Reductions on the Kepler problem

(1)

faculty of science and engineering

Reductions on the Kepler problem

Master Project Mathematics

June 2018

Student: S. de Vries

First supervisor: K. Efstathiou, PhD Second supervisor: Prof. dr. H. Waalkens

(2)

1 Introduction

In this thesis we will consider the Kepler Problem and its reductions. The Kepler problem is a classical problem originating from the 17-th century. Its initial application was to describe the motion of the planets around the sun. The Kepler problem can be fully described by Newton’s second law and Newton’s gravitational law. Several laws for the motion of this system are attributed to the German scientist Johannes Kepler. The Kepler laws can be proven by elementary vector calculus, as done in [1]. Surprisingly, the study of the Kepler system still continues today and even new insights have been obtained in recent years.

In the twentieth century several regularisations of the Kepler system have been introduced in order to remove the singularity coming from the collision orbit. These regularisations all transform the Kepler system to a system that is easier to analyse. In 1906 Levi-Civita introduced the two-dimensional Levi- Civita transformation in [2]. Partly because of its elegance, we will outline this transformation in Appendix C, other reasons to outline this transformation are to give newer readers a clear outline of the Levi-Civita transformation and to show that the Levi-Civita transformation indeed generalizes to the Kustaanheimo-Stiefel transformation. In 1965 Kustaanheimo and Stiefel [3] gen- eralised the Levi-Civita transformation to three dimensions, which is the transformation that will be used in this thesis and will be called KS-transformation for short. Five years later, in 1970, Moser [4] introduced a different regularisation for the Kepler system. This regularisation maps the level sets of the Hamilto- nian to the tangent bundle of the sphere, where the flow of the Hamiltonian turns out to be represented by geodesics. In 1982 Kummer [5] showed a relation between the KS-transformation and the Moser transformation by constructing an isomorphism between the KS phase space and the phase space of the Ke- pler system as obtained after the Moser transformation. Another regularisation method has been proposed by Ligon and Schaaf in 1976 [6]. A more understand- able characterisation of this method is written by Cushman and Duistermaat [7] in 1996, where they show that the phase space of the Kepler problem can be identified with the geodesics of the sphere. In 2012 Heckman and De Laat showed that the Ligon-Schaaf regularisation can be understood as an adaptation of the Moser regularisation [8].

In recent years, the Kustaanheimo-Stiefel transformation has been used for example by Efstathiou and Sadovskii [9] in order to study the perturbed Kepler system for certain fibrations. An important application of the Kepler problem is describing the physical system of the hydrogen atom, as for example is done in [9]. In quantum mechanics the principal quantum number n and the magnetic quantum number m can be defined. In [10], Dullin and Waalkens shown by that for sufficiently large, but still negative, values of the Hamiltonian, the Kepler system has monodromy. From this it can be concluded that for certain fibrations a third quantum number cannot be globally defined. In this thesis we will aim to reproduce this result and in addition to consider the positive energy case.

Monodromy is a property of integrable systems. The concept of monodromy

(4)

is related to non-trivial torus bundles. Take a 2N-dimensional symplectic manifold, where N is the number of degrees of freedom and take N independent integrals, whose mutual Poisson brackets equal zero. Under these conditions, the Arnold-Liouville theorem assures that the level set for regular values of the integral map is diffeomorphic to the N-dimensional torus, if these level sets are compact and connected. Suppose that we take a simple loop in the space of regular values of the integral map. Under the assumptions of the Arnold-Liouville theorem, each regular value on this loop corresponds to a N-dimensional torus.

Geometrically, the system has non-trivial monodromy, if the entire loop does not lift to an N + 1-dimensional torus.

In order to quantify the concept of monodromy for integrable systems, we will use the monodromy vector k. Consider the case for N = 2 and a two- dimensional torus T with a point p on T . Suppose that one of the integrals J generates a periodic action defined on T with period 2π, whose flow ψJ thus defines a closed path γ. Let H be a Hamiltonian function. If we consider its possibly non-periodic Hamiltonian action on T , then its flow ψH will cross γ at some point p⁰. Consider the level set H = h and J = j, where (h, j) is a regular value of the vector valued map (H, J). Define θ(h, j) to be the time in which the flow ψJ sends p to p⁰. The quantity θ(h, j) is independent of which point p is picked. If we consider a loop Γ on the space of regular values of the integral map, we observe that θ(h, j) might increase or decrease with 2kπ for integer k along Γ. If and only if k is nonzero the system has non-trivial monodromy.

The Kepler system is superintegrable i.e. the number of independent integrals of motion of the system is higher than the amount of degrees of freedom.

The Kepler system has 5 independent integrals of motion. These independent integrals of motion can be obtained by taking the Hamiltonian, the components of the angular momentum vector and the Laplace-Runge-Lenz vector. These integrals of motion are not independent as the angular momentum vector is per- pendicular to the Laplace-Runge-Lenz vector and the absolute values of these two vectors can be expressed in terms of each other by using the Hamiltonian.

The superintegrability of the Kepler system has as a result that the orbits can be joined in different ways to form different toric fibrations. Different fibrations may have different properties, such as their geometry and the presence of non-trivial monodromy.

In order to show monodromy, we will first apply the Kustaanheimo-Stiefel transformation. We will reduce the system with respect to the Hamiltonian H, one of the components of the angular momentum called l3( there is no particular reason to reduce with respect to the third component of the angular momentum, reducing with any other component of the angular momentum should give similar results), and an artificial constant ζ obtained through the KS-transformation.

