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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.

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    $\begingroup$ 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. $\endgroup$ – Geoff Robinson Jan 2 '16 at 23:32
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    $\begingroup$ @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 $S_{n-1}$. $\endgroup$ – Nicholas Proudfoot Jan 3 '16 at 0:23
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    $\begingroup$ @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 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). $\endgroup$ – darij grinberg Jan 3 '16 at 0:29
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    $\begingroup$ 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 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). $\endgroup$ – darij grinberg Jan 3 '16 at 1:19
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    $\begingroup$ 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 $S_{n-1}$. $\endgroup$ – Geoff Robinson Jan 3 '16 at 1:39
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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.)

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  • $\begingroup$ 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. $\endgroup$ – darij grinberg Jan 3 '16 at 1:52
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    $\begingroup$ This all said, I still want to know my isomorphism $\operatorname{Lie}_n \otimes \mathbb{C}^n \to \mathbb{C}\left[S_n\right]$ ! $\endgroup$ – darij grinberg Jan 3 '16 at 1:54
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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 $S_n\subset S_{n+1}$).

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  • $\begingroup$ Thank you! This is more than I can chew off right now, of course. Does your book, or Loday-Valette, bring me anywhere close? $\endgroup$ – darij grinberg Jan 3 '16 at 2:59
  • $\begingroup$ @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! $\endgroup$ – Vladimir Dotsenko Jan 3 '16 at 12:26

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