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For each positive integer $i$, let $A_i$ be a fixed representation of the symmetric group $S_i$. I won't tell you exactly what $A_i$ is, but let's say that I have a very explicit description of its character.

Let $\lambda$ be a partition of a fixed integer $n$, and let $\ell$ be its length. Let $W_\lambda \subset S_n$ be the stabilizer of $\lambda$. This group $W_\lambda$ is a semidirect product of the Young subgroup $S_\lambda = S_{\lambda_1}\times\ldots\times S_{\lambda_\ell}$ and a subgroup $T_\lambda \subset S_\ell$, which shuffles pieces of $\lambda$ of the same size.

How can one compute the character of $\operatorname{Ind}_{W_\lambda}^{S_n}(A_{\lambda_1}\otimes\ldots \otimes A_{\lambda_\ell})$? Here $S_{\lambda_i}\subset S_\lambda$ acts on the $i^\text{th}$ factor of the tensor product, and $T_\lambda$ acts by permuting the factors of the tensor product.

In theory, I know that I have all of the necessary information to compute this character, but I wonder if there are any useful combinatorial tricks to organize the data.

Regarding the motivation for this question, I would like to apply the recursive formula in Remark 3.6 of this paper http://pages.uoregon.edu/njp/ukl-m1.pdf when $W=S_n$ and $\mathcal{A}$ is the braid arrangement.

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Let $\mathrm{ch}(A_i)$ denote the Frobenius characteristic of $A_i$ (as defined e.g. in Enumerative Combinatorics, vol. 2, page 351). Thus $\mathrm{ch}(A_i)$ is a symmetric function. If $\lambda$ has $m_i$ parts equal to $i$, then the Frobenius characteristic of your induced character is given by $\prod_i h_{m_i}[\mathrm{ch}(A_i)]$, where the brackets denote plethysm. See for instance page 449 of the previous reference for a statement of a somewhat more general result. The proof is a fairly straightforward computation. Since plethysm is notoriously difficult to work with, this interpretation will probably not be of much use for your question about computing the character.

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  • $\begingroup$ Thanks! This is exactly what I needed (including the generalization). $\endgroup$ – Nicholas Proudfoot Jul 22 '15 at 16:19

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