# Question on Ball Quotients

Let $X$ be a compact Kahler surface which is a ball quotient. Can such $X$ contain a torus $T$ such that the fudamental class of $T$ is non-trivial? I expect this is false as $\pi_{1}(X)$ is a hyperbolic group, thus $\mathbb{Z} \times \mathbb{Z}$ can not occur as a subgroup of $\pi_{1}(X)$ (here I consider the subgroup generated by the loops of the torus $T$). I can also handle the case when the subgroup is $\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$, but I am not able to exclude the cases when the subgroup is $0$, $\mathbb{Z}$, or $\mathbb{Z}/n\mathbb{Z}$ for $n \geq 2$. Can these cases occur?

-
Yes, I assume $X$ is compact. The torus $T$ is a submanifold of $X$, and I consider the subgroup of $\pi_{1}(X)$ generated by the loops on $T$. – kla Jul 29 '12 at 14:08
In a different vein, when people study the birational classification of <i>non-compact</i> ball quotients, they often use "toroidal compactifications", i.e. compactify $B/\Gamma$ by adjoining elliptic curves. – Peter Dalakov Jul 29 '12 at 16:55
Do you mean "complex torus, embedded holomorphically"? Then no, from looking at the universal covers: every holomorphic map from $\mathbb C$ to a complex ball is constant. – inkspot Jul 29 '12 at 19:47
@inkspot Thanks. Very good point! No, I assume the torus $T$ is a symplectic submanifold of $X$. – kla Jul 29 '12 at 21:14

Fundamental group of $X$ is torsion-free, so the image of $\eta: \pi_1(T^2)\to \pi_1(X)$ is either trivial or infinite cyclic. In any case, you can realize $\eta$ by a composition of maps $$T^2\to S^1\to X.$$ The first map will kill the fundamental class of the torus, since $H_2(S^1)=0$.
@Misha Sorry, it is not clear to me where you used the fact that $X$ is a ball quotient when you relaize $\eta$ as a composition. Consider the following example: Let $X = E(1)$, an elliptic surface obtained by blowing up a pencil of cubics in $CP^2$ at $9$ base points of the pencil. Note that the class of the fiber torus is non-zero, and the loops of the fiber torus are nullhomotopic. $X$ is not a ball quotient in this case. – kla Jul 29 '12 at 19:35
Since the universal cover of X is contractible, its homotopy groups vanish in dimensions $\ge 2$, hence homotopy class of a map to X is determined by map of fundamental groups (Whitehead's theorem). – Misha Jul 30 '12 at 3:55
@kla: You have group homomorphisms $Z^2\to Z\to \pi_1(X)$. Each is realized by a continuous map of the respective spaces $T^2\to S^1\to X$ by the same Whitehead's theorem, since circle has contractible universal cover. – Misha Jul 30 '12 at 5:41