The two-dimensional complex unit ball $B$ has group of biholomorphic automorphisms $PU(2,1)$.


If $Γ$ is an arithmetic subgroup of $PU(2,1)$, the quotient $Γ\text{\\}B$ is an orbifold.


Taking its minimal resolution gives a complex manifold $X$.


**Question:** How to compute the Hodge numbers of $X$, given the group $Γ$ (in matrix form or generators-relation form)?



Computing the invariants of ball quotient surfaces
I'm assuming that you know "where" in the commensurability class your lattice is. By this, I mean you perhaps have $\Gamma$ as a subgroup of some principal arithmetic lattice $\Lambda$ of known index, e.g., as a subgroup of $\mathrm{PU}(h, \mathcal{O}\_k)$ where $h$ is a hermitian form on $k^3$ for an appropriate number field $k$. You then know the orbifold Euler characteristic of $X = \Gamma \backslash \mathrm{B}$, using Prasad's formula for the orbifold Euler characteristic of $Y = \Lambda \backslash \mathrm{B}$.


To start, you need to know the conjugacy classes of finite subgroups of $\Gamma$. This can be a bit subtle. You cannot, for example, read these all off of a presentation. You could, for example, have a computer find a fundamental domain for the action of $\Gamma$ on $\mathrm{B}$ (if you have matrix generators), then find the stabilizers of the lower-dimensional faces. Alternately, you can sometimes figure out the torsion in $\Lambda$ by hand, but you still need to convert these elements to words in $\Gamma$ if you're just working with a presentation. Long story short, if you don't already know $X$ as a normal variety (along with the orbifold locus), this step can be a gigantic pain.


Assuming we now have the conjugacy classes of finite subgroups, you can figure out all the singularities of $X$ (as a normal variety). In essence, you care about the subgroups with an isolated fixed point for their action on $\mathrm{B}$. These are all classical singularities from subgroups of $\mathrm{U}(2)$, so you're now ready to resolve them to get the smooth resolution $\overline{X}$. Also, from these finite subgroups, you can figure out the topological Euler characteristic of $X$ from the orbifold Euler characteristic (this is, for instance, explained in the book of Barthel, Hirzebruch, and Hofer), and hence you can find $c\_2(\overline{X})$. There is an additional subtle point here: you need to know the genus of each curve in the orbifold locus. These are arithmetic Fuchsian subgroups of $\Gamma$ associated with stabilizers of certain sub-balls of $\mathrm{B}$, so you need to figure these out as well - this is possible, but again quite annoying.


We now figure out the irregularity $q(\overline{X}) = h^{1,0}(\overline{X})$. Take the quotient of $\Gamma$ by its subgroups generated by complex reflections, you get a presentation for $\pi\_1(X)$. You can then abelianize to find $H\_1(X, \mathbb{Z})$.


From all the standard relations between Hodge numbers and characteristic numbers, all we need now is the holomorphic Euler characteristic $\chi(\mathcal{O}\_{\overline{X}})$. This you can deduce from Hirzebruch proportionality. To be a bit careful, let $Z \to X$ be a cover by a smooth ball quotient (e.g., you can find an explicit congruence subgroup of $\Gamma$ that is torsion-free), and we then have
$$
c\_1^2(Z) = 3 c\_2(Z) = 9 \chi(\mathcal{O\_Z}),
$$
and we know $c\_2(Z)$ from the orbifold Euler characteristic of $Y$ and $[\Lambda : \pi\_1(Z)]$. Now use properties of the holomorphic Euler characteristic under finite maps to figure out $\chi(\mathcal{O}\_{\overline{X}})$. We now have $p\_g(\overline{X}) = h^{2,0}(\overline{X})$, and can get $h^{1,1}(\overline{X})$ from $c\_2(\overline{X})$.


You could really argue that everything I've described comes from finding an explicit Galois cover $Z \to Y$ by a smooth ball quotient with finite Galois group $F$. You know $q(Z)$ from abelianizing $\pi\_1(Z) \le \Gamma$, then Hirzebruch proportionality gives you the other Hodge numbers of $Z$. Now it's all about computing invariants for resolutions of quotient surfaces. Knowing the number field $k$, one can find a completely explicit $F$ by taking $\pi\_1(Z) < \Lambda$ to be a torsion-free congruence subgroup (and only only needs to avoid certain explicitly computable congruence subgroups to ensure torsion-free).


You could probably go for the canonical divisor instead of the holomorphic Euler characteristic, but I think that's more work. Also, when $X$ has cusps, there is additional fuss about resolving the cusp singularities, but the effect on the characteristic numbers can be found in the literature (Holzapfel's book, work of Hirzebruch, Feustel, ...).


While this all sounds impossible in practice, people have done this for small Picard modular groups (i.e., where $k$ is an imaginary quadratic field). However, the problem does get out of hand pretty quickly.