To be able to reduce the system, we will use several coordinate transformations to make the equations of motion diagonal and obtain a set of harmonic oscilla- tors. The equations that describe the geometry of the reduced system depend on the value of the Hamiltonian. The reduced space will be compact for negative energies, but it will be unbounded for positive energies. We will thereafter study the bifurcation diagram of the system with respect to another integral, that can

(5)

obtained through separation of variables in prolate spheroidal coordinates. In order to show monodromy, we will examine which points in the reduced space have non-trivial (S¹) isotropy under the action of H, l3 and ζ. We will be able to show that the system in Kustaanheimo-Stiefel space has monodromy for negative values of the Hamiltonian that are sufficiently large. By using the result obtained earlier in this thesis, we will be able to apply a theorem outlined in [11], which shows us that the system has monodromy for negative values of the Hamiltonian. In order to show monodromy for positive energies, we will reduce the space by a rescaled Hamiltonian N and ζ in order to apply this theorem.

We will need to do these extra reductions since the orbits are not compact for positive energies. The geometry of this reduced system is diffeomorphic to the tangent space of the sphere. From the S¹isotropy under the action of l3, we will be able to conclude that the reduced system has non-trivial monodromy. Since reducing a system does not cause monodromy, the monodromy should also be present in the original system.

(6)

2 The Kepler problem and the Kustaanheimo- Stiefel transformation

Consider the Kepler problem in three dimensions with two point masses M1and M₂. Let M1 be fixed at the origin. Let P and Q be the 3-dimensional cartesian coordinates and describing the position and momenta of M2. The corresponding Hamiltonian function for the Kepler problem is given by

H = 1

2P²− 1

pQ², (1)

where P = (P1, P₂, P₃)^T, Q = (Q1, Q₂, Q₃)^T and E is the energy of the system.

2.1 The Kustaanheimo-Stiefel transformation

We will apply the Kustaanheimo-Stiefel transformation (KS transformation for short) introduced in [3]. The KS transformation maps the coordinates in the KS space (q1, q2, q3, q4, p1, p2, p3, p4)^T ∈ R⁸ to the original Euclidean space R⁶ and is given by

KS(q, p) = (M_KS· q)^T, 1

2q²(M_KS· p)^T = Q, 0, P,2ζ q²

T

, for some ζ and where

M_KS =







q₁ −q₂ −q₃ q₄ q₂ q₁ −q₄ −q₃ q₃ q₄ q₁ q₂ q4 −q3 q2 −q1







, q = (q₁ q₂ q₃ q₄)^T, p = (p₁ p₂ p₃ p₄)^T.

In terms of qi and pi the KS transformation is given by

KS(q, p) =







q₁²− q₂²− q²₃+ q²₄ 2q1q2− 2q3q4

2q1q3+ 2q2q4

0

q₁p₁−q2p₂−q3p₃+q₄p₄ 2q²

q2p1+q1p2−q4p3−q3p4

2q²

q₃p₁+q₄p₂+q₁p₃+q₂p₄ 2q²

q₄p₁−q3p₂+q₂p₃−q1p₄ 2q²







T

.

We claim that the quantity

ζ = 1

2(q4p1− q3p2+ q2p3− q1p4)

is invariant in time under the flow induced by any time-independent Hamiltonian function on the original Cartesian coordinates in R⁶. This result can be obtained

(7)

by studying Poisson brackets. For a quick overview of the theory of Poisson brackets, we refer to Appendix A.

By computing the Poisson brackets of ζ and any of the original coordinates, we observe that ζ is an integral of motion. In order to turn the KS transformation into a canonical transformation, we will set ζ = 0.

If we want to study Hamiltonian dynamics on the KS space, we should determine whether the KS transformation is canonical. The following proposition assures that this is indeed the case.

Proposition 2.1. The KS transformation preserves Poisson brackets on the original coordinates.

Proof. To prove this theorem, it is sufficient to show that the coordinates obtained after the KS-transformation satisfy {Qi, Pj} = δ_jⁱ , {Qi, Qj} = 0 and {P_i, P_j} = −_q^ζ₆P

k_ijkQ_k, where ζ = 0. By computing the Poison brackets, one observes that this is indeed the case.

As a result of the KS-transformation, we can obtain the following useful formulas

|Q| = q², (2)

P²+ ζ 2q²

²

= p²

4q². (3)

To obtain these formulas, it is helpful to know that M^T = q²M⁻¹. We can use equations (2) and (3) to write the Hamiltonian as

H = 1

2P²− 1

|Q| = p² 8q² − ζ²

8q⁴− 1 q².

This Hamiltonian function can be regularised. To do so, we slow down time near the origin, which we achieve by multiplying the Hamiltonian by 4q². In order to avoid extra terms in the equations of motion due to the re-scaling of time, we consider the level set H = h. We can use this to obtain another Hamiltonian function, which corresponds to the same dynamics. Furthermore we can add a real number to the Hamiltonian without changing the dynamics.

Therefore we can introduce the following Hamiltonian function

N = 4q²(H − h) + 4 = 1 2

p²− 8hq²−ζ² q²

. (4)

(8)

2.2 Keplerian integrals of motion in the KS space

Throughout the rest of this thesis, we will require several integrals of motions.

In this subsection we will determine an expression for these Keplerian integrals of motion in KS-coordinates.

The first integrals of motion that we will express in KS-coordinates will be the components of the angular momentum l

l₁= P₂Q₃− P₃Q₂= 1 2

q₁p₄+ q₂p₃− q₃p₂− q₄p₁ , l2= P1Q3− P3Q1=1

2

− q1p3+ q2p4+ q3p1− q4p2

,

l3= P1Q2− P2Q2= 1 2

q1p2− q2p1+ q3p4− q4p3

.

