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Let $X$ be a compact complex surface of general type which a ball quotient. Is it true that $\pi_{1}(X)$ can not contain ${\mathbb{Z}}^{2}$ as a subgroup? What kind of infinite abelian groups can occur as a subgroup of $\pi_{1}(X)$?

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up vote 5 down vote accepted

If I understand your question, you ask if a cocompact torsion free subgroup of $PU(2,1)$ (namely $\Gamma=\pi_{1}(X)$) can contain a ${\mathbb Z}^2$.

This is not the case, because $\Gamma$ is a Gromov-hyperbolic group, due to the fact that $X$ has a negatively curved riemannian metric (quotient of that of the complex hyperbolic plane, aka 4-ball).

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Thank you very much for your answer. Do you know if the subgroups of the form $\mathbb{Z} \oplus \mathbb{Z}_{n}$ can occur, where $n \geq 2$? – David Dec 11 '11 at 17:56
Let me also ask a related question. Can $X$ as above contain a homologically essential real two dimensional submanifold that realizes the subgroup $\mathbb{Z} \oplus \mathbb{Z}_{n}$? I suspect the answer is no, but I am not able to prove it. – David Dec 11 '11 at 18:22
They can appear as subgroups of $PU(2,1)$, but not as a fundamental group, because the finite order elements have fixed points in the complex hyperbolic plane. – BS. Dec 11 '11 at 18:26

I don't know what a ball quotient is, but the decisive value is the genus of $X$. If the genus is zero, $X$ is the projective space, if the genus is two, $X$ is a torus and the fundamental group is ${\mathbb Z}^2$ if the genus is $\ge 2$, then $X$ is a quotient of the upper half plane and the fundamental group $\pi$ is a uniform torsion-free lattice in $G=PSL_2({\mathbb R})$. Therefore every element is semisimple and the centralizer of any element $g\ne 1$ in $G$ is a torus, so the centralizer in $\pi$ is isomorphis to $\mathbb Z$. This means that in the case of genus $\ge 2$ the answer is $\mathbb Z$ only.

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Anton, you are thinking of Riemann surfaces, which have complex dimension 1. The question was about complex surfaces, which have complex dimension 2. – Angelo Dec 11 '11 at 9:12

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