Let $G$ be a finite group, and $\{\chi_i\}$ the set of characters of its complex irreducible representations. I'm trying to compute
$$\sum_{g \in G} \chi_i (g) \chi_j (g) \chi_k (g)$$
for any three irreducible characters $\chi_i, \chi_j, \chi_k$.
According to this question, it can be interpreted as
the multiplicity of the trivial rep of $G$ in the (tensor) product rep $\chi_1\otimes \chi_2\otimes \chi_3$ of $G\times G\times G$ restricted to $G$ via $g \mapsto (g,g,g)$.
From this, if the restricted representation is irreducible, then the sum should be $|G|$ if the restricted representation is trivial and $0$ otherwise.
However, this is not really simpler than computing the above sum from the character table. Is there a simple way of computing the above sum?
Best Answer
Indeed, this sum is $|G|$ times the multiplicity of the trivial representation inside the tensor product $\chi_1 \otimes \chi_2 \otimes \chi_3$ of the three representations. It is also the multiplicity of $\bar{\chi}_3$ (i.e. the dual character to $\chi_3$) inside $\chi_1 \otimes \chi_2$.
These are called Clebsch-Gordan coefficients. As far as I understand while we have this nice formula as a character sum (so you can just compute them for any particular group), they are otherwise pretty difficult to understand in general. In the case where $G$ is a symmetric group these give the so-called Kronecker coefficients, which are famously difficult to understand combinatorially.