Nash embedding theorem states that every smooth Riemannian manifold can be smoothly isometrically embedded into some Euclidean space $E^N$. This result is of fundamental importance, for it unifies the intrinsic and extrinsic points of view of Riemannian geometry, however, it is less clear that it is also useful. Most if not all results in differential geometry that I am aware of seem to have been obtained using the intrinsic point of view so avoiding any recurse to Nash embedding theorem.

Can you mention results that used in their proof the Nash embedding in an essential way, or results whose proof was considerably simplified by the Nash embedding result? If not can you explain why Nash theorem is less useful and powerful than expected?

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    $\begingroup$ If everything is smooth, it is indeed never necessary to use the Nash theorem to prove a theorem about intrinsic Riemannian geometry. Even if the proof were somehow easier, using the embedding would probably obscure the essential geometric intuition of what's going on. It is, however, used to define maps between Riemannian manifolds of weak regularity, since it's not so clear to how define such a map directly and intrinsically. If, however, the manifold is isometrically embedded, then it's a weak map into $\mathbb{R}^N$ whose image lies almost everywhere in the target manifold. $\endgroup$
    – Deane Yang
    Oct 6, 2019 at 16:55
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    $\begingroup$ As Deane says, the Nash embedding theorem is useful in geometric analysis especially in the study of low regularity objects. Another example is if one wants to study minimal surfaces in an arbitrary Riemannian manifold one can isometrically embedded the ambient manifold into Euclidean space and consider the surface as a submanifold of prescribed mean curvature of Euclidean space (the regularity of such objects is well understood). $\endgroup$
    – RBega2
    Oct 6, 2019 at 17:07
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    $\begingroup$ @DeaneYang: I've read this statement about defining maps of low regularity to manifolds using isometric embeddings before, but it has always confused me. At least if the target manifold is compact, why does it matter that it is isometrically embedded? Can we really not just use any smooth embedding into $\mathbb R^N$? $\endgroup$ Oct 6, 2019 at 21:32
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    $\begingroup$ @JohnPardon, if the embedding is isometric, then the Riemannian metric on the target manifold is induced by the Euclidean metric, so the formulas for the gradient and Hessian of the map into the target manifold can be written nicely in terms of the gradient and Hessian of the map into Euclidean space. $\endgroup$
    – Deane Yang
    Oct 7, 2019 at 2:04
  • $\begingroup$ @DeaneYang: Thank you for mentioning the application to geometry under weak regularity. Can you please provide some references? Perhaps with them your comment could be posted as an answer and voted. $\endgroup$ Oct 7, 2019 at 7:20

5 Answers 5


The Nash embedding theorem is an existence theorem for a certain nonlinear PDE ($\partial_i u \cdot \partial_j u = g_{ij}$) and it can in turn be used to construct solutions to other nonlinear PDE. For instance, in my paper

Tao, Terence, Finite-time blowup for a supercritical defocusing nonlinear wave system, Anal. PDE 9, No. 8, 1999-2030 (2016). ZBL1365.35111.

I used the Nash embedding theorem to construct discretely self-similar solutions to a supercritical defocusing nonlinear wave equation $-\partial_{tt} u + \Delta u = (\nabla F)(u)$ on (a backwards light cone in) ${\bf R}^{3+1}$ that blew up in finite time. Roughly speaking, the idea was to first construct the stress-energy tensor $T_{\alpha \beta}$ and then find a field $u$ that exhibited that stress-energy tensor; the stress-energy tensor $T_{\alpha \beta} = \partial_\alpha u \cdot \partial_\beta u -\frac{1}{2} \eta_{\alpha \beta} ( \partial^\gamma u \cdot \partial_\gamma u + F(u))$ was close enough to the quadratic form $\partial_i u \cdot \partial_j u$ that shows up in the isometric embedding problem that I was able to use the Nash embedding theorem (applied to a backwards light cone, quotiented by a discrete scaling symmetry) to resolve the second step of the argument. The field $u$ had to take values in quite a high dimensional space - I ended up using ${\bf R}^{40}$ - because of the somewhat high dimension needed in the target Euclidean space for the Nash embedding theorem to apply.

