# Riemannian 2-manifolds not realized by surfaces in $\mathbb{R}^3$?

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.

Thanks!

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 $\mathbb{R}^3$?

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I think the canonical examples of 2-manifolds that can be immersed but not embedded in $\mathbb{R}^3$ are the en.wikipedia.org/wiki/Klein_bottle, en.wikipedia.org/wiki/Boy%27s_surface and various half-way models of en.wikipedia.org/wiki/Sphere_eversion – 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

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– Joseph O'Rourke Oct 17 '10 at 16:02

## 3 Answers

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: http://www.springerlink.com/content/m775p64w64351260/

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: http://arxiv.org/abs/1009.6214 The bibliography for that last one has no shortage of other relevant sounding titles.

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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

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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
– 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.

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– Joseph O'Rourke Oct 16 '10 at 17:42
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