Conformal transformations, curvature, and energy

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Conformal transformations, curvature, and energy Conformal transformations, curvature, and energy

Richard G. Ligo University of Iowa

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Recommended Citation Recommended Citation Ligo, Richard G.. "Conformal transformations, curvature, and energy." PhD (Doctor of Philosophy) thesis, University of Iowa, 2017. https://doi.org/10.17077/etd.99g6qj3i

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CONFORMAL TRANSFORMATIONS, CURVATURE, AND ENERGY

by

Richard G. Ligo

A thesis submitted in partial fulfillment of therequirements for the Doctor of Philosophy

degree in Mathematicsin the Graduate College of

The University of Iowa

May 2017

Thesis Supervisor: Associate Professor Oguz Durumeric

Copyright byRICHARD G. LIGO

2017All Rights Reserved

Graduate CollegeThe University of Iowa

Iowa City, Iowa

CERTIFICATE OF APPROVAL

PH.D. THESIS

This is to certify that the Ph.D. thesis of

Richard G. Ligo

has been approved by the Examining Committee for thethesis requirement for the Doctor of Philosophy degreein Mathematics at the May 2017 graduation.

Thesis committee:

Oguz Durumeric, Thesis Supervisor

Charles Frohman

Keiko Kawamuro

Walter Seaman

Maggy Tomova

ACKNOWLEDGEMENTS

Thank you to my advisor, Oguz Durumeric, a man with whom I have found

myself brilliantly compatible in many ways. I am deeply appreciative of his guidance

in all avenues—mathematical, professional, and personal.

Thank you to all the other mathematicians who have played a role in this

journey. To David Stewart: our nine-minute conversation regarding MATLAB was

the gateway to nearly a third of my work. To Charlie Frohman: in those times when

Oguz and I wanted another opinion, your mathematical wisdom was most useful.

To Maggy Tomova: your professional advice has been invaluable, and I credit you

with giving me the initiative to work with Oguz. To Keiko Kawamuro and Walter

Seaman: I am thankful for your willingness to serve on my review panel alongside

Oguz, Charlie, and Maggy. To my colleagues: you have always been willing to listen

to my bursts of geometric ideas, despite not being students of the subject.

Thank you also to my family. My father’s patient confidence and mother’s

unwavering support have served as great motivators. For those times I needed to step

away from my work, my brother’s tremendous energy and endless creativity provided

welcome respites.

Finally, I am thankful to God for having given me this wonderful opportunity

and the resilience to bring it to completion. Mathematics will always be one of my

favorite parts of His creation.

ii

ABSTRACT

Space curves have a variety of uses within mathematics, and much attention

has been paid to calculating quantities related to such objects. The quantities of

curvature and energy are of particular interest to us. While the notion of curvature is

well-known, the Mobius energy is a much newer concept, having been first defined by

Jun O’Hara in the early 1990s. Foundational work on this energy was completed by

Freedman, He, and Wang in 1994, with their most important result being the proof

of the energy’s conformal invariance. While a variety of results have built upon those

of Freedman, He, and Wang, two topics remain largely unexplored: the interaction of

curvature and Mobius energy and the generalization of the Mobius energy to curves

with a varying thickness. In this thesis, we investigate both of these subjects.

We show two fundamental results related to curvature and energy. First, we

show that any simple, closed C2 curve can be transformed in an energy-preserving and

length-preserving way that allows us to make the pointwise curvature arbitrarily large

at a point. Next, we prove that the total absolute curvature of a C2 curve is uniformly

bounded with respect to conformal transformations. This is accomplished mainly via

an analytic investigation of the effect of inversions on total absolute curvature.

In the second half of the thesis, we define a generalization of the Mobius

energy for simple curves of varying thickness that we call the “nonuniform energy.”

We call such curves “weighted curves,” and they are defined as the pairing of a

curve parametrization and positive, continuous weight function on the same domain.

iii

We then calculate the first variation formulas for several different variations of the

nonuniform energy. Variations preserving a closed curve’s shape and total weight are

shown to have no minimizers exactly when the curve has a point of zero curvature.

Variations that “slide” the weight along a closed curve are shown to preserve energy

is special cases.

iv

PUBLIC ABSTRACT

Curves in n-dimensional space can be described mathematically, which allows

us to calculate various quantities related to these objects. The curvature tells us how

much a curve is bending, whereas the energy provides a measure of the complexity of

the curve. Conformal transformations of n-dimensional space rearrange curves in n-

dimensional space in a predictable way, and it can be shown that this rearrangement

does not change a curve’s energy. We address two main questions related to the ideas

of curvature, energy, and conformal transformations.

First, how are the curvature and energy related to each other? This question

ultimately boils down to understanding how conformal transformations affect curva-

ture. We show that while the effects can appear exotic, the total amount of bending

can only become so large.

Second, is there a way to describe the energy of a curve with varying thickness?

Following this question we investigate how energy varies as the thickness of a curve is

adjusted. We show that certain examples behave is nonintuitive ways; for example,

the energy of a certain family of curves does not change when the thickness is changed

in a specific way.

v

TABLE OF CONTENTS

LIST OF FIGURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii

CHAPTER

1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1 Conformal Transformations . . . . . . . . . . . . . . . . . . . . . 11.2 Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 The Mobius Energy . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 POINTWISE CURVATURE AND CONFORMAL TRANSFORMA-TIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 TOTAL ABSOLUTE CURVATURE AND CONFORMAL TRANS-FORMATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.1 Total Curvature of Polygonal Curves . . . . . . . . . . . . . . . . 253.2 Total Curvature of C2 Curves . . . . . . . . . . . . . . . . . . . . 33

3.2.1 Curvature Formulas and Consequences . . . . . . . . . . . 333.2.2 Analysis of Inversions of Closed C2 Curves . . . . . . . . 40

3.2.2.1 Inversion Centers on the Curve . . . . . . . . . . 423.2.2.2 Inversion Centers off the Curve . . . . . . . . . . 56

3.2.3 Analysis of Inversions of Open, Piecewise C2 Curves . . . 113

4 THE NONUNIFORM ENERGY . . . . . . . . . . . . . . . . . . . . . 117

4.1 Definition of the Nonuniform Energy . . . . . . . . . . . . . . . . 1174.2 Properties of the Nonuniform Energy . . . . . . . . . . . . . . . 1184.3 Properties of Minimizers of the Nonuniform Energy . . . . . . . 1254.4 Computing the Nonuniform Energy . . . . . . . . . . . . . . . . 130

5 VARIATIONS OF THE NONUNIFORM ENERGY . . . . . . . . . . 134

5.1 Curve Variations . . . . . . . . . . . . . . . . . . . . . . . . . . . 1375.1.1 The Free Curve Variation . . . . . . . . . . . . . . . . . . 1385.1.2 The Equilength Variation . . . . . . . . . . . . . . . . . . 142

5.2 Weight Variations . . . . . . . . . . . . . . . . . . . . . . . . . . 1605.2.1 The Equitotal Variation . . . . . . . . . . . . . . . . . . . 161

vi

5.2.2 The Translatory Variation . . . . . . . . . . . . . . . . . 1735.2.3 The Compromisal Variation . . . . . . . . . . . . . . . . . 193

REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

vii

LIST OF FIGURES

Figure

1.1 The intrinsic distance is at least the Euclidean distance. . . . . . . . . . 6

1.2 A square’s total absolute curvature changes from 2π to 6π if invertedthrough a circle whose center lies in the square’s interior. . . . . . . . . . 8

1.3 An ellipse’s total absolute curvature changes from 2π to 8.65452 if invertedthrough a circle whose center is the origin. . . . . . . . . . . . . . . . . . 9

2.1 Example of an osculating circle. . . . . . . . . . . . . . . . . . . . . . . . 11

2.2 An illustration of Proposition 2.3 with an ellipse. . . . . . . . . . . . . . 18

3.1 The inversion of λ to Ir,c(λ). . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.2 The inversion of p to Ir,c(p), where c is in the interior of p . . . . . . . . . 28

3.3 The inversion of p to Ir,c(p), where c is on p, but not a corner. . . . . . . 29

3.4 The inversion of p to Ir,c(p), where c is on a corner of p. . . . . . . . . . . 30

3.5 The inversion of p to Ir,c(p), where c is in the exterior of p. . . . . . . . . 31

3.6 Inversion through a distant center is approximately a reflection. . . . . . 40

3.7 Diagram for Lemma 3.19. . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3.8 Diagram for Proposition 3.21. . . . . . . . . . . . . . . . . . . . . . . . . 59

3.9 An approach for analyzing the total curvature of γ when the inversioncenter is close to γ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

3.10 The segment of γ analyzed in Lemmas 3.25 and 3.26. . . . . . . . . . . . 75

3.11 The segment of γ analyzed in Lemma 3.27. . . . . . . . . . . . . . . . . . 92

3.12 The segments from Lemmas 3.26 and 3.27 reunite to form γ in Proposi-tion 3.28. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

viii

3.13 An illustration of the construction of W and Bnd (c0). . . . . . . . . . . . 110

4.1 Examples of weighted curves. . . . . . . . . . . . . . . . . . . . . . . . . 118

4.2 The weighted knots (γ, µ) and (I1,(0,0)(γ), µ). . . . . . . . . . . . . . . . . 122

5.1 An illustration of the two ways to obtain the first variation formula forthe equilength variation. . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

5.2 While they have different distributions, each weighted knot has the sametotal weight. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

5.3 The graph of µn, as used in Proposition 5.24. . . . . . . . . . . . . . . . 169

5.4 The division of [0, `]× [0, `] into regions I, II, and III for Proposition 5.24. 170

5.5 Each weighted knot has the same weight profile, but they are translationsof each other. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

5.6 A non-torus knot satisfying Properties (a) and (b) of Proposition 5.35. . 190

ix

1

CHAPTER 1INTRODUCTION

We shall start by laying out the definitions and descriptions related to three

fundamental ideas: conformal transformations, curvature, and Mobius energy.

1.1 Conformal Transformations

Conformal transformations are the common thread by which all our work is

connected, and so we begin with them.

Definition 1.1. Let U ⊆ Rn be an open set. Then any transformation T : U → Rn

is called a conformal transformation if it is angle-preserving.

Liouville’s Theorem for conformal transformations gives us an extremely useful

characterization of conformal transformations.

Theorem 1.1 (Liouville [3]). Let n ≥ 3 and say T : U → Rn is a conformal trans-

formation. Then f is the restriction to U of a composition of translations, rotations,

reflections, dilations, and inversions, at most one of each.

Liouville’s theorem does not account for the case in which n = 2. The Riemann

mapping theorem states that any two simply connected open subsets of C∪ {∞} are

conformally equivalent provided that neither is the whole complex plane [1]. However,

Liouville’s theorem still applies if we have a transformation of R2 ∪ {∞} = C∪ {∞},

since (i) a map of the extended complex plane onto itself is conformal if and only if it

is a linear fraction transformation or its conjugates [6], and (ii) the linear fractional

2

transformations with conjugates can be decomposed exactly as stated in Liouville’s

theorem.

This leads us to the definition of a Mobius transformation. While this defini-

tion is nonstandard, it coincides with that used by Freedman, He, and Wang [4].

Definition 1.2. We call a map T : Rn ∪ {∞} → Rn ∪ {∞} a Mobius transformation

if it is generated by a composition of at most one each of translations, rotations,

reflections, dilations, and inversions.

The first three of these elementary transformations are isometries; that is,

transformations that preserve distance. The isometries together with the dilations

form the generating set for another set of transformations, which we define here.

Definition 1.3. Let n ≥ 2. Then any transformation T : Rn → Rn generated by at

most one each of translations, rotations, reflections, and dilations of Rn is called an

affine similarity.

It is important to note that affine similarities are different from affine trans-

formations. Specifically, affine transformations allow for shear transformations of Rn,

whereas affine similarities do not.

Inversions are arguably the most interesting of the five types of transformations

listed in Theorem 1.1. We define them as follows:

Definition 1.4. Let n ≥ 2. Then we define inversion in the unit sphere as the map

I : Rn ∪ {∞} → Rn ∪ {∞} given by

I(x) =x

|x|2.

3

This formula can be generalized to inversion through an n-sphere of radius r centered

at c ∈ Rn via

Ir,c(x) = c+r2(x− c)|x− c|2

.

The generalized inversion formula will be very useful in Chapters 2 and 3.

1.2 Curvature

Note: Recall that a simple curve is one without self-intersection.

Throughout the entirety of this thesis, every curve considered will be

simple or simple closed unless otherwise noted.

The term “curvature” has various meanings throughout the whole of geometry,

but we wish to focus only on how it relates to curves. In particular, we will examine

the pointwise curvature and total absolute curvature of curves.

Definition 1.5. For a curve γ parametrized by arclength, the pointwise curvature is

given by the formula

κabsγ (s) =

∣∣∣∣dTds (s)

∣∣∣∣ ,where T (s) is the unit tangent of γ.

Remark. This is a measurement of how fast a curve is turning while a point moves

on it with unit speed. If a curve is not parametrized by arclength, we can still define

its pointwise curvature via a general formula—see Lemma 2.4.

There are a variety of notations used throughout literature to denote pointwise

curvature. We shall use κsgnγ (t) and κabs

γ (t) to represent the signed and absolute point-

4

wise curvatures of γ, respectively. It is important to note that the signed pointwise

curvature only exists when γ ⊆ R2.

In addition, many of the curves we will consider are so-called “closed curves.”

More formally, a curve γ : [0, `] → Rn is a Ck closed curve if γ has k continuous

derivatives and γ can be extended to a Ck curve γ : R→ Rn that is `-periodic (that

is, ` is the smallest positive number such that γ(t) = γ(t + `) for all t ∈ R). When

doing analysis around γ(0) = γ(`), we will work with γ|[0−ε,`−ε].

As mentioned, we also explore the notion of the total curvature of a curve. This

is obtained by integrating the pointwise curvature with respect to arclength along the

curve—exact formulas are discussed in Chapter 3. Even more so than for the pointwise

curvature, there is no canonical notation for total curvature. Let κsgntot (γ) and κabs

tot (γ)

represent the signed and absolute total curvatures of γ, respectively. Again, note that

the total signed curvature only exists when γ ⊆ R2.

1.3 The Mobius Energy

Let γ : [0, `] → R3 be an embedding of a curve, and let K be the space of all

such embeddings.

Definition 1.6. Then a knot energy functional is a real-valued functional e on K

that satisfies the following:

(i) e is bounded from below.

(ii) e is continuous with respect to a suitable topology.

(iii) e blows up as a knot degenerates to an immersion with a double point.

5

In 1990 Jun O’Hara proposed a knot energy functional that would later come

to be known as the Mobius energy [7], shown in Definition 1.7.

Definition 1.7. Let γ : [0, `] → Rn be a C1 curve. Then the Mobius energy is the

functional E : K → R defined as

E(γ) =

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]|γ′(s)||γ′(t)| ds dt.

Dγ(s, t) is called the intrinsic distance; specifically, Dγ(s, t) is the shortest distance

between γ(s) and γ(t) along γ.

Notice immediately that |γ(s)−γ(t)| ≤ Dγ(s, t), with equality only in the case

where the segment of γ between γ(s) and γ(t) is linear. (Illustrated in Figure 1.1.)

This tells us right away that E(γ) ≥ 0 for all γ, so the Mobius energy satisfies

Property (i) of Definition 1.6. The dynamic between the Euclidean and intrinsic

distances also allows us to recognize that Property (iii) is satisfied. We can see from

the formula given in Definition 1.7 that the Mobius energy will vary continuously

with respect to ambient isotopies of γ, so we know that Property (ii) is also satisfied.

Many results have been built out of O’Hara’s definition and the foundational

work presented by Freedman, He, and Wang’s 1994 paper, Mobius Energy of Knots

and Unknots [4]. Theorem 1.2 is arguably the most important of their results; in fact,

it is the reason that the functional possesses the name “Mobius.”

Theorem 1.2 (Freedman, He, and Wang). Let γ : [0, `]→ Rn be a closed curve and

let T be a Mobius transformation of Rn ∪ {∞}. Then we have two cases:

(i) If T (γ) ⊆ Rn, then E(T (γ)) = E(γ).

6

Figure 1.1: The intrinsic distance is at least the Euclidean distance.

(ii) If ∞ ∈ T (γ), then E(T (γ) \ {∞}) = E(γ)− 4.

Recall that every Mobius transformation of Rn ∪ {∞} (where n ≥ 2) can be

written as a composition of translations, rotations, reflections, dilations, and inver-

sions. This means that Theorem 1.2 gives us tremendous freedom when working with

the Mobius energy, as we can perform various transformations and leave the energy

unchanged.

1.4 Overview

Our work is comprised of two main parts. Chapters 2 and 3 make up the first

part, and Chapters 4 and 5 are the second.

In Chapter 2, we initiate an investigation of the connections between the

Mobius energy and curvature of a curve—this chapter focuses only on pointwise cur-

vature. In particular, we show that any noncircular close curve can be conformally

transformed to have arbitrarily high pointwise curvature without changing its length

7

or Mobius energy, seen in Proposition 2.3.

Proposition 2.3. Let γ : [0, `] → Rn be a non-circular, closed, C2 curve. Then for

any K ∈ R+ there exists a length and energy-preserving transformation T such that

the maximum absolute pointwise curvature of T (γ) is greater than K.

To conclude Chapter 2, we compute formulas for the pointwise curvature of

the image of a curve under inversion in R2, R3, and Rn.

In order to understand the connections between the Mobius energy and total

absolute curvature, we focus our attention on the interaction between total curvature

and Mobius transformations in Chapter 3. As this interaction is trivial for isometries

and straightforward for dilations, the majority of our work is devoted to understanding

the effects of inversions on the total absolute curvature. Consider Examples 1.1

and 1.2, and notice that in both cases convexity is not preserved.

Example 1.1. Let p be a square in R2. Then p has total absolute curvature 2π.

If c is in the interior of p, then Ir,c(p) has total absolute curvature 6π, as shown in

Figure 1.2. Notice that the choice of r has no effect on the total absolute curvature. If

one takes C2-approximations of the square with total curvature 2π (such as rounding

at the corners) their inversions will have total curvature close to 6π.

Example 1.2. Let γ be the ellipse defined as γ(t) = (2 cos(t), sin(t)). Then γ has

total absolute curvature 2π. If c = (0, 0), then Ir,c(γ) has total absolute curvature

approximately 8.65452, as shown in Figure 1.3. As in Example 1.1, the choice of r

has no effect.

8

Figure 1.2: A square’s total absolute curvature changes from 2π to 6π if inverted

through a circle whose center lies in the square’s interior.

Somewhat suprisingly, we found that the set of total absolute curvatures of

all Mobius transformations of a given curve γ have a finite supremum. This behavior

is a consequence of Theorem 3.33. Notice the parallels between this result and The-

orem 1.2: both have reasonable behavior when inversion centers are away from the

curve and a correction term when inversion centers coincide with the curve.

Theorem 3.33. Let γ : [0, `] → Rn be a closed C2 curve parametrized by arclength,

and define the map Φ : R+ × Rn → R by

Φ(r, c) =

κabstot (Ir,c(γ)) + 2π c ∈ γ([0, `])

κabstot (Ir,c(γ)) c /∈ γ([0, `])

Then Φ is continuous and bounded.

9

Figure 1.3: An ellipse’s total absolute curvature changes from 2π to 8.65452 if inverted

through a circle whose center is the origin.

It is important to note that κabstot (Ir,c(γ)) is constant with respect to r (Corol-

lary 3.6). We can then obtain Theorem 3.33 by dividing the plane into four parts:

points on the curve, points in a neighborhood of the curve, points in a neighborhood

of infinity, and the remaining points. Inversions with centers in each of these re-

gions are analyzed separately, and then these individual results are synthesized into

Theorem 3.33.

In Chapter 4 we begin an investigation of a generalization of the Mobius energy.

Called the “nonuniform energy,” it is specifically designed to be compatible with

curves of nonuniform thickness. We describe a number of its elementary properties,

many of which are referential to the work completed by Freedman, He, and Wang [4].

We then turn to describing the properties of minimizers of the nonuniform energy.

Finally, we give a brief description of our approach for computing the nonuniform

energy through the use of MATLAB.

In Chapter 5 we investigate several variations of the nonuniform energy. Sec-

10

tion 5.1 provides first variation formulas for two curve variations—a general variation

and a variation that preserves arclength. Section 5.2 describes three different weight

variations and calculates their first and second variations. In addition, counterintu-

itive behaviors within the equitotal and translatory variations are investigated, cul-

minating in proofs. Specifically, we show that the equitotal variation of closed curves

has no absolute minimizer exactly when the curve has a point of zero curvature and

that specific classes of weighted knots are invariant under translatory variation.

11

CHAPTER 2POINTWISE CURVATURE AND CONFORMAL

TRANSFORMATIONS

Our goal in this section is twofold. First, we will show that the maximum of

the pointwise curvature can be made arbitrarily large by a length-preserving inversion

for any non-circular closed curve (Proposition 2.3). Secondly, we will derive formulas

for the pointwise curvature of the image of a curve under inversion in R2, R3, and Rn

(Proposition 2.5).

The pointwise curvature is most easily visualized from the perspective of os-

culating circles—that said, we make the following definition:

Definition 2.1. Let γ : [0, `] → Rn be a curve. If γ is C2 at a point t ∈ [0, `], then

the osculating circle at γ(t) is the unique circle η (see Figure 2.1) given by

η(s) =

(γ(t) +

γ′′(t)

κabsγ (t)2

)+−γ′′(t)κabsγ (t)2

cos(κabsγ (t)s

)+

γ′(t)

κabsγ (t)

sin(κabsγ (t)s

).

In the case where κabsγ (t) = 0, we simply define η to be the line tangent to γ at γ(t).

Figure 2.1: Example of an osculating circle.

12

Remark. Notice that the radius of the osculating circle at γ(t) is the reciprocal of the

absolute pointwise curvature at γ(t).

Lemma 2.1. A Mobius tranformation of Rn ∪ {∞} sends circles and lines to circles

and lines.

Proof. Recall that any Mobius transformation can be decomposed into at most one

each of translations, rotations, reflections, dilations, and inversions. As the first four

types of transformations all preserve circles and lines, we need focus only on the

behavior of circles and lines with respect to inversions. We have two cases:

Case 1: n = 2.

This is proved for C ∪ {∞} in [6], which equivalent to this case. Specifically,

a circle C [alternatively, a line L] is sent to a line when the center of the “inverting

circle” is on C [L], and C [L] is sent to a circle otherwise.

Case 2: n ≥ 3.

Let C be a circle in Rn. We first must prove a claim:

Claim 2.1.1. An inversion sends (n−1)-spheres to (n−1)-spheres or (n−1)-planes.

Proof of Claim 2.1.1: Let S ⊆ Rn be an (n − 1)-sphere, and let c be the center of

our “inverting (n− 1)-sphere” We have two cases:

Case A: c ∈ S.

Choose q ∈ S. Now choose ~v ∈ TcRn such that ~v ⊥ c− q. Then it follows that

{~v, c− q} defines a 2-plane P ⊆ Rn containing c and q. Let D be the circle given by

P ∩S, and note that the image D of D under inversion must be a line by Case 1. As

13

this is true for all ~v, we have that the union of all possible D will be an (n−1)-plane.

To see the final statement, consider the following argument: Let c1 be the

antipodal point of c on S, and define c2 be the image of c1 under inversion. Then c,

c1, and c2 are colinear, and note that ←→c1c2 ⊥ S at both c and c1. Then every great

circle of S through c and c1 is inverted to a line, perpendicular to ←→c1c2 at c2 and vice

versa.

Case B: c /∈ S.

As c /∈ S, we can choose p, q ∈ S so that c, p, q are collinear and unique. Now

choose ~v ∈ TcRn such that ~v ⊥ p− q. Then it follows that {~v, p− q} defines a 2-plane

P ⊆ Rn containing c, p, and q. Let D be the circle given by P ∩ S, and note that

the image D of D under inversion must be a circle by Case 1. As this is true for all

~v, we have that the union of all possible D will be an (n− 1)-sphere.

To see the final statement, consider the following argument: Let p1 and p2 be

the closest and the furthest points of S from c, respectively. (Note that this tells us

that p1 and p2 will be anitpodal on S, and ←→p1p2 ⊥ S at p1 and p2 and c ∈ ←→p1p2.)

Let p3 and p4 be the images of p1 and p2 under inversion, respectively, and define

p5 = 12(p3 + p4). All c, p1, p2, p3, p4, and p5 are colinear. Every great circle of S

through p1 and p2 is inverted to a circle through p3 and p4 perpedicular to ←→p1p2, and

centered at p5, and vice versa. The union of all such circles is the image of S under

inversion, which is another (n− 1)-sphere.

As both cases A and B hold, we have proved our claim. �

Recall that every circle can be defined as a transverse intersection of a collec-

14

tion n − 1 many (n − 1)-spheres and (n − 1)-planes, and vice-versa. Hence, we can

choose n− 1 many (n− 1)-spheres S1, . . . , Sn−1 such that

C = S1 ∩ · · · ∩ Sn−1.

Let Si represent the image of Si under inversion—it follows from our claim that each

Si will be either an (n− 1)-sphere or (n− 1)-plane. Hence, we know that

C = S1 ∩ · · · ∩ Sn−1.

is indeed a circle (or a line in the case that each Si is a plane), which completes the

proof.

We now prove a lemma critical to the proof of Proposition 2.3.

Lemma 2.2. A smooth Mobius transformation sends osculating circles and lines to

osculating circles and lines.

Proof. Lemma 2.1 tells us that a Mobius transformation sends circles and lines to

circles and lines. Hence, the image of any osculating circle or line is a circle or line

under a Mobius transformation—it then only remains to show that this circle or line

is a second degree approximation of the curve.

Now recall that smooth maps preserve any second degree approximation. As

osculating circles and lines are second degree approximations of a curve, our image

circle or line will be a second degree approximation of the curve, which completes the

proof.

15

Say that γ is a curve, c ∈ Rn, and r ∈ R+. Let Snr (c) represent the n-sphere

of radius r centered at c, let Bnr (c) represent the open n-ball of radius r centered at

c, and let length(γ) represent length of γ. An example of Proposition 2.3 is shown in

Figure 2.2.

Proposition 2.3. Let γ : [0, `] → Rn be a non-circular, closed, C2 curve. Then for

any K ∈ R+ there exists a length-preserving and energy-preserving transformation T

such that the maximum absolute pointwise curvature of T (γ) is greater than K.

Proof. Since γ is not a circle, there exists some point s ∈ [0, `] such that the osculating

circle at γ(s) does not coincide with every other point on γ. Set q = γ(s), and let

p = γ(t) be a point on γ not coinciding with the osculating circle at q, which we call

Oq. Since p /∈ Oq, there exists some open n-ball Bnε (p) such that

Bnε (p) ∩Oq = ∅ and

1

ε> K.

Note that both of these conditions simply require that ε is sufficiently small, so we

immediately know that such ε exists.

We now claim that there exists some c /∈ γ and r ∈ R+ such that

(i) Ir,c(Oq) = S1δ (x), where 1

δ> K.

(ii) length(Ir,c(γ)) = length(γ).

To prove the claim we proceed as follows. Choose c0 /∈ γ and r0 ∈ R+ so that

Sn−1r0

(c0) ⊆ Bnε (p). As Oq is disjoint from Bn

ε (p), we know then that Ir0,c0(Oq) will be

an osculating circle to Ir0,c0(γ) (refer to Lemma 2.2) at Ir0,c0(q) inside Sn−1r0

(c0), which

we call S1δ (x). Since S1

δ (x) ⊆ Bnr0

(c0) ⊆ Bnε (p) and Sn−1

r0(c0) ⊆ Bn

ε (p), we then have

16

that

1

δ>

1

r0

>1

ε> K.

This follows from comparing the curvature of S1δ (x) to the curvature of circle obtained

from the intersection of the 2-plane containing S1δ (x) and Sn−1

r0(c0). Hence, we have

that Condition (i) is satisfied. Note now that

limc→p

length(Ir0,c(γ)) =∞ and limr→0

length(Ir,c0(γ)) = 0.

The first limit comes from the fact that as the center of the inverting circle approaches

γ, the image of γ under inversion has points sent further away from p, since r0 is

fixed. In the second limit, c0 is fixed. Observe then that for all choices of r ∈ R+, the

images Ir,c0(γ) will all be scale copies of each other. Additionally, since c0 /∈ γ, we can

choose r sufficiently small so that no point of γ is in Bnr (c0). Thus, Ir,c0(γ) ⊆ Bn

r (c0).

Combining this with the statement about scale copies gives us the second limit above.

Given the c0, r0 we have in hand at this point, we have three cases.

Case 1: length(Ir0,c0(γ)) = length(γ).

Since Conditions (i) and (ii) are satisfied, we need not change our choice of c0

and r0.

Case 2: length(Ir0,c0(γ)) > length(γ).

We can continuously decrease length(Ir,c0(γ)) by continuously decreasing r.

(Here the continuity comes from our above statement about scale copies.) Hence, we

simply decrease r until Condition (ii) is satisfied. Note that any decrease in r will

always preserve Sn−1r (c0) ⊆ Bn

ε (p), which ensures that Condition (i) remains satisfied.

17

Case 3: length(Ir0,c0(γ)) < length(γ).

We can increase length(Ir0,c(γ)) indefinitely by moving c closer to p. Our goal

here is to choose c sufficiently close to p to obtain length(Ir,c(γ)) > length(γ), which

reduces this case to case 2. Note that any movement of c closer to p will always

preserve Sn−1r (c) ⊆ Bn

ε (p), which ensures that Condition (i) remains satisfied. As γ

is a regular curve, we know that γ has measure zero in Rn. Hence, γ cannot be dense

in any neighborhood of p, so we may always choose some c /∈ γ sufficiently close to p.

Now set T (x) = Ir,c(x), and observe the following:

• length(T (γ)) = length(γ) from Condition (ii).

• E(T (γ)) = E(γ), since T is a Mobius transformation, and T (γ) ⊆ Rn.

• The maximum absolute pointwise curvature of T (γ) is at least K, as the oscu-

lating circle at T (q) has radius δ satisfying 1δ> K.

So then T is indeed the desired map, completing the proof.

The result proved in Proposition 2.3 can be generalized for the nonuniform

energy, as will be shown in Corollary 4.5.

We now work toward the goal of obtaining a formula for the pointwise cur-

vature of the image of a curve under inversion, which is Proposition 2.5. First, we

provide a proof of a basic result.

Lemma 2.4. Let γ(t) be a regular curve not necessarily parametrized by arclength,

and let ν(t) = |γ′(t)| > 0. Then

κabsγ (t) =1

ν3(t)|γ′′(t)ν(t)− γ′(t)ν ′(t)| .

18

Figure 2.2: An illustration of Proposition 2.3 with an ellipse. As the inversion center

approaches the ellipse, we shrink the inversion radius to preserve length—the inverted

curve will still have a point whose curvature is arbitrarily large.

Proof. The arclength parameter s is traditionally defined as

s(t) =

∫ t

0

|γ′(u)| du.

Then it follows that dsdt

(t) = |γ′(t)|, so we have that dtds

(s) = 1|γ′(s)| = 1

ν(s).

Recall then that if s is the arclength parameter then

κabsγ (s) =

∣∣∣∣dTds (s)

∣∣∣∣ =

∣∣∣∣dTdt (t(s)) · dtds

(s)

∣∣∣∣ =1

ν(s)

∣∣∣∣dTdt (t(s))

∣∣∣∣ =⇒ κabsγ (t) =

1

ν(t)|T ′(t)|,

where T (t) = γ′(t)ν(t)

(the unit tangent). Observe also that

γ′(t) = ν(t)T (t) =⇒ γ′′(t) = ν ′(t)T (t)+ν(t)T ′(t) =⇒ T ′(t) =1

ν(t)(γ′′(t)−ν ′(t)T (t)).

We can then substitute this identity for T ′(t) into our above equation for the curva-

ture, which gives

κabsγ (t) =

1

ν(t)

∣∣∣∣ 1

ν(t)(γ′′(t)− ν ′(t)T (t))

∣∣∣∣ =1

ν3(t)|γ′′(t)ν(t)− ν ′(t)γ′(t)|,

19

as claimed.

With Lemma 2.4 in hand, we can proceed to the calculation of formulas for

the pointwise curvature of the image of curve under inversion. These formulas are

immediately useful in Chapter 3, as integrating them along the curve will provide a

formula for total curvature.

