Product of a class function with a conjugate of an irreducible character

Question

Let $\chi$ be an irreducible complex character of a finite group $G$ and define $f:G \to \mathbb{C}$ by $f(g)=|\{h \in G:h^2=g\}|$. From a question which I am trying to solve it appears that $f(g)\overline{\chi(g)}=\chi(g^2)$, but I have no idea how one should go about showing that this is actually true (assuming that it is indeed correct). Any ideas?

Answer 1

Consider the set $S$ of pairs $(a,b) \in G\times G$ satisfying $a^2 = b$. Now consider the expression:

$$\frac{1}{|G|}\sum_{(a,b) \in S} \chi(b) $$

On the one hand we could just sum over the first coordinate and since each $a$ value appears with exactly one such $b$ this is just:

$$ \frac{1}{|G|} \sum_{a\in G} \chi(a^2)$$

On the other hand we could sum over the second coordinate, this time the different $b$ values occur different numbers of times, but this number of times is just $f(b)$ by definition. Therefore we get:

$$ \frac{1}{|G|} \sum_{b\in G} f(b)\chi(b)$$

Therefore:

$$ \frac{1}{|G|} \sum_{a\in G} \chi(a^2) = \frac{1}{|G|} \sum_{b\in G} f(b)\chi(b)$$

But wait that's not what you wanted! You wanted the $\overline{\chi(b)}$ on the right hand side, how can this be?

Well notice that $f(b)$ is a real function so

$$\overline{\frac{1}{|G|} \sum_{b\in G} f(b)\chi(b)} = \frac{1}{|G|} \sum_{b\in G} f(b)\overline{\chi(b)}$$

But $\frac{1}{|G|} \sum_{b\in G} f(b)\chi(b)$ is rational as it is a rational virtual character multiplicity (in fact it is 0,1, or -1 but that's a story for another time) so it's equal to its own conjugate.

Answer 2

For any representation $\pi: G\to GL(V)$ of a finite group let $$i_\pi = \sum_{x \in G} \pi(x)$$

Then $i_\pi = i_{\pi^{-1}}$ and $\pi(g) = \pi^{-1}(g^{-1})$ so $$\pi(g)i_\pi = \pi^{-1}(g^{-1}) i_{\pi^{-1}} = \sum_{x \in G} \pi^{-1}(g^{-1})\pi^{-1}(x^2)= \sum_{x \in G} \pi^{-1}(g^{-1}x^2)=\sum_{x \in G} \pi(x^{-2}g)=i_\pi \pi(g)$$

Thus $\pi|_{i_\pi V}$ is a representation $G \to GL(i_\pi V)$.

With $\pi$ the regular representation then $tr(\pi(g)) = |G| 1_{g = 1}$ and

the basis of $i_\pi V$ are the different cosets $y G^2 \in G/G^2$.

$G$ acts by permutation on those and $tr(\pi(g)|_{i_\pi V}) =\sum_{yG^2, gyG^2=yG^2}1$.

For $V$ a $k$-vector space with $char(k) \nmid |G|$, decomposing the regular representation in a sum of irreducible representations $\pi = \sum_j \rho_j$ then $\pi i_\pi = \sum_j \rho_j i_{\rho_j}$ where $\rho_j|_{i_{\rho_j}}$, being a subrepresentation of $\rho_j$, is $0$ or $\rho_j$.