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…)

I’m going to keep the same notation as before.  In particular, we’re studying how the energy integral behaves with respect to variations of curves.  Now I want to prove the second variation formula when ${c}$ is a geodesic.Now to compute ${\frac{d^2}{d^2 u} E(u)|_{u=0}}$, for further usage. We already showed$\displaystyle E'(u) = \int_I g\left( \frac{D}{dt} \frac{\partial}{\partial u} H(t,u), \frac{\partial}{\partial t} H(t,u) \right)$ Differentiating again yields the messy formula for ${E''(u)}$:

$\displaystyle \int_I g\left( \frac{D}{du} \frac{D}{dt} \frac{\partial}{\partial u} H, \frac{\partial}{\partial t} H \right) + \int_I g\left( \frac{D}{dt} \frac{\partial}{\partial u} H, \frac{D}{du}\frac{\partial}{\partial t} H(t,u) \right).$

Call these ${I_1(u), I_2(u) }$.

${I_2}$

Now ${I_2(0)}$ is the easiest, since by symmetry of the Levi-Civita connection we get$\displaystyle I_2(0) = \int g\left( \frac{D}{dt} \frac{\partial}{\partial u} H(t,u), \frac{D}{dt}\frac{\partial}{\partial u} H(t,u) \right) = \int g\left( \frac{D}{dt} V, \frac{D}{dt} V \right).$ For vector fields along ${c}$ ${E,F}$ with ${E(a)=F(a)=E(b)=F(b)=0}$, we have$\displaystyle \int g\left( \frac{D}{dt} E, \frac{D}{dt} F \right) = - \int g\left( \frac{D^2}{dt^2} E , F \right).$ This is essentially a forum of integration by parts. Indeed, the difference between the two terms is

$\displaystyle \frac{d}{dt} g\left( \frac{D}{dt} E, F \right).$

So if we plug this in we get

$\displaystyle \boxed{ I_2(0) = -\int g\left( \frac{D^2}{dt^2} V , V \right).}$

${I_1}$

Next, we can write

$\displaystyle I_1(0) = \int_I g\left( \frac{D}{du} \frac{D}{dt} \frac{\partial}{\partial u} H(t,u) |_{u=0}, \dot{c}(t) \right)$ Now ${R}$ measures the difference from commutation of ${\frac{D}{dt}, \frac{D}{du}}$. In particular this equals

$\displaystyle \int_I g\left( \frac{D}{dt} \frac{D}{du} \frac{\partial}{\partial u} H(t,u) |_{u=0}, \dot{c}(t) \right) + \int_I g\left( R(V(t), \dot{c}(t)) V(t), \dot{c}(t)) \right).$

By antisymmetry of the curvature tensor (twice!) the second term becomes

$\displaystyle \int_I g\left( R( \dot{c}(t), V(t)) \dot{c}(t), V(t), \right).$

Now we look at the first term, which we can write as

$\displaystyle \int_I \frac{d}{dt} g\left( \frac{D}{du} \frac{\partial}{\partial u} H(t,u), \dot{c}(t)\right)$

since ${\ddot{c} \equiv 0}$. But this is clearly zero because ${H}$ is constant on the vertical lines ${t=a,t=b}$. If we put everything together we obtain the following “second variation formula:”

Theorem 1 If ${c}$ is a geodesic, then$\displaystyle \boxed{\frac{d^2}{du^2}|_{u=0} E(u) = \int_I g\left( R( \dot{c}(t), V(t)) \dot{c}(t) - \frac{D^2 V}{Dt^2}, V(t) \right).}$

Evidently that was some tedious work, and the question arises: Why does all this matter? The next goal is to use this to show when a geodesic cannot minimize the energy integral—which means, in particular, that it doesn’t minimize length. Then we will obtain global comparison-theoretic results.

Now, let’s finish the proof of the Hopf-Rinow theorem (the first one) started yesterday. We need to show that given a Riemannian manifold ${(M,g)}$ which is a metric space ${d}$, the existence of arbitrary geodesics from ${p}$ implies that ${M}$ is complete with respect to ${d}$. Actually, this is slightly stronger than what H-R states: geodesic completeness at one point ${p}$ implies completeness.

