I have come across a sequence of representations $V_n$ of the symmetric group $S_{n+2}$ which has the property that restricting the action $S_n \subset S_{n+2}$ gives the regular representation: $$ Res^{S_{n+2}}_{S_n} V_n = \mathbb{Q}S_n. $$ In other words, there is some natural way to give the regular rep of $S_n$ an action of $S_{n+2}$. This (to me) is surprising, but I imagine this has already been observed.

For concreteness, here are the first few terms in the sequence, written as a sum of partitions (using the usual indexing of representations of $S_{n+2}$ by partitions of $n+2$):


I have two questions:

  1. Is there already work on unrestrictions of the regular representation of a symmetric group? Is my particular sequence of representations $V_n$ well-known?
  2. In general, are there circumstances under which a representation of $S_n$ has a canonical way to extend the action to $S_{n+1}$?
  • $\begingroup$ 2 seems unlikely. When certain doubly transitive subgroups exist, we can find representations with the same description as yours, but that are permutation representations containing trivial subrepresentations. So sometimes there are multiple, presumably sometimes there is one, and sometimes none. $\endgroup$
    – Will Sawin
    Feb 2, 2012 at 7:19
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    $\begingroup$ I interpret 2 as asking two questions. (i) Can we characterize those $S_n$-representations that unrestrict to an $S_{n+1}$ representation in an essentially unique way? (ii) Can we find a class of $S_n$-representations that unrestrict to an $S_{n+1}$ representations in a somewhat canonical way? Incidentally, +1 for the question, and I hadn't seen the word "unrestrict" before and I like it. $\endgroup$ Feb 2, 2012 at 8:16
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    $\begingroup$ Upvoted for the word "unrestriction." $\endgroup$
    – JSE
    Feb 2, 2012 at 17:26

2 Answers 2

  1. Yes your series of representations looks (except for the first term - but there must be the law of small numbers lurking around) like the sign representation times the Whitehouse module, see, e.g. these slides by Richard Stanley. Basically, the Whitehouse module can be viewed as the respective component of the cyclic operad Lie.

  2. The last sentence of the previous paragraph gives an operadchik's view on your second question. If your representation of $S_n$ appears as a component of arity $n$ of a certain operad, then the most "natural" situation in which it can be extended to $S_{n+1}$ is when your operad is cyclic or anticyclic (see, e.g. this paper of Getzler and Kapranov, and this paper of Chapoton for some flavour of the story).


When $n+2$ is a prime $p,$ it is relatively easy to construct such a representation. For then the symmetric group $S_{p}$ contains a Frobenius group $F$ of order $p(p-1).$ Every element of $F$ either has order $p$ or has order dividing $p-1.$ Furthermore, each non-identity element of $F$ of order dividing $p-1$ has exactly one fixed point in the natural permutation action of $S_{p}.$ Consequently, if we induce the trivial module for $F$ to $S_{p},$ and restrict that module back to the natural copy of $S_{p-2},$ then we obtain the regular module for $S_{p-2}$ (using for example, Mackey's formula and the fact that $[S_{p}:F] = (p-2)!,$ and noting that $F^{x} \cap S_{p-2} = 1$ for all $x \in S_{p}).$ However, I do not see offhand an easy way to generalize this argument when $n+2$ is not prime

Note also that a rather simpler argument using the cyclic subgroup $X$ generated by an $n+1$ cycle shows that (for general $n$), the permutation module afforded by the action of $S_{n+1}$ on the (say right) cosets of $X$ restricts to the regular module of $S_{n}.$

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    $\begingroup$ More generally, starting with any one-dimensional representation of that cyclic subgroup of $S_{n+1}$ you can induce up to $S_{n+1}$ to get a representation that restricts to the regular rep of $S_n$. If instead of the trivial one you start with a faithful one, then you get the one which is probably coming up here, namely the restriction of the Whitehouse rep of $S_{n+2}$, mentioned in Dotsenko's answer. $\endgroup$ Feb 2, 2012 at 14:38
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    $\begingroup$ Yes, I did think about that, but for some reason (incorrectly) got the idea that the original question was phrased in terms of rational representations. $\endgroup$ Feb 2, 2012 at 19:08
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    $\begingroup$ The funny thing is that even though other faithful one-dimensional reps are defined over the cyclotomic field, the induced ones are honest rational representations. $\endgroup$ Feb 2, 2012 at 19:25
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    $\begingroup$ Yes, they are rational representations-up to equivalence-but the matrices givn by the induced representation construction are not rational matrices (at least for $S_{n+1}$), though the restriction of the induced representation back to $S_n$ is a genuine rational representation. $\endgroup$ Feb 2, 2012 at 20:01

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