Proposition 2.5. Let γ be a C2 curve in Rn parametrized by arclength, and fix

c ∈ Rn and r ∈ R+. Then the pointwise curvature of γ = Ir,c(γ) is given by any of the

following formulas. (Note that we frequently omit input variables to avoid unnecessary

complexity.)

(i) Let n = 2 and write γ(s) = (x(s), y(s)) and c = (c1, c2). Then

κsgnγ (s) =1

r2

[−(x′y′′− x′′y′)((x− c1)2+ (y − c2)2) + 2((x− c1)y′− (y − c2)x′)

].

(ii) Let n = 3 and write γ(s) = (x(s), y(s), z(s)), c = (c1, c2, c3), and ∆ = (x −

c1)2 + (y − c2)2 + (z − c3)2. Then

κabsγ (s) =1

r2

[[(x′y′′ − x′′y′)∆− 2((x− c1)y′ − (y − c2)x′)]

2

+ [(x′z′′ − x′′z′)∆− 2((x− c1)z′ − (z − c3)x′)]2

+ [(y′z′′ − y′′z′)∆− 2((y − c2)z′ − (z − c3)y′)]2] 1

2.

(iii) Write γ(s) = (x1(s), . . . , xn(s)), c = (c1, . . . , cn), and ∆ =∑n

i=1(xi−ci)2. Then

κabsγ (s) =1

r2∆

∣∣γ′′∆2 − γ′∆∆′ − (γ − c)∆∆′′ + (γ − c)(∆′)2∣∣ .

Proof. We shall prove each part individually.

20

Proof of Part (i): Define γ(t) = Ir,c(γ(t)), and write

γ(t)=(x(t), y(t))=

(r2(x(t)− c1)

(x(t)−c1)2 + (y(t)−c2)2+ c1,

r2(y(t)− c2)

(x(t)−c1)2 + (y(t)−c2)2+ c2

).

Then recall that the signed curvature of any curve γ(t) = (x(t), y(t)) in R2 will be

given as

κsgnγ (t) =

x′(t)y′′(t)− x′′(t)y′(t)[x′(t))2 + (y′(t))2]3/2

.

In the following calculation we make two notational omissions to preserve the read-

ability of our work:

• Omitting input variables.

• Omitting c = (c1, c2).

We will return c to our expression once we are closer to the final result. With this in

mind, consider the following:

x = r2x[x2 + y2]−1

x′ = r2[x′[x2 + y2]−1 − 2x[(xx′ + yy′)(x2 + y2)−2]

]x′′ = r2

[x′′(x2 + y2)−1 − 4x′[(xx′ + yy′)(x2 + y2)−2]

+ x[−2(1 + xx′′ + yy′′)(x2 + y2)−2 + 8(xx′ + yy′)2(x2 + y2)−3]]

Of course, y and its derivatives are obtained similarly.

Carrying out a considerable amount of algebra then tells us that the numerator

of κsgnγ will be given as

r4[−(x′y′′ − x′′y′)(x2 + y2)−2 + 2(xy′ − yx′)(x2 + y2)−3

].

With a shorter calculation, we can obtain that the denominator will be r6(x2 +y2)−3.

During these calculations, it is important to remember that γ is parametrized by

21

arclength, as we end up using that 1 = (x′)2 + (y′)2. So then it follows that

κsgnγ (t) =

1

r2

[−(x′y′′ − x′′y′)(x2 + y2) + 2(xy′ − yx′)

].

Rewriting c in its proper places then gives the formula in Part (i).

Proof of Part (ii): This will be carried out similarly to the proof of Part (i), and we

will again omit input variables and c until the end of the proof. Define ∆ = x2+y2+z2

and γ = Ir,c(γ), so then

γ = (x, y, z) =

(r2x

∆,r2y

∆,r2z

∆

).

Recall that we can define the curvature of γ in terms of the cross product. That is,

κabsγ (t) =

|γ′ × γ′′||γ′|3

The cross product in question can be calculated to be

|γ′ × γ′′| =√

(y′z′′ − y′′z′)2 + (x′z′′ − x′′z′)2 + (x′y′′ − x′′y′)2 (2.1)

We then make the following calculation:

x = r2x∆−1

x′ = r2(x′∆−1 − x∆−2∆′

)x′′ = r2

(x′′∆−1 − 2x′∆−2∆′ + 2x∆−3(∆′)2 − x∆−2∆′′

)Of course, these formulas are similar for y and z. We can then expand Equation (2.1)

22

as

= r4[(

∆−2(x′y′′ − x′′y′) + ∆−3∆′′(xy′ − x′y) + ∆−3∆′(x′′y − xy′′))2

+(∆−2(x′z′′ − x′′z′) + ∆−3∆′′(xz′ − x′z) + ∆−3∆′(x′′z − xz′′)

)2

+(∆−2(y′z′′ − y′′z′) + ∆−3∆′′(yz′ − y′z) + ∆−3∆′(y′′z − yz′′)

)2] 1

2.

(2.2)

Next, we calculate the derivatives of ∆, remembering that (x′)2 + (y′)2 + (z′)2 = 1

because γ is parametrized by arclength.

∆ = x2 + y2 + z2

∆′ = 2xx′ + 2yy′ + 2zz′

∆′′ = 2(1 + xx′′ + yy′′ + zz′′)

Before we proceed further with the numerator, we observe that the denominator can

be calculated as follows:

|γ′|3 =[√

(x′)2 + (y′)2 + (z′)2]3

= r6[(x′∆−1 − x∆−2∆′)2 + (y′∆−1 − y∆−2∆′)2 + (z′∆−1 − z∆−2∆′)2

] 32

= r6[∆−2

((x′)2 + (y′)2 + (z′)2︸︷︷︸

=1

)−∆−3∆′ (2xx′ + 2yy′ + 2zz′)︸︷︷︸

=∆′

+∆−4(∆′)2 (x2 + y2 + z2)︸︷︷︸=∆

] 32

= r6∆−3

We can then obtain a simplified expression for κabsγ by dividing Equation (2.2) by this

23

expression for the denominator:

|γ′ × γ′′||γ′|3

=1

r2

[(∆(x′y′′ − x′′y′) + ∆′′(xy′ − x′y) + ∆′(x′′y − xy′′))2

+ (∆(x′z′′ − x′′z′) + ∆′′(xz′ − x′z) + ∆′(x′′z − xz′′))2

+ (∆(y′z′′ − y′′z′) + ∆′′(yz′ − y′z) + ∆′(y′′z − yz′′))2] 1

2.

Expanding this expression makes a tremendous mess, but wading through all the

related algebra that we have chosen to omit tells us that κabsγ will be given as

=1

r2

[[(x′y′′ − x′′y′)∆− 2(xy′ − yx′)]2

+ [(x′z′′ − x′′z′)∆− 2(xz′ − zx′)]2

+ [(y′z′′ − y′′z′)∆− 2(yz′ − zy′)]2] 1

2.

Rewriting c in its proper place then gives the formula in Part (ii).

Proof of Part (iii): This will bear some similarities to the proofs of Parts (i) and (ii),

and we will again omit input variables and c for the duration of the proof. Let

∆ = |γ − c|2 and γ = Ir,c(γ). We can then write

γ =r2

∆γ

γ′ =r2

∆2(γ′∆− γ∆′)

γ′′ =r2

∆3

(γ′′∆2 − γ∆∆′′ − 2γ′∆∆′ + 2γ(∆′)2

)Define ν(t) = |γ′(t)|. Then we can observe that

ν2 = γ′ • γ′ =r4

∆4(γ′∆− γ∆′) • (γ′∆− γ∆′)

=r4

∆4(γ′ • γ′︸︷︷︸

=1

∆2 − 2γ′ • γ︸︷︷︸=∆′

∆∆′ + γ • γ︸︷︷︸=∆

(∆′)2) =r4

∆2.

24

Hence, we may conclude that

ν =r2

∆, which implies ν ′ =

−r2∆′

∆2.

Recall from Lemma 2.4 that

κabsγ (t) =

1

ν3|νγ′′ − ν ′γ′| .

Substitution of the above work then gives that κabsγ equals

∆3

r6

∣∣∣∣r2

∆

(r2

∆3

(γ′′∆2 − γ∆∆′′ − 2γ′∆∆′ + 2γ(∆′)2

))+r2∆′

∆2

(r2

∆2(γ′∆− γ∆′)

)∣∣∣∣ .As this quickly simplifies to Part (iii), the proof is complete.

25

CHAPTER 3TOTAL ABSOLUTE CURVATURE AND CONFORMAL

TRANSFORMATIONS

The nature of the interaction between the total absolute curvature and Mobius

energy is of great interest to us. After initial investigations, it became clear that to

study this question we would need an understanding of how Mobius transformations

affect the total absolute curvature of a curve. As affine similarities have no effect on

total curvature (see Proposition 3.4), we devote nearly all of our focus to inversions.

In the following results, we briefly examine this relationship for convex polyg-

onal curves in R2 (Section 3.1) before moving to the total curvatures of closed C2

curves in Rn in Section 3.2.

3.1 Total Curvature of Polygonal Curves

As inversions send lines to either circles or lines, polygonal curves have rela-

tively predictable behavior under inversions. Hence, we use convex polygonal curves

as a starting point for our investigation of the effect of inversions on the total absolute

curvature. Before we begin, we must prove a lemma about how inversions affect line

segments.

Lemma 3.1. Let λ : [0, `]→ Rn be a line segment and c ∈ Rn and r ∈ R+.

(i) If c /∈ λ, then κabstot (Ir,c(λ)) = 2∠(λ(0)− c, λ(`)− c).

(ii) If c ∈ λ, then κabstot (Ir,c(λ)) = 0.

Proof. We shall prove Part (i) first. Let α = ∠(λ(0) − c, λ(`) − c). Since c /∈ λ,

26

we know that Ir,c(λ) will be a subarc of a circle through c. Let β equal the angle

represented by this subarc, and recognize that then β = κabstot (Ir,c(λ)). Note that Ir,c(λ)

will be a circle of infinite radius (that is, a line) in the case that α = 0; we will handle

this case separately. The typical case gives rise to Figure 3.1.

Figure 3.1: The inversion of λ to Ir,c(λ).

An elementary result from Euclidean geometry tells us that β = 2α in the

α 6= 0 case, which implies that κabstot (Ir,c(λ)) = 2∠(λ(0)− c, λ(`)− c), as desired.

Now recall that the image under inversion of any line through an inversion’s

center will simply be the same line. This takes care of both the special case α = 0

and Part (ii), as both cases are subsegments of a line through the inversion center.

In both of these circumstances, since the image is a line, κabstot (Ir,c(λ)) = 0. (For the

27

case when α = 0, this completes the proof because κabstot (Ir,c(λ)) = 0 = 2α.)

With Lemma 3.1 in hand, we can now move to our main result. Notice that

its five different parts correspond to different regions of R2; a similar approach will

be used when working with C2 curves in Section 3.2.

Theorem 3.2. Let p : [0, `]→ R2 be a closed, convex, polygonal curve of n segments.

• Let λ1, . . . , λn be the segments of p.

• Let p(ti) be the initial point of λi.

• Let βi be the exterior angle at the joint p(ti).

Then for r ∈ R+, we have the following:

(i) If c ∈ R2 is in the interior of the curve, then κabstot (Ir,c(p)) = 6π.

(ii) If c ∈ R2 is such that c ∈ p(t) but t 6= ti for any i, then κabstot (Ir,c(p)) = 4π.

(iii) If c ∈ R2 is such that c = p(ti) for some i, then κabstot (Ir,c) = 4π − 3βi.

(iv) If c ∈ R2 is in the exterior of the curve, then κabstot (Ir,c(p)) < 6π.

(v) lim|c|→∞ κabstot (Ir,c(p)) = 2π, where c /∈ p.

Proof. We shall prove each part individually.

Proof of Part (i): Consider Figure 3.2. Notice that

n∑i=1

βi = 2π andn∑i=1

αi = 2π,

where αi = ∠ (p(ti)− c, p(ti+1)− c).

Let βi represent the turning angle at the joint Ir,c(p(ti)). Since inversion is

28

Figure 3.2: The inversion of p to Ir,c(p), where c is in the interior of p

conformal, we then know that βi = βi for all i, so

n∑i=1

βi = 2π.

It follows from Lemma 3.1 that each Ir,c(λi) will contribute 2αi to κabstot (Ir,c(p)), so we

know that the total contribution of all the Ir,c(λi) will be

n∑i=1

κabstot (Ir,c(λi)) =

n∑i=1

2αi = 4π.

Hence, we have that κabstot (Ir,c(p)) = 2π + 4π = 6π. �

Proof of Part (ii): Without loss of generality, say that c ∈ λ1 (but not an endpoint),

and consider Figure 3.3. Notice that

n∑i=1


αi = π.

29

Figure 3.3: The inversion of p to Ir,c(p), where c is on p, but not a corner.



n∑i=1

βi = 2π.

Lemma 3.1 tells us that each Ir,c(λi) for i 6= 1 will contribute 2αi to κabstot (Ir,c(p)), so

we know that the total contribution of all the Ir,c(λi) for i 6= 1 will be

n∑i=2


n∑i=2

2αi = 2π.

Since λ1 is a line through the inversion center, its image will be a line, so they

contribute nothing to the total curvature (again, see Lemma 3.1).

Hence, we have that κabstot (Ir,c(p)) = 2π + 2π = 4π. �

Proof of Part (iii): Without loss of generality say that c = p(t1), and consider

30

Figure 3.4. Notice that

n∑i=2

βi = 2π − β1 andn−1∑i=2

αi = π − β1.

Figure 3.4: The inversion of p to Ir,c(p), where c is on a corner of p.

Let βi represent the turning angle at the joint Ir,c(p(ti)) for i 6= 1. (The angle at

p(t1) “disappears,” since it will be an angle at infinity.) Since inversion is conformal,

we then know that βi = βi for i 6= 1, so

n∑i=2

βi = 2π − β1.

It follows from Lemma 3.1 that each Ir,c(λi) for i 6= 1, n will contribute 2αi to

31

κabstot (Ir,c(p)), so we know that the total contribution of all the Ir,c(λi) for i 6= 1 will be

n−1∑i=2


n∑i=2

2αi = 2π − 2β1.

Since λ1 and λn are lines through the inversion center, their images will be lines, so

they contribute nothing to the total curvature (again, see Lemma 3.1).

Hence, we have that κabstot (Ir,c(p)) = 2π − β1 + 2π − 2β1 = 4π − 3β1. �

Proof of Part (iv): Consider Figure 3.5. Notice that

n∑i=1


αi < 2π.

Figure 3.5: The inversion of p to Ir,c(p), where c is in the exterior of p.

32


n∑i=1

βi = 2π.


know that the total contribution of all the Ir,c(λi) will satisfy

n∑i=1


n∑i=1

2αi < 4π.

Hence, we have that κabstot (Ir,c(p)) < 2π + 4π = 6π. �

Proof of Part (v): We will refer to Figure 3.5. Notice that as |c| → ∞, αi → 0 for

all i. This means that for every ε > 0 there exists some R ∈ R+ such that αi <ε

2n

for all c satisfying |c| > R. As a result, we get

n∑i=1

αi <n∑i=1

ε

2n=ε

2.

Let βi represent the turning angle at the joint Ir,c(p(ti)). Since inversion is conformal,

we then know that βi = βi for all i, so

n∑i=1

βi = 2π.


know that the total contribution of all the Ir,c(λi) will satisfy

n∑i=1


n∑i=1

2αi < ε.

Hence, we have for any ε > 0 that there exists R ∈ R+ such that κabstot (Ir,c(p)) < 2π+ ε

when |c| > R, so we may conclude that lim|c|→∞ κabstot (Ir,c(p)) = 2π. �

This completes the proof.

33

3.2 Total Curvature of C2 Curves

Our investigation of total curvature is considerably more in-depth than our

work with the pointwise curvature, and will be separated into three subsections. In

Subsection 3.2.1, we will derive formulas for the total curvature of an inverted curve

in R2, R3, and Rn and examine the immediate consequences of these formulas. In

Subsection 3.2.2, we will use analysis to determine the behavior of the total curvature

when the inversion center is near or on the curve. Our ultimate goal is Theorem 3.33.

In Subsection 3.2.3 we generalize some of our more important results to the case in

which our curve is only piecewise C2 and not necessarily closed.



Φ(r, c) =

κabstot (Ir,c(γ)) + 2π c ∈ γ([0, `])


.


3.2.1 Curvature Formulas and Consequences

In this section, we begin by deriving formulas for the total curvature of an

inverted curve (Corollary 3.5). We use these formulas to show that total curvature

is invariant with respect to inversion radius (Corollary 3.6) and describe its behavior

as the inversion center goes to infinity (Proposition 3.7).

We begin by providing a formula for the total absolute curvature of a curve

with an arbitrary parametrization.

34

Lemma 3.3. Let γ : [0, `]→ Rn be a regular C2 curve with an arbitrary parametriza-

tion in t. Then

κabstot (γ) =

∫ `

0

|κabsγ (t)||γ′(t)| dt,

and κabstot (γ) is independent of the regular parametrization chosen.

Proof. Define

s(t) =

∫ t

0

|γ′(u)| du and L = s(`),

and notice then that

ds

dt= |γ′(t)|, which implies ds = |γ′(t)|dt.

Let γ be the reparametrization of γ by arclength, that is γ(s) = γ(t(s)). Then we

know that κabsγ (s) = κabs

γ (t(s)) (cf. Lemma 2.4), and

κabstot (γ) =

∫ L

0

|κabsγ (s)| ds =

∫ `

0

|κabsγ (s(t))||γ′(t)| dt =

∫ `

0

|κabsγ (t)||γ′(t)| dt = κabs

tot (γ).


As mentioned in the introduction to this chapter, the behavior of the total

absolute curvature with respect to affine similarities is trivial. We prove this in

Proposition 3.4.

Proposition 3.4. Say that γ : [0, `] → Rn is a regular C2 curve with an arbitrary

parametrization in t. Then

κabstot (A(γ)) = κabstot (γ)

for any affine similarity A of Rn.

35

Proof. Recall that the affine similarities of Rn are transformations that can be decom-

posed into isometries and dilations of Rn. As we know that isometries will not change

the curvature of γ, it follows that isometries will not change the total curvature of γ.

Hence, it remains to show that the total curvature is invariant under dilations.

Remember that any dilation β of γ can be written as β = a · γ, where a ∈ R+.

Define νγ(t) = |γ′(t)| and νβ(t) = |β′(t)|. Notice that

β′(t) = aγ′(t), β′′(t) = aγ′′(t), νβ(t) = aνγ(t), and ν ′β(t) = aν ′γ(t).

Then substitution into the curvature formula from Lemma 2.4 yields the following:

κabsβ (t) =

1

ν3β(t)|β′′(t)νβ(t)− β′(t)ν ′β(t)|

=1

a3ν3γ(t)|a2γ′′(t)νγ(t)− a2γ′(t)ν ′γ(t)|

=1

a

(1

ν3γ(t)|γ′′(t)νγ(t)− γ′(t)ν ′γ(t)|

)=

1

aκabsγ (t)

We can then substitute into the result from Lemma 3.3 to acquire

κabstot (β) =

∫ `

0

|κabsβ (t)||β′(t)| dt

=

∫ `

0

∣∣∣∣1aκabsγ (t)

∣∣∣∣ |aγ′(t)| dt=

∫ `

0

|κabsγ (t)||γ′(t)| dt = κabs

tot (γ).

As a result, we know that total curvature is invariant under dilations. Thus, total

curvature is invariant under affine similarities, as desired.

As promised, we now provide formulas for the total curvature of the image of

a curve under inversion.

36

Corollary 3.5. Let γ : [0, `] → Rn be a regular C2 curve parametrized by arclength,

and fix c ∈ Rn and r ∈ R+. Then the total absolute curvature of γ = Ir,c(γ) is given

by any of the following formulas. (Note that we frequently omit input variables to

avoid unnecessary complexity.)

(i) Let n = 2 and write γ(s) = (x(s), y(s)) and c = (c1, c2). Then

κabstot (γ) =

∫ `

0

∣∣∣∣(x′y′′ − x′′y′)− 2(x− c1)y′ − (y − c2)x′

(x− c1)2 + (y − c2)2

∣∣∣∣ ds.(ii) Let n = 3 and write γ(s) = (x(s), y(s), z(s)), c = (c1, c2, c3), and ∆ = (x −

c1)2 + (y − c2)2 + (z − c3)2. Then

κabstot (γ) =

∫ `

0

[[(x′y′′ − x′′y′)− 2

(x− c1)y′ − (y − c2)x′

∆

]2

+

[(x′z′′ − x′′z′)− 2

(x− c1)z′ − (z − c3)x′

∆

]2

+

[(y′z′′ − y′′z′)− 2

(y − c2)z′ − (z − c3)y′

∆

]2] 1

2

ds.

(iii) Write γ(s) = (x1(s), . . . , xn(s)), c = (c1, . . . , cn), and ∆ =∑n

i=1(xi−ci)2. Then

κabstot (γ) =

∫ `

0

1

∆2

∣∣γ′′∆2 − γ′∆∆′ − (γ − c)∆∆′′ + (γ − c)(∆′)2∣∣ ds.

Proof. Notice that none of the formulas from Proposition 2.5 are given with respect

to arclength. This is not a problem, as we can simply apply Lemma 3.3, multiplying

each formula by |γ′(t)| and integrating.

Fortunately, we can use the Rn case to prove Parts (i) to (iii) simultaneously.

Recall from the proof of Part (iii) of Proposition 2.5 that

|γ′|2 = γ′ • γ′ =r4

∆2.

37

Hence, we may conclude that |γ′| = r2

∆. Multiplying each of the formulas in Propo-

sition 2.5 by this and integrating then gives the desired result. Note: We write the

expanded form of ∆ in Part (i); that is, ∆ = (x− c1)2 + (y − c2)2.

Although short, Corollary 3.6 is an extremely useful result. Specifically, it tells

us that the total absolute curvature of an inverted curve is not affected by the radius

of the inversion sphere. It is important to note that Corollary 3.6 refers only to the

total absolute curvature of γ, not its pointwise curvature.

Corollary 3.6. Let γ be a regular C2 curve in Rn. Then for a fixed c we have that

κabstot (Ir,c(γ)) = κabstot (I1,c(γ))

for all r ∈ R+.

Proof. Notice that Part (iii) of Corollary 3.5 has no dependence on r.

It follows from Corollary 3.6 that we need only focus on inversion centers, and

Proposition 3.7 is the first region of inversion centers we address. Specifically, this

result characterizes the effect of inversions with centers in a neighborhood of infinity.

Proposition 3.7. Let γ : [0, `]→ Rn be a regular C2 curve parametrized by arclength.

Then

lim|c|→∞

κabstot (Ir,c(γ)) = κabstot (γ)

for any choice of r.

Proof. Let ε > 0. Since γ has a compact domain, there exists some R ∈ R+ such that

γ ⊆ BnR(0). Set M = 8`

ε+R and let |c| > M . Using the result from Corollary 3.5, we

38

may now observe the following:

|κabstot (γ)− κabs

tot (γ)|

=

∣∣∣∣∫ `

0

1

∆2

∣∣γ′′∆2 − γ′∆∆′ − (γ − c)∆∆′′ + (γ − c)(∆′)2∣∣ ds− ∫ `

0

|γ′′| ds∣∣∣∣

≤∫ `

0

∣∣∣∣∣∣∣∣∣∣γ′′ − γ′∆′∆

− (γ − c)∆′′

∆+ (γ − c)

(∆′

∆

)2∣∣∣∣∣− |γ′′|

∣∣∣∣∣ dsRecall that ∆ = |γ − c|2. Then we can write

∆ = (γ − c) • (γ − c), ∆′ = 2γ′ • (γ − c), and ∆′′ = 2 + 2γ′′ • (γ − c).

So then we can continue as follows:


tot (γ)|

≤∫ `

0

∣∣∣∣∣∣∣∣∣γ′′ − γ′2γ′ • (γ − c)

|γ − c|2− (γ − c)2 + 2γ′′ • (γ − c)

|γ − c|2+ (γ − c)(2γ′ • (γ − c))2

|γ − c|4

∣∣∣∣− |γ′′|

∣∣∣∣∣ ds=

∫ `

0

∣∣∣∣∣∣∣∣∣∣ 1

|γ − c|

[−2

(γ − c)|γ − c|

− 2γ′(γ′ •

γ − c|γ − c|

)+ 4

γ − c|γ − c|

(γ′ •

γ − c|γ − c|

)2]

+

[γ′′ − 2

γ′′ • (γ − c)(γ − c) • (γ − c)

(γ − c)] ∣∣∣∣∣− |γ′′|

∣∣∣∣∣ dsRecall that the reflection of a vector v in a hyperplane through the origin orthogonal

to a vector a is given by the formula

Refa(v) = v − 2v • a

a • aa.

Hence, we have that

γ′′ − 2γ′′ • (γ − c)

(γ − c) • (γ − c)(γ − c) = Refγ−c(γ

′′).

39

In addition, note that |γ′′| = |Refγ−c(γ′′)|. So we continue with the following:


tot (γ)|

≤∫ `

0

∣∣∣∣∣∣∣∣∣∣ 1

|γ − c|

[−2

(γ − c)|γ − c|

− 2γ′(γ′ •

γ − c|γ − c|

)+ 4

γ − c|γ − c|

(γ′ •

γ − c|γ − c|

)2]

+ [Refγ−c(γ′′)]

∣∣∣∣∣− |Refγ−c(γ′′)|

∣∣∣∣∣ ds≤∫ `

0

1

|γ − c|

∣∣∣∣∣−2(γ − c)|γ − c|

− 2γ′(γ′ •

γ − c|γ − c|

)+ 4

γ − c|γ − c|

(γ′ •

γ − c|γ − c|

)2∣∣∣∣∣

+ |Refγ−c(γ′′)− Refγ−c(γ

′′)| ds

≤∫ `

0

1

|γ − c|

[2

∣∣∣∣(γ − c)|γ − c|

∣∣∣∣+ 2|γ′||γ′|∣∣∣∣ γ − c|γ − c|

∣∣∣∣+ 4

∣∣∣∣ γ − c|γ − c|

∣∣∣∣ |γ′|2 ∣∣∣∣ γ − c|γ − c|

∣∣∣∣2]

+ |0| ds

=

∫ `

0

8

|γ − c|ds

Since |c| > M , we can then observe that

∫ `

0

8

|γ − c|ds ≤

∫ `

0

8

|c| − |γ|ds ≤

∫ `

0

8

|c| −Rds =

8`

|c| −R<

8`

M −R= ε.

Hence, we have that lim|c|→∞ κabstot (γ) = κabs

tot (γ), as desired.

Remark. Proposition 3.7 can be seen immediately for the R2 and R3 cases by simply

inspecting the formulas for κabstot given in Parts (i) and (ii) of Corollary 3.5.

In effect, Proposition 3.7 is saying that inversion with respect to a distant

center is nearly a reflection. Recall from Corollary 3.6 that the total curvature is

invariant with respect to the inversion radius. This means that we can always adjust

the radius so that the inversion circle will intersect our curve in some point. If the

center is far away, the curvature of the inversion circle will very small—this means

40

that arc of the circle intersecting our curve looks nearly linear, which gives us an

“almost reflection.” An example of this idea is shown in Figure 3.6.

Figure 3.6: Inversion through a distant center is approximately a reflection.

3.2.2 Analysis of Inversions of Closed C2 Curves

We will now work toward obtaining an understanding of the behavior of the

total absolute curvature of Ir,c(γ) when c is close to or on γ and γ is closed and C2.

This dichotomy for c allows us to easily separate our work into two main components.

We will first address the case c ∈ γ (Subsubsection 3.2.2.1), then move to c /∈ γ

41

(Subsubsection 3.2.2.2). Somewhat surprisingly, the former case is the more easily

handled one.

First, we take a moment to clarify some terminology.

Definition 3.1. Let γ be a closed curve in Rn and say that c ∈ γ. Notice that for this

choice c of inversion center, Ir,c(γ) is no longer be a closed curve in Rn. Specifically,

the image of the point c ∈ γ is the point at infinity. We call such curve a broken

inverted curve.

Proposition 3.8 is a well-known result (for a recent exposition, see [5] and [2]),

which we have reframed here in the context of our particular problem. If c ∈ Rn, let

dist(c, γ) = infs γ(s).

Proposition 3.8. Let γ : [0, `]→ Rn be a closed C2 curve parametrized by arclength.

Then there exists some d > 0 such that for any c ∈ Rn satisfying 0 < dist(c, γ) < d

we have the following:

(i) There exists unique t ∈ [0, `] such that |γ(t)− c| = dist(c, γ).

(ii) There exists unique ~ωt ∈ Tγ(t)Rn such that |~ωt| = 1, ~ωt ⊥ γ′(t), and ~ωt is in the

same direction as c− γ(t).

(iii) c = γ(t) + dist(c, γ) • ~ωt.

This proposition immediately allows us to define an important quantity.

Definition 3.2. The value r(γ) = sup{d ∈ R+ | d satisfies Proposition 3.8} is called

the normal injectivity radius of γ, or thickness of γ.

42

With a foundation laid by Proposition 3.8, we can now state the ultimate goal

of this subsection: Proposition 3.28. This proposition characterizes the behavior of

the total curvature for inversions close to a curve.

Proposition 3.28. Let γ : [0, `] → Rn be a regular closed C2 curve parametrized by

arclength. For all ε > 0 there exists d > 0 for all t ∈ [0, `] and ~ωt satisfying |~ωt| = 1

and ~ωt ⊥ γ′(t) such that

∣∣κabstot (Ir,λt(0)(γ)) + 2π − κabstot (Ir,λt(u)(γ))∣∣ < ε,

whenever 0 < u < d, where λt(u) = γ(t) + u~ωt.

In particular, Proposition 3.28 can be thought of as saying “the total absolute

curvature of the inverted curve is approximately 2π plus the curvature of the broken

inverted curve given by inversion centered at the point on the curve closest to the

inversion center.” It is helpful to note that we need not worry about when u is

negative—indeed, for every ~ωt satisfying |~ωt| = 1 and ~ωt ⊥ γ′(t), we know that −~ωt

also satisfies | − ~ωt| = 1 and −~ωt ⊥ γ′(t).

3.2.2.1 Inversion Centers on the Curve

To begin our work for c ∈ γ, it is helpful to have a view of the main goal of

this subsubsection: Proposition 3.14.


arclength. Then the function F : [0, `]→ R defined by

F (t) = κabstot (I1,γ(t)(γ))

43

is uniformly continuous.

Notice that the curve I1,γ(t)(γ) is a broken inverted curve, since the inversion

center is necessarily a point on the curve. Hence, Proposition 3.14 also tells us that

the set of total absolute curvatures of every broken inverted curve is a bounded set.

Indeed, this is formally stated as Corollary 3.15.

Lemma 3.9 is a result from Manfredo do Carmo’s Riemannian Geometry [3].

As the proof in this text omits a few important points, we present a more complete

version.

Lemma 3.9. Let h : (−δ, δ)× U → Rn be a Ck function that satisfies h(0, t) = 0 for

all t ∈ U . Then there exists some Ck−1 function g : (−δ, δ)× U → Rn satisfying the

following:

(i) h(s, t) = sg(s, t), and

(ii) g(0, t) = ∂h∂s

(s, t)

∣∣∣∣s=0

.

Proof. Define `(s, t) = ∂h∂s

(s, t) and then define

g(s, t) =

∫ 1

0

`(su, t) du.

We immediately know that ` is Ck−1, so then g is Ck−1. It remains to show Equali-

ties (i) and (ii).

We show Equality (i) first. When s = 0, it follows from the given information

that

h(0, t) = 0 = 0 · g(0, t),

44

since g(0, t) is finite. Hence, we have h(s, t) = sg(s, t) when s = 0. To see that this

equality holds when s 6= 0, observe the following:

sg(s, t) = s

∫ 1

0

`(su, t) du Definition of g

=

∫ 1

0

`(su, t) d(su) s is constant w.r.t. the integration

=

∫ s

0

`(v, t) dv Change of variables with su = v

=

∫ s

0

∂h

∂s(s, t)

∣∣∣∣s=v

dv Definition of `

= h(s, t)− h(0, t) Evaluate

= h(s, t)− 0 = h(s, t) Given h(0, t) = 0 for all t

Hence, we also have that h(s, t) = sg(s, t) when s 6= 0, so we know h(s, t) = sg(s, t)

for all s.

