Hamiltonian Vector Fields on a Space of Curves on the 3-Sphere

Hamiltonian vector ﬁelds on a space of curves on the 3-sphere

Md Ismail Hossain

A thesis submitted to the Faculty of Graduate Studies of The University of Manitoba in partial fulﬁlment of the requirements of the degree of

MASTER OF SCIENCE

Department of Mathematics University of Manitoba Winnipeg

This thesis reviews aspects related to the integrability of a Hamiltonian system on a space of arc length parametrized curves on the unit sphere S3 in R4 of a fixed length L. In particular, we find that the flow of the Hamiltonian vector field Z L corresponding to the total torsion function X(s) 7→ τ(s)ds generates the curve o shortening equation. Additionally, we show that the total torsion function belongs to a hierarchy of Poisson commuting functions. Acknowledgements

I would first like to thank my thesis supervisor Dr. Derek Krepski. I am and will forever be grateful for his valuable guidance, support and encouragement throughout my studies. He steered me in the right direction when I needed. I would also like to thank Dr. Adam Clay and Dr. Saman Muthukumarana for being a part of my thesis committee and providing valuable comments on this thesis. I would also like to thank the Department of Mathematics and my supervisors’ NSERC grants for their financial support during my studies here in the University of Manitoba. Finally, I would like to express my profound gratitude to my parents for their love, affection and support throughout my life. I would also like to thank my wife Lubana Yasmin for motivating me throughout my years of study. This thesis is dedicated to my beloved parents

i Contents

Introduction 1

1 Mathematical Background 8 1.1 Symplectic manifolds ...... 8 1.2 Hamiltonian vector ﬁelds ...... 13 1.3 Brackets ...... 15 1.4 Integrable systems ...... 17

2 A Space of Curves on the 3-Sphere 20 2.1 Preliminaries ...... 20

2.2 A Space of Curves in SU2 ...... 23

3 Hamiltonian vector fields and Complete integrability 30 Z L 3.1 Hamiltonian vector field of τds ...... 32 0 Z L 1 2 3.2 Hamiltonian vector field of 2 κ ds ...... 36 0 Z L 1 1 1 1 3.3 Hamiltonian vector field of (κ0)2 + κ2τ 2 + κ2τ + κ2 − κ4 ds 40 0 2 2 8 8 3.4 Poisson commuting functions ...... 46 3.5 Future research ...... 48

Appendices 48

A Lemmas in Chapter 2 49

ii B Computations in Chapter 3 51 B.1 Calculations in Section 3.1 ...... 52 B.2 Calculations in Section 3.3 ...... 57 B.3 Calculations in Section 3.4 ...... 64

Bibliography 69

iii Introduction

Let us begin with the notion of Hamiltonian mechanics. Classical mechanics has two main points of view, Lagrangian formalism and Hamiltonian formalism. Here, we intend to focus only on Hamiltonian formalism. Consider n particles A1,...,An

3 moving in R under the inﬂuence of respective forces F1,...,Fn. Recall, Newton’s second law of motion states that

Fi = miai, i = 1, . . . , n,

where mi denotes the mass of the i-th particle and ai denotes its acceleration due to the force Fi acting on it. For simplicity, we assume the force is conservative, that is,

3 there exist functions Vi : R → R, i = 1, . . . , n typically called potentials such that

Fi = −∇Vi. Letting qi denote the position of the i-th particle, Newton’s second law takes the form d2q −∇V = m i , i = 1, . . . , n, (1) i i dt2 d2q where i = a , acceleration of the i-th particle. Equation (1) represents a system dt2 i of second order diﬀerential equations and a unique solution to these equations can be obtained given the initial position and velocity of the particles. A convenient way to dq deal with the system (1) is to introduce momentum coordinates p = m i , where i i dt dq i denotes the velocity of the i-th particle. Let q = (q , . . . , q ), p = (p , . . . , p ) dt 1 n 1 n

1 and introduce a function

n 2 X kpik H(q, p) = + Vi(qi) i=1 2mi called the Hamiltonian, which represents the sum of kinetic energy and potential energy of the system. Then Newton’s second law (1) is equivalent to Hamilton’s equations  dq p ∂ kp k2 ∂H  i i i  = = + Vi(qi) =  dt mi ∂pi 2mi ∂pi (2)  2 dpi d qi ∂H   = mi 2 = −∇Vi(qi) = −  dt dt ∂qi which is a system of ﬁrst order diﬀerential equations. In the discussion above, the

3n parameter space of possible positions (q1, . . . , qn) ∈ R of the n particles is known as the conﬁguration space, while the position-momentum space (q1, . . . , qn, p1, . . . , pn) ∈

R3n × R3n is known as the phase space of the system. The dynamical system given by (2) is known as a Hamiltonian system with the Hamiltonian H. There are a few advantages of the Hamiltonian formulation of the system (1). One of these advantages can readily be seen as Hamilton’s equations are first order differential equations and hence relatively easier to solve than second order differential equations given by the system (1). Next, we will discuss how the Hamiltonian approach generalizes to other phase spaces. Let us choose the configuration space N of a dynamical system to be an m- dimensional differentiable manifold. Then the phase space is given by the cotangent bundle T ∗N of the manifold N. In the discussion above, N = R3n and T ∗N = R3n × R3n. The cotangent bundle T ∗N naturally has a symplectic structure, that is, T ∗N is a 2m-dimensional differentiable manifold equipped with a closed non- degenerate differential 2-form. See Example 1.1.2. This is the starting point of symplectic geometry, which generalizes the above discussion to ‘phase spaces’ that are not necessarily the cotangent bundle of some configuration space manifold.

2 Solutions to the system (2) can be viewed as curves γ(t) = (q(t), p(t)) on the phase space R3n × R3n, whose tangent vectors must agree with the vector ﬁeld

∂H ∂H XH = , − . ∂pi ∂qi

Notice that the function H determines the vector ﬁeld XH on the right side of the equation (2). In the more general setting of symplectic geometry, how does a

Hamiltonian H : M → R determine a vector ﬁeld on a “phase space” given by a symplectic manifold M? The answer to this question is precisely what a symplectic structure achieves. The Hamiltonian vector ﬁeld XH corresponding to the function H is determined via the relation

dH(V ) = ω(XH ,V ), ∀ vector ﬁelds V, (3) where ω is the symplectic form on M and dH(V ) denotes the directional derivative of H in the direction of V . In this way, one may write down Hamilton’s equations as d (γ(t)) = X (γ(t)). (4) dt H

Then the solutions to (4) are integral curves of the Hamiltonian vector ﬁeld XH . A Hamiltonian system (4) may incorporate conserved quantities. For example, the function H itself is conserved, i.e. H remains constant along the integral curves of XH . This is easy to see in case of (2), since

d n ∂H dq ∂H dp H(q, p) = X i + i dt i=1 ∂qi dt ∂pi dt

n ∂H ∂H ∂H ∂H = X + − i=1 ∂qi ∂pi ∂pi ∂qi

= 0.

3 Also H is conserved in the more general setting as it is immediate from equation (3)

dH(XH ) = ω(XH ,XH ) = 0.

More generally, for any other function f, df(XH ) represents the directional derivative of f in the direction XH of the evolution of the system. Hence, f is conserved along the flow of the system if and only if df(XH ) = 0. In this case, we say the functions f and H Poisson commute (see section 1.4). Other conserved quantities, also known as integrals of motion, may exist and if there are sufficiently many (see Definition 1.17) of these, then we say that the system (4) is completely integrable. Existence of enough integrals of motion allows one to introduce new coordinates in which the system is solvable by quadrature, where quadrature stands for integrals, definite or indefinite. Sometimes it is not possible to obtain an analytic solution to the system. In this case, numerical methods are often used to solve the system and the system is still considered integrable. In case of infinite dimensional phase spaces, we can use a similar approach as in the finite dimensional case to write down Hamilton’s equations and to look for integrals of motion. The focus of this thesis is to study the complete integrability of a Hamiltonian system defined on a particular infinite dimensional symplectic manifold.

Literature review

For the purpose of this thesis, we closely follow the paper [6] authored by V. Jurdjevic. In [6], V. Jurdjevic showed that the integral curves of Hamiltonian vector ﬁelds corresponding to respective integrals of motion generate some well known equations in physics and also discussed their integrability. More precisely, in [6], V. Jurdjevic used a symplectic structure on an inﬁnite dimensional manifold M of arc length parametrized curves X(s) on the 3-sphere S3 = {x ∈ R4 : kxk = 1} to compute

4 flows of Hamiltonian vector fields. We will return to the precise description of M in Chapter 2. In what follows, we describe the formalism that lead to the equations under consideration. Z L 1 2 Consider the function f2 : M → R given by X(s) 7→ κ (s)ds, where κ(s) 2 0 denotes the curvature of X(s), and let Xf2 denote its Hamiltonian vector field. The

ﬂow of Xf2 deﬁnes a “curve” in M, which is a variation of curves Y (s, t), with Y (s, 0) = X(s). As we will see in more detail in Chapter 2, the tangent vectors

Λ(s, t) along each curve s 7→ Y (s, t) are, roughly speaking, in R3. In [6], it is shown that Λ(s, t) evolves according to the equation

∂Λ ∂2Λ = × Λ, (5) ∂t ∂s2 which is known as Heisenberg’s magnetic equation (HME) in the literature (see for instance [17], [8], [6]). Jurdjevic also computes the Hamiltonian vector ﬁeld Z L 1 2 1 2 corresponding to the function f3 = κ (s)τ(s) + κ ds (see Theorem 7 in [6]), 0 2 4 where τ(s) denotes torsion, and claims that the integral curves of the Hamiltonian Z L vector ﬁeld associated with the total torsion function f1 = τ ds generate the 0 equation ∂Λ = κN (6) ∂t where N denote the unit normal vectors to the curves in Y (s, t).

As shown in [6], the functions f1, f2, f3 and

Z L 1 0 2 1 2 2 1 2 1 2 1 4 f4 = (κ ) + κ τ (s) + κ τ(s) + κ − κ ds 0 2 2 2 8 8 pairwise Poisson commute, which suggests they may be a part of an inﬁnite list of Poisson commuting functions, that is, a completely integrable system. Functions similar to the above appear in the paper [10] by J. Langer and R.

5 Perline, which considers a similar space of curves in Euclidean space R3. The authors of [10] showed that the families of curves γ(s, t) in R3 evolve according to the filament ∂γ equation = κB may be viewed as a completely integrable Hamiltonian system. ∂t Complete integrability of the filament equation implies that the space of curves possesses infinitely many functions which are pairwise Poisson commuting. The successive functions in this infinite list are obtained through the repeated application of a recursion operator. The recursion operator is constructed using two compatible Poisson structures on the space of curves.

The functions f1, f4 are also mentioned in [16], where the authors show the nonlinear Schrödinger equation is a completely integrable Hamiltonian system. In this thesis, we consider an infinite dimensional manifold M, a space of arc length parametrized curves on S3. The manifold M admits a symplectic structure, and we compute the Hamiltonian vector fields corresponding to the functions f1, f2, f4 defined on M and show that f1, f2, f3, f4 pairwise Poisson commute. It seems plausible to have an infinite list of Poisson commuting functions on M. How- ever, at this point, we do not have any mechanism like the recursion operator mentioned in [10] to produce successive functions in this infinite list.

Outline of the thesis

This thesis consists of three chapters. In Chapter 1, we summarize the overall con- cepts that are necessary to understand this thesis. Introduction to symplectic manifolds, Hamiltonian vector ﬁelds and integrable systems are provided in this chapter. Chapter 2 describes the space of curves of our interest along with some basic properties of the Lie group SU2, the set of all 2 × 2 unitary matrices having determinant 1. Chapter 3 contains all the results of this thesis. In particular, we establish the

6 following result: Z L Proposition 3.1. Let f1 : M → R be the function given by f1 = τ ds and let 0

Xf1 denote its Hamiltonian vector ﬁeld. If g(t, s) denote the integral curves of Xf1 ∂g with tangent vectors (s, t) = g(s, t)Λ(s, t), then Λ(s, t) evolves according to the ∂s Curve Shortening equation (CSE),

∂Λ (s, t) = κ(s, t)N(s, t). (7) ∂t

Similarly, in Proposition 3.3, we give the evolution equation for the Hamiltonian vector ﬁeld corresponding to the function f4.

Finally, we show that the functions f1, f2, f3, f4 on M pairwise Poisson commute. Suggestions for further research are described.

7 1

Mathematical Background

In this chapter, we present some required background for this study. The reader is assumed to be familiar with introductory differential geometry. In Section 1.1, we focus on discussing finite dimensional and infinite dimensional symplectic manifolds along with some basic definitions. In Section 1.2, we discuss Hamiltonian vector fields and their integral curves with examples. Section 1.3 reviews the notion of Lie bracket, Lie algebra and Poisson bracket. Section 1.4 contains the notion of complete integrability of a Hamiltonian system. We followed [1], [2], [9], [12] [14], [15] for the material presented in this chapter.

1.1 Symplectic manifolds

Let V be an N-dimensional vector space over R, and let ω : V × · · · × V → R be a | {z } k k-linear map. That is,

ω(v1, . . . , αvi, . . . , vk) = αω(v1, . . . , vi, . . . , vk)

0 0 ω(v1, . . . , vi + vi, . . . , vk) = ω(v1, . . . , vi, . . . , vk) + ω(v1, . . . , vi, . . . , vk)

8 0 for α ∈ R and v1, . . . , vi, vi, . . . , vk ∈ V . The k-linear map ω is said to be skew- symmetric if for any pair of arguments

ω(v1, . . . , vi, . . . , vj, . . . , vk) = −ω(v1, . . . , vj, . . . , vi, . . . , vk)

Deﬁnition 1.1. Let M be a smooth manifold. Let k > 0 be an integer. A diﬀerential k-form ω is a smooth assignment p 7→ ωp, where p ∈ M and ωp is a skew-symmetric k-linear map

ωp : TpM × · · · × TpM → R | {z } k where TpM denotes the tangent space to M at p.

Definition 1.2. Let M be a smooth manifold and ω a differential k-form. The exterior derivative dω is a differential (k + 1)-form defined by

k+1 X i+1 ˆ dω(X1,...,Xk+1) = (−1) Xi ω(X1,..., Xi,...,Xk+1) i=1

X i+j ˆ ˆ + (−1) ω [Xi,Xj],X1,..., Xi,..., Xj,...,Xk+1 1≤i

where X1,...,Xk+1 are smooth vector ﬁelds on M, and the entries with a hat sign are absent in the expression. [14],[15].

Remark 1.1.1. For finite dimensional manifolds, the definition of exterior derivative as in Definition 1.2 is known to be equivalent to the local form in coordinates.

That is, given a chart (U, x1, . . . , xm) on M, the exterior derivative of the k-form

X ∞ ω = fI dxI where I denotes multi-index 1 ≤ i1 < ··· < ik ≤ m and fI ∈ C (U) I are functions is deﬁned as

m X X ∂fI dω = dxl ∧ dxI . I l=1 ∂xl

9 See Theorem 13 on page 213 in [14] for the proof of equivalence. On infinite dimensional manifolds, the local formula is inadequate and we use the global formula as a suitable substitute. Section 33 in Chapter 7 in [9] has a careful analysis of differential forms on infinite dimensional manifolds. As noted in Remark 33.22 therein (see also the Lemma following Remark 33.22) Definition 1.1 is compatible with the global definition given in Definition 1.2.

Let ω be a k-form and ξ be a l-form on M. Then the exterior derivative operator d satisﬁes the following properties,

(a) d is linear.

(b) d(ω ∧ ξ) = dω ∧ ξ + (−1)k ω ∧ dξ.

Deﬁnition 1.3. A diﬀerential k-form ω is said to be exact if ω = dα, where α is a (k − 1) form and it is said to be closed if dω = 0. All exact forms are closed but not all closed forms are exact.

Deﬁnition 1.4. A diﬀerential 2-form ω is said to be symplectic if it is closed and non-degenerate. Here, non-degeneracy means that for all p ∈ M, if there exists a vector u ∈ TpM such that ωp(u, v) = 0 for all v ∈ TpM, then u = 0.

Deﬁnition 1.5. A symplectic manifold is a pair (M, ω) where M is a manifold and ω is a symplectic form on M. Sometimes, we refer to the symplectic form as a symplectic structure in this thesis.

On a ﬁnite dimensional manifold M, the form ω being symplectic implies that dim TpM= dim M must be even. This is due to the fact that non-degenerate skew-symmetric bilinear forms can exist only on even dimensional vector spaces.

10 2n Example 1.1.1. The manifold M = R with coordinates x1, . . . , xn,y1, . . . , yn and n X the form ω = dxi ∧ dyi is a symplectic manifold. i=1 To check ω is closed, notice that

n X ω = dxi ∧ dyi i=1 n X = d xi dyi i=1

n X where xi dyi is a 1-form and hence ω is exact. Therefore ω is closed. i=1

n ∂ ∂ X To prove the non-degeneracy of ω, let u = ai +bi be a vector in i=1 ∂xi p ∂yi p TpM, the tangent space to M at p. The form ω is non-degenerate at p if ωp(u, v) = 0

n ∂ ∂ X for all v = fi + gi in TpM, then u = 0. Now, i=1 ∂xi p ∂yi p

  n ∂ ∂ ∂ ∂ X ωp(u, v) = dxi ai + bi dyi fi + gi  i=1 ∂xi p ∂yi p ∂xi p ∂yi p

  n ∂ ∂ ∂ ∂ X − dyi ai + bi dxi fi + gi  i=1 ∂xi p ∂yi p ∂xi p ∂yi p

n X = aigi − bifi i=1

n n X X 2 2 So, aigi − bifi = 0, ∀fi, gi. Let gi = ai and fi = −bi. Then ai + bi = 0 i=1 i=1 implies ai, bi = 0 and consequently u = 0. Thus if ωp(u, v) = 0 for any choice of v, then u = 0. Therefore, ω is non-degenerate.

Example 1.1.2. Let N be an n-dimensional manifold. Then its cotangent bundle M = T ∗N naturally has a symplectic structure. Let π : T ∗N → N be the projection map and θ be a co-vector in T ∗N. Then the

∗ 1-form α at θ is the map αθ : Tθ(T N) → R given by αθ(v) = θ (Tθπ)(v) , where

∗ v ∈ Tθ(T N) and Tθπ denotes the tangent map of π at θ.

11 ∗ ∗ Again, let us choose a chart T U, q1, . . . , qn, p1, . . . , pn) on T N. Then the 1- n X form in terms of coordinates is deﬁned as β = pi ∧ dqi. The deﬁnition of 1-form i=1 is actually independent of the choice of charts on T ∗N. This can be seen from the n ∂ ∂ o following calculation as both α and β agree on a basis , , i = 1, . . . , n for ∂qi ∂pi the tangent space of T ∗N at any point θ. So,

n ∂ ∂ X ∂ (i) αθ = θ Tθπ( ) = pi ∧ dqi = pi, since θ itself is a co-vector ∂qi ∂qi i=1 ∂qi n ∂ X ∂ and β = pi ∧ dqi = pi. ∂qi i=1 ∂qi

∂ ∂ ∂ ∂ (ii) αθ = θ Tθπ( ) = 0, since Tθπ sends to zero and β = ∂pi ∂pi ∂pi ∂pi n X ∂ pi ∧ dqi = 0. i=1 ∂pi

n X Set ω = −dβ = dqi ∧ dpi. Then ω is closed and non-degenerate as shown i=1 in the previous example. Such a form ω is called the canonical symplectic form on T ∗N.

In this thesis, we will work with an infinite dimensional manifold. In general, infinite dimensional manifolds are manifolds modeled on an infinite dimensional locally convex vector space as the finite dimensional manifolds are modeled on Rn. There are various models of infinite dimensional manifolds in the literature such as Ba- nach manifold, Hilbert manifold, Fréchet manifold etc. We will be using the Fréchet manifold here. The focus of this thesis is to work on a particular Fréchet manifold which will be defined in Chapter 2. The differential forms on an infinite dimensional Fréchet manifold are defined in the same manner as defined on the finite dimensional manifolds. As a result, the formulas associated with the differential forms (such as, exterior derivative, Cartan’s magic formula) on finite dimensional manifolds still hold on Fréchet manifold.

12 ∗ For every p ∈ M the symplectic form ω induces a map ω˜p : TpM → (TpM) which is injective, by non-degeracy of ωp : TpM × TpM → R. When this map is also surjective, ω is said to be strongly non-degenerate. In ﬁnite dimensions, these are equivalent; however in inﬁnite dimensions, this is not the case. Thus sometimes ω is called ‘weakly’ non-degenerate.

1.2 Hamiltonian vector ﬁelds

Let (M, ω) be a ﬁnite dimensional symplectic manifold and let h : M → R be a smooth function. The diﬀerential dh of h is a 1-form. By non-degeneracy of ω there

exists a unique vector field Xh on M such that ιXh ω = ω(Xh, −) = dh. For infinite dimensional symplectic manifolds, it is possible that the 1-form dh of a function h does not equal ω(X, −) for any vector field X. This is due to the weak non-degeneracy of the symplectic form. In spite of this fact, Hamiltonian vector fields are defined in the same manner as in the finite dimensional case.

Deﬁnition 1.6. A vector ﬁeld Xh on (M, ω) for which ω(Xh, −) = dh is called a

Hamiltonian vector ﬁeld for the function h : M → R.

Example 1.2.1. Let (R2, ω) be a symplectic manifold with the symplectic form ω = dx ∧ dy. Also let the Hamiltonian function h : R2 → R be given by h(x, y) =

2 2 x + y . We will compute the Hamiltonian vector ﬁeld Xh at a point p = (a, b) on

2 R . We need to show that ω(Xh,V ) = dh(V ), where Xh and V are vector ﬁelds of ∂ ∂ ∂ ∂ the form X = f(x, y) + g(x, y) and V = v (x, y) + v (x, y) respectively. h ∂x ∂y 1 ∂x 2 ∂y

Let γ(t) = (a + tv1, b + tv2) be the ﬂow generated by V . Then

dhp(V ) = h(a + tv1, b + tv2) dt 0

d 2 2 = [(a + tv1) + (b + tv2) ] dt 0

13 2 2 = 2av1 + 2tv + 2bv2 + 2tv 1 2 t=0

= 2av1 + 2bv2 (1.1)

On the other hand,

∂ ∂ ∂ ∂ ω(X ,V ) = (dx ∧ dy) f + g , v + v h ∂x ∂y 1 ∂x 2 ∂y

∂ ∂ ∂ ∂ = dx(f + g ) dy(v + v ) ∂x ∂y 1 ∂x 2 ∂y

∂ ∂ ∂ ∂ − dx(v + v ) dy(f + g ) 1 ∂x 2 ∂y ∂x ∂y

= fv2 − gv1 (1.2)

From equation (1.1) and (1.2), we have f = 2b and g = −2a. Therefore, the ∂ ∂ Hamiltonian vector ﬁeld is X = 2b − 2a corresponding to the Hamiltonian h. h ∂x ∂y

Definition 1.7. Let X be a vector field on M. For every point p ∈ M, suppose d there exists a unique curve Φ (p) such that Φ (p) = p and [Φ (p)] = X[Φ (p)] for t 0 dt t t all t. Then the curve Φt(p) is said to be an integral curve or local flow of X. Then

Φt(Φs(p)) = Φt+s(p) whenever both sides are deﬁned.

∂ ∂ Example 1.2.2. Let X = 2x − 2y be a vector ﬁeld on M = 2. Then the ∂x ∂y R system of diﬀerential equations generated by X is,

dx   = 2x  dt  dy  = −2y  dt

Solving the above equations by separation of variables with initial condition p =

2t −2t (x0, y0), we obtain x(t) = x0e and y(t) = y0e . Thus an integral curve of X is

14 2t −2t Φt(p) = (x0e , y0e ).

1.3 Brackets

We can view a vector ﬁeld on M as a linear map X : C∞(M) → C∞(M) that satisﬁes the product rule,

X(fg) = X(f)g + fX(g) for f, g ∈ C∞(M). That is, X is a derivation of the algebra C∞(M) of smooth functions. The vector ﬁeld X acting on a smooth function f is given by,

X(f) = df(X) where df(X) is the directional derivative of f in the direction of X.

Example 1.3.1. For a pair of vector ﬁelds X1,X2 on M, the commutator

∞ ∞ [X1,X2] = X1 ◦ X2 − X2 ◦ X1 : C (M) → C (M)

is again a vector field. To show [X1,X2] is a vector field, we verify that it satisfies the product rule stated above by direct calculation. We obtain,

(X1 ◦ X2)(fg) = X1 X2(f)g + fX2(g)

= X1(X2(f))g + fX1(X2(g)) + X2(f)X1(g) + X1(f)X2(g) (1.3)

Similarly,

(X2 ◦ X1)(fg) = X2(X1(f))g + fX2(X1(g)) + X1(f)X2(g) + X2(f)X1(g) (1.4)

15 Subtracting equation (1.4) from (1.3) we get,

[X1,X2](fg) = X1(X2(f))g + fX1(X2(g)) − X2(X1(f))g − fX2(X1(g))

= (X1 ◦ X2 − X2 ◦ X1)(f)g + f(X1 ◦ X2 − X2 ◦ X1)(g)

= [X1,X2](f)g + f[X1,X2](g)

Therefore, [X1,X2] is a vector ﬁeld.

Definition 1.8. The vector field [X1,X2] = X1 ◦ X2 − X2 ◦ X1 is called the Lie bracket of the vector fields X1,X2.

Definition 1.9. The Lie derivative of a differential k-form ω is defined as

LX ω = ιX dω + dιX ω (1.5)

This identity is known as Cartan’s magic formula. See Theorem (4.3.3) in [12].

Proposition 1.10. If X1 and X2 are Hamiltonian vector ﬁelds on a symplectic manifold (M, ω), then their Lie bracket [X1,X2] is a Hamiltonian vector ﬁeld with

Hamiltonian ω(X2,X1). [2].

Proof. For the symplectic form ω we have,

ι[X1,X2]ω = LX1 ιX2 ω − ιX2 LX1 ω

= dιX1 ιX2 ω + ιX1 dιX2 ω − ιX2 dιX1 ω − ιX2 ιX1 dω [Using (1.5)]

= d ω(X2,X1)

The second and third term vanish since ιX2 ω and ιX1 ω are exact forms and thus their exterior derivatives are zero. The fourth term also vanishes since ω is a symplectic

16 form.

Deﬁnition 1.11. A Lie algebra is a vector space g together with a bilinear map [·, ·]: g × g → g called the Lie bracket such that it satisﬁes the following properties:

(i) [X1,X2] = −[X2,X1], ∀ X1,X2 ∈ g,

(ii) [X1, [X2,X3]] + [X2, [X3,X1]] + [X3, [X1,X2]] = 0, ∀ X1,X2,X3 ∈ g

That is, the Lie bracket is skew-symmetric and satisﬁes the Jacobi identity.

Deﬁnition 1.12. Let (M, ω) be a symplectic manifold. Let f, g : M → R be two smooth functions on M. Then the Poisson bracket of f and g is deﬁned as,

{f, g} = ω(Xf ,Xg).

1.4 Integrable systems

Deﬁnition 1.13. A Hamiltonian system is a triple (M, ω, h), where (M, ω) is a symplectic manifold and h : M → R is a smooth function, called the Hamiltonian. [2].

