# Segre embedding and intersections by hyperplanes

Consider the Segre embedding $$\mathbb{P}^2 \times \mathbb{P}^2 \to \mathbb{P}^8.$$ Denote by $$V$$ the image of the Segre embedding and by $$B$$ the locus of triples $$(H_1, H_2, H_3)$$ with $$H_i \in H^0(\mathcal{O}_{\mathbb{P}^8}(1))$$ a hyperplane section such that $$V \cap H_1 \cap H_2 \cap H_3$$ is a proper intersection i.e., a curve. Clearly, $$B$$ is an open subset in $$H^0(\mathcal{O}_{\mathbb{P}^8}(1))^{\oplus 3}$$. Denote by $$B' \subset B$$ the subloci consisting of triples $$(H_1, H_2, H_3)$$ such that the intersection $$V \cap H_1 \cap H_2 \cap H_3$$ is a singular curve. My question is: Is $$B'$$ necessarily of codimension $$1$$ in $$B$$ or can it be of higher codimension in $$B$$? Any hint/reference will be most welcome.

• I guess, $B$ lies in the direct sum of three copies, not in the tensor product. Commented May 17, 2023 at 12:21
• @Sasha Yes, sorry, I have made the correction. Commented May 17, 2023 at 12:22

Let $$W = H^0(\mathcal{O}_{\mathbb{P}^8}(1))$$. Consider the variety $$X$$ of tuples $$(P,H_1,H_2,H_3) \in V \times W^{\oplus 3}$$ such that $$V \cap H_1 \cap H_2 \cap H_3$$ is singular at $$P$$. Then it is easy to see that the projection $$X \to V, \qquad (P,H_1,H_2,H_3) \mapsto P$$ is flat and its fibers have dimension $$22$$ (indeed, there are 4 parameters for a 2-dimensional subspace in the tangent space $$T_PV$$ and $$3 \cdot 6 = 18$$ parameters for the $$H_i$$ passing through $$P$$ and containing this tangent space), hence $$\dim(X) = 4 + 22 = 26$$. It is also easy to see that $$X$$ is irreducible.
On the other hand, the general fiber of the map $$X \to W^{\oplus 3}, \qquad (P,H_1,H_2,H_3) \mapsto (H_1,H_2,H_3)$$ over its image is finite (to see this it is enough to find just one triple $$(H_1,H_2,H_3)$$ such that $$V \cap H_1 \cap H_2 \cap H_3$$ has a finite number of singular points), hence the dimension of the image of $$X$$ is 26, hence it is a divisor in $$W^{\oplus 3}$$.
• Thanks. I had two questions. 1) I understand that $T_pV$ is $4$-dimensional. How are there "$4$ parameters for a 2-dimensional subspace in the tangent space"? 2) Why do we not consider the case of 3-dimensional subspace in the tangent space? Commented May 17, 2023 at 13:25
• 1) $\dim(\mathrm{Gr}(2,4)) = 4$. 2) If a 3-dimensional subspace is contained, then a 2-dimensional subspace is contained as well, so this already is taken into account. Commented May 17, 2023 at 13:46