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이학석사 학위논문

Hamiltonian mechanics andSymmetries

(해밀턴 역학과 대칭)

2014년 8월

서울대학교 대학원

수리과학부

김 성 찬


(해밀턴 역학과 대칭)

지도교수 Urs Frauenfelder

이 논문을 이학석사 학위논문으로 제출함

2014년 4월

서울대학교 대학원

수리과학부

김 성 찬

김 성 찬의 이학석사 학위논문을 인준함

2014년 6월

위 원 장 (인)

부 위 원 장 (인)

위 원 (인)


A dissertation

submitted in partial fulfillment

of the requirements for the degree of

Master of Science

to the faculty of the Graduate School ofSeoul National University

by

Seongchan Kim

Dissertation Director : Professor Urs Frauenfelder

Department of Mathematical SciencesSeoul National University

August 2014

c© 2014 Seongchan Kim

All rights reserved.

Abstract

The aim of this thesis is to understand the relation of symmetries and inte-

grals of Hamiltonian system and their importance. We also treat an amazing

structure of an integrable system and moment maps to investigate equiva-

lent Hamiltonian systems. Through the paper, we study the Kepler problem

in detail as the example to see how these concepts work in a concrete sys-

tem. We show that the Kepler problem can be figured out by the angular

momentum of unit-speed geodesic circles on the unit-sphere.

Key words: Hamiltonian system, Integrable system, Kepler problem, An-

gular momentum, Moment map

Student Number: 2012-20245

i

Contents

Abstract i

1 Preliminaries on symplectic geometry 1

1.1 Symplectic manifolds . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Hamiltonian mechanics . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Noether theorem . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Invariant tori 11

2.1 Arnold-Liouville-Jost theorem . . . . . . . . . . . . . . . . . . 11

2.2 Mischenko-Fomenko theorem . . . . . . . . . . . . . . . . . . . 13

3 Kepler problem 15

3.1 Delaunay variables . . . . . . . . . . . . . . . . . . . . . . . . 15

3.2 The Laplace-Runge-Lenz vector . . . . . . . . . . . . . . . . . 17

3.3 Moser regularization . . . . . . . . . . . . . . . . . . . . . . . 22

3.4 Integrals after regularization . . . . . . . . . . . . . . . . . . . 24

3.5 Lie algebra structure . . . . . . . . . . . . . . . . . . . . . . . 27

4 Equivalent Hamiltonian systems 29

4.1 Group action . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.1.1 Lie group action . . . . . . . . . . . . . . . . . . . . . . 29

4.1.2 Adjoint and Coadjoint representations . . . . . . . . . 30

4.2 Moment maps . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

Abstract (in Korean) 41

ii

Chapter 1

Preliminaries on symplectic

geometry

In Hamiltonian mechanics, we treat a phase space, which carries a canonical

symplectic structure. Symplectic geometry is the mathematical theory which

plays a crucial role in the formulation of Hamiltonian mechanics. We start

this thesis with the following basic notions of symplectic geometry.

Throughout this paper, we assume that M is a 2n-dimensional smooth

closed manifold.

1.1 Symplectic manifolds

Definition 1.1.1. A symplectic structure on M is a nondegenerate closed

2-form ω ∈ Ω2(M). A pair (M,ω) is called a symplectic manifold.

It follows from nondegeneracy of ω that n-fold wedge product ω ∧ · · · ∧ ωis a nonvanishing 2n-form and hence it defines a volume form on M . There-

fore, every symplectic manifold is orientable.

Example 1.1.2. The easiest example of a symplectic manifold is an Eu-

clidean space. Let M = R2n with coordinates q1, · · · , qn, p1, · · · , pn. Consider

a closed 2-form

ω0 =n∑j=1

dqj ∧ dpj.

1

CHAPTER 1. PRELIMINARIES ON SYMPLECTIC GEOMETRY

The n-fold wedge product is given by

ω0 ∧ · · · ∧ ω0 = dq1 ∧ dp1 ∧ · · · ∧ dqn ∧ dpn,

which is a nonvanishing 2n form on R2n. Therefore, (R2n, ω0) is a symplectic

manifold.

Lemma 1.1.3. The cotangent bundle T ∗M of a smooth manifold Mn carries

a canonical symplectic structure.

Proof. Consider the natural projection

π : T ∗M →M

(x, v∗) 7→ x.

The differential of π at (x, v∗) is the linear map dπ(x,v∗) : T(x,v∗)T∗M → TxM.

Define the canonical 1-form λ on T ∗M by

λ(x,v∗) = v∗x dπ(x,v∗) : T(x,v∗)T∗M → R.

Let (T ∗U, q1, · · · qn, p1, · · · , pn) be a local chart on T ∗M , where (U, q1, · · · , qn)

is a local chart on M . Then on T ∗U , we obtain

v∗x dπ(x,v∗)(ξ, η) = v∗(ξ) =∑j

pjξj =∑j

pjdqj(ξ, η),

for (ξ, η) ∈ T ∗U , i.e., locally the canonical 1-form λ is given by∑

j pjdqj.

Thus we have a symplectic form ω = −dλ on T ∗M which is given locally by

ω =∑n

j=1 dqj ∧ dpj. This proves the Lemma.

Definition 1.1.4. A diffeomorphism ψ : (M1, ω1)→ (M2, ω2) between two

symplectic manifolds is called a symplectomorphism if it preserves the sym-

plectic form, i.e.,

ψ∗ω2 = ω1.

In studying symplectic geometry, it is important to study invariants un-

der symplectomorphisms. The following theorem shows that dimension is

2


the only local invariant of symplectic manifolds up to symplectomorphisms,

which is a contrast to Riemannian geometry where the curvature is such a

local invariant.

Theorem 1.1.5. (Darboux) Let (M,ω) be a 2n-dimensional symplectic man-

ifold, and let x be any point in M . Then there exists a local coordinate chart

(U, q1, · · · , qn, p1, · · · , pn) centered at x such that on U

ω =n∑j=1

dqj ∧ dpj.

A chart (U, q1, · · · , qn, p1, · · · , pn) is called a Darboux chart.

We conclude that every symplectic manifold (M,ω) is locally symplec-

tomorphic to (R2n, ω0). Thus in order to construct invariants of symplectic

manifolds, one has to go beyond local considerations.

1.2 Hamiltonian mechanics

Because of nondegeneracy of ω, we can identify the tangent bundle TM with

the cotangent bundle T ∗M via a map

TM → T ∗M

X 7→ ιXω

Then for any smooth function H on M , we obtain a unique vector field XH

which is defined by

ιXHω = dH.

Such a vector field is called a Hamiltonian vector field associated with a

Hamiltonian H. A pair (M,ω,H) is called a Hamiltonian system. A flow φtHof XH is called a Hamiltonian flow, i.e.,

dφtHdt

= XH φtH , φ0H = idM .

Lemma 1.2.1. (Conservation of Energy) Let p ∈M . ThenH φtH(p) = H(p)

for all t.

3


Proof.d

dtH(φtH(p)) = dH(φtH(p))

d

dtφtH(p)

= dH(φtH(p))XH(φtH(p))

= ωφtH(p)(XH , XH) = 0,

i.e., H φtH(p) is time-independent. It follows that

H φtH(p) = H φ0H(p) = H(p).

The previous lemma implies that the Hamiltonian flow moves along an

energy hypersurface. In other words, a Hamiltonian vector field XH is tangent

to the level surfaces of H.

Lemma 1.2.2. The Hamiltonian flow φtH is a symplectomorphism, i.e.,

(φtH)∗ω = ω for all t.

Proof. By Cartan formula, we have

LXHω = dιXHω + ιXHdω = ddH = 0.

By definition of Lie bracket, it implies that (φtH)∗ω = ω.

