Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Define a circle as a geometric circle of positive, finite radius: a set of points in $\mathbb{R}^3$ congruent to the set $x^2 + y^2 = r^2$ in the $xy$-plane. [Edited as per BMann's comment.] I am interested in classifying surfaces $S$ embedded in $\mathbb{R}^3$ in two categories:

  1. Define $S$ to be a circled surface if every point of $S$ is contained in a circle that lies in $S$. (I choose the name "circled" in analogy with ruled surfaces.) Thus, $S$ is a union of circles; it is covered by circles.
  2. Define $S$ to be a hoop surface if it may be partitioned into circles, i.e., it is a disjoint union of circles (hoops). In such a partition, every point of $S$ is contained in a unique circle of the partition that lies in $S$. [Edited as per Charles Siegel's comment.]

A specialization would be to restrict the circles in either class to be congruent, unit-radius circles. Generalizations include: (a) relaxing embedded to immersed; (b) replacing circles by spheres in $\mathbb{R}^d$; (c) replacing geometric circles with topological circles.

Setting aside these variants, let me list the hoop surfaces I know:
--Tubular Spline Image--

  1. Tubular surfaces, topological cylinders, either infinite in both directions, or with a circular boundary at one or both ends. These are sometimes called tubular splines in computer graphics.
  2. Tubular surfaces capped at either or both ends but with one point missing from the cap.
  3. Torus surfaces; genus 1.
  4. Disk-like surfaces (composed of nested circles) with one interior point removed. [Edited as per J.C. Ottem's comment.]
  5. A surface homemorphic to the plane with one point removed.
  6. [Due to Cristi Stoica, explained in his answer below.] A surface homeomorphic to two planes connected by a wormhole.

Is this list complete?

Every hoop surface is a circled surface, but planes and spheres are circled surfaces. I cannot see how to form a circled surface of genus greater than 1.

Have these concepts been explored before, and if so, under what names? My searches have been unsuccessful. Thanks for any ideas or literature pointers!

Addendum. The variant of immersed hoops surfaces partitioned into topological circles has been thoroughly analyzed in a paper by Gábor Moussong and Nándor Simányi, "Circle Decompositions of Surfaces":

"The main result in this note is Corollary 3 ... stating that any surface with a circle decomposition is homeomorphic to either a torus, a Klein bottle, an annulus, a Möbius band, an open annulus, a half-open annulus, or an open Möbius band."

share|improve this question
2  
Why do you bring homeomorphism into the picture? The square is homeomorphic to a disk, but is not covered by circles in the sense you want.. –  J.C. Ottem Nov 27 '10 at 1:56
    
@J.C.: Good point! I was struggling to describe those surfaces succinctly. They are not planar. But you are right, I shouldn't mention homeomorphism, for just the reason you articulate. –  Joseph O'Rourke Nov 27 '10 at 2:04
    
You definition of a hoop surface doesn't imply what you say it does. You mention that every point lies on a unique circle, which is false, for instance, on the torus, each point lies on two obvious circles, a vertical and a horizontal one. You could fix this by choosing a circling, and saying that it lies on a unique circle in it. –  Charles Siegel Nov 27 '10 at 2:56
    
The tube formed from a lemniscate does not count as a genus-2 surface? –  J. M. Nov 27 '10 at 3:31
    
@Charles-Siegel, if you look at the Villarceau circles referenced below by Michael Hardy, there are also two more (non-obvious) circles that also going through each point on a torus. There's an excellent image at www-fourier.ujf-grenoble.fr/~bkloeckn/images/villarceau.png on Benoît Kloeckner's web site. –  sleepless in beantown Nov 27 '10 at 8:30

6 Answers 6

Here's how to get circled surfaces of arbitrary genus:

First: Any open subset of $S^2$ is circlable. If the boundary is smooth, it is circlable with circles of constant size: the size just needs to be less than the minimum distance of the boundary from the cut locus (the cut locus is the set of points with more than one shortest arc to the boundary).

Therefore, if you take any finite union of balls with no triple intersections and whose boundaries are not tangent, the boundary of the union is circlable by circles of constant (small) size. These surfaces can have arbitrary genus.

There are also $C^\infty$ smooth circlable genus $g$ surfaces, but they require circles of varying radii. Just take one of the surfaces above, and replace neighborhoods of the circles of intersection by short tubes that are smoothly tangent to the pair of intersecting spheres.

share|improve this answer
    
@Bill: Ah, very nice to invoke the cut locus! The union of balls is a clean construction. Thanks for these insights! –  Joseph O'Rourke Nov 27 '10 at 19:54

We can simplify the problem by first thinking at it from topological viewpoint, then adding the metric.

We are looking for surfaces in $\mathbb R^3$ which are obtained as $I\times S^1$, where $S^1$ is the circle. Now, $I$ can be homeomorphic with an interval from $\mathbb R$, or with $S^1$. So, from topological viewpoint, you can have open and closed tubes - cylinders and tori. We can exclude Klein's bottle because the surfaces are embedded in $\mathbb R^3$.

Adding now the metric, we see that we can get various shapes of tori, and cylinders with none, one or both ends infinite, as you said. Then, the radius being allowed to vary, you can get the punctured caps, which are not necessarily spheres, and may even be conical. They can even be flat, obtaining in particular the punctured disk or the punctured plane you mentioned. Or combinations: the caps can be partially flat. If you do not impose the surface to be smooth, you can have something like a cylinder with both its bases, provided they are punctured.

