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Is there a classification of the equivalent of a "developable surface" in $\mathbb{R}^4$? Analogous to: planes, cylinders, cones, and tangent developables in $\mathbb{R}^3$? Edit: Here I am imagining "developing" a 3-dimensional manifold embedded in $\mathbb{R}^4$ into $\mathbb{R}^3$. (Apologies for the earlier misleading version!)

I would appreciate any suggestions for source materials here. My only source (Edit: now evidently misleading) is one page (p.342) in Hilbert and Cohn-Vossen (Geometry and the Imagination), in which they say: in $\mathbb{R}^4$

there are surfaces that are isometric to the Euclidean plane in the small but are not ruled.

But now I see from the comments that this must mean a two-dimensional surface embedded in $\mathbb{R}^4$, which is not exactly what I seek.

A precise definition of developable 3-manifold in $\mathbb{R}^d$ would also be much appreciated. Thanks for pointers!

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Isn't the definition of a developable surface that the Gauss curvature is zero? That's independent of the ambient space. – Ryan Budney Jul 28 '11 at 23:46
"Flat" tori cut out by the equations $x_1^2+y_1^2=a^2$, $x_2^2+y_2^2=b^2$ are zero Gauss curvature surfaces in $\mathbb{R}^4$ which are not ruled as they are compact. See e.g. – j.c. Jul 29 '11 at 0:40
@Ryan & jc: OK, I see now where the confusion lies (of course with me). I want to "develop" a 3-manifold embedded in $\mathbb{R}^4$ into $\mathbb{R}^3$. I can see I am misusing terminology, in particular the word "surface," which I was thinking as 3-dimensional rather than 2-dimensional. Sorry to trigger a tutorial! – Joseph O'Rourke Jul 29 '11 at 0:52
see… apparently you need the Ricci curvature of your 3-manifold to vanish. – Will Jagy Jul 29 '11 at 5:08
It's not so much abstract 3-manifolds as, say, orientable ones in $\mathbb R^4.$ For your general concerns you would enjoy "Elementary Topics in Differential Geometry" by John A. Thorpe. He does everything for hypersurfaces of Euclidean space that one ordinarily sees for abstract manifolds. I admit, he does not get to the full curvature tensor. But you do need more information than you have on the second fundamental form, etc. As usual, if Rob Kusner has some time, a discussion in person is worth any number of MO answers. – Will Jagy Jul 29 '11 at 18:19

This is really just an amplification of Deane's answer. As he points out, it is indeed true that a flat submanifold $M^n\subset \mathbb{R}^{n+1}$ whose second fundamental form is nonvanishing is canonically ruled by $(n{-}1)$-planes. However, just being ruled in this way does not imply that the submanifold is flat. For example, the hyperboloid of one sheet in $\mathbb{R}^3$ is ruled by lines (in two ways, in fact), and it is not flat.

The explicit description of the local flat hypersurfaces in $\mathbb{R}^{n+1}$ is, of course, classical, and it goes like this: Start with a regular curve $p:(a,b)\to S^n\subset\mathbb{R}^{n+1}$, choose a smooth function $g:(a,b)\to\mathbb{R}$, and solve the differential equation $dq = g\ dp$, yielding a smooth curve $q:(a,b)\to\mathbb{R}^{n+1}$. Now consider the set $M$, consisting of the vectors of the form $q(t) + v$ where $v$ satisfies $v\cdot p(t) = v\cdot p'(t) = 0$. This will be a (ruled) hypersurface most places, and, where it is, the induced metric will be flat. Note that this says that such hypersurfaces depend on $n$ functions of 1 variable (the first $n{-}1$ of them come from the curve $p$ and one more comes from the function $g$).

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Thanks, Robert. I completely forgot that ruled does not imply flat. – Deane Yang Jul 30 '11 at 16:20

EDIT: I explain below why a flat hypersurface implies ruled, but the converse is not true. See Robert Bryant's answer for a more precise explanation.

It is a theorem that any 2-dimensional open surface in $R^3$ that has zero Gauss curvature everywhere (and therefore is a flat 2-dimensional Riemannian manifold) is a ruled surface (i.e., is foliated by straight line segments). Such surfaces apparently are called developable surfaces.

I have not bothered to check the details but I am betting that everything generalizes in a straightforward manner to an $n$-dimensional hypersurface in $R^{n+1}$. Such a hypersurface has a flat Riemannian metric if and only if the sectional curvature vanishes on all tangent 2-planes. But the sectional curvature of any tangent 2-plane is equal to the determinant of the restriction of the second fundamental form at that point to the 2-plane. Therefore, the hypersurface is flat if and only if the rank of the second fundamental form is at most 1. The proof that a 2-dimensional surface is ruled probably extends to show that any flat hypersurface in higher dimensions is foliated by codimension 2 planes. So these could be called developable hypersurfaces. In other words, a hypersurface is flat if and only if it consists of a 1-parameter family of codimension 2 planes.

