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Jun 19, 2017 at 23:12 answer added Avi Steiner timeline score: 3
Jun 16, 2017 at 9:57 comment added C. Dubussy @Avi Steiner : Thanks for your comment. I was aware of these equivalences but the definition with filtrations was the first to appear in Kashiwara's papers about the Riemann-Hilbert problem. And he said that this definition "clearly allows to recover the usual definition for differential equations". So perhaps it is the "best" definition to make the link with the old one.
Jun 15, 2017 at 18:49 comment added Avi Steiner You might want to look into the other equivalent conditions. E.g.: 1) Every pullback to a smooth curve has all r.h. cohomology; 2) the analytic and formal power series solution complexes agree.
Jun 15, 2017 at 13:59 history edited C. Dubussy CC BY-SA 3.0
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Jun 15, 2017 at 13:57 comment added C. Dubussy You're right, I have been too quick, my filtration is not a good candidate. I will edit the post with your suggestion. The main problem for me is to prove the equality $f\cdot \text{gr}(M)=0$.
Jun 15, 2017 at 13:54 comment added Simon Wadsley Also note that your statement says that there exists a good filtration with some property not that every good filtration has that property.
Jun 15, 2017 at 13:54 comment added Simon Wadsley I think you are mistaken that your filtration is good. You want something like $M_n=D_n.(1+D_\mathbb{C}P)$ where $D_n$ denotes the differential operators of order at most $n$. You will only get $M_n=M$ for some $n$ if $M$ is an integrable connection ie a f.g. $O_X$-module.
Jun 15, 2017 at 13:19 history asked C. Dubussy CC BY-SA 3.0