How do you switch between representations of an algebraic group and its Lie algebra?

Question

I'm interested in the structures of categories like $Rep(GL_n), Rep(SL_n)$, etc. of algebraic representations of an algebraic group. I understand that there should be some relation between these and the categories of representations of the corresponding Lie algebras. However, it's not as intuitive to me what's going on here as with the case of, say, a Lie group, perhaps because the notion of a "tangent vector" is somewhat different.

So, how does one switch between the categories $Rep(G)$ and $Rep(\mathfrak{g})$ for $G$ an algebraic group and $\mathfrak{g}$ its Lie algebra---are there functors in each direction? Can this be used to prove that $Rep(G)$ is semisimple when $G$ is reductive? In another direction, can the structure of $Rep(\mathfrak{g})$ as known from the representation theory of, say, semisimple Lie algebras give the structure of $Rep(G)$?

Answer 1

If $G$ is semisimple simply connected in characteristic zero, the differential at $1$ gives an equivalence of (tensor) categories $Rep(G)\to Rep({\mathfrak g})$. If $G$ is not semisimple, this is not the case, but this functor is always fully faithful (i.e. an equivalence onto a full subcategory) if $G$ is connected. The essential image of this functor can be described explicitly. Namely, consider the Levi decomposition ${\mathfrak g}={\mathfrak l}\ltimes {\mathfrak u}$, where ${\mathfrak l}$ is reductive and ${\mathfrak u}=Lie(U)$, where $U$ is the unipotent radical of $G$. Then the image is those finite dimensional representations of ${\mathfrak g}$ for which ${\mathfrak u}$ acts nilpotently, and the weights for ${\mathfrak l}$ are integral (i.e. descend to chartacters of the maximal torus).

Answer 2

Suppose that the algebraic group $G$ over $k$ acts on the vector space $V$, i.e. that there is map of algebraic groups $G \to GL(V).$ Passing to Lie algebras (= Zariski tangent space to the identity) is a functor, and so we get a map on Lie algebras $\mathfrak g \to End(V),$ which is the corresponding Lie algebra representation.

Another way to do this, closer to the differential point of view (and which you will need anyway to identify the Lie algebra of $GL(V)$ with $End(V)$) is as follows: $G(k[\epsilon])$ acts on $V[\epsilon]$ (take dual number-valued points of the original morphism). In particular, the Zariski tangent space at the identity (which on the one hand is $\mathfrak g$, by definition, and on the other hand is the subgroup of $G(k[\epsilon])$ consisting of elements mapping to the identity under the specialization $\epsilon \mapsto 0$) acts on $V[\epsilon]$ by endomorphisms which reduce to the identity after setting $\epsilon = 0$. One checks that such a map is of the form $v \mapsto v + L(v)\epsilon,$ where $L \in End(V)$.
Sending it to $L$ gives the required map $\mathfrak g \to End(V)$.

(Note: we are taking $g \in G(k[\epsilon])$ lying over the identity, applying the representation $\rho$ to get $\rho(g)$, then forming the difference quotient $(\rho(g) - 1)/\epsilon.$ Hopefully the connection with differentiation is clear.)

One has to be slightly cautious about going back from $\mathfrak{g}$ to $G$, since there are the following subtleties which any approach has to take into account: the field $k$ had better have char. 0; the group $G$ had better be linear algebraic, and furthermore either nilpotent, or simply connected semi-simple; and the representation has better be finite-dimensional.

One could try the following: take a finite dimensional representation $V$ of $\mathfrak{g}$; extend it to a rep. of the universal enveloping algebra $U(\mathfrak{g})$; use the fact that $V$ is finite-dimensional to extend the rep'n to a certain completion of $U(\mathfrak{g})$; inside this completion, look at the group-like elements under the canonical co-multiplication on $U(\mathfrak{g})$; show that these elements form a linear algebraic group $G$ with Lie algebra $\mathfrak g$. (The intuition is that we can map $\mathfrak{g}$ into a well-chosen completion of $U(\mathfrak{g})$ via a formal version of the exponential map.)

[This last suggestion is based on a discussion in Serre's Lie algebras/Lie groups book, but I don't remember if he carefully treats this algebraic group context; it may be that he is rather focussing on the Lie group setting.]