Lemma 1.2.2 shows that a Hamiltonian flow is a volume preserving and

it is impossible that a Hamiltonian system has asymptotically stable equilib-

rium positions.

Note that a Hamiltonian flow φtH is a solution of x = XH(x). Choose local

coordinates (q, p) on (M,ω). Then it is equivalent to the system of first order

differential equations

q = Hp, p = −Hq.

These equations are called Hamiltonian equations.

Lemma 1.2.3. Given a Hamiltonian H : M → R and a symplectomoprhism

ψ : (M,ω)→ (M,ω), we have XHψ = ψ∗XH . In particular, a symplectomor-

phism preserves Hamiltonian systems.

4


Given two Hamiltonians H,G ∈ C∞(M), the Poisson bracket is defined

by

H,G := ω(XH , XG).

Locally, it is expressed by

H,G =n∑j=1

HqjGpj −HpjGqj .

Moreover, one can see that

XH,G = [XG, XH ].

This implies that a map H 7→ XH is a Lie algebra anti-homomorphism

C∞(M)→ X(M)

H 7→ XH

·, · → −[·, ·].

The next lemma provides a simple criterion for a diffeomorphism to be a

symplectomorphism.

Lemma 1.2.4. Let (M1, ω1) and (M2, ω2) be symplectic manifolds. A diffeo-

morphism φ : M1 →M2 is a symplectomorphism if and only if it preserves

Poisson brackets, i.e.,

F,G φ = F φ,G φ ,

for all F,G ∈ C∞(M2).

Definition 1.2.5. A smooth function F : M → R is called an integral, or a

preserved quantity of XH if it is constant along a Hamiltonian flow φtH , i.e.,

F φtH = F. In particular, H itself is an integral of XH .

Lemma 1.2.6.d

dtφ∗tα = φ∗tLXtα,

where α ∈ Ωk(M) and φ is the isotopy generated by a vector field Xt.

5


Proof. We will prove this lemma by induction on k. First, suppose that

k = 0, i.e., α = f ∈ C∞(M). Then we compute that

d

dtφ∗tf =

d

dt(f φt) = df(φt)

d

dtφt

= df(φt)Xt(φt) = Xtf(φt)

= LXtf(φt) = φ∗t (LXtf).

Locally, any 1-form α can be written as

α =∑j

fjdgj,

where fj, gj ∈ C∞(M). Thus, it suffices to consider the case α = fdg. Then

d

dtφ∗tfdg =

d

dt(f φt · d(g φt))

=d

dt(f φt)d(g φt) + (f φt)

d

dt(d(g φt))

=d

dt(f φt)d(g φt) + (f φt)d(

d

dt(g φt))

= (d

dtφ∗tf)d(φ∗tg) + (φ∗tf)d(

d

dtφ∗tg)

= φ∗t (LXtf)φ∗tdg + (φ∗tf)d(φ∗t (LXtg))

= φ∗t (LXtf · dg + fdLXtg)

= φ∗t (LXtf · dg + fLXtdg)

= φ∗tLXt(fdg).

Now assume that k > 1. Suppose that lemma holds true for every α ∈ Ωl(M),

where l < k. Let α ∈ Ωk(M). Locally, it can be written by

α =∑

fi1,··· ,ikdxi1 ∧ · · · ∧ dxik .

Writing λ = fi1,··· ,ikdxi1 and µ = dxi2 ∧ · · · ∧ dxik , we see that α can be writ-

ten as a sum of differential forms of the form λ ∧ µ. So, it remains to prove

6


the lemma for λ ∧ µ.

d

dtφ∗t (λ ∧ µ) =

d

dt(φ∗tλ ∧ φ∗tµ)

=d

dtφ∗tλ ∧ φ∗tµ+ φ∗tλ ∧

d

dtφ∗tµ

= φ∗tLXtλ ∧ φ∗tµ+ φ∗tλ ∧ φ∗tLXtµ= φ∗t (LXtλ ∧ µ+ λ ∧ LXtµ)

= φ∗tLXt(λ ∧ µ).

This proves the lemma.

It follows from

d

dt(F φtH) =

d

dt(φtH)∗F = (φtH)∗LXHF

= (φtH)∗ιXHdF = (φtH)∗ιXH ιXFω

= (φtH)∗ω(XF , XH) = (φtH)∗ F,H ,

that F is an integral of XH if and only if the Poisson bracket of F and H is

zero: F,H = 0.

Definition 1.2.7. A Hamiltonian system (M2n, ω,H) is called integrable if

it possesses n independent integrals F1, · · ·Fn which are in involution, i.e.,

Fi, Fj = 0 for all i, j.

By independence of F1, · · · , Fn, we mean that dF1, · · · , dFn are linearly in-

dependent.

Example 1.2.8. (Kepler problem) A Hamiltonian of the planar Kepler prob-

lem is of the form

H(q, p) =1

2|p|2 − 1

|q|,

where q = (q1, q2) and p = (p1, p2). Consider the angular momentum

~L = q × p = (0, 0, q1p2 − q2p1).

7


Write L = q1p2 − q2p1. We claim that L is an integral. Indeed, we compute

that

H,L =q1√

q21 + q223 (−q2)− p1p2 +

q2√q21 + q22

3 q1 + p1p2 = 0.

Moreover, it follows from

dH =q1|q|3

dq1 +q2|q|3

dq2 + p1dp1 + p2dp2

dL = p2dq1 − p1dq2 − q2dp1 + q1dp2

that dH and dL are linearly dependent only if

q1|q|3

= kp2

q2|q|3

= −kp1

p1 = −kq2p2 = kq1

for some constant k. Observe that k should be nonzero. Then

q1|q|3

= k2q1 ⇒ q1(1

|q|3− k2) = 0

q2|q|3

= k2q2 ⇒ q2(1

|q|3− k2) = 0

Since (q1, q2) 6= (0, 0), it follows that |q| is constant. In short, H and L are

not independent on the circular Kepler orbit.

1.3 Noether theorem

Definition 1.3.1. Given F ∈ C∞(M), the flow φtF of a Hamiltonian vector

field XF is called a symmetry of a Hamiltonian system (M,ω,H) if the

following equality holds:

H φtF = H for all t.

8


In classical mechanics, It is important to figure out preserved quantities

of the system. The Noether theorem proves that various laws of conserva-

tion, such as momentum and angular momentum, are particular cases of the

theorem.

Theorem 1.3.2. (Noether) For a Hamiltonian system (M,ω,H) and a smooth

function F : M → R, the following are equivalent.

(i) H φtF = H for all t, i.e., φtF is a symmetry.

(ii) F φtH = F for all t, i.e., F is a preserved quantity.

Proof. H φtF = H ⇔ H,F = 0⇔ F,H = 0⇔ F φtH = F.

Thus, we conclude that there is a one-to-one correspondence between

integrals and symmetries. Observe that in the proof of Noether theorem,

we just use the definitions of integrals, symmetries, and Poisson brackets.

However, the theorem is more difficult to prove if one uses the method of

Lagrangian formalism. For example, see [2]. This is a power of Hamiltonian

mechanics.

Example 1.3.3. Consider L(q, p) = q1p2 − q2p1. By definition of Hamilto-

nian vector field, we get

XL =∂L

∂p1

∂

∂q1+∂L

∂p2

∂

∂q2− ∂L

∂q1

∂

∂p1− ∂L

∂q2

∂

∂p2

= −q2∂

∂q1+ q1

∂

∂q2− p2

∂

∂p1+ p1

∂

∂p2.

