# Is the Nash Embedding Theorem a special case of the Whitney Embedding Theorem?

The Whitney Embedding Theorem states that every smooth manifold can be embedded in Euclidean space.

The Nash Embedding Theorem states that every Riemannian manifold can be embedded in Euclidean space.

So I wonder: Can I regard Nash’s theorem as a special case of Whitney’s theorem?

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Nash's theorem says something stronger: every Riemannian manifold can be isometrically embedded in Euclidean space. –  Zhen Lin Nov 13 '12 at 8:10

Just to add something to Jesse's answer, the idea behind the proof of the Easy Whitney Embedding Theorem is to place different pieces of the given $n$-dimensional smooth manifold in 'general position' in $\mathbb{R}^{2n + 1}$. The proof is not very hard to follow; I think that Munkres does a pretty good job in his book Topology. The Hard Whitney Embedding Theorem, which tries to embed a smooth $n$-dimensional manifold in $\mathbb{R}^{2n}$, requires a more technical proof. A clever idea, called 'Whitney's trick' nowadays, is the main idea behind the proof. Notice that we have no notion of distance on a general smooth manifold $M$ unless some metric on $M$ is specified. Hence, both versions of the Whitney Embedding Theorem do not talk about preserving distances between points when constructing the required smooth embedding.

The Nash Embedding Theorem, however, is much harder. Not only must you embed the given Riemannian manifold in Euclidean space, you must do so isometrically, i.e., in a way that preserves distances between points. This requires the solution of a formidable system of partial differential equations that yields the required isometric embedding. Nash solved this PDE system using a special version of Newton's iteration method, called Newton's method with post-conditioning. When unmodified, Newton's iteration method, in general, fails to converge to a solution because each step of the iteration might result in the loss of derivatives, i.e., the order of differentiability is reduced. Nash recovered the lost derivatives by applying smoothing operators (defined via convolution) at each step of the iteration. This ensures that Newton's iteration method does actually converge to a solution. The application of a smoothing operator at each step is called post-conditioning. As you can see, Nash's result is definitely much harder and requires more technology to prove than Whitney's results.

These two results also have different natures. The Whitney Embedding Theorem is more topological in character, while the Nash Embedding Theorem is a geometrical result (as it deals with metrics). However, the structure of smooth manifolds is sufficiently rigid to ensure that they are also geometrical objects (cf. my comment below Jesse's answer), to which the Nash Embedding Theorem can be applied.

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If you want embeddings (as opposed to immersions) you need to add $1$ to the dimensions you state: easy Whitney embeds into $\mathbb{R}^{2n+1}$ and as you say you can push the dimension down by one to $2n$. The example to keep in mind is the real projective plane which cannot be embedded into $\mathbb{R}^3$. The first chapter of Hirsch's book on differential topology contains a very good exposition of these ideas, too. –  commenter Nov 13 '12 at 9:19
Yes, I accidentally stated the immersion result, which is not what I intended. Thanks for pointing that out! –  Haskell Curry Nov 13 '12 at 9:21
Your explanation is quite heuristic, thanks very much! –  hxhxhx88 Nov 13 '12 at 16:15
Very nice: I can't help feeling that the film A Beautiful Mind would have been much better if the above had been incorporated into the screenplay... –  Georges Elencwajg Mar 13 '13 at 21:26

It's really the other way around: Whitney's Theorem is (in some sense) a special case of Nash's Theorem.

Whitney's Theorem says that every smooth manifold can be smoothly embedded in some euclidean space $\mathbb{R}^N$.

Nash's Theorem says that every Riemannian manifold can be isometrically in some euclidean space $\mathbb{R}^N$, and hence smoothly embedded.

While it's true that every Riemannian manifold is (by definition) a smooth manifold, it's also true that every smooth manifold can be equipped with a Riemannian metric (and thereby becomes a Riemannian manifold).

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To see why every smooth $n$-dimensional manifold $M$ can be equipped with a Riemannian metric, first form an open cover of $M$ consisting of subsets of $M$ that are homeomorphic to open subsets of $\mathbb{R}^{n}$. From these open subsets of $\mathbb{R}^{n}$, pull back (via the individual homeomorphisms) the standard Euclidean inner product. Each member of the open cover is now equipped with an inner product. Next, construct a partition-of-unity subordinate to the open cover to consistently sum up all these inner products to get a global inner product on $M$. –  Haskell Curry Nov 13 '12 at 8:55
Got it. Thank you for answer!. –  hxhxhx88 Nov 13 '12 at 16:17