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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.

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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 $$g_\phi(\alpha'(s),\alpha'(s)) = e^{-2\phi(\alpha(s))} \implies {\rm d}_\phi(x,x_0) \leq \int_0^{{\rm d}(x,x_0)} e^{-\phi(\alpha(s))}\,{\rm d}s \leq \int_0^{+\infty}e^{-s}\,{\rm d}s = 1.$$

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

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