once again get a contradiction. Hence the left adjoint functor for
$R$ does not exist.
Suppose $K$ is right adjoint for $R$. Then
we have a natural bijection $$\mathbf{Grp}(R(\Gamma), H) \to \widetilde{\mathbf{GrAlg}_F}(\Gamma, K(H)).$$
Considering projections of $\bigoplus_{\lambda \in \Lambda} A$ on different components (now we assume $|\Lambda| > |\Hom(G_{K(H)}, H)|$),
we again prove that $K(H)$ is a zero grading for every group $H$, each $\widetilde{\mathbf{GrAlg}_F}(\Gamma, K(H))$ must contain at most one element, but $\mathbf{Grp}(R(\Gamma), H)$ can contain more than one element. We get a contradiction and the right adjoint functor for
$R$ does not exist.
\end{proof}
\begin{proposition}\label{PropositionAbsenceOfAdjointsUGGF1}
The functor $R_1$ has neither left nor right adjoint.
\end{proposition}
\begin{proof} Suppose $K$ is left adjoint for $R_1$. Then
we have a natural bijection $$\widetilde{\mathbf{GrAlg}^1_F}(K(H), \Gamma) \to \mathbf{Grp}(H, R_1(\Gamma)).$$
We claim that for any group $H$ the grading $K(H)$ is the grading
on an algebra isomorphic to $F$.
Indeed, let $H$ be a group and let $K(H) \colon A=\bigoplus_{g\in G} A^{(g)}$.
Denote by $\Upsilon$ the grading on $F$ by the trivial group. Then $\mathbf{Grp}(H, R_1(\Upsilon))$
and $\widetilde{\mathbf{GrAlg}^1_F}(K(H), \Upsilon)$ both
consist of one element. In particular, there exists a unital homomorphism
$\varphi \colon A \to F$. Hence there exists an ideal $\ker \varphi \subsetneqq A$ of codimension $1$.
Now denote by $\Xi$ the grading on $A$ by the trivial group.
If $\ker \varphi\ne 0$, then $\widetilde{\mathbf{GrAlg}_F^1}(K(H), \Xi)$
consists of at least two different elements: the identity map $A\to A$
and the composition of $\varphi$
and the embedding $F \cdot 1_A \to A$. Since $\mathbf{Grp}(H, R_1(\Xi))$
consists of a single element, we get $\ker \varphi=0$ and $A \cong F$.
In particular, $\widetilde{\mathbf{GrAlg}_F^1}(K(H), \Gamma)$ contains exactly one element.
Again, considering a group $H$ that admits a non-trivial automorphism
and its group algebra $FH$ with the standard grading $\Gamma$, we obtain that $\mathbf{Grp}(H, R(\Gamma))=\Hom(H,H)$ contains more than one element and we get a contradiction. Hence the left adjoint functor for $R_1$ does not exist.
Suppose $K$ is right adjoint for $R_1$. Then
we have a natural bijection $$\mathbf{Grp}(R_1(\Gamma), H) \to \widetilde{\mathbf{GrAlg}^1_F}(\Gamma, K(H)).$$
Note that the set $\mathbf{Grp}(R_1(\Gamma), H)$ is always
non-empty since it contains at least the homomorphism that maps everything to $1_H$. Fix a group $H$. Let $K(H) \colon A=\bigoplus_{g\in G} A^{(g)}$.
Let $B$ be an $F$-algebra with the cardinality $|B|$
that is greater than the cardinality $|A^{(1_G)}|$. For example,
$B = \End_F V$ where $V$ is a vector space with a basis that has a cardinality greater than $|A^{(1_G)}|$. Let $\Gamma$ be the grading on $B$ by the trivial group. Then there exist no injective maps $B \to A^{(1_G)}$
and therefore the set $\widetilde{\mathbf{GrAlg}^1_F}(\Gamma, K(H))$
is empty. Again we get a contradiction and the right adjoint functor for
$R$ does not exist.
\end{proof}
The observations above show that, in order to get indeed an adjuction, one must restrict the category of algebras
to the algebras that are well determined by their universal grading group.
\subsection{Adjunction in the case of group algebras}
Let $F$ be a field and let $\mathbf{Grp}'_F$ be the category where the objects
are groups $G$ that do not have non-trivial one dimensional representations (in other words, $H^1(G, F^{\times}) = 0$) and the morphisms are all group homomorphisms. Let $\mathbf{GrpAlg}'_F$ be the category
where the objects are group algebras $FG$ of groups $G$ from $\mathbf{Grp}'_F$
with the standard grading
and the morphisms are all non-zero graded algebra homomorphisms.
Let $U$ be the functor $\mathbf{GrpAlg}'_F \to \mathbf{Grp}'_F$
defined by $U(FG)=G$ and $\varphi\left(FG_1^{(g)}\right)\subseteq FG_1^{\left(U(\varphi)(g) \right)}$
for $\varphi \colon FG_1 \to FG_2$.
Denote by $F-$ the functor which associates to each group its group algebra over the field $F$.
\begin{proposition}\label{PropositionEquivGroupAlgGroups}
There exists a natural bijection $\theta_{G, A} \colon \mathbf{GrpAlg}'_F(FG, A)
\to \mathbf{Grp}'_F(G, U(A))$ where $G \in \mathbf{Grp}'_F$
and $A\in \mathbf{GrpAlg}'_F$.
Furthermore, $FU(-)= 1_{\mathbf{GrpAlg}'_F}$
and $U(F-)= 1_{\mathbf{Grp}'_F}$, i.e. the categories
$\mathbf{Grp}'_F$ and $\mathbf{GrpAlg}'_F$ are isomorphic.
\end{proposition}
\begin{proof}
Let $\varphi\colon FG \to FH$ be a non-zero graded algebra homomorphism.
Then $\varphi(u_{g_0}) \ne 0$ for some $g_0\in G$ and
$\varphi(u_{g_0})=\varphi(u_{g_0} u_g^{-1}) \varphi(u_g)\ne 0$
implies $\varphi(u_g)\ne 0$ for all $g\in G$. Hence $\varphi$ is graded injective. Therefore $\varphi$ is determined by group homomorphisms $\psi \colon G \to H$ and $\alpha \colon G \to F^\times$ such that $\varphi(u_g) = \alpha(g)u_{\psi(g)}$ for $g\in G$.
Since $G$ does not have non-trivial one dimensional representations,
$\alpha$ is trivial and we have the natural bijection $\theta_{G, A}$. The equalities $FU(-)\cong 1_{\mathbf{GrpAlg}'_F}$
and $U(F-)\cong 1_{\mathbf{Grp}'_F}$ are verified directly.
\end{proof} | 0.026318 |
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\begin{document}
\title{Isomorphisms of Poisson systems over locally compact groups}
\author{Amanda Wilkens}
\thanks{This research was supported in part by NSF grant DMS-1937215.}
\address{Department of Mathematics\\ University of Texas at Austin\\ 2515 Speedway, PMA \indent 8.100\\ Austin, Texas 78712\\ United States}
\email{amanda.wilkens@math.utexas.edu}
\urladdr{http://web.ma.utexas.edu/users/amandawilkens}
\keywords{}
\subjclass[2020]{}
\begin{abstract}
A Poisson system is a Poisson point process and a group action, together forming a measure-preserving dynamical system.
Ornstein and Weiss proved Poisson systems over many amenable groups were isomorphic in \cite{MR910005}.
We consider Poisson systems over non-discrete, non-compact, locally compact Polish groups, and
we prove by construction all Poisson systems over such a group are finitarily isomorphic,
producing examples of isomorphisms for non-amenable group actions.
As a corollary, we prove Poisson systems and products of Poisson systems are finitarily isomorphic.
\end{abstract}
\maketitle
\section{Introduction}\label{intro}
In what follows, we construct an isomorphism between any two Poisson systems over a non-discrete, non-compact, locally compact Polish group.
We define Poisson systems and isomorphisms between them before formally stating our main result.
Let $\gamma>0$. A random variable $Z$ taking values on $\mathbb N:=\{0,1,2,\ldots\}$ with $$\mathbb P(Z =m) = \frac{e^{-\gamma}\gamma^m}{m!}$$ is a \textit{Poisson random variable with mean $\gamma$.}
Let $G$ be a group with $\sigma$-finite Haar measure $\lambda$, and let $\alpha>0$. A \textit{Poisson point process on $G$ with intensity $\alpha$} is a random process $X$, which takes values on the space $\mathbb M:=\mathbb M(G,\mathcal B(G))$ of Borel simple point measures on $G$,
such that: For every Borel subset $A \in \mathcal B(G)$ with $\lambda(A)<\infty$, the number of points of $X$ in $A$, denoted by $N(A)$, is a Poisson random variable with mean $\alpha\cdot\lambda(A)$. For pairwise disjoint Borel sets $A_1, \ldots, A_{\ell}$, the random variables $N(A_1), \ldots, N(A_{\ell})$ are independent.
Let $P_\alpha$ be the law of a Poisson point process on $G$ with intensity $\alpha$, and let $\mu\in\mathbb M$ and $g\in G$.
The triple $(\mathbb M, P_\alpha, G)$ is a measure-preserving dynamical system, which we refer to as a \textit{Poisson system} and denote as $\mathbb X_\alpha$,
where the action of $G$ on $\mathbb M$ is given by $g(\mu)(A)=\mu(g^{-1}A)$ for all $A\in\mathcal B(G)$.
It may be helpful to think of a Poisson system as a random collection of points in a state space, along with actions inherited from the state space which left-translate the collection of points around the space.
A map $\phi:\mathbb X_\alpha\rightarrow \mathbb X_\beta$ is a \textit{factor from $\mathbb X_\alpha$ to $\mathbb X_\beta$} if, on a set of $P_\alpha$ full-measure, it behaves with respect to the group action and relevant measures, so that $\phi\circ g=g\circ\phi$ and $P_\alpha\circ\phi^{-1}=P_\beta$. A factor is an \textit{isomorphism} if it is a bijection and its inverse is a factor from $\mathbb X_\beta$ to $\mathbb X_\alpha$.
We impose some additional restrictions on $G$ and, by the end of Section \ref{group}, will have detailed each choice. Let $G$ be non-discrete, non-compact, and locally compact Polish.
\begin{thm}\label{iso}
All Poisson systems over $G$ are isomorphic.
\end{thm}
Ornstein and Weiss proved all Poisson systems over any group with ``good entropy theory'' were isomorphic \cite{MR910005}.
Groups with good entropy theory are locally compact, amenable, and unimodular, with some further restrictions. There are many new isomorphism examples as a result of Theorem \ref{iso}; some in particular are for Poisson systems over the Euclidean group $E(n)$ for $n\geq 3$ and the special linear group $SL(n,\mathbb R)$ for $n\geq 2$. Any real Lie group with countably many connected components satisfies the conditions on $G$. So does the field $\mathbb Q_p$ of $p$-adic numbers for any prime $p$.
Our construction does not distinguish between amenable and non-amenable groups, or unimodular and non-unimodular.
We prove the next theorem over the course of proving Theorem \ref{iso}.
\begin{thm}\label{factor}
Let $\alpha, \beta >0$.
There exists a factor from $\mathbb X_\alpha$ to $\mathbb X_\beta$.
\end{thm}
Holroyd, Lyons, and Soo construct a factor from $\mathbb X_\alpha$ to $\mathbb X_\beta$ in the case $G=\mathbb R^n$, for all $n\geq 1$, in their paper on Poisson splitting \cite{splitting}. The Ornstein and Weiss isomorphisms of \cite{MR910005}, while applicable to a large class of groups, are non-constructive. Via an explicit construction, Holroyd, Lyons, and Soo gain some additional properties. The factor maps of \cite{splitting} extend the action from translations to isometries of $\mathbb R^n$ and are finitary, which we define shortly. Our proof of Theorem \ref{factor} extends the construction in \cite{splitting}.
In \cite{finisoppp}, Soo and the author construct an isomorphism between any two Poisson systems over $\mathbb R^n$. The isomorphism builds on the factor map from \cite{splitting} and carries its additional properties. Our proof of Theorem \ref{iso} extends this construction.
Recall, a locally compact space is Polish if and only if it is second countable.
Birkhoff and Kakutani proved a topological group is metrizable if and only if it is first countable (\cite{metgeo}, Theorem 2.B.2).
A metric $d$ on a topological space $(X,T)$ is \textit{proper} if spheres with respect to $d$ are relatively compact, and is \textit{compatible} if $d$ defines $T$.
Struble proved
a locally compact topological group is metrizable with a proper, left-invariant, compatible metric if and only if it is second countable \cite{struble}.
To define a finitary factor, we fix
$d$ as a proper, left-invariant, compatible metric on $G$ and
denote the open sphere centered at $g\in G$ of radius $r$, with respect to the metric $d$, as $B(g,r)$ (although, for most of the paper, we use a more specific metric and a different definition of $B(g,r)$, beginning in Section \ref{marker}).
We use either the random variable $X$, which takes values on $\mathbb M$, or the collection of Borel simple point measures $\mu\in\mathbb M$ to refer to a Poisson point process on $G$, depending on context. The restriction of $\mu\in\mathbb M$ to a Borel set $A$ is $\mu|_A(B):=\mu(B\cap A)$ for $B\in\mathcal B(G)$ and is itself a Poisson process on $A$ (\cite{kingman}, Chapter 2).
Let $\phi$ be a factor from $\mathbb X_\alpha$ to $\mathbb X_\beta$, and let $\mu,\mu'\in\mathbb M$. We always denote the identity element of a group as $e$. A \textit{coding window} of $\phi$ is a map $\omega:\mathbb M\rightarrow \mathbb N\cup \{\infty\}$ such that if
$$\mu|_{B(e,\omega(\mu))}=\mu'|_{B(e,\omega(\mu))}$$ then $\phi(\mu)|_{B(e,1)}=\phi(\mu')|_{B(e,1)}$. The factor map $\phi$ is \textit{finitary} if there exists a $P_\alpha$ almost-surely finite coding window of $\phi$, or in other words, if $P_\alpha$ almost-surely the sphere $B(e,1)$ under $\mathbb X_\beta$ is completely determined by a sphere at $e$ of finite radius under $\mathbb X_\alpha$.
The choice of $e$ is arbitrary due to $G$-equivariance of $\phi$.
A finitary factor is a finitary isomorphism if it is a bijection and its inverse is a finitary factor from $\mathbb X_\beta$ to $\mathbb X_\alpha$.
\begin{thm}\label{finitary}
All Poisson systems over $G$ are finitarily isomorphic.
\end{thm}
Theorem \ref{finitary} is a direct consequence of the proofs of Theorems \ref{iso} and \ref{factor}, wherein we extend the construction of the finitary maps between Poisson systems over $\mathbb R^n$ from \cite{splitting, finisoppp}.
\subsection{The isomorphism problem for countable groups}
Theorems \ref{iso}, \ref{factor}, and \ref{finitary} have a rich history in the setting of countable groups, where Poisson systems take the form of Bernoulli shifts. Let $D$ be a countable group and $(K,\kappa)$ a standard Borel probability space. Endow
$K^D:=\{x:D\rightarrow K\}$
with the product $\sigma$-algebra generated by cylinder sets, and let $\kappa^D$ be the product measure on $K^D$.
The group $D$ acts on $K^D$ by $(dx)(g)=x(d^{-1}g)$ for all $x\in K^D$ and $d,g\in D$. The system $(K^D,\kappa^D,D)$ is called a \textit{Bernoulli shift over $D$.} It was an open question as to whether the Bernoulli shifts over the integers with outcome sets $\{0,1\}$ and $\{0,1,2\}$ were isomorphic as measure-preserving dynamical systems for about twenty years, until the late 1950's.
Around that time, Kolmogorov formalized the concept of entropy for ergodic theory \cite{MR2342699}. Entropy measures the randomness of a system and is invariant under isomorphism in Kolmogorov's construction. Entropy does not apply to the isomorphism question in our setting, so we do not give a formal definition. But we note the entropies of the Bernoulli shifts referenced above are different, implying the shifts are not isomorphic. (Poisson systems over groups with good entropy theory have infinite entropy \cite{MR910005}, and one might conjecture the same holds for Poisson systems over $G$.)
Kolmogorov further proved entropy is non-increasing under factor maps between Bernoulli shifts over $\mathbb Z$.
More results followed, including Sinai's on the existence of a factor map between shifts with non-increasing entropy \cite{MR2342699}, and Ornstein's proof that shifts are isomorphic if and only if they share the same entropy \cite{Ornstein}.
Keane and Smorodinsky later constructed isomorphisms between shifts with finite coding windows \cite{keanea, keaneb}. The finitary definition for Poisson systems first appears in \cite{splitting} under the terminology ``strongly finitary'' and is based on the finitary definition for Bernoulli shifts, which can be found in \cite{MR2306207}.
In 1975, Kieffer proved entropy must be non-increasing for any factor map between shifts with group action from a countable amenable group \cite{MR393422}, and Stepin introduced a method called coinduction, which we use as well, to extend Ornstein's result to all countable groups containing $\mathbb Z$ as a subgroup and some further groups \cite{stepin}. In their 1987 paper, Ornstein and Weiss, as a precursor to proving Poisson systems over groups with good entropy theory were isomorphic, proved Bernoulli shifts with equal entropy over such groups were isomorphic \cite{MR910005}.
More recently, Bowen \cite{MR2552252} and Kerr and Li \cite{MR2813530} extended entropy theory to countable sofic groups.
Bowen additionally proved shifts with equal entropy over any countable group, as long as both shifts have outcome sets of cardinality at least $3$, were isomorphic \cite{almostornstein}. Seward removed the cardinality restriction and added the finitary property in \cite{seward}, thus closing the open question for the countable sofic setting. In \cite{MR3959054}, Seward states a conjecture which, if true, closes the open question in full generality for countable groups.
\subsection{Paper outline}
We first push the constructions on $\mathbb R^n$ in \cite{splitting, finisoppp} through to non-discrete, non-compact, and locally compact Polish groups that are compactly generated (Section \ref{cg}); see Sections \ref{group} and \ref{marker} for more information on the compactly generated requirement. From here, we drop the compactly generated condition with two coinduction arguments, one for the case in which $G$ contains a non-compact, compactly generated open subgroup (Section \ref{noncompact}), and one for the case in which $G$ does not (Section \ref{compact}).
We discuss technical details of Poisson point processes and systems on groups, obstructions to lifting any requirements on $G$, and a further question
in Section \ref{group}.
We begin construction of an isomorphism between Poisson systems over compactly generated $G$ in Section \ref{marker} by building a $G$-equivariant partition deterministically from information in the source system.
The idea goes back to Keane and Smorodinsky's finitary factor maps for Bernoulli shifts in the $\mathbb Z$ setting, as found in \cite{keanea}. Here, Keane and Smorodinsky introduce markers and fillers; markers are strings of symbols defined to identify potential sources of randomness, and fillers occupy the remaining space. To be useful, markers must be defined carefully enough as to not disrupt inherent independence conditions of the source system. We prove independence properties and harness the randomness made available by our marker system in Section \ref{random}.
To convert pieces of randomness into uniform random variables, we rely on the Borel isomorphism theorem for measures (see Remark \ref{bitform}) and the uncountable property of $G$ (since $G$ is non-discrete and second countable, it must be uncountable). We apply the isomorphism theorem again to convert the uniform random variables into Poisson point processes on bounded Borel sets determined by the marker partition. As long as we are careful with independence properties, the processes sum to a Poisson point process on $G$ (see Remark \ref{cutpaste}). This is the idea behind the proof of Theorem \ref{factor}.
The construction up until this point builds on the one in \cite{splitting}. But towards proving Theorem \ref{iso} we apply tools from \cite{finisoppp}. We note the method described in the above paragraph is not bijective; when generating a Poisson point process on a bounded Borel set, the unique empty process occurs with non-zero probability. Our path around this obstruction is an intermediate isomorphism from a Poisson system to a particular product of Poisson systems, which we state in Proposition \ref{preproduct} after further definitions.
Let $\alpha,\beta_1,\beta_2>0$ and $\psi:\mathbb X_\alpha \rightarrow \mathbb X_{\beta_1}\times \mathbb X_{\beta_2}$. Denote $\psi(\cdot)=(\psi(\cdot)_1,\psi(\cdot)_2)$. If $\psi\circ g=(g\circ \psi_1,g\circ\psi_2)$ on a set of $P_\alpha$ full-measure, and
for a Poisson point process $X$ on $G$ with intensity $\alpha$, the outputs $\psi(X)_i$ are independent Poisson processes on $G$ with intensities $\beta_i$
for $i=1,2$, then we say $\psi$ is a \textit{factor from $\mathbb X_\alpha$ to $(\mathbb X_{\beta_1},\mathbb X_{\beta_2})$}. The map $\psi$ is \textit{finitary} if each coordinate mapping is finitary.
Likewise, let $\zeta:X_{\beta_1} \times X_{\beta_2} \rightarrow X_\alpha$.
If $\zeta\circ g=g\circ \zeta$ and $(P_{\beta_1}\times P_{\beta_2})\circ\zeta^{-1}=P_\alpha$ on a set of $P_{\beta_1}\times P_{\beta_2}$ full-measure, then $\zeta$ is a \textit{factor from $(\mathbb X_{\beta_1},\mathbb X_{\beta_2})$ to $\mathbb X_\alpha$}.
Let $\mu_1,\mu_2,\nu_1,\nu_2\in\mathbb M$. A \textit{coding window} of $\zeta$ is a function $\omega:\mathbb M\times\mathbb M\rightarrow \mathbb N\cup\{\infty\}$, such that if
$$(\mu_1,\mu_2)|_{B(e,\omega(\mu_1,\mu_2))}=(\nu_1,\nu_2)|_{B(e,\omega(\mu_1,\mu_2))},$$
then
$\zeta(\mu_1,\mu_2)|_{B(e,1)}=\zeta(\nu_1,\nu_2)|_{B(e,1)}.$
A factor from $(\mathbb X_{\beta_1},\mathbb X_{\beta_2})$ to $\mathbb X_\alpha$ is \textit{finitary} if there exists a coding window of $\zeta$ that is finite $P_{\beta_1}\times P_{\beta_2}$ almost-surely.
A factor from $\mathbb X_\alpha$ to $(\mathbb X_{\beta_1},\mathbb X_{\beta_2})$ is an (finitary) isomorphism if its inverse is a (finitary) factor from $(\mathbb X_{\beta_1},\mathbb X_{\beta_2})$ to $\mathbb X_\alpha$.
\begin{prop}\label{preproduct}
Let $G$ be a non-discrete, non-compact, locally compact and compactly generated Polish group, and let $\alpha,\beta >0$. There exists a finitary isomorphism from $\mathbb X_\alpha$ to $(\mathbb X_\alpha,\mathbb X_\beta)$.
