# Sum of product of characters of $S_{n+m}$ over $S_n$

I asked this in MSE first, here, but it is not getting any attention.

Let $$S_n$$ be the group of permutations acting on the set $$\{1,...,n\}$$.

Given $$R_1,R_2$$ two irreps of $$S_n$$ with characters $$\chi_{R_i}$$, I know that $$\sum_{a\in S_n}\chi_{R_1}(a)\chi_{R_2}(ab)=n!\frac{\chi_{R_1}(b)}{\chi_{R_1}(1)}\delta_{R_1R_2},$$ when $$b\in S_n$$.

What I want to know is what is the result of $$\sum_{a\in S_n}\chi_{R_1}(a)\chi_{R_2}(ab),$$ where now $$R_1$$ is an irrep of $$S_n$$ but $$R_2$$ is an irrep of $$S_{n+m}$$, with $$b\in S_{n+m}$$ and $$a$$ having $$\{n+1,...,n+m\}$$ as fixed points. Can this be expressed in terms of the representation induced from $$R_1$$, for example?

• Are you sure about the first formula you wrote? I tried calculating it using the inner product of characters and I get that the sum that you wrote is equal to $\chi_{R_1}(b)|S_n|\delta_{R_1,R_2}$ May 25, 2022 at 16:27
• @EhudMeir there was a $n!$ missing, yes May 25, 2022 at 16:40
• Interesting question! Have you tried some examples? May 25, 2022 at 18:34

I would rather write $$\chi_{R_1}(a)=\chi_{R_1^*}(a^{-1})$$. This allows us to see the sum $$\frac{1}{n!}\sum_{a\in S_n}\chi_{R_1^*}(a^{-1})\chi_{R_2}(ba)$$ as the trace on $$\mathrm{Hom}_{\mathbb{C}}(R_1,R_2)$$ of the map $$f\mapsto \frac{1}{n!}\sum_{a\in S_n}bafa^{-1}=b\frac{1}{n!}\sum_{a\in S_n}afa^{-1}$$. If $$b=1$$, this is projection to the space of invariants, so its trace is the dimension of this space $$\delta_{R_1,R_2}$$. If $$b\neq 1$$, then it's projection to invariants, followed by composition with $$b$$. If you choose a basis consisting of $$b$$ (which is sent to $$\chi_{R_1}(b)/\chi_{R_1}(1)$$ times itself), and then a basis of the other isotypic components in $$\mathrm{Hom}_{\mathbb{C}}(R_1,R_2)$$ (which are all killed), you can see this gives the factor of $$\chi_{R_1}(b)/\chi_{R_1}(1)$$ (incidentally, this seems to be a restatement of the proof that this ratio is an algebraic integer).
For the generalization, we still have that $$\frac{1}{n!}\sum_{a\in S_n}\chi_{R_1^*}(a^{-1})\chi_{R_2}(ba)$$ as the trace on $$\mathrm{Hom}_{\mathbb{C}}(R_1,R_2)$$ of the map $$f\mapsto \frac{1}{n!}\sum_{a\in S_n}bafa^{-1}=b\frac{1}{n!}\sum_{a\in S_n}afa^{-1}$$. However, this projection might have a much larger image, since there might be many homomorphisms in $$\mathrm{Hom}_{S_n}(R_1,R_2)$$, since $$R_2$$ isn't irreducible as an $$S_n$$-module. My impulse is that there's no easy expression for this trace, but I can't say I'm sure.