2.9. Lecture 9

2.9.1. Schur Orthogonality

The proof of Theorem 2.8.7 is split into two parts. First, in this section we show that the irreducible characters form an orthonormal set. Later on we will show that this set spans $CF(G)$ .

Given two representations $\pi,\rho$ , we can create a new representation:

Lemma 2.9.1.

Let $(\pi,V)$ and $(\rho,W)$ be two representations of a group $G$ . The pair $\big{(}c_{\pi}^{\rho},\operatorname{Hom}(V,W)\big{)}$ is a representation of $G$ given by

c_{\pi}^{\rho}(g)(T):=\rho(g)T\pi(g^{-1}),

for $T\in\operatorname{Hom}(V,W)$ .

Proof.

This is left as an exercise (see Problem 2.9.3). ∎

Lemma 2.9.2.

(i)

$\operatorname{Hom}_{G}(V,W)=\{T\in\operatorname{Hom}(V,W)\,|\,c_{\pi}^{\rho}(g% )T=T,\,\forall g\in G\}$ .
(ii)

$\chi_{c_{\pi}^{\rho}}=\chi_{\rho}\,\overline{\chi_{\pi}}$ .

Proof.

Part (i) follows directly from the definition of $\operatorname{Hom}_{G}(V,W)$ :

	$\displaystyle T\in\operatorname{Hom}_{G}(V,W)$	$\displaystyle\Leftrightarrow\rho(g)T=T\pi(g)\;\forall g\in G$
		$\displaystyle\Leftrightarrow\rho(g)T\pi(g^{-1})=T\;\forall g\in G$
		$\displaystyle\Leftrightarrow c_{\pi}^{\rho}(g)T=T\;\forall g\in G,$

as claimed.

For part (ii), given $g\in G$ , let $\{{\bf v}_{i}\}$ be an eigenbasis of $V$ for $\pi(g)$ (with $\pi(g){\bf v}_{i}=\lambda_{i}{\bf v}_{i}$ ) and $\{{\bf w}_{j}\}$ an eigenbasis of $W$ for $\rho(g)$ (with $\rho(g){\bf w}_{j}=\mu_{j}{\bf w}_{j}$ ). Then $\{T_{i,j}\}$ forms a basis of $\operatorname{Hom}(V,W)$ , where $T_{i,j}$ is the linear map defined by

T_{i,j}({\bf v}_{k})=\begin{cases}{\bf 0}\quad&\mathrm{if\;}k\neq i,\\ {\bf w}_{j}\quad&\mathrm{if\;}k=i.\end{cases}

The set $\{T_{i,j}\}$ forms an eigenbasis of $\operatorname{Hom}(V,W)$ for $c_{\pi}^{\rho}(g)$ :

	$\displaystyle\big{(}c_{\pi}^{\rho}(g)T_{i,j}\big{)}({\bf v}_{k})=\rho(g)T_{i,j% }\big{(}\pi(g^{-1}){\bf v}_{k}\big{)}=\rho(g)T_{i,j}(\overline{\lambda_{k}}{% \bf v}_{k})$	$\displaystyle=\overline{\lambda_{k}}\rho(g)T_{i,j}({\bf v}_{k})$
		$\displaystyle=\begin{cases}\overline{\lambda_{k}}\rho(g){\bf 0}\quad&\mathrm{% if\;}k\neq i,\\ \overline{\lambda_{i}}\rho(g){\bf w}_{j}\quad&\mathrm{if\;}k=i\end{cases}$
		$\displaystyle=\begin{cases}{\bf 0}\quad&\mathrm{if\;}k\neq i,\\ \overline{\lambda_{i}}\mu_{j}{\bf w}_{j}\quad&\mathrm{if\;}k=i\end{cases}=% \overline{\lambda_{i}}\mu_{j}T_{i,j}({\bf v}_{k}).$

Thus,

\chi_{c_{\pi}^{\rho}}(g)=\sum_{i,j}\overline{\lambda_{i}}\mu_{j}=\left(\sum_{i% }\overline{\lambda_{i}}\right)\left(\sum_{j}\mu_{j}\right)=\chi_{\rho}(g)% \overline{\chi_{\pi}(g)}.

∎

We make another useful definition:

Definition 2.9.3.

Let $(\pi,V)$ be a representation of $G$ . Define

V^{G}=\{{\bf v}\in G\,|\,\pi(g){\bf v}={\bf v},\,\forall g\in G\},

to be the set of fixed points of $(\pi,V)$ .

Note that $(\pi,V^{G})$ is a subrepresentation of $G$ , and in fact $(\pi,V^{G})=({\,\mathrm{Id}},V^{G})$ .

Lemma 2.9.4.

