# Homology generated by lifts of simple curves

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?

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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 – Daniel Valenzuela Oct 12 '15 at 15:21

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.

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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.

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An example providing a negative answer to your question is discussed in Appendix A of this paper of Putman-Wieland (http://www.ams.org/mathscinet-getitem?mr=3092259), and it was previously introduced in the Teichmueller dynamics literature under the name of "Eierlegende Wollmilchsau".

As it turns out, I learned from Looijenga that your question is related to some conjectures of Ivanov and Putman-Wieland (which are discussed for example in this blog post here (https://matheuscmss.wordpress.com/2015/04/24/some-comments-on-the-conjectures-of-ivanov-and-putman-wieland/#comment-1658)).

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It doesn't look like a counterexample to me. – Ben Wieland Aug 28 '15 at 19:39
You're right that there is a bit of information lacking in my answer. In a work of Boggi and Looijenga (still in progress), they show among other things that the primitive homology of $\tilde{\Sigma}$ relative to $p$ (=the submodule generated by $\Gamma$ in the OP's notation) is non-trivial (i.e., not finite-index in $H_1(\tilde{\Sigma},\mathbb{Z})$) whenever the corresponding higher-Prym representation has a non-trivial subspace of vectors with finite orbits. Thus, I think that the Eierlegende Wollmilchsau is a counterexample thanks to the "non-trivial finite orbits property". – Matheus Aug 29 '15 at 16:08
Alternatively, I think that you can check directly this property in the case of the Eierlegende Wollmilchsau: if I'm not mistaken, one can compute with the definition of the Eierlegende Wollmilchsau covering $p$ to check that the kernel $K$ of $p_*$ (i.e., the homology classes on the genus 3 surface $\tilde{\Sigma}$ projecting to zero on the torus under $p$) is non-trivial, and we have a direct sum decomposition $K\oplus P = H_1(\tilde{\Sigma},\mathbb{Q})$ where $P$ is the submodule generated by $\Gamma$ (in the OP's notation). – Matheus Aug 29 '15 at 16:16
Just draw some pictures and see that the OP's curves generate the rational homology. I think integral, too, but I'm not sure. Are you sure Boggi and Looijenga are talking about the same group? – Ben Wieland Aug 29 '15 at 21:31
Maybe I misunderstood both the OP's question and the statement of Boggi and Looijenga draft, but what I had in mind about the features of the Eierlegende Wollmilchsau is also contained in the following answer (mathoverflow.net/questions/136415/…) to a related question. – Matheus Aug 30 '15 at 22:14