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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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…)