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Let $Q_i(i=1,2,3)$ be quadric hypersurface in $\mathbb{P}^4$. Consider a net of quadrics $\Lambda=(Q_1,Q_2,Q_3)$.

I can't understand some part of proof of Corollary 2.8(p.11) in Stability of genus 5 canonical curves.

Corollary 2.8 If a net of quadrics in $\mathbb{P}^4$ is semistable, then the corresponding intersection is connected and purely 1-dimensional.

My question is:

(1) They said "Fulton-Hansen connectedness theorem gives the first statement." But to use the theorem, I need irreducibility of quadrics. How can I use the theorem? Can I choose a basis $\{Q_1,Q_2,Q_3\}$ of the net such that each $Q_i$ is irreducible?

(2) Assume that we can choose a basis $\{Q_1,Q_2,Q_3\}$ of the net, such that $S:=Q_1\cap Q_2$ is a quartic surface and there is an irreducible component $S'$ of $S$ which is contained in $Q_3$. If $deg(S')=2$, then why the span of $S'$ is a hyperplane?

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  • $\begingroup$ what is the corresponding net of a rectangular prism $\endgroup$
    – user66811
    Commented Feb 6, 2015 at 19:03

2 Answers 2

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(1) Yes.

Consider a net of reducible quadrics (here the dimension of the projective space is not relevant): since all the associate quadratic forms have rank at most 2, and since they vary linearly, then there is a vector subspace of codimension 2 contained in the kernel of all the quadratic forms in the net. Then, in suitable coordinates, it is the net $ \langle x_0^2, x_0x_1, x_1^2 \rangle$ which is not semistable.

(2) Every nondegenerate (=not contained in any hyperplane) variety in ${\mathbb P}^r$ has degree strictly bigger than its codimension (so a nondegenerate surface in ${\mathbb P}^4$ has degree strictly at least 2+1=3).

This can be easily proved by induction on the codimension: it is trivial in codimension 1, and the induction step can be easily performed by projecting the variety from a smooth point, reducing both the codimension and the degree by 1.

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  • $\begingroup$ Thank. In (2), I know that S′ is contained in a hyperplane. What's the meaning of "span"? Why a pencil of quadrics in the net contains this hyperplane? $\endgroup$
    – Sasga
    Commented Mar 14, 2014 at 11:50
  • $\begingroup$ "Span" means "the smallest linear space containing it". As an irreducible quadric can't be contained in a plane, its span must be that hyperplane, that has dimension 3. About your second question: let $H$ be this hyperplane, then $S'$ is contained in both $Q_1 \cap H$ and $Q_2 \cap H$. Since $S'$ is defined by a single quadratic relation in $H$, $Q_1$ and $Q_2$ are proportional modulo the equation of $H$, so a suitable combination of them vanishes along $H$. $\endgroup$
    – Roberto Pignatelli
    Commented Mar 14, 2014 at 16:26
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An ample divisor on a variety of dimension $\ge 2$ is always connected, see Cor. III.7.9 in Hartshorne. For the second see Ex. I.7.8.

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  • $\begingroup$ Do you mean each $Q_i$ is connected? What about the intersection? And in (2), I know $S'$ is contained in a hyperplane ($S'\subset Q_3\cap$ a hyperplane). What's the meaning of "span"? $\endgroup$
    – Sasga
    Commented Mar 13, 2014 at 15:13
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    $\begingroup$ Apply this 3 times. $\endgroup$
    – Sasha
    Commented Mar 13, 2014 at 15:42
  • $\begingroup$ They may not meet transversally. I can't apply that. $\endgroup$
    – Sasga
    Commented Mar 14, 2014 at 11:55
  • $\begingroup$ It does not matter how they meet. Still each time it is an ample divisor. $\endgroup$
    – Sasha
    Commented Mar 14, 2014 at 12:01

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