Timeline for Is there a Nash-type theorem for symplectic manifolds?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 26, 2016 at 13:03 | comment | added | Danny Ruberman | If $M^k$ is closed, then the integral of $(f^*\omega)^k$ is zero. But $(g\alpha)^k$ is a volume form, so its integral is positive. So I don't think that version works either. | |
| Dec 26, 2016 at 12:52 | comment | added | Ali Taghavi | @DannyRuberman What about the following weaker version: For every symplectic manifold $(M, \alpha)$ there is an embedding $f: M \to \mathbb{R}^{2n}, \omega=\sum dx_{i} \wedge dy_{i}$ such that $f*{w}= g\alpha$ for a positive smooth fumction $g$? | |
| Feb 1, 2016 at 9:17 | vote | accept | Alex M. | ||
| Feb 1, 2016 at 2:59 | comment | added | user21574 | see also e-collection.library.ethz.ch/eserv/eth:24332/eth-24332-02.pdf | |
| Feb 1, 2016 at 2:56 | comment | added | user21574 | If $M$ be a Kahler manifold with $c_1(M)>0$, then you can embed $M$ in $\mathbb CP^N$ | |
| Jan 31, 2016 at 22:28 | comment | added | Danny Ruberman | Well, it depends on what you're trying to accomplish by this embedding. I believe that one can approximate an arbitrary form by one with rational periods (integrals over a basis of homology classes) and then take a multiple of the form to make it integral. | |
| Jan 31, 2016 at 21:11 | comment | added | Alex M. | Apparently, for those embedding results one needs $\omega$ to be integral. I cannot visualize this, so I cannot tell how restrictive this is. | |
| Jan 31, 2016 at 20:42 | history | answered | Danny Ruberman | CC BY-SA 3.0 |