7
$\begingroup$

It is true that for any $n$ there exists a compact complex manifold which Frolicher spectral sequence does not degenerate at the $n$-th page(https://arxiv.org/pdf/0709.0481.pdf).

Does the analogous statement hold for smooth proper varieties in positive characteristics? That is, for a field $k$ of characteristics $p>0$ is it true that for any $n\geq 1$ there exists a smooth proper variety $X$ such that its Hodge to de Rham spectral sequence has a non-zero differential on the $n$-th page?

It is certainly true for $n=1$.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ I was wondering about that too! I think this is not known, but I might be wrong. $\endgroup$
    – Piotr Achinger
    Jan 14, 2017 at 23:09
  • $\begingroup$ 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.) $\endgroup$
    – R. van Dobben de Bruyn
    Jan 15, 2017 at 1:12
  • 1
    $\begingroup$ Could there exist a variety $X$ where $X^n$ has a nonzero differential on the $n$th page for all $n$? $\endgroup$
    – Will Sawin
    Jan 15, 2017 at 3:31
  • $\begingroup$ @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 $X \times Y$. $\endgroup$
    – R. van Dobben de Bruyn
    Jan 20, 2017 at 18:48
  • $\begingroup$ 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. $\endgroup$
    – R. van Dobben de Bruyn
    Jan 20, 2017 at 18:53

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.