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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.

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  • $\begingroup$ 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
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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).

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  • $\begingroup$ 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. $\endgroup$ – YCor Mar 25 '19 at 19:37
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@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.

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  • $\begingroup$ 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
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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 ($\simeq (\mathbb Z/2\mathbb Z)^n)$.

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

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