Timeline for Reference for de Rham cohomology in positive characteristic
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Nov 2, 2015 at 21:51 | comment | added | David E Speyer | Those Deligne notes are great, thanks! I learned a bunch anyway. I have often thought that the absence of algebraic de Rham from standard AG courses is a real shame; one consequence of this is that there aren't a lot of good references. But, of course, it is quite possible that someone will come along tomorrow and give us one. | |
| Nov 2, 2015 at 20:34 | comment | added | R. van Dobben de Bruyn | Thanks. I am aware of this approach. I have added some explanation to the original question to clarify what it is that I am looking for. Sorry if it was a bit unclear what I am looking for. | |
| Nov 2, 2015 at 14:07 | comment | added | David E Speyer | Serre duality shows $1/3$ of the terms are dual, induction shows another $1/3$ of the terms are dual. If I could build compatible pairings between the sequences, I could deduce by formal linear algebra that the last $1/3$ are dual. But I couldn't figure out how to build the pairings without looking inside the definition of hypercohomology, at which point the above route became faster. If I really cared about the proper case, I might go back and try harder. | |
| Nov 2, 2015 at 14:05 | comment | added | David E Speyer | I'll also mention that I had a much more conceptual way that I wanted to do it. Let $\Omega^{\leq p}$ be the de Rham complex truncated to the first $p$ terms, and define $\Omega^{\geq p}$ likewise. I wanted to show, by induction on $p$, that $\mathbb{H}^q(\Omega^{\leq p})$ and $\mathbb{H}^{n-q}(\Omega^{\geq n-p})$ are dual. The idea would be to use the short exact sequences of complexes $0 \to \Omega^p[p] \to \Omega^{\leq p} \to \Omega^{\leq p-1} \to 0$ and $0 \to \Omega^{\geq n-p+1}[1] \to \Omega^{\geq n-p} \to \Omega^{n-p} \to 0$. This gives two long exact sequences in hypercohomology ... | |
| Nov 2, 2015 at 14:01 | comment | added | David E Speyer | I was using gemetrically integral, although that should be removable: Properly fromulated, the statement plays well with change of base field and with disjint union. Not so sure about the proper case. I wanted to have $\int$ defined on all of $A^{nn}$ in order to have the pairings defined on the whole double complex. My guess is yes, but I don't really know. | |
| Nov 2, 2015 at 8:37 | comment | added | R. van Dobben de Bruyn | Thanks; this is great! I will take some time to parse all that is written. In the meantime, I have two questions. Are you assuming that $X$ is geometrically integral (to get $H^{nn} = k$, and not some finite extension)? (I am perfectly happy to stick to this assumption.) More importantly, do you think this could be generalised to the proper case? In that case we don't have access to a cover by $n+1$ opens (I think), so something else is needed. | |
| Nov 2, 2015 at 4:24 | comment | added | pro | and thanks to permalinks we now also have a reference | |
| Nov 2, 2015 at 4:03 | history | answered | David E Speyer | CC BY-SA 3.0 |