Theorem 1.14. Let (M, ω, h) be a Hamiltonian system. Let f ∈ C∞(M). Then

{f, h} = 0 if and only if f is constant along integral curves of Xh. [2].

Proof. From the deﬁnition of Poisson bracket we have,

{f, h} = − ω(Xf ,Xh)

= − df(Xh)

That is, {f, h} is the directional derivative of f along the integral curve of Xh. Hence

{f, h} = 0, if and only if f is constant along the integral curves of Xh.

17 Deﬁnition 1.15. Let (M, ω) be a symplectic manifold. Two functions f, g ∈ C∞(M) are said to Poisson commute if {f, g} = 0.

Deﬁnition 1.16. A function f as in Theorem 1.14 is said to be an integral of motion for the Hamiltonian system (M, ω, h). [2].

A symplectic vector space is a pair (V, ϕ), where V is a vector space over R and ϕ : V × V → R is a non-degenerate bilinear skew-symmetric form. Let U be subspace of V and U ⊥ = {v ∈ V : ϕ(v, u) = 0, ∀ u ∈ U} be the orthogonal complement subspace of U. Then V = U ⊕ U ⊥. The subspace U is called isotropic

⊥ ⊥ if ϕ|U = 0, i.e. U ⊆ U . By non-degeneracy, dim V = dim U + dim U . Since V is 2n dimensional, the dimension of U can be at most n. 1 The functions f , f , ··· , f on M, where n = dim(M) are said to be indepen- 1 2 n 2 dent if their diﬀerentials (df1)p, (df2)p, ··· , (dfn)p are linearly independent at every point p in some open dense subset of M. Let f1, f2, ··· , fn be pairwise Poisson commuting integrals of motion on the symplectic manifold (M, ω). Let Up ⊆ TpM be a subspace at p spanned by the vectors X1|p , ··· ,Xn|p. For any two vectors X X u1 = ai Xi|p and u2 = bj Xj|p in Up, i j

X X ωp(u1, u2) = aibj ω(Xi|p ,Xj|p) = aibj{fi, fj}(p) = 0 i,j i,j

Therefore, Up is an isotropic subspace of the tangent space TpM at p and Up can be at most n dimensional. See [2] for details.

Deﬁnition 1.17. A Hamiltonian system (M, ω, h) is completely integrable if there 1 exists n = dim(M) independent integrals of motion f = h, f , ··· , f on M, 2 1 2 n which are pairwise Poisson commuting with respect to the Poisson bracket. That is,

{fi, fj} = 0, for all i, j. [2]. If M is inﬁnite dimensional, we will say that (M, ω, h) is completely integrable whenever there exist inﬁnitely many independent functions f1, f2, f3, ··· which are pairwise Poisson commuting.

18 The existence of n independent Poisson commuting functions allows one to write down Hamilton’s equations in a new coordinate system. In this case, Hamilton’s equations are linear and solutions to these equations can be obtained analytically. For further details, please see Liouville’s Theorem on page 272 in [1].

19 2

A Space of Curves on the 3-Sphere

In this chapter, we develop some theory that will enable us to carry out future computations. We begin with recalling some elementary facts about the Lie group

SU2, the set of all 2 × 2 unitary matrices of determinant 1. The covariant derivative

3 on SU2 obtained through the identification of SU2 with S will be used later in this chapter to describe the tangent space of the manifold of our interest M. The manifold M is defined as a certain space of arc length parametrized curves. Finally, we define the symplectic form on M as used in [6]. The contents of this chapter are relying upon [4], [5], [6], [7], [13].

2.1 Preliminaries

A Lie group G is a smooth manifold with a group structure so that the group operations are smooth maps, i.e. the multiplication (g, h) 7→ gh and inverse g 7→ g−1 are smooth maps. Here are few examples of matrix Lie groups:

(i) The general linear group, GLn(C) = {A ∈ Mn(C): det(A) 6= 0}.

(ii) The special linear group, SLn(C) = {A ∈ Mn(C): det(A) = 1}.

20 ∗ (iii) The special unitary group, SUn = {A ∈ Mn(C): det(A) = 1, AA = I}.

In this thesis, we are interested in the group SU2, the set of all 2×2 unitary matrices    u v having determinant 1. Any matrix X ∈ SU2 is of the form X =  , where −v¯ u¯ u, v ∈ C and uu¯ + vv¯ = 1. To any Lie group G, one can associate a Lie algebra g. See Deﬁnition 1.11. For

G = SU2 the corresponding Lie algebra will be denoted by g = su2. It consists of all 2 × 2 skew-Hermitian matrices of trace zero with bracket [A, B] = BA − AB.

Deﬁnition 2.1. For any two matrices A, B in su2, let hA, Bi = −2 Trace(AB).

Proposition 2.2. hA, Bi deﬁnes an inner product on su2.

     ia1 a2 + ia3  ib1 b2 + ib3 Proof. Let A =   ,B =   and     −a2 + ia3 −ia1 −b2 + ib3 −ib1    ic1 c2 + ic3 C =   be matrices in su2. Also let α ∈ be a scalar. We will   R −c2 + ic3 −ic1 show that hA, Bi satisﬁes the properties of an inner product.

(i) Symmetry:

hB,Ai = −2 Trace(BA) = −2 Trace(AB) = hA, Bi

(ii) Linearity:

hαA, Bi = 4α(a1b1 + a2b2 + a3b3) = −α 2 Trace(AB) = αhA, Bi

and

hA + B,Ci = −2 Trace (A + B)C

21 = 4(a1c1 + a2c2 + a3c3) + 4(b1c1 + b2c2 + b3c3)

= −2 Trace(AC) − 2 Trace(BC)

= hA, Ci + hB,Ci

(iii) Positive deﬁniteness:

2 2 2 hA, Ai = −2 Trace(AA) = 4(a1 + a2 + a3) ≥ 0

and hA, Ai = 0 if and only if A is a zero matrix.

3 Lemma 2.3. The Lie algebra su2 is isomorphic to R with Lie bracket given by cross product. Moreover, under this isomorphism, the inner product in Deﬁnition 2.1 corresponds to the dot product.

3 Proof. Let Φ: su2 → R be the linear map given by Φ(A) = a , where A =   ia1 a2 + ia3   3   ∈ su2 and a = 2(a1, a2, a3) ∈ .   R −a2 + ia3 −ia1

   ib1 b2 + ib3 Let B =   be a matrix in su2 and b = 2(b1, b2, b3) be a vector   −b2 + ib3 −ib1 in R3. Then,

   i2(a3b2 − a2b3) 2(a1b3 − a3b1) + i2(a2b1 − a1b2) [A, B] =     −2(a1b3 − a3b1) + i2(a2b1 − a1b2) −i2(a3b2 − a2b3)

On the other hand, b × a = 2 2(a3b2 − a2b3), 2(a1b3 − a3b1), 2(a2b1 − a1b2) , which shows that Φ([A, B]) = b × a .

Additionally, hA, Bi = 4(a1b1 + a2b2 + a3b3) = a · b.

22 Lemma 2.4. For any matrices A, B, C ∈ su2, we have hA, [B,C]i = h[A, B],Ci.

3 Proof. Under the isomorphism between su2 and R given by Lemma 2.3, the above Lemma coincides with the formula a · (b × c) = (a × b) · c in R3.

As with any Lie group G, the tangent bundle of SU2, denoted by TSU2 is triv- ∼ ializable. That is, TSU2 = SU2 × su2. So for any g ∈ SU2 and A ∈ su2, the map

(g, A) 7→ (TI Lg)(A) is a diﬀeomorphism, where Lg : G → G is the map given by

Lg(h) = gh, ∀g, h ∈ G and TI Lg denotes its tangent map at the identity I. Any tangent vector P in TgSU2 will be denoted by P = gA, for some A ∈ su2.

2.2 A Space of Curves in SU2

Let X(s) in SU2 be an arc length parametrized curve defined on a fixed interval [0,L]. A curve X(s) satisfying X(0) = I will be called anchored. Suppose the dX tangent vectors of X(s) are defined by (s) = X(s)Λ(s), where Λ(s) ∈ su satisfy ds 2 the conditions hΛ(s), Λ(s)i = 1 and Λ(0) = Λ(L), where Λ(0) = A for some fixed

A ∈ su2.

Deﬁnition 2.5. The space of arc length parametrized anchored curves satisfying the above conditions will be denoted by M.

The space of all smooth maps

Map [0,L],SU2 = {smooth maps X : [0,L] → SU2} is a Fréchet manifold by Corollary 2.3.2 in [5]. As argued in section 3 in [6], M is a smooth Fréchet submanifold of Map [0,L],SU2 .    u v Remark 2.2.1. Consider a matrix, X =   ∈ SU2 for u, v ∈ C. Then by −v¯ u¯ deﬁnition det(X) = uu¯ + vv¯ = 1.

23 Set u = x0 + ix1 and v = x2 + ix3 for x0, x1, x2, x3 ∈ R. Then (x0 + ix1)(x0 −

2 2 2 2 ix1) + (x2 + ix3)(−x2 + ix3) = x0 + x1 + x2 + x3 = 1. Hence we may view SU2 as the 3-sphere S3 ⊂ R4.

Lemma 2.6. For any matrices A, B, C in su2, we have

[A, [B,C]] = hA, CiB − hA, BiC.

3 Proof. Under the isomorphism between su2 and R given by Lemma 2.3, the above Lemma coincides with the vector triple product formula in R3.

Now we will deﬁne the covariant derivative on SU2 as follows.

Deﬁnition 2.7. The covariant derivative of a curve of tangent vectors v(s) =

X(s)U(s) along a curve X(s) in SU2 is given by

D dU 1 v(s) = X(s) (s) + [U(s), Λ(s)] (2.1) ds ds 2

dX where Λ(s) = X−1(s) (s) is in su . [6], [7]. ds 2

Proposition 2.8. The covariant derivative on SU2 corresponds to the covariant derivative on S3 ⊂ R4.

3 3 Proof. Let x(s) be a curve in S and v(s) ∈ Tx(s)S a ﬁeld of tangent vectors along the curve x(s). Then the covariant derivative of v(s) is given by

D dv dv (v) = − · x x. ds ds ds

For any two matrices P,Q in SU2 corresponding to p = (p0, p1, p2, p3), q = 1 (q , q , q , q ) in S3 ⊂ 4, we can write p · q = Trace(PQ−1). See Lemma A.1 0 1 2 3 R 2 in Appendix A.

24 3 Under this correspondence between S and SU2 we can write

D dV 1 dV (V ) = − Trace X−1 X ds ds 2 ds

dV 1 dV = X X−1 − Trace X−1 I (2.2) ds 2 ds

3 where X ∈ SU2 given in Remark 2.2.1 corresponds to x ∈ S and V = XU ∈ TX SU2, where U ∈ su2 corresponds to v. Now the right hand side of equation (2.1) yields

dU 1 dV 1 dX 1 dX (s) + [U(s), Λ(s)] = X−1 − X−1 X−1V − X−1VX−1 (2.3) ds 2 ds 2 ds 2 ds

dX where U = X−1V and Λ = X−1 . Comparing equations (2.2) and (2.3), it suﬃces ds to show dX dX dV X−1V + VX−1 = Trace X−1 X (2.4) ds ds ds

Now the right hand side of equation (2.4) can be written as

dV dX dU Trace X−1 X = Trace U + X X−1 X ds ds ds

dU = Trace XΛUX−1 + X X−1 X ds

dU = Trace (XΛUX−1) + Trace (X X−1) X ds

= Trace (ΛU)X

dU because Trace = 0. Expressing the left hand side of equation (2.4) in terms of ds Λ and U we get

dX dX X−1V + VX−1 = XΛU + XUΛ ds ds

25 = X(ΛU + UΛ)

= Trace (ΛU)X

because Λ,U ∈ su2. See Lemma A.2 in Appendix A. Therefore, equation (2.4) holds. This proves the proposition.

Lemma 2.9. Suppose that X(s, t) is a ﬁeld of curves in SU2 with its inﬁnitesimal directions

∂X ∂X Z(s, t) = X−1(s, t) (s, t) and W (s, t) = X−1(s, t) (s, t). ∂s ∂t

Then ∂Z ∂W − + [Z,W ] = 0. (2.5) ∂t ∂s

[6], [7].

D ∂X Proof. Let be the covariant derivative along the curve t 7→ X(s, t) and dt ∂s D ∂X be the covariant derivative along the curve s 7→ X(s, t). Since the covariant ds ∂t derivatives on SU2 are symmetric [13], we have

D ∂X D ∂X = dt ∂s ds ∂t

∂Z 1 ∂W 1 X + [Z,W ] = X + [W, Z] ∂t 2 ∂s 2

Simplifying we get ∂Z ∂W − + [Z,W ] = 0 ∂t ∂s

The tangent space of M is deﬁned by the following proposition.

26 Proposition 2.10. The tangent space TX M at an arc length parametrized curve dX X(s) with Λ(s) = X(s)−1 (s) consists of curves v(s) = X(s)V (s) with V (s) the ds solution of dV (s) = [Λ(s),V (s)] + U(s) (2.6) ds such that V (0) = 0, where U(s) is a curve in su2 subject to the conditions that U(0) = 0 and hΛ(s),U(s)i = 0. [6], [7].