To see Equality (ii), set s = 0 and observe the following:

g(0, t) =

∫ 1

0

`(0 · u, t) du Definition of g

=

∫ 1

0

`(0, t) du Simplify

= `(0, t)

∫ 1

0

1 du `(0, t) is constant w.r.t. u

=∂h

∂s(0, t) Definition of `

Thus, we have that g(0, t) = ∂h∂s

(0, t) when s = 0.

Remark. Note that Lemma 3.9 defines g continuously as

g(s, t) =

h(s,t)s

s 6= 0

∂h∂s

(0, t) s = 0

.

45

We now use Lemma 3.9 to construct a lemma critical to the main result of

this subsubsection.

Lemma 3.10. Let γ : [0, `] → Rn be a regular closed C2 curve and define L :

[0, `]× [0, `]→ Rn by

L(s, t) = γ(s+ t)− γ(t)− γ′(t)s− γ′′(t)

2s2.

Then

L(s, t)

s2,

∂L∂s

(s, t)

s, and

∂2L

∂s2(s, t)

are all continuously extendable to [0, `]× [0, `].

Proof. Since γ is C2, we immediately know that L(s, t) is C2 on [0, `] × [0, `], so

∂2L∂s2

(s, t) is C0 on [0, `]× [0, `].

Now notice for all t that s = 0 gives

L(0, t) = γ(0 + t)− γ(t)− γ′(t) · 0− γ′′(t)

2· 02 = 0.

Hence, Lemma 3.9 tells us that there exists some G1(s, t) such that G1 is C1 on

[0, `]× [0, `] and L(s, t) = sG1(s, t). Now notice further for all t that s = 0 gives

G1(0, t) = γ′(0 + t)− γ′(t)− γ′′(t) · 0 = 0.

Hence, Lemma 3.9 tells us that there exists some G2(s, t) such that G2 is C0 on

[0, `]× [0, `] and G1(s, t) = sG2(s, t). As a result, we have for s 6= 0 that

L(s, t) = s2G2(s, t) =⇒ L(s, t)

s2= G2(s, t).

46

Since G2(s, t) is C0 on [0, `]× [0, `], we know that L(s,t)s2

is continuously extendable to

[0, `]× [0, `].

Finally, observe for all t that s = 0 gives

∂L

∂s(0, t) = γ′(0 + t) + γ′(t)− γ′′(t)0 = 0.

So then Lemma 3.9 tells us that there exists some G3(s, t) such that K is C0 on

[0, `]× [0, `] and ∂L∂s

(s, t) = sG3(s, t). As a result, we have for s 6= 0 that

∂L∂s

(s, t)

s= G3(s, t).

Since G3 = (s, t) is C0 on [0, `]× [0, `], we know∂L∂s

(s,t)

sis continuously extendable to

[0, `]× [0, `]. This completes the proof.

Lemma 3.11 allows us to rewrite rational expressions containing a function in

a useful way.

Lemma 3.11. Let g(s) be a continuous function such that 1 + sg(s) > 0 for all s in

an open interval containing 0. Then it follows that there exists a function g such that

g is continuous and satisfies

1

1 + sg(s)= 1 + sg(s).

Proof. Define g by

g(s) =−g(s)

1 + sg(s).

Since g is continuous and 1 + sg(s) > 0, it immediately follows that g is continuous.

In addition, observe the following:

1 + sg(s) = 1 + s

(−g(s)

1 + sg(s)

)=

1 + sg(s)

1 + sg(s)+−sg(s)

1 + sg(s)=

1

1 + sg(s).

47


With Lemmas 3.10 and 3.11 in hand, we are now prepared to show the conti-

nuity of the integrand of κabstot (Ir,γ(t)(γ)) on the square [0, `]× [0, `].

Lemma 3.12. Let γ : [0, `] → Rn be a regular closed C2 curve parametrized by

arclength and define ∆(s, t) = |γ(s)− γ(t)|2. Then

f(s, t) =

∣∣∣∣γ′′(s+ t)− γ′(s+ t)∆′(s+ t, t)

∆(s+ t, t)

−(γ(s+ t)− γ(t))∆′′(s+ t, t)

∆(s+ t, t)+ (γ(s+ t)− γ(t))

(∆′(s+ t, t)

∆(s+ t, t)

)2∣∣∣∣∣

is continuously extendable to [0, `]× [0, `], and it is bounded on that same set.

Proof. Using L as defined in Lemma 3.10, we can write

γ(s+ t) = γ(t) + γ′(t)s+γ′′(t)

2s2 + L(s, t).

Now fix t, and set B = γ′(t) and C = γ′′(t)2

. We abbreviate the s-derivatives of L as

L′(s, t) = ∂L∂s

(s, t) and L′′(s, t) = ∂2L∂s2

(s, t). Also omitting the input variables for L,

we set B = γ′(t) and C = γ′′(t)2

and write

γ(s+ t)− γ(t) = Bs+ Cs2 + L,

γ′(s+ t) = B + 2Cs+ L′, and

γ′′(s+ t) = 2C + L′′.

Let ∆ = |γ(s+ t)− γ(t)|2. Then we can also write out ∆ and its s-derivatives below,

which we abbreviate as ∆′ and ∆′′. As part of this process, we define W , U , and V

48

as shown. In addition, we omit the input variables on ∆, L, and their derivatives.

∆ = |γ(s+ t)− γ(t)|2

= (γ(s+ t)− γ(t)) • (γ(s+ t)− γ(t))

= (Bs+ Cs2 + L) • (Bs+ Cs2 + L)

= B •Bs2 + 2B • Cs3 + 2B • Ls+ C • Cs4 + 2C • Ls2 + L • L

Since γ is parametrized by arclength, it follows that B •B = 1 and B •C = 0. Applying

these facts allows us to continue with the following for s 6= 0:

∆ = s2 + 2B • Ls+ C • Cs4 + 2C • Ls2 + L • L

= s2 + s3

(2B •

L

s2+ C • Cs+ 2C •

L

s+L

s2•L

s

)

We know from Lemma 3.10 that Ls2

is continuously extendable to [0, `]× [0, `], so let

L be such a continuous extension. Note that this substitution also allows us to write

Ls

= Ls. Then we can continue as follows:

∆ = s2 + s3(

2B • L+ C • Cs+ 2C • Ls+ L • Ls)

︸︷︷︸W

= s2(1 + sW )

Note that all the terms of W are continuous on [0, `]× [0, `], so (1+sW ) is continuous

on [0, `] × [0, `], and we can continuously extend the above representation of ∆ to

[0, `]× [0, `]. Further, we know that ∆ ≥ 0 for all s, with ∆ = 0 exactly when s = 0.

This implies that 1 + sW > 0 for all s. These facts will later be used in conjunction

with Lemma 3.11.

49

We apply similar ideas to calculate ∆′, using L to represent the continuous

extension of L′

swe know exists from Lemma 3.10. Observe the following for s 6= 0:

∆′ = 2γ′(s+ t) • (γ(s+ t)− γ(t))

= 2(B + 2Cs+ L′) • (Bs+ Cs2 + L)

= 2B •Bs+ 6B • Cs2 + 2B • L+ 2B • L′s+ 4C • Cs3 + 4C • Ls+ 2C • L′s2

+ 2L • L′

= 2s+ 2B • L+ 2B • L′s+ 4C • Cs3 + 4C • Ls+ 2C • L′s2 + 2L • L′

= 2s+ 2s2

(B •

L

s2+B •

L′

s+ 2C • Cs+ 2C •

L

s+ C • L′ +

L

s2• L′)

= 2s+ 2s2(B • L+B • L+ 2C • Cs+ 2C • Ls+ C • L′ + L • L′

)︸︷︷︸

U

= 2s(1 + sU)

Note that all of the terms of U are continuous on [0, `] × [0, `]. Hence, (1 + sU) is

continuous on [0, `]× [0, `], and we can continuously extend the above representation

of ∆′ to [0, `]× [0, `].

The following will be similar to the above calculations. Recall that γ′ • γ′ = 1

because of our parametrization. Then for s 6= 0:

∆′′ = 2γ′′(s+ t) • (γ(s+ t)− γ(t)) + 2γ′(s+ t) • γ′(s+ t)

= 2(2C + L′′) • (Bs+ Cs2 + L) + 2

= 4B • Cs+ 4C • Cs2 + 4C • L+ 2B • L′′s+ 2C • L′′s2 + 2L • L′′ + 2

= 2 + 4C • Cs2 + 4C • L+ 2B • L′′s+ 2C • L′′s2 + 2L • L′′

50

= 2 + 2s

(2C • Cs+ 2C •

L

s+B • L′′ + C • L′′s+

L

s• L′′)

= 2 + 2s(

2C • Cs+ 2C • Ls+B • L′′ + C • L′′s+ Ls • L′′)

︸︷︷︸V

= 2(1 + sV )

Note that all of the terms of V are continuous on [0, `] × [0, `]; hence, (1 + sV ) is

continuous on [0, `]× [0, `], and we can continuously extend the above representation

of ∆′′ to [0, `]× [0, `].

We noted above that 1 + sW > 0 and W is continuous, both for all s. Thus,

Lemma 3.11 tells us that there exists some continuous function such that

1

1 + sW= 1 + sW .

Hence, we can write the following, with X, Y , and Z defined as shown:

∆′

∆=

2s(1 + sU)

s2(1 + sW )

=2

s(1 + sU)(1 + sW )

=2

s

(1 + s (U + W + sUW )︸︷︷︸

X

)=

2

s(1 + sX)

∆′′

∆=

2(1 + sV )

s2(1 + sW )

=2

s2(1 + sV )(1 + sW )

=2

s2

(1 + s (V + W + sV W )︸︷︷︸

Y

)=

2

s2(1 + sY )

(∆′

∆

)2

=

(2

s(1 + sX)

)2

=4

s2

(1 + s (2X + sX2)︸︷︷︸

Z

)=

4

s2(1 + sZ)

51

Note that the continuity of U , V , and W on [0, `] × [0, `] ensures that X, Y , and Z

are continuous on [0, `]× [0, `].

All of the above setup now allows us to write the following for s 6= 0:

f(s, t) =

∣∣∣∣2C + L′′

− (B + 2Cs+ L′) · 2

s(1 + sX)

− (Bs+ Cs2 + L) · 2

s2(1 + sY )

+ (Bs+ Cs2 + L) · 4

s2(1 + sZ)

∣∣∣∣=

∣∣∣∣2C + L′′

+

(−2

B

s− 4C − 2L

)+ sX

(−2

B

s− 4C − 2L

)+

(−2

B

s− 2C − 2L

)+ sY

(−2

B

s− 2C − 2L

)+

(4B

s+ 4C + 4L

)+ sZ

(4B

s+ 4C + 4L

) ∣∣∣∣=

∣∣∣∣4Bs − 2B

s− 2

B

s+ 2C − 2C + 4C − 4C + 4L− 2L− 2L+ L′′

+X(−2B − 4Cs− 2Ls

)+ Y

(−2B − 2Cs− 2Ls

)+ Z

(4B + 4Cs+ 4Ls

) ∣∣∣∣=

∣∣∣∣2L− 2L+ L′′

+X(−2B − 4Cs− 2Ls

)+ Y

(−2B − 2Cs− 2Ls

)+ Z

(4B + 4Cs+ 4Ls

) ∣∣∣∣Note that each of the terms in the above representation of f is continuous on [0, `]×

[0, `], so we may continuously extend f to [0, `] × [0, `]. The crucial point here was

the cancellation of the terms containing Bs

.

52

As this extension is continuous on a compact set, it is also bounded.

Lemma 3.13 is a basic and standard fact about integration we need to complete

our proof of the main result.

Lemma 3.13. Let f : [0, `]× [0, `]→ R be continuous. Then it follows that

F (t) =

∫ `

0

f(s, t) ds

is a continuous function in t.

Proof. Let ε > 0 and fix some t0 ∈ [0, `]. We shall show that F is continuous at t0.

Since f is a continuous function on a compact domain, we know f is uniformly

continuous. Hence, we have some δ > 0 such that |f(s1, t1)− f(s2, t2)| < ε`

whenever

|(s1, t1) − (s2, t2)| < δ. Note that this implies that |f(s, t) − f(s, t0)| < ε`

whenever

|(s, t)− (s, t0)| < δ. Note further that |(s, t)− (s, t0)| = |t− t0|.

Let t be such that |t− t0| < δ, and observe the following:

|F (t)− F (t0)| =∣∣∣∣∫ `

0

f(s, t) ds−∫ `

0

f(s, t0) ds

∣∣∣∣ Definition of F

=

∣∣∣∣∫ `

0

f(s, t)− f(s, t0) ds

∣∣∣∣ Property of the integral

≤∫ `

0

|f(s, t)− f(s, t0)| ds Integral triangle inequality

<

∫ `

0

ε

`ds Shown above

=ε

`

∫ `

0

1 ds =ε

`· ` = ε Simplify

Hence, we have for any ε > 0 that there exists some δ > 0 such that |F (t) −

F (t0)| < ε when |t − t0| < δ, so F is continuous at t0. Since our choice of t0 was

arbitrary, we know that F is continuous for all t.

53

As we have now obtained Lemma 3.12 and the supporting Lemma 3.13, we

can finally turn our attention to the proof of Proposition 3.14.


arclength. Then the function F : [0, `]→ R defined by

F (t) = κabstot (I1,γ(t)(γ))

is uniformly continuous.

Proof. Note that the function in Lemma 3.12 can be integrated in s over [0, `] to

obtain κabstot (I1,γ(t)(γ)). Then combining the results from Lemmas 3.12 and 3.13 and

Part (iii) of Corollary 3.5 gives us the continuity of F . As the domain of F is [0, `]

(a compact set), it follows that the continuity of F is uniform.

A few simple corollaries can be immediately derived from Proposition 3.14. In

particular, Corollary 3.17 will be useful for deriving results for c near γ. The first tells

us that the set of total absolute curvatures of all broken inverted curves is always a

bounded set.

Corollary 3.15. Let γ : [0, `] → Rn be a regular closed C2 curve parametrized by

arclength. Then there exists M ∈ R+ such that κabstot (Ir,c(γ)) ≤ M for all r ∈ R+ and

c ∈ γ.

Proof. Proposition 3.14 tells us that κabstot (I1,γ(t)(γ)) is continuous for t ∈ [0, `]. Since

[0, `] is compact and κabstot (Ir,c(γ)) does not depend on r, this immediately completes

the proof.

54

The result of Corollary 3.15 is then extended in Corollary 3.16 to tell us that

the total absolute curvature of any inverted curve (including broken inverted curves)

is finite.


arclength. Then κabstot (Ir,c(γ)) <∞ for all r ∈ R+ and c ∈ Rn.

Proof. Recall from Corollary 3.6 that total absolute curvature is independent of the

choice of r, so we know that κabstot (Ir,c(γ)) will be defined for all r if it is defined for a

single value of r. We thus have two cases with respect to c.

Case 1: c ∈ γ.

Proposition 3.14 tells us that κabstot (I1,c(γ)) is continuous for c ∈ γ, so we know

that κabstot (Ir,c(γ)) <∞ for all c ∈ γ.

Case 2: c /∈ γ.

Note that the integrand for κabstot (Ir,c(γ)) given by Part (iii) of Corollary 3.5 is

continuous for all s, and the integration domain is compact. Hence, we know that

κabstot (Ir,c(γ)) <∞ for all c /∈ γ.

Corollary 3.17 does not immediately appear impactful, but it will be used in

Subsubsection 3.2.2.2. This corollary simply tells us that a very small segment of a

broken inverted curve will have very small total absolute curvature.


arclength. Then for all ε > 0, there exists δ > 0 independent of t such that

κabstot

(Ir,γ(t)(γ|[t−δ,t+δ])

)< ε.

55

Proof. Let ε > 0 and without loss of generality fix t ∈ [0, `]. Recall from Propo-

sition 3.14 that the integrand of κabstot (I1,γ(t)(γ)) will be continuous in both s (the

integration variable) and t. As this continuity is on a compact domain, there exists

some M ∈ R+ bounding the integrand above. Recall that κabstot

(Ir,γ(t)(γ|t−δ,t+δ)

)does

not depend on r, then choose δ < ε2M

and observe

κabstot

(Ir,γ(t)(γ|[t−δ,t+δ])

)= κabs

tot

(I1,γ(t)(γ|[t−δ,t+δ])

)≤∫ t+δ

t−δM ds = 2Mδ < 2M

ε

2M= ε.

Note that our choice of t has no effect on this bound, which completes the proof.

Although Corollary 3.18 may seem redundant, its summative nature can help

us to obtain a wider perspective on what we have shown thus far. Notice that the

function F used in Proposition 3.14 is a restriction of Φ as defined in Corollary 3.18.


arclength, and define the function Φ : R+ × Rn → R by Φ(r, c) = κabstot (Ir,c(γ)). Then


(i) Φ(r, c) is constant in r.

(ii) Φ(r, c)|R+×(Rn\γ) is continuous.

(iii) Φ(r, c)|R+×γ is continuous.

Proof. Property (i) is a restatement of Corollary 3.6. Property (ii) follows from the

fact that the integrand given in Part (iii) of Corollary 3.5 is continuous on R+×Rn\γ.

Property (iii) follows from Proposition 3.14.

56

3.2.2.2 Inversion Centers off the Curve

We now turn our attention to the case in which c /∈ γ, but where c is still

close to γ. This work will build toward the proof of Proposition 3.28, which describes

the behavior of the total absolute curvature as the center of inversion approaches the

curve.

Before we directly attack the lemmas required for Proposition 3.28, we take

a quick detour that culminates in Corollary 3.22. While not as strong of a result as

Proposition 3.28, Corollary 3.22 provides an unique perspective on our problem.

Corollary 3.22. Let γ be a closed C2 curve in Rn parametrized by arclength. Let

c ∈ γ, and let {cn} be a sequence of points in Rn converging to c satisfying cn /∈ γ for

all n. Then

limn→∞

κabstot (Ir,cn(γ)) ≥ κabstot (Ir,c(γ)) + 2π.

Corollary 3.22 tells us that the function representing the total absolute curva-

ture of the inversion of γ is semicontinuous with respect to the inversion center—our

total absolute curvature must fall at least 2π when our inversion center touches γ.

We need a few preliminary results before moving to Corollary 3.22. The first

of these is Lemma 3.19, which is essentially a reformulation of a classic inequality.

Lemma 3.19. Let u and v be two unit vectors. Then the following hold:

(i) |u− v| = 2 sin(∠(u,v)

2

)(ii) ∠(u, v) = 2 arcsin

(|u−v|

2

)Proof. Consider Figure 3.7.

57

Figure 3.7: Diagram for Lemma 3.19.

Hence, we can obtain via trigonometry:

|u− v|2

= sin

(∠(u, v)

2

)=⇒ |u− v| = 2 sin

(∠(u, v)

2

).

This completes the proof of Part (i); to see Part (ii), simply solve Part (i) for ∠(u, v).

Next, we prove a lemma that describes the relationship between the “turning

angle” (that is, the deviation of the tangent vector of a curve) and the total absolute

curvature. This is a well-known, basic result.

Lemma 3.20. Let γ : [0, `] → Rn be a C2 curve parametrized by arclength, and let

a, b ∈ [0, `]. Then we have that

κabstot (γ∣∣[a,b]

) ≥ ∠(γ′(b), γ′(a)).

Proof. Recall that the tantrix T : [0, `]→ Sn−1 of γ is the map defined by

T (s) =γ′(s)

|γ′(s)|= γ′(s),

58

with the second equality due to the fact that γ is parametrized by arclength. Notice

then that T (s) forms a curve on Sn−1, and recall also that

κabstot (γ

∣∣[a,b]

) =

∫ b

a

|γ′′(s)| ds =

∫ b

a

|T ′(s)| ds.

Hence, we can calculate the total curvature of γ∣∣[a,b]

by calculating the length of the

curve T (s) on Sn−1. Of course,∫ ba|T ′(s)| ds must be bounded below by the minimum

distance between T (a) and T (b); keep in mind that this is the distance along the

sphere.

This means that the minimum distance will be the arclength of a great circle

(in this case unit) through the points T (a) and T (b). Recall that the length d of an

arc with a central angle θ is given by d = rθ. But r = 1, so we can conclude that the

minimum spherical distance between T (a) and T (b) will be θ = ∠(T (a), T (b)).

Remembering that γ′(s) = T (s), we may then say that

κabstot (γ

∣∣[a,b]

) =

∫ b

a

|T ′(s)| ds ≥ ∠(γ′(a), γ′(b)),

which is the desired result.

The final preliminary result we need is Proposition 3.21. In particular, this

result tells us that the total absolute curvature of the inverted image of a very small

arc is bounded below when the inversion center is sufficiently close to the arc.

Proposition 3.21. Let γ : [0, `]→ Rn be a closed C2 curve parametrized by arclength,

and let t ∈ [0, `]. Then for all ε > 0 there exists δ > 0 and d > 0 such that

κabstot (Ir,c(γ|[t−δ,t+δ])) > 2π − ε

59

when c ∈ Bnd (γ(t)).

Proof. Let ε > 0. Without loss of generality, say that r = 1 and t = 0. We will

use the abbreviation γ = I1,c(γ). While we have yet to define some of the variables

shown, it can be helpful to keep Figure 3.8 in mind.

Figure 3.8: Diagram for Proposition 3.21.

We will obtain the result by proving two different statements, specifically

κabstot (I1,c(γ|[0,δ])) > π − ε

2and κabs

tot (I1,c(γ|[−δ,0])) > π − ε

2.

As the two have similar proofs, we show only the +δ case. Our approach requires a

handful of inequalities, which are built in the following claims.

60

Claim 3.21.1. There exists some δ > 0 such that

∠(γ′(0), γ′(s)) <ε

8for all s ∈ [−δ, δ]. (3.1)

Proof of Claim 3.21.1: Since γ is C2, there exists some δ > 0 such that Inequal-

ity (3.1) is immediately satisfied by also using Lemma 3.19. �

Claim 3.21.2. Define wδ = γ(δ)− c. Then for sufficiently small δ there exists some

d > 0 such that

∠(γ′(δ), wδ) <ε

8for all c ∈ Bn

d (γ(0)). (3.2)

Proof of Claim 3.21.2: Claim 3.21.2 will be shown through the assembly of three

different statements, which we show in turn below.

As γ is C2, we know that limδ→0γ(δ)−γ(0)|γ(δ)−γ(0)| = γ′(0). Hence, δ can be chosen

small enough to ensure that∣∣∣∣γ′(δ)− γ(δ)− γ(0)

|γ(δ)− γ(0)|

∣∣∣∣ < ε

32.

Note that arcsin(x2

)≤ x for all 0 ≤ x ≤ 2. Combining this with Lemma 3.19 gives

ε

16> 2

∣∣∣∣γ′(δ)− γ(δ)− γ(0)

|γ(δ)− γ(0)|

∣∣∣∣≥ 2 arcsin

∣∣∣γ′(δ)− γ(δ)−γ(0)

|γ(δ)−γ(0)|

∣∣∣2

= ∠

(γ′(δ),

γ(δ)− γ(0)

|γ(δ)− γ(0)|

).

Hence, we have that

∠

(γ′(δ),

γ(δ)− γ(0)

|γ(δ)− γ(0)|

)<

ε

16. (3.3)

As d must be chosen after δ, we proceed from this point of the proof with δ fixed.

Recall that arcsin(x2

)≤ x for all 0 ≤ x ≤ 2 and combine this with Lemma 3.19

61

to observe the following. We will use the abbreviations D0 = |γ(δ) − γ(0)| and

Dc = |γ(δ)− c|. Obviously D0 > 0, Dc > 0 and fixed since δ > 0.

∠

(γ(δ)− γ(0)

|γ(δ)− γ(0)|,γ(δ)− c|γ(δ)− c|

)≤ 2

∣∣∣∣γ(δ)− γ(0)

D0

− γ(δ)− cDc

∣∣∣∣= 2

∣∣∣∣γ(δ)Dc − γ(0)Dc − γ(δ)D0 + cD0

D0Dc

∣∣∣∣= 2

∣∣∣∣γ(δ)Dc − γ(δ)D0 + cD0 − cDc + cDc − γ(0)Dc

D0Dc

∣∣∣∣≤ 2|γ(δ)||Dc −D0|+ |c||D0 −Dc|+Dc|c− γ(0)|

D0Dc

Via reverse triangle inequality we can obtain

|D0−Dc| =∣∣|γ(δ)−γ(0)|−|γ(δ)−c|

∣∣ ≤ |γ(δ)−γ(0)−γ(δ)+c| = |γ(0)−c| = |c−γ(0)|.

Using this fact, we substitute as follows:

∠

(γ(δ)− γ(0)

|γ(δ)− γ(0)|,γ(δ)− c|γ(δ)− c|

)≤ 2|γ(δ)||c− γ(0)|+ |c||c− γ(0)|+Dc|c− γ(0)|

D0Dc

= 2|c− γ(0)|[|γ(δ)|+ |c|+Dc

D0Dc

]=

(|c− γ(0)||γ(δ)− c|

)︸︷︷︸

C1

2|γ(δ)|D0

+

(|c||c− γ(0)||γ(δ)− c|

)︸︷︷︸

C2

2

D0

+ |c− γ(0)|︸︷︷︸C3

2

D0

Hence, it follows that

∠

(γ(δ)− γ(0)

|γ(δ)− γ(0)|,γ(δ)− c|γ(δ)− c|

)≤ C1

2|γ(δ)|D0

+ C22

D0

+ C32

D0

. (3.4)

Note that D0 is independent of our choice of d, since it does not depend on c. Hence,

we need only know that making d sufficiently small makes C1, C2, and C3 sufficiently

62

small. Note that |c − γ(0)| decreases as d decreases. This takes care of C3. Com-

pactness ensures that we can bound c away from γ(δ) using sufficiently small d, so

we also have taken care of C1. Since c will be close to γ(0), we know that |c| will be

bounded, which takes care of C2. Hence, we may conclude that d > 0 can be selected

to ensure that

(|c− γ(0)||γ(δ)− c|

)2|γ(δ)|D0

<ε

48,(

|c||c− γ(0)||γ(δ)− c|

)2

D0

<ε

48, and

(|c− γ(0)|) 2

D0

<ε

48.

Inequality (3.4) then allows us to conclude that

∠

(γ(δ)− γ(0)

|γ(δ)− γ(0)|,γ(δ)− c|γ(δ)− c|

)<

ε

16. (3.5)

Recall that wδ = γ(δ)− c, which tells us that

∠

(γ(δ)− c|γ(δ)− c|

, wδ

)= 0, (3.6)

since the two vectors are parallel. Assembling Inequalities (3.3) and (3.5) and Equa-

tion (3.6) then allows us to observe the following:

∠(γ′(δ), wδ) ≤ ∠(γ′(δ),

γ(δ)− γ(0)

|γ(δ)− γ(0)|

)+ ∠

(γ(δ)− γ(0)

|γ(δ)− γ(0)|,γ(δ)− c|γ(δ)− c|

)+ ∠

(γ(δ)− c|γ(δ)− c|

, wδ

)<

ε

16+

ε

16+ 0 =

ε

8

This completes the proof of Inequality (3.2) and the proof of Claim 3.21.2. �

63

Claim 3.21.3. If Claim 3.21.2 is satisfied, then

∠(−γ′(δ), wδ) <ε

8, where γ = Ir,c(γ). (3.7)

Proof of Claim 3.21.3: To see that this claim holds, consider the following:

ε

8> ∠(γ′(δ), wδ) Inequality (3.2)

= ∠(γ′(δ), wδ) Inversion is conformal

= ∠(γ′(δ),−wδ) wδ = −wδ, since wδ is in the radial direction of the inversion

= ∠(−γ′(δ), wδ) Negate both angles

Hence, we have proved our claim. �

Claim 3.21.4. For every c ∈ Bnd (γ(0)) there exists some s0 ∈ (−δ, δ) such that

∠(γ′(s0), γ′(s0)) = 0. (3.8)

Proof of Claim 3.21.4: Recall that vectors perpendicular to the radial direction

of an inversion will have their direction maintained under inversion. Hence, it will

suffice to show for every c ∈ Bnd (γ(0)) that there exists some s0 ∈ (−δ, δ) such

that γ′(s0) ⊥ (γ(s0) − c). Since γ|[−δ,δ] is a C1 curve on a compact domain, it has

some normal tube N with radius d0 > 0. Since every point in N \ γ satisfies the

perpendicularity property above, we first ensure that d is small enough to satisfy

Bnd (γ(0)) ⊆ N . However, this alone does not guarantee that s0 6= ±δ.

We must also make sure that

0 < d ≤ 1

3min (|γ(0)− γ(δ)|, |γ(0)− γ(−δ)|) .

64

Observe that

|c− γ(0)| < d ≤ 1

3|γ(0)− γ(δ)| =⇒ |c− γ(δ)| ≥ 2

3|γ(δ)− γ(0)|.

This second condition then tells us that γ(δ) is not the closest point of γ|[−δ,δ] to c.

We can similarly show that γ(−δ) is not the closest point of γ|[−δ,δ] to c. Hence, there

exists s0 ∈ (−δ, δ) such that γ′(s0) ⊥ (γ(s0) − c), which completes the proof of our

claim. �

To summarize, the above work allows us to conclude that there exists δ, d > 0

such that when c ∈ Bnd (γ(0)) we have the following:

• Equation (3.8): ∠(γ′(s0), γ′(s0)) = 0

• From Inequality (3.1): ∠(γ′(s0), γ′(0)) < ε8

• From Inequality (3.1): ∠(γ′(0), γ′(δ)) < ε8

• Inequality (3.2): ∠(γ′(δ), wδ) <ε8

• Inequality (3.7): ∠(wδ,−γ′(δ)) < ε8

We can then assemble these facts to obtain

∠(γ′(s0),−γ′(δ)) ≤ ∠(γ′(s0), γ′(s0)) + ∠(γ′(s0), γ′(0))

+ ∠(γ′(0), γ′(δ)) + ∠(γ′(δ), wδ) + ∠(wδ,−γ′(δ))

< 0 +ε

8+ε

8+ε

8+ε

8=ε

2.

Using supplementary angles, we can then determine that

∠(γ′(s0),−γ′(δ)) + ∠(γ′(s0), γ′(δ)) = π

∠(γ′(s0), γ′(δ)) = π − ∠(γ′(s0),−γ′(δ))

65

∠(γ′(s0), γ′(δ)) > π − ε

2

It then follows from Lemma 3.20 that

κabstot (γ|[s0,δ]) ≥ ∠(γ′(s0), γ′(δ)) > π − ε

2.

The proof for the −δ case is similar, so we may also conclude that

κabstot (γ|[−δ,s0]) ≥ ∠(γ′(s0), γ′(−δ)) > π − ε

2.

As total curvature is additive on C2 curves, we can then conclude that

κabstot (γ|[−δ,δ]) = κabs

tot (γ|[−δ,s0]) + κabstot (γ|[s0,δ]) > 2π − ε.

Since this was shown without loss of generality for r = 1 and t = 0, we have the

desired result for all values of r ∈ R+ and t ∈ [0, `].

With Proposition 3.21 in hand, we can finally move to the proof of Corol-

lary 3.22. As most of the work is buried in the proof of the former, the proof of the

latter is straightforward.

Corollary 3.22. Let γ be a closed C2 curve in Rn parametrized by arclength. Let

c ∈ γ, and let {cn} be a sequence of points in Rn converging to c satisfying cn /∈ γ for

all n. Then

lim infn→∞

κabstot (Ir,cn(γ)) ≥ κabstot (Ir,c(γ)) + 2π.

Proof. Without loss of generality say that c = γ(0). Let ε > 0. Recall the total cur-

vature is additive over segments of a C2 curve. Let δ be as specified in Corollary 3.17,

66

and choose d (and possibly smaller δ) as in Proposition 3.21. If |cn − γ(0)| < d, then


κabstot (Ir,cn(γ)) = κabs

tot (Ir,cn(γ|[δ,`−δ])) + κabstot (Ir,cn(γ|[−δ,δ]))

=⇒ κabstot (Ir,cn(γ)) ≥ κabs

tot (Ir,cn(γ|[δ,`−δ])) + 2π − ε

=⇒ lim infn→∞

κabstot (Ir,cn(γ)) ≥ lim inf

n→∞κabs

tot (Ir,cn(γ|[δ,`−δ])) + 2π − ε

Notice that c /∈ γ([δ, `− δ]), so we have that

lim infn→∞

κabstot (Ir,cn(γ|[δ,`−δ])) = κabs

tot (Ir,c(γ|[δ,`−δ])).

