# Does every smooth manifold of infinite topological type admit a complete Riemannian metric?

To elaborate a bit, I should say that the question of the existence of a complete metric is only of interest in the case of manifolds of infinite topological type; if a manifold is compact, any metric is complete, and if a noncompact manifold has finite topological type(ie is diffeomorphic to the interior of a compact manifold with boundary,) one can contruct a complete metric on the manifold with boundary via a partition of unity, and then divide by the square of a defining function to get an complete asymptotically metric on the interior.

I have absolutely no intuition for how "wild" these manifolds can be. The only examples I can think of are infinite connected sums and quotients of negatively curved symmetric spaces by sufficiently complicated groups, but I'd imagine that one can construct some pathological examples by limiting arguments.

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By Whitney embedding theorem any smooth manifold embeds into some Euclidean space as a closed subset. The induced metric is complete.

In fact, a good exercise is to show that any Riemannian metric is conformal to a complete metric.

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We can even prove more: every smooth manifold can be equipped with a Riemannian metric of bounded geometry (positive injectivity radius, bounded curvature tensor and all derivatives of the curvature tensor are also bounded).

This is Theorem 2' in "R. E. Greene, Complete metrics of bounded curvature on noncompact manifolds, Archiv der Mathematik 31 (1978), no. 1, 89-95".

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It's not my field, so I am really out on a limb here. Maybe someone can tell my why the following naïve idea wouldn't work: Make a Riemannian metric by partition of unity. For any point $p$ in the manifold, let $h(p)$ be the infimum of the lengths of all paths starting at $p$ which are not contained in any compact subset of the manifold. Now $h$ won't be smooth, so you need to smooth it without changing it substantially. Divide the metric by $h^2$.

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The idea is right I think but your $h$ is unnecessary complicated; there are easier ways to find a proper smooth positive function on a manifold. –  Igor Belegradek Mar 20 '10 at 16:12
Yeah, but is it enough for it to be proper? It seems to me it has to have some relationship with the metric to insure that paths that stray outside of any compact will be infinitely long in the rescaled metric. –  Harald Hanche-Olsen Mar 20 '10 at 16:44
Definition of $h$ uses "length", which gives relationship with the metric. I am not claiming $h$ is proper but it could be. –  Igor Belegradek Mar 20 '10 at 17:39
On a somewhat related note, Sha and Yang have shown the for $n$ and $m \geq 2$, the connect sum of $S^n\times S^m$ with itself an infinite number of times carries a metric of positive Ricci curvature. I don't think these metrics are Einstein, though. –  Jason DeVito Mar 20 '10 at 17:02