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Let $\mathbb{C}[S_n]$ be the regular representation of the symmetric group $S_n$, and let $\mathbb{C}^n$ be the vector representation.

Question: Does there exist a representation $V$ (of dimension $(n-1)!$) such that $V\otimes\mathbb{C}^n\cong \mathbb{C}[S_n]$? If so, does $V$ admit any particularly nice description?

For example, when $n=4$, we can take $V = V(4)\oplus V(2,1,1) \oplus V(2,2)$.

This problem has a nice solution if we only ask the isomorphism to be equivariant for the subgroup $S_{n-1}\subset S_n$. We can realize $\mathbb{C}[S_n]$ as the cohomology of the configuration space of $n$ points in $\mathbb{R}^k$ for any odd $k$ (even $k=1$ is okay). Then forgetting the $n$th point gives us a fiber bundle that behaves like a product on cohomology, and we obtain an isomorphism $\mathbb{C}[S_n]\cong \mathbb{C}[S_{n-1}]\otimes\mathbb{C}^n$. But we have broken the symmetry by choosing which point to forget, so this is only an isomorphism of $S_{n-1}$ representations. I want to do everything $S_n$-equivariantly, which is indeed possible when $n=4$, as the above example demonstrates.

Update: As Geoff Robinson observes in the comments below, an $S_n$-representation $V$ is a solution to my problem if and only if the restriction of $V$ to $S_{n-1}$ is isomorphic to the regular representation.

YCor
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Nicholas Proudfoot
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  • The $S_{n-1}$-equivariant version boils down to the fact that the left $S_{n-1}$-set $S_n$ is the disjoint union of $n$ copies of the regular left $S_{n-1}$-set $S_{n-1}$, right? (And the fact that $\mathbb{C}\left[S_m\right] \otimes V \cong \mathbb{C}\left[S_m\right]^{\times \dim V}$ for any $m$ and any $S_m$-module $V$.) – darij grinberg Jan 02 '16 at 23:24
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    You seem to be asking for a (not necessarily irreducible) character of $S_{n}$ of degree $(n-1)!$ which vanishes on all non-identity permutations ( in the natural permutation action) which have a fixed point. This character won't be irreducible for $n > 3$. You have demonstrated the existence for $n =4$ and there is such a character (irreducible of degree $2$) when $n = 3$. It seems tricky when $n >4$, but $n=5$ should be easy to check with a computer. – Geoff Robinson Jan 02 '16 at 23:32
  • @GeoffRobinson: Yes, that's a nice way to reformulate the question. I actually have a conjectural candidate for such a $V$ is in general, and I do intend to check my conjecture for some higher $n$ with a computer. I just wanted to find out here whether or not this problem is already known to have a solution. – Nicholas Proudfoot Jan 02 '16 at 23:38
  • I haven't seen such a character before but I don't specialize in reps of the symmetric group, so others may know of one ( hopefully Richard Stanley will see this question at some point and may shed some light on it). I would certainly be interested to know if such character exists in general. – Geoff Robinson Jan 02 '16 at 23:42
  • Oops, of course it works for $n=2$. Anyway, I'd suspect that something like the Lie representations (cf. http://arxiv.org/abs/1505.04196 ) could be relevant. – darij grinberg Jan 02 '16 at 23:43
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    @GeoffRobinson: Just to rephrase your comment, you have observed that my question is equivalent to asking for a representation $V$ of $S_n$ with the property that the restriction of $V$ to $S_{n-1}$ is isomorphic to the regular representation of $S_{n-1}$. – Nicholas Proudfoot Jan 03 '16 at 00:23
  • Yes, equivalently ( more or less back to where you started), does the regular module for $S_{n-1}$ extend to a module for $S_{n}$? – Geoff Robinson Jan 03 '16 at 00:25
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    @GeoffRobinson: That's equivalent? The answer is positive, and for a rather simple reason: The module $\operatorname{Lie}n$ as defined on page 3 of http://www-math.mit.edu/~rstan/transparencies/whouse.pdf restricts to $S{\left{2,3,\ldots,n\right}}$ as a regular representation (permuting the given basis). – darij grinberg Jan 03 '16 at 00:29
  • @darijgrinberg : Yes, it's equivalent. A permutation of $S_{n}$ fixes a point if and only if some conjugate of it is in $S_{n-1}$. Hence a character of $S_{n}$ vanishes on all non-identity elements fixing a point if and only if it vanishes on alll non-identity elements of $S_{n-1}.$ I am not too familiar with those Whitehouse representations, though I have seen the before. They seem to do the job. – Geoff Robinson Jan 03 '16 at 00:37
  • @GeoffRobinson: And why is the fixed-point version equivalent to the original? – darij grinberg Jan 03 '16 at 00:38
  • @darij grinberg :Consider the product of the characters of the two representations. The natural permutation character takes strictly positive values on permutations with fixed points, so the character afforded by $V$ must take value $0$ on all non-identity permutations with fixed points, an in particular vanishes on all non-identity elements of $S_{n-1}$. – Geoff Robinson Jan 03 '16 at 00:43
  • OK, I guess I'll have to sleep over it; either way, thanks for transforming the question into something I could answer. An explicit isomorphism $\operatorname{Lie}_n \otimes \mathbb{C}^n \to \mathbb{C}\left[S_n\right]$, of course, would be the best answer. – darij grinberg Jan 03 '16 at 00:51
  • @darijgrinberg: Thanks, I did not know that $Lie_n$ had this property! I think my question is now fully answered. (If you want to post a reference to this fact as an answer, I will accept it.) – Nicholas Proudfoot Jan 03 '16 at 01:04
  • @NicholasProudfoot: It seems that each of Geoff and myself currently understands only their own half of the argument, which makes an answer tricky to write :) – darij grinberg Jan 03 '16 at 01:05
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    Actually, Corollary 8.7 of Reutenauer's Free Lie Algebras says that $\operatorname{Lie}n \cong \operatorname{Ind}^{S_n}{C_n} E$, where $C_n$ is the cyclic group generated by an $n$-cycle in $S_n$, and where $E$ is any faithful $1$-dimensional representation of $C_n$ (that is, the representation which sends said cycle to a primitive $n$-th root of unity). Combining this with the projection formula that is item 3 in http://mathoverflow.net/q/18799/ , we obtain $\operatorname{Lie}_n \otimes \mathbb{C}^n \to \mathbb{C}\left[S_n\right]$ (though, annoyingly, this gives us complex coefficients). – darij grinberg Jan 03 '16 at 01:19
  • (Yet more annoyingly, Reutenauer's proof uses characters, so it won't give us an explicit isomorphism either way.) – darij grinberg Jan 03 '16 at 01:23
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    It is actually immediate (with the benefit of hindsight) that the permutation module afforded by the action of the cosets of $\langle (12 \ldots n-1 n) \rangle$ will do for $V$, which is related to the Corollary in Reutenaur's book mentioned by Darij Grinberg. It suffices by Mackey's formula to note that no non-identity power of an $n$-cycle is conjugate to an element in $S_{n-1}$. – Geoff Robinson Jan 03 '16 at 01:39
  • Typo: action on the cosets of $\langle (12 \ldots n-1 n) \rangle$. – Geoff Robinson Jan 03 '16 at 01:49

