Timeline for Is the quantum algebra unique (up to isomorphism) in deformation quantization ?
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6 events
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| Jan 28, 2013 at 16:04 | comment | added | Adrien | @Damien> That's what I meant, your point 2) is definitely true at the universal level, but that being true or not for actual Poisson structures is open and probably a hard problem, isn'it ? | |
| Jan 28, 2013 at 15:17 | comment | added | DamienC | @Adrien. I think that point 2 is not the content of Dolgushev's paper (though related to it). But your last sentence is correct (and THIS is the main point of 2.). @Alexander 1. I was just saying that if you make the choice of a local universal formality morphism (i.e. given by weights associated to graphs) then the globalization is essentially unique. @Alexander 2. this is not what I am saying BUT in the context of the class of a star-product, the Poisson structures you are looking at are of the form $\hbar\pi+\cdots$. wiht a given fixed $\pi$. If $\pi$ is ND then they are all equivalent. | |
| Jan 28, 2013 at 12:44 | comment | added | Adrien | Damien> Isn't your point 2 the content of Dolgushev's paper arXiv:1109.6031, at least at the universal level ? It states that the action of grt on homotopy classes of stable formality isomorphisms is free and transitive (hence highly non trivial). However, as noticed in arXiv:1211.4230 (and if I understand correctly, which is far from being granted) there is no known example of an actual Poisson structure for which the realization of this action is non trivial. | |
| Jan 28, 2013 at 10:38 | comment | added | Alexander Chervov | Consider R^2n. It seems what you write imply that GRT will act trivially if Poisson structure is symplectic, and may be non-trivivially if Poisson structure is non symplectic... Do I understand correctly ? It would be very unexpected for me... I am not great expert, of course, but still... | |
| Jan 28, 2013 at 10:35 | comment | added | Alexander Chervov | Damien, thank you very much for your answer. But I am not clear about some points. " concerning the symplectic case, the main reason why the classification map doesn't depend on any choice is because the quantization is unique. Let me explain further: the choices involved in the formality isomorphism appear in the local case. " Do you mean globalization kills additional choices ? Probably no. | |
| Jan 28, 2013 at 8:38 | history | answered | DamienC | CC BY-SA 3.0 |