I work entirely over a field of characteristic $0$, in case it matters.

Recall that a Poisson algebra is a commutative algebra $A$ with a bracket $\lbrace,\rbrace : A^{\wedge 2} \to A$ which is (1) a Lie bracket, i.e. it satisfies a Jacobi identity, and (2) a derivation in each variable. Or, maybe even better is that $(A,\lbrace,\rbrace: A^{\wedge 2} \to A)$ is a Poisson algebra if $A$ is a commutative algebra and $\forall a\in A$ $\lbrace a,-\rbrace : A \to A$ is a derivation of $(A,\lbrace,\rbrace)$.

A coisotrope in $A$ is a vector subspace $I \subseteq A$ which is (1) an ideal for the multiplication on $A$ and (2) Lie subalgebra for $\lbrace,\rbrace$. In particular, I do not demand that $I$ be an ideal for the bracket. The most important examples of Poisson algebras are $A = \mathcal C^\infty(M)$, where $M$ is a symplectic manifold; then an embedded submanifold $N \hookrightarrow M$ is coisotropic ($\mathrm T^\perp N \subseteq \mathrm T N$) iff the vanishing ideal of $N$ is a coisotrope. For more details and equivalent characterizations, see:

One reason to invent Poisson algebras is that the arise naturally as "deformation" or "quantization" problems: a Poisson algebra is the linear or infinitesimal data for a noncommutative algebra. In the most studied symplectic case, the Poisson algebra is in an important sense "maximally Poisson-noncommutative": the Poisson-center of $A$ consists only of (locally) constant functions. After quantization, the corresponding algebras are maximally noncommutative, and so should be algebras of (bounded) operators on some Hilbert space. In this sense, the quantization of a symplectic manifold $M$ is some Hilbert space $H$, or maybe its projectivization $\mathbb P H = H / \mathbb C^\times$.

Then the general dictionary says that lagrangian, i.e. minimal coisotropic, submanifolds of a symplectic manifold $M$ should correspond to elements of $H$ (or maybe elements of $\mathbb P H$, i.e. lines in $H$). My question is to understand a generalization of this that relaxes two things:

  1. I am interested in algebras that are Poisson but not symplectic, and so might have center; then I do not expect to have as good a "Hilbert-space" description of the quanization. Rather, my quantizations of Poisson algebras $(A,\lbrace,\rbrace)$ are nothing more nor less than (flat) deformations of $A$ in the $\lbrace,\rbrace$ direction.

  2. In the non-symplectic setting, one loses a good theory of "lagrangian" submanifold, and the best stand-ins are the coisotropes.

My question is then something along the following:

Suppose I have a Poisson algebra $(A,\lbrace,\rbrace)$ with a coisotrope $I \subseteq A$, and I have a reasonably good quantization of $A$ in the $\lbrace,\rbrace$-direction. What should I expect/hope to see at the quantum level that corresponds to $I$?

Asking the same question in the opposite direction:

Suppose I have an associative algebra $B$ with a formal parameter $\hbar$ (and satisfying some strong flatness/topological freeness conditions), such that $B/(\hbar B)$ is commutative. Then the associated graded algebra $A = \bigoplus (\hbar B)^n / (\hbar B)^{n+1}$ is Poisson. What structures (e.g. ideals, left-modules, etc.) on $B$ become coisotropes in $A$ upon taking associated-graded?

  • $\begingroup$ This question retains its interest---and perhaps is even more interesting---if I (viewed as a submanifold) is not a linear subspace, but rather a embedded (nonlinear) Kronecker join. A great deal of modern quantum simulation theory arises from the (largely empirical) observation that some Poisson algebras are respected by pullback. Here too there are many unanswered questions; the one I asked yesterday on MathOverflow "Is an immersed Kronecker join always a multilinear variety on a Hilbert space?" is one such. In that same post, Theorem 1 may be a (nonlinear) answer to your question. $\endgroup$ May 10, 2011 at 9:45
  • $\begingroup$ You might want to remove zero from $H$ before taking the quotient. $\endgroup$
    – S. Carnahan
    Aug 4, 2011 at 3:00

3 Answers 3


Concerning your first question I have a couple of suggestions: first coisotropic is in some sense the best we can have in a truely Poisson situation: there is nothing like lagrangian (unfortunately).

As you already said, lagrangian sometimes is associated to pure states in the quantum regime, here the main argument is coming from the WKB approximation in physics, which has some very interesting mathematical formulations: in the booklet of Bates&Weinstein (you probably know) you can find a lot of these ideas.

