Two Classical Examples of Integrable Systems

Two classical examples of integrable systems

Florian Beck 37th day in quarantine

String Math Seminar (Summer 2020) Recap Equations of motion in Darboux coordinates (pi , qj ) ∂H ∂H q˙j = , p˙i = − . ∂pj ∂qi The Hamiltonian system ((M, ω), H) is called Liouville/completely integrable if there are n independent, Poisson commuting

conserved quantities F1,..., Fn, {H, Fj } = 0.

Liouville theorem: let Mf be a level set of n F := (F1,..., Fn): M → R .

In Darboux coordinates around m ∈ Mf deﬁne the action Z q X S(F , q) := pi (F , q)dqi , pi = pi (F , q). q0 i

Recap (I)

Phase space of a mechanical / Hamiltonian system is a symplectic 2n manifold (M , ω) with Hamiltonian H : M → R.

1 ∂H ∂H q˙j = , p˙i = − . ∂pj ∂qi The Hamiltonian system ((M, ω), H) is called Liouville/completely integrable if there are n independent, Poisson commuting

conserved quantities F1,..., Fn, {H, Fj } = 0.

Liouville theorem: let Mf be a level set of n F := (F1,..., Fn): M → R .

In Darboux coordinates around m ∈ Mf deﬁne the action Z q X S(F , q) := pi (F , q)dqi , pi = pi (F , q). q0 i

Recap (I)

Phase space of a mechanical / Hamiltonian system is a symplectic 2n manifold (M , ω) with Hamiltonian H : M → R.

Equations of motion in Darboux coordinates (pi , qj )

1 The Hamiltonian system ((M, ω), H) is called Liouville/completely integrable if there are n independent, Poisson commuting

conserved quantities F1,..., Fn, {H, Fj } = 0.

Liouville theorem: let Mf be a level set of n F := (F1,..., Fn): M → R .

In Darboux coordinates around m ∈ Mf deﬁne the action Z q X S(F , q) := pi (F , q)dqi , pi = pi (F , q). q0 i

Recap (I)

Phase space of a mechanical / Hamiltonian system is a symplectic 2n manifold (M , ω) with Hamiltonian H : M → R.

Equations of motion in Darboux coordinates (pi , qj ) ∂H ∂H q˙j = , p˙i = − . ∂pj ∂qi

1 Liouville theorem: let Mf be a level set of n F := (F1,..., Fn): M → R .

In Darboux coordinates around m ∈ Mf deﬁne the action Z q X S(F , q) := pi (F , q)dqi , pi = pi (F , q). q0 i

Recap (I)

Phase space of a mechanical / Hamiltonian system is a symplectic 2n manifold (M , ω) with Hamiltonian H : M → R.

Equations of motion in Darboux coordinates (pi , qj ) ∂H ∂H q˙j = , p˙i = − . ∂pj ∂qi The Hamiltonian system ((M, ω), H) is called Liouville/completely integrable if there are n independent, Poisson commuting

conserved quantities F1,..., Fn, {H, Fj } = 0.

1 In Darboux coordinates around m ∈ Mf deﬁne the action Z q X S(F , q) := pi (F , q)dqi , pi = pi (F , q). q0 i

Recap (I)

Phase space of a mechanical / Hamiltonian system is a symplectic 2n manifold (M , ω) with Hamiltonian H : M → R.

conserved quantities F1,..., Fn, {H, Fj } = 0.

Liouville theorem: let Mf be a level set of n F := (F1,..., Fn): M → R .

1 Recap (I)

Phase space of a mechanical / Hamiltonian system is a symplectic 2n manifold (M , ω) with Hamiltonian H : M → R.

conserved quantities F1,..., Fn, {H, Fj } = 0.

Liouville theorem: let Mf be a level set of n F := (F1,..., Fn): M → R .

In Darboux coordinates around m ∈ Mf deﬁne the action Z q X S(F , q) := pi (F , q)dqi , pi = pi (F , q). q0 i 1 F1,..., Fn, ψ1 := ∂S/∂F1, . . . , ψn := ∂S/∂Fn

in a neighborhood of m ∈ Mf . Equations of motion in these coordinates:

Recap (II)

Then the following are conjugate coordinates

2 Equations of motion in these coordinates:

Recap (II)

Then the following are conjugate coordinates

F1,..., Fn, ψ1 := ∂S/∂F1, . . . , ψn := ∂S/∂Fn

in a neighborhood of m ∈ Mf .

