4
$\begingroup$

A well-known problem is to classify all covering spaces of a topological space $X$. For example, if $X$ is a semi-locally simply connected space, then each equivalent class of a covering space of $X$ is corresponding to conjugacy class of a subgroup of $\pi_1 (X)$. Now my question is that:

Is there any classification of all spaces for which $X$ is a covering space?

For instance, for which spaces is $\mathbb{S}^n$ a covering space, up to homeomorphism or up to homotopy equivalence?

$\endgroup$
  • 5
    $\begingroup$ For the specific problem you ask about the sphere, the term is "spherical space form problem". See Wolf, Spaces of Constant Curvature, part III, and indiana.edu/~jfdavis/books/Spherical_space_forms.pdf $\endgroup$ – Gro-Tsen Nov 19 '17 at 10:33
  • $\begingroup$ @Gro-Tsen Thank you for the comment and thank you for the great link. $\endgroup$ – M.Ramana Nov 19 '17 at 13:16
10
$\begingroup$

In general, I would expect this to be a quite intractable problem. For instance, let's assume we are only interested in the category of manifolds, and we ask the question which $3$-manifolds are covered by $\mathbb{R}^{3}.$ Here, by the solution to the geometrization conjecture for $3$-manifolds, every closed, orientable, irreducible and atoroidal $3$-manifold with infinite fundamental group is hyperbolic, and is therefore covered by $\mathbb{R}^{3}.$ In fact, most of the $8$ model geometries in $3$-dimensions are diffeomorphic to $\mathbb{R}^{3}$, with the exceptions being $S^{2}\times \mathbb{R}$ and $S^{3},$ and therefore in a sense which I won't make precise, the building blocks of "most" $3$-manifolds are covered by $\mathbb{R}^{3}.$ The sense in which you call this a classification of $3$-manifolds covered by $\mathbb{R}^{3}$ is up for debate.

In general, every closed aspherical manifold has contractible universal cover, of which $\mathbb{R}^{n}$ is the usual candidate, so hoping to classify the topological spaces covered by $\mathbb{R}^{n}$ would include a classification of "most" aspherical manifolds.

You might ask for a much coarser classification, but this at least shows that the problem is very complicated.

$\endgroup$
  • $\begingroup$ Your explanation is great and complete. I didn't know that this is an old and complicated problem. From your answer, I understood that my question is too general. Thank you so much for your help. My question, in fact, arises from some concept in shape theory which proposed by Karol Borsuk concerning the capacity of compacta. But now I understood that the relationship between them is also complicated. Thank you very much. $\endgroup$ – M.Ramana Nov 19 '17 at 13:27
  • $\begingroup$ As mentioned above, the classification all spaces with the $X$ as covering space (up to homoemorphism) is so complicated. But is it a hard problem up to homotopy equivalence? $\endgroup$ – M.Ramana Nov 19 '17 at 17:08
  • $\begingroup$ It’s an extremely hard problem up to homotopy equivalence. If $X$ is simply connected the problem is equivalent to classifying all pairs consisting of a group and a homotopy coherent action of that group on $X$. $\endgroup$ – Qiaochu Yuan Nov 19 '17 at 18:28
  • $\begingroup$ (When I say "up to homotopy equivalence" I mean including the possibility of replacing $X$ with a homotopy equivalent space. This changes the classification problem, e.g. all of the $\mathbb{R}^n$ cover different, even non-homotopy equivalent, spaces, but they are all homotopy equivalent. If you don't want to allow this then I don't think allowing the covered space to vary up to homotopy equivalence makes things any easier.) $\endgroup$ – Qiaochu Yuan Nov 19 '17 at 18:43
  • $\begingroup$ @QiaochuYuan Thank you very much for your explanation and your help. $\endgroup$ – M.Ramana Dec 8 '17 at 4:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.