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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?

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    $\begingroup$ 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 $\chi_{R_1}(b)|S_n|\delta_{R_1,R_2}$ $\endgroup$
    – Ehud Meir
    May 25, 2022 at 16:27
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    $\begingroup$ @EhudMeir there was a $n!$ missing, yes $\endgroup$
    – thedude
    May 25, 2022 at 16:40
  • $\begingroup$ Interesting question! Have you tried some examples? $\endgroup$
    – LSpice
    May 25, 2022 at 18:34

1 Answer 1

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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.

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  • $\begingroup$ "Isotopic" should be "isotypic", right? $\endgroup$
    – LSpice
    May 26, 2022 at 14:25

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