Relationship between the character group and cotangent bundle

Question

Given a manifold $ M $ , the cotangent bundle $ T^*(M) $ carries a symplectic structure given by the wedge-product. In classical mechanics, if $ M $ is taken to be the state space of positions, then $ T^*(M) $ is regarded as the phase space which captures both positions and momenta. In the case when $M\cong \mathbb R^n$, then $ T^*(\mathbb R^n) \cong \mathbb R^n \times \mathbb (\mathbb R^n)^* \cong \mathbb R^n \times \mathbb R^n \cong \mathbb R^{2n} $.

However, in quantum mechanics the story is a bit different. Given a group $ G $ which encodes some discrete physical symmetries of the quantum system, the state space of a discrete variable system is the Hilbert space of square summable functions on $ G $:

$$ \ell^2(G) := \left\{\phi:G\to \mathbb C \ \middle|\ \sum_{g \in G} |\phi (g) |^2 < \infty \right\}. $$

Similarly, given a group $G$ with Lebesgue measure, capturing some continuous symmetry of the quantum system, the state space is the space of square integrable functions on $ G $:

$$ L^2(G) := \left\{\phi:G\to \mathbb C \ \middle|\ \int_{g \in G} |\phi (g) |^2 dg < \infty \right\}. $$

Given a quantum state with state space generated by the group $ G $, the phase space is taken to be the group $G\times\hat G$ where $\hat G \:= \mathsf{Grp}(G, {\mathbb S}^1) $ is the group of homorphisms into the circle. In the case when $ G = \mathbb R ^n $, then as groups, $ \mathbb R \times \hat {\mathbb R} \cong \mathbb{R}^{2n} $, which is the underlying group of the cotangent bundle $T^*(\mathbb R^n) \cong \mathbb R^{2n}$.

However, for non-euclidean state spaces, the quantum and classical mechanical pictures diverge.

For example, a classical rotor spins around the state space $ \mathbb{S}^1 $, and a quantum rotor has the state space $ L^2(\mathbb S ^1)$. However, the cotangent bundle $T^* (\mathbb{S}^1) \cong \mathbb R \times \mathbb{S}^1$ is the cylinder, whereas the quantum phase-space $ \mathbb S^1 \times \hat {\mathbb S^1} \cong \mathbb Z\times \mathbb S^1$ has “discretized momentum”. As groups, $\mathbb Z \times \mathbb S^1 \hookrightarrow \mathbb R \times \mathbb S^1 $ the quantum state space can be regarded as a subgroup of the classical state space.

Is there something more general going on? Does anyone know if there is some general mathematical fact which connects the character group to the cotangent bundle?

Please forgive me if I am missing some extra conditions on the group, I am not an expert in this domain.

Answer 1

Suppose that $M$ is a non-zero connected Abelian Lie group (i.e., $M \cong \mathbb{R}^p \times \mathbb{T}^q$ for $p,q \in \mathbb{N}_0$ with $p+q > 0$); let $\mathfrak{m}$ denote the Lie algebra of $M$. By connectedness of $M$, the exponential map $\exp : \mathfrak{m} \to M$ is a surjective homomorphism, so that you can identify the Pontrjagin dual $\hat{M}$ of $M$ with an additive subgroup of $\mathfrak{m}^\ast$ by identifying $\alpha \in \hat{M}$ with its differential $\alpha_\ast \in \operatorname{Hom}(\mathfrak{m},\mathfrak{u}(1)) \cong \mathfrak{m}^\ast$. In the concrete case of $M = \mathbb{R}^p \times \mathbb{T}^q$, so that $\mathfrak{m} \cong \mathbb{R}^p \oplus \mathbb{R}^q$ and $\exp : (x,y) \mapsto (x,y \bmod \mathbb{Z}^q)$, the resulting map $\mathbb{R}^p \times \mathbb{Z}^q \cong \hat{M} \hookrightarrow \mathfrak{m}^\ast \cong \mathbb{R}^p \oplus \mathbb{R}^q$ is the obvious inclusion.