most recent 30 from http://mathoverflow.net 2013-05-26T00:07:44Z http://mathoverflow.net/feeds/user/3058 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92316/equivariant-version-of-a-spectral-sequence-in-beilinson-ginzburg-soergel Daniel Moseley 2012-03-26T22:16:22Z 2012-03-26T22:16:22Z

<p>In Beilinson, Ginzburg, and Soergel, "Koszul Duality Patterns in Representation Theory" (comment 3.4), the authors outline a spectral sequence as follows:</p> <p>Given a filtered complex algebraic variety $X=X_0 \supset \cdots \supset X_r = \varnothing$ and $\mathcal{F} \in D(X)$, $\mathbb{H}^{\bullet} (\mathcal{F})$ is the limit of the spectral sequence with $E_1$-term $E_1 ^{p,q} = \mathbb{H}_{X_p - X_{p+1}} ^{p+q} (\mathcal{F})$.</p> <p>I believe that the analogous result is true in the T-equivariant derived category. Does anyone know of a reference where I might find such a statement?</p>

http://mathoverflow.net/questions/51239/representations-of-semidirect-products-of-symmetric-groups Daniel Moseley 2011-01-05T19:56:48Z 2011-01-05T20:26:21Z

<p>This is sort of a vague (I apologize in advance) question, but I'm interested in the representation theory of the following group</p> <p>$A \rtimes B$, where $A = (S_1)^{m_1} \times (S_2)^{m_2} \times \ldots \times (S_r)^{m_r}$, $B = S_{m_1} \times S_{m_2} \times ... \times S_{m_r}$, and $B$ acts on $A$ by permuting the factors. Is something nice known about the representation theory of these groups? Does anyone know a good reference for something like this?</p>

http://mathoverflow.net/questions/11061/computer-program-to-solve-a-system-of-polynomial-equations-over-a-finite-field Daniel Moseley 2010-01-07T19:17:08Z 2010-01-07T20:53:21Z

<p>I have a set of polynomial equations for which I want to know the solutions (actually really the number of solutions). It would be great if I could get a computer to do it, but I'm not sure exactly which programs will do things over finite fields. I surmise that Macaulay 2 may be able to do something like this, but I'm not quite sure how.</p>