# Real points of reductive groups and connected components

Let $$\mathbf G$$ be a connected reductive group over $$\mathbb R$$, and let $$G = \mathbf G(\mathbb R)$$. Then $$G$$ is not necessarily connected as a Lie group, e.g. $$\mathbf G = \operatorname{GL}_n$$. Does $$G$$ have finitely many connected components? I heard it's true when $$\mathbf G$$ is semisimple, by a theorem of Cartan.

• Yes, it does, see the answers below. Moreover, if $\mathbf{G}$ is semisimple and simply connected, then $\mathbf{G}(\mathbb{R})$ is connected, see this answer. – Mikhail Borovoi Mar 22 '19 at 20:07
• Moreover, if $\mathbf{G}$ is reductive and anisotropic (compact), then $\mathbf{G}(\mathbb{R})$ is connected, because then $\mathbf{G}(\mathbb{R})$ is homotopically equivalent to $\mathbf{G}(\mathbb{C})$ – Mikhail Borovoi Mar 22 '19 at 20:33

For every $$\mathbb{R}$$-scheme $$X$$ of finite type, $$\pi_0(X(\mathbb{R}))$$ is finite. This follows e.g. from Theorem 2.3.6 in Bochnak, Coste, Roy, Real Algebraic Geometry (basic structure theorem for semi-algebraic sets).

• In Borel-Serre (CMH, 1964, §6.7), the authors attribute this result to Whitney. This might be: H. Whitney, Elementary Structure of Real Algebraic Varieties, Ann. of Math. (2) 66 1957 545-556. – YCor Mar 25 '19 at 19:37

@LaurentMoret-Bailly gives an algebraic-geometry reference, which is much better than the one I'm about to give. My argument applies only to the quasi-split case. Its only (negligible) virtue is that it lets us exercise some structure theory of quasi-split groups.

Let $$\mathbb B$$ be a Borel subgroup of $$\mathbb G$$, with maximal torus $$\mathbb T$$. Then $$\mathbb G$$ is the (disjoint) union of the finitely many (indexed by the Weyl group $$W \mathrel{:=} \mathrm W(\mathbb G, \mathbb T)$$) double cosets of $$\mathbb B$$, so it's enough to show that each of those has finitely many connected components. Since, for $$w \in W$$, the double coset $$\mathbb B w\mathbb B$$ is isomorphic as a variety to $$\prod_{\substack{\alpha > 0 \\ w\alpha < 0}} \mathbb U_\alpha \times \mathbb T \times \prod_{\alpha > 0} \mathbb U_\alpha$$, where $$\mathbb U_\alpha \cong \mathfrak{gl}_1$$ denotes the root subgroup corresponding to $$\alpha$$ and positivity is taken with respect to $$\mathbb B$$, it suffices to show that $$\mathbb T(\mathbb R)$$ has finitely many connected components. Since $$\mathbb T$$ is the almost-direct product of an $$\mathbb R$$-split torus $$\mathbb T_d$$ and an $$\mathbb R$$-anisotropic torus $$\mathbb T_a$$, we have that $$\mathbb T_d(\mathbb R) \times \mathbb T_a(\mathbb R)$$ maps into $$\mathbb T(\mathbb R)$$ with finite cokernel $$\operatorname H^1(\mathbb R, \mathbb T_d \cap \mathbb T_a)$$, and since $$\mathbb T_d(\mathbb R) \cong (\mathbb R^\times)^{\operatorname{rank} \mathbb T_d}$$, it suffices to prove that $$\mathbb T_a(\mathbb R)$$ has finitely many connected components.

Now maybe a real real-group theorist would know that $$\mathbb T_a$$ is a product of circle groups $$\mathbb S^1 \mathrel{:=} \ker(\operatorname{Res}_{\mathbb C/\mathbb R}\operatorname{GL}_1 \to \operatorname{GL}_1)$$, but I am a $$p$$-adicist, so I have to notice that $$\operatorname{conj} + 1$$ maps $$\mathrm X^*(\mathbb T_a)$$ into $$\mathrm X^*(\mathbb T_a)^{\operatorname{conj}} = \{0\}$$, whence $$\mathrm X^*(\mathbb T_a)$$ is, not just as a group but as a $$\mathrm{Gal}(\mathbb C/\mathbb R)$$-module, a direct sum of copies of $$\mathrm X^*(\mathbb S^1)$$; so $$\mathbb T_a$$ is (as an $$\mathbb R$$-torus) a product of copies of $$\mathbb S^1$$, so $$\mathbb T_a(\mathbb R) \cong \prod \mathbb S^1(\mathbb R) = \prod \mathrm S^1$$, which is connected.

• Actually I guess the cokernel is a priori only a subgroup of, not necessarily all of, $\operatorname H^1(\mathbb R, \mathbb T_d \cap \mathbb T_a)$. To make this argument work in general, one would need to understand the structure of anisotropic $\mathbb R$-groups. As a $p$-adicist, I find it hard to come to terms with the sheer variety (no pun intended) of such groups. – LSpice Mar 22 '19 at 22:42

Suppose $$G$$ is a complex reductive group (not necessarily connected), defined over $$\mathbb R$$.

1) $$G(\mathbb R)=K\text{exp}(\mathfrak p)$$ where $$K$$ is a maximal compact subgroup of $$G(\mathbb R)$$. This is the Cartan decomposition; for disconnected $$G$$ this is due to Mostow. So $$G(\mathbb R)$$ is homotopically equivalent to $$K$$.

2) Every compact group has a canonical (reductive, algebraic) complexification (Chevalley) with the same component group.

3) Every algebraic complex group has finitely many components.

This gives a conceptual, though not necessarily elementary, proof, and it isn't necessary that $$G$$ is connected. If $$G$$ is connected one can say a bit more: $$G(\mathbb R)/G(\mathbb R)^0$$ is an elementary two-group ($$\simeq (\mathbb Z/2\mathbb Z)^n)$$.

• Thanks for your answer. Can you say a little more about (2)? What is meant by a complexification of a compact real Lie group? – D_S Mar 25 '19 at 4:14
• The complexification of an arbitrary compact Lie group is defined using Hopf algebras. For a nice exposition see Claudio Procesi's book Lie Groups, An Approach through Invariants and Representations, Chapter 8, Section 7.2, Theorem 3 (and the Corollary). – Jeffrey Adams Mar 25 '19 at 14:52
• Surely you mean something other than "$G(\mathbb R)$ is diffeomorphic to $K$" in (1)? If it were so, then every real algebraic group would have compact real points. – LSpice Mar 25 '19 at 15:04
• Thanks Loren, of course, it now correctly says homotopically equivalent. – Jeffrey Adams Mar 25 '19 at 19:24