advanced31 min studyLesson 445 in Mathematics

Lagrangian submanifolds and action–angle coordinates

Recognize half-dimensional zero-pairing geometry and build action–angle coordinates for a single regular oscillator.

On this page

What you’ll understand

  • Test whether a submanifold is Lagrangian using dimension and restricted form.
  • Relate Lagrangian graphs to closed one-forms and generating potentials.
  • Compute an oscillator action with a consistent angular orientation.

A Lagrangian submanifold is a maximal-dimensional place where the symplectic pairing vanishes on tangent directions. This condition does not make the ambient form degenerate; it restricts attention to a carefully chosen half-dimensional slice. Graphs of generating potentials, zero sections of cotangent bundles and regular invariant tori provide important examples. Action–angle coordinates organize the latter into phase and conserved magnitude. We first construct them explicitly for one oscillator, leaving the multi-integral existence theorem and its compactness hypotheses for the later Liouville-torus lesson.

Zero pairing needs the correct dimension

In a symplectic vector space of dimension 2n, a subspace is isotropic when the form vanishes on every pair of its vectors. Such a subspace has dimension at most n. It is Lagrangian when its dimension is exactly n, equivalently when it equals its symplectic orthogonal. A smooth embedded submanifold L is Lagrangian when every tangent space has this property, or when dim L=n and the pulled-back form is zero. Checking only the zero pullback is insufficient: a curve in a four-dimensional phase space is automatically isotropic but has too small a dimension to be Lagrangian.

In T*Q, the zero section and each fiber are Lagrangian by direct canonical-coordinate calculation. For the graph of a one-form β, pullback of θ is β and pullback of ω is −dβ. Thus the graph is Lagrangian exactly when β is closed. Locally a closed β is dS, so a local scalar potential describes the graph. Globally, closedness does not imply one single-valued potential; the circle translation example persists. A Lagrangian condition is a geometric restriction, while exactness adds global information about one-form periods.

dim⁡L=n,ιL∗ω=0⟺L Lagrangian,(graph⁡β)∗ω=−dβ\dim L=n,\quad\iota_L^*\omega=0\quad\Longleftrightarrow\quad L\text{ Lagrangian},\qquad (\operatorname{graph}\beta)^*\omega=-d\beta
The ambient manifold has dimension 2n. A zero restricted two-form and the half-dimension requirement must both be checked.↔ Scroll wide equations horizontally.

Action is an oriented loop integral

For H=(q²+p²)/2 on the punctured phase plane, write q=√(2I)sin θ and p=√(2I)cos θ, with I>0 and θ modulo 2π. Direct differentiation gives dq∧dp=dθ∧dI. This angular orientation matters: our Hamilton equations give θ̇=1 and Ī=0, matching q̇=p,ṗ=−q. The action is one over 2π times the integral of θ_canonical=p dq around the orbit in the direction of the flow. The same symbol θ often names both a one-form and an angle, so we identify which meaning is intended in each formula.

For the oscillator, the integral is the area π times the squared radius, equal to 2πI, so the normalized action is I=H. The angle coordinate is periodic rather than a global real number, and the transformation is singular at I=0 where the whole circle collapses to an equilibrium. Consequently no action–angle chart of this type covers the origin. The energy circle is a one-dimensional Lagrangian submanifold of the plane, but its neighboring circles are needed to supply the full two-dimensional coordinate system.

q=2Isin⁡ϑ,p=2Icos⁡ϑ,ω=dϑ∧dI,I=12π∮p dq,ϑ˙=∂IHq=\sqrt{2I}\sin\vartheta,\quad p=\sqrt{2I}\cos\vartheta,\quad\omega=d\vartheta\wedge dI,\quad I=\frac1{2\pi}\oint p\,dq,\quad\dot\vartheta=\partial_IH
The angle ϑ is periodic with period 2π. This choice makes the action positive along our clockwise Hamiltonian oscillator orbit.↔ Scroll wide equations horizontally.

Lagrangian geometry is not automatically integrability

A single Lagrangian submanifold need not be invariant under a given Hamiltonian. Invariance requires X_H to be tangent; for a connected Lagrangian L this is equivalent to H restricted to L being constant. The equivalence follows because ω(X_H,v)=dH(v) and the symplectic orthogonal of TL equals TL. Even an invariant Lagrangian graph does not automatically provide a neighborhood filled by compact invariant tori. The action–angle theorem requires enough commuting independent functions plus a compact connected regular joint fiber.

Lagrangian graphs also appear in Hamilton–Jacobi theory: a potential S solving H(q,dS)=E makes its graph lie in one energy level, and hence invariant wherever it is a smooth Lagrangian graph. Caustics can make projection to configuration space fail, although the phase-space submanifold remains smooth. A single generating potential may then require several charts. This explains why a geometrically simple phase-space object can acquire multivalued momentum branches when viewed only through position coordinates.

A coupled momentum graph

In T*R², consider p₁=2q₁+q₂ and p₂=q₁+4q₂. Show the graph is Lagrangian and find a generating potential.

  1. The graph has dimension two inside four-dimensional phase space. Its momentum Jacobian is [[2,1],[1,4]], which is symmetric.

  2. Pull back ω: dq₁∧(2dq₁+dq₂)+dq₂∧(dq₁+4dq₂)=dq₁∧dq₂+dq₂∧dq₁=0.

  3. Integrating p₁ in q₁ gives S=q₁²+q₁q₂+g(q₂). Matching S_{q₂}=q₁+4q₂ gives g=2q₂² plus an arbitrary constant.

The result

The graph is the exact Lagrangian graph of dS for S=q₁²+q₁q₂+2q₂². Symmetric cross derivatives are the essential compatibility check.

Explore the model

Interactive lab

Interactive lab · oscillator

Measure action from an undamped orbit

Pause and Reset, then set damping ratio ζ=0 and natural frequency ω₀=1. Predict H=(position²+velocity²)/2=0.5 and action I=0.5. Step to several times and verify that value. For a fresh comparison Reset and set ζ=0,ω₀=2; the physical Hamiltonian is p²/2+2q², with energy 2 and action E/ω₀=1. This is a fixed unit-mass oscillator, not a general action-integral calculator.

Measure action from an undamped orbitUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. 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 forFor a harmonic oscillator the action equals energy divided by angular frequency, and damping must be absent for invariant energy loops.

Check your understanding

A curve in a four-dimensional symplectic manifold has zero restricted two-form. What follows?

Try it yourself

For the graph p₁=−q₂,p₂=q₁ in T*R², compute the restricted symplectic form and decide whether a local potential S with p=dS can exist.

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

Give me a hint

Compare the two mixed derivatives, or substitute dp₁=−dq₂ and dp₂=dq₁ into the canonical form.

Compare with the solution

The restricted form is −dq₁∧dq₂+dq₂∧dq₁=−2dq₁∧dq₂, which is nonzero. Although the graph has the required half-dimension, it is not Lagrangian. A potential would require S_{q₁q₂}=−1 and S_{q₂q₁}=1, contradicting equality of mixed derivatives for a smooth S. Thus no local generating potential exists on an open configuration neighborhood.

Take it with you

  • A Lagrangian submanifold is isotropic and exactly half-dimensional.
  • Closed covector graphs are Lagrangian; global potentials require exactness.
  • Action–angle coordinates need regular noncollapsed loops and a consistent orientation.

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