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The Nash embedding theorem tells us that every smooth Riemannian m-manifold can be embedded in $R^n$ for, say, $n = m^2 + 5m + 3$. What can we say in the special case of 2-manifolds? For example, can we always embed a 2-manifold in $R^3$?

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Certainly one can't embed any compact non-orientable manifold in $\mathbb{R}^3$. It's well-known that one can't embed the hyperbolic plane isometrically in $\mathbb{R}^3$, but I don't know a convenient reference for this off-hand. –  Robin Chapman Sep 4 '10 at 10:35
    
See also this related MO question: mathoverflow.net/questions/31222/… –  Joseph O'Rourke Sep 4 '10 at 12:19
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BTW, the local version of your question is still open. Namely, is it true that any point on a surface has a nbhd which admits a smooth isometric embedding into $\mathbb R^3$? –  Anton Petrunin Sep 5 '10 at 2:57
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Anton's point is a very good one. –  Deane Yang Sep 7 '10 at 21:28
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@Robin Chapman: I think it is proved in the last chapter of Do Carmo's book "The differential Geometry of curves and surfaces" –  Daniel Barter Sep 12 '10 at 5:51
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7 Answers

up vote 16 down vote accepted

The Nash-Kuiper embedding theorem states that any orientable 2-manifold is isometrically ${\cal C}^1$-embeddable in $\mathbb{R}^3$. A theorem of Thompkins [cited below] implies that as soon as one moves to ${\cal C}^2$, even compact flat $n$-manifolds cannot be isometrically ${\cal C}^2$-immersed in $\mathbb{R}^{2n-1}$. So the answer to your question for smooth embeddings is: No, as others have pointed out. I believe Gromov reduced the dimension you quote of the space needed for any compact surface to 5, but I don't have a precise reference for that.

Tompkins, C. "Isometric embedding of flat manifolds in Euclidean space," Duke Math.J. 5(1): 1939, 58-61.

Edit. Both Deane Yang and Willie Wong were correct that the Gromov result is in Partial Differential Relations. I believe this is it, on p.298: "We construct here an isometric $\cal{C}^\infty$ ($\cal{C}^{\mathrm{an}}$)-imbedding of $(V,g) \rightarrow \mathbb{R}^5$ for all compact surfaces $V$." $g$ is a Riemannian metric on $V$.

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I think Gromov's result may be in his Partial Differential Relations book (I am not sure: I am travelling right now and don't have it with me). –  Willie Wong Sep 4 '10 at 23:00
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If you allow immersions, then Poznyak proved that $\mathbb{R}^4$ works for any compact part of a complete surface. In particular, regarding Anton's comment above, this implies that surfaces admit local embeddings into $\mathbb{R}^4$. Compactness is important for Gromov's result - apparently it's not known whether the hyperbolic plane has a smooth isometric embedding in $\mathbb{R}^5$, though Blanusa constructed one in $\mathbb{R}^6$. The book by Han and Hong cited by BS in his answer is a good source for many of these questions. –  j.c. Sep 8 '10 at 15:38
    
My above comment was written too carelessly; see mathoverflow.net/questions/78026/… for more detailed discussion. –  j.c. Oct 13 '11 at 18:40
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Lately I've been designing and making collections of pieces, cut from foam with a computer-controlled cutter, that can be joined to one another in an interlocking way to approximate an arbitrary surface, so I've become more aware of some of the obstructions to smooth isometric embeddings. (A few of some earlier versions of the modelling sytem are shown on the home page of Kelly Delp, with whom I've been collaborating). Here are two examples:

alt text alt text

It seems like an interesting question to give good sufficient conditions beyond the non-negatively curved case and good obstructing conditions, beyond conditions that I will outline below. (I'm not an expert in this stuff, so there may be much more that is known.)

In the first place, any smooth closed surface M^2 in R^3, if you think about the Gauss map (the point of the unit normal vector, translated to be based at the origin), it is clear that the closure of the subset of M^2 that has positive Gaussian curvature map surjectively --- it suffices to take the point on the surface that maximizes inner product with a vector in that direction. In other words, there must be at least 4 pi worth of positive curvature. (For the construction kits, this gives a linear inequality for how many seams of various types are needed to construct a closed surface in space).

The above condition is not sufficient. If you start from a standard torus obtained as a circle of revolution, the Gauss map covers the sphere precisely twice, once with positive orientation from the outer shell, and once in the negative sense from the inner part. The positively curved part meets the negatively curved part on two round circles, along which the normal vectors are parallel and so the Gauss map is constant. For any slight perturbation of this surface in space, the Gauss map changes, but the area of the two regions is constant up to first order: the area of the image changes by the area swept out by the evolving boundary curve, but the Gauss image boundary curves of length zero have to grow before they can begin to capture area.

It's easy to perturb the metric of a round torus to make the total positive part of Gaussian curvature increase to first order. These perturbations of the metric can't be extended to perturbations of the embedding. If you do this perturbation on a portion of the torus, half a bagel so to speak, it increases its angle of curling.

