2 Character theory 2.3 Example: S₄ 2.5 Universal projections

2.4 Inner products and homomorphisms

In this and the next section we complete the proof of character orthogonality!

If $(\rho,V)$ and $(\sigma,W)$ are two complex representations of a finite group $G$ , then we can define a representation $\operatorname{Hom}(\rho,\sigma)$ on the vector space

\operatorname{Hom}(V,W)=\{\text{linear maps $T:V\rightarrow W$}\}

with $G$ -action ’by conjugation’:

(g\cdot T)(v)=\sigma(g)T(\rho(g)^{-1}v).

Lemma 2.27.

1.

If $\rho$ and $\sigma$ have characters $\chi$ and $\psi$ , then $\operatorname{Hom}(\rho,\sigma)$ has character $\overline{\chi}\sigma$ .
2.

We have

$\operatorname{Hom}_{G}(V,W)=\operatorname{Hom}(V,W)^{G};$

in other words, the $G$ -homomorphisms are the $G$ -fixed points of $\operatorname{Hom}(V,W)$ .

Proof.

1.

Let $g\in G$ and let $v_{1},\ldots,v_{n}$ and $w_{1},\ldots,w_{m}$ be bases of $V$ and $W$ consisting of eigenvectors for $\rho(g)$ and $\sigma(g)$ , with $\lambda_{i}$ the eigenvalue attached to $v_{i}$ and $\mu_{i}$ that attached to $w_{i}$ . Let $T_{ij}\in\operatorname{Hom}(V,W)$ send $v_{i}$ to $w_{j}$ and all other basis vectors to zero; in matrix terms, this is the matrix with 1 in column $i$ , row $j$ , and 0 everywhere else. Then the $T_{ij}$ are a basis for $\operatorname{Hom}(V,W)$ . Moreover, recalling that the $\lambda_{i}$ are roots of unity, we have

$gT_{ij}=\lambda_{i}^{-1}\mu_{j}T_{ij}=\overline{\lambda_{i}}\mu_{j}T_{ij}.$

So the trace of $\operatorname{Hom}(\rho,\sigma)(g)$ is

$\sum_{i,j}\overline{\lambda_{i}}\mu_{j}=\overline{\sum_{i}\lambda_{i}}\sum_{j}% \mu_{j}=\overline{\chi}(g)\psi(g)$

as required.
2.

If $T\in\operatorname{Hom}(V,W)$ , then $gT=T$ means

$\sigma(g)T(\rho(g)^{-1}v)=T(v)$

for all $v$ . Replacing $v$ by $w=\rho(g)v$ , we get

$\sigma(g)T(w)=T(\rho(g)w)$

for all $w\in W$ . So $T\in\operatorname{Hom}(V,W)^{G}$ if and only if this holds for all $g$ and all $w$ , i.e. if and only if $T$ is a $G$ -homomorphism.∎

The key theoretical result is:

Theorem 2.28.

If $V$ and $W$ are two representations of $G$ with characters $\chi$ and $\psi$ respectively, then

\left\langle\chi,\psi\right\rangle=\dim\operatorname{Hom}_{G}(V,W).

Proof.

Step 1: We first consider the case that $(\rho,V)$ is the trivial representation. Then we have to prove that

\dim W^{G}=\frac{1}{|G|}\sum_{g\in G}\psi(g).

Note that the operator

\pi=\frac{1}{|G|}\sum_{g\in G}\rho(g)

maps $W$ to $W^{G}$ and is the identity on $W^{G}$ . This means that $\pi$ must be a projection onto $W^{G}$ . It follows that, in some basis, its matrix is diagonal with $\dim W^{G}$ ‘1’s and the remaining entries ‘0’,¹¹ 1 See also problem [[prob-projection]]. so

	$\displaystyle\dim W^{G}$	$\displaystyle=\operatorname{tr}(\pi)$
		$\displaystyle=\operatorname{tr}\left(\frac{1}{\|G\|}\sum_{g\in G}\rho(g)\right)$
		$\displaystyle=\frac{1}{\|G\|}\sum_{g\in G}\psi(g)$

as required.

Step 2: Note that $\operatorname{Hom}_{G}(V,W)=\operatorname{Hom}(V,W)^{G}$ . The character of $\operatorname{Hom}(V,W)$ is $\overline{\chi}\psi$ . So, by the first part,

\dim\operatorname{Hom}_{G}(V,W)=\frac{1}{|G|}\sum_{g\in G}\overline{\chi}(g)% \psi(g)

which is exactly what we have to show. ∎

Corollary 2.29.

Suppose that $V$ and $W$ are irreducible representations with characters $\chi$ and $\psi$ . Then

\left\langle\chi,\psi\right\rangle=\begin{cases}1&\text{if $V\cong W$}\\ 0&\text{otherwise}.\end{cases}

Proof.

This is immediate from Theorem 2.28 and Schur’s lemma. ∎

We have therefore proved part (1) of Theorem 2.17, as well as Theorem 2.18 and Theorem 2.19. In other words, we know that representations are determined by their characters and that the rows of the character table are orthonormal (with an appropriately weighted inner product). We now only have to show that the character table is square.