# Universal hyperplane section and nondegeneracy of general hyperplane section

I have a question about Exercise 18.11 In Harris' book Algebraic Geometry, on page 231:

Give a proof of the nondegeneracy of the general hyperplane section of an projective irreducible nondegenerated variety $$X \subset \mathbb{P}^n$$ of degree $$\ge 2$$ (...that's just to avoid the case $$X$$ linear; see also EDIT below) over $$k= \mathbb{C}$$ without invoking the notion of degree (or Bezout's theorem), as follows. First, show that if the general hyperplane section of $$X$$ spans a $$k$$-plane, we have a rational map

$$\varphi: (\mathbb{P}^{n})^* \dashrightarrow \mathbb{G}(k,n)$$

defined by sending a general hyperplane $$H \in (\mathbb{P}^{n})^*$$ to the span of $$H \cap X$$. Next, use the fact that the universal hyperplane section $$\Omega_X$$ (for definition see p 43 or below) of $$X$$ is irreducible to deduce that for any $$H \in (\mathbb{P}^{n})^*$$ and any point $$\Lambda \in \varphi(H)= \Gamma_{\varphi} \cap (\{H\} \times \mathbb{G}(k,n))$$ (that is, any point in the image of the fiber of the graph $$\Gamma_{\varphi}$$, over $$H$$), the hyperplane section $$H \cap X$$ lies on the $$k$$-plane $$\Lambda$$. It follows that if the general hyperplane section of $$X$$ is degenerate, then all are - but any $$n$$ independent points of $$X$$ will span a hyperplane $$H$$ with $$X \cap H$$nondegenerate, contradiction.

Remark: The universal hyperplane section $$\Omega_X \subset X \times (\mathbb{P}^{n})^*$$ is defined as subvariety $$\{(p, H) \ \vert \ p \in H \cap X \}$$ und is irreducible (Theorem 5.8, page 53).

Question: I not understand how the irreducibility of $$\Omega_X$$ can be related to this problem, ie to use it here to conclude that $$H \cap X \subset \Lambda$$ for every $$\Lambda \in \varphi(H)$$.

Some thoughts: Let $$U \subset (\mathbb{P}^{n})^*$$ the open dense subset where $$\varphi$$ is regular. Since $$\Gamma_{\varphi}$$ is closure of the graph $$\Gamma_U:=\{(H, \varphi(H) \ \vert \ H \in U \}$$, it follows that $$\Gamma_{\varphi}$$ is irreducible and it suffice to find a closed subset of $$\Gamma_{\varphi}$$ which contains $$\Gamma_U$$.
A natural choice seems to take the closed subset $$A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$$ and intersect it with $$\Gamma_{\varphi}$$. By contruction it contains $$\Gamma_U$$.

But this approach nowhere makes use of the universal hyperplane section $$\Omega_X$$. So my question is not how to prove the claim somehow, but how to argue as Harris suggested using explicitly the irreducibility of the universal hyperplane section.

EDIT 1: as Libli noticed the exercise is flawed if we not exclude the case that $$X$$ is linear, let's do it and add the additional assumption that the degree of $$X$$ is $$\ge 2$$.

EDIT 2: actually I noticed that it could happen that that the set $$A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$$ might be not always closed inside $$(\mathbb{P}^{n})^* \times \mathbb{G}(k,n)$$; so my approach is flawed, but the concern of the question stays the same, how is it possible to deduce the exercise's Claim exploiting irreducibility of $$\Omega_X$$?

My guess is that presumably the argument I'm seeking for might work like that there exist certain dense subset $$B \subset \Omega_X$$ carrying informations about $$\Gamma_{\varphi}$$ and showing equality $$B = \Omega_X$$ would imply consequently the claim that $$H \cap X \subset \Lambda$$ for all $$\Lambda \in \varphi(H)$$. And the task becomes to show that this $$B$$ is closed...

