# A proof of the Nomizu-Ozeki theorem

Assume that $$M$$ is a differentiable manifold such that, when equipped with any Riemannian metric, it is always complete. Prove that $$M$$ is compact. Hint: Prove that any succession $$(p_{n})_{ n \in \mathbf{N}} \subset M$$ has an accumulation point, start by choosing minimum geodesics $$\gamma _{n}$$ that join a fixed point $$p\in M$$ with $$p_{n}$$. I tried to show that it's sequentially compact, with the help of the hint, but nothing has worked for me. I would appreciate a help or a comment if you have had this problem or have thought about it.

This result is known as the Nomizu-Ozeki theorem. See The Existence of Complete Riemannian Metrics, Proceedings of the American Mathematical Society, Vol. 12, No. 6, pp. 889-891, 1961.

I will summarize what happens: they prove that

(i) any Riemannian metric on $$M$$ is conformally equivalent to a complete metric, and;

(ii) any Riemannian metric is conformally equivalent to a metric which makes $$M$$ bounded, that is, bounded with respect to the Riemannian distance.

If $$M$$ is bounded with respect to a complete metric, it is compact. So this means that combining the two results mentioned above, your claim follows. I will not reproduce the proof of (i), but will explain (ii).

By (i), assume that the initial metric $$g$$ is complete, and fix $$x_0 \in M$$. The function $$M \ni x \mapsto {\rm d}(x,x_0) \in \Bbb R$$ is continuous, but not necessarily smooth, so we take $$\phi\colon M \to \Bbb R$$ smooth such that $$\phi(x) > {\rm d}(x,x_0)$$ for all $$x \in M$$ instead. Then consider the conformal metric $$g_\phi = e^{-2\phi}g$$, and the induced distance function $${\rm d}_\phi$$. We claim that $${\rm d}_\phi(x,x_0) < 1$$ for all $$x \in M$$, which concludes the argument. By Hopf-Rinow, take a minimizing $$g$$-geodesic $$\alpha\colon \Bbb R \to M$$ joining $$x_0$$ to $$x$$, parametrized by arc-length measured by $$x_0$$ (i.e., $$\alpha(0)=x_0$$ and $$\alpha({\rm d}(x,x_0)) = x$$). Then we have that, in general, $${\rm d}(\alpha(s), x_0) = s$$ for all $$s$$, and so $$\phi(\alpha(s))>s$$ for all $$s$$. With this, we compute $$g_\phi(\alpha'(s),\alpha'(s)) = e^{-2\phi(\alpha(s))} \implies {\rm d}_\phi(x,x_0) \leq \int_0^{{\rm d}(x,x_0)} e^{-\phi(\alpha(s))}\,{\rm d}s \leq \int_0^{+\infty}e^{-s}\,{\rm d}s = 1.$$

• Thanks Ivo, I had read a Nomizu article but I didn't find the relationship of the conformal metrics. – Kevin Oct 19 '19 at 15:57