Another surface which is homeomorphic to a tube is obtained by taking two parallel planes, removing two disks, and smoothly joining the holes with a tube - a wormhole.

In the case of tori, there are two main subcases, depending whether you choose to deform the "big circle" or the "small circle".

share|improve this answer
    
@Cristi: The wormhole is a great idea! Definitely I missed that possibility! Thanks for the clarifying reformulation! –  Joseph O'Rourke Nov 27 '10 at 12:31
    
@Cristi-Stoica, very nice construction of the wormhole. –  sleepless in beantown Nov 29 '10 at 22:44

Perhaps I am confused by your question, most probably by the definition of "geometric." I would be willing to bet there are ways to embed tori in $\mathbb{R}^3$ which do not admit a hoop surface structure by geometric circles (i.e. put a "dent" in the torus).

On the other hand, if we are just talking about writing a surface as the disjoint union of topological circles, then you are correct that one cannot have a closed surface of genus $\geq 2$ which is a hoop surface as you defined it. This is because a surface with non-zero Euler characteristic cannot admit a continuous vector field which is non-vanishing at every point, and writing the surface as a disjoint union of circles would give us this.

share|improve this answer
    
@BMann: By a geometric circle I mean a set in $R^3$ congruent to $x^2 + y^2 = r^2$ in the plane. Nice argument that even topological hoop surfaces cannot have genus more than 1! –  Joseph O'Rourke Nov 27 '10 at 1:45
    
Ah, I fully understand your question now. My first paragraph is irrelevant. Also, if we allow topological circles, as you suggested in your question, every surface is a circled surface, simply by taking the circles small enough to lie in patches homeomorphic to $\mathbb{R}^2$. For hooped surfaces, I believe the list is pretty much the same, but with the Moebius strip added. –  BMann Nov 27 '10 at 1:59

I recently spent an unhealthy amount of time thinking about whether or not you can cover R3 with circles. Turns out you can. For your own search it might be worth looking at Szulkin, Andrzej, 'R 3 is the Union of Disjoint Circles', Amer. Math. Monthly 90 (1983) and then move on to more recent papers which cite it, to get some idea of where people went next.

I don't know how well known this is, but for those who are dying to know: Observe that you can cover a two-punctured sphere (Falls under 2. In Joseph's list). Now consider a family of circles lying in the xy plane, radii 1, centred at the points (4k+1,0,0). Each sphere about the origin intersects this family in exactly two places.

share|improve this answer
    
I realise you are primarily interested in surfaces but the R3 result surprised me so much I felt compelled to tell the world (the MO Community). –  Spencer Nov 27 '10 at 9:43
1  
@Spencer: Actually, this was the topic of an earlier MO question: Is it possible to partition $R^3$ into unit circles mathoverflow.net/questions/28647/… . –  Joseph O'Rourke Nov 27 '10 at 12:15
    
Ah yes. Thank you. This post ought to be there. –  Spencer Nov 27 '10 at 14:58

Since you want to "explore" an idea rather than seeking an answer to a well-defined question, maybe you'd be interested in Wikipedia's article titled Villarceau circles.

share|improve this answer
    
"Given an arbitrary point on a torus, four circles can be drawn through it"---Nice! –  Joseph O'Rourke Nov 27 '10 at 2:07

If you had a torus in $\mathbb{R}^3$ created by sweeping a circle of radius $1$ perpendicular to the path defined by the circle $x^2+y^2=1$ (or equivalently the parametrized version of the circle as $x=sin(\theta), y=cos(\theta), 0 \le \theta \le 2\pi$), then the point $(0,0,0)$ is contained in an infinite number of circles.

I believe you are asking about the outer hull of a swept surface defined by sweeping an outline/path object along another outline or path object. As Michael Hardy pointed out, there is more than one way to specify a torus as the sweep of a circle. In fact, there are four ways with the Villarceau circles. Sometimes, specifying one of the non-obvious circles is a better way to solve certain problems (see Advanced view of the napkin ring problem? about the volume of napkin rings)

There's a great image of the non-obvious Villarceau circles at Benoît Kloeckner's web site:

http://www-fourier.ujf-grenoble.fr/~bkloeckn/images/villarceau.png

If the radius of curvature of the sweep-path becomes smaller than the radius of the circle being swept, then you get strangeness unless your meshing algorithm remembers to remove the extraneous mesh-points that are within the swept-volume's outer hull.

I am not sure I'm completely following your differentiation between circled vs. hoop surfaces, unless you mean that in a circled surface the swept circle is maintained perpendicular to the sweep path, whereas in a hoop surface, it is not. Is that what you meant?

It's also possible to define a sweep surface where the radius of the swept circle also varies parametrically along the sweep path, and if you do that in two-dimensions, you can emulate the behaviour of an ink-pen or a marker where increasing the pressure (or slowing the speed of the pen) leads to a thicker weight line being drawn at certain regions of the "writing path".

share|improve this answer
    
@sleepless: The difference between circled and hoop surfaces is that the former are unions of (generally overlapping) circles, whereas each of the latter may be partitioned into disjoint circles. –  Joseph O'Rourke Nov 27 '10 at 12:37

Your Answer

 
discard

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.