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Mar 2, 2021 at 2:04 history edited Carlos Esparza CC BY-SA 4.0
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May 14, 2020 at 15:51 vote accept Carlos Esparza
Nov 12, 2019 at 22:36 answer added Igor Khavkine timeline score: 7
Nov 11, 2019 at 23:31 comment added Carlos Esparza @Qfwfq Well, Peetre's theorem says a lot more - namely that every differential operator locally is a polynomial with $C^\infty$ coefficients in the partial derivatives. Q1 is whether that is true in the holomorphic category. For example on $\mathbb{C}$ this would mean that locally $D = f_0 + f_1 \partial + f_2 \partial^2 + \cdots + f_n \partial^n$.
Nov 11, 2019 at 16:09 comment added Qfwfq Ok, I see. - Assume we're on a space $(X,O_X)$ locally ringed in algebras over a char zero field $K$, and $E$ and $F$ are two (finite rank) locally free sheaves on $X$. We have: 1) the set $Hom_K (E,F)$ of $K$-linear maps of sheaves of $K$-vector spaces. 2) the set $Diff_K(E,F)$ of the $D\in Hom_K(E,F)$ locally satisfying the recursive $[D,M_f]\in Diff^{n-1}$ type relation. So, if I've understood well, smooth Peetre's theorem says the sets 1) and 2) are equal in this case? and your Q.1 was whether that was also true in the holomorphic category?
Nov 11, 2019 at 15:36 comment added Carlos Esparza @Qfwfq on a (paracompact) manifold that's equivalent to saying that $D$ is a morphism if sheaves. But that's just a particularity if the real case, in general you should define differential operators as morphisms of sheaves (e.g. the support of an nonzero holomorphic function on a connected domain is the entire domain)
Nov 11, 2019 at 15:32 comment added Qfwfq According to Wikipedia, Peetre's theorem needs a support-decreasing hypothesis (in the real differentiable case).
Nov 11, 2019 at 13:50 history edited Carlos Esparza CC BY-SA 4.0
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Nov 11, 2019 at 9:57 answer added Michael Bächtold timeline score: 1
S Nov 9, 2019 at 20:03 history suggested Amir Sagiv CC BY-SA 4.0
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Nov 9, 2019 at 19:34 comment added Carlos Esparza Éléments de Géometrie Algébrique by Grothendieck
Nov 9, 2019 at 19:32 comment added Amir Sagiv Welcome to Math Overflow. What is EGA?
Nov 9, 2019 at 19:32 review Suggested edits
S Nov 9, 2019 at 20:03
Nov 9, 2019 at 18:39 history edited Carlos Esparza CC BY-SA 4.0
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Nov 9, 2019 at 18:33 history edited Carlos Esparza CC BY-SA 4.0
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Nov 9, 2019 at 18:30 review First posts
Nov 9, 2019 at 19:32
Nov 9, 2019 at 18:25 history asked Carlos Esparza CC BY-SA 4.0