Central idempotent of a group representation Given a finite group $G$ with representation $\rho: G \mapsto V$, and let $\chi$ be the character of an irreducible representation. It is well known that the following map
$$\pi = \frac{\chi(1)}{|G|} \sum_g \chi(g) \rho(g)$$
is the projection into the isotypical component of $\rho$ corresponding to $\chi$, but I cannot find a self-contained proof of this fact. In particular, is there a direct way to show that $\pi^2 = \pi$?
 A: Note $\mathrm{End}(V)\cong V\otimes V^{\ast}$ as $G$-bimodules, and $\cong V^{\oplus\dim V}$ as a $G$-module (aka rep).
The map $\mathbb{C}[G]\to\bigoplus\limits_{\small V\in\mathrm{Irr}(G)}\mathrm{End}(V)$ extending $g\mapsto\rho_{\small V}(g)$ linearly is an algebra isomorphism.
Thus, the regular representation $\mathbb{C}[G]\cong\bigoplus V^{\oplus \dim V}$ as left $G$-modules.
Suppose $e_{\small V}$ is the idempotent of $\mathbb{C}[G]$, corresponding to the identity map on the irrep $V$ and to the zero endomorphism on other irreps. Write $e_{\small V}=\sum a_g g$. Multiply by $h^{-1}$, and apply $\mathrm{tr}_{\mathbb{C}[G]}$.
To take the trace on the RHS, write multiplication-by-$h^{-1}g$ as a matrix using $G$ as a basis, and note the diagonals are either all $0$ (if $h^{-1}g$ is nontrivial) or all $1$ (if $h^{-1}g$ is the identity element of $G$), giving a trace of $0$ or $|G|$ respectively. More generally, if $X$ is a $G$-set, then multiplication-by-$g$'s trace as an endomorphism is the number of fixed points of $g$ acting on $X$.
To take the trace on the LHS, note $\mathrm{tr}_{\mathbb{C}[G]}(x)=\sum (\dim V)\mathrm{tr}_{\small V}(x)$ because of how the regular representation decomposes as a left $G$-module. Therefore, we get
$$ (\dim V)\chi_{\small V}(h^{-1})=a_h|G|. $$
Solving for $a_h$ above we can write $e_{\small V}$ in full:
$$ e_{\small V}=\frac{\dim V}{|G|}\sum_{g\in G}\chi_{\small V}(g^{-1})g $$
