3
$\begingroup$

Let $M$ be a hyperbolic 3-manifold with finitely generated fundamental group. Assume $E$ is a geometrically infinite end (not of geometrically finite type, i.e. the convex core can not be separated of the end for a small neighborhood of $E$). Then there exists a sequence of closed geodesics approaching $E$. Call this family $\mathcal{F}$.

Question: Is it true that there are closed geodesics (in the above family) that are arbitrarily far apart? To be more precise: for a closed geodesic $\gamma$ define $l_\gamma$ as the minimum distance from $\gamma$ to a geodesic in $\mathcal{F}$. Is it true that $\{l_\gamma\}_{\gamma\in\mathcal{F}}$ is unbounded?

If the answer is no, do people know hypothesis that could imply such a statement?.

$\endgroup$

1 Answer 1

3
$\begingroup$

You seem to be asking two different questions:

  1. The first question appears to be: "Given a family of closed geodesics ${\mathcal F}$ which exits an end $E$, is it true that there are members of ${\mathcal F}$ which are arbitrarily far apart?"

This is true and follows directly from the definition :

A sequence of closed geodesics exits an end $E$ if for every compact $K\subset E$ all but finitely many members of the sequence are disjoint from $K$.

Now use metric neighborhoods of radius $i$ of your $\gamma$ as the compacts $K$: This yields geodesics $\gamma_i\in {\mathcal F}$ such that $d(\gamma, \gamma_i)\ge i$.

  1. You seem to be also asking:

"For a closed geodesic $\gamma$ define $l_\gamma$ as the minimum distance from $\gamma$ to a geodesic in ${\mathcal F}$ (I assume, different from $\gamma$). Is it true that $\{l_\gamma: \gamma\in {\mathcal F}\}$ is unbounded?"

The answer to this question is negative. An example is given by $M$ which is an infinite cyclic cover of a manifold $N$ fibered over the circle. Let $g: M\to M$ denote the generator of the deck-group of the covering $M\to N$. Pick a geodesic closed $\gamma$ in $M$ and set ${\mathcal F}=\{\gamma_i= g^i(\gamma): i\in {\mathbb N}\}$. Then it is immediate that for $\gamma_i\in {\mathcal F}$ the number $l_{\gamma_i}$ is independent of $i$ and, hence, your set is bounded.

Of course, it is entirely possible that you meant to ask yet another question (I can think of a couple question along the lines of the above), in which case you should think carefully what your question really is.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.