User chris bowman - MathOverflow

The Jantzen-Schaper theorem

2013-05-15T05:14:22Z

Does anybody have an electronic copy of Schaper's PhD thesis:

K.D. SCHAPER, ‘Charakterformeln fur Weyl-Moduln und Specht-Moduln in Primcharacteristik’, Diplomarbeit, Bonn, 1981.

I would like to understand how to pass between Jantzen's formulation of the "Jantzen sum formula" for ${\rm GL}_n$ in terms of reflections in a type $A$ geometry, to James and Mathas' combinatorial treatment given in terms of hooks in the Young diagram.

Reference for Clifford theory (of algebras)

2013-01-23T10:19:31Z

Clifford theory relates the representation theory of a group to that of a normal subgroup. A good reference for this is Curtis and Reiner's "Methods in Representation theory II" Theorem 11.1.

Theorem: Clifford theory

Le $N $ be a normal subgroup of a finite group $G$. Let $M$ be a simple $kG$-module and $L$ a simple $kN$-submodule of ${\rm res}^G_N(M)$. Then the following statements hold:

(i) ${\rm res}^G_N(M)$ is a semisimple $kN$-module, and is isomorphic to a direct sum of conjugates of $L$

(ii) the $kN$-homogenous components of ${\rm res}^G_N(M)$ are permuted transitively by $G$.

(iii) Let $\hat{L}$ be a $kN$-homogenous component of ${\rm res}^G_N(M)$ containing $L$, and let $\hat{N}={ x\in G: x \hat{L}=\hat{L} }$. Write $G$ as a disjoint union $G=\cup_{i=1}^n g_i \hat{N}$. Then ${g_iL: 1 \leq i \leq n}$ is a complete set of non-isomorphic conjugates of $L$, and each appears with equal multiplicity in ${\rm res}^G_N(M)$.

$ \ $

I am looking for a reference which generalises this theorem to other algebras. In particular the "skew-group ring" situation where a finite group, $G$, acts by automorphisms on an algebra $A$. We then get that Clifford theory relates the representation theories of $A$ and $$A \rtimes G = \textbraceleft \sum_{x \in G} a_x x : a \in A \textbraceright.$$

One possible reference for this is Ram and Rammage "Affine Hecke algebras, cyclotomic Hecke algebras and Clifford theory", however they focus on induction and restriction between $A \rtimes G $ and $A \rtimes H$ where $H$ is the inertia group of a given simple module. I would prefer a reference to a theorem of the above form, directly relating $A \rtimes G$ and $A$, so that I can just write "please see...." without going into any more detail. This is lazy of me, I know, but I think that such a reference should exist.

Littlewood Richardson rule and seminormal basis of Specht modules

2013-01-03T11:07:40Z

Background

Seminormal Basis of Specht modules of $\mathfrak{S}_n$

Let $\lambda$ be a partition of $n$. A $\lambda$-tableau is a bijection $\mathfrak{t}:\lambda \to {1,2...,n}$. We say a tableau, $\mathfrak{t}$, s standard if the entries are increasing along the rows and columns. We let $\mathcal{T}_{\lambda}$ denote the set of standard $\lambda$-tableaux.

Let $1\leq i < j\leq n$, we define the axial distance, $a(i,j)$, as follows: if $i$ occurs in row $i_0$ and column $i_1$ and $j$ occurs in row $j_0$ and column $j_1$, then $a(i,j)=(i_0-i_1) -(j_0-j_1)$.

If $\mathfrak{t}$ is a $\lambda$-tableau and $w \in \Sigma_n$ let $w\mathfrak{t}$ be the tableau obtained from $\mathfrak{t}$ by replacing each entry in $\mathfrak{t}$ by its image under $w$. If $\mathfrak{t}$ is a standard $\lambda$-tableau, we set $\mathfrak{t}_{i \leftrightarrow i+1}$ equal to $w \mathfrak{t}$ if this is still a standard $\lambda$-tableau, and 0 otherwise.

For a given partition $\lambda$ of $n$, the Specht module ${\mathbf{S}(\lambda)}$ has a basis given by the set of standard $\lambda$-tableaux. With respect to this basis the generators act as follows \begin{align*}
{\rho_{\lambda}}(s_{i,i+1})\mathfrak{t} = \frac{1}{a(i,i+1)} \mathfrak{t} + \left(1 + \frac{1}{a(i,i+1)}\right) \mathfrak{t}_{i \leftrightarrow i+1}
\end{align*}

This basis is very compatible with induction and restriction rules (see \emph{Seminormal representations of Weyl groups and Iwahori-Hecke algebras}, Arun Ram).

