Explicit computation of induced modules of semidirect products with the symmetric group - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T07:53:02Z http://mathoverflow.net/feeds/question/9756 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9756/explicit-computation-of-induced-modules-of-semidirect-products-with-the-symmetric Explicit computation of induced modules of semidirect products with the symmetric group Akhil Mathew 2009-12-25T23:00:00Z 2009-12-31T04:29:12Z <p>I've gotten stuck in a project I have been working on, essentially on the following combinatorial question about the symmetric group.</p> <p>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.<br /> 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 <code>$$M^{\pi}_c = \mathrm{Ind} ( M^{n_1}_{c_1} \otimes \dots \otimes M^{n_k}_{c_k})$$</code> where the induction is from the subalgebra $T_{n_1} \otimes \dots \otimes T_{n_k} \subset T_n$ to $T_n$.</p> <p>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 $S_{n_1} \times \dots \times S_{n_k}$.)</p> <p>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.</p> <p>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). </p> <p>Is there an approach to compute these matrices? </p> <p>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 <a href="http://www.newton.ac.uk/programmes/ALT/seminars/032716301.html" rel="nofollow">program</a>. (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. </p> <p><strong>Edit</strong> (12/27) I added a bounty today and here is some additional information that may be useful:</p> <p>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.</p> <p>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 $Rep(S_t)$. </p> http://mathoverflow.net/questions/9756/explicit-computation-of-induced-modules-of-semidirect-products-with-the-symmetric/10183#10183 Answer by David Jordan for Explicit computation of induced modules of semidirect products with the symmetric group David Jordan 2009-12-30T21:41:49Z 2009-12-31T04:29:12Z <p>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.</p> <blockquote> <p><strong>Fact:</strong> 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 $V\otimes H\cong H^{\oplus dim(V)}$.</p> </blockquote> <p>The proof of this fact is given as follows (see <a href="http://www-math.mit.edu/~etingof/tenscat.pdf" rel="nofollow">http://www-math.mit.edu/~etingof/tenscat.pdf</a>, or Akhil's comments below):</p> <p>$Hom_H(V\otimes H,W)=Hom_H(H,^\ast V\otimes W) = \widetilde{^\ast V\otimes W}$</p> <p>On the other hand, $Hom_H(\tilde{V}\otimes H,W) = \tilde{V}\otimes Hom_H(H,W) = \widetilde{V\otimes W}$,</p> <p>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)</p> <p>(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).</p> <p>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.</p> <p>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.</p> <p>[edited an error from preceding paragraph]</p> <p>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 $\mathbb{C}[S_n/S_\pi]\otimes \tilde{\mathfrak{h}^\ast}$.</p> <p>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.</p> <p>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.</p> <p>Note that it doesn't seem to matter how $\mathbb{C}[S_n/S_{\pi}]$ decomposes into simples, since they all get lumped together.</p>