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.


Thanks in advance, Miel.



Maximal compact subgroups of a semisimple Lie group are conjugate
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.


