Central idempotent of a group representation

Question

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 $\pi^2 = \pi$?

Answer 1

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:

$$ e_{\small V}=\frac{\dim V}{|G|}\sum_{g\in G}\chi_{\small V}(g^{-1})g $$

Answer 2

While there is already an excellent answer, I wanted to add how the statement follows from the Schur orthogonality relations in an elementary way. Let $\lambda$ be an irreducible unitary representation of $G$ and $\chi:= \operatorname{tr}\circ \lambda$ be the corresponding character. The Schur orthogonality relations for matrix elements of such representations read (in part)

\begin{equation*} \sum_{g\in G} \overline{\lambda}_{ij}(g) \lambda_{kl}(g) = \frac{|G|}{\chi(e)} \delta_{ik}\delta_{jl}. \end{equation*}

This statement is sometimes called "Great Orthogonality Theorem". From here it is straightforward to show the relation indicated by Ben Grossmann in his comment:

\begin{align*} \sum_{g\in G}\chi(g)\chi(g^{-1}h) &=\sum_{g\in G} \operatorname{tr}(\lambda(g)) \operatorname{tr}(\lambda(g^{-1})\lambda(h))\\ &=\sum_{g\in G}\sum_{i,j,k} \lambda_{ii}(g) \overline{\lambda}_{jk}(g) \lambda_{jk}(h)\\ &=\sum_{i,j,k}\left( \sum_{g\in G}\overline{\lambda}_{jk}(g) \lambda_{ii}(g) \right) \lambda_{jk}(h)\\ &=\sum_{i,j,k}\frac{|G|}{\chi(e)} \delta_{ji}\delta_{ki}\lambda_{jk}(h)\\ &= \frac{|G|}{\chi(e)}\chi(h). \end{align*}

Showing $\pi^2 = \pi$ is now a routine computation.