advanced32 min studyLesson 446 in Mathematics

Momentum maps and Noether conservation

Convert infinitesimal symmetries into Hamiltonian generators and track exactly why invariant energies conserve their momentum components.

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What you’ll understand

  • Compute a momentum component from a symmetry vector field.
  • Derive conservation using the fixed Poisson-bracket convention.
  • Distinguish a symplectic action, a Hamiltonian action and equivariance of its momentum map.

A continuous symmetry moves phase-space points without changing the geometric structure. A momentum map asks for a Hamiltonian generator for every infinitesimal symmetry, packaging those generators into one map valued in the dual Lie algebra. If the chosen energy shares the symmetry, the corresponding momentum remains constant. This is the Hamiltonian form of Noether’s principle. The argument requires both geometric compatibility and invariance of the Hamiltonian; a symmetric phase space by itself does not make every possible energy conserve angular momentum or linear momentum.

Package the generators of a group action

Let a Lie group G act smoothly by symplectomorphisms on M. For a Lie-algebra element ξ, let ξ_M be the vector field obtained by differentiating exp(tξ)·x. A momentum map J:M→g* satisfies dJ_ξ=ι_{ξ_M}ω, where J_ξ(x)=⟨J(x),ξ⟩. In our convention this means X_{J_ξ}=ξ_M. Equivariance additionally asks J(g·x)=g·J(x), where the coadjoint action is defined by (g·μ)(ξ)=μ(Ad_{g^{-1}}ξ). We will explicitly assume equivariance when using reduction.

Every ξ_M preserves ω, so its contraction is closed, but it need not be exact. Translation on a symplectic two-torus gives the closed nonexact form dp from the earlier lesson, preventing a global real momentum component for that direction. Even when individual generators exist, arranging their group-level compatibility can introduce additional conditions. Adding a constant covector to J preserves its differential defining equations, but preserves equivariance only when the constant is fixed by the coadjoint action. Thus the generator equation and equivariance should not be treated as identical assertions.

Jξ=⟨J,ξ⟩,dJξ=ιξMω,XJξ=ξM,J(gx)=g⋅J(x) for an equivariant mapJ_\xi=\langle J,\xi\rangle,\qquad dJ_\xi=\iota_{\xi_M}\omega,\qquad X_{J_\xi}=\xi_M,\qquad J(gx)=g\cdot J(x)\ \text{for an equivariant map}
The infinitesimal Hamiltonian sign matches ι_XHω=dH. Equivariance is an additional group-level compatibility condition used by the reduction theorem.↔ Scroll wide equations horizontally.

Energy invariance supplies the conservation step

If H is invariant under the group, then ξ_MH=0 for every infinitesimal generator. Since ξ_M=X_{J_ξ}, our observable convention gives ξ_MH={H,J_ξ}. Antisymmetry then gives dJ_ξ/dt={J_ξ,H}=0 along the H flow. This is the complete local conservation calculation. It applies where the trajectories exist and to autonomous generators; an explicitly time-dependent symmetry or Hamiltonian requires its corresponding modified statement. A momentum map supplies candidate quantities, while energy invariance decides whether those quantities are conserved by the selected dynamics.

For a cotangent-lifted action on a configuration space Q, the tautological form is invariant. Cartan’s identity and ω=−dθ give the natural component J_ξ(q,p)=p(ξ_Q(q)). Translation along a configuration axis therefore has its conjugate momentum as generator. Planar rotation has angular momentum q₁p₂−q₂p₁. These formulas follow from covector evaluation and do not require a Euclidean kinetic energy. Whether the mechanical Hamiltonian respects the translation or rotation is a separate test on its potential and other terms.

