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

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 http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pjm/1102991730. 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.

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Jim, I recently went back to some of those old posts to insert warnings about lack of Levi in char > 0 (in case anyone looks at those again). Is there published reference for Chev's examples? (In Borel's alg gps book he mentions examples exist but nothing more.) Gopal and I didn't know ref, so we put discussion of G(W_2(k)) for reductive G in appendix to the p-red book. We may have chance to insert literature reference in final version, so please let me know about possible ref. asap. The next comment will address your question. –  BCnrd Apr 21 '10 at 22:13
This is definitely a direction where exploration is needed, though I can't supply it. The older work on linear groups doesn't help (I only got the example in my ancient paper from Tits, who may have gotten it from Chevalley). In particular, my 1967 paper doesn't lead onward at all, though it is freely available online as other Pacific Journal papers are. Where I tried to use Lie algebras, the Frobenius kernels would be the obvious substitute to exploit for arbitrary powers of the prime. –  Jim Humphreys Apr 21 '10 at 22:56
Over $k$ admitting a degree $p^2$ extension $K$ with height $\le 1$ (such as finitely generated extn of F_p with trdeg $> 1$) Ofer found likely examples of nontrivial central extension $G$ of Res_{K/k}(SL_n) by $\mathbf{G}_ a$. That Weil restriction is perfect (by hand or because SL_n is simply connected) so the central extension structure cannot split over $\overline{k}$. It would be worthwhile to see if such an example (if it really exists) may fail to have Levi over $\overline{k}$; I have not thought about it seriously. – BCnrd 11 secs ago –  BCnrd Apr 21 '10 at 23:27

[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.

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I agree that examples unrelated to Witt vectors might occur. There may even be examples which look entirely artificial, but of course it would be much nicer if instead they all arose naturally in a scheme-theoretic framework. Your last comment is important to keep in mind, since work from about 1989 by Gerry Schwarz and others has made it clear that some reductive group actions on affine space can't be linearized. Anyway, low degree cohomology is a natural tool if one can learn enough about it. –  Jim Humphreys Apr 23 '10 at 12:28
@Jim: right - when I wrote yesterday I couldn't remember "non-linearizable" examples. –  George McNinch Apr 23 '10 at 14:09