Let $(\pi,V)$ be a representation of $G$ . Then

\operatorname{dim}(V^{G})=\frac{1}{|G|}\sum_{g\in G}\chi_{\pi}(g).

Proof.

Let $\langle\cdot,\cdot\rangle$ be a $G$ -invariant inner product on $V$ . We write

(\pi,V)=(\pi,V^{G})\oplus\big{(}\pi,(V^{G})^{\perp}\big{)},

and let ${\bf v}_{1},\ldots,{\bf v}_{\operatorname{dim}(V^{G})}$ be a basis of $V^{G}$ , and ${\bf w}_{1},\ldots,{\bf w}_{\operatorname{dim}\big{(}(V^{G})^{\perp}\big{)}}$ a basis of $(V^{G})^{\perp}$ . Then ${\bf v}_{1},\ldots,{\bf v}_{\operatorname{dim}(V^{G})},{\bf w}_{1},\ldots,{\bf w% }_{\operatorname{dim}\big{(}(V^{G})^{\perp}\big{)}}$ is a basis of $V$ , which we denote by ${\mathcal{B}}$ .

Defining $S=\frac{1}{|G|}\sum_{h\in G}\pi(h)\in\operatorname{Hom}(V,V)$ , observe that for any ${\bf v}\in V$ , we have

\pi(g)S{\bf v}=\pi(g)\frac{1}{|G|}\sum_{h\in G}\pi(h){\bf v}=\frac{1}{|G|}\sum% _{h\in G}\pi(gh){\bf v}=\frac{1}{|G|}\sum_{h\in G}\pi(h){\bf v}=S{\bf v},

hence $S{\bf v}\in V^{G}$ . Moreover, if ${\bf v}\in V^{G}$ , then

S{\bf v}=\frac{1}{|G|}\sum_{h\in G}\pi(h){\bf v}=\frac{1}{|G|}\sum_{h\in G}{% \bf v}=\frac{1}{|G|}|G|{\bf v}={\bf v},

hence $S|_{V^{G}}={\,\mathrm{Id}}$ . On the other hand, note that since $(\pi,(V^{G})^{\perp})$ is a subrepresentation, we have $\pi(g){\bf v}\in(V^{G})^{\perp}$ for all ${\bf v}\in(V^{G})^{\perp}$ and $g\in G$ , hence $S{\bf v}\in(V^{G})^{\perp}$ for all ${\bf v}\in(V^{G})^{\perp}$ . Since $V^{G}\cap(V^{G})^{\perp}=\{0\}$ , $S|_{(V^{G})^{\perp}}=0$ , so writing the matrix of $S$ with respect to the basis ${\mathcal{B}}$ gives

[S]_{{\mathcal{B}}}=\operatorname{diag}\big{(}1,1,\ldots,1,0,0\ldots,0\big{)},

with $\operatorname{dim}(V^{G})$ ones, and $\operatorname{dim}\big{(}(V^{G})^{\perp}\big{)}$ zeroes. In conclusion,

\operatorname{dim}(V^{G})=\operatorname{tr}\big{(}[S]_{{\mathcal{B}}}\big{)}=% \operatorname{tr}\left(\frac{1}{|G|}\sum_{h\in G}\pi(h)\right)=\frac{1}{|G|}% \sum_{h\in G}\operatorname{tr}(\pi(h))=\frac{1}{|G|}\sum_{h\in G}\chi_{\pi}(h),

as claimed. ∎

Remark 2.9.5.

The main part of the proof of Lemma 2.9.4 shows that $\frac{1}{|G|}\sum_{g\in G}\pi(g)$ is an orthogonal projection onto $V^{G}$ ; this is an important result that we will generalise further next time.

Lemma 2.9.6.

Let $(\pi,V)$ and $(\rho,W)$ be two representations of a group $G$ . Then

\langle\chi_{\rho},\chi_{\pi}\rangle_{G}=\operatorname{dim}\operatorname{Hom}_% {G}(V,W).

Proof.

We will apply Lemma 2.9.4 to the representation $\big{(}c_{\pi}^{\rho},\operatorname{Hom}(V,W)\big{)}$ . By Lemma 2.9.2(i) and Lemma 2.9.4,

\operatorname{dim}\big{(}\operatorname{Hom}_{G}(V,W)\big{)}=\operatorname{dim}% \big{(}\operatorname{Hom}(V,W)^{G}\big{)}=\frac{1}{|G|}\sum_{g\in G}\chi_{c_{% \pi}^{\rho}}(g).

Now applying Lemma 2.9.2(ii), we have $\chi_{c_{\pi}^{\rho}}(g)=\chi_{\rho}(g)\overline{\chi_{\pi}(g)}$ , and so

\operatorname{dim}\big{(}\operatorname{Hom}_{G}(V,W)\big{)}=\frac{1}{|G|}\sum_% {g\in G}\chi_{\rho}(g)\overline{\chi_{\pi}(g)}=\langle\chi_{\rho},\chi_{\pi}% \rangle_{G}.

