0
$\begingroup$

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)

$\endgroup$
1
  • 1
    $\begingroup$ You should be able to adapt the answer here. $\endgroup$
    – jgon
    Commented Jun 14, 2019 at 6:37

1 Answer 1

1
$\begingroup$

Here is a solution, mostly constructed by my Riemann geometry professor, for anyone seeking further questions regarding similar problems. Any remarks or corrections are welcome:

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .