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I am aware of classification theorems for principal bundles, vector bundles, and covering spaces $\pi:E\to B$ over a fixed base space $B$. Principal and vector bundles over $B$ are classified by homotopy classes of maps from $B$ into the base space of suitable universal bundles, while covering spaces over $B$ are classified by (conjugacy classes of) subgroups of the fundamental group of $B$.

Rather than taking $B$ to be fixed, suppose we take the total space $E$ to be fixed.

Question: are there classifications of principal bundles, and/or vector bundles, and/or covering spaces having a fixed total space $E$?

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    $\begingroup$ It's pretty clear that any naive version of such "classification" is entirely hopeless: topological groups having decent homotopy type (at least all discrete groups) act freely on a contractible space; taking products allows you to shove any number of actions into a single contractible space. // There's a more reasonable version of your question: identifying a subset of bundles over fixed base B, such that total space is some E. This task is usually very hard and seldom has any tractable answer. $\endgroup$
    – Denis T
    Commented Jan 31 at 20:46
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    $\begingroup$ On the other hand, there are plenty of results of the form "groups of X type cannot act freely on a space of type Y". Finite groups with a noncyclic Sylow subgroup cannot act without fixed points on a sphere of any dimension; circle does not act freely on a space with nonzero Euler characteristic (there are a lot of papers about free circle actions). $\endgroup$
    – Denis T
    Commented Jan 31 at 20:51
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    $\begingroup$ There are many examples of non-equivalent free $G$-actions on the same total space, especially when $G$ is a circle or a discrete group. For example, if two closed manifolds are h-cobordant, their products with a circle are diffeomorphic (except maybe in low dimensions), which gives free circle actions on the same $E$ with different orbit spaces. While having the same total space ties your hands somewhat, there is enough room to do topology and get different quotients of free $G$-actions. $\endgroup$
    – Igor Belegradek
    Commented Feb 1 at 0:01
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    $\begingroup$ @MatthewKvalheim: I don't have a good answer (of why there isn't a classification). I thought a lot about vector bundles with the same total space and I don't have any ideas what the classification might look like in that case. Is the equivalence relation a $G$-bundle isomorphism? People in transformation groups certainly study (not necessarily free) $G$-actions on specific manifolds such as spheres, tori, low-dimensional manifolds. Even in these cases fixing the manifold on which $G$ acts does not lead very far. Even the study of quotients of $S^n$ by a free involution is quite complex. $\endgroup$
    – Igor Belegradek
    Commented Feb 5 at 22:54
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    $\begingroup$ @MatthewKvalheim: if $n\neq 4$, then any manifold homeomorphic to $\mathbb R^n$ is also diffeomorphic to it. As far as I know there is no single example of closed manifold whose universal cover is an exotic $\mathbb R^4$. On the other hand, some exotic $\mathbb R^4$s do admit group actions, see arxiv.org/pdf/1705.06644.pdf; I am not sure if any of these actions are free and properly discontinuous. For many manifolds we know their universal cover is diffeomorphic to $\mathbb R^n$, e.g. nilmanifold, nonpositively curved manifold. $\endgroup$
    – Igor Belegradek
    Commented Feb 17 at 20:52

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