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Timeline for Algebraic cycles

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Jul 31, 2012 at 12:30 history edited upd CC BY-SA 3.0
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Jul 25, 2012 at 12:03 vote accept upd
Jul 25, 2012 at 9:45 answer added diverietti timeline score: 3
Jul 25, 2012 at 7:05 comment added diverietti Maybe I shall give a more detailed answer. I'll do that.
Jul 24, 2012 at 12:16 comment added Francesco Polizzi Take a numerical Godeaux surface (i.e. a minimal surface of general type with $p_g=q=0$, $K^2=1$) without $-2$ curves. Then $K_X$ is ample and $K^2=1$, but the cycle $[K_X]$ cannot be represented by the class of a complex submanifold, since $h^0(X, K_X)=p_g=0$, i.e. there are no holomorphic sections at all. Of course if $A$ is very ample then what you want is true, since $h^0(X, A) >0$.
Jul 24, 2012 at 12:07 comment added upd Thanks. If I understood correctly, the class $[A]$ itself can not necessarily represented by a fundamental class of an analytic subvarity? Is there a condition on the line bundle that tells when such meromorphic section is holomorphic? Probably this holds true if the line bundle is very ample? If the line bundle $L$ is just ample and say the section has self-intersection $1$ (or say $L$ generates $H^2(X)$), is it true that the section you mentioned is holomorphic?
Jul 24, 2012 at 10:09 comment added diverietti In fact, it follows from Lefschetz theorem on $(1,1)$-classes plus the fact that every line bundle on an projective manifold does have a meromorphic section. The cycle in question is then given as the zero-poles divisor of the meromorphic section.
Jul 24, 2012 at 9:45 comment added Francesco Polizzi Yes, any such a class can be represented by a (linear combination of) fundamental classes of analytic subvarieties of $X$, and this follows from Lefschetz Theorem on $(1,1)$-classes. See Griffiths-Harris "Principles of Algebraic Geometry" p. 163
Jul 24, 2012 at 9:30 history asked upd CC BY-SA 3.0