Noether's Theorem (Symplectic Version)

The symplectic version of Noether's theorem establishes a precise correspondence between symmetries of a Hamiltonian system and conserved quantities, mediated by the moment map.

Statement

Theorem4.2Noether's Theorem (Symplectic)

Let $(M, \omega, H)$ be a Hamiltonian system with a Hamiltonian action of a Lie group $G$ with moment map $\mu: M \to \mathfrak{g}^*$ . If $H$ is $G$ -invariant, then $\mu$ is conserved along the Hamiltonian flow of $H$ : $\frac{d}{dt}\mu(\phi_t^H(x)) = 0 \quad \text{for all } x \in M.$ Conversely, if $F: M \to \mathbb{R}$ is conserved ( $\{F, H\} = 0$ ), then the Hamiltonian flow of $F$ is a one-parameter group of symmetries of $H$ .

Proof

For $\xi \in \mathfrak{g}$ , let $\mu_\xi = \langle \mu, \xi \rangle$ . The Hamiltonian vector field of $\mu_\xi$ is $X_\xi$ (the infinitesimal generator of the $G$ -action). Since $H$ is $G$ -invariant: $X_\xi(H) = 0$ . Therefore: $\{H, \mu_\xi\} = X_{\mu_\xi}(H) = X_\xi(H) = 0.$ This means $\mu_\xi$ is constant along the flow of $X_H$ , i.e., $\mu$ is conserved. $\square$

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ExampleAngular Momentum Conservation

For a particle in a central force field on $\mathbb{R}^3$ , the $SO(3)$ -symmetry (rotational invariance) yields the moment map $\mu(q, p) = q \times p \in \mathfrak{so}(3)^* \cong \mathbb{R}^3$ , which is the angular momentum. By Noether's theorem, all three components of angular momentum are conserved.

RemarkMomentum Map Formulation

The moment map $\mu: M \to \mathfrak{g}^*$ encodes all conserved quantities simultaneously. The conservation law $\{H, \mu_\xi\} = 0$ for all $\xi$ is equivalent to $H$ being constant on the orbits of $G$ . This elegant reformulation unifies Noether's theorem with the theory of symplectic reduction.