Hence, we may continue our string of implications as follows:

=⇒ lim infn→∞

κabstot (Ir,cn(γ)) ≥ κabs

tot (Ir,c(γ|[δ,`−δ])) + 2π − ε

=⇒ lim infn→∞

κabstot (Ir,cn(γ)) ≥

[κabs

tot (Ir,c(γ))− κabstot (Ir,c(γ|[−δ,δ]))

]+ 2π − ε

=⇒ lim infn→∞

κabstot (Ir,cn(γ)) ≥

[κabs

tot (Ir,c(γ))− ε)]

+ 2π − ε

=⇒ lim infn→∞

κabstot (Ir,cn(γ)) ≥ κabs

tot (Ir,c(γ)) + 2π − 2ε

As this is true for all ε > 0 and c = γ(0) was taken without loss of generality, we have

the desired conclusion.

While obtaining semicontinuity is nice, we can do better. Notice that Propo-

sition 3.28 even provides a uniformity condition. With that in mind, we will now

work directly toward its proof. This is done in three main steps: First, we must

quantify the behavior of the total absolute curvature coming from points of γ near

the inversion point (we will show that it is approximately equal to 2π). Second, we

67

must quantify the behavior of the total absolute curvature coming from the points of

γ away from the inversion point (we will show it is approximately equal to the curva-

ture of a broken inverted curve). Finally, we will combine these estimates to derive

an estimate for the total absolute curvature of the entirety of the inverted image of

γ. Note also that the following work requires γ to be closed.

To understand the curvature of the inverted curve when the inversion center

c is nearby (but not on) the curve γ, we use the following strategy (illustrated in

Figure 3.9):

1. For a point c sufficiently close to the curve γ, let t be such that γ(t) be the

unique point on γ closest to c.

2. Specify a line λt through c and γ(t), and parametrize it by arclength in u.

3. Specify cut points γ(t + δ(u)) and γ(t − δ(u)) to separate the curve into two

segments—one is an interval about γ(t), and the other is its complement.

4. Set u so that λt(u) = c. (Notice that cut points will depend on dist(c, γ)!)

5. Analyze the total absolute curvature of each segment separately, then combine

the two results.

It is critical to note that our choice of δ will not matter once the reassembly of the

total absolute curvature of γ is complete. The choice only affects the total absolute

curvatures of the individual segments, not the entire curve.

With our strategy before us, we move to proving an array of lemmas. The first

is Lemma 3.23, which tells us that the image under inversion of an appropriately-

sized arc of circle or segment of line with a sufficiently close inversion will have total

68

Figure 3.9: An approach for analyzing the total curvature of γ when the inversion

center is close to γ.

absolute curvature approximately equal to 2π. The essential idea: an inversion with

an inversion center very close to a circular arc transforms the arc into circle minus an

a small arc.

Lemma 3.23. Let γ : [0, `]→ Rn be a closed C2 curve parametrized by arclength. For

all ε > 0 there exists d > 0 for all t ∈ [0, `] and ~ωt satisfying |~ωt| = 1 and ~ωt ⊥ γ′(t)

69

such that one has ∣∣κabstot (Ir,λt(u)(ηt|[−δ(u),δ(u)]))− 2π∣∣ < ε

whenever 0 0.

Case 1: ηt is a circle with radius r.

We will abbreviate ηt, λt and ~ωt as η, λ, and ~ω, respectively. Since total

curvature is invariant with respect to affine similarities by Proposition 3.4, we may

say without loss of generality that η(s) = (sin(s), cos(s)− 1, 0, . . . , 0). We then have

that η′(0) = (1, 0, 0, . . . , 0), so we know ~ω is of the form ~ω = 1r(0, ω2, ω3, . . . , ωn).

(Note that rescaling the circle to ensure that it has radius one also necessarily scales

~ω.) We can compute the following:

η(s)− λ(u) =(

sin(s), cos(s)− 1− uω2

r,−uω3

r, . . . ,−uωn

r

)η′(s) = (cos(s),− sin(s), 0, . . . , 0)

η′′(s) = (− sin(s), cos(s), 0, . . . , 0)

∆(s, u) = |η(s)− λ(u)|2 = 2 + 2uω2

r− 2

(1 + u

ω2

r

)cos(s) +

u2

r2

∆′(s, u) = 2(

1 + uω2

r

)sin(s)

∆′′(s, u) = 2(

1 + uω2

r

)cos(s)

We can then substitute all of the above work into Part (iii) of Corollary 3.5 to obtain

κabstot (Ir,λ(u)(η|[−δ(u),δ(u)])) =

∫ δ(u)r

− δ(u)r

u√

4r2 + 4ruω2 + u2

2r2 + 2ruω2 + u2 − 2r(r + uω2) cos(s)ds.

70

Integration then yields the following:

κabstot (Ir,λ(u)(η|[−δ(u),δ(u)])) = 2 arctan

[tan(s2

)√4r2 + 4ruω2 + u2

u

] ∣∣∣∣∣δ(u)r

− δ(u)r

= 2 arctan

tan(δ(u)2r

)√4r2 + 4ruω2 + u2

u

− 2 arctan

tan(−δ(u)

2r

)√4r2 + 4ruω2 + u2

u

= 4 arctan

tan(δ(u)2r

)√4r2 + 4ruω2 + u2

u

Recall that |~ω| = 1, so then |ω2| ≤ 1. As γ is compact, we know that maxs κ

absγ (s)

exists; call it κm > 0, and note that r ≥ 1κm

for every osculating circle ηt. Note

additionally that tan(x) ≥ x for sufficiently small x, so we have that tan(δ(u)2r

)≥ δ(u)

2r

for sufficiently small u. Now consider the following for u ≤ 1κm

:

tan(δ(u)2r

)√4r2 + 4ruω2 + u2

u≥

δ(u)2r

√4r2 + 4ruω2 + u2

u

=δ(u)

u·√

4r2 + 4ruω2 + u2

2r

=u

19

u

√4r2 + 4ruω2 + u2

4r2

=1

u89

√1 +

u

rω2 +

u2

4r2

Recall that |ω2| ≤ 1, which allows us to continue with the following:

tan(δ(u)2r

)√4r2 + 4ruω2 + u2

u≥ 1

u89

√1− u

r+

u2

4r2

=1

u89

√(1− u

2r

)2

=1

u89

(1− u

2r

)

71

Note that u ≤ 1κm≤ r, so we have that

r ≥ u =⇒ 1 ≥ u

r=⇒ 2 ≥ 1 +

u

r=⇒ 1 ≥ 1

2+

u

2r=⇒ 1− u

2r≥ 1

2.

Hence, we may conclude that

tan(δ(u)2r

)√4r2 + 4ruω2 + u2

u≥ 1

u89

· 1

2=

1

2u89

.

It is important to notice that this reduction removes any dependence on t. Recall

that arctan is increasing and bounded above by π2. We then have the following:

|κabstot (Ir,λ(u)(η|[−δ(u),δ(u)]))− 2π|

=

∣∣∣∣∣∣4 arctan

tan(δ(u)

2

)√4r2 + 4ruω2 + u2

u

− 2π

∣∣∣∣∣∣≤∣∣∣∣4 arctan

[1

2u89

]− 2π

∣∣∣∣We then have for all t where ηt is a circle that there exists some d1 > 0 such that

∣∣κabstot (Ir,λ(u)(η|[−δ(u),δ(u)]))− 2π

∣∣ ≤ ∣∣∣∣4 arctan

[1

2u89

]− 2π

∣∣∣∣ < ε

when 0 < u < d1.

Case 2: η is a line.

Since total curvature is invariant with respect to isometries, we may say with-

out loss of generality that η(s) = (s, 0, . . . , 0). We then have that η′(0) = (1, 0, . . . , 0),

72

so we know ~ω is of the form ~ω = (0, ω2, . . . , ωn). Hence, we can calculate the following:

η(s)− λ(u) = (s,−uω2, . . . ,−uωn)

η′(s) = (1, 0, . . . , 0)

η′′(s) = (0, 0, . . . , 0)

∆(s, u) = |η(s)− λ(u)|2 = s2 + u2

∆′(s, u) = 2s

∆′′(s, u) = 2

We can then substitute all of the above work into Part (iii) of Corollary 3.5 to obtain

κabstot (Ir,λ(u)(η|[−δ(u),δ(u)])) =

∫ δ(u)

−δ(u)

2u

s2 + u2ds.

Integration then yields the following:

κabstot (Ir,λ(u)(η|[−δ(u),δ(u)])) = 2 arctan

[ su

] ∣∣∣∣δ(u)

−δ(u)

= 2 arctan

[δ(u)

u

]− 2 arctan

[−δ(u)

u

]= 4 arctan

[δ(u)

u

]We then have for all t where ηt is a line that there exists some d2 > 0 such that

∣∣κabstot (Ir,λ(u)(η|[−δ(u),δ(u)]))− 2π

∣∣ =

∣∣∣∣4 arctan

[1

u89

]− 2π

∣∣∣∣ < ε

when 0 < u < d2.

Picking d = min{d1, d2} completes the proof.

Lemma 3.24 is similar to Lemma 3.10, with only a few exceptions. This will

be useful for proving another one of our main lemmas.

73

Lemma 3.24. Let γ : [0, `]→ Rn be a C2 curve and for t ∈ [0, `] let σt be its parabolic

approximation at γ(t). That is,

σt(s) = γ(t) + γ′(t)s+1

2γ′′(t)s2.

Now define Lt : [0, `]→ Rn by

Lt(s) = γ(s+ t)− σt(s).

Then

Lt(s)

s2,

L′t(s)

s, and L′′t (s)

are all continuously extendable to [0, `]. Additionally, each of these is bounded uni-

formly on [0, `]× [0, `].

Proof. Since both γ and σ are C2, we immediately know that Lt(s) is C2 on [0, `], so

L′′t (s) is C0 on [0, `].

Now notice that s = 0 gives

Lt(0) = γ(0 + t)− σt(0) = γ(t)− σt(0) = 0.

Hence, Lemma 3.9 tells us that there exists some Gt(s) such that Gt is C1 on [0, `]

and Lt(s) = sGt(s). Now notice further for all t that s = 0 gives

Gt(0) = γ′(0 + t)− σ′t(0) = γ′(t)− σ′t(0) = 0.

Hence, Lemma 3.9 tells us that there exists some Ht(s) such that Ht is C0 on [0, `]

and Gt(s) = sHt(s). As a result, we have for s 6= 0 that

Lt(s) = s2Ht(s) =⇒ Lt(s)

s2= Ht(s).

74

Since Ht is C0 on [0, `], we know that Lt(s)s2

is continuously extendable to [0, `]. Finally,

observe for all t that s = 0 gives

L′t(0) = γ′(0 + t)− σ′t(0) = γ′(0)− σ′t(0) = 0.

So then Lemma 3.9 tells us that there exists some Kt(s) such that Kt is C0 on [0, `]

and L′t(s) = sKt(s). As a result, we have for s 6= 0 that

L′t(s)

s= Kt(s).

Since Kt is C0 on [0, `], we know thatL′t(s)s

is continuously extendable to [0, `].

Note that the definition of σ allows us to write the following:

Lt(s)

s2=γ(s+ t)− γ(t)− γ′(t)s− γ′′(t)

2s2

s2

L′t(s)

s=γ′(s+ t)− γ′(t)− γ′′(t)s

s

L′′t (s) = γ′′(s+ t)− γ′′(t).

It is clear from this representation that each of these terms is continuous in t. Fur-

thermore, the above work shows that each of these is at least continuously extendable

in s; therefore, each is continuously extendable to [0, `] × [0, `]. As this domain is

compact, we may then conclude that each is uniformly bounded in s and t. This

completes the proof.

Lemma 3.25 shows that an inverted segment of the parabolic approximation

of a curve γ at a point provides an approximation of the total absolute curvature of

an inverted segment of γ at that point for a sufficiently close inversion center and

sufficiently small segments.

75

Figure 3.10: The segment of γ analyzed in Lemmas 3.25 and 3.26.



such that one has

∣∣κabstot (Ir,λt(u)(γ|[t−δ(u),t+δ(u)]))− κabstot (Ir,λt(u)(σt|[−δ(u),δ(u)]))∣∣ < ε

whenever 0 0. We will first prove the case where t = 0, then show that the proof for

t = 0 can be extended to work for any t. That said, set t = 0, and abbreviate σt and λt

as σ and λ, respectively. Let ∆γ(s, u) = |γ(s)− λ(u)|2 and ∆σ(s, u) = |σ(s)− λ(u)|2.

We begin with the following, omitting all input variables and defining Ni and D as

shown below:

∣∣κabstot (Ir,λ(γ|[−δ,δ]))− κabs

tot (Ir,λ(σ|[−δ,δ]))∣∣

76

=

∣∣∣∣∣∫ δ

−δ

∣∣∣∣γ′′ − γ′∆′γ∆γ

− (γ − λ)∆′′γ∆γ

+ (γ − λ)(∆′γ)

2

∆2γ

∣∣∣∣ ds−∫ δ

−δ

∣∣∣∣σ′′ − σ′∆′σ∆σ

− (σ − λ)∆′′σ∆σ

+ (σ − λ)(∆′σ)2

∆2σ

∣∣∣∣ ds∣∣∣∣∣

≤∫ δ

−δ| γ′′ − σ′′︸︷︷︸

N1

| ds

+

∫ δ

−δ

[∣∣∣− (γ′∆′γ∆γ∆2σ − σ′∆′σ∆σ∆2

γ

)︸︷︷︸N2

−((γ − λ)∆′′γ∆γ∆

2σ − (σ − λ)∆′′σ∆σ∆2

γ

)︸︷︷︸N3

+((γ − λ)(∆′γ)

2∆2σ − (σ − λ)(∆′σ)2∆2

γ

)︸︷︷︸N4

∣∣∣/ (∆2γ∆

2σ)︸︷︷︸

D

]ds

(3.9)

Over the remainder of this proof we shall derive individual bounds on N1, N2, N3,

N4, and D to show that∫ δ

−δ|N1| ds+

∫ δ

−δ

| −N2 −N3 +N4|D

ds ≤ 2Mu19 +

18Mu

u89 − 12Mu

. (3.10)

Define L : [0, `]→ Rn by L(s) = γ(s)− σ(s), which gives γ(s) = σ(s) + L(s).

Recall that σ is given explicitly as σ(s) = γ(0) + γ′(0)s+ γ′′(0)2s2. Defining A = γ(0),

B = γ′(0), and C = γ′′(0)2

, we may write

γ(s) = A+Bs+ Cs2 + L and σ(s) = A+Bs+ Cs2.

Note that our definition of A also allows us to write λ(u) = A+ u~ω. Hence, we have

that

γ(s)− λ(u) = Bs+ Cs2 + L− u~ω and σ(s)− λ(u) = Bs+ Cs2 − u~ω.

We can also compute

γ′(s) = B + 2Cs+ L′, σ′(s) = B + 2Cs, γ′′(s) = 2C + L′′, and σ′′(s) = 2C.

77


and L′

sare continuously extendable to [0, `]. Call

these extensions L and L, respectively, and note that then s2L = L and sL = L′. We

then rewrite γ, σ, and their derivatives as follows:

γ(s)− λ(u) = sB + s2C + s2L− u~ω σ(s)− λ(u) = sB + s2C − u~ω

γ′(s) = B + s2C + sL σ′(s) = B + s2C

γ′′(s) = 2C + L′′ σ′′(s) = 2C

Now recall that

B • C = γ′(0) • γ′′(0) = 0 and B •B = γ′(0) • γ′(0) = 1,

since γ was parametrized by arclength. Also, remember that

B • ~ω = γ′(0) • ~ω = 0 and ~ω • ~ω = 1

by our choice of ~ω. Using all of this in conjunction with the above expansions allows

us to make the following calculations.

∆γ = (γ − λ) • (γ − λ)

= s2 + u2 + s3(2B • L) + s4(C • C) + s4(2C • L) + s4(L • L)

− s2u(2C • ~ω)− s2u(2L • ~ω)

∆′γ = 2γ′ • (γ − λ)

= s · 2 + s2(2B • L) + s2(2B • L) + s3(4C • C)

+ s3(2C • L) + s3(2L • L) + s3(4C • L)

− su(4C • ~ω)− su(2L • ~ω)

78

∆′′γ = 2γ′ • γ′ + 2γ′′ • (γ − λ)

= 2 + s(2B • L′′) + s2(4C • C) + s2(4C • L) + s2(2C • L′′) + s2(2L • L′′)

− u(4C • ~ω)− u(2L′′ • ~ω)

We can make similar calculations for σ.

∆σ = (σ − λ) • (σ − λ) = s2 + u2 + s4(C • C)− s2u(2C • ~ω)

∆′σ = 2σ′ • (σ − λ) = s · 2 + s3(4C • C)− su(4C • ~ω)

∆′′σ = 2σ′ • σ′ + 2σ′′ • (σ − λ) = 2 + s2(4C • C)− u(4C • ~ω)

In order to simplify our calculations, we will use the following consolidated versions

of each of these ∆-terms, with the Xi, Yi being defined appropriately. Note that while

the Xi, Yi might contain variables s and u, they can each be bounded for all s and

u when t = 0. (In fact, notice that they can actually be uniformly bounded for all

t. This will be used later when extending the proof of the t = 0 case to the general

case.)

∆γ = s2 + u2 + s3X1 + s2uX2 ∆σ = s2 + u2 + s3Y1 + s2uY2

∆′γ = s · 2 + s2X3 + suX4 ∆′σ = s · 2 + s2Y3 + suY4

∆′′γ = 2 + sX5 + uX6 ∆′′σ = 2 + sY5 + uY6

Also observe for appropriately defined Xi, Yi that

∆2γ = s4 + s2u2 · 2 + u4 + s5X7 + s4uX8 + s3u2X9 + s2u3X10 and

∆2σ = s4 + s2u2 · 2 + u4 + s5Y7 + s4uY8 + s3u2Y9 + s2u3Y10.

79

We now proceed to the calculation of the Ni, starting with N1. Observe that

N1 = γ′′ − σ′′ = L′′,

so we have that

N1 = L′′ .

We now focus our efforts on the calculation each of the remaining cross terms in

Expression (3.9), appropriately defining all Zi,Wi as the calculation proceeds. As

with Xi, Yi, note that each of the Zi and Wi can all be bounded for all s and u for

t = 0. We begin by calculating the factors of N2.

∆′γ∆γ = (s · 2 + s2X3 + suX4)(s2 + u2 + s3X1 + s2uX2)

= s3 · 2 + su2 · 2 + s4Z1 + s3uZ2 + s2u2Z3 + su3Z4

∆′σ∆σ = (s · 2 + s2Y3 + suY4)(s2 + u2 + s3Y1 + s2uY2)

= s3 · 2 + su2 · 2 + s4W1 + s3uW2 + s2u2W3 + su3W4

Hence, we can observe the following:

∆′γ∆γ∆2σ = (s3 · 2 + su2 · 2 + s4Z1 + s3uZ2 + s2u2Z3 + su3Z4)

· (s4 + s2u2 · 2 + u4 + s5Y7 + s4uY8 + s3u2Y9 + s2u3Y10)

= s7 · 2 + s5u2 · 6 + s3u4 · 6 + su6 · 2

+ s8Z5 + s7uZ6 + s6u2Z7 + s5u3Z8

+ s4u4Z9 + s3u5Z10 + s2u6Z11 + s1u7Z12

∆′σ∆σ∆2γ = (s3 · 2 + su2 · 2 + s4W1 + s3uW2 + s2u2W3 + su3W4)

· (s4 + s2u2 · 2 + u4 + s5X7 + s4uX8 + s3u2X9 + s2u3X10)

80

= s7 · 2 + s5u2 · 6 + s3u4 · 6 + su6 · 2

+ s8W5 + s7uW6 + s6u2W7 + s5u3W8

+ s4u4W9 + s3u5W10 + s2u6W11 + s1u7W12

We then multiply by γ′ and σ′, respectively; as well, we also define Vi and Ui appro-

priately. Unlike Xi, Yi, Zi, and Wi, our Vi and Ui will be vector quantities, but as

with those definitions above, Vi and Ui will still be bounded uniformly in s and u for

t = 0.

γ′∆′γ∆γ∆2σ = (B + s2C + sL)

(s7 · 2 + s5u2 · 6 + s3u4 · 6 + su6 · 2

+ s8Z5 + s7uZ6 + s6u2Z7 + s5u3Z8

+ s4u4Z9 + s3u5Z10 + s2u6Z11 + s1u7Z12

)= s7(2B) + s5u2(6B) + s3u4(6B) + su6(2B)

+ s8V1 + s7uV2 + s6u2V3 + s5u3V4

+ s4u4V5 + s3u5V6 + s2u6V7 + s1u7V8

σ′∆′σ∆σ∆2γ = (B + s2C)

(s7 · 2 + s5u2 · 6 + s3u4 · 6 + su6 · 2

+ s8W5 + s7uW6 + s6u2W7 + s5u3W8

+ s4u4W9 + s3u5W10 + s2u6W11 + s1u7W12

)= s7(2B) + s5u2(6B) + s3u4(6B) + su6(2B)

+ s8U1 + s7uU2 + s6u2U3 + s5u3U4

+ s4u4U5 + s3u5U6 + s2u6U7 + s1u7U8

Finally, we can take the difference of these two terms, defining appropriate Ti, which,

81

again, we know can all be bounded uniformly in s and u for t = 0.

N2 = γ′∆′γ∆γ∆2σ − σ′∆′σ∆σ∆2

γ

=(s7(2B) + s5u2(6B) + s3u4(6B) + su6(2B)

+ s8V1 + s7uV2 + s6u2V3 + s5u3V4

+ s4u4V5 + s3u5V6 + s2u6V7 + s1u7V8

)−(s7(2B) + s5u2(6B) + s3u4(6B) + su6(2B)

+ s8U1 + s7uU2 + s6u2U3 + s5u3U4

+ s4u4U5 + s3u5U6 + s2u6U7 + s1u7U8

)= s8T1 + s7uT2 + s6u2T3 + s5u3T4 + s4u4T5 + s3u5T6 + s2u6T7 + s1u7T8

Hence, we have that

N2 = s8T1 + s7uT2 + s6u2T3 + s5u3T4 + s4u4T5 + s3u5T6 + s2u6T7 + s1u7T8 .

We now embark on a similar calculation of N3, starting with its factors.

∆′′γ∆γ = (2 + sX5 + uX6)(s2 + u2 + s3X1 + s2uX2)

= s2 · 2 + u2 · 2 + s3Z13 + s2uZ14 + su2Z15 + u3Z16

∆′′σ∆σ = (2 + sY5 + uY6)(s2 + u2 + s3Y1 + s2uY2)

= s2 · 2 + u2 · 2 + s3W13 + s2uW14 + su2W15 + u3W16


∆′′γ∆γ∆2σ = (s2 · 2 + u2 · 2 + s3Z13 + s2uZ14 + su2Z15 + u3Z16)


82

= s6 · 2 + s4u2 · 6 + s2u4 · 6 + u6 · 2

+ s7Z17 + s6uZ18 + s5u2Z19 + s4u3Z20

+ s3u4Z21 + s2u5Z22 + su6Z23 + u7Z24

∆′′σ∆σ∆2γ = (s2 · 2 + u2 · 2 + s3W13 + s2uW14 + su2W15 + u3W16)


= s6 · 2 + s4u2 · 6 + s2u4 · 6 + u6 · 2

+ s7W17 + s6uW18 + s5u2W19 + s4u3W20

+ s3u4W21 + s2u5W22 + su6W23 + u7W24

We then multiply by γ − λ and σ − λ, respectively.

(γ − λ)∆′′γ∆γ∆2σ = (sB + s2C + s2L− u~ω)

·(s6 · 2 + s4u2 · 6 + s2u4 · 6 + u6 · 2

+ s7Z17 + s6uZ18 + s5u2Z19 + s4u3Z20

+ s3u4Z21 + s2u5Z22 + su6Z23 + u7Z24

)= s7(2B) + s5u2(6B) + s3u4(6B) + su6(2B)

− s6u(2~ω)− s4u3(6~ω)− s2u5(6~ω)− u7(2~ω)

+ s8V9 + s7uV10 + s6u2V11 + s5u3V12 + s4u4V13

+ s3u5V14 + s2u6V15 + s1u7V16 + u8V17

83

(σ − λ)∆′′σ∆σ∆2γ = (sB + s2C − u~ω)

·(s6 · 2 + s4u2 · 6 + s2u4 · 6 + u6 · 2

+ s7W17 + s6uW18 + s5u2W19 + s4u3W20

+ s3u4W21 + s2u5W22 + su6W23 + u7W24

)= s7(2B) + s5u2(6B) + s3u4(6B) + su6(2B)

− s6u(2~ω)− s4u3(6~ω)− s2u5(6~ω)− u7(2~ω)

+ s8U9 + s7uU10 + s6u2U11 + s5u3U12 + s4u4U13

+ s3u5U14 + s2u6U15 + s1u7U16 + u8U17

Finally, we can take the difference of these two terms:

N3 = (γ − λ)∆′′γ∆γ∆2σ − (σ − λ)∆′′σ∆σ∆2

γ

=(s7(2B) + s5u2(6B) + s3u4(6B) + su6(2B)

− s6u(2~ω)− s4u3(6~ω)− s2u5(6~ω)− u7(2~ω)

+ s8V9 + s7uV10 + s6u2V11 + s5u3V12 + s4u4V13

+ s3u5V14 + s2u6V15 + s1u7V16 + u8V17

)−(s7(2B) + s5u2(6B) + s3u4(6B) + su6(2B)

− s6u(2~ω)− s4u3(6~ω)− s2u5(6~ω)− u7(2~ω)

+ s8U9 + s7uU10 + s6u2U11 + s5u3U12 + s4u4U13

+ s3u5U14 + s2u6U15 + s1u7U16 + u8U17

)= s8T9 + s7uT10 + s6u2T11 + s5u3T12 + s4u4T13

+ s3u5T14 + s2u6T15 + s1u7T16 + u8T17

84

Hence, we have that

N3 = s8T9 + s7uT10 + s6u2T11 + s5u3T12 + s4u4T13

+ s3u5T14 + s2u6T15 + s1u7T16 + u8T17

.

And now we finally calculate N4.

(∆′γ)2 = (s · 2 + s2X3 + suX4)(s · 2 + s2X3 + suX4)

= s2 · 4 + s3Z25 + s2uZ26

(∆′σ)2 = (s · 2 + s2Y3 + suY4)(s · 2 + s2Y3 + suY4)

= s2 · 4 + s3W25 + s2uW26


(∆′γ)2∆2

σ = (s2 · 4 + s3Z25 + s2uZ26)


= s6 · 4 + s4u2 · 8 + s2u4 · 4

+ s7Z27 + s6uZ28 + s5u2Z29 + s4u3Z30 + s3u4Z31 + s2u5Z32

(∆′σ)2∆2γ = (s2 · 4 + s3W25 + s2uW26)


= s6 · 4 + s4u2 · 8 + s2u4 · 4

+ s7W27 + s6uW28 + s5u2W29 + s4u3W30 + s3u4W31 + s2u5W32

85

And so we then multiply by γ − λ and σ − λ, respectively.

(γ − λ)(∆′γ)2∆2

σ = (sB + s2C + s2L− u~ω)

·(s6 · 4 + s4u2 · 8 + s2u4 · 4

+ s7Z27 + s6uZ28 + s5u2Z29

+ s4u3Z30 + s3u4Z31 + s2u5Z32

)= s7(4B) + s5u2(8B) + s3u4(4B)

− s6(4u~ω)− s4u3(8u~ω)− s2u5(4u~ω)

+ s8V18 + s7uV19 + s6u2V20 + s5u3V21

+ s4u4V22 + s3u5V23 + s2u6V24

(σ − λ)(∆′σ)2∆2γ = (sB + s2C − u~ω)

·(s6 · 4 + s4u2 · 8 + s2u4 · 4

+ s7W27 + s6uW28 + s5u2W29

+ s4u3W30 + s3u4W31 + s2u5W32

)= s7(4B) + s5u2(8B) + s3u4(4B)

− s6(4u~ω)− s4u3(8u~ω)− s2u5(4u~ω)

+ s8U18 + s7uU19 + s6u2U20 + s5u3U21

+ s4u4U22 + s3u5U23 + s2u6U24

At last, we can take the difference of these two terms:

N4 = (γ − λ)(∆′γ)2∆2

σ − (σ − λ)(∆′σ)2∆2γ

86

=(s7(4B) + s5u2(8B) + s3u4(4B)

− s6(4u~ω)− s4u3(8u~ω)− s2u5(4u~ω)

+ s8V18 + s7uV19 + s6u2V20 + s5u3V21

+ s4u4V22 + s3u5V23 + s2u6V24

)−(s7(4B) + s5u2(8B) + s3u4(4B)

− s6(4u~ω)− s4u3(8u~ω)− s2u5(4u~ω)

+ s8U18 + s7uU19 + s6u2U20 + s5u3U21

+ s4u4U22 + s3u5U23 + s2u6U24

)= s8T18 + s7uT19 + s6u2T20 + s5u3T21 + s4u4T22 + s3u5T23 + s2u6T24

Hence, we have that

N4 = s8T18 + s7uT19 + s6u2T20 + s5u3T21 + s4u4T22 + s3u5T23 + s2u6T24 .

We can then combine N2, N3, and N4 to obtain the numerator of the second integrand

in Expression (3.9).

−N2 −N3 +N4 = −(s8T1 + s7uT2 + s6u2T3 + s5u3T4 + s4u4T5

+ s3u5T6 + s2u6T7 + s1u7T8

)−(s8T9 + s7uT10 + s6u2T11 + s5u3T12 + s4u4T13

+ s3u5T14 + s2u6T15 + s1u7T16 + u8T17

)+(s8T18 + s7uT19 + s6u2T20 + s5u3T21

+ s4u4T22 + s3u5T23 + s2u6T24

)

87

= s8T25 + s7uT26 + s6u2T27 + s5u3T28 + s4u4T29

+ s3u5T30 + s2u6T31 + s1u7T32 + u8T33

We still need to acquire an expression for D, which is the following calculation:

D = ∆2γ∆

2σ

=(s4 + s2u2 · 2 + u4 + s5X7 + s4uX8 + s3u2X9 + s2u3X10

)·(s4 + s2u2 · 2 + u4 + s5Y7 + s4uY8 + s3u2Y9 + s2u3Y10

)= s8 + s6u2 · 4 + s4u4 · 6 + s2u6 · 4 + u8

+ s9R1 + s8uR2 + s7u2R3 + s6u3R4 + s5u4R5 + s4u5R6 + s3u6R7 + s2u7R8

Hence, we have that

D = s8 + s6u2 · 4 + s4u4 · 6 + s2u6 · 4 + u8

+ s9R1 + s8uR2 + s7u2R3 + s6u3R4 + s5u4R5 + s4u5R6 + s3u6R7 + s2u7R8

.

We shall now proceed to determining a bound for | −N2 −N3 + N4|/D. As each of

the Ti and Ri are finite products and sums of bounded terms, we know there exists

some M ∈ R+ such that M ≥ |Ti| and M ≥ |Ri| for all i when t = 0. Hence, we have

that

| −N2 −N3 +N4| ≤ |s|8M + |s|7uM + |s|6u2M + |s|5u3M + |s|4u4M

+ |s|3u5M + |s|2u6M + |s|1u7M + u8M.

Our bounds of integration then tell us that |s| ≤ δ, but we also have that δ = u19 ,

88

which gives the following for sufficiently small u:

| −N2 −N3 +N4| ≤ (u19 )8M + (u

19 )7uM + (u

19 )6u2M + (u

19 )5u3M

+ (u19 )4u4M + (u

19 )3u5M + (u

19 )2u6M + (u

19 )1u7M + u8M

= u89M + u

169 M + u

83M + u

329 M

+ u409 M + u

163 M + u

569 M + u

649 M + u8M

≤ 9Mu89

Since |s| ≤ δ = u19 , for sufficiently small u we also have the following:

D ≥ (u19 )8 −

((u

19 )6u2M + (u

19 )4u4M + (u

19 )2u6M + u8

+ (u19 )9M + (u

19 )8uM + (u

19 )7u2M + (u

19 )6u3M

+ (u19 )5u4M + (u

19 )4u5M + (u

19 )3u6M + (u

19 )2u7R8

)= u

89 −

(u

83M + u

409 M + u

569 M + u8

+ uM + u179 M + u

259 M + u

113 M + u

419 M + u

499 M + u

193 M + u

659 M

)≥ u

89 − 12Mu

This allows us to bound the second integral in Expression (3.9). Also recalling that

δ = u19 , we can obtain

∫ δ

−δ

| −N2 −N3 +N4|D

ds ≤∫ δ

−δ

9Mu89

u89 − 12Mu

ds = 2δ9Mu

89

u89 − 12Mu

=18Mu

u89 − 12Mu

.