\end{prop}
With Proposition \ref{preproduct}, we modify the construction used in the proof of Theorem \ref{factor} in order to prove Theorem \ref{iso}. Instead of using randomness from a source system to populate a single Poisson process on $G$, we populate a pair of Poisson processes on $G$, bijectively. We make sure members of the pair are independent so that we may send the permuted pair back through the inverse factor map. All these operations together constitute an isomorphism between Poisson systems.
As a corollary to Proposition \ref{preproduct} and Theorem \ref{iso}, we gain the next result, going back to the usual requirements on $G$.
\begin{cor}\label{product}
Poisson systems over $G$ are finitarily isomorphic to products of Poisson systems over $G$.
\end{cor}
We prove Theorems \ref{iso}, \ref{factor}, and \ref{finitary}, Proposition \ref{preproduct}, and Corollary \ref{product} in stages (first for compactly generated groups, then for groups with compactly generated subgroups, then in general) in Section \ref{construct}.
The main difficulties of the paper are ensuring the collection and distribution of randomness performed by factor maps uphold independence properties, and securing bijectivity. Tools for dealing with the latter port easily from the $\mathbb R^n$ setting and \cite{finisoppp}, while dealing with the former is not as direct and informs the restrictions on $G$. We point out some of the differences between our construction and that of \cite{splitting} as they arise.
\subsection*{Acknowledgments}
The author thanks Lewis Bowen for helpful comments and discussions provided throughout the writing of this paper, particularly those on word metrics, coinduction, and generating sets of compact groups, and Brandon Seward for inspiring the project by suggesting the construction in \cite{finisoppp} might work for Lie groups.
\section{Poisson systems over groups}\label{group}
Although Poisson point processes are most commonly defined on $\mathbb R^n$, all that is needed for existence is a Borel space with measurable singletons and a non-atomic measure satisfying a weak finiteness condition; see \cite{kingman}, Chapter 2. Considering a Poisson point process as a measure-preserving dynamical system restricts the possibilities somewhat.
Recall $\mathbb M$ is the space of Borel simple point measures on $G$. An element of $\mathbb M$ is a collection of ``turned on'' points (points with simple point measure value $1$, simple meaning a point may only be turned on once) scattered throughout $G$. The scattering obeys the law $P_\alpha$ for some $\alpha>0$. For $g\in G$ to act as a measure-preserving transformation on $\mathbb M$, the group $G$ must have a left-invariant measure. Haar's theorem states any locally compact group has such a measure, and a later converse proved in stages by Weil and Mackey states for a left-invariant measure to exist on a group, the group must be locally compact (\cite{metgeo}, Chapter 1). Hence, for a Poisson system over $G$ to be defined, we cannot weaken the locally compact condition on $G$.
Further, the intensity of the process must be a constant value. Poisson point processes with constant intensity are called \textit{homogeneous} or, for reasons we will address in this section, \textit{uniform}.
If Haar measure on $G$ is $\sigma$-finite, then it
satisfies the necessary conditions to host a Poisson process. It may be possible to remove the $\sigma$-finite condition and still have examples of groups where Poisson systems are valid, but this situation is less clear than the locally compact one. We note if a locally compact group is not $\sigma$-compact, its Haar measure is not $\sigma$-finite (\cite{harmonic}, Chapter 2).
\subsection{Construction specific requirements}\label{specific}
Our approach to prove Theorem \ref{iso} is by construction. In order to construct a factor from $\mathbb X_\alpha$ to $\mathbb X_\beta$, we create an output Poisson point process from a source process. To do so, we need plenty of independent randomness from the source, which we collect with a marker system. The construction must be $G$-equivariant, limiting how we can define markers. The definition is based on distances between and densities of clustered points in the source process. Thus we require that $G$ have a left-invariant metric.
Recall Birkhoff and Kakutani's theorem that a group is metrizable if and only if it is first countable.
Struble proved a locally compact group is first countable and $\sigma$-compact if and only if it is second countable \cite{struble}.
We require $G$ to be locally compact and $\sigma$-compact in order for its Haar measure to be left-invariant and $\sigma$-finite. Since we need a left-invariant metric, we reasonably require $G$ to be locally compact and second countable (in other words, locally compact Polish).
We often directly use the second countable property (which implies hereditarily Lindel\"{o}f) in proofs to find countable open covers of $G$ and of open subsets of $G$, replacements for the $\mathbb Z^n$ lattices used in \cite{splitting}.
The marker system in \cite{splitting} depends on a weak connectedness property of $\mathbb R^n$, stated at the start of Section \ref{marker}. A topological group $T$ is said to be \textit{compactly generated} if there exists a compact set $K\subseteq T$ such that
$$T=\bigcup_{i\in\mathbb N}\left(K\cup K^{-1}\cup \{e\}\right)^i.$$
In order for a locally compact Polish group to satisfy the marker property, it suffices that it be compactly generated.
We discuss some nuances in Sections \ref{marker} and \ref{compact}, and remark that
the compactly generated condition is related to the notion of connectedness; see \cite{metgeo}, Proposition 4.B.8.
A locally compact Polish group may not be compactly generated, but
it must contain a compactly generated open subgroup $H$ of countable index (\cite{metgeo}, Corollary 2.C.6).
Then we may apply a version of Theorem \ref{iso} for compactly generated groups (Theorem \ref{preiso}) to Poisson systems over $H$, as well as the other copies of $H$ in $G$ given by left cosets.
This is Stepin's method of coinduction: A $G$-action is induced from the existing $H$-actions on cosets.
Bowen uses coinduction in \cite{weakiso} to prove there exists a factor from any Bernoulli shift to any other one for countable groups containing a non-abelian free subgroup and includes a proof of Stepin's theorem from \cite{stepin}.
The proof applies to our setting with minimal adjustments, which we make in Section \ref{noncompact}, for $H$ non-compact. We make further adjustments if there is no non-compact, compactly generated open subgroup in Section \ref{compact}.
As an example of a locally compact Polish group that is not compactly generated but contains a non-compact, compactly generated open subgroup, consider
$H\times\mathbb R,$ where
$$H:=\bigoplus_{i\in\mathbb N}\mathbb Z/2\mathbb Z.$$
On the other hand, the group $H\times S^1,$ where $S^1$ is the circle group, is locally compact Polish, but has no non-compact, compactly generated open subgroup.
At the core of our construction is a liberal use of the Borel isomorphism theorem for measures, stated in Remark \ref{bitform} as found in \cite{kechris}, Theorem 17.41. We need Borel subsets of $G$ to be standard Borel spaces (that is, isomorphic to a Polish space in which measurable sets are Borel) in order to apply the isomorphism theorem to extract randomness and populate an output Poisson process. Such is the case when $G$ is Polish (\cite{kechris}, Corollary 13.4).
We say a probability measure $\kappa$ on a measurable space $(A,\mathcal B(A))$ is \textit{continuous} if $\kappa(\{a\})=0$ for all $a\in A$. Note Haar measure on a discrete, locally compact group is a counting measure. Thus $G$ being non-discrete is integral to our construction. Certainly, Theorem \ref{iso} is false for countable sofic groups.
\begin{rem}[Borel isomorphism theorem for measures]\label{bitform}
Let $A$ be a standard Borel space and $\kappa$ a continuous probability measure on $(A,\mathcal B(A))$. Then there exists a Borel isomorphism $S:A\rightarrow [0,1]$ with $\kappa \circ S^{-1}$ Lebesgue measure on $[0,1]$.
\end{rem}
Occasionally, we must integrate over the
space $\mathbb M$ equipped with
the smallest $\sigma$-algebra such that the projection maps
$$\pi_A:\mathbb M\rightarrow\mathbb N\cup\{\infty\}$$
sending $\mu\in\mathbb M$ to $\mu(A)$ are measurable for all $A\in\mathcal B(G)$. We denote this $\sigma$-algebra as $\mathcal M(G,\mathcal B(G))$. For this purpose we want Poisson processes on $G$, that is, random variables taking values on $\mathbb M$, to act as well-behaved as they do on $\mathbb R^n$ and $\mathbb M(\mathbb R^n,\mathcal B(\mathbb R^n))$. This is the case when $G$ is a complete separable metric space (equivalently, Polish). Then $\mathbb M$ is such a space as well, and any probability measure on $(\mathbb M,\mathcal M(G,\mathcal B(G)))$ is $\sigma$-finite (\cite{reiss}, Chapter 8). In particular, Fubini's theorem applies, useful for proving independence properties.
For a homogeneous Poisson point process $X$ on a complete separable metric space, given a bounded Borel set $A$ of the space, the points of $X|_A$ are uniformly distributed throughout $A$ (\cite{reiss}, Theorem 1.2.1). We use this property many times throughout the paper and state its most useful form to us in Remark \ref{findist}.
Throughout the paper, we fix $\lambda$ as Haar measure on $G$. A uniform random variable without reference to a set is assumed to be on the interval $[0,1]$. We denote the Dirac measure with mass at $g\in G$ as $\delta(g)$.
\begin{rem}[Distribution on finite measure sets]\label{findist}
Let $X$ be a Poisson point process on $G$ with intensity $\alpha$, and let $A\in\mathcal B(G)$ such that $\lambda(A)$ is finite. Let $Z$ be a Poisson random variable with mean $\alpha\cdot\lambda(A)$, and let $\{U_i\}_{i\in\mathbb N}$ be a sequence of independent random variables, each uniformly distributed on $A$ and independent of $Z$. Then
$$X|_A\overset{d}{=}\sum_{i=1}^Z\delta(U_i).$$
\end{rem}
It remains to address the non-compact requirement on $G$. A Poisson point process of intensity $\alpha$ on a compact group $K$ is the empty process $\emptyset\in\mathbb M(K)$ with probability $e^{-\alpha\cdot\lambda (K)}>0$. Under our construction, it is impossible to produce an output process from an empty source process. But as long as $G$ is non-compact, by definition we are guaranteed, almost-surely, a Poisson point process on $G$ with countably infinitely many points.
\subsection{Non-unimodular case}
Suppose $G$ is locally compact and $\sigma$-compact (so Poisson systems are well-defined). Benjy Weiss proposed the following isomorphism when $G$ is non-unimodular and the modular homomorphism
$\Delta:G\rightarrow (0,\infty)$
is surjective. (The modular homomorphism measures the difference between left and right Haar measures on $G$; see \cite{harmonic}, Chapter 2 for a definition.)
Fix $\alpha>0$, and let $X$ be a Poisson point process on $G$ with intensity $\alpha$. Let $g\in G$. The map sending $X$ to $Xg$ is a finitary isomorphism from $\mathbb X_\alpha$ to $\mathbb X_{\alpha\cdot\Delta(g)}$. Since $\Delta$ is surjective, all Poisson systems over $G$ are isomorphic.
This gives an incredibly short proof of Theorem \ref{iso} under non-unimodular and surjective modular homomorphism assumptions.
\subsection{Extending the group action}
While we maintain the finitary property belonging to the factor maps in \cite{splitting, finisoppp} in our construction, we do not extend the group action from the base group. The referenced factor maps are between systems $(\mathbb M(\mathbb R^n),P_\alpha,\isom\mathbb R^n)$ and $(\mathbb M(\mathbb R^n),P_\beta,\isom\mathbb R^n)$, where $\isom\mathbb R^n$ is the isometry group of $\mathbb R^n$. Our factors will be between $(\mathbb M(G),P_\alpha,G)$ and $(\mathbb M(G),P_\beta,G)$.
Beyond extending the group action to an isometry group, one may think of extensions more generally. For a locally compact group $G$ and a closed subgroup $H$ with modular homomorphisms $\Delta_G$ and $\Delta_H$, there exists a $G$-invariant Radon measure on $G/H$ if and only if $\Delta_G|H=\Delta_H$ (\cite{harmonic}, Theorem 2.49). Lewis Bowen posed the following question.
\begin{q}\label{extend}
Let $G$ be locally compact, let $H$ be a closed subgroup of $G$ such that $\Delta_G|H=\Delta_H$, and let $\alpha,\beta >0$. Under what further conditions are $(\mathbb M(G/H),P_\alpha,G)$ and $(\mathbb M(G/H),P_\beta,G)$ isomorphic?
\end{q}
Evans shows in \cite{Evans} that for the Poisson splitting constructed in \cite{splitting} (related to but not precisely the finitary factor map of the same paper), it is impossible to extend equivariance further than $\isom \mathbb R^n$.
\section{Markers}\label{marker}
Our first step towards an isomorphism is to construct markers.
To do so, we generalize the ones on $\mathbb R^n$ found in \cite{splitting,finisoppp}. These $\mathbb R^n$ markers rely on some connected-like behavior of $\mathbb R^n$ equipped with the Euclidean metric; namely, for all $g,h\in \mathbb R^n$ and $m\in \mathbb N$ with $|g-h|=m$ and $p,q\in \mathbb N$ such that $p+q=m$, there exists $k\in \mathbb R^n$ such that $|g-k|=p$ and $|h-k|=q$.
A word metric on any group always exhibits such behavior and is left-invariant, but depending on the group, may lack any number of desirable properties, such as measurability, properness, or compatibility.
However, a locally compact and compactly generated group has a measurable, proper word metric (\cite{metgeo}, Remark 4.B.3) with other serviceable properties, which we identify as we use.
We fix $G$ as a non-discrete, non-compact, locally compact and compactly generated Polish group throughout this section, Section \ref{random}, and Section \ref{cg}, and
$S$ as a compact generating subset of $G$. Since $G$ is uncountable, so is $S$.
The \textit{word metric defined by $S$ on $G$} is the map $d_S:G\times G\rightarrow \mathbb N$ such that for any $g,h\in G$, we have
$$d_S(g,h):=\min
\left\{ n\geq 0 : \begin{aligned}
& \text{ there exist } s_1,\ldots,s_n\in S\cup S^{-1}\\
& \text{ such that } g^{-1}h=s_1\ldots s_n\\
\end{aligned}\right\}.
$$
We define the sphere centered at $g\in G$ of radius $r\in\mathbb N$ as
$$B(g,r):=\{h\in G:d_S(g,h)\leq r\},$$
replacing our earlier definition.
Note we include elements at distance $r$ from $g$.
Spheres are compact with respect to the underlying topology of $G$, since $B(e,m)=\left(S\cup S^{-1}\right)^m$ for any $m\in\mathbb N$.
The closed set $$A(g,a,b):=\{h\in G : a\leq d_S(g,h)\leq b\}$$ is the \textit{shell centered at $g$ from $a$ to $b$}.
Shells will be used to ``mark'' sources of randomness.
Recall $N(A)$ is the number of points of $X$ in $A$, for any $A\in\mathcal B(G)$ with finite measure.
We say $g\in G$ is a \textit{seed} if $B(g,60)$ satisfies the following properties:
\begin{enumerate}
\item For every sphere $B$ of radius $1$ such that $B\subset A(g,16,20)$, $N(B)\geq 1$.
\item The shell $A(g,20,60)$ is empty; that is, $N(A(g,20,60))=0$.
\end{enumerate}
For any seed $g$, we call $A(g,16,20)$ its \textit{dense shell} and $A(g,20,60)$ its \textit{empty shell}. Seeds are defined so that we may place an equivalence relation on those within a small enough distance of each other, and the remaining ones must be far enough apart to avoid an intersection of dense and empty shells. In particular, we have chosen radii so seeds cannot spawn inside empty shells of other seeds (since $60-20>2\cdot 16$) and so we have a sizable enough gap between $g$ and $A(g,16,60)$ to prove independence properties. The particular constants are unimportant beyond meeting these requirements.
Our isomorphism construction deeply depends upon Lemma \ref{seed}---it is either referenced directly or lurks in the background of most subsequent proofs.
\begin{lem}[Seed distances]\label{seed}
Let $X$ be a Poisson point process on $G$ with intensity $\alpha$, and let $g,h\in G$ be seeds of $X$. Then $d_S(g,h)\notin [2,78]$.
\end{lem}
\begin{proof}
Note spheres of radius $1$ centered at elements in $A(g,17,19)$ contain at least one point of $X$, and $A(h,20,60)$ contains no points of $X$. We split the proof into two cases.
First, let $m,n\in\mathbb N$ such that $17 \leq m \leq 19$ and $21 \leq n \leq 59$, and
suppose $d_S(g,h)=m+n$. Let $s_1,\ldots ,s_{m+n}\in S\cup S^{-1}$ such that $g^{-1}h=s_1\ldots s_{m+n}$. Then $d_S(g,gs_1\ldots s_i)=i$ for $1\leq i\leq m+n$. Thus, we may suppose $g'\in B(gs_1\ldots s_m,1)$ with $\delta(g')=1$. Consider that $h^{-1}gs_1\ldots gs_m=(s_{m+1}\ldots s_{m+n})^{-1}$, and
$$d_S(h,gs_1\ldots gs_m)\geq d_S(g,h)-d_S(g,gs_1\ldots gs_m)=n,$$ so $d_S(h,gs_1\ldots gs_m)=n$ by the triangle inequality. But then $g' \in A(h,20,60)$. We have a contradiction via a ``turned on'' point in the dense shell of $g$ belonging also to the empty shell of $h$, so $d_S(g,h)\notin [38,78].$
For the next case, fix $m=19$, and let $n\in\mathbb N$ such that $21\leq n \leq 57$. Suppose $d_S(g,h)=n-m$.
Let $h'\in G$ such that $d(h,h')=n$, meaning $h'\in A(h,21,59)$.
We may choose $s_1'\ldots s'_n\in S\cup S^{-1}$ such that $h^{-1}h'=s_1'\ldots s'_n$ and
$s_1,\ldots ,s_{n-m}\in S\cup S^{-1}$ such that $g^{-1}h=s_1\ldots s_{n-m}$. Then $g^{-1}h'=s_1\ldots s_{n-m}s_1'\ldots s'_n$, and
$$d_S(g,h')\geq d_S(h,h')-d_S(g,h)=m,$$
so $d_S(g,h')= m$, yielding another contradiction. So $d_S(g,h)\notin [2,38]$. Combined with the first case, we are done.
\end{proof}
In light of Lemma \ref{seed}, we define an equivalence relation $\sim$ on seeds by setting $g\sim h$ whenever $d_S(g,h)< 2$.
Given a seed $g$, we denote its equivalence class of seeds under $\sim$ as $[g]$.
The \textit{core} of a seed class is the non-empty, closed intersection
$$\mathcal C[g]:=\bigcap_{g\in [g]} B(g,2).$$
Some cores contain a unique point of $X$; we call such cores \textit{identifiable}.
The unique $X$-point in an identifiable core is its \textit{landmark} $\ell[g]:=X|_{\mathcal C[g]}.$
We denote the union or sum (depending on our perspective of a landmark as a point in $G$ or a random variable, which we switch between) of landmarks as $\ell(X)$.
For any subset $A$ of $G$, set
$$\diam A:=\sup_{g,h\in A}d_S(g,h).$$
We say $A$ is \textit{bounded} if $\diam A<\infty$. Since $G$ is compactly generated, the boundedness of $A$ is independent of a choice between the metric $d_S$ and a ``nice enough'' compatible metric (meaning, a geodesically adapted metric, see \cite{metgeo}, Corollary 4.B.11).
Note $\diam \mathcal C[g] \leq 4$ for any seed $g$, by Lemma \ref{seed}. This informs our next definition.
The \textit{fitted shell} of a landmark is the uncountable Polish set $$F[g]:=A(\ell[g],6,10),$$
and $F(X)$ refers to the union of fitted shells under $X$.
Each $X|_{F[g]}$ is a potential source of randomness. Lemma \ref{indlem1} helps to ensure independence conservation when collecting and distributing randomness from $X|_{F(X)}$.
\begin{lem}[Fitted shell independence property]\label{indlem1}
Let $\mu\in\mathbb M$, and define $$\hat F(\mu):=\bigcup_{\ell[g]\in\ell(\mu)}A(\ell[g],5,11).$$
Suppose $\mu'\in\mathbb M$ such that $\mu|_{\hat F(\mu)^c}=\mu'|_{\hat F(\mu)^c}$.
Then $F(\mu)=F(\mu')$.
\end{lem}
This property states that given two Poisson processes which agree everywhere except on the fitted shells of one of the processes and relatively small ``bumper shells'' around them, both processes have the same fitted shells. In other words, the set $F(X)$ depends only on $X|_{\hat F(X)^c}$.
If two processes agree just on their cores, dense shells, and empty shells, both processes should have the same fitted shells. Indeed, the cores of two processes do not even need to agree beyond containing a unique point. We have no need to prove these stronger statements, but we do use the tolerance provided by the bumper shells in the proof of Lemma \ref{indlem2}.
\begin{proof}[Proof of Lemma \ref{indlem1}]
Let $g\in G$ be a seed under $\mu$ with identifiable core.
Consider that $\mathcal C[g]\subseteq \hat F(\mu)^c$ by Lemma \ref{seed}, so $\ell[g]$ is the landmark of $[g]$ under both $\mu$ and $\mu'$, as long as $g$ is a seed under $\mu'$. Thus
in order to show $F(\mu)=F(\mu')$, it suffices to restrict our attention to dense and empty shells and show $$\mu|_{A(g,16,60)}=\mu'|_{A(g,16,60)}$$
to imply $g$ is a seed under $\mu'$.
Again by Lemma \ref{seed}, we know if $h$ is an identifiable seed under $\mu$ such that $h\notin[g]$, then $A(\ell[h],5,11)\cap B(g,60)=\emptyset$. This, along with $\mu|_{\hat F(\mu)^c}=\mu'|_{\hat F(\mu)^c}$, gives us that $g$ is a seed under $\mu'$.
\end{proof}
Only those $X|_{F[g]}$ such that $N(X|_{F[g]})\geq 1$ are isomorphic to uniform random variables. If $N(X|_{F[g]})=1$, then $X|_{F[g]}$ is uniformly distributed on $F[g]$, and the Borel isomorphism theorem from Remark \ref{bitform} applies. If $N(X|_{F[g]})>1$, we easily have an isomorphism still by the same theorem. But when $N(X|_{F[g]})=0$, the process $X|_{F[g]}$ is empty, and we cannot extract any useful randomness. In the language of the isomorphism theorem, the probability measure is not continuous.
To construct an output Poisson point process on $G$ given the input process $X$,
we construct Poisson point processes on cells of a partition of $G$. The new processes as well as the partition both need to come from randomness within $X$, but the information sources should be independent of each other in order for the new processes to coalesce into a process on $G$.
Thus, we want to place a partition on $G$ based on information independent of those fitted shells containing at least one point of $X$. To do so we use a type of Voronoi tessellation.
The statement in Remark \ref{cutpaste} is a corollary of the Superposition Theorem in \cite{kingman} and justifies piecing processes together. Remark \ref{cutpaste} holds for any complete separable metric space.
\begin{rem}[Poisson point process cut and paste]\label{cutpaste}
Let $X$ be a Poisson point process on $G$ with intensity $\alpha$, $N\subseteq \mathbb N$, and $\{A_i\}_{i\in N}$ a measurable partition of $G$. Then
$$\sum_{i\in N}X|_{A_i}\overset{d}{=}X.$$
\end{rem}
We introduce the Voronoi partition. Let $\mu\in\mathbb M$ and $x$ be a point of $\mu$. We use $x$ to express both an element of $G$ and a point measure. The \textit{Voronoi cell} of $x$ is $$\mathcal V(x):=\{g\in G:d_S(x,g)\leq d_S(y,g) \text{ for } y\in \mu\}.$$
In this context we refer to each $x\in\mu$ as a \textit{site}. The \textit{Voronoi tessellation} of $\mu$ is the collection $\mathcal V(\mu):=\{\mathcal V(x)\}_{x\in\mu}$. Recall $d_S$ is measurable, so each Voronoi cell is closed. Some elements of $G$ could belong to multiple Voronoi cells, although this cannot happen in $\mathbb R^n$. Ab\'ert and Mellick identify this distinction in \cite{cost} and introduce a tie-breaking function in pursuit of a true partition; we do the same.
Being uncountable Polish, the group $G$ is Borel isomorphic to $\mathbb R$. Fix a Borel isomorphism $T:G\rightarrow \mathbb R$. The \textit{Voronoi cell under} $T$ of $x$ is
$$\mathcal V(x,T):=\{g\in G:d_S(x,g)\leq d_S(y,g) \text{ and } T(g^{-1}x)\leq T(g^{-1}y) \text{ for } y\in \mu\}$$
and $\mathcal V(\mu,T):=\{\mathcal V(x,T)\}_{x\in\mu}$ is the \textit{Voronoi tessellation under} $T$ of $\mu$. The collection of cells $\mathcal V(\mu,T)$ is a measurable partition of $G$, and the intersection of Voronoi cells under $T$ has measure zero.
Given a seed $g$ with identifiable core, if its fitted shell is non-empty, we denote its fitted shell as $F^*[g]$, its landmark as $\ell^*[g]$, and its core as $\mathcal C^*[g]$. We say $g$ and its associated class, fitted shell, landmark, and core are \textit{harvestable}, with \textit{harvest} $\mu|_{F^*[g]}$.
The union of harvestable fitted shells is $F^*(\mu)$, and the union (or sum) of harvestable landmarks is $\ell^*(\mu)$. Any $\mu\in\mathbb M$ has ``enough'' harvestable landmarks for our construction $P_\alpha$ almost-surely, which we prove in Lemma \ref{infland} with a Borel-Cantelli argument.
In Lemma \ref{infland} and elsewhere, we use open covers made up of sets related to $d_S$-spheres. The spheres themselves are compact, but not necessarily open with respect to the underlying topology of $G$. If, for any $r\in\mathbb N$, we take the interior of $B(e,r+1)$, we end up with an open set which contains $B(e,r)$, since $G$ is completely and perfectly normal, and so have some constraint on the size of the interior.
For any $A\in\mathcal B(G)$, we denote the interior of $A$ as $\inter A$ and the closure of $A$ as $\cl A$, both with respect to the topology of $G$.
\begin{lem}[Infinitely many landmarks]\label{infland}
Let $X$ be a Poisson point process on $G$ with intensity $\alpha$, and let $A\in\mathcal B(G)$ such that $\lambda(A)=\infty$. Then $A$ contains infinitely many harvestable landmarks of $X$, $P_\alpha$ almost-surely.
\end{lem}
\begin{proof}
Fix $r\in\mathbb N$ such that $r>202$. The collection $\{\inter B(g,r)\}_{g\in A}$ is an open cover of $A$. It is also an open cover of the open set
$$\bigcup_{g\in A} \inter B(g,r).$$
Since $G$ is second countable, it is also hereditarily Lindel\"{o}f, and we may let the collection $\{\inter B(g_i,r)\}_{i\in\mathbb N}$ be a countable subcover of $\{\inter B(g,r)\}_{g\in A}$ for some $\{g_i\}_{i\in\mathbb N}\subseteq A$. By Zorn's Lemma, we may let $\{\inter B(g_{i_j},r)\}_{j\in\mathbb N}$ be a maximal disjoint subcollection of $\{\inter B(g_i,r)\}_{i\in\mathbb N}$ for some $\{g_{i_j}\}_{j\in\mathbb N}\subseteq\{g_i\}$.
Let $E_j$ be the event that $\inter B(g_{i_j},r/2)$ contains a harvestable landmark of $X$. It follows from the definition of a harvestable landmark that $\mathbb P(E_j)>0$. By Lemma \ref{indlem1}, the events $\{E_j\}_{j\in\mathbb N}$ are disjoint (the restriction on $r$ forces a distance of at least $100$ between any $\inter B(g_{i_j},r/2)$). Furthermore, the $\mathbb P(E_j)$ are equal for all $j$. Then Borel-Cantelli implies the events $E_j$ occur infinitely often $P_\alpha$ almost-surely.
\end{proof}
Finally, we
define the \textit{restricted Voronoi cell} of $\ell^*[g]$, understood to be under $T$, as
$$\mathcal V^*(\ell^*[g]):=
\left\{f\in G: \begin{aligned}
& d_S(\ell^*[g],f)\leq d_S(\ell^*[h],f) \text{ and } \\
& T(f^{-1}\ell^*[g])\leq T(f^{-1}\ell^*[h]) \text{ for } \ell^*[h]\in\mathcal \ell^*(\mu)\\
\end{aligned} \right\}.
$$
The \textit{restricted Voronoi tessellation} of $\mu$, $$\mathcal V^*(\mu):=\{\mathcal V^*(\ell^*[g])\}_{\ell^*[g]\in\ell^*(\mu)}$$
is our desired, measurable $G$-partition.
It is simple to check that restricted Voronoi tessellations, like the other Voronoi tessellations defined, are $G$-equivariant.
Later, we use the randomness from each harvest to populate a Poisson point process inside the restricted Voronoi cell of its landmark site. In a corollary to Lemma \ref{infland}, we prove restricted Voronoi cells are bounded, key in showing our construction is finitary.
\begin{cor}[Bounded Voronoi cells]\label{bounded}
Let $X$ be a Poisson point process on $G$ with intensity $\alpha$ and restricted Voronoi tessellation $\mathcal V^*(X)$. Each restricted Voronoi cell is bounded, $P_\alpha$ almost-surely.
\end{cor}
\begin{proof}
Let $V$ be a restricted Voronoi cell in $\mathcal V^*(X)$. By its definition, $V$ contains precisely one harvestable landmark of $X$, so Lemma \ref{infland} implies $\lambda (V)<\infty$. Suppose for a contradiction $V$ is unbounded.
Fix $r\in\mathbb N\setminus\{0\}$. The collection $\{\inter B(v,r)\}_{v\in V}$ covers $V$, and
$$\bigcup_{v\in V}\inter B(v,r)$$
is open. We take a countable subcover $S:=\{\inter B(v_i,r)\}_{i\in\mathbb N}$, which must be countably infinite: else $V$ would be bounded. Let $M$ be a maximal disjoint subcollection of $S$ (Zorn's Lemma guarantees the existence of such a subcollection). The subcollection $M$ must be countably infinite as well, so $M$ consists of infinitely many disjoint spheres of equal Haar measure. Let $\mathcal O$ be the union of sets in $M$. We have $\lambda (\mathcal O)=\infty$.
Lemma \ref{infland} implies $\mathcal O$ contains infinitely many harvestable landmarks of $X$. Each must belong to some sphere in $M$. Say $L:=\{\inter B(v_{i_j},r)\}_{j\in\mathbb N}$ is the set of spheres in $M$ which contain a harvestable landmark, $\mathcal L$ is the union of sets in $L$, and $\ell_j$ is the landmark inside $\inter B(v_{i_j},r)$.
Suppose $\mathcal L$ is bounded. Then $\mathcal O\setminus \mathcal L$ must contain infinitely many harvestable landmarks, but this contradicts the definition of $\mathcal L$. So suppose $\mathcal L$ is unbounded. Let $\ell$ be the unique harvestable landmark in $V$. Then for all $v\in V$ and $j\in \mathbb N$, we have $d_S(v,\ell)\leq d_S(v,\ell_j)$. Certainly for some $j$, we have
$d_S(v,\ell_j)\leq 2r$. But then $V$ must be bounded, a contradiction.
\end{proof}
\section{Independent randomness}\label{random}
In this section, we explore the independence properties necessary to our construction.
For any Poisson point process $X$ on $G$ and $A\in\mathcal B(G)$, the restrictions $X|_A$ and $X|_{A^c}$ are independent Poisson processes. Further, their sum has the same law as $X$ (see Remark \ref{cutpaste}). Introducing a Poisson point process $Y$ on $G$, independent of $X$ and of the same intensity, allows us to rephrase this fact in the following way, more conducive to proving independent properties:
$$Y|_A+X|_{A^c}\overset{d}{=}X.$$
We want a version of this fact for a random Borel set in $G$, in particular the union of fitted shells $F(X)$ which depends on $X|_{\hat F(X)^c}$. Lemma \ref{indlem1} in Section \ref{marker} is a first step in this direction. We need Lemma \ref{indlem2} as well to prove our desired version, which we state first in Proposition \ref{indprop}.
\begin{prop}[Fitted shell cut and paste]\label{indprop}
Let $X$ and $Y$ be independent Poisson point processes on $G$, both with intensity $\alpha$. The process $Z:=Y|_{F(X)}+X|_{F(X)^c}$ is equal in distribution to $X$, and $F(Z)=F(X)$.
\end{prop}
Proposition \ref{indprop} restates Proposition 16 in \cite{splitting}, and our proof adapts the proofs of Proposition 16 and Lemma 18 in the same paper from Poisson processes on $\mathbb R^n$ to Poisson processes on $G$.
As a corollary (Corollary \ref{indcor}) we get a similar statement for the union of harvestable fitted shells $F^*(X)$, which is really what we are working towards.
Ignoring technical details, the main tools in the proof of Proposition \ref{indprop} are Lemma \ref{indlem1}, which tells us $F(X)$ depends only on $X|_{\hat F(X)^c}$, and Fubini's theorem.
Lemma \ref{indlem2}, adapted from Lemma 17 in \cite{splitting}, is used in the proof of Proposition \ref{indprop} and addresses some of those technical details.
\begin{lem}\label{indlem2}
Let $X$ be a Poisson point process on $G$ with intensity $\alpha$ and $A$ be a bounded Borel set. There exists a finite set $S$, a collection of bounded Borel sets $\{F_s\}_{s\in S}$, and a collection of disjoint events $\{E_s\}_{s\in S}$ such that:
\begin{enumerate}
\item The union over $S$ of the events $E_s$ occurs $P_\alpha$ almost-surely.
\item For all $s\in S$, conditioned on $E_s$, we have $F(X)\cap A\subseteq F_s\subseteq \hat F(X)$.
\item Let $\sigma(X|_{F_s^c})$ be the $\sigma$-algebra generated by $X|_{F_s^c}$. For all $s\in S$, the event $E_s$ is $\sigma(X|_{F_s^c})$-measurable.
\end{enumerate}
\end{lem}
It would be more straightforward to apply Lemma \ref{indlem2} in the proof of Proposition \ref{indprop} if, instead of relying on bounded $A\in\mathcal B(G)$, we could replace $A$ directly with $G$. We cannot do so because of the first property. The cardinality of the set $S$ matches the cardinality of some cover of $A$. Requiring $A$ to be bounded lets us find a finite subcover. If $S$ were even countably infinite, the probability of the union over $S$ of any events indexed by $S$ would be zero almost-surely.
In the proof, we use the fact that since $d_S$ is a proper metric quasi-isometric to a proper, compatible metric on $G$, subsets of finite diameter in $(G,d_S)$ are relatively compact (\cite{metgeo}, Remark 4.B.3, Corollary 4.B.11, and Definition 2.A.7).
\begin{proof}[Proof of Lemma \ref{indlem2}]
The collection $\{B(g,10)\}_{g\in A}$ contains fitted shells of landmarks belonging to $X|_A$. Let $\mathcal O$ be the union of sets in the collection, so $F(X|_A)\subseteq\mathcal O$. We consider $\mathcal O$ since $F(X|_A)$ may not be contained in $A$ itself.
Note $\cl \mathcal O$ is compact.
The collection $\{\inter B(g,1)\}_{g\in\cl\mathcal O}$ is an open cover of $\cl \mathcal O$;
let $\{\inter B(g_i,\rho)\}_{i=1}^N$ be a finite subcover, and let $\{D_i\}_{i=1}^N$ be the disjointification of the finite subcover.
Let $E_i$ be the event that $D_i$ contains a landmark of $X|_A$. Although the $D_i$ are disjoint, the events $E_i$ are not, due to the nature of landmarks. To obtain disjoint events, for all $s\in \{0,1\}^{N}$, define
$$E_s:=\left(\bigcap_{1 \leq i\leq N:s_i=1}E_i\right)\cap\left(\bigcap_{1 \leq i\leq N:s_i=0}E_i^c\right)$$
and set $S:=\{s\in\{0,1\}^{N}:\mathbb P(E_s)>0\}$. Now for all $s\in S$, the events $E_s$ are disjoint. Furthermore, the $E_s$ satisfy (1).
To prove (2), we want to define $F_s$ based on $E_s$ so that $F(X)\cap A\subseteq F_s\subseteq \hat F(X)$.
Suppose $\ell[h]$ is a landmark of $X|_A$ in $D_i$: Then $d(g_i,\ell[h])\leq 1$. For any $f$ in the fitted filler of $\ell[h]$, we have $5 \leq d(g_i,f)\leq 11$,
so set
$$F_s:=\bigcup_{1 \leq i\leq N:s_i=1}A(g_i,5,11)$$
which satisfies our conditions.
Property (3) assures us $E_s$ does not depend on any information from $X|_{F_s}$ and follows from Property (2) and Lemma \ref{indlem1}.
\end{proof}
We are ready to prove Proposition \ref{indprop}. Let $\Omega$ be a set. For any subset $A\subseteq\Omega$, we use $\mathds{1}_{A}$ to denote the indicator random variable of $A$.
\begin{proof}[Proof of Proposition \ref{indprop}]
By Remark \ref{cutpaste}, for any fixed $B\in\mathcal B(G)$,
we know $$Y|_B+X|_{B^c}\overset{d}{=}X.$$ Our goal is to prove the same for the random union of fitted fillers $F(X)$.
First note, since $Z|_{F(X)^c}=X|_{F(X)^c}$, it is immediate from Lemma \ref{indlem1} that $F(Z)=F(X)$.
Fix a bounded $A\in\mathcal B(G)$ and let $S$, $\{F_s\}$, and $\{E_s\}$ be defined as in the proof of Lemma \ref{indlem2}, under $X$. Fix $s\in S$ and let $\mathcal A$ be any measurable set of point measures in $\mathcal M$. We claim
\begin{equation}\tag{$\ast$}
\mathbb P\left(\left\{X|_{F_s\cap A}+X|_{F_s^c\cap A}\in\mathcal A\right\}\cap E_s\right)
\end{equation}
is equal to
\begin{equation}\tag{$\ast\ast$}
\mathbb P\left(\left\{Y|_{A\cap F(X)}+X|_{A\setminus F(X)}\in\mathcal A\right\}\cap E_s\right).
\end{equation}
Note $(\ast)=\mathbb P\left(\left\{X|_A\in\mathcal A\right\}\cap E_s\right)$, because $F_s$ is not a random set, and $(\ast\ast)=\mathbb P\left(\left\{Z|_A\in\mathcal A\right\}\cap E_s\right)$ by Remark \ref{cutpaste}.
Supposing the claim were true and taking a sum over $S$ yields
$\mathbb P(Z|_A\in\mathcal A)=\mathbb P(X|_A\in\mathcal A)$ by Property (1) of Lemma \ref{indlem2}. To obtain $\mathbb P(Z\in\mathcal A)=\mathbb P(X\in\mathcal A)$, that is,
$$Z\overset{d}{=}X,$$
fix $r\in \mathbb N\setminus\{0\}$ and consider the open $G$-cover $\{\inter B(g,r)\}_{g\in G}$. Let $\{\inter B(g_i,r)\}_{i\in\mathbb N}$ be a countable subcover and $\{D_i\}_{i\in\mathbb N}$ its disjointification. The argument applied to $A$ applies in turn to each $D_i$, and Remark \ref{cutpaste} implies $$\bigcup_{i=0}^\infty Z|_{D_i}\overset{d}{=}Z;$$
similarly for $X$. Thus, all that remains is to prove the claim.
Let $\mu$ be the law of $X|_{F_s\cap A}$ and $\nu$ the law of $\left( X|_{F_s^c\cap A}, \mathds{1}_{E_s}, A\cap F(X) \right)$. The processes $X|_{F_s\cap A}$ and $X|_{F_s^c\cap A}$ are independent by definition, while
$X|_{F_s\cap A}$ is independent of
$( X|_{F_s^c\cap A}, \mathds{1}_{E_s}, A\cap F(X) )$
by Lemma \ref{indlem2}.
Let $x,y\in(\mathbb M,\mathcal M)$ under $\mu$ and $(w,e,b)\in(\mathbb M,\mathcal M)\times(\mathbb M,\mathcal M)\times (G,\mathcal B(G))$ under $\nu$. Recall any probability measure on $(\mathbb M,\mathcal M)$ is $\sigma$-finite. We have
\begin{align*}
(\ast)&=\iint \mathds{1}_{[x+w\in\mathcal A]} \mathds{1}_{E_s}\, d\mu(x)\, d\nu(w,e,b)\\
&= \iint \mathds{1}_{\left[x|_{F\left(x\right)}+x|_{F\left(x\right)^c}+w\in\mathcal A\right]} \mathds{1}_{E_s}\, d\mu(x) \,d\nu(w,e,b)\\
&= \int \left(\int \mathds{1}_{\left[x|_{F\left(x\right)}+x|_{F\left(x\right)^c}+w\in\mathcal A\right]}\, d\mu(x) \right) \mathds{1}_{E_s}\, d\nu(w,e,b)\\
&= \iiint \mathds{1}_{\left[y|_{F\left(x\right)}+x|_{F\left(x\right)^c}+w\in\mathcal A\right]} \mathds{1}_{E_s}\, d\mu(x)\, d\mu(y)\, d\nu(w,e,b)= (\ast\ast)
\end{align*}
by Remark \ref{cutpaste} and Fubini's theorem.
\end{proof}
With Proposition \ref{indprop}, we have shown $X|_{F(X)}$ and $X|_{F(X)^c}$ are independent Poisson point processes, conditioned on $F(X)$. We want to have a restricted Voronoi tessellation in place so that we may populate each restricted Voronoi cell $\mathcal V^*(\ell^*[g])$ with a Poisson point process via randomness from the harvest $X|_{F^*[g]}$. Such a series of actions requires more information than we can get from $F(X)$ alone; we also need to know which fitted shells contain at least $1$ point of $X$ and so are harvestable. That is, we need to know $F^*(X)$.
If we know $F^*(X)$, we still do not gain any information regarding locations of points in harvestable fitted shells. We do gain information on the process restricted to $F(X)\setminus F^*(X)$ (we learn it is empty), but this does not disrupt the independence properties we ultimately require, which we prove in Corollaries \ref{indcor} and \ref{source}, extensions of Lemma 24 in \cite{splitting} and Corollaries 9 and 10 in \cite{finisoppp}.
For convenience, we enumerate the sets of landmarks, harvestable landmarks, fitted shells, and harvestable fitted shells under $X$ as $\{\ell_i\}_{i\in\mathbb N}$, $\{\ell^*_i\}_{i\in\mathbb N}$, $\{F_i\}_{i\in\mathbb N}$, and $\{F^*_i\}_{i\in\mathbb N}$, respectively.
\begin{cor}[Harvestable fitted shell cut and paste]\label{indcor}
Let $X$ and $Y$ be independent Poisson point processes on $G$, both with intensity $\alpha$. The process $Z:=Y|_{F^*(X)}+X|_{F^*(X)^c}$ is equal in distribution to $X$, and $F^*(Z)=F^*(X)$.
\end{cor}
\begin{proof}
Since $Z|_{F^*(X)^c}=X|_{F^*(X)^c}$, it must be true that $Z|_{F(X)^c}=X|_{F(X)^c}$, which implies $F(Z)=F(X)$ by Lemma \ref{indlem1}. Then of course $F^*(Z)=F^*(X)$. It remains to show $X|_{F^*(X)}$ is independent of $X|_{F^*(X)^c}$.
Note $$F^*(X)^c=F(X)^c\cup \left( F(X)\setminus F^*(X) \right).$$
From Proposition \ref{indprop}, we know the random variables $\{N(F_i)\}_{i\in\mathbb N}$ are independent of each other
and of $X|_{F(X)^c}$.
It follows that the processes $X|_{F(X)\setminus F^*(X)}$ and $X|_{F^*(X)}$ are independent.
Thus $X|_{F(X)^c}$ and $X|_{F^*(X)}$ are independent as well.
The set $F(X)\setminus F^*(X)$ depends only on $(X|_{F(X)^c}, \{N(F_i)\}_{i\in\mathbb N})$, so we are done.
\end{proof}
With Corollary \ref{indcor}, we prove Corollary \ref{source}, which we use to prove that given a Poisson point process $X$ with intensity $\alpha$, we may generate processes on cells of $\mathcal V^*(X)$ which sum to a Poisson process on $G$ with intensity $\beta$. Harvests of $X$ generate the new processes, while $\mathcal V^*(X)$ depends on $\ell^*(X)\subseteq X|_{F^*(X)^c}$, independent of $X|_{F^*(X)}$.
\begin{cor}[Source independence]\label{source}
Let $X$ be a Poisson point process on $G$ with intensity $\alpha$, and let $\{U_i\}_{i\in\mathbb N}$ be a sequence of independent uniform random variables, independent of $X$. There exists a measurable map $\xi:\mathbb M\times \mathcal B(G)\rightarrow [0,1]$ such that
$$\left(X|_{F^*(X)^c}, F^*(X), \left\{\xi\left(X,F^*_i\right)\right\}_{i\in\mathbb N}\right)\overset{d}{=} \left(X|_{F^*(X)^c}, F^*(X),\{U_i\}_{i\in\mathbb N}\right).$$
\end{cor}
\begin{proof}
Each harvest $X|_{F^*_i}$ is a Poisson point process with intensity $\alpha$ on $F^*_i$. By Corollary \ref{indcor},
elements of the set of all harvests of $X$
are independent as well as independent of $X|_{F^*(X)^c}$, and by Lemma \ref{infland},
the cardinality of the set of all harvests is $P_\alpha$ almost-surely countably infinite.