However, I would like to draw your attention to some other point: Having a coisotropic $I$ in some Poisson algebra $A$ (so we left geometry, purely algebraic setting) then the thing you can do classically is reduction: You take the idealizer $B \subseteq A$ of $I$ with respect to the Poisson bracket and this turns out to be the largest Poisson subalgebra of $A$ having $I$ as Poisson ideal. So the quotient $B/I$ is again a Poisson algebra. In your favorite geometric setting with nice assumption this corresponds precisely to the Poisson algebra of functions of the (Marsden-Weinstein) quotient: But we see something more:

$A/I$ is a $A$-left module (sure) and it becomes also a $B/I$ right module (just check that things are well-defined). In fact, a little exercise shows that if $A$ has a unit (let's assume that) then $B/I = End_A(A/I)^{op}$. So we are in some sense even very close to a Morita context (it is not, though...)

Usually I don't like to do that: but to make a little advertisement for some own work, I have a quite detailed paper with Simone Gutt on the above reduction proceedure where we investigate the relations of the representation theories of the big algebra $A$ and the reduced algebra $B/I$ :)

Now the remarkable thing is that this (still classical) bimodule structure might have good chances to survive quantization. In fact (and here one should quote Martin Bordemann's long french preprint as well as the works of Cattaneo&Felder) under certain geometric conditions deformation quantization gives indeed such a quantization.

So the noncommutative version is a left ideal $I$ and then the above quotient proceedure just works the same on the algebraic level. Of course, in DQ the trickey question is whether $B/I$ is still something like the quantized functions on the classical Marsden-Weinstein quotient and even more trickey: whether the classical coisotropic ideal can be quantized into a left ideal at all. For this there are obstructions, even local ones, in the Poisson setting, while it works locally in the symplectic setting by taking an adapted Darboux chart. Globally, it is also trickey in the symplectic setting: Martin Bordemann discusses this in detail...

OK: the conclusion is something like coisotropic ideals are used for reduction and their quantization will be left ideals used in the same way. Both lead to the above bimodule structures on $A/I$ which is geometrically the coisotropic submanifold itself.

As a small warning: it is not true in deformation quantization that all modules (of interest) arise this way. There are other modules which have their support say on points: one can use $\delta$-functionals as positive functional (after some correction terms) and get a GNS like construction also in formal deformation quantization. Then in this case, the module has sort of support on that point...

Ah, the second question: never thought about that in detail, but perhaps the above picture gives some ideas on "reverse engineering"..?


I guess the canonical answer is "modules" (either left or right). The claim is that the associated graded of a module has coisotropic support is usually called "Gabber's theorem"- it is quite non-trivial, but true. Of course, actual modules correspond to coisotropic subvarieties with some additional structure (because the coisotropic subvariety is just the support of the associated graded module) -- what is that structure is not really known in general but there are many cases when some sort of answer to this question is known.

  • 5
    $\begingroup$ A reference for this very general theorem is here: jstor.org/stable/2374101 (I have heard Knop gave a simplified proof, but I don't know a reference.) My understanding (correct me if I'm wrong) is that if $I$ is the annihilator of $gr(M)$ it is easy to see that $I$ is a Poisson subalgebra of $A$, and the (hard) theorem is that the radical of $I$ is also a Poisson subalgebra. $\endgroup$ May 10, 2011 at 15:05

This question was already positively answered. I'd like to add a remark on a situation where things go in a very clear way supporting Waldmann's statementes. Say $G$ is a Poisson Lie group.

A (closed) subgroup $H$ is coisotropic iff the annhilator of its Lie algebra $\mathfrak h$ is a Lie subalgebra in $\mathfrak g^*$.

Quantization of such subgroups are exactly two-sided coideal (this is the subgroup part) which are also one sided ideal (this is the coisotropy assumption).

A two-sided coideal and one sided ideal inside a Hopf algebra is exactly what is needed to define a subalgebra of coinvariants which is a one sided coideal, i.e. a quantum homogeneous spaces seen as a subalgebra of the quantum algebra. This fact reflects the property that modding out a Poisson-Lie group by a coisotropic subgroup you get a Poisson homogeneous space (of quotient type, i.e. such that the quotient map is Poisson).

The idea that one sided ideal is the non commutative counterpart of coisotropy was stated in a paper in the beginning of the 90's by Jiang-Hua Lu: the statement was named coisotropic creed.


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