Since we can always take M larger, ensure that M ≥ |L′′|. Then we can bound the

first integral of Expression (3.9) via

∫ δ

−δ|N1| ds =

∫ δ

−δ|L′′| ds ≤

∫ δ

−δM ds ≤= 2δM = 2Mu

19 .

89

Combining these two integral bounds then gives us that


tot (Ir,λ(σ|[−δ,δ]))∣∣ ≤ 2Mu

19 +

18Mu

u89 − 12Mu

.

Note that this is exactly the bound specified in Inequality (3.10), so our claim is

proved. It then follows that there exists some d > 0 such that

2Mu19 +

18Mu

u89 − 12Mu

< ε

whenever 0 < u < d for t = 0. Hence, we have that


tot (Ir,λ(σ|[−δ,δ]))∣∣ < ε

whenever 0 < u < d for t = 0, as desired. This completes the proof for t = 0.

This proof can easily be extended to all t. First, notice that the bounds

developed above are only written in terms of M and u. As u is obviously independent

of t, it remains only to show that our choice of M is independent of our choice of t.

Recall that M was chosen so that M ≥ |Ti|, M ≥ |Ri|, and M ≥ |L′′|. Fortunately,

each of these terms is comprised of the variables s and u, as well as terms coming from

γ, specifically, A, B, C, L, L, and L′′. The s and u are on compact domains, so M

can be chosen independently of them. The curve quantities A, B, and C depend only

on t, but since the domain of our curve is compact and the curve is C2, we can choose

M independently of these as well. The functions L, L, and L′′ are all continuous on

the square [0, `]× [0, `], so then M may also be chosen independently of them, as well.

Thus, we can choose some M entirely independent of t, and our proof of the t = 0

case uniformly extends to the case for all t.

90

Lemma 3.26 combines Lemmas 3.23 and 3.25 to give us a result similar to

Lemma 3.23 for the general curve γ.



such that one has ∣∣κabstot (Ir,λt(u)(γ|[t−δ(u),t+δ(u)]))− 2π∣∣ < ε

whenever 0 0, let σt(s) be the parabolic approximation of γ at t, and let ηt(s)

be the osculating circle parametrized with respect to arclength at γ(t). (It will be

a line in the case that κabsγ (t) = 0.) Without loss of generality, say that t = 0, and

abbreviate σt, ηt, and λt as σ, η, and λ, respectively. Note that

γ(0) = σ(0) = η(0), γ′(0) = σ′(0) = η′(0), and γ′′(0) = σ′′(0) = η′′(0),

so we know that σ is also a parabolic approximation of η at γ(0). Now observe the

following, omitting all input variables:

∣∣κabstot (Ir,λ(γ|[−δ,+δ]))− 2π

∣∣ =∣∣∣κabs

tot (Ir,λ(γ|[−δ,+δ]))− κabstot (Ir,λ(σ|[−δ,+δ]))

+ κabstot (Ir,λ(σ|[−δ,+δ]))− κabs

tot (Ir,λ(η|[−δ,+δ]))

+ κabstot (Ir,λ(η|[−δ,+δ]))− 2π

∣∣∣≤∣∣κabs

tot (Ir,λ(γ|[−δ,+δ]))− κabstot (Ir,λ(σ|[−δ,+δ]))

∣∣+∣∣κabs

tot (Ir,λ(σ|[−δ,+δ]))− κabstot (Ir,λ(η|[−δ,+δ]))

∣∣+∣∣κabs

tot (Ir,λ(η|[−δ,+δ]))− 2π∣∣

91

Lemma 3.25 tells us that there exists d1, d2 > 0 such that the first and second terms

are less than ε3

for 0 0 such that the third term is less than ε3

for 0 < u < d3

for all t. By defining d = min{d1, d2, d3}, we then have that

∣∣κabstot (Ir,λ(γ|[−δ,+δ]))− 2π

∣∣ < ε

3+ε

3+ε

3= ε

whenever u such that 0 < u < d for all t, which is the desired result.

Notice that the proof of Lemma 3.26 and its supporting results will still hold

when defining δ(u) = cu19 , where c ∈ R+.

Figure 3.11 describes the approximation given by Lemma 3.27. This time, we

show that the total absolute curvature of the image of almost all of our curve under

inversion is approximated by the total absolute curvature of a broken inverted curve.



such that one has

∣∣κabstot (Ir,λt(0)(γ|[t+δ(u),`+t−δ(u)]))− κabstot (Ir,λt(u)(γ|[t+δ(u),`+t−δ(u)]))∣∣ < ε

whenever 0 0. We will first prove the case where t = 0, then show that the proof

for t = 0 can be extended to work for any t. That said, set t = 0, and abbreviate

λt as λ. Let ∆γ(s, u) = |γ(s) − λ(u)|2 and use the abbreviations ∆0 = ∆γ(s, 0) and

92

Figure 3.11: The segment of γ analyzed in Lemma 3.27.

∆u = ∆γ(s, u). We begin with the following, omitting all input variables and defining

Ni and D as shown below:

∣∣κabstot (Ir,λ(0)(γ|[δ,`−δ]))− κabs

tot (Ir,λ(u)(γ|[δ,`−δ]))∣∣

=

∣∣∣∣∣∫ `−δ

δ

∣∣∣∣γ′′ − γ′∆′0∆0

− (γ − λ(0))∆′′0∆0

+ (γ − λ(0))(∆′0)2

∆20

∣∣∣∣ ds−∫ `−δ

δ

∣∣∣∣γ′′ − γ′∆′u∆u

− (γ − λ(u))∆′′u∆u

+ (γ − λ(u))(∆′u)

2

∆2u

∣∣∣∣ ds∣∣∣∣∣

93

≤∫ `−δ

δ

| γ′′ − γ′′︸︷︷︸N1

| ds

+

∫ `−δ

δ

[∣∣∣− (γ′∆′0∆0∆2u − γ′∆′u∆u∆

20

)︸︷︷︸N2

−((γ − λ(0))∆′′0∆0∆2

u − (γ − λ(u))∆′′u∆u∆20

)︸︷︷︸N3

+((γ − λ(0))(∆′0)2∆2

u − (γ − λ(u))(∆′u)2∆2

0

)︸︷︷︸N4

∣∣∣/ (∆20∆2

u)︸︷︷︸D

]ds

(3.11)

Obviously we have that

N1 = γ′′ − γ′′ = 0.

We shall use a representation of γ in terms of its parabolic approximation at t (recall

that we assumed without loss of generality t = 0.) That is, we write γ as

γ(s) = A+Bs+ Cs2 + L,

where A = γ(0), B = γ′(0), and C = γ(0)2

. Note that L will be the same L given

in Lemma 3.24. Additionally, note that our definition of A means that we can also

write λ(u) = A+ u~ω. Thus, we have that

γ(s)− λ(0) = Bs+ Cs2 + L and γ(s)− λ(u) = Bs+ Cs2 + L− u~ω.

We can also compute

γ′(s) = B + 2Cs+ L′ and γ′′(s) = 2C + L′′.


and L′

sare continuously extendable to [0, `]. Call

these extensions L and L, respectively, and note that then s2L = L and sL = L′. We

94

then rewrite γ and its derivatives as follows:

γ(s)− λ(0) = sB + s2C + s2L γ′(s) = B + s2C + sL

γ(s)− λ(u) = sB + s2C + s2L− u~ω γ′′(s) = 2C + L′′

Now recall that

B • C = γ′(0) • γ′′(0) = 0 and B •B = γ′(0) • γ′(0) = 1,

since γ was parametrized by arclength. Also, remember that

B • ~ω = γ′(0) • ~ω = 0 and ~ω • ~ω = 1

by our choice of ~ω. Using all of this in conjunction with the above expansions allows

us to make the following calculations, as in Lemma 3.25.

∆u = (γ − λ(u)) • (γ − λ(u))

= s2 + u2 + s3(2B • L) + s4(C • C) + s4(2C • L) + s4(L • L)

− s2u(2C • ~ω)− s2u(2L • ~ω)

(3.12)

∆′u = 2γ′ • (γ − λ(u))

= s · 2 + s2(2B • L) + s2(2B • L) + s3(4C • C)

+ s3(2C • L) + s3(2L • L) + s3(4C • L)

− su(4C • ~ω)− su(2L • ~ω)

(3.13)

∆′′u = 2γ′ • γ′ + 2γ′′ • (γ − λ(u))

= 2 + s(2B • L′′) + s2(4C • C) + s2(4C • L) + s2(2C • L′′) + s2(2L • L′′)

− u(4C • ~ω)− u(2L′′ • ~ω)

(3.14)

95

We can obtain ∆0 and its derivatives by setting u = 0 in the above calculations, as

follows:

∆0 = (γ − λ(0)) • (γ − λ(0))

= s2 + s3(2B • L) + s4(C • C) + s4(2C • L) + s4(L • L) (3.15)

∆′0 = 2γ′ • (γ − λ(0))

= s · 2 + s2(2B • L) + s2(2B • L) + s3(4C • C)

+ s3(2C • L) + s3(2L • L) + s3(4C • L)

(3.16)

∆′′0 = 2γ′ • γ′ + 2γ′′ • (γ − λ(0))

= 2 + s(2B • L′′) + s2(4C • C) + s2(4C • L) + s2(2C • L′′) + s2(2L • L′′) (3.17)

In order to simplify our calculations, we will use the following massively con-

solidated versions of each of the ∆-terms (Expressions (3.12) to (3.17)), with the

X, Y , Z, P , Q, and R being defined appropriately. Note that although consolida-

tion variables are not constant, they will all be bounded in s and u for t = 0. The

consolidated representations are written as

∆0 = X, ∆u = X+uP, ∆′0 = Y, ∆′u = Y +uQ, ∆′′0 = Z, and ∆′′u = Z+uR.

We now begin the calculation of the cross terms in Expression (3.11). We will define

Ti (notice that these are different than those used in the proof of Lemma 3.25) as we

proceed to further consolidate our expressions. These Ti will all be bounded in s and

t for t = 0. First, the cross coefficients:

∆′0∆0∆2u = XY (X + uP )2 = X3Y + uT1

96

∆′u∆u∆20 = (X + uP )(Y + uQ)X2 = X3Y + uT2

∆′′0∆0∆2u = ZX(X + uP )2 = X3Z + uT3

∆′′u∆u∆20 = (Z + uR)(X + uP )X2 = X3Z + uT4

(∆′0)2∆2u = Y 2(X + uP )2 = X2Y 2 + uT5

(∆′u)2∆2

0 = (Y + uQ)2X2 = X2Y 2 + uT6

Note that γ − λ(0) = γ − A and γ − λ(u) = γ − A− u~ω. We can then calculate the

following:

N2 = γ′∆′0∆0∆2u − γ′∆′u∆u∆

20

= γ′(X3Y + uT1)− γ′(X3Y + uT2) = uT7

N3 = (γ − λ(0))∆′′0∆0∆2u − (γ − λ(u))∆′′u∆u∆

20

= (γ − A)(X3Z + uT3)− (γ − A− u~ω)(X3Z + uT4) = uT8

N4 = (γ − λ(0))(∆′0)2∆2u − (γ − λ(u))(∆′u)

2∆20

= (γ − A)(X2Y 2 + uT5)− (γ − A− u~ω)(X2Y 2 + uT6) = uT9

As a result, we may write

−N2 −N3 +N4 = −uT7 − uT8 + uT9 = uT10.

Since T10 is a sum and product of terms bounded in s and u for t = 0, we know that

there exists some M ∈ R+ such that |T10| ≤ M for all s and u when t = 0. Hence,

we have that

| −N2 −N3 +N4| ≤ uM.

We now address the last term in Expression (3.11), D.

97

Claim 3.27.1. Let κm be the maximum curvature of γ, which we know exists, since

γ is C2 and has a compact domain. If |s| ≤ 1κm

, we have that

|D| ≥ u89

256.

Proof of Claim 3.27.1: As γ is closed, recall that γ(`−s) = γ(−s). (See Figure 3.11.)

In order to complete this proof we split |s| ≤ 1κm

into the cases −1κm≤ s ≤ 0 and

0 ≤ s ≤ 1κm

. As the proofs are similar, we will only show the latter case.

Let |s| ≤ 1κm

, and define P = (γ′(0))⊥, the normal hyperplane at γ(0). Recall

that γ is parametrized by arclength and define f :[−1κm, 1κm

]→ R by f(s) = (γ(s) −

γ(0)) • γ′(0). Observe that

dist(γ(s), P ) =∣∣projγ′(0)(γ(s)− γ(0))

∣∣=

∣∣∣∣(γ(s)− γ(0)) • γ′(0)

γ′(0) • γ′(0)γ′(0)

∣∣∣∣= |(γ(s)− γ(0)) • γ′(0)| = |f(s)|.

We can then observe that

f ′(s) = γ′(s) • γ′(0) and f ′′(s) = γ′′(s) • γ′(0).

Hence, we have the following:

f(0) = (γ(0)− γ(0)) • γ′(0) = 0 (3.18)

f ′(0) = γ′(0) • γ′(0) = 1 (3.19)

The definition of f and Cauchy-Schwarz imply

|f ′′(s)| = |γ′′(s) • γ′(0)| ≤ |γ′′(s)||γ′(0)| = |γ′′(s)| ≤ κm =⇒ f ′′(s) ≥ −κm.

98

Integrating this inequality and applying Equation (3.19) then gives

f ′(s) ≥ 1− κms.

We then integrate this inequality and apply Equation (3.18) to obtain

f(s) ≥ s− κm2s2. (3.20)

For 0 ≤ s ≤ 1κm

we have

1

κm≥ s =⇒ 1

2≥ κm

2s =⇒ 1− κm

2s ≥ 1

2=⇒ s− κm

2s2 ≥ s

2.

Substitution of this inequality into Inequality (3.20) implies

f(s) ≥ s

2

for 0 ≤ s ≤ 1κm

. With a similar proof, one can conclude that |f(s)| ≥ |s|2

for −1κm≤

s ≤ 1κm

. As |f(s)| = dist(γ(s), P ) and λ(0), λ(u) ∈ P , we may then conclude that

|γ(s)− λ(0)| ≥ |s|2

and |γ(s)− λ(u)| ≥ |s|2.

Recall the definition of D, which allows us to write

|D| = ∆20∆2

u = |γ(s)− λ(0)|4|γ(s)− λ(u)|4 ≥(s

2

)4 (s2

)4

=s8

256.

As |s| ≥ δ(u) = u19 , we may conclude that |D| ≥ u

89

256, as desired. �

We shall use Claim 3.27.1 to prove a second claim.

Claim 3.27.2. There exists sufficiently small u so that

|D| ≥ u89

256.

99

Proof of Claim 3.27.2: We have already shown this for |s| ≤ 1κm

in Claim 3.27.1. As

such, we may assume that |s| ≥ 1κm

.

Since γ is compact and C2 and |s| ≥ 1κm

, there exists some K1 ∈ R+ such that

|γ(s)− λ(0)| = |γ(s)− γ(0)| ≥ K1.

Now choose u small enough so that λ(u) is contained the normal neighborhood

of γ guaranteed to exist by Proposition 3.8. Again, since γ is compact and C2 and

|s| ≥ 1κm

, there exists some K2 ∈ R+ such that |γ(s)− λ(u)| ≥ K2. We can then use

the definition of D to write

|D| = ∆20∆2

u = |γ(s)− λ(0)|4|γ(s)− λ(u)|4 ≥ K41K

42 .

Hence, we simply choose u to be small enough so that K41K

42 ≥ u

89

256to obtain the

desired result. �

Having proved Claim 3.27.2, we now have for sufficiently small u that∫ `−δ

δ

|N1| ds+

∫ `−δ

δ

| −N2 −N3 +N4||D|

ds ≤∫ `−δ

δ

0 ds+

∫ `−δ

δ

256uM

u89

≤ 256`Mu19 .

It then follows that there exists some d > 0 such that

256`Mu19 < ε

whenever 0 < u < d for t = 0. Thus, we have that

∣∣κabstot (Ir,λ(0)(γ|[δ,`+δ]))− κabs

tot (Ir,λ(u)(γ|[δ,`+δ]))∣∣ < ε

whenever 0 < u < d for t = 0, as desired. This completes the proof for t = 0.

This proof can easily be extended to all t. First, notice that the bounds

developed above are only written in terms of M and u. As u is obviously independent

100

of t, it remains only to show that our choice of M is independent of our choice of

t. Recall that M was chosen so that M ≥ |Ti| for i = 4, 5, 15. Fortunately, each of

these terms is comprised of the variables s and u, as well as terms coming from γ,

specifically, A, B, C, L, L, and L′′. The s and u are on compact domains, so M can

be chosen independently of them. The curve quantities A, B, and C depend only on

t, but since the domain of our curve is compact and the curve is C2, we can choose

M independently of these as well. The functions L, L, and L′′ are all continuous on

the square [0, `] × [0, `], so M may also be chosen independently of them, as well.

Thus, we can choose some M entirely independent of t, so our proof of the t = 0 case

uniformly extends to the case for all t.

The combination of Lemmas 3.26 and 3.27 is enough for us to finally prove

Proposition 3.28. It can be helpful to keep Figure 3.12 in mind.

Proposition 3.28. Let γ : [0, `]→ Rn be a closed C2 curve parametrized by arclength.

For all ε > 0 there exists d > 0 for all t ∈ [0, `] and ~ωt satisfying |~ωt| = 1 and ~ωt ⊥ γ′(t)

such that one has

∣∣κabstot (Ir,λt(0)(γ)) + 2π − κabstot (Ir,λt(u)(γ))∣∣ < ε,

whenever 0 0 and define δ : R→ R+ by δ(u) = u19 . As total curvature is additive,

we know that

κabstot (Ir,λt(u)(γ)) = κabs

tot (Ir,λt(u)(γ|[t+δ,`+t−δ])) + κabstot (Ir,λt(u)(γ|[t−δ,t+δ])).

102

≤ |κabstot (Ir,λt(0)(γ))− κabs

tot (Ir,λt(0)(γ|[t+δ,`+t−δ]))|

+ |κabstot (Ir,λt(0)(γ|[t+δ,`+t−δ]))− κabs

tot (Ir,λt(u)(γ|[t+δ,`+t−δ]))|

+ |κabstot (Ir,λt(u)(γ|[t−δ,t+δ]))− 2π|

(3.21)

Lemma 3.12 and Corollary 3.17 tell us that the integrand of κabstot (Ir,λt(0)(γ)) will be

bounded, so there exists some d1 > 0 such that

|κabstot (Ir,λt(0)(γ))− κabs

tot (Ir,λt(0)(γ|[t+δ,`+t−δ]))| <ε

3

when 0 0 such

that

|κabstot (Ir,λt(0)(γ|[t+δ,`+t−δ]))− κabs

tot (Ir,λt(u)(γ|[t+δ,`+t−δ]))| <ε

3

when 0 0 such

that

|κabstot (Ir,λt(u)(γ|[t−δ,t+δ]))− 2π| < ε

3

when 0 < u < d3 for all t. Set d = min{d1, d2, d3}. Then the substitution of these

three inequalities into Expression (3.21) yields


tot (Ir,λt(u)(γ))| < ε

3+ε

3+ε

3= ε

when 0 < u < d for all t, which is the desired result.

Remark. Proposition 3.28 can then quickly lead us to a pair of corollaries and a pair

of theorems, one of them being Theorem 3.33.

Corollary 3.29 is the first of several results appearing as the fruit of Proposi-

tion 3.28. This particular result provides a bound for the total absolute curvature of

103

an inverted image of a curve, provided the inversion center is in a sufficiently small

neighborhood of the original curve.

Corollary 3.29. Let γ : [0, `]→ Rn be a closed C2 curve parametrized by arclength,

and define

M1 = maxt∈[0,`]

κabstot (Ir,γ(t)(γ)).

Then for every ε > 0 there exists some d ∈ R+ such that for all r ∈ R+ and all c ∈ Rn

such that dist(c, γ) < d,

0 ≤ κabstot (Ir,c(γ)) ≤M1 + 2π + ε.

Proof. Notice that we immediately have 0 ≤ κabstot (Ir,c(γ)), as the total absolute cur-

vature cannot be negative. Further, recall from Corollary 3.6 that κabstot (Ir,c(γ)) will

not depend on r.

Let ε > 0. Note that M1 as defined above is guaranteed to exist by the fact

that [0, `] is compact and the integrand giving κabstot (Ir,γ(t)(γ)) is bounded on [0, `]×[0, `]

by Lemma 3.12.

We will have two cases.

Case 1: c ∈ γ.

If c ∈ γ, then we immediately have that κabstot (Ir,c(γ)) ≤ M1, so this case is

complete.

Case 2: c /∈ γ.

Since γ is a C2 curve with a compact domain, Proposition 3.8 tells us that

there exists some d1 > 0 such that Parts (i) to (iii) of Proposition 3.8 are satisfied

104

when 0 < dist(c, γ) < d1. Since d1 is being chosen so that Proposition 3.8 holds,

we know that we may define λt(u) = γ(t) + u~ωt, where |~ωt| = 1 and ~ωt ⊥ γ′(t).

Additionally, setting uc = dist(c, γ) implies λt(uc) = c; note also that λt(0) = γ(t).

It also follows from Proposition 3.28 that there exists some d2 > 0 such that

∣∣κabstot (Ir,λt(0)(γ)) + 2π − κabs

tot (Ir,λt(u)(γ))∣∣ < ε

when 0 < u < d2 for all t.

Set d = min{d1, d2}, let c be such that 0 < dist(c, γ) < d, and notice that

0 < uc < d. Then we have the following for all t:

∣∣κabstot (Ir,λt(0)(γ)) + 2π − κabs

tot (Ir,λt(uc)(γ))∣∣ < ε From Proposition 3.28

=⇒∣∣κabs

tot (Ir,γ(t))(γ)) + 2π − κabstot (Ir,c(γ))

∣∣ < ε Use λt(0) = γ(t), λt(uc) = c

=⇒∣∣κabs

tot (Ir,c(γ))− (κabstot (Ir,γ(t))(γ)) + 2π)

∣∣ < ε

=⇒ κabstot (Ir,c(γ))− (κabs

tot (Ir,γ(t)(γ)) + 2π) < ε

=⇒ κabstot (Ir,c(γ)) < κabs

tot (Ir,γ(t)(γ)) + 2π + ε

It then follows from the definition of M1 that

κabstot (Ir,c(γ)) ≤M1 + 2π + ε,

when 0 < dist(c, γ) < d for all t, as desired.

The bound developed in Corollary 3.29 can then be united with the bound for

inversion center in a neighborhood of infinity (Proposition 3.7) to obtain a bound for

inversion centers anywhere in Rn.

105

Theorem 3.30. Let γ : [0, `] → Rn be a closed C2 curve parametrized by arclength.

Then there exists some M0 ∈ R+ such that for all r, c we have κabstot (Ir,c(γ)) ≤M0.

Proof. Recall from Proposition 3.7 that lim|c|→∞ κabstot (Ir,c(γ)) = κabs

tot (γ) for any r ∈

R+. Hence, we have that there exists some R ∈ R+ such that

κabstot (Ir,c(γ)) ≤ κabs

tot (γ) + 1

for all c ∈ X, where X = {c ∈ Rn | |c| > R}.

Corollary 3.29 tells us that there exists some d > 0 and M1 such that

κabstot (Ir,c(γ)) ≤M1

when c ∈ N , where Y = {c ∈ Rn | dist(c, γ) < d}.

Note that the set Z = Rn \ (X ∪Y ) is a compact set upon which the integrand

of κabstot (Ir,c(γ)) is continuous. Hence, there exists some M2 such that

κabstot (Ir,c(γ)) ≤M2

when c ∈ Z.

Then the desired M0 is defined as M0 = max{κabstot (γ) + 1,M1,M2}, which


We will now finally work toward addressing the connection between the total

absolute curvature and Mobius energy. We begin by proving an equality about sets

of total absolute curvatures and affine similarites.

106

Lemma 3.31. Let γ be a C2 curve in Rn and let A be an affine similarity of Rn.

Then

{κabstot (Ir,c(γ)) | r ∈ R+, c ∈ Rn} = {κabstot (Ir,c(A(γ))) | r ∈ R+, c ∈ Rn}.

Proof. Recall that any affine similarity A of Rn can be written as A(x) = aBx + y,

where a ∈ R+, y ∈ Rn and B is an orthogonal linear transformation of Rn. Now

observe the following:

Iar,A(c)(A(x)) = A(c) +(ar)2(A(x)− A(c))

|A(x)− A(c)|2

= aBc+ y +a2r2(aBx+ y − aBc− y)

|aBx+ y − aBc− y|2

= aBc+a3r2B(x− c)a2|B(x− c)|2

+ y

= aBc+aBr2(x− c)|x− c|2

+ y

= aB

(c+

r2(x− c)|x− c|2

)+ y

= A(Ir,c(x))

Hence, Iar,A(c)(A(γ)) = A(Ir,c(γ)), implying κabstot (Iar,A(c)(A(γ))) = κabs

tot (A(Ir,c(γ))). We

may then combine this with the fact that A is in affine similarity and Proposition 3.4

to obtain

κabstot (Iar,A(c)(A(γ))) = κabs

tot (A(Ir,c(γ))) = κabstot (Ir,c(γ)).

This implies that

{κabstot (Iar,A(c)(A(γ))) | r ∈ R+, c ∈ Rn} = {κabs

tot (Ir,c(γ)) | r ∈ R+, c ∈ Rn}. (3.22)

107

Recognize that if a ∈ R+, we have {ar | r ∈ R+} = {r | r ∈ R+}. Additionally, since

A is a bijection, we have that {A(c) | c ∈ Rn} = {c | c ∈ Rn}. As a result, we may

say that

{κabstot (Iar,A(c)(A(γ))) | r ∈ R+, c ∈ Rn} = {κabs

tot (Ir,c(A(γ))) | r ∈ R+, c ∈ Rn}.

We then combine this with Equation (3.22) to obtain

{κabstot (Ir,c(γ)) | r ∈ R+, c ∈ Rn} = {κabs

tot (Ir,c(A(γ))) | r ∈ R+, c ∈ Rn},

as desired.

The previous lemma then allows us to prove that the set of total absolute

curvatures of the family of Mobius-equivalent curves is uniformly bounded, seen in

Proposition 3.32.

Proposition 3.32. Let γ : [0, `] → Rn be a closed C2 curve parametrized by ar-

clength. Then there exists some M ∈ R+ such that κabstot (T (γ)) ≤ M0 for all Mobius

transformations T .

Proof. Recall that any Mobius transformation T can be decomposed into a sequence of

at most one each of translations, rotations, reflections, dilations, and inversions. Any

Mobius transformation generated without an inversion is simply an affine similarity

and has no effect on total absolute curvature (making the desired result immediate).

Hence, we assume that the decomposition of T contains an inversion. We can then

write T as T = A2 ◦ I ◦ A1, where A2 and A1 are (possibly trivial) affine similarities

and I is an inversion.

108

Let C and A denote the sets of Mobius transformations and affine similarities,

respectively, and observe the following:

{κabstot (T (γ)) | T ∈ C }

= {κabstot (A2(Ir,c(A1(γ)))) | r ∈ R+; c ∈ Rn;A1, A2 ∈ A } Shown above

= {κabstot (Ir,c(A1(γ))) | r ∈ R+, c ∈ Rn, A1 ∈ A } Proposition 3.4

=⋃

A1∈A

{κabstot (Ir,c(A1(γ))) | r ∈ R+, c ∈ Rn} Equivalent definition

=⋃

A1∈A

{κabstot (Ir,c(γ)) | r ∈ R+, c ∈ Rn} Lemma 3.31

= {κabstot (Ir,c(γ)) | r ∈ R+, c ∈ Rn} Simplify

It follows from Theorem 3.30 that there exists some M0 such that κabstot (Ir,c(γ)) ≤M0

for all r ∈ R+ and c ∈ Rn. Hence we have for any Mobius transformation T that

κabstot (T (γ)) ≤M0,

as desired.

At long last, we can finally prove Theorem 3.33. This theorem tells us the the

total curvature as a function of inversion center is only removably discontinuous, and

we can remove the discontinuity in a straightforward way.



Φ(r, c) =

κabstot (Ir,c(γ)) + 2π c ∈ γ([0, `])


.

109


Proof. It immediately follows from Theorem 3.30 that Φ will be bounded, so it only

remains to show that Φ is continuous.

We know from Corollary 3.18 that Φ(r, c) is continuous at every (r, c) ∈ R+×

(Rn \ γ). Since Φ is constant in r, it only remains to show that Φ is continuous at

each c0 ∈ γ.

Let ε > 0 and t0 be such that c0 = γ(t0). Choose d1 > 0 so that it satisfies the

following conditions, (i)-(iii):

(i) |κabstot (Ir,γ(t0)(γ)) − κabs

tot (Ir,γ(t)(γ))| < ε2

for |t0 − t| < d1, possible by Proposi-

tion 3.14.

(ii) Proposition 3.8 is satisfied for d1.

Since d1 is being chosen so that Proposition 3.8 holds, we know that we can define

λt(u) = γ(t) + u~ωt, where |~ωt| = 1 and ~ωt ⊥ γ′(t). This allows us to state a final

condition:

(iii)∣∣κabs

tot (Ir,λt(0)(γ)) + 2π − κabstot (Ir,λt(u)(γ))

∣∣ < ε2

when 0 < u < d1 for all t, which is

possible via Proposition 3.28.

Now define the open set

W = {γ(t) + u~ωt | t ∈ (t0 − d1, t0 + d1), ~ωt ⊥ γ′(t), |~ωt| = 1, 0 ≤ u < d1}.

Since W is open, there exists some d ∈ R+ such that 0 < d < d1 and Bnd (c0) ⊆ W . It

may be helpful here to refer to Figure 3.13.

Now say c ∈ Bnd (c0) ⊆ W . The definition of W and Part (i) of Proposition 3.8

110

Figure 3.13: An illustration of the construction of W and Bnd (c0). Note that the

particular c shown in the diagram corresponds with case 2.

tell us that if t ∈ [0, `] is such that |c − γ(t)| = dist(c, γ), then t ∈ (t0 − d1, t0 + d1).

Consequently, c = γ(t) + u~ωt = λt(u) for some t ∈ (t0 − d1, t0 + d1) and ~ωt ⊥ γ′(t)

with |~ωt| = 1, and

0 ≤ u = dist(c, γ) ≤ dist(c, c0) < d < d1.

(Hence, all of Condition (i), Condition (ii) for all c satisfying |c − c0| < d, and

112

Condition (i) tells us that

|κabstot (Ir,γ(t0)(γ))− κabs

tot (Ir,γ(t)(γ))| < ε

2.

Since uc = dist(c, γ), we have that uc < d. We also know that c /∈ γ, so we have that

uc > 0, which gives 0 < uc < d. Hence, Condition (iii) tells us that


tot (Ir,λt(uc)(γ))| < ε

2.

We can then substitute both of these inequalities into Expression (3.23) to obtain

|κabstot (Ir,c0(γ)) + 2π − κabs

tot (Ir,c(γ))| < ε

2+ε

2= ε.

This concludes the second case.

As both cases hold, we have that Φ is continuous at every c0 ∈ γ, so Φ is

continuous.

We now move to an example of Theorem 3.33.

Example 3.1. While somewhat trivial, the circle provides a simple illustration of

Theorem 3.33. Let γ : [0, `] → Rn be a circle parametrized by arclength. Then we

actually have that Φ(r, c) ≡ 2π, so Φ is obviously continuous and bounded. We can

see this by considering two cases.

Case 1: c ∈ γ.

Then we know that Ir,c(γ) will be a line, so κabstot (Ir,c(γ)) = 0. As c ∈ γ, it

follows from the definition of Φ that Φ(r, c) = 0 + 2π = 2π.

Case 2: c /∈ γ.

113

Then we know that Ir,c(γ) will be a circle, so κabstot (Ir,c(γ)) = 2π. As c /∈ γ, it

follows from the definition of Φ that Φ(r, c) = 2π.

