Timeline for Elements of graded algebra associated with the algebra of differential operators as smooth sections
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Jul 1, 2018 at 5:28 | comment | added | Michael Bächtold | You’re welcome. Thanks for accepting my answer. | |
| Jun 30, 2018 at 23:55 | comment | added | truebaran | Many thanks for your answer: as you can see I accepted your answer as well awarded you with the bounty. I'm going to analyze all the details, thank you once again! | |
| Jun 30, 2018 at 21:27 | history | edited | Michael Bächtold | CC BY-SA 4.0 |
typo
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| Jun 30, 2018 at 21:25 | comment | added | Michael Bächtold | @truebaran I gave a few more details about the isomorphism with symmetric powers in the answer. Concerning your last question, maybe you're missing that the symmetric algebra over vector fields embeds canonically into $C^\infty (T^*M)$ as fiberwise polynomial functions on the cotangent. Hence a symbol in the sense of my answer becomes a section in your sense. | |
| Jun 30, 2018 at 21:05 | history | edited | Michael Bächtold | CC BY-SA 4.0 |
addresse question in comments.
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| Jun 30, 2018 at 16:20 | history | bounty ended | truebaran | ||
| Jun 30, 2018 at 16:20 | vote | accept | truebaran | ||
| Jun 29, 2018 at 19:53 | comment | added | truebaran | ...thing take a section $t \in \Gamma(M,E)$, produces a smooth function on which we act then by differential operator. This produces another smooth function and finally we want to compose with $s$ but does this composition make sens?). I would be also very happy if you could please explain me the following issue: each symbol may be viewed as a section acting $T^*M \to Hom(p^*E,p^*F)$ where $p$ is the projection or as a morphism of bundles $p^*E \to p^*F$. In both these descriptions the symmetric power does not appear. What am I missing? | |
| Jun 29, 2018 at 19:46 | comment | added | truebaran | Thank you for an explanation! It seems that once you have identified our module $\mathcal{S_k}$ as $\Gamma(M,U_k)$ for some bundle then we have a short exact sequence of the form $0 \to I_x \mathcal{S}_k \to \mathcal{S}_k \to (U_k)_x \to 0$. Therefore I'm more interested in the second identification of the bundle $U_k$ in terms of symmetric powers. I would be happy if you can also adress the following issues: how to define the inverse isomorphism (is this map which you described well defined? We compose $r$ with succesive vector fields but at the end we compose with $s$ thus the whole ...) | |
| Jun 29, 2018 at 9:08 | comment | added | Michael Bächtold | @truebaran I tried to simplify my reply to your comment. Have a look and let me know if you see something missing. | |
| Jun 29, 2018 at 9:07 | history | edited | Michael Bächtold | CC BY-SA 4.0 |
shortened the argument
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| Jun 26, 2018 at 8:13 | comment | added | Michael Bächtold | @truebaran I tried to address your comment in my edit, which I rewrote now. | |
| Jun 26, 2018 at 8:13 | history | edited | Michael Bächtold | CC BY-SA 4.0 |
tried to be more explicit and address directly the concern in the comment.
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| Jun 25, 2018 at 10:03 | history | edited | Michael Bächtold | CC BY-SA 4.0 |
expanded
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| Jun 1, 2018 at 20:24 | comment | added | truebaran | Still I don't see how to finish the argument. Serre-Swan'ts theorem gives you that $\mathcal{S}^k$ can be identified with $\Gamma^{\infty}(M,V^k)$ for some vector bundle $V^k$ (provided we can show that $\mathcal{S}^k$ is finitely generated and projective). On the other hand, here we would like to have a concrete description of $V^k$ as $U^k$ where the fiber over $x$ is $\mathcal{S}^k/ I_x \mathcal{S}$. As far as I know, somehow we have to use the fact that $C^{\infty}(M)$ is central in $\mathcal{S}$. | |
| Jun 1, 2018 at 9:34 | comment | added | Michael Bächtold | If i recall correctly, you can define a natural iso $\mathcal{S}^d\to S^d X\otimes \mathcal{S}^0$. The first factor of the tensor product is the symmetric power of derivations on $C^\infty(M)$. | |
| May 31, 2018 at 15:03 | comment | added | Michael Bächtold | I think you can also find a proof of that in the Nestruev book. | |
| May 31, 2018 at 14:49 | comment | added | truebaran | You are right about the finitness of dimension. However, I still don't see why $\mathcal{S}^k$ need to be projective and finitely generated. | |
| May 31, 2018 at 14:16 | history | answered | Michael Bächtold | CC BY-SA 4.0 |