The first thing to notice is that ${\exp: T_p(M) \rightarrow M}$ is smooth by the global smoothness theorem and the assumption that arbitrary geodesics from ${p}$ exist. Moreover, it is surjective by the second Hopf-Rinow theorem.

Now fix a ${d}$-Cauchy sequence ${q_n \in M}$. We will show that it converges. Draw minimal geodesics ${\gamma_n}$ travelling at unit speed with

$\displaystyle \gamma_n(0)=p, \quad \gamma_n( d(p,q_n)) = q_n.$  (more…)

Ok, yesterday I covered the basic fact that given a Riemannian manifold ${(M,g)}$, the geodesics on ${M}$ (with respect to the Levi-Civita connection) locally minimize length. Today I will talk about the phenomenon of “geodesic completeness.”

Henceforth, all manifolds are assumed connected.

The first basic remark to make is the following. If ${c: I \rightarrow M}$ is a piecewise ${C^1}$-path between ${p,q}$ and has the smallest length among piecewise ${C^1}$ paths, then ${c}$ is, up to reparametrization, a geodesic (in particular smooth). The way to see this is to pick ${a,b \in I}$ very close to each other, so that ${c([a,b])}$ is contained in a neighborhood of ${c\left( \frac{a+b}{2}\right)}$ satisfying the conditions of yesterday’s theorem; then ${c|_{[a,b]}}$ must be length-minimizing, so it is a geodesic. We thus see that ${c}$ is locally a geodesic, hence globally.

Say that ${M}$ is geodesically complete if ${\exp}$ can be defined on all of ${TM}$; in other words, a geodesic ${\gamma}$ can be continued to ${(-\infty,\infty)}$. The name is justified by the following theorem:

Theorem 1 (Hopf-Rinow)

The following are equivalent:

• ${M}$ is geodesically complete.
• In the metric ${d}$ on ${M}$ induced by ${g}$ (see here), ${M}$ is a complete metric space (more…)

Fix a Riemannian manifold with metric ${g}$ and Levi-Civita connection ${\nabla}$. Then we can talk about geodesics on ${M}$ with respect to ${\nabla}$. We can also talk about the length of a piecewise smooth curve ${c: I \rightarrow M}$ as

$\displaystyle l(c) := \int g(c'(t),c'(t))^{1/2} dt .$

Our main goal today is:

Theorem 1 Given ${p \in M}$, there is a neighborhood ${U}$ containing ${p}$ such that geodesics from ${p}$ to every point of ${U}$ exist and also such that given a path ${c}$ inside ${U}$ from ${p}$ to ${q}$, we have

$\displaystyle l(\gamma_{pq}) \leq l(c)$

with equality holding if and only if ${c}$ is a reparametrization of ${\gamma_{pq}}$.

In other words, geodesics are locally path-minimizing.   Not necessarily globally–a great circle is a geodesic on a sphere with the Riemannian metric coming from the embedding in $\mathbb{R}^3$, but it need not be the shortest path between two points. (more…)

Ok, we know what connections and covariant derivatives are. Now we can use them to get a map from the tangent space ${T_p(M)}$ at one point to the manifold ${M}$ which is a local isomorphism. This is interesting because it gives a way of saying, “start at point ${p}$ and go five units in the direction of the tangent vector ${v}$,” in a rigorous sense, and will be useful in proofs of things like the tubular neighborhood theorem—which I’ll get to shortly.

Anyway, first I need to talk about geodesics. A geodesic is a curve ${c}$ such that the vector field along ${c=(c_1, \dots, c_n)}$ created by the derivative ${c'}$ is parallel. In local coordinates ${x_1, \dots, x_n}$, here’s what this means. Let the Christoffel symbols be ${\Gamma^k_{ij}}$. Then using the local formula for covariant differentiation along a curve, we get

$\displaystyle D(c')(t) = \sum_j \left( c_j''(t) + \sum_{i,k} c_i'(t) c_k'(t) \Gamma^j_{ij}(c(t)) \right) \partial_j,$

so ${c}$ being a geodesic is equivalent to the system of differential equations

$\displaystyle c_j''(t) + \sum_{i,k} c_i'(t) c_k'(t) \Gamma^j_{ij}(c(t)) = 0, \ 1 \leq j \leq n.$ (more…)