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.

$\begingroup$ 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. $\endgroup$ – Mikhail Borovoi Mar 22 '19 at 20:07

$\begingroup$ 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 $\mathbf{G}(\mathbb{C})$ $\endgroup$ – 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 semialgebraic sets).

$\begingroup$ In BorelSerre (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 545556. $\endgroup$ – YCor Mar 25 '19 at 19:37
@LaurentMoretBailly gives an algebraicgeometry reference, which is much better than the one I'm about to give. My argument applies only to the quasisplit case. Its only (negligible) virtue is that it lets us exercise some structure theory of quasisplit 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 almostdirect 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 realgroup 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.

$\begingroup$ 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. $\endgroup$ – 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 twogroup ($\simeq (\mathbb Z/2\mathbb Z)^n)$.

$\begingroup$ Thanks for your answer. Can you say a little more about (2)? What is meant by a complexification of a compact real Lie group? $\endgroup$ – D_S Mar 25 '19 at 4:14

$\begingroup$ 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). $\endgroup$ – Jeffrey Adams Mar 25 '19 at 14:52

1$\begingroup$ 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. $\endgroup$ – LSpice Mar 25 '19 at 15:04

1$\begingroup$ Thanks Loren, of course, it now correctly says homotopically equivalent. $\endgroup$ – Jeffrey Adams Mar 25 '19 at 19:24