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?


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!

  • 3
    $\begingroup$ +1, but this doesn't seem to cover the non-semisimple case. For example, universal covers of tori are not representable in algebraic groups. $\endgroup$ – 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).

  • 4
    $\begingroup$ 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. $\endgroup$ – Pierre-Yves Gaillard Aug 24 '10 at 13:20
  • $\begingroup$ 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. $\endgroup$ – Pierre-Yves Gaillard Aug 24 '10 at 13:52
  • $\begingroup$ Thanks, Pierre. Do you know if a similar argument works over C (with a different group)? $\endgroup$ – Qiaochu Yuan Aug 24 '10 at 14:25
  • $\begingroup$ No, I'm sorry, I don't. But I'm sure many users know this kind of things. $\endgroup$ – Pierre-Yves Gaillard Aug 24 '10 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.