Answer 3

Hmm. There exists an easy-to-define functor $\mathrm{Rep}G\to\mathrm{Rep}\left(\mathrm{Lie}G\right)$, but I am not sure whether it is the "standard" one and whether it is new to you.

Let $G$ be an algebraic group, represented by a Hopf algebra $A$ (actually, bialgebra is enough). It is known that the Lie algebra $\mathrm{Lie}G$ can be described as the Lie algebra

$\mathrm{Der}_{\varepsilon}\left(A,k\right)=\lbrace d:A\to k\mid d\text{ is a }k\text{-linear map satisfying }d\left(xy\right)=\varepsilon\left(x\right)d\left(y\right)+d\left(x\right)\varepsilon\left(y\right)\text{ for all }x,y\in A\rbrace$.

The Lie bracket on this vector space $\mathrm{Der}_{\varepsilon}\left(A,k\right)$ is given by $\left[d,e\right]=d\ast e-e\ast d$, where $\ast$ means the convolution product on $\mathrm{Hom}_k\left(A,k\right)$.

Now, a representation of the algebraic group $G$ is a right $A$-comodule $V$. We want to make $V$ a left $\mathrm{Der}_{\varepsilon}\left(A,k\right)$-module. This is done by

$dv=v_{\left(0\right)}\otimes d\left(v_{\left(1\right)}\right)$ for every $d\in\mathrm{Der}_{\varepsilon}\left(A,k\right)$ and $v\in V$.

We are using Sweedler notation here.

This turns every representation of $G$ into one of $\mathrm{Der}_{\varepsilon}\left(A,k\right)=\mathrm{Lie}G$. As for the other direction, I don't think it can always be done.

Answer 4

One point which seems not to have been mentioned (but is implicit in Pavel's answer) is that the small question posed at the end of the main question, whether $\mathrm{Rep}(\mathfrak{g})$ semisimple implies $\mathrm{Rep}(G)$ semisimple, has a simple answer regardless of most technical conditions. In fact, in characteristic zero it is true.

The relevant theorem is:

Theorem: Suppose $k$ has characteristic zero, let $G$ be a connected affine algebraic $k$-group, and let $\mathfrak{g}$ be its Lie algebra. If $V$ is a $G$-representation and $W \subset V$ is a subspace, then $W$ is $G$-stable if and only if it is $\mathfrak{g}$-stable.

The proof for faithful representations is in Humphreys' "Linear Algebraic Groups", Theorem 13.2, and to extend it to any representation one need only show that the Lie algebra is compatible with fibered products, which is tautological. (Possibly this result must be stated over an algebraically closed field; this is so in the book, and I am badly acquainted with rationality properties.)

As a consequence, $V$ is irreducible or completely reducible for $G$ if and only if it is for $\mathfrak{g}$.

Answer 5

First off, let's assume from the outset that we're working in characteristic zero because Lie algebras fail to capture the proper information about the group otherwise.

The functor $\operatorname{Lie}: \{\text{affine groups}\} \to \{\text{Lie algebras}\}$ has a left adjoint. This adjoint sends a Lie algebra $\mathfrak{g}$ to the group $G(\mathfrak{g})$ obtained by Tannaka duality from the tensor category of representations of $\mathfrak{g}$ equipped with the forgetful functor to $k$-vector spaces. I believe that in general, this functor can behave badly, but for $\mathfrak{g}$ semisimple, $G(\mathfrak{g})$ is the universal connected group with Lie algebra $\mathfrak{g}$, in the sense that any other connected group with Lie algebra $\mathfrak{g}$ is a nice quotient of it.

Thus, we see that the category of representations of a semisimple Lie algebra $\mathfrak{g}$ will be equivalent to the category of representations of some group whose Lie algebra is $\mathfrak{g}$. In general, however, representations of $G$ will only be a full subcategory of representations of $\operatorname{Lie}(G)$ (think of $SO(3)$, for example). So even in the semisimple case, there is not a functor from $\operatorname{Rep}(\operatorname{Lie}(G)) \to \operatorname{Rep}(G)$ in general.

A good overview of the Tannakian viewpoint is in this short article by Milne, which discusses the results I mentioned above.