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I'm rather ignorant in both fields, but I would still like to endeavor asking this question. I've just learned that any Lie group is diffeomorphic to a compact Lie group cross $\mathbb{R}^n$. While there is no (to my knowledge) equivalent to diffeomorphic in the group-schemes language, this does have obvious implications about the cohomology of Lie groups (which has an analog in the group-scheme language.)

So: Is there an analog of this theorem for group-schemes? What is it? What can we say?

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For algebraic groups over a perfect field $k$, one has Chevalley's Theorem. It says that every algebraic group $G$ over $k$ contains a unique closed normal linear subgroup $H$ such that $G/H$ is an abelian variety. The abelian variety is the analogue of the compact Lie group, and the linear group $H$ is the analogue of affine space.

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You mean to say that every (connected) linear group over $\mathbb{R}$ is diffeomorphic to $\mathbb{R}^n$? This seems like a stronger version of the theorem that every linear group is rational, but I've never heard of it before. Also, in the theorem you state, is it implied that $H$ is connected? – James D. Taylor Oct 2 '10 at 16:21
@James D. Taylor---what is a "(connected) linear group over $\mathbf{R}$" and what does it mean for such a thing to be diffeomorphic to $\mathbf{R}^n$? For me, GL_1 is a connected linear algebraic group over the reals. It's not diffeomorphic to anything though, because it's not a sensible topological space. Its real points can be given the structure of a sensible topological space---but this space is not connected. So there are issues here. Another example is the restriction of scalars from $\mathbf{C}$ to $\mathbf{R}$ of $GL(1)$; this has real points isomorphic to $\mathbf{C}^\times$. – Kevin Buzzard Oct 2 '10 at 20:54
...[ran out of space]...and $\mathbf{C}^\times$ isn't diffeo to $\mathbf{R}^n$ because it isn't simply connected. Perhaps these examples above may help to clarify what your question is. – Kevin Buzzard Oct 2 '10 at 20:57
Yes, thanks Kevin. My question is: in what way does Chevalley's theorem answer the question? It's portrayed as a more general theorem than the one I stated in the question. I'm trying to understand why. In order to relate the two, I assumed that cfranc had it in his mind that $H$ is diff. to $\mathbb{R}^n$. Indeed you're right that there are problems with the definition of "diff." here, and even with "connected". Still -- for his answer to make sense he must have had an idea of how these two things are related in the classical case. – James D. Taylor Oct 2 '10 at 23:02
Dear James D. Taylor: It doesn't really answer your question. If $G$ is a conn'd lin. alg. group over field $k$ of char. 0 then qt map $G \rightarrow G/U$ for unipotent radical $U$ admits splitting: $G = L \rtimes U$ over $k$ with conn'd reductive $L$ ("Levi factor"). Now $U$ is $k$-isomorphic to affine space as a variety, and if $G$ does not contain ${\rm{GL}}_1$ as $k$-subgroup and $k = \mathbf{R}$ then $L(\mathbf{R})$ is maximal compact subgp of $G(\mathbf{R})$. A pseudo-answer to your question, perhaps. For conn'd reductive $G$ it says nothing. But what would we want for $G = {\rm{SL}}_2$? – BCnrd Oct 2 '10 at 23:23

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