Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is there a cusped (non-compact) orientable hyperbolic 3-manifold $M$ which has Heegaard genus $g$, and has a finite-sheeted cover with Heegaard genus $<g$? Moreover, if the cover is index $n$, then it should have fewer than $kn$ cusps, where $k$ is the number of cusps of $M$ (so it does not induce the trivial $n$-fold cover of each cusp component).

Remarks: I believe this is an open question. To clarify, by Heegaard genus $g$ of a cusped hyperbolic manifold $M$, I mean that $M$ is the interior of a compact manifold $N$ with torus boundary components, so that $N$ has Heegaard genus $g$, i.e. there is a surface of genus $g$ in $N$ splitting it into two compression bodies. For closed manifolds, there are examples of Alan Reid. One may modify these examples to get cusped examples which are the trivial cover of the cusps. For more discussion in the closed case, see this paper of Peter Shalen.

I would also be happy with a 1-cusped manifold of Heegaard genus 3, and which has a $k$-fold cover with a maximal embedded horocusp of volume $\leq \pi$ and $<k$ cusps (e.g. an irregular 3-fold cover with 2 cusps). (although Nathan's answer answers the original version of this part of the question, I realized I hadn't given the correct formulation).

share|improve this question
@Agol: Do you mean that the Heegaard genus is the minimal Heegaard genus of $M$? –  yanqing Jul 11 '13 at 2:11
Yes, that is the usual meaning of the Heegaard genus of a manifold. –  Ian Agol Jul 11 '13 at 3:29
add comment

1 Answer

Here's an answer to your alternate question: The manifold m135, which is the punctured torus bundle $-LLRR$, has Heegaard genus 3 and max cusp volume < 2.83. There are plenty of examples of this kind, even if you want the lower bound on Heegaard genus to be certified by the rank of the homology, e.g. m136, m306, m307, s298, s299, s595 which have max cusp volumes in the range [2.8,3.12].

share|improve this answer
Hey Nathan, thanks for the examples - I was aware of that one. However, I realized that I wanted a stronger condition, that there is an irregular cover which is bicuspid. In the case that the rank of $H_1$ is 3, then there is a regular bicuspid cover, which has number of cusps equal to the index of the cover. So I don't want genus to be certified by rank of $H_1$, which makes things trickier. Marc found many manifolds in the snappea census which have a 3-fold irregular cover with a cusp of volume < 3; however, they appear to be Heegaard genus 2. –  Ian Agol Jul 18 '13 at 15:41
To clarify, you want the base manifold to have cusp volume < 3 or the cover to have that property? –  Nathan Dunfield Jul 19 '13 at 3:07
Yes, the cover should have cusp volume $<3$ (which will imply the base will too). –  Ian Agol Jul 19 '13 at 4:31
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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