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Let $X$ be a smooth projective complex algebraic variety of general type. Suppose that the (topological) fundamental group of $X$ is an infinite abelian group and that $\pi_2(X^{an})$ is finite.

What can we say about $X$?

I am mainly interested in the case where $\dim X = 2$, and $\Omega^1_X$ satisfies some positivity properties such as ampleness.

What is "known" about surfaces $X$ with infinite abelian fundamental group, finite $\pi_2$ and ample $\Omega^1_X$?

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There is a Hurewicz homomorphism from $\pi_2$ to $H_2$. For every projective complex manifold of dimension $>0$, $H_2$ is infinite: intersection against the first Chern class of an ample divisor defines a nonzero homomorphism from $H_2$ to $\mathbb{Z}$. –  Jason Starr Jul 26 '13 at 14:53
    
By a theorem of Gurjar, see "Two remarks on the topology of projective surfaces." Math. Ann. 328 (2004), no. 4, 701–706, $\pi_2$ is always torson free if Shafarevich's conjecture is true; this is now known in many cases. So if $\pi_2$ is finite, it should be trivial. –  ulrich Jul 27 '13 at 10:44
    
It seems likely that there is no such surface. If the universal cover is Stein (as claimed by user37314), then it would have to be contractible (since $\pi_1$ and $\pi_2$ are tivial). So $\pi(X)$ determines the cohomology of $X$ which must then be that of an abelian surface. This easily implies that $X$ must itself be an abelian surface since the map to the Albanese is forced to be of degree 1. –  ulrich Jul 27 '13 at 11:04
    
@ulrich: Shafarevich conjecture is true for $X$ with nilpotent fundamental group (Katzarkov, 1997). –  Misha Jul 27 '13 at 11:39
    
My comments and posted answer were wrong! –  Jason Starr Jul 27 '13 at 11:43

3 Answers 3

up vote 7 down vote accepted

I combine user37314's answer and my comments; the claim is that any smooth projective complex algebraic surface with $\pi_1$ abelian and $\pi_2$ finite has a finite cover which must be an abelian surface; in particular, $X$ cannot be of general type.

By replacing $X$ by a finite cover, we may assume that $\pi_1(X)$ is free abelian. The map $X \to Alb(X)$ induces an isomomorphism on $\pi_1$ and the universal cover of $Alb(X)$ is a complex Euclidean space, so by taking the fibre product we get a proper map from the universal cover $\tilde{X}$ of $X$ to a complex Euclidean space. The fundamental class of any positive dimensional fibre of this map would give a non-torsion class in $H_2(\tilde{X}) = \pi_2(\tilde{X})$ (since $\pi_1$ is trivial) so this map must be finite.

It follows that $\tilde{X}$ is Stein and moreover $\pi_2(\tilde{X}) = 0$ (see Gurjar, "Two remarks on the topology of projective surfaces." Math. Ann. 328 (2004), no. 4, 701–706; we do not need any results about the Shafarevich conjecture.) Since $\tilde{X}$ is a Stein surface it follows (from Morse theory) that $H_i(\tilde{X}) = 0$ for all $i > 2$ and so by Hurewicz, $\pi_i(\tilde{X}) = 0$ for all $i$. It follows that $\tilde{X}$ is contractible, so the cohomology of $X$ must be the group cohomology of $\pi_1(X)$. Since $X$ is a projective surface this can only happen if $\pi_1(X)$ has rank $4$ and the cohomology ring of $X$ must be isomorphic (by pullback) to that of its Albanese. This implies that the Albanese map has degree $1$ and all fibres are finite so it must be an isomorphism.

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With regard to surfaces: it is a simple theorem that $\pi_2$ of any closed $4$-manifold is torsion free; so one doesn't need any complex geometry to see that there are no such projective surfaces. The argument I will give was observed by the late Jerry Levine and me in 2004 after some correspondence with Gurjar following his paper referred to in ulrich's answer. The idea is to apply the argument that follows to the universal cover of a closed 4-manifold. It seems that the more interesting question, addressed by Gurjar for complex surfaces, is whether $\pi_2$ is actually free; our attempts to resolve this ran into some well-known problems related to group cohomology and ends of manifolds.

Let X be a simply connected $4$-manifold with empty boundary. Then $H_2 (X)$ is isomorphic to $H^2_c (X)$, cohomology with compact supports. This is the direct limit of $H^2 (X,X-C)$, where $C$ ranges over compact subsets of X. we have an exact sequence $$ 0 \to H^1 (X-C) \to H^2 (X,X-C) \to H^2 (X) $$ Now $H^1 (X-C)$ is torsion-free because $H^1$ always is. $H^2 (X)=Hom(H_2 (X),Z)$ is also torsion-free since $X$ is simply connected, and so $H^2 (X,X-C)$ is torsion-free.

But now it is easy to see that the direct limit of a family of torsion-free groups is torsion-free.

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Your assumption implies that there is a finite holomorphic map of the universal cover of X into complex euclidean space .In particular the universal cover is Stein.

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Which assumption implies that there is a finite holomorphic map of the universal cover into complex euclidean space? –  Uiterloo Jul 26 '13 at 19:20
    
your assumptions on the fundamental group implies that the universal cover admits a proper holomorphic map to complex Euclidean space .The assumption on the –  Mohan Ramachandran Jul 26 '13 at 19:25
    
:contd second homotopy group implies the Kahler form is exact on the universal cover so by Stokes's theorem the universal cover has no positive dimensional compact complex subvarieties. –  Mohan Ramachandran Jul 26 '13 at 19:28

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