2 Hence completely integrable systems are solvable ‘by quadrature’. Plan for today: • Two classical examples (Kepler and Neumann problem) with direct application of Liouville theorem. • Separation of variables. • Example of Hamiltonian reduction. • Relation to algebraic curves.

Recap (II)

Then the following are conjugate coordinates

F1,..., Fn, ψ1 := ∂S/∂F1, . . . , ψn := ∂S/∂Fn

in a neighborhood of m ∈ Mf . Equations of motion in these coordinates:

F˙i = {H, Fi } = 0, ψ˙j = {H, ψj } =: Ωj , Ωj = Ωj (F ).

2 Plan for today: • Two classical examples (Kepler and Neumann problem) with direct application of Liouville theorem. • Separation of variables. • Example of Hamiltonian reduction. • Relation to algebraic curves.

Recap (II)

Then the following are conjugate coordinates

F1,..., Fn, ψ1 := ∂S/∂F1, . . . , ψn := ∂S/∂Fn

in a neighborhood of m ∈ Mf . Equations of motion in these coordinates:

F˙i = {H, Fi } = 0, ψ˙j = {H, ψj } =: Ωj , Ωj = Ωj (F ). Hence completely integrable systems are solvable ‘by quadrature’.

2 • Two classical examples (Kepler and Neumann problem) with direct application of Liouville theorem. • Separation of variables. • Example of Hamiltonian reduction. • Relation to algebraic curves.

Recap (II)

Then the following are conjugate coordinates

F1,..., Fn, ψ1 := ∂S/∂F1, . . . , ψn := ∂S/∂Fn

in a neighborhood of m ∈ Mf . Equations of motion in these coordinates:

F˙i = {H, Fi } = 0, ψ˙j = {H, ψj } =: Ωj , Ωj = Ωj (F ). Hence completely integrable systems are solvable ‘by quadrature’. Plan for today:

2 • Separation of variables. • Example of Hamiltonian reduction. • Relation to algebraic curves.

Recap (II)

Then the following are conjugate coordinates

F1,..., Fn, ψ1 := ∂S/∂F1, . . . , ψn := ∂S/∂Fn

in a neighborhood of m ∈ Mf . Equations of motion in these coordinates:

2 • Example of Hamiltonian reduction. • Relation to algebraic curves.

Recap (II)

Then the following are conjugate coordinates

F1,..., Fn, ψ1 := ∂S/∂F1, . . . , ψn := ∂S/∂Fn

in a neighborhood of m ∈ Mf . Equations of motion in these coordinates:

2 • Relation to algebraic curves.

Recap (II)

Then the following are conjugate coordinates

F1,..., Fn, ψ1 := ∂S/∂F1, . . . , ψn := ∂S/∂Fn

in a neighborhood of m ∈ Mf . Equations of motion in these coordinates:

2 Recap (II)

Then the following are conjugate coordinates

F1,..., Fn, ψ1 := ∂S/∂F1, . . . , ψn := ∂S/∂Fn

in a neighborhood of m ∈ Mf . Equations of motion in these coordinates:

F˙i = {H, Fi } = 0, ψ˙j = {H, ψj } =: Ωj , Ωj = Ωj (F ). Hence completely integrable systems are solvable ‘by quadrature’. Plan for today: • Two classical examples (Kepler and Neumann problem) with direct application of Liouville theorem. • Separation of variables. • Example of Hamiltonian reduction. • Relation to algebraic curves. 2 Kepler problem Historically, the ﬁrst (higher-dimensional) integrable system introduced by Kepler (∼ 1600) and ﬁrst solved by Newton (1687). Hamiltonian system: 3 6 X • (M := R , ω) where ω = dxi ∧ dyi i=1 (x1, x2, x3) ∈ M center of mass coordinates. • Hamiltionian: 3 1 X 2 H = 2 yi + V(r) , r = k(x1, x2, x3)k. i=1 In two-body problem: V (r) = C/r. For us: any potential dependent only on r.

Statement (I)

3 Motion of two (point) particles in R only interacting with each other.

3 Hamiltonian system: 3 6 X • (M := R , ω) where ω = dxi ∧ dyi i=1 (x1, x2, x3) ∈ M center of mass coordinates. • Hamiltionian: 3 1 X 2 H = 2 yi + V(r) , r = k(x1, x2, x3)k. i=1 In two-body problem: V (r) = C/r. For us: any potential dependent only on r.

