A simple formula in finite group representation

Question

Let $G$ be a finite group and ${\rm Fun}(G)$ be the set of complex valued functions over $G$. There is an interesting algebra structure to be put on ${\rm Fun}(G)$, namely the convolution:

$$ (a \ast b )(x) := \Sigma_{g\in G} \,\, a(g)b(g^{-1}x)$$

One can show that the set of class functions ${\rm Class}(G)$ is a subalgebra under convolution. Therefore, by basic group representation theory, the convolution of two irreducible characters is a linear combination of irreducible characters.

I have computed for $\Bbb Z/\Bbb Z_2$ and $S_3$. And it seems that the general formula should be

$$ \chi_i \ast \chi_j = \frac{|G| \delta_{ij} \chi_i}{\dim(V_i)}.\tag{$\ast$}$$ However, I cannot prove it.

Attempts

1. I have computed for $\Bbb Z/\Bbb Z_2$ and $S_3$.

2. I have tried to use Fourier transform and inversion, but it amounts to prove the same thing at the end.

3. I have also skimmed through the basic texts in this field including Serre and Fulton+Harris, but fail to find a formula for this.

Question

Is the formula $(\ast)$ true in general? How do you prove if that's the case? It seems very simple and I apologize if I have overlooked a proof in basic texts. Thank you in advance.

Answer 1

Thanks to @Whatsup's useful comment in the original post, I now have a proof. I write it up here, with some references to Serre's Linear Representations of Finite Groups.

The algebra $(\operatorname{Fun}(G),\ast)$ is isomorphic to the group algebra $(\mathbb{C}[G],\cdot)$ simply by sending $\delta_x$ to $[x]$. Proposition 2.5.5 shows that $\mathbb{C}[G]$ decomposed to $\bigoplus_i V_i^{d_i}$ as a $G$-module, where $d_i = \dim V_i$. Theorem 2.6.8 further shows that the projector $p_i$ from $\mathbb{C}[G]$ to $V_i^{d_i}$ is given as an element in $\mathbb{C}[G]$ by

$$ p_i = \frac{d_i}{|G|} \sum_{g\in G} \,\, \overline{\chi_i(g)} [g] .$$

The corresponding element in $\operatorname{Fun}(G)$ is then $\frac{d_i}{|G|}\overline{\chi_i}$. Therefore, since $p_i^2=p_i$, we have

$$ \frac{d_i^2}{|G|^2} \overline{\chi_i} \ast \overline{\chi_i} = \frac{d_i}{|G|}\overline{\chi_i}.$$

Finally, we get $$\chi_i\ast\chi_i = \frac{|G|}{d_i}\chi_i,$$ completing the proof.