Another well-known integral of the system is the eccentricity vector. This vector is given by

E|H=h= P × l − Q

|Q| = P × l − h −1

2P²Q. (5)

To obtain this equation, we used the original equation (1) for the Hamilto- nian to substitute |Q|. By using the last expression from equation (5), we can write the eccentricity vector in KS-coordinates as

E|H=h= e + ζ q²

l − ζ

2q²Q , where the vector e is given by

e1=1 8

− p²₁+ p²₂+ p²₃− p²₄+ 8h(q₁²− q₂²− q²₃+ q²₄) , e2=1

4

− p1p2+ p3p4+ 8h(q1q2− q3q4) , e3=1

4

− p1p3− p2p4+ 8h(q1q3+ q2q4) .

(9)

3 Algebraic reduction of the Kepler system

In this section we will reduce the Kepler system by the actions of ζ and l3. This can be done using the Hamilton equations to determine the equations of motion.

We will use the following notation in the rest of this section.

˙ p_i= ∂ζ

∂q_i, ˙q_i= −∂ζ

∂p_i, p⁰_i= ∂l3

∂q_i, q⁰_i= −∂l3

∂p_i.

3.1 Reduction of the Kepler system with respect to ζ and l

₃

To study the action of ζ and l3, we will introduce the following coordinates, which will conveniently make the equations of motion diagonal.

u1= q1+ iq2+ q3+ iq4, u₂= q₁− iq2− q3+ iq₄, v₁= p₁+ ip₂+ p₃+ ip₄, v₂= p₁− ip₂− p₃+ ip₄.

Using the Hamilton equations we find that the equations of motion in the new coordinates are given by

˙

u1= iu1, u⁰₁= iu1,

˙

u2= iu2, u⁰₂= −iu2,

˙v1= iv1, v⁰₁= iv1,

˙v2= iv2, v⁰₂= −iv2.

Let T^k be the k-dimensional torus. We will call a homomorphism a T^k (respectively an R^k) action on a manifold M if it is a homomorphism from T^k 7→ diff(M) (respectively R^k 7→diff(M)), where diff(M) is the space of diffeomorphism on M. We will discuss the definition of homomorphisms briefly in Appendix D. Using the equations of motion, the T² action induced by the action of ζ and l3is given by

T²×C⁴→ C⁴: (s, t), (u1, u2, v1, v2) 7→ (eî(s+t)u1, eî(s−t)u2, eî(s+t)v1, eî(s−t)v2).

This observation allows us to make the following statement. (6)

Lemma 3.1. Let j = 1, 2. The algebra of invariant polynomials under the action of ζ and l3 is generated by

ρj =1 8

uju¯j+ vjv¯j

, χj= 1 8

uju¯j− vj¯vj

,

(10)

ψj =1 8

ujv¯j+ ¯ujvj

, ιj= 1 8

ujv¯j− ¯ujvj

. Furthermore, the following relations hold

ρj ≥ 0,

ρ²_j− χ²_j− ψ_j²= ι²_j. (7) Proof. Stating that ρj, χ_j, ψ_j, ι_jare the generators of the the algebra of invariant polynomials is equivalent to saying that the algebra of invariant polynomials is generated by uju¯j, vjv¯j, ujv¯j and ¯ujvj. We observe that these quantities are indeed invariant under the action given by equation (6).

The first observation that we make is that if a polynomial is invariant, it necessarily must be the sum of invariant monomials. Therefore we will consider an arbitrary monomial given by uⁿ1¹u¯1n₂uⁿ₂³u¯2n₄v₁ⁿ⁵v¯1n₆v₂ⁿ⁷v¯2n₈.

The action of ζ and l3on such an arbitrary monomial is given by uⁿ₁¹u¯1n₂

uⁿ₂³u¯2n₄

v₁ⁿ⁵v¯1n₆

vⁿ₂⁷v¯2n₈

7→

eⁱ⁽ⁿ¹⁺ⁿ⁵⁻ⁿ²⁻ⁿ⁶^)(s+t)+i(n³⁺ⁿ⁷⁻ⁿ⁴⁻ⁿ⁸^)(s−t)uⁿ₁¹u¯₁ⁿ²uⁿ₂³u¯₂ⁿ⁴vⁿ₁⁵v¯₁ⁿ⁶vⁿ₂⁷v¯₂ⁿ⁸. If we suppose that this monomial is invariant, we obtain the relations

n1+ n5= n2+ n6, n3+ n7= n4+ n8.

From these relations it can be concluded that every monomial can be expressed in terms of uju¯_j, v_jv¯_j, u_jv¯_j, ¯u_jv_j. The statement ρj≥ 0follows directly from the definition of ρj. Equation (7) is just an algebraic relation and can be obtained by expressing ρj, χ_j, ψ_j and ιj in terms of u1, u₂, v₁and v2. This completes the proof.

The integrals of motion of the Kepler system can be expressed in terms of the generators of the algebra of invariant polynomials. We will express the following integrals in terms of the generators.

N = (1 − 8h)(ρ1+ ρ2) − (1 + 8h)(χ1+ χ2), (8)

ζ = ι1+ ι2, (9)

l₃= ι₁− ι₂. (10)

Demanding that ζ = 0 allows us to rewrite the equation for l3 as l3= 2ι1= −2ι2.

Since l3is an integral of motion, we observe that ι1, ι2are invariant. Substituting ι1 and ι2 for ¹₂l3, allows us to express equation (7) as

ρ²_j− χ²_j− ψ²_j = 1

4l₃², with ρj ≥ 0.

