User chris bowman - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T12:43:57Z http://mathoverflow.net/feeds/user/19113 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130668/the-jantzen-schaper-theorem The Jantzen-Schaper theorem Chris Bowman 2013-05-15T05:14:22Z 2013-05-15T23:17:10Z <p>Does anybody have an electronic copy of Schaper's PhD thesis:</p> <p>K.D. SCHAPER, ‘Charakterformeln fur Weyl-Moduln und Specht-Moduln in Primcharacteristik’, Diplomarbeit, Bonn, 1981.</p> <p>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.</p> http://mathoverflow.net/questions/119640/reference-for-clifford-theory-of-algebras Reference for Clifford theory (of algebras) Chris Bowman 2013-01-23T10:19:31Z 2013-01-23T13:23:45Z <p>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.</p> <p>Theorem: Clifford theory</p> <p>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:</p> <p>(i) ${\rm res}^G_N(M)$ is a semisimple $kN$-module, and is isomorphic to a direct sum of conjugates of $L$</p> <p>(ii) the $kN$-homogenous components of ${\rm res}^G_N(M)$ are permuted transitively by $G$.</p> <p>(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 ${\rm res}^G_N(M)$.</p> <p>$\$</p> <p>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 $$A \rtimes G = \textbraceleft \sum_{x \in G} a_x x : a \in A \textbraceright.$$ </p> <p>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.</p> http://mathoverflow.net/questions/117943/littlewood-richardson-rule-and-seminormal-basis-of-specht-modules Littlewood Richardson rule and seminormal basis of Specht modules Chris Bowman 2013-01-03T11:07:40Z 2013-01-03T16:00:58Z <p>Background </p> <p><strong>Seminormal Basis of Specht modules of $\mathfrak{S}_n$</strong></p> <p>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.</p> <p>Let $1\leq i &lt; j\leq n$, 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 $a(i,j)=(i_0-i_1) -(j_0-j_1)$.</p> <p>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. </p> <p>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*}<br> {\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}<br> \end{align*}</p> <p>This basis is very compatible with induction and restriction rules (see \emph{Seminormal representations of Weyl groups and Iwahori-Hecke algebras}, Arun Ram).</p> <p><strong>The Littlewood--Richardson rule</strong></p> <p>The LR rule describes the coefficients in the restriction $$\mathbf{S}(\nu)\downarrow_{\mathfrak{S}_{r_1}\times \mathfrak{S}_{r_2}} \cong \oplus c^{\nu}_{\lambda,\mu} \mathbf{S}(\lambda) \boxtimes \mathbf{S}(\mu)$$</p> <p>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 $\mathfrak{t}$.</p> <p>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.</p> <p><strong>Question:</strong></p> <p>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.</p> http://mathoverflow.net/questions/95499/cell-modules-for-type-d-weyl-group Cell modules for type D Weyl group Chris Bowman 2012-04-29T11:16:36Z 2012-04-29T11:16:36Z <p>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 $B$?</p> <p>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). </p> <p>Thanks!</p> http://mathoverflow.net/questions/86520/cohomology-hg-k-of-wreath-products Cohomology $H^*(G,K)$ of wreath products Chris Bowman 2012-01-24T09:53:58Z 2012-01-24T16:25:47Z <p>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 $p$.</p> <p>What is $H^*(G,k)$? </p> <p>If $i \leq p-3$ and we're in the symmetric group case, then $H^i(Sym(n), k)=0$. </p> <p>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.</p> <p>Is anything else possible to say? What I REALLY want is that the symmetric group result generalises so that:</p> <p>If $i \leq p-3$, then $H^i(G, k)=0$ for $i \leq p-3$, for $G$ a wreath product as above.</p> <p>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.</p> http://mathoverflow.net/questions/80394/quasi-hereditary-covers Quasi-hereditary covers Chris Bowman 2011-11-08T15:19:42Z 2011-11-08T15:19:42Z <p>What is the quasi-hereditary cover of a quasi-hereditary algebra?</p> <p>Is the algebra A somehow (always) a quasi-hereditary cover of itself? Or is their a relation to Ringel duality in some way?</p> http://mathoverflow.net/questions/130668/the-jantzen-schaper-theorem/130773#130773 Comment by Chris Bowman Chris Bowman 2013-05-17T09:17:38Z 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? http://mathoverflow.net/questions/130668/the-jantzen-schaper-theorem/130773#130773 Comment by Chris Bowman Chris Bowman 2013-05-16T01:37:58Z 2013-05-16T01:37:58Z Sorry $\Delta=V$, I swapped notation half-way through http://mathoverflow.net/questions/130668/the-jantzen-schaper-theorem/130773#130773 Comment by Chris Bowman Chris Bowman 2013-05-16T01:37:12Z 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 $\sum_{i&gt;0}V^i(4,1)$ 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? http://mathoverflow.net/questions/119640/reference-for-clifford-theory-of-algebras/119656#119656 Comment by Chris Bowman Chris Bowman 2013-01-24T11:31:08Z 2013-01-24T11:31:08Z That's perfect. Thank you very much. http://mathoverflow.net/questions/95499/cell-modules-for-type-d-weyl-group Comment by Chris Bowman Chris Bowman 2012-05-01T18:31:07Z 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 &quot;internal construction&quot; of the &quot;Specht&quot; modules is in Gecke's paper &quot;Hecke algebras of finite type are cellular&quot;. http://mathoverflow.net/questions/95499/cell-modules-for-type-d-weyl-group Comment by Chris Bowman Chris Bowman 2012-04-30T07:42:48Z 2012-04-30T07:42:48Z Oops. No, I just want a basis of the &quot;Specht modules&quot; - or cellular basis of the Hecke algebra (at q=1). http://mathoverflow.net/questions/80394/quasi-hereditary-covers Comment by Chris Bowman Chris Bowman 2011-11-08T17:00:00Z 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 &quot;reputation points&quot; but don't understand how this works.