∎

2.9.2. Consequences of the inner product formula

We collect a few important consequences of Lemma 2.9.6:

Proof of orthogonality in Theorem 2.8.7.

By Schur’s Lemma (Corollary 1.4.2) and Lemma 2.9.6: if $\pi$ and $\rho$ are irreducible representations of $G$ , then

\langle\chi_{\pi},\chi_{\rho}\rangle_{G}=\operatorname{dim}\big{(}% \operatorname{Hom}_{G}(V,W)\big{)}=\begin{cases}1\quad&\mathrm{if\;}(\pi,V)% \cong(\rho,W),\\ 0\quad&\mathrm{otherwise}.\end{cases}

∎

Theorem 2.9.7.

(i)

Two representations are isomorphic if and only if they have the same character.
(ii)

A representation $(\pi,V)$ is irreducible if and only if

$\|\chi_{\pi}\|_{G}^{2}=\langle\chi_{\pi},\chi_{\pi}\rangle_{G}=1.$

Proof.

The fact that isomorphic representations have the same character has been proved already (Lemma 2.7.5), so for (i), we need to show the other direction. Letting $(\pi,V)$ be any representation of $G$ , we can decompose $\pi$ into irreducibles by Theorem 1.5.12:

(\pi,V)\cong\bigoplus_{\rho\in\mathrm{Irr}(G)}(\rho,W_{\rho})^{n_{\rho}},

(2.9.8)

where the $n_{\rho}$ are non-negative integers. Writing $\chi_{\pi}=\sum_{\rho}n_{\rho}\chi_{\rho}$ , for any $\sigma\in\mathrm{Irr}(G)$ , we then have

\langle\chi_{\pi},\chi_{\sigma}\rangle_{G}=\sum_{\rho\in\mathrm{Irr}(G)}n_{% \rho}\langle\chi_{\rho},\chi_{\sigma}\rangle_{G}=n_{\sigma}.

The representation $\pi$ is thus completely determined by its character.

For part (ii), by (2.9.8) we have that $\|\chi_{\pi}\|_{G}^{2}=\sum_{\rho}n_{\rho}^{2}$ , which (being the sum of squares of non-negative integers) is equal to $1$ if and only if a single $n_{\rho}$ equals one, and the rest are zero.

∎

2.9.3. Exercises

Problem 50. Verify that $\big{(}c_{\pi}^{\rho},\operatorname{Hom}(V,W)\big{)}$ is a representation of $G$ .

The next few problems give an alternative proof of Theorem 2.8.7 (actually a slightly stronger result is proved).

Problem 51. Let $(\pi,V)$ be a finite-dimensional representation of a group $G$ . Suppose that $V$ is equipped with a $G$ -invariant inner product $\langle\cdot,\cdot\rangle$ . Show that

\chi_{\pi}(g)=\sum_{i=1}^{\operatorname{dim}V}\langle\pi(g){\bf v}_{i},{\bf v}% _{i}\rangle,

where $\{{\bf v}_{i}\}_{i=1,\ldots,\operatorname{dim}(V)}$ is any orthonormal basis for $V$ with respect to the invariant inner product.

Remark: A function $\Phi:G\rightarrow{\mathbb{C}}$ of the form $\Phi(g)=\langle\pi(g){\bf v}_{1},{\bf v}_{2}\rangle$ , where ${\bf v}_{1},{\bf v}_{2}\in V$ for some unitary representation $(\pi,V)$ of $G$ is called a matrix coefficient.

Problem 52. Let $(\pi,V)$ and $(\rho,W)$ be two representations of a group $G$ . For $T\in\operatorname{Hom}(V,W)$ , define $T^{G}$ by

T^{G}=\frac{1}{|G|}\sum_{g\in G}\pi_{2}(g)T\pi_{1}(g^{-1}).

Show that $T^{G}\in\operatorname{Hom}_{G}(V,W)$ .

Problem 53. Show that if $(\pi,V)$ and $(\rho,W)$ are two non-isomorphic irreducible unitary representations of $G$ , then their matrix coefficients are orthogonal with respect to the standard $L^{2}$ -inner product on ${\mathbb{C}}$ -valued functions on $G$ , i.e.

\frac{1}{|G|}\sum_{g\in G}\big{\langle}\pi(g){\bf v}_{1},{\bf v}_{2}\big{% \rangle}_{V}\overline{\big{\langle}\rho(g){\bf w}_{1},{\bf w}_{2}\big{\rangle}% _{W}}=0\qquad\forall\,{\bf v}_{1},{\bf v}_{2}\in V,\,{\bf w}_{1},{\bf w}_{2}% \in W.

