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Note that $M$ of dimension $n$ can be imbedded differentiably as a closed submanifold of ${\bf R}^{N=2n+1}$.

Here Let $f$ be an imbedding. $f$ is one-to-one immersion, that is, rank $n$, which is homeomorphic.

Question : Here I have a question : If we allowed $N$ to be large, then $f : M\rightarrow f(M)$ can be isometric ?

Thank you in advance.

HK Lee
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    http://en.wikipedia.org/wiki/Nash_embedding_theorem – Moishe Kohan Mar 21 '14 at 11:11
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    If a manifold has a metric is it always induced by a riemannian structure? – Seth Mar 21 '14 at 11:41
  • I mean : Consier fixed Riemannian manifold $M$ with a Riemannian metric $g$. Then is there isometric $smooth$ map $f$ where $f : (M^n,g) \rightarrow (f(M),G)\subset {\bf R}^m$. Here $G$ is a metric induced from cannonical metric in ${\bf R}^m$. That is, point is that any Riemannian manifold with $g$ can be a submanifold of ${\bf R}^m$ ? – HK Lee Mar 21 '14 at 11:43

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First, to clear the terminological confusion: There is a theory of metric spaces and a theory of Riemannian manifolds (Riemannian geometry). Both have their own notion of a metric and an isometry and these two notions are quite different. See the discussion here about which distance functions are defined via Riemannian metrics. It is clear, however, from the question asked that it is about Riemannian metrics and Riemannian isometric embeddings. The answer is given by

Nash embedding theorem. If $g$ is a Riemannian metric of class $C^k$ on an $m$-dimensional $C^k$-smooth manifold, then there exists a $C^k$-smooth isometric embedding $f: (M,g)\to R^N$, $N = m(m+1)(3m+11)/2$. Here $R^N$ is equipped with the standard flat Riemannian metric.

The most subtle (and amazing) part of this theorem is that $f$ is as smooth as $g$ is.

See here for the detailed discussion, including improvements and generalizations of this theorem. While the result itself, I believe, is not particularly useful in Riemannian geometry, Nash's proof of his theorem was of tremendous importance in geometry and analysis (see here).

Moishe Kohan
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