Is it true that $\langle \psi, \omega\rangle$ is the dimension of $Hom(V,W)$? If $(\psi,V)$ and $(\omega,W)$ are two characters of two representations of a  group $G$, then
$$
\langle \psi, \omega\rangle = \dim \operatorname{Hom}_G(V,W)?
$$
Here $\langle\cdot,\cdot\rangle$ is the standard inner product of characters of representations and $\operatorname{Hom}_G(V,W)$ is the vector space of intertwining operators from $V$ to $W$.
I am guessing this is true from examples I have seen. I think that Schur's lemma in fact says that this is true when $V,W$ are irreducible. But is it true in general? If so, how might one go about proving it? (I am not asking for a complete proof, just the basic idea of it.)
I should maybe add that as a definition the character of a representation is the trace thing.
 A: If $\newcommand{\Hom}{\text{Hom}}\Hom(V,W)$ denotes the vector space
homomorphisms from $V$ to $W$ then $\Hom(V,W)$
is a  $G$-module with character $\overline\psi\omega$ and $\Hom_G(V,W)=\Hom(V,W)^G$. Now if $U$ is a $G$-module with character $\chi$ then
$U^G$ has dimension $\frac1{{|G|}}\sum_{g\in G}\chi(g)$. Putting all
this together gives the inner product formula.
A: Just expanding a bit, partially for my own sake.
Note that $\mathrm{Hom}(V,W) \cong V^* \otimes W$ via the isomorphism $v^* \otimes w \mapsto (x \mapsto v^*(x)w)$ which is an iso of representations as well. Using the fac that characters are multiplicative over tensor product, we can check that $\chi_{V^* \otimes W}=\overline{\chi_V} \cdot \chi_W$.
From this, we can see that $\mathrm{dim} (\mathrm{Hom}(V,W)^G)=\mathrm{dim} (V^* \otimes W)^G=\frac{1}{|G|} \sum_{g \in G}\overline{\chi_V} \cdot \chi_W$
where the last equality comes from the fact that $$\frac{1}{|G|} \sum_{g \in G}\overline{\chi_V} \cdot \chi_W$$
is a projection onto $(V^* \otimes W)^G$, so it acts by identity on the subspace, and hence the trace agrees with the dimension.