I don't at the moment know a rigorous proof that there are no smooth embeddings of these perturbed metrics, but I suspect a proof could be given. (If one had a sequence of eapproximating metrics, one technical issue is that they might admit embeddings having greater and greater variation of the 2nd derivative, so the limit might be a non-C^2 embedding of the Nash-Kuiper embeddings as described by Deane Yang.)

In practice, constructions made from pieces of foam that approximate a torus of revolution are surprisingly rigid --- there is very little tolerance to modify the embedding in a way that will make it close if it doesn't want to.

On the other hand, if you change the torus shape to add a corkscrew effect, the Guass map on the boundary betwen positive and negative is no longer constant. These shapes have much greater ability to accomodate a change in the metric.

It's easy to come up with many other examples of surfaces where there are components of the positively curved part bounded by curves that have parallel normal vectors --- people often draw these instinctively. These surfaces are all limited by the same kind of rigidity.

There is another class of obstructions I'm aware of that from Hilbert's theorem that for any hyperbolic surface with a real analytic isometric immersion in space, the total |Gaussian curvature enclosed| by a quadrilateral of asymptotic lines is less than 2 pi (the upper bound for areas of quadrilaterals in the hyperbolic plane). This has subsequently been generalized in several ways, but I'm not up-to-date on what's known. The trouble is that one doesn't easily know ahead of time what the asymptotic lines will be. However, this gives some qualitative obstrutions, so I suspect one could prove that if you take two copies of a large disk in the hyperbolic plane and bridge between them by an annulus of positive curvature, the resulting metric on the sphere has no C^2 isometric embedding. If you make paper models, in practice they get riffly edges that are qualitatively incompatible with a C^2 isometric approximation, because the asymptotic line fields turn in a homotopically non-trivial way.

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Bill, I don't have anything to add to what you say (except that it was Joseph O'Rourke who mentioned Nash-Kuiper and not me), but your thoughts reminded me of a different question I've thought about: Is the isometric embedding of a smooth closed surface in $R^3$ necessarily locally rigid? Convex ones are. Connelly has a counterexample for a nonconvex polyhedron (which can be seen in the metal at IHES). At one point I thought about trying to create smooth version of Connelly's example by replacing the edges and vertices by equivalent smooth models. Do you have any thoughts about this? –  Deane Yang Sep 8 '10 at 2:24
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Fixed the link to Delp's homepage, somehow the ~ had gotten messed up. –  Noah Snyder Sep 8 '10 at 3:13
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I'd like to summarize and elaborate on what's been said:

First, closed non-orientable surfaces such as the Klein bottle have no topological embedding in $R^3$ and therefore no isometric embedding.

The question is more reasonable if restricted to, say, closed orientable surfaces. Without the topological obstruction, this is a question about global solutions to the system of PDE's given by the isometry condition. But not any global solution. A global solution is necessarily an immersion but not necessarily an embedding. Moreover, the system of PDE's is a rather nasty one. It can be rewritten as a single Monge-Ampere equation, whose type is determined by the sign of the Gauss curvature, so the equation is elliptic where the Gauss curvature is positive and hyperbolic where the curvature is negative. Near a point where the Gauss curvature changes sign but has nonzero gradient, the equation is called "Tricomi type".

As pointed out by José, there is a geometric obstruction to global solutions: any isometrically embedded surface has at least two points with positive Gauss curvature (given by the two points of contact of an osculating sphere). So, for example, orientable surfaces with nonpositive curvature have no isometric embedding in $R^3$.

At this point, if you're still focused on isometric embeddings in $R^3$, then the question is whether any oriented surface with "enough" positive Gauss curvature is isometrically embeddable. The only result I know in this direction is Nirenberg's solution to the Weyl problem, which states that any closed surface with positive Gauss curvature has a unique isometric embedding in $R^3$. This and Nirenberg's solution to the 2-d Minkowski problem launched a golden era of using nonlinear elliptic PDE theory to solve many hard global problems in differential geometry, as exemplified by the work of Yau and Schoen.

Or you could go to higher dimensions. By now I don't remember the details, but I think that it is correct that Gromov showed that any 2-d surface can be isometrically embedded in $R^5$. I also don't remember where the proof is given, but the first place to look is his book, Partial Differential Relations, where he shows how to solve systems of underdetermined PDE's using "soft" techniques such as the h-principle. The ideas in this book sit somewhere between the hard analysis of PDE theory (he actually provides a proof of the Nash-Moser implicit function theorem) and the soft or "flabby" approach of topology. Either the proof or a reference to it should be in the book somewhere.

