Suppose that the smooth manifold $ M $ has the n-sphere for its universal cover (in the topological sense). Does there exist a Riemannian metric on $ M $ (not necessarily compatible with the covering map) for which all sectional curvatures of $ M $ are constant, equal to 1?



Must a manifold covered by $ S^n $ admit a metric of constant positive sectional curvature?
Hitchin showed ([here](https://www.sciencedirect.com/science/article/pii/0001870874900218), I think) that there is an exotic sphere which admits no metric of positive scalar curvature. This manifold certainly has the sphere as its topological universal cover, but its sectional curvatures can't all be positive let alone 1.




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The OP is also interested in the smooth case. I think there still are counterexamples, sometimes called "fake lens spaces" - this is a useful term to google, though many of the hits focus on the topological category.


The first step is to classify spherical space forms (i.e. manifolds with constant sectional curvature 1). By various classical results in Riemannian geometry this amounts to classifying all finite groups which act freely on the sphere by isometries. This has been done, and so your question becomes "are there any other finite groups which act freely and smoothly on the sphere?" The answer is "no" in even dimensions because the only nontrivial group which can act on an even dimensional sphere is $\mathbb{Z}/2\mathbb{Z}$ (the Euler characteristic is multiplicative for coverings). The answer is also "no" in dimension 3 by the geometrization theorem.


But more interesting things can happen in higher dimensions. The results inevitably use a lot of heavy duty surgery theory, so I'll point you to a survey paper by [Hambleton](https://arxiv.org/pdf/1412.8187.pdf) which goes into more detail. He credits Lee and Petrie for producing the first smooth counterexamples.