Further,
$$\mathcal S(X):=\left\{(\ell^*_i)^{-1}X|_{F^*_i}\right\}_{i\in\mathbb N}$$
is a sequence of independent and identically distributed Poisson processes on the set $A(e,6,10)$.
Each process in $\mathcal S(X)$ forgets its starting location, so $\mathcal S(X)$ is independent of $(X|_{F^*(X)^c}, F^*(X)).$
Set $n_i:=N\left(F^*_i\right)$. Note $0<n_i<\infty$ almost-surely for all $i$. Let $\{V_j\}_{j=1}^{n_i}$ be a sequence of independent uniform random variables on $A(e,6,10)$, independent of $X$.
By Remark \ref{findist},
$$(\ell^*_i)^{-1}X|_{F^*_i}\overset{d}{=}\sum_{j=1}^{n_i}\delta(V_j).$$
For $k\in\mathbb N\setminus \{0\}$, fix Borel isomorphisms $R_k:A(e,6,10)^k\rightarrow [0,1]$ such that
$R_k(V_1,\ldots,V_k)$ is a uniform random variable on $[0,1]$.
Denote the shifted process points of $\left(\ell^*_i\right)^{-1}X|_{F^*_i}$ as $v_1^i,v_2^i,\ldots, v_{n_i}^i$, and let $B\in\mathcal B(G)$.
Define $\xi:\mathbb M\times\mathcal B(G)\rightarrow [0,1]$ by
\[ \xi\left(X, B\right)=\begin{cases}
R_{n_i}(v_1^i,\ldots,v_{n_i}^i) & \text{ if } B=F^*_i \\
0 & \text{ otherwise. }
\end{cases}
\]
Then $\xi\left(X, F^*_i\right)$ is uniformly distributed on $[0,1]$ and $\{\xi (X,F^*_i )\}_{i\in\mathbb N}$ is a sequence of independent uniform random variables, independent of $(X|_{F^*(X)^c}, F^*(X)).$
Thus
$$\left(X|_{F^*(X)^c}, F^*(X), \{\xi (X,F^*_i )\}_{i\in\mathbb N}\right) \overset{d}{=} \left(X|_{F^*(X)^c}, F^*(X),\{U_i\}_{i\in\mathbb N}\right)$$
as desired.
\end{proof}
We have proven all necessary independence properties for the proof of Theorem \ref{factor}. One further independence property is required to prove Proposition \ref{preproduct} and Theorem \ref{iso}. We need to know the Voronoi partition and the output system are independent.
\begin{lem}[Partition and output independence]\label{partition}
Let $X$ be a Poisson point process on $G$ with intensity $\alpha$ and restricted Voronoi tessellation $\mathcal V^*(X)$, and suppose $\{U_i\}_{i\in\mathbb N}$ is a sequence of independent uniform random variables, independent of $X$. Let $\kappa:\mathbb M\times[0,1]\rightarrow\mathbb M$ be a measurable map such that $\kappa(X,U_i)$ is a Poisson point process on $\mathcal V^*\left(\ell^*_i\right)$ with intensity $\beta>0$. Then
$$\sum_{i\in\mathbb N}\kappa (X,U_i)$$
is a Poisson point process on $G$ with intensity $\beta$, independent of $X$.
\end{lem}
Lemma \ref{partition} is a version of Lemma 16 in \cite{finisoppp}, and in fact is implied from the same lemma, which relies on Fubini's theorem, along with Corollaries \ref{indcor} and \ref{source}. We need only recall that $P_\alpha$ for any $\alpha >0$ on $\mathbb M$ is $\sigma$-finite.
\section{Factor map and isomorphism}\label{construct}
We start in Section \ref{cg} with a version of Theorem \ref{factor} for compactly generated groups. We then prove Proposition \ref{preproduct} and a version of Theorem \ref{iso}.
Once the compactly generated case is complete, we drop the extra condition in Section \ref{subgroups} and prove Theorem \ref{iso} via coinduction in a series of steps; first for non-discrete, non-compact, locally compact Polish groups with a non-compact, compactly generated open subgroup in Section \ref{noncompact}, then for those without in Section \ref{compact}, and putting everything together in Section \ref{general}.
\subsection{Compactly generated case}\label{cg}
To prove Theorem \ref{factor} for compactly generated $G$, we construct a factor from $\mathbb X_\alpha$ to $\mathbb X_\beta$ via the harvests and restricted Voronoi tessellation of the source system $\mathbb X_\alpha$. That the factor is finitary is immediate from construction and Corollary \ref{bounded}.
\begin{thm}[Finitary factor for compactly generated groups]\label{prefactor}
Let $G$ be a non-discrete, non-compact, locally compact and compactly generated Polish group, and let $\alpha, \beta >0$.
There exists a finitary factor from $\mathbb X_\alpha$ to $\mathbb X_\beta$.
\end{thm}
\begin{proof}
Suppose $X$ and $Y$ are Poisson point processes on $G$ with intensities $\alpha$ and $\beta$, respectively. We construct a factor map $\phi:\mathbb X_\alpha \rightarrow \mathbb X_\beta$ so that
$$\phi(X)\overset{d}{=}Y.$$
Let $\xi:\mathbb M\times\mathcal B(G)\rightarrow [0,1]$
be defined as in the proof of Corollary \ref{source}.
Recall the restricted Voronoi tessellation of $X$ is
$$\mathcal V^*(X)=\{\mathcal V^*(\ell^*_i)\}_{\ell^*_i\in\ell^*(X)},$$
a measurable $G$-partition $P_\alpha$ almost-surely.
Cells of $\mathcal V^*(X)$ are standard Borel spaces.
We populate a Poisson point process with intensity $\beta$ on each shifted cell $\mathcal S_i:=\left(\ell^*_i\right)^{-1}\mathcal V^*\left(\ell^*_i\right)$ (also a standard Borel space) with randomness from $X|_{F^*_i}$. Cells are shifted to avoid disrupting $G$-equivariance of our eventual factor map.
Remark \ref{bitform} implies we may fix Borel maps $\pi_i:[0,1]\rightarrow \mathbb M(\mathcal S_i,\mathcal B(\mathcal S_i))$ such that if $U$ is a uniform random variable, then $\pi_i(U)$ is a process of intensity $\beta$ on $\mathcal S_i$, for $i\in\mathbb N$.
Define $\phi:\mathbb X_\alpha \rightarrow \mathbb X_\beta$ as
$$\phi(X):=\sum_{i\in\mathbb N} \ell^*_i\pi_i\left(\xi(X,F_i^*)\right).$$
Recall $\xi$ extracts randomness from an input process restricted to its shifted harvestable fitted shells
$\left(\ell^*_i\right)^{-1}F^*_i.$
By Corollary \ref{source}, $\left\{\xi\left(X,F^*_i\right)\right\}_{i\in\mathbb N}$ is a sequence of independent uniform random variables, independent of $X_{F^*(X)^c}$. The landmarks $\ell_i^*(X)$ and tessellation $\mathcal V^*(X)$ depend only on $X_{F^*(X)^c}$ by definition. Hence Remark \ref{cutpaste} implies $\phi(X)$ is a Poisson point process of intensity $\beta$ on $G$. To finish, we check $\phi$ is $G$-equivariant.
Let $g\in G$.
If $g$ acts on $X$, then shifted shells remain the same:
$$\left(g\ell^*_i\right)^{-1}gF^*_i=\left(\ell^*_i\right)^{-1}F^*_i.$$ Similarly, shifted restricted Voronoi cells $\mathcal S_i$ are $g$-invariant. Thus
$$\phi\circ g(X)=\sum_{i\in\mathbb N} (g\ell^*_i)\pi_i\left(\xi(X,F_i^*)\right)=g\circ \phi(X).$$
Corollary \ref{bounded} implies $\phi$ is finitary, since each Voronoi cell under $X$ is completely determined by $X$ restricted to some bounded set containing the cell and adjacent cells, and each $F_i^*\subset \mathcal V^*(\ell^*_i)$.
\end{proof}
The factor map in Theorem \ref{prefactor} fails to be bijective. The law of a Poisson point process on a bounded Borel set is not a continuous measure (the empty process is an atom), and so we cannot expect an isomorphism between such a process and a uniform random variable. The problem and its solution become clearer if we break the application of the Borel isomorphism theorem for measures in the proof down into a more manageable size.
\begin{lem}[Maps for construction]\label{maps}
Let $A\in\mathcal B(G)$ be bounded, $U$ a uniform random variable, and $\alpha >0$. There exists a measurable map $\pi:[0,1]\rightarrow \mathbb M(A,\mathcal B(A))$ such that $\pi(U)$ is a Poisson point process on $A$ with intensity $\alpha\cdot\lambda(A)$.
\end{lem}
\begin{proof}
The existence of such a map follows from Remark \ref{bitform}, but we take a slightly more constructive approach. Let $Z$ be a Poisson random variable with mean $\alpha\cdot\lambda(A)$. For $m\in\mathbb N$, set $p_m:=\mathbb P(Z<m)$. The intervals $[p_m,p_{m+1})$ partition $[0,1)$. Define piecewise maps $q:[0,1)\rightarrow \mathbb N$ and $r:[0,1)\rightarrow [0,1)$ by $q(x):=m$ and
$$r(x):=\frac{x-p_m}{p_{m+1}-p_m}$$
for $x\in [p_m,p_{m+1})$.
Conditioned on $q(U)=m$, $r(U)$ is a uniform random variable independent of $q(U)$.
Starting from one uniform random variable, applying the maps $q$ and $r$ effectively gives us a Poisson and uniform random variable independent of each other.
Whenever $q(U)$ returns a value greater than $1$, we want to split $r(U)$ into a corresponding number of uniform random variables on $A$.
Now we appeal to the Borel isomorphism theorem in Remark \ref{bitform}.
For $k\in\mathbb N\setminus\{0\}$, let $R_k:[0,1]\rightarrow A^k$, where $[0,1]$ and $A$ carry uniform measures and $A^k$ the product measure, be an isomorphism. Define $\pi:[0,1]\rightarrow \mathbb M(A,\mathcal B(A))$ so that
$$ \pi(x):=
\begin{cases}
\emptyset & \text{ whenever } q(x)=0 \\
\displaystyle\sum_{j=1}^{q(x)} \delta\left( R_{q(x)}(r(x))_j \right) & \text{ otherwise.}\\
\end{cases} $$
By Remark \ref{findist}, $\pi(U)$ is a Poisson process on $A$ with intensity $\alpha\cdot\lambda(A)$.
\end{proof}
The map $\pi$ in Lemma \ref{maps} is still not bijective, but is as close to bijective as possible, and what goes wrong is easier to see.
A Poisson point process on a finite measure Borel set $A$ is empty with probability $e^{-\alpha\cdot\lambda(A)}$. At any such event, the map $\pi$ is
uncountably infinite to one.
But whenever the process contains at least one point, $\pi$ is one to one.
To construct a bijective factor map, we must preserve the information otherwise lost when a Poisson point process on a bounded Voronoi cell is determined to be empty.
Our solution is the same as in \cite{finisoppp} for the $\mathbb R^n$ case. Along with generating an output Poisson point process from an input one, we generate an additional output process using the ``extra'' information. We need to do some further work to end up with a final output of a single Poisson process, but this motivates Proposition \ref{preproduct}.
\begin{proof}[Proof of Proposition \ref{preproduct}]
Suppose $X$ and $Y$ are independent Poisson point processes on $G$ with intensities $\alpha$ and $\beta$, respectively. We build a factor map $\phi: \mathbb X_\alpha \rightarrow (\mathbb X_\alpha,\mathbb X_\beta)$. We denote $\phi (X)$ as $(\phi(X)_1,\phi(X)_2)$ (and use similar notation for all coordinate maps) and construct $\phi$ so that
$$\phi(X)\overset{d}{=}(X,Y).$$
Let the shifted Voronoi cells $\mathcal S_i$ under $X$ be defined as in the proof of Theorem \ref{prefactor}.
Let $b_n$ be defined as in the proof of Lemma \ref{maps}; likewise, for $i\in\mathbb N$, let $q_i$ and $r_i$ be defined as $q$ and $r$, but adjust the definitions to match a Poisson random variable with mean $\beta\cdot\lambda(\mathcal S_i)$.
For $i\in\mathbb N$ and $k\in\mathbb N\setminus\{0\}$, fix a Borel isomorphism $R_{i,k}:[0,1]\rightarrow \mathcal S_i^k$, where $[0,1]$ and $\mathcal S_i$ carry uniform measures and $\mathcal S_i^k$ the product measure. Set $N_i:=N(X|_{F_i^*})$ and similarly fix Borel isomorphisms $T_i:[0,1]\rightarrow A(e,6,10)^{N_i}$. Note each $N_i>0$.
Define
$$\pi_i:[0,1]\rightarrow \mathbb M(A(e,6,10), \mathcal B(A(e,6,10)))\times \mathbb M(\mathcal S_i, \mathcal B(\mathcal S_i))$$
so that
$$ \pi_i(x):=
\begin{cases}
\left( \displaystyle\sum_{j=1}^{N_i} \delta(T_i(r(x))_j), \emptyset \right) & \text{ whenever } q_i(x)=0 \\[4ex]
\left(\left(\ell_i^*\right)^{-1} X|_{F^*_i}, \displaystyle\sum_{j=1}^{q_i(x)} \delta\left( R_{i,q_i(x)}(r(x))_j \right)\right) & \text{ otherwise.}\\
\end{cases} $$
For $x\in [0,1]$ and each $i$, if $q_i(x)\neq 0$, the uniform random variable $r(x)$ generates a non-empty Poisson point process of intensity $\beta$ on the shifted Voronoi cell $\mathcal S_i$, by Remark \ref{findist}. When $q_i(x)=0$, the uniform random variable $r(x)$ generates, point for point, a resampled harvest on its fitted filler shifted to the identity. No information goes to waste.
Let $\xi:\mathbb M\times\mathcal B(G)\rightarrow [0,1]$
be defined as in the proof of Corollary \ref{source}, so $\left\{\xi\left(X,F^*_i\right)\right\}_{i\in\mathbb N}$ is a sequence of independent uniform random variables, independent of $X_{F^*(X)^c}$. Set $\xi_i:=\xi(X,F^*_i)$, and define
$$\phi(X)_2:=\sum_{i\in\mathbb N} \ell^*_i\pi_i(\xi_i)_2.$$
The arguments in the proofs of Theorem \ref{prefactor} and Lemma \ref{maps} imply $\phi(X)_2$ is a Poisson point process of intensity $\beta$ on $G$.
Each $\ell_i^*\pi_i(\xi_i)_1$ is a Poisson point process of intensity $\alpha$ on $F_i^*$: If $q_i(\xi_i)\neq 0$, then $\ell_i^*\pi_i(\xi_i)_1=X|_{F^*_i}$. If $q_i(\xi_i) = 0$, then
$$\ell_i^*\displaystyle\sum_{j=1}^{N_i} \delta(T_i(r(\xi_i))_j)\overset{d}{=}X|_{F^*_i}$$ by Remark \ref{findist} and Corollary \ref{source}. Define
$$\phi(X)_1:=X|_{F^*(X)^c}+ \displaystyle\sum_{i\in\mathbb N} \ell_i^*\pi_i(\xi_i)_1,$$
so by the same statements, $\phi(X)_1$ is a Poisson point process of intensity $\alpha$ on $G$.
By Corollary \ref{source} and Lemmas \ref{partition} and \ref{maps}, the processes $\phi(X)_1$ and $\phi(X)_2$ are independent. Set $\phi(X):=(\phi(X)_1,\phi(X)_2)$. By construction, the map $\phi$ is bijective. It is straightforward to check $\phi$ is $G$-equivariant in each coordinate, since the set $\ell^*(X)$ is $G$-equivariant. Thus $\phi$ is a factor from $\mathbb X_\alpha$ to $(\mathbb X_\alpha,\mathbb X_\beta)$. Each coordinate mapping is finitary by Corollary \ref{bounded}, so $\phi$ is finitary.
The inverse map $\phi^{-1}$ serves as a finitary factor from $(\mathbb X_\alpha,\mathbb X_\beta)$ to $\mathbb X_\alpha$, so $\phi$ is a finitary isomorphism from $\mathbb X_\alpha$ to $(\mathbb X_\alpha,\mathbb X_\beta)$.
\end{proof}
A version of the main theorem for compactly generated groups follows from Proposition \ref{preproduct}.
\begin{thm}[Finitary isomorphism for compactly generated groups]\label{preiso}
Let $G$ be a non-discrete, non-compact, locally compact and compactly generated Polish group. All Poisson systems on $G$ are finitarily isomorphic.
\end{thm}
\begin{proof}
Let $X$ be a Poisson point process on $G$ with intensity $\alpha$. Let $\phi: \mathbb X_\alpha \rightarrow (\mathbb X_\alpha,\mathbb X_\beta)$ be the map from the proof of Proposition \ref{preproduct}, so that $\phi(X)_1$ is a Poisson point process on $G$ with intensity $\alpha$ independent of $\phi(X)_2$, a Poisson point process on $G$ with intensity $\beta$.
To construct a bijective factor map from $\mathbb X_\alpha$ to $\mathbb X_\beta$, we compose $\phi^{-1}$ with a permuted version of $\phi$.
First, we examine $\phi^{-1}$ on its own. The following algorithm describes $\phi^{-1}$, starting with the input $(\phi(X)_1,\phi(X)_2):=(X',Y)$ and ending with the output $Z$:
\begin{enumerate}
\item Set $Z_c=X'|_{F^*(X')^c}$.
\item Consider $\mathcal V^*(X')$, a $G$-partition, superimposed on $Y$. For all cells $V_i$ in the partition:
\begin{enumerate}
\item If $N(V_i)\neq 0$ under $Y$, reverse the instructions from $\phi$ to generate a Poisson process on $F^*_i$ (under $X'$) with intensity $\alpha$ via the randomness in $V_i$, call it $Z_i$.
\item If $N(V_i)=0$ under $Y$, use this information in addition to $X'|_{F^*_i}$ and instructions from $\phi$ to generate a Poisson process on $F_i^*$ with intensity $\alpha$, call it $Z'_i$.
\end{enumerate}
\item Set $$Z=Z_c + \displaystyle\sum_{i\in\mathbb N:N(V_i)\neq 0} Z_i + \displaystyle\sum_{i\in\mathbb N:N(V_i)= 0}Z'_i.$$
\end{enumerate}
Note $Z$ is equal to $X$ on the set $F^*(X')^c=F^*(X)^c$ (by Lemma \ref{indlem1}) as well as on those harvestable fitted fillers associated with non-empty Voronoi cells (by construction of $\phi$). On the remaining harvestable fitted fillers, $Z$ contains the same number of points as $X$. By Remark \ref{cutpaste}, as expected, we have
$$Z\overset{d}{=}X.$$
Evidently, if we apply $\phi^{-1}$ to the permuted pair $(Y,X')$, the output will be a Poisson point process on $G$ with intensity $\beta$, equal to $Y$ on the set $F^*(Y)$ and on some harvestable fitted fillers (under $Y$), while on the remaining fillers, the output contains the same number of points as $Y$.
Such an application is well-defined since, by Proposition \ref{preproduct}, $X'$ and $Y$ are independent. Hence
$$\phi^{-1}(Y,X')\overset{d}{=}Y.$$
By Proposition \ref{preproduct}, the composition of $\phi^{-1}$ with a permuted $\phi(X)$ is $G$-equivariant, bijective, and finitary. Thus it is a finitary isomorphism from $\mathbb X_\alpha$ to $\mathbb X_\beta$.
\end{proof}
\subsection{Extension via coinduction}\label{subgroups}
Now we fix $G$ as a non-discrete, non-compact, locally compact Polish group that is not compactly generated.
Let $K$ be some compact, symmetric (so $K=K^{-1}$) neighborhood of $e\in G$. The \textit{subgroup generated by $K$} is
$$\langle K \rangle:=\bigcup_{i\in\mathbb N} K^i,$$
and is compactly generated and $\sigma$-compact. It is simple to show $\langle K \rangle$ is open and closed (see, for example, Proposition 2.4 in \cite{harmonic}) as well as of countable, possibly finite, index in $G$ (\cite{metgeo}, Corollary 2.C.6). Further, $\langle K \rangle$ is locally compact and second countable, so $\langle K \rangle$ is uncountable Polish. It is either non-compact or compact (both instances occur; see Section \ref{specific} for examples).
In either case, the method of coinduction applies to extend the isomorphism for compactly generated groups to $G$. The non-compact case is much more direct, so we begin there after realizing Stepin's method for our setting.
Fix $H$ as some non-discrete, locally compact and compactly generated open subgroup of $G$. Then $H$ has countable index in $G$. We denote the set of left cosets of $H$ in $G$ as $G/H$. Since $H$ is open, it is Polish; thus $\mathbb M(H)$ (the space of Borel simple point measures on $H$) and $\mathbb M(H)^{G/H}$ with the product topology are as well.
Choose coset representatives $\{g_i\}_{i\in N}\subset G$ for $G/H$, where $N\subseteq \mathbb N$ enumerates the index of $H$ in $G$.
Let $(\nu (g_i H))_{i\in N} \in \mathbb M(H)^{G/H}$.
Towards obtaining a $G$-action on $\mathbb M(H)^{G/H}$,
define $s:G/H\rightarrow G$ so that $s$ sends a coset to its representative, and $c:G\times G/H\rightarrow H$ so that
$$c(f,gH):=s(fgH)^{-1}fs(gH)$$
for $f\in G$ and $gH\in G/H$.
For $f_1,f_2\in G$ and $gH\in G/H$, it follows from the above definition that $c(f_1f_2,gH)=c(f_1,f_2gH)c(f_2,gH)$ (so $c$ is a cocycle).
The \textit{action of $H\curvearrowright\mathbb M(H)$ coinduced to $G$} is given by
$$f(\nu(g_iH))_{i\in N}:=(c(f^{-1},g_iH)^{-1}\nu(f^{-1}g_iH))_{i\in N}.$$
The cocycle condition can be used to show
$$f_1f_2(\nu(g_iH))_{i\in N}=f_1(f_2(\nu(g_iH))_{i\in N}),$$
so the action is well-defined.
Each $c(f^{-1},g_iH)^{-1}\in H$, and recall we have a natural action $H\curvearrowright\mathbb M(H)$.
Applying $f\in G$ to $(\nu(g_iH))_{i\in N}\in\mathbb M(H)^{G/H}$ permutes the sequence and shifts individual elements in the sequence by some element in $H$.
