Timeline for Why is there no symplectic version of spectral geometry?
Current License: CC BY-SA 3.0
4 events
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| Sep 9, 2018 at 20:45 | comment | added | Qfwfq | @B K: reading the answer by @Dan Fox, it would seem that a metaplectic structure is of "topological" type rather than properly "geometrical". | |
| Oct 19, 2017 at 17:12 | comment | added | B K | Thank you for the reference. While it looks very interesting, one needs to assume that the symplectic manifold admits a metaplectic structure in order to study symplectic spinors. This is not exactly in the spirit of my question because assuming an extra structure besides the symplectic form can of course lead the question ad absurdum: For example, assume that M is Kaehler. Then it is both symplectic and Riemannian and one could argue that there is indeed a spectral geometry of symplectic manifolds, as long as they are Kaehler :-) | |
| Oct 19, 2017 at 16:42 | comment | added | Vít Tuček | I was about to suggest this. Even if there is no analogue for the Laplace operator, there is a symplectic version of the Dirac operator. But beware! The symplectic spinor representation is infinite-dimensional! | |
| Oct 19, 2017 at 15:43 | history | answered | Igor Rivin | CC BY-SA 3.0 |