Some elementary questions about deformation quantization

Question

I am interested in deformations of affine Poisson algebras, and so this is the setting in which I shall write out the elementary definitions involved. All algebras and vector spaces shall be over $\mathbb{C}$ although most of my questions make sense over any field. Any answers to any part of my question will be very welcome indeed, thanks in advance!

An affine Poisson algebra $A$ is a finitely generated, reduced, commutative algebra with a bracket $\{ \cdot, \cdot\} : A \otimes A \rightarrow A$ which is skew-symmetric, satisfies the Jacobi identity and such that for all $f \in A$ the map $\{f, \cdot\} : A \rightarrow A$ is a derivation.

I am aware of two broad notions of deformation. Filtered deformations arise when $A$ is also graded $A = \bigoplus_{i \geq 0} A^i$ and the Poisson bracket is homogeneous of some fixed degree $d$, meaning $\{ \cdot, \cdot\} : A^i \otimes A^j \rightarrow A^{i + j - d}$ for all $i, j$. If $B$ is a non-commutative, filtered algebra with $B = \bigcup_{i \geq 0} B_i$ such that $[B_i, B_j] \subseteq B_{i + j-d}$ then $\operatorname{gr} B$ inherits a Poisson structure by setting $$\{ f, g\} := [\overline{f}, \overline{g}] + B_{\deg f + \deg g - d - 1}$$ for homogeneous elements $f, g \in \operatorname{gr} B$ where $\overline{f}, \overline{g}$ are lifts of $f,g$ to $B$. If $\operatorname{gr} B \cong A$ as Poisson algebras then we say that $B$ is a filtered deformation of $A$.

A closely related notion of deformation involves star products. Suppose that $A$ is a Poisson algebra as above. Then a $\ast$-product is an associative $\mathbb{C}[[\hbar]]$-linear product on $A[[\hbar]]$ which is written $$f \ast g = fg + \hbar B_1(f, g) + \hbar^2 B_2(f,g) + \cdots$$ where $f, g \in A$ and $\ast$ satisfies $f \ast g - g \ast f = \hbar\{f, g\} + \mathcal{O}(\hbar^2)$. The algebra $(A[[\hbar]], \ast)$ is then referred to as a deformation quantisation of the Poisson algebra $(A, \{ \cdot, \cdot\})$.

Kontsevich famously defined a $\ast$-product on the $\mathcal{C}^\infty(M)$ where $M$ is a Poisson manifold, thus showing that deformation quantisations always exist in this case. Smooth affine Poisson varieties are special cases of these and I believe that the ring of regular functions can be endowed with a $\ast$-product à la Kontsevich.

I hope everything that I've written thus far is clear and correct.

Question 1: what is the precise connection between these two definitions of quantisation?

Subquestion 1a: When $B_i(f,g) = 0$ for $i \gg 0$ we can define $f \ast g|_{\hbar = 1}$ and if this holds for all $(f,g) \in A\times A$ we can define a product $\ast_{\hbar = 1} : A\otimes A \rightarrow A$. Does Kontsevich's $\ast$-product have this vanishing property (when $\operatorname{Spec} A$ is smooth)?

Subquestion 1b: If $(A, \{\cdot,\cdot\})$ is graded then is there automatically a filtration on $(A, \ast_{\hbar = 1})$ such that the associated graded is $(A, \{\cdot, \cdot\})$?

Subquestion 1c: If the Poisson structure on $A$ is not graded then how is the $\hbar = 1$ product on $A$ related to the Poisson structure? Is there something more general that a filtered deformation coming from this construction and can it be axiomatised?

Kontsevich's $\ast$-product is a consequence of his formality theorem, and so the choice of $\ast$-product is only well-defined up to so-called gauge equivalence, but gauge equivalent products subtend isomorphic algebras.

Question 2: Is isomorphisms a strictly finer invariant than gauge equivalence?

This can be asked in two settings:

Subquestion 2a: If $(A[[\hbar]], \ast) \cong (A[[\hbar]], \ast')$ then does it follow that the structures are gauge equivalent?

Subquestion 2b: If $(A, \ast_{\hbar = 1}) \cong (A, \ast'_{\hbar = 1})$ then does it follow that the structures are gauge equivalent?

