# Climbing Mount Bourbaki

Thoughts on mathematics

November 27, 2009

## Proof of the fixed point theorem

Posted by Akhil Mathew under differential geometry, MaBloWriMo | Tags: Cartan fixed point theorem, compact Lie groups, cosine inequality, fixed points, negative curvature |

Today we will prove the fixed point theorem, which I restate here for convenience:

Theorem 1 (Elie Cartan) Let ${K}$ be a compact Lie group acting by isometries on a simply connected, complete Riemannian manifold ${M}$ of negative curvature. Then there is a common fixed point of all ${k \in K}$.

There is a Haar measure on ${K}$. In fact, we could even construct this by picking a nonzero alternating ${n}$-tensor (where ${n=\dim K}$) at ${T_e(K)}$, and choosing the corresponding ${K}$-invariant ${n}$-form on ${K}$. This yields a functional ${C(K) \rightarrow \mathbb{R}}$, which we can assume positive by choosing the orientation appropriately. This yields the Haar measure ${d \mu}$ by the Riesz representation theorem.

Now define ${J(q) := \int_K d^2(q,kp) d \mu(k).}$ This is a continuous function ${M \rightarrow \mathbb{R}}$ which has a minimum, because ${J(q)>J(p)}$ for ${q}$ outside some compact set containing ${p}$. Let the minimum occur at ${q_0}$. I claim that the minimum is unique, which will imply that it is a fixed point of ${K}$.

It can be checked that ${J}$ is continuously differentiable; indeed, let ${q_t}$ be a curve. Then ${d^2(q_t,kp)}$ can be computed as in yesterday when ${kp \neq q_t}$; when they are equal, it is still differentiable with zero derivative because of the ${d^2}$. (I am sketching things here because I don’t currently want to dive into the technical details; see Helgason’s book for them.)

So now take ${q_t}$ to be a geodesic joining the minimal point ${q_0}$ to some other point ${q_1}$. Now

$\displaystyle \frac{d}{dt} J(q_t)|_{t=0} = \int_K \frac{d}{dt} d^2(q_t, kp) |_{t=0} d\mu(k) = 0.$

Then we get

$\displaystyle \int_K d(q_0,kp) \cos \alpha d\mu(k) = 0$

where ${\alpha}$ is an appropriate angle as in yesterday’s post. When ${q_0=kp}$, this ${\alpha}$ is not well-defined, but ${d(q_0,kp)=0}$, so it is ok. Now

$\displaystyle \int_K d^2(q_1, k.p) d\mu \geq \int_K \left( d^2(q_0, kp) + d^2(q_0,q_1) - 2 d(q_0, kp) \cos \alpha \right) d \mu(k) .$

This is because of the cosine inequality. But the cosine part vanishes, so this is strictly greater than ${J(q_0)}$. In particular, since ${q_1}$ was arbitrary, ${q_0}$ was a global minimum for ${J}$—and it is thus a fixed point.
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