Let $(M^n,g)$ be a Riemannian manifold with non-empty smooth boundary $\partial M$. For any two points $x,y\in M$, the distance between $x$ and $y$ may be defined as $$ d(x,y)=\inf_\gamma Length(\gamma), $$ where the infimum is taken over all $C^1$ curves lying in $M$. Can we prove there exists a path in the closure $\bar{M}$ which achieves $d(x,y)$? And the length-minimizing path is piecewise $C^1$? Note that since $\partial M$ is non-empty, the length-minimizing path (if exists) may intersect the boundary. Any reference for this question?
2
-
1$\begingroup$ Interpreting the closure as $M \cup \partial M$, it seems to me that, in general, minimizers can have corners. Any non-convex region of $\mathbb{R}^2$ is gives an example. $\endgroup$– RazielCommented Aug 2, 2018 at 16:28
-
$\begingroup$ @Raziel You are right. The minimizer may only be $C^0$ but piecewise $C^1$. $\endgroup$– Changwei XiongCommented Aug 2, 2018 at 21:57
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
9
As stated the question is unclear. What do you mean by the "closure"? Metric completion? If so, what do you mean by piecewise $C^1$? Do you know the answer when the boundary is empty? The boundary does not seem to matter here.
The standard result of this type is the Hopf-Rinow theorem which implies that any two points in locally compact complete length space can be joined by a minimizing geodesic if they can be joined by a finite length curve. See Theorem 2.5.23 in Course of metric geometry, by Burago-Burago-Ivanov.
Note that the metric completion of the non-complete Riemannian manifold need not be locally compact (think of the universal cover of the once-punctured Euclidean plane).
answered Aug 2, 2018 at 12:55
-
1$\begingroup$ Although, the metric completion of the universal cover of the once-punctured plane does have a minimizing geodesic between any two points, despite the failure of local compactness. $\endgroup$ Commented Aug 2, 2018 at 13:20
-
1$\begingroup$ @LeeMosher: yes, in this case the completion is a complete CAT(0) space. It would be nice to have an example of a completion with no minimal geodesic between some points. $\endgroup$ Commented Aug 2, 2018 at 13:32
-
$\begingroup$ @IgorBelegradek Many thanks for the answer and the correction. Yes by closure I meant metric completion. I thought that the shortest path should be $C^0$ and piecewise $C^1$. But it seems that it should be at least $C^1$ if we are in smooth manifolds with smooth boundary. And by definition the manifold with smooth boundary has local chart along the boundary which is mapped onto a half space in Euclidean space, and so $\bar{M}$ may be locally compact. The Hopf-Rinow theorem in this form is what I need. Thank you. $\endgroup$ Commented Aug 3, 2018 at 4:17
-
$\begingroup$ @ChangweiXiong: you didn't answer my question on what is meant by $C^1$ if $\bar M$ is interpreted as the metric completion. Conceivably the minimizer can spend a lot of time in the nonsmooth part of $\bar M$ where the notion of $C^1$ makes no sense. The boundary is a separate issue. $\endgroup$ Commented Aug 3, 2018 at 13:08
-
$\begingroup$ @IgorBelegradek: I am sorry for not answering your question. In fact I am not familiar with the delicate definition of $C^1$ in metric geometry. I think by $C^1$ I only mean a $C^1$ map from an interval to $\bar{M}$. I don't know if there is a problem on this notion. And may I draw your attention to Theorem 1 in the paper "Alexander, Ralph and Alexander, S., Geodesics in Riemannian manifolds-with-boundary. Indiana Univ. Math. J. 30 (1981), no. 4, 481--488"? There they used the $C^1$ notion. Maybe I can follow their definition. Thanks a lot. $\endgroup$ Commented Aug 4, 2018 at 0:52
$\begingroup$
$\endgroup$
0
Alexander, Berg, Bishop: The Riemannian obstacle problem. https://projecteuclid.org/download/pdf_1/euclid.ijm/1255989406
