# Mod-p cohomology of $GL(n,p^d)$

In the classic paper On the Cohomology and K-Theory of the General Linear Groups Over a Finite Field, Quillen proved (Theorem 6):

$H^i(GL(n,p^d),\mathbb{F}_p)=0$ for $0 < i < d(p-1)$ and all $n$.

On the other hand, the cohomology of a finite group doesn't completely vanish (this is nicely discussed on MO: Non-vanishing of group cohomology in sufficiently high degree). So there is a minimal integer $i_0=i_0(n,d) \ge d(p-1)$ satisfying $$H^{i_0}(GL(n,p^d),\mathbb{F}_p) \neq 0$$ Is Quillen's lower bound $d(p-1)$ sharp ? (I couldn't find any information about sharpness in Quillen's paper). If not, is the precise value of $i_0$ known ?

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There are some results known in this direction; see the paper On the vanishing ranges for the cohomology of finite groups of Lie type, by Christopher Bendel, Daniel Nakano, and Cornelius Pillen (Int. Math. Res. Not. (2012), 2817-2866). In particular, they show that if $p \geq n+2$, then the first degree in which nonzero cohomology appears is $d(2p-3)$.
Thanks for your answer. I don't have access to the journal, but I there is a paper with same title and result on the arxiv: arxiv.org/pdf/0906.0026.pdf. There the role of $n$ is unclear to me: Isn't $SL_n$ a simple (i.e. finite center) simply connected algebraic group scheme of type $A_{n-1}$ ? Then, according to 6.2 Corollary, $i_0(SL(n,p^d))= d(2p-3)$, provided $p\ge (n-1)+2 = n+1$ and from the Hochschild-Serre spectral sequence, it would follow $i_0(GL(n,p^d))=d(2p-3)$ if $p \ge n+1$ (instead of $p\ge n+2$ in the paper). Am I missing something ? – Demin Hu Feb 24 '13 at 13:08
@Demin: In Corollary 6.2 (both arXiv and IMRN) they deal with Lie type $A_n$ and the bounds $n \geq 2, p \geq n+2$; this applies to $G=\mathrm{SL}_{n+1}$. It's confusing, but probably consistent throughout. I haven't checked the fine points, but did note that their misprint "indentifies" at the start of Section 6 got changed to "identifies" in IMRN. There may be other minor changes like that. @Christopher: Your use of $n$ is misleading, since the paper uses this for the Lie rank of a simple, simply connected group. – Jim Humphreys Feb 24 '13 at 20:46
Jim, thanks for your help. @Christopher: Maybe it's a typo, but it's peculiar that $p \ge n+2$ is required in 6.15 in the proposition and in the theorem as well. But of course, the assumption $p \ge n+2$ doesn't contradict my guess that $p\ge n+1$ is sufficient. – Demin Hu Feb 25 '13 at 0:49