3.2.3 Analysis of Inversions of Open, Piecewise C2 Curves

Thus far, our discussion has focused primarily on closed C2 curves. If we

generalize to the case where our curve is possibly open and only piecewise C2, we lose

some of our strongest results from previous sections. In particular, we no longer have

continuity of the function Φ, as described in Theorem 3.33. However, we can still

show that the total absolute curvature of our inverted curve is uniformly bounded

over all possible choices of inversion center. This result is the subject of Theorem 3.35,

which is a generalization of Theorem 3.30.

We first provide two definitions.

Definition 3.3. γ : [0, `]→ Rn is a piecewise C2 curve if there exists a set {t0, . . . , tk}

such that the following hold:

(i) 0 = t0 < t1 < · · · < tk = `.

(ii) γi = γ|[ti−1,ti] : [ti−1, ti] → Rn is C2 on [ti−1, ti]. Notice that this requires

κabsγi

(t) ≤ ci <∞ for t ∈ [ti−1, ti] and ci ∈ R+.

(iii) ∠(γ′i(ti), γ′i+1(ti)) is well defined for all i. Notice that if the curve is closed that

∠(γ′k(`), γ′1(0)) must also be well defined.

Now that piecewise C2 curves have been defined, we must formally describe

how to calculate their total absolute curvature.

Definition 3.4. For a piecewise C2 curve γ with k segments, its total absolute cur-

114

vature is given according to the formula

κabstot (γ) =

k∑i=1

κabstot (γi) +

m∑i=1

∠(γ′i(ti), γ′i+1(ti)),

where m = k if γ is closed, and m = k − 1 otherwise.

With these two definitions in hand, we move to a generalization of Proposi-

tion 3.7.

Proposition 3.34. Let γ : [0, `] → Rn be a piecewise C2 curve parametrized by

arclength. Then

lim|c|→∞

κabstot (Ir,c(γ)) = κabstot (γ)

for any choice of r.

Proof. Let γi be the C2 segments of γ, and note that Proposition 3.7 tells us that

lim|c|→∞

κabstot (Ir,c(γi)) = κabs

tot (γi).

(It may also be helpful to observe that Proposition 3.7 does not require the curve be

closed.) As inversions are conformal transformations, we also know for all i that

∠(Ir,c(γ′i(ti)), Ir,c(γ

′i+1(ti))) = ∠(γ′i(ti), γ

′i+1(ti)).

(The previous statement is only true when c 6= γ(ti) for all i, so simply force |c| to

be sufficiently large.) Let m = k if γ is closed, and m = k − 1 otherwise. Since total

curvature is additive, it then follows that

lim|c|→∞

κabstot (Ir,c(γ)) = lim

|c|→∞

n∑i=1

κabstot (Ir,c(γi)) +

m∑i=1


′i+1(ti)))

=n∑i=1

κabstot (γi) +

m∑i=1

∠(γ′i(ti), γ′i+1(ti)) = κabs

tot (γ),


115

Next, we move to a generalization of Theorem 3.30, which tells us that the

image under inversion of any piecewise C2 curve will have bounded total absolute

curvature. Notice that this also allows the possibility that γ is open, which was not

true in Theorem 3.30.

Theorem 3.35. Let γ : [0, `]→ Rn be a piecewise C2 curve parametrized by arclength.

Then there exists some M ∈ R+ such that for all r ∈ R+ and c ∈ Rn we have

κabstot (Ir,c(γ)) < M .

Proof. Let γ1, . . . , γk be the C2 segments of γ. (If we have the special case in which γ

is closed and only fails to be C2 at one point t0, split γ into two segments by selecting

some other point t1 6= t0 to use as a second break point.)

Since γ is a piecewise C2 curve, we know from Definition 3.3 that each γi has

bounded pointwise curvature on [ti−1, ti], so we can construct some simple closed C2

curve Γ : [0, Li]→ Rn for each i such that Γi(t) = γi(t) for t ∈ [ti−1, ti].

Theorem 3.30 implies that there exists some Mi such that κabstot (Ir,c(Γi)) ≤ Mi

for all r ∈ R+ and c ∈ Rn. As total absolute curvature is additive, this implies

κabstot (Ir,c(γi)) ≤Mi for all r ∈ R+ and c ∈ Rn.

As inversions are conformal transformations, we know for all r ∈ R+ and

c ∈ Rn that


′i+1(ti))) ≤ ∠(γ′i(ti), γ

′i+1(ti)),

with equality holding except when the inversion center c satisfies c = γ(ti). (In that

case we have that ∠(Ir,c(γ′i(ti)), Ir,c(γ

′i+1(ti))) = 0, as the angle is eliminated when the

curve is broken by the inversion.)

116

Let m = k if γ is closed, and m = k − 1 otherwise. It then follows for all

r ∈ R+ and c ∈ Rn that

κabstot (Ir,c(γ)) =

k∑i=1

κabstot (Ir,c(γi)) +

m∑i=1


′i+1(ti)))

≤k∑i=1

Mi +m∑i=1

∠(γ′i(ti), γ′i+1(ti)).

Hence, choosing

M =k∑i=1

Mi +m∑i=1

∠(γ′i(ti), γ′i+1(ti))


117

CHAPTER 4THE NONUNIFORM ENERGY

4.1 Definition of the Nonuniform Energy

The Mobius energy has both intrinsic value and value to applied problems.

That being said, we believe it appropriate to define a generalization of the Mobius

energy compatible with what we will call “weighted curves.” This concept is described

formally in Definition 4.1 and illustrated in Figure 4.1. As this definition is dependent

on the lengths of subarcs of our curves, we will always require that our curves be

rectifiable—hence, we shall assume that γ is rectifiable in Chapters 4 and 5 unless

otherwise stated. In addition, we shall continue to assume that γ is simple or simple

closed unless otherwise stated.

Definition 4.1. Let γ : [0, `] → Rn be a curve parametrized by arclength and

µ : [0, `] → R+ be a continuous function. Then the pair (γ, µ) is called a weighted

curve and µ is called a weight function, see Figure 4.1.

Remark. Recall that any closed curve γ can be extended to an `-periodic function

γ : R→ Rn, as described in Section 1.2. Similarly, µ can be extended to µ : R→ R+,

where ` is the least period.

Definition 4.2. For a simple C0 rectifiable curve γ : [0, `] → Rn parametrized by

arclength and weight function µ, we define a functional E(γ, µ) called the nonuniform

118

Figure 4.1: Examples of weighted curves.

energy of γ with respect to µ via

E(γ, µ) =

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(s)µ(t) ds dt. (4.1)

Remark. Note that choosing µ(t) ≡ 1 makes E(γ, µ) = E(γ), the standard Mobius

energy originally given by O’Hara. Also observe that E(γ, µ) is invariant if the ori-

entation is reversed via γ(s) = γ(`− s) and µ(s) = µ(`− s).

4.2 Properties of the Nonuniform Energy

We now extend the definition of E(γ, µ) to arbitrary regular parametrizations

of C1 curves. Specifically, for a C1 curve γ : [0, `]→ Rn satisfying |γ′(t)| > 0 for all t

and µ : [0, `]→ R+, we have

E(γ, µ) =

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(s)µ(t)|γ′(s)||γ′(t)| ds dt. (4.2)

This will be shown to be consistent with Definition 4.2 as a consequence of Prop-

erty (iii) of Lemma 4.1.

119

Although Equation (4.2) is defined for piecewise C1 curves, we will see in

Corollary 4.4 that if γ is not entirely C1 then it has infinite nonuniform energy.

We now describe some elementary properties of the nonuniform energy.

Lemma 4.1. The following hold for any regular closed curve γ and weight function

µ:

(i) E(γ, µ) is invariant under isometries of γ.

(ii) E(γ, µ) is invariant under rescaling of γ.

(iii) E(γ, µ) is invariant under reparametrizations (γ ◦ φ, µ ◦ φ) of (γ, µ) for any

φ : [0, `]→ [0, `], where φ is bijective and |φ′(t)| > 0 for all t.

Proof. Note that the distances (both Euclidean and intrinsic) between points on γ

will be unchanged under isometries. In addition, the lengths of tangent vectors of γ

will be unchanged under isometries. Hence, Property (i) holds.

Let c ∈ R+. Then we may observe that

E(cγ, µ) =

∫ `

0

∫ `

0

[1

|cγ(s)− cγ(t)|2− 1

Dcγ(s, t)2

]µ(s)µ(t)|cγ′(s)||cγ′(t)| ds dt

=

∫ `

0

∫ `

0

[1

c2|γ(s)− γ(t)|2− 1

c2Dγ(s, t)2

]µ(s)µ(t)|γ′(s)||γ′(t)|c2 ds dt

=

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(s)µ(t)|γ′(s)||γ′(t)| ds dt

= E(γ, µ).

Hence, Property (ii) holds.

With respect to Property (iii): When reparametrizing γ, we must also corre-

spondingly reparametrize µ. Let φ be a bijective function mapping the domain of γ

120

to itself with |φ′(x)| > 0 for all x; note then that either φ(0) = 0 and φ(`) = ` (the

case φ′(x) > 0 for all x) or φ(0) = ` and φ(`) = 0 (the case φ′(x) < 0 for all x).

Without loss of generality, we assume that φ′(x) > 0 for all x. γ ◦ φ will then be a

reparametrization of γ, and µ ◦ φ will have the same weights along the points of the

curve in Rn. Then we may use a change of variables by replacing φ(x) and φ(y) with

s and t respectively to observe that

E(γ ◦ φ, µ ◦ φ) =

∫ `

0

∫ `

0

[1

|γ(φ(x))− γ(φ(y))|2− 1

Dγ◦φ(x, y)2

]· µ(φ(x))µ(φ(y))|(γ(φ(x)))′||(γ(φ(y)))′| dx dy

=

∫ `

0

∫ `

0

[1

|γ(φ(x))− γ(φ(y))|2− 1

Dγ(φ(x), φ(y))2

]· µ(φ(x))µ(φ(y))|γ′(φ(x))||γ′(φ(y))|φ′(x)dx φ′(y)dy

=

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(s)µ(t)|γ′(s)||γ′(t)| ds dt

= E(γ, µ).

Note that the proof is similar in the case that φ′(x) < 0 for all x. Hence, Property (iii)

holds in the sense described above.

We now describe the relationship between E and E. Let γ : [0, `] → Rn be a

C1 curve satisfying |γ′(t)| > 0 for all t, and let µ : [0, `] → R+ be a weight function.

Define

s = ψ(s) =

∫ s

0

|γ′(u)| du and t = ψ(t) =

∫ t

0

|γ′(u)| du.

Then it follows that

ds

ds= |γ′(s)| and

dt

dt= |γ′(t)| =⇒ ds = |γ′(s)|ds and dt = |γ′(t)|dt.

121

In addition, define φ = ψ−1, which implies s = φ(s) and t = φ(t). Notice now that

then γ ◦ φ will be parametrization with respect to arclength. Combining this with

Property (iii) of Lemma 4.1 then allows us to write the following:

E(γ, µ) = E(γ ◦ φ, µ ◦ φ)

=

∫ `

0

∫ `

0

[1

|γ(φ(s))− γ(φ(t))|2− 1

Dγ◦φ(s, t)2

]· µ(φ(s))µ(φ(t))|(γ(φ(s)))′||(γ(φ(t)))′| ds dt

=

∫ `

0

∫ `

0

[1

|γ(φ(s))− γ(φ(t))|2− 1

Dγ◦φ(s, t)2

]µ(φ(s))µ(φ(t)) · 1 · 1 ds dt

= E(γ ◦ φ, µ ◦ φ)

This allows us to take E = E on regular C1 curves. Hence, we shall simplify

our notation by writing E for C1 regular curves for the remainder of the text.

Example 4.1. Consider the weighted knot (γ, µ), where γ is the ellipse parametrized

by γ(t) = (2 cos t, sin t) and µ is the weight function µ(t) = 2 + cos 2t. Then we can

calculate that

E(γ, µ) ≈ 27.6901 and E(I(γ), µ) ≈ 30.3376.

Notice that this immediately implies that the nonuniform energy is not Mobius in-

variant, as it has failed to be invariant with respect to inversions. An illustration of

the weighted knots (γ, µ) and (I1,(0,0)(γ), µ) is provided in Figure 4.2.

Some regularity of γ is required in order for E(γ, µ) to make sense. E(γ, µ) is

not even defined on discontinuous curves, as the intrinsic distance is no longer well-

defined. In the case that γ is continuous but is not C1, we can show that the energy

122

Figure 4.2: The weighted knots (γ, µ) and (I1,(0,0)(γ), µ).

will be infinite. First, we show a bound for the nonuniform energy in terms of the

standard Mobius energy.

Proposition 4.2. Let µ be a weight function and set m = mint µ(t) and M =

maxt µ(t). Then for any curve γ parametrized with respect to arclength, the nonuni-

form energy, E(γ, µ), obeys the inequality

m2E(γ) ≤ E(γ, µ) ≤M2E(γ),

where E(γ) is the standard Mobius energy.

Proof. Note that m,M ∈ R always exist and are positive, since µ is a weight function

(therefore continuous). Then we may see the following, since |γ(s)− γ(t)| ≤ Dγ(s, t):

E(γ, µ) =

∫ ∫ [1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(s)µ(t) ds dt

≥∫ ∫ [

1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]m ·mds dt

= m2

∫ ∫ [1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]ds dt

= m2E(γ)

123

We can similarly obtain E(γ, µ) ≤M2E(γ), which completes the proof.

From this comes an immediate corollary.

Corollary 4.3. For any curve γ and any weight function µ, E(γ, µ) =∞ if and only

if E(γ) =∞.

Then this corollary leads to another one.

Corollary 4.4. If γ 6∈ C1, then E(γ, µ) is infinite.

Proof. It is a result from O’Hara’s work that E(γ) =∞ when γ 6∈ C1 [7], so combining

this with Corollary 4.3 completes the proof.

We can also use Proposition 4.2 to generalize Proposition 2.3 to the case of

the nonuniform energy.

Corollary 4.5. Let γ : [0, `] → Rn be a closed C2 curve that is not a circle, let

µ : [0, `] → R+ be a weight function, and set M = maxt∈[0,`] µ(t). Then for any

K ∈ R+ there exists a length-preserving transformation T such that the maximum

absolute pointwise curvature of T (γ) is greater than K, and E(T (γ), µ) ≤M2E(γ).

Proof. Recall from Proposition 4.2 that E(γ, µ) ≤ M2E(γ). From Proposition 2.3

we have that there exists some T such that T (γ) has maximum absolute pointwise

curvature greater than K and E(T (γ)) = E(γ). Then it immediately follows that

E(T (γ, µ)) ≤M2E(T (γ)) = M2E(γ), which completes the proof.

The use of Proposition 4.2 and Corollary 4.3 allow us to extend several of the

results proved by Freedman, He, and Wang to the nonuniform energy.

124

Proposition 4.6. Let γ be a closed curve in R3, and say that its derivative is L2+δ-

integrable for some δ > 0. Then E(γ, µ) is finite for any weight function µ.

Proof. This is immediate when applying Corollary 4.3 to proposition 1.5 in [4].

Proposition 4.7. Let µ be any weight function, and let γ be a closed curve in R3,

where c(γ) represents the average crossing number of γ. In the case that γ is closed,

let c([γ]) be the crossing number of the knot type of γ. Then the following hold:

(i) If γ is a proper rectifiable line, then

c(γ) ≤ 1

2πm2E(γ, µ).

(ii) If γ is a simple closed curve, then

2πm2c([γ]) + 4m2 ≤ E(γ, µ).

(iii) If γ is a simple rectifiable curve, then

c(γ) ≤ 11

12πm2E(γ, µ) +

1

π.

(iv) If γ is a unit speed rectifiable arc with domain [−5δ, 5δ] and γδ = γ|[−δ,δ], then

c(γδ) ≤2

3πm2E(γ, µ).

Proof. All of these follow from the application of the bounds of Proposition 4.2 to

Theorems 3.2, 3.3, 3.6, and 3.7 of [4], respectively.

Proposition 4.8. Let µ be a weight function and γ be a curve. Then γ is tame if

E(γ, µ) is finite.

Proof. This immediately follows from the application of Corollary 4.3 to theorem 4.1

of [4].

125

4.3 Properties of Minimizers of the Nonuniform Energy

Now that we have defined the nonuniform energy and investigated its basic

properties, we seek to learn more about its minimizers. Following the lead of Freed-

man, He, and Wang, we were able to generalize several of their results for the Mobius

energy to our nonuniform case. Many of the proofs of [4] are easily extendable by sim-

ple modifications. They also give a sketch of how to prove Propositions 4.11 and 4.12

in the case of the Mobius energy—we provide a complete proof for the nonuniform

case.

Corollary 4.9. Let µ be a weight function and γ be a closed curve. If m = mint µ(t),

then E(γ, µ) ≥ 4m2.

Proof. Let γ0 be a circle. Now recall from the work of Freedman, He, and Wang

that γ0 is the absolute minimizer of the Mobius energy among closed curves, and

E(γ0) = 4 [4]. Then it follows from Proposition 4.2 that

E(γ, µ) ≥ m2E(γ) ≥ m2E(γ0) = 4m2.


Propositions 4.11 and 4.12 describe necessary conditions for curves minimizing

the nonuniform energy. A similar idea for the Mobius energy is mentioned in [4], but

we provide a more complete proof here. First, we need to state Lemma 4.10.

Lemma 4.10. Let ~x, ~y ∈ Rn, and write ~x = (x1, . . . , xn) and ~y = (y1, . . . , yn). Then

if x1, y1 ≥ 0 we have the following:

126

(i)n∑i=1

(xi − yi)2 ≤ (−x1 − y1)2 +n∑i=2

(xi − yi)2.

(ii) Equality in Property (i) holds if and only if x1 = 0 or y1 = 0; that is, when

either ~x or ~y is on the hyperplane x1 = 0.

(iii) In Rn, dist((x1, . . . , xn), (y1, . . . , yn)) > dist((−x1, . . . , xn), (y1, . . . , yn)), pro-

vided neither ~x or ~y is on the hyperplane x1 = 0.

We begin by describing the case in R2.

Proposition 4.11. Fix a weight function µ : [0, `] → R+. Then any closed, planar

curve γ : [0, `] → R2 parametrized by arclength that absolutely minimizes E(γ, µ) is

convex.

Proof. Fix some weight function µ, and let γ be a simple, closed, planar curve mini-

mizing E(γ, µ). By way of contradiction say that γ is not convex. We shall show that

there exists γ : [0, `]→ R2 parametrized by arclength for which E(γ, µ) > E(γ, µ).

Since γ is not convex, then there exists a closed half-plane H ⊆ R2 such that

the following are satisfied:

(i) γ is disjoint from the interior of H.

(ii) γ intersects ∂H in at least two points. We can parametrize γ so that these

points are given by γ(0) and γ(b) for some b ∈ (0, `). Define γ1 = γ|[0,b] and

γ2 = γ|[b,`].

(iii) Neither γ1 or γ2 are entirely contained in ∂H.

127

We can then expand the definition of the nonuniform energy as follows:

E(γ, µ) =

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(s)µ(t) ds ds

=

∫ b

0

∫ b

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(s)µ(t) ds ds

+

∫ b

0

∫ `

b

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(s)µ(t) ds ds

+

∫ `

b

∫ b

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(s)µ(t) ds ds

+

∫ `

b

∫ `

b

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(s)µ(t) ds ds

Using symmetry, we can reduce this expression to

E(γ, µ) =

∫ b

0

∫ b

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(s)µ(t) ds ds

+

∫ `

b

∫ `

b

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(s)µ(t) ds ds

+ 2

∫ b

0

∫ `

b

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(s)µ(t) ds ds.

Define γ2 to be the reflection of γ2 through ∂H, and then define

γ(t) =

γ1(t) t ∈ [0, b]

γ2(t) t ∈ [b, `]

.

Note that γ is still parametrized with respect to arclength, as reflection is an isometry.

(It can also be helpful to note that γ′(0) = γ′(0) and γ′(b) = γ′(b).) This will not

affect any of the intrinsic distance calculations in the integrals above. Furthermore,

it will not change the Euclidean distance calculation |γ(s)− γ(t)|, provided that s, t

128

are either both in [0, b] or both in [b, `]. We can formally see this invariance as

∫ b

0

∫ b

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]|γ′(s)||γ′(t)|µ(s)µ(t) ds ds

=

∫ b

0

∫ b

0

[1

|γ1(s)− γ1(t)|2− 1

Dγ1(s, t)2

]|γ′1(s)||γ′1(t)|µ(s)µ(t) ds ds

=

∫ b

0

∫ b

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]|γ′(s)||γ′(t)|µ(s)µ(t) ds ds,

and

∫ `

b

∫ `

b

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2


=

∫ `

b

∫ `

b

[1

|γ2(s)− γ2(t)|2− 1

Dγ2(s, t)2

]|γ′2(s)||γ′2(t)|µ(s)µ(t) ds ds

=

∫ `

b

∫ `

b

[1

|γ2(s)− γ2(t)|2− 1

Dγ2(s, t)2

]|γ′2(s)||γ′2(t)|µ(s)µ(t) ds ds

=

∫ `

b

∫ `

b

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]|γ′(s)||γ′(t)|µ(s)µ(t) ds ds.

Next we will prove that

∫ b

0

∫ `

b

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]|γ′(s)|γ′(t)|µ(s)µ(t) ds ds

>

∫ b

0

∫ `

b

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]|γ′(s)||γ′(t)|µ(s)µ(t) ds ds.

Observe that the if the inequality

|γ(s)− γ(t)| ≤ |γ(s)− γ(t)| for s ∈ [0, b], t ∈ [b, `]

holds and becomes strict for some subsets of [0, b] and [b, `] of nonzero measure, then

our claim is proved. We can rewrite this statement as

|γ1(s)− γ2(t)| ≤ |γ1(s)− γ2(t)| for s ∈ [0, b], t ∈ [b, `],

129

which we know holds and is strict on subsets of nonzero measure as a result of

Lemma 4.10. (Note that although the hyperplane (a line in this case, as we are

in R2) we are reflecting through is not the line x1 = 0, we may use an isometry to

make it so.)

The combination of this inequality with the above work yields the following:

E(γ, µ) =

∫ b

0

∫ b

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2


+

∫ `

b

∫ `

b

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2


+ 2

∫ b

0

∫ `

b

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2


>

∫ b

0

∫ b

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2


+

∫ `

b

∫ `

b

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2


+ 2

∫ b

0

∫ `

b

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2


= E(γ, µ)

Hence, E(γ, µ) > E(γ, µ), which is a contradiction. Thus, γ must be convex.

The notions of this proof can also be extended n dimensions.

Proposition 4.12. Fix a weight function µ : [0, `] → R+. Let γ : [0, `] → Rn be

a closed curve parametrized by arclength, and say that there exists a half-hyperspace

H ⊆ Rn satisfying the following:

(i) γ is disjoint from the interior of H.

(ii) γ intersects ∂H is at least two points. We can parametrize γ so that these points

are given by γ(0) and γ(b) for some b ∈ (0, `). Define γ1 = γ|[0,b] and γ2 = γ|[b,`].

130

(iii) Neither γ1 or γ2 are entirely contained in ∂H.

Then γ cannot be an absolute minimizer for E(γ, µ).

Proof. Our proof here will be nearly identical to that of Proposition 4.11. Ultimately,

it is contingent upon the proof that

|γ1(s)− γ2(t)| ≤ |γ1(s)− γ2(t)| for s ∈ [0, b], t ∈ [b, `], (4.3)

with strictness on some subsets of [0, b] and [b, `] with nonzero measure. As in the

proof of Proposition 4.11, this follows immediately from Lemma 4.10.

4.4 Computing the Nonuniform Energy

The ability to compute the nonuniform energy of specific examples was critical

to much of the work completed in Chapter 5, and so we feel it is worthwhile to outline

our approach to this task.

The integrand of the nonuniform energy presents two very significant chal-

lenges:

1. The antiderivative of the integrand often does not have an easily obtainable

closed form.

2. The integrand contains terms with singularities.

The first problem can be dealt with by performing the calculation numerically. Un-

fortunately, this solution gives rise to the second problem, as we cannot simply handle

the singularities using an antiderivative. Recall that the singularities in the integrand

occur only on the diagonal (that is, where s = t) and are intended to be regularizing;

131

that is, they should cancel each other out, ultimately leaving the integrand with a

value of 112

(κabsγ )2 on the diagonal (see Lemma 5.1).

However, this value arises from the subtraction of two fractions with extremely

small denominators, making it easy for the calculation to lose accuracy near the

diagonal. Furthermore, the calculation is not even defined on the diagonal, since that

would require division by zero. One could write a piecewise-defined function replacing

the diagonal with the correct values, but we realized a more elegant solution that

would also prevent problems nearby the diagonal. As MATLAB takes the steps to

cover the double domain, we offset the steps for the second copy of the domain by a

half-step. This allows us to write simple code while still maintaining accuracy and

avoiding the diagonal.

Our work in MATLAB serves as a powerful calculator with the ability to

perform iterative calculations. Ultimately, we would really like to have the ability to

calculate

E(γ(s), µ(s)),

where s is an arclength parameter. This leads us to two problems for which we must

still explain our resolution:

1. γ is not parametrized by arclength.

2. The domain of µ does not correspond to that of γ.

Let L be the length of the parametrization domain of µ (which is 2π) in all of our

132

experiments. Then what we really want to calculate is

E

(γ(s), µ

(L

`s

)).

Unfortunately, we do not have an arclength parametrization for γ, nor is one easy

to come by. However, we can perform a workaround using a change of variables.

Rather than finding an arclength parametrization for γ, we put the inverse of such a

reparametrization in µ. To be explicit, we define

ψ(t) =

∫ t

0

|γ′(u)| du.

Then it follows from the basic properties of E that

E

(γ(s), µ

(L

`s

))= E

(γ(t), µ

(L

`ψ(t)

)).

As a result, we can acquire the value on the left by calculating the value on the right,

which is considerably easier to acquire.

From here, the remainder of our programming is the implementation of rela-

tively straightforward calculations, runtime optimizations, or convenience features.

In order to optimize the program, many of the values requires for the energy

calculation are programmed to be pre-calculated and stored (as they are calculated

many times). This alone dramatically reduced the running time of the program,

although it did make some of the calculations slightly more complicated to implement.

The function ψ we defined above was the main example of this change. As even a

single output of ψ is expensive to compute, we compiled a table of possible output

values for ψ to use as a reference instead. While this makes ψ much faster, it does

restrict our input values to only a specific set.

133

The convenience features manifested themselves in many forms. First, we

wanted a way to compare a large number of E values while translating the weight

function along γ. We created a function calculate multiple E values as we varied the

translation distance; this function also makes and saves graphs of the results, curve,

and weight function. On top of this, we implemented functionality to run many of

these for a specific family of weight functions with torus knots. Since some of these

combinations were constant with respect to the weight translation, we implemented

a method for the program to detect this constancy and save the data regarding it to

an Excel file.

134

CHAPTER 5VARIATIONS OF THE NONUNIFORM ENERGY

There are a two main ways we can vary a weighted curve: varying the curve

itself and varying the distribution of the weight. Section 5.1 addresses variations of

the curve, whereas Section 5.2 addresses variations of the weight. Before we proceed

to either, we must first prove a useful lemma.

Lemma 5.1. Let γ : [0, `]→ Rn be a C∞ curve parametrized by arclength and define

g(s, t) =1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2.

Then g(s, t), ∂g∂s

(s, t), and ∂g∂t

(s, t) are all continuously extendable to [0, `] × [0, `] by

taking the following:

(i) g(s, s) = 112

(κabsγ (s))2

(ii) ∂g∂s

(s, s) = ∂g∂t

(s, s) = 112κabsγ (s)(κabsγ (s))′

Proof. We first write the Taylor approximation for γ(t) centered at s, specifically

γ(t) = γ(s) +B(t− s) + C(t− s)2 +D(t− s)3 + E(t− s)4 + R4(t− s)5,

where

B = γ′(s), C =γ′′(s)

2, D =

γ′′′(s)

6, E =

γ′′′′(s)

24, and R4 =

∣∣∣∣ R4

|t− s|5

∣∣∣∣ .In addition, we choose M ∈ R+ such that M ≥ |B|, |C|, |D|, |E|, |R4| for all s. (We

know that R4 is bounded by Taylor’s theorem because γ is C∞ [8].) We shall use the

135

abbreviation κ(s) = κabsγ (s). Next, we calculate the following:

B •B = 1 C • C =κ2(s)

4B • E = −κ(s)κ′(s)

8

B • C = 0 B •D = −κ2(s)

6C •D =

κ(s)κ′(s)

12

Appropriately defining H1 then allows us to calculate the following:

g(s, t) =1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

=1

(t− s)2 − κ2(s)12

(t− s)4 − κ(s)κ′(s)12

(t− s)5 +H1(t− s)6− 1

(s− t)2

=κ2(s)

12+ κ(s)κ′(s)

12(t− s) +H1(t− s)2

1− κ2(s)12

(t− s)2 − κ(s)κ′(s)12

(t− s)3 +H1(t− s)4

Hence, we can take that

g(s, s) =1

12(κabs

γ (s))2.

Note that H1 will be the sum of products of B, C, D, E, R4, and powers of (t− s).

As a result, we know that H1 is bounded for t close to s, so we may assume that M

is also large enough to also bound H1. In addition, note that this proves that g(s, t)

is continuously extendable in t if we set g(s, s) = 112κ2(s).

Now let ε > 0, fix s0 ∈ [0, `], and choose δ > 0 such that

|g(s, t)− g(s, s)| < ε

2and

∣∣∣∣κ2(s)

12− κ2(s0)

12

∣∣∣∣ < ε

2

when |s − s0| < δ and |t − s| < 3δ. The second bound is possible because γ is C2.

The first bound follows from the fact that

|g(s, t)− g(s, s)| ≤ 25M |t− s|1− 25M |t− s|2

,

136

in other words, the bound of |g(s, t)− g(s, s)| is independent of our choice of s. Note

then that when |(s, t)− (s0, s0)| < δ we have

δ > |(s, t)− (s0, s0)| =√

(s− s0)2 + (t− s0)2 ≥√

(s− s0)2 = |s− s0|.

We can then also see the following:

δ > |(s, t)− (s0, s0)|

= |(s, t)− (s, s) + (s, s)− (s0, s0)|

≥ |t− s| − |(s, s)− (s0, s0)|

= |t− s| −√

2|s− s0|

=⇒ δ +√

2|s− s0| > |t− s|

As we know from above that |(s, t) − (s0, s0)| < δ implies |s − s0| < δ, we may then

conclude that

3δ > δ +√

2δ > δ +√

2|s− s0| > |t− s|.

Then for our choice of δ we have see that

|g(s, t)− g(s0, s0)| ≤ |g(s, t)− g(s, s)|+ |g(s, s)− g(s0, s0)|

= |g(s, t)− g(s, s)|+∣∣∣∣κ2(s)

12− κ2(s0)

12

∣∣∣∣ < ε

2+ε

2= ε

whenever |(t, s) − (s0, s0)|. Hence, we may conclude that g is continuous at (s0, s0).

As g is already continuous away from the diagonal of [0, `]× [0, `], it follows that g is

continuously extendable to all of [0, `]× [0, `]

Next, we calculate

∂

∂sg(s, t) =

2γ′(s) • (γ(t)− γ(s))

|γ(s)− γ(t)|4− 2

(t− s)3.

137

We can then use appropriately defined H2 and H3 and the Taylor approximation

constructed above to obtain

∂g

∂s(s, t) =

2γ′(s) • (γ(t)− γ(s))

|γ(s)− γ(t)|4− 2

(t− s)3

=2[(t− s) + −κ2(s)

6(t− s)3 + −κ(s)κ′(s)

8(t− s)4 +H2(t− s)5

](t− s)4 + −κ2(s)

6(t− s)6 + κ(s)κ′(s)

6(t− s)7 +H3(t− s)8

− 2

(t− s)3

=κ(s)κ′(s)

12+H2(t− s)

1− κ2(s)6

(t− s)2 − κ(s)κ′(s)8

(t− s)3 +H3(t− s)4.


∂g

∂s(s, s) =

κ(s)κ′(s)

12

It follows similarly that ∂g∂s

(s, s) = κ(s)κ′(s)12

. The proof of the continuous extendability

of these derivatives to [0, `]× [0, `] is similar to the proof above for g.

Remark. Notice that

∂

∂ug(u, u) =

∂g

∂s(u, u) +

∂g

∂t(u, u),

which corresponds to the results given in Lemma 5.1

5.1 Curve Variations

We will consider two types variations of E defined by varying the underlying

curve. In Subsection 5.1.1 we generalize the variation investigated in [4]. After that,

Subsection 5.1.2 turns our attention to a special case in which the length of the

underlying curve is preserved.

