\chapter{Characters} Characters are basically the best thing ever. To every representation $V$ of $A$ we will attach a so-called character $\chi_V : A \to k$. It will turn out that the characters of irreps of $V$ will determine the representation $V$ completely. Thus an irrep is just specified by a set of $\dim A$ numbers. \section{Definitions} \begin{definition} Let $V = (V, \rho)$ be a finite-dimensional representation of $A$. The \vocab{character} $\chi_V : A \to k$ attached to $A$ is defined by $\chi_V = \Tr \circ \rho$, i.e.\ \[ \chi_V(a) \defeq \Tr\left( \rho(a) : V \to V \right). \] \end{definition} Since $\Tr$ and $\rho$ are additive, this is a $k$-linear map (but it is not multiplicative). Note also that $\chi_{V \oplus W} = \chi_V + \chi_W$ for any representations $V$ and $W$. We are especially interested in the case $A = k[G]$, of course. As usual, we just have to specify $\chi_V(g)$ for each $g \in S_3$ to get the whole map $k[G] \to k$. Thus we often think of $\chi_V$ as a function $G \to k$, called a character of the group $G$. Here is the case $G = S_3$: \begin{example} [Character table of $S_3$] Let's consider the three irreps of $G = S_3$ from before. For $\CC_{\text{triv}}$ all traces are $1$; for $\CC_{\text{sign}}$ the traces are $\pm 1$ depending on sign (obviously, for one-dimensional maps $k \to k$ the trace ``is'' just the map itself). For $\refl_0$ we take a basis $(1,0,-1)$ and $(0,1,-1)$, say, and compute the traces directly in this basis. \[ \begin{array}{|r|rrrrrr|} \hline \chi_V(g) & \id & (1\;2) & (2\;3) & (3\;1) & (1\;2\;3) & (3\;2\;1) \\ \hline \Ctriv & 1 & 1 & 1 & 1 & 1 & 1 \\ \CC_{\mathrm{sign}} & 1 & -1 & -1 & -1 & 1 & 1 \\ \refl_0 & 2 & 0 & 0 & 0 & -1 & -1 \\ \hline \end{array} \] \end{example} The above table is called the \vocab{character table} of the group $G$. The table above has certain mysterious properties, which we will prove as the chapter progresses. \begin{enumerate}[(I)] \ii The value of $\chi_V(g)$ only depends on the conjugacy class of $g$. \ii The number of rows equals the number of conjugacy classes. \ii The sum of the squares of any row is $6$ again! \ii The ``dot product'' of any two rows is zero. \end{enumerate} \begin{abuse} The name ``character'' for $\chi_V : G \to k$ is a bit of a misnomer. This $\chi_V$ is not multiplicative in any way, as the above example shows: one can almost think of it as an element of $k^{\oplus |G|}$. \end{abuse} \begin{ques} Show that $\chi_V(1_A) = \dim V$, so one can read the dimensions of the representations from the leftmost column of a character table. \end{ques} \section{The dual space modulo the commutator} For any algebra, we first observe that since $\Tr(TS) = \Tr(ST)$, we have for any $V$ that \[ \chi_V(ab) = \chi_V(ba). \] This explains observation (I) from earlier: \begin{ques} Deduce that if $g$ and $h$ are in the same conjugacy class of a group $G$, and $V$ is a representation of $\CC[G]$, then $\chi(g) = \chi(h)$. \end{ques} Now, given our algebra $A$ we define the \vocab{commutator} $[A,A]$ to be the $k$-vector subspace spanned by $xy-yx$ for $x,y \in A$. Thus $[A,A]$ is contained in the kernel of each $\chi_V$. \begin{definition} The space $A\ab \coloneqq A / [A,A]$ is called the \vocab{abelianization} of $A$. Each character of $A$ can be viewed as a map $A\ab \to k$, i.e.