(11)

For a particular value of l3, we can visualise this space embedded in R⁶ as the product of two two-dimensional manifolds. Rearranging terms might help to recognise that these two-dimensional manifolds are hyperboloids. The result of this rearrangement is

ρ²_j= 1

4l²₃+ χ²_j+ ψ_j², with ρj ≥ 0.

We will introduce the following notation to describe the structure of the space of invariant polynomials under the action of ζ and l3. Let Cl₃ be defined as

Cl₃= {(ρ, χ, ψ) ∈ R³|ρ ≥ 0, ρ²= 1

4l₃²+ χ²+ ψ²}.

Using this notation, we can identify the reduced space under the action of ζ and l3 with Cl₃× Cl₃.

The reduced space is homeomorphic to the Euclidean plane and for l3 6= 0 this space is even a manifold. In order to obtain a mapping from (χ, ψ) ∈ R²to (ρ, χ, ψ) ∈ Cl₃, we can pick the mapping

(χ, ψ) 7→r 1

4l₃²+ χ²+ ψ², χ, ψ .

Since the value of l3 is an integral of motion, this mapping allows us indeed to identify Cl3× C_l₃ with R²× R². We note that this identification is differentiable if and only if l36= 0

3.2 Fibers of the reduced space and the reconstruction to KS space

From equation (6), we can deduce that most points in Cl3× Cl3 lift to a 2-torus in KS space. When a point is not lifted to a 2-torus, the Marsden-Weinstein- Meyer theorem implies that this point reduces to a singular point in the reduced space. The Marsden-Weinstein-Meyer theroem is stated in Appendix B. In our case this singularity is the non-differentiability of equation (7) for ιi= ¹₂l₃= 0. Since the space given by Cl3 has a singularity for l3= 0, the reconstruction to KS space is not trivial for the particular value of l3 = 0. We will do the reconstruction for l3 = 0 by considering two cases. Let j = 1, 2. The first observation that we make is that ρj = 0implies that uj = vj = 0. If ρj 6= 0 then uju¯j = 4(ρj+ χj)and vjv¯j= 4(ρj− χj). Similarly we observe that uj¯vj = 4(ψj− iιj)and ¯ujvj = 4(ψj+ iιj). The last two equations are equivalent. We can use these equations to obtain the following relation for the angles between uj and vj

4(ψj− iιj) = ujv¯j = |uj||vj| exp i arg(uj) − arg(vj).

Thus, if ρj, χ_j, ψ_j, ι_j are given, we find that |uj| and |vj| are fixed and there is a relation between the arguments of both complex quantities. This allows

(12)

us to conclude that for ρj, χj, ψj, ιj, with ρj 6= 0, the corresponding values of uj, vj ∈ C²are homeomorphic to S¹.

Hence we can make the follwoing reconstruction to KS space. Points in Cl3× Cl3 with ρ1 = ρ2= 0correspond to a single point in C⁴. Points with for j, k = 1, 2with j 6= k and such that ρj 6= ρk = 0correspond to a S¹ in C⁴ and points with ρ16= 0 6= ρ2 correspond to a T²in C⁴.

3.3 Reduction of the Kepler system with respect to N

In this subsection we will reduce the system with respect to N. In order to determine the action of N, we will recall the equation (8), which gives the following expression for N

N = (1 − 8h)(ρ1+ ρ2) − (1 + 8h)(χ1+ χ2).

Furthermore we can calculate the Poisson brackets between ρj, ψ_j, χ_j for j = 1, 2. The non-zero Poisson brackets between ρj, ψ_j, χ_j are given by

{ρ_j, ψ_j} = χ_j, {ρ_j, χ_j} = −ψ_j, {ψ_j, χ_j} = −ρ_j.

The Poisson brackets are useful for obtaining the equation of motion. The equation of motion of induced by the flow of N are given by

˙

ρj= {ρj, N } = (1 + 8h)ψj,

˙

χj= {χj, N } = (1 − 8h)ψj,

ψ˙j = {ψj, N } = (1 + 8h)ρj− (1 − 8h)χj.

Lemma 3.2. The flow of N on Cl3× Cl3 induces a S¹ action for h < 0 and a R action for h > 0.

Furthermore, the algebra of invariant polynomials of the action of N is generated by

α₁= (1 − 8h)ρ₁− (1 + 8h)χ₁, α₂= (1 − 8h)ρ₂− (1 + 8h)χ₂, γ =

(1 + 8h)ρ1− (1 − 8h)χ1

(1 + 8h)ρ2− (1 − 8h)χ2

− 32hψ1ψ2,

δ =

(1 + 8h)ρ1− (1 − 8h)χ1

ψ2−

(1 + 8h)ρ2− (1 − 8h)χ2

ψ1. These generators satisfy the relation

γ²− 32hδ²= (α²₁+ 32hι₁²)(α²₂+ 32hι²₂). (11) For h > 0, additionally γ ≥ 0 and for h < 0, 0 ≤ αj ≤ 4for i = 1, 2.

(13)

Proof. Let us define for j = 1, 2

w_k^±= (1 + 8h)ρk− (1 − 8h)χk±√ 32hψk. The time derivatives of these quantities are given by

˙

w_k^±= (1 + 8h)²ψk− (1 − 8h)²ψk±√ 32h

(1 + 8h)ρk− (1 − 8h)χk

= ±√ 32hw^±_k. Furthermore we note that indeed ˙αj= 0. Hence we will make the linear trans-(12) formation

(ρ₁, χ₁, ψ₁, ρ₂, χ₂, ψ₂) 7→ (α₁, w⁺₁, w⁻₁, α₂, w₂⁺, w⁻₂).

