Timeline for How can I see the "missing" Poisson center when the rank of the Poisson structure drops?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 21, 2011 at 3:53 | vote | accept | Theo Johnson-Freyd | ||
| Oct 21, 2011 at 3:53 | history | bounty ended | Theo Johnson-Freyd | ||
| Oct 16, 2011 at 9:39 | comment | added | Nicola Ciccoli | A number of side remarks which may convince that the Poisson module over distrbutions can be a good choice: 1. Smooth functions embed into distributions, via this embedding Casimirs go into distributional Casimirs. 2. We have the support of a distribution, and may therefore define the concept of support for the Casimirs submodule, a generalization of the zero set of usual Casimirs. Ddistributions detect the support and the order of singularity. 3. It is quite easy to show that in the extreme easy cases you get exactly what you would expect (0-Poisson, symplectic Poisson, regular Poisson). | |
| Oct 15, 2011 at 18:41 | comment | added | Theo Johnson-Freyd | Great. So my comment to my question was of course nonsense — $H^0$ is precisely the space of Casimirs. (Somehow I had gotten myself confused, and thought I had a homology theory, so that $H_0$ was "all functions modulo the vanishing ideal of the Poisson bivector". I do get something like this if I move to $H^1$.) If I think the answer is to use distributions, then I can formalize them as a Poisson module. Indeed, it is the module dual to the module of smooth functions. (Of course, "module" really means "sheaf of modules over the corresponding algebroid.) | |
| Oct 15, 2011 at 18:36 | comment | added | Theo Johnson-Freyd |
I have made no edits except to fix some TeX. As always, there is a problem that Markdown sometimes strips backslashes off of special characters. I have fixed this by adding <p> tags around the problematic paragraph — Markdown is instructed to do no processing to things inside such tags.
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| Oct 15, 2011 at 18:30 | history | edited | Theo Johnson-Freyd | CC BY-SA 3.0 |
TeX fix
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| Oct 15, 2011 at 9:09 | comment | added | Nicola Ciccoli | One point about which I was not clear. Casimirs are indeed $0$-Poisson cohomology (and Casimir distributions are $0$-cohomology with coefficients in the Poisson module of distributions). But to "see the singular leaves" one may need to go to $1$-Poisson cohomology. | |
| Oct 15, 2011 at 8:47 | comment | added | Mariano Suárez-Álvarez | Ask on tea.mathoverflow.net or flag your answer for moderator attention. | |
| Oct 15, 2011 at 8:42 | comment | added | Nicola Ciccoli | I wonder whether there is a way to merge this user which I wrongly created for this answer, with the user writing this comment (i.e. merging me into me... ) | |
| Oct 15, 2011 at 8:25 | history | answered | Nicola Ciccoli | CC BY-SA 3.0 |