Statement (I)

3 3 6 X • (M := R , ω) where ω = dxi ∧ dyi i=1 (x1, x2, x3) ∈ M center of mass coordinates. • Hamiltionian: 3 1 X 2 H = 2 yi + V(r) , r = k(x1, x2, x3)k. i=1 In two-body problem: V (r) = C/r. For us: any potential dependent only on r.

Statement (I)

3 (x1, x2, x3) ∈ M center of mass coordinates. • Hamiltionian: 3 1 X 2 H = 2 yi + V(r) , r = k(x1, x2, x3)k. i=1 In two-body problem: V (r) = C/r. For us: any potential dependent only on r.

Statement (I)

3 • Hamiltionian: 3 1 X 2 H = 2 yi + V(r) , r = k(x1, x2, x3)k. i=1 In two-body problem: V (r) = C/r. For us: any potential dependent only on r.

Statement (I)

3 In two-body problem: V (r) = C/r. For us: any potential dependent only on r.

Statement (I)

3 Motion of two (point) particles in R only interacting with each other. Historically, the ﬁrst (higher-dimensional) integrable system introduced by Kepler (∼ 1600) and ﬁrst solved by Newton (1687). Hamiltonian system: 3 6 X • (M := R , ω) where ω = dxi ∧ dyi i=1 (x1, x2, x3) ∈ M center of mass coordinates. • Hamiltionian: 3 1 X 2 H = 2 yi + V(r) , r = k(x1, x2, x3)k. i=1

3 For us: any potential dependent only on r.

Statement (I)

3 Statement (I)

3 The rotational symmetry corresponds to the conserved angular momentum

J = (J1, J2, J3), εijk Jk = Jij := xi yj − xj yi .

Note: compontents of J do not Poisson commute. Instead consider

2 2 2 2 H, J3 = J12, J := J12 + J23 + J13.

These are three independent Poisson commuting conserved quantities =⇒ completely integrable.

Statement (II)

∂V (r) Equations of motions:x ¨i = − . ∂xi

4 εijk Jk = Jij := xi yj − xj yi .

Note: compontents of J do not Poisson commute. Instead consider

2 2 2 2 H, J3 = J12, J := J12 + J23 + J13.

These are three independent Poisson commuting conserved quantities =⇒ completely integrable.

Statement (II)

∂V (r) Equations of motions:x ¨i = − . ∂xi The rotational symmetry corresponds to the conserved angular momentum

J = (J1, J2, J3),

4 Note: compontents of J do not Poisson commute. Instead consider

2 2 2 2 H, J3 = J12, J := J12 + J23 + J13.

These are three independent Poisson commuting conserved quantities =⇒ completely integrable.

Statement (II)

∂V (r) Equations of motions:x ¨i = − . ∂xi The rotational symmetry corresponds to the conserved angular momentum

J = (J1, J2, J3), εijk Jk = Jij := xi yj − xj yi .

4 Instead consider

2 2 2 2 H, J3 = J12, J := J12 + J23 + J13.

These are three independent Poisson commuting conserved quantities =⇒ completely integrable.

Statement (II)

∂V (r) Equations of motions:x ¨i = − . ∂xi The rotational symmetry corresponds to the conserved angular momentum

J = (J1, J2, J3), εijk Jk = Jij := xi yj − xj yi .

Note: compontents of J do not Poisson commute.

4 These are three independent Poisson commuting conserved quantities =⇒ completely integrable.

Statement (II)

∂V (r) Equations of motions:x ¨i = − . ∂xi The rotational symmetry corresponds to the conserved angular momentum

J = (J1, J2, J3), εijk Jk = Jij := xi yj − xj yi .

Note: compontents of J do not Poisson commute. Instead consider

2 2 2 2 H, J3 = J12, J := J12 + J23 + J13.

4 =⇒ completely integrable.

Statement (II)

∂V (r) Equations of motions:x ¨i = − . ∂xi The rotational symmetry corresponds to the conserved angular momentum

J = (J1, J2, J3), εijk Jk = Jij := xi yj − xj yi .

Note: compontents of J do not Poisson commute. Instead consider

2 2 2 2 H, J3 = J12, J := J12 + J23 + J13.

These are three independent Poisson commuting conserved quantities

4 Statement (II)

∂V (r) Equations of motions:x ¨i = − . ∂xi The rotational symmetry corresponds to the conserved angular momentum

J = (J1, J2, J3), εijk Jk = Jij := xi yj − xj yi .

Note: compontents of J do not Poisson commute. Instead consider

2 2 2 2 H, J3 = J12, J := J12 + J23 + J13.

