# Is the quantum algebra unique (up to isomorphism) in deformation quantization ?

Consider a Poisson algebra A (i.e. commutative algebra with Poisson bracket).

Let $\hat A$ be a deformation quantization of the algebra A. We know that construction of deformation quantization and more generally "formality isomorphism of Kontsevich" depends on choices of certain data (in Kontsevich approach we can change "propagator" and coordinates on manifold, in Tamarakin's approach we can choose arbitrary associator.)

Question Is it known/true/expected that for different choices of "that datum" we nevertheless obtain isomorphic quantum algebras $\hat A$?

May be one needs certain restrictions on setup (e.g. only smooth algebras A, only "generic" quantization morphism, whatever...) to guarantee uniqueness ?

Kontsevich also mentions that Grothendieck-Teichmuller group should act on the set of all deformation quantizations. Is it at least true that two quantizations living in the same orbit of that group give isomorphic quantum algebras ?

Question in formally precise form Consider a Poisson algebra. Choose two different formality isomorphisms (e.g. with different propagators or associators).

Define two star-products $\star$ and $\star'$ with the help of these two formality isomorphisms.

Question Are the algebras defined by these two star-products isomorphic ? More strongly - are these star products "equivalent" ? (See definition of equivalence in Stefan Waldmann's answer below or in Kontsevich paper).

If our manifold is R^2n we canonical Poisson bracket { p_i q_j } = delta_ij, then undoubtly the quantum algebra should be unique and isomorphic to Heisenberg algebra: $[ \hat p_i , \hat q_j ] = delta_{ij}$. But I am not sure that even in this case it is that much obvious - if we change coordinates we can make Poisson bracket arbitrary weird, it would be non-obvious that we get isomorphism with Heisenberg algebra. And moreover it might depend on category we are working with (polynomial or smooth functions).

More general example is Lie algebra $g$ - corresponding quantization should be isomorphic to universal enveloping algebra, but again it is not that much obvious. (In Kontsevich paper he devoted some special (small) arguments to prove that his quantum algebra is isomorphic to U(g)).

Concerning choices of different coordinates in classical algebra A - it already states in Kontsevich paper that obtained algebras will be isomorphic. More strongly star-products will be "equivalent".
See the last formula on the page 3 of his paper. However nothing is said about choices of different propagators

I had discussed this question with some experts some years ago, but there was no clear answer.

The motivation to ask partly comes from MO-discussions here: Quantization of a classical system (e.g. the case of a billard)

-
Why say undoubtly if you cannot prove it? :-) –  Mariano Suárez-Alvarez Jan 27 '13 at 6:02
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. –  Theo Johnson-Freyd Jan 27 '13 at 6:47
@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 –  Alexander Chervov Jan 27 '13 at 13:12
...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 ? –  Alexander Chervov Jan 27 '13 at 13:16
@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. ... –  Theo Johnson-Freyd Jan 30 '13 at 5:27

In deformation quantization there is a full classification available: let us first focus on the symplectic case which is easier. If $(M, \omega)$ is a symplectic manifold (like the $\mathbb{R}^2$ in your example) then the equivalence classes of star products are classified by formal series in the second deRham cohomology of $M$, a purely topological quantity.

Here equivalence means that $\star$ and $\star'$ are called equivalent if there is a formal series $T = \mathrm{id} + \sum_{n = 1}^\infty \hbar^r T_r$ of differential operators $T_r$ starting with the identity in zeroth order such that $$T(f \star g) = Tf \star'Tg$$ This implies clearly that the deformations yield isomorphic algebras. But it is sligtly stronger, as one requires $T$ to start with the identity in zeroth order. This means that $T$ is invisible in the classical limit. If you are interested in a physical interpretation, this is very important as the interpretation of the elements of $C^\infty(M)$ should not be changed under quantization: the same function should correspond to the same physical observable, both in the classical and the quantum world.

There is a good reason that the above notion of equivalence can be interpreted as the freedom of having possible different ordering descriptions as they show up in other quantization schemes.

So in your example, there is only one deformation up to equivalence since the second deRham cohomology of $\mathbb{R}^2$ is trivial.

If $M$ admits a non-trivial deRham cohomology then the higher order terms in the classifying "characteristic class" $c(\star)$ of the star product have the physical interpretation of a magnetic field. Their non-triviality then gives a magnetic monopol. So from that point of view (which can be justified physically more striclty) quantization requires to be told what magnetic charges are around and then it is unique up to equivalence, which means physically, up to the choice of an ordering description.

Names involved in this classification results are Bertelson-Cahen-Gutt, Nest-Tsygan, Fedosov, Weinstein-Xu, and many others.

Let me add here the precise definition of the characteristic class. On a symplectic manifold, given a star product $\star$ there is an intrinsic way to construct a formal series $$c(\star) \in \frac{[\omega]}{i\hbar} + \mathrm{H}^2_{dR}(M, \mathbb{C})[[\hbar]]$$ of second de Rham cohomology classes in such a way that $\star$ is equivalent to $\star'$ iff their classes coincide. Moreover, for every such formal series there is also a star product $\star$ having this series as characteristic class. The choice of the origin in this affine space is by convention, and given by the class of the symplectic form itself.

There are several ways to construct $c(\star)$, all leading to the same result. The easiest is probably the construction in Gutt-Rawnsley link text. As a result: there is always the way to choose the class $c(\star) = \frac{[\omega]}{i\hbar}$ without higher order contributions. This is the somehow canonical choice on a symplectic manifold. However, having this choice does of course not mean to have a unique quantization (up to equivalence): depending on the non-triviality of the deRham cohomology, there are hiogher orders possible and then one get's non-trivial equivalence classes of quantizations.

