Maximal compact subgroups of a semisimple Lie group are conjugate

I'm trying to go through the proof that all maximal compact subgroups of a semisimple Lie group $G$ are conjugate. I know that a possible proof follows the following steps:

• Take one maximal compact subgroup $K$ of $G$ (which I know that exists). Consider $G/K$. This is a $CAT(0)$ space.
• Take any other maximal compact subgroup $L$ of $G$. $L$ acts on $G/K$ so by Cartan's fixed point theorem (or something similar), $L$ has a fixed point in $G/K$, call it $x_0$.
• The stabilizer of $x_0$ in $G$ is a conjugate of $K$, so $L\subseteq gKg^{-1}$ for some $g\in G$. By maximality of $L$ we get an equality, concluding the proof.

To prove the first stage one defines a Riemannian structure on $G/K$, which turns out to have non-negative sectional curvature $K(X,Y)=-||[X,Y]||^2$, so it is locally $CAT(0)$. However, I could not see why $G/K$ is simply connected.

I could not find a reference giving a neat (or any) proof of that fact. I would be very grateful if someone could provide me with a reference for this fact.

• The group of $\mathbf{R}$-points of a Zariski-connected semisimple matrix group over $\mathbf{R}$ (e.g., ${\rm{SO}}(q)$ for $(V,q)$ with indefinite signature and $\dim V>2$) can be disconnected, but has finite $\pi_0$. The case of finite $\pi_0$ works well (e.g, maximal compacts meet each connected component), but doesn't follow formally from the connected case. A nice treatment of $G$ with $\pi_0(G)$ finite, giving vector subgroups $V_1,\dots,V_n\subset G$ making multiplication $K \times \prod V_i\rightarrow G$ a diffeomorphism, is Ch. XV of Hochschild's "Structure of Lie groups" (see 3.1). – user27920 Jul 26 '14 at 18:45
• $G/K$ is contractible... If $\frak g =\frak k +\frak p$ is a Cartan decomposition of $\frak g$, then $(X,k)\mapsto \exp(X)k$ is a diffeomorphism ${\frak p}\times K\to G$, and as a result $\mathrm{Exp}$ (of the Riemannian connection) is a diffeomorphism ${\frak p}\to G/K$. This is in Cartan, and e.g. in Helgason, Chap. VI, Thm 1.1. – Francois Ziegler Jul 26 '14 at 19:31

Consider the fiber bundle \begin{align} K\rightarrow G\rightarrow G/K. \end{align}

Herein we assume $G$ is connected.

Since $G/K$ is non-positively curved, it is aspherical (i.e. all its higher homotopy groups vanish). Thus, the long exact sequence of homotopy groups induced by the above fiber bundle implies the short exact sequence \begin{align} 1\rightarrow \pi_1(K)\rightarrow \pi_1(G)\rightarrow \pi_1(G/K)\rightarrow 1. \end{align} The inclusion of $K\rightarrow G$ is a homotopy equivalence (this requires proof but is standard and for matrix groups comes from the polar decomposition), and hence the induced map \begin{align} \pi_1(K)\rightarrow \pi_1(G) \end{align} is an isomorphism. Hence, by the previous short exact sequence $\pi_1(G/K)=1$ and $G/K$ is simply connected.

• First of all, thanks! I would be happy to see why $K\rightarrow G$ is a homotopy equivalence in the general case, if it fine. Also, where did $\pi_0(K)$ go? – Miel Sharf Jul 26 '14 at 18:48
• @Andy Sanders: I second Miel's question: how do you prove that $K \rightarrow G$ is a homotopy equivalence without already having the Euclidean space "complement" to $K$ in $G$ (with which one sees directly that $G/K$ is a Euclidean space and $\pi_0(K)=\pi_0(G)$, so no need for consideration of homotopy groups in the argument)? – user27920 Jul 26 '14 at 18:53
• That $K\rightarrow G$ is a homotopy equivalence is a consequence of the Iwasawa decomposition en.wikipedia.org/wiki/Iwasawa_decomposition though this certainly requires some work to prove that there is such a decomposition. I agree with user52824 comment though, once you have this it's pretty clear that $G/K$ is Euclidean. Perhaps I should say that the 'proof' I give makes clear what needs to be shown to see that $G/K$ is simply connected. At this point, I'll leave the rest as an "exercise." – Andy Sanders Jul 26 '14 at 18:58