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

This is related to my question Adelic description of moduli of $G$-bundles on a curve.

Let $X$ be a smooth, projective, and geometrically connected curve over a finite field $\mathbb{F}_q$ and $G$ a smooth connected affine algebraic group over $\mathbb{F}_q$. Under what conditions on $G$ does any $G$-bundle on $X$ trivialize over a Zariski open subset of $X$? A priori this can only be done over an étale cover of $X$.

In the accepted answer to the aforementioned question it is claimed that this holds for simply connected and semisimple $G$ by a theorem of Harder. Does anyone have a reference for this result, preferably in English or French? What about e.g. $G = PGL_2$? Non-split tori? Unipotent groups? I would love to see either positive results or counterexamples.

share|cite|improve this question
Any smooth connected unipotent group over a perfect field is filtered by $\mathbf{G}_a$'s, so the unipotent case is OK by additive Hilbert 90 for the function field $K$ of $X$. Since Br($X$) vanishes (CFT for $K$), the map $H^1(X,GL_n) \rightarrow H^1(X,PGL_n)$ is surjective and identifies the target with the quotient of the set of (isom. classes of) rank-$n$ vector bundles modulo line-bundle twisting, so $H^1(X,PGL_n)$ is infinite for $n > 1$, and $H^1(K,PGL_n)={\rm{Br}}(K)[n]$ is too (CFT). But $H^1(X,PGL_n)\rightarrow H^1(K,PGL_n)$ vanishes since $H^1(K,GL_n)=1$, so $PGL_n$ is OK too. – user29283 Mar 30 '13 at 18:27
Serre's Galois Cohomology textbook. Read the chapter on "Conjecture II". – Jason Starr Mar 30 '13 at 20:04
@Jason: That chapter doesn't seem to provide a non-German reference for the prof of Harder's theorem, and also doesn't seem to address whether the map $H^1(X,G) \rightarrow H^1(K,G)$ is nontrivial (which is what it seems the OP is asking about, weaker than triviality of $H^1(K,G)$). What sort of relevant information can be gleaned from that part of Serre's book? For example, knowing $H^1(K,G) \ne 1$ for some $G$ arising from the constant field doesn't seem to help to "spread out" a nontrivial $G$-torsor over $K$ to a $G$-torsor over the entirety of $X$ (e.g., even for $G$ a torus). – user29283 Mar 30 '13 at 21:10
@Xuhan. Regarding Zariski local triviality versus rational points, this follows from Nisnevich's work on the Grothendieck-Serre conjecture. – Jason Starr Mar 30 '13 at 21:29
@Jason:Ah, OK, so by Nisnevich, the question posed is exactly asking about the non-triviality of $H^1_{\rm{Zar}}(X,G)$. But I don't think Serre's book addresses this sort of thing (or am I mistaken?), and there are so few Zariski-exact sequences of interesting algebraic groups that it isn't clear (to me) how one gets a handle on this $H^1$. Do you know any connected reductive $G$ over $k = \mathbf{F}_q$ for which $H^1_{\rm{Zar}}(X,G)$ is nontrivial? – user29283 Mar 31 '13 at 3:58

Clearly Harder's theorem implies that $H^1(K,G) = 1$ for any reductive $G$ such that $[G,G]$ is simply connected.

One can combine this with the argument in xuhan's comment to see that the result holds for any split reductive $G$, given the following: there exists a central extension $\widetilde{G}$ of $G$ by a connected (split) torus $T$ such that $[\widetilde{G},\widetilde{G}]$ is simply connected. This can be proved using root data.

Then, following xuhan, we see that $H^1(X,G) = H^1(X,\widetilde{G})/H^1(X,T)$ because $H^2(X,T) = 0$ by class field theory. But, as noted above, $H^1(K,\widetilde{G}) = 1$, so the map $H^1(X,G) \to H^1(K,G)$ vanishes as desired.

I guess the remaining cases at this point are non-smooth unipotent groups and nonsplit tori.

share|cite|improve this answer

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.