# Explicit computation of induced modules of semidirect products with the symmetric group

I've gotten stuck in a project I have been working on, essentially on the following combinatorial question about the symmetric group.

One can obtain a 1-dimensional representation $M^n_c$ of the algebra $T_n := S_n \rtimes \mathbb{C}[y_1, \dots, y_n]$ by letting each $y_i$ act by $c$ and $S_n$ act trivially.
Given a partition $\pi$ of $n =n_1 + \dots + n_k$ and $c_1, \dots, c_k \in \mathbb{C}$, we can consider the standard induced module $$M^{\pi}_c = \mathrm{Ind} ( M^{n_1}_{c_1} \otimes \dots \otimes M^{n_k}_{c_k})$$ where the induction is from the subalgebra $T_{n_1} \otimes \dots \otimes T_{n_k} \subset T_n$ to $T_n$.

As far as I know, the simple quotients of these standard modules form a complete class for the simple representations of $T_n$. (I think this is a general fact about semidirect products of a finite group with a commutative algebra, together with the fact that representations of the symmetric group $S_n$ can be obtained by taking quotients induction of the trivial representation of a Young subgroup $S_{n_1} \times \dots \times S_{n_k}$.)

My question is how to represent this explicitly in terms of the simple objects in the semisimple category $Rep(S_n)$. First of all, the $S_n$-module $M^{\pi}_c$ can be described in a combinatorial manner in terms of the simple objects in $Rep(S_n)$ using the Kostka numbers. The action of the $y_i$'s on $M^{\pi}_c$ can be given in terms of a suitable morphism $y: \mathfrak{h} \otimes M^{\pi}_c \to M^{\pi}_c$, where $\mathfrak{h} \in Rep(S_n)$ is the regular representation.

The Pieri rule allows one to compute the decomposition in irreducibles (as parametrized by Young diagrams, of course) of $\mathfrak{h} \otimes M^{\pi}_c$, so we can view $y$ as a bunch of matrices based upon this decomposition (matrices w.r.t. the simple objects in $Rep(S_n)$, not as vector spaces).

Is there an approach to compute these matrices?

I am interested in this because it may allow a way to directly interpolate the construction of these modules to complex rank, via Pavel Etingof's program. (I believe one can interpolate the construction in another way, by reasoning more directly on the definition of the category $Rep(S_t)$ given by Deligne, but this seems to be useless as far as explicit computations--which might be helpful to study the "degeneracy phenomena" that Etingof has suggested might exist--are concerned.) In this case we have tensor categories $Rep(S_t)$ for $t$ not necessarily an integer, and while the interpretation in terms of vector spaces fails, the one in terms of Young diagrams does not.

Edit (12/27) I added a bounty today and here is some additional information that may be useful:

It should come out that the matrices representing the $y$-morphism $\mathfrak{h} \otimes M^{\pi}_c \to M^{\pi}_c$ are polynomials in the dimension $n$. When increasing $n$, we change the partition $\pi$ by adding to the first (largest) element and leaving the rows below fixed. Since a simple object in $Rep(S_t)$ for $t$ not an integer can be represented as a normal Young diagram (of size, say, $N$) with a "very long line" of "size" $t-N$ at the top, this kind of a polynomial interpolation will allow for an interpolation of the $M^{\pi}_c$ to complex rank. My claim that it should come out as a polynomial was based upon studying induction directly on these categories and finding it was interpolable. However, I don't know how to compute the $y$'s directly as matrices via the simple object decomposition. My hope was that there is a clean not-too-computationally-intensive way to do this, but unfortunately I'm not yet sufficiently comfortable with the theory of the symmetric group to have any ideas as to how to proceed.

I am also interested in the degenerate affine Hecke algebra of type A, where these kinds of standard induced modules can be defined similarly. Their simple quotients form a complete collection of irreducible modules for the Hecke algebra according to a theorem of Zelevinsky, and I know that these, too, can be interpolated by reasoning on the definitions in Deligne's paper (so one gets objects in the interpolated category $Rep(H_t)$, which is defined in Etingof's talk). But I am curious here too how it is possible to compute the $y$-morphisms as matrices using the decomposition into irreducibles in $Rep(S_t)$.

• Hi Akhil, this is an interesting question. I have some thoughts on it but I am going to take a few days before I post an answer so I can be clear. In the meantime, am I correct that M_c^\pi, as a representation of S_n, is isomorphic to Ind^{S_n}_{S_\pi}(C), or in other words C[S_n] / C[S_\pi]? Then the y_i act on this representation? This seems to follow from the fact that the subalgebra T_\pi is really S_{\pi}C[Y] (right?). I just want to make sure I haven't confused the setup. Dec 29, 2009 at 21:55
• Yes, that's correct. (I assume in Ind^{S_n}_{S_\pi}(C), C refers to the trivial representation of S_\pi.) I think you're also right about the subalgebra T_\pi being S_\pi C[Y] (i.e. the image in T_n, or intrinsically S_\pi \rtimes C[Y]). The statement is also true for the dAHA case, though it requires the analog of the PBW theorem for it. Dec 30, 2009 at 1:09

I suspect I know the answer, but I don't yet have a proof (not because I think it would be hard to prove, but because I didn't try really; when you see my guess, you'll likely want to believe it). The answer is stated not in the basis of simples, because I didn't compute the decomposition of $\mathbb{C}[S_n/S_{pi}]$. However, it is stated in the tensor category S_n-mod, so that given that decomposition, you can easily adjust what I write here.