If $(\mathbb M(H),P_\alpha,H)$ is a Poisson system over $H$, then
$$\mathbb X^H_\alpha:=(\mathbb M(H)^{G/H},P_\alpha^{G/H},G)$$
is the \textit{Poisson system coinduced from $H\curvearrowright (\mathbb M(H),P_\alpha)$}.
With coinduction, we prove some systems of interest are isomorphic in Lemma \ref{coin1}, which applies to both the non-compact and compact subgroup cases.
Factor maps between $(\mathbb M(G),P_\alpha,G)$ and $\mathbb X^H_\alpha$ are defined as expected.
Such factors are finitary if each of their coordinate mappings are.
\begin{lem}[Coset copy isomorphism]\label{coin1}
Let $(\mathbb M(H),P_\alpha,H)$ be a Poisson system over $H$. The Poisson system coinduced from $H\curvearrowright (\mathbb M(H),P_\alpha)$ is finitarily isomorphic to $(\mathbb M(G),P_\alpha,G)$.
\end{lem}
\begin{proof}
Let $(\nu (g_i H))_{i\in N} \in \mathbb M(H)^{G/H}$. Define $\psi:\mathbb M(H)^{G/H}\rightarrow \mathbb M(G)$ by
$$\psi ((\nu(g_i H))_{i\in N} ) := \sum_{i\in N}g_i\nu(g_i H).$$
The map $\psi$ takes a sequence in $\mathbb M(H)^{G/H}$ and sends each element of the sequence to the copy of $H$ in $G$ identified by the element's index. If each element in the input sequence were a Poisson point process on $H$ with intensity $\alpha$, then the output would be a Poisson point process on $G$ with intensity $\alpha$ by Remark \ref{cutpaste}, since $G/H$ is countable.
The map $\psi$ is measurable, bijective, and finitary with finitary inverse.
To show $\psi$ is a finitary isomorphism, it remains to check $\psi$ and its inverse are $G$-equivariant on a set of full measure. In fact, $\psi$ is $G$-equivariant on $\mathbb M(H)^{G/H}$. Let $f\in G$. We have
$$f\psi((\nu (g_i H))_{i\in N})=f\sum_{i\in N}g_i\nu(g_i H)$$
as well as
$$\psi(f(\nu (g_i H))_{i\in N})=\psi((c(f^{-1},g_iH)^{-1}\nu(f^{-1}g_iH))_{i\in N}),$$
where for each $i$, $(c(f^{-1},g_iH)^{-1}=h_i^{-1}$ for some $h_i^{-1}\in H$ and $g_{k_i}\in\{g_i\}$ such that $f^{-1}g_i=g_{k_i}h_i$. So we may write
\begin{align*}
\psi(f(\nu (g_i H))_{i\in N})&=\psi((\nu(f^{-1}g_iH)\circ h_i)_{i\in N})\\
&=\sum_{i\in N}g_i(\nu(f^{-1}g_iH)\circ h_i)\\
&=\sum_{i\in N}fg_{k_i}h_i(\nu(g_{k_i}H)\circ h_i)\\
&=f\sum_{i\in N}g_{k_i}\nu(g_{k_i} H)=f\psi((\nu (g_i H))_{i\in N}).
\end{align*}
A similar argument shows $\psi^{-1}$ is $G$-equivariant.
Hence $\mathbb X^H_\alpha$ is finitarily isomorphic to $(\mathbb M(G),P_\alpha,G)$.
\end{proof}
\subsubsection{Non-compact and compactly generated subgroup case}\label{noncompact}
Within this subsection, we assume $H$ is non-compact, in addition to being a non-discrete, locally compact and compactly generated open subgroup of $G$.
Suppose we have a factor map $\phi:(\mathbb M(H),P_\alpha,H)\rightarrow (\mathbb M(H),P_\beta,H)$, for some $\alpha,\beta >0$. Keeping the same notation for $(\nu (g_i H))_{i\in N} \in \mathbb M(H)^{G/H}$ as before, define $\Psi:\mathbb M(H)^{G/H}\rightarrow \mathbb M(H)^{G/H}$ by
$$\Psi((\nu(g_iH))_{i\in\mathbb N}):=(\phi(\nu(g_i H)))_{i\in\mathbb N}.$$
We say $\Psi$ is the \textit{factor map coinduced by $\phi$}. Such a factor is finitary if each coordinate mapping is.
\begin{lem}[Coinduced isomorphism]\label{coin2}
Let $\alpha,\beta >0$, and let $(\mathbb M(H),P_\alpha,H)$ and $(\mathbb M(H),P_\beta,H)$ be Poisson systems over $H$. The Poisson systems coinduced from $H\curvearrowright (\mathbb M(H),P_\alpha)$ and $H\curvearrowright (\mathbb M(H),P_\beta)$ are finitarily isomorphic.
\end{lem}
\begin{proof}
By Theorem \ref{preiso}, there exists a finitary isomorphism from $(\mathbb M(H),P_\alpha,H)$ to $(\mathbb M(H),P_\beta,H)$. Let $\phi$ be such an isomorphism. Let $\Psi$ be the factor map coinduced by $\phi$. Then $\Psi$ is measurable, bijective, finitary with finitary inverse, and by Remark \ref{cutpaste},
$$P_\alpha^{G/H}\circ \Psi^{-1}=P_\beta^{G/H}.$$
Let $f\in G$ and $(\nu(g_iH))_{i\in N}$. With $f$ acting as in the proof of Lemma \ref{coin1}, here we have
$$f\Psi(\nu(g_iH))=f\phi(\nu(g_iH))=\phi(\nu(f^{-1}g_iH)\circ h_i)$$
for some $h_i\in H$, since $\phi$ is $H$-equivariant and $\phi(\nu(g_iH))$ is an element of $\mathbb M(H)$ indexed by $G/H$, so the $f$-action is well-defined. On the other hand,
$$\Psi(f\nu(g_iH))=\Psi(\nu(f^{-1}g_iH)\circ h_i)=\phi(\nu(f^{-1}g_iH)\circ h_i).$$
Thus $\Psi$ is a finitary isomorphism between the Poisson systems coinduced from $H\curvearrowright (\mathbb M(H),P_\alpha)$ and $H\curvearrowright (\mathbb M(H),P_\beta)$.
\end{proof}
A version of Theorem \ref{iso} follows for $G$ containing such an $H$ as defined in this subsection. We prove Theorem \ref{iso} for the general case in Section \ref{general}.
\subsubsection{Compact subgroup case}\label{compact}
Here we assume $G$ is non-discrete, non-compact, locally compact Polish, not compactly generated, and does not contain any non-compact, compactly generated open subgroups. It is easy to see by the argument at the start of Section \ref{subgroups} that $G$ must contain a compact open subgroup. (In fact, $G$ contains infinitely many compact open subgroups.) Let $H$ be a compact open subgroup of $G$, although we will define $H$ a bit more specifically shortly. Our goal is to construct an isomorphism between the coinduced systems $\mathbb X^H_\alpha$ and $\mathbb X^H_\beta$ for any $\alpha,\beta >0$, then lift the isomorphism to $\mathbb X_\alpha$ and $\mathbb X_\beta$ with Lemma \ref{coin1}.
Unlike in Section \ref{noncompact}, we cannot utilize prior results for non-compact, compactly generated groups. Instead, we modify the results in Sections \ref{marker} and \ref{random} to work for our current needs. We again appeal to a word metric to define markers on $H$. The subgroup $H$ is trivially compactly generated, but assigning $H$ as its own generating set yields a rather useless word metric. To find a generating set so that $H$ is spacious enough under the corresponding word metric to allow marker independence properties, we use Lemma \ref{space}.
\begin{lem}\label{space}
Let $G$ be a non-compact, locally compact Polish group without a non-compact, compactly generated open subgroup, and fix $n\in\mathbb N\setminus \{0\}$. Then $G$ contains a compact, symmetric neighborhood $S$ of $e$ such that $S^{n}\neq S^{n+1}$.
\end{lem}
\begin{proof}
Let $V$ be a compact, symmetric neighborhood of $e$ in $G$ and set $J:=\langle V \rangle$.
Certainly $J$ is a compactly generated open subgroup of $G$, so by the assumptions on $G$, it must be compact. Fix $m\in\mathbb N\setminus \{0\}$, and let $W$ be another compact, symmetric neighborhood of $e$ with $V\subset W$ such that $J$ has finite index at least $m$ in the compact open subgroup
$L:=\langle W \rangle.$
Note, such a $W$ must exist since $G$ is non-compact without any non-compact, compactly generated open subgroups.
We want a subgroup of $J$ that is normal in $L$. Set
$$K:=\bigcap_{g\in L} gJg^{-1}\leq J.$$
Then $K \trianglelefteq L$. Since $J$ has finite index in $L$, the normalizer subgroup $$N_{L}(J):=\{g\in L : gJg^{-1}=J\}$$ of $J$ in $L$ has finite index in $L$. The index of $N_L(J)$ in $L$ is the number of conjugacy classes of $J$ in $L$, so the intersection in the definition of $K$ is finite, implying $K$ is open, hence compact.
Consider $F:=L/K$, a finite group. Any finite group $D$ has a minimal, symmetric generating set $R$ with order at most $\log_2|D|$. For large enough $|D|$, the final inequality in the chain of inequalities
$$|R^n|\leq |R|^n\leq \left(\log_2|D|\right)^n < |D|$$
holds.
We have a group $F$ such that $m\leq |F|$ for any $m\in\mathbb N\setminus\{0\}$. Let $S$ be a minimal, symmetric generating set for $F$ with $|S|\leq \log_2|F|$. Thus for any $n$ we can find $F$ and $S$ such that
$|S^n| < |F|$, meaning $S^n\subsetneq S^{n+1}$.
If $S$ generates $F$ so that $S^n\subsetneq S^{n+1}$, then $S\cup K$ generates $L$ and $(S\cup K)^n\subsetneq (S\cup K)^{n+1}$. Note $S\cup K$ is a compact, symmetric neighborhood of $e$.
\end{proof}
Now fix $S$ as a compact, symmetric neighborhood of $e$ in $G$ such that $S^{60}\neq S^{61}$, and set
$H:=\langle S \rangle,$
so $H$ is a compact open subgroup of $G$. We proceed to redefine concepts from Section \ref{marker}. Although we use the same terminology, some definitions are slightly different.
Let $d_S:H\times H\rightarrow \mathbb N$ be the word metric defined by $S$ on $H$, so that for any $g,h\in H$,
$$d_S(g,h)=\min
\left\{ n\geq 0 : \begin{aligned}
& \text{ there exist } s_1,\ldots,s_n\in S\\
& \text{ such that } g^{-1}h=s_1\ldots s_n\\
\end{aligned}\right\}.
$$
Note the diameter of $H$ with respect to $d_S$, denoted as $\diam H$, is finite and at least $60$.
Spheres are again defined so as to contain their boundary. Let $X$ be a Poisson point process on $H$ with intensity $\alpha$.
We say $h\in H$ is a \textit{seed} if $B(h,\diam H)=H$ satisfies the following under $X$:
\begin{enumerate}
\item For every sphere $B$ of radius $1$ such that $B\subset A(h,16,20)$, $N(B)\geq 1$.
\item The shell $A(h,20,\diam H)$ is empty.
\end{enumerate}
With non-zero probability, $H$ does not contain any seeds. However, we are ultimately interested in a sequence of Poisson point processes on $H$ indexed by $G/H$, in which case infinitely many contain seeds (Lemma \ref{infland2}). Because $H$ is compact, the index of $H$ in $G$ is countably infinite.
The proof of Lemma \ref{seed2} follows the same argument as the proof of Lemma \ref{seed}.
\begin{lem}[Seed distances for a compact group]\label{seed2}
Let $X$ be a Poisson point process on $H$ with intensity $\alpha$, and let $g,h\in H$ be seeds under $X$. Then $d_S(g,h)\leq 2.$
\end{lem}
As before, we define an equivalence relation by setting seeds $g,h$ equivalent whenever $d_S(g,h)\leq 2$. Given a seed $g$, its equivalence class of seeds is $[g]$. The \textit{core} of $[g]$ is
$$\mathcal C[g]:=\bigcap_{g\in[g]}B(g,2).$$
Cores containing a unique point of $X$ are \textit{identifiable}, and the unique $X$-point in an identifiable core is its \textit{landmark} $\ell[g]:=X|_{\mathcal C[g]}$. The \textit{fitted shell} of a landmark is
$$F[g]:=A(\ell[g],6,10).$$
Lemma \ref{seed2} implies Lemma \ref{indlem1} applies here, with $\mu\in\mathbb M(H)$ such that $\mu$ contains a landmark (necessarily unique in $H$). If $X|_{F[g]}\neq\emptyset$, we denote $F[g]$ as $F^*[g]$ and use similar notation for the landmark and core of $[g]$. The associated seed $g$ with its class, fitted shell, landmark, and core are \textit{harvestable}, and $X|_{F^*[g]}$ is their \textit{harvest}.
Let $\{g_i\}_{i\in\mathbb N}$ be coset representatives of $G/H$ with fixed enumeration,
and let
$$\nu:=(\nu(g_i H))_{i\in\mathbb N}\in \mathbb M(H)^{G/H}.$$
An element $\nu(g_i H)$ of the sequence is a \textit{harvestable process} if it contains a harvest.
Let $\psi$ be the map from the proof of Lemma \ref{coin1}. Then $\psi(\nu)\in \mathbb M(G)$. Denote the set of harvestable landmarks of $\nu$ in $G$ via $\psi$ as $\ell^*(\nu):=\{\ell^*_j\}_{j\in \mathbb N}$.
We think of $\ell^*(\nu)$ as points in $G$
and use it to define a Voronoi tessellation on $G$.
Fix a Borel isomorphism $T:G\rightarrow \mathbb R$ and a proper, left-invariant, compatible metric $d$ on $G$. The \textit{restricted Voronoi cell} of $\ell^*_j$ is the set
$$\mathcal V^*(\ell^*_j):=
\left\{f\in G: \begin{aligned}
& d(\ell^*_j,f)\leq d(\ell^*_k,f) \text{ and } \\
& T(f^{-1}\ell^*_j)\leq T(f^{-1}\ell^*_k) \text{ for } \ell^*_k\in\mathcal \ell^*(\nu)\\
\end{aligned} \right\}
$$
and the \textit{restricted Voronoi tessellation} of $\nu$ is
$$\mathcal V^*(\nu):=\{\mathcal V^*(\ell_j^*)\}_{j\in \mathbb N}.$$
Cosets of $H$ are bounded in $G$, so the next lemma implies
restricted Voronoi cells are bounded almost-surely.
\begin{lem}[Harvests on a sequence]\label{infland2}
Let $X:=(X_i)_{i\in\mathbb N}$ be a sequence of independent Poisson point processes on $H$ with intensities $\alpha$, indexed by $G/H$, and let $A\in\mathcal B(G/H)$ such that $|A|=\infty$. Then $A$ contains infinitely many harvestable processes $P_\alpha^A$ almost-surely.
\end{lem}
\begin{proof}
Let $I\subseteq \mathbb N$ such that $I$ enumerates $A$.
Fix $i\in I$ and let $E_i$ be the event that $X_i$ is harvestable. By the definition of a harvestable process, $\mathbb P(E_i)>0$ and $\mathbb P(E_i)=\mathbb P(E_j)$ for all $j$. The events $(E_i)_{i\in\mathbb N}$ are independent. Thus Borel-Cantelli implies the statement.
\end{proof}
As before, harvestable fitted shells are sources of randomness for populating output processes. We need a version of Proposition \ref{indprop} for a sequence of independent Poisson point processes $X:=(X_i)_{i\in \mathbb N}$ on $H$ with intensities $\alpha$.
Denote the fitted shell of $X_i$ as $F_i$, if it exists, and the collection of fitted shells under $X$ as $F(X)$. If $X_i$ does not contain a fitted shell, set $F_i=H$, so
$$X_i|_{F_i}= \begin{cases}
X_i|_{F_i} & \text{ if } X_i \text{ contains a fitted shell}\\
X_i & \text{ otherwise. }
\end{cases}$$
Set
$X|_{F(X)}:=(X_i|_{F_i})_{i\in\mathbb N}$; define $X_i|_{F_i^{c}}$, $X_i|_{F_i^{*}}$, and $X_i|_{F_i^{*c}}$ similarly.
Lemma \ref{indlem2} holds on each $H$ copy with $G=H=A$ by Lemma \ref{seed2}, and independence between sequence elements is by assumption. Thus we have the following.
\begin{prop}[Fitted shells on a sequence]\label{indprop2}
Let $X:=(X_i)_{i\in \mathbb N}$ and $Y:=(Y_i)_{i\in \mathbb N}$ be sequences of independent Poisson point processes on $H$ with intensities $\alpha$. The sequence $Z:=Y|_{F(X)}+X|_{F(X)^c}$ is equal in distribution to $X$, and $F(Z)=F(X)$.
\end{prop}
Versions of Corollaries \ref{indcor} and \ref{source} for sequences follow from Proposition \ref{indprop2}. We explicitly redefine $\xi$ from Corollary \ref{source} for future reference. For each $\ell^*_j\in \ell^*(X)$, set $A_j:=A(\ell^*_j,6,10)$, so that $A_j\subset G$ corresponds to a harvestable fitted filler of some Poisson process on $H$. Then
$\{(\ell_j^*)^{-1}\psi(X)|_{A_j}\}_{j\in \mathbb N}$
is a sequence of independent and identically distributed Poisson processes on $A(e,6,10)\subset G$. For $k\in\mathbb N\setminus\{0\}$, let
$$R_k:A(e,6,10)\rightarrow [0,1]$$
be the Borel isomorphisms from the proof of Corollary \ref{source}, and as in the same proof, denote the shifted process points of $(\ell_j^*)^{-1}\psi(X)|_{A_j}$ as $$v_1^j,v_2^j,\ldots v_{n_j}^j,$$ where $n_j:=N(A_j)$. Define
$\xi:\mathbb M(G)\times\mathcal B(G)\rightarrow [0,1]$
so that for a sequence $X$ of independent and identically distributed Poisson processes on $H$ indexed by ${G/H}$,
$$\xi(\psi(X),B)=\begin{cases}
R_{n_j}(v_1^j,\ldots, v_{n_j}^j) & \text{ if } B=A_j\\
0 & \text{ otherwise.}
\end{cases}$$
Then $\{\xi(\psi(X),A_j)\}_{j\in\mathbb N}$ is a sequence of independent uniform random variables almost-surely.
The restricted Voronoi tessellation of $X$ depends on $X|_{F^*(X)^c}$ and the fixed enumeration $\{g_i\}_{i\in\mathbb N}$ of coset representatives, so it is independent of $X|_{F^*(X)}$. Thus a version of Lemma \ref{partition} holds, and we are prepared to prove a version of Proposition \ref{preproduct}, which implies a version of Theorem \ref{iso}.
\begin{prop}[Intermediate product isomorphism]\label{preproduct2}
Let $\alpha,\beta >0$. There exists a finitary isomorphism from $\mathbb X^H_\alpha$ to $(\mathbb X_\alpha^H,\mathbb X_\beta^H)$.
\end{prop}
\begin{proof}
Let $X:=(X(g_i H))_{i\in\mathbb N}$ be a sequence of independent Poisson point processes on $H$ with intensities $\alpha$. Recall
$$\psi(X)=\sum_{i\in\mathbb N}g_iX(g_iH)$$
is a Poisson point process on $G$ with intensity $\alpha$
and restricted Voronoi tessellation $\mathcal V^*(X)$.
Set $$\mathcal S_j:=(\ell^*_j)^{-1}\mathcal V^*(\ell^*_j) \text{ and } N_j:=N(\psi(X)|_{A_j})$$
for $j\in \mathbb N.$ Here $\mathcal S_j$ is a Borel subset of $G$ ``centered'' at $e\in G$. Let the maps $q_j$, $r_j$, $R_{j,k}$, $T_j$, and $\pi_j$ be defined as in the proof of Proposition \ref{preproduct} for $j\in \mathbb N$ and $k\in \mathbb N\setminus \{0\}$.
We have a sequence $\{\xi(\psi(X),A_j)\}_{j\in\mathbb N}$ of independent uniform random variables almost-surely. Set $\xi_j:=\xi(\psi(X),A_j)$ for $j\in\mathbb N$. The sum
$$W_2:=\sum_{j\in\mathbb N}\ell_j^*\pi_j(\xi_j)_2$$
is a Poisson process on $G$ with intensity $\beta$
by Remark \ref{cutpaste}, Proposition \ref{indprop2}, and versions of Corollary \ref{source} and Lemma \ref{partition}.
We define the second coordinate of our desired factor map as
$\phi(X)_2:=\psi^{-1}(W_2).$
Note $\phi(X)_2$ is a sequence of independent Poisson processes on $H$ with intensities $\beta$, indexed by $G/H$.
Consider that
$$W_1:=\psi(X)|_{\left(\bigcup_{j\in\mathbb N} A_j\right)^c}+\sum_{j\in\mathbb N}\ell^*_j\pi_j(\xi_j)_1$$
is a Poisson process of intensity $\alpha$ on $G$, equal to $\psi(X)$ outside of harvestable fitted fillers, as well as inside fillers associated with non-empty Voronoi cells, and resampled point for point inside fillers associated with empty Voronoi cells via the leftover information from those $\xi_j$, as in the proof of Proposition \ref{preproduct}.
Define
$\phi(X)_1:=\psi^{-1}(W_1)$
and
$$\phi(X):=(\phi(X)_1,\phi(X)_2).$$
The coordinates $\phi(X)_1$ and $\phi(X)_2$ are independent by a version of Lemma \ref{partition},
so $\phi$ is a measurable, bijective map from $\mathbb X^H_\alpha$ to $(\mathbb X_\alpha^H,\mathbb X_\beta^H)$.
Recall from the proof of Lemma \ref{coin1} that $\psi$ and $\psi^{-1}$ are $G$-equivariant. Further, the extraction map $\xi$ is $G$-invariant, as is each population map $\pi_j$. Let $f\in G$, and recall the action of $H\curvearrowright \mathbb M(H)$ coinduced to $G$. We have
$$\phi(fX)_2=\psi^{-1}\left(\sum_{j\in\mathbb N}f\ell_j^*\pi_j(\xi_j)_2\right)=f\psi^{-1}\left(W_2\right)= f\phi(X)_2.$$
Moreover, the set $\cup A_j$ is $G$-equivariant, so by a similar argument, $$\phi(fX)_1=f\phi(X)_1.$$
We have shown $\phi$ is a bijective factor from $\mathbb X_\alpha^H$ to $(\mathbb X_\alpha^H,\mathbb X_\beta^H)$. It is finitary, since Lemma \ref{infland2} implies restricted Voronoi cells are bounded almost-surely. It is straightforward to check $\phi^{-1}$ is a finitary factor from $(\mathbb X_\alpha^H,\mathbb X_\beta^H)$ to $\mathbb X_\alpha^H$.
