Timeline for Is the quantum algebra unique (up to isomorphism) in deformation quantization ?
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Feb 1, 2013 at 18:35 | comment | added | Alexander Chervov | @Theo Okay, I am sorry seems I misunderstand your claim. | |
| Jan 30, 2013 at 5:38 | comment | added | Theo Johnson-Freyd | And I meant to say that my comments are largely beside the point. Because there are deep questions about "universal" quantization formulas, and these are the ones you now emphasize in your revised question. Unfortunately, off the top of my head I don't know precise statements about the relation between outputs of different universal quantization procedures. | |
| Jan 30, 2013 at 5:35 | comment | added | Theo Johnson-Freyd | But all I really meant to point out is that you should expect the space of deformations to be roughly the "size" of the space of $\hbar$-dependent Poisson structures (Kontsevich gives an isomorphism between these spaces, but naive "size" estimates are for free), whereas the usual word "Poisson algebra" does not allow $\hbar$-dependence, as $\hbar$ is not part of the classical algebra. So they really cannot be the same size, and you should expect many deformations for the same Poisson structure. | |
| Jan 30, 2013 at 5:32 | comment | added | Theo Johnson-Freyd | ... Sorry. I mean $A = $ functions in two variables $x_1,x_2$, and $a\star b = \exp( \hbar^2 \partial_{x_1} \partial_{y_2}) a(x_1,x_2) b(y_1,y_2) |_{y_i = x_i}$. In any case this certainly deserves to correspond to the $\hbar$-dependent Poisson structure that is the canonical one in two variables rescaled by $\hbar$. But making the correspondence precise is subtle, and one of the main results of Kontsevich is how to do so — indeed, the point is that the only way to make it precise requires choosing an associator (or similar gauge fixing). | |
| Jan 30, 2013 at 5:27 | comment | added | Theo Johnson-Freyd | @Alexander: The question is now clearer, thank you. But I don't understand what you mean in your comment. I think I understand Kontsevich's work on deformation quantization reasonably well. The classical definition of a deformation quantization of a Poisson algebra $(A,\pi)$ is an associative multiplication on $A[[\hbar]]$ that begins $a\star b = ab + \frac\hbar2 \pi(a,b) + O(\hbar^2)$. Thus in particular the deformation $a\star b = \exp( \hbar^2 \partial_x \partial_y) a(x) b(y) |_{x=y}$ of functions in two variables is a deformation corresponding to the trivial Poisson structure. ... | |
| Jan 28, 2013 at 8:38 | answer | added | DamienC | timeline score: 3 | |
| Jan 27, 2013 at 13:16 | comment | added | Alexander Chervov | ...continued ||| but it will give you Poisson bracket which start with h^n , but it will be non-trivial, and hence it should NOT be considered as "Poisson structure on A to be trivial,..." Is my point clear ? | |
| Jan 27, 2013 at 13:12 | comment | added | Alexander Chervov | @Theo I added some clarifications. Hope you agree question is well-posed now ? PS "many nontrivial deformations, which you can "turn on" with speed much slower than your deformation parameter." It seems to me many many people are making the same mistake: that Poisson bracket = first order of commutator /h . IT IS NOT correct way of thinking. Correct way is use Kontsevich or Fedosov (see SW's answer) and it gives for any star product the Poisson bracket which MAY DEPEND ON "h" !!! So in that way if you "turn on..." you will get that you are quantizing NOT the trivial Poisson bracket|||continue | |
| Jan 27, 2013 at 13:05 | history | edited | Alexander Chervov | CC BY-SA 3.0 |
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| Jan 27, 2013 at 8:57 | answer | added | Stefan Waldmann | timeline score: 13 | |
| Jan 27, 2013 at 6:47 | comment | added | Theo Johnson-Freyd | I take your question to be about "universal" quantizations (like those of Kontsevich and Tamarkin) rather than ones that happen to occur. But the opening is phrased a bit confusingly. I could take the Poisson structure on $A$ to be trivial, for example. Then universal quantizations should "quantize" $A$ by making no change at all. But there are generically many nontrivial deformations, which you can "turn on" with speed much slower than your deformation parameter. On the other hand, it is not immediate to me how best to clarify your question into one that is well-posed. | |
| Jan 27, 2013 at 6:02 | comment | added | Mariano Suárez-Álvarez | Why say undoubtly if you cannot prove it? :-) | |
| Jan 27, 2013 at 5:54 | history | asked | Alexander Chervov | CC BY-SA 3.0 |