These are three independent Poisson commuting conserved quantities =⇒ completely integrable.

4 • canonical one-form α = pr dr + pθdθ + pϕdϕ • conserved quantities: H = 1 p2 + 1 p2 + 1 p2 + V (r), 2 r r 2 θ r 2 sin2 θ ϕ J2 = p2 + 1 p2 , θ sin2 θ ϕ

J3 = pϕ. 2 3 Fix a (regular) value f of (H, J , J3): M → R . On Mf , we can separate variables:

pr = pr (r), pθ = pθ(θ), pϕ = pϕ(ϕ). (1)

Then (r, θ, ϕ) are called separated variables on Mf . In particular, α is closed along Mf .

Solution (I)

Work in spherical coordinates (r, θ, ϕ):

5 • conserved quantities: H = 1 p2 + 1 p2 + 1 p2 + V (r), 2 r r 2 θ r 2 sin2 θ ϕ J2 = p2 + 1 p2 , θ sin2 θ ϕ

J3 = pϕ. 2 3 Fix a (regular) value f of (H, J , J3): M → R . On Mf , we can separate variables:

pr = pr (r), pθ = pθ(θ), pϕ = pϕ(ϕ). (1)

Then (r, θ, ϕ) are called separated variables on Mf . In particular, α is closed along Mf .

Solution (I)

Work in spherical coordinates (r, θ, ϕ):

• canonical one-form α = pr dr + pθdθ + pϕdϕ

5 2 3 Fix a (regular) value f of (H, J , J3): M → R . On Mf , we can separate variables:

pr = pr (r), pθ = pθ(θ), pϕ = pϕ(ϕ). (1)

Then (r, θ, ϕ) are called separated variables on Mf . In particular, α is closed along Mf .

Solution (I)

Work in spherical coordinates (r, θ, ϕ):

• canonical one-form α = pr dr + pθdθ + pϕdϕ • conserved quantities: H = 1 p2 + 1 p2 + 1 p2 + V (r), 2 r r 2 θ r 2 sin2 θ ϕ J2 = p2 + 1 p2 , θ sin2 θ ϕ

J3 = pϕ.

5 On Mf , we can separate variables:

pr = pr (r), pθ = pθ(θ), pϕ = pϕ(ϕ). (1)

Then (r, θ, ϕ) are called separated variables on Mf . In particular, α is closed along Mf .

Solution (I)

Work in spherical coordinates (r, θ, ϕ):

• canonical one-form α = pr dr + pθdθ + pϕdϕ • conserved quantities: H = 1 p2 + 1 p2 + 1 p2 + V (r), 2 r r 2 θ r 2 sin2 θ ϕ J2 = p2 + 1 p2 , θ sin2 θ ϕ

J3 = pϕ. 2 3 Fix a (regular) value f of (H, J , J3): M → R .

5 pr = pr (r), pθ = pθ(θ), pϕ = pϕ(ϕ). (1)

Then (r, θ, ϕ) are called separated variables on Mf . In particular, α is closed along Mf .

Solution (I)

Work in spherical coordinates (r, θ, ϕ):

• canonical one-form α = pr dr + pθdθ + pϕdϕ • conserved quantities: H = 1 p2 + 1 p2 + 1 p2 + V (r), 2 r r 2 θ r 2 sin2 θ ϕ J2 = p2 + 1 p2 , θ sin2 θ ϕ

J3 = pϕ. 2 3 Fix a (regular) value f of (H, J , J3): M → R . On Mf , we can separate variables:

5 Then (r, θ, ϕ) are called separated variables on Mf . In particular, α is closed along Mf .

Solution (I)

Work in spherical coordinates (r, θ, ϕ):

• canonical one-form α = pr dr + pθdθ + pϕdϕ • conserved quantities: H = 1 p2 + 1 p2 + 1 p2 + V (r), 2 r r 2 θ r 2 sin2 θ ϕ J2 = p2 + 1 p2 , θ sin2 θ ϕ

J3 = pϕ. 2 3 Fix a (regular) value f of (H, J , J3): M → R . On Mf , we can separate variables:

pr = pr (r), pθ = pθ(θ), pϕ = pϕ(ϕ). (1)

5 In particular, α

is closed along Mf .

