# 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.

• Yes, this is the whole point of the inner product. Dec 14 '17 at 18:28
• @LordSharktheUnknown: Ok, I had just never seen this. Feel free to write an answer that I can accept. Dec 14 '17 at 18:29
• You need $G$ to be finite and you need to be working over a field of characteristic not dividing $|G|$ (or else the inner product of characters is not well-defined). Dec 14 '17 at 21:41

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.

• Ok, so one needs to kow about induced representations? Dec 14 '17 at 18:37
• @JohnDoe No. $U^G$ is the $G$-invariant elements of $U$. Dec 14 '17 at 18:38
• Ok, that makes more sense. Dec 14 '17 at 18:39
• The dimension formula for $U^G$ comes from the observation that 1) the element $\frac{1}{|G|} \sum_{g \in G} g$ is projection onto the fixed points and 2) the trace of a projection is the dimension of its image. Dec 14 '17 at 21:42

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.

• Oh sorry, replace the $\chi$'s floating around with $\psi, \omega$. Dec 14 '17 at 19:01