# A Singular Foliation of $\mathbb{C}P^2$ which does not admit a global transverse submanifold

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)?

## 1 Answer

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

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