Solution (I)

Work in spherical coordinates (r, θ, ϕ):

• canonical one-form α = pr dr + pθdθ + pϕdϕ • conserved quantities: H = 1 p2 + 1 p2 + 1 p2 + V (r), 2 r r 2 θ r 2 sin2 θ ϕ J2 = p2 + 1 p2 , θ sin2 θ ϕ

J3 = pϕ. 2 3 Fix a (regular) value f of (H, J , J3): M → R . On Mf , we can separate variables:

pr = pr (r), pθ = pθ(θ), pϕ = pϕ(ϕ). (1)

Then (r, θ, ϕ) are called separated variables on Mf .

5 Solution (I)

Work in spherical coordinates (r, θ, ϕ):

• canonical one-form α = pr dr + pθdθ + pϕdϕ • conserved quantities: H = 1 p2 + 1 p2 + 1 p2 + V (r), 2 r r 2 θ r 2 sin2 θ ϕ J2 = p2 + 1 p2 , θ sin2 θ ϕ

J3 = pϕ. 2 3 Fix a (regular) value f of (H, J , J3): M → R . On Mf , we can separate variables:

pr = pr (r), pθ = pθ(θ), pϕ = pϕ(ϕ). (1)

Then (r, θ, ϕ) are called separated variables on Mf . In particular, α is closed along Mf .

5 Z r Z θ Z ϕ 0 0 0 S(r, θ, ϕ) = pr dr + pθ dθ + pϕ dϕ . r0 θ0 ϕ0 2 The conjugate coordinates ψH = ∂S/∂H, ψJ2 = ∂S/∂J ,

ψJ3 = ∂S/∂J3 satisfy ˙ ˙ ˙ ψH = 1, ψJ2 = 0, ψJ3 = 0. (2)

HOMEWORK: check these equations.

Hence we obtain the standard solution( ψH (t0) = 0) : Z r dr 0 t − t0 = √ . 2(H−V (r 0))−J2/r 02) r0

Solution (II)

Let us apply Liouville’s theorem and deﬁne the action

6 2 The conjugate coordinates ψH = ∂S/∂H, ψJ2 = ∂S/∂J ,

ψJ3 = ∂S/∂J3 satisfy ˙ ˙ ˙ ψH = 1, ψJ2 = 0, ψJ3 = 0. (2)

HOMEWORK: check these equations.

Hence we obtain the standard solution( ψH (t0) = 0) : Z r dr 0 t − t0 = √ . 2(H−V (r 0))−J2/r 02) r0

Solution (II)

Let us apply Liouville’s theorem and deﬁne the action

Z r Z θ Z ϕ 0 0 0 S(r, θ, ϕ) = pr dr + pθ dθ + pϕ dϕ . r0 θ0 ϕ0

6 ˙ ˙ ˙ ψH = 1, ψJ2 = 0, ψJ3 = 0. (2)

HOMEWORK: check these equations.

Hence we obtain the standard solution( ψH (t0) = 0) : Z r dr 0 t − t0 = √ . 2(H−V (r 0))−J2/r 02) r0

Solution (II)

Let us apply Liouville’s theorem and deﬁne the action

Z r Z θ Z ϕ 0 0 0 S(r, θ, ϕ) = pr dr + pθ dθ + pϕ dϕ . r0 θ0 ϕ0 2 The conjugate coordinates ψH = ∂S/∂H, ψJ2 = ∂S/∂J ,

ψJ3 = ∂S/∂J3 satisfy

6 HOMEWORK: check these equations.

Hence we obtain the standard solution( ψH (t0) = 0) : Z r dr 0 t − t0 = √ . 2(H−V (r 0))−J2/r 02) r0

Solution (II)

Let us apply Liouville’s theorem and deﬁne the action

Z r Z θ Z ϕ 0 0 0 S(r, θ, ϕ) = pr dr + pθ dθ + pϕ dϕ . r0 θ0 ϕ0 2 The conjugate coordinates ψH = ∂S/∂H, ψJ2 = ∂S/∂J ,

ψJ3 = ∂S/∂J3 satisfy ˙ ˙ ˙ ψH = 1, ψJ2 = 0, ψJ3 = 0. (2)

6 Solution (II)

Let us apply Liouville’s theorem and deﬁne the action

Z r Z θ Z ϕ 0 0 0 S(r, θ, ϕ) = pr dr + pθ dθ + pϕ dϕ . r0 θ0 ϕ0 2 The conjugate coordinates ψH = ∂S/∂H, ψJ2 = ∂S/∂J ,

ψJ3 = ∂S/∂J3 satisfy ˙ ˙ ˙ ψH = 1, ψJ2 = 0, ψJ3 = 0. (2)

HOMEWORK: check these equations.