\ an element of $(A\ab)^\vee$. \end{definition} \begin{example} [Examples of abelianizations] \listhack \begin{enumerate}[(a)] \ii If $A$ is commutative, then $[A,A] = \{0\}$ and $A\ab = A$. \ii If $A = \Mat_k(d)$, then $[A,A]$ consists exactly of the $d \times d$ matrices of trace zero. (Proof: harmless exercise.) Consequently, $A\ab$ is one-dimensional. \ii Suppose $A = k[G]$. Then in $A\ab$, we identify $gh$ and $hg$ for each $g,h \in G$; equivalently $ghg\inv = h$. So in other words, $A\ab$ is isomorphic to the space of $k$-linear combinations of the \emph{conjugacy classes} of $G$. \end{enumerate} \end{example} \begin{theorem} [Character of representations of algebras] Let $A$ be an algebra over an algebraically closed field. Then \begin{enumerate}[(a)] \ii Characters of pairwise non-isomorphic irreps are linearly independent in $(A\ab)^\vee$. \ii If $A$ is finite-dimensional and semisimple, then the characters attached to irreps form a basis of $(A\ab)^\vee$. \end{enumerate} In particular, in (b) the number of irreps of $A$ equals $\dim A\ab$. \end{theorem} \begin{proof} Part (a) is more or less obvious by the density theorem: suppose there is a linear dependence, so that for every $a$ we have \[ c_1 \chi_{V_1}(a) + c_2 \chi_{V_2}(a) + \dots + c_r \chi_{V_r} (a) = 0\] for some integer $r$. \begin{ques} Deduce that $c_1 = \dots = c_r = 0$ from the density theorem. \end{ques} For part (b), assume there are $r$ irreps. We may assume that \[ A = \bigoplus_{i=1}^r \Mat(V_i) \] where $V_1$, \dots, $V_r$ are the irreps of $A$. Since we have already showed the characters are linearly independent we need only show that $\dim ( A / [A,A] ) = r$, which follows from the observation earlier that each $\Mat(V_i)$ has a one-dimensional abelianization. \end{proof} Since $G$ has $\dim \CC[G]\ab$ conjugacy classes, this completes the proof of (II). \section{Orthogonality of characters} Now we specialize to the case of finite groups $G$, represented over $\CC$. \begin{definition} Let $\Classes(G)$ denote the set conjugacy classes of $G$. \end{definition} If $G$ has $r$ conjugacy classes, then it has $r$ irreps. Each (finite-dimensional) representation $V$, irreducible or not, gives a character $\chi_V$. \begin{abuse} From now on, we will often regard $\chi_V$ as a function $G \to \CC$ or as a function $\Classes(G) \to \CC$. So for example, we will write both $\chi_V(g)$ (for $g \in G$) and $\chi_V(C)$ (for a conjugacy class $C$); the latter just means $\chi_V(g_C)$ for any representative $g_C \in C$. \end{abuse} \begin{definition} Let $\FunCl(G)$ denote the set of functions $\Classes(G) \to \CC$ viewed as a vector space over $\CC$. We endow it with the inner form \[ \left< f_1, f_2 \right> = \frac{1}{|G|} \sum_{g \in G} f_1(g) \ol{f_2(g)}. \] \end{definition} This is the same ``dot product'' that we mentioned at the beginning, when we looked at the character table of $S_3$. We now aim to prove the following orthogonality theorem, which will imply (III) and (IV) from earlier. \begin{theorem}[Orthogonality] For any finite-dimensional complex representations $V$ and $W$ of $G$ we have \[ \left< \chi_V, \chi_W \right> = \dim \Homrep(W, V). \] In particular, if $V$ and $W$ are irreps then \[ \left< \chi_V, \chi_W \right> = \begin{cases} 1 & V \cong W \\ 0 & \text{otherwise}. \end{cases} \] \end{theorem} \begin{corollary}[Irreps give an orthonormal basis] The characters associated to irreps form an \emph{orthonormal} basis of $\FunCl(G)$. \end{corollary} In order to prove this theorem, we have to define the dual representation and the tensor representation, which give a natural way to deal with the quantity $\chi_V(g)\ol{\chi_W(g)}$. \begin{definition} Let $V = (V, \rho)$ be a representation of $G$. The \vocab{dual representation} $V^\vee$ is the representation on $V^\vee$ with the action of $G$ given as follows: for each $\xi \in V^\vee$, the action of $g$ gives a $g \cdot \xi \in V^\vee$ specified by \[ v \xmapsto{g \cdot \xi} \xi\left( \rho(g\inv)(v) \right). \] \end{definition} \begin{definition} Let $V = (V, \rho_V)$ and $W = (W, \rho_W)$ be \emph{group} representations of $G$. The \vocab{tensor product} of $V$ and $W$ is the group representation on $V \otimes W$ with the action of $G$ given on pure tensors by \[ g \cdot (v \otimes w) = (\rho_V(g)(v)) \otimes (\rho_W(g)(w)) \] which extends linearly to define the action of $G$ on all of $V \otimes W$. \end{definition} \begin{remark} Warning: the definition for tensors does \emph{not} extend to algebras. We might hope that $a \cdot (v \otimes w) = (a \cdot v) \otimes (a \cdot w)$ would work, but this is not even linear in $a \in A$ (what happens if we take $a=2$, for example?). \end{remark} \begin{theorem} [Character traces] If $V$ and $W$ are finite-dimensional representations of $G$, then for any $g \in G$. \begin{enumerate}[(a)] \ii $\chi_{V \oplus W}(g) = \chi_V(g) + \chi_W(g)$. \ii $\chi_{V \otimes W}(g) = \chi_V(g) \cdot \chi_W(g)$. \ii $\chi_{V^\vee}(g) = \ol{\chi_V(g)}$. \end{enumerate} \end{theorem} \begin{proof} Parts (a) and (b) follow from the identities $\Tr(S \oplus T) = \Tr(S) + \Tr(T)$ and $\Tr(S \otimes T) = \Tr(S) \Tr(T)$. However, part (c) is trickier. As $(\rho(g))^{|G|} = \rho(g^{|G|}) = \rho(1_G) = \id_V$ by Lagrange's theorem, we can diagonalize $\rho(g)$, say with eigenvalues $\lambda_1$, \dots, $\lambda_n$ which are $|G|$th roots of unity, corresponding to eigenvectors $e_1$, \dots, $e_n$. Then we see that in the basis $e_1^\vee$, \dots, $e_n^\vee$, the action of $g$ on $V^\vee$ has eigenvalues $\lambda_1\inv$, $\lambda_2\inv$, \dots, $\lambda_n\inv$. So \[ \chi_V(g) = \sum_{i=1}^n \lambda_i \quad\text{and}\quad \chi_{V^\vee}(g) = \sum_{i=1}^n \lambda_i\inv = \sum_{i=1}^n \ol{\lambda_i} \] where the last step follows from the identity $|z|=1 \iff z\inv = \ol z$. \end{proof} \begin{remark} [Warning] The identities (b) and (c) do not extend linearly to $\CC[G]$, i.e.\ it is not true for example that $\chi_V(a) = \ol{\chi_V(a)}$ if we think of $\chi_V$ as a map $\CC[G] \to \CC$. \end{remark} \begin{proof} [Proof of orthogonality relation] The key point is that we can now reduce the sums of products to just a single character by \[ \chi_V(g) \ol{\chi_W(g)} = \chi_{V \otimes W^\vee} (g). \] So we can rewrite the sum in question as just \[ \left< \chi_V, \chi_W \right> = \frac{1}{|G|} \sum_{g \in G} \chi_{V \otimes W^\vee} (g) = \chi_{V \otimes W^\vee} \left( \frac{1}{|G|} \sum_{g \in G} g \right). \] Let $P : V \otimes W^\vee \to V \otimes W^\vee$ be the action of $\frac{1}{|G|} \sum_{g \in G} g$, so we wish to find $\Tr P$. \begin{exercise} Show that $P$ is idempotent. (Compute $P \circ P$ directly.) \end{exercise} Hence $V \otimes W^\vee = \ker P \oplus \img P$ (by \Cref{prob:idempotent}) and $\img P$ is the subspace of elements which are fixed under $G$. From this we deduce that \[ \Tr P = \dim \img P = \dim \left\{ x \in V \otimes W^\vee \mid g \cdot x = x \; \forall g \in G \right\}. \] Now, consider the natural isomorphism $V \otimes W^\vee \to \Hom(W, V)$. \begin{exercise} Let $g \in G$. Show that under this isomorphism, $T \in \Hom(W, V)$ satisfies $g \cdot T = T$ if and only if $T(g \cdot w) = g \cdot T(w)$ for each $w \in W$. (This is just unwinding three or four definitions.) \end{exercise} Consequently, $\chi_{V \otimes W^\vee}(P) = \Tr P = \dim \Homrep(W,V)$ as desired. \end{proof} The orthogonality relation gives us a fast and