Friday, 8 July 2011

at.algebraic topology - Is the universal covering of an open subset of $mathbb{R}^n$ diffeomorphic to an open subset of $mathbb{R}^n$ ?

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.

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