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Let $X$ be a smooth complex projective variety of dimension $d$. Then the Hilbert scheme of $n$ points $X^{[n]}$ is not irreducible in general, but it has always the main component $X^{[n]}_{sm}$ of smoothable subschemes: these are the finite subschemes $\xi\subseteq X$ of length $n$ that can be deformed to a collection of $n$ distinct points on $X$. Equivalently, $X^{[n]}_{curv}$ is the closure in $X^{[n]}$ of the set of $n$ distinct points in $X$.The main component $X^{[n]}_{sm}$ is irreducible and of dimension $d\cdot n$.

My question is about an open subset of this component: the subset of curvilinear subschemes. A finite subscheme $\xi \subseteq X$ is called curvilinear if it is contained in a smooth curve $C\subseteq X$ or, equivalently, if for each point $P\in \xi$ the tangent space has dimension at most one: $\dim_{\mathbb{C}} T_p\xi \leq 1$.

The curvilinear locus $X^{[n]}_{curv} \subseteq X^{[n]}$ is an open subset, and moreover it is contained in the main component $X^{[n]}_{sm}$.

Question: is it true that the curvilinear locus is a large open subset in the main component? Meaning that the complement $X^{[n]}_{sm} - X^{[n]}_{curv}$ has codimension at least two in $X^{[n]}_{sm}$.

For example, this is true if $X$ is either a curve or a surface, and it should also be true in arbitrary dimension if the number of points $n$ is at most $3$. I imagine that this has been already studied before, and I would be perfectly happy with a reference.

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  • $\begingroup$ Once you get to the range where $X^{[n]}_{\text{sm}}$ is not a connected component of $X^{[n]}$, I believe that $X^{[n]}_{\text{sm}}$ must have codimension $1$ intersection with the union of the remaining irreducible components by Hartshorne's Connectedness Theorem . . . $\endgroup$
    – Jason Starr
    Commented Feb 26, 2018 at 18:21
  • $\begingroup$ Thank you, that would answer the question. But, how would you apply Hartshorne's Connectedness Theorem to prove that? Also, I guess you mean the range where $X^{[n]}_{sm}$ is not an irreducible component of $X^{[n]}$, since $X^{[n]}$ is always connected. $\endgroup$
    – Daniele A
    Commented Feb 27, 2018 at 2:23
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    $\begingroup$ I mean the range where $X^{[n]}_{\text{sm}}$ is not a connected component, i.e., the range where there exists at least one additional irreducible component of $X^{[n]}$ that intersects $X^{[n]}_{\text{sm}}$. Hartshorne's Connectedness Theorem says that a Cohen-Macaulay scheme is connected in codimension $1$. If the intersection of $X^{[n]}_{\text{sm}}$ with the remaining irreducible components is codimension $2$ in $X^{[n]}_{\text{sm}}$, then removing that codimension $2$ intersection disconnects $X^{[n]}$. $\endgroup$
    – Jason Starr
    Commented Feb 27, 2018 at 2:50
  • $\begingroup$ Ah, thanks! I see what you mean by $X^{[n]}_{sm}$ being a connected component. Also, my comment of before makes no sense, since $X^{[n]}_{sm}$ is always an irreducible component. What I probably wanted to write is that, since $X^{[n]}$ is connected, then $X^{[n]}_{sm}$ is not a connected component if and only if $X^{[n]}$ is not irreducible. $\endgroup$
    – Daniele A
    Commented Feb 27, 2018 at 3:25
  • $\begingroup$ However, about Hartshorne's Connectedness Theorem: do we know that $X^{[n]}$ is Cohen-Macaulay? I had the impression that it can have very bad singularities. For example K. Lee proves in "The singularities of the principal components of the Hilbert scheme of points" that $X^{[n]}_{sm}$ is not Cohen-Macaulay in general. $\endgroup$
    – Daniele A
    Commented Feb 27, 2018 at 3:30

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