ADDED: I have confirmed that the above is right. In particular, any neighborhood of a point where the second fundamental form is not identically zero can be foliated uniquely into codimension 2 planes. This is a nice and rather easy exercise in the local geometry of a hypersurface in Euclidean space using the Frobenius theorem.

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Thanks so much, Deane! – Joseph O'Rourke Jul 30 '11 at 1:58

This is only a partial answer, and it only addresses the compact case.

If you take developable to have the original sense of the word (provided by jc's Wikipedia link above), a developable 3-manifold in $\mathbb R^4$ would be (in the language of geometrization) a Euclidean 3-manifold. Meaning, with the Riemann metric it inherits from $\mathbb R^4$, it is flat.

I believe the question of which compact Euclidean 3-manifolds embed in $\mathbb R^4$ was settled by Crisp and Hillman. The answer is only $S^1 \times S^1 \times S^1$ and the bundle over $S^1$ whose fibre is $S^1 \times S^1$ with the monodromy (of order two) acting by rotation by $\pi$ on the torus $S^1 \times S^1$.

  • J. Crisp, J. Hillman, Embedding Seifert fibred 3-manifolds and $Sol^3$-manifolds in 4-space, Proc. Lond. Math. Soc. (3) (1998) no. 3 685--710.

Of course, these are simply smooth embeddings. I do not know whether or not you can position these manifolds so that they inherit a flat metric from a particular embedding. I suspect you can't find such positions but off the top of my head I don't have a proof.

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@Ryan: Great! "Euclidean 3-manifold" is a useful keyphrase, new to me. I appreciate your help. – Joseph O'Rourke Jul 29 '11 at 1:22

This is discussed thoroughly in section 42 of "Differential Geometry of Two Dimensional Surfaces in Hyperspace" by E.B. Wilson and C.L.E Moore. The entire book "Proceedings of the American Academy of Arts and Sciences" (volume 52) is available from Google Books or from the archive.

I think it makes sense to define a developable hypersurface as the envelope of a one parameter family of hyperplanes, this way it seems clear that in codimension 1 there is a foliation by hyperplanes. (See exercise 4.63 in the notes "Intorduction to Differential Geometry" by Robbin and Salamon)

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A link to read the article online is here: – j.c. Jul 29 '11 at 0:42
@Gjergji: Thanks, this is new to me and will be useful. I apologize for changing the question out from under your feet, so to speak! – Joseph O'Rourke Jul 29 '11 at 1:16

Edit: First, as in comment above, see When is a Riemannian metric equivalent to the flat metric on $\mathbb R^n$?

Getting there. What you want is Ricci flat 3-manifolds in $\mathbb R^4.$ Plus you want to write this in terms of the second fundamental form. So, what you are looking for is at least two out of three principal curvatures equal to 0. I found this nice three page item by Hicks, LINK, what you have is the case $n=3$ in Corollary 6. The terminology of Hicks is not up to date necessarily.

As I commented, you would like a book for undergraduates by John A. Thorpe called Elementary Topics in Differential Geometry. His whole point is to write everything for hypersurfaces of $\mathbb R^{n+1}.$ Given your work in discrete geometry, I think this viewpoint will always be useful to you.

This is also an exercise on the final page in Unit 12 of LINK Exercise 12-3. Express the Ricci curvature of a hypersurface in $\mathbb R^{n+1}$ in a principal direction in terms of the principal curvatures.

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Thanks, Will, I just ordered the Thorpe book. For the record, Hick's Cor.6: "A hypersurface is Ricci flat if and only if it is Einstein with total curvature zero. If $n=3$ and $M$ is Ricci flat, then the second mean curvature is also zero so at points $m$ on $M$ that are not flat points ($L_m \neq 0$), the multiplicity of the nonzero principle curvature is unity." – Joseph O'Rourke Jul 29 '11 at 20:52
For 3-manifolds the Riemann curvature tensor is recoverable from the Ricci tensor, so "Ricci flat" implies its a Euclidean manifold -- locally isometric to open subsets of $\mathbb R^3$. – Ryan Budney Jul 29 '11 at 21:40
@Ryan: Great! Terminological variations are converging... – Joseph O'Rourke Jul 29 '11 at 21:46

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