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I am interested in properly discontinuous cocompact subgroups of the group $PU(n,1)$ of automorphisms of the complex hyperbolic space $H^n_{\mathbb{C}}$, says for $n=2,3$. Is there such a lattice $G$ whose abelianisation $G' = G/[G,G]$ has positive rank? How large can the rank of $G'$ be?

In dimension $n=1$, the situation is that any such a lattice is the fundamental group of a hyperbolic surface, so the rank of its abelianisation, i.e the rank of the first homology group of the surface, is twice the genus. So, it can be arbitrary large.

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  • $\begingroup$ @YCor: Thank you for editing the title. $\endgroup$ Jul 30, 2018 at 15:03

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This is going to be loo long for comment and hence posted as an answer.

[1] There are some arithmetic lattices in $SU(p,1)$ which arise as follows. Let $E$ be a totally imaginary quadratic extension of a totally real number field $F$ of degree $d$ over $\mathbb Q$. Let $V=E^{p+1}$ and $h:V\times V \rightarrow E$ a form linear in the first variable and "conjugate linear" (with respect to the non-trivial automorphism of $E/F$; the unitary group $U(h)$ of the Hermitian form is an algebraic group over $F$. WE asume $h$ is so chosen that $U(h)F\otimes _{\mathbb Q } {\mathbb R}\simeq U(p,1)\times U(p+1)^{d-1}$. Then $U(h)(O_F)$ contains a $congruence \quad subgroup$ with nonvanishing and arbitrarily high first Betti number. This is a result due to Kazhdan and a proof is given in the book by Borel Wallach referred to in one of the answers.

[2] There are other kinds of arithmetic groups arising from unit groups of (e.g.) division algebras over $F$ with an involution of the second kind. When $p=2$ (complex hyperbolic surfaces) it is a result of Rapoport and Rogawski that there are no $congruence$ $subgroups$ with non-vanishing first Betti number. However, it is possible that the group has non-congruence subgroups with non-vanishing first betti number since, being a lattice in a real rank one group, congruence subgroup property is not conjectured to be true.

[3] It can be shown that once the first Betti number is non-zero for some arithmetic group, there exists a finite cover with arbitrarily high first betti number (this is an argument due essentially to Borel (some papers of Alan Reid deal with this question).

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Studying Betti numbers of lattices of $SU(p, q)$ is a classical subject and I barely know its history so let me just give some pointers to the literature focusing on $q=1$.

  1. Some examples of lattices in $SU(p,1)$ with nonzero first Betti number appear in [Kazhdan, D., Some applications of the Weil representation. J. Analyse Mat. 32 (1977), 235–248]. There were then works by A. Borel and N. Wallach, and for example, Walalch in [Square Integrable Automorphic Forms and Cohomology of Arithmetic Quotients of $SU(p, q)$, Math. Annalen, (266) 1984, pp 261–278 showed that the first Betti number of lattices in $SU(p,1)$ can be made arbitrary large.

  2. There is also a much studied class of complex hyperbolic manifolds that are fake projective planes (i.e. they have they same Betti number as $CP^2$). See e.g. the survey by S.-K. Yeung Classification of fake projective planes.

  3. For most recent results on nonuniform lattices see Cusp and $b_1$ growth for ball quotients and maps onto $\mathbb Z$ with finitely generated kernel by M. Stover.