138

5.1.1 The Free Curve Variation

Freedman, He, and Wang define a certain variation of the Mobius energy in

[4] by varying the curve in a simple and natural way. In this section, we generalize

their result to the nonuniform energy.

Say that γ : [0, `] → Rn is a simple closed, rectifiable curve with an arbitrary

regular parametrization, and let L = length(γ). Additionally, let µ : [0, λ]→ R be a

closed C∞ weight function. (By closed, we mean that µ and each of its derivatives

are equal at 0 and λ. See the remark following Definition 4.1. Additionally, note that

λ need not be equal to `.) Then we can make the following definition.

Definition 5.1. Let h : [0, `] → Rn be a closed C∞ curve with an arbitrary regular

parametrization, and define γε(s) = γ(s)+ εh(s). Then we call E(γε, µ) the free curve

variation of E in the h direction, and we let Lε = length(γε).

Definition 5.1 allows us to state the following proposition; note that it does

not require that γ be parametrized by arclength. Additionally, this formula allows µ

to be λ-periodic (while γ is `-periodic).

We begin with a lemma.

Lemma 5.2. Let γ : [0, `] → Rn be a simple closed C∞ curve parametrized by ar-

clength. Then we have the following:

(i)∂

∂ε|γ′ε(s)| = 〈γ′(s), h′(s)〉

(ii)d

dεLε

∣∣∣∣ε=0

=

∫ `

0

〈γ′(u), h′(u)〉 du

(iii)d

dε

L

Lε

∣∣∣∣ε=0

=−1

L

∫ `

0

〈γ′(u), h′(u)〉 du

139

Proof. Proof of Part (i):

Recall that |γ′(s)| = 1 and observe the following:

∂

∂ε|γ′ε(s)|

∣∣∣∣ε=0

=∂

∂ε〈γ′ε(s), γ′ε(s)〉

12

∣∣∣∣ε=0

=1

2〈γ′ε(s), γ′ε(s)〉

−12 · 2〈γ′ε(s), h′(s)〉

∣∣∣∣ε=0

=〈γ′ε(s), h′(s)〉|γ′ε(s)|

∣∣∣∣ε=0

= 〈γ′(s), h′(s)〉

As this is the desired result, we are done.

Proof of Part (ii):

Observe that

d

dεLε

∣∣∣∣ε=0

=

∫ `

0

∂

∂ε|γ′ε(u)| du

∣∣∣∣ε=0

=

∫ `

0

〈γ′(u) + εh′(u), h′(u)〉|γ′(u) + εh′(u)|

du

∣∣∣∣ε=0

.

Setting ε = 0 implies that |γ′(u)+εh′(u)| = |γ′(u)| = 1 (since γ was parameterized by

arclength), and this immediately simplifies the third expression to the desired result.

Proof of Part (iii):

Observe that

d

dε

L

Lε

∣∣∣∣ε=0

=

(−LL2ε

d

dεLε

) ∣∣∣∣ε=0

=−1

L

∫ `

0

〈γ′(u), h′(u)〉 du.

Proposition 5.3. The variation of γ in the h direction gives the first variation of

140

E(γ, µ) as

dE

dε(γε, µ)

∣∣∣∣ε=0

=

∫ `

0

∫ `

0

[〈γ′(s), h′(s)〉|γ′(s)|

− 〈γ(s)− γ(t), h(s)− h(t)〉|γ(s)− γ(t)|2

]· |γ

′(s)||γ′(t)||γ(s)− γ(t)|2

µ

(λ

L

∫ s

0

|γ′(u)| du)µ

(λ

L

∫ t

0

|γ′(u)| du)ds dt

+ 2

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]|γ′(s)||γ′(t)|

· µ(λ

L

∫ t

0

|γ′(u)| du)µ′(λ

L

∫ s

0

|γ′(u)| du)

·

[λ∫ s

0〈γ′(u),h′(u)〉|γ′(u)| du

L−λ∫ `

0〈γ′(u),h′(u)〉|γ′(u)| du

∫ s0|γ′(u)| du

L2

]ds dt.

Proof. Beginning with the generalized formula for the nonuniform energy, we have

the following:

d

dεE(γε, µ)

∣∣∣∣ε=0

=d

dε

∫ `

0

∫ `

0

[1

|γε(s)− γε(t)|2− 1

Dγε(s, t)2

]|γ′ε(s)||γ′ε(t)|

· µ(λ

Lε

∫ s

0

|γ′ε(u)| du)µ

(λ

Lε

∫ t

0

|γ′ε(u)| du)ds dt

∣∣∣∣ε=0

=

∫ `

0

∫ `

0

∂

∂ε

([1

|γε(s)− γε(t)|2− 1

Dγε(s, t)2

]|γ′ε(s)||γ′ε(t)|

) ∣∣∣∣ε=0

· µ(λ

L

∫ s

0

|γ′(u)| du)µ

(λ

L

∫ t

0

|γ′(u)| du)ds dt

+

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]|γ′(s)||γ′(t)|

· ∂∂ε

(µ

(λ

Lε

∫ s

0

|γ′ε(u)| du)) ∣∣∣∣

ε=0

µ

(λ

L

∫ t

0

|γ′(u)| du)ds dt

+

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]|γ′(s)||γ′(t)|

· µ(λ

L

∫ s

0

|γ′(u)| du)∂

∂ε

(µ

(λ

Lε

∫ t

0

|γ′ε(u)| du)) ∣∣∣∣

ε=0

ds dt

141

The derivative of the factor in the second term of ddεE(γε, µ) can be directly calculated

as

d

dεµ

(λ

Lε

∫ s

0

|γ′ε(u)| du) ∣∣∣∣

ε=0

= µ′(λ

L

∫ s

0

|γ′(u)| du)d

dε

(λ

Lε

∫ s

0

|γ′ε(u)| du) ∣∣∣∣

ε=0

= µ′(λ

L

∫ s

0

|γ′(u)| du)[

d

dε

(λ

Lε

) ∣∣∣∣ε=0

∫ s

0

|γ′(u)| du+λ

L

d

dε

(∫ s

0

|γ′ε(u)| du) ∣∣∣∣

ε=0

]= µ′

(λ

L

∫ s

0

|γ′(u)| du)

·

−λ ∫ `0 〈γ′(u),h′(u)〉|γ′(u)| du

∫ s0|γ′(u)| du(∫ `

0|γ′(u)| du

)2 +λ∫ s

0〈γ′(u),h′(u)〉|γ′(u)| du∫ `

0|γ′(u)| du

= µ′

(λ

L

∫ s

0

|γ′(u)| du)

·

[−λ∫ `

0〈γ′(u),h′(u)〉|γ′(u)| du

∫ s0|γ′(u)| du

L2+λ∫ s

0〈γ′(u),h′(u)〉|γ′(u)| du

L

].

The derivative of the factor in the third term is similar. By using a symmetry ar-

gument to exchange the roles of s and t in the third integral, we may combine the

second and third integrals into one. The derivative of the factor in the first term

has already been calculated by Freedman, He, and Wang in [4]. Using their work in

combination with ours then gives the desired result.

Remark. While it is tempting to consider the case in which γ is parametrized by

arclength, notice that variation in the h direction would immediately “ruin” this

parametrization. As a result, this case ends up being much more difficult to handle,

and is the subject of much discussion in Subsection 5.1.2.

142

5.1.2 The Equilength Variation

A question of great interest is the following: “What is the first variation for-

mula for the nonuniform energy when the curve is varied in such a way that its length

is preserved and its parametrization is always according to arclength?” In partic-

ular, this variation interests us because it most closely matches that of a real-life

scenario—one in which we vary some closed loop of fixed length with nonuniform

thickness.

Unfortunately, the road to acquiring this solution is somewhat complex, and

requires careful setup. That said, consider the following descriptions and definitions.

We let γ : [0, `]→ Rn be the C∞ arclength parametrization of a simple closed

curve, and we call its length L. (Notice here that L = `; we merely use separate letters

for emphasis.) Additionally, we µ : [0, `]→ R+ be a closed C∞ weight function.

We shall vary our curve in the h : [0, `]→ Rn direction; note that this requires

that h is closed, and we also assume that h is C∞. We then define γε(s) = γ(s)+εh(s)

to be the variation of γ in the h direction, and we let Lε = length(γε).

Next, we define γε = LLεγε, which makes length(γε) = L. After that, define

φε(s) =∫ s

0|γ′ε(u)| du, and set x = φε(s), which makes s = φ−1

ε (x). Finally, let γ∗ε (x) =

γε(φ−1ε (x)), and notice that this is a parametrization proportional to arclength.

At this point, we make a few remarks:

• Note that then φε is a function mapping [0, `] to itself.

• Note that s corresponds to the original domain, whereas x is in the new domain.

• Note that γ∗ε (x) = LLε

(γ(φ−1ε (x))+εh(φ−1

ε (x))) is a parametrization with respect

143

to arclength.

We intend to use this notation to construct a variation of E(γ, µ) in the h

direction that preserves the length of γ and the distribution of µ (that is, the variation

must also be parameterized by arclength). To this end, we consider the diagram shown

in Figure 5.1.

E(γ(s), µ) E(γ∗(x), µ) E(γ∗ε (x), µ)

E(γ(s), µ ◦ φε) E(γε(s), µ ◦ φε)

(1)

(2)

(3)

Figure 5.1: An illustration of the two ways to obtain the first variation formula for

the equilength variation.

Our eventual goal will be to calculate first variation formulas for the two cases

displayed in Figure 5.1; in particular:

i. The direct traversal of path (1).

ii. The consecutive traversal of paths (2) and (3).

But first, we must build up a collection of several basic formulas though some straight-

forward calculations. In order to assist with these calculations, we have compiled a

list of the relevant notation discussed above, shown in Definition 5.2.

144

Definition 5.2. The following notation will be used throughout Subsection 5.1.2:

• γε(s) = γ(s) + εh(s)

• Lε = length(γε) =

∫ `

0

|γ′ε(u)| du

• γε(s) =L

Lεγε

• φε =

∫ s

0

|γ′ε(u)| du

• x = φε(s)

• γ∗ε (x) = γε(φ−1(x)

)• γ∗ε (x) =

L

Lεγ∗ε =

L

Lε

(γ(φ−1(x)

)+ εh

(φ−1ε (x)

))Lemma 5.4.

∂φε∂s

(s) =L

Lε|γ′(s) + εh′(s)|.

Proof. This follows immediately from the definition

φε(s) =

∫ s

0

∣∣∣∣ LLε (γ′(u) + εh′(u))

∣∣∣∣ duand the second FTC.

Lemma 5.5.

∂φε∂ε

(s)

∣∣∣∣ε=0

=

∫ s

0

〈γ′(u), h′(u)〉 du− s

L

∫ `

0

〈γ′(u), h′(u)〉 du.

Proof. Observe the following:

∂φε∂ε

(s)

∣∣∣∣ε=0

=

∫ s

0

∂

∂ε|γ′ε(u)| du

∣∣∣∣ε=0

=

∫ s

0

∂

∂ε

(L

Lε〈γ′(u) + εh′(u), γ′(u) + εh′(u)〉

12

)du

∣∣∣∣ε=0

145

=

((d

dε

(L

Lε

))∫ s

0

|γ′(u) + εh′(u)| du

+L

Lε

∫ s

0

∂

∂ε〈γ′(u) + εh′(u), γ′(u) + εh′(u)〉

12 du

)∣∣∣∣ε=0

We can use Part (i) of Lemma 5.2 to obtain the derivative in the first term at ε = 0.

Computing the second term’s derivative yields the following:

∂φε∂ε

(s)

∣∣∣∣ε=0

=−1

L

∫ `

0

〈γ′(u), h′(u)〉 du∫ s

0

|γ′(u)| du

+

(L

Lε

∫ s

0

〈γ′(u) + εh′(u), h′(u)〉|γ′(u) + εh′(u)|

du

) ∣∣∣∣ε=0

Note that setting ε = 0 then gives Lε = L. Additionally, it tells us that

|γ′(u) + εh′(u)|∣∣ε=0

= |γ′(u)| = 1

(since γ was parameterized by arclength), so the second integral in the first term will

simply become s. Hence, we have

∂φε∂ε

(s)

∣∣∣∣ε=0

=−sL

∫ `

0

〈γ′(u), h′(u)〉 du+

∫ s

0

〈γ′(u), h′(u)〉 du,

which can be rearranged to obtain the desired result.

Lemma 5.6.

∂φ−1ε

∂ε(x) =

−∂φε∂ε

(φ−1ε (x))

∂φε∂s

(φ−1ε (x))

.

Proof. Say that s and x are the variables for the domain and codomain of φε, respec-

tively. That is, φε(s) = x. As our function is invertible, we may write

φε(φ−1ε (x)) = x.

146

Differentiating both sides of this equation with respect to ε then yields

∂φε∂ε

(φ−1ε (x)) +

∂φε∂s

(φ−1ε (x))

∂φ−1ε

∂ε(x) = 0.

We can then solve this for ∂φ−1ε

∂ε(x), which gives the desired result.

Lemma 5.7.

∂φ−1ε

∂ε(x)

∣∣∣∣ε=0

=x

L

∫ `

0

〈γ′(u), h′(u)〉 du−∫ x

0

〈γ′(u), h′(u)〉 du.

Proof. Apply Lemmas 5.4 to 5.6, then set ε = 0. Note that when ε = 0 we will see

φε become the identity and LLε

= 1. Combining these with the fact that |γ′(s)| = 1


Lemma 5.8.

∂γ∗ε∂ε

(x)

∣∣∣∣ε=0

= (xγ′(x)− γ(x))1

L

∫ `

0

〈γ′(u), h′(u)〉 du− γ′(x)

∫ x

0

〈γ′(u), h′(u)〉 du+h(x).

Proof. First observe that

∂γ∗ε∂ε

(x)

∣∣∣∣ε=0

=d

dε

(L

Lε

(γ(φ−1ε (x)

)+ εh

(φ−1ε (x)

))) ∣∣∣∣ε=0

=

((d

dε

(L

Lε

))(γ∗ε ) +

L

Lε

∂

∂ε

(γ(φ−1

ε (x)) + εh(φ−1ε (x))

)) ∣∣∣∣ε=0

.

The first term is familiar from Lemma 5.2. Note that ε = 0 implies that φε will be

the identity and γ∗0 = γ. Then the above may be continued as

∂γ∗ε∂ε

(x)

∣∣∣∣ε=0

= γ(x)−1

L

∫ `

0

〈γ′(u), h′(u)〉 du

+L

Lε

[γ′(φ−1

ε (x))∂φ−1

ε

∂ε(x) + h(φ−1

ε (x)) + εh′(φ−1ε (x))

∂φ−1ε

∂ε(x)

] ∣∣∣∣ε=0

.

147

Setting ε = 0 simplifies the second and third terms and vanishes the fourth. Combin-

ing this with Lemma 5.7 yields

∂γ∗ε∂ε

(x)

∣∣∣∣ε=0

= −γ(x)1

L

∫ `

0

〈γ′(u), h′(u)〉 du

+ γ′(x)

(x

L

∫ `

0

〈γ′(u), h′(u)〉 du−∫ x

0

〈γ′(u), h′(u)〉 du)

+ h(x).

This can then be rearranged to obtain the desired result.

Lemma 5.9.

∂

∂ε

(1

|γ∗ε (x)− γ∗ε (y)|2− 1

|x− y|2

) ∣∣∣∣ε=0

=−2

|γ(x)− γ(y)|4

·

⟨γ(x)− γ(y),

((xγ′(x)− γ(x))

1

L

∫ `

0

〈γ′(u), h′(u)〉 du−

γ′(x)

∫ x

0

〈γ′(u), h′(u)〉 du+ h(x)

)−(

(yγ′(y)− γ(y))1

L

∫ `

0

〈γ′(u), h′(u)〉 du

−γ′(y)

∫ y

0

〈γ′(u), h′(u)〉 du+ h(y)

)⟩.


∂

∂ε

1

|γ∗ε (x)− γ∗ε (y)|2

∣∣∣∣ε=0

=∂

∂ε〈γ∗ε (x)− γ∗ε (y), γ∗ε (x)− γ∗ε (y)〉−1

∣∣∣∣ε=0

= −1 〈γ∗ε (x)− γ∗ε (y), γ∗ε (x)− γ∗ε (y)〉−2

(2

⟨γ∗ε (x)− γ∗ε (y),

∂

∂ε

(γ∗ε (x)− γ∗ε (y)

)⟩) ∣∣∣∣ε=0

=−2⟨γ∗ε (x)− γ∗ε (y), ∂

∂ε(γ∗ε (x)− γ∗ε (y))

⟩|γ∗ε (x)− γ∗ε (y)|4

∣∣∣∣ε=0

148

Note that setting ε = 0 gives γ∗ε (x) = γ(x). Then we combine this with the application

of Lemma 5.8 to obtain

∂

∂ε

1

|γ∗ε (x)− γ∗ε (y)|2

∣∣∣∣ε=0

=−2

|γ(x)− γ(y)|4

·

⟨γ(x)− γ(y),

((xγ′(x)− γ(x))

1

L

∫ `

0

〈γ′(u), h′(u)〉 du−

γ′(x)

∫ x

0

〈γ′(u), h′(u)〉 du+ h(x)

)−(

(yγ′(y)− γ(y))1

L

∫ `

0

〈γ′(u), h′(u)〉 du

−γ′(y)

∫ y

0

〈γ′(u), h′(u)〉 du+ h(y)

)⟩,

which is exactly the desired result.

Before we proceed, let p(ε, t) be implicitly defined by∫ t+p(ε,t)t

|γ′ε(u)| du = Lε2

.

Lemma 5.10.

∂p

∂ε(ε, t)

∣∣∣∣ε=0

=1

2

∫ `

0

〈γ′(u), h′(u)〉 du−∫ t+`/2

t

〈γ′(u), h′(u)〉 du.

Proof. Recall the implicit definition of p, that is∫ t+p(ε,t)

t

|γ′ε(u)| du =Lε2.

Then computing the ε-derivative of this expression at ε = 0 yields[|γ′ε(t+ p(ε, t))| ∂

∂ε(t+ p(ε, t))

] ∣∣∣∣ε=0

+

[∫ t+p(ε,t)

t

〈γ′ε(u), h′(u)〉|γ′ε(u)|

du

] ∣∣∣∣ε=0

=1

2

∂

∂εLε

∣∣∣∣ε=0

.

Note that p(0, t) = `2

for all t and recall that |γ′(u)| = 1. By also applying Part (i) of

Lemma 5.2 to the right side, we can simplify the previous expression to

∂p

∂ε(ε, t)

∣∣∣∣ε=0

+

∫ t+`/2

t

〈γ′(u), h′(u)〉 du =1

2

∫ `

0

〈γ′(u), h′(u)〉 du.

149

Rearranging this then gives the desired result.

Remark. Although

1

|γ∗ε (x)− γ∗ε (y)|2− 1

|x− y|2

is C1 extendable to the diagonal x = y for all ε > 0 (see Lemma 5.1), its ε-derivative

evaluated at ε = 0 (appearing in Proposition 5.11) may or may not be C0 extendable

to the diagonal. We do not provide a proof or counterexample—this motivates our

second approach to the first variation (see Figure 5.1), shown in Proposition 5.17.

Proposition 5.11. The first variation formula for E(·, µ) along (1) in Figure 5.1

will be the following integral, provided that the integrand is continuously extendable

to the diagonal x = y and the integral converges:

dE

dε(γ∗ε , µ)

∣∣∣∣ε=0

=

∫ `

0

∫ y+`/2

y−`/2µ(x)µ(y)

[−2

|γ(x)− γ(y)|4

⟨γ(x)− γ(y),(

(xγ′(x)− γ(x))1

L

∫ `

0

〈γ′(u), h′(u)〉 du

− γ′(x)

∫ x

0

〈γ′(u), h′(u)〉 du+ h(x)

)−(

(yγ′(y)− γ(y))1

L

∫ `

0

〈γ′(u), h′(u)〉 du

− γ′(y)

∫ y

0

〈γ′(u), h′(u)〉 du+ h(y)

)⟩]dx dy.

Proof. Recall that we can write the nonuniform energy of γ∗ε with respect to µ as

E(γ∗ε , µ) =

∫ `

0

∫ `

0

[1

|γ∗ε (x)− γ∗ε (y)|2− 1

Dγ∗ε (x, y)2

]µ(x)µ(y)|(γ∗ε )′(x)||(γ∗ε )′(y)| dx dy.

150

We can then combine the definition of p with the fact that γ∗ε is parametrized with

respect to arclength to acquire

E(γ∗ε , µ) =

∫ `

0

∫ y+p(ε,y)

y−(`−p(ε,y))

[1

|γ∗ε (x)− γ∗ε (y)|2− 1

|x− y|2

]µ(x)µ(y) dx dy.

Note that this leaves the only dependences on ε in the first term and the bounds of

the inside integral. Now observe the following:

d

dεE(γ∗ε , µ)

∣∣∣∣ε=0

=d

dε

∫ `

0

∫ y+p(ε,y)

y−`+p(ε,y)

[1

|γ∗ε (x)− γ∗ε (y)|2− 1

|x− y|2

]µ(x)µ(y) dx dy

∣∣∣∣ε=0

=

∫ `

0

[([1

|γ∗ε (y + p(ε, y))− γ∗ε (y)|2− 1

|y + p(ε, y)− y|2

]

·µ(y + p(ε, y))µ(y)∂p

∂ε(ε, y)

) ∣∣∣∣∣ε=0

−([

1

|γ∗ε (y − `+ p(ε, y))− γ∗ε (y)|2− 1

|y − `+ p(ε, y)− y|2

]· µ(y − `+ p(ε, y))µ(y)

∂p

∂ε(ε, y)

) ∣∣∣∣∣ε=0

+

(∫ y+p(ε,y)

y−`+p(ε,y)

∂

∂ε

([1

|γ∗ε (x)− γ∗ε (y)|2− 1

|x− y|2

]µ(x)µ(y)

)dx

)∣∣∣∣∣ε=0

]dy

=

∫ `

0

[([1∣∣γ(y + `

2)− γ(y)

∣∣2 − 1∣∣ `2

∣∣2]µ

(y +

`

2

)µ(y)

(∂p

∂ε(ε, y)

∣∣∣∣∣ε=0

))

−

([1∣∣γ(y − `+ `2)− γ(y)

∣∣2 − 1∣∣−`+ `2

∣∣2]

·µ(y − `+

`

2

)µ(y)

(∂p

∂ε(ε, y)

∣∣∣∣∣ε=0

))

+

(∫ y+`/2

y−`/2

∂

∂ε

([1

|γ∗ε (x)− γ∗ε (y)|2− 1

|x− y|2

]µ(x)µ(y)

) ∣∣∣∣∣ε=0

dx

)]dy

=

∫ `

0

[([1∣∣γ(y + `

2)− γ(y)

∣∣2 − 4

`2

]µ

(y +

`

2

)µ(y)

(∂p

∂ε(ε, y)

∣∣∣∣∣ε=0

))

151

−

([1∣∣γ(y − `

2)− γ(y)

∣∣2 − 4

`2

]µ

(y − `

2

)µ(y)

(∂p

∂ε(ε, y)

∣∣∣∣∣ε=0

))

+

(∫ y+`/2

y−`/2

∂

∂ε

(1

|γ∗ε (x)− γ∗ε (y)|2

) ∣∣∣∣∣ε=0

µ(x)µ(y) dx

)]dy

Notice that the `-periodicity of γ and µ cause the first and second lines in the final

expression above to cancel out, as they will be the negatives of each other. Since

Lemma 5.9 gives us an expression for the derivative in the third line of the final

expression, we obtain

d

dεE(γ∗ε , µ)

∣∣∣∣ε=0

=

∫ `

0

∫ y+`/2

y−`/2µ(x)µ(y)

[−2

|γ(x)− γ(y)|4

⟨γ(x)− γ(y),(

(xγ′(x)− γ(x))1

L

∫ `

0

〈γ′(u), h′(u)〉 du

− γ′(x)

∫ x

0

〈γ′(u), h′(u)〉 du+ h(x)

)−(

(yγ′(y)− γ(y))1

L

∫ `

0

〈γ′(u), h′(u)〉 du

− γ′(y)

∫ y

0

〈γ′(u), h′(u)〉 du+ h(y)

)⟩]dx dy.

which is exactly the desired result.


will be

dE

dε(γ, µ ◦ φε)

∣∣∣∣ε=0

= 2

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(t)µ′(s)

·(∫ s

0

〈γ′(u), h′(u)〉 du− s

L

∫ `

0

〈γ′(u), h′(u)〉 du)ds dt.

152


d

dεE(γ, µ ◦ φε)

∣∣∣∣ε=0

=d

dε

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(φε(s))µ(φε(t)) ds dt

∣∣∣∣ε=0

=

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]∂

∂ε

(µ(φε(s))µ(φε(t))

)∣∣∣∣ε=0

ds dt

=

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

](µ′(φε(s))

∂φε∂ε

(s)

∣∣∣∣ε=0

µ(φε(t))

)ds dt

+

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

](µ(φε(s))µ

′(φε(t))∂φε∂ε

(t)

∣∣∣∣ε=0

)ds dt

Note that we can acquire an expression for ∂φε∂ε

via Lemma 5.5. Further, recognize

that φε will be the identity when ε = 0. Due to the symmetry of the term in brackets,

we can exchange of the roles of s and t in the last line of the above expression to

obtain

d

dεE(γ, µ ◦ φε)

∣∣∣∣ε=0

= 2

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(t)µ′(s)

·(∫ s

0

〈γ′(u), h′(u)〉 du− s

L

∫ `

0

〈γ′(u), h′(u)〉 du)ds dt.

Since

1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

continuous extends to the diagonal s = t by Lemma 5.1, this integral exists. As this

is exactly the desired result, we are finished.

Lemma 5.13.

∂

∂ε

1

|γε(s)− γε(t)|2

∣∣∣∣ε=0

=−2〈γ(s)− γ(t), h(s)− h(t)〉

|γ(s)− γ(t)|4.

153


∂

∂ε|γε(s)− γε(t)|−2

∣∣∣∣ε=0

=∂

∂ε〈γε(s)− γε(t), γε(s)− γε(t)〉−1

∣∣∣∣ε=0

= −1〈γε(s)− γε(t), γε(s)− γε(t)〉−2 · 2〈γε(s)− γε(t), h(s)− h(t)〉∣∣∣∣ε=0

=−2〈γ(s)− γ(t), h(s)− h(t)〉

|γ(s)− γ(t)|4


Lemma 5.14.

∂

∂ε

1(∫ ts|γ′ε(u)| du

)2

∣∣∣∣ε=0

=−2∫ ts〈γ′(u), h′(u)〉 du(t− s)3

.


∂

∂ε

1(∫ ts|γ′ε(u)| du

)2

∣∣∣∣ε=0

=∂

∂ε

[∫ t

s

|γ′ε(u)| du]−2 ∣∣∣∣

ε=0

= −2

[∫ t

s

|γ′ε(u)| du]−3

∂

∂ε

∫ t

s

|γ′ε(u)| du∣∣∣∣ε=0

=−2∫ ts〈γ′ε(u),h′(u)〉|γ′ε(u)| du(∫ t

s|γ′ε(u)| du

)3

∣∣∣∣∣ε=0

Recall that |γ(s)| = 1. Then setting ε = 0 gives

−2∫ ts〈γ′(u),h′(u)〉

1du(∫ t

s1 du

)3 =−2∫ ts〈γ′(u), h′(u)〉 du(t− s)3

,



154

will be

dE

dε(γε, µ)

∣∣∣∣ε=0

= 2

∫ `

0

∫ t+ `2

t− `2

[[∫ ts〈γ′(u), h′(u)〉 du

(t− s)3− 〈γ(s)− γ(t), h(s)− h(t)〉

|γ(s)− γ(t)|4

]|γ′(t)|

+

[1

|γ(s)− γ(t)|2− 1

|s− t|2

]〈γ′(t), h′(t)〉

]|γ′(s)|µ(s)µ(t) ds dt.

Proof. Note that when traversing path (3), our weight function does not change. As

a result, an ε-derivative of the change along (3) only sees the change of γ to γε. With

that in mind, we recall the definition of E(·, µ) to see

dE

dε(γε, µ)

∣∣∣∣ε=0

=d

dε

∫ `

0

∫ `

0

[1

|γε(s)− γε(t)|2− 1

Dγε(s, t)2

]µ(s)µ(t)|γ′ε(s)||γ′ε(t)| ds dt

∣∣∣∣ε=0

.

Recalling the definition of γε and remembering that E(·, µ) is invariant under rescal-

ing, we can then reformulate the above expression as

dE

dε(γε, µ)

∣∣∣∣ε=0

=d

dε

∫ `

0

∫ `

0

[1

|γε(s)− γε(t)|2− 1

Dγε(s, t)2

]µ(s)µ(t)|γ′ε(s)||γ′ε(t)| ds dt

∣∣∣∣ε=0

.

Using the periodicity of the terms in the integrand, we may rewrite the bounds of

the interior integral using the function p defined above. This allows us to substitute

using

Dγε(s, t) =

(∫ t

s

|γε(u)| du),

which gives the following:

dE

dε(γε, µ)

∣∣∣∣ε=0

=d

dε

∫ `

0

∫ t+p(ε,t)

t−`+p(ε,t)

1

|γε(s)− γε(t)|2− 1(∫ t

s|γ′ε(u)| du

)2

155

· µ(s)µ(t)|γ′ε(s)||γ′ε(t)| ds dt

∣∣∣∣∣ε=0

=

∫ `

0

[( 1

|γε(t+ p(ε, t))− γε(t)|2− 1(∫ t

t+p(ε,t)|γ′ε(u)| du

)2

·

(µ(t+ p(ε, t))µ(t)|γ′ε(t+ p(ε, t))||γ′ε(t)|

∂p

∂ε(ε, t)

))∣∣∣∣∣ε=0

−

( 1

|γε(t+ `− p(ε, t))− γε(t)|2− 1(∫ t

t+`−p(ε,t) |γ′ε(u)| du)2

·

(µ(t+ `− p(ε, t))µ(t)|γ′ε(t+ `− p(ε, t))||γ′ε(t)|

∂p

∂ε(ε, t)

))∣∣∣∣ε=0

+

∫ t+ `2

t− `2

∂

∂ε

1

|γε(s)− γε(t)|2− 1(∫ t

s|γ′ε(u)| du

)2

|γ′ε(s)||γ′ε(t)|∣∣∣∣∣

ε=0

· µ(s)µ(t) ds

]dt

=

∫ `

0

[( 1

|γ(t+ `2)− γ(t)|2

− 1(∫ tt+ `

2|γ′(u)| du

)2

· µ(t+

`

2

)µ(t)

∣∣∣∣γ′(t+`

2

)∣∣∣∣ |γ′(t)|∂p∂ε (ε, t)

∣∣∣∣∣ε=0

)

−

( 1

|γ(t− `2)− γ(t)|2

− 1(∫ tt− `

2|γ′(u)| du

)2

· µ(t− `

2

)µ(t)

∣∣∣∣γ′(t− `

2

)∣∣∣∣ |γ′(t)|∂p∂ε (ε, t)

∣∣∣∣∣ε=0

)

+

∫ t+ `2

t− `2

∂

∂ε

1

|γε(s)− γε(t)|2− 1(∫ t

s|γ′ε(u)| du

)2

|γ′ε(s)||γ′ε(t)|∣∣∣∣∣

ε=0

· µ(s)µ(t) ds

]dt

156

Recall that |γ′(s)| = 1. Then we can obtain

dE

dε(γε, µ)

∣∣∣∣ε=0

=

∫ `

0

[([1

|γ(t+ `2)− γ(t)|2

− 1(− `

2

)2

]µ

(t+

`

2

)µ(t)

∂p

∂ε(ε, t)

∣∣∣∣∣ε=0

)

−

([1

|γ(t− `2)− γ(t)|2

− 1(`2

)2

]µ

(t− `

2

)µ(t)

∂p

∂ε(ε, t)

∣∣∣∣∣ε=0

)

+

∫ t+ `2

t− `2

∂

∂ε

1

|γε(s)− γε(t)|2− 1(∫ t

s|γ′ε(u)| du

)2

|γ′ε(s)||γ′ε(t)|∣∣∣∣∣

ε=0

· µ(s)µ(t) ds

]dt.

