Solving the geodesic equation with Sobolev space techniques

Question

I was wondering if there is a proof that every Riemannian metric on $\mathbb{R}^n$ is complete using Sobolev space techniques. Here is the start of an argument. Classically, one can reduce the geodesic problem to the problem of finding minimizers of the energy functional $$E(\gamma) = \int_a^b ||\gamma'(t)||^2 \, dt,$$ where $\gamma$ is a smooth curve over $[a, b].$

Specifically, if $g$ is our Riemannian metric and $d = d_g(p,q),$ we are looking to prove that $E$ attains a minima on the space of all functions $\gamma : [0, d]\to \mathbb{R}^n$ which have $\gamma(0) = p, \gamma(d) =q.$

Using standard ideas, I can prove the following:

1. With the trace operator, interpret this space of paths as a subspace of $W^{1,2}([0, d], \mathbb{R}^n).$
2. Prove that $E$ is weakly lower semicontinuous (the crucial point beings it convexity).
3. Using the weak lower semicontinuity, show that $E$ has to attain a minima (this is not immediate, but if one uses the fact that the Sobolev norm is just $L^2+E^2 \leq dE + E^2$ for $E$ the length functional, then it is not too hard).

This minima is, however, not guaranteed to be a classical function. Is it possible to show that this minima is a classical function, thereby solving the geodesic problem?

I am well aware of the standard way to solve this: use Euler-Lagrange to derive the geodesic equation, and then just apply classical ODE theory. However, I was interested if there was a way of finishing this approach, mostly as a way of testing my knowledge of Sobolev space techniques towards a problem which hopefully should be easy to solve.

Answer 1

There are several mistakes in your understandings.

• there are Riemannian metrics on $\mathbb R^n$ that is not complete: as suggested in the comment, take a diffeomorphism $\phi : \mathbb R^n \to B$ from $\mathbb R^n$ to a ball $B$ and let $g = \phi^*g_0$, where $g_0$ is the Euclidean metric on $B$. This $g$ is not complete.

• there seems to be a misunderstanding about the definition of completeness. The fact that every two points can be joined by a shortest geodesic does not imply that the metric is complete. The statement is a corollary of geodesically completeness, but not equivalent. Again, the ball with the Euclidean metric is a counter-example.

• Even if you want just this weaker corollary that any two points on $(\mathbb R^n, g)$ can be joined by a shortest geodesic, this is false: take a diffeomorphism between $\varphi: \mathbb R^n \to C$, where $C$ is a non-convex open subset in $\mathbb R^n$ and consider $g= \varphi^* g_0$.

So what's wrong with your argument?

• The space $W^{1,2} ([0, d] , \mathbb R^n)$ is not suitable for the length minimizing problem on $(\mathbb R^n, g)$. You will need to consider the space $$ W^{1,2}_g ([0, d], \mathbb R^n) = \left\{ \gamma : [0, d] \to \mathbb R^n : \int_0^d \sqrt{g(\gamma', \gamma')} dt <\infty\right\},$$ so that the length functional is at least well-defined. (note that the length of a curve $\gamma \in W^{1,2} ([0, d], \mathbb R^n)$ measured using $g$ might be $+\infty$).

• Existence of minimizers in $W^{1, 2}_g([0, d], \mathbb R^n)$ is hard, if not impossible in general, since we don't have the compactness theorem for this "Sobolev space". Take the simpler example where $g$ is conformal, that is $g = e^{-f} g_0$ where some $f \in C^\infty (\mathbb R^n)$. Then $$ \int_0^ d \sqrt{g(\gamma', \gamma')} dt = \int_0 ^d |\gamma'| e^{-f} dt.$$ (In this case, this is called a weighted sobolev space). Take e.g. $f(x) = |x|^2$, there is not even $C^0$ estimates of the form: $$\sup _{[0,d]} |\gamma (t) - \gamma (0)| \le C \int_0^d |\gamma'| e^{-|x|^2} dt.$$ There might be some weight version of this, but that's not sufficient for the minimization process.