Is there a singular holomorphic foliation $F$ of $\mathbb{C}P^2$ which does not admit a global transverse holomorphic curve? More precisely there is no an immersed holomorphic submanifold of $\mathbb{C}P^2$ which intersect all regular leaves, transversely? If there exist such an example $F$, does this foliation admit a smooth(but not necessarily holomorphic) global transverse submanifold (of real dimension 2)?


The answer to this question is negative. If we require the transversal surface to be an immersed holomorphic curve then the only foliation for which such a surface exists is the pencil of lines. This is proven the Claim below. I will consider more generally the second part of your question where we require the transversal surface to be smooth but not necessarily holomorphic.

I will produce examples of foliations that don't have such a smooth transversal surface at the end of this answer. And I guess in reality that the only foliation that possesses a transversal surface is the pencil of lines (though I can not prove it so far).

Note that each singular holomorphic foliation on $\mathbb CP^2$ defines a rank one subsheaf of $T\mathbb CP^2$ and its first Chern class is at most $1$. I claim the following statement concerning transversal pairs $(\cal F,S)$ where $S$ is smooth (but not necessarily holomorphic). The case when $S$ is holomorphic immersed is proven in the same way.

Claim. There exists at most two types of transversal pairs $(S,\cal F)$ with $S$ smooth.

1)The first type is when $\cal F$ is a pencil of lines and $S$ is a sphere of degree $1$.

2) The second type is when $S$ is a torus in $\mathbb CP^2$ of degree $0$. Let us first prove this claim, first giving a remark. In particular $S$ can not be holomorphic.

Remark. Is $S$ is transversal to $\cal F$, it is orientable. Moreover we can choose its orientation in such a way, that it intersects the leafs of $\cal F$ positively.

Proof of Claim. Let $\cal F$ be a singular holomorphic foliation and let $S$ be a surface transversal to $\cal F$. Let $L$ be the line sub-bundle of the restriction $T\mathbb CP^2$ to $S$ such that $L$ is tangent to $\cal F$ along $S$. Let us choose an orientation on $S$ so that the homology class of $S$ in $H_2(\mathbb CP^2)$ is $d\ge 0$. Note $S$ is orientable, because it is transversal to $\cal F$). Then since $\cal F$ is transversal to $S$, over $S$ $T\mathbb CP^2|_S$ splits into the sum $L\oplus TS$, so we have

$$3d=c_1(T\mathbb CP^2|_S)=c_1(L\oplus TS)=c_1(L)+2-2g=dc_1({\cal F})+2-2g\;\;\;\;(1)$$

$$d(3-c_1({\cal F}))=2(1-g).$$

Since $c_1({\cal F})\le 1$ we conclude that only two possibilities can occur

1) $d=1$, $g=0$, and $c_1({\cal F})=1$.

2) $d=0$ and $g=1$. $\square$

Now, I don't know if examples of second type exist, when $S$ is a torus and $deg(S)=0$. It is hard to believe that they exist. But at least let me give some examples of foliations of degree $0$ that don't have a transversal surface.

Example. Let $\cal F$ be a foliation on $\mathbb CP^2$ tangent to the orbits of action of some $\mathbb C^*$ action on $\mathbb CP^2$. This foliation has degree $0\ne 1$, so by the above claim the only surface transversal to $\cal F$ can be a degree $0$ torus. I claim that such tori don't exist.

Indeed, suppose by contradiction that $S$ is such a torus. Let's take a point $x\in S$. Then there is a $\mathbb C^*$-orbit $O$ passing through $x$. The closure of $O$ is a complex curve $\overline O$ in $\mathbb CP^2$. By the Remark in the beginning, there is an orientation of $S$ such that all points of intersection of $S$ with $\overline O$ are positive, i.e. $S\cdot \overline O\ge 1$. This contradicts to the fact that $\deg(S)=0$. $\square$.

  • $\begingroup$ Thank you very much for your answer. I try to understand its details. $\endgroup$ – Ali Taghavi May 5 at 18:27
  • $\begingroup$ Sure, Ali, please ask if you want to clarify something $\endgroup$ – Dmitri Panov May 6 at 16:26

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