Let $X$ be a compact complex manifold equipped with a holomorphic symplectic form $\omega$. Let $D$ be a smooth divisor on $X$. At each point of $D$, the restriction of $\omega$ to $D$ has one-dimensional kernel. This gives a non-singular foliation $F$ on $D$. Is it possible that some leaves of $F$ are algebraic while some are not?
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$\begingroup$ Is this form skew-symmetric? closed? $\endgroup$– nikitamarkarianCommented May 17, 2016 at 9:14
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$\begingroup$ @nikitamarkarian Sure, sorry, I forgot to specify this. $\endgroup$– lks8271Commented May 17, 2016 at 11:32
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1 Answer
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Suppose $X$ is projective manifold endowed with holomorphic symplectic form. Let $D$ be smooth divisor on $X$.
Characteristic foliation with algebraic and non-algebraic leaves. Suppose that $X$ is a product of two abelian varieties of dimension $n$ ($n\ge 2$) $A_1 \times A_2$ in such a way that the fibers of the natural projection $\pi: X \to A_2$ are Lagrangian. Let $D$ be the pull-back under $\pi$ of an ample divisor $E$ on $A_2$. The characteristic foliation on $D$ will be everywhere tangent to $\pi$ and over each fiber is a linear foliation. The slope of the foliation on the fiber over a a point $p \in E$ is determined by the tangent of $E$ at $p$. But since $E$ is an ample divisor, it has non-degenerate Gauss map and every possible linear foliation on $A_1$ will appear among the restriction of the characteristic foliation $D$ to fibers of $\pi$. If $A_1$ itself is a product of elliptic curves (or isogeneous to the product of an elliptic curve and an abelian variety of dimension $n-1$) then $A_1$ carries linear foliations with all leaves algebraic. This shows the existence of divisors with the requested property.
Characteristic foliation with all leaves algebraic. If every leaf of the characteristic foliation is algebraic then it was proved by Amerik and Campana (refining previous result by Hwang and Viehweg) that either every leaf is rational or $X$ is, up to étale coverings, the product of a symplectic surface $S$ with a symplectic manifold $Y$ and $D$ is the product of a curve $C\subset S$ with $Y$. In the latter case the leaves of the characteristic foliation are the fibers of the projection to $Y$. In particular, the Kodaira dimension of $D$ is at most $1$.
Characteristic foliation on ample divisors. One interesting problem on the subject (already raised by Hwang and Viehweg) is whether or not the characteristic foliation on an ample divisor can have one algebraic leaf. To the best of my knowledge this problem is wide open.
answered May 18, 2016 at 2:59
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2$\begingroup$ Just to mention one more thing about this question: the uniruledness conjecture (and thus the abundance conjecture) predicts that whenever $\mathcal{O}_X(D)|_D$ is not (eventually) effective, then the characteristic foliation is algebraically integrable with rational curves as leaves. For small dimensions this is known, but I believe this is wide open starting at dimension $6$. $\endgroup$ Commented May 18, 2016 at 11:05
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$\begingroup$ @Jorge Vitório Pereira Thank you for the answer! And can something specific be said about the algebraic leaves if it is known that some of them are algebraic and some are not(not in your example but in general)? And, sorry for the off-topic, but can this situation(algebraic and non-algebraic leaves) occur when the dimension $n$ is $1$ and the foliation is genereated by a global holomorphic 1-form on a surface? $\endgroup$– lks8271Commented May 18, 2016 at 16:32
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$\begingroup$ Concerning your first question not much comes to mind. For the second question the answer is no (one algebraic leaf implies every leaf is algebraic ) if you assume that the $1$-form has no zeros. You have just to observe that you can find a primitive for $1$-form at an analytic neighbourhood of any algebraic leaf and one can choose these neighbourhoods foliated by algebraic leaves. If you allow singularities then you can easily produce examples with algebraic and non-algebraic leaves by blowing-up. $\endgroup$ Commented May 18, 2016 at 22:44