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Is it true that the fundamental groups of compact Kahler surfaces have property T if and only if it they are finite? I am having trouble finding counterexamples to this, but maybe that's just me...

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Igor, if you add the assumption that the fundamental group is Gromov-hyperbolic, then it becomes a very interesting question to which currently there are no counter-examples. –  Misha Mar 25 '12 at 13:55
    
@Misha: Isn't it hard just to find a hyperbolic group with prop T? –  Igor Rivin Mar 25 '12 at 17:54
    
@Igor Rivin: Igor, there are several ways to construct hyperbolic groups with property T. The oldest: (1) Uniform lattices in quaternionic hyperbolic space. More recent: (2) Fundamental groups of 2-dimensional simplicial complexes where links of vertices have smallest eigenvalue $>1/2$. (3) Uniform lattices acting on some hyperbolic buildings. (4) Random groups (in certain regimes) are infinite hyperbolic with property T. Very recent: (5) Oppenheim's constructions. However, it is conjectured that 2-dimensional (hyperbolic) groups are never Kahler (except for surface groups). –  Misha Mar 25 '12 at 19:54
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1 Answer

up vote 8 down vote accepted

According to this survey by Donu Arapura, Toledo proved that many arithmetic lattices in higher rank algebraic $\mathbb{Q}$-groups (with hermitian symmetric space) are fundamental groups of smooth projective surfaces.

In particular $Sp(2n,\mathbb{Z})$ for $n>2$, is such a group, and has property (T).

Note that once you get a group as fundamental group of a smooth projective variety you obtain a smooth projective surface with the same fundamental group by intersecting with some generic hyperplanes.

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Ah, thanks! That certainly does it... –  Igor Rivin Mar 25 '12 at 17:54
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