Jξ(q,p)=p(ξQ(q)),H invariant⟹{H,Jξ}=0⟹ddtJξ={Jξ,H}=0J_\xi(q,p)=p(\xi_Q(q)),\qquad H\text{ invariant}\Longrightarrow\{H,J_\xi\}=0\Longrightarrow\frac{d}{dt}J_\xi=\{J_\xi,H\}=0
The two bracket orders differ by a sign, but both vanish under invariance. The cotangent formula uses the canonical lifted action.↔ Scroll wide equations horizontally.

Conserved components lead toward reduction

Fixing a conserved momentum value selects an invariant subset of phase space. Quotienting a symmetry that acts inside that subset can then remove redundant motion, but not every level or quotient is smooth. Equivariance identifies which subgroup preserves a nonzero value, while regularity and freeness determine whether the level and orbit space are manifolds. Those conditions enter the next lesson. Simply counting one conserved scalar and deleting one coordinate is not a proof that the remaining dynamics has a nondegenerate symplectic form.

Several momentum components can also fail to Poisson commute for nonabelian symmetries, so symmetry alone does not automatically produce a Liouville-integrable system. Commuting circle symmetries are easier to combine into independent integrals, but their rank can still drop at fixed points. The useful diagnostic is therefore explicit: compute the infinitesimal generator, integrate its contraction if possible, verify equivariance when required, and then check the bracket with the actual H. This sequence separates topological obstructions, algebraic compatibility and dynamical conservation instead of collapsing them into one appeal to symmetry.

Angular momentum for the isotropic oscillator

On T*R² with canonical ω, use simultaneous counterclockwise rotation of q and p. Find its generator J and verify conservation for H=(|p|²+|q|²)/2.

  1. The infinitesimal rotation is (−q₂,q₁,−p₂,p₁). Contracting it with Σdq_j∧dp_j gives d(q₁p₂−q₂p₁), so J=q₁p₂−q₂p₁.

  2. Its bracket with H is p₂p₁−p₁p₂−(−q₂q₁+q₁q₂)=0. Hence J remains constant along the oscillator flow.

  3. At q=(2,0),p=(0,3), angular momentum is 6 and energy is (4+9)/2=13/2. Both remain constant, although each individual position and momentum coordinate changes.

The result

The conserved rotation generator comes from the lifted action and the isotropic energy’s invariance, not from a claim that all oscillators conserve angular momentum.

Explore the model

Interactive lab

Interactive lab · vectors

A rotation generator is tangent to a circle

Pause and Reset, then set angle=0° and magnitude=2 to represent q=(2,0). Predict the counterclockwise infinitesimal generator (−q₂,q₁)=(0,2). For a fresh comparison Reset and set angle=90°,magnitude=2; the new q=(0,2) has generator (−2,0). Compare the displayed projections with the quarter-turn predictions. This one-vector instrument does not contain momenta or compute angular momentum.

A rotation generator is tangent to a circlePlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe geometric symmetry direction is a tangent vector; the momentum map supplies the Hamiltonian whose field equals that direction.

Check your understanding

A Hamiltonian group action has momentum map J. Which additional condition makes J_ξ conserved under the chosen autonomous H?

Try it yourself

Keep J=q₁p₂−q₂p₁ but change the energy to H=(p₁²+p₂²+q₁²+4q₂²)/2. Compute J̇ and evaluate it at q=(1,2).

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The momentum derivatives of H still cancel in the first bracket terms; inspect the unequal position coefficients.

Compare with the solution

Using J_q=(p₂,−p₁), J_p=(−q₂,q₁) and H_q=(q₁,4q₂), the bracket is J̇={J,H}=−(−q₂q₁+4q₁q₂)=−3q₁q₂. At q=(1,2), it is −6, independently of the momentum at that instant. The anisotropic potential breaks rotational invariance, so the same valid rotation momentum map no longer gives a conserved quantity for this H.

Take it with you

  • Momentum components generate infinitesimal symmetries with the module’s fixed Hamiltonian sign.
  • Conservation requires invariance of the chosen Hamiltonian.
  • Global existence, equivariance and regular reduction are separate requirements.

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