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A celebrated theorem of Nash is that every $C^k$ ($k\geq 3$) Riemannian manifold $(M,g)$ can be isometrically embedded into some Euclidean space $\mathbb{R}^d$ for some $d\in \mathbb{N}$. However, I have not been able to track down any results when $M$ is a manifold with boundary. Is there a result of the same ilk when $M$ has nonempty boundary?

Perhaps more specifically, is it true that any Riemannian manifold with boundary can be isometrically and properly embedded into some $\mathbb{R}^d$? (Properly is included so that in particular, the image of the boundary corresponds to the topological closure of the embedded manifold in $\mathbb{R}^d$.)

This seems like a natural question to ask, and I was wondering if it has already been studied.

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    $\begingroup$ I believe it's straightforward to embed $M$ smoothly into an open manifold $N$ (maybe by attaching an open collar to $\partial M$?). It is then straightforward to extend the Riemannian metric smoothly to $N$. Now you can apply that Nash theorem to $N$. $\endgroup$
    – Deane Yang
    Commented Nov 2, 2019 at 20:01
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    $\begingroup$ Or do the reverse of what Deane suggests: remove the boundary, embed that Riemann manifold, then check to see if you can ensure that extends to the manifold with boundary. $\endgroup$
    – Ryan Budney
    Commented Nov 2, 2019 at 20:02
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    $\begingroup$ @RyanBudney, that's a good point. You can indeed do it, but, in doing so, you're effectively showing that the standard proof of the Nash theorem for an open manifold is easily adapted to a manifold with boundary. $\endgroup$
    – Deane Yang
    Commented Nov 3, 2019 at 20:56
  • $\begingroup$ For applications, one might want the manifold to lie in a halfspace with the boundary of $M$ mapping to the boundary of that halfspace. Is that possible as well? $\endgroup$
    – Sebastian Goette
    Commented Nov 4, 2019 at 10:52
  • $\begingroup$ @DeaneYang could you make an answer from your comment (so the question would disappear from unanswered). $\endgroup$
    – Anton Petrunin
    Commented Dec 18, 2022 at 17:32

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Anton asked me to post an answer. I think I never did, because it's too easy for me to get even simple topological arguments wrong.

If I'm not mistaken, Tom Goodwillie's answer here shows that double of a manifold $M$ with boundary can be made into a smooth manifold $\widetilde{M}$ and there is a smooth embedding of $M$ into $\widetilde{M}$. The Riemannian metric can now be extended smoothly to $\widetilde{M}$ using an appropriate partition of unity and the appropriate theorem that says a smooth function on the closed half ball can be extended to the closed ball. Since $\widetilde{M}$ has no boundary, the Nash theorem says that for $N > 0$ large enough, there is an isometric embedding of $\widetilde{M}$ into $\mathbb{R}^N$. The restriction of this to $M$ is of course an isometric embedding of $M$ in $\mathbb{R}^N$.

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