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This question may be trivial to people with the right background, but I do not see the answer.

Let $\Bbbk$ be an algebraically closed field. Can any one-dimensional group variety (over $\Bbbk$) act transitively on $\mathbb{P}^1_{\Bbbk}$?

Here is my reasoning so far:
-Every $g \in G$ fixes a point of $\mathbb{P}^1$ since the associated linear map has an eigenvector.
-Let $G_0$ be the identity component of $G$. If $G_0$ fixes a point $p$, then $G/G_0$ surjects onto the orbit of $p$, so $p$ has finite $G$-orbit. Hence $G$ does not act transitively. So, it suffices to assume $G$ is connected and show that it fixes a point.
-I think (but am not confident here) that the only one-dimensional connected group varieties are $(\Bbbk, +)$, $(\Bbbk^*, \cdot)$, and abelian varieties (i.e., elliptic curves). The last are not an issue since any morphism $E \to PGL_2$ is constant, where $E$ is complete and $PGL_2$ is affine.
-Since $G$ is one-dimensional irreducible and the isotropy subgroup of a point is closed, to show that $G$ fixes $p$, it suffices to show that infinitely many elements of $G$ fix $p$.

The last step, I think I can do for the multiplicative group, and for the additive group if $\Bbbk$ has characteristic zero, but my basic line of approach does not seem to work for the additive group in characteristic $p$. I also am less than completely confident of the classification I've stated for one-dimensional connected group varieties; if someone could point me to a good reference for this, I would appreciate it.

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Humphreys-Linear Algebraic Groups Ch. 20 is called "One dimensional Groups" (affine only). G commutative with transitive orbit on X implies X=G/H, H comm, so H normal so X is comm. alg. group. P^1 is not a comm. alg. gp. –  Peter McNamara Mar 24 '11 at 0:29

1 Answer 1

up vote 9 down vote accepted

The case of a 1-dimensional connected affine group (which in particular is solvable) is settled most easily by invoking the Borel Fixed Point Theorem: a connected solvable (affine) group acting as an algebraic group on an irreducible projective variety has a fixed point. If the given "group variety" is an abelian variety, of course, you have to argue separately. Similarly for a non-connected affine group, in which the connected component of the identity has a fixed point.

ADDED: The question here involves only the 1950s work on algebraic groups ("group varieties") by Borel, Chevalley, Rosenlicht, ... By now parts of the theory have been streamlined, but anyway it's a little artificial to extract from the structure theory just the minimal amount of information you need for your purpose. However it's done you probably need to distinguish somewhere between characteristic 0 and characteristic $p$. Here is a more detailed outline of what might be argued:

1) It's useful to reduce the problem first to the case where $G$ is connected, which is an elementary step.

2) To handle the case when $G$ is not affine (thus an abelian variety of dimension 1), you implicitly rely on Chevalley's deep structure theorem on algebraic groups which was recently revisited by Brian Conrad: A modern proof of Chevalley’s theorem on algebraic groups, J. Ramanujan Math. Soc. 17 (2002), no. 1, 1–18.

3) When $G$ is an abelian variety, it has no everywhere-defined regular functions (like other complete varieties) and thus can't have a nontrivial algebraic group morphism to an affine group such as $PGL_2$, which is shown concretely to be the automorphism group of the projective line (6.4 in my book).

4) If $G$ is connected and affine of dimension 1, it is commutative: see 20.1 in my book, which doesn't yet get into the precise classification of such groups.

5) To finish the argument as Peter suggests you need something like Lemma 21.1 in my book, which is the key step toward the Borel Fixed Point Theorem in 21.2. (That theorem is relatively easy to prove after some preparation, but is the key to further deep structure theory for reductive groups.) In Peter's sketch, you get only a bijective morphism from $G/H$ to the projective line, not necessarily an isomorphism; so 21.1 is needed in characteristic $p$. For me the best conceptual viewpoint is found in the fixed point theorem, which of course applies much more generally to actions of connected solvable groups on projective (or complete) varieties.

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