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

In the representation theory of Lie groups (say, over $\mathbb{R}$ or $\mathbb{C}$), one can show that a Lie algebra homomorphism between the Lie algebras of two algebraic groups $G$ and $H$ always results in a Lie group homomorphism if $G$ is simply connected.

It seems like one might be able to make this argument over a general field. In place of fundamental group, we could ask that the etale fundamental group of $G$ be trivial. Does this allow us to show that a homomorphism of Lie algebras results in a homomorphism of groups? Then one could prove that the semisimplicity of the algebra and of the group are equivalent using this argument.

share|improve this question
2  
In nonzero char., ss of Lie alg. is bad notion, so consider char. 0 (with conn'dness). Etale fundamental gp is red herring. Structure theory of linear alg. gps in char. 0 (e.g., Levi decomposition, "algebraicity" of subalg. of $\mathfrak{gl}_n$ that are own derived algebras) proves algebraically that a smooth conn'd affine gp is ss iff Lie alg. is ss with trivial center. There's a good notion of simply conn'd ss gp in any char (but SL$_2(\mathbf{R})$ not simply conn'd, SL$_2$ is simply conn'd as $\mathbf{R}$-gp). With this concept, can promote Lie alg. maps to gp maps for such gps in char 0. –  BCnrd Dec 2 '10 at 6:21
    
@BCnrd, how can I bribe you so that you start writing actual answers? :) –  Mariano Suárez-Alvarez Dec 2 '10 at 14:30
    
You could try cutting and pasting his comments into the answer box! –  Allen Knutson Dec 2 '10 at 15:04

2 Answers 2

No, it does not. The additive $G$ and the multiplicative $H$ groups have isomorphic Lie algebras but only trivial group homomorphisms between them.

The Lie algebra of $G$ has a restricted structure! You may wonder what happens if there is a homomorphism of restricted Lie algebras. This would uniquely lift to a homomorphism of the first Frobenius kernels but not whole groups.

The easiest example does not have trivial $\pi_1$: consider a semidirect product of multiplicative and additive groups with the action of the mulitplicative group twisted by Frobenius. It will have an abelian Lie algebra, isomorphic as a restricted Lie algebra to the Lie algebra of the direct product. There is no corresponding homomorphism of groups.

With trivial $\pi_1$, I can think of $SL_2 (K)$ where $K$ is algebarically closed of characteristic 2. The Lie algebra will have a nontrivial homomorphism to the Lie algebra of the additive group $K_a$ but no such homomorphism exists for groups.

share|improve this answer
    
Okay, but does characteristic 0 imply that Lie algebra homomorphisms correspond to group homomorphisms? This is probably trivial, since any variety over characteristic $0$ can be defined over $\mathbb{C}$, and I would imagine that the topological fundamental group of $\mathbb{C}$ and the algebraic fundamental group over an arbitrary field are the same, and that furthermore the property that Lie algebra homomorphisms correspond to group homomorphisms is invariant of characteristic $0$ algebraically closed base field (essentially this is mostly Lefschetz). –  David Corwin Dec 3 '10 at 15:19
    
But the "additive/multiplicative group" example Bugs gives is a counterexample already in characteristic 0. And on the other hand, BCnrd's comment above explains (albeit trsly) that in char. 0 the desired conclusion is valid for simply connected semisimple groups (note that neither $\mathbf{G}_a$ nor $\mathbf{G}_m$ is semisimple). Concerning your "Lefschetz principle" comments, the argument that BCnrd outlines doesn't involve C and works even without stipulating "alg closed", but requires the notion of simply connected defined "algebraically" for semisimple linear algebraic groups. –  George McNinch Dec 3 '10 at 18:27
    
Yes, you need to be careful in characteristic zero too: the map $G_a\rightarrow G_m$ is the exponent. It is not a homomorphism of algebraic groups, only of holomorphic groups. In particular, you are rosted over an arbitrary field of characteristic zero:-)) –  Bugs Bunny Dec 4 '10 at 12:31
    
The right statement is that in characteristic zero there is a category equivalence between Lie algebras and formal groups. Thus, you can lift only to formal groups, in general. –  Bugs Bunny Dec 4 '10 at 12:34

Just to make explicit BCnrd's comment that the Lie algebra is not-so-great in non-zero characteristic, consider the special linear group $G=\operatorname{SL}_2$ of $2 \times 2$ matrices of det 1 in char. 2, a simply connected semisimple group. The Lie algebra $L = \mathfrak{sl}_2$ contains a 1-dimensional ideal $I$ spanned by the identity matrix, and the quotient algebra $L/I$ is isomorphic to the (restricted) Lie algebra $M = \operatorname{Lie}(\mathbf{G}_a \times \mathbf{G}_a)$.

But the natural quotient mapping $L \to M$ does not result in a non-trivial homomorphism of alg groups $\operatorname{SL}_2 \to \mathbf{G}_a \times \mathbf{G}_a$.

For what it is worth, consider $G = \operatorname{SL}_p$ in char p>0. Again the Lie algebra $L$ of $G$ contains a 1 dimensional ideal $I$ spanned by the identity matrix. For $p>2$, I'm not aware that the Lie algebra $L/I$ is the Lie algebra of any algebraic group, though $L/I$ is isomorphic to an invariant subalgebra of the Lie algebra of the adjoint group $G_1 = \operatorname{PGL}_p$, and the mapping $L \to L/I \subset \operatorname{Lie}(G_1)$ is the tangent mapping of the standard (inseparable) isogeny $G \to G_1$.

[oops: just noticed that much of this is redundant w/ last bit of Bugs Bunny's answer...]

share|improve this answer

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.