# Relationship between the character group and cotangent bundle

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.

• Character groups only really make sense for abelian groups, for what it's worth. Also, I don't really understand why you are expecting anything for groups that are not isomorphic to their dual groups Commented Jun 23 at 20:24
• @YemonChoi The reason I care about groups which are not isomorphic to their dual groups, is because the duality between the circle and the integers is useful for quantum error correction. Already in this case, the product of the group with its character embeds within the cotangent bundle, so I am curious if this always happens, for example. Commented Jun 23 at 21:16
• What if your group is the integers? What is the "cotangent bundle" supposed to be in this case? If you are only interested in connected Lie groups, then the abelian connected Lie groups are just products of copies of ${\bf R}$ with copies of ${\bf T}$, and then you may as well just analyze them by hand Commented Jun 23 at 22:54
• Thank you for your comments, Yes, this doesn't seem to work if you start with quantum angular momentum and try to build the classical phase space. But it is not just that I care about this specific example... it is only the most well-studied. I hope to recover the structure of "Lagrangian subgoups" of the quantum phase-space from Lagrangian submanifolds of the cotangent bundle. At least in this example, there is a clear analogy, but I wonder if the analogy can be made more formal. Commented Jun 24 at 8:12

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.
• I wish I did. I think you might get some intriguing possible context in the literature around Heisenberg groups, quantum harmonic analysis, and the symplectic Fourier transform. There's a definite idea out there that a locally compact Abelian group (LCA) equipped with an alternating non-degenerate $U(1)$-valued bicharacter (e.g., if $M$ is an LCA group, then $\widehat{M} \times M$ with $\sigma((\alpha,x),(\beta,y)) := (\alpha,y) \overline{(\beta,x)}$) is a "quantum" analogue of a symplectic vector space. Maybe try this preprint and its references? Commented Jun 26 at 19:43