Now for the Poisson case: here the situation is similar in so far as in the symplectic case you can interprete the higher order terms of the class as perturbation of the symplectic form. Kontsevich proved using his formality theorem that the possible star products are classified by formal deformations of the Poisson tensor $\pi = \hbar \pi_1 + \hbar^2 \pi_2 + \cdots$ up to the action of the formal diffeomorphism group. This is given by exponentiating the formal vector fields $X = \hbar X_1 + \hbar^2 X_2 + \cdots$.

In the symplectic case, this of course matches with the "characteristic class" classification. One can even show that, under correct identification, the classes are equal (this was done by Bursztyn, Dolgushev, and myself).

In particular, not all deformations will be isomorphic in general, this depends now really very much on the underlying manifold and the Poisson tensor.

Surprisingly, and this point I don't really understand well, in the Poisson case, the classification, i.e. the identification of the class of $\star$ and the class of $\pi$ depends to some extend on the choice of the formality map (and on those, GT acts). This is funny, because in the symplectic case, the classification can be done quite intrinsically, not depending on any sort of choices. Here the construction of Deligne, explained later also in Gutt-Rawnsley, is very nice.

Note added: I fear that for the refined question I have no answer. The situation is the following: one has two types of data needed to construct a quantization (in the sense of star products) of a Poisson manifold together with a meaningful classification:

First is to choose a quantization machinery. In the generic Poisson case, the only one we know is a formality map, which can be obtained by e.g. Kontsevich's or Tamarkin's methods. These constructions require certain choices, like the ones mentioned by Alexander Chervov_ a propagator in Kontsevich's construction or an associator for Tamarkin.

Second, having a formality we can choose a classical deformation $\pi = \hbar \pi_1 + \hbar^2 \pi_2 + \cdots$ of the given Poisson structure $\pi_1$ and plug it into the formality. This will give a star product and any two obtained this way will be equivalent iff the classical deformations $\pi$ and $\pi'$ are equivalent in the sense of the action of the formal diffeos.

So the question, how the notion of equivalence depends on the choice of the formality seems to be much more subtle. I have no idea about that. ONe knows that there are really different formalities (in a sense of being non-homotopic) but whether and how they yield a really different classification scheme seems to be not known (please correct me). I had some discussions about that with Vasiliy Dolgushev some time ago, but we didn't come to a real conclusion...

The somehow surprising thing is that in the symplectic case, the classification is independent of any construction, and done by this characteristic class. Since the classifying space is the deRham cohomology $H_{dR}^2(M, \mathbb{C})[[\hbar]]$, we have a canonical choice for the trivial class. So in this sense, we really have a canonical quantization up to equivalence. Note however, that there are good reasons for considering also nontrivial classes in the symplectic case.

From that point if view, the original question in the Poisson case can also be formulates as folows. given $\pi = \hbar \pi_1$ and a formality yielding $\star$, is there another formality such that $\star$ has a corresponding class $\pi' = \hbar \pi_1 + \hbar^2 \pi_2 + \cdots$ which is not equivalent to $\pi$ by means of the formal diffeos?

OK, not much of a help, but just a refomulation of the problem.

-
Thank you very much! Let me ask for some clarification later. –  Alexander Chervov Jan 27 '13 at 9:59
" then the equivalence classes of star products are classified by formal series in the second deRham cohomology of M" would you be so kind to clarify what kind of star-products you consider ? I mean you need to put requirement that they "quantize" $\omega$, e.g. not $\omega + h\omega$ . –  Alexander Chervov Jan 27 '13 at 13:21
"One can even show that, under correct identification, the classes are equal (this was done by Bursztyn, Dolgushev, and myself)." What out of many formality isomorphism you are using ? (I mean what propagator, or what associator) ? PS By the way in that paper you write: "and the fact that this element agrees with the Fedosov class of the star product ∗ has been conjectured by A. Chervov and L. Rybnikov in [13..." :):) It is very nice to discuss with you ! –  Alexander Chervov Jan 27 '13 at 13:29
I added some clarifications to my question. Does your answer imply that in symplectic case the algebras are isomorphic ? –  Alexander Chervov Jan 27 '13 at 13:30
@Alexander Chervov: yeah, it's funny to meet this way. I added a couple of remarks (too long for a comment) Essentially, I have no idea about you refined question in the Poisson case :( –  Stefan Waldmann Jan 28 '13 at 7:36

Let me give a conjectural answer.

Point 1. $GRT$ acts non-trivially on poly-vector fields by $L_\infty$-isomorphisms (see T.Willwacher, M. Kontsevich’s graph complex and the Grothendieck-Teichm¨uller Lie algebra, arXiv:1009.1654 for an explicit description of this actions).

Point 2. I guess that $GRT$ actually acts non-trivially on the set of equivalence classes of Maurer-Cartan elements (I haven't checked this).

Point 3. composing Kontsevich's formality map with the action of GRT gives a negative answer to your question if Point 2 is correct. .

Point 4. 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. By itself, the globalization procedure does not require any additional choice.

-
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. –  Alexander Chervov Jan 28 '13 at 10:35
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... –  Alexander Chervov Jan 28 '13 at 10:38
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. –  Adrien Jan 28 '13 at 12:44
@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. –  DamienC Jan 28 '13 at 15:17
@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 ? –  Adrien Jan 28 '13 at 16:04