The situation is the following.

A finite-index subgroup $\Gamma$ of $SL_2(\mathbb Z)$ acts on the upper-half plane $\mathcal H$. It has a fundamental domain, obtained by a union of translates of the fundamental domain for $SL_2(\mathbb Z)$. The quotient $\mathcal H/\Gamma$ can be compactified by adding a finite number of cusps(I have checked that this can indeed be done, in some book). We call it $X$.

Now I want to compute the de Rham cohomology or singular (co)homology of $X$. I am unable to do it in the general case. Any hints on how to proceed would be appreciated.

The difficulty I am facing is that I am given a group to work with, and the standard examples of computations are with simple spaces through Mayer-Vietories. I do not a priori have a nice Mayer-Vietories decomposition of the space. Or perhaps the best method is not through Mayer-Vietories?

More generally, if $\Gamma$ is a discrete subgroup of $SL_2(\mathbb R)$, firstly, 1) How would one construct a fundamental domain? and, 2) How would one compute the homology?

Re to Sam Nead: I had only the computation of $H_1$ or $H^1$ in mind.

  • 2
    $\begingroup$ In any concrete instance: simply triangulate the fundamental domain, and compute. Slightly more theoretically, use the theory of modular symbols (as described in papers of Mazur and Manin from the 1970s, in Cremona's book, and in many other places). $\endgroup$ – Emerton May 19 '10 at 21:51
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    $\begingroup$ Wouldn't it me best to just abelianise the fundamental group eg. using generators and relations as described here: en.wikipedia.org/wiki/Fundamental_polygon. $\endgroup$ – Dan Piponi May 19 '10 at 21:53
  • $\begingroup$ The first (co)homology group of the genus $g$ surface is $Z^{2g}$. The zeroth and second are both $Z$. The ring structure is a direct sum of g copies of the matrix [[0 1],[1 0]]. If you want an answer more sensitive to your problem, you'll have to be more precise. $\endgroup$ – Sam Nead May 23 '10 at 17:46
  • $\begingroup$ To echo algori and Sam, you need to explain how the fundamental group is being specified. Without that information it's impossible to provide a practical way to compute $H^1$. $\endgroup$ – Deane Yang May 23 '10 at 19:08
  • $\begingroup$ What I had in mind is the following: The choice of a nice triangulation is an art. Is there a more systematic way to go about it? In fact, I wanted to delete this question. But after I announced the intention in another question, algori answered this question and Greg Kuperberg suggested that I better not delete this. So I am keeping it up. If it were not for the discomfort of troubling Emerton, sigfpe and algori, I would have deleted this question(since as you say I also think no purpose is served by keeping it open). $\endgroup$ – Akela May 24 '10 at 0:03

It is hard to answer the question without actually knowing what the group is. Here are some remarks:

  1. For classical congruence subgroups of $PSL_2(\mathbf{Z})$ there is a formula for the number of cusps and the genus. It can be found in many books on modular forms and related things. If memory serves, it is given in Diamond-Shurman (and probably in Shimura's book too).

  2. For a general $\Gamma$ of finite index or not the quotient $H/\Gamma$ is homotopy equivalent to the graph $X_{comb}$ constructed as follows. Let $\Gamma\setminus PSL_2(\mathbf{Z})$ be the set of the right cosets; there is a natural right action of $PSL_2(\mathbf{Z})$ on it. The set of the vertices of $X_{comb}$ is the disjoint union of the sets of orbits of the standard elements of order 2 and 3 (tried to put the matrices here, but these won't show properly). Two vertices of $X_{comb}$ are joined with an edge iff the corresponding orbits intersect (and there are as many edges joining the vertices as there are elements in the intersection). This gives the cohomology and more.

  • $\begingroup$ For the classical congruence subgroups all the relevant formulas should be at modular.math.washington.edu/books/modform/modform/… . $\endgroup$ – Qiaochu Yuan May 23 '10 at 20:00
  • $\begingroup$ This comment is just a confirmation that Shimura's book has a formula (Prop 1.40) for genus assuming you have the index and an enumeration of special points, and Shimura calculates precise inputs for $\Gamma_0(N)$ and $\Gamma(N)$. Katz-Mazur calculates the relevant inputs for several other congruence groups. $\endgroup$ – S. Carnahan May 23 '10 at 20:06
  • $\begingroup$ Various people noted that the question is naive. It was my original intention to delete it. But now that you have answered it, I am in a quandary as it would mean your answer also would vanish. So what do you think about deleting the question? If you would let me I would go ahead with it. $\endgroup$ – Akela May 24 '10 at 2:57

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