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Deane Yang
Deane Yang
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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 restrictiomnrestriction of this to $M$ is of course an isometric embedding of $M$ in $\mathbb{R}^N$.

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 restrictiomn of this to $M$ is of course an isometric embedding of $M$ in $\mathbb{R}^N$.

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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Ben McKay
Ben McKay
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Anton asked me to post an answer. I think I never dieddid, 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 restrictiomn of this to $M$ is of course an isometric embedding of $M$ in $\mathbb{R}^N$.

Anton asked me to post an answer. I think I never died, 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 restrictiomn of this to $M$ is of course an isometric embedding of $M$ in $\mathbb{R}^N$.

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 restrictiomn of this to $M$ is of course an isometric embedding of $M$ in $\mathbb{R}^N$.

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Deane Yang
Deane Yang
  • 27.5k
  • 5
  • 89
  • 180

Anton asked me to post an answer. I think I never died, 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 restrictiomn of this to $M$ is of course an isometric embedding of $M$ in $\mathbb{R}^N$.