2 Answers2

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Let $H$ be a regular subgroup of $S_n$, for instance a transitive cyclic subgroup of order $n$. Then the permutation module $V$ of the action on the coset space $S_n/H$ has the requested property, as only the identity element of $S_n$ lies in a conjugate of $H$ and fixes a point in the natural action at the same time.

(Remark: Just noticed that in the very same minute Geoff Robinson gave the identical answer in a comment. So I marked my answer CW.)

Peter Mueller
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  • This is soooo nice! (@both of you.) Let me just spell out the exact isomorphism. Let $\left[n\right]$ denote the left $S_n$-set $\left{1,2,\ldots,n\right}$. Then, the map $S_n \to \left(S_n/H\right)\times \left[n\right], \ \sigma \mapsto \left(\sigma H, \sigma\left(1\right)\right)$ is an isomorphism of left $S_n$-sets (since it is $S_n$-equivariant and injective and its domain has the same size as its target). By linearization, it becomes an isomorphism $\mathbb{C}\left[S_n/H\right] \otimes \mathbb{C}^n \to \mathbb{C}\left[S_n\right]$ of left $S_n$-modules. – darij grinberg Jan 03 '16 at 01:52
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    This all said, I still want to know my isomorphism $\operatorname{Lie}_n \otimes \mathbb{C}^n \to \mathbb{C}\left[S_n\right]$ ! – darij grinberg Jan 03 '16 at 01:54
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The fact that the Lie module (as proposed by Darij Grinberg) works, as well, as an explicit isomorphism of modules, follows from the theory of cyclic operads: see Corollary 6.9 in http://sites.math.northwestern.edu/~getzler/Papers/cyclic.pdf (to be precise, restrict the statement of that Corollary to $S_n\subset S_{n+1}$).

Vladimir Dotsenko
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  • Thank you! This is more than I can chew off right now, of course. Does your book, or Loday-Valette, bring me anywhere close? – darij grinberg Jan 03 '16 at 02:59
  • @darijgrinberg Neither LV nor my book with Bremner discusses enough of cyclic operads (well, we don't discuss them at all, and LV discuss them just a little bit). Section 13.1.10 of LV would help in unraveling bits on Harrison homology which may be missing in Getzler-Kapranov, but as for cyclic operads and appropriate cyclic homology complexes, I kind of feel that GK can't be avoided! – Vladimir Dotsenko Jan 03 '16 at 12:26