Question

Harold Williams, Pablo Solis, and I were chatting and the following question came up.

In Lie group land (where you're doing differential geometry), given a finite-dimensional Lie algebra g, you can find a faithful representation g → End(V) by Ado's theorem. Then you can take the group generated by the exponentiation of the image to get a Lie group G⊆GL(V) whose Lie algebra is g. I think this is correct, but please do tell me if there's a mistake.

This argument relies on the exponential map, which we don't have an the algebraic setting. Is there some other argument to show that any finite-dimensional Lie algebra g is the Lie algebra of some algebraic group (a closed subgroup of GL(V) cut out by polynomials)?

Answer 1

A Lie subalgebra of gl(n,k) which is the Lie algebra of an algebraic subgroup of GL(n,k) is called an algebraic subalgebra. Apparently there are Lie subalgebras which are not algebraic, even in characteristic zero. If g is the Lie algebra of an affine algebraic group then it must be ad-algebraic, ie. its image in End(g) under the adjoint representation must be an algebraic subalgebra. An example of a non-ad-algebraic Lie algebra is given on pg. 385 of Lie Algebras and Algebraic Groups, by Tauvel and Yu.

Answer 2

If a Lie subalgebra of gl(V) is the Lie algebra of an algebraic group, then it contains the semisimple and nilpotent factors of any element. There is a five-dimensional Lie algebra for which this fails, which you can find in Bourbaki (or on p153 of my notes www.jmilne.org/math/CourseNotes/ala.html )

Answer 3

I recommend http://eom.springer.de/l/l058380.htm .

Answer 4

My suspicion is yes, at least over C, and that the thing to do is take the Zariski closure in GL(V) of the exponentials of the Lie algebra elements. Of course, over random fields, one doesn't have this trick.

Might a trick like looking at the subgroup of GL(V) fixing all invariant polynomials for the Lie algebra work?

Answer 5

Over a field that's not C --- well, at least in non-zero characteristic --- , the problem is deeper than Ben suggests: the proof of Ado's theorem that I know requires characteristic zero. I think if the theorem were true in non-zero characteristic Mark Haiman would have said so --- he seemed to suggest in his class that it was not.

Incidentally, you don't really need Ado's theorem, which includes data about the action of the nilpotency ideal, just to find a faithful action. Levi's theorem splits any lie algebra as semisimple semi-direct solvable, and this is enough to find a faithful finite-dimensional representation.

Also, even with Ado's theorem, there's a warning. The Zariski closure, and indeed even the analytic closure, of the image of the exponential might have higher dimension. E.g. the irrational line in the torus.