About the proof of Proposition 6.45 on Ziller's notes

Question

I'm currently going through W. Ziller's notes on symmetric spaces, and I've come across one argument he makes which I can't seem to wrap my head around.

Suppose $(G,K)$ is a symmetric pair of the noncompact type with Cartan decomposition $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$ (that is, the Killing form $B$ is negative definite in $\mathfrak{p}$). I wish to prove that $K$ is a maximal compact subgroup of $G$.

I'll briefly describe Ziller's proof as follows: given that $f\colon \mathfrak{p}\times K\to G$, $f(X,g)=\operatorname{Exp}(X)g$ is a diffeomorphism, where $\operatorname{Exp}$ stands for the Lie exponential map, we suppose a compact subgroup $K\subseteq L\subseteq G$. Since $L$ is compact, we can define an inner product over $\mathfrak{g}$ such that for every $X\in \mathfrak{l}=\operatorname{Lie}(L)$, $\operatorname{ad}_{X}$ is skew-symmetric. This implies that $B|_{\mathfrak{l}}$ is negative semidefinite, and in reality it is negative definite, since its kernel is $\mathfrak{z}(\mathfrak{g})\cap\mathfrak{l}=0$. Because of this, we must have $\mathfrak{k}=\mathfrak{l}$, so that $K=L^{0}$, the identity component of $L$ (hence, $K$ is normal in $L$). Therefore, $L/K$ is a $0$-dimensional compact Lie group (that is, a finite group).

Now the part that I don't understand

Take a nontrivial element $gK\in L/K$, which corresponds to some $g\in L\setminus K$. Since $f$ is a diffeomorphism, we can write $g=\operatorname{Exp}(X)y$ for some uniquely determined $X\in \mathfrak{p}$, $y\in K$. Then, since $L/K$ is finite, we get that for some $n>0$, $g^{n}=\operatorname{Exp}(nX)y'\in K$, so that $\operatorname{Exp}(nX)y'=\operatorname{Exp}(0)z$ for $z\in K$, contradicting that $f$ is injective.

Here is my question: how can we be sure that $g^{n}$ admits an expression as above? Since elements of $G$ don't commute, it doesn't seem obvious to me that we can make such a claim.

Thank you in advance!

Answer 1

Since $K$ is the identity component of $L$, $K$ is normal in $L$. Thus, $g$ normalizes $K$.

This implies that $\exp(X)$ also normalizes $K$. Indeed, if $k\in K$, then $ \exp(X) k \exp(X)^{-1} = \exp(X)y (y^{-1} k y) (\exp(X) y)^{-1} = g(y^{-1} k y)g^{-1}$. Since both $y,k\in K$, and $g$ normalizes $K$, the result follows.

The normalizing condition can equivalently stated by saying that for any $k\in K$, there is a $k'\in K$ with $\exp(X) k = k' \exp(X)$. In other words, we can pass $\exp(X)$ across elements of $K$ as long as we're willing to change the element of $K$. Applying this to $g^n = (\exp (X)y)(\exp(X)y)...(\exp(X)y)$, we can rewrite this as $\exp(x)^n y' = \exp(nx) y'$ for some $y'\in K$.

Answer 2

Recently, I've been taking another look at these basic results on symmetric spaces of the noncompact type. With this proposition in particular, I came up with a more geometric proof using the Cartan fixed point theorem. I'm posting it as an answer to my question in case someone is interested.

Let $(G,K)$ be a symmetric pair of the noncompact type, $M=G/K$ the resulting symmetric space, and $o=eK$. Just before proving the result discussed in the post, Ziller proves that the Riemannian exponential map $\operatorname{exp}_{o}\colon \mathfrak{p}\to M$ is a diffeomorphism, so $M$ is a Hadamard manifold.

We will apply the Cartan fixed point theorem, which states that if $N$ is a Hadamard manifold and $H$ is a group acting by isometries on $N$, then $H$ has a fixed point if and only if $H$ has a bounded orbit. This can be found for example in Patrick Eberlein's ''Geometry of Nonpositively Curved Manifolds'', in which the author derives it from the law of cosines.

Now, assume $L$ is a compact subgroup of $G$ containing $K$. Since $L$ is compact, all $L$-orbits are bounded subsets of $M$, so by the Cartan fixed point theorem we can find a point $q\in M$ such that $L\cdot q = \{q\}$. In particular, $L\subseteq G_{q}$, where $G_{p}$ denotes the isotropy subgroup of $p$.

Write $q=g\cdot o$ for some $g\in G$. Then $G_{q}=gG_{o}g^{-1}=gKg^{-1}$, so $K\subseteq L \subseteq gKg^{-1}$. By compactness of $K$, one has $K=gKg^{-1}$, which means $K=L$. Thus, $K$ is a maximal compact subgroup.