# Building a hypersurface from hyperplane sections

Let $$P_0,\ldots,P_n$$ denote the coordinate hyperplanes in $$\mathbb P^n$$, and suppose that in each $$P_i$$ I have a degree $$d$$ hypersurface $$V_i$$. I am trying to understand what is the obstruction to the existence of a degree $$d$$ hypersurface $$V \subset \mathbb P^n$$ such that $$V \cap P_i = V_i$$.

There are some obvious obstructions: for example, my prescribed hyperplane sections $$V_i$$ must agree on intersections so that $$V_i \cap P_j = V_j \cap P_i$$. I don't think this alone is quite sufficient. In the case $$n=3$$ for example, let $$p = P_1 \cap P_2 \cap P_3$$ and suppose that $$V_1,V_2,V_3$$ all go through this point. Then $$T_p V_1 + T_p V_2 + T_p V_3$$ had better have dimension $$2$$, not $$3$$.

I am sure there is some exact sequence I'm missing - what is it?

(note: of course $$V$$ does not have to be unique, at least if $$d \geq n+1$$. You can add monomials containing all of the variables to the defining equation, and it doesn't change the intersection of the resulting surface with the coordinate hyperplanes)

The condition $$V_i \cap P_j = V_j \cap P_i$$ (if understood properly) is the only obstruction. To see this denote by $$Z$$ the union of $$P_i$$. Then we have an exact sequnce $$0 \to \mathcal{O}_Z \to \bigoplus \mathcal{O}_{P_i} \to \bigoplus \mathcal{O}_{P_i \cap P_j} \to \bigoplus \mathcal{O}_{P_i \cap P_j \cap P_k} \to \dots$$ (such an exact sequence exists for any union of transverse hypersurfaces and can be proved by induction on the number of components). Tensoring it by $$\mathcal{O}(d)$$ one deduces an isomorphism $$H^0(Z,\mathcal{O}_Z(d)) = \mathrm{Ker}\Big(\bigoplus H^0(P_i,\mathcal{O}_{P_i}(d)) \to \bigoplus H^0(P_i \cap P_j, \mathcal{O}_{P_i \cap P_j}(d)) \Big).$$ Thus, to give a section of $$\mathcal{O}_Z(d)$$ one needs to give a collection of sections of $$\mathcal{O}_{P_i}(d)$$ that agree on pairwise intersections (this is the right way to state the condition).
On the other hand, $$Z$$ is a hypersurface of degree $$n+1$$, hence there is an exact sequence $$0 \to \mathcal{O}(d-n-1) \to \mathcal{O}(d) \to \mathcal{O}_Z(d) \to 0$$ on $$\mathbb{P}^n$$, and since $$H^1(\mathbb{P}^n,\mathcal{O}(d-n-1)) = 0$$ (I assume that $$n \ge 2$$), it follows that the morphism $$H^0(\mathbb{P}^n, \mathcal{O}(d)) \to H^0(Z, \mathcal{O}_Z(d))$$ is surjective, hence any such section lifts to an equation of a hypersurface in $$\mathbb{P}^n$$.
• Thanks! This is a big help, but I'm still missing something. I really get the $V_i$ as elements of $\mathbb PH^0(P_i,\mathcal O(d))$, so I still need to be able to check whether they have lifts to $H^0(P_i,\mathcal O(d))$ that agree in the sense you describe. Is there any easy way to see this? – Mark Jan 4 '20 at 14:23
• @Mark: Consider the simplest example: $n = 2$, $d = 1$. You have a triangle of lines on $\mathbb{P}^2$. Imagine you have a point on each of the lines, away from their intersection points. This configuration of $V_i$ automatically satisfies your compatibility condition in the projectivized spaces of sections. But of course, not every such triple is collinear. – Sasha Jan 4 '20 at 21:17