# Exact sequence on $\mathbb{P}^3$ obtained from Euler sequence

In the article of title: On the singular scheme of codimension one holomorphic foliations in $$\mathbb{P}^3$$ the author states that the sequence on $$\mathbb{P}^3$$ $$0\longrightarrow \mathcal{F}\oplus\mathscr{O}_{\mathbb{P}^3}\longrightarrow\mathscr{O}_{\mathbb{P}^3}^{\oplus4}(1)\longrightarrow\mathcal{I}_Z(d+2)\longrightarrow0$$is exact, where $$\mathcal{F}$$ is a reflexive sheaf of rank two.

The author states that the sequence above is obtained of the two exact sequences below: $$0\longrightarrow\mathscr{O}_{\mathbb{P}^3}\longrightarrow\mathscr{O}_{\mathbb{P}^3}^{\oplus4}(1)\longrightarrow\mathcal{T}_{\mathbb{P}^3}\longrightarrow0$$ and $$0\longrightarrow\mathcal{F}\longrightarrow\mathcal{T}_{\mathbb{P}^3}\longrightarrow\mathcal{I}_Z(d+2)\longrightarrow0$$where $$\mathcal{I}_{Z}$$ is the ideal sheaf of the singular scheme $$Z$$ of foliation.

How to get the first exact sequence from the last two?

Suggestions will be welcome.

• It seems to me that there are is a some very relevant information in the paper that you're not sharing. Isn't the proof you refer to assuming $\mathcal{F}$ is split (i.e. a sum of line bundles)? – user347489 Aug 16 '19 at 1:40
• @user347489. Yes !! I read the result in the article. Good article .$\mathcal{F}$ is split. – Allan Ramos Aug 16 '19 at 1:56
• The fact that $\mathcal{F}$ is split helps in annulling cohomology $H^{1}(\mathbb{P}^{3}, \mathcal{F}(q) \oplus \mathcal{O}_{\mathbb{P}^{3}}(q))$. – Allan Ramos Aug 16 '19 at 2:03

First note that the sheaf $$\mathcal{F}$$ is the kernel of a map $$\omega \colon \mathcal{T}_{\mathbb{P}^3}\rightarrow \mathscr{O}_{\mathbb{P}^3}(d+2)$$ defined by a degree $$d+1$$ homogeneous $$1$$-form $$\omega = F_0dx_0 +F_1dx_1 +F_2dx_2+F_3dx_3$$, such that $$\sum x_iF_i = 0$$. Clearly the image is the ideal sheaf of the singular scheme twisted by $$\mathscr{O}_{\mathbb{P}^3}(d+2)$$.

From the Euler sequence we see that every (local) vector field is given by the class four (local) sections $$(p_0,p_1,p_2,p_3)$$ of $$\mathscr{O}_{\mathbb{P}^3}(1)$$ modulo the relation $$(p_0,p_1,p_2,p_3)\sim (q_0,q_1,q_2,q_3) \Leftrightarrow (p_0-q_0,p_1-q_1,p_2-q_2,p_3-q_3) = \lambda\cdot (x_0,x_1,x_2,x_3)$$ for some (local) holomorphic function $$\lambda$$.

Then lift the map given by $$\omega$$ to the representatives of such vector fields. This gives $$\omega \colon \mathscr{O}_{\mathbb{P}^3}(1)^{\oplus4}\rightarrow \mathscr{O}_{\mathbb{P}^3}(d+2)$$ which is just $$(p_0,p_1,p_2,p_3)\mapsto F_0p_0 +F_1p_1+F_2p_2+F_3p_3$$ . Denote the kernel of this map by $$\mathcal{F}_0$$. Then $$0 \rightarrow \mathcal{F}_0 \rightarrow\mathscr{O}_{\mathbb{P}^3}(1)^{\oplus4}\rightarrow \mathscr{I}_{Z}(d+2) \rightarrow 0$$

We have that $$\mathcal{F}_0$$ is spanned by the representatives of the elements of $$\mathcal{F}$$ and we have an exact sequence $$0 \rightarrow \mathscr{O}_{\mathbb{P}^3}\rightarrow \mathcal{F}_0 \rightarrow \mathcal{F} \rightarrow 0$$ where the copy of $$\mathscr{O}_{\mathbb{P}^3}$$ inside $$\mathcal{F}_0$$ is spanned by $$(x_0,x_1,x_2,x_3)$$ which projects to the zero vector field.

Therefore, to prove that $$\mathcal{F}_0 = \mathcal{F}\oplus\mathscr{O}_{\mathbb{P}^3}$$ one must prove that this sequence splits.

In the case of the theorem 3.5 of the paper, $$\mathcal{F}$$ is locally free. Hence a sufficient condition is $$H^1(\mathcal{F}^\vee) = \{0\}$$. This condition holds if $$\mathcal{F}$$ is split.

If the ideal $$(F_0,F_1,F_2,F_3)$$ is saturated, we have from the sequence $$0 \rightarrow \mathcal{F}_0 \rightarrow\mathscr{O}_{\mathbb{P}^3}(1)^{\oplus4}\rightarrow \mathscr{I}_{Z}(d+2) \rightarrow 0$$ that $$H^1(\mathcal{F}_0(k))=0$$, for every $$k$$. Using that $$\mathcal{F}^\vee = \det(\mathcal{F}^\vee)\otimes \mathcal{F}$$ and the sequence $$0 \rightarrow \mathscr{O}_{\mathbb{P}^3}\rightarrow \mathcal{F}_0 \rightarrow \mathcal{F} \rightarrow 0$$ twisted by $$\det(\mathcal{F}^\vee) = \mathscr{O}_{\mathbb{P}^3}(d-2)$$, we see that $$H^1(\mathcal{F}^\vee) = \{0\}$$ also holds.

• +1. Just note that the existence of the sequence $0\to O\to F_0\to F\to 0$ doesn't depend in the geometry. It actually holds for any two exact sequences that relate in the way the two above do. In this sense the real content is the splitting of this sequence. – user347489 Aug 17 '19 at 3:48
• @user347489 For sure! I thought that there was something natural about the sequences involved to give the answer. Then I realized that that kernel does not split naturally. Fortunately, it is true under the hypotheses of the theorem in question. – Alan Muniz Aug 17 '19 at 12:08