Let $X$ be a smooth protective variety, or just a smooth Kahler manifold. Is it possible to have two curves $C_1$ and $C_2$ in $X$ such that their difference in $H_2(X,\mathbb{Z})$ is a non-trivial torsion class ?
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asked Oct 25, 2012 at 4:15
Mohammad Farajzadeh-TehraniMohammad Farajzadeh-Tehrani
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3 Answers
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Yes, this is possible. For an example of a Calabi-Yau threefold with such differences of curves, see my paper with Pavanelli http://arxiv.org/pdf/math/0512182.pdf. I am sure there are much simpler examples, however.
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$\begingroup$ Thanks Mark, and do these two homology classes have non-trivial GW invariants? I am asking this just for curiosity. $\endgroup$ Commented Oct 25, 2012 at 14:59
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$\begingroup$ Yes, they do. This Calabi-Yau is a small resolution of a (2,2,2,2) complete intersection in ${\bf P}^7$, and the curves in question are exceptional curves for this small resolution. Each such curve contributes $1$ to the Gromov-Witten invariants of its homology class. $\endgroup$ Commented Oct 25, 2012 at 15:48
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This is off the top of my head, but I think that the canonical class of the Enriques surface is a torsion class given by the difference of curves. Every Enriques surface is obtained from a rational elliptic surface by performing log-transforms on two of the elliptic fibers. The class $F_1 + F_2 - F$, where $F_i$ are the transformed fibers and $F$ is a generic fiber, is then 2-torsion.
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Every divisor class $D$ on a surface is the difference of two smooth, connected curves. Choose a very ample divisor $A$ and an $n$ so that $D+nA$ is also very ample. Then $(D+nA)-nA=D$ so $D$ is the difference of two curves. They may be chosen smoothly by Bertini's Theorem.
ADDED LATER: They may also be chosen to be connected. The Lefschetz hyperplane theorem shows that hyperplane sections of surfaces are connected.
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$\begingroup$ It seems to me that your $S'$ is just the product $Y\times E$. To get multiple fibers you have to make some more complicated construction. $\endgroup$– ritaCommented Oct 25, 2012 at 10:38
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$\begingroup$ Yeah, dunno what I was thinking... I totally just got it backwards. $\endgroup$ Commented Oct 25, 2012 at 11:11
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$\begingroup$ In dim>1, an ample divisor is always connected. This follows by Lefschetz theorem. $\endgroup$– HenriCommented Oct 25, 2012 at 15:53
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$\begingroup$ what is torsion here! $\endgroup$ Commented Oct 29, 2012 at 3:02
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$\begingroup$ This works whether or not $D$ is torsion, for any $(1,1)$-class in $H^2(X,\mathbb{Z})$. $\endgroup$ Commented Nov 1, 2012 at 3:25