# The Classification of all spaces for which $X$ is a covering space

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?

• 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 – Gro-Tsen Nov 19 '17 at 10:33
• @Gro-Tsen Thank you for the comment and thank you for the great link. – M.Ramana Nov 19 '17 at 13:16

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.
• 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? – M.Ramana Nov 19 '17 at 17:08
• 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 $X$. – Qiaochu Yuan Nov 19 '17 at 18:28
• (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.) – Qiaochu Yuan Nov 19 '17 at 18:43