# Is the universal cover of an algebraic group an algebraic group?

Here algebraic group means affine algebraic group in both instances. Also I'm mainly interested in groups over $\mathbb{C}$. In fact I'm taking $\pi_1(G)$ to mean the fundamental group of $G_{an}$, the analytification. So I guess my question only applies to the base field being either $\mathbb{C}$ or $\mathbb{R}$.

In this case these groups are also Lie groups with Lie algebras. If $\mathfrak{g}$ is a semisimple Lie algebra then there is a connection with the weight and root lattices: there is a 1-1 correspondence between connected Lie groups with Lie algebra $G$ and lattices $\Lambda$ with $\Lambda_W \supset \Lambda \supset \Lambda_R$.

The group corresponding to $\Lambda = \Lambda_R$ is always algebraic because it is the adjoint group. A slightly more general question is then, for $\Lambda_W \supset \Lambda \supset \Lambda_R$ is the corresponding group $G_\Lambda$ affine algebraic?

-

## 2 Answers

Over $\mathbf{C}$ I believe the answer is yes (the universal cover is algebraic), although I'm not really an expert. Here's the story as I understand it.

A connected[*] semisimple linear algebraic group over a field $k$ is called simply connected if it admits no nontrivial isogeny from another connected group. (An isogeny is a surjective, flat homomorphism of algebraic $k$-groups with finite kernel.) Now just like in the Lie group story, nice algebraic groups are classified by combinatorial data. Precisely, a reductive $k$-group $G$ together with a split maximal torus $T$, is classified by a "root datum", which is roughly the roots and coroots of $G$ with respect to $T$, plus the lattices of characters and cocharacters of $T$. ("Roughly" because when $k$ is not algebraically closed, you need to keep track of the action of the Galois group of $k$ on the lattices, too.) Correspondingly, isogenies between such groups match bijectively with appropriate "morphisms" between the respective root data. These theorems are a bit involved, but can certainly be found in Borel's book on linear algebraic groups, for the case you care about. (Algebraically closed field of characteristic zero, namely $\mathbf{C}$.)

A consequence of this theory is that for any connected semisimple linear algebraic $\mathbf{C}$-group, there exists an "algebraic universal cover", i.e. an isogeny $\tilde{G}\to G$ from a simply connected $\mathbf{C}$-group $\tilde{G}$. (In fact, we don't need to be working over $\mathbf{C}$ for this to be true.)

Now here's the crux. I claim that if $G$ is a simply connected $\mathbf{C}$-group, then the closed points $G(\mathbf{C})$ are simply connected in the classical topology. The hypothesis that we are over $\mathbf{C}$ is crucial: $\mathrm{Sp}(2n)$ is a simply connected $\mathbf{R}$-group, but the universal cover of $\mathrm{Sp}(2n)(\mathbf{R})$ is the non-algebraic metaplectic group.

Let me sketch the proof of the claim, which is suprisingly hard. (This was explained to me by Brian Conrad; any errors I introduce are, of course, my own.) First, it's a fact, although not a tautology, that since $G$ is connected, $G(\mathbf{C})$ is connected in the classical topology. So that's a good start. Next, by classical Lie theory, the complex Lie group $G(\mathbf{C})$ is homotopy equivalent to a maximal compact real Lie-subgroup $K$. Since $K$ is a compact manifold, this implies that $H^1(G(\mathbf{C}),\mathbf{Z})$ is finitely generated. But this group is the abelianization of $\pi_1(G(\mathbf{C}))$ (Hurewicz); and the fundamental group is already abelian because this is true for all topological groups. So $\pi_1(G(\mathbf{C}))$ is finitely generated (and abelian). In particular, it has a finite quotient. So if $G(\mathbf{C})$ were not (classically) simply connected, there would be a finite covering map of complex Lie groups $G'\to G(\mathbf{C})$. Now a hard theorem of Grauert (relating $\pi_1(G(\mathbf{C}))$ to the "\'etale fundamental group" of $G$, which classifies the algebraic analogue of finite covering maps) implies that $G'$, as well as its analytic group structure, can uniquely be given an algebraic structure. In other words, there is an algebraic $\mathbf{C}$-group $G_0'$ with $G'=G_0'(\mathbf{C})$, and an isogeny $G_0'\to G$, such that the induced map on $\mathbf{C}$-points is $G'\to G(\mathbf{C})$. In particular $G_0'\to G$ is a nontrivial isogeny, since on $\mathbf{C}$-points it is a nontrivial finite convering map. And this contradicts the (algebraic) simple-connectedness of $G$.

Whew! So in summary, for connected semisimple linear algebraic $\mathbf{C}$-groups $G$, the universal cover of $G(\mathbf{C})$ is precisely $\tilde G(\mathbf{C})$ where $\tilde G$ is the simply connected form of $G$.

[*] Warning: In this answer, "connected" means Zariski-connected. For algebraic groups over $\mathbf{R}$, this is VERY different from the $\mathbf{R}$-points being connected in the classical topology!

-
+1, but this doesn't seem to cover the non-semisimple case. For example, universal covers of tori are not representable in algebraic groups. –  Scott Carnahan Jun 16 '11 at 3:53

For groups over $\mathbb{R}$ the answer is no. The group of real points of an affine algebraic group has a faithful finite-dimensional linear representation over $\mathbb{R}$, but it is well-known that the universal cover of $\text{SL}_2(\mathbb{R})$ has no faithful finite-dimensional linear representations (unfortunately I don't know a reference for this).

-
One knows the finite dim. reps of the Lie algebra. Each of them lifts to G := SL(2,R). Thus, any nontrivial cover of G has no faithful finite dim. rep. –  Pierre-Yves Gaillard Aug 24 '10 at 13:20
To be a little more precise, the irreducible finite dimensional representations, of the Lie algebra sl(2,R), and thus of the group SL(2,R), are the symmetric powers of the standard representation. –  Pierre-Yves Gaillard Aug 24 '10 at 13:52
Thanks, Pierre. Do you know if a similar argument works over C (with a different group)? –  Qiaochu Yuan Aug 24 '10 at 14:25
No, I'm sorry, I don't. But I'm sure many users know this kind of things. –  Pierre-Yves Gaillard Aug 24 '10 at 15:35