Hence we obtain the standard solution( ψH (t0) = 0) : Z r dr 0 t − t0 = √ . 2(H−V (r 0))−J2/r 02) r0

6 Neumann problem • harmonic forces (‘harmonic oscillators’) in each direction with frequencies a1 < ··· < aN , ai ∈ R, • motion is constrained to the sphere N N−1 N X 2 S := {(x1,..., xN ) ∈ R : xi = 1}. i=1 Resulting Lagrangian

N  N  X 1 2 2 1 X 2 L(x, x˙) = 2 (x ˙i − ai xi ) + 2 Λ  xj − 1 i=1 j=1

• harmonic oscillators • Lagrange multiplier

Lagrangian statement (I)

N Motion of a particle in R with the following constraints:

7 • motion is constrained to the sphere N N−1 N X 2 S := {(x1,..., xN ) ∈ R : xi = 1}. i=1 Resulting Lagrangian

N  N  X 1 2 2 1 X 2 L(x, x˙) = 2 (x ˙i − ai xi ) + 2 Λ  xj − 1 i=1 j=1

• harmonic oscillators • Lagrange multiplier

Lagrangian statement (I)

N Motion of a particle in R with the following constraints: • harmonic forces (‘harmonic oscillators’) in each direction with frequencies a1 < ··· < aN , ai ∈ R,

7 Resulting Lagrangian

N  N  X 1 2 2 1 X 2 L(x, x˙) = 2 (x ˙i − ai xi ) + 2 Λ  xj − 1 i=1 j=1

• harmonic oscillators • Lagrange multiplier

Lagrangian statement (I)

N Motion of a particle in R with the following constraints: • harmonic forces (‘harmonic oscillators’) in each direction with frequencies a1 < ··· < aN , ai ∈ R, • motion is constrained to the sphere N N−1 N X 2 S := {(x1,..., xN ) ∈ R : xi = 1}. i=1

7 • harmonic oscillators • Lagrange multiplier

Lagrangian statement (I)

N  N  X 1 2 2 1 X 2 L(x, x˙) = 2 (x ˙i − ai xi ) + 2 Λ  xj − 1 i=1 j=1

7 • Lagrange multiplier

Lagrangian statement (I)

N  N  X 1 2 2 1 X 2 L(x, x˙) = 2 (x ˙i − ai xi ) + 2 Λ  xj − 1 i=1 j=1

• harmonic oscillators

7 Lagrangian statement (I)

N  N  X 1 2 2 1 X 2 L(x, x˙) = 2 (x ˙i − ai xi ) + 2 Λ  xj − 1 i=1 j=1

• harmonic oscillators • Lagrange multiplier 7 Lagrangian statement (I)

N  N  X 1 2 2 1 X 2 L(x, x˙) = 2 (x ˙i − ai xi ) + 2 Λ  xj − 1 i=1 j=1

• harmonic oscillators • Lagrange multiplier 7 In our example, they are equivalent to N X 2 x¨i = (−ai + Λ)xi , xi = 1. (3) i=1

HOMEWORK: let

L0 := diag(a1,..., aN ), t X := (x1,..., xN ) , t Y := (y1,..., yN ) , N h•, •i : standard Euclidean product on R . Then (3) are equivalent to 2 X¨ = −L0X − kX˙ k − hX , L0X i X

Lagrangian statement (II)

Euler–Lagrange equations: ∂L − d ∂L = 0. ∂xi dt ∂x˙i

8 HOMEWORK: let

L0 := diag(a1,..., aN ), t X := (x1,..., xN ) , t Y := (y1,..., yN ) , N h•, •i : standard Euclidean product on R . Then (3) are equivalent to 2 X¨ = −L0X − kX˙ k − hX , L0X i X

Lagrangian statement (II)

Euler–Lagrange equations: ∂L − d ∂L = 0. ∂xi dt ∂x˙i In our example, they are equivalent to N X 2 x¨i = (−ai + Λ)xi , xi = 1. (3) i=1

8 Then (3) are equivalent to 2 X¨ = −L0X − kX˙ k − hX , L0X i X

Lagrangian statement (II)

Euler–Lagrange equations: ∂L − d ∂L = 0. ∂xi dt ∂x˙i In our example, they are equivalent to N X 2 x¨i = (−ai + Λ)xi , xi = 1. (3) i=1

HOMEWORK: let

L0 := diag(a1,..., aN ), t X := (x1,..., xN ) , t Y := (y1,..., yN ) , N h•, •i : standard Euclidean product on R .