The Littlewood--Richardson rule

The LR rule describes the coefficients in the restriction $$\mathbf{S}(\nu)\downarrow_{\mathfrak{S}_{r_1}\times \mathfrak{S}_{r_2}} \cong \oplus c^{\nu}_{\lambda,\mu} \mathbf{S}(\lambda) \boxtimes \mathbf{S}(\mu)$$

There are many formulations of this rule. For example, the Jeu de Taquin version maps standard skew-tableaux of shape $\nu/\lambda$ to those of shape $\mu$. The LR coef, $c^{\nu}_{\lambda, \mu}$ is the cardinality of the fiber $f^{-1}(\mathfrak{t})$ for any $\mu$-tableau $\mathfrak{t}$.

So we have a map from $\nu$-tableaux to $\lambda \times \mu$-tableaux. The fibers give the LR coefficients. However, this map is not a homomorphism of Specht modules.

Question:

Is there a reference for an explicit construction of such a homomorphism? I.e. a formulation of the LR rule which is compatible with the seminormal bases of Specht modules.

Cell modules for type D Weyl group

2012-04-29T11:16:36Z

Does anyone know a reference for the construction of cell modules for the Weyl group of type $D$, without any reference to the Weyl group of type $B$?

What would be even better is an idempotent construction along the lines of: take the product of the row and column stabilisers of the partition (as in the type $A$ case) with some other idempotent (as in the type $B$ case).

Thanks!

Cohomology $H^*(G,K)$ of wreath products

2012-01-24T09:53:58Z

Let $G = Sym(a) \wr Sym(b)$ be a wreath product of symmetric groups - I'm particularly interested in the Weyl group of type $B$, $Sym(2) \wr Sym(n)$. Let $k$ be a field of characteristic $p$.

What is $H^*(G,k)$?

If $i \leq p-3$ and we're in the symmetric group case, then $H^i(Sym(n), k)=0$.

If $i=1$ and $G$ is as above then $H^1(G, k)=0$, for all $a$ and $b$, unless we're in characteristic 2.

Is anything else possible to say? What I REALLY want is that the symmetric group result generalises so that:

If $i \leq p-3$, then $H^i(G, k)=0$ for $i \leq p-3$, for $G$ a wreath product as above.

Any ideas if this is true? It might be that it holds with fewer restrictions on $i$ and $G$ - the proof I know for the symmetric group case uses the Schur functor and tilting modules for $GL_n$. However, the result proved is FAR more general (it concerns all Specht modules) - so maybe this $GL_n$ approach isn't needed.

Quasi-hereditary covers

2011-11-08T15:19:42Z

What is the quasi-hereditary cover of a quasi-hereditary algebra?

Is the algebra A somehow (always) a quasi-hereditary cover of itself? Or is their a relation to Ringel duality in some way?

Comment by Chris Bowman

2013-05-17T09:17:38Z

So, is there some obvious way of passing from non-dominant weights to dominant weights? Is this similar to the cancellation one sees close to the walls when using the generic patterns for Weyl modules? For example, in the SL_3 case I think these are mentioned in papers of Doty (and perhaps Sullivan). Would it be fair to say that James and Mathas' formulation is easier to work with (for a given example) in the GL_n case, than Jantzen's own formulation?

Comment by Chris Bowman

2013-05-16T01:37:58Z

Sorry $\Delta=V$, I swapped notation half-way through

Comment by Chris Bowman

2013-05-16T01:37:12Z

Thanks Jim! I'll have a look at that. I was looking at $V(4,1)$ for $SL_3$ in characteristic 3. I know that I should be able to rewrite the sum $\sum_{i>0}V^i(4,1)$ as V(3,0)+V(0,3)+V(0,0) But I couldn't get from Delta(4,1) to Delta(0,0) using a single reflection in the geometry, I had to take 3 reflections and compose. Is my problem that I'm focussing on SL_3, rather than GL_3?

Comment by Chris Bowman

2013-01-24T11:31:08Z

That's perfect. Thank you very much.

Comment by Chris Bowman

2012-05-01T18:31:07Z

I've mostly been reading the work of Ariki, Mathas, Hu, Ram, Michel, and Marin. It seems that there is no easy construction (without reference to the type B case). I've since been told that the only place I'm likely to find an "internal construction" of the "Specht" modules is in Gecke's paper "Hecke algebras of finite type are cellular".

Comment by Chris Bowman

2012-04-30T07:42:48Z

Oops. No, I just want a basis of the "Specht modules" - or cellular basis of the Hecke algebra (at q=1).

Comment by Chris Bowman

2011-11-08T17:00:00Z

Thanks Ben (and Martin) that's what I thought it should be - very helpful. I'd like to give you the "reputation points" but don't understand how this works.