I am a little rusty on my Riemannian geometry. In addressing a problem in PDE's I came across a situation that I cannot reconcile with the Hopf-Rinow Theorem. If $\Omega \subset \mathbb{R}^n$ is a bounded, open set with smooth boundary, then $\mathbb{R}^n - \Omega$ is a Riemannian manifold with smooth boundary. Since $\mathbb{R}^n - \Omega$ is closed in $\mathbb{R}^n$, it follows that $\mathbb{R}^n - \Omega$ is a complete metric space. However, the Hopf-Rinow Theorem seems to indicate that $\mathbb{R}^n - \Omega$ (endowed with the usual Euclidean metric) is not a complete metric space since not all geodesics $\gamma$ are defined for all time. Am I missing something here? Do the hypotheses of the Hopf-Rinow theorem have to be altered to accommodate manifolds with boundary?

  • $\begingroup$ For manifolds with boundary, completeness as a metric space and completeness in the sense that all geodesic curves extend forever, are no longer equivalent. Clearly, the boundary can ensure that all Cauchy sequences converge (possibly to a boundary point), and just as clearly, there is going to be some geodesic curves that "run into" the boundary and cannot be continued beyond it. $\endgroup$ – Jeppe Stig Nielsen Jul 13 at 1:44

Hopf-Rinow concerns, indeed, Riemannian manifolds with no boundary.

  • $\begingroup$ Thanks! I thought that was the case, but I wanted to check. Just out of curiosity, if one allows for say broken geodesics (allowing geodesics to be reflected off the boundary) using the metric, do you know if you can extend the Hopf-Rinow theorem to manifolds with boundary? $\endgroup$ – John May 27 '13 at 21:33
  • $\begingroup$ Hmm, I don't know about this. It starts to sound like various billiard-ball problems ... $\endgroup$ – Ted Shifrin May 27 '13 at 21:35
  • $\begingroup$ @John I do not know precisely what you ask, but if I consider $M = \{ (x,y)\in\mathbb{R}^2 \mid y \le x^2 \}$ with the usual metric (inherited from $\mathbb{R}^2$), then starting from the point $p=(0,-1)$, or from any point of $M$, I guess, there is clearly going to be points in $M$ that I cannot hit. Each geodesic curve will reflect at most once on the boundary (parabola). But maybe you thought of compact manifolds with boundary? $\endgroup$ – Jeppe Stig Nielsen Jul 13 at 2:02

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