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

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

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 be a manifold of negative curvature, and let be a **normal neighborhood** of ; this means that is a diffeomorphism of some neighborhood of onto , and any two points in 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 by Cartan-Hadamard.)

So take . Draw the geodesics between the respective pairs of points, and let be the inverse images in under . Note that are straight lines, but is not in general. Let be the points in corresponding to respectively. Let be the angle between ; it is equivalently the angle at the origin between the lines , which is measured through the inner product structure.

Now from the figure and since geodesics travel at unit speed, and similarly for . Moreover, we have , where the first inequality comes from the fact that has negative curvature and 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

In particular, all this yields

So we have a **cosine inequality**.

There is in fact an ordinary plane triangle with sides , since these satisfy the appropriate inequalities (unless lie on the same geodesic, which case we exclude). The angles of this plane triangle satisfy

by the boxed equality. In particular, if we let (resp. ) be the angles between the geodesics (resp. ), then by symmetry and

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

**Derivative of the distance function **

The next ingredient for the proof of Cartan’s fixed point theorem is the derivative of the distance function as one point varies along a curve.

*Henceforth I will assume that is simply connected, complete, and of negative curvature.*

Let be a curve in , parametrized by arc length, not containing the point . Let .

Proposition 2We have

for the angle between the curve and the unique geodesic from to .

First of all, it’s enough to compute ; this will fit in better with the law of cosines.

Now is a diffeomorphism. So we can consider the curve in corresponding to . By Hopf-Rinow, we have

where the latter distance is in .

Next, we compute

as shown in the figure.

As , the first term is and contributes nothing to the derivative. Now is approximately for small, so putting this together we get

as in the figure

and therefore

Now is the length of the projection of onto the line from to . Let the components be , where is the component in the direction of to , and is orthogonal to it, so and . In the curve , we have orthogonal to and in the direction of the geodesic . Also, ; straight lines through the origin get sent to geodesics in , which travel at constant speed.

Note that by the assumption on .

In particular, is the length —which we have just shown is .

I have followed Helgason here.

November 27, 2009 at 9:56 pm

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March 16, 2011 at 11:35 pm

This is an interesting paper. Thanks!