Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A smooth surface $S$ embedded in $\mathbb{R}^3$ whose metric is inherited from $\mathbb{R}^3$ (i.e., distance measured by shortest paths on $S$) is a Riemannian 2-manifold: differentiable because smooth and with the metric just described. Two questions:

  1. Are such surfaces a subset of all Riemannian 2-manifolds? Are there Riemannian 2-manifolds that are not "realized" by any surface embedded in $\mathbb{R}^3$? I assume _Yes_.
  2. If so, is there any characterization of which Riemannian 2-manifolds are realized by such surfaces? In the absence of a characterization, examples would help.


Edit. In light of the useful responses, a sharpening of my question occurs to me: 3. Is the only impediment embedding vs. immersion? Is every Riemannian 2-manifold realized by a surface immersed in $\mathbb{R}^3$?

share|cite|improve this question
I think the canonical examples of 2-manifolds that can be immersed but not embedded in $\mathbb{R}^3$ are the, and various half-way models of –  kahen Oct 16 '10 at 14:23
It seems like the flat torus should present problems beyond embedding/immersing...? –  Aaron Mazel-Gee Oct 17 '10 at 3:55
Aaron: Good point! The flat torus has a $C^1$ embedding. See this MO question:… . –  Joseph O'Rourke Oct 17 '10 at 16:02

3 Answers 3

up vote 2 down vote accepted

There are pointers to a wealth of information on this question in the responses to the Math Overflow question, which you mentioned in your comment to Paul VanKoughnett's response.

In particular, Deane Yang's response gives a nice summary of the situation, and Bill Thurston's response seems to give a good perspective on the problem of trying to find a characterization of Riemannian manifolds that admit such an embedding.

Regarding the third question you mention in an edit. This is essentially a local problem. All this from the same MO question:

From BS's response: there is not even a local isometric embedding in general:

from Will Jaggy's response (and Deane Yang's comments on it): If the metric is analytic, then you can construct a local isometric embedding. Some recent progress on characterizing the requirements when the degree of smoothness is relaxed: The bibliography for that last one has no shortage of other relevant sounding titles.

share|cite|improve this answer
Thanks, yasmar, this is very useful! Somehow I was not connecting the two topics directly in my mind... –  Joseph O'Rourke Oct 16 '10 at 23:08

The hyperbolic plane cannot be smoothly isometrically embedded in $\mathbb{R}^3$. It can be so in $\mathbb{R}^5$. It is open (as far as I know) if it can be embedded in $\mathbb{R}^4$. I believe this is mentioned in Do Carmo's book on curves and surfaces.

Edit: Not a complete characterization, but Amsler has shown (see below for reference) that any Riemannian surface with constant negative curvature, if attempted to be imbedded in $\mathbb{R}^3$, must have singularities.

Amsler, M.H., Des surfaces a courbure negative constante dans l'espace a trois dimensions et de leurs singularites, Math. Ann. 130, 1955, 234-256

share|cite|improve this answer
Thanks for the example and the reference! I have do Carmo's book in my office; will look up. Thanks! –  Joseph O'Rourke Oct 16 '10 at 14:21
See also… which applies to $C^2$ immersions of complete Riemannian 2-manifolds with constant negative curvature. –  Willie Wong Oct 17 '10 at 0:30

The Whitney embedding theorem says you can always embed a smooth $n$-manifold in $\mathbb{R}^{2n}$, and immerse it in $\mathbb{R}^{2n-1}$. Nonorientable Riemann surfaces, for example, don't embed in $\mathbb{R}^3$, but there are some pretty good immersions (the typical picture of the Klein bottle is a good example).

For a Riemannian manifold, Nash and Kuiper proved that there's a $C^1$ globally isometric embedding into $\mathbb{R}^{2n+1}$ (and, in fact, that you can arbitrarily closely approximate any metric $C^\infty$ embedding into at least $\mathbb{R}^{n+1}$ by a global isometric $C^1$ embedding). For a global isometric $C^\infty$ embedding, it looks like the current lower bound is max$(n(n+1)/2+2n,n(n+1)/2+n+5)$. For a local one, you can do it into $n(n+1)/2+n$-space.

This means that for a globally isometric and analytic embedding of a surface, you might have to go up to $\mathbb{R}^{10}$. Ew.

share|cite|improve this answer
Yes, thanks! See also this MO question:… –  Joseph O'Rourke Oct 16 '10 at 17:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.