When we say (see pag. 9 of Lectures on the Geometry of Quantization) that

the image of the differential of the phase function lies in the level set of the classical Hamiltonian

is it simply equivalent to reading the following equation?

$$H(x,S'(x)) = E$$

i.e. to applying the inverse Hamiltonian $H^{-1}$ operator to both members?

In which way is relevant that the cotangent bundle (or the dual of the tangent bundle) $T^*M$ is isomorphic to $\mathbb{R}^2$?

Is that isomorphism (together with $dS: \mathbb{R} \rightarrow T^*\mathbb{R}$) taken into consideration to replace $p$ with $S'$ or has it to do with the exponential map between the Lie algebra (as tangent space) and the Lie group or maybe with the de Rahm cohomology: when we say

"the Chern class of the line bundle $L$"

is $L={\rm Im}(dS)$ an $n$-dimensional submanifold of $H^{-1}(E)$ and so is it the so-called "level manifold/set" or something else unrelated?

This is what I've tried to study more from a mathematical standpoint, of course I'd like to put it into a physical perspective.

  • $\begingroup$ $L = \mathrm{im}(dS)$ stands for a Lagrangian submanifold. It has a dimension of $\dim(T^*M)/2= \dim(M)$. The line bundle $L$ in this context, usually denotes a line bundle over the cotangent bundle $T^*M$ it has a dimension of $2\dim(M)+1$. They are not the same object. $\endgroup$ – David Bar Moshe May 12 '19 at 16:14
  • $\begingroup$ Not exactly the same object, but, for example, in the quantization context, I think we can define the Maslov line bundle of a Lagrangian submanifold... $\endgroup$ – Xlsx2020 May 14 '19 at 23:38

First of all, the importance of the diffeomorphism on points in configuration space is that the cotangent lift (see 6.3 of Cotangent Bundles) $T^*f$ is guaranteed to be symplectic. If $M$ is a cotangent bundle $T^*Q$ then the obvious choice for the pre quantum bundle $B$ is $M \times\mathbb{C}$. Indeed in the prequantization we introduce the polarization by requiring that $q$ goes to $\frac {\partial} {\partial p}$

A symplectic manifold (M,ω) is said to be prequantizable if the integral of $\omega$ around any 2-cycle lies in $2\pi\mathbb{Z}$

Now given the Lie Algebra $\mathfrak{h}$ of a compact and connected Lie group $H$, the character group $\hat H$ can be identified with $H^1(H,\mathbb{Z})$. The Maslov class of a lagrangian immersion $\iota: L → T^∗M$ is defined as the degree-1 cohomology class and determines via the exponential map $\mathbb{R} \to U(1)$ an isomorphism class of flat hermitian line bundles over $L$ (see pag. 56 of the pdf linked in the question).

Another way of looking at this is that if we impose the strong Legendre condition, that $\mathscr L$ is globally a diffeomorphism, then Lagrange’s equations on $TQ$ are transfromed by $\mathscr L$ into Hamilton’s equations on $T^∗Q$ with canonical symplectic structure.

Finally, given a Lagrangian submanifold $\Lambda$ of the cotangent bundle, contained in the energy shell $H^{-1}(E)$, it is desirable to attach the WKB expansion to the manifold $\Lambda$ (see Semi-classical approximations)

The WKB method applies when the semi-classical density in phase space is supported by a Lagrangian submanifold: for the levels of a separable system this leads to the Bohr-Sommerfeld rules corrected by the Maslov index.

And that corresponds to the Maslov quantization condition, described in the already mentioned Lectures on the Geometry of Quantization at pag. 45

Returning to the harmonic oscillator of Example 4.5, we see that the level set $H^{−1}(E)$ satisfies the Maslov condition provided that for some integer n, $E = (n + 1/2)$. Allowable energy levels in this case therefore correspond to the Bohr-Sommerfeld condition, which actually gives the precise energy levels for the quantum harmonic oscillator.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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