Proof. Let Y (s, t) be a family of anchored arc length parametrized curves such that

∂Y

Y (s, 0) = X(s). Then v(s) = (s, t) is a tangent vector at X(s) for which ∂t t=0 v(0) = 0 since the curves in Y (s, t) are anchored. Let Z(s, t) and W (s, t) be two tangent vectors deﬁned by

∂Y ∂Y Z(s, t) = Y (s, t)−1 (s, t) and W (s, t) = Y (s, t)−1 (s, t) ∂s ∂t

Let Λ(s) = Z(s, 0) and V (s) = W (s, 0). Then from equation (2.5) we have

∂Z ∂W (s, 0) − (s, 0) + [Z(s, 0),W (s, 0)] = 0 ∂t ∂s

dV h i ∂Z which implies (s) = Λ(s),V (s) + U(s), where U(s) = (s, 0). ds ∂t Since the curves s 7→ Y (s, t) are arc length parametrized for each t, we have hZ(s, t),Z(s, t)i = 1 and Z(0, t) = A. Therefore

∂Z ∂Z Z(s, 0), (s, 0) = 0 and (0, t) = 0 ∂t ∂t which implies hΛ(s),U(s)i = 0 and U(0) = 0.

Now, suppose that V (s) is a solution of (2.6) generated by a curve U(s) ∈ su2

27 with U(0) = 0 satisfying hΛ(s),U(s)i = 0 and U(0) = 0. Let us deﬁne,

1 Z(s, t) = q Λ(s) + tU(s) (2.7) 1 + t2hU(s),U(s)i

Then we have

D E 1 h Z(s, t),Z(s, t) = hΛ(s), Λ(s)i + 2thΛ(s),U(s)i+ 1 + t2hU(s),U(s)i i t2hU(s),U(s)i

1 h i = 1 + t2hU(s),U(s)i 1 + t2hU(s),U(s)i

= 1

Therefore Y (s, t), the solution of the initial value problem

∂Y (s, t) = Y (s, t)Z(s, t),Y (0, t) = I ∂s

∂Y corresponds to the family of anchored arc length parametrized curves. Let (s, t) = ∂t Y (s, t)W (s, t). Then we get

∂Z ∂W (s, 0) − (s, 0) + [Z(s, 0),W (s, 0)] = 0 (2.8) ∂t ∂s

Now diﬀerentiating Z(s, t) with respect to t we obtain

∂Z −1 −3 (s, t) = 1 + t2hU(s),U(s)i 2 2thU(s),U(s)i Λ(s) ∂t 2

q −1 1 + t2hU(s),U(s)i + t2 1 + t2hU(s),U(s)i 2 hU(s),U(s)i + U(s) 1 + t2hU(s),U(s)i

∂Z which implies (s, t) = U(s). Thus equation (2.8) reduces to ∂t t=0

dW h i (s, 0) = Λ(s),W (s, 0) + U(s) ds

Therefore, W (s, 0) solves the above equation and thus is equivalent to V (s).

In this thesis we will use the following symplectic form that is used by V. Jurdjevic in [6] on the space of curves M.

Deﬁnition 2.11. Let v1(s) = X(s)V1(s) and v2(s) = X(s)V2(s) be two tangent vectors at a curve X(s). Then according as in equation (2.6) write

dV (s) U (s) = i − [Λ(s),V (s)], i = 1, 2 i ds i

such that Ui(0) = 0 and hΛ(s),Ui(s)i = 0. Let ω be the 2-form on M deﬁned by

Z L D E ωX (V1,V2) = − Λ(s), [U1(s),U2(s)] ds. (2.9) 0

29 3

Hamiltonian vector ﬁelds and Complete integrability

dX Consider a curve X(s) on the space of curves M with unit tangent vector = XΛ. ds D dX Then the curvature κ(s) of X(s) is deﬁned by ( ) . According to the deﬁnition ds ds of covariant derivative given by the equation (2.1), we have

D dX dΛ 1 dΛ (s) = X(s) + [Λ(s), Λ(s)] = X(s) (s) ds ds ds 2 ds

dΛ since [Λ(s), Λ(s)] = 0. Therefore κ(s) = . ds Now the expression for the torsion τ(s) corresponding to X(s) depends on how one chooses the frame above X(s). For our purpose, we choose the Serret-Frenet frame. The Serret-Frenet frame along the curve X(s) is deﬁned via the following relations:

D D D (XΛ) = κXN, (XN) = −κXΛ + τXB, (XB) = −τXN (3.1) ds ds ds where, Λ,N and B are skew-Hermitian matrices of trace zero that correspond to the tangent, normal and bi-normal unit vectors to the curve X(s). The derivation of the

30 above Serret-Frenet formulas is similar to the derivation in R3. Let us consider the following four functions on M. Here, the ordering of these functions follows from the literature.

Z L 1. f1 = τ(s)ds 0 Z L 1 2 2. f2 = κ (s)ds 2 0 Z L 1 2 1 2 3. f3 = κ (s)τ(s) + κ ds 0 2 4 Z L 1 0 2 1 2 2 1 2 1 2 1 4 4. f4 = (κ ) (s) + κ (s)τ (s) + κ (s)τ(s) + κ (s) − κ (s) ds 0 2 2 2 8 8

The ﬂow of the Hamiltonian vector ﬁeld corresponding to the function f2 generates ∂Λ ∂2Λ ‘Heisenberg’s magnetic equation’ given by (s, t) = (s, t), Λ(s, t) and the ∂t ∂2s

flow of the Hamiltonian vector field corresponding to f1 generates the ‘curve short- ∂Λ ening equation’ given by (s, t) = κ(s, t)N(s, t). Also, as we will show, the above ∂t functions pairwise Poisson commute. It seems plausible to have a list of infinitely many pairwise Poisson commuting functions on M, in which f1, f2, f3, f4 are the top four functions. The subsequent functions in such an infinite list are yet to be determined, and this raises the question of complete integrability of this Hamiltonian system. A similar problem has been handled in [10]. The authors of [10] have shown that the filament equation defined on a space of curves in R3 is a completely integrable Hamiltonian system. That is, the space of curves possesses infinitely many functions which are pairwise Poisson commuting. The successive functions in this infinite list are obtained through the repeated application of a recursion operator. The recursion operator is constructed using two compatible Poisson structures on the space of curves.

In this chapter, we will compute the Hamiltonian vector ﬁelds Xfi , i = 1, 2, 4 corresponding to the functions f1, f2 and f3 via the symplectic structure on M. The computations of the Hamiltonian vector ﬁelds associated with the functions f1 and

31 f4 are new, while those with f2 have been reproduced here from [6]. On the other hand, computation of the Hamiltonian vector ﬁeld corresponding to the function f3 can be found in [6].

For computational convenience we will rewrite the functions f1, f2, f4 in terms of the tangent vector Λ using the Serret-Frenet equations. Thus the functions take the following forms

Z L 0 0 −1 00 0 1. f1 = hΛ , Λ i hΛ , [Λ(s), Λ ]ids 0 Z L 1 0 0 2. f2 = hΛ , Λ ids 2 0 Z L 1 00 0 2 0 0 −1 1 00 0 2 0 0 −1 1 0 0 2 3. f4 = hΛ , Λ i hΛ , Λ i + hΛ , [Λ, Λ ]i hΛ , Λ i − hΛ , Λ i ds 0 2 2 8 where Λ0, Λ00 denote ﬁrst and second derivative of Λ respectively with respect to s. See Appendix B for detailed calculation.

Towards the end of this chapter, we will show that the functions f1, f2, f3, f4 pairwise Poisson commute, suggesting they may belong to a list of inﬁnitely many Poisson commuting functions, which would show the complete integrability of the corresponding Hamiltonian system. Section 3.5 discusses the potential of further research in this topic.

Z L 3.1 Hamiltonian vector ﬁeld of τds 0 Z L Proposition 3.1. Let f1 : M → R be the function given by f1 = τ ds and let 0

Xf1 denotes its Hamiltonian vector ﬁeld. If g(s, t) denote the integral curves of Xf1 ∂g with tangent vectors (s, t) = g(s, t)Λ(s, t), then Λ(s, t) evolves according to the ∂s Curve Shortening equation (CSE)

∂Λ (s, t) = κ(s, t)N(s, t). (3.2) ∂t

32 Proof. Let Y (s, t) be a family of anchored arc length parametrized curves that sat-

∂Y isﬁes Y (s, 0) = X(s) and v(s) = (s, t) = X(s)V (s). ∂t t=0 ∂Y Also let Z(s, t) denote the matrices deﬁned by (s, t) = Y (s, t)Z(s, t). It ∂s dV follows that Λ(s) = Z(s, 0) and that V (s) solves = [Λ(s),V (s)] + U(s) with ds ∂Z U(s) = (s, 0). ∂t

Then the directional derivative of f1 at X(s) along V is given by

∂ Z L ∂Z ∂Z −1 ∂2Z ∂Z df1X (V ) = (s, t), (s, t) 2 (s, t), [Z, (s, t)] ds ∂t 0 ∂s ∂s ∂s ∂s t=0

Z L ∂ ∂Z ∂Z −1 ∂2Z ∂Z

= (s, t), (s, t) 2 (s, t), [Z, (s, t)] ds 0 ∂t ∂s ∂s ∂s ∂s t=0

Z L ∂Z ∂Z −1 ∂ ∂2Z ∂Z

+ (s, t), (s, t) 2 (s, t), [Z, (s, t)] ds 0 ∂s ∂s ∂t ∂s ∂s t=0

Z L dΛ dΛ −2 dΛ dU d2Λ dΛ = −2 (s), (s) (s), (s) (s), [Λ(s), (s)] ds 0 ds ds ds ds ds2 ds

Z L dΛ dΛ −1 d2U dΛ + (s), (s) (s), [Λ(s), (s)] ds 0 ds ds ds2 ds

Z L dΛ dΛ −1 d2Λ dU + (s), (s) (s), [Λ(s), (s)] ds 0 ds ds ds2 ds

Z L dΛ dΛ −1 d2Λ dΛ + (s), (s) (s), [U(s), (s)] ds 0 ds ds ds2 ds

= I1 + I2 + I3 + I4 (say) (3.3)

Using integration by parts multiple times and considering the fact that Λ(0) = Λ(L) we ﬁnd

Z L D E df1X (V ) = − G(s),U(s) ds (3.4) 0

See Appendix B.1 for detailed calculation regarding equation (3.4). The Hamiltonian

33 vector ﬁeld Xf1 associated with f2 is of the form

Xf1 (X(s)) = X(s)F1(s)

for some curve F1(s) ∈ su2.

Since Xf1 (X) is in TX M, F1(s) is the solution of

dF h i 1 (s) = Λ(s),V (s) + U (s),F (0) = 0 (3.5) ds f1 1

for some curve Uf1 (s) ∈ su2 satisfying the conditions Uf1 (0) = 0 and hΛ(s),Uf1 (s)i = 0.

Then the curve Uf1 satisﬁes the symplectic form

Z L D E ωX (F1,F ) = − Λ(s), [Uf1 (s),U(s)] ds (3.6) 0

where, U(s) is an arbitrary curve in su2 that satisﬁes U(0) = 0 and hΛ(s),U(s)i = 0. Now by deﬁnition from equation (3.4) and(3.6) we obtain

Z L D E Z L D E G(s),U(s) ds = Λ(s), [Uf1 (s),U(s)] ds 0 0

Which implies

Z L D E D E 0 = G(s),U(s) − Λ(s), [Uf1 (s),U(s)] ds 0

Z L D E D E = G(s),U(s) − [Λ(s),Uf1 (s)],U(s) ds [Using Lemma 2.4] 0

Z L D E = G(s) − [Λ(s),Uf1 (s)],U(s) ds (3.7) 0

Substituting U(s) = [Λ(s),C(s)] in equation (3.28) for some curve C(s) that

34 satisﬁes C(0) = 0,

Z L

0 = G(s) − [Λ(s),Uf1 (s)], [Λ(s),C(s)] ds 0

Z L h i = G(s) − [Λ(s),Uf1 (s)], Λ(s) ,C(s) ds 0

Z L h i h i = G(s), Λ(s) − [Λ(s),Uf1 (s)], Λ(s) ,C(s) ds 0

Recall, if f(x)g(x) = 0 for all x ∈ [a, b] and for any arbitrary continuous function g(x) then f(x) = 0 on [a, b]. Thus since C(s) is arbitrary,

h i h i 0 = G(s), Λ(s) − [Λ(s),Uf1 (s)], Λ(s)

h i h i = G(s), Λ(s) − hΛ(s), Λ(s)iUf1 (s) − hUf1 (s), Λ(s)iΛ(s) [Using Lemma 2.6]

h i = G(s), Λ(s) − Uf1 (s)[since hUf1 (s), Λi = 0]

and therefore h i Uf1 = − Λ(s),G(s) (3.8)

Now expressing G(s) in terms of curvature and torsion using the Serret-Frenet equations we obtain

G(s) = −2κ−2κ0τN − 2τΛ + 2κ−1τ 2B + 2k−1τ 0N + 2κ−3(κ0)2B − κ−2κ00B

+ 2κ−2κ0τN + τΛ − κ0τN − k−1τ 2B − 2κ−3(κ0)2B + κ−2κ00B − κ−1τ 0N

− κ−1τ 2B + τΛ + κB + τΛ

= κB + τΛ

35 Thus

Uf1 = −[Λ, κB + τΛ] = κN

The integral curves t 7→ X(s, t) of Xf1 are the solutions of

∂X ∂X = X(s, t)F (s, t) and = X(s, t)Λ(s, t) ∂t 1 ∂s

where F1(s, t) is the solution of

∂F 1 (s, t) = [Λ(s, t),F (s, t)] + κN (3.9) ∂s 1

According to the Lemma 2.9 we have

∂Λ ∂F (s, t) − 1 (s, t) + [Λ(s, t),F (s, t)] = 0 (3.10) ∂t ∂s 1

Using the equation (3.9) in (3.10) we get,

∂Λ (s, t) − [Λ(s, t),F (s, t)] − κ(s, t)N(s, t) + [Λ(s, t),F (s, t)] = 0 ∂t 1 1

Therefore, ∂Λ (s, t) = κ(s, t)N(s, t) ∂t which is known as Curve shortening equation (CSE). See [3], [6].

