3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ I guess, $B$ lies in the direct sum of three copies, not in the tensor product. $\endgroup$
    – Sasha
    Commented May 17, 2023 at 12:21
  • $\begingroup$ @Sasha Yes, sorry, I have made the correction. $\endgroup$
    – user45397
    Commented May 17, 2023 at 12:22

1 Answer 1

5
$\begingroup$

This is a standard projective duality argument.

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 $W^{\oplus 3}$.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – user45397
    Commented May 17, 2023 at 13:25
  • $\begingroup$ 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. $\endgroup$
    – Sasha
    Commented May 17, 2023 at 13:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .