Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose I have a symplectic manifold $M$, and have a deformation quantization of it, i.e. an associative product $\ast:C(M)[[\hbar]]\otimes C(M)[[\hbar]]\to C(M)[[\hbar]]$ so that $f\ast g=fg+\{f,g\}\hbar+O(\hbar^2)$, where $\{\cdot,\cdot\}:C(M)\otimes C(M)\to C(M)$ is the Poisson bracket coming from the symplectic structure.

According to Lu (http://www.ams.org/mathscinet-getitem?mr=1244874 page 395), a Lagrangian submanifold $L\subseteq M$ should correspond to a left-module over $C(M)[[\hbar]]$, that is, we have a deformed module structure $\ast:C(M)[[\hbar]]\otimes C(L)[[\hbar]]\to C(L)[[\hbar]]$, which reduces to the standard module structure when $\hbar=0$, and is compatible with the start product on $C(M)[[\hbar]]$. (Actually, Lu thought of the left-ideal $I_L\subseteq C(M)[[\hbar]]$ so that $C(L)[[\hbar]]=C(M)[[\hbar]]/I_L$, but I perfer to think of the deformation of the module structure.)

Then we can write, for $f\in C(M)$ and $g\in C(L)$, that $f\ast g=fg+\{f,g\}\hbar+O(\hbar^2)$, for some brackets $\{\cdot,\cdot\}:C(M)\otimes C(L)\to C(L)$.

What are these brackets? How are the related to the symplectic structure?

It seems that in general the existence of a deformed module structure on $C(L)[[\hbar]]$ is nontrivial (perhaps not even known for arbitrary $L$?), but I am hoping that perhaps the deformation to first order is something that one can understand more easily.

share|improve this question

2 Answers 2

up vote 4 down vote accepted

Well, in the symplectic case, the situation is somehow much simpler as in the general Poisson case where you only can speak about coisotropic (there is no good meaning of minimal coisotropic as the rank may vary). In the symplectic case you have a theorem of Weinstein which states that a there is a tubular neighbourhood of $L$ which is symplectomorphic to a neighbourhood of the zero section of $T^*L$. Thus the question of a module structure is reduced to the case of a cotangent bundle since star products are local. For cotangent bundles there is a good understanding whether you can have a module structure on the functions on the configuration space $L$: the characteristic class of $\star$ has to be trivial. In fact, together with Martin Bordemann and Nikolai Neumaier we constructed such module structures in a series of papers in the end of the nineties. Also Markus Pflaum has some papers on this. Thus the global statement is that on $L$ you have a module structure for $\star$ iff the characteristic class of $\star$ is trivial in an tubular neighbourhood of $L$.

The module structures have (for particular star products) a very nice interpretation as global symbol calculus for differential operators on $L$. Moreover, if the char. class is not trivial but at least integral (up to some $2\pi$'s) then there is a module structure on the sections of some line bundle over $L$, coming quite close to the functions on $L$. Physically, this is important for the quantization of Dirac's magnetic monopole.

As DamienC already said, the situation in the general Poisson case or even in the general coisotropic case on symplectic manifolds is much more involved. Here my answer to In the dictionary between Poisson and Quantum, what corresponds to Coisotropic? might also be of interest for you.

Oh, I forgot: the first order term can be obtained as in the flat case, at least morally. On the configuration space $L$ you chose your favorite connection then the first order term of the module structure is something like "half the Poisson bracket" which means that \begin{equation} f \bullet \psi = \iota^* f \psi + i \hbar \iota^* \frac{\partial f}{\partial p_i} \frac{\partial \psi}{\partial q^i} + \cdots \end{equation} (modulo some constants I forgot) where $q^i$ are coordinates on $L$ and $p_i$ are the corresponding fiber coordinates on the cotangent bundle. This has an intrinsic meaning. Here $\iota: L \longrightarrow T^*L$ is the zero section...

share|improve this answer

One has to choose an identification of a tubular neighborhood $U$ of $L$ in $M$ with a tubular neighborhood $V$ of the zero section of the normal bundle $NL$.

Once one has done this, it means that we have a projection $p:U\to L$. Then your bracket is "simply", for $f\in C^\infty(M)$ and $g\in C^\infty(L)$, $\{f,g\}:=\{f_{|U},p^*(g)\}_{|L}$.

But $C^\infty(C)$ is, to my opinion, not the right object to consider (see e.g. the work of Cattaneo-Felder).

Remark: another approach is via Tsygan's Oscillatory modules.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.