Z L 3.2 Hamiltonian vector ﬁeld of 1 κ2 ds 2 0 Z L 1 2 Proposition 3.2. Let f2 : M → R be the function given by f2 = 2 κ ds and 0 let Xf2 denote its Hamiltonian vector ﬁeld. If g(t, s) denote the integral curves of ∂g X with tangent vectors (s, t) = g(s, t)Λ(s, t), then Λ(s, t) evolves according to f2 ∂s

36 Heisenberg’s magnetic equation(HME)

∂Λ ∂2Λ (s, t) = (s, t), Λ(s, t) . (3.11) ∂t ∂s2

Proof. Suppose that a family of anchored arc length parametrized curves denoted

∂Y by Y (s, t) satisﬁes Y (s, 0) = X(s) and v(s) = (s, t) = X(s)V (s). ∂t t=0 ∂Y Let Z(s, t) denote the matrices deﬁned by (s, t) = Y (s, t) Z(s, t). It fol- ∂s dV lows from Proposition 2.10 that Λ(s) = Z(s, 0) and V (s) is the solution of = ds ∂Z(s, t)

[Λ(s),V (s)] + U(s) with U(s) = . ∂t t=0 Then the directional derivative of f2 at X(s) in the direction of V is given by

1 ∂ Z L ∂Z ∂Z

df2X (V ) = (s, t), (s, t) ds 2 ∂t 0 ∂s ∂s t=0

1 Z L ∂Z ∂ ∂Z

= · 2 (s, t), (s, t) ds 2 0 ∂s ∂s ∂t t=0

Z L ∂Z ∂ ∂Z = (s, 0), (s, 0) ds 0 ∂s ∂s ∂t

Z L dΛ dU = (s), (s) ds 0 ds ds

s=L dΛ Z L d2Λ

= (s),U(s) − 2 (s),U(s) ds ds s=0 0 ds

s=L dΛ

The boundary term (s),U(s) equals zero due to the assumption Λ(0) = ds s=0 Λ(L). Consequently

Z L d2Λ df2X (V ) = − (s),U(s) ds (3.12) 0 ds2

The Hamiltonian vector ﬁeld Xf2 corresponding to f2 is of the form

Xf2 X(s) = X(s)F2(s)

37 for some curve F2(s) ∈ su2.

Since Xf2 (X) ∈ TX M,F (s) is the solution of

dF h i 2 (s) = Λ(s),F (s) + U (s),F (0) = 0 (3.13) ds 2 f2 2

for some curve Uf2 (s) ∈ su2 satisfying the conditions Uf2 (0) = 0 and hΛ(s),Uf2 (s)i = 0.

Then the curve Uf2 satisﬁes the following equation

Z L D E ωX (F2(s),F (s)) = − Λ(s), [Uf2 (s),U(s)] ds (3.14) 0

where, U(s) ∈ su2 is an arbitrary curve that satisﬁes U(0) = 0 and hΛ(s),U(s)i = 0. From equation (3.12) and(3.14) we obtain

Z L d2Λ Z L D E (s),U(s) ds = Λ(s), [Uf2 (s),U(s)] ds 0 ds2 0

Which implies

Z L d2Λ D E 0 = (s),U(s) − [Λ(s),Uf2 (s)],U(s) ds [Using Lemma 2.4] 0 ds2

Z L d2Λ = (s) − [Λ(s),Uf2 (s)],U(s) ds (3.15) 0 ds2

Replacing U(s) by [Λ(s),C(s)] for some curve C(s) that satisﬁes C(0) = 0, equation (3.15) becomes

Z L d2Λ 0 = (s) − [Λ(s),Uf2 (s)], [Λ(s),C(s)] ds 0 ds2

Z L d2Λ = (s) − [Λ(s),Uf2 (s)], Λ(s) ,C(s) ds [Using Lemma 2.4] 0 ds2

38 Z L d2Λ h i = (s), Λ(s) − [Λ(s),Uf2 (s)], Λ(s) ,C(s) ds. 0 ds2

Since C(s) is arbitrary, we have

d2Λ h i 0 = (s), Λ(s) − [Λ(s),U (s)], Λ(s) ds2 f2

d2Λ h i = (s), Λ(s) − hΛ(s), Λ(s)iU (s) − hU (s), Λ(s)iΛ(s) [Using Lemma 2.6] ds2 f2 f2

d2Λ = (s), Λ(s) − U (s)[since hU (s), Λ(s)i = 0] ds2 f2 f2 and therefore d2Λ U (s) = − Λ(s), (s) (3.16) f2 ds2

The integral curves t 7→ X(s, t) of Xf2 are the solutions of

∂X ∂X = X(s, t) F (s, t) and = X(s, t) Λ(s, t) ∂t 2 ∂s

where F2(s, t) is the solution of

∂F ∂2Λ 2 (s, t) = [Λ(s, t),F (s, t)] − Λ(s, t), (s, t) (3.17) ∂s 2 ∂2s

Now according to Lemma 2.9 we have

∂Λ ∂F (s, t) − 2 (s, t) + [Λ(s, t),F (s, t)] = 0 (3.18) ∂t ∂s 2

Using the equation (3.17) in (3.18) we get

∂Λ ∂2Λ (s, t) − [Λ(s, t),F (s, t)] + Λ(s, t), (s, t) + [Λ(s, t),F (s, t)] = 0 ∂t 2 ∂2s 2

39 After the cancellation we are left with

∂Λ ∂2Λ (s, t) = (s, t), Λ(s, t) ∂t ∂2s which is known as Heigenberg’s magnetic equation. See [6], [8],[17]. Therefore, Λ(s, t) evolves according to (HME).

Z L 1 3.3 Hamiltonian vector ﬁeld of (κ0)2 + κ2τ 2 + 0 2 1 1 1 κ2τ + κ2 − κ4 ds 2 8 8

Proposition 3.3. Let f4 : M → R be the function given by

Z L 0 2 1 2 2 1 2 1 2 1 4 f4 = (κ ) (s) + κ (s)τ (s) + κ (s)τ(s) + κ (s) − κ (s) ds 0 2 2 8 8

and let Xf4 denote its Hamiltonian vector ﬁeld. If g(t, s) denote the integral curves ∂g of X with tangent vectors (s, t) = g(s, t)Λ(s, t), then Λ(s, t) evolves according f4 ∂s to,

∂Λ 3 3 (s, t) = (−κτ 3 + 3κ00τ + 3κ0τ 0 + κ3τ + κτ 00)N + (−κ000 + 3κττ 0 + 3κ0τ 2 − κ2κ0)B ∂t 2 2 (3.19) where the primes denote derivatives with respect to the arc length s.

Proof. Let Y (s, t) be a family of anchored arc length parametrized curves that sat-

∂Y isﬁes Y (s, 0) = X(s) and v(s) = (s, t) = X(s)V (s). ∂t t=0 ∂Y Also let Z(s, t) be the matrices deﬁned by (s, t) = Y (s, t)Z(s, t). It follows ∂s dV that Λ(s) = Z(s, 0) and that V (s) is the solution of = [Λ(s),V (s)] + U(s) with ds ∂Z U(s) = (s, 0). ∂t

For computational convenience, we split the function f4 into three diﬀerent func-

40 tions, namely

Z L 1 00 0 2 0 0 −1 (i) I1 = hΛ , Λ i hΛ , Λ i ds 0 2 Z L 1 00 0 2 0 0 −1 (ii) I2 = hΛ , [Λ, Λ ]i hΛ , Λ i ds 0 2 Z L 1 0 0 2 (iii) I3 = − hΛ , Λ i ds 0 8

The directional derivative corresponding to I1, dI1X (V ) at X in the direction of V is given by

∂ Z L 1∂2Z ∂Z 2∂Z ∂Z −1 dI1X (V ) = 2 (s, t), (s, t) (s, t), (s, t) ds ∂t 0 2 ∂s ∂s ∂s ∂s t=0

Z L ∂Z ∂Z −2∂Z ∂ ∂Z ∂2Z ∂Z 2

= − (s, t), (s, t) (s, t), (s, t) 2 (s, t), (s, t) 0 ∂s ∂s ∂s ∂t ∂s ∂s ∂s t=0

Z L ∂2Z ∂Z ∂ ∂2Z ∂Z ∂Z ∂Z −1

+ 2 (s, t), (s, t) 2 (s, t), (s, t) (s, t), (s, t) 0 ∂s ∂s ∂t ∂s ∂s ∂s ∂s t=0

Z L ∂2Z ∂Z ∂2Z ∂ ∂Z ∂Z ∂Z −1

+ 2 (s, t), (s, t) 2 (s, t), (s, t) (s, t), (s, t) 0 ∂s ∂s ∂s ∂t ∂s ∂s ∂s t=0

Z L dΛ dΛ −2 dΛ dU d2Λ dΛ 2 = − (s), (s) (s), (s) (s), (s) ds 0 ds ds ds ds ds2 ds

Z L d2Λ dΛ −2d2U dΛ dΛ dΛ −1 + (s), (s) (s), (s) (s), (s) ds 0 ds2 ds ds2 ds ds ds

Z L d2Λ dΛ −2 d2Λ dU dΛ dΛ −1 + (s), (s) (s), (s) (s), (s) ds 0 ds2 ds ds2 ds ds ds

= I11 + I12 + I13 (say) (3.20)

Using integration by parts and Λ(0) = Λ(L), we obtain

Z L dI1X (V ) = − G1(s),U(s) ds (3.21) 0

See Appendix B.2 for calculations associated with the above equation.

41 Similarly, the directional derivative corresponding to I2 at X along V is given by

∂ Z L 1 ∂2Z h ∂Z i2 ∂Z ∂Z −1 dI2X (V ) = 2 (s, t), Z(s, t), (s, t) (s, t), (s, t) ds ∂t 0 2 ∂s ∂s ∂s ∂s t=0

Z L dΛ dΛ −2 dΛ dU d2Λ h dΛ i2 = − (s), (s) (s), (s) (s), Λ(s), (s) ds 0 ds ds ds ds ds2 ds

Z L dΛ dΛ −1d2Λ h dΛ id2U h dΛ i + (s), (s) (s), Λ(s), (s) (s), Λ(s), (s) ds 0 ds ds ds2 ds ds2 ds

Z L dΛ dΛ −1d2Λ h dΛ id2Λ h dU i + (s), (s) (s), Λ(s), (s) (s), Λ(s), (s) ds 0 ds ds ds2 ds ds2 ds

Z L dΛ dΛ −1d2Λ h dΛ d2Λ h dΛ i + (s), (s) (s), Λ(s), (s)] (s), U(s), (s) ds 0 ds ds ds2 ds ds2 ds

= I21 + I22 + I23 + I24 (say) (3.22)

Using integration by parts few times, we obtain

Z L dI2X (V ) = − G2(s),U(s) ds (3.23) 0

See Appendix B.2 for calculations regarding the above equation.

Finally, we compute that the directional derivative corresponding to I3 is given by

∂ Z L 1∂Z ∂Z 2

dI3X (V ) = − (s, t), (s, t) ds ∂t 0 8 ∂s ∂s t=0

1 Z L dΛ dΛ dΛ dU = − (s), (s) (s), (s) ds 2 0 ds ds ds ds

Again, using integration by parts we obtain

Z L DdΛ d2Λ E dΛ dI3X (V ) = (s), (s) (s),U(s) ds 0 ds ds2 ds

42 1 Z L DdΛ dΛ E d2Λ + (s), (s) (s),U(s) ds 2 0 ds ds ds2

Z L = G3(s),U(s) ds (3.24) 0

See Appendix B.2 for calculations regarding the above equation. Therefore, the directional derivative corresponding to f4 at X along V is given by

df4X (V ) = dI1X (V ) + dI2X (V ) + dI3X (V )

Z L D E = − G1(s) + G2(s) − G3(s),U(s) ds 0

Z L D E = − R(s),U(s) ds (3.25) 0

where, R(s) = G1(s) + G2(s) − G3(s). See Appendix B.2 for details.

The Hamiltonian vector ﬁeld Xf4 associated with f4 is of the form

Xf4 (s) = X(s)F4(s)

for some curve F4(s) ∈ su2.

Since Xf4 (X) is in TX M, F4(s) is the solution of

dF h i 4 (s) = Λ(s),V (s) + U (s),F (0) = 0 (3.26) ds f4 4

for some curve Uf4 (s) ∈ su2 that satisﬁes the conditions Uf4 (0) = 0 and hΛ(s),Uf4 (s)i = 0.

