Is the universal covering of an open subset of $\mathbb{R}^n$ diffeomorphic to an open subset of $\mathbb{R}^n$ ?

Question

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?

Answer 1

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:

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

where $\chi$ is the Euler characteristic of the surface.

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 $S^4$.

Answer 2

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.

Answer 3

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.

Answer 4

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 $\mathbb R^n$.

Answer 5

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 $\mathbb R^n$.