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Yes, thanks, the $R^5$ result must be as you say in here: Gromov, M. Partial Differential Relations, Springer-Verlag, Ergeb. der Math. 3 Folge, Bd. 9, Berlin-Heidelberg-New-York, 1986. I don't have that book. It would be useful to have a precise reference. –  Joseph O'Rourke Sep 4 '10 at 14:32
    
Dear Prof. Yang, There are preferred isometric embeddings for the 2-sphere in $R^3 $ as the round sphere, the flat Clifford torus in $R^4$. Ignoring the idea of best possible dimension, are there preferred isometric embeddings of the oriented higher genus constant curvature compact surfaces in $R^k$? Here the concept of "preferred" could mean anything appropriate. Will. –  Will Jagy Sep 7 '10 at 23:25
    
Actually, some nice embedding in $S^k$ or $H^k$ would also be more than I know at the moment. –  Will Jagy Sep 7 '10 at 23:26
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Will, I am unaware of any particularly nice isometric embeddings of complete hyperbolic surfaces in a higher dimensional Euclidean space. Even the hyperbolic plane is problematic. The best isometric embedding of the hyperbolic plane I know is into 2+1 Minkowski space. –  Deane Yang Sep 8 '10 at 1:11
    
Thank you, Deane. –  Will Jagy Sep 8 '10 at 1:22
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I'm not convinced this is an appropriate question, but giving the benefit of the doubt and since the OP is probably not a geometer...

The flat torus cannot be smoothly embedded in $\mathbb{R}^3$ isometrically. Since the torus is compact, any embedded torus lies in the interior of some sphere centered at the origin. Decrease the radius of the sphere until it just touches the torus for the first time. At the point of contact, the curvature of the torus agrees with that of the sphere, and is hence positive.

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@Jose: I believe your statement is correct only with the additional qualification that the embedding is smooth? Zalgaller has constructed an isometric embedding of the flat torus. See my answer to this MO question: mathoverflow.net/questions/31222/… . I think in your proof sketch the conclusion that the curvatures agree at the contact point requires the embedding to be smooth. –  Joseph O'Rourke Sep 4 '10 at 12:19
    
@Joseph: Guilty as charged! But aren't we in the smooth category? At least that is what I understood from the way the question was phrased. –  José Figueroa-O'Farrill Sep 4 '10 at 12:42
    
@Jose: Yes, what you say holds in the smooth category. I guess that was implied; sorry. So consider my comment just a remark that without that assumption, the situation alters. –  Joseph O'Rourke Sep 4 '10 at 12:49
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There are interesting material on embedding surfaces in $R^3$ in the book

Isometric Embedding of Riemannian Manifolds in Euclidean Spaces by Qing Han and Jia-Xing Hong, AMS, 2006. See this link.

Moreover, it seems that even embedding a small neighborhood of a point can be problematic, but I don't know if there is a smooth counterexample.

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I had a conversation today with Christopher Croke of Penn., he pointed out that there was progress, extremely recent progress, on the local version, isometrically embedding a piece of surface in $\mathbf R^3.$ This is all in the $C^\infty$ category, it is always possible that more is known in $C^\omega$ and very likely less is known in, say $C^2.$

Anyway, if the curvature does not vanish in the neighborhood, then it can be isometrically embedded in $\mathbf R^3.$ The same is true if the curvature vanishes at a point but some directional derivative does not.

Well, from a new item on the arXiv by Qing Han and Marcus Khuri,

It is shown that if the Gaussian curvature vanishes to finite order and its zero set consists of two Lipschitz curves intersecting transversely at a point, then local sufficiently smooth isometric embeddings exist.

The title of the preprint is "On the Local Isometric Embedding in $\mathbf R^3$ of Surfaces with Gaussian Curvature of Mixed Sign."

http://arxiv.org/abs/1009.6214

Both Han and Khuri have published on this topic in the Journal of Differential Geometry, Han in 2003, Khuri in 2007.

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The local existence of an isometric embedding in $R^3$ of a real analytic $2$-dimensional Riemannian metric is classical and easy to prove by applying Cauchy-Kovalevski to the Monge-Ampere equation that arises. The generalization to higher dimensions is known as the Cartan-Janet theorem and is a bit trickier, because the system of PDE's is highly degenerate. However, it can be proved by induction of the dimension of the domain, where you use Cauchy-Kovalevski at each step. Cartan formulated the question as an exterior differential system and used the Cartan-Kahler theorem. –  Deane Yang Oct 12 '10 at 0:55
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On the other hand, in the $C^\infty$ category, the isometric embedding equation is the nastiest naturally occurring system of PDE's that I know. Bryant, Griffiths, and I showed that in some sense the linearized system is a generic first order $n$-by-$n$ system of PDE's. So in general it is not elliptic, hyperbolic, or parabolic. Either very clever ad hoc techniques specific to the system need to be developed, or microlocal analysis is used to analyze the linearized operator. –  Deane Yang Oct 12 '10 at 0:59
    
Thanks, Deane. I thought $C^\omega$ would be known but I was not sure. For Han and Khuri, it would appear that the next case would be a monkey saddle, where the curvature function behaves as the imaginary part of $(x + i y)^3$ near the origin. –  Will Jagy Oct 12 '10 at 2:55
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Another counter example, and somehow more exotic, is the Klein bottle.

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but that is probably a lot easier to prove since the Klein bottle doesn't embed in R^3 at all based upon some algebraic topology considerations. –  Daniel Pomerleano Sep 4 '10 at 20:47
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