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I am trying to multiply two generators of center $Z(\mathbb{C}[S_n])$ of ring algebra of symmetric group of $n$ elements. We know that these generators are given by sums of conjugacy classes in $S_n,$ which are given by various partitions. E.g. for $S_3$ one gets $$Z(\mathbb{C}[S_3])=\mathbb{C}\langle (1)(2)(3), (12)(3)+(23)(1)+(13)(2), (123)+(132) \rangle.$$ Now the question is: Giving two partitions $p_1$ and $p_2$ of $n$ how to get the product of the associated generators $x_{p_1} \cdot x_{p_2}$ of $Z(\mathbb{C}[S_n])?$
E.g, in the given example $S_3$ we get $$x_{(2,1)}\cdot x_{(2,1)}=3x_{(1,1,1)}+3x_{(3)}.$$

By saying ''to get'' I mean an algorithm that takes two partitions and spits out the partitions that appear in the product, together with the corresponding coefficients.

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  • $\begingroup$ You mean $3x_{(1,1,1)}+6x_{(3)}$? $\endgroup$
    – user43326
    Commented May 16, 2019 at 16:59
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    $\begingroup$ I just made a quick google search: it's not the solution (you are looking for the structure constants of $Z[\mathbb{C}S_n]$ with respect to that set of algebra generators) but maybe it has some info: core.ac.uk/download/pdf/82470256.pdf - Look e.g. on page 727 (page 3 of the pdf doc) $\endgroup$
    – Qfwfq
    Commented May 16, 2019 at 17:04
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    $\begingroup$ This is the same as mathoverflow.net/questions/62088. $\endgroup$
    – Richard Stanley
    Commented May 16, 2019 at 17:08
  • $\begingroup$ @user43326 No - multiplying you get 9 terms, 3 of them go to neutral = $x_{(111)}$, and 6 others are 3 times $x_{(3)}=(123)+(132).$ <br/> $\endgroup$
    – Filip
    Commented May 16, 2019 at 19:57
  • $\begingroup$ @Richard Stanley - true, thanks for pointing that out! $\endgroup$
    – Filip
    Commented May 16, 2019 at 20:01

1 Answer 1

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This is just a small comment. Let $G$ be a finite group and let $K_1$, $\ldots$, $K_n$ be the class sums that form a basis of $Z(\mathbb{C}[G])$. For $K_i$ let $k_i$ be the size and $g_i \in G$ a representative of the corresponding conjugacy class.

Then in general $$K_i K_j = \sum_{t} C_{i,j,t} K_t$$ where $C_{i,j,t}$ are non-negative integers and $$C_{i,j,t} = \frac{k_ik_j}{|G|} \sum_{\chi \in Irr(G)} \frac{\chi(g_i)\chi(g_j)\overline{\chi(g_t)}}{\chi(1)}$$

This is very well known, see for example exercise 3.9 in the character theory book of Isaacs. Anyway, this shows that we have a formula for the coefficients in terms of character values. It might be useless in practice for computations though.

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