Is there a continuous partition of space into circles?

Question

Question 1. Is there a continuous partition of space $\mathbb{R}^3$ into circles?

I strongly suspect not.

It is well-known by diverse arguments that space can be partitioned into circles. There is an elegant simple argument from the axiom of choice, which works generally with circles, ellipses, triangles, squares, and so forth, almost any kind of shape. One simply proceeds in a transfinite recursion, placing the next circle so as to hit a desired point, while avoiding the previous (fewer than continuum many) circles. (See the excellent article Jonsson, M.; Wästlund, J., Partitions of $\mathbb{R}^3$ into curves, Math. Scand. 83, No. 2, 192–204 (1998). ZBL0951.52018. JStor.) The argument can be modified to make the circles all of different radii, or all of the same radii, or realizing every radius exactly once, or making no two circles coplanar nor even in parallel planes, or making every circle come within 1 of the origin, and so forth. It is very flexible.

There are also explicit constructions that do not rely on the axiom of choice, producing a partition of space into circles. For example, see theorem 1.1 of the Jonsson-Wästlund article. It does remain open, however, how to achieve several of the further properties, such as all-circles-same-size, by an explicitly described partition.

Meanwhile, none of the explicit constructions that I know of are continuous, and I suspect that there is no continuous partition of space into circles.

A continuous partition of space into circles is a partition assigning to each point in space a circle, in such a way that the function mapping each point to the center of the corresponding circle and the line perpendicular is a continuous function. We do not need to give the circles an orientation.

I am here in Madison, having spoken at the seminar, and several of us (myself, Joe Miller, Steffen Lempp, Uri Andrews, Mariya Soskova) spent the afternoon discussing it.

We can prove that there is no continuous partition of space into circles, if there would be a pair of linked circles present. The reason is that if circles C and D are linked, then we can continuously deform circle C to any desired circle C', simply by following a path mapping any point of C to a point on C' and considering the circles that arise for the points on that path. All these continuous deformations of C will remain linked with D, and so C' is linked with D (and consequently, by also deforming D to any D', we see that any two circles are linked with each other). But now consider the points in the disc spanning D. The function mapping those points to the radius of the corresponding circle on which those points lie is continuous, and so it achieves a maximum by compactness. So none of the circles linked with D can reach points very far away, a contradiction.

The link-free case, however, remains unsettled.

Question 2. Is there a continuous partition of space $\mathbb{R}^3$ into loops?

That is, we weaken the requirements from geometric circles to topological circles. For the notion of continuity, we can say that two loops are close if they have parameterizations whose images remain always close.

Again, I suspect the answer is negative.

The linked case argument works just as well for topological circles, so the open case occurs when all loops are unlinked.

Answer 1

As a partial answer, there is no smooth partition of $\mathbb{R}^3$ into smooth loops. For, consider the quotient space; it is necessarily a two-orbifold (locally a surface, possibly having isolated points each having a neighbourhood modelled on $\mathbb{R}^2$ modulo a finite-order rotation).

Since $\mathbb{R}^3$ is non-compact and simply connected, so is the quotient. The only such orbifold is $\mathbb{R}^2$. Thus the partition gives $\mathbb{R}^3$ the structure of an $S^1$-bundle over $\mathbb{R}^2$.

Now, $\mathbb{R}^2$ is contractible. So any $S^1$-bundle over it is isomorphic to $S^1 \times \mathbb{R}^2$. This has fundamental group $\mathbb{Z}$ and we are done.

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Smoothness is not really necessary. Instead we need that every loop has a small neighbourhood which is homeomorphic to a "standard fibering of a solid torus". If this holds, then the partition gives $\mathbb{R}^3$ the structure of a Seifert fibered space. The non-compact case is more difficult than the compact case, but there is still a rich theory.

The extra structure (of namely having good neighbourhoods) holds when the parts of the partition are (continuously varying) round circles. So the answer to Question 1 is no. Question 2 feels more subtle; after all, continuous loops in space can be pretty badly behaved! So obtaining the bundle structure seems more difficult.

Answer 2

Yes, there is a topological foliation of $\mathbb R^3$ by smooth circles. A foliation by topological circles was constructed in by Vogt in

Vogt, Elmar, A foliation of ${\mathbb{R}}^3$ and other punctured 3-manifolds by circles, Publ. Math., Inst. Hautes Étud. Sci. 69, 215-232 (1989). ZBL0688.57016.

In the subsequent paper he improved to the result, extending it to higher dimensions and getting smooth circles (but still, just a topological foliation):

Vogt, Elmar, Existence of foliations of Euclidean spaces with all leaves compact, Math. Ann. 296, No. 1, 159-178 (1993). ZBL0816.57018.

But these foliations correspond to decompositions of $\mathbb R^3$ which are not upper semicontinuous. There are no upper semicontinuous decompositions of $\mathbb R^3$ in round circles, see

Cobb, John, Nice decompositions of (R^ n) entirely into nice sets are mostly impossible, Geom. Dedicata 62, No. 1, 107-114 (1996). ZBL0854.54015.