# Algebraic groups without torsors

If $G$ is an algebraic group such that $H^1(S, G) = 0$ for all schemes $S$, must $G$ be the trivial group?

My original motivation for the question is the rationale I always give students for studying stacks: when moduli problems involve nontrivial automorphisms, a fine moduli space cannot exist. The reasoning is that if $X$ is an object of a moduli problem $M$ admitting an automorphism group $G$ then one gets an embedding of the classifying stack $\mathrm{B}G$ of $G$-torsors inside of $M$. Two distinct $G$-torsors over a base $S$ give distinct maps to $\mathrm{B}G$ and hence to $M$ that are pointwise identical and constant, hence must be the same in any purported fine moduli space for $M$. This argument requires a nontrivial $G$-torsor to work.

It is not difficult to show that a nontrivial $G$-torsor exists for any disconnected group, and there are many familiar examples for connected groups. On the other hand, the analogous question for torsors on $C^\infty$ manifolds under Lie groups has a negative answer, since maps from $X$ to the Lie group $\mathbf R$ correspond to sections over $X$ of $C^\infty(X)$, and $H^1(X, C^\infty(X)) = 0$ because of partitions of unity.

• If you want to produce a nontrivial torsor, it suffices to produce a nontrivial torsor after base change to the algebraic closure of a residue field of the base scheme. So now you can use the structure theory of algebraic groups. For multiplicative groups, $\mathbb{A}^{n+1}\setminus\{0\}\to \mathbb{P}^n$ is a nontrivial torsor. For additive groups, there is a nontrivial torsor over $\mathbb{A}^2\setminus\{0\}$. For semisimple groups, the $G$-torsor induced over $\mathbb{P}^1$ by a nontrivial torsor for the maximal torus is nontrivial, cf. Grothendieck, et al. – Jason Starr Oct 30 '16 at 19:28
• The notion of "automorphism group" and "automorphism scheme" seem to be conflated in the question. Anyway, a way to convey the relevance of rigidity is not to bring in BG (using a stack to motivate the need to consider stacks?), but to observe that the utility of universal objects over fine moduli schemes rests on cartesian diagrams between objects being uniquely determined up to unique isomorphism (upstairs) by maps along the base. In the presence of non-trivial automorphisms over the base that breaks down, so the framework is bad and must be replaced with something better, ergo stacks. – nfdc23 Oct 30 '16 at 20:27
• @nfdc23: I certainly agree, but what I really want is to explain why it is important to distinguish "unique up to isomorphism" from "unique up to unique isomorphism", the point being that non-unique isomorphisms allow you to construct twists by torsors. Of course, I wouldn't actually invoke $BG$ to motivate algebraic stacks to the uninitiated, but it seemed harmless to use it here---the question wasn't actually about pedagogy. – Jonathan Wise Oct 30 '16 at 20:44

I am just posting my comment as an answer. If you want to produce a nontrivial torsor, it suffices to produce a nontrivial torsor after base change to the algebraic closure of a residue field of the base scheme. So now you can use the structure theory of algebraic groups. For multiplicative groups, $\mathbb{A}^{n+1}\setminus\{0\} \to \mathbb{P}^n$ is a nontrivial torsor. For additive groups, there is a nontrivial torsor over $\mathbb{A}^2\setminus\{0\}$. For semisimple groups, the $G$-torsor induced over $\mathbb{P}^1$ by a nontrivial torsor for the maximal torus is nontrivial, cf. Grothendieck, et al.

Edit. At the request of Jonathan Wise, here are some references.

MR0087176 (19,315b)
Grothendieck, A.
Sur la classification des fibrés holomorphes sur la sphère de Riemann. (French)
Amer. J. Math. 79 (1957), 121–138.
53.3X

MR0263826 (41 #8425)
Harder, Günter
Halbeinfache Gruppenschemata über vollständigen Kurven. (German)
Invent. Math. 6 (1968), 107–149.
14.50

Jonathan Wise also asks how to deduce nontriviality of torsors from nontriviality for factor torsors. Let $\phi:G\to L$ be a homomorphism of algebraic groups. For every $G$-torsor $E_G$, the quotient of $E_G \times L$ by the "diagonal" action of $G$ is an $L$-torsor $E_L$. If $E_G$ has a section, then so does $E_L$.

First, consider the case that $\phi$ is a surjective (thus flat) homomorphism from a solvable group to a multiplicative group over an algebraically closed field. By the structure theory of algebraic groups, there exists a closed subgroup $L'\subset G$ such that $\phi:L'\to L$ is an isogeny. The group of $L$-torsors over $\mathbb{P}^1$ is a free Abelian group. Up to multiplying the torsor by a suitably positive and divisible integer, it is induced from an $L'$-torsor. For the inclusion homomorphism $L'\subset G$, there is an associated $G$-torsor $E_G$ whose $L$-torsor $E_L$ is a multiple of the original $L$-torsor over $\mathbb{P}^1$. In particular, if the original $L$-torsor is nontrivial, then so is the $G$-torsor.

Next, consider the case that $\phi:G\to L$ is a surjective (thus flat) homomorphism from $G$ to a reductive group whose kernel is solvable, over an algebraically closed field. Let $T\subset L$ be a maximal torus. Let $\phi^{-1}(T)\subset G$ be the inverse image. By the previous case, after multiplying by a sufficiently positive and divisible integer, every nontrivial $T$-torsor over $\mathbb{P}^1$ lifts to a $\phi^{-1}(T)$-torsor, which then induces a $G$-torsor over $\mathbb{P}^1$. By Grothendieck, if the original $T$-torsor is nontrivial, then so is the $L$-torsor associated to this $G$-torsor. Thus the $G$-torsor is also nontrivial.

The only remaining case is when $G$ itself is a unipotent group over an algebraically closed field. A bit more generally, assume that $G$ is any non-reductive group. Let $G\hookrightarrow L$ be any closed embedding of $G$ into a reductive group. Then $L/G$ is not affine by Matsushima-Richardson. If the $G$-torsor $L$ over $L/G$ were split, then the affine variety $L$ would be isomorphic to $G\times (L/G)$, and this is not affine.

Edit. I should also say, most of what I know about this I learned from the excellent articles of Johan Martens and Michael Thaddeus. Their proof of the Grothendieck-Harder theorem is very clear and geometric.

• What is the reference to Grothendieck, et al.? Is this SGA3? What about extension groups: if G is an extension of G'' by G', and both G' and G'' possess nontrivial torsors over some bases, is there a reason G should have a nontrivial torsor? – Jonathan Wise Oct 31 '16 at 18:23
• Terrific! Thanks for the great answer. – Jonathan Wise Nov 1 '16 at 14:57