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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 distinguishes his manifolds from $\mathbb{R}^n$ by constructing them so that they are not simply connected at infinity. 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

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

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 distinguishes his manifolds from $\mathbb{R}^n$ by constructing the so that they are not simply connected at infinity. 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

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 distinguishes his manifolds from $\mathbb{R}^n$ by constructing them so that they are not simply connected at infinity. 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

I think not. In any dimension $n$ 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 distinguishes his manifolds from $\mathbb{R}^n$ by constructing the so that they are not simply connected at infinity. 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

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 distinguishes his manifolds from $\mathbb{R}^n$ by constructing the so that they are not simply connected at infinity. 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