Let $M$ be a Riemannian manifold with boundary. Can it isometrically embed into a Riemannian manifold without boundary of the same dimension?
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2$\begingroup$ You can double $M$ by gluing two copies of it. The smoothness of the double was discussed on the site before, for example here. $\endgroup$– Ivan IzmestievCommented Oct 16, 2016 at 18:31
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4$\begingroup$ Any smooth manifold $M$ with boundary can be smoothly embedded into an open manifold $\widetilde{M}$, in fact the interior of $M$ itself. Extending the Riemannian metric from $M$ to $\widetilde{M}$ can be done in essentially the same way a smooth function on $M$ can be extended to $\widetilde{M}$. $\endgroup$– Deane YangCommented Oct 16, 2016 at 21:13
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3 Answers
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This paper by Pigolla and Veronelli contains a proof of that, with the additional constraint that the larger manifold is complete (which is not difficult to achieve).
Let me note that the main point of the paper is to consider this problem when we ask the extension to satisfy some curvature bound which hold on the manifold with boundary. There are many entertaining questions of this kind, and I recommend to to take a look at the paper to every intrigued Riemannian geometer.
answered Oct 17, 2016 at 7:43
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A simple direct proof is as follows: Let $M^o = M\setminus \partial M$ be the interior of $M$. Choose a vector field $X$ on $M$ which along the boundary is non-zero and point into the interior, and which is 0 off some neighborhood of the boundary. Use the flow of $X$ to flow $M$ into $M^o$. It remains to extend the Riemannian metric from the image of $M$ in $M^o$ to the whole of $M^o$.
This proof also works for manifolds with corners.
answered Oct 17, 2016 at 19:16
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$\begingroup$ @PeterMichnor, a similar vector field and a partition of unity. $\endgroup$ Commented Oct 18, 2016 at 3:22
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This paper at the annals describes such an embedding on page 1097 for compact Riemannian manifolds, although the main results are embeddings between manifolds with boundaries.
answered Oct 16, 2016 at 17:55
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$\begingroup$ The article places the problem in a very restrictive context: the manifolds are $2$-dimensional, simply-connected, their geodesics have no conjugate points and the boundaries are strictly convex. Well, with enough convenient hypotheses everything can be proven... $\endgroup$– Alex M.Commented Jun 13, 2018 at 17:25