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Let $X \subset \mathbb{P}^m$ be a smooth subvariety (e.g. a surface or some hypersurface) and $X^{[n]}$ be its Hilbert scheme of $n$ points. Which elements of $X^{[n]}$ determine an $(n - 1)$-dimensional linear subspace of $\mathbb{P}^m$? Is the collection of such elements locally closed/constructible? If not, is there a more "reasonable" way to think about this? I'm mostly interested in the case $n = 3$.

This may seem like a strange question since we can answer the question of when $n$ points in $\mathbb{P}^m$ determine an $(n - 1)$-dimensional linear subspace of $\mathbb{P}^m$ just using some linear algebra. The motivation for this question is that these "points" come from intersections of a variety with a linear subspace and can come with some kind of multiplicity. The simplest example is a point with a direction in $X^{[2]}$ for a curve $X$ which comes from the tangent line to a point in the plane $\mathbb{P}^2$ (keep track of point and direction of tangent line).

Two simple cases/examples with $X = \mathbb{A}^2$:

$n = 2$: We either have two distinct points or a point with a given direction. In each case, the point of $X^{[2]}$ determines a line.

$n = 3$: The situation here is a bit more complicated if we want a plane.

  1. If the support is on three distinct points, it is enough to say that they are not collinear.

  2. If the support is at two distinct points with one point and different point with a direction attached to it, it is enough to say that this direction is not the same as the line connecting these two points in order to determine a plane.

  3. There is the case with support at one point. After analyzing the ideal, it turns out that there is a point with two different tangent directions attached to it or a point with a "triple nilpotent" (including both first and second order). The first part gives a plane and I don't think the second part would work.

Even with being able to do casework "geometrically", it's not clear to me how the cases for $n = 3$ above translate into algebraic conditions. Is there an $X$ (e.g. some surface) where such an analysis is feasible?

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    $\begingroup$ This should be an open condition, by the semi continuity theorem for coherent sheaf Cohomology, or more easily because the map from the universal family of $\operatorname{Hilb}^n$s of the universal family of $n-1$-dimensional linear sub spaces. $\endgroup$
    – Will Sawin
    Commented Nov 5, 2019 at 8:23

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For $Z\in X^{[n]}$, let $\mathcal{I}_Z$ be the ideal sheaf of $Z$ in $\mathbb{P}^m$; you are asking about the locus where $h^0(\mathcal{I}_Z(1))$ attains its minimal dimension $m-n+1$. Now $\mathcal{I}_Z$ is the restriction to $\{Z\} \times \mathbb{P}^m$ of a universal ideal sheaf $\mathcal{I}$ on $X^{[n]}\times \mathbb{P}^m$, flat over $X^{[n]}$; by the semicontinuity theorem (Hartshorne, III, Theorem 12.8), this locus is open.

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    $\begingroup$ Just to clarify, does this mean that specifying some arbitrary dimension other than $m - n + 1$ gives a locally closed condition? $\endgroup$
    – modnar
    Commented Nov 5, 2019 at 15:21
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    $\begingroup$ Yes. More precisely perhaps, imposing any upper bound on that dimension gives an open condition. $\endgroup$
    – abx
    Commented Nov 5, 2019 at 16:54
  • $\begingroup$ If we want $Z \in X^{[n]}$ to determine some fixed linear subspace $\Lambda$ of dimension $n - 1$, is it enough to repeat the steps above with $X \cap \Lambda$ replacing $X$ to find the subset of elements of $X^{[n]}$ determining $\Lambda$? $\endgroup$
    – modnar
    Commented Nov 5, 2019 at 21:09
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    $\begingroup$ Yes, but you don't even need to repeat: just take the intersection of $(X\cap\Lambda)^{[n]}\subset X^{[n]}$ with the previous open subset. $\endgroup$
    – abx
    Commented Nov 6, 2019 at 5:24

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