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Let $Y$ be a projective scheme. The naive definition of a Hilbert scheme of subschemes $X$ of $Y$ would require us to projectively embed $Y$, then ask that $X$ have a fixed Hilbert polynomial $p$.

However, this space is usually disconnected. Cheap example: $Y$ is two points, $p=1$, and the Hilbert scheme is $Y$ itself. More expensive example: $Y = {\mathbb P}^1 \times {\mathbb P}^1$ in the Segre embedding, $p(d) = d+1$. Then $X$ is a line from one of the two rulings, so this naive Hilbert scheme is a disjoint union of two $\mathbb P^1$s.

If $F\to X$ is a flat family of subschemes of $Y$, i.e. $F \subseteq Y\times X$, then the class in $K_\bullet(Y)$ of a fiber -is- isn't locally constant on $X$. (EDIT: Oops! Jason Starr suggests a related invariant that is, in a comment.) So the proper definition of "the Hilbert scheme of subschemes of $Y$" should include "with fixed $K$-class $p \in K_\bullet(Y)$". (If $Y$ is projective space, the $K$-class is exactly the same data as the Hilbert polynomial.) This finer invariant suffices to break apart the disconnected spaces in the examples above.

I presume that, in general, fixing the $K$-class is not enough to get a connected Hilbert scheme. (Certainly Hartshorne's proof of connectivity in the case $Y={\mathbb P}^n$ does not go through, as it uses a big group action.)

Is there an easy example of a projective scheme $Y$ and two subschemes $X_1,X_2$ defining the same class in $K$-homology of $Y$, but there is no flat connected family of subschemes of $Y$ including $X_1,X_2$ as fibers?

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    $\begingroup$ What do you mean by $K_*(Y)$, and by the class of $Y$ in $K_*(Y)$? It is topological K-theory? For algebraic K-theory, it is certainly false that the class of the structure sheaf of a fiber is locally constant. $\endgroup$
    – Angelo
    Nov 29, 2012 at 16:07
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    $\begingroup$ @Angelo: Allen can answer for himself, but I bet he means the image of the map $K_\ast(X) \to \text{Hom}_{\mathbb{Z}}(K^\ast(X),\mathbb{Z})$ induced by the pairing $K^\ast(X) \times K_\ast(X) \to \mathbb{Z}$ sending $([\mathcal{E}],[\mathcal{F}])$ to $\chi(X,\mathcal{E}\otimes_{\mathcal{O}_X}\mathcal{F})$. This is the sort of thing that Martin Olsson and I used in our work to try to pin down the numerical invariants of a quasi-compact, open and closed piece of the Hilbert scheme. $\endgroup$
    – Jason Starr
    Nov 29, 2012 at 18:23
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    $\begingroup$ No, I was just wrong, but I like Jason's fix. $\endgroup$
    – Allen Knutson
    Nov 30, 2012 at 2:03

1 Answer 1

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Take a generic quintic in $\mathbb CP^4$ and consider lines on it

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  • $\begingroup$ In this example I assumed that the question was about topological $K$-theory. $\endgroup$
    – aglearner
    Nov 29, 2012 at 16:22
  • $\begingroup$ This would certainly be an interesting answer. Is the claim that it has lines, they're rigid, but they're all in the same K-class? $\endgroup$
    – Allen Knutson
    Nov 30, 2012 at 2:20
  • $\begingroup$ @Allen, I have to look on the fix of Jason, in fact I can only claim the obvious: that on the quintic all lines give the same integral homology classes since $H_2$ of the quintic is $\mathbb Z$. Since I don't understand yet the definition of Jason, I can not say anything else. But I would imagine that all "discreet" invariants of these lines would coincide, but any "finer" might not (and probably will not). $\endgroup$
    – aglearner
    Nov 30, 2012 at 8:42
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    $\begingroup$ @aglearner: It is indeed true that the structure sheaves of any two lines on a quintic threefold give the same image in $\text{Hom}_{\mathbb{Z}}(K^\ast(X),\mathbb{Z})$. For $\mathcal{F} = \mathcal{O}_L$, the structure sheaf of a line $L$ in $X$, for every locally free sheaf $\mathcal{E}$ on $X$, $\chi(X,\mathcal{E}\otimes_{\mathcal{O}_X}\mathcal{F})$ equals $\chi(L,\mathcal{E}|_L)$. By Riemann-Roch (or the classification of locally free sheaves on $\mathbb{P}^1$), this is just $\text{deg}(\mathcal{E}) + \text{rank}(\mathcal{E})$, independent of $L$. $\endgroup$
    – Jason Starr
    Nov 30, 2012 at 13:17

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