In these new coordinates the flow of N is given by

(α1, w⁺₁, w⁻₁, α2, w₂⁺, w₂⁻) 7→ (α1, e

√

32htw⁺₁, e⁻

√

32htw⁻₁, α2, e

√

32htw⁺₂, e⁻

√

32htw⁻₂), which shows that the flow of N is an S¹ action for h < 0 and an R action for h > 0.

The standard pairing argument (as seen in the proof of lemma 3.1) then im-

plies that the space of invariant polynomials is generated by α1, α2, w₁⁺w₁⁻, w₂⁺w₂⁻, w⁺₁w₂⁻, w⁻₁w₂⁺. Furthermore we observe the following identities

γ = 1

2(w₁⁺w₂⁻+ w⁻₁w⁺₂), δ = 1 2√

32h(w⁺₁w⁻₂ − w₁⁻w⁺₂) and

w⁺₁w⁻₁ = α²₁+ 8hl₃², w₂⁺w₂⁻= α²₂+ 8hl²₃.

Hence we can conclude that the algebra of invariant polynomials is generated by α1, α₂, γ, δ

Equation (11) can be verified by writing γ²− 32hδ²= (γ +√

32hδ)(γ −√ 32hδ)

= w⁻₁w⁺₂w⁺₁w⁻₂ = w₁⁺w₁⁻w₂⁺w₂⁻= (α²₁+ 32hι²₁)(α²₂+ 32hι²₂).

For h < 0, the equations 0 ≤ αj ≤ 4are satisfied since equation (7) implies that αj ≥ 0as ρj ≥ χj and thus (1 − 8h)ρj≥ (1 + 8h)χj. By demanding that N = 4we obtain αj≤ α1+ α2= N = 4.

In order to show that γ > 0 for h > 0, we define

γ₀=p

(ρ₁− χ₁)(ρ₂− χ₂) − 8hp

(ρ₁+ χ₁)(ρ₂+ χ₂)2

≥ 0.

This allows us to show γ ≥ γ0≥ 0, as we obtain the following inequality

(14)

γ − γ₀= h

2(|u₁|²|v₂|²+ |u₂|²|v₁|²− 4Re(u₁¯v₁)Re(u₂v¯₂) + 2|u₁||u₂||v₁||v₂|)

≥ h

2(|u1|²|v2|²+ |u2|²|v1|²− 4|u1||u2||v1||v2| + 2|u1||u2||v1||v2|)

= h

2 |u1||v2| − |u2||v1|2

≥ 0.

3.4 Geometry of the reduced system

We will study equation (11) to some more extent in this subsection. In order to do so, we first recall that α1+ α2= N = 4and thus we can solve α2= 4 − α1. We will reduce the number of variables in equation (11) even more by omitting δ². This results in the following equation for h < 0

γ²≤ (α²₁+ 8hl²₃)((4 − α₁)²+ 8hl²₃). (13) For h > 0, this procedure will result in the following equation

γ²≥ (α²₁+ 8hl²₃)((4 − α₁)²+ 8hl²₃). (14) The sets described by (13) and (14) can be visualised in a two-dimensional figure, as is done in Figure 1. Both for positive and negative energies, the set of points satisfying equation (13) or equation (14) decreases in size as l3² increases. For the case with negative energy, there is only a single feasible point for l²3= −_2h¹ and if l²3> −_2h¹, there is no single point that satisfies equation (13). For positive energies, there will always be points that satisfy equation (14).

It is important to note that for all h 6= 0 and l3 = 0, the surface given by equation (11) is not differentiable at the points (α, γ, δ) = (0, 0, 0) and (α, γ, δ) = (4, 0, 0).

4 Isotropy groups

In this section, we will determine the isotropy induced by the flow of N, ζ and l3

in KS-space. Before we introduce the concept of isotropy, the notion of a group action will be recalled once more.

Definition 4.1. A group action on a manifold M is a homomorphism that maps a group G to the space of diffeomorphisms on M.

For a short overview of homomorphisms see Appendix D. In the context of Hamiltonian mechanics we often encounter a group action as a flow ψg of a certain integral of motion H, where g is some quantity that represents time.

Hence we have for every time g a diffeomorphism ψg, which represents the flow of H. Using this notion of group action, we can make definitions about isotropy.

Let Diff(M) be the space of diffeomorphisms on M.

(15)

Figure 1: On the left side, the lemon shaped area represents the values of α, γ for which there is a δ such that equation (11) holds for h < 0 and l3 = 0with 0 ≤ α1≤ 4. The area above this lemon shaped area shows the values of α, γ for which there is a δ such that equation (11) holds with γ ≥ 0 for h > 0 and l3= 0. On the right side, the lower area represents the values of α, γ for which there is a δ such that equation (11) holds with 0 ≤ α1≤ 4for h < 0 and 8hl²3= −1/4. The upper area shows the values of α, γ for which there is a δ such that equation (11) holds with γ ≥ 0 for h > 0 and l3= 8hl²₃= 1/4.

Definition 4.2. Take a group action ψg defined on a manifold M by G 7→

Diff(M). The isotropy group JU of a set U in M is defined as JU = {g ∈ G | ∀x ∈ U : ψg(x) = x}.

We say that the isotropy group of a set U is trivial if the isotropy group J_U only consists of the identity of G. We say that an action is effective on a manifold M if the isotropy group of the entire manifold is trivial. The action is free on a manifold M if the isotropy group of any set U ⊂ M is trivial. We say that a set S has S¹isotropy, if the set JS is homeomorphic to S¹.

4.1 Action of N in the KS space.

In order to determine the isotropy group induced by the flow of N, ζ and l3, we will first determine the flow of N in KS-space. We will recall that the expression for N in KS-coordinates is given by

N = 1

2(p²− 8hq²).