Here is also the most naive candidate: set $$B:= \{(p,H) \in \Omega_X \ \vert \ \forall \Lambda \in \varphi(H): p \in \Lambda \}$$. As I said, that's a very naive guess :) But it is of course dense in $$\Omega_X$$ and the big question becomes if this subset is closed?

What I can show that the set $$\{(p,H) \in \Omega_X \ \vert \ \exists \Lambda \in \varphi(H): p \in \Lambda \}$$ is closed as image under proper projection map $$p_{12}: X \times (\mathbb{P}^{n})^* \times \mathbb{G}(k,n) \to X \times (\mathbb{P}^{n})^*$$ of closed subset

$$p_{12}^{-1}(\Omega_X) \cap p_{13}^{-1}(\{ (p,\Lambda) \ \vert \ p \in \Lambda \}) \cap p_{23}^{-1}(\Gamma_{\varphi})$$

but the conclusion that this set indeed equals $$\Omega_X$$ implies just that for every $$H$$ there exist some $$\Lambda \in \varphi(H)$$ with $$H \cap X \subset \Lambda$$, but that's not enough.

The result of this exercice is correct (and well-known). I am however not sure that the intermediate result used in the exercice is true. Take $$X \subset \mathbb{P}^2$$ be a line. Then for generic $$H \in \left(\mathbb{P}^2\right)^*$$, the intersection $$X \cap H$$ is degenerate (it is a point in a $$\mathbb{P}^1$$). But for $$H = X$$, the intersection $$H \cap X$$ is not degenerate in $$H$$. I guess the problem comes from the definition of $$\{(H,\Lambda), \ H \cap X \subset \Lambda \}$$ as a closed subset. You can always define it as the closure of some set, but then, at the boundary, things will get messy. In the example above, then all couples $$(X,x)$$ for $$x \in X$$ will be included in the closure, but of course, $$X \cap X \not\subset x$$.
I believe a better approach to prove the result of the exercice is the following. Let $$J_X$$ be the ideal sheaf of $$X$$ in $$\mathbb{P}^n$$. Then $$h^0(\mathbb{P}^n, J_X(1))$$ is the number of independant points in $$\left(\mathbb{P}^n\right)^*$$ which contains $$X$$. If $$X \subset \mathbb{P}^n$$ is not degenerate, then for all $$H \in \left(\mathbb{P}^n\right)^*$$ the intersection $$X \cap H$$ is proper (that is has the right codimension) and so we have $$J_{X \cap H} = J_X \otimes \mathcal{O}_H$$, where $$J_{X \cap H}$$ is the ideal sheaf of $$X \cap H$$ in $$H$$.
Assume now that for generic $$H \in \left(\mathbb{P}^n \right)^*$$, the section $$X \cap H$$ is degenerate, that is $$h^0(H, J_X \otimes \mathcal{O}_H(1)) > 0$$. By the semi-continuity Theorem for $$h^0$$, we get that for all $$H \in \left(\mathbb{P}^n \right)^*, \ h^0(H, J_X \otimes \mathcal{O}_H(1)) > 0$$, that is for all $$H \in \left(\mathbb{P}^n \right)^*$$, the section $$X \cap H$$ is degenerate. And now you can conclude as in the exercice.
• you are right, the statement of exercise is wrong if $X$ has degree $1$ , ie is linear subspace. But if we add the assumpion that $X$ has degree $\ge2$, could we argue exploiting irreducibility of $\Omega_X$ as Harris suggested? Feb 27 at 12:19
• I understand, even if $X$ is NOT linear, there could occure troubles with the set $A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$. Actually I not pretty sure that it is aways closed in $(\mathbb{P}^{n})^* \times \mathbb{G}(k,n)$ Feb 27 at 12:36
• @user7391733 : I am not sure the irreducibility assumption is helpful in any way. The result is still true when $X$ is reducible. This result is more of a cohomological nature than a topological one, so I guess one should not focus too much on the topological assumptions. Feb 27 at 14:51