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...
$\begingroup$
$\endgroup$
3
asked Mar 25, 2012 at 4:52
-
$\begingroup$ 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. $\endgroup$– MishaCommented Mar 25, 2012 at 13:55
-
$\begingroup$ @Misha: Isn't it hard just to find a hyperbolic group with prop T? $\endgroup$– Igor RivinCommented Mar 25, 2012 at 17:54
-
1$\begingroup$ @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). $\endgroup$– MishaCommented Mar 25, 2012 at 19:54
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
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.
-
$\begingroup$ Ah, thanks! That certainly does it... $\endgroup$ Commented Mar 25, 2012 at 17:54