Solving the system of ODEs, we obtain

φtL(q, p) =

cos t − sin t 0 0

sin t cos t 0 0

0 0 cos t − sin t

0 0 sin t cos t

q1q2p1p2

where (q1, q2, p1, p2) = (q1(0), q2(0), p1(0), p2(0)). Observe that in a configu-

ration space R2, the flow φtL appears as the rotation about the origin. By

9


Noether theorem, the flow φtL is a symmetry. Indeed, we compute that

H φtL(q, p) =1

2|p(t)|2 − 1

|q(t)|

=1

2(p1(t)

2 + p2(t)2)− 1√

q1(t)2 + q2(t)2

=1

2|p|2 − 1

|q|= H(q, p).

Example 1.3.4. Consider the unit-sphere S2. Define a Hamiltonian

G : T ∗S2 → R

(ξ, η) 7→ 1

2|ξ|2|η|2,

where ξ = (ξ0, ξ1, ξ2) and η = (η0, η1, η2). Then its Hamiltonian equations are

given byξ = η

η = −|η|2ξ.Note that these equations represent the geodesic circles on S2, i.e., the Hamil-

tonian flow is a geodesic flow. Consider an angular momentum

ξ × η = (ξ1η2 − ξ2η1, ξ2η0 − ξ0η2, ξ0η1 − ξ1η0).

Define F (ξ, η) = ξ1η2 − ξ2η1. One can easily check that F,H = 0. More-

over, its Hamiltonian flow is given by

φtF (ξ, η) =

1 0 0 0 0 0

0 cos t − sin t 0 0 0

0 sin t cos t 0 0 0

0 0 0 1 0 0

0 0 0 0 cos t − sin t

0 0 0 0 sin t cos t

ξ0ξ1ξ2η0η1η2

Observe that G φtF = G, i.e., φtF is a symmetry. For the other two compo-

nents, we obtain the similar results. Consequently, this Hamiltonian system

has SO(3)-symmetry.

10

Chapter 2

Invariant tori

Chapter 2 is based on integrable systems.We start the chapter with presenting

a remarkable theorem, which is proved in [5], of Arnold-Liouville-Jost which

explains that every integrable system has invariant tori. By the theorem, an

integrable Hamiltonian system can be transformed into an extremely simple

system and then the solution can be easily integrated. So it is very important

to know how many independent integrals a system has. An orbit of the system

is contained in the invariant torus in which it forms a closed orbit or a dense

subset. As a generalization of Arnold-Liouville-Jost theorem, we see briefly

Mischenko-Fomenko theorem, see [3].

2.1 Arnold-Liouville-Jost theorem

Assume that a Hamiltonian system (M2n, ω,H) is integrable with n in-

volutive independent integrals F1, · · · , Fn. Define a map F : M → Rn by

F (x) = (F1(x), · · · , Fn(x)). Since dF1, · · · , dFn are linearly independent, the

rank of F is equal to n. So, the level set F−1(c) is n-dimensional submani-

fold of M for every regular value c of F . Now consider the tangent space at

x ∈ F−1(c),TxF

−1(c) = v ∈ TxM : dF (x)v = 0 .

Since

0 = Fi, Fj = ω(XFi , XFj) = dFi(XFj),

11

CHAPTER 2. INVARIANT TORI

we get XFj(x) ∈ TxF−1(c) for all j and XF1(x), · · · , XFn(x) are linearly in-

dependent. So they form a basis of the tangent space TxF−1(c). Since every

Fj is an integral of H, by similar argument the vector field XH is tangent

to F−1(c), i.e., the flow of XH leaves F−1(c) invariant. The following theo-

rem provides more significant results for integrable systems with an additive

condition.

Theorem 2.1.1. (Arnold-Liouville-Jost) Let (M2n, ω,H) be an integrable

system with n integrals F1, · · · , Fn. Assume that one of the level sets, say

F−1(0) ⊂M , is compact and connected. Then

(i) F−1(0) is an embedded n-dimensional torus T n.

(ii) There is an open neighborhood U of F−1(0) in M which can be de-

scribed by the action-angle variables (x, y) as follows: Let D1 and D2 be open

neighborhoods of the origin in Rn. If x = (x1, · · · , xn) are the variables on

the torus T n = Rn/Zn and y = (y1, · · · , yn) ∈ D1, then there exists a diffeo-

morphism

ψ : T n ×D1 → U = ∪c∈D2

(F−1(c) ∩ U

)and a diffeomorhpism µ : D2 → D1, with µ(0) = 0, such that

ψ∗ω = ω0 , µ F ψ(x, y) = y.

In particular, ψ maps the torus T n × 0 diffeomorphically onto F−1(0) and

the torus T n × y onto F−1(c) ∩ U where y = µ(c).

Corollary 2.1.2. Assume as in Theorem 2.1.1. Then by the symplectomor-

phism ψ in the theorem, the given Hamiltonian system is transformed into

the new Hamiltonian system associated with the Hamiltonian

H ψ(x, y) = h(y)

on (T n ×D1, ω0), where the Hamiltonian h depends only on the integrals and

not on the angle variables x.

The above corollary implies that on ψ−1(U) the transformed Hamiltonian

system is of the form

x = hy, y = 0.

12


The solution of this system is given by

x(t) = x(0) + thy(y(0)), y(t) = y(0).

Geometrically, every torus T n × y ⊂ T n ×D1 is invariant under the flow

φth and the restriction onto such a torus is linear:

φth∣∣Tn×y

: (x, y) 7→ (x+ thy(y), y).

Here, hy(y) describes the frequency of this torus. We conclude that in U ⊂M

all the solutions of x = XH(x) are quasi-periodic.

Example 2.1.3. Consider H and L in the example 1.2.8. Define a map

F = (H,L) : T ∗(R2 \ (0, 0))→ R2. Choose (a, b) ∈ R2 with a < 0 and b 6= 0.

Then we have

F−1(a, b) = H−1(a) ∩ L−1(b)

=

(q, p) :

1

2|p|2 − 1

|q|= a

∩ L−1(b).

Since (1/2)|p|2 is always nonnegative, we get−1/|q| ≤ a and hence−1/a ≥ |q|,which implies that the motion is bounded. It follows that |p| is also bounded.

Since H and L are continuous, we conclude that F−1(a, b) is closed and

bounded, i.e., compact. Now by Theorem 2.1.1, each connected component

of the set F−1(a, b) is an embedded 2-dimensional torus T 2 for any (a, b) ∈ R2

such that a < 0 and b 6= 0.

2.2 Mischenko-Fomenko theorem

Let (M,ω) be a 2n-dimensional symplectic manifold. For any positive integer

d ≤ n, we obtain the following theorem.

Theorem 2.2.1. (Mischenko-Fomenko) Suppose that a map

F = (F1, · · · , F2n−d) : M → F (M) ⊂ R2n−d

is a submersion with compact and connected fibers. Assume that

13


(1) There exist functions Pij : F (M)→ R such that

Fi, Fj = Pij F, i, j = 1, · · · , 2n− d.

(2) The matrix P with entries Pij has rank 2n− 2d at all points of F (M).

Then:

(i) Every fiber of F is diffeomorphic to Td.(ii) Every fiber of F has a neighborhood U endowed with a diffeomorphism

β × α : U → β(U)× Td, β(U) ⊂ R2n−d,

such that the level sets of F coincide with the level sets of β and, writing

β = (q1, · · · , qn−d, p1, · · · , pn−d, a1, · · · , ad), we obtain

ω∣∣U

=n−d∑j=1

dqj ∧ dpj +d∑j=1

daj ∧ dαj.

14

Chapter 3

Kepler problem

We discuss Kepler problem in detail. The Kepler problem has long been a

subject of theoretical interest since it is an important example of an inte-

grable mechanical system. A major advance in the classical treatment of the

Kepler problem was made by Laplace when he discovered the existence of

new integrals other than the angular momentum. We will find these new inte-

grals and map into cotangent bundle of unit-sphere via Moser regularization.

