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The following theorem is well known in the literature:

Let $M$ and $N$ be riemannian manifolds and let $f : M \to N$ be a local isometry. If $M$ is complete and $N$ is connected, then $f$ is a covering map.

My question is: does the same theorem hold when we assume that $M$ and $N$ are now riemannian manifolds with boundary?

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  • $\begingroup$ Just consider the inclusion $[0,1]\subset [0,2]$. $\endgroup$ – YCor Dec 10 '19 at 13:14
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I think it does not hold: For example, let $M$ be any complete Riemannian manifold with connected, totally geodesic boundary, and let $N$ be its double—the disjoint union of two copies of $M$ glued along their boundary. The inclusion map $M\to N$ is a local isometry, but is not a covering map.

A Riemannian manifold with boundary should be complete just when it is complete as a metric space.

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  • $\begingroup$ But in this case $N$ does not have a boundary, which is assumed in my question. $\endgroup$ – Eduardo Longa Dec 10 '19 at 2:15
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    $\begingroup$ @EduardoLonga: could one not use the same idea for a riemannian manifold with two isomorphic boundary components, gluing just one? More specifically, I'm thinking of a cylinder $[0, 1] \times \mathbb{S}^1$, and gluing two of these: one along $\{0\} \times \mathbb{S}^1$ and the other along $\{1\} \times \mathbb{S}^1$. $\endgroup$ – Eric Canton Dec 10 '19 at 2:18
  • $\begingroup$ Or, less complicated: take $f: M \to N$ to be the cylinder $M = [0, 1/2] \times \mathbb{S}^1$ including into $N = [0, 1] \times \mathbb{S}^1$. $\endgroup$ – Eric Canton Dec 10 '19 at 2:19
  • $\begingroup$ Related question: is a local isometry a local diffeomorphism, even in the boundary case? If so, a local isometry must send boundary to boundary. $\endgroup$ – Eduardo Longa Dec 10 '19 at 2:20

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