Timeline for Arbitrarily non-degenerate Hodge to de Rham spectral sequence
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 1, 2017 at 2:24 | comment | added | R. van Dobben de Bruyn | Footnote: the reason that my statement about the Hodge-de Rham spectral sequence of $X \times Y$ is not a complete formality is that the differentials of the de Rham complex are not $\mathcal O_X$-linear. | |
| Jan 20, 2017 at 18:53 | comment | added | R. van Dobben de Bruyn | To prove the first statement about $E_r^{pq}(X \times Y)$, use the Čech + de Rham double complex to compute the spectral sequences on $X$ and $Y$. Then prove that their tensor product (suitably totalised to make it a double complex again) computes the spectral sequence on $X \times Y$. Finally, over a field, the spectral sequence associated to the (totalised) tensor product of two double complexes has $E_r(C\otimes D) = E_r(C)\otimes E_r(D)$ on each page. This is proved by setting up a map of spectral sequences from right to left, which is an isomorphism by Künneth + induction. | |
| Jan 20, 2017 at 18:48 | comment | added | R. van Dobben de Bruyn | @WillSawin: Nice idea! Unfortunately, is doesn't work. Indeed, the spectral sequence of $X\times Y$ is given by $E_r^{pq}(X \times Y) = \bigoplus_{i,j} E_r^{ij}(X) \otimes E_r^{p-i,q-j}(Y)$, with the obvious differentials (for a general treatise of sign conventions, see SGA 4$_3$, Exp. XVII, 1.1.4). Thus, if the spectral sequences of $X$ and $Y$ degenerate on the $E_r$ page, then so does that of $X \times Y$. | |
| Jan 15, 2017 at 3:31 | comment | added | Will Sawin | Could there exist a variety $X$ where $X^n$ has a nonzero differential on the $n$th page for all $n$? | |
| Jan 15, 2017 at 1:12 | comment | added | R. van Dobben de Bruyn | Trivial remark: a $d$-dimensional variety does not have any differentials past the $d$-th page. Thus, such examples should have really large dimension. (This for example shows that we cannot hope to mimic Mumford's example of a non-closed $1$-form: his example is on a surface and uses resolution of singularities.) | |
| Jan 14, 2017 at 23:09 | comment | added | Piotr Achinger | I was wondering about that too! I think this is not known, but I might be wrong. | |
| Jan 14, 2017 at 21:33 | history | asked | SashaP | CC BY-SA 3.0 |