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?

  • $\begingroup$ General note: Ingrid Irmer uploaded a paper addressing this question last week to the arxive. It contains some nice results about this subspace and conditions for it to generate. arxiv.org/abs/1508.04815 $\endgroup$ Oct 12 '15 at 15:21
  • $\begingroup$ @DanielValenzuela: I should warn people that there are serious issues with Irmer's paper. $\endgroup$ Aug 6 '17 at 18:41

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.

  • $\begingroup$ Thanks for both answers. It is amazing that such a simple question is open! $\endgroup$ Jan 29 '12 at 17:28
  • $\begingroup$ Yeah. Please let me know if you ever get anywhere on it or find it in the literature. $\endgroup$ 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.


I just posted a paper with Justin Malestein entitled "Simple closed curves, finite covers of surfaces, and power subgroups of $\text{Out}(F_n)$" that gives a nearly complete answer to this question. It can be downloaded here.

In this paper, we construct examples of finite covers of punctured surfaces where the first rational homology is not spanned by lifts of simple closed curves. We can deal with any surface whose fundamental group is a nonabelian free group, though we do not know how to deal with closed surfaces.

In fact, we prove a much more general result. Let $\mathcal{O}$ be a subset of a nonabelian free group $F_n$ that is contained in the union of finitely many $\text{Aut}(F_n)$-orbits. For instance, if $F_n$ is the fundamental group of a punctured surface then $\mathcal{O}$ could be the set of all simple closed curves (or even for some fixed $m \geq 0$ the set of all closed curves with at most $m$ self-intersections). Then there exists a finite-index normal subgroup $R$ of $F_n$ such that $H_1(R;\mathbb{Q})$ is not spanned by the homology classes of powers of elements of $\mathcal{O}$.

I also should point out two related papers (see the introduction to my paper above for a more complete summary of the literature). The paper

T. Koberda, R. Santharoubane, Quotients of surface groups and homology of finite covers via quantum representations, Invent. Math. 206 (2016), no. 2, 269–292.

constructs examples of finite covers $\widetilde{S}$ of surfaces $S$ such that $H_1(\widetilde{S};\mathbb{Z})$ is not spanned by lifts of simple closed curves on $S$. They can deal with any orientable surface whose fundamental group is nonabelian (including closed surfaces). However, unlike us they cannot ensure that lifts of simple closed curves do not span a finite-index subgroup of $H_1(\widetilde{S};\mathbb{Z})$. This is why they only deal with integral homology as opposed to rational homology.

Another interesting paper along these lines is

B. Farb, S. Hensel, Finite covers of graphs, their primitive homology, and representation theory, New York J. Math. 22 (2016), 1365–1391.

Among other things, this paper proves that the "rational simple closed curve homology" is everything for abelian covers. It also constructs counterexamples to this for genus $1$ surfaces with $1$ or $2$ punctures.


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