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May 12, 2021 at 11:05 comment added Z. M I did not claim that I see the precise link. I would like to replace the Deligne-Beilinson by de Rham because it is simpler (which could be understood as an "enriched" version of de Rham). What it seems to me is that, by taking filtered colimit of "models" (I don't know whatever it is), the result for de Rham cohomology might be very similar to Hodge-completed derived de Rham cohomology, and in that sense, Beilinson's isomorphism seems related.
May 11, 2021 at 19:06 comment added Mikhail Bondarko Why?:) Do you have any references or explanations for this association?
May 10, 2021 at 14:19 comment added Z. M What if you replace Deligne-Beilinson cohomology by de Rham cohomology? Apparently, this looks a bit like Beilinson's isomorphism $B_{\operatorname{dR}}^+\simeq\widehat{\operatorname{dR}}_{\overline{\mathbb Q_p}/\mathbb Q_p}$.
May 10, 2021 at 11:32 history edited Mikhail Bondarko CC BY-SA 4.0
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Jul 28, 2014 at 17:09 history edited Mikhail Bondarko CC BY-SA 3.0
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Jul 27, 2014 at 7:39 history asked Mikhail Bondarko CC BY-SA 3.0