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](http://www.newton.ac.uk/programmes/ALT/seminars/032716301.html). (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 $Rep(S\_t)$. 



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


