Suppose that $M$ is a non-compact manifold of finite topological type with one end which is the universal cover of some closed manifold $N$.
Is $M $ necessarily homeomorphic to the total space of some vector bundle over a compact manifold?
In fact the only examples I can think up are much more limited, just of the form $M = \Sigma \times \mathbb{R}^n$ where $\Sigma$ is a closed simply connected manifold.
Cross posted on stack exchange https://math.stackexchange.com/questions/4417368/universal-covers-with-one-end.
Best Answer
I think not. In any dimension $n\geq 4$ there are examples, constructed by Mike Davis, of contractible manifolds $M^n$ that are not homeomorphic to $\mathbb{R}^n$, and yet are the universal cover of a compact manifold $N$. The only way that $M$ could be a vector bundle over a compact closed manifold $X$ is if $X$ were a point and the fiber Euclidean space, which is evidently not true. (I suppose that the base space of the vector bundle would have to be closed, or else $M$ would itself have boundary.)
Davis constructs his manifolds so that they are not simply connected at infinity; this distinguishes them from $\mathbb{R}^n$. At least if $n=4$ this would imply that they are not vector bundles over a non-compact manifold either.
M. Davis, Groups generated by reflections and aspherical manifolds not covered by Euclidean space. Ann. of Math. (2) 117 (1983), no. 2, 293–324