Then the curve Uf4 satisﬁes the following equation

Z L D E ωX (F4(s),F (s)) = − Λ(s), [Uf4 (s),U(s)] ds (3.27) 0

43 where, U(s) is an arbitrary curve in su2 that satisﬁes U(0) = 0 and hΛ(s),U(s)i = 0. From equation (3.25) and(3.27) we obtain

Z L D E Z L D E R(s),U(s) ds = Λ(s), [Uf4 (s),U(s)] ds 0 0

Which implies

Z L D E 0 = R(s),U(s) − Λ(s), [Uf4 (s),U(s)] ds 0

Z L D E = R(s),U(s) − [Λ(s),Uf4 (s)],U(s) ds [Using Lemma 2.4] 0

Z L

= R(s) − [Λ(s),Uf4 (s)],U(s) ds (3.28) 0

Substituting U(s) = [Λ(s),C(s)] for some curve C(s) that satisﬁes C(0) = 0, equation (3.28) becomes

Z L

0 = R(s) − [Λ(s),Uf4 (s)], [Λ(s),C(s)] ds 0

Z L h = R(s) − [Λ(s),Uf4 (s)], Λ(s) ,C(s) ds 0

Z L h i h i = R(s), Λ(s) − [Λ(s),Uf4 (s)], Λ(s) ,C(s) ds 0

Since C(s) is arbitrary

h i h i 0 = R(s), Λ(s) − [Λ(s),Uf4 (s)], Λ(s)

h i h i = R(s), Λ(s) − hΛ(s), Λ(s)iUf4 (s) − hUf4 (s), Λ(s)iΛ(s) [Using Lemma 2.6]

h i = R(s), Λ(s) − Uf4 (s)[since hUf4 (s), Λ(s)i = 0]

44 and therefore h i Uf4 (s) = − Λ(s),R(s)

Expressing R(s) in terms of curvature and torsion, we get

R(s) = κ000N − κκ00Λ + 3κ00τB − 3κ0τ 2N + 2κ2τ 2Λ − κτ 3B − 3κττ 0N+ (3.29) 3 1 3 +3κ0τ 0B + κτ 00B + κ2κ0N + κ4Λ + κ3τB 2 2 2

See Appendix B.2 for detail calculations. Therefore

Uf4 (s) = −[Λ(s),R(s)] 3 = −(−3κ00τ + κτ 3 − κ3τ − 3κ0τ 0 − κτ 00)N 2 3 − (κ000 − 3κ0τ 2 − 3κττ 0 + κ2κ0)B (3.30) 2

The integral curves t 7→ X(s, t) of Xf4 are the solutions of

∂X ∂X = X(s, t)F (s, t) and = X(s, t)Λ(s, t) ∂t 4 ∂s

where F4(s, t) is the solution of

∂F 3 4 (s, t) = [Λ(s, t),F (s, t)] + (3κ00τ − κτ 3 + κ3τ + 3κ0τ 0 + κτ 00)N ∂s 4 2

3 + (−κ000 + 3κ0τ 2 + 3κττ 0 − κ2κ0)B (3.31) 2

From Lemma 2.9 we know

∂Λ ∂F (s, t) − 4 (s, t) + [Λ(s, t),F (s, t)] = 0 (3.32) ∂t ∂s 4

45 From equations (3.30), (3.31) and (3.32) we obtain,

∂Λ (s, t) − [Λ(s, t),F (s, t)] − U (s) + [Λ(s, t),F (s, t)] = 0 ∂t 4 f4 4

After cancellation we get,

∂Λ 3 (s, t) = − κτ 3 + 3(κ0τ)0 + κ3τ + κτ 00 N ∂t 2 3 + − κ000 + 3κ0τ 2 + 3κττ 0 − κ2κ0 B 2

This concludes the proof.

3.4 Poisson commuting functions

Theorem 3.4. The functions f1, f2, f3, f4 on M pairwise Poisson commute.

Proof. Here, we will show that the functions f2 and f4 Poisson commute.

The Poisson bracket of the functions f2 and f4 is deﬁned as

Z L D E {f2, f4} = ωX (F2(s),F4(s)) = − Λ(s), [Uf2 (s),Uf4 (s)] ds (3.33) 0

where Uf2 (s) and Uf4 (s) are Hamiltonian vector ﬁelds associated with f2 and f4 respectively. We need to show that {f2, f4} = 0. From equation (3.16) we have

00 0 2 0 Uf2 = [Λ , Λ] = [κ N − κ Λ + κτB, Λ] = −κ B + κτN and from equation (3.30) we have

3 3 U = (−κτ 3 + 3κ00τ + 3κ0τ 0 + κ3τ + κτ 00)N + (−κ000 + 3κττ 0 + 3κ0τ 2 − κ2κ0)B f4 2 2

46 Taking the Lie bracket of these two vector ﬁelds we obtain

3 [U (s),U (s)] = κκ000τ − 3κ2τ 2τ 0 − 3κκ0τ 3 + κ3κ0τ Λ + κκ0τ 3 − 3κ0κ00τ f2 f4 2

3 − 3(κ0)2τ 0 − κ3κ0τ − κκ0τ 00 Λ. 2

Then

3 hΛ(s), [U (s),U (s)]i = κκ000τ − 3κ2τ 2τ 0 − 3κκ0τ 3 + κ3κ0τ + κκ0τ 3 − 3κ0κ00τ f2 f4 2

3 − 3(κ0)2τ 0 − κ3κ0τ − κκ0τ 00 . (3.34) 2

From equation (3.33) and (3.34) we conclude

Z L 000 2 2 0 0 3 3 3 0 0 3 0 00 {f2, f4} = − [κκ τ − 3κ τ τ − 3κκ τ + κ κ τ] + [κκ τ − 3κ κ τ 0 2

3 − 3(κ0)2τ 0 − κ3κ0τ − κκ0τ 00] ds 2

Z L d h 3 1 i = − − κ2τ 3 − (κ0)2τ − κκ0τ 0 − (κ0)2τ + κκ00τ ds 0 ds 2 2

= 0

Similar proofs for the remaining pair of functions are given in the Appendix B.3.

Therefore, f2 Poisson commutes with f4.

Remark 3.4.1. Since the functions f1, f2, f3 and f4 Poisson commute with each other, it suggests that they may be a part of an inﬁnite list of Poisson commuting functions. See [10]. If this was the case, then the Hamiltonian systems corresponding to the above functions would be completely integrable.

47 3.5 Future research

As we have mentioned earlier, the authors of [10] have shown that the filament equation is a completely integrable system on the space of curves in R3. That is, there exist an infinite list of pairwise Poisson commuting functions on the space of curves under consideration. In [10], the authors constructed a recursion operator using two compatible Poisson structures on the space of curves to compute the successive functions in this infinite list. In this thesis, we have used the following symplectic structure given by equation (26) in [6],

Z L D E ωX (V1,V2) = − Λ(s), [U1(s),U2(s)] ds 0 on the space of curves M. There might exist other symplectic structures on M which are compatible with ω. One such candidate is

Z L D E ΩX (V1,V2) = Λ(s), [V1(s),V2(s)] ds 0 as indicated by V. Jurdjevic in [6]. Perhaps the Poisson brackets resulting from these symplectic structures can be used to construct a recursion operator to obtain the complete integrability for the systems which are Hamiltonian with respect to both structures [11], similar to the approach of Joel Langer and Ron Perline in [10].

48 Appendix A

Lemmas in Chapter 2

−1 Lemma A.1. For any two matrices P,Q in SU2, 2(p · q) = Trace(PQ ), where p, q are vectors in R4.

4 Proof. Let p = (p0, p1, p2, p3) and q = (q0, q1, q2, q3) be vectors in R . Then

p · q = p0q0 + p1q1 + p2q2 + p2q2 + p3q3

Let      p0 + ip1 p2 + ip3  q0 + iq1 q2 + iq3 P =   ,Q =       −p2 + ip3 p0 − ip1 −q2 + iq3 q0 − iq1

  q0 − iq1 −q2 − iq3 −1   be matrices in SU2. Then Q =   and   q2 − iq3 q0 + iq1

−1 Trace(PQ ) = (p0 + ip1)(q0 − iq1) + (p2 + ip3)(q2 − iq3)

+ (−p2 + ip3)(−q2 − iq3) + (p0 − ip1)(q0 + iq1)

= 2(p0q0 + p1q1 + p2q2 + p2q2 + p3q3)

49 Lemma A.2. For any two matrices Λ and U in su2, ΛU + UΛ = Trace(ΛU)I.

Proof. Let

     iλ1 λ2 + iλ3  iu1 u2 + iu3 Λ =   ,U =       −λ2 + iλ3 −iλ1 −u2 + iu3 −iu1

be matrices in su2. Then

  −λ1u1 − λ2u2 − λ3u3 + i(λ2u3 − λ3u2) −λ1u3 + λ3u1 + i(λ1u2 − λ2u1)        ΛU =           λ1u3 − λ3u1 + i(λ1u2 − λ2u1) −λ1u1 − λ2u2 − λ3u3 + i(λ3u2 − λ2u3) and

  −λ1u1 − λ2u2 − λ3u3 + i(λ3u2 − λ2u3) λ1u3 − λ3u1 + i(λ2u1 − λ1u2)        UΛ =           λ3u1 − λ1u3 + i(λ2u1 − λ1u2) −λ1u1 − λ2u2 − λ3u3 + i(λ2u3 − λ3u2)

Adding ΛU and UΛ, we obtain

  −2(λ1u1 + λ2u2 + λ3u3) 0       ΛU + UΛ =         0 −2(λ1u1 + λ2u2 + λ3u3)

= −2(λ1u1 + λ2u2 + λ3u3)I

On the other hand, Trace(ΛU) = −2(λ1u1 + λ2u2 + λ3u3). Therefore, ΛU + UΛ = Trace(ΛU)I.

50 Appendix B

Computations in Chapter 3

The functions f1, f2, f3, f4 in terms of Λ: To rewrite the functions f1, f2, f3, f4 in terms of the tangent vector Λ, we need the expressions for κ, τ and κ0 in terms of Λ. To do so we will make use of the Serret-Frenet formulas given in terms of ordinary derivative of Λ,N,B. The formulas are given by

dΛ dN 1 dB 1 = κN, = −κΛ + (τ + )B, = −(τ + )N ds ds 2 ds 2

1 The expression for torsion in the above formulas is in fact τ + 2 , whereas the torsion in the Serret-Frenet formulas using covariant derivative is τ. The expression for κ is obtained by

hΛ0, Λ0i = hκN, κNi = κ2

00 0 00 0 2 1 0 1 00 0 hΛ , Λ i Now, Λ = κ N − κ Λ + κτB + κB. So κ = hΛ , Λ i = q . By deﬁnition 2 κ hΛ0, Λ0i of torsion we have

1 1 1 hN 0,Bi = Λ00 − κ0Λ0, [Λ, Λ0] κ κ2 κ

51 1 1 1 τ + = Λ00, [Λ, Λ0] 2 κ κ

1 D E 1 τ = Λ00, [Λ, Λ0] − κ2 2

Using the above information we obtain

Z L Z L 0 0 −1 00 0 1. f1 = τ(s)ds = hΛ , Λ i hΛ , [Λ(s), Λ ]ids. 0 0 Z L Z L 1 2 1 0 0 2. f2 = 2 κ (s)ds = hΛ , Λ ids. 0 2 0 Z L 1 Here we ignore the term − ds, because it does not aﬀect the computation 0 2 of Hamiltonian vector ﬁeld.

Z L Z L h1 2 1 2i 1 00 0 3. f3 = κ (s)τ(s) + κ ds = hΛ , [Λ, Λ ]i ds. 0 2 4 0 2 Z L 1h 0 2 1 2 2 1 2 1 2 1 4 i 4. f4 = (κ ) (s) + κ (s)τ (s) + κ (s)τ(s) + κ (s) − κ (s) ds 0 2 2 2 8 8  Z L 1 00 0 2 0 0 −1 1 00 0 2 0 0 −1 1D 00 0 E =  hΛ , Λ i hΛ , Λ i + hΛ , [Λ, Λ ]i hΛ , Λ i − Λ , [Λ, Λ ] + 0 2 2 2

 1 0 0 1D 00 0 E 1 0 0 1 0 0 0 0 1 0 0 2 hΛ , Λ i + Λ , [Λ, Λ ] − hΛ , Λ i + hΛ , Λ ihΛ , Λ i − hΛ , Λ i ds 8 2 4 8 8

Z L 1 1 1 = hΛ00, Λ0i2hΛ0, Λ0i−1 + hΛ00, [Λ, Λ0]i2hΛ0, Λ0i−1 − hΛ0, Λ0i2 ds. 0 2 2 8

B.1 Calculations in Section 3.1

In this section we provide the calculations related to the Proposition 3.1. Steps for obtaining equation (3.4): We will use integration by parts to evaluate the integrals I1,I2,I3.

I1: Let

dU DdΛ dΛE−2 dΛ Dd2Λ dΛ E dV = ds and U = −2 , , [Λ, ] ds ds ds ds ds2 ds

52 Then

h iL Z L I1 = UV − V dU 0 0

Z L DdΛ dΛE−3 DdΛ d2Λ E dΛ Dd2Λ dΛ E = 0 − 8 (s), (s), (s) (s) (s), [Λ(s), (s)] , 0 ds ds ds ds2 ds ds2 ds

Z L DdΛ dΛ E−2 d2Λ Dd2Λ dΛ E U(s) ds + 2 (s), (s) (s) (s), [Λ(s), (s)] ,U(s) ds 0 ds ds ds2 ds2 ds

Z L DdΛ dΛ E−2 dΛ Dd3Λ dΛ E + 2 (s), (s) (s) (s), [Λ(s), (s)] ,U(s) ds 0 ds ds ds ds3 ds

Z L DdΛ dΛ E−2 dΛ Dd2Λ d2Λ E + 2 (s), (s) (s) (s), [Λ(s), (s)] ,U(s) ds 0 ds ds ds ds2 ds2

= I11 + I12 + I13 + I14 (say) (B.1)

h iL The term UV vanishes since Λ(0) = Λ(L). 0

I2: Let d2U dΛ dΛ−1 dΛ dV = ds and U = , Λ, ds2 ds ds ds

Then

Z L *dΛ dΛ −2 dΛ d2Λ dΛ dU + I2 = 2 (s), (s) (s), (s) Λ(s), (s) , (s) ds 0 ds ds ds ds2 ds ds

Z L *dΛ dΛ −1 d2Λ dU + − (s), (s) Λ(s), (s) , (s) ds 0 ds ds ds2 ds

= I21 + I22 (say) (B.2)

Using integration by parts again we obtain

Z L DdΛ dΛ E−3 DdΛ d2Λ E2 h dΛ i I21 = 8 (s), (s) (s), (s) Λ(s), (s) ,U(s) ds 0 ds ds ds ds2 ds

53 Z L DdΛ dΛ E−2 Dd2Λ d2Λ E h dΛ i − 2 (s), (s) (s), (s) Λ(s), (s) ,U(s) ds 0 ds ds ds2 ds2 ds

Z L DdΛ dΛ E−2 DdΛ d3Λ E h dΛ i − 2 (s), (s) (s), (s) Λ(s), (s) ,U(s) ds 0 ds ds ds ds3 ds

