Question

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.

Answer 1

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.

Answer 2

This is not an answer to your question. I just wanted to point out that there are plenty of examples of complete Einstein manifolds of infinite topological type. I am aware of the following two papers at least:

1. Complete Ricci-flat Kåhler manifolds of infinite topological type, by Anderson, Kronheimer and LeBrun

2. Continued fractions and Einstein manifolds of infinite topological type, by Calderbank and Singer

Answer 3

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 $h^2$.