Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
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
show 1 more comment

2 Answers

up vote 6 down vote accepted

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).

I would propose an Enriques surface, but I'm not at all convinced it is "simplest".

share|improve this answer
My answer crossed Donu's comments. Sorry. –  BS. Jan 6 '11 at 15:45
No problem. Perhaps evidence that Enriques surfaces are the most obvious examples, although perhaps not the simplest. –  Donu Arapura Jan 6 '11 at 15:50
add comment

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 $(x,y)\mapsto (-x,-y)$.

share|improve this answer
Thanks Tom, yes if you don't take the involution then this is the classical example of an Hopf surface! –  Hugo Chapdelaine Jan 6 '11 at 21:26
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.