# 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

Yes, see e,g, http://en.wikipedia.org/wiki/Nash_embedding_theorem

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 $M^n\to\mathbf{R}^{n^2+5n+3}$.

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Indeed, any Riemannian manifold can be realized as a submanifold of Euclidean space, but in practice you never want to study a Riemannian manifold that way. The Nash isometric embedding theorem is actually rarely used by Riemannian geometers. Its impact in mathematics has been primarily the analytic ideas introduced by Nash in the proof. –  Deane Yang Mar 7 '10 at 3:15
@Deane. It is true, BUT there are some theorems which were first proved for embeddable manifolds and generalized later (it was before Nash). Say presentation of Euler characteristic through the curvature tensor. –  Anton Petrunin Mar 7 '10 at 3:31
@Anton: Historically, you're right. Back then people were a lot more comfortable working with first and second fundamental forms than with the Riemann curvature tensor. But by now we've all become much more comfortable working intrinsically. If anything, today it seems cleaner and simpler to working with a Riemannian manifold than a submanifold of Euclidean space. –  Deane Yang Mar 7 '10 at 3:37