Z L DdΛ dΛ E−2 DdΛ d2Λ E h d2Λ i − 2 (s), (s) (s), (s) Λ(s), (s) ,U(s) ds 0 ds ds ds ds2 ds2 (B.3) and

Z L DdΛ dΛ E−2 DdΛ d2Λ E h d2Λ i I22 = −2 (s), (s) (s), (s) Λ(s), (s) ,U(s) ds 0 ds ds ds ds2 ds2

Z L DdΛ dΛE−1 h d3Λ i + (s), Λ(s), (s) ,U(s) ds 0 ds ds ds3

Z L DdΛ dΛ E−1 hdΛ d2Λ i + (s), (s) (s), (s) ,U(s) ds (B.4) 0 ds ds ds ds2

I3 returns the same outcome as I22 when evaluated using integration by parts. So

I3 = I22 (B.5)

and the integral I4 takes the form

* Z L dΛ dΛ −1 hd2Λ dΛ i I4 = − (s), (s) (s), (s) ,U(s) ds (B.6) 0 ds ds ds2 ds

Therefore

df1X (V ) = I11 + I12 + I13 + I14 + I21 + 2I22 + I4

Z L D E = − G(s),U(s) ds (B.7) 0

54 where,

DdΛ dΛE−3 DdΛ d2Λ E dΛ Dd2Λ dΛ E G(s) = 8 (s), (s), (s) (s) (s), [Λ(s), (s)] ds ds ds ds2 ds ds2 ds

DdΛ dΛ E−2 d2Λ Dd2Λ dΛ E + 2 (s), (s) (s) (s), [Λ(s), (s)] ds ds ds2 ds2 ds

DdΛ dΛ E−2 dΛ Dd3Λ dΛ E + 2 (s), (s) (s) (s), [Λ(s), (s)] ds ds ds ds3 ds

DdΛ dΛ E−2 dΛ Dd2Λ d2Λ E + 2h (s), (s) (s) (s), [Λ(s), (s)] ds ds ds ds2 ds2

DdΛ dΛ E−3 DdΛ d2Λ E2 h dΛ i + 8 (s), (s) (s), (s) Λ(s), (s) ds ds ds ds2 ds

DdΛ dΛ E−2 Dd2Λ d2Λ E h dΛ i − 2 (s), (s) (s), (s) Λ(s), (s) ds ds ds2 ds2 ds

DdΛ dΛ E−2 DdΛ d3Λ E h dΛ i − 2 (s), (s) (s), (s) Λ(s), (s) ds ds ds ds3 ds

DdΛ dΛ E−2 DdΛ d2Λ E h d2Λ i − 4 (s), (s) (s), (s) Λ(s), (s) ds ds ds ds2 ds2

DdΛ dΛE−1 h d3Λ i + 2 (s), Λ(s), (s) ds ds ds3

DdΛ dΛ E−1 hdΛ d2Λ i + 2 (s), (s) (s), (s) ds ds ds ds2

dΛ dΛ −1 hd2Λ dΛ i − (s), (s) (s), (s) ds ds ds2 ds

Now we will express G(s) in terms of curvature and torsion using Serret-Frenet equations. Before doing so here we provide a list of computations that will help us rewrite expressions like G(s) with respect to curvature and torsion.

1. Λ0 = κN

2. Λ00 = κ0N − κ2Λ + κτB

55 3. Λ000 = κ00N − 3κκ0Λ + 2κ0τB − κ3N + κτ 0B − κτ 2N

4. Λiv = κ000N − 4κκ00Λ + 3κ00τB − 3(κ0)2Λ − 6κ2κ0N + 3κ0τ 0B − 3κ0τ 2N

+ κ4Λ − κ3τB + κτ 00B − 3κττ 0N + κ2τ 2Λ − κτ 3B

5. hΛ0, Λ0i = κ2

6. hΛ0, Λ00i = κκ00

7. hΛ0, Λ000i = κκ00 − κ4 − κ2τ 2

8. hΛ0, Λivi = κκ000 − 6κ3κ0 − 3κκ0τ 2 − 3κ2ττ 0

9. hΛ00, Λ00i = (κ0)2 + κ4 + κ2τ 2

10. hΛ00, Λ000i = κ0κ00 + 2κ3κ0 + κκ0τ 2 + κ2ττ 0

11. [Λ, Λ0] = κB

12. [Λ, Λ00] = κ0B − κτN

13. [Λ, Λ000] = κ00B − 2κ0τN − κ3B − κτ 0N − κτ 2B

14. [Λ0, Λ00] = κ3B + κ2τΛ

15. hΛ00, [Λ, Λ0]i = κ2τ

16. hΛ00, [Λ, Λ00]i = 0

17. hΛ00, [Λ0, Λ00]i = 0

18. hΛ00, [Λ, Λ000]i = −2(κ0)2τ − κκ0τ 0 + κκ00τ − κ4τ − κ2τ 3

19. hΛ000, [Λ, Λ0]i = 2κκ0τ + κ2τ 0

20. hΛ000, [Λ, Λ00]i = 2(κ0)2τ + κκ0τ 0 − κκ00τ + κ4τ + κ2τ 3

21. hΛiv, [Λ, Λ0]i = 3κκ00τ + 3κκ0τ 0 − κ4τ + κ2τ 00 − κ2τ 3

56 Therefore

G(s) = −8κ−2κ0τN + 2κ−2τκ0N − 2τΛ + 2κ−1τ 2B + 4κ−2κ0τN + 2κ−1τ 0N

+ 4κ−3(κ0)2B − 2κ−2κ00B + 2κ−2κ0τN − 2κ−3(κ0)2B + κ−2κ00B − κ−1τ 0N

− κ−1τ 2B + τΛ + κB + τΛ

B.2 Calculations in Section 3.3

The calculations related to the Proposition 3.3 are provided in this section. Steps to obtain equation (3.21): Using integration by parts as before we obtain,

I11: Let

dU dΛ dΛ −2 dΛ d2Λ dΛ 2 dV = ds and U = − (s), (s) (s) (s), (s) ds ds ds ds ds2 ds

Then

Z L *dΛ dΛ −3 dΛ d2Λ 3 dΛ + I11 = −4 (s), (s) (s), (s) (s),U(s) ds 0 ds ds ds ds2 ds

Z L *dΛ dΛ −2 d2Λ d2Λ dΛ 2 + + (s), (s) (s) (s), (s) ,U(s) ds 0 ds ds ds2 ds2 ds

* + Z L dΛ dΛ −2 dΛ d2Λ dΛ d3Λ dΛ E + 2 (s), (s) (s) (s), (s) (s), (s) ,U(s) ds 0 ds ds ds ds2 ds ds3 ds

Z L *dΛ dΛ −2 dΛ d2Λ dΛ d2Λ d2Λ + + 2 (s), (s) (s) (s), (s) (s), (s) ,U(s) ds 0 ds ds ds ds2 ds ds2 ds2

(B.8)

57 I12: Let

d2U dΛ2Λ dΛ dΛ dΛ dΛ −1 dV = ds and U = (s), (s) (s) (s), (s) ds2 ds2 ds ds ds ds

Then we get

Z L *d3Λ dΛ dΛ dΛ dΛ −1 dU + I12 = − (s), (s) (s) (s), (s) , (s) ds 0 ds3 ds ds ds ds ds

Z L *d2Λ d2Λ dΛ dΛ dΛ −1 dU + − (s), (s) (s) (s), (s) , (s) ds 0 ds2 ds2 ds ds ds ds

Z L *d2Λ dΛ d2Λ dΛ dΛ −1 dU + − (s), (s) (s) (s), (s) , (s) ds 0 ds2 ds ds2 ds ds ds

Z L *d2Λ dΛ 2 dΛ dΛ dΛ −2 dU + + 2 (s), (s) (s) (s), (s) , (s) ds 0 ds2 ds ds ds ds ds

= I121 + I122 + I123 + I124 (B.9)

Using integration by parts again for each of the above integrals in (B.9) we get

Z L Dd4Λ dΛ E dΛ DdΛ dΛ E−1 I121 = (s), (s) (s) (s), (s) ,U(s) ds 0 ds4 ds ds ds ds

Z L Dd3Λ d2Λ E dΛ DdΛ dΛ E−1 + (s), (s) (s) (s), (s) ,U(s) ds 0 ds3 ds2 ds ds ds

Z L Dd3Λ dΛ E d2Λ DdΛ dΛ E−1 + (s), (s) (s) (s), (s) ,U(s) ds 0 ds3 ds ds2 ds ds

Z L Dd3Λ dΛ E dΛ DdΛ dΛ E−2 DdΛ d2Λ E − 2 (s), (s) (s) (s), (s) (s), (s) ,U(s) ds 0 ds3 ds ds ds ds ds ds2

Z L Dd2Λ d3Λ E dΛ DdΛ dΛ E−1 I122 = 2 (s), (s) (s) (s), (s) ,U(s) ds 0 ds2 ds3 ds ds ds

58 Z L Dd2Λ d2Λ E d2Λ DdΛ dΛ E−1 + (s), (s) (s) (s), (s) ,U(s) ds 0 ds2 ds2 ds2 ds ds

Z L Dd2Λ d2Λ Ed2Λ DdΛ dΛ E−2 DdΛ d2Λ E − 2 (s), (s) (s) (s), (s) (s), (s) ,U(s) ds 0 ds2 ds2 ds2 ds ds ds ds2

Z L Dd3Λ dΛ E d2Λ DdΛ dΛ E−1 I123 = (s), (s) (s) (s), (s) ,U(s) ds 0 ds3 ds ds2 ds ds

Z L Dd2Λ d2Λ E d2Λ DdΛ dΛ E−1 + (s), (s) (s) (s), (s) ,U(s) ds 0 ds2 ds2 ds2 ds ds

Z L Dd2Λ dΛ E d3Λ DdΛ dΛ E−1 + (s), (s) (s) (s), (s) ,U(s) ds 0 ds2 ds ds3 ds ds

Z L Dd2Λ dΛ E2 d2Λ DdΛ dΛ E−2 − 2 (s), (s) (s) (s), (s) ,U(s) ds 0 ds2 ds ds2 ds ds and

Z L Dd2Λ dΛ EDd3Λ dΛ EdΛ DdΛ dΛ E−2 I124 = −4 (s), (s) (s), (s) (s) (s), (s) ,U(s) ds 0 ds2 ds ds3 ds ds ds ds

Z L Dd2Λ dΛ EDd2Λ dΛ EdΛ DdΛ dΛ E−2 − 4 (s), (s) (s), (s) (s) (s), (s) ,U(s) ds 0 ds2 ds ds2 ds ds ds ds

Z L Dd2Λ dΛ E3 dΛ DdΛ dΛ E−3 + 8 (s), (s) (s) (s), (s) ,U(s) ds (B.10) 0 ds2 ds ds ds ds

I13: Let

dU d2ΛΛ dΛ d2Λ dΛ dΛ −1 dV = ds and U = (s), (s) (s) (s), (s) ds ds2 ds ds2 ds ds

Consequently I13 becomes

Z L *d3Λ dΛ d2Λ dΛ dΛ −1 + I13 = − (s), (s) (s) (s), (s) ,U(s) ds 0 ds3 ds ds2 ds ds

59 Z L *d2Λ d2Λ d2Λ dΛ dΛ −1 + − (s), (s) (s) (s), (s) ,U(s) ds 0 ds2 ds2 ds2 ds ds

Z L *d2Λ dΛ d3Λ dΛ dΛ −1 + − (s), (s) (s) (s), (s) ,U(s) ds 0 ds2 ds ds3 ds ds

Z L *d2Λ dΛ 2 d2Λ dΛ dΛ −2 + + 2 (s), (s) (s) (s), (s) ,U(s) ds (B.11) 0 ds2 ds ds2 ds ds

Thus we obtain

dI1Λ(V ) = I11 + I12 + I13

Z L D E = − G1(s),U(s) ds 0

where G1(s) is the sum of all the terms in each of the above integrals excluding U(s).

