For $G$ any finite group and $V$ any irreducible complex representation of $G$ with character $\chi\_V$, is it always true that


$$ \frac{1}{\left| G \right|} \displaystyle\sum\_{g\in G} \chi\_V(g)\chi\_V(g^{-1}h) = \frac{\chi\_V(h)}{\dim(V)}?$$


I see already that if $V$ is one-dimensional, then $\chi\_V$ is a group homomorphism from $G$ to $\mathbb C^\*$ and the statement is immediate. Also, if $h=e$, then this is just part of the orthogonality relations for characters of irreducible representations.


Also, this identity seems to be equivalent to saying that the element
$e\_V=\frac{\dim(V)}{\left|G\right|}\displaystyle\sum\_{g\in G} \overline{\chi\_V(g)}g$ in the group algebra of $G$ is an idempotent, so if the only proof leads from this fact (if it is a fact), so be it, but I was hoping for something more direct, perhaps along the lines of proofs of the orthogonality relations themselves.



A "shifted" orthogonality relation for characters of irreducible representations of finite groups?
Consider the linear transformation $\rho(e\_V)$ to which $e\_V$ maps in the representation $\rho$ on $V$. The element $e\_V$ is in the center of the group algebra. So its image commutes with $\rho(g)$ for all $g\in G.$ But since $V$ is irreducible, by Schur's lemma, $\rho(e\_V)=\lambda \textit{Id}$ for some constant $\lambda.$ And we know that the trace of $\rho(e\_V)$ is $\dim(V)$ by the orthogonality relation $\left<\chi,\chi\right>=1.$ The only way this can happen is if $\rho(e\_V)=\mathrm{Id}$. To finish, look at the trace of $\rho(e\_Vh).$


[Benjamin](https://mathoverflow.net/questions/413432/a-shifted-orthogonality-relation-for-characters-of-irreducible-representations/413516#comment1059919_413432)'s and [Mark](https://mathoverflow.net/questions/413432/a-shifted-orthogonality-relation-for-characters-of-irreducible-representations/413516#comment1060059_413432)'s remarks actually made this answer pop out at me once I was able to stare at $e\_V$ written properly.


