# G-invariant projector of irreducible representation

In the proof of Theorem 3.8 Etingof's notes (page 37, orthogonality of characters) there is the following claim. Let $G$ be a finite group, and let $P = \frac{1}{|G|}\sum_{g \in G} g \in \mathbb C[G]$. Let $X$ be an irreducible representation of $G$. Then $$P|_X = \begin{cases} \text{Id} & \text{if } X = \mathbb C \\ 0 & \text{otherwise}. \end{cases}$$

Why this is true?

In particular, why is the following not a counterexample? Let $G = \mathbb Z / 2 = \left< t | t^2=1 \right>$ act on $X = \mathbb C$ by $\rho(t) : z \mapsto -z$. Clearly $X$ is irreducible since it is one-dimensional, but it would appear that for the representation $X$, we have $P(z) = \frac12(z+(-z)) = 0$.

• Note that when talking about the complex representation theory of a group $G$, referring to "the" representation $\mathbb{C}$ generally means you should view it as the trivial 1-dimensional $\mathbb{C}G$-module. – Nephry Nov 29 '15 at 15:11
• Thanks! I was wondering why the dimension of $X$ mattered; that clears it up. I think I can get it from there. – Evan Chen Nov 29 '15 at 18:01

I finally got it thanks to Nephry's comment. The text "if $X = \mathbb C$" refers to if $X$ is the trivial representation corresponding to the trivial homomorphism $G \to \operatorname{GL}(X)$. With that, here's the proof:
Since $P$ commutes with all elements of $g$, it follows that $P : X \to X$ is intertwining. By Schur's Lemma, $P$ is either an isomorphism or the zero map. Observe also that $P^2 = P$, so $P$ is either the identity or zero. Now the image $\operatorname{img} P$ is a $G$-invariant subspace, so it must be $\{0\}$ or the entire space $X$, and the latter case corresponds to $X$ being the trivial representation.