algebraic groups and their Lie algebras

Question

I have probably a stupid question about representations of algebraic groups:

Let $G$ be an algebraic group and $L$ be a Lie algebra of $G$. What is the connection between categories of representations of $G$ and $L$ (are they equivalent)?

Now, let $V$ be an irreducible representation of $G$, how to prove that $V$ is also an irreducible representation of $L$ (in case it is true)? It should be true and I believe, proof is easy, but still, I do not see that and I am looking for a nice proof of that.

And vice versa, let $V$ be an irreducible representation of $L$, is it also irreducible representation of $G$ ?

Answer 1

I suggest the following lecture notes of Bruhat:

www.math.tifr.res.in/~publ/ln/tifr14.pdf

Chapter 3 & 4 should answer most of your questions.

For example, there are statements like this :

Proposition 1(pg.19). To every analytic representation h : G −→ G′ there corresponds a map dh : U(G) → U(G′) which is a representation of algebras such that ( f ◦ h) = (dh() f ) ◦ h.

Corollary (pg.36). Let G and H be two Lie groups having g and J as their Lie algebras. If G is connected and simply connected, to every representation π of g in J, there corresponds one and only one representation f of G → H such that d f = π.

If you are interested in semisimple, connected, simply connected groups only, both give you an isomorphism between the categories of complex representations. Equivalence of categories is weaker than isomorphism. Moreover, the categories of representations are both semisimple in this case, i.e., reps decompose into irreducible ones. Thus, also your second and third question can be answered affirmative in this case.

Answer 2

The question is out of focus, I think. If it's really about algebraic groups rather than Lie groups, that should be made clear. (And a tag should be added either way.) As Marc Palm indicates, there are well understood analytic pathways connecting representations of Lie groups and representations of their Lie algebras, though of course the details are quite nontrivial and restrictions are needed.

For linear algebraic groups in characteristic 0, the connections are much more problematical. Chevalley realized this in his early book (1951), where he took some steps using a sort of formal exponentiation process to get back from the Lie algebra to the algebraic group. In modern form some of this is written down in later textbooks (Borel for instance). Probably the most comprehensive source is Section 6 of Chapter II in Demazure-Gabriel Groupes algebriques (optimistically labelled "tome I"). Even here one sees that not so much can be said outside the framework of semisimple groups and their Lie algebras. Note that the concept of "simply connected" isn't even meaningful for most algebraic groups.

In prime characteristic, almost nothing works along the lines of your question, even if you limit to simply connected semisimple groups. Here there is lots of literature, but for example it's long been known that the irreducible representations of the group and its Lie algebra diverge drastically.

Answer 3

A graph argument settles this issue very nicely, as follows.

Consider a linear algebraic group $G$ over a field $k$ of characteristic 0, and let $\mathfrak{g}$ be its Lie algebra. For a finite-dimensional $k$-vector space $V$, let $f:\mathfrak{g} \rightarrow \mathfrak{gl}(V)$ be a representation. Provided that $G$ is connected, since ${\rm{char}}(k) = 0$ clearly there is at most one $k$-homomorphism $\rho:G \rightarrow {\rm{GL}}(V)$ such that ${\rm{Lie}}(\rho) = f$, so assuming $G$ is connected we seek conditions under which such a $\rho$ always exists.

Let $\mathfrak{h} \subset \mathfrak{g} \times \mathfrak{gl}(V)$ be the graph of $f$. Clearly if $\rho$ is to exist and the $k$-subgroup $H \subset G \times {\rm{GL}}(V)$ is its graph then $H$ is connected and $\mathfrak{h} = {\rm{Lie}}(H)$, so $H$ is uniquely determined (as a $k$-subgroup of $G \times {\rm{GL}}(V)$) since ${\rm{char}}(k) = 0$. So we seek conditions under which the Lie subalgebra $\mathfrak{h}$ of $\mathfrak{g} \times \mathfrak{gl}(V)$ "exponentiates" to a connected closed $k$-subgroup $H$ (and then we need conditions to ensure ${\rm{pr}}_1:H \rightarrow G$ is an isomorphism).

Assume $G$ is its own derived group, so $\mathfrak{g}$ is its own derived subalgebra (as ${\rm{char}}(k)=0$) and hence the same for $\mathfrak{h}$. Now comes the crucial point: it is a general fact over fields $k$ of char. 0 (see Cor. 7.9 in Ch. II of Borel's textbook on algebraic groups) that the derived subalgebra of any Lie subalgebra of a linear algebraic group over $k$ "exponentiates" to a connected closed $k$-subgroup. So $\mathfrak{h} = {\rm{Lie}}(H)$ for a unique connected closed $k$-subgroup $H \subset G \times {\rm{GL}}(V)$. The necessary and sufficient condition for $f$ to arise from some $\rho$ is that ${\rm{pr}}_1:H \rightarrow G$ is an isomorphism.

This projection has Lie algebra map $\mathfrak{h} \rightarrow \mathfrak{g}$ that is the analogous projection which is visibly an isomorphism (due to the definition of $\mathfrak{h}$ as the graph of $f$), so $H \rightarrow G$ is an isogeny. As such, its kernel is etale (since ${\rm{char}}(k) = 0$) and hence central (since $H$ is connected), so it is a finite central $k$-subgroup of $H$. Thus, we just need that $G$ admits no nontrivial isogenous (smooth) connected central extension. This is automatic when $G$ is assumed to be semisimple and simply connected (in the sense of algebraic groups).

So we win whenever $G$ is a connected semisimple $k$-group that is simply connected. We also win whenever $G$ is a unipotent $k$-group (by entirely different arguments), but presumably you're not interested in that case.