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Prove the following theorem:

For $3\leq r\leq \infty$ let $(M; g)$ be a Riemannian $C^r$-manifold. Then there exists an isometric $C^r$-embedding of $(M; g)$ into a Euclidean space $\mathbb{R}^n$.

I have no idea how to prove this. Can someone help me?

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    This is an extremely deep theorem. Even the proof of the local result involves substantial partial differential equations. Look up the Nash Embedding Theorem. – Ted Shifrin Feb 08 '14 at 00:29
  • @TedShifrin: I know that the Nash embedding theorem states that if $M$ is a given $m$-dimensional Riemannian manifold (analytic or of class $C^k, 3 \leq k \leq \infty$), then there exists a number $n$ and an injective map $f: M \to \mathbb{R}^n$ such that for every point $p\in M$, the derivative $df_p$ is a linear map from the tangent space $T_pM$ to $\mathbb{R}^n$ which is compatible with the given inner product on $T_pM$ and the standard dot product of $\mathbb{R}^n$. I don't understand how one can show that $f$ is an isometric $C^r$-embedding. Could you shed light on your thought process? –  Feb 08 '14 at 00:50
  • @SanathDevalapurkar: The word "compatible" is synonymous with "isometric" in the sense of Riemannian geometry. This should answer your question. Note that there is also another meaning for the word "isometry" which means that the map is distance-preserving. Sometimes such maps are called "strongly isometric". Of course, a strongly isometric map will not exist to $R^n$ for any Riemannian manifold whose metric is not flat. – Moishe Kohan Feb 12 '14 at 11:02
  • @studiosus I agree with that. Now I understand what Ted Shifrin meant. Thanks for that! –  Feb 12 '14 at 23:37
  • Good. Then you should just write your own answer and accept it, so that it is not listed as open. – Moishe Kohan Feb 13 '14 at 03:55
  • @Studiousus O.K. I'll do that. –  Feb 13 '14 at 05:20

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