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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.

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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
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1 Answer

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.

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