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  


\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.