Skip to main content
6 events
when toggle format what by license comment
Apr 2, 2014 at 5:20 vote accept Jared
Mar 26, 2014 at 7:36 comment added Jeremy Rickard The fact that the trivial module has maximum complexity is fairly elementary, since a (not necessarily minimal) projective resolution of any module $M$ can be obtained by taking the tensor product of $M$ with a minimal projective resolution of $k$. The fact that the complexity of $k$ is the Krull dimension of the cohomology ring probably does need at least the beginnings of the theory of support varieties.
Mar 25, 2014 at 19:42 answer added Jim Humphreys timeline score: 3
Mar 25, 2014 at 19:07 comment added Jared @JeremyRickard: Do your statements follow from the fact that the complexity of $M$ is the dimension of its support variety, and the support variety of $M=k$ is the full space $\mathrm{Spec}(H^{ev}(G_{(r)},k))$? I'm just now learning about support varieties, so I'm not quite yet comfortable with them.
Mar 25, 2014 at 15:54 comment added Jeremy Rickard There's a recent paper on the arxiv by Ngo that looks at similar questions, not for $\mathrm{GL}_n$ but for simple classical group schemes, and gets lower bounds. Maybe you can adapt the methods. By the way, as you probably know, the maximum complexity is attained for $M=k$, the trivial module, and your question is equivalent to asking for the Krull dimension of the cohomology ring $H^*(G_{(r)},k)$.
Mar 25, 2014 at 2:21 history asked Jared CC BY-SA 3.0