

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

Is the universal covering of a connected open subset $U$ of ℝn diffeomorphic to an open subset of ℝn (standard differentiable structure)?

If not true in general, is there any condition on $U$ which guarantees a positive answer?

share|cite|improve this question
If you don't require the differentiable structure on the first copy of $\mathbb{R}^n$ to be standard, then the answer is no, due to the existence of large exotic structures on $\mathbb{R}^4$. – S. Carnahan Mar 14 '10 at 19:26
I wonder, what happens if you take the complement of a knot in $R^3$? Torus knot in $R^4$? – Ilya Grigoriev Mar 15 '10 at 3:30
Even crazier: complement of a Hawaiian earring in $R^3$. Cool question, by the way. – Ilya Grigoriev Mar 15 '10 at 3:37
Complement of a knot won't work: it's usually a hyperbolic manifold, so the universal cover is the hyperbolic 3-space, which is diffeomorphic to $R^3$. – Ilya Grigoriev Mar 15 '10 at 3:46
Complement of a knot in $S^3$ (and hence $\mathbb{R}^3$) will never work (even if it fails to be hyperbolic), as such a manifold is Haken, and it is a theorem of Waldhausen that the universal cover of a Haken manifold is $\mathbb{R}^3$. So you can avoid Geometrization. – Richard Kent May 12 '11 at 1:54

The answer is no, and there is a counter-example in dimension $4$.

A theorem of Whitney and Massey states that the total space of a disc-bundle over a non-orientable surface $\Sigma$ embeds in $S^4$ if and only if the normal Euler class of the disc bundle is one of the integers:

$$\{2\chi -4, 2\chi, \cdots, 4-2\chi\}$$


So for example, if $\Sigma = \mathbb RP^2$, $\chi = 1$. So normal Euler classes $-2$ and $+2$ appear for embeddings $\mathbb RP^2 \to S^4$. These come from the standard embeddings of $\mathbb RP^2$ in $S^4$.

The universal cover of this total space is the pull-back of that bundle along the covering map $S^2 \to \mathbb RP^2$. But the total space of this bundle is orientable so it can't embed in $S^4$ as it's Euler class is not zero -- the pull-back bundle must be isomorpic to the tangent bundle of $S^2$. And this does not embed in $S^4$.

share|cite|improve this answer

The open subset $U$ is parallelizable and hence so is its universal cover. A classical theorem of Morris Hirsch says that any open parallelizable $n$-manifold can be immersed into $\mathbb R^n$. Now one could ask whether any open parallelizable $n$-manifold embeds into $\mathbb R^n$. This is formally more general than the original question, so it might be easier to produce a counterexample in this case. Also this more general question strikes me as more natural.

share|cite|improve this answer
Universal cover of a subset of ${\mathbb R}^n$ could be immersed to ${\mathbb R}^n$ by trivial reason (just by the projection). – Petya Mar 15 '10 at 2:22
Here is the point I was trying to make: suppose you have an open parallelizable $n$-manifold; how does one decide whether it embeds into $\mathbb R^n$? The original question describes a formally more special situation but it is unclear to me whether it is really more special. – Igor Belegradek Mar 15 '10 at 2:35
If one takes figure-eight-immersion of cylinder $S^1\times(0,1)$ then you can not deform it into an embedding. – Anton Petrunin Mar 16 '10 at 17:56
@Igor, I guess that the result on parallelizable n-manifold which you mention is proved by h-principle --- one simply deforms a frame into a coordinate frame. I just want to say that the same thing can not work here in principle. (In other words: It should be a counterexample and if NOT then proof will be VERY hard.) – Anton Petrunin Mar 29 '10 at 3:19
A punctured 2-torus is an open parallelizable manifold, and it does not embed in $\mathbb{R}^2$, because of non-trivial intersection on $H_1$. Maybe adding "simply connected" works. – BS. Jul 13 '10 at 17:13

I don't have an answer, only a heuristically inspired hunch. If we think of the figure eight, we can thicken it slightly to an open connected set in the plane. The universal cover is the universal TV antenna times an open interval. But this can be put into the plane by narrowing the branches of the thickened UTVA as one moves out from the center, and since one can do this arbitrarily fast, even tiny branches very far out can be prevented from colliding.