Expression of G1(s) in terms of curvature and torsion becomes

000 00 00 G1(s) = κ N − κκ Λ + κ τB

Steps to obtain equation (3.23):

Z L DdΛ dΛE−3 DdΛ d2ΛE dΛ Dd2Λ h dΛiE2 I21 = −4 , , , Λ, ,U ds 0 ds ds ds ds2 ds ds2 ds

Z L DdΛ dΛE−2 d2Λ Dd2Λ h dΛiE2 + , , Λ, ,U ds 0 ds ds ds2 ds2 ds

Z L DdΛ dΛE−2 dΛ Dd3Λ h dΛiE Dd2Λ h dΛiE + 2 , , Λ, , Λ, ,U ds 0 ds ds ds ds3 ds ds2 ds

Z L DdΛ dΛE−2 dΛ Dd2Λ h d2ΛiE Dd2Λ h dΛiE + 2 , , Λ, , Λ, ,U ds 0 ds ds ds ds2 ds2 ds2 ds

60 Z L DdΛ dΛE−2 DdΛ d2ΛEDd2Λ h dΛiE h dΛi dU I22 = 2 , , , Λ, Λ, , ds 0 ds ds ds ds2 ds2 ds ds ds

Z L DdΛ dΛE−1 Dd3Λ h dΛiE h dΛi dU − , , Λ, Λ, , ds 0 ds ds ds3 ds ds ds

Z L DdΛ dΛE−1 Dd2Λ h d2ΛiE h dΛi dU − , , Λ, Λ, , ds 0 ds ds ds2 ds2 ds ds

Z L DdΛ dΛE−1 Dd2Λ h dΛiE h d2Λi dU − , , Λ, Λ, , ds 0 ds ds ds2 ds ds2 ds

= I221 + I222 + I223 + I224 (say) (B.12)

dU Again, we evaluate each of the integrals in equation (B.12) by letting dV = ds ds and the rest as U. Thus we get

Z L DdΛ dΛE−3 DdΛ d2ΛE2 Dd2Λ h dΛiE h dΛi I221 = 8 , , , Λ, Λ, ,U ds 0 ds ds ds ds2 ds2 ds ds

Z L DdΛ dΛE−2 Dd2Λ d2ΛEDd2Λ h dΛiE h dΛi − 2 , , , Λ, Λ, ,U ds 0 ds ds ds2 ds2 ds2 ds ds

Z L DdΛ dΛE−2 DdΛ d3ΛEDd2Λ h dΛiE h dΛi − 2 , , , Λ, Λ, ,U ds 0 ds ds ds ds3 ds2 ds ds

Z L DdΛ dΛE−2 DdΛ d2ΛEDd3Λ h dΛiE h dΛi − 2 , , , Λ, Λ, ,U ds 0 ds ds ds ds2 ds3 ds ds

Z L DdΛ dΛE−2 DdΛ d2ΛEDd2Λ h d2ΛiE h dΛi − 2 , , , Λ, Λ, ,U ds 0 ds ds ds ds2 ds2 ds2 ds

Z L DdΛ dΛE−2 DdΛ d2ΛEDd2Λ h d2ΛiE h d2Λi − 2 , , , Λ, Λ, ,U ds 0 ds ds ds ds2 ds2 ds2 ds2 (B.13)

61 Z L DdΛ dΛE−2 DdΛ d2ΛEDd3Λ h dΛiE h dΛi I222 = −2 , , , Λ, Λ, ,U ds 0 ds ds ds ds2 ds3 ds ds

Z L DdΛ dΛE−1 Dd4Λ h dΛiE h dΛi − , , Λ, Λ, ,U ds 0 ds ds ds4 ds ds

Z L DdΛ dΛE−1 Dd3Λ h d2ΛiE h dΛi − , , Λ, Λ, ,U ds 0 ds ds ds3 ds2 ds

Z L DdΛ dΛE−1 Dd3Λ h d2ΛiE h d2Λi − , , Λ, Λ, ,U ds (B.14) 0 ds ds ds3 ds2 ds2

Z L DdΛ dΛE−2 DdΛ d2ΛEDd2Λ h d2ΛiE h dΛi I223 = −2 , , , Λ, Λ, ,U ds 0 ds ds ds ds2 ds2 ds2 ds

Z L DdΛ dΛE−1 Dd3Λ h d2ΛiE h dΛi + , , Λ, Λ, ,U ds 0 ds ds ds3 ds2 ds

Z L DdΛ dΛE−1 Dd2Λ h d3ΛiE h dΛi + , , Λ, Λ, ,U ds 0 ds ds ds2 ds3 ds

Z L DdΛ dΛE−1 Dd2Λ hdΛ d2ΛiE h dΛi + , , , Λ, ,U ds 0 ds ds ds2 ds ds2 ds

Z L DdΛ dΛE−1 Dd2Λ h d2ΛiE h d2Λi + , , Λ, Λ, ,U ds (B.15) 0 ds ds ds2 ds2 ds2 and

Z L DdΛ dΛE−2 DdΛ d2ΛEDd2Λ h dΛiE h d2Λi I224 = −2 , , , Λ, Λ, ,U ds 0 ds ds ds ds2 ds2 ds ds2

Z L DdΛ dΛE−1 Dd3Λ h dΛiE h d2Λi + , , Λ, Λ, ,U ds 0 ds ds ds3 ds ds2

62 Z L DdΛ dΛE−1 Dd2Λ h d2ΛiE h d2Λi + , , Λ, Λ, ,U ds 0 ds ds ds2 ds2 ds2

Z L DdΛ dΛE−1 Dd2Λ h d2ΛiE h d3Λi + , , Λ, Λ, ,U ds 0 ds ds ds2 ds2 ds3

Z L DdΛ dΛE−1 Dd2Λ h dΛiE hdΛ d2Λi + , , Λ, , ,U ds (B.16) 0 ds ds ds2 ds ds ds2

Similarly I23 gives us

Z L DdΛ dΛE−2 DdΛ d2ΛEDd2Λ h dΛiE hd2Λ i I23 = 2 , , , Λ, , Λ ,U ds 0 ds ds ds ds2 ds2 ds ds2

Z L DdΛ dΛE−1 Dd3Λ h dΛiE hd2Λ − , , Λ, , Λ],U ds 0 ds ds ds3 ds ds2

Z L DdΛ dΛE−1 Dd2Λ h d2ΛiE hd2Λ − , , Λ, , Λ],U ds 0 ds ds ds2 ds2 ds2

Z L DdΛ dΛE−1 Dd2Λ h dΛiE hd2Λ dΛi − , , Λ, , ,U ds 0 ds ds ds2 ds ds2 ds

Z L DdΛ dΛE−1 Dd2Λ h dΛiE hd3Λ − , , Λ, , Λ],U ds (B.17) 0 ds ds ds2 ds ds3 and ﬁnally

Z L DdΛ dΛE−1 Dd2Λ h dΛiE Dd2Λ hdΛ iE I24 = − , , Λ, , ,U ds 0 ds ds ds2 ds ds2 ds

Z L DdΛ dΛE−1 Dd2Λ h dΛiE Dhd2Λ dΛi E = − , , Λ, , ,U ds 0 ds ds ds2 ds ds2 ds

Z L DdΛ dΛE−1 Dd2Λ h dΛiE hd2Λ dΛi = − , , Λ, , ,U ds (B.18) 0 ds ds ds2 ds ds2 ds

Therefore, we obtain

dI2Λ(V ) = I21 + I22 + I23 + I24

63 Z L D E = − G2(s),U(s) ds 0

where G2(s) is the sum of all the terms in each of the above integrals excluding U(s).

Now, G2(s) in terms of curvature and torsion takes the following form

0 2 2 2 3 0 3 0 0 G2(s) = −3κ τ N + 2κ τ Λ − κτ B − 3κττ N + κ τB + 3κ τ B

+ 2κ00τB + κτ 00B

Steps to obtain equation (3.24):

DdΛ d2Λ E dΛ G (s) = (s), (s) (s) 3 ds ds2 ds

1DdΛ dΛ E d2Λ + (s), (s) (s) 2 ds ds ds2

Expressing G3(s) in terms of curvature and torsion we get

3 1 1 G (s) = κ2κ0N + κ4Λ + κ3τB (B.19) 3 2 2 2

B.3 Calculations in Section 3.4

In this section we discuss the Poisson commuting functions mentioned in the Theorem 3.4.

Z L Theorem B.1. The function f1 = τ(s)ds Poisson commutes with the functions 0 Z L 1 2 (a) f2 = κ (s) ds. 2 0 Z L 1 2 (b) f3 = κ (s)τ(s) ds. 2 0 Z L 1 0 2 1 2 2 1 4 (c) f4 = (κ ) (s) + κ (s)τ (s) − κ (s) ds. 0 2 2 8

64 Proof. The Poisson bracket of two functions fa and fb is given by the formula

Z L

{fa, fb} = ωX (Fa(s),Fb(s)) = − hΛ(s), [Ufa (s),Ufb (s)]ids 0

where Ufa (s) and Ufb (s) are Hamiltonian vector ﬁelds corresponding to fa and fb respectively.

The Hamiltonian vector ﬁelds of f1, f2 and f3 are Uf1 = κ(s, t)N(s, t), Uf2 =

00 000 000 3 00 0 0 00 000 [Λ , Λ] and Uf3 = [Λ −hΛ , ΛiΛ]− 2 hΛ, Λ iΛ respectively. Here, Λ , Λ , Λ denote the ﬁrst, second and third derivatives of Λ with respect to s. Proof of (a): We are required to show that

Z L

{f2, f1} = ωX (F2(s),F1(s)) = − hΛ(s), [Uf2 (s),Uf1 (s)]ids = 0 0

Now

00 0 2 0 Uf2 = [Λ , Λ] = [κ N − κ Λ + κτB, Λ] = −κ B + κτN, so

0 0 [Uf2 (s),Uf1 (s)] = [−κ B + κτN, κN] = κκ Λ.

Then

Z L 0 {f2, f1} = − hΛ, κκ Λids 0

Z L = − κκ0ds 0

Z L d 1 = − ( κ2)ds 0 ds 2 = 0.

Therefore, the function f2 Poisson commutes with f1.

65 Proof of (b): The proof is similar to the proof in part(a). Here we need to show that Z L

{f1, f3} = ωX (F1(s),F3(s)) = − hΛ(s), [Uf1 (s),Uf3 (s)]ids = 0 0

Now Λ000 = κ00N − 3κκ0Λ + 2κ0τB − κ3N + κτ 0B − κτ 2N,

hΛ000, ΛiΛ = −3κκ0Λ,

hΛ, Λ00i = hΛ, κ0N − κ2Λ + κτBi = −κ2,

3 3 hΛ, Λ00iΛ0 = −κ2 · κNi = − κ3N. 2 2

000 000 3 00 0 U = Λ − hΛ , ΛiΛ − hΛ, Λ iΛ f3 2

1 = κ00N + 2κ0τB + κ3N + κτ 0B − κτ 2N. 2

Then the Lie bracket of Uf1 and Uf3 gives

0 2 0 [Uf1 ,Uf3 ] = −2κκ τΛ − κ τ Λ.

Consequently

Z L 0 2 0 {f3, f1} = − hΛ, (−2κκ τ − κ τ )Λids 0

Z L = − (−2κκ0τ − κ2τ 0)ds 0

66 Z L d = (κ2τ)ds 0 ds

= 0.

Therefore, the function f1 Poisson commutes with f3. Proof of (c): We need to show that

Z L

{f1, f4} = ωX (F1(s),F4(s)) = − hΛ(s), [Uf1 (s),Uf4 (s)]ids = 0 0

Now

3 3 U = (−κτ 3 + 3κ00τ + 3κ0τ 0 + κ3τ + κτ 00)N + (−κ000 + 3κττ 0 + 3κ0τ 2 − κ2κ0)B. f4 2 2

3 [U (s),U (s)] = [(−κτ 3 + 3κ00τ + 3κ0τ 0 + κ3τ + κτ 00)N + (−κ000 + 3κττ 0 + 3κ0τ 2 f1 f4 2

3 − κ2κ0)B, κN] 2

3 = (−κκ000 + 3κ2ττ 0 + 3κκ0τ 2 − κ3κ0)Λ 2

Then taking the inner product with Λ we get

3 hΛ, [U ,U ]i = −κκ000 + 3κ2ττ 0 + 3κκ0τ 2 − κ3κ0. f1 f4 2

As a consequence

Z L 000 2 0 0 2 3 3 0 {f1, f4} = − (−κκ + 3κ ττ + 3κκ τ − κ κ )ds 0 2

Z L d 1 d 3 d 3 = − − κκ00 + (κ0)2 + κ2τ 2 + κ4 ds 0 ds 2 ds 2 ds 8

67 = 0.

Therefore, the function f1 Poisson commutes with f4.

Z L 1 2 Theorem B.2. The function f2 = κ (s) ds Poisson commutes with the func- 2 0 Z L 1 2 tion f3 = κ (s)τ(s) ds. 2 0

Proof. See Theorem 7 in [6].

Z L 1 2 Theorem B.3. The function f2 = κ (s) ds Poisson commutes with the func- 2 0 Z L h1 0 2 1 2 2 1 4 i tion f4 = (κ ) (s) + κ (s)τ (s) − κ (s) ds. 0 2 2 8

Proof. See Theorem 3.4.

Z L 1 2 Theorem B.4. The function f3 = κ (s)τ(s) ds Poisson commutes with the 2 0 Z L h 0 2 1 2 2 1 4 i function f4 = (κ ) (s) + κ (s)τ (s) − κ (s) ds. 0 2 8

Proof. We need to show that the Poisson bracket of the functions f3 and f4 is equal to zero. That is,

{f4, f3} = 0 which is deﬁned by the symplectic form.

The Hamiltonian vector ﬁelds associated with f3 and f4 in terms of curvature and torsion are

1 U (s) = κ00N + 2κ0τB + κ3N + κτ 0B − κτ 2N f3 2 and

3 3 U (s) = (−κτ 3 + 3κ00τ + 3κ0τ 0 + κ3τ + κτ 00)N + (−κ000 + 3κττ 0 + 3κ0τ 2 − κ2κ0)B f4 2 2

68 The inner product of the Lie bracket of these two vector ﬁelds with Λ(s) generates

D E 3 1 3 Λ, [U (s),U (s)] = − κ2κ0κ00 − κ3κ000 − 3κ0κ00τ 2 + κκ000τ 2 − κ5κ0 − κ00κ000 f3 f4 2 2 4

− 3κκ0(τ 0)2 − κ2τ 0τ 00 − 6(κ0)2ττ 0 − κκ0τ 4 − 2κ2τ 3τ 0 − 2κκ0ττ 00

Thus the Poisson bracket

Z L 3 2 0 00 1 3 000 0 00 2 000 2 3 5 0 00 000 {f4, f3} = − − κ κ κ − κ κ − 3κ κ τ + κκ τ − κ κ − κ κ 0 2 2 4

− 3κκ0(τ 0)2 − κ2τ 0τ 00 − 6(κ0)2ττ 0 − κκ0τ 4 − 2κ2τ 3τ 0 − 2κκ0ττ 00 ds

Z L d 1 1 1 1 = − − κ3κ00 + κκ00τ 2 − κ6 − (κ00)2 − κ2(τ 0)2 0 ds 2 8 2 2

1 − 2τ 2(κ0)2 − κ2τ 4 − 2κκ0ττ 0 ds 2

= 0.

Therefore, the function f3 Poisson commutes with the function f4.

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