MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $k$ be a finite field and $G$ a finite type smooth $k$-group scheme. Let $G^0$ and $\Gamma$ be the connected component of identity and the component group of $G$, so there is an exact sequence $1 \rightarrow G^0 \rightarrow G \rightarrow \Gamma \rightarrow 1$. Is the induced $H^1(k, G) \rightarrow H^1(k, \Gamma)$ surjective?

By Lang's theorem, $H^1(k, G^0) = *$, so the map in question is injective. It is surjective if $G^0$ is central (in particular, if $G$ is commutative), since for commutative $G^0$ one has $H^2(k, G^0) = 0$ for cohomological dimension reasons. It is also surjective if the short exact sequence splits as a semi-direct product. Thus, my question is: does the surjectivity hold in general for noncommutative smooth $G$?

share|cite|improve this question
Showing fibers have $\le 1$ point reduces (by cocycle-twisting) to ${\rm{H}}^1(k,\mathscr{G})=1$ for every $k$-form $\mathscr{G}$ of $G^0$; just for $G^0$ is insufficient. Classifying $K$-forms of ${\rm{SL}}_n$ for a general $K$ uses $G={\rm{Aut}}_{{\rm{SL}}_n/K}$, a semi-direct product of $\mathbf{Z}/(2)$ against $G^0 = {\rm{PGL}}_n$. Here, ${\rm{H}}^1(K,G^0)=1$ but the fiber of ${\rm{H}}^1(K,G)\rightarrow {\rm{H}}^1(K,\mathbf{Z}/(2))$ over separable quadratic $K'/K$ is ${\rm{SU}}(h)$'s for $(-1)^{n+1}$-hermitian $h:{K'}^n\times{K'}^n\rightarrow K'$; one is quasi-split but more can arise! – user27920 Jun 29 '14 at 21:22
As an explicit example for the end of the above comment, if $K = \mathbf{R}$ then there's the $K$-anisotropic form corresponding to the compact Lie group ${\rm{SU}}(n)$ (so the fiber of ${\rm{H}}^1(\mathbf{R}, {\rm{Aut}}_{{\rm{SL}}_n/\mathbf{R}}) \rightarrow {\rm{H}}^1(\mathbf{R},\mathbf{Z}/(2))$ over the nontrivial class in the target has more than one point even though the kernel is trivial). – user27920 Jun 29 '14 at 21:27
up vote 5 down vote accepted

This is proved when $G$ is a linear algebraic group in Corollary III.2.4.3 of Serre's book on Galois cohomology (p. 135 in my version). I believe it is only stated here for linear algebraic groups as Serre is at that point looking at general fields of cohomological dimension at most $1$. In the special case of finite fields where one has Lang's theorem, it seems that the proof generalises to your setting.

share|cite|improve this answer
Serre's Corollary 2 in 2.4 (as opposed to Corollary 3 which you mention) provides surjectivity in complete generality: for smooth groups of finite type (not just linear algebraic) over perfect fields with cohomological dimension $\le 1$. His proof of the main theorem from which the corollaries are deduced use Chevalley's theorem on the structure of the identity component beyond the affine case (over perfect fields) and the simple structure of connected unipotent groups over perfect fields. Throughout Chapter III it is tacitly assumed (as noted at the start) that the ground field is perfect. – user27920 Jun 29 '14 at 21:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.