8 2 X¨ = −L0X − kX˙ k − hX , L0X i X

Lagrangian statement (II)

Euler–Lagrange equations: ∂L − d ∂L = 0. ∂xi dt ∂x˙i In our example, they are equivalent to N X 2 x¨i = (−ai + Λ)xi , xi = 1. (3) i=1

HOMEWORK: let

L0 := diag(a1,..., aN ), t X := (x1,..., xN ) , t Y := (y1,..., yN ) , N h•, •i : standard Euclidean product on R . Then (3) are equivalent to

8 Lagrangian statement (II)

Euler–Lagrange equations: ∂L − d ∂L = 0. ∂xi dt ∂x˙i In our example, they are equivalent to N X 2 x¨i = (−ai + Λ)xi , xi = 1. (3) i=1

HOMEWORK: let

L0 := diag(a1,..., aN ), t X := (x1,..., xN ) , t Y := (y1,..., yN ) , N h•, •i : standard Euclidean product on R . Then (3) are equivalent to 2 X¨ = −L0X − kX˙ k − hX , L0X i X 8 Angular momentum:

Jkl = xk yl − xl yk , J := (Jkl ). Hamiltonian N N 1 X 2 1 X 2 H := 4 Jkl + 2 ak xk . k,l k=1 Hamiltonian equations ˙ ∂H ˙ ∂H X = ∂Y = −JX , Y = − ∂X = −JY − L0X .

Hamiltonian statement

2N Let (M := R , ω) be the standard symplectic manifold with 2N X ω = dxi ∧ dyi . i=1

9 Hamiltonian N N 1 X 2 1 X 2 H := 4 Jkl + 2 ak xk . k,l k=1 Hamiltonian equations ˙ ∂H ˙ ∂H X = ∂Y = −JX , Y = − ∂X = −JY − L0X .

Hamiltonian statement

2N Let (M := R , ω) be the standard symplectic manifold with 2N X ω = dxi ∧ dyi . i=1 Angular momentum:

Jkl = xk yl − xl yk , J := (Jkl ).

9 Hamiltonian equations ˙ ∂H ˙ ∂H X = ∂Y = −JX , Y = − ∂X = −JY − L0X .

Hamiltonian statement

2N Let (M := R , ω) be the standard symplectic manifold with 2N X ω = dxi ∧ dyi . i=1 Angular momentum:

Jkl = xk yl − xl yk , J := (Jkl ). Hamiltonian N N 1 X 2 1 X 2 H := 4 Jkl + 2 ak xk . k,l k=1

9 ˙ ∂H = −JX , Y = − ∂X = −JY − L0X .

Hamiltonian statement

2N Let (M := R , ω) be the standard symplectic manifold with 2N X ω = dxi ∧ dyi . i=1 Angular momentum:

Jkl = xk yl − xl yk , J := (Jkl ). Hamiltonian N N 1 X 2 1 X 2 H := 4 Jkl + 2 ak xk . k,l k=1 Hamiltonian equations ˙ ∂H X = ∂Y

9 ˙ ∂H Y = − ∂X = −JY − L0X .

Hamiltonian statement

2N Let (M := R , ω) be the standard symplectic manifold with 2N X ω = dxi ∧ dyi . i=1 Angular momentum:

Jkl = xk yl − xl yk , J := (Jkl ). Hamiltonian N N 1 X 2 1 X 2 H := 4 Jkl + 2 ak xk . k,l k=1 Hamiltonian equations ˙ ∂H X = ∂Y = −JX ,

9 = −JY − L0X .

Hamiltonian statement

2N Let (M := R , ω) be the standard symplectic manifold with 2N X ω = dxi ∧ dyi . i=1 Angular momentum:

Jkl = xk yl − xl yk , J := (Jkl ). Hamiltonian N N 1 X 2 1 X 2 H := 4 Jkl + 2 ak xk . k,l k=1 Hamiltonian equations ˙ ∂H ˙ ∂H X = ∂Y = −JX , Y = − ∂X

9 Hamiltonian statement

2N Let (M := R , ω) be the standard symplectic manifold with 2N X ω = dxi ∧ dyi . i=1 Angular momentum:

Jkl = xk yl − xl yk , J := (Jkl ). Hamiltonian N N 1 X 2 1 X 2 H := 4 Jkl + 2 ak xk . k,l k=1 Hamiltonian equations ˙ ∂H ˙ ∂H X = ∂Y = −JX , Y = − ∂X = −JY − L0X .

