Geodesics are locally length-minimizing November 13, 2009

Posted by Akhil Mathew in differential geometry, MaBloWriMo.
Tags: , ,

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.

To prove this will require a bit of work. Here is a warm-up lemma we shall need.

Lemma 2

Let ${c: I \rightarrow M}$ be a curve, and ${V,W}$ vector fields along ${c}$. Then$\displaystyle \frac{d}{dt} g(V,W) = g\left( \frac{D}{dt} V,W\right) + g\left( V,\frac{D}{dt} W\right).$

To prove this, write ${V = \sum v^i(t) Q^i(t), W =\sum w^i(t) Q^i(t)}$ where the ${ Q^i}$ are parallel and orthonormal at ${c(0)}$ (hence along ${c}$). Then

$\displaystyle g(V,W) = \sum v^i w^i$

while

$\displaystyle \frac{D}{dt} V = \sum \dot{v}^i Q^i, \frac{D}{dt} W = \sum \dot{w}^i Q^i$

by the rules for connections. Then the statement of the lemma becomes merely the product rule.

A corollary of this lemma is that geodesics ${\gamma}$ have constant speed ${| \dot{\gamma}|_g}$.

Now we move on to proving the theorem. First of all, let’s choose ${U}$ such that it is the diffeomorphic image of the unit ball ${B_r(0) \subset T_p(M)}$ under the exponential map ${\exp_p}$; this is because the exponential map’s differential at zero is the identity. Then every path ${c}$ as above can be written in the form ${c(t) = \exp_p( r(t) s(t))}$, where ${s(t) \in T_p(M)}$ and ${|{s(t)}| = 1}$ (with the norm on ${T_p(M)}$ coming from the inner product ${g}$). Now this is a geodesic up to reparametrization iff ${s}$ is constant.

We have

$\displaystyle c'(t) = d\exp_p( r'(t)s(t)) + d\exp_p( r(t) s'(t)) = A+B.\ \ \ \ \ (1)$

Motivated by this, consider the map ${F: J \times S^{n-1} \rightarrow M}$ where ${J}$ is a small interval containing the origin, with

$\displaystyle (u,w) \rightarrow \exp( u w).$

Lemma 3 (Gauss) We have ${g(A,B) = 0}$.

Let ${v}$ trace out a path in ${S^{n-1}}$ which is also a 1-dimensional closed submanifold with tangent vector ${s'(t)}$ at ${t}$. Now ${g(A,B)}$ is a function of ${t}$, which is

$\displaystyle G(u,v) := g\left( \frac{\partial}{\partial u} F(u,v), \frac{\partial}{\partial v} F(u,v) \right)$

evaluated at ${u=r(t),v=t}$. So it will be enough to prove that the vector field ${G(u,v)}$ along the surface ${F(u,v)}$ vanishes. Take the partial derivative $\frac{\partial}{\partial u} G(u,v)$ with respect to ${u}$, using the first lemma:

$\displaystyle g\left( \frac{D}{\partial u} \frac{\partial}{\partial u} F(u,v), \frac{\partial}{\partial v} F(u,v) \right) + g\left( \frac{\partial}{\partial u} F(u,v), \frac{D}{\partial u} \frac{\partial}{\partial v} F(u,v) \right) .\ \ \ \ \ (2)$

The first term vanishes by definition of a geodesic—the image of a line in ${J}$ with a fixed point of ${S^{n-1}}$ gets sent via ${F}$ to a geodesic. As for the second, we can use the Clairaut-like theorem for symmetric connections to get that this equals

$\displaystyle g\left( \frac{\partial}{\partial u} F(u,v), \frac{D}{\partial v} \frac{\partial}{\partial u} F(u,v) \right) .$

Now in here we can pull out the ${\frac{D}{\partial v}}$ to get

$\displaystyle \frac{1}{2} \frac{\partial}{\partial v} g\left( \frac{\partial}{\partial u} F(u,v), \frac{\partial}{\partial u} F(u,v)\right) = \frac{1}{2} \frac{\partial}{\partial v} 1 = 0$

(Recall that geodesics move at constant speed $|v|$.) Going back a few equations to (2) shows that ${G}$ is constant in ${u}$. Since ${G(0,v) = 0}$ for all ${v}$ because of the term ${\frac{\partial}{\partial v} F(0,v) = 0}$ (all geodesics start at ${p}$!), we find that ${G(u,v) \equiv 0}$. This implies the lemma.

With notation as in (1), we have by the Gauss lemma

$\displaystyle l(c) = \int g(A+B,A+B)^{1/2} = \int g(A,A)^{1/2} + \int g(B,B)^{1/2}.$

This will be minimized precisely when ${B \equiv 0}$, which is when ${c}$ is a geodesic.