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Jun 28, 2014 at 11:07 vote accept Alex Gavrilov
Jun 27, 2014 at 1:47 history edited Ben Wieland
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Jun 27, 2014 at 1:32 answer added Ben Wieland timeline score: 3
Jun 23, 2014 at 12:39 comment added Alex Gavrilov It makes sense. This trick does not work for divisors, because $H^1$ has no torsion.
Jun 22, 2014 at 16:49 comment added Jason Starr Here is something worth considering. Let $Y$ and $Z$ be complex, projective manifolds each of which has $p$-torsion in $H_1$, e.g., $Y$ and $Z$ are both Enriques surfaces so $H_1=\mathbb{Z}/2\mathbb{Z}$. Now let $X$ be the product $Y\times Z$. By Kunneth's theorem, there is torsion in $H_2(X,\mathbb{Z})$. What algebraic cycle gives rise to this torsion?
Jun 22, 2014 at 3:34 history asked Alex Gavrilov CC BY-SA 3.0