# Are all Riemannian metrics induced by Euclidean metrics? [Nash Embedding Theorem]

Let $M$ be a smooth manifold. We can get a Riemannian metric on $M$ by at least two methods: first by partitions of unity and second by the Whitney embedding theorem: we can embed $M$ into a Euclidean space of sufficiently large dimension, and we thus get a Riemannian metric on $M$ by restricting the Euclidean metric on the ambient space.

Can all Riemannian metrics on $M$ be constructed in the second way above?

[Edited for for punctuation, grammar and clarity -- PLC]

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Always check the Wikipedia article! The Nash embedding theorem is cited in the very first section ("Overview") on Riemannian manifolds. –  Qiaochu Yuan Mar 7 '10 at 3:31

Briefly, every $C^1$--metric on a $C^1$-manifold is induced by a $C^1$-embedding $M^n\to\mathbf{R}^{2n+1}$; every $C^\infty$ metric on a $C^{\infty}$ manifold is induced by $C^\infty$-embedding $M^n\to\mathbf{R}^{n^2+5n+3}$.