The existence of complete Riemannian metric

If $M$ is a differential manifold, can we necessarily find a complete Riemannian metric on $M$? (I know we can find a Riemannian metric without completeness assumption.)

Solution 1:

The answer is yes, altought the construction is not simple. Given any metric on a differential manifold, it can be shown that it is conformally equivalent to a complete riemannian manifold.

Reference: Nomizu, Katsumi, and Hideki Ozeki. "The existence of complete Riemannian metrics." Proceedings of the American Mathematical Society 12.6 (1961): 889-891.

http://www.jstor.org/discover/10.2307/2034383?uid=2&uid=4&sid=21105114015163is

Solution 2:

For a proof of Problem 13-17 [p.346] in Lee's Intro. to Smooth Manifolds (2nd ed.), see https://www.ams.org/journals/proc/1961-012-06/S0002-9939-1961-0133785-8/S0002-9939-1961-0133785-8.pdf. The following remarks may help readers read the above paper.

Remark 1. For the information about conformal metrics, read On conformal metrics notation. Conformal metrics $d,d'$ generate the same topology.

Remark 2. For all $x,y\in M$, if $S(x,r(x))\cap S(y,r(y))\neq\varnothing$, then $|r(x)-r(y)|=d(x,y)$; otherwise, $|r(x)-r(y)|<d(x,y)$.

Remark 3. $\omega (x)$ can be constructed by using partitions of unity. That is, $\omega (x)$ is constructed by the formula given in p.46, l.$-$10, Lee's Intro. to Smooth Manifolds (2nd ed.) except that $j$ [the coefficient of $\psi_j$] should be replaced with the maximum of $r^{-1}$ on an appropriate compact set.

Remark 4. $(M,d')$ is complete.

Proof. Let $\{x_n\}$ be a Cauchy sequence in $(M,d')$.\ We may assume $\{x_n\}\subset S'(x_1,1/3)$. Since $S'(x_1,1/3)\subset S(x_1, r(x)/2)$ and the latter set is relatively compact in $(M,d)$, $\exists x\in \bar{S}(x_1, r(x)/2): x_n\to x$. Note that $x\in \bar{S}'(x_1,1/3)$.

This answer is taken from my paper http://www.lcwangpress.com/papers/qm.pdf.