Timeline for Families of ideal sheaves: What's the correct definition?
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12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 22, 2012 at 7:26 | comment | added | Sasha | The argument in Lemma 5.3, Prop. 5.4 and Thm. 5.5 with $Y$ replaced by $X$ and $L$ eplaced by $O_X(1)$ proves that $M_I$ is a scheme. | |
| Sep 22, 2012 at 4:55 | comment | added | 36min | Maybe you misunderstood my question. If $M_I(X)$ is a scheme, then $P-Hilb$ is a projective scheme. I've no problem with that. But why is $M_I(X)$ a scheme? If it is merely a functor then the fiber product doesn't necessarily exists as a scheme. (In the paper it is said to be a scheme.) | |
| Sep 21, 2012 at 8:00 | comment | added | Sasha | $P$-$Hilb(X)$ is a fiber product of $Hilb$ and $M_{PI}$ ove $M_I$. Therefore it is a closed subscheme in the product $Hilb\times M_{PI}$. Both $Hilb$ and $M_{PI}$ are projective. hence so is $P$-$Hilb$. | |
| Sep 20, 2012 at 18:39 | comment | added | 36min | Now I think I'm fine with your definition of $M_I(X)$, but why is that a projective scheme? This is really needed since we want to show that $P-Hilb(X)$ is a projective scheme and it turns out to be a fiber product of the Hilbert scheme and $M_{PI}(X)$ over $M_I(X)$. | |
| Sep 20, 2012 at 6:30 | comment | added | 36min | @MartinG What's this codimension 2 result? Any 2 ideals that are abstractly isomorphic are set-theoretically the same ideal? | |
| Sep 19, 2012 at 9:19 | comment | added | Sasha | @MartinG: Codimension 2 is automatic as soon as numeric class of the sheaf is fixed. Also, see my comment to your answer --- the way you suggest to define $M_I$ is not quite good. | |
| Sep 19, 2012 at 8:34 | comment | added | MartinG | @Sasha: This viewpoint makes sense, except that for the embedding of $I$ into $\mathcal{O}_X$ to be unique, you need to assume codimension at least 2. This is what 36min complained about with maximal ideals in $\mathbb{Z}$. So we have to restrict to codimension at least 2 to get a bijection. (What I wrote in my answer, then, is basically that this $M_I(X)$ can be defined without reference to ideals, and that the morphism from the Hilbert scheme is in fact an isomorphism, although that is not entirely obvious and probably independent from what Bridgeland is doing, as you say.) | |
| Sep 19, 2012 at 7:53 | comment | added | Sasha | What is the problem with functoriality? Once again, $M_I(X)(S)$ is the equivalence classes of objects $F$ on $S\times X$ such that for each point $s\in S$ the derived pullback $j_s^*F$ is isomorphic to an ideal sheaf, ok? If $u:T \to S$ is a morphism and $u^*F$ is the pullback then for any $t\in T$ you have $j_t^*u^*F \cong j_{u(t)}^*F$ is isomorphic to an ideal sheaf. | |
| Sep 19, 2012 at 6:32 | comment | added | 36min | You always need remember the embedding. What you said is sometimes we can just remember the image. That's what you mean by up to a constant. And I don't think my question is answered. E.g. What is $M_I(X)$? What's the functoriality? | |
| Sep 19, 2012 at 6:26 | comment | added | Sasha | These are just different questions. Sometimes you need to remember the embedding, sometimes you don't. | |
| Sep 19, 2012 at 6:18 | comment | added | 36min | Well, he said if the derived restrictions are sheaves, then you do get a flat family of sheaves before definition 3.7. How on earth can one forget the inclusion in the definition of an ideal sheaf? Any maximal ideal of $\mathbb{Z}$ are abstractly isomorphic as $\mathbb{Z}$-modules, so it is essential to remember the embedding. It is not just a subtle thing. | |
| Sep 19, 2012 at 6:09 | history | answered | Sasha | CC BY-SA 3.0 |