Let $H_1$ and $H_2$ be two planes in $\mathbb{P}^3$.Let $P$ be a set of $9$ points such that no three lie on a line. Suppose $H_1$ contains 4 of them and $H_2$ contains remaining $5$ points. Is it true that $P$ imposes independent conditions on quadrics ?
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1$\begingroup$ No, because the points lie on the quadric $Q = H_1 \cap H_2$. $\endgroup$– SashaCommented Feb 15, 2020 at 18:31
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$\begingroup$ Thanks for the answer. But if $Q= H_1\cap H_2$ is the only quadric , then it imposes 9 independent conditions. Thus it does not give any contradiction. $\endgroup$– user130022Commented Feb 15, 2020 at 19:20
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$\begingroup$ Sorry, I misunderstood the question. Indeed, if you take 5 and 4 points in general position on the planes they impose independent conditions on quadrics. $\endgroup$– SashaCommented Feb 16, 2020 at 8:21
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$\begingroup$ Thank you very much for the reply. Could you please give a sketch of the proof or at least a reference? If no three points lie in a line, then i hope they are in general position in plane. $\endgroup$– user130022Commented Feb 16, 2020 at 8:57
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1 Answer
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Let $C'' \subset H_2$ be the unique conic on $H_2$ containing the 5 points and let $$ \{C'_t\}_{t \in \mathbb{P}^1} \subset H_1 $$ be the pencil of conics containing the 4 points. Let $$ L = H_1 \cap H_2. $$ The points impose independent conditions on quadrics if and only if there is no $t$ such that $$ C'_t \cap L = C'' \cap L.\tag{*} $$ Indeed, any quadric passing through these points (except for $H_1 \cap H_2$) intersects the planes along conics $C'_t \cup C''$ for some $t$, and it intersects the line $L$ along a 2-point scheme (jr contains $L$), so $(*)$ follows.
The converse follows from the exact sequences $$ 0 \to \mathcal{O}_{\mathbb{P}^3} \to \mathcal{O}_{\mathbb{P}^3}(2) \to \mathcal{O}_{H_1 \cup H_2}(2) \to 0 $$ and $$ 0 \to \mathcal{O}_{H_1 \cup H_2}(2) \to \mathcal{O}_{H_1}(2) \oplus \mathcal{O}_{H_2}(2) \to \mathcal{O}_{L}(2) \to 0. $$
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$\begingroup$ Thank you very much for the answer. I have a silly question: why there is no $t$ such that (*) hols ? $\endgroup$ Commented Feb 16, 2020 at 10:26
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$\begingroup$ Assume for simplicity that the pencil has no conics containing $L$. Then $t \mapsto C'_t \cap L$ defines a map $\mathbb{P}^1 \to S^2L = \mathbb{P}^2$ (its image is, in fact, a line). On the other hand $C'' \cap L$ defines a point of $S^2L$. So, the extra generality assumption that you need is that the above line in $S^2L$ does not contain the above point. $\endgroup$– SashaCommented Feb 16, 2020 at 10:32
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$\begingroup$ Now i have understood. Thank you very much. $\endgroup$ Commented Feb 16, 2020 at 10:37
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$\begingroup$ Can we impose that generality assumption from the generality assumption on points in the plain ? sorry for bothering you again $\endgroup$ Commented Feb 16, 2020 at 10:41
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$\begingroup$ Sorry I got the answer. Probably we can move the point in the 2nd plane or in 1st plane to get that generality hypothesis. thanks. $\endgroup$ Commented Feb 16, 2020 at 10:43