Let $C$ be a curve on a smooth projective three-fold $M$ equipped with the restriction of the Fubini-Study metric $\omega$. I'd like to know if there exists a surface $S$ such that for every closed $(1,1)$-form $\alpha$, $\displaystyle \int_C \alpha = k \int_{S} \alpha \wedge \omega$ for some $k>0$ (possibly depending on $C$).
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$\begingroup$ Is your question asking whether the class of $C$ (in cohomology) is equal to the class of $S$ cupped with the class of $\omega$? If so, I don't think this is true: the point is that given $S$, curves obtained by intersecting $S$ with various hyperplanes will fill out all of $S$, but there exist $M$ and curves $C$ on $M$ so that any (reduced) curve on $M$ whose cohomology class is a multiple of the class of $C$ is actually equal to $C$. This happens when there is a morphism $M \to M'$ which is an isomorphism onto its image outside $C$ but maps $C$ to a point. $\endgroup$– nafCommented Dec 15, 2023 at 12:44
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$\begingroup$ I guess I am asking whether given a C, there is an S, such that a multiple of C is S cupped with omega. $\endgroup$– VamsiCommented Dec 15, 2023 at 16:41
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$\begingroup$ That is not true. Let $M$ be a general ample hypersurface in $\mathbb{CP}^2 \times \mathbb{CP}^2$ of bidegree $(d,e)$ for $1\leq d \leq 3$ and $e\gg 0$. There exist curves $C$ in $M$ of bidegree $(1,0)$, but there are no surfaces in $M$ of bidegree $(r,0)$. $\endgroup$– Jason StarrCommented Dec 15, 2023 at 18:43
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$\begingroup$ I see. Sorry can you elaborate? By ampleness and Mori, the line bundle of every surface is presumably ample and hence no (r,0) ones I suppose. But why are there (1,0) curves? Sorry if I am making a stupid mistake but the ample hypersurface is CP3 and for that this property appears to be true. $\endgroup$– VamsiCommented Dec 16, 2023 at 1:50
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1$\begingroup$ If $S$ exists for such a curve $C$, then by intersecting $S$ with various hyperplanes you would get curves not equal to $C$ having the same homology class (up to a scalar). The homology class of $C$ maps to zero in the homology of $M'$, but the class of no other curve can map to zero (assuming $M'$ is projective) since the class of any curve is always nonzero in the homology of a projective variety $\endgroup$– nafCommented Dec 18, 2023 at 10:13
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