Let $k$ be a field, and $G$ a finite group such that the characteristic of $k$ does not divide $|G|$. Then $kG$ is a semisimple $k$-algebra, and the representation theory of $G$ over $k$ is semisimple.
Fix an irreducible representation $V$ of $kG$. Then for any representation $W$, define the isotypic component of $W$ as the sum of all subspaces isomorphic to $V$, denoted $W[V]$. $$ W[V] = \sum_{\substack{U \subseteq W \\ U \cong V}} U$$ One can check that the assignment $W \mapsto W[V]$ is functorial. Define the element $\psi_V \in kG$ by the formula $$ \psi_V := \frac{\dim V}{|G|} \sum_{g \in G} \chi_V(g) g^{-1}$$ where $\chi_V: G \to k$ is the character of $V$, defined as $\chi_V(g) = \operatorname{tr} \rho_V(g)$. Then $\psi_V$ acts in $W$ as the $G$-equivariant projector onto $W[V]$, in the sense that $\rho_W(\psi_V) \in \mathrm{End}_k(W)$ is $G$-equivariant, and projects to the subspace $W[V]$.
Question: Is there a "nice" explanation of the formula defining $\psi_V$?
Here is the direction I would like "nice" to go in. The functor $W \mapsto W^G = W[\text{trivial}]$ taking a representation to its subspace of invariants can equivalently be thought of as the map $W \mapsto \psi_1 W$, where $\psi_1 \in kG$ is the element $$ \psi_1 = \frac{1}{|G|} \sum_{g \in G} g $$ and checking this fact is very easy. We also have that the "$V$-isotypic component functor" $W \mapsto W[V]$ is isomorphic to the functor $$ W \mapsto V \otimes_k \operatorname{Hom}_{kG}(V, W)$$ where we consider the $k$-vector space $\operatorname{Hom}_{kG}(V, W)$ to have trivial $G$-action. We can follow this second functor up with the evaluation map $$ \operatorname{ev}: V \otimes_k \operatorname{Hom}_{kG}(V, W) \to W, \quad v \otimes f \mapsto f(v) $$ to get an actual subspace of $W$ as the image of $\operatorname{ev}$. Moreover, using the various isomorphisms/identities:
- $\operatorname{Hom}_k(V, W)$ is a $G$-representation via $(g \cdot f)(v) = g f(g^{-1} v)$.
- $\operatorname{Hom}_k(V, W)^G = \operatorname{Hom}_{kG}(V, W)$
- $\operatorname{Hom}_k(V, W) \cong V^* \otimes W$ as $G$-representations.
We can get to the fact that the image of $\operatorname{ev}$ must be the image of
$$ V \otimes V^* \otimes W \twoheadrightarrow V \otimes (V^* \otimes W)^G \hookrightarrow V \otimes V^* \otimes W \xrightarrow{\operatorname{ev} \otimes 1} W$$ which on a pure tensor $v \otimes f \otimes w$ is (by using $\psi_1$) $$v \otimes f \otimes w \mapsto \frac{1}{|G|} v \otimes \sum_{g \in G} gf \otimes gw \mapsto \frac{1}{|G|} \sum_{g \in G} f(g^{-1} v) (gw)$$
This looks so close to $\psi_V$ that I feel like this must be on the right track, but I'm quite stuck taking this further. I would very much like the map $W \mapsto V \otimes \operatorname{Hom}_{kG}(V, W)$ to just lead to a formula for $\psi_V$, without having to know a bunch of orthogonality-of-characters stuff. (I am happy to accept the category is semisimple).