\end{proof}
\subsubsection{General case}\label{general}
Finally, fix $G$ as a non-discrete, non-compact, locally compact Polish group. Then $G$ has a compactly generated open subgroup. We use the results from Sections \ref{cg}, \ref{noncompact}, and \ref{compact} to prove Theorem \ref{iso}.
\begin{proof}[Proof of Theorem \ref{iso}]
Let $H\leq G$ be a compactly generated open subgroup, and let $\alpha,\beta >0$. First suppose $H=G$. Then Theorem \ref{preiso} implies the statement. If $H$ is non-compact,
by Lemma \ref{coin2}, the Poisson systems
coinduced from $H\curvearrowright (\mathbb M(H),P_\alpha)$ and $H\curvearrowright (\mathbb M(H),P_\beta)$ are finitarily isomorphic, and by Lemma \ref{coin1}, they are finitarily isomorphic to
$(\mathbb M(G),P_\alpha,G)$ and $(\mathbb M(G),P_\beta,G)$,
respectively.
Thus the systems $(\mathbb M(G),P_\alpha,G)$ and $(\mathbb M(G),P_\beta,G)$ are finitarily isomorphic.
Now suppose $H$ is compact.
Let $X$ be a sequence of independent Poisson point processes with intensities $\alpha$ on $H$, indexed by $G/H$.
Let $\phi$ be defined as in the proof of Proposition \ref{preproduct2}. The same argument in the proof of Theorem \ref{preiso} works here, and the composition of $\phi^{-1}$ with the permuted outputs of $\phi(X)$ is a finitary isomorphism between the coinduced systems $\mathbb X^H_\alpha$ and $\mathbb X^H_\beta$. By Lemma \ref{coin1}, the systems $\mathbb X_\alpha$ and $\mathbb X_\beta$ are finitarily isomorphic.
\end{proof}
Theorems \ref{factor} and \ref{finitary} follow immediately.
\begin{proof}[Proof of Corollary \ref{product}]
Let $H\leq G$ be a compactly generated open subgroup, and let $\alpha,\beta,\theta >0$.
Again we consider cases on $H$, and first suppose $H=G$. Let $X,Y,$ and $Z$ be Poisson point processes on $H$ with intensities $\alpha$, $\beta$, and $\theta,$ respectively.
The map $\phi$ from Proposition \ref{preproduct} depends on the intensity of the output $\phi(X)_2$. To avoid confusion, denote the map $\phi$ such that $\phi(X)_2$ is a Poisson point process with intensity $\alpha$ as $\phi_\alpha$. The map sending $X$ to
$\phi^{-1}_\beta(\phi_\beta(X)_2,\phi_\beta(X)_1)$ is a finitary isomorphism from $\mathbb X_\alpha$ to $\mathbb X_\beta$; call this map $\zeta_\beta$. Consider that $$\phi_\theta(\zeta_\beta(X))\overset{d}{=}(Y,Z).$$
By the proofs of Proposition \ref{preproduct} and Theorem \ref{preiso}, the composition $\phi_\theta(\zeta_\beta)$ is a finitary isomorphism from $\mathbb X_\alpha$ to $(\mathbb X_\beta,\mathbb X_\theta)$.
Next assume $H\neq G$. Our argument does not depend on whether $H$ is compact. We merely note Proposition \ref{preproduct2} still holds when $H$ is non-compact. Let $X',Y',$ and $Z'$ be sequences of Poisson point processes on $H$ with intensities $\alpha$, $\beta$, and $\theta,$ respectively, indexed by $G/H$. In contrast with the prior case, we redefine $\phi$ to be the map from Proposition \ref{preproduct2}. But as before, denote $\phi$ such that $\phi(X')_2$ is a sequence of Poisson processes with intensities $\alpha$ as $\phi_\alpha$. Redefine $\zeta_\beta$ to be the map sending $X'$ to $\phi^{-1}_\beta(\phi_\beta(X')_2,\phi_\beta(X')_1)$, so that
$$\phi_\theta(\zeta_\beta(X'))\overset{d}{=}(Y',Z').$$
The proofs of Proposition \ref{preproduct2} and Theorem \ref{iso} imply $\phi_\theta(\zeta_\beta)$ is a finitary isomorphism from $\mathbb X^H_\alpha$ to $(\mathbb X^H_\beta,\mathbb X^H_\theta)$.
By Lemma \ref{coin1}, the systems $\mathbb X^H_\alpha$ and $\mathbb X_\alpha$ are finitarily isomorphic, so the relevant product map is a finitary isomorphism between $(\mathbb X^H_\beta,\mathbb X^H_\theta)$ and $(\mathbb X_\beta,\mathbb X_\theta)$. Therefore $\mathbb X_\alpha$ and $(\mathbb X_\beta,\mathbb X_\theta)$ are finitarily isomorphic.
\end{proof}
\bibliography{embedding}
\bibliographystyle{amsalpha}
\end{document} | 0.001544 |
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TITLE: Lifting automorphisms on a Blow-up surface
QUESTION [3 upvotes]: Let $X$ be a projective surface and let $x\in X$ be a smooth point. Consider the blow up $Bl_{x}X$ of $X$ in $x$, and let $E$ be the exceptional divisor. Suppose we know that $E$ is the only (-1)-curve in $Bl_{x}X$, so that $\alpha(E) = E$ for any $\alpha\in Aut(Bl_{x}X)$. Then any automorphisms $\alpha\in Aut(Bl_{x}X)$ induces an automorphism of $X$ fixing $x$, let us denote $Aut_{x}(X)\subseteq Aut(X)$ the subgrup of this automorphisms. We get a injective morphism
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Over 100 nearby golf courses are sure to thrill even the most seasoned player. Soak in the hot tub after a long day on the links or by the pool in the sun. Tennis, hiking, mountain biking, Jeep tours, and horseback riding are just a few of the many great activities around the area. Explore the Andreas and Tahquitz canyons on one day, or head into town for a bit of culture, modern shopping and comfort.
Learn more about Palm Springs | 0.020395 |
I'm auditioning now for scholarships and all, and I'm doing Carnival of Venice.
I gotta do lyrical and technical. I can double tongue well at fast speeds, but triple tongue is poor so I aint doing variation II. The variation for lyrical of course is Introduction.
Then for technical, which would impress the auditioners most, variation III, or IV? I find IV easier than III, but I've really been working at variation III and it sounds smooth and alright. But what's the best choice here that shows off my skill best assuming I play them well?
Edit: I'm talking about the song in Arban book just to make things clear. | 0.116891 |
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Explanations of EVA, MVA and NPV and
Extracts from this document...
Introduction
Explanations of EVA, MVA and NPV and their relationship with each other The concept of EVA is a measure of economic profit and was popularised and originally trade-marked by Stern Stewart Consulting Company in the 1980's. Economic Value Added (EVA) can be defined as the difference between net operating profit after taxes and the monetary value of a company's total cost of capital. Should a company's profit exceed the overall costs of funds they create EVA. It can be so important because EVA is the most efficient internal measure of the true economic profit of a company. Managers within any company can use this measure in order to obtain any crucial information they may need when making crucial decisions. When broken up EVA can simply be defined as an estimate of the amount by which earnings exceed or fall short of the required lowest rate of return that shareholders could receive by investing in the company. ...read more.
Middle
In order to see whether a value has been added, or destroyed, to the company over a period of time the difference in MVA from one date to a the next should be calculated. With MVA representing the stock markets assessment of a company it can be derived that the higher the MVA the better as this would represent greater wealth for the shareholders. The finance decision model Net Present Value (NPV) can be defined as the difference between the present value of an investment's future net cash flows less the initial investment. The result of NPV expresses how much value an investment will result in which is done by measuring over a period of time all cash flows, and back towards the present time. Should the result of the NPV method be positive, should there not be a better investment anywhere else, it can be concluded that an investment should be made. ...read more.
Conclusion
If debt is used in the right way, it can help to lower the WACC, so it must be made sure that the WACC calculation has taken account of all factors that could affect it. A company's WACC is a very important figure,. These corporate investments should result in an increase in stock prices. These are the projects that should be invested in. Investments that earn less than the firm's WACC will result in a decrease in stockholder value and should be avoided by the company. So, in other words the WACC must be looked at to see if the current capital structure of the company is in good shape and will benefit any investors. .. | 0.898011 |
The Contender
Pro (for)
Losing
0 Points
Stricter Gun Control
Post Voting Period
The voting period for this debate has ended.
after 1 vote the winner is...
Varrack
Debate Rounds (4)
Votes (1)
4 comments have been posted on this debate. Showing 1 through 4 records.
Posted by Varrack 1 year ago
If you post then the number of rounds per person would be uneven, 3-2, so you have to skip.
Report this Comment
Posted by InnovativeEphemera 1 year ago
So am I posting my last round now or skipping?
Report this Comment
Posted by Varrack 1 year ago
Dang that sucks. Almost happened to me last round - I submitted my argument with 9 seconds left. Fortunately conduct is only worth 1 point so it probably won't butcher the debate.
Report this Comment
Posted by InnovativeEphemera 1 year ago
Damn. Halfway through typing. Sorry bro.
Report this Comment
1 votes has been placed for this debate.
Reasons for voting decision: Both sides FF'd. But Pro's FF is more important. It causes him to drop Con's final responses. This debate boiled down to something where I have read most of the literature: how firearms effect the crime rate. If guns reduce crime or have no effect, there is no reason to restrict guns. If they harm our society, there is justifiable reason to ban them. Pro in many cases says the UK restrictions are lax--something Con showed was untrue. Even assuming the UK has more access to firearms, I don't see how it is relevant. Pro offers one really good argument, though: cross-sectional data proves guns increase the homicide rate. Con responds with a study published in the Harvard JLLP (Kates and Mauser 2010). The study, as Con tells us, finds little relationship between firearms and crime, and the relationship may entail that guns reduce crime. This means any restrictions would increase the crime rate and be unjust. Had Pro not FF'd, he may have won. But from hat we had, Con won. Good debate. | 0.00336 |
TITLE: Problem 1.5.11 in Aluffi's Algebra: Chapter 0
QUESTION [3 upvotes]: I am trying to prove the following:
Let $A$ and $B$ be sets endowed with the equivalence relations $\sim_A$ and $\sim_B$, respectfully. Define the relation $\sim$ on $A \times B$ by setting
$$(a_1,b_1) \sim (a_2,b_2) \; \text{if and only if} \; a_1\sim_A a_2 \; \text{and} \; b_1\sim_B b_2.$$
Use the universal property of quotients to establish that there are functions $(A\times B)/\sim \rightarrow A/\sim_A$, $(A\times B)/\sim\rightarrow B/\sim_B$.
Prove that $(A\times B)/\sim$ with these two functions satisfies the
universal property for the product of $(A/\sim_A)\times (B/\sim_B)$.
Conclude without further work that $(A\times B)/\sim \cong (A\sim_A)\times (B/\sim_B)$.
First of all, sorry that I cannot draw the diagrams here. Hopefully, this isn't too terrible.
I worked out the first two parts without much difficulty: let $\varphi_A$ and $\varphi_B$ denote the maps which sends $(a,b)$ to $[a]_{\sim_A}$ and $(a,b)$ to $[b]_{\sim_B}$, respectively. Then, by the universal property of quotient spaces, there exists unique functions $\psi_A:(A\times B)/\sim \rightarrow A/\sim_A$ and $\psi_B:(A\times B)/\sim \rightarrow B/\sim_B$ such that $\varphi_A=\psi_A\circ \pi$ and $\varphi_B=\psi_B\circ \pi$, where $\pi:A\times B\rightarrow (A\times B)/\sim$ is the projection map. Thus, by the universal property for $(A/\sim_A)\times (B\sim_B)$ we can conclude there exists a unique function $\Psi:A\times B\sim\rightarrow (A/\sim_A)\times(B/\sim_B)$ such that $\pi_A\circ \Psi=\psi_A$ and $\pi_B\circ \Psi=\psi_B$, where $\pi_i$ represents the projection onto the $i$ coordinate. Namely we get $\Psi=\psi_A\times\psi_B$, where the uniqueness of $\Psi$ follows from the uniqueness of $\psi_A$ and $\psi_B$.
From here I don't see how to conclude this is an isomorphism. Any ideas? Have I made a mistake?
REPLY [4 votes]: Any two objects satisfying the universal property of the product are isomorphic.
More generally, a limit of a diagram, if it exists, is unique up to a unique isomorphism.
A product of $X$ and $Y$ is the limit of the discrete diagram with objects $X$ and $Y$ (discrete means no morphisms in the diagram). | 0.047471 |
In an effort to expose young people to the music of Bob Dylan, Reimagine Music is offering Subterranean Homesick Blues: A Tribute to Bob Dylan’s Bringing It All Back Home featuring modern, independent artists, according to the Associated Press.
Jim Sampas, who has done similar projects with Bruce Springsteen’s Nebraska and the Beatles’ Rubber Soul, produced the compilation.
Artists like Castanets, The Helio Sequence and Asobi Seksu interpret 1965’s Bringing It All Back Home. The tribute also includes seven bonus tracks.
Tracklisting:
1. “Subterranean Homesick Blues”— Peter Moren
2. “She Belongs to Me” — Ane Brun
3. “Maggie’s Farm” — Castanets
4. “Love Minus Zero” — Mirah
5. “Outlaw Blues” — The Morning Benders
6. “On the Road Again” — Julie Doiron
7. “Bob Dylan’s 115th Dream” — Asobi Seksu
8. “Mr. Tambourine Man” — The Helio Sequence
9. “Gates of Eden” — DM Stith
10. “It’s Alright, Ma (I’m Only Bleeding)” — Franz Nicolay
11. “It’s All Over Now, Baby Blue” — Sholi
12. “If You’ve Gotta Go, Go Now” (Bonus Track) — J. Tillman
13. “Sitting on a Barbed Wire Fence” (Bonus Track) — Sea Wolf
14. “I’ll Keep it With Mine” (Bonus Track) — Denison Witmer
15. “Mama, You’ve Been on My Mind” (Bonus Track) — Laura Veirs
16. “Farewell Angelina” (Bonus Track) — William Fitzsimmons
17. “California” (Bonus Track) — Will Dailey
18. “It’s All Over Now, Baby Blue” (Bonus Track) [Electric Version] — Sholi
You can pick up a digital copy here.
Sign up for the Paste newsletter Get our daily summary of the day's top articles and new items. Sign Up Thank you! Your email address has been added to our list. You will begin receiving our newsletter within 48 hours. | 0.025742 |
Athlete | Skeleton
Abbotsford Sports Hall of Fame Class of 2015
Amy Gough first drew notice playing rugby at Abbotsford Senior Secondary, where she was part of a provincial championship-winning Panthers squad.
Gough tried skeleton for the first time in 2002 at a “Discover Skeleton” event at Canada Olympic Park in Calgary. That experience ignited a passion for the sport, and four years later, she was on Canada’s World Cup team, finishing seventh in the overall points standings. Gough spent the next two seasons predominantly on the Intercontinental Cup circuit (the second tier of international skeleton competition), winning three races in 2008-09. She was back on the World Cup Circuit in 2009-10, earning a silver medal at Park City, Utah along with three other top-10 finishes.
The signature moment of Gough’s career was making the Canadian team for the 2010 Vancouver Olympics. She gave her hometown fans a huge thrill by clocking the second-fastest time in the first of four Olympic runs at the Whistler Sliding Centre, trailing only eventual gold medalist Amy Williams of Great Britain. Gough ended up finishing seventh, just 0.65 seconds away from a podium spot. She completed two more World Cup campaigns – in 2010-11 she earned a trio of bronze medals to finish fifth in the overall world standings, and in 2011-12 she won her first World Cup race at Winterberg, Germany then added another bronze medal en route to a sixth-place overall finish. | 0.747096 |
\begin{document}
\maketitle
\begin{abstract}
In this paper, we study the set of homogeneous geodesics of a
left-invariant Finsler metric on Lie groups. We first give a simple
criterion that characterizes geodesic vectors. As an application, we
study some geometric properties of bi-invariant Finsler metrics on
Lie groups. In particular a necessary and sufficient condition that
left-invariant Randers metrics are of Berwald type is given. Finally
a correspondence of homogeneous geodesics to critical points of
restricted Finsler metrics is given. Then results concerning the
existence homogeneous geodesics are obtained.
\end{abstract}
\textbf{Keywords}: Invariant Finsler metrics, Homogeneous geodesics,
Geodesic vectors, Randers spaces.
\\
\textbf{PACS numbers}: 02.40.Ky, 02.40.Sf, 4520J\\
\textbf{Mathematics Subject Classifications}: 53C60; 53C35; 53C30;
53C22
\\
\section{Introduction}
\
A classical problem of differential geometry is to study geodesics
of Riemannian manifolds $(M,g)$. Of particular interest are
geodesics with some special properties, for example homogeneous
geodesics. A geodesic of a Riemannian manifold $(M,g)$ is called
homogeneous if
it is an orbit of a one-parameter group of isometries of $M$. For results on homogeneous geodesics in
homogeneous Riemannian manifolds we refer to [8], [14], [12], [11].\\
Homogeneous geodesics have important applications to mechanics. For
example, the equation of motion of many systems of classical
mechanics reduces to the geodesic equation in an appropriate
Riemannian manifold $M$.\\ Geodesics of left-invariant Riemannian
metrics on Lie groups were studied by V. I. Arnold extending Euler's
theory of rigid-body motion [1]. A major part of V. I. Arnold's
paper is devoted to the study of homogeneous geodesics. Homogeneous
geodesics are called by V. I. Arnold "relative equilibriums ". The
description of such relative equilibria is important for qualitative
description of the behavior of the corresponding mechanical system
with symmetries. There is a big literature in mechanics devoted to
the investigation of relative equilibria. Homogeneous geodesics are
interesting also in pseudo-Riemannian geometry and light-like
homogeneous geodesics are of particular interest. For results on
homogeneous geodesics in homogeneous pseudo-Riemannian manifolds we
refer for example to [18], [19], [21], [7], [3], [6]. In [18], [21]
and [7], the authors study plane-wave limits(Penrose limits )of
homogeneous spacetimes along light-like homogeneous geodesics.\\
About the existence of homogeneous geodesics in a general
homogeneous Riemannian manifold, we have, at first, a result due to
V. V. Kajzer who proved that a Lie group endowed with a
left-invariant Riemannian metric admits at least one homogeneous
geodesic [10]. More recently O. Kowalski and J. Szenthe extended
this result to all homogeneous Riemannian manifolds [13]. An
extension of result of [13] to reductive homogeneous
pseudo-Riemannian manifolds has been also obtained [21], [6].
Homogeneous geodesics of left-invariant Lagrangians on Lie groups
were studied by J. Szenthe [24]. In this paper, we study the set of
homogeneous geodesics of a left-invariant Finsler metric on Lie
groups.
\section{Preliminaries}
\subsection{Finsler spaces}
In this section, we recall briefly some known facts about Finsler
spaces. For details, see [2], [23], [4]. \
Let $M$ be a n-dimensional $C^{\infty}$ manifold and
$TM=\bigcup_{x\in M}
T_{x}M$ the tangent bundle. If the continuous function $F:TM\longrightarrow
R_{+}$ satisfies the conditions that it is $C^{\infty}$ on $TM\setminus
\{0\}$; $F(tu)=tF(u)$ for all $t\geq 0$ and $u\in TM$, i.e, $F$ is
positively homogeneous of degree one; and for any tangent vector $y\in T_{x}M\setminus \{0\}$, the following bilinear symmetric form $g_{y}:T_{x}M \times T_{x}M\longrightarrow R$ is positive definite :
$$g_{y}(u,v)=\frac{1}{2}\frac{\partial^{2}}{\partial s\partial t}[F^{2}(x,y+su+tv)]|_{s=t=0},$$
then we say that $(M,F)$ is a Finsler manifold.\\
Let $$g_{ij}(x,y)=(\frac{1}{2}F^{2})_{y^{i}y^{j}}(x,y).$$ By the
homogeneity of $F$, we have
$$g_{y}(u,v)=g_{ij}(x,y)u^{i}v^{j},\hskip.5cm F(x,y)=\sqrt{g_{ij}(x,y)y^{i}y^{j}}.$$
Let $\gamma:[0,r]\longrightarrow M$ be a piecewise $C^{\infty}$
curve. Its integral length is defined as
$$L(\gamma)=\int_{0}^{r}F(\gamma(t),\dot{\gamma}(t))dt.$$ For $x_{0},x_{1}\in
M$ denote by $\Gamma(x_{0},x_{1})$ the set of all piecewise
$C^{\infty}$ curve $\gamma:[0,r]\longrightarrow M$ such that
$\gamma(0)=x_{0}$ and $\gamma(r)=x_{1}$. Define a map $d_{F}:M\times
M\longrightarrow [0,\infty)$ by \[d_{F}(x_{0},x_{1})=\inf_{\gamma\in
\Gamma(x_{0},x_{1})} L(\gamma).\] Of course, we have
$d_{F}(x_{0},x_{1})\geq 0$, where the equality holds if and only if
$x_{0}=x_{1}$; $d_{F}(x_{0},x_{2})\leq
d_{F}(x_{0},x_{1})+d_{F}(x_{1},x_{2})$. In general, since $F$ is
only a positive homogeneous function, $d_{F}(x_{0},x_{1})\neq
d_{F}(x_{1},x_{0})$, therefore $(M,d_{F})$ is only a non-reversible
metric space.
Let $\pi^{\ast}TM$ be the pull-back of the tangent bundle $TM$ by
$\pi: TM\setminus\{0\}\longrightarrow M$. Unlike the Levi-Civita
connection in Riemannian geometry, there is no unique natural
connection in the Finsler case. Among these connections on
$\pi^{\ast}TM$, we choose the \emph{Chern connection} whose
coefficients are denoted by $\Gamma^{i}_{jk}$(see[2,p.38]). This
connection is almost $g-$compatible and has no torsion. Here
$g(x,y)=g_{ij}(x,y)dx^{i}\otimes
dx^{j}=(\frac{1}{2}F^{2})_{y^{i}y^{j}}dx^{i}\otimes dx^{j}$ is the
Riemannian metric on the pulled-back bundle $\pi^{\ast}TM$.
The Chern connection defines the covariant derivative $D_{V}U$ of a
vector field $U\in \chi(M)$ in the direction $V\in T_{p}M$. Since,
in general, the Chern connection coefficients $\Gamma^{i}_{jk}$ in
natural coordinates have a directional dependence, we must say
explicitly that $D_{V}U$ is defined with a fixed reference vector.
