# Complete is equivalent to Misner-complete for Riemannian manifolds.

I'm trying to prove that, if $$M$$ is a smooth Riemannian manifold, then completeness of $$M$$ is equivalent to Misner-completeness.

A pesudo-Riemannian (or semi-Riemannian) manifold $$M$$ is Misner-complete if and only if every geodesic $$\gamma: [0,b) \to M,$$ $$b < \infty,$$ lies in a compact subset of $$M.$$

Now one way is simple, I think. If $$M$$ is complete, then for a geodesic $$\gamma: [0,b) \to M,$$ let $$p = \gamma(0), v = \gamma'(0).$$ Since $$M$$ is complete, there is the maximal geodesic starting at $$p$$ with initial velocity $$v,$$ $$\gamma_v : \mathbb R \to M$$ with $$\gamma_v |_{[0,b)} = \gamma.$$ But then $$\gamma_v([0,b])$$ is compact and contains $$\gamma([0,b)).$$

Now for the other way I am completely lost. It would seen some of the equivalences for a Riemann manifold to be complete via Hopf-Rinow theorem would make things easier, but I've had no success so far.

Any hints are appreciated, thanks.

(for reference, this is problem 13, chapter 5 of O'Neill, Semi-Riemannian Geometry, also problem 5.9.6 of Riemannian Geometry by Petersen, third edition)

– jgon
Jun 14 '19 at 6:37

Let $$M$$ be Riemannian and Misner-complete. Take $$\gamma : [0,b) \to M$$ a geodesic, which can be reparameterized as to be an unit speed geodesic. Then $$\gamma([0,b)) \subseteq K$$ a compact subset of $$M.$$ Also, since $$b< \infty,$$ the length of $$\gamma$$ is finite. For a sequence $$\{t_n\} \in [0,b)$$ converging to $$b,$$ since $$K$$ is compact, the sequence $$\gamma(t_n)$$ has a converging subsequence, so we can as well assume $$\{t_n\}$$ is such that $$\gamma(t_n) \to p \in K.$$ We now prove this $$p$$ is the extension of $$\gamma.$$ Since it is a geodesic, extension as a continuous curve is equivalent to extension as a geodesic, so by proving this we are done.
Assume there is another sequence $$\{s_n\}$$ in $$[0,b)$$ converging to $$b$$ with $$\gamma(s_n)$$ converging to $$q\neq p,$$ so we have $$d(p,q) > 0$$ (where $$d$$ is the Riemann distance). Take $$\epsilon > 0$$ such that $$0 < \epsilon < \frac{1}{3}d(p,q).$$ For $$n \in \mathbb N$$ large enough, $$\gamma(t_n) \in B(p,\epsilon )$$ and $$\gamma(s_n) \in B(q,\epsilon)$$ (the open balls for Riemann distance). By the triangle inequality, $$d(p,q) \leq d(p, \gamma(t_n)) +d(\gamma(t_n),\gamma(s_n)) + d(q, \gamma(s_n)) < 2\epsilon +d(\gamma(t_n),\gamma(s_n))$$ which gives $$\epsilon < d(\gamma(t_n),\gamma(s_n)),$$ for $$n$$ large enough.
By a standard process we can construct two increasing subsequences of $$\{t_n\}$$ and $$\{s_n\}$$ such that $$t_{n_k} < s_{n_k},$$ for all $$k \in \mathbb{N}.$$
To finish, take an $$l \in \mathbb N$$ with $$\epsilon < d(\gamma(t_{n_k}),\gamma(s_{n_k}))$$ for $$k > l$$ and such that $$b < l\cdot \epsilon.$$ We have $$\epsilon < d(\gamma(t_{n_k}), \gamma(s_{n_k})) \leq L(\gamma|_{[t_{n_k}, s_{n_k}]}),$$ where $$L(\gamma|_{[t_{n_k}, s_{n_k}]})$$ is the length of $$\gamma$$ over the interval $${[t_{n_k}, s_{n_k}]}.$$ Summing over $$k,$$ $$b < l\cdot \epsilon < \sum_{k=1}^l L(\gamma|_{[t_{n_k}, s_{n_k}]}) \leq b,$$ a contradiction.