Moser regularization provides an interesting perspective to see Kepler prob-

lem in different view, i.e., we can see Kepler problem on unit-sphere instead

of an Euclidean space. Finally, we connect integrals of Kepler problem to

angular momentum of unit-speed geodesic on unit-sphere.

3.1 Delaunay variables

In this section, we see briefly the explicit way to find invariant tori for the

Kepler problem. For details, see [1].

Let a = −1/2H be semimajor axis and let e be an eccentricity of the Ke-

pler orbit. Define the domain D ⊂ T 2 × R2 ∼= T (T 2) by D = T 2 × S, where

S =

(a, e) ∈ R2 : a > 0, e ∈ (0, 1),

and E ⊂ T (R2 \ 0) by

E =

(q, p) ∈ T (R2 \ 0) : H(q, p) < 0 < e(q, p) < 1, L(q, p) > 0.

15

CHAPTER 3. KEPLER PROBLEM

Note that the conditions H < 0 < e < 1 and a > 0 implies that every Kepler

orbit is an ellipse in the setsD and E and the condition L > 0 of E determines

an orientation of the orbits.

Now the Kepler map K : E → D is defined as follows. At (q, p) ∈ E,

α(q, p) denotes the polar angle and β(q, p) denotes the argument of the per-

ihelion, see Fig. 3.1. Then the Kepler map K is defined by

Figure 3.1

K : E → D (q, p) 7→ (α(q, p), β(q, p), a(q, p), e(q, p)).

It is well-known that this map is a C∞ diffeomorphism. Denote the coordi-

nates of T (T 2) ∼= T 2 × R2 by (x1, x2, y1, y2). Note that x1, x2 ∈ C∞(T (T 2), S1)

are angular variables, not real-valued functions. Define γ, λ ∈ C∞(D,S1) and

16


Γ,Λ ∈ C∞(D) by

γ(x1, x2, y1, y2) = x2

λ(x1, x2, y1, y2) = x1 − x2

Γ(x1, x2, y1, y2) =√y1(1− y22)

Λ(x1, x2, y1, y2) =√y1

Then the Delaunay variables of the Kepler problem are given by

g = K∗γ, l = K∗λ, G = K∗Γ, L = K∗Λ,

where g, l ∈ C∞(E, S1) and G,L ∈ C∞(E). Define F : D → F (D) ⊂ T (T 2)

by F (α, β, a, e) = (γ, λ,Γ,Λ). It is also well-known that this map is a C∞

diffeomorphism and the map ∆ = F K : E → F (D) defined by

∆(q, p) = (γ, λ,Γ,Λ)

is a symplectomorphism. This implies that E can be identified with a subset

of T (T 2) and hence it has invariant tori T 2 × (Γ,Λ). Note that we did not

use Theorem 2.1.1.

3.2 The Laplace-Runge-Lenz vector

In this section, we study new integrals of the Kepler problem and their prop-

erties. Recall that the Kepler orbits of the negative energy are bounded.

Since in a mechanical problem which has a potential energy of the form

U(q) = −k/|q| all bounded orbits are elliptical, see [2], we conclude that ev-

ery Kepler orbit of the negative energy is an ellipse. Since the eccentricity is

constant on an ellipse, one can guess that the new integrals would be related

to the eccentricity of the Kepler orbit.

In order to derive the new integral, we follow [4]. Note that the Kepler

problem is in a central field F (q) = (−1/|q|2)(q/|q|) and hence we can write

p = (− 1

|q|2)q

|q|.

17


We compute that

p× ~L = (− 1

|q|2)q

|q|× (q × p)

= (− 1

|q|2)q

|q|× (q × q)

= − 1

|q|3[q × (q × q)]

= − 1

|q|3[q(q · q)− |q|2q]

and

q · q =1

2(q · q + q · q)

=1

2

d

dt(q · q)

=1

2

d|q|2

dt= |q|d|q|

dt.

Then we haved

dt(p× ~L) = p× ~L

=q

|q|− d|q|

dt

q

|q|2

=d

dt(q

|q|).

The first equality follows from the fact that the angular momentum is con-

stant on the orbit. Now define the vector

A = (A1, A2, A3) = p× ~L− q

|q|,

which is called the Laplace-Runge-Lenz(LRL) vector. Explicitly, it is of the

form

A =

(p2(q1p2 − q2p1)−

q1|q|,−p1(q1p2 − q2p1)−

q2|q|, 0).

Observe that ~L and A are perpendicular, i.e., the LRL vector A lies in the

same plane with the Kepler orbits. We claim that A1 and A2 are integrals.

18


Indeed, we compute that

H,A1 =q1|q|3

(−q2p2)− p1(p22 −1

|q|+

q21|q|3

)

+q2|q|3

(2q1p2 − q2p1)− p2(−p1p2 +q1q2|q|3

) = 0.

Similirly, we get H,A2 = 0. Therefore, on the Kepler orbit the LRL vector

A is constant.

Next, we will show that the absolute value of A is an eccentricity. We

compute that

A21 + A2

2 =

(p22L

2 +q21|q|2− 2q1p2L

|q|

)+

(p21L

2 +q22|q|2

+2q2p1L

|q|

)= |p|2L2 +

q21 + q22|q|2

− 2L2

|q|

= 2L2

(1

2|p|2 − 1

|q|

)+ 1

= 2HL2 + 1.

Since√

2HL2 + 1 represents an eccentricity of the Kepler orbit, see [7], this

proves the claim.

Finally, we claim that the LRL vector A points in the direction of perihe-

lion of the Kepler orbit, i.e., the point on the orbit where it is nearest to the

origin. Since the vector A is constant on the orbit, it suffices to check at one

point to find a direction of A. Choose the perihelion on the orbit. Observe

that both p× ~L and q/|q| point in the direction of the perihelion. Since the

eccentricity of an ellipse is nonzero, the LRL vector is nonzero. This proves

the claim. see Fig. 3.2.

Remark 3.2.1. Note that H,L,A1, and A2 are not linearly independent

integrals. Indeed, since A21 + A2

2 = 2HL2 + 1 we have

2L2dH + 4HLdL− 2A1dA1 − 2A2dA2 = 0,

which implies that dH, dL, dA1, and dA2 are linearly dependent.

19


Figure 3.2

Remark 3.2.2. In fact, the Kepler problem has a lower dimensional invari-

ant tori than T 2. Indeed, with three integrals H,L and A1 Kepler problem

satisfies the conditions of Theorem 2.2.1 as follows: write F = (H,L,A1).

Since A21 + A2

2 = 2HL2 + 1 and L,A1 = A2, which will be proved in sec-

tion 3.5, we get

L,A1 = 2HL2 − A21 + 1 6= 0.

(Here, we excluded the set (q, p) : A2(q, p) = 0 from the domain of F .)

That is, P23 and P32 are the only nonzero components of the matrix P which

is of the form

P =

0 0 0

0 0 P23

0 P32 0

Obviously, the rank of the matrix P is two. Since in this case we have n = 2

and d = 1, this implies that all conditions are satisfied. Therefore, there is

an invariant tori T 1 = S1. Observe that the projection of this torus onto the

20


configuration space R2 \ (0, 0) is the Kepler orbit.

Remark 3.2.3. In fact, the integral A1 = p2(q1p2 − q2p1)− q1/|q| can be

obtained as the limit of the integral of the Euler problem.