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  • $\begingroup$ You seem to imply that the first Betti numbers of finite covers of fake projective planes are unbounded. Is it right ? $\endgroup$
    – BS.
    Jul 29, 2018 at 15:12
  • $\begingroup$ @BS.: I did not say that. The question asked for examples when the first Betti number is nonzero. Fake projective plane provide a well-understood class of examples with $b_1=0$. $\endgroup$ Jul 29, 2018 at 15:16
  • $\begingroup$ For $p,q\geq 2$, the group $G={\rm SU}(p,q)$ has Kazhdan's property (T). By Kazhdan's theorem, the same is true of any uniform lattice $\Gamma$ in $G$. As an easy corollary, the abelianization of $\Gamma$ is finite. $\endgroup$ Jul 30, 2018 at 2:24
  • $\begingroup$ @VictorProtsak: a sample result (due to Wallach cited above) for $SU(p,q)$ is that $b_q=\dim H^q(\Gamma;\mathbb R)$ can be made arbitrary large for a suitable $\Gamma$. I just thought it is illuminating not to restrict to $q=1$. $\endgroup$ Jul 30, 2018 at 2:55
  • $\begingroup$ @Igor Belegradek: Thank you for the references to the papers of Kazdan and Borel-Wallach. Cartwright and Steger have recently found a way to classify all fake projective plane (in fact, it is an algorithm). $\endgroup$ Jul 30, 2018 at 15:10
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There are sequences of congruence covers of certain arithmetic complex hyperbolic surfaces with unbounded first Betti number, as proven in this paper of Simon Marshall: https://arxiv.org/abs/1301.7244 (the proof unfortunately uses advanced automorphic forms techniques). In higher dimensions I don't know what happens, there are some speculations in more recent work of Marshall with Sug-Woo Shin https://arxiv.org/abs/1804.05047. Note that in general it is not known whether a complex hyperbolic lattice has a finite-index subgroup with positive first Betti number.

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The other answers give fine examples, but I found a reference to the 1981 thesis of Livné which constructs complex hyperbolic lattice which admits a surjective holomorphic map to a Riemann surface. This is detailed in chapter 16 of

Deligne, Pierre; Mostow, George Daniel, Commensurabilities among lattices in $\text{PU}(1,n)$, Annals of Mathematics Studies. 132. Princeton, NJ: Princeton University Press. 183 p. $ 49.95/ hbk; $ 19.95/pbk; £ 33.50/hbk; £ 15.00/pbk (1993). ZBL0826.22011.

Since Riemann surfaces have covers with arbitrarily large betti numbers, so do the corresponding ball quotients. The point is here that if one has a holomorphic map $\phi: X \to Y$, $X$ and $Y$ compact, $Y$ a Riemann surface, then $\pi_1(X)$ must surject a finite-index subgroup of $Y$. Otherwise $\phi_\#(\pi_1(X))$ would induce an infinite cover $\tilde{Y}\to Y$ and lift $\tilde{\phi}: X\to \tilde{Y}$ so that $\phi$ factors through $\tilde{\phi}$. But the map $\tilde{\phi}$ must be constant, since it maps a compact complex manifold to a non-compact Riemann surface (by the open mapping theorem, restricted to 1-dimensional complex subspaces of $X$, $\tilde{\phi}$ is open if non-constant, and hence $\tilde{\phi}(X)$ is both open and compact in $\tilde{Y}$, a contradiction). Thus, $\pi_1(X)$ must surject $\pi_1(\tilde{Y})$ for $\tilde{Y}\to Y$ a finite-sheeted cover. Then covers of $\tilde{Y}$ with arbitrarily large betti number induce such covers of $X$.

Some generalizations to other examples are given by Deraux using a forgetful map.

If the lattice is arithmetic, then covers induced from a map to a Riemann surface will usually not be congruence covers.

In the arithmetic case, once one has a (congruence) cover with positive betti number, one can find further (congruence) covers with arbitrarily large betti numbers, as hinted at in Venkataramana's answer. Another perspective on this is given here.

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  • $\begingroup$ Thank you for your answer. There are recent constructions of non-arithmetic lattices due to Deraux, Paupert, Parker which are refection groups with explicit representations and very detailed descriptions of the fundamental domains: Martin Deraux, John R. Parker, Julien Paupert, New non-arithmetic complex hyperbolic lattices. Invent. Math. 203 (2016), 681-771. $\endgroup$ Jul 30, 2018 at 15:22

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