4
$\begingroup$

Let $M$ be a complete hyperbolic manifold of dimension $n$, let $\varepsilon=\varepsilon_n$ be the Margulis constant. Let $M_{[\varepsilon,\infty)}$ be the thick part of $M$ with respect to $\varepsilon$. Suppose that $\pi_1(M)$ is infinite. Is it true that $\pi_1(M_{[\varepsilon,\infty)})$ is also infinite. Or, in case $M_{[\varepsilon,\infty)}$ is not connected, whether there exist a connected component of $M_{[\varepsilon,\infty)}$ such that its fundamental group is infinite.

PS: I am reading a paper, where this fact seems to be an important point in the proof. This is totally not my area, so it might as well be trivial for anyone familiar with these notions, however, I will be much obliged for an expalnation why this is (or is not) true.

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
6
$\begingroup$

Yes, because $\pi_1$ of the boundary of a component of the $\epsilon$-thin part always surjects to the fundamental group of that thin part, except in one special 2-dimensional case: if there is a short orientation-reversing curve on a surface, then its component of the thin part is a Moebius band, and the boundary generates a subgroup of index 2. This can be dealt with by passing to the orientable double cover, which still would have infinite fundamental group.

Except for the mild exception above, $\pi_1(M_{[\epsilon,\infty)})$ maps surjectively to $\pi_1(M)$, so if the latter is infinite the former is infinite.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you, that settles it for $n>2$ for me, since then $M_{[\varepsilon,\infty)}$ is connected, as I understand. Is it still connected if $n=2$? (Why for instance a short curve cannot separate 2 components?) $\endgroup$
    – Rostyslav Kravchenko
    Commented Feb 5, 2011 at 18:28
  • $\begingroup$ Good point: for $n=2$, the fundamental group of the thick part isn't even well-defined, you have to decide what component the base point is in. However, the conclusion still holds, because you could take the covering space of $M$ corresponding to one particular component of the thick part. Really, every non-trivial subgroup of $\pi_1(M)$ is infinite when $M$ is a hyperbolic manifold. $\endgroup$
    – Bill Thurston
    Commented Feb 5, 2011 at 18:41
3
$\begingroup$

By Margulis lemma, components of the $\varepsilon$-thin part are cusps or $\varepsilon$-tubes, so the interior of the $\varepsilon$-part is $M$ with cusps chopped off, and a finite (possibly empty) collection of embedded closed geodesics deleted. Thus the thick part is connected, and by general position the inclusion of the thick part into $M$ is $\pi_1$-surjective when $n=3$, and is a $\pi_1$-isomorphism when $n>3$. In particular, the fundamental group of the thick part is infinite, because it surjects onto an infinite group $\pi_1(M)$.

EDIT: in the above I assumed that $n>2$.

Improve this answer
$\endgroup$

You must log in to answer this question.

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