The corresponding equations of motion are

˙ q = p,

˙ p = 8hq.

We are interested in the flow induced by N on the following coordinates u1= q1+ iq2+ q3+ iq4,

(16)

u2= q1− iq2− q3+ iq4, v₁= p₁+ ip₂+ p₃+ ip₄, v₂= p₁− ip₂− p₃+ ip₄.

This allows us to formulate the equations of motion on these coordinates as

˙

u1= p1+ ip2+ p3+ ip4= v1,

˙

u₂= q₁− iq2− q3+ iq₄= v₂,

˙v1= 8h(q1+ iq2+ q3+ iq4) = 8hu1,

˙v2= 8h(q1− iq2− q3+ iq4) = 8hu2.

In order to diagonalize the equations of motion, we introduce the following coordinates

ν1=√

8hu1+ v1, ν2=√

8hu2+ v2, µ1= −√

8hu1+ v1, µ2= −√

8hu2+ v2. In these new coordinates the flow of N is diagonal and its equations of motion(15) are given by

˙ ν₁=√

8hν₁, ν˙₂=√

8hν₂, , µ˙₁= −√

8hµ₁, µ˙₂= −√ 8hµ₂.

The action induced by ¹₂(ζ + l3),¹₂(ζ − l3)on KS space can be deduced from equation (6). Using equation (6), we can express how the flow of ¹₂(ζ +l₃),¹₂(ζ − l3)

acts on ν1, ν2, µ1 and µ1. Flowing for a time (t, s) along the flow induced by ¹₂(ζ + l₃),¹₂(ζ − l₃)can be described as

T²× C⁴→ C⁴: (s, t), (ν1, ν2, µ1, µ2) → (eîtν1, eîsν2, eîtµ1, eîsµ2).

If we now consider the total action of ¹₂(ζ + l₃),¹₂(ζ − l₃)and N, we obtain the following T³ action

(ν₁, ν₂, µ₁, µ₂) → (e^it+

√8hzν₁, e^is+

√8hzν₂, e^it−

√8hzµ₁, e^is−

√8hzµ₂). (16) This action is not effective for h < 0, as the element (s, t, z) = (π, π,^√_−8h^π ) gives rise to a trivial action. By using different generators, we can obtain an effective action. Throughout the rest of this subsection, we will assume that h < 0. An effective action is generated by

(J₁, J₂, J₃) = 1

2(ζ + l₃),1

2(ζ − l₃),1 2

(l₃+ ζ) + (l₃− ζ) + 1

√−8hN , (17) whose action is given by

(ν1, ν2, µ1, µ2) → (eît+izν1, eîs+izν2, eîtµ1, eîsµ2). (18)

(17)

This action is indeed effective as the identity, given by (s, t, z) = (0, 0, 0), is the only element in (R/2πZ)³ that results in a trivial action.

While the entire manifold has trivial isotropy, there are submanifolds in {(ν1, ν2, µ1, µ2)} ∈ C⁴ with non-trivial isotropy. These submanifolds can be characterised as sets

S_I = {(a₁, a₂, a₃, a₄), where ai∈ C for i ∈ I and aj= 0for i 6∈ I} ∈ C⁴, (19) where I is a subset of {1, 2, 3, 4} with at most two elements. All sets of this form have non-trivial isotropy. Hence we can state the following lemma as a summary of this subsection.

Lemma 4.1. Let h < 0 and consider the T³action generated by the map given in (17), then its action is given by equation (18) and this action is effective.

Furthermore, for any I ⊂ {1, 2, 3, 4} with at most two elements, the sets SI

has non-trivial isotropy.

The proof of this lemma comes down to repeating the steps done at the start of this section to obtain the flow of

1

2(ζ + l3),1

2(ζ − l3),1 2

(l3+ ζ) + (l3− ζ) + 1

√−8hN .

From equation (18), we conclude that only the element (s, t, z) = (0, 0, 0) results in a trivial action. Hence the group action is effective. The isotropy of any of the submanifolds can be determined by demanding that the exponents in front of the non-zero coordinates should equal 1. Doing so gives two independent equations for tree unknowns, which implies that the solution set is a 1-dimensional set.

4.2 Non-trivial isotropy in the Kepler system

Though all of the sets SIhave non-trivial isotropy, not all of these sets satisfy all equations that must necessarily be satisfied by any physically relevant solution to the Kepler system. In this section we will check which sets with non-trivial isotropy correspond to a physically relevant solutions. In order to do so, we recall some formulas. Let i = 1, 2 in the following. We will recall from the definition of µi and νi as in equation (15) that

u_i= 1 2√

8h(ν_i− µi), vi= 1

2(νi+ µi).

We furthermore recall that

ζ = ι1+ ι2, l₃= ι₁− ι2,

(18)

where

ι_i= i 8

u_i¯v_i− ¯u_iv_i . We also require the following formulas

α₁= 1

4(v₁¯v₁− 8hu1u¯₁), α2= 1

4(v2¯v2− 8hu2u¯2), γ = h

2|¯u2v1− u1v¯2|²+ 1

16|v1v¯2− 8hu1u¯2|²,

δ = 1 32

(¯u2v1− u1¯v2)(¯v1v2− 8h¯u1u2) + (u2v¯1− ¯u1v2)(v1v¯2− 8hu1u¯2) . To obtain these formulas for γ and δ, we used the expression for w^±k with k = 1, 2 as given in equation (12). In order to determine which sets SI correspond to a physically realistic situation for h > 0, the easiest approach is to observe that one can obtain the following implications

ν₁= 0 ⇒ α₁= 0, ν2= 0 ⇒ α2= 0, µ₁= 0 ⇒ α₁= 0, µ2= 0 ⇒ α2= 0.

(20)

Since we demand that N = α1+ α2= 4, a situation where α1= α2= 0does not correspond to a physically relevant solution. Hence, for h > 0, there are at most two physically relevant submanifolds with non-trivial isotropy, namely S_{1,3}and S{2,4}. In order to be a physically relevant solution, a solution needs to satisfy N = 4 and ζ = 0.

Since the equations describing S{1,3}and S{2,4} are symmetric, we will only deal with the set S{2,4}. For points in S{2,4}the coordinates ν1= µ1= 0, which implies that u1= v₁ = 0 and therefore ι1= 0 (as well as ρ1 = χ₁ = ψ₁ = 0).

Using the relations for ζ and l3 and setting ζ = 0, we find that l3= 0.

The only remaining equation which needs to be checked in order to claim that there are points in S{2,4}that correspond to a physically relevant solution is the equation 4 = α2, or equivalently

16 = 4α2= |v2|²− 8h|u2|². (21) By using a geometric point of view, we see that if we pick any value for u2in C², then there is a circle of values for v2such that the equation (21) is satisfied.

Hence, the geometry of the solution set is isomorphic to C × S¹. Thus we can conclude that this case indeed corresponds to a physical relevant solution.

For h < 0, the implications given by (19) are no longer valid. Hence we will need to treat all sets. These sets are:

(19)

• (Case I) If we consider S{1,3} (respectively S{2,4}), then we will be in a similar case as in the positive energy case where there is a physically relevant solution. This group corresponds to γ, δ, l3 = 0 and α1 = 4 (respectively α2= 4).

• (Case II) If we consider S{1,2} (resp. S{3,4}), then by expressing vi in terms of uiby using µ1= µ2= 0(resp. ν1= ν2= 0), we can see that the equation ζ = 0, implies that u1u¯1+ u2u¯2 = 0, thus ui = vi = 0. Hence this case does not correspond to a physical solution.

• (Case III) If we consider S{2,3} (resp. S{1,4}). If we follow the same approach as in Case II, then we find that ζ = 0 gives u1u¯1− u2u¯2 = 0, which can be satisfied. This relation implies α1 = α2 = 2. Furthermore we find that γ = δ = 0 and

l3= 2ι1= −4i 8

√

8hu1u¯1= − 4

√−8h8 8α 32h = −

√−8h 4h =

√ 8 4√

−h (resp. l3= −

√8 4√

−h), which gives h = −_2l¹2 3.

The sets corresponding to an index set I with fewer than 2 elements, will satisfy also all the equations which are satisfied in Case II, hence they do not correspond to a physically relevant solution.

4.3 Concluding remarks

While the results in this section might seem computational, these results can be visually linked to figure 1.

For negative values of h, there are four physically relevant submanifolds SI

with non-trivial isotropy.

Two of these submanifolds, S{1,3} and S{2,4} satisfy νi = µ_i = 0 for some i ∈ {1, 2}, which implies that γ = δ = l3= 0and 0 = αi6= αj= 4for i, j = 1, 2 with i 6= j. Thus S{1,3}and S{2,4} correspond to the non-smooth points of the lemon-like shape in figure 1. As mentioned earlier, for h < 0 the set of points satisfying equation (11) shrinks as |l3| increases until the value of l3² = _−2h¹ is reached after which there are no more feasible points. The remaining two physically relevant submanifolds with non-trivial isotropy are given by the sets S_{2,3} and S{1,4}. These sets satisfy νj = µi = 0 for i, j = 1, 2 with i 6= j.

The corresponding value for l3 is given by l3 = ±^√¹

−2h. These submanifolds correspond to the situation in which only a single point satisfies equation (11).

For positive values of h, there are only two physical relevant submanifolds SI

with non-trivial isotropy, namely S{1,3} and S{2,4}. As in the negative energy case, these two submanifolds correspond to the points where equation (11) is not differentiable, or equivalently to the non-smooth points shown in figure 1.

Hence we can say that for both positive and negative values of h, the physical relevant submanifolds with non-trivial isotropy correspond to singularities of the surface given by equation (11).

(20)

From equation (18) the generators of the isotropy groups for the physically relevant submanifolds with non-trivial isotropy can be determined. All these submanifolds posses S¹isotropy. Let XJ_i be the vector field generated by Jifor i = 1, 2, 3. For the submanifolds SI, we have the following results

• The action induced by equation (17) is trivial on S{1,3} if and only if t = z = 0. Hence the corresponding isotropy group is generated by the flow of (XJ1, 0, 0).

• The action induced by equation (17) is trivial on S{1,4} if and only if t = −z and s = 0. Hence the corresponding isotropy group is generated by the flow of (0, XJ₂, −XJ₃).

• The action induced by equation (17) is trivial on S{2,3} if and only if s = −z and s = 0. Hence the corresponding isotropy group is generated by the flow of (XJ₁, 0, −XJ₃).

• The action induced by equation (17) is trivial on S{2,4} if and only if s = z = 0. Hence the corresponding isotropy group is generated by the flow of (0, XJ₂, 0).

5 Level sets of G

In order to show monodromy, we will analyse one more integral of motion, which comes from separation in spheroidal coordinates. The bifurcation diagram that we will obtain in this section, plays a crucial role in the next section where we will show that the Kepler system has monodromy for particular reductions and for a range of values of h. The integral of motion that we will study in this section is described in [10] and is given by

G = l²− 2e3= γ − α₁α₂ 16h +1

2l²₃−1

2(α2− α1).