The Euler problem is a kind of three-body problem with two fixed centers

at (0, 0) and (1, 0) and its Hamiltonian is given by

H(q, p) =1

2|p|2 − 1− µ

|q|− µ

|q − (1, 0)|where µ is the reduced mass. Observe that the Kepler problem is the limit case

of the Euler problem, i.e., the Euler problem for µ = 0. The Euler problem

is integrable with two integrals H and

A = −(q1p2 − q2p1)2 + (q1p2 − q2p1)p2 −(1− µ)q1|q|

− µ(1− q1)|q − (1, 0)|

.

Geometrically, it represents the following quantity

|q|2 · |q − (1, 0)|2 · dθ1dt· dθ2dt− 2((1− µ) cos θ1 + µ cos θ2),

where θj is the angle between the lines connecting the center mj and the

particle and the line connecting two centers, for j = 1, 2. See, Fig. 3.3.

Figure 3.3

As µ→ 0, we obtain

A = −(q1p2 − q2p1)2 + (q1p2 − q2p1)p2 −q1|q|

21


and then this is an integral for the Kepler problem. Since L = q1p2 − q2p1 is

already an integral, so we may remove the first term. Then we get

A = (q1p2 − q2p1)p2 −q1|q|,

which is the first component of LRL vector.

3.3 Moser regularization

In this section, we recall Moser regularization in [9].

For the first step of the regularization, we introduce a new Hamiltonian

K which is defined by

K = |q|(H +1

2) =

1

2(|p|2 + 1)|q| − 1,

where −1/2 is the energy level of H which we want to regularize. Changing

the time variable t to s by

s =

∫dt

|q|,

we see that on K = 0

Kq = |q|Hq, Kp = |q|Hp.

It follows fromdq

dt=∂H

∂p,dp

dt= −∂H

∂q

thatdq

ds=dq

dt

dt

ds=dq

dt|q| = |q|Hp = Kp

dp

ds=dp

dt

dt

ds=dp

dt|q| = −|q|Hq = −Kq.

Therefore, the flow of K at K = 0 with the time variable s coincides with

the flow of H at H = −1/2 with the time variable t. In other words, the

Hamiltonian system with the Hamiltonian H on H = −1/2 is transformed

into the Hamiltonian system with the new Hamiltonian K on K = 0.

22


The second step is to apply a symplectomorphism ψ : R4 → R4 defined

by ψ(q, p) = (y,−x). Then the Hamiltonian K becomes

K(x, y) =1

2(|x|2 + 1)|y| − 1.

Consider a mapping

K(x, y) =1

2(K + 1)2 =

(|x|2 + 1)2|y|2

8.

Since Hamiltonian equations depend only on the gradient of K on K = 0, we

can replace the Hamiltonian system with K on K = 0 by the Hamiltonian

system with K on K = 1/2.

Finally, we perform the stereographic projection Φ : T ∗S2 → R4 which is

defined by

xj =ξj

1− ξ0, yj = ηj(1− ξ0) + ξjη0 for j = 1, 2.

Note that this map Φ is a symplectomorphism. Then our Hamiltonian system

is transformed into the new Hamiltonian system

ξ′ = Gη , η′ = −Gξ.

on T ∗S2 with a Hamiltonian

G(ξ, η) =1

2|ξ|2|η|2.

Recall that this system of equations expresses geodecis flows on T ∗S2 and

unit-speed geodesics, i.e., geodesics with |η| = 1 on G = 1/2, correspond to

the Kepler orbits on H = −1/2.

So far, we have excluded the north pole (1, 0, 0), which corresponds to

the collision states. We now include this point to compactify the energy

hypersurface. More precisely, we transform the geodesics through the north

pole into collision orbits. Then our regularized energy hypersurface becomes

G−1(1

2) =

(ξ, η) ∈ T ∗S2 : |η| = 1

,

23


i.e., a unit cotangent bundle T ∗1S2. Recall that the unit cotangent bundle is

diffeomorphic to RP 3 and hence it is compact.

Note that during the regularization, we interchange the roles of the po-

sition and the momentum variables so that the north pole on the sphere

corresponds to collision states at which velocities explode. So far we have

considered the only energy value H = −1/2. The general case of a nega-

tive energy surface H = −1/2c2 can be reduced to this case by a scaling of

variables according to

q 7→ c2q , p 7→ c−1p , t 7→ c3t.

We have proven the following theorem.

Theorem 3.3.1. For a negative value c, the energy surface H = c can be

mapped one-to-one and into the unit cotangent bundle T ∗1S2 of S2. This map-

ping is onto T ∗1 S2, which is the unit cotangent bundle punctured at one point

corresponding to the collision states. Moreover, the Kepler orbit is mapped

into the unit-speed geodesic on the punctured sphere after a change of the

independent variable. Finally, the Kepler problem has SO(3)-symmetry.

3.4 Integrals after regularization

As we saw in the previous section, in Kepler problem the symmetries are

continuous operations that map one orbit onto another without changing

the energy H. Since L,A1 and A2 are integrals, there are corresponding

symmetries by Noether theorem. In Example 1.3.3 we saw that the symmetry

corresponding to L is a rotation of the orbit. It does not effect on the energy

of the orbit. However, the symmetries corresponding to A1 and A2 are not

that simple. Their symmetries are more subtle since the symmetry operations

take place in a higher-dimensional space. These symmetries take place in a

phase space, not in a configuration space. It implies that we cannot see

the symmetry in a physical world. We map these integrals into T ∗1S2 via

stereographic projection.

After interchanging the roles of the position and the momentum and

applying stereographic projection, we obtain

24


L =ξ1

1− ξ0(η2(1− ξ0) + ξ2η0

)− ξ2

1− ξ0(η1(1− ξ0) + ξ1η0

)=ξ1η2 − ξ2η1

A1 =− (ξ1η2 − ξ2η1)ξ2

1− ξ0− η1(1− ξ0) + ξ1η0

(1− ξ0)|η|

=−ξ1ξ2η2 + ξ22η1

1− ξ0− η1(1− ξ0) + ξ1η0

(1− ξ0)|η|

A2 =(ξ1η2 − ξ2η1)ξ1

1− ξ0− η2(1− ξ0) + ξ2η0

(1− ξ0)|η|

=−ξ1ξ2η1 + ξ21η2

1− ξ0− η2(1− ξ0) + ξ2η0

(1− ξ0)|η|Since

(ξ × η)× ξ = (ξ22η0 − ξ0ξ2η2 − ξ0ξ1η1 + ξ21η0,

ξ20η1 − ξ0ξ1η0 − ξ1ξ2η2 + ξ22η1,

ξ21η2 − ξ1ξ2η1 − ξ0ξ2η0 + ξ20η2),

using 〈ξ, η〉 = 0 and |ξ| = 1, we obtain

ξ22η0 − ξ0ξ2η2 − ξ0ξ1η1 + ξ21η0 = η0

and then

(ξ × η)× ξ − η1− ξ0

+ (ξ × η)× (1, 0, 0)

=1

1− ξ0(0,ξ20η1 − ξ0ξ1η0 − ξ1ξ2η2 + ξ22η1 − η1 + (ξ0η1 − ξ1η0)(1− ξ0),

ξ21η2 − ξ1ξ2η1 − ξ0ξ2η0 + ξ20η2 − η2 + (ξ0η2 − ξ2η0)(1− ξ0))

=1

1− ξ0(0,− ξ1ξ2η2 + ξ22η1 − η1(1− ξ0)− ξ1η0,

− ξ1ξ2η1 + ξ21η2 − η2(1− ξ0)− ξ2η0).

Finally, with |η| = 1 we get

25


(ξ × η)× ξ − η1− ξ0

+ (ξ × η)× (1, 0, 0)

= (0,−ξ1ξ2η2 + ξ22η1

1− ξ0− η1(1− ξ0) + ξ1η0

(1− ξ0)|η|,−ξ1ξ2η1 + ξ21η2

1− ξ0− η2(1− ξ0) + ξ2η0

(1− ξ0)|η|)

= (0, A1, A2).

