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 $Rep(G)$?

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    The Lie algebra will not see non-infinitesimal data: quotients of $G$ by discrete subgroups, for example, or, more sillily (?), the case where $G$ is not connected... – Mariano Suárez-Álvarez Feb 3 '10 at 21:31
  • Any restrictions on the characteristic of the ground field? In any case, taking the differential of a rational representation gives a representation of the Lie algebra. In characteristic 0 there should be a nice correspondence. In characteristic p > 0 all I know is that the irreducible >restricted< representations of the Lie algebra of a simply-connected semisimple group which is defined over the prime field are precisely the modules S(lambda) for p-restricted weights lambda. – user717 Feb 3 '10 at 21:34
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    If you like the idea of Lie groups more than algebraic groups, then use the unitary trick! This works for reductive groups, I guess, and reduces things to Lie groups. The representation theory of split connected reductive algebraic groups is "the same in char 0 and char p", so to understand e.g. the algebraic reps of the algebraic group GSp_{12} over a finite field is "the same" as understanding it over C. And over C it's again "the same" as C^infty Lie group reps of the maximal compact subgroup, so e.g. to understand the algebraic reps of GL_n you only need to understand the rep thy of U(n). – Kevin Buzzard Feb 3 '10 at 22:30
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    @Kevin: what do you mean the representation theory "is the same in char 0 and char p"? The category of representations is pretty different in the two cases in lots of ways -- one is semisimple the other isn't, for example. The unitary trick will tell you that the rep theory of the reductive group over $\mathbb C$ is the same as that over the maximal compact subgroup, the compact form of the group, but that has nothing to do with char p. – Kevin McGerty Feb 4 '10 at 1:44
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    @Kevin M: thanks for putting me right about this. The theories are "the same" only (if I've understood correctly) in the highly weak sense that the irreducible modules for G are parametrised by dominant integral weights. On the other hand, your example shows that (a) extension groups might be different and (b) dimensions might be different if you move between char 0 and char p, so in fact the rep theory of the two cases is hugely different. – Kevin Buzzard Feb 4 '10 at 14:34
up vote 20 down vote accepted

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).

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.]

  • @Emerton: as you recall, in his (excellent) book Lie Algebras and Lie Groups, Serre works only in the analytic category, although much of what he says can be carried over immediately to linear algebraic groups. This came up in mathoverflow.net/questions/10730/… – Pete L. Clark Feb 3 '10 at 22:26

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.

  • I'm somewhat confused--what do you mean by "$G$ is represented by a Hopf algebra $A$"? – Akhil Mathew Feb 4 '10 at 0:22
  • I mean that $A$ is the coordinate Hopf algebra of $G$, or, equivalently, that the functor $G$ is given by $G\left(R\right)=\mathrm{Hom}_{\mathrm{Alg}}\left(A,R\right)$. Okay, I was assuming that "algebraic group" means "affine algebraic group" here, but wasn't the author assuming this too? – darij grinberg Feb 4 '10 at 10:20

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.

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 $\mathfrak{g}$.

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