Timeline for hilbert quot stacks vs schemes
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Mar 8, 2016 at 23:29 | comment | added | usr0192 | What threw me off was (when I quickly scanned it) Hall-Rydh said their results were new even for X, S schemes. But reading it more carefully, I guess what they are saying is new is that if X \to S is not separated, their result is that the HIlbert Stack is algebraic, but we know aprori (in the case when X, S are schemes) it can't have nontrival automorphism groups, so their Hilbert Stack is an algebraic space. | |
| Mar 8, 2016 at 23:29 | comment | added | usr0192 | @JasonStarr Yes, those functors. I mean, what confusing me was a paper by Hall and Rydh where they talk about the Hilbert stack HIlb_{X/S}, where $X, S$ are stacks - but there I guess since stacks form a 2-category, morphisms (such as closed immersions in the case of HIlbert schemes) can have automorphisms, so that's why they need to consider the Hilbert stack. | |
| Mar 8, 2016 at 18:44 | comment | added | Jason Starr | Are you asking about the Hilbert and Quot functors from my article with Olsson? Those functors manifestly take values in sets, rather than in groupoids. So there is no question that, when they happen to be representable, they are representable by algebraic spaces rather than general algebraic stacks. | |
| Mar 8, 2016 at 16:48 | comment | added | Sasha | Because they classify not just objects, but quotient objects. The morphism from $O_X$ (in case of Hilbert schemes) does the rigidification. | |
| Mar 8, 2016 at 16:46 | comment | added | usr0192 | @sasha yes but I'm asking why the don't have automorphisms? | |
| Mar 8, 2016 at 16:41 | comment | added | Sasha | The main reason is that the objects they classify do not have automorphisms. | |
| Mar 8, 2016 at 15:54 | history | asked | usr0192 | CC BY-SA 3.0 |