Consequently, on the geodesics with unit speed we have

A =(ξ × η)× ξ − η

1− ξ0+ (ξ × η)× (1, 0, 0),

where A = (0, A1, A2). Note that (ξ × η)× ξ and η are the same vectors on

the unit speed geodesics and hence the first term of A vanishes at ξ 6= (1, 0, 0),

i.e., the north pole which corresponds to the collision states. Therefore, we

obtain

A = (0, A1, A2) = (ξ × η)× (1, 0, 0) = (0, ξ0η1 − ξ1η0, ξ0η2 − ξ2η0)

and then

ξ × η = (L,−A2, A1)

at all points ξ ∈ S2 but north pole. In other words, on the unit cotangent bun-

dle T ∗1S2 three integrals of the Kepler problem correpond to the components

of the angular momentum ξ × η of the unit speed geodesics at ξ 6= (1, 0, 0). It

is reasonable that A1 and A2 are not defined at the north pole since the north

pole corresponds to the collision states and the absolute value of A coincides

with the eccentricity of the orbit. However, since ξ × η is defined on the whole

sphere, we can extend the definition of A by A1 = L3 and A2 = −L2, where

Lj is the jth component of ξ × η. We have proven the following proposition.

Proposition 3.4.1. The angular momentum of unit-speed geodesic on the

unit-sphere S2 provides

(i) eccentricity of the Kepler orbit, i.e., magnitude of LRL vector

(ii) direction of perihelion of the Kepler orbit, i.e., direction of LRL vector

(iii) Kepler’s second law, i.e., conservation of the angular momentum

(iv) Kepler problem has three independent integrals.

26


Therefore, we can investigate the Kepler orbits via angular momentum of

unit-speed geodesic on unit-sphere.

3.5 Lie algebra structure

In this section, we figure out Lie algebra structure of the Kepler problem.

First recall that the Lie algebra so(3) of the Lie group SO(3) is a collection

of anti-symmetric matrices and has basis elements

A1 =

0 0 0

0 0 −1

0 1 0

, A2 =

0 0 1

0 0 0

−1 0 0

, A3 =

0 −1 0

1 0 0

0 0 0

These basis elements have the relation

[Ai, Aj] = εijkAk, 1 ≤ i, j, k ≤ 3,

where εijk is the Levi-Civita symbol.

Now consider our three integrals L,A1 and A2. Recall that after regular-

ization, we have ξ × η = (L,−A2, A1), i.e.,

L = ξ1η2 − ξ2η1,A1 = ξ0η1 − ξ1η0,A2 = ξ0η2 − ξ2η0.

One can easily see thatL,A1 = A2,

A1, A2 = L,

A2, L = A1.

Thus, a collection of all integrals of the regularized Kepler problem on S2

carries a Lie algebra structure of so(3), where the Lie bracket of so(3) cor-

responds to the Poisson bracket. Moreover, by argument in section 3.4 and

Lemma 1.2.4, on the energy hypersurface the original Kepler problem has

27


the same Lie algebra structure. Indeed, we compute that

L,A1 =p2(−q2p2) + q2(p22 −

1

|q|+

q21|q|3

)

− p1(2q1p2 − q2p1)− q1(−p1p2 +q1q2|q|3

)

=− q1p1p2 + q2p21 −

q2|q|

=− p1(q1p2 − q2p1)−q2|q|

= A2,

A2, L =(−p1p2 +q1q2|q|3

)(−q2)− (−q1p2 + 2q2p1)p2

+ (p21 −1

|q|+

q22|q|3

)q1 − (−q1p1)(−p1)

=q1p22 − q2p1p2 −

q1|q|

=p2(q1p2 − q2p1)−q1|q|

= A1,

A1, A2 =(p22 −1

|q|+

q21|q|3

)(−q1p2 + 2q2p1)− (−q2p2)(−p1p2 +q1q2|q|3

)

+ (−p1p2 +q1q2|q|3

)(−q1p1)− (2q1p2 − q2p1)(p21 −1

|q|+

q22|q|3

)

=2q1p2|q|− 2q2p1|q|− q1p21p2 − q1p32 + q2p

31 + q2p1p

22

=2(q1p2 − q2p1)

|q|− (q1p2 − q2p1)(p21 + p22) = −2HL = L.

28

Chapter 4

Equivalent Hamiltonian

systems

Chapter 4 treats equivalent Hamiltonian systems via moment maps. We start

with the basic notions of Lie group action and adjoint and coadjoint repre-

sentations. These notions provide a way to define a moment map. Together

with arguments in Chapter 3, we find a moment map of the Kepler problem

and study how to see equivalent Kepler problems via moment map.

4.1 Group action

This section is designed for review of Lie group action, presenting some fun-

damental results that will be needed to define a moment map in the next

section. Through this section, we assume that G is a Lie group and g is the

corresponding Lie algebra.

4.1.1 Lie group action

Given a Lie group action ψ : G×M →M , define a map φξ : R×M →M

for ξ ∈ g by

φξ(t, x) = ψ(exp(tξ), x).

29

CHAPTER 4. EQUIVALENT HAMILTONIAN SYSTEMS

By definition of group action, we compute that

φξ(0, x) = ψ(e, x) = x,

φξ(t, φξ(s, x)) = φξ(t, ψ(exp(sξ), x))

= ψ(exp(tξ), ψ(exp(sξ), x))

= ψ(exp(t+ s)ξ, x)

= φξ(t+ s, x),

i.e., φξ is the flow on M . Note that t is defined on the whole R since every

ξ ∈ g is complete and exp(tξ) is the one-parameter group generated by ξ.

The infinitesimal generator of this flow is defined by

Xξ =d

dt

∣∣∣∣t=0

ψexp(tξ),

where ψexp(tξ)(x) = ψ(exp(tξ), x).

4.1.2 Adjoint and Coadjoint representations

Any Lie group G acts on itself by conjugation

G→ Diff(G)

g 7→ Cg, Cg(a) = g · a · g−1.The derivative at the identity of Cg is an invertible linear map Adg : g→ g.

Here, we used the fact that TeG and g are canonically isomorphic. Explicitly,

Adg is given by

Adg(v) =d

dt

∣∣∣∣t=0

(g · exp(tv) · g−1).

Letting g vary, we get a map

Ad : G→ GL(g)

g 7→ Adg

Since Cgh = Cg Ch for all g, h ∈ G, we have Adgh = Adg Adh, i.e., Ad is a

group homomorphism. The map Ad : G→ GL(g) is called an adjoint repre-

sentation of G on g. The derivative of Ad at the identity is a map

ad : g→ gl(g),

which is called an adjoint endomorphism.

30


Lemma 4.1.1. adv(w) = [v, w] for all v, w ∈ g, where [·, ·] is a Lie bracket

of g.

Proof. By definition of ad and Ad, we have

adv(w) =d

ds

∣∣∣∣s=0

(Adexp(sv)w)

=d

ds

∣∣∣∣s=0

d

dt

∣∣∣∣t=0

(exp(sv) exp(tw) exp(−sv))

=d

ds

∣∣∣∣s=0

d

dt

∣∣∣∣t=0

(Rexp(−sv)(exp(sv) exp(tw))

)=

d

ds

∣∣∣∣s=0

(Texp(sv)Rexp(−sv)

) (TeLexp(sv)

)w.

The last equality follows from the chain rule. Let V,W be the left-invariant

vector fields on G such that Ve = v and We = w. We claim that the flow

ψ : R×M →M of V is given by ψt = Rexp tV . Define a map ψ(p) : R→M

by ψ(p)(t) = ψ(t, p). We now compute that

(Lg ψ(p))′(t) = dLg(ψ(p)(t))(ψ(p))′(t)

= dLg(ψ(p)(t))Vψ(p)(t) = VLgψ(p)(t).

