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

Let $\Sigma$ be a compact connected oriented surface and $p:\tilde{\Sigma}\to\Sigma$ a finite regular cover.

Consider the set $\Gamma$ of simple closed curves on $\tilde{\Sigma}$ obtained as a connected component of $p^{-1}(\gamma)$ where $\gamma$ is a simple curve in $\Sigma$.

My question is: is it true that $\Gamma$ generates $H_1(\tilde{\Sigma},\mathbb{Z})$ and if not, can we identify the subspace it generates?

share|improve this question

2 Answers 2

up vote 12 down vote accepted

As far as I know, this is open.

In fact, I think the following weaker question is open.

Let $\Theta$ be the set of loops $\gamma$ in $\widetilde \Sigma$ such that the image of $\gamma$ in $\Sigma$ is not a filling curve. If $\Sigma$ is not a pair of pants, is $H_1(\widetilde \Sigma ; \mathbb{Z})$ generated by $\Theta$?

This weaker statement would give a new proof of the congruence subgroup problem for the mapping class group of a genus two surface (which is a theorem of Boggi).

I don't know the answer even when $\Sigma$ is the 5-punctured sphere. I do know that if $\Sigma$ is the 5-punctured sphere, then the answer to the weak question is yes provided the deck group is generated by a subgroup carried by an embedded pair of pants.

share|improve this answer
Thanks for both answers. It is amazing that such a simple question is open! –  Julien Marché Jan 29 '12 at 17:28
Yeah. Please let me know if you ever get anywhere on it or find it in the literature. –  Richard Kent Jan 29 '12 at 20:08

If $\Sigma$ is a sphere, disk, annulus or torus, then this is true.

If $\Sigma$ is a thrice-punctured sphere, then this is false. The only simple closed curves on a thrice-punctured sphere are parallel to the three boundary curves. In the preimage of a finite-sheeted cover, one obtains only boundary parallel curves, and therefore the subspace of homology is the subspace generated by boundary components.

I'm not sure what happens for more complicated surfaces in general. It seems to work for index two covers and planar covers.

share|improve this answer

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.