In particular, let $\sigma :[0,r]\longrightarrow M$ be a smooth
curve with velocity field $T=T(t)=\dot{\sigma}(t)$. Suppose that $U$
and $W$ are vector fields defined along $\sigma$. We define $D_{T}U$
with \emph{reference vector} $W$ as
$$D_{T}U=\left[\frac{dU^{i}}{dt}+U^{j}T^{k}(\Gamma^{i}_{jk})_{(\sigma ,W)}\right]\frac{\partial}{\partial x^{i}}\mid_{\sigma(t)}.$$
A curve $\sigma:[0,r]\longrightarrow M$, with velocity
$T=\dot{\sigma}$ is a Finslerian \emph{geodesic} if\\
$D_{T}\left[\frac{T}{F(T)}\right]=0$ ,\hskip.5cm with reference vector $T$.\\
We assume that all our geodesics $\sigma (t)$ have been
parameterized to have constant Finslerian speed. That is, the length
$F(T)$ is constant. These geodesics are characterized by the
equation \\
$D_{T}T=0$ , \hskip.5cm with reference vector $T$.\\ \\Since
$T=\frac{d\sigma^{i}}{dt}\frac{\partial}{\partial x^{i}}$, this
equation says that
$$\frac{d^{2}\sigma^{i}}{dt^{2}}+\frac{d\sigma^{j}}{dt}\frac{d\sigma^{k}}{dt}(\Gamma^{i}_{jk})_{(\sigma ,T)}=0.$$
If $U,V$ and $W$ are vector fields along a curve $\sigma$, which has
velocity $T=\dot{\sigma}$, we have the derivative rule
$$\frac{d}{dt}g_{_W}(U,V)=g_{_W}(D_{T}U,V)+g_{_W}(U,D_{T}V)$$
whenever $D_{T}U$ and $D_{T}V$ are with reference vector $W$ and one
of the following conditions holds:
\begin{description}
\item[i)] U or V is proportional to W, or
\item[ii)] W=T and $\sigma$ is a geodesic.
\end{description}
\subsection{Left-invariant Finsler metrics on Lie groups}
Let $G$ be a connected Lie group with Lie algebra $\frak{g}=T_{e}G$.
We may identify the tangent bundle $TG$ with $G\times \frak{g}$ by
means of the diffeomorphism that sends $(g,X)$ to
$(L_{g})_{\ast}X\in T_{g}G$.
\begin{Def}
A Finsler function $F:TG\longrightarrow R_{+}$ will be called
$G$-invariant if $F$ is constant on all $G$-orbits in $TG=G\times
\frak{g}$; that is, $F(g,X)=F(e,X)$ for all $g\in G$ and $X\in
\frak{g}$.
\end{Def}
The G-invariant Finsler functions on $TG$ may be identified with the
Minkowski norms on $\frak{g}$. If $F:TG\longrightarrow R_{+}$ is an
G-invariant Finsler function, then we may define
$\widetilde{F}:\frak{g}\longrightarrow R_{+}$ by
$\widetilde{F}(X)=F(e,X)$, where $e$ denotes the identity in $G$.
Conversely, if we are given a Minkowski norm
$\widetilde{F}:\frak{g}\longrightarrow R_{+}$, then $\widetilde{F}$
arises from an G-invariant Finsler function $F:TG\longrightarrow
R_{+}$ given by $F(g,X)=\widetilde{F}(X)$ for all $(g,X)\in G\times
\frak{g}$.\\
Let $G$ be a connected Lie group, $L:G\times G\longrightarrow G$ the
action being defined by the left-translations
$L_{g}:G\longrightarrow G$, $g\in G$ and $TL:G\times
TG\longrightarrow TG$ the action given by the tangent linear maps
$TL_{g}:TG\longrightarrow TG$, $g\in G$ of the left-translations. \\
A smooth vector field $X:TG-\{0\}\longrightarrow TTG$ is said to be
left-invariant if
$$TTL_{g}\circ X\circ TL^{-1}_{g}=X \hskip1cm \forall g \in G.$$
By a classical argument of calculus of variation we have the
following proposition.
\begin{prop}
If $F:TG\longrightarrow R_{+}$ is a left-invariant Finsler metric
then its geodesic spray $X$ is left-invariant as well.
\end{prop}
\section{Homogeneous geodesics of left invariant Finsler metrics}
\begin{Def}
Let $G$ be a connected Lie group, $\frak{g}=T_{e}G$ its Lie algebra
identified with the tangent space at the identity element,
$\widetilde{F}:\frak{g} \longrightarrow R_{+}$ a Minkowski norm and
$F$ the left-invariant Finsler metric induced by $\widetilde{F}$ on
$G$. A geodesic $\gamma :R_{+}\longrightarrow G$ is said to be
\emph{homogeneous} if there is a $Z\in \frak{g}$ such that
$\gamma(t)=exp(tZ)\gamma(0)$, $t\in R_{+}$ holds. A tangent vector
$X\in T_{e}G-\{0\}$ is said to be a \emph{geodesic vector} if the
1-parameter subgroup $t\longrightarrow exp(tX)$, $t\in R_{+}$, is a
geodesic of $F$.
\end{Def}
The geodesic defined by a geodesic vector is obviously a homogeneous
one. Conversely, let $\gamma$ be a geodesic with $\gamma(0)=g$ which
is homogeneous with respect to a 1-parameter group of
left-translations, namely $$\gamma(t)=exp(tY)g,\hskip.5cm t\in
R_{+},$$ then a homogeneous geodesic $\widetilde{\gamma}$ is given
by
\begin{eqnarray*}
\widetilde{\gamma}(t) &=& L^{-1}_{g}\circ \gamma(t)=L^{-1}_{g}\circ R_{g}\circ exp(tY)\\
&=&exp(Ad(g^{-1})tY).e=exp(Ad(g^{-1})tY)\widetilde{\gamma}(0),
\end{eqnarray*}
which means that $X=Ad(g^{-1})Y$ is a geodesic vector.\\
For results on homogeneous geodesics in homogeneous Finsler
manifolds we refer to [16]. The basic formula characterizing
geodesic vector in the Finslerian case was derived in [16], Theorem
3.1. In the following theorem we present a new elementary proof of
this theorem for left invariant Finsler metrics on Lie groups.
\begin{thm}
Let $G$ be a connected Lie group with Lie algebra $\frak{g}$, and
let $F$ be a left-invariant Finsler metric on $G$. Then $X\in
\frak{g}-\{0\}$ is a geodesic vector if and only if
$$g_{X}(X,[X,Z])=0$$ holds for every $Z\in
\frak{g}$.
\end{thm}
Proof: Following the conventions of [9] a left-invariant vector
field associated to an element $X$ in $T_{e}G$ is denoted by
$\widetilde{X}:G\longrightarrow TG$; that is
$\widetilde{X}_{x}=L_{x\ast}X$. For any left invariant vector fields
$\widetilde{X}, \widetilde{Y}, \widetilde{Z}$ on $G$, we have
\begin{equation}
\widetilde{Y}g_{\widetilde{X}}(\widetilde{Z},\widetilde{X})=g_{\widetilde{X}}(D_{\widetilde{Y}}\widetilde{Z},\widetilde{X})+g_{\widetilde{X}}(\widetilde{Z},D_{\widetilde{Y}}\widetilde{X})
\qquad \mbox{with reference $\widetilde{X}$}
\end{equation}
Similarly,
\begin{equation}
\widetilde{Z}g_{\widetilde{X}}(\widetilde{Y},\widetilde{X})=g_{\widetilde{X}}(D_{\widetilde{Z}}\widetilde{Y},\widetilde{X})+
g_{\widetilde{X}}(\widetilde{Y},D_{\widetilde{Z}},\widetilde{X})
\end{equation}
\begin{equation}
\widetilde{X}g_{\widetilde{X}}(\widetilde{Z},\widetilde{X})=g_{\widetilde{X}}(D_{\widetilde{X}}\widetilde{Z},\widetilde{X})+
g_{\widetilde{X}}(\widetilde{Z},D_{\widetilde{X}},\widetilde{X})
\end{equation}
All covariant derivatives have $\widetilde{X}$ as reference
vector.\\
Subtracting (2) from the summation of (1) and (3) we get
\begin{eqnarray*}
g_{\widetilde{X}}(\widetilde{Z},D_{\widetilde{X}+\widetilde{Y}}\widetilde{X})+ g_{\widetilde{X}}(\widetilde{X}-\widetilde{Y},D_{\widetilde{Z}}\widetilde{X})&=& \widetilde{Y}g_{\widetilde{X}}(\widetilde{Z},\widetilde{X})-\widetilde{Z}g_{\widetilde{X}}(\widetilde{Y},\widetilde{X})+\widetilde{X}g_{\widetilde{X}}(\widetilde{Z},\widetilde{X})\\
&&-g_{\widetilde{X}}([\widetilde{Y},\widetilde{Z}],\widetilde{X})-g_{\widetilde{X}}([\widetilde{X},\widetilde{Z}],\widetilde{X}),
\end{eqnarray*}
where we have used the symmetry of the connection, i.e.,
$D_{\widetilde{Z}}\widetilde{X}-D_{\widetilde{X}}\widetilde{Z}=[\widetilde{Z},\widetilde{X}]$.
Set $\widetilde{Y}=\widetilde{X}-\widetilde{Z}$ in the above
equation, we obtain
\begin{equation}
2g_{\widetilde{X}}(\widetilde{Z},D_{\widetilde{X}}\widetilde{X})=2\widetilde{X}g_{\widetilde{X}}(\widetilde{Z},\widetilde{X})-\widetilde{Z}g_{\widetilde{X}}(\widetilde{X},\widetilde{X})-2g_{\widetilde{X}}([\widetilde{X},\widetilde{Z}],\widetilde{X}).
\end{equation}
Since $F$ is left-invariant, $dL_{x}$ is a linear isometry between
the spaces $T_{e}G=\frak{g}$ and $T_{x}G$, $\forall x\in G$.
Therefore for any left-invariant vector field $\widetilde{X},
\widetilde{Z}$ on $G$, we have
$$g_{\widetilde{X}}(\widetilde{Z},\widetilde{X})=g_{X}(Z,X)$$
i.e., the functions $g_{\widetilde{X}}(\widetilde{Z},\widetilde{X})$
, $g_{\widetilde{X}}(\widetilde{X},\widetilde{X})$ are constant.
Therefore from (4) the following is obtained
$$g_{\widetilde{X}}(\widetilde{Z},D_{\widetilde{X}}\widetilde{X})\mid_{e}=-g_{\widetilde{X}}([\widetilde{X},\widetilde{Z}],\widetilde{X})\mid_{e}=-g_{X}([X,Z],X).$$
Consequently the assertion of the theorem follows.$\Box$ \\\\
The following Proposition is well known for left-invariant
Riemannian metrics.
\begin{prop}
Let $G$ be a connected Lie group furnished with a left-invariant
Finsler metric $F$. Then the following are equivalent,
\begin{enumerate}
\item $F$ is right-invariant, hence bi-invariant.
\item $F$ is $Ad(G)-$invariant.
\item$g_{Y}([X,U],V)+g_{Y}(U,[X,V])+2C_{Y}([X,Y],U,V)=0$,\hskip.4cm$\forall \hskip.2cm Y\in \frak{g}-\{0\}, X, U, V
\in\frak{g}$, where $C_{y}$ is the Cartan
tensor of $F$ at $Y$.\\
If the Finsler structure $F$ is absolutely homogeneous, then one
also has.
\item The inversion map $g\longrightarrow g^{-1}$ is an
isometry
of $G$.
\end{enumerate}
\end{prop}
Proof: The equivalence of the first two assertion is routine, and we
omit the details. The equivalence between (1) and (3) is a result of
S. Deng and Z. Hou [5]. If $F$ is absolutely homogeneous, one can
check quite easily that (4) is equivalent to (1).$\Box$
\begin{cor}
If $G$ is a Lie group endowed with a bi-invariant Finsler metric,
then the geodesics through the identity of $G$ are exactly
one-parameter subgroups.
\end{cor}
Proof: Since $F$ is bi-invariant, we have
$$g_{Y}([X,U],V)+g_{Y}(U,[X,V])+2C_{Y}([X,Y],U,V)=0$$$\forall \hskip.2cm Y\in \frak{g}-\{0\}, X, U, V
\in\frak{g}$. It follows from the
homogeneity of $F$ that $C_{Y}(Y,V,W)=0$. So we have
$$g_{Y}([X,Y],Y)=0.$$ The result now follows from the
Theorem 3.2.$\Box$\\\\
A connected Finsler space $(M,F)$ is said to be symmetric [15] if to
each $p\in M$ there is associated an isometry
$s_{p}:M\longrightarrow M$ which is \begin{description}
\item[(i)] involutive ($s_{p}^{2}$ is the identity).
\item[(ii)]has $p$ as an isolated fixed point, that is, there
is a neighborhood $U$ of $p$ in which $p$ is the only fixed
point of $s_{p}$.
\end{description}
$s_{p}$ is called the symmetry of the point $p$.
\begin{thm}
Suppose $G$ is a Lie group with a bi-invariant absolutely
homogeneous Finsler metric, then $G$ is a symmetric Finsler space.
\end{thm}
Proof: Consider the smooth mapping $f:G\longrightarrow G$,
$f:x\longrightarrow x^{-1}$. Then $f_{\ast}:T_{e}G\longrightarrow
T_{e}G$ maps a vector $\xi \in T_{e}G$ to $-\xi$; in particular,
$df$ is an isometry of $\frak{g}=T_{e}G$. Clearly,
$f=R_{g^{-1}}fL_{g^{-1}}$. Therefore, $df_{g}:T_{g}G\longrightarrow
T_{g^{-1}}G$ is an isometry for any $g\in G$.\\ Let
$s_{g}(x)=gx^{-1}g,\hskip.3cm g,x \in G$. The mapping $s_{g}$ is an
isometry, because $s_{g}=R_{g}fR_{g^{-1}}$. Thus, $s_{g}$ is an
isometry of $G$, obviously fixing the point $g$. Furthermore,
$s_{g}^{2}(x)=g(gx^{-1}g)^{-1}g=x.$ To show that $s_{g}$ is the
symmetry used in the definition of a symmetric Finsler space, it
suffices to show that $(s_{g})_{\ast}\xi=-\xi$ whenever $\xi \in
T_{g}G$. \\ Let us start with the case $g=e$. Let
$\xi=\frac{d}{dt}\gamma (t)|_{t=0}\in T_{e}G$ where $\gamma (t)$ is
a one-parameter subgroup of $G$. Then $\gamma (t)^{-1}=\gamma (-t)$,
and $(s_{e})_{\ast}\xi =\frac{d}{dt}\mid_{t=0}\gamma (-t)=-\xi$.
Now, if $\xi$ is in $T_{e}G$ for an arbitrary $g\in G$, then
$ds_{g}=dR_{g}dfdR_{g^{-1}}$, so
$ds_{g}(\xi)=dR_{g}(df(dR_{g^{-1}}(\xi)))=dR_{g}(-dR_{g^{-1}}(\xi))=-\xi$.\\$\Box$
\
Let $M$ be a smooth n-dimensional manifold, a Randers metric on $M$
consists of a Riemannian metric
$\widetilde{a}=\widetilde{a}_{ij}dx^{i}\otimes dx^{j}$ on $M$ and a
1-form $b=b_{i}dx^{i}$, [2], [22]. Here $\widetilde{a}$ and $b$
define a function $F$ on $TM$ by
$$F(x,y)=\alpha(x,y)+\beta(x,y) \hskip2cm x\in M ,y\in T_{x}M$$
where $\alpha(x,y)=\sqrt{\widetilde{a}_{ij}y^{i}y^{j}}$ ,
$\beta(x,y)=b_{i}(x)y^{i}$. $F$ is Finsler structure if
$\|b\|=\sqrt{b_{i}b^{i}}<1$ where $b^{i}=\widetilde{a}^{ij}b_{j}$,
and $(\widetilde{a}^{ij})$ is the inverse of $(\widetilde{a}_{ij})$.
The Riemannian metric $\widetilde{a}=\widetilde{a}_{ij}dx^{i}\otimes
dx^{j}$ induces the musical bijections between 1-forms and vector
fields on $M$, namely $\flat :T_{x}M\longrightarrow T_{x}^{\ast}M$
given by $y\longrightarrow \widetilde{a}_{x}(y,\circ)$ and its
inverse $\sharp :T_{x}^{\ast}M\longrightarrow T_{x}M$. In the local
coordinates we have $$(y^{b})_{i}=\widetilde{a}_{ij}y^{j}\hskip1cm
y\in
T_{x}M$$$$(\theta^{\sharp})^{i}=\widetilde{a}^{ij}\theta_{j}\hskip1cm
\theta\in T_{x}^{\ast}M$$ Now the corresponding vector field to the
1-form $b$ will be denoted by $b^{\sharp}$, obviously we have
$\|b\|=\|b^{\sharp}\|$ and
$\beta(x,y)=(b^{\sharp})^{\flat}(y)=\widetilde{a}_{x}(b^{\sharp},y)$.
Thus a Randers metric $F$ with Riemannian metric
$\widetilde{a}=\widetilde{a}_{ij}dx^{i}\otimes dx^{j}$ and 1-form
$b$ can be showed by
$$F(x,y)=\sqrt{\widetilde{a}_{x}(y,y)}+\widetilde{a}_{x}(b^{\sharp},y)\hskip1cm x\in M ,y\in T_{x}M$$
where $\widetilde{a}_{x}(b^{\sharp},b^{\sharp})<1 \hskip1cm \forall
x\in M$.
\\\\ Let
$F(x,y)=\sqrt{\widetilde{a}_{x}(y,y)}+\widetilde{a}_{x}(X,y)$ be a
left invariant Randers metric. It is easy to check that the
underlying Riemannian metric $\widetilde{a}$ and the vector field
$X$ are also left invariant.
\begin{thm}
Let $G$ be a Lie group with a left-invariant Randers metric $F$
defined by the Riemannian metric
$\widetilde{a}=\widetilde{a}_{ij}dx^{i}\otimes dx^{j}$ and the
vector field $X$. Then the Randers metric $F$ is of Berwald type if
and only if $ad_{X}$ is skew-adjoint with respect to $\widetilde{a}$
and $\widetilde{a}(X,[\frak{g},\frak{g}])=0$.
\end{thm}
Proof: For all $Y,Z\in \frak{g}$,
\begin{equation}
2\widetilde{a}(Y,\nabla_{Z}X)=\widetilde{a}(Z,[Y,X])+\widetilde{a}(X,[Y,Z])-\widetilde{a}(Y,[X,Z]).
\end{equation}
where $\nabla$ is the Levi-Civita connection of
$(M,\widetilde{a})$.\\
If $ad_{X}$ is skew-adjoint, then first and last terms of
(5) sum to $0$. If additionally
$\widetilde{a}(X,[\frak{g},\frak{g}])=0$, then the middle term is
also $0$. So $\nabla_{Z}X=0$ for all $Z\in \frak{g}$, which means
that $X$ is parallel. By theorem 11.5.1. of [2] the Randers metric
is of Berwald type if
and only if $X$ is parallel with respect to $\widetilde{a}$.\\
Conversely, assume that the Randers metric is of Berwald type, so
the left side of (5) equals $0$ for all $Y,Z \in \frak{g}$. When
$Y=Z$, this yields $2\widetilde{a}(Y,[Y,X])=0$ for all $Y\in
\frak{g}$, which implies that $ad_{X}$ is skew-adjoint. This
property makes the first and third terms of (5) sum to zero, so
$\widetilde{a}(X,[Y,Z])=0$ for all $Y,Z \in \frak{g}$. In other
words
$\widetilde{a}(X,[\frak{g},\frak{g}])=0$.$\Box$\\\\
By a simple modification of the previous procedure, we can easily
obtain the following.
\begin{thm}
Let $(M=\frac{G}{H},F)$ be a homogeneous Randers space with $F$
defined by the Riemannian metric $\widetilde{a}$ and the vector
field $X$. Let $\frak{m}$ be the orthogonal complement of $\frak{h}$
in $\frak{g}$ with respect to the inner product induced on
$\frak{g}$ by $\widetilde{a}$. Then the Randers metric $F$ is of
Berwald type if and only if $\left(ad_{X}\right)_{\frak{m}}$ is
skew-adjoint and
$\widetilde{a}\left(X,[\frak{m},\frak{m}]_{\frak{m}}\right)=0$,
where $\left(ad_{X}\right)_{\frak{m}}$ denotes
$\left(ad_{X}\right)_{\frak{m}}:\frak{m}\longrightarrow \frak{m}$,
$\left(ad_{X}\right)_{\frak{m}}(y)=[X,y]_{\frak{m}}$.
\end{thm}
\begin{thm}
Let $G$ be a Lie group with a bi-invariant Randers metric $F$
defined by the Riemannian metric
$\widetilde{a}=\widetilde{a}_{ij}dx^{i}\otimes dx^{j}$ and the
vector field $X$. Then the Randers metric $F$ is of Berwald type.
\end{thm}
Proof: Let
$F(p,y)=\sqrt{\widetilde{a}_{p}(y,y)}+\widetilde{a}_{p}(X,y)$.\\ Now
for $s,t \in R$
\begin{eqnarray*}
F^{2}(y+su+tv) &=& \widetilde{a}(y+su+tv,y+su+tv)+\widetilde{a}^{2}(X,y+su+tv) \\
& & +2\sqrt{\widetilde{a}(y+su+tv,y+su+tv)} \widetilde{a}(X,y+su+tv)
\end{eqnarray*}
By definition
$$g_{y}(u,v)=\frac{1}{2}\frac{\partial^{2}}{\partial r \partial
s}F^{2}(y+ru+sv)\mid_{r=s=0}.$$ So by a direct computation we get
\begin{eqnarray*}
g_{y}(u,v) &=& \widetilde{a}(u,v)+
\widetilde{a}(X,u)\widetilde{a}(X,v)\\\\
&& +\frac{\widetilde{a}(u,v)\widetilde{a}(X,y)}{\sqrt{\widetilde{a}(y,y)}}-\frac{\widetilde{a}(v,y)\widetilde{a}(u,y)\widetilde{a}(X,y)}{\widetilde{a}(y,y)\sqrt{\widetilde{a}(y,y)}}
\\\\
&&+\frac{\widetilde{a}(X,v)\widetilde{a}(u,y)}{\sqrt{\widetilde{a}(y,y)}}+\frac{\widetilde{a}(X,u)\widetilde{a}(v,y)}{\sqrt{\widetilde{a}(y,y)}}.\\
\end{eqnarray*}
So for all $y,z \in \frak{g}$ we have
\begin{eqnarray*}
g_{y}(y,[y,z]) &=& \widetilde{a}(y,[y,z])+\widetilde{a}(X,y)\widetilde{a}(X,[y,z])
\\ \\
&&
+\frac{\widetilde{a}(y,[y,z])\widetilde{a}(X,y)}{\sqrt{\widetilde{a}(y,y)}}+\widetilde{a}(X,[y,z])\sqrt{\widetilde{a}(y,y)}\\\\
&=&\widetilde{a}(y,[y,z])\left(1+\frac{\widetilde{a}(X,y)}{\sqrt{\widetilde{a}(y,y)}}\right)\\\\
&&+\widetilde{a}(X,[y,z])\left(\widetilde{a}(X,y)+\sqrt{\widetilde{a}(y,y)}\right).
