Timeline for twisted Poisson structures, degenerate metrics and integrability properties of (2,0)-tensors
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 30, 2012 at 19:44 | vote | accept | issoroloap | ||
| Dec 30, 2012 at 19:45 | |||||
| Dec 30, 2012 at 15:37 | comment | added | issoroloap | I was also wandering why you needed symmetry. I'll wait for your upgrde then! Thank you! | |
| Dec 30, 2012 at 15:25 | comment | added | Robert Bryant | @issoroloap: You're welcome. Actually, I realized that my construction of an operator doesn't need symmetry or skew-symmetry either, and it doesn't require choosing any metric $g$, so I'll add that to my answer above. It gives an alternative to Peter's construction. Of course, his obstruction tensor and mine are `equivalent' in some sense, but the transformation between them will need to use the extra choices that Peter makes in his construction. | |
| Dec 30, 2012 at 14:50 | comment | added | issoroloap | Thank you Robert, thank you Peter. I also started with Peter's idea precisely from that paper and got stuck with the fact that in the symmetric case one always gets zero from the bracket. However the second proposal with the two curvatures for the two distributions seems really conclusive. In the general case I was interested in perturbing a genuine Poisson structure with a symmetric tensor such that the integrability of the annihilator of the kernel is preserved. That would be a antisymmetric-symmetric version of bihamiltonian systems with hamiltonians given by the perturbed casimirs. | |
| Dec 29, 2012 at 20:59 | history | edited | Peter Michor | CC BY-SA 3.0 |
added 1116 characters in body
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| Dec 29, 2012 at 19:06 | comment | added | Peter Michor | @Robert. Oh yes, in the skew case the grading is just what one needs, My mistake! Thanks. | |
| Dec 29, 2012 at 18:34 | comment | added | Robert Bryant | @Peter: I thought about this, but the problem is that one only has the cometric $g\in C^\infty(S^2(TM))$ as data. Of course, one can regard $g$ as a function on $T^\ast M$ that is quadratic on the fibers, but the Poisson bracket of $g$ with itself will vanish no matter what $g$ is. Moreover, because there are no first-order invariants if $g$ has full rank (by the Fundamental Lemma of Riemannian geometry), it's hard to see how one can avoid a construction of a first-order invariant that could detect the desired integrability without making an assumption of constant rank. | |
| Dec 29, 2012 at 17:34 | history | answered | Peter Michor | CC BY-SA 3.0 |