# Lattices of PU(n,1) with large abelianization

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.

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

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).

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.

• You seem to imply that the first Betti numbers of finite covers of fake projective planes are unbounded. Is it right ?
– BS.
Jul 29, 2018 at 15:12
• @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 $b_1=0$. Jul 29, 2018 at 15:16
• 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. Jul 30, 2018 at 2:24
• @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 $q=1$. Jul 30, 2018 at 2:55
• @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). Jul 30, 2018 at 15:10

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.

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.

• 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. Jul 30, 2018 at 15:22