By substituting α2= 4 − α1in the previous equation, we obtain

G(α₁, γ, δ, l₃) =γ − 4α1+ α²₁

16h +1

2l²₃− (2 − α₁). (22) Equation (22) can be used obtain an expression for γ. Denote the value of G by g, then the level set of G = g is given by

γ = 16h g −1

2l²₃+ 2

+ (4 − 16h)α₁− α²₁. (23) We will analyse the level sets of G on the surface given by equation (11) i.e.

γ²− 32hδ²= (α₁²+ 8hl₃²)((4 − α1)²+ 8hl²₃), where γ ≥ 0 for h > 0 and for h < 0, we demand that 0 ≤ α1≤ 4.

(21)

The level sets going trough the singular points are of special interest, since these points have been shown to possess non-trivial isotropy. These singular points are

x0= (α1, γ, δ, l3) = (0, 0, 0, 0), x4= (α1, γ, δ, l3) = (4, 0, 0, 0).

We will discuss the level sets of these points in the next subsections.

5.1 The level sets of G through singular points for h > 0

Let i ∈ {0, 4}. In this subsection we will determine which points x satisfy both G(x) = G(xi)and equation (11), with γ ≥ 0.

Let h > 0, the problem can be simplified by noting that δ does not appear in the equation for G. Hence having an equality in equation (11), with γ ≥ 0, is equivalent to satisfying equation (14), i.e.

γ²≥ (α²₁+ 8hl²₃)((4 − α₁)²+ 8hl²₃).

Since both x0and x4satisfy l3= 0, we can simplify the equations. Imposing that l3= 0 simplifies equation (14) to

γ ≥ |α²₁− 4α1|. (24)

The level sets of x0 and x4 are given by the equations

γx₀ = (4 − 16h)α1− α²₁, (25) γx₄= 64h + (4 − 16h)α1− α²₁. (26) We notice that these equations are both parabolas. For 0 < α1 < 4we can easily compare γx₀ and γx₄ to equation (24), which gives the following relation between γx₀ and γx₄

γx₀ = (4 − 16h)α1− α²₁< 4α1− α₁²< 4α1− α²₁+ 16h(4 − α1) = γx₄. From this equation and the fact that equation (26) gives a parabola, we can conclude that there are some functions γ(α1), δ(α1)and a number t ≤ 0 such that (α1, γ(α1), δ(α1))is in the level set of x4 for t ≤ α1 ≤ 4. For the level set of x0 there are no points in the level set of x0 with α1 > 0, which can be seen from the relation γx0 ≤ 4α1− α²₁≤ |4α1− α₁²|.

It remains to determine the sign of γx0− |4α1− α²₁|for negative α1, which we will do be considering the sign and roots of

(4 − 8h)α₁− α²₁. (27)

The roots of this equation are given by α1= 0and α1= 4 − 8h. Note that our calculations are only correct for α1 ≤ 0. Hence we must have 4 − 8h < 0 in order to find two distinct roots, or equivalently h >¹₂. Recalling that equation (27) gives rise to a parabola, we conclude that the level set of x0contains solely x0 if and only if h ≤¹₂.

(22)

5.2 The level sets of G through singular points for h < 0

Let i ∈ {0, 4}, in this subsection we will determine all points x which satisfy both G(x) = G(xi)and equation (11) with 0 ≤ α1≤ 4.

We will apply a similar trick as in the previous subsection. The existence of solutions for equation (11) with 0 ≤ α1 ≤ 4 for h < 0 is equivalent with a solution for (13), i.e.

γ²≤ (α²₁+ 8hl²₃)((4 − α1)²+ 8hl²₃).

Since both x0 and x4 satisfy l3 = 0, we can simplify the equations. Imposing that l3= 0 simplifies equation (14) to

|γ| ≤ 4α1− α²₁. (28)

The level sets of x0 and x4 are given by equation (25) and equation (26). For the level set γx0, we obtain the following relation for 0 ≤ α1≤ 4

|γx0| ≥ (4 − 16h)α1− α²₁≥ 4α1− α²₁. (29) Therefore, if we restrict to values such that 0 ≤ α1 ≤ 4 and equation (11) is satisfied, the level set of x0 consist only of x0 for h < 0. For the level set of x4, we observe

γx₄ = 4α1− α²₁+ 16h(4 − α1) ≤ 4α1− α²₁. Hence equation equation (28) is satisfied if and only if

γx₄ = 4α1− α²₁+ 16h(4 − α1) ≥ α²₁− 4α1.

Solving 4α1− α²₁+ 16h(4 − α1) − (α²₁− 4α1) = 0gives α1= 4and α1= −8h. Hence the expression is positive for α1 between 4 and −8h. We require that 0 ≤ −8h ≤ 4to have a solution satisfying equation (11) with 0 ≤ α1≤ 4. Thus, for negative h, equation (11) and G = 2 have one common solution if and only if h < ¹₂.

5.3 Concluding remarks on the level sets of G

Using the results of the previous subsections, we can determine the shape of the bifurcation diagram of G.

We first notice that the level set of G = 2 (respectively G = −2) splits the α₁, γ−plane in two parts. In one of these parts G ≤ 2 (respectively G ≤ −2) and in the other part G ≥ 2 (respectively G ≥ −2). Note that G is polynomial and thus if we consider G evaluated on a compact set, its minimum and maximum are attained. Combining this with the information obtained for the level sets of Gfor l3= 0, we conclude the following:

• For h ≤ −¹₂, we find that −2 ≤ G ≤ 2 .

• For −¹₂ < h < 0, we find that there exists some c > 2 such that 2 ≤ G ≤ c.

Reductions on the Kepler problem