mechanical way to check whether a finite-dimensional representation $V$ is irreducible. Namely, compute the traces $\chi_V(g)$ for each $g \in G$, and then check whether $\left< \chi_V, \chi_V \right> = 1$. So, for example, we could have seen the three representations of $S_3$ that we found were irreps directly from the character table. Thus, we can now efficiently verify any time we have a complete set of irreps. \section{Examples of character tables} \begin{example} [Dihedral group on $10$ elements] Let $D_{10} = \left< r,s \mid r^5 = s^2 = 1, rs = sr\inv \right>$. Let $\omega = \exp(\frac{2\pi i}{5})$. We write four representations of $D_{10}$: \begin{itemize} \ii $\Ctriv$, all elements of $D_{10}$ act as the identity. \ii $\Csign$, $r$ acts as the identity while $s$ acts by negation. \ii $V_1$, which is two-dimensional and given by $r \mapsto \begin{bmatrix} \omega & 0 \\ 0 & \omega^4 \end{bmatrix}$ and $s \mapsto \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. \ii $V_2$, which is two-dimensional and given by $r \mapsto \begin{bmatrix} \omega^2 & 0 \\ 0 & \omega^3 \end{bmatrix}$ and $s \mapsto \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. \end{itemize} We claim that these four representations are irreducible and pairwise non-isomorphic. We do so by writing the character table: \[ \begin{array}{|c|rccr|} \hline D_{10} & 1 & r, r^4 & r^2, r^3 & sr^k \\ \hline \Ctriv & 1 & 1 & 1 & 1 \\ \Csign & 1 & 1 & 1 & -1 \\ V_1 & 2 & \omega+\omega^4 & \omega^2+\omega^3 & 0 \\ V_2 & 2 & \omega^2+\omega^3 & \omega+\omega^4 & 0 \\ \hline \end{array} \] Then a direct computation shows the orthogonality relations, hence we indeed have an orthonormal basis. For example, $\left< \Ctriv, \Csign \right> = 1 + 2 \cdot 1 + 2 \cdot 1 + 5 \cdot (-1) = 0$. \end{example} \begin{example} [Character table of $S_4$] We now have enough machinery to to compute the character table of $S_4$, which has five conjugacy classes (corresponding to cycle types $\id$, $2$, $3$, $4$ and $2+2$). First of all, we note that it has two one-dimensional representations, $\Ctriv$ and $\Csign$, and these are the only ones (because there are only two homomorphisms $S_4 \to \CC^\times$). So thus far we have the table \[ \begin{array}{|c|rrrrr|} \hline S_4 & 1 & (\bullet\;\bullet) & (\bullet\;\bullet\;\bullet) & (\bullet\;\bullet\;\bullet\;\bullet) & (\bullet\;\bullet)(\bullet\;\bullet) \\ \hline \Ctriv & 1 & 1 & 1 & 1 & 1 \\ \Csign & 1 & -1 & 1 & -1 & 1 \\ \vdots & \multicolumn{5}{|c|}{\vdots} \end{array} \] Note the columns represent $1+6+8+6+3=24$ elements. Now, the remaining three representations have dimensions $d_1$, $d_2$, $d_3$ with \[ d_1^2 + d_2^2 + d_3^2 = 4! - 2 = 22 \] which has only $(d_1, d_2, d_3) = (2,3,3)$ and permutations. Now, we can take the $\refl_0$ representation \[ \left\{ (w,x,y,z) \mid w+x+y+z=0 \right\} \] with basis $(1,0,0,-1)$, $(0,1,0,-1)$ and $(0,0,1,-1)$. This can be geometrically checked to be irreducible, but we can also do this numerically by computing the character directly (this is tedious): it comes out to have $3$, $1$, $0$, $-1$, $-1$ which indeed gives norm \[ \left< \chi_{\refl_0}, \chi_{\refl_0} \right> = \frac{1}{4!