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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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 http://en.wikipedia.org/wiki/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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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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