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Suppose $n$ quadric hypersurfaces cut out $2^n$ distinct points $p_1,\ldots,p_{2^n}$ in $\mathbb{P}^n$. What is the maximal number of points $p_i$ a quadric can contain without containing all of them?

For n=2, there is of course a quadric going through any three points and avoiding the last point. In $\mathbb{P}^3$, it is easy to exhibit a quadric going through 6 points (and avoiding the last two). This is however not possible with 7 points (as can be seen using a projection to $\mathbb{P}^2$).

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up vote 4 down vote accepted

To complement the answers of Sasha and Damiano I would like to give one (completley non-generic) example of a very specific configuration when a quadric contains exactly $3\cdot 2^{n-2}$ points.

Take in $\mathbb C^n$ the collection of $2^n$ points $(\pm 1,..., \pm 1)$. It is clear that this collection is the intersection of n quadrics. Indeed, we can obtain these points by taking intersection of $n$ degenerate quadrics $Q_i:=(x_i+1)(x_i-1)=0$, then we perturb a bit each quadric by adding a little multiple of $(\sum_i x_i^2-n)$. This way we get $n$ smooth quadrics that intersect transversally in the above $2^n$ points. Finally take one more singular quadric $(x_1-1)(x_2-1)=0$ it contains exactly $3\cdot 2^{n-2}$ points.

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This and damiano's answer finishes the problem (modulo the proof off the EGH-conjecture). Thanks! – J.C. Ottem Dec 29 '10 at 20:20

I do not know how much progress has been made on this, but what you ask is part of a conjecture appearing in a paper of Eisenbud, Green and Harris (see Cayley-Bacharach Theorems and Conjectures, Conjecture CB10). The authors make the following conjecture.

Conjecture CB10. (Eisenbud, Green, Harris) Let $\Gamma$ be a complete intersection of $n$ quadrics in $\mathbb{P}^n$. If $X \subset \mathbb{P}^n$ is any hypersurface of degree $k$ containing a subscheme $\Gamma_0 \subset \Gamma$ of degree strictly greater than $2^n − 2^{n−k}$, then $X$ contains $\Gamma$.

If $X$ is a quadric, this says that if $X$ contains more than $3 \cdot 2^{n-2}$ of the points, then it must contain them all.

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Thanks Damiano. I googled around and found a reference to a paper by Lazarsfeld, containing a proof of this conjecture. Still I wonder how sharp this bound is, e.g., can a quadric contain all the $3\cdot 2^{n-2}$ points? – J.C. Ottem Dec 29 '10 at 17:28

I think the answer is the stupid estimate $(n^2+n)/2$. On one hand, for any $(n^2 + n)/2$ points there is at least $n+1$ quadrics passing through it, just by counting the parameters. On the other hand, if you pick up $c_n := (n^2+n)/2 + 1$ points in generic position there are precisely $n$ quadrics through them. Now let $H_n$ be the irreducible component of the Hilbert scheme $(P^n)^{[c_n]}$ containing the subschemes consisting of $c_n$ distinct points. We have a rational map $H_n \to Gr(n,S^2V^*)$, $Z \mapsto H^0(P^n,I_Z(2))$ (here $V$ is the vector space of simension $n+1$ such that $P^n = P(V)$). It is clearly finite over generic point. Since $H_n$ is irreducible it follows that for generic $n$-tuple of quadrics any $c_n$ points in there intersection are generic on $H_n$, hence there are only $n$ quadrics passing through them.

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But for n=4, it is possible to find a quadric containing 12 of the points, not only 10. – J.C. Ottem Dec 28 '10 at 23:33
Why? Can you explain? – Sasha Dec 29 '10 at 4:10
@Sasha: I haven't checked the numbers you give, but I think that the issue is that the points in a complete intersection of quadrics are not general. You are starting with $2^n$ special points and then want to see how many conditions subsets of them impose on quadrics. Presumably, this means that no subset of $(n^2+n)/2$ points in a complete intersection of quadrics is general. – damiano Dec 29 '10 at 9:38
@Damiano: Certainly I thought about this, but it seems that the argument I gave shows that although the $2^n$ points are special, the subsets of $c_n$ points are general. – Sasha Dec 29 '10 at 10:43

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