Also, there is a major indirect use of the Nash embedding theorem: the Nash-Moser iteration scheme that was introduced in order to prove this theorem has since proven to be a powerful tool to establish existence theorems for several other nonlinear PDE, though in many cases it turns out later that with some trickery one can avoid this scheme. For instance the original proof by Hamilton of the local existence for Ricci flow in

Hamilton, Richard S., Three-manifolds with positive Ricci curvature, J. Differ. Geom. 17, 255-306 (1982). ZBL0504.53034.

relied on Nash-Moser iteration, though a later trick of de Turck in

DeTurck, Dennis M., Deforming metrics in the direction of their Ricci tensors, J. Differ. Geom. 18, 157-162 (1983). ZBL0517.53044.

allowed one to avoid using this scheme. (For Nash embedding itself, a somewhat similar trick of Gunther in

Günther, Matthias, Isometric embeddings of Riemannian manifolds, Proc. Int. Congr. Math., Kyoto/Japan 1990, Vol. II, 1137-1143 (1991). ZBL0745.53031.

can be used to also avoid applying Nash-Moser iteration.)

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    $\begingroup$ Terry, that's a pretty cool and clever use of the Nash theorem. As for the Nash-Moser iteration scheme, in almost all cases, as you point out, there have been approaches found that avoid the need for Nash-Moser. Another example of this is Klainerman's thesis on global solutions to a nonlinear wave equation. I would say that, except for Gunther's trick (which I still find rather mysterious), the proof without Nash-Moser is more enlightening than the one using it. As far as I know, the only application, in addition to yours, that still requires Nash-Moser is KAM theory. $\endgroup$
    – Deane Yang
    Oct 7, 2019 at 2:10
  • $\begingroup$ @DeaneYang is there a construction of nonsymmetric Zoll surfaces without Nash--Moser? $\endgroup$ Nov 16 at 16:25
  • $\begingroup$ @AntonPetrunin, I’ve never thought about this. Do you mean looking one near the standard metric? $\endgroup$
    – Deane Yang
    Nov 16 at 17:56
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    $\begingroup$ @DeaneYang In "The Radon transform on Zoll surfaces", Guillemin uses Nash--Moser, to construct a one-parameter family of functions $\phi_t\colon\mathbb{S}^2\to\mathbb{R}$ such that $f=\tfrac\partial{\partial t}\phi_t$ is a given odd function on the sphere and $\mathbb{S}^2$ with the conformal factor $\phi_t$ is a Zoll surface. It would be nice avoid Nash--Moser here. $\endgroup$ Nov 16 at 18:51

In an influential paper, Li and Yau introduced the notion of conformal volume of a Riemannian manifold.

Li, Peter; Yau, Shing-Tung, A new conformal invariant and its applications to the Willmore conjecture and the first eigenvalue of compact surfaces, Invent. Math. 69, 269-291 (1982). ZBL0503.53042, MR674407, eudml.

See also El Soufi and Ilias for the generalization of the application to first eigenvalues in all dimensions.

For a Riemannian manifold $M$, and $\phi: M\to S^n$ a (branched) conformal immersion, $$V_c(n, \phi) = \sup_{\gamma \in G} V(M,(\gamma\circ\phi)^* can),$$ where $G$ is the group of conformal (Möbius) transformations of $S^n$, and $can$ is the canonical round metric on $S^n$. Then $V_c(n,M) =\underset{ \phi:M\to S^n}{\inf} V_c(n,\phi)$, where the infimum is taken over all conformal immersions into $S^n$. Moreover, $V_c(M)=\lim_{n\to \infty} V_c(n,M)$.

Then $V_c(M)$ is well-defined because of the Nash embedding theorem: there is an isometric embedding of $M$ into $\mathbb{R}^n$ for $n$ sufficiently large, and hence an conformal immersion into $S^n$.

El Soufi and Ilias prove the theorem:

Theorem Let $(M,g)$ be a compact Riemannian manifold of dimension $m$. Then $$\lambda_1(M,g) V(M,g)^{2/m}\leq V_c(M)^{2/m}.$$

Equality holds iff $(M,g)$ admits an isometric immersion into $S^n$ (up to scaling) by first eigenfunctions.


Sobolev mappings between Riemannian manifolds. Let $N$ be closed, and $M$ is compact, possibly with boundary. A natural definition of Sobolev mappings between Riemannian manifolds $W^{1,p}(M,N)$ requires the isometric embedding ot $N$ into an Euclidean space $N\subset\mathbb{R}^\nu$ which is the Nash theorem. Then the space is defined as $$ W^{1,p}(M,N)=\{u\in W^{1,p}(M,\mathbb{R}^\nu):\, u(x)\in N \text{ a.e.}\} $$ The space is equipped with the metric inherited from $W^{1,p}(M,\mathbb{R}^\nu)$.

