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Question: is it true that every finite-dimensional connected complete Riemannian manifold of non-positive sectional curvature is a Polish space, i.e. homeomorphic to a complete separable metric space?

My reasoning is that by the Cartan-Hadamard theorem, a universal covering space of a connected complete Riemannian manifold of non-positive sectional curvature is diffeomorphic, and thus also homeomorphic, to a Euclidean space. The space itself is a universal covering space by taking the identity map as a universal cover, and the Euclidean space is a complete separable metric space.

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    $\begingroup$ The space itself need not be a universal cover (think of $S^ 1$). However, complete Riemannian manifolds are always complete separable metric spaces. $\endgroup$ – Teri Mar 3 '17 at 10:30
  • $\begingroup$ Whitney's theorem shows that any connected smooth manifold $M$ can be properly imbedded in an Euclidean space $E$. In other words you can view $M$ as a closed subset of an Euclidean space with the induced topology. As such it is a Polish space. For details on Whitney's imbedding see Chapter IV Sect 1 of Whitney's Geometric Integration Theory. $\endgroup$ – Liviu Nicolaescu Mar 3 '17 at 11:14
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    $\begingroup$ I have to amend my comment with another. Whitney's definition of manifold includes the assumption that it admits a countable atlas. $\endgroup$ – Liviu Nicolaescu Mar 3 '17 at 11:21
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    $\begingroup$ In the context of the question asked: Riemannian implies metrizable implies paracompact + connected implies second countable and so the countable atlas assumption is not an obstruction. $\endgroup$ – Willie Wong Mar 3 '17 at 15:17
  • $\begingroup$ It may be a question of definitions. Some people might consider an uncountable disjoint union of, say, circles, to be a smooth manifold, since it's locally diffeomorphic to $\mathbb{R}^1$, and then you could equip it with a Riemannian metric. This isn't Polish because it's not separable. In that context, the long line might also be a Riemannian manifold. But most people impose a second-countability assumption, in which case it's true that every second-countable smooth manifold is Polish. $\endgroup$ – Nate Eldredge Mar 3 '17 at 18:46

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