Proofs of a character identity?

Question

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?

Answer 1

$\newcommand\card[1]{\lvert#1\rvert}$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, 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.

Answer 2

Here is another way to view things: it uses the fact that for each complex irreducible character $\chi$ of $G$, there is an algebra homomorphism $\omega_{\chi} : Z(\mathbb{C}G) \to \mathbb{C}$ defined by $\omega_{\chi}(X) = \frac{\chi(X)}{\chi(1)} .$ This is the underlying root (from a modern viewpoint) of the formula of Frobenius quoted in the question, and the formula beloved of Paul Sally quoted in L. Spice's answer.

For $S$ a subset of $G$ invariant under conjugation, let $S^{+} = \sum_{s \in S} s$ in $\mathbb{C}G.$

Hence for each $\lambda$ we find that $$\frac{\chi_{\lambda}(C_{1}^{+}C_{2}^{+} \ldots C_{k}^{+})}{\chi_{\lambda}(1)} = \prod_{i=1}^{k} |C_{i}|\frac{\chi_{\lambda} (c_{i})}{\chi_{\lambda}(1)}.$$

Expressing the element $C_{1}^{+} \ldots C_{k}^{+}$ as a sum of products of group elements explains the formula of the question.