\end{eqnarray*}
So we have
\begin{equation}
g_{y}(y,[y,z]) = \widetilde{a}(y,[y,z])\left(\frac{F(y)}{\sqrt{\widetilde{a}(y,y)}}\right)+\widetilde{a}(X,[y,z])F(y)
\end{equation}
Since $\widetilde{a}$ is bi-invariant, $\widetilde{a}(y,[y,z])=0$
and $ad(x)$ is skew-adjoint for every $x\in \frak{g}$. Since $F$ is
bi-invariant, $g_{y}(y,[y,z])=0$. So From (6) we get
$\widetilde{a}(X,[y,z])=0$ for all $y,z \in \frak{g}$. Therefore, by
Theorem 3.6, we see that $(G,F)$ is of Berwald type.$\Box$
\begin{cor}
Let $G$ be a Lie group with a left-invariant Randers metric $F$
defined by the Riemannian metric
$\widetilde{a}=\widetilde{a}_{ij}dx^{i}\otimes dx^{j}$ and the
vector field $X$. If the Randers metric $F$ is of Berwald type then
$X$ is a geodesic vector.
\end{cor}
The following lemma can be found in [20, p.301].
\begin{lem}\emph{(Milnor)}
Let $G$ be a Lie group endowed with a left-invariant Riemannian
metric $\widetilde{a}$. If $x \in \frak{g}$ is
$\widetilde{a}-$orthogonal to the commutator ideal $[\frak{g},
\frak{g}]$, then $Ricci(x)\leq 0$, with equality if and only if
$ad_{x}$ is skew-adjoint with respect to $\widetilde{a}$.
\end{lem}
\begin{cor}
Let $G$ be a Lie group with a left-invariant Randers metric $F$
defined by the Riemannian metric
$\widetilde{a}=\widetilde{a}_{ij}dx_{i}\otimes dx_{j}$ and the
vector field $X$. If the Randers metric $F$ is of Berwald type then
the Ricci curvature of $\widetilde{a}$ in the direction
$u=\frac{X}{\sqrt{\widetilde{a}(X,X)}}$ is zero.
\end{cor}
Proof: The corollary is a direct consequence of Theorem 3.6 and
Lemma 3.10 .\\$\Box$
\section{Homogeneous geodesics and the critical points of the restricted Finsler function}
Let $G$ be a connected Lie group, $\frak{g}=T_{e}G$ its Lie algebra,
$Ad:G\times \frak{g} \longrightarrow \frak{g}$ the adjoint action,
$G(X)=\{Ad(g)X\mid g\in G\}\subset \frak{g}$ the orbit of an element
$X\in \frak{g}$ and $G_{X}<G$ the isometry subgroup at $X$. The set
$\frac{G}{G_{X}}$ of left-cosets of $G_{X}$ endowed with its
canonical smooth manifold structure admits the canonical left-action
$$\Lambda:G\times \frac{G}{G_{X}}\longrightarrow \frac{G}{G_{X}}
\hskip1cm (g,aG_{X})\longrightarrow gaG_{X},$$ which is also smooth.
Moreover, a smooth bijection $\rho:\frac{G}{G_{X}}\longrightarrow
G(X)$ is defined by $\rho(aG_{X})=Ad(a)X$ which thus yields an
injective immersion into $\frak{g}$ which is equivariant
with respect to the actions $\Lambda$ and $Ad$.\\
Now consider a Minkowski norm $\widetilde{F}:\frak{g}\longrightarrow
R$, then $F$ defines a left-invariant Finsler metric on $G$ by
$$F(x,U)=\widetilde{F}(dL_{x^{-1}}U), \hskip.5cm U\in T_{x}G,$$ where $L_{x}:G\longrightarrow
G$ is the left translation by $x\in G$. Let
$Q(Z)=\widetilde{F}^{2}(Z)$, \hskip.2cm $Z\in \frak{g}$. Using the
formula $\widetilde{F}(Z)=\sqrt{g_{Z}(Z,Z)}$, we have $Q(Z)=g_{Z}(Z,Z)$.\\
The smooth function $q=Q\circ\rho:\frac{G}{G_{X}}\longrightarrow R$
will be called the restricted Minkowski norm on $\frac{G}{G_{X}}$.\\
In the following, we give an extension of results of [25] to
left-invariant Finsler metrics. We use some ideas from [25], [26] in
our proofs.
\begin{thm}
Let $G$ be a connected Lie group and $\widetilde{F}$ a Minkowski
norm on its Lie algebra $\frak{g}$. For $X\in \frak{g}-\{0\}$ let
$U\in \frak{g}$ be such that $X\in G(U)$ for the corresponding
adjoint orbit and let $gG_{U}\in \frac{G}{G_{U}}$ be the unique
coset with $\rho(gG_{U})=X$. Then $X$ is a geodesic vector if and
only if $gG_{U}$ is a critical point of $q=Q\circ \rho$ the
restricted Minkowski norm on $\frac{G}{G_{U}}$.
\end{thm}
Proof: The coset $gG_{U}$ is a critical point of $q$ if and only if
$vq=0$ for $v\in T_{gG_{U}}(\frac{G}{G_{U}})$. But as
$\frac{G}{G_{U}}$ is homogeneous, for each $v$ there is a $Z\in
\frak{g}$ such that $v=\widetilde{Z}(gG_{U})$ where
$\widetilde{Z}:\frac{G}{G_{U}}\longrightarrow T(\frac{G}{G_{U}})$ is
the infinitesimal generator of the action $\Lambda$ corresponding to
$Z$. Consider also the infinitesimal generator $\widehat{Z}:
\frak{g}\longrightarrow T\frak{g}$ of the adjoint action
corresponding to $Z$. Since the injective immersion $\rho$ is
equivariant with respect to the action $\Lambda$ and $Ad$ the
following holds: $\widehat{Z}\circ\rho=T\rho\circ\widetilde{Z}$. But
then the following is valid:
\begin{eqnarray*}
v(q) &=& \widetilde{Z}(q)\mid_{gG_{U}}=\widetilde{Z}(Q\circ p)\mid_{gG_{U}}
\\\\
&=& \left(T\rho\widetilde{Z}\right)\mid_{gG_{U}}Q=(\widehat{Z}\circ\rho)\mid_{gG_{U}}Q \\
&=& \left(\frac{d}{dt}\mid_{t=0}(Ad(exptZ)X)\right)Q \\
&=& \frac{d}{dt}\mid_{t=0}Q(Ad(exptZ)X) \\
&=&
\frac{d}{dt}\mid_{t=0}g_{_{Ad(exptZ)X}}(Ad(exptZ)X,Ad(exptZ)X)\\\\
&=& g_{X}([Z,X],X)+g_{X}(X,[Z,X])+2C_{X}([Z,X],X,X) \\\\
&=& 2g_{X}([Z,X],X),
\end{eqnarray*}
where $C_{X}$ is the Cartan tensor of $F$ at $X$. It follows from
the homogeneity of $F$ that $C_{X}([Z,X],X,X)=0$. Since the map
$\alpha:\frak{g}\longrightarrow T_{gG_{U}}(\frac{G}{G_{U}})$,
$Z\longrightarrow \widetilde{Z}(gG_{U})$ is an epimorphism, the
assertion of the theorem follows.$\Box$
\begin{cor}
Let $G$ be a compact connected semi-simple Lie group and
$\widetilde{F}$ a Minkowski norm on its Lie algebra $\frak{g}$. Then
each orbit of the adjoint action $Ad:G\times \frak{g}\longrightarrow
\frak{g}$ contains at least two geodesic vectors.
\end{cor}
Proof: Consider an orbit $G(X)$ of the adjoint action, the
corresponding coset manifold $\frac{G}{G_{X}}$ and the injective
immersion $\rho:\frac{G}{G_{X}}\longrightarrow \frak{g}$. Since $G$
is compact and semi-simple then the manifold $\frac{G}{G_{X}}$
becomes compact, and the restricted Minkowski norm
$q=Q\circ\rho:\frac{G}{G_{X}}\longrightarrow R$ has at least two
critical points.$\Box$ \\\\
The following corollary is a consequence of the preceding corollary.
Two geodesics are considered different if their images are
different.
\begin{cor}
Let $G$ be compact connected semi-simple Lie group of
\hskip.1cm$rank\geq2$ and $\widetilde{F}$ a Minkowski norm on its
Lie algebra. Then the left-invariant Finsler metric $F$ induced by
$\widetilde{F}$ on $G$ has infinitely many homogeneous geodesic
issuing from the identity element.
\end{cor}
Proof: The proof is similar to the Riemannian case, so we omit it
[25].$\Box$
\\\\
\section{Some examples}
\begin{ex}
\end{ex}
Let $G$ be a three-dimensional connected Lie group endowed with a
left-invariant Riemannian metric $\widetilde{a}$.
\begin{enumerate}
\item Let $G$ be an unimodular Lie group. According to a result
due to J. Milnor (see[20,Theorem 4.3, p.305],[17]) there exist an
orthonormal basis $\{e_{1}, e_{2}, e_{3}\}$ of the Lie algebra
$\frak{g}$ such that $$[e_{1}, e_{2}]=\lambda_{3}e_{3}, \hskip.5cm [e_{2}, e_{3}]=\lambda_{1}e_{1},\hskip.5cm[e_{3}, e_{1}]=\lambda_{2}e_{2}.$$
Let $F$ be a left invariant Randers metric on $G$ defined by the
Riemannian metric $\widetilde{a}$ and the vector field $X=\epsilon
e_{1}$, $0 <\epsilon<1$ i.e.
$$F(p,y)=\sqrt{\widetilde{a}_{p}(y,y)}+\widetilde{a}_{p}(X,y).$$
We note, by using Theorem 3.6, that $(G,F)$ is not of the
Berwald type. We want to describe all geodesic vectors of
$(G,F)$.\\ For $s,t \in R$ \begin{eqnarray*}
F^{2}(y+su+tv) &=& \widetilde{a}(y+su+tv,y+su+tv)+\widetilde{a}^{2}(X,y+su+tv) \\
& & +2\sqrt{\widetilde{a}(y+su+tv,y+su+tv)} \widetilde{a}(X,y+su+tv)
\end{eqnarray*}
By definition
$$g_{y}(u,v)=\frac{1}{2}\frac{\partial^{2}}{\partial r \partial
s}F^{2}(y+ru+sv)\mid_{r=s=0}.$$ So by a direct computation we get
\begin{eqnarray*}
g_{y}(u,v) &=& \widetilde{a}(u,v)+
\widetilde{a}(X,u)\widetilde{a}(X,v)\\\\
&& +\frac{\widetilde{a}(u,v)\widetilde{a}(X,y)}{\sqrt{\widetilde{a}(y,y)}}-\frac{\widetilde{a}(v,y)\widetilde{a}(u,y)\widetilde{a}(X,y)}{\widetilde{a}(y,y)\sqrt{\widetilde{a}(y,y)}} \mbox{\qquad
}\\\\
&&+\frac{\widetilde{a}(X,v)\widetilde{a}(u,y)}{\sqrt{\widetilde{a}(y,y)}}+\frac{\widetilde{a}(X,u)\widetilde{a}(v,y)}{\sqrt{\widetilde{a}(y,y)}}.\\
\end{eqnarray*}
So for all $z\in \frak{g}$ we have
\begin{equation}
g_{y}(y,[y,z])=
\widetilde{a}\left(X+\frac{y}{\sqrt{\widetilde{a}(y,y)}} \hskip.1cm,
[y,z]\right)F(y)
\end{equation}
Using Theorem 3.2 and (7) we can check easily that $e_{1}$ is a
geodesic vector.\\ By using Theorem 3.2 and (7) a vector
$y=y_{1}e_{1}+y_{2}e_{2}+y_{3}e_{3}$ of $\frak{g}$ is a geodesic
vector if and only if $$\widetilde{a}\left(\epsilon
e_{1}+\frac{y_{1}e_{1}+y_{2}e_{2}+y_{3}e_{3}}{\sqrt{y_{1}^{2}+y_{2}^{2}+y_{3}^{2}}}
\hskip.1cm, [y_{1}e_{1}+y_{2}e_{2}+y_{3}e_{3} , e_{j}]\right)=0$$
for each $j=1, 2, 3.$\\ So we get :
$$(\lambda_{2}-\lambda_{3})y_{2}y_{3}=0,$$
$$-\epsilon y_{3}\lambda_{1}-\frac{1}{\sqrt{y_{1}^{2}+y_{2}^{2}+y_{3}^{2}}}y_{1}y_{3}\lambda_{1}+\frac{1}{\sqrt{y_{1}^{2}+y_{2}^{2}+y_{3}^{2}}}y_{1}y_{3}\lambda_{3}=0,$$
$$-\epsilon y_{2}\lambda_{1}+\frac{1}{\sqrt{y_{1}^{2}+y_{2}^{2}+y_{3}^{2}}}y_{1}y_{2}\lambda_{1}-\frac{1}{\sqrt{y_{1}^{2}+y_{2}^{2}+y_{3}^{2}}}y_{1}y_{2}\lambda_{2}=0.$$
As a special case, if $\lambda_{1}=\lambda_{2}=\lambda_{3}\neq 0$ we
conclude that all geodesic vectors $y$ are those from the set
$Span\{e_{1}\}$. Consequently, there is only one homogeneous
geodesic.
\item Let $G$ be a non-unimodular Lie group. According to a
result due to J. Milnor (see[20, Lemma 4.10, p.309],[17]) there
exists an orthogonal basis $\{e_{1}, e_{2}, e_{3}\}$ of the Lie
algebra $\frak{g}$ such that
$$[e_{1},e_{2}]=\alpha e_{2}+\beta e_{3},\hskip.5cm[e_{2},e_{3}]=0,\hskip.5cm [e_{1},e_{3}]=\gamma e_{2}+\delta e_{3}, $$
where $\alpha , \beta, \gamma, \delta $ are real numbers such
that the matrix
$$
\left(
\begin{array}{cc}
\alpha & \beta \\
\gamma & \delta \\
\end{array}
\right)
$$
has trace $\alpha+\delta =2$ and $\alpha \gamma + \beta\delta =0.$
Let $F$ be a left invariant Randers metric on $G$ defined by the
Riemannian metric $\widetilde{a}$ and the vector field $X=\epsilon
e_{1}, \hskip.2cm 0<\epsilon <1.$\\
By using Theorem 3.2 and (7), a vector
$y=y_{1}e_{1}+y_{2}e_{2}+y_{3}e_{3}$ of $\frak{g}$ is a geodesic
vector if and only if $$\widetilde{a}\left(\epsilon
e_{1}+\frac{y_{1}e_{1}+y_{2}e_{2}+y_{3}e_{3}}{\sqrt{y_{1}^{2}+y_{2}^{2}+y_{3}^{2}}}
\hskip.1cm, [y_{1}e_{1}+y_{2}e_{2}+y_{3}e_{3} , e_{j}]\right)=0$$
for each $j=1, 2, 3.$\\
This condition leads to the system of equations\\\\
$$y_{2}(-\alpha y_{2}-\gamma y_{3})+y_{3}(-y_{2}\beta -\delta y_{3})=0,$$
$$y_{1}y_{2}\alpha+y_{1}y_{3}\beta=0,$$
$$y_{1}y_{2}\gamma+y_{1}y_{3}\delta=0.$$
Putting $\alpha=2, \delta=0, \gamma=0$ the above equations take the
form $$2y_{2}\left(y_{2}+\frac{\beta}{2}y_{3}\right)=0,$$
$$2y_{1}\left(y_{2}+\frac{\beta}{2}y_{3}\right)=0.$$
So a vector $y$ of $\frak{g}$ is a geodesic vector if and only if :
\begin{description}
\item[-] $y\in Span(e_{1}, e_{3})$ for $\beta=0.$
\item[-] $y\in Span(e_{1})\bigcup Span(e_{3})\bigcup
Span(\frac{\beta}{2}e_{2}-e_{3})$ for $\beta \neq 0$
\end{description}
\end{enumerate} | 0.044522 |
When I first came to SAL groups, I had been attending a different recovery group for a year. While I had never heard the term of surrender before, it was skittered throughout many of the comments of SAL women each week in their shares. Embarrassed that I was unfamiliar with something that seemed so fundamental, I kept trying to figure out what they were talking about.
As I became brave enough to ask directly for other group members to explain surrender to me, and I began to work with my own sponsor, I quickly began to realize that surrender is something that you only begin to understand when you begin to practice it. Here’s a quick run-down of what surrender has come to mean to me:
What Surrender Is
“Surrender is the simple but profound wisdom of yielding to rather than opposing the flow of life…it is to relinquish inner resistance to what is.” -Eckhart Tolle, The Power of Now, p206
In Rhyll Croshaw’s book, What Can I Do About Me? she explains that the first 3 steps are a daily process of learning to surrender her own will by believing in God’s great power. In SAL, we often talk about the “surrender process” which gives us tangible, concrete actions to take to help us let go of our inner resistance.
On my knees-we give our fear, resentment, pride, control…whatever…to God through prayer.
On the phone-we reach out and call our sponsor or another group member who understands the Steps and the surrender process, and speak our surrender out loud to her.
In the box-we write our surrender on a scrap of paper and put it in a “Surrender Box”, “God Jar”, or even in the garbage can.
Surrender is the process by which I let go of my will and seek to align myself with God’s will. It is most effective when done in the moment I become aware of my lack of serenity…not at the end of a resentful day or week.
Surrender requires the awareness and humility to be open to the idea that the way I am currently approaching the situation is not working. I must be willing to let go and try things His way.
When I truly surrender, I can feel a tangible lifting in my heart and my body: a weight off my shoulders. I am acknowledging that I am no longer in charge of something or someone. God is. I feel light. I feel free. As one SAL woman put it, “I don’t worry about it anymore.”
I may need to surrender over and over again if I keep finding myself returning mentally or physically to a troubling situation. Surrender is a practice I can use every single time I lose my serenity. It is perhaps the most important tool I have to keep God at my center.
I have learned that His will for me may change moment to moment in different situations, so it is less effective to white-knuckle my way through a tension-filled day, only to follow it up with a big surrender. Neither can I expect some big blanket list of boundaries to always produce a desired outcome. Rather, if I am willing to take one step at a time, practice awareness and accountability for my own negative emotions, and surrender in the moment, He will reveal His will for me one step at a time. I will experience greater serenity in my relationships and more stillness in my soul.
What Surrender is Not
Surrender is NOT a license to vent and complain. Sometimes I have caught myself saying, “I just need to surrender that…” and then I go on and on complaining about how terrible someone is.
This is not a surrender. This is me using the term “surrender” to dump all my criticisms of someone. Surrender always comes back to letting go of MY underlying fear or coping strategy. The focus is always on me and what I need to let go of.
For example, if someone else is highly critical and nothing I do is ever good enough for them, it has been much more effective for me to focus on surrendering MY fear that I am not enough, than to surrender that THEY are critical and difficult. I can surrender MY need to be validated by the approval of others, and feel peace no matter what the other person decides to do. With surrender, my serenity is no longer a slave to someone else’s behavior.
As I practice this, it also becomes clear when a boundary is needed. I can apply the serenity prayer to any situation: what do I have the power to change? This is where I need to create or hold a boundary.
What do I not have the power to change? This is what I need to accept through surrender.
This type of surrender always brings the promised peace of the program. The license-to-criticize kind just feeds my resentment.
I have been practicing surrender as an integral part of my daily life for the past 2 years and I am still figuring it out. Recovery is a process and not a destination. But I can say, that when I get it right and I truly surrender in the moment, that is where the peace is. That is where the miraculous process starts, and that is the key that unlocks the indescribable magic of Steps 6 & 7, and He comes in and takes away the junk that is keeping me from connecting with Him and others. It Works When I Work It.
We would all love to learn from your insights and experiences with Surrender.
Looking forward to reading your comments this week!
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16 thoughts on “What is Surrender?”
Surrender is amazingly beautiful!! It totally feels like magic! I for ever be greatful for learning how to access Christ power and grace; I am in aw everytime I surrender I find peace, forgiveness, LOVE, mercy and reasutance!
Eva–you have a gift of being able to get right to the heart of recovery. It is the gift of grace, peace, forgiveness, LOVE, and mercy that comes when we surrender. Thank you for sharing your strength, hope, and experience!
I love the quote by Tolle -seriously so good. And I really can relate to what you added at the end… surrender is such a learning process.
This morning, I surrendered my will about several situations, and it was very freeing. The rest of my day went better for sure. Surrender is such a vital truth -I need it badly, daily. Sometimes hourly.
Yes–I love that word “freeing.” Surrender frees me from the weight of negative emotions, and the false belief that I am supposed to know how or be able to fix things. Surrender is freedom!…and it is learning to depend upon the One who can fix everything.
I LOVE the surrender process! Thank you so much for sharing your experience and knowledge about it. I too am so grateful for this amazing tool in my life. I have to be honest for myself it can be really hard to humble myself and choose to surrender. It takes a great amount of courage and humility to surrender. One of the reason I love being a Sponsor is seeing so many women who are humble and courageous as they work thier 1, 2 and 3 to surrender and be open to knowing Gods will for them. These women are examples to me everyday. I love the feeling of clarity and peace I feel as I surrender to God. What a beautiful tool the 12 step program has given me.
Thanks Tosha. I love how you point out that it takes courage and humility to surrender. I am also so grateful for the constant strength I find in sponsees and other group members as I attend my meetings. Seeing surrender work in the lives of others definitely inspires me to surrender more often! | 0.920729 |
Child of Light - The World of Lemuria Trailer
Featuring a plethora of locations players will visit during their time playing, The World of Lumeria showcases some beautiful graphics.
From caves to forests and several other locations all visited by the floating Aurora. These are no doubt just a showcase and there will be others in the full game, as Aurora quests to return Light to Lumeria.
Child of Light comes out 30th April for PS3, PS4, Xbox 360, Xbox One, Wii U and PC. | 0.005608 |