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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.

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    $\begingroup$ 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. $\endgroup$
    – BCnrd
    Commented Dec 2, 2010 at 6:21
  • $\begingroup$ @BCnrd, how can I bribe you so that you start writing actual answers? :) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Dec 2, 2010 at 14:30
  • $\begingroup$ You could try cutting and pasting his comments into the answer box! $\endgroup$
    – Allen Knutson
    Commented Dec 2, 2010 at 15:04

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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.

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  • $\begingroup$ 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). $\endgroup$
    – David Corwin
    Commented Dec 3, 2010 at 15:19
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    $\begingroup$ 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. $\endgroup$
    – George McNinch
    Commented Dec 3, 2010 at 18:27
  • $\begingroup$ 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:-)) $\endgroup$
    – Bugs Bunny
    Commented Dec 4, 2010 at 12:31
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    $\begingroup$ 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. $\endgroup$
    – Bugs Bunny
    Commented Dec 4, 2010 at 12:34
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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...]

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