Cartan fixed point theorem | Climbing Mount Bourbaki

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.

I am now aiming to prove an important fixed point theorem:

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 are several ingredients in the proof of this result. These will provide examples of the techniques that I have discussed in past posts.

Geodesic triangles

Let ${M}$ be a manifold of negative curvature, and let ${V}$ be a normal neighborhood of ${p \in M}$ ; this means that ${\exp_p}$ is a diffeomorphism of some neighborhood of ${0 \in T_p(M)}$ onto ${V}$ , and any two points in ${V}$ are connected by a unique geodesic. (This always exists by the normal neighborhood theorem, which I never proved. However, in the case of Cartan’s fixed point theorem, we can take ${V=M}$ by Cartan-Hadamard.)

So take ${a=p,b,c\in V}$ . Draw the geodesics ${\gamma_{ab}, \gamma_{ac}, \gamma_{bc}}$ between the respective pairs of points, and let ${\Gamma_{ab}, \Gamma_{bc}}$ be the inverse images in ${T_p(M) = T_a(M)}$ under ${\exp_p}$ . Note that ${\Gamma_{ab}, \Gamma_{ac}}$ are straight lines, but ${\Gamma_{bc}}$ is not in general. Let ${a',b',c'}$ be the points in ${T_p(M)}$ corresponding to ${a,b,c}$ respectively. Let ${A}$ be the angle between ${\gamma_{ab}, \gamma_{ac}}$ ; it is equivalently the angle at the origin between the lines ${\Gamma_{ab}, \Gamma_{ac}}$ , which is measured through the inner product structure.

Now ${d(a,b) = l(\gamma_{ab}) = l(\Gamma_{ab}) = d(a',b')}$ from the figure and since geodesics travel at unit speed, and similarly for ${d(a,c)}$ . Moreover, we have ${d(b,c) = l(\gamma_{bc}) \geq l(\Gamma_{bc}) \leq d(b',c')}$ , where the first inequality comes from the fact that ${M}$ has negative curvature and ${\exp_p}$ then increases the lengths of curves; this was established in the proof of the Cartan-Hadamard theorem.

We have evidently by the left-hand-side of the figure

$\displaystyle d(b',c')^2 = d(a',c')^2 + d(a',b')^2 - 2d(a',b')d(a',c') \cos A.$

In particular, all this yields

$\displaystyle \boxed{d(b,c)^2 \geq d(a,c)^2 + d(a,b)^2 - 2d(a,b)d(a,c) \cos A.}$

So we have a cosine inequality.

There is in fact an ordinary plane triangle with sides ${d(a,b), d(b,c),d(a,c)}$ , since these satisfy the appropriate inequalities (unless ${a,b,c}$ lie on the same geodesic, which case we exclude). The angles ${A',B',C'}$ of this plane triangle satisfy

$\displaystyle A \leq A'$

by the boxed equality. In particular, if we let ${B}$ (resp. ${C}$ ) be the angles between the geodesics ${\gamma_{ab}, \gamma_{bc}}$ (resp. ${\gamma_{ac}, \gamma_{bc}}$ ), then by symmetry and ${A'+B'+C=\pi}$

$\displaystyle \boxed{A + B+ C \leq \pi.}$

This is a fact which I vaguely recall from popular-math books many years back. The rest is below the fold. (more…)

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Climbing Mount Bourbaki

Proof of the fixed point theorem

Towards Cartan’s fixed point theorem: Cosine inequalities and derivative of the distance

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