Now there is more room in higher dimensions, so the above kind of argument should actually be easier to carry through than in the plane. Perhaps if one excludes torsion in the fundamental group at least, the countability would be enough if the dimension is at least 3. Just visualize the countably many generating loops, and wiggle them very slightly (there is room enough) so they don't intersect. Then hopefully one can proceed as with the figure eight above. That there could be countably many branches at the forks does not seem to be an essential difficulty.

The above argument does not work generally in the plane, but for the plane the desired statement follows from (a special case of) the uniformization theorem of complex analysis: Every simply connected open Riemann surface is conformally equivalent (and thus diffeomorphic) to the whole plane or the open upper half plane.

EDIT: I think it must be more complicated than this. Otherwise any open connected subset with torsionfree fundamental group, of a manifold of dimension at least three, would have a universal cover diffeomorphic to an open connected set in the same manifold. Surely this is wrong? (By the way, there are open connected sets in Euclidean space with torsion in their fundamental groups).

EDIT: I doubt there is room enough to make this work in dimension 3, maybe in dimension 4.

share|cite|improve this answer
Dear Marius, you write: "Every open connected set in the plane is conformally equivalent (and thus diffeomorphic) to the whole plane or the open upper half plane." This is clearly false: you probably mean the universal cover of this open set. – Georges Elencwajg Mar 15 '10 at 12:37
Dear Georges, of course you are right! I forgot the condition of simple connectedness in the uniformization theorem. – Marius Overholt Mar 15 '10 at 13:09

Consider the standard embedding of the unit interval in $\mathbb R^2$ viz. $I=[0,1]\times \{0\} \subset \mathbb R^2$. Let $C$ denote the Cantor subset $C \subset I$ and define $U= \mathbb R^2 - C$, an open subset of $\mathbb R^2$.

I seem to remember that $\pi_1(U)$ has cardinality at least the continuum and so the fibers of the universal covering $\tilde{U} \to U$ are such big discrete sets that I would guess that $\tilde{U} $ can't be embedded in $\mathbb R^2$.

EDIT Thanks to Petya and Ryan for explaining that $\pi_1(U)$ is actually countable and that what "I seem to remember" is false. Sincere apologies to all for my misleading answer.

For the sake of atonement, here is another argument for the countability of $\pi_1(U)$. Since $U$ is locally connected, locally compact and second countable, any connected covering (or even étalé space) of $U$ is second countable by the theorem of Poincaré-Volterra. Hence the fibers of the covering, being discrete, are countable. But these fibers are equipotent to $\pi_1(U)$ , which must thus be countable. This argument seems to be valid for any open subset of $\mathbb R^n$.

share|cite|improve this answer
It seems to me that each loop could be approximated (in the same homotopy class) by a polygonal loop with finite number of rational vertexes. Hence, $\pi_1$ is countable. – Petya Mar 15 '10 at 0:57
Another argument that gives Petya's conclusion would be to triangulate your Cantor set complement (which can be done with countably-many triangles). So $\pi_1$ is countably-presented. – Ryan Budney Mar 15 '10 at 1:28
I think you are remembering something like an infinite binary tree doubled over a Cantor set. – Douglas Zare Mar 15 '10 at 2:25

An open subset of (standard) $\mathbb R^n$ has a flat metric, so its universal covering space is a simply connected Euclidean space form. The only one such thing is $\mathbb R^n$.

share|cite|improve this answer
What is if the subset is the complement to a point? – Petya Mar 14 '10 at 21:00
I think maybe you're assuming complete? – Ryan Budney Mar 14 '10 at 21:02
Ahhh. Right: completion is needed... So much for the answer! – Mariano Suárez-Alvarez Mar 14 '10 at 21:08

Your Answer