Fact: Let H be a finite dimensional semi-simple Hopf algebra (e.g. H=\mathbb{C}[S_n]), and let $V\in H$-mod be an irrep. Let us regard H as an H-module via the left action. Then $V\otimes H\cong H^{\oplus dim(V)}$.

The proof of this fact is given as follows (see http://www-math.mit.edu/~etingof/tenscat.pdf, or Akhil's comments below):

$Hom_H(V\otimes H,W)=Hom_H(H,^\ast V\otimes W) = \widetilde{^\ast V\otimes W}$

On the other hand, $Hom_H(\tilde{V}\otimes H,W) = \tilde{V}\otimes Hom_H(H,W) = \widetilde{V\otimes W}$,

where $\tilde{M}$ means we forget the module $M$ down to a vector space, which we use as a multiplicity space (just because the direct sum decomposition I asserted originally isn't canonically given, you just know that there's this multiplicity space)

(above we took right duals since I didn't assume $H$ is commutative or co-commutative; for $C[G]$ there is no need to distinguish.) One could (and should) be uncomfortable that we got duals on the one hand and not on the other. However, the standard representation for $S_n$ is special in that it is isomorphic to its own dual, by sending $e_i$ to $e^i$ (the point is that the standard rep for $S_n$ has a basis build into its definition).

The general fact above about Hopf algebras is used to relate Frobenius-Perron dimension for representations of Hopf algebras to ordinary dimension of the underlying vector space; indeed the regular representation is the unique eigenvector which realizes the Frobenius Perron dimension as an eigenvalue.

Okay so now we are considering $\mathfrak{h}\otimes \mathbb{C}[S_n/S_{\pi}]\to \mathbb{C}[S_n/S_{\pi}]$. This is then isomorphic to $(\mathfrak{h}\otimes \mathbb{C}[S_n])\otimes_{S_\pi}\mathbf{1}$, where we tensor the trivial $S_\pi$-module on the right. This is because $C[S_n]$ is a $S_n-S_\pi$ bi-module, so that the map $\mathfrak{h}\otimes \mathbb{C}[S_n] \to \mathfrak{h}\otimes \mathbb{C}[S_n/S_\pi]$ given by right multiplying with the symmetrizer $a_\pi=\sum_{g\in S_\pi} g$ is an $S_n$-morphism, and allows us to identify $\mathfrak{h}\otimes \mathbb{C}[S_n]\otimes_{S_\pi}\mathbf{1}$ with $\mathfrak{h}\otimes \mathbb{C}[S_\pi]$. Morally, this is just because $S_\pi$ acts on the right, while the other action is on the left.

[edited an error from preceding paragraph]

Together with the Fact, this implies that $\mathfrak{h}\otimes \mathbb{C}[S_n/S_\pi]$ is in fact just isomorphic to $\mathbb{C}[S_n/S_\pi]^{\oplus dim(\mathfrak{h})},$ which I should really write as $\mathbb{C}[S_n/S_\pi]\otimes \tilde{\mathfrak{h}^\ast}$.

Well, now we have this function $c: \mathfrak{h}\to \mathbb{C}$. We will project $\mathbb{C}[S_n/S_\pi]^{\oplus dim(\mathfrak{h})}$ (or rather $\mathbb{C}[S_n/S_\pi]\otimes \tilde{\mathfrak{h}}$) to $\mathbb{C}[S_n/S_\pi]$ by just applying $c$ to the multiplicity space.

I haven't really proved that this last paragraph is what happens, but once one has applied "Fact" above, this seems like the only natural guess. I imagine verifying it would be pretty straightforward.

Note that it doesn't seem to matter how $\mathbb{C}[S_n/S_{\pi}]$ decomposes into simples, since they all get lumped together.

• Thanks for the response! I do have a few questions. I thought about the result you mentioned about Hopf algebras, taking $H = C[G]$ for $G$ a finite group. Then if $V$ is a nontrivial irreducible representation of $G$ and $W$ is any representation, $Hom_G(C[G] \otimes V, W) \simeq Hom_G(C[G], V^{\vee} \otimes W) \simeq V^{\vee} \otimes W$. Also $Hom_G( C[G]^{\oplus dim(V)}, W) \simeq W^{\oplus dim(V)}$. I don't think these are isomorphic as $G$-modules. I'll add another comment for the other question I had because of the 500-character limit. Dec 31, 2009 at 1:57
• Actually I answered the other question myself. I'm just a bit confused about why $\mathfrak{h} \otimes C[S_n/S_\pi]$ is isomorphic to $C[S_n/S_\pi]^{dim(\mathfrak{h})}$. Dec 31, 2009 at 2:12
• Actually Akhil, I think you have provided the proof of the "Fact" I mentioned. Note that Hom spaces aren't representations, but rather they are just vector spaces. When you write $Hom_G(C[G],V^\vee\otimes W)\cong V^\vee\otimes W$, you mean as a vector space (this is true, as such a map is determined by where 1 goes). As a vector space, $V^\vee\otimes W\cong W^{\oplus dim(V)}$, as I claimed. So you have proven the "fact". Dec 31, 2009 at 3:56
• I found a reference for you, from some of Pavel's notes: www-math.mit.edu/~etingof/tenscat.pdf page 81. Note that the proof is the same as the one you proposed. Dec 31, 2009 at 4:05
• I'll edit the post to include the new information. Dec 31, 2009 at 4:07