2
$\begingroup$

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 $P(z) = \frac12(z+(-z)) = 0$.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Nephry Nov 29 '15 at 15:11
  • 1
    $\begingroup$ Thanks! I was wondering why the dimension of $X$ mattered; that clears it up. I think I can get it from there. $\endgroup$ – Evan Chen Nov 29 '15 at 18:01
1
$\begingroup$

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.

$\endgroup$

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.