This is Theorem 4.5.1 in the text Introduction to Representation Theory by Etingof et al. and is Theorem 3.8 in the PDF lecture notes. The statement of the theorem is:

Theorem 4.5.1. For any representations $V, W$ $$ (\chi_V, \chi_W) = \dim \operatorname{Hom}_G(W, V),$$ and $$ (\chi_V, \chi_W) = \begin{cases} 1, & \text{if $V \cong W,$} \\ 0, & \text{if $V \not\cong W$} \end{cases} $$ if $V, W$ are irreducible.

Here $(\cdot, \cdot)$ denotes the inner product defined as follows:

$$ ( f_1, f_2 ) = \frac{1}{|G|} \sum_{g \in G} f_1(g) \overline{f_2(g)}$$

In the proof of the theorem, it is argued that $(\chi_V, \chi_W) = \operatorname{tr}|_{V \otimes W^*}(P)$ where $P = \frac{1}{|G|}\sum_{g\in G} g$ (thus $P$ lives in the group algebra $\mathbb{C}[G]$) and that $P$ acts as the identity in the trivial representation and zero in all other irreducible representations. So far I understand.

Then comes the part I don't understand:

Therefore, for any representation $X$ the operator $P|_X$ is the $G$-invariant projector onto the subspace $X^G$ of $G$-invariants in $X$. Thus, \begin{align*} \operatorname{tr}|_{V \otimes W^*}(P) &= \dim \operatorname{Hom}_G(\mathbb{C}, V \otimes W^*) \\ &= \dim(V \otimes W^*)^G = \dim \operatorname{Hom}_G(W, V).\end{align*}

I'm guessing the implied reason why $P$ projects onto the $G$-invariant subspace is that $\mathbb{C}[G]$ is semisimple. OK, fine. And that does seem to explain why the first and third quantities should be equal. But I don't understand why the first and second, or second and third, or third and fourth quantities should be equal.

  • 1
    $\begingroup$ You don't need semisimplicity to see that $P$ projects onto the $G $-invariants. This is what $P $ always does, acting on any representation of $G $; check it by direct computation. $\endgroup$ – darij grinberg Dec 23 '17 at 4:40
  • 3
    $\begingroup$ The equalities you are wondering about come from the facts that $\operatorname{Hom}_G (V,W) = \left(\operatorname{Hom}(V,W)\right)^G$ and $\operatorname{Hom}_G (\mathbb{C}, M) \cong M^G $ for any $G $-module $M $. $\endgroup$ – darij grinberg Dec 23 '17 at 4:43

It's unfortunate that Etingof is so terse here because this is really a very beautiful proof and it's worth understanding carefully. First, a general fact:

If $E : V \to V$ is an idempotent linear operator, then it projects onto its image, which is the same thing as its subspace of fixed points.

This has nothing to do with semisimplicity. Just compute that if $Ev = v$ then $v$ is in the image; conversely if $Ev = w$ then $Ew = E^2 v = Ev = w$ so if $w$ is in the image then it's fixed. We don't even need $V$ to be finite-dimensional here.


$$P = \frac{1}{|G|} \sum_{g \in G} g$$

is an idempotent (exercise), its image acting on any representation $V$ is contained in the $G$-invariant subspace of $V$, and it fixes all $G$-invariant elements. So the subspace it projects onto contains all $G$-invariants and is contained in all $G$-invariants, and hence must be exactly all $G$-invariants.


The functor $\text{Hom}_G(\mathbb{C}, V)$ takes the $G$-fixed points of $V$, so that we have a natural isomorphism

$$\text{Hom}_G(\mathbb{C}, V) \cong V^G.$$

Said another way, $\mathbb{C}$ represents the fixed point functor. This is where the second equality comes from.


The dimension of the image of an idempotent linear operator on a finite-dimensional vector space is its trace. Hence

$$\dim \text{Hom}_G(\mathbb{C}, V) = \dim V^G = \text{tr} \left( \frac{1}{|G|} \sum_{g \in G} \rho(g) \right) = \langle 1, \chi_V \rangle.$$

This is because idempotents only have eigenvalues $1$ and $0$ and are diagonalizable, so the multiplicity of the eigenvalue $1$ is exactly the dimension of the fixed point subspace. This is where the first equality comes from.


If $V, W$ are finite-dimensional there is a natural isomorphism

$$\text{Hom}_G(V, W) \cong (V^{\ast} \otimes W)^G.$$

This is mostly a matter of chasing through the definitions, although you can also take a more abstract approach using the axioms of a closed monoidal category. $V^{\ast} \otimes W$ corresponds to the inner hom $[V, W]$ describing all linear maps $V \to W$, and the $G$-action on this has the property that its fixed points are exactly all $G$-equivariant linear maps $V \to W$. This is where the third equality comes from.

Putting it all together, we get

$$\dim \text{Hom}_G(V, W) \cong \dim (V^{\ast} \otimes W)^G = \text{tr}(P : V^{\ast} \otimes W \to V^{\ast} \otimes W) = \langle \chi_V, \chi_W \rangle$$

as desired. This result can be interpreted as saying that $\text{Hom}_G(V, W)$ categorifies the inner product on characters, which is a fun idea to play with. For example, the fact that inner products are conjugate-linear in the first variable and linear in the second corresponds to the fact that the Hom functor is contravariant in the first variable and covariant in the second (and also preserves finite direct sums in both variables).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.