The space does not depend on the isometric embedding of $N$. If $\iota_1:N\to\mathbb{R}^{\nu_1}$ and $\iota_2:N\to\mathbb{R}^{\nu_2}$ are two isometric embeddings and we denote by
$$ W^{1,p}_{\iota_1}(M,N) \quad \text{and} \quad W^{1,p}_{\iota_2}(M,N) $$ the spaces obtained with respect to these embeddings, then for a mapping $u:M\to N$ we have that $$ \iota_1\circ u\in W^{1,p}_{\iota_1}(M,N) \quad \text{if and only if} \quad \iota_2\circ u\in W^{1,p}_{\iota_2}(M,N). $$ However, the metric in the space depends on the embedding, but the map $$ W^{1,p}_{\iota_1}(M,N)\ni u\mapsto \iota_2\circ\iota_1^{-1}\circ u\in W^{1,p}_{\iota_2}(M,N) $$ is a homeomorphism of spaces.

This class of mappings appear in a natural way in the study of geometric variational problems for mappings between manifolds. Like for example, the theory of harmonic mappings. One of the early problems was the question whether smooth mappings are dense. That led to a very fruitful research showing deep connections to algebraic topology.

You can find some basic information about this space as well as references in the survey paper (available on my website):

P. Hajłasz, Sobolev mappings between manifolds and metric spaces. In: Sobolev Spaces in Mathematics I. Sobolev type Inequalities pp. 185-222. International Mathematical Series. Springer 2009.

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    $\begingroup$ Thank you for the answer and reference. Is the space $W^{1,p}(M,N)$ dependent on just $N$, or does it also depend on the embedding of $N$ into $\mathbb{R}^\nu$? From what I get from the equation in display I would say the latter. I that case wouldn't be the notation $W^{1,p}(M,N)$ a sort of abuse of notation? $\endgroup$ Oct 7, 2019 at 13:33
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    $\begingroup$ @EttoreMinguzzi I added more details showing that the space does not depend on the embedding. $\endgroup$ Oct 7, 2019 at 14:48

An attempt to answer the question "What can we do with Nash's embedding theorem?" is given in the paper http://mathlab.math.scu.edu.tw/mp/pdf/S30N35.pdf Very broad perspective on Nash's imbedding theorem is provided by Gromov's article "Geometric, algebraic, and analytic descendants of Nash isometric embedding theorems" https://www.ams.org/journals/bull/2017-54-02/S0273-0979-2016-01551-5/

One may continue with questions about relations between regularity classes + the topology/geometry of the source manifolds with the dimensions q of the ambient spaces, but the most compelling problems raised by the results of Nash are not about these.

Nash, like Columbus, unwillingly discovered a new land. Refining and improving Nash’s isometric imbedding results would be like building bigger and faster ships than those in which Columbus had crossed the Atlantic.

But what is this new land? What is its geography, geology, ecology? How can one explore and cultivate this land? What can one build on this land? What is its future? It may be hard to decide what this land is but it is easy to say what it is not:

what Nash discovered is not any part of the Riemannian geometry, neither it has much (if anything at all) to do with the classical PDE.

Nash’s theorems are only superficially similar to the existence (and non-existence) results for isometric embeddings that rely on PDE and/or on relations between intrinsic, i.e. induced Riemannian, and extrinsic geometries of submanifolds in Euclidean spaces. (The primary instance of the latter is the proof of the existence of isometric immersions of surfaces with positive curvatures into the Euclidean 3-space $R^3$ by means of elliptic a priori estimates, that are certain bounds on the extrinsic curvature of a locally convex surface $X⊂R^3$ in terms of the intrinsic Gauss curvature of X.)

Nash’s results points in the opposite direction:

typically, the geometry of a Riemannian manifold X has no significant influence on its isometric embeddings to $R^q$.

In order to get an idea of what kind of mathematics may lie in this "opposite direction" we shall look at Nash’s theorems and his proofs from a variety of different perspectives.

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    $\begingroup$ The quote is nice but it doesn’t answer the question. Saying that Nash’s theorem is not part of Riemannian geometry, not much to do with classical PDE, and shows no significant influence of the geometry on the imbeddings — all that does not suggest why the theorem is useful in a positive sense. $\endgroup$
    – Matt F.
    Oct 6, 2019 at 21:25
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    $\begingroup$ @MattF. To be fair, it seemed to me that the original question also asked/allowed for negative answers: "If not can you explain why Nash theorem is less useful and powerful than expected?" $\endgroup$
    – Yemon Choi
    Oct 7, 2019 at 3:23

Let me add that the generalized Gauss–Bonnet formula was first proved for embedded Riemannian manifolds. It was done independently in Weyl's "On the volume of tubes" (1939) and Allendoerfer's "The Euler number of a Riemannian manifold" (1940). (By the way, the proof is nice and simple.)

Together with Nash's theorem, it gives the formula for all Riemannian manifolds. So Nash's theorem can be applied to prove the generalized Gauss–Bonnet theorem.


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