Timeline for Differential operators between modules, $\mathcal{D}_A(M, M)$ necessarily a filtered, almost commutative ring?
Current License: CC BY-SA 3.0
5 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 7, 2015 at 10:42 | history | edited | Giovanni Moreno | CC BY-SA 3.0 |
added 145 characters in body
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| Jul 7, 2015 at 9:44 | comment | added | Michael Bächtold | Here's an explicit counterexample: $B=C^\infty(\mathbb{R})$, $M=B\oplus B$ and consider the first order operators $D_1: (f,g)\mapsto (\partial f,g)$ and $D_2: (f,g)\mapsto (\partial g, \partial f)$. Now $[D_1,D_2]$ is of order 2 and not 1, hence $\mathcal{D}(M,M)$ is not almost commutative. So finitely generated an projective is not enough. Maybe invertible $M$ will do. | |
| Jul 7, 2015 at 8:14 | comment | added | Michael Bächtold | Wouldn't almost commutative require that any two zero order operators $D_1,D_2:M\to M$ commute? | |
| Jul 7, 2015 at 7:54 | history | answered | Giovanni Moreno | CC BY-SA 3.0 |