Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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)?

share|improve this question
2  
A very nice fact over fields $k$ of char. 0: for any linear algebraic $k$-group G and Lie $k$-subalgebra h in g = Lie(G), [h,h] = Lie(G') for a (unique) connected closed $k$-subgroup $G'$ in $G$. In particular, if h is a semisimple Lie $k$-subalgebra of g (so h = [h,h]) then it is the Lie algebra of a connected closed $k$-subgroup of $G$. See 7.9 in Borel's book on linear algebraic groups (and 7.7 for a nec/sufficient condition in general, in char. 0). So as always, it's the commutative/solvable stuff that creates all the headaches. –  BCnrd May 5 '10 at 16:47

8 Answers 8

up vote 15 down vote accepted

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.

share|improve this answer

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 )

share|improve this answer

Sorry to tune in so late to this conversation, but I think it's worth pointing out some of the paper trail on the question (supplementing BCnrd's comment). This goes back to Chevalley's initial work in the 1950s, especially in volume II (in French) of his projected six volume work Theorie des groupes de Lie. The first volume (in English) was published by Princeton Press, then II and III followed but no more; his 1956-58 Paris seminar changed the whole approach to linear algebraic groups and largely ignored the Lie algebras. In Section 14 of II, working over an arbitrary field of characteristic 0, Chevalley asks which Lie subalgebras $\mathfrak{g} \subset \mathfrak{gl}(V)$ (with $\dim V < \infty$) can be Lie algebras of closed subgroups of the general linear group. He worked out a number of nice features of the unique smallest algebraic subalgebra containing $\mathfrak{g}$: it has the same derived algebra as $\mathfrak{g}$, for instance. In fact, the derived algebra of any $\mathfrak{g}$ is algebraic.

Some of these ideas were written down by Borel (Section 7) and by me (Chap. V) in our Springer graduate texts, working over an algebraically closed field of characteristic 0 (my treatment came from the earlier Bass/Borel notes). These sources include further references to papers by Hochschild and others along with the more scheme-theoretic treatment in the 1970 book by Demazure and Gabriel: II, Section 6, no. 2. They assume $k$ is a field and $\mathfrak{G}$ is a " $k$-groupe localement algebrique" with an appropriate Lie algebra attached, then study possible algebraic subalgebras.

In prime characteristic the notion of algebraic Lie algebra becomes much more problematic. See Seligman's 1965 book Modular Lie Algebras, VI.2, for some discussion. For example, the Lie algebra $\mathfrak{sl}(p,k)$ is usually simple modulo its one-dimensional center, but the quotient algebra can't be algebraic since then it would be the Lie algebra of a known simple algebraic group. Even more extreme are the extra simple Lie algebras of "Cartan type" (and others for small primes), for which there are no corresponding groups. `

share|improve this answer

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

share|improve this answer

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?

share|improve this answer
    
I'll have to think about it more, but I really like the trick you're proposing. –  Anton Geraschenko Oct 10 '09 at 20:12

There is a very interesting way in which this statement can fail. If K is a number field and G is an algebraic group over K having good reduction away from N, then a Lie subalgebra h of Lie(G) is going to be algebraic if and only if the reduction mod P for primes P not dividing N is closed under p-th powers.

For example, take G = G_m x G_m over a number field K. and consider the subalgebra of Lie(G) defined by the graph of the map multiplication by a in K. Then this is going to be algebraic if and only if a is in Q.

The above result is proven in Bost's paper "Algebraic Leaves of Algebraic Foliations over Number Fields".

N

share|improve this answer
1  
This isn't right: any abelian Lie algebra is surely algebraic! –  Victor Protsak May 5 '10 at 7:47
    
Yes, but it won't be of the form Lie(H) for a K-algebraic subgroup H of G unless it satisfies the criterion. –  Nicolás May 5 '10 at 8:35
    
The word some (algebraic group) in the formulation is crucial –  Victor Protsak May 6 '10 at 0:19

Let ${\mathbb R}$ act on ${\mathbb R}^2$ by rotation at unit speed, and on a second ${\mathbb R}^2$ by rotation at some irrational speed. Let $G$ be the (solvable) semidirect product ${\mathbb R} \ltimes {\mathbb R}^4$. Then the coadjoint orbits of $G$ may not be locally closed. I don't think that should happen in the algebraic case.

share|improve this answer
    
Hmmmm.......why not? –  Dr Shello Jun 30 '11 at 3:06
    
The orbit through $x\in X$ is the image of the composite map $G \to G \times \{x\} \to G \times X \to X$. Images of algebraic maps are constructible sets. Constructible sets (are nasty but) have an open set on which they are locally closed. So $G\cdot x$ has an open dense set on which it's locally closed. And $G\cdot x$ is homogeneous, so it's locally closed everywhere. –  Allen Knutson Jul 4 '11 at 15:12

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.

share|improve this answer
1  
<a href="en.wikipedia.org/wiki/Ado's_theorem">Wikipedia</…; says Ado is true in all characteristics. –  Ben Webster Oct 7 '09 at 21:32
    
same link as Ben posted: en.wikipedia.org/wiki/Ado%27s_theorem –  Anton Geraschenko Oct 8 '09 at 5:14
    
Ah, great. Except it wasn't Ado who proved it, but Iwasawa (1948) and Harish-Chandra (1949), according to Wikipedia. We skipped these theorems in Mark's class, which is why I don't know them. I wonder why Mark didn't mention them... –  Theo Johnson-Freyd Oct 8 '09 at 6:07
    
Well, so, a glance at the paper shows that Harish-Chandra considers only the case when the characteristic is 0, but emphasizes that his proof is entirely algebraic. Iwasawa's paper isn't available online via MathSciNet. –  Theo Johnson-Freyd Oct 8 '09 at 6:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.