# Torsion in the Betti cohomology of complex surfaces

Q1. So what is the "simplest example" of a compact complex manifold of dimenension $2$, say $X$, for which $H_B^2(X,\mathbb{Z})$ has non-trivial torsion.

Q2. How do we think about these torsion elements? What is the geometrical content behind it?

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An Enriques surface. Torsion in $H^2$ equals torsion in $H_1$ which comes from etale covers with abelian Galois group. –  Donu Arapura Jan 6 '11 at 15:19
Thnaks Donu. But I guess that in general there is no reason why the torsion in $H^1$ should inject in $H^2$. May be I should also require $X$ to be simply connected. In any case, thanks for your example. –  Hugo Chapdelaine Jan 6 '11 at 15:31
Hugo, I realize I was bit too concise (but I'm a bit rushed). The identification with $H^2(X,\Z)_{torsion}\cong H_1(X,\Z)_{torsion}$ comes from he universal coefficient theorem. So for simply connected surfaces, there is no torsion. One can also see that $Pic$ surjects onto the torsion in $H^2$, and then take the Kummer cover associated to the corresponding line bundle. –  Donu Arapura Jan 6 '11 at 15:42
Another nice example is the Godeaux surface. See book by Barth, Peters and Van de Venn. –  Donu Arapura Jan 6 '11 at 15:44
Donu is saying that the torsion in $H^2$ is isomorphic to the torsion in $H_1$. More precisely, by universal coefficients the torsion in $H^2$ is isomorphic to the torsion in $Ext(H_1,\mathbb Z)$, which if $H_1$ is finitely generated is $Hom((H_1)_{tors},\mathbb Q/\mathbb Z)$ and isomorphic to $(H_1)_{tors}$. If $X$ is simply connected then $H^2$ is torsion-free. –  Tom Goodwillie Jan 6 '11 at 16:09

The torsion of $H_B^2(X,\mathbb{Z})$ is that of $H_1(X,\mathbb{Z})=\pi_1(X)^{ab}$ (universal coefficient theorem for cohomology), so the simplest case should be a simply connected complex surface quotiented by a fixed point free holomorphic involution (or a prime order automorphism).
Are you interested in a complex manifold which is not Kaehler? $\mathbb RP^3\times S^1$ admits a complex structure. Remove the origin from $\mathbb C^2$, and then divide by the free action of $\mathbb Z\times \mathbb Z/2$ generated by $(x,y)\mapsto (2x,2y)$ and $(x,y)\mapsto (-x,-y)$.