# Holomorphic Foliations having transverse sections

In the introduction to the paper "On the Geometry of Holomorphic Flows and Foliations Having Transverse Sections" by Ito and Scardua, one reads the following "a holomorphic codimension one foliation on a compact manifold is not necessarily transverse to some compact Riemann surface. Indeed, the existence of such a compact transverse section often implies several restrictions on the foliation..."

My question is simply "What restrictions?" This is not really discussed in their paper nor the references, as they concentrate on real transversals. Also, most results I have seen are local, I'm interested in the global statement as phrased by Scardua.

I'd be happy to even concentrate on the case of smooth projective surfaces and smooth transverse algebraic curves. The only general result of which I am aware is the result by Bogomolov/McQuillan in "Rational Curves on Foliated Varieties" and the follow-up article by Kebekus/Conde/Toma "Rationally connected foliations after Bogomolov and McQuillan", that if the transverse curve has positive self-intersection (i.e. is an ample divisor), then the foliation is a fibration with rational leaves. (I'm phrasing it in the case of dimension 2, similar results also hold in higher dimension, giving rational connectedness of the leaves, and with weaker hypotheses as to the nature of the curve and foliation singularities, amount of tangency, etc).

But how about everywhere transverse curves with non-positive self-intersection? What restrictions does this give on the foliation?

• Do you want the curve $C$ in $X$ to be everywhere transverse to the foliation $\mathcal{F}$? If so, then there is an obvious topological condition: the natural map from the direct sum $\mathcal{F}|_C \oplus T_C$ to the restricted tangent bundle $T_X|_C$ must be an isomorphism. This can easily fail, e.g., if $X$ is a complex torus containing no elliptic curves and $\mathcal{F}$ is any codimension one foliation on $X$. – Jason Starr Jul 14 '14 at 18:29
• @JasonStarr Yes, everywhere transverse. So yes, the conormal sequence splits and satisfies the condition that you mentioned. You say this can easily fail. I see it in some simple situations, like the one you mentioned, but in general, to make it interesting let's say on surfaces of general type with Picard rank not equal to one, I don't see how we can use this condition. – Jordan Thomas Jul 14 '14 at 21:11
• Do you have a specific general type surface you would like to understand? Probably, using deformation theory, I could give one example of a foliation on a general type surface of Picard number 2 that has no transversal, however, I doubt that is what you are looking for. – Jason Starr Jul 15 '14 at 17:18
• @JasonStarr Yes, I was really looking for something more along the lines of a classification, such as what jvp provided below. Also, I was wondering if there was any result which would imply that a foliation with infinitely many everywhere transverse curves is a fibration, but it seems to me that there is probably no known result of this form. – Jordan Thomas Jul 17 '14 at 23:01

If a smooth curve $C$ has zero self-intersection and is everywhere transverse to a foliation on a surface then there are also strong restrictions.

If $C$ is rational then the foliation is a Riccati foliation (i.e. is birationally equivalent to the projectivization of a meromorphic flat connection on rank two bundle over a curve).

For an arbitrary smooth curve $C$ everywhere transverse to a foliation $\mathcal F$ on a projective surface we can argue as follows. If $T^* \mathcal F$ is not pseudo-effective then by Miyaoka's Theorem (Bogomolov-McQuillan is a generalization of Miyaoka's Theorem) then $\mathcal F$ is a foliation by rational curves.

If instead $T^*\mathcal F$ is pseudo-effective then the Zariski decomposition $P+N$ ($P$ nef $\mathbb Q$-divisor and $N$ effective contractible $\mathbb Q$-divisor) of $T^* \mathcal F$ satisfies $P \cdot C = 0$. Hodge index theorem implies that $P$ is numerically equivalent to a rational multiple of $C$. This already imposes strong restriction on $\mathcal F$. It must be a foliation of special type (Kodaira dimension is not maximal).

Synthesis. Assume there exists a smooth curve C of zero self-intersection which is everywhere transverse to a foliation $\mathcal F$. I believe that going through the classification of foliations of special type we can proof that at least one of the following assertions holds true:

• The foliation $\mathcal F$ is a foliation by rational curves.
• The curve $C$ is a fiber of rational fibration and the foliation is transverse to the general fiber of it (Riccati foliation).
• The curve $C$ is a fiber of an isotrivial elliptic fibration and the foliation is transverse to the general fiber of it (turbulent foliation).
• The curve $C$ is a fiber of an isotrivial fibration with fibers of genus $\ge 2$ and the leaves of the foliation are all algebraic.

The main references here are McQuillan's paper Canonical models of foliations; and Brunella's book Birational geometry of foliations.

Examples. In contrast, it is rather easy to produce examples of foliations tranverse to curves of negative self-intersection. Start with a reduced foliation of general type on any surface and take a sufficiently general smooth curve. The tangencies between the foliation and the curve will be simple tangencies. Blowing-up each of the tangencies points twice will give rise to a curve everywhere transverse to the resulting foliation.