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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)?

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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 $\deg(S)=0$. $\square$.

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  • $\begingroup$ 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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