Finally I would like to know something about quantising Poisson subvarieties. First comes the general question:

Question 3: Is Kontsevich's quantisation functorial in any sense?

Now comes the more specific version of this question. Suppose that $I \unlhd A$ is a Poisson ideal, so that $A/ I$ is a Poisson algebra and the projection $A \rightarrow A/I$ is a Poisson map. Kontsevich's quantisation gives products $(A[[\hbar]], \ast)$ and $(A/I[[\hbar]], \ast')$.

Subquestion 3a: Do we have a surjection $(A[[\hbar]], \ast) \twoheadrightarrow (A/I[\hbar]], \ast')$?

And finally the $\hbar = 1$ version:

Subquestion 3b: Do we have a surjection $(A, \ast_{\hbar = 1}) \twoheadrightarrow (A/I, \ast'_{\hbar=1})$?

Again, thanks in advance for any comments, answers, advice... anything at all really.

Answer 1

a lot of questions, let me try on some of them :)

The bad news is that in most of the interesting situations the higher order terms of the star product, the $B_i$ will not vanish. Heuristically this can be understood as follows (and can be made precise in many cases): $B_1$ is a bidifferential operator of order at least one, since it has to produce the Poisson bracket as antisymmetric term. Now the associativity failure if you only have $B_1$ is a composition of two $B_1$'s hence generically a tridifferential operator of order $2$. Thus to compensate this you need the $B_2$ (since you want something associative!) which then enters the associativity condition together with $B_0$, i.e. the multiplication. But $B_0$ is bidifferential of order zero, so you need a bidifferential operator $B_2$ in order to produce terms with $2$ derivatives, in order to compensate the $B_1$ terms. This continues by induction. So whenever the bidifferential operators you have already found up to $B_k$ are of order at least $k$, then $B_{k+1}$ tends to be of order $k+1$. "Tends to be" can be made precise, say in the case of symplectic Poisson structures, where the bidifferential operators are sufficiently nondegenerate. The zero Poisson bracket $B_1 = 0$ of course needs no higher order to fix associativity :) But you see that you will typically need infinitely many contributions to get associativity.

This makes 1b and 1c sort of obsolete. Of course there might be situations where you only have finitely many $B_i$ different from zero, but I have not yet seen examples (being, of course, only a proof of my limited experience).

However: one can of course require honest convergence of the formal series. This is an ongoing huge project of several people (including myself) and by far not well-understood. One enters the realm of strict deformation quantization, rewarding, challenging, but also quite trickey...

For 2: the difference between gauge transformations (or equivalence transformations) and isomorphisms is that an isomorphism is of the form \begin{equation} \Phi = \Phi_0 + \hbar \Phi_1 + \hbar^2\Phi_2 + \cdots \end{equation} with linear maps $\Phi_k\colon A \longrightarrow A$, extended $\hbar$-linearly. In this case, $\Phi_0$ is easily shown to be an isomorphism of the underformed algebra $A$. For an equivalence transformation you require $\Phi_0 = id$, thus having slightly less freedom. In case of, say, symplectic star products, one can furthermore show that $\Phi_0$ is the pullback with a symplectic diffeo. But if this diffeo is large enough, it can act nontrivially on the second deRham cohomology. If your characteristic class of the star product is now not invariant under $\Phi_0$ then such an isomorphism really changes the class with respect to equivalence transformations.

For 3: this is a long long story. In short: no. In geometric situation the question is probably not even that relevant since there are not too many Poisson ideals. They would correspond to Poisson submanifolds and your functoriality would sort of imply that the big star product is tangent to the submanifold. This is known to be false in most cases, say for linear Poisson structures (see a paper by Cahen, Gutt, Rawnsley?)

More important is the question of a Poisson subalgebra which is a associative ideal, i.e. a coisotropic submanifold. They are used to phase space reduction and produce reduced Poisson algebras. Now it is a quite nontrivial question to characterize those coisotropic submanifolds for which you can perform reduction. In general: no, but there are many favorable positive situations. Here papers pf Bordemann or Cattaneo-Felder might give you some inspiration. In the symplectic case there is also a very old approach by Bordemann, Herbig and myself using some BRST like construction.

Again 3b is beyond hope.