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Oct 4, 2013 at 16:00 history edited KoopaTroopa CC BY-SA 3.0
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Oct 4, 2013 at 15:43 comment added KoopaTroopa I'm sorry, I should have written isomorphic as groups or the corresponding finite groups of Lie-Type over $\mathbb{F}_2$ are isomorphic.
Oct 4, 2013 at 15:42 history edited KoopaTroopa CC BY-SA 3.0
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Sep 28, 2013 at 16:54 answer added Jim Humphreys timeline score: 5
Sep 28, 2013 at 15:01 comment added Marguax SO$_{2n+1}$ and Sp$_{2n}$ are not isomorphic as algebraic groups in characteristic 2 (the natural isogeny between them in characteristic 2 has nontrivial unipotent infinitesimal kernel). Their groups of rational points (over a perfect field) are identified via that isogeny, but that is weaker than algebraic isomorphism.
Sep 28, 2013 at 14:16 comment added KoopaTroopa There is this paper from Frank Lübeck link and I have computed the composition factors of the tensorproducts of the natural module. If $n \leq 6$, I got the desired representation this way. however, if $n \geq 7$, the dimensions get to big, to compute the factors. I think, the representations I'm looking for correspond somehow, becaus of the duality in characteristic $2$ of $B_n$ and $C_n$, to Spin-representations of the orthogonal groups $\mathrm{SO}_{2n+1}$ wich are isomorphic to $\mathrm{Sp}_{2n}$ in even char..
Sep 28, 2013 at 13:44 comment added Marguax In positive characteristic it is generally difficult to determine the dimension of higher-weight representations (could be smaller than in characteristic 0); how do you know for that particular $\lambda$ that the dimension is $2^n$ for characteristic 2?
Sep 28, 2013 at 11:25 review First posts
Sep 28, 2013 at 11:28
Sep 28, 2013 at 11:06 history asked KoopaTroopa CC BY-SA 3.0