So, Lg ψ(p) is an integral curve of V starting at Lg ψ(p)(0) = Lg(p). By

uniqueness of integral curves, we have

Lg ψt(p) = Lg ψ(p)(t) = ψLg(p)(t) = ψt Lg(p),

i.e., Lg ψt = ψt Lg. Then it follows that

Rexp(tV )(g) = g exp(tV ) = Lg(exp(tV ))

= Lg(ψt(e)) = ψt(Lg(e)) = ψt(g).

This proves the claim. We conclude that

d

ds

∣∣∣∣s=0


) (TeLexp(sv)

)w =

d

ds

∣∣∣∣s=0


)Wexp(sv)

=d

ds

∣∣∣∣s=0

(Texp(sv)ψ−s

)Wexp(sv)

= Lvw = [v, w].

31


Let 〈·, ·〉 be the natural pairing between g∗ and g

〈·, ·〉 : g∗ × g→ R(ξ,X) 7→ 〈ξ,X〉 = ξ(X)

Given ξ ∈ g∗, define Ad∗gξ by⟨Ad∗gξ,X

⟩= 〈ξ, Adg−1X〉 for any X ∈ g.

Then we obtain the coadjoint representation which is defined by

Ad∗ : G→ GL(g∗)

g 7→ Ad∗g

Observe that Ad∗g Ad∗h = Ad∗gh.

Lemma 4.1.2. Given an action ψ : G×M →M , define ψg(x) = ψ(g, x).

Then XAdg−1 (ξ) = ψ∗gXξ for ξ ∈ g and g ∈ G.

Proof. It suffices to show that their flows coincide with each other. Since

ψ is a group action, we have

ψ−1g ψexp(tξ)ψg = ψg−1 exp(tξ)g.

Put γ(t) = ψexp(tξ)(p). Then γ′(0) = Xξ(p) and γ(0) = p. Choose q ∈M such

that ψg(q) = p. Then for f ∈ C∞(M)

(ψ∗gXξ)qf = (Xξ)p(f ψ−1g )

= γ′(0)(f ψ−1g )

=d

dt

∣∣∣∣t=0

(f ψ−1g γ)(t)

=d

dt

∣∣∣∣t=0

(f ψ−1g ψexp(tξ))(p)

=d

dt

∣∣∣∣t=0

(f ψ−1g ψexp(tξ) ψg)(q)

= (ψ−1g ψexp(tξ) ψg)′(0)f,

i.e., ψ−1g ψexp(tξ)ψg is the flow of ψ∗gXξ.

32


Next, we show that ψg−1 exp(tξ)g is the flow of XAdg−1 (ξ). We claim that for

Lie groups G,H and for a Lie group homomorphism F : G→ H, we have

exp(tdF (e)X) = F (exp(tX)) for every X ∈ g. Put γ(t) = F (exp(tX)). Since

γ(s+ t) = F (exp(s+ t)X)

= F (exp(sX) exp(tX))

= F (exp(sX))F (exp(tX))

= γ(s)γ(t),

a map γ : R→ H is a group homomorphism. Moreover,

γ′(0) =d

dt

∣∣∣∣t=0

F (exp(tX))

= dF (e)d

dt

∣∣∣∣t=0

exp(tX)

= dF (e)X.

Thus, F (exp(tX)) is the flow of dF (e)X. On the other hand, exp(tdF (e)X)

is the flow of dF (e)X. This proves the claim. Now put F = Cg−1 : G→ G

and X = ξ and then we have

exp(tAdg−1(ξ)) = exp(tdCg−1(e)ξ) = Cg−1(exp(tξ)) = g−1 exp(tξ)g.

Therefore, ψg−1 exp(tξ)g is the flow of XAdg−1 (ξ). This proves the Lemma.

By the previous lemma, we see that a map φ : g→ X(M) : ξ 7→ Xξ is a

Lie algebra anti-homomorphism. Indeed, we compute that

d

dt

∣∣∣∣t=0

ψexp(t(aξ+η)) =d

dt

∣∣∣∣t=0

ψexp(taξ)ψexp(tη)

=d

dt

∣∣∣∣t=0

ψexp(taξ)ψe + ψed

dt

∣∣∣∣t=0

ψexp(tη)

= ad

dt

∣∣∣∣t=0

ψexp(tξ) +d

dt

∣∣∣∣t=0

ψexp(tη)

= aXξ +Xη,

33


i.e., the correspondence ξ 7→ Xξ is linear. It follows that

[Xξ, Xη] = LXξXη =d

dt

∣∣∣∣t=0

ψ∗exp(tξ)Xη

=d

dt

∣∣∣∣t=0

XAdexp(−tξ)(η) = Xad−ξ(η)

= X[−ξ,η] = X[η,ξ].

The fourth equality follows from the linearity of φ,

d

dt

∣∣∣∣t=0

XAdexp(−tξ)η =d

dt

∣∣∣∣t=0

φ(Adexp(−tξ)η)

= φ(d

dt

∣∣∣∣t=0

Adexp(−tξ)η)

= φ(ad−ξ(η))

= Xad−ξ(η).

4.2 Moment maps

Definition 4.2.1. Let G be a compact Lie group with Lie algebra g. A group

action ψ : G×M →M of G on a symplectic manifold (M,ω) is said to be

symplectic if ψ∗gω = ω for each g ∈ G , where ψg(x) = ψ(g, x).

Definition 4.2.2. A map µ : M → g∗ is called equivariant with respect

to the action ψ of G on M and the coadjoint action Ad∗g of G on g∗ if

µ ψg = Ad∗g µ for all g ∈ G.

We are now in position to define a moment map.

Definition 4.2.3. Assume that the group action ψ of G on M is symplectic.

An equivariant map µ : M → g∗ is called a moment map for the G-action ψ

if it satisies the following: given ξ ∈ g and x ∈M , define

Xξ(x) :=d

dt

∣∣∣∣t=0

ψexp(tξ)(x) ∈ TxM.

Then

d 〈µ, ξ〉 = ιXξω for all ξ ∈ g,

34


i.e., a vector field Xξ is the Hamiltonian vector field corresponding to a

function 〈µ, ξ〉.

Remark 4.2.4. The condition d 〈µ, ξ〉 = ιXξω implies that the correspon-

dence ξ 7→ Hξ = 〈µ, ξ〉 defines a Lie algebra homomorphism. Indeed, we com-

pute thatHξ, Hη = ω(Xξ, Xη) = d 〈µ, ξ〉 (Xη)

=d

dt

∣∣∣∣t=0

〈µ, ξ〉 exp(tη)

=d

dt

∣∣∣∣t=0

〈µ(exp(tη)), ξ〉

=d

dt

∣∣∣∣t=0

⟨Ad∗tη µ, ξ

⟩=

d

dt

∣∣∣∣t=0

〈µ,Ad−tηξ〉

= 〈µ, ad−η(ξ)〉 = 〈µ, [−η, ξ]〉= 〈µ, [ξ, η]〉 = H[ξ,η].

For the first equality, we used XHξ = Xξ. Indeed, since

ω(XHξ , ·) = dHξ = d 〈µ, ξ〉 = ω(Xξ, ·),

it follows from nondegeneracy of ω.

Proposition 4.2.5. Let ψ : G×M →M be symplectic action on (M,ω)

and µ : M → g∗ be a corresponding moment map. Suppose that H : M → Ris invariant under the action ψ, i.e., H(x) = H(ψg(x)) for all x ∈M and

g ∈ G. Then the moment map µ is constant along the integral curves of XH ,

i.e., µ(φtH(x)) = µ(x) for all x and t, where φtH is the flow of XH .

