Let $G$ be a finite group, $g \geq 0, k\geq 1 $ integers, and $(C\_1,...,C\_k)$ a tuple of conjugacy classes of $G$. I am interested in proofs of the following identity


$$
\sum\_{(c\_1,...,c\_k) \in C\_1 \times \cdots \times C\_k} \sum\_{V\_\lambda} d(V\_\lambda)^{1-2g} \chi\_\lambda(c\_1^{-1}\cdots c\_k^{-1}) = |C\_1|\cdots|C\_k|\sum\_{V\_\lambda} d(V\_\lambda)^{2-2g-k}\chi\_\lambda(c\_1)\cdots \chi\_\lambda(c\_k)
$$


where the sums with $V\_\lambda$ are over all irreducible characters of $G$ and $d(V\_\lambda) = \chi\_\lambda(1)$ is the dimension of the representation. I obtained this from messing around with a theorem due to Frobenius (for $g=0$ and later generalized originally using a quantum topology approach - hence the $g$ for genus).


Is there some nice (ideally purely character-theoretic) proof of this identity?



Proofs of a character identity?
$\newcommand\card[1]{\lvert#1\rvert}$I use the functional identity $\card{C\_1}\chi(c\_1)\chi(c\_2) = \chi(1)\sum\_{c\_1' \in C\_1} \chi(c\_1' c\_2)$, which identifies multiples of irreducible characters $\chi$, and which I'll re-write in the form
$$
\sum\_{c\_1 \in C\_1} \chi(c\_1)\chi(c\_2)
= \chi(1)\sum\_{c\_1 \in C\_1} \chi(c\_1 c\_2).
$$
This was my advisor (Paul Sally)'s all-time favourite result, and I am delighted to have a chance to use it.


I replace the cardinality on the RHS by a sum over $C\_1 \times \dotsb \times C\_k$. I also re-index the LHS by $\overline\lambda$ instead of $\lambda$, so that there's no need for inverses. Finally, I work with the $\chi$-by-$\chi$ version, as suggested by @FedorPetrov in the [comments](https://mathoverflow.net/questions/388863/proofs-of-a-character-identity#comment990319_388863), in the sense that
$$
\chi(1)^{1 - 2g}\sum\_{(c\_1, \dotsc, c\_k) \in C\_1 \times \dotsb \times C\_k} \chi(c\_1\dotsm c\_k)
= \chi(1)^{2 - 2g - k}\sum\_{(c\_1, \dotsc, c\_k) \in C\_1 \times \dotsb \times C\_k} \chi(c\_1)\dotsm\chi(c\_k),
$$
or rather that
$$
\chi(1)^{k - 1}\sum\_{(c\_1, \dotsc, c\_k) \in C\_1 \times \dotsb \times C\_k} \chi(c\_1\dotsm c\_k)
= \sum\_{(c\_1, \dotsc, c\_k) \in C\_1 \times \dotsb \times C\_k} \chi(c\_1)\dotsm\chi(c\_k).
$$


We prove this modified result by induction. It's obvious for $k = 1$. (Most inductions that I know that start at $k = 1$ have an interesting degenerate case at $k = 0$, but I don't know what the proper degenerate case is here! Plugging in $k = 0$ naïvely doesn't work.) In general, we have
\begin{align\*}
& \sum\_{c\_1 \in C\_1} 
\chi(c\_1)\Bigl(
\sum\_{(c\_2, \dotsc, c\_{k + 1}) \in C\_2 \times \dotsb \times C\_{k + 1}} \chi(c\_2\dotsm c\_{k + 1})
\Bigr)
\\
={} & \sum\_{c\_1 \in C\_1}
\chi(c\_1)\Bigl(
\chi(1)^{k - 1}\sum\_{(c\_2, \dotsc, c\_{k + 1}) \in C\_2 \times \dotsb \times C\_{k + 1}} \chi(c\_2\dotsm c\_{k + 1})\Bigr) \\
={} & \chi(1)^k\sum\_{(c\_1, \dotsc, c\_{k + 1}) \in C\_1 \times \dotsb C\_{k + 1}} \chi(c\_1 c\_2\dotsm c\_{k + 1}),
\end{align\*}
as desired.