Hint: Define the map $T\in\operatorname{Hom}(V,W)$ by

T(v)=\langle{\bf v},{\bf v}_{1}\rangle_{V}{\bf w}_{1}\qquad\forall{\bf v}\in V.

Then use Problem 2.9.3 and Schur’s lemma to study $\langle{\bf w}_{2},T^{G}{\bf v}_{2}\rangle_{W}$ .

Problem 54. Let $(\pi,V)$ be an irreducible unitary representation of $G$ . Show that

\frac{1}{|G|}\sum_{g\in G}\big{\langle}\pi(g){\bf v}_{1},{\bf v}_{2}\big{% \rangle}_{V}\overline{\big{\langle}\pi(g){\bf u}_{1},{\bf u}_{2}\big{\rangle}_% {V}}=\frac{\langle{\bf v}_{1},{\bf u}_{1}\rangle_{V}\overline{\langle{\bf v}_{% 2},{\bf u}_{2}\rangle_{V}}}{\operatorname{dim}V}\qquad\forall\,{\bf v}_{1},{% \bf v}_{2},{\bf u}_{1},{\bf u}_{2}\in V.

Hint: Define the map $T\in\operatorname{Hom}(V)$ by

T({\bf v})=\langle{\bf v},{\bf v}_{1}\rangle_{V}{\bf u}_{1}\qquad\forall\,{\bf v% }\in V.

Then compute $\langle{\bf u}_{2},T^{G}({\bf v}_{2})\rangle_{V}$ using Problem 2.9.3 and Schur’s lemma (compare with the proof of Lemma 2.9.4; show that $\operatorname{tr}T^{G}=\operatorname{tr}T$ ).

Problem 55. Use Problems 2.9.3 to 2.9.3 to give an alternative proof of the fact that the characters of irreducible representations form an orthonormal set in $CF(G)$ .

The following problems give a generalisation of Fourier analysis to non-commutative finite groups:

Problem 56. For each $(\sigma,W_{\sigma})\in\mathrm{Irr}(G)$ , let $\{{\bf w}_{\sigma,i}\}_{i=1,\ldots,\operatorname{dim}\sigma}$ be an orthonormal basis of $W_{\sigma}$ with respect to an invariant inner product $\langle\cdot,\cdot\rangle_{W_{\sigma}}$ . Show that $\{\Phi_{\sigma,i,j}\}_{\sigma\in\mathrm{Irr}(G),1\leq i,j\leq\operatorname{dim% }\sigma}$ is an orthonormal basis of $L^{2}(G)=\{f:G\rightarrow{\mathbb{C}}\}$ with respect to the inner product $\langle f_{1},f_{2}\rangle_{G}=\frac{1}{|G|}\sum_{g\in G}f_{1}(g)\overline{f_{% 2}(g)}$ , where

\Phi_{\sigma,i,j}(g):=\sqrt{\operatorname{dim}\sigma}\langle\sigma(g){\bf w}_{% \sigma,i},{\bf w}_{\sigma,j}\rangle_{W_{\sigma}}.

Problem 57. (Challenging!) For each $(\sigma,W_{\sigma})\in\mathrm{Irr}(G)$ and $f\in L^{2}(G)=\{f:G\rightarrow{\mathbb{C}}\}$ , define $\widehat{f}(\sigma)\in\operatorname{Hom}(W_{\sigma})$ by

\widehat{f}(\sigma)=\frac{1}{|G|}\sum_{g\in G}f(g)\sigma(g^{-1}).

(a)

Show that $f\mapsto\widehat{f}(\sigma)$ is an intertwining operator between $(\rho,L^{2}(G))$ , and $\big{(}\widetilde{\sigma},\operatorname{Hom}(W_{\sigma})\big{)}$ , where $(\rho(g)f)(h)=f(hg)$ , $\forall g,h\in G$ $f\in L^{2}(G)$ , and $\widetilde{\sigma(g)}(T)=\sigma(g)T$ for all $T\in\operatorname{Hom}(W_{\sigma})$ , $g\in G$ .
(b)

Show the Plancherel identity

$\langle f_{1},f_{2}\rangle_{G}=\sum_{\sigma\in\mathrm{Irr}(G)}\operatorname{% dim}\sigma\,\operatorname{tr}\big{(}\widehat{f_{1}}(\sigma)\widehat{f_{2}}(% \sigma)^{*}\big{)},$

and use it to deduce the Fourier inversion formula

$f(g)=\sum_{\sigma\in\mathrm{Irr}(G)}\operatorname{dim}\sigma\,\operatorname{tr% }\big{(}\widehat{f}(\sigma)\sigma(g)\big{)}.$