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Let F be a field of characteristic $p > 0$. Let $g$ be a linear lie algebra, that is there exists a natural number $n$ such that $g \subset M_n(F)$. Does there exist a condition involving $n$ and $p$ such that $g$ is semisimple if and only if its killing form is non-degenerate?

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2 Answers 2

The short answer is no. In prime characteristic, the Killing form sometimes behaves badly even for simple Lie algebras. If "semisimple" means that the solvable radical is zero, there is no way to obtain the classical equivalences with non-degeneracy of the Killing form and with the direct sum decomposition into simples. Moreover, the simple Lie algebras have only recently been classified when $p=5$ (Premet-Strade), while for $p=2,3$ little is known and for $p>5$ the classification takes an enormous amount of work (by Block-Wilson and others). Conditions on the dimension of a faithful representation or on the prime are not enough to sort out the concept of semisimplicity. Still, a lot is known. For example, Seligman and others explored in the 1960s the class of modular Lie algebras for which the Killing form is non-degenerate.

ADDED: Much more could be said along these lines, but for older results see the book Modular Lie Algebras by G.B. Seligman (Springer, 1967). Modular Lie algebras have been of much less importance overall than linear algebraic groups, whose Lie algebras are "restricted" (have a nice $p$th power operation) but still don't reflect precisely the group structure or representation theory. Moreover, the representations of simple or more generally semisimple Lie algebras in prime characteristic are poorly understood even for those arising from Lie algebras of semisimple algebraic groups. So working with a fixed $p$ and fixed $n$ will usually not be illuminating.

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Thanks a lot for your reply Jim. Although it was bit disheartening to know that very little is known about this problem. –  Pooja Nov 21 '10 at 19:17
    
In some ways a lot is actually known, but your original formulation doesn't get at that. While enormous progress has been made on finding all simple Lie algebras (starting with the restricted ones), far less is known about their linear representations. The notion of semisimple in prime characteristic still looks too opaque to be explored systematically, using the range of tools now available. –  Jim Humphreys Nov 22 '10 at 23:03
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This started out as a comment on Jim's answer, but it got too long (EDIT : It looks like Jim edited his answer to say some of this while I wrote this up).

First, I recommend Seligman's book "Modular Lie Algebras" for a nice account of what happens to Lie algebras in positive char if you assume that the Killing form is nondegenerate.

It's maybe also worth giving an example to show what can happen (there might be easier ones, but this came up in a paper I wrote a while ago and gave me no end of headaches). Consider $\mathfrak{sp}_{2g}(\mathbb{Z}/p\mathbb{Z})$. It is not hard to show that this is simple for $p > 2$. For instance, an argument is contained in the paper

N. Jacobson, Classes of restricted Lie algebras of characteristic p. I, Amer. J. Math. 63 (1941), 481–515.

By the way, this is false for $p=2$. Anyway, one can calculate the Killing form, and it turns out to be degenerate exactly when $p$ divides $g+1$. One interesting observation, however, is that we are really looking at the wrong bilinear form. Elements of $\mathfrak{sp}_{2g}(\mathbb{Z}/p\mathbb{Z})$ are matrices, so we can define a blinear form

$$(A,B) = \text{Tr}(AB).$$

This is NOT the Killing form, as we are not looking at the adjoint representation. One can then show that this bilinear form is nondegenerate for all $p>2$.

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In characteristic 0, this form would just be off by a factor from the Killing form, right? Something like 4(g+1), maybe. The factor of 4 would also explain why it always fails for p=2. –  Keerthi Madapusi Pera Nov 21 '10 at 16:49
    
That's right. The joy of characteristic 0 is that annoying things like that don't happen. –  Andy Putman Nov 21 '10 at 16:52
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Let's take semisimple to mean that the solvable radical is zero. Then the usual proof in characteristic $0$ goes through in characteristic $p$ to show that if a Lie algebra $L$ has a nondegenerate bilinear form with appropriate invariance properties (similar to the Killing form), then $L$ is semisimple. The converse, alas, is false, but in many examples you can find good bilinear forms like the one I mentioned above. Seligman's book has good information on this (though since it is old I expect that more is known). –  Andy Putman Nov 21 '10 at 20:09
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(continued) I don't have Seligman's book handy, but in the paper I mentioned in my answer I refer to page 47 of it, so that is probably a good place to start looking. Be warned, however, that semisimplicity doesn't work in char p like it does in char 0; for instance, semisimple Lie algebras in char p are not necessarily direct sums of simple Lie algebras. –  Andy Putman Nov 21 '10 at 20:13
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For Lie algebras with a nondegenerate trace form, the last two sections of Seligman's Chapter II summarize what was known at the time based mainly on work by Zassenhaus and Block. Typically such a Lie algebra over an algebraically closed field of characteristic >3 is close to being classical, while exceptions in classical cases are spelled out. I don't think much else is known even now, though there was at the time an open question about existence of a nondegenerate trace form for type $E_8$ when $p=5$. (Is it still open?) "Semisimple" remains hard to pin down completely –  Jim Humphreys Nov 21 '10 at 23:03
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