# Nash isometric embedding for noncompact manifolds

It seems that the smooth isometric embedding theorem by Nash is true also for noncompact manifolds.

Is it true that any (complete, connected) Riemannian manifold $$(M^n,g)$$ admits a proper smooth isometric embedding $$\iota:M^n\hookrightarrow\mathbb R^N$$ into some Euclidean space?

This is equivalent to ask that $$\iota(M)$$ is a closed subset of $$\mathbb R^N$$.

Remark. It is pretty trivial to modify the proof of Whitney's embedding theorem to obtain that any manifold admits a proper embedding into some Euclidean space (if we don't care about obtaining the best dimension for which this is possible...). But of course completeness is a necessary condition to find a proper isometric embedding $$\iota$$. Is completeness also sufficient?

Are weaker results known? E.g. is it true if $$\operatorname{inj}(M)>0$$ and/or the sectional curvature $$\operatorname{sec}(M)$$ is bounded?

Nash's theorem states that any smooth embedding $$f$$ with Lipschitz constant less than 1 can be approximated by smooth isometric embedding (if the dimension of Euclidean space is sufficiently large). So you only need to find a proper embedding $$f$$.

Take any embedding with Lipshitz constant $$<\tfrac13$$, and add one more coordinte $$z$$ that is a proper function with Lipshitz constant $$<\tfrac13$$. You get a smooth proper embedding $$f$$ with Lipschitz constant at most $$\tfrac23$$.

To construct $$z$$, you may smooth a distance function $$\tfrac14\cdot \mathrm{dist}_p$$; for sure you need completeness.

• Very nice! I also added a similar argument along the lines of the paper mentioned by Igor Belegradek. – Mizar Sep 29 '18 at 11:47
• @Mizar: Edits so vast as the one that you attempted to perform on Anton Petrunin's answer are not really "edits", because they change the original post too much. Better add that argument as a separate answer, yourself. – Alex M. Sep 29 '18 at 12:03

A slight modification of Anton Petrunin's answer, namely Müller's idea brought up in Igor Belegradek's comment, allows one to deduce the desired result starting from the classical statement of the isometric embedding theorem, i.e.

Any Riemannian manifold $$(M^n,g)$$ admits a smooth isometric embedding into a Euclidean space.

Fix $$p\in M$$ and take again a smoothing $$\phi$$ of $$\frac{1}{4}\operatorname{dist}_p$$ with $$|\phi-\frac{1}{4}\operatorname{dist}_p|\le 1$$ and $$|\nabla\phi|_g\le\frac{1}{2}$$ (see below for the details). Now apply the classical Nash embedding theorem to the Riemannian manifold $$(M^n,g-d\phi\otimes d\phi)$$, obtaining an isometric embedding $$F$$. Then $$(F,\phi)$$ is the desired proper isometric embedding of $$(M^n,g)$$, since $$\operatorname{dist}_p$$ (and thus $$\phi$$) is a proper map by completeness.

Construction of $$\boldsymbol{\phi}$$. Take a locally finite open cover $$\{U_j\}$$ of $$M$$ such that $$\overline{U}_j$$ is compact and lies in a coordinate chart, together with a subordinated partition of unity $$\{\rho_j\}$$. Since $$|\nabla\operatorname{dist}_p|_g\le 1$$ a.e., we can approximate $$\frac{1}{4}\operatorname{dist}_p$$ on $$U_j$$ by a smooth function $$\phi_j$$ with $$|\nabla\phi_j|_g\le\frac{3}{8},\qquad|\phi_j-\frac{1}{4}\operatorname{dist}_p|\le\min\bigg\{1,\frac{1}{8N_j'\|\nabla\rho_j\|_{L^\infty}}\bigg\},$$ where $$N_k:=\#\{\ell:U_k\cap U_\ell\neq\emptyset\}$$ and $$N_j':=\max\{N_k\mid U_j\cap U_k\neq\emptyset\}$$. Finally, $$\phi:=\sum\rho_j\phi_j$$ has $$|\phi-\frac{1}{4}\operatorname{dist}_p|\le 1$$ and, being $$\nabla\phi=\sum_k\rho_k\nabla\phi_k+\sum_k\nabla\rho_k(\phi_k-\frac{1}{4}\operatorname{dist}_p),$$ on each $$U_j$$ (and thus on $$M$$) we have $$|\nabla\phi|_g\le\frac{3}{8}+\sum_{k:U_j\cap U_k\neq\emptyset}\frac{1}{8N_k'}\le\frac{1}{2}$$ (as $$N_k'\ge N_j$$).

• Hi, I wonder about the proneness, initial question, for instance is it true that we can assume that the embedding get a uniform tubular neighbourhood? In general no, but assuming that $M$ is homogenously regular is that correct, do you have a precise reference? where do you use the completness in your argument? Thx – Paul Oct 10 '18 at 14:25