9 • Lagrangian picture: Lagrange multiplier. • Hamiltonian picture: Hamiltonian reduction / Noether’s theorem. In our example:

• the group (R, +) acts on M via (X , Y ) 7→ (X , Y + λX ), λ ∈ R. • To this symmetry corresponds the conserved quantity N 1 X 2 1 2 µ: M → R, µ(X ) = 2 xi = 2 kX k , i=1 i.e. the value of the moment map µ remains constant along the trajectories of the Hamiltonian ﬂow.

Hamiltonian reduction (I)

How to incorporate the constraint kX k = 1 along trajectories?

10 Lagrange multiplier. • Hamiltonian picture: Hamiltonian reduction / Noether’s theorem. In our example:

Hamiltonian reduction (I)

How to incorporate the constraint kX k = 1 along trajectories? • Lagrangian picture:

10 • Hamiltonian picture: Hamiltonian reduction / Noether’s theorem. In our example:

Hamiltonian reduction (I)

How to incorporate the constraint kX k = 1 along trajectories? • Lagrangian picture: Lagrange multiplier.

10 Hamiltonian reduction / Noether’s theorem. In our example:

Hamiltonian reduction (I)

How to incorporate the constraint kX k = 1 along trajectories? • Lagrangian picture: Lagrange multiplier. • Hamiltonian picture:

10 In our example:

Hamiltonian reduction (I)

How to incorporate the constraint kX k = 1 along trajectories? • Lagrangian picture: Lagrange multiplier. • Hamiltonian picture: Hamiltonian reduction / Noether’s theorem.

10 In our example:

Hamiltonian reduction (I)

How to incorporate the constraint kX k = 1 along trajectories? • Lagrangian picture: Lagrange multiplier. • Hamiltonian picture: Hamiltonian reduction / Noether’s theorem.

10 • the group (R, +) acts on M via (X , Y ) 7→ (X , Y + λX ), λ ∈ R. • To this symmetry corresponds the conserved quantity N 1 X 2 1 2 µ: M → R, µ(X ) = 2 xi = 2 kX k , i=1 i.e. the value of the moment map µ remains constant along the trajectories of the Hamiltonian ﬂow.

Hamiltonian reduction (I)

How to incorporate the constraint kX k = 1 along trajectories? • Lagrangian picture: Lagrange multiplier. • Hamiltonian picture: Hamiltonian reduction / Noether’s theorem. In our example:

10 • To this symmetry corresponds the conserved quantity N 1 X 2 1 2 µ: M → R, µ(X ) = 2 xi = 2 kX k , i=1 i.e. the value of the moment map µ remains constant along the trajectories of the Hamiltonian ﬂow.

Hamiltonian reduction (I)

How to incorporate the constraint kX k = 1 along trajectories? • Lagrangian picture: Lagrange multiplier. • Hamiltonian picture: Hamiltonian reduction / Noether’s theorem. In our example:

• the group (R, +) acts on M via (X , Y ) 7→ (X , Y + λX ), λ ∈ R.

10 i.e. the value of the moment map µ remains constant along the trajectories of the Hamiltonian ﬂow.

Hamiltonian reduction (I)

How to incorporate the constraint kX k = 1 along trajectories? • Lagrangian picture: Lagrange multiplier. • Hamiltonian picture: Hamiltonian reduction / Noether’s theorem. In our example:

• the group (R, +) acts on M via (X , Y ) 7→ (X , Y + λX ), λ ∈ R. • To this symmetry corresponds the conserved quantity N 1 X 2 1 2 µ: M → R, µ(X ) = 2 xi = 2 kX k , i=1

10 Hamiltonian reduction (I)

How to incorporate the constraint kX k = 1 along trajectories? • Lagrangian picture: Lagrange multiplier. • Hamiltonian picture: Hamiltonian reduction / Noether’s theorem. In our example:

the trajectories of the Hamiltonian ﬂow. 10 −1 1 2 N := µ ( 2 ) = {(X , Y ): kX k = 1} ⊂ M. Caveat: not a symplectic manifold. But: preserved by the full symmetry group R. To get the equations of motion we need to consider ∼ 2 N/R = {(X , Y ): kX k = 1, hX , Y i = 0 }

(non-canonical) gauge condition

Hamiltonian reduction: the symplectic form ω descends to a symplectic formω ˆ on N/R. What are the equations of motion on N/R?

Hamiltonian reduction (II)