Timeline for Torsion in the Betti cohomology of complex surfaces
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 6, 2011 at 21:21 | answer | added | Tom Goodwillie | timeline score: 1 | |
| Jan 6, 2011 at 16:33 | comment | added | Hugo Chapdelaine | Yes, I just worked it out for myself. I had forgotten about the universal coefficient theorem for chain complexes. I also just amused myself to construct explicitly all the $n$ extensions in $Ext(\mathbb{Z}/n,\mathbb{Z})$. | |
| Jan 6, 2011 at 16:09 | comment | added | Tom Goodwillie | 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. | |
| Jan 6, 2011 at 15:54 | vote | accept | Hugo Chapdelaine | ||
| Jan 6, 2011 at 15:44 | comment | added | Donu Arapura | Another nice example is the Godeaux surface. See book by Barth, Peters and Van de Venn. | |
| Jan 6, 2011 at 15:42 | comment | added | Donu Arapura | 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. | |
| Jan 6, 2011 at 15:38 | answer | added | BS. | timeline score: 7 | |
| Jan 6, 2011 at 15:31 | comment | added | Hugo Chapdelaine | 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. | |
| Jan 6, 2011 at 15:19 | comment | added | Donu Arapura | An Enriques surface. Torsion in $H^2$ equals torsion in $H_1$ which comes from etale covers with abelian Galois group. | |
| Jan 6, 2011 at 15:09 | history | asked | Hugo Chapdelaine | CC BY-SA 2.5 |