Geometric quantization is a marvelous tool for understanding the relation between classical physics and quantum physics. However, it's a bit like a power tool — you have to be an expert to operate it without running the risk of seriously injuring your brain. Here's a brief sketch of how it goes. This is pretty terse; for the details you'll have to read the series of articles on geometric quantization on the sci.physics.research archive.
Warning: we can only do this step if \(\omega\) satisfies the Bohr–Sommerfeld condition which says that \(\omega/2\pi\) defines an integral cohomology class. If this condition holds, the bundle \(L\) is determined up to isomorphism, but not canonically. After choosing \(L\), the curvature \(\omega\) determines the connection up to a gauge transformation locally, but not globally: we also need to choose its holonomies around noncontractible loops. So, it is best to consider \(L\) and \(\nabla\) as extra choices required for geometric quantization.
Warning: for \(P\) to be a polarization, there are some crucial technical conditions we impose on the subspaces \(P_x\). First, they must be isotropic: the complexified symplectic form \(\omega\) must vanish on them. Second, they must be Lagrangian they must be maximal isotropic subspaces. Third, they must vary smoothly with \(x\). And fourth, they must be integrable.
Modulo some technical trickery, we get this example when we run the above machinery and use a certain god-given real polarization on \(X = T^*M\), namely the one given by the vertical vectors.
Here are some definitions of important terms. Unfortunately they are defined using other terms that you might not understand. If you are really mystified, try Wikipedia, or some books on differential geometry and the math of classical mechanics.
$$ \omega(\cdot,v_f) = df $$
In other words, for any vector field \(u\) on \(X\) we have
$$ \omega(u,v_f) = df(u) = u(f) $$
The vector field \(v_f\) is guaranteed to exist and be unique by the fact that \(\omega\) is nondegenerate.
This trick works as follows: given any smooth function \(f\) we can take its differential \(df\), which is a 1-form. Then there is a unique vector field \(v(f)\), the Hamiltonian vector field associated to \(f\), such that
$$ \omega(\cdot,v_f) = df $$
Using this we define
$$ \{f,g\} = \omega(v_f, v_g) $$
It's easy to check that we also have $$ \{f,g\} = dg(v(f)) = v(f)(g) $$
so \(\{f,g\}\) says how much \(g\) changes as we differentiate it in the direction of the Hamiltonian vector field generated by \(f\).
In the familiar case where \(M\) is \(\mathbb{R}^{2n}\) with momentum and position coordinates \(p_i, q_i \), the Poisson brackets of \(f\) and \(g\) work out to be
$$ \{f,g\} = \sum_i \frac{\partial f}{\partial p_i} \frac{\partial g}{\partial q_i} - \frac{\partial f}{\partial q_i} \frac{\partial g}{\partial p_i} $$
© 2018 John Baez
baez@math.removethis.ucr.andthis.edu