Timeline for Geometric interpretation of filtered rings and modules
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 11, 2021 at 22:16 | comment | added | Z. M | I heard that it is due to Simpson and Deligne that filtered objects are simple objects over the stack $\mathbb A^1/\mathbb G_m$, which corresponds precisely to graded objects over $\mathbb Z[t]$ where $\deg t=1$ in this answer. A modern treatment could be found in Moulinos' paper arxiv.org/abs/1907.13562 | |
| May 16, 2019 at 21:29 | comment | added | Tim Campion | This question and answer is an old favorite of mine. Maybe it's worth adding that if $\mathcal R$ is the Rees algebra of a filtered ring $R$. then a filtered module over $R$ is a $\mathbb G_m$-equivariant quasicoherent sheaf over $Spec \mathcal R$, perhaps with some torsionfreeness condition. | |
| May 27, 2010 at 9:54 | comment | added | Leonid Positselski | No, one does not need $F_m=A$ for $m$ nonnegative at all. My answer simply presumes the notation in which $F_{i-1} A$ is contained in $F_i A$, and of course 1 is in $F_0 A$, so 1 is in $F_1 A$, to. E.g., the filtration can be increasing, with $F_iA=0$ for $i<0$, or it can be decreasing, with $F_iA=A$ for $i\geq0$, or it can extend nontrivially in both directions. All these cases are covered. | |
| May 27, 2010 at 0:12 | comment | added | Allen Knutson | In order for 1 to be in F_1 A, we need all F_m to equal R for m nonnegative, and for the OP's A_i to be Leonid's F_{-i}. Then the fiber over 0 is indeed gr A, supported in negative degrees. Anyway, to answer the question "What is the filtered ring A?" the Rees construction answers "A deformation of the graded ring gr A" / "Something that degenerates to the graded ring gr A". | |
| May 25, 2010 at 17:03 | vote | accept | Jan Weidner | ||
| May 25, 2010 at 15:26 | comment | added | Jan Weidner | Thanks, this is exactly the kind of answer I was looking for! | |
| May 25, 2010 at 15:11 | history | edited | Leonid Positselski | CC BY-SA 2.5 |
The condition on R (that it has to be torsion-free over C[t]) added.
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| May 25, 2010 at 14:39 | history | answered | Leonid Positselski | CC BY-SA 2.5 |