Proof. The condition implies that H(x) = H(ψexp(tξ)(x)) for all ξ ∈ g and

t ∈ R. Differentiating at t = 0, we get

35


0 = dH(x)Xξ(x)

= ω(XH , Xξ)

= −ω(Xξ, XH)

= −ιXξω(XH)

= −d 〈µ, ξ〉 (XH)

= −XH(〈µ, ξ〉).

It implies that 〈µ, ξ〉 is constant along the integral curves of XH for all ξ ∈ g.

This proves the proposition.

Example 4.2.6. Recall that Kepler problem has SO(3)-symmetry. Define

an action ψ : SO(3)× T ∗S2 → T ∗S2 by

ψA(ξ, η) = (Aξ,Aη) for A ∈ SO(3) .

A symplectic form is given by

ω =2∑j=0

dξj ∧ dηj = −dλ,

where λ =∑2

j=0 ηjdξj. Since ψ∗Aλ = λ, this action is symplectic. Recall that

the Lie algebra so(3) of SO(3) is a collection of all anti-symmetric matrices.

We can identify R3 and so(3) via a map R3 → so(3) which is defined by

x = (x1, x2, x3) 7→ Φx =

0 −x3 x2x3 0 −x1−x2 x1 0

.

In our argument we use the following equalities, where proofs are straight-

forward.(i) Φxy = x× y(ii) [Φx,Φy] = Φx×y

(iii) ΦAx = AΦxA−1 for A ∈ so(3)

(iv) trace(ΦTxΦy) = 2(x · y).

36


The last equality defines an inner product on so(3) and then via this inner

product we can identify so(3)∗ and so(3) by

so(3)→ so(3)∗

A 7→ 1

2trace(AT ·)

Then for A,B ∈ so(3) the natural pairing becomes

〈A,B〉 =

⟨1

2trace(AT ·), B

⟩=

1

2trace(ATB).

Now define a map µ : T ∗S2 → so(3) ∼= so(3)∗ by

µ(ξ, η) = Φξ×η.

We will show that this map is a moment map for the action ψ. We first

compute that

µ ψA(ξ, η) = µ(Aξ,Aη) = ΦAξ×Aη

= [ΦAξ,ΦAη] = A[Φξ,Φη]A−1

= AΦξ×ηA−1.

Also, for V ∈ so(3)

〈Ad∗A(Φξ×η), V 〉 = 〈Φξ×η, AdA−1V 〉=⟨Φξ×η, A

−1V A⟩

=⟨AΦξ×ηA

−1, V⟩.

We conclude that Ad∗A µ = µ ψA, i.e., the map µ is equivariant.

Recall that for V = (vij) ∈ so(3) and (ξ, η) ∈ T ∗S2

XV (ξ, η) =d

dt

∣∣∣∣t=0

ψexp(tV )(ξ, η)

=d

dt

∣∣∣∣t=0

(exp(tV )ξ, exp(tV )η)

= (V ξ, V η)

37


Then

ιXV ω∣∣(ξ,η)

=(v21η1 + v31η2)dξ0 + (v12η0 + v32η2)dξ1 + (v13η0 + v23η1)dξ2

+ (v12ξ1 + v13ξ2)dη0 + (v21ξ0 + v23ξ2)dη1 + (v31ξ0 + v32ξ1)dη2

On the other hand,

d 〈µ(ξ, η), V 〉 =d 〈Φξ×η, V 〉

=1

2dtrace(ΦT

ξ×ηV )

=− 1

2dtrace(Φξ×ηV )

=− 1

2d(v21(−ξ0η1 + ξ1η0) + v31(ξ2η0 − ξ0η2) + v12(ξ0η1 − ξ1η0)

+ v32(−ξ1η2 + ξ2η1) + v13(−ξ2η0 + ξ0η2) + v23(ξ1η2 − ξ2η1))=(v21η1 + v31η2)dξ0 + (v12η0 + v32η2)dξ1 + (v13η0 + v23η1)dξ2

+ (v12ξ1 + v13ξ2)dη0 + (v21ξ0 + v23ξ2)dη1 + (v31ξ0 + v32ξ1)dη2

Therefore,

ιXV ω∣∣(ξ,η)

= d 〈µ(ξ, η), V 〉

for all V ∈ so(3) and for all (ξ, η) ∈ T ∗S2. This shows that the map µ is a

moment map.

We now see the meaning of this moment map. We first observe that

G ψA(ξ, η) = G(ξ, η), where G is the regularized Hamiltonian in section

3.3. Identifying of so(3) and R3, the moment map µ becomes

µ : T ∗S2 → R3

(ξ, η) 7→ ξ × η

Then it follows from propositon 4.2.5 that each component of angular mo-

mentum ξ × η is an integral of XG as shown before. Hence, we conclude that

we can study all integrals of the Kepler problem via the moment map µ.

Moreover, it follows from µ ψA(ξ, η) = Ad∗A µ(ξ, η) = Aµ(ξ, η)A−1 and

ψ∗Aω = ω that we can interchange the roles of integrals by SO(3)-action and

the resultant Hamiltonian system is equivalent to the original one. This pro-

cedure is just interchanging the rotational axes of S2 by SO(3)-action so

38


that the corresponding components of the angular momentum ξ × η are in-

terchanged. For example, we can compose Kepler problem which takes −A2

as the angular momentum. Moreover, the Hamiltonian system corresponding

to H and −A2 is equivalent to the Hamiltonian system corresponding to H

and L, which is the one we have considered so far. Therefore, we can find

invariant tori corresponding to H and −A2 from invariant tori corresponding

to H and L which are expressed by Delaunay variables.

Figure 4.1

39

Bibliography

[1] R. Abraham and J.E. Marsden, Foundations of mechanics, 2nd ed,

Addison Wesley, New York (1978).

[2] V.I. Arnold, Mathematical methods of classical mechanics, Springer-

Verlag, Berlin (1978).

[3] F. Fasso, Superintegrable Hamiltonian systems: geometry and perturba-

tions, Acta Appl. Math. 87 (2005) 93-121.

[4] H. Goldstein, C. Poole, and J. Safko, Classical mechanics, 3rd ed, Ad-

dison Wesley (2002).

[5] H. Hofer and E. Zehnder, Symplectic invariants and Hamiltonian Dy-

namics, Birkhauser (1994).

[6] J.M. Lee, Introduction to smooth manifolds, Springer, New York (2003).

[7] D.O. Mathuna, Integrable systems in celestial mechanics, Birkhauser

(2007).

[8] D. Mcduff and D. Salamon, Introduction to symplectic topology, 2nd ed,

Clarendon press, Oxford (1998).

[9] J. Moser, Regularization of Kepler’s problem and the averaging method

on a manifold, Comm. Pure Appl. Math. 23 (1970) 609-636.

[10] C. Silva, Lectures on symplectic geometry, Springer, no.1764 (2008).

40

국문초록

이 논문은 해밀턴계의 대칭과 운동 상수의 관계와 그 중요성 대해 다룬다.

우리는또한적분가능계가가지고있는놀라운구조와,동등한해밀턴계를

살피기 위한 운동량 사상에 대해서도 살펴볼 것이다. 이런 모든 개념들이

실제의 예에서는 어떻게 작용하는지 보기 위해 고전역학에서 매우 중요한

케플러 문제를 자세히 다룰 것이다. 마지막으로 케플러 문제는 2차원 단위

구면 상의 단위 속력 측지선들의 각운동량에 의해 표현될 수 있다는 것을

보일 것이다.

주요어휘: 해밀턴계, 적분가능계, 케플러 문제, 각운동량, 운동량 사상

학번: 2012-20245

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