} \left( \underbrace{3^2}_{\id} + \underbrace{6\cdot(1)^2}_{(\bullet\;\bullet)} + \underbrace{8\cdot(0)^2}_{(\bullet\;\bullet\;\bullet)} + \underbrace{6\cdot(-1)^2}_{(\bullet\;\bullet\;\bullet\;\bullet)} + \underbrace{3\cdot(-1)^2}_{(\bullet\;\bullet)(\bullet\;\bullet)} \right) = 1. \] Note that we can also tensor this with the sign representation, to get another irreducible representation (since $\Csign$ has all traces $\pm 1$, the norm doesn't change). Finally, we recover the final row using orthogonality (which we name $\CC^2$, for lack of a better name); hence the completed table is as follows. \[ \begin{array}{|c|rrrrr|} \hline S_4 & 1 & (\bullet\;\bullet) & (\bullet\;\bullet\;\bullet) & (\bullet\;\bullet\;\bullet\;\bullet) & (\bullet\;\bullet)(\bullet\;\bullet) \\ \hline \Ctriv & 1 & 1 & 1 & 1 & 1 \\ \Csign & 1 & -1 & 1 & -1 & 1 \\ \CC^2 & 2 & 0 & -1 & 0 & 2 \\ \refl_0 & 3 & 1 & 0 & -1 & -1 \\ \refl_0 \otimes \Csign & 3 & -1 & 0 & 1 & -1 \\\hline \end{array} \] \end{example} \section\problemhead \begin{dproblem} [Reading decompositions from characters] Let $W$ be a complex representation of a finite group $G$. Let $V_1$, \dots, $V_r$ be the complex irreps of $G$ and set $n_i = \left< \chi_W, \chi_{V_i} \right>$. Prove that each $n_i$ is a non-negative integer and \[ W = \bigoplus_{i=1}^r V_i^{\oplus n_i}. \] \begin{hint} Obvious. Let $W = \bigoplus V_i^{m_i}$ (possible since $\CC[G]$ semisimple) thus $\chi_W = \sum_i m_i \chi_{V_i}$. \end{hint} \end{dproblem} \begin{problem} Consider complex representations of $G = S_4$. The representation $\refl_0 \otimes \refl_0$ is $9$-dimensional, so it is clearly reducible. Compute its decomposition in terms of the five irreducible representations. \begin{hint} Use the previous problem, with $\chi_W = \chi_{\refl_0}^2$. \end{hint} \begin{sol} $\Csign \oplus \CC^2 \oplus \refl_0 \oplus (\refl_0\otimes\Csign)$. \end{sol} \end{problem} \begin{problem} [Tensoring by one-dimensional irreps] Let $V$ and $W$ be irreps of $G$, with $\dim W = 1$. Show that $V \otimes W$ is irreducible. \begin{hint} Characters. Note that $|\chi_W| = 1$ everywhere. \end{hint} \begin{sol} First, observe that $|\chi_W(g)|=1$ for all $g \in G$. \begin{align*} \left< \chi_{V \otimes W}, \chi_{V \otimes W} \right> &= \left< \chi_V \chi_W, \chi_V \chi_W \right> \\ &= \frac{1}{|G|} \sum_{g \in G} \left\lvert \chi_V(g) \right\rvert^2 \left\lvert \chi_W(g) \right\rvert^2 \\ &= \frac{1}{|G|} \sum_{g \in G} \left\lvert \chi_V(g) \right\rvert^2 \\ &= \left< \chi_V, \chi_V \right> = 1. \end{align*} \end{sol} \end{problem} \begin{problem} [Quaternions] Compute the character table of the quaternion group $Q_8$. \begin{hint} There are five conjugacy classes, $1$, $-1$ and $\pm i$, $\pm j$, $\pm k$. Given four of the representations, orthogonality can give you the fifth one. \end{hint} \begin{sol} The table is given by \[ \begin{array}{|c|rrrrr|} \hline Q_8 & 1 & -1 & \pm i & \pm j & \pm k \\ \hline \Ctriv & 1 & 1 & 1 & 1 & 1 \\ \CC_i & 1 & 1 & 1 & -1 & -1 \\ \CC_j & 1 & 1 & -1 & 1 & -1 \\ \CC_k & 1 & 1 & -1 & -1 & 1 \\ \CC^2 & 2 & -2 & 0 & 0 & 0 \\\hline \end{array} \] The one-dimensional representations (first four rows) follows by considering the homomorphism $Q_8 \to \CC^\times$. The last row is two-dimensional and can be recovered by using the orthogonality formula. \end{sol} \end{problem} \begin{sproblem} [Second orthogonality formula] \label{prob:second_orthog} \gim Let $g$ and $h$ be elements of a finite group $G$, and let $V_1$, \dots, $V_r$ be the irreps of $G$. Prove that \[ \sum_{i = 1}^r \chi_{V_i}(g) \ol{\chi_{V_i}(h)} = \begin{cases} |C_G(g)| & \text{if $g$ and $h$ are conjugates} \\ 0 & \text{otherwise}. \end{cases} \] Here, $C_G(g) = \left\{ x \in G : xg = gx \right\}$ is the centralizer of $g$. \begin{hint} Write as \[ \sum_{i=1}^r \chi_{V_i \otimes V_i^\vee} (gh\inv) = \chi_{\bigoplus_i V_i \otimes V_i^\vee}(gh\inv) = \chi_{\CC[G]}(gh\inv). \] Now look at the usual basis for $\CC[G]$. \end{hint} \end{sproblem}