Are there "reasonable" criteria for existence/non-existence of Levi factors or their conjugacy in prime characteristic?

Question

Classical theorems attributed to Levi, Mal'cev, Harish-Chandra for a finite dimensional Lie algebra over a field of characteristic 0 state that it has a Levi decomposition (semisimple subalgebra plus solvable radical) and that all such semisimple subalgebras (Levi factors) are conjugate in a strong sense: see Jacobson, Lie Algebras, III.9, for example. This carries over to connected linear algebraic groups, but in prime characteristic there are counterexamples going back perhaps to Chevalley that involve familiar group schemes like $SL_2$ over rings of Witt vectors. Recent posts here have somewhat ignored that difficulty, having just characteristic 0 in mind. Borel and Tits redefined "Levi factor" to be a reductive complement to the unipotent radical, which is makes no real difference in characteristic 0 but allows them to concentrate on positive answers for parabolic subgroups of reductive groups in general. Other familiar subgroups of reductive groups like the identity component of the centralizer of a unipotent element require much more subtle treatment, as in work of George McNinch.

Whether or not the characteristic $p$ question is important, it has remained open for many decades (say over an algebraically closed field). I gave up after one forgettable paper (Pacific J. Math. 23, 1967). The problem is still easy to state:

Are there effective necessary or sufficient conditions for existence or uniqueness of Levi factors in a connected linear algebraic group over an algebraically closed field of prime characteristic?

It's clear that a scheme-theoretic viewpoint may be needed. Possibly the known counterexamples using Witt vectors suggest in some way all possible counterexamples? (Or is the question hopeless to resolve completely?)

EDIT: For online access to my 1967 paper, via Project Euclid, see Link. Here Chevalley's counterexample is mentioned only in the abstract, but in remarks later on it is noted that Borel-Tits (III.15) gave an example involving two Levi subgroups which fail to be conjugate; see NUMDAM link to PDF version of Publ. Math. IHES 27 (1965) at http://www.numdam.org:80/?lang=en

In April 1967 Tits responded to my inquiry with a letter outlining the behavior of the group scheme $SL_2$ over the ring of Witt vectors of length 2, which gives a 6-dimensional algebraic group over the underlying field with unipotent radical of dimension 3 but no Levi factor. He remarked that he got this counterexample from P. Roquette but had also been told about Chevalley's counterexample.

ADDED: The question as formulated probably doesn't have a neat answer, but meanwhile George McNinch has delved much deeper (over more general fields) in his new arXiv preprint 1007.2777. Some technical steps rely on the forthcoming book Pseudo-reductive groups (Cambridge, 2010) by Conrad-Gabber-Prasad.

Answer 1

[See Edit below.]

This isn't really an answer, but I believe it is relevant.

Work geometrically, so $k$ is alg. closed. Let $G$ reductive over $k$, and let $V$ be a $G$-module (linear representation of $G$ as alg. gp.).

If $\sigma$ is a non-zero class in $H^2(G,V)$, there is a non-split extension $E_\sigma$ of $G$ by the vector group $V$ -- a choice of 2-cocyle representing $\sigma$ may be used to define a structure of alg. group on the variety $G \times V$. Here "non-split" means "$E_\sigma$ has no Levi factor".

And if $H^2(G,V) = 0$, then any $E$ with reductive quotient $G$ and unipotent radical that is $G$-isomorphic to $V$ has a Levi factor.

You can look at the $H=\operatorname{SL}_2(W_2(k))$ example from this viewpoint; $H$ is an extension of $\operatorname{SL}_2$ by the first Frobenius twist $A = (\mathfrak{sl}_2)^{[1]}$ of its adjoint representation. Of course, this point of view doesn't really help to see that $H$ has no Levi factor; the fact that $H^2(\operatorname{SL}_2,A)$ is non-zero only tells that it might be interesting (or rather: that there is an interesting extension). The extension $H$ determines a class in that cohomology group, and the argument in the pseudo-reductive book of Conrad Gabber and Prasad -- or a somewhat clunkier representation theoretic argument I gave some time back -- shows this class to be non-zero, i.e. that $H$ has no Levi factor.

So stuff you know about low degree cohomology of linear representations comes up. And this point of view can be used to give examples that don't seem to be related to Witt vectors.

A complicating issue in general is that there are actions of reductive $G$ on a product of copies of $\mathbf{G}_a$ that are not linearizable, so one's knowledge of the cohomology of linear representations of $G$ doesn't help...

Edit: It isn't clear I was correct last April about that "complicating issue". See this question.

Also: the manuscript arXiv:1007.2777 includes a "cohomological" construction of an extension $E$ of SL$_3$ by a vector group of dim $(3/2)(p-1)(p-2)$ having no Levi factor in char. $p$, and an example of a group having Levi factors which aren't geometrically conjugate.