Then the `-periodicity of µ and γ imply that the first two lines are the negatives of

each other, so our expression reduces simply to the third line. We then apply Part (ii)

of Lemma 5.2 and Lemmas 5.13 and 5.14 as follows:

dE

dε(γε, µ)

∣∣∣∣ε=0

=

∫ `

0

∫ t+ `2

t− `2

∂

∂ε

1

|γε(s)− γε(t)|2− 1(∫ t

s|γ′ε(u)| du

)2

|γ′ε(s)||γ′ε(t)|∣∣∣∣∣

ε=0

· µ(s)µ(t) ds dt

=

∫ `

0

∫ t+ `2

t− `2

[∂

∂ε

1

|γε(s)− γε(t)|2− 1(∫ t

s|γ′ε(u)| du

)2

∣∣∣∣∣

ε=0

|γ′(s)||γ′(t)|

+

1

|γ(s)− γ(t)|2− 1(∫ t

s|γ′(u)| du

)2

∂

∂ε

(|γ′ε(s)|

)∣∣∣∣ε=0

|γ′(t)|

+

1

|γ(s)− γ(t)|2− 1(∫ t

s|γ′(u)| du

)2

|γ′(s)| ∂∂ε

(|γ′ε(t)|

)∣∣∣∣ε=0

]

· µ(s)µ(t) ds dt

157

=

∫ `

0

∫ t+ `2

t− `2

[[2∫ ts〈γ′(u), h′(u)〉 du

(t− s)3− 2〈γ(s)− γ(t), h(s)− h(t)〉

|γ(s)− γ(t)|4

]|γ′(s)||γ′(t)|

+

1

|γ(s)− γ(t)|2− 1(∫ t

s|γ′(u)| du

)2

〈γ′(s), h′(s)〉|γ′(t)|+

1

|γ(s)− γ(t)|2− 1(∫ t

s|γ′(u)| du

)2

|γ′(s)|〈γ′(t), h′(t)〉]

· µ(s)µ(t) ds dt

We can then replace the definition of Dγ(s, t) and use symmetry to obtain the follow-

ing:

dE

dε(γε, µ)

∣∣∣∣ε=0

=

∫ `

0

∫ t+ `2

t− `2

[2∫ ts〈γ′(u), h′(u)〉 du

(t− s)3− 2〈γ(s)− γ(t), h(s)− h(t)〉

|γ(s)− γ(t)|4

]

· |γ′(s)||γ′(t)|µ(s)µ(t) ds dt

+

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)

]〈γ′(s), h′(s)〉|γ′(t)|µ(s)µ(t) ds dt

+

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)

]|γ′(s)|〈γ′(t), h′(t)〉µ(s)µ(t) ds dt

=

∫ `

0

∫ t+ `2

t− `2

[2∫ ts〈γ′(u), h′(u)〉 du

(t− s)3− 2〈γ(s)− γ(t), h(s)− h(t)〉

|γ(s)− γ(t)|4

]

· |γ′(s)||γ′(t)|µ(s)µ(t) ds dt

+ 2

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)

]|γ′(s)|〈γ′(t), h′(t)〉µ(s)µ(t) ds dt

Since γ is parametrized by arclength, Dγ(s, t) = |s − t| if the bounds of the integral

are changed to t− `2, t+ `

2. This along with some simple algebra then gives

dE

dε(γε, µ)

∣∣∣∣ε=0

158

= 2

∫ `

0

∫ t+ `2

t− `2

[[∫ ts〈γ′(u), h′(u)〉 du

(t− s)3− 〈γ(s)− γ(t), h(s)− h(t)〉

|γ(s)− γ(t)|4

]|γ′(t)|

+

[1

|γ(s)− γ(t)|2− 1

|s− t|2

]〈γ′(t), h′(t)〉

]|γ′(s)|µ(s)µ(t) ds dt,

which completes the proof.

Before proceeding to our second representation of the first variation formula,

we must prove a lemma similar to Lemma 5.1.

Lemma 5.16. Let γ : [0, `] → Rn and h : [0, `] → Rn be arclength parametrizations

of closed C∞ curves and define

G(s, t) =

∫ ts〈γ′(u), h′(u)〉 du

(t− s)3− 〈γ(s)− γ(t), h(s)− h(t)〉

|γ(s)− γ(t)|4.

Then G is continuously extendable to [0, `]× [0, `] by taking

G(s, s) =1

12〈γ′′(s), h′′(s)〉 − |γ

′′(s)|2

6〈γ′(s), h′(s)〉.

Proof. As in Lemma 5.1, we shall prove the result through the use of Taylor’s theorem

[8]. We begin by writing the Taylor approximation for γ(t) centered at s, specifically

γ(t) = γ(s) +B(t− s) + C(t− s)2 +D(t− s)3 + E(t− s)4 + R4(t− s)5,

where

B = γ′(s), C =γ′′(s)

2, D =

γ′′′(s)

6, E =

γ′′′′(s)

24, and R4 =

∣∣∣∣ R4

|t− s|5

∣∣∣∣ .We also write a similar approximation for h(t) centered at s, specifically

h(t) = γ(s) + b(t− s) + c(t− s)2 + d(t− s)3 + e(t− s)4 + r4(t− s)5,

159

where b, c, d, e, r4 are defined similarly to their capital counterparts. This allows us

to write the following for appropriately defined H1:

〈γ′(v + s), h′(v + s)〉 = B • b+ (2B • c+ 2b • C)v + (3B • d+ 3b •D + 4C • c)v2

+ (B • e+ 4b • E + 6C • d+ 6c •D)v3 + 5H1v4

We can use a change of variables to obtain the following for appropriately defined H2:∫ t

s

〈γ′(u), h′(u)〉 du

=

∫ t−s

0

〈γ′(v + s), h′(v + s)〉 dv

=

[B • bv + (B • c+ b • C)v2 +

(B • d+ b •D +

4

3C • c

)v3

+

(B • e+ E • b+

3

2C • d+

3

2D • c

)v4 +H2v

5

]∣∣∣∣∣t−s

0

= B • b(t− s) + (B • c+ b • C)(t− s)2 +

(B • d+ b •D +

4

3C • c

)(t− s)3

+

(B • e+ E • b+

3

2C • d+

3

2D • c

)(t− s)4 +H2(t− s)5

Abbreviating κabsγ (s) = κ(s) and appropriately defining H3, we can obtain

|γ(s)− γ(t)|4 = (t− s)4 − κ2(s)

6(t− s)6 +

−κ(s)κ′(s)

6(t− s)7 +H3(t− s)8.

We can then compute the following for appropriately defined H4 and H5:

G(s, t) =

∫ ts〈γ′(u), h′(u)〉 du

(t− s)3− 〈γ(s)− γ(t), h(s)− h(t)〉

|γ(s)− γ(t)|4

=13C • c− κ2(s)

6B • b+H4(t− s)

1 +H5(t− s)2


G(s, s) =1

3C • c− κ2(s)

6B • b =

1

12〈γ′′(s), h′′(s)〉 − |γ

′′(s)|2

6〈γ′(s), h′(s)〉.

160

The proof of the continuous extendability of G to [0, `]× [0, `] is similar to that shown

in Lemma 5.1.

Proposition 5.17. The first variation formula for E(·, µ) along the concatenation

of paths (2) and (3) in the Figure 5.1 is

dE

dε(γε, µ ◦ φε)

∣∣∣∣ε=0

= 2

∫ `

0

∫ t+ `2

t− `2

[ [1

|γ(s)− γ(t)|2− 1

|s− t|2

]µ′(s)

·(∫ s

0

〈γ′(u), h′(u)〉 du− s

L

∫ `

0

〈γ′(u), h′(u)〉 du)

+

[∫ ts〈γ′(u), h′(u)〉 du

(t− s)3− 〈γ(s)− γ(t), h(s)− h(t)〉

|γ(s)− γ(t)|4

]|γ′(t)||γ′(s)|µ(s)

+

[1

|γ(s)− γ(t)|2− 1

|s− t|2

]〈γ′(t), h′(t)〉|γ′(s)|µ(s)

]µ(t) ds dt.

Proof. Simply add the results of Propositions 5.12 and 5.15 and perform minor al-

gebra. Note that Lemmas 5.1 and 5.16 are needed to show the continuity of the

integrand. Consequently, the formula for dEdε

(γε, µ ◦ φε)∣∣ε=0

has a convergent inte-

gral, and it is valid for all variations.

5.2 Weight Variations

In this subsection we shall cover three different weight variations. The equi-

total variation preserves the total amount of weight while changing its arrangement

and is explored in Subsection 5.2.1. The translatory variation preserves the weight

profile, and simply “slides” the weight along the curve—this is investigated in Sub-

section 5.2.2. Lastly, we use Subsection 5.2.3 to discuss the compromisal variation,

which blends two distributions into a single weight function.

161

Definition 5.3. Throughout Section 5.2 we shall use the abbreviations

g(s, t) =

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]|γ′(s)||γ′(t)| and F (s) =

∫ `

0

g(s, t)µ(t) dt.

Remark. Recall Freedman, He, and Wang’s definition of the energy relative to a point

[4]. The function we call F (s) in Definition 5.3 is analogous to this definition in the

case of the nonuniform energy. Following their convention, we write E(γ, µ, s) = F (s).

5.2.1 The Equitotal Variation

Let us consider the variation of E(γ, µ) in which the curve γ is fixed but the

weight described by µ is allowed to “reallocate.” Note that this requires the total

weight to remain fixed throughout the variation, which is why we force η to have an

integral of zero in Definition 5.5.

Definition 5.4. Let µ : [0, `] → R+ be a weight function. Then we call W =∫ `0µ(s) ds the total weight.

We can easily derive a simple bound for E(γ, µ) in terms of the total weight.

Proposition 5.18. Let γ : [0, `]→ Rn be a closed C∞ curve, let µ : [0, `]→ R+ be a

weight function, and define

g0 = min(s,t)

g(s, t) and W =

∫ `

0

µ(t) dt.

Then E(γ, µ) ≥ g0W2.

Proof. Recall that g0 must exist and be nonnegative as a result of Lemma 5.1 (g

is continuous on the square [0, `] × [0, `], a compact set). As both g and µ are

162

nonnegative, we must simply observe that

E(γ, µ) =

∫ `

0

∫ `

0

g(s, t)µ(s)µ(t) ds dt

≥∫ `

0

∫ `

0

g0µ(s)µ(t) ds dt

= g0

∫ `

0

µ(s) ds

∫ `

0

µ(t) dt = g0W2.

Hence E(γ, µ) ≥ g0W2, as desired.

Definition 5.5. For a closed C∞ curve γ : [0, `] → Rn parametrized by arclength

and a weight function µ : [0, `]→ R+, let η : [0, `]→ R be such that

∫ `

0

η(s) ds = 0.

We then define µε(s) : [0, `] → R by µε(s) = µ(s) + εη(s), and we call E(γ, µε) the

equitotal variation of E in the η direction. An example is shown in Figure 5.2.

Figure 5.2: While they have different distributions, each weighted knot has the same

total weight.

163

Remark. Notice that Definition 5.5 implies

∫ `

0

µε(s) ds = W =

∫ `

0

µ(s) ds,

for all ε, which is why we call Definition 5.5 the equitotal variation.

Proposition 5.19. Let γ be a closed C∞ curve parameterized with respect to arclength

and let µ be a weight function. The equitotal variation of E in the η direction gives

the first variation of E(γ, µ) as

dE

dε(γ, µε)

∣∣∣∣ε=0

= 2

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(t)η(s) ds dt.

Proof. We can obtain this result from direct calculation, which is detailed in the

following:

dE

dε(γ, µε)

∣∣∣∣ε=0

=d

dε

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µε(s)µε(t) ds dt

∣∣∣∣ε=0

=

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]∂

∂ε(µε(s)µε(t))

∣∣∣∣ε=0

ds dt

=

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

](η(s)µ(t) + µ(s)η(t)) ds dt

=

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(s)η(t) ds dt

+

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(t)η(s) ds dt

Consider the first double integral; since its limits of integration are identical, we may

exchange the labels s and t in its integrand without changing its value. Further,

note that two terms inside the brackets within the integrand are symmetric in s

and t. Hence, we may return s and t to their “original” places within the brackets.

Ultimately, this manipulation causes the first double integral to become identical to

164

the second double integral, so we may simply combine the two integrals to get the

desired result.

A final note: the exchange of the integral and derivative above is possible

because the derivative of the integrand with respect to ε is continuous in each s, t,

and ε.

We can now use this formula derive some conclusions about critical points.

Proposition 5.20. µ is a critical point of the equitotal variation of E(γ, µ) if and

only if F (s) ≡ C, where C ∈ R≥0.

Proof. Note that the definition of F (s) and the statement of Proposition 5.19 allow

us to write

dE

dε(γ, µε)

∣∣∣∣ε=0

= 2

∫ `

0

∫ `

0

g(s, t)µ(t)η(s) dt ds = 2

∫ `

0

F (s)η(s) ds. (5.1)

Proof of (⇒): Let µ be a critical point of E(γ, µ). If µ is a critical point, then

Equation (5.1) is equal to 0 for every choice of η. Now define

A =1

`

∫ `

0

F (s) ds.

Then we have that∫ `

0

(F (s)− A) ds =

∫ `

0

F (s) ds− À = À− À = 0.

Hence, we may choose η0(s) = F (s)− A. Now observe the following:∫ `

0

(F (s)− A)2 ds =

∫ `

0

(F (s)− A)η0(s) ds

=

∫ `

0

F (s)η0(s) ds− A∫ `

0

η0(s) ds

= 0− A · 0 = 0

165

Notice that the first integral in the second line because µ was assumed to be a critical

point. Of course, (F (s)− A)2 ≥ 0, so then∫ `

0

(F (s)− A)2 ds = 0 =⇒ F (s)− A ≡ 0 =⇒ F (s) ≡ A.

Thus, way may conclude that F (s) is constant when µ is a critical point.

Proof of (⇐): Let F (s) ≡ C, where C ∈ R≥0. Then using Equation (5.1) we may

write

dEµεdε

(γ)

∣∣∣∣ε=0

= 2

∫ `

0

Cη(s) ds = 2C

∫ `

0

η(s) ds = 0,

for all choices of η, since∫ `

0η(s) ds = 0 for any η. Hence, µ is a critical point of

E(γ, µ).

Remark. The remark following Definition 5.3 allows us to restate Proposition 5.20 as

“µ is a critical point of E(γ, µ) ⇐⇒ the nonuniform energy relative to a point is

constant.”

Corollary 5.21. If µ0 is a critical point of the equitotal variation of E(γ, µ) then

E(γ, µ0) = CW , where C is the constant to which F (s) is equal.

Proof. As µ0 is a critical point, we have that F (s) ≡ C from Proposition 5.20. We

can then use the definition of F (s) to write

E(γ, µ0) =

∫ `

0

F (s)µ0(s) ds =

∫ `

0

Cµ0(s) ds = C

∫ `

0

µ0(s) ds = CW,


Lemma 5.22. If γ is C∞ and parametrized by arclength, we may write

F ′(s) = 2

∫ s+ `2

s− `2

[1

(s− t)3− 〈γ

′(s), γ(s)− γ(t)〉|γ(s)− γ(t)|4

]µ(t) dt.

166

Proof. The result is derived from the following calculation; note the simplifications

afforded by the fact that γ is parametrized by arclength.

d

dsF (s) =

d

ds

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]|γ′(s)||γ′(t)|µ(t) dt

=d

ds

∫ s+ `2

s− `2

[1

|γ(s)− γ(t)|2− 1

(s− t)2

]µ(t) dt

=

[1∣∣γ(s)− γ(s+ `

2

)∣∣2 − 1

(s−(s+ `

2

))2

]µ

(s+

`

2

)

−

[1∣∣γ(s)− γ(s− `

2

)∣∣2 − 1

(s−(s− `

2

))2

]µ

(s− `

2

)

+

∫ s+ `2

s− `2

∂

∂s

[1

|γ(s)− γ(t)|2− 1

(s− t)2

]µ(t) dt

=

[1∣∣γ(s)− γ(s+ `

2

)∣∣2 − 1(− `

2

)2

]µ

(s+

`

2

)

−

[1∣∣γ(s)− γ(s− `

2

)∣∣2 − 1(`2

)2

]µ

(s− `

2

)

+

∫ s+ `2

s− `2

[−2〈γ′(s), γ(s)− γ(t)〉|γ(s)− γ(t)|4

+2

(s− t)3

]µ(t) dt

The `-periodicity of γ and µ allow the first two terms to cancel, and the third terms

can be immediately rearranged to obtain the desired result.

Of course, the Leibniz integral rule demands that the partial derivative of the

integrand with respect to s be continuous in both s and t in order for us to perform

the above calculation. This continuity can be shown via the third degree Taylor

approximation of γ [8]. (Also see Lemma 5.1.)

Corollary 5.23. If µ is a critical point of E(γ, µ), then

0 =

∫ s+ `2

s− `2

[1

(s− t)3− 〈γ

′(s), γ(s)− γ(t)〉|γ(s)− γ(t)|4

]µ(t) dt.

167

Proof. Proposition 5.20 tells us that F (s) will be constant when µ is a critical point,

so F ′(s) = 0 when µ is a critical point. Combining this with Lemma 5.22 and dividing

by 2 then gives the desired result.

Proposition 5.24. Let γ : [0, `] → Rn be a closed C∞ curve parametrized by ar-

clength, and define

M1 =

{µ : [0, `]→ R+ | µ ∈ C0,

∫ `

0

µ(t) dt = 1

}.

Then it follows that

infµ∈M1

E(γ, µ) = 0

if and only if there exists t0 ∈ [0, `] such that κabsγ (t0) = 0.

Proof. Proof of (⇒): Assume that infµ∈M1 E(γ, µ) = 0. As the total weight of any

µ ∈M1 equals 1, it follows from Proposition 5.18 that E(γ, µ) ≥ g0 for all µ ∈M1,

where

g0 = min(s,t)

g(s, t).

Hence, we know that g0 = 0. There are two situations in which we could obtain

g0 = 0:

Case 1: There exists s0, t0 ∈ [0, `] with s0 6= t0 such that g(s0, t0) = 0.

Recalling the definition of g allows us to write

0 =1

|γ(s0)− γ(t0)|2− 1

(s0 − t0)2=⇒ |γ(s0)− γ(t0)| = |s0 − t0|.

This last inequality only occurs when the segment γ([s0, t0]) is a line, which implies

that κabsγ (s) = 0 for all s ∈ [s0, t0].

168

Case 2: There exists s0 ∈ [0, `] such that g(s0, s0) = 0.

Recall from Lemma 5.1 that g(s, s) =(κabsγ (s))2

12. As a result, g(s0, s0) = 0

implies(κabsγ (s0))2

12= 0, which implies κabs

γ (s0) = 0, as desired.

Proof of (⇐): Assume that there exists t0 ∈ [0, `] such that κabsγ (t0) = 0. Without

loss of generality, we may assume that t0 = `2.

To prove that infµ∈M1 E(γ, µ) = 0, we shall describe a sequence {µn} such

that {E(γ, µn)} → 0 and µn ∈M1 for all n. For each n ∈ N for which n > ` define

hn = n2− `

2+ 1

nand µn : [0, `]→ R+ (illustrated in Figure 5.3) by

µn(s) =

hn s ∈[`2− 1

2n, `

2+ 1

2n

]n2−`n

2s+ −`n2+(3+`2)n2−3`n+4

4ns ∈ ( `

2− 3

2n, `

2− 1

2n)

`n−n2

2s+ `n2+(3−`2)n2−3`n+4

4ns ∈ ( `

2+ 1

2n, `

2+ 3

2n)

1n

s 6∈(`2− 3

2n, `

2+ 3

2n

).

It is important here to note that for all values of n > ` we have∫ `

0

µn(s) ds = ` · 1

n+

1

n

(hn −

1

n

)+ 2

[1

2· 1

n

(hn −

1

n

)]=`

n+ 2

1

n

(n

2− `

2

)=`

n+n− `n

= 1,

which confirms that each µn is indeed an element of M1.

We now define three subsets of the [0, `]× [0, `] square as follows; this partition

is illustrated in Figure 5.4.

I =

(`

2− 3

2n,`

2+

3

2n

)×(`

2− 3

2n,`

2+

3

2n

)II =

(((`

2− 3

2n,`

2+

3

2n

)× [0, `]

)∪(

[0, `]×(`

2− 3

2n,`

2+

3

2n

)))\ I

169

Figure 5.3: The graph of µn, as used in Proposition 5.24.

III =([0, `]× [0, `]

)\ (I ∪ II)

Using Figure 5.4 as a guide, we shall derive an estimate for E(γ, µn). Recall

that g(s, t) is continuously extendable to [0, `] × [0, `], so then there exists some

M ∈ R such that |g(s, t)| < M for all s, t. Note that g(`2, `

2

)=

(κabsγ ( `2

))2

12= 0, and let

εn represent the least upper bound of g(s, t) in I; that is, |g(s, t)| ≤ εn when (s, t) ∈ I.

Then the continuity of g will ensure that as n→∞ we will have εn → 0. (All these

facts about g follow from Lemma 5.1.)

Notice that the integrand representing E(γ, µn) can be thought of as a product

170

Figure 5.4: The division of [0, `]× [0, `] into regions I, II, and III for Proposition 5.24.

with three factors; that is, we can write

E(γ, µn) =

∫ `

0

∫ `

0

g(s, t)µn(s)µn(t) ds dt.

Then an estimate of E(γ, µn) in each region can be made by multiplying three factors

with a bound for the area of the region, as follows:

Region I:

(εn · hn · hn) ·(

3

n· 3

n

)=

(3

2− 3`

2n+

3

n2

)2

εn.

Region II:

(M · hn ·

1

n

)· 2(` · 3

n

)=

(3− 3`

n+

6

n2

)`M

n.

171

Region III: (M · 1

n· 1

n

)· (` · `) =

`2M

n2.

Note that each of these estimates will go to 0 as n→∞, so we have that E(γ, µn)→ 0

as n→∞. It then follows that

infµ∈M

E(γ, µ) = 0,

which completes the proof.

This result immediately leads to a corollary about the existence of minimizers

for the equitotal variation.

Corollary 5.25. Let γ : [0, `]→ Rn be a closed C∞ curve parametrized by arclength

such that there exists t0 ∈ [0, `] with κabsγ (t0) = 0. Then the equitotal weight variation

does not have an absolutely minimizing weight function for γ.

Proof. Let MW be the set of all weight functions parametrized on [0, `] with total

weight W , and without loss of generality take W = 1. It then follows from Proposi-

tion 5.24 that infµ∈M1 E(γ, µ) = 0. Hence, it only remains to show that there does

not exist µ0 ∈M1 such that E(γ, µ0) = 0.

Since γ is not a line (as it is closed), there exists s0, t0 ∈ [0, `] such that

|γ(s0)− γ(t0)| < |s0 − t0|,

which implies that g(s0, t0) > 0. As g is continuously extendable to [0, `] × [0, `] by

Lemma 5.1, we then have that there exists some δ1, δ2 > 0 such that g(s, t) > 0 for all

(s, t) ∈ (s0− δ1, s0 + δ1)× (t0− δ2, t0 + δ2). As it is required that any weight function

172

µ is positive everywhere, we know regardless of our choice of µ that the integrand of

E(γ, µ) must make a positive contribution on (s0 − δ1, s0 + δ1) × (t0 − δ2, t0 + δ2),

which is a region of nonzero measure. Since g is never negative, this ultimately forces

E(γ, µ) to be positive. Hence, no µ exists such that E(γ, µ) = 0.

Remark. Proposition 5.24 and Corollary 5.25 can be extended for non-closed curves.

However, the proof needs to be adjusted to handle the case in which the point of zero

curvature is an endpoint.

Proposition 5.26. The equitotal variation of E in the η direction gives the second

variation of E(γ, µ) as

d2E

dε2(γ, µε)

∣∣∣∣ε=0

= 2

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]η(s)η(t) ds dt

when γ is rectifiable, closed, C∞, and parametrized with respect to arclength.


following:

d2E

dε2(γ, µε)

∣∣∣∣ε=0

=d2

dε2

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µε(s)µε(t) ds dt

∣∣∣∣ε=0

=

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]∂2

∂ε2(µε(s)µε(t))

∣∣∣∣ε=0

ds dt

=

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]· ∂

2

∂ε2(µ(s)µ(t) + εµ(s)η(t) + εµ(t)η(s) + ε2η(s)η(t)

) ∣∣∣∣ε=0

ds dt

=

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]2η(s)η(t) ds dt

173

Corollary 5.27. If γ(t) = (cos t, sin t) and µ = c for some c ∈ R+, then (γ, µ) is a

critical point of the equitotal weight variation.

Proof. It follows from calculation that F (s) ≡ C. Indeed, note for the circle that

F (s) =

∫ `

0

c · g(s, t) dt

is constant since the circle is invariant under rotation about the origin, and F (s) is

independent of s. As Proposition 5.20 tells us that µ is a critical point of the equitotal

variation if and only if F (s) ≡ C, this completes the proof.

Remark. We conjecture that (γ, µ) as described in Corollary 5.27 is neither a relative

minimum or relative maximum of the equitotal weight variation. There is numerical

evidence for this fact, but we do not have a proof.

5.2.2 The Translatory Variation

Let us consider a variation of E(γ, µ) in which the curve γ is fixed but the

weight described by µ is allowed to “slide.” Note that this means that the profile of

the weight function will remain unchanged.

Definition 5.6. For a closed curve γ : [0, `] → Rn parametrized by arclength and a

closed weight function µ : [0, `]→ R+, define µτ (s) : [0, `]→ R+ by µτ (s) = µ(s+ τ)

for τ ∈ R and call E(γ, µτ ) the translatory variation of E. Note that we must use

the extension µ of µ described in Chapter 4 for µτ to make sense; specifically, we

174

use µ : R → R+, where ` is the least period. An example of translatory variation is

shown in Figure 5.5.

Figure 5.5: Each weighted knot has the same weight profile, but they are translations

of each other.

Notice that Definition 5.6 does not change the shape of µ (since µ is periodic),

which is why we call Definition 5.6 the translatory variation.

Remark. The translatory variation of any weighted knot (γ, µ) always has an absolute

minimizer—notice that E(γ, µτ ) changes continuously with τ and the domain of τ is

compact (as γ was assumed to be closed).

We now calculate the first variation formula for the translatory variation.

Proposition 5.28. Let γ be a closed curve parametrized with respect to arclength,

and let µ be a closed C1 weight function. Then the translatory variation of E in the

τ direction gives the first variation of E(γ, µ) as

dE

dτ(γ, µτ )

∣∣∣∣τ=0

= 2

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ(t)µ′(s) ds dt.

175


following:

dE

dτ(γ, µτ )

∣∣∣∣τ=0

=d

dτ

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µτ (s)µτ (t) ds dt

∣∣∣∣τ=0

=

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]∂

∂τ(µτ (s)µτ (t))

∣∣∣∣τ=0

ds dt

=

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

](µ′(s)µ(t) + µ(s)µ′(t)) ds dt

=

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ′(s)µ(t) ds dt

+

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]µ′(t)µ(s) ds dt

Now consider the second double integral. Since the limits of integration are identical,

we may exchange the labels s and t in its integrand without changing its value.

Further, note that two terms inside the brackets within the integrand are symmetric

in s and t. Hence, we may return s and t to their “original” places within the brackets.

Ultimately, this manipulation causes the second double integral to become identical

to the first double integral, so we may simply combine the two integrals to get the

desired result.

Unfortunately, most curves are not easily parametrized with respect to ar-

clength. This leads us to the following corollary, which is in practice a much more

useful form of Proposition 5.28.

Corollary 5.29. Say γ : [0, `] → Rn is a closed curve with an arbitrary regular

parametrization, and set L = length(γ). If µ : [0, λ] → R+ is a closed C1 weight

176

function, the translatory variation of E in the τ direction gives the first variation of

E(γ, µ) as

dE

dτ(γ, µτ )

∣∣∣∣τ=0

= 2

∫ `

0

∫ `

0

[1

|γ(s)− γ(t)|2− 1

Dγ(s, t)2

]|γ′(s)||γ′(t)|

· µ(λ

L

∫ t

0

|γ′(u)| du)µ′(λ

L

∫ s

0

|γ′(u)| du)(

λ

L

)ds dt.

Note: The shift in τ is done so that it is proportional to arclength, not the parameter

of γ.

Remark. Using the notation outlined above, we can rewrite the result of Proposi-

tion 5.28 for curves parametrized with respect to arclength as

dE

dτ(γ, µτ )

∣∣∣∣τ=0

= 2

∫ `

0

∫ `

0

g(s, t)µ(t)µ′(s) ds dt = 2

∫ `

0

F (s)µ′(s) ds.

The following proposition describes a sufficient condition for a critical point

of the translatory variation.

Proposition 5.30. Let γ : [0, `] → Rn be a closed curve parametrized by arclength

and let µ : [0, `]→ R+ be a closed C1 weight function. If F (s) is constant for a given

(γ, µ), then µ is a critical point of the translatory variation of E(γ, µ).

Proof. Say that F (s) is constant, so then F (s) ≡ C for some C. The first variation

formula for the translatory variation can be written as follows:

dE

dτ(γ, µ) = 2

∫ `

0

F (s)µ′(s) ds

= 2

∫ `

0

Cµ′(s) ds

= 2C(µ(`)− µ(0)) = 0

177

Hence, µ is indeed a critical point.

Proposition 5.31. Let γ : [0, `] → Rn be a closed curve parametrized by arclength

and let µ : [0, `] → R+ be a closed weight function. If g(s, t) is diagonally constant

for a given γ; that is, if g(s + a, t + a) = g(s, t) for all a, then E(γ, µ) is invariant

under the translatory variation, regardless of the choice of µ.

Proof. Say that g(s, t) is diagonally constant, as described above. Then the nonuni-

form energy of (γ, µ0) is given as

E(γ, µ0) =

∫ `

0

∫ `

0

g(s, t)µ(s)µ(t) ds dt

Now perform a change of variables with s = x + a and t = y + a. This leads to the

following manipulation:

E(γ, µ0)

=

∫ `−a

−a

∫ `−a

−ag(x+ a, y + a)µ(x+ a)µ(y + a) dx dy

=

∫ `−a

−a

∫ `−a

−ag(x, y)µ(x+ a)µ(y + a) dx dy

=

∫ `

0

∫ `

0

g(s, t)µ(s+ a)µ(t+ a) ds dt

= E(γ, µa)

As each step never changed the value of the expression and a is arbitrary, we have

that diagonal invariance implies translatory constancy.

Remark. Notice that if g(s, t) is diagonally constant, then µ will be a critical point of

the translatory variation of E, regardless of our choice of µ.

178

It is unknown if g(s, t) diagonally constant for some γ implies that F (s) is

constant. However, pure diagonal constancy (and no relation to a curve) is known to

be insufficient to make F constant, as seen in Example 5.1.

Example 5.1. Let g : [0, `] × [0, `] → R+ and µ : [0, `] → R+ both be `-periodic in

all input variables. Then g(s, t) diagonally constant does not imply that

F (s) =

∫g(s, t)µ(t) ds

is constant. To see this, consider the following counterexample:

Let g(s, t) = sin2(s− t) and µ(t) = 2 + sin2(t). We can see immediately that

g(s+ a, t+ a) = sin2(s+ a− t− a) = sin2(s− t) = g(s, t) ∀a,

so g is indeed diagonally constant. However,

F (s) =

∫ 2π

0

g(s, t)µ(t) dt =π(cos 2s+ 10)

4

is not constant.

Most surprisingly, experiments revealed a family of knots and weight functions

whose nonuniform energy does not change when the weight function is translated—

Figure 5.5 is an illustration of such an example, as each weighted knot gives E ≈

305.99. Specifically, we saw this behavior in (p, q)-torus knots and sinusoidal weight

functions given by

γ(t) =

(2 + cos qt) cos pt(2 + cos qt) sin pt

sin qt

and µ(t) = b+ cos(cs),

where p, q, b, c ∈ N, and p, q are relatively prime. This discovery led to Proposi-

tion 5.35 and is further described in Example 5.2.

179

Proposition 5.35. Say ` = 2π, and let µb,c = b + cos(cs) for b, c ∈ N. Then let

q ∈ N, and choose γ parametrized by arclength so that g satisfies the following:

(a) g(s, t) = g(−s,−t)

(b) g(s, t) = g(s+ 2πq, t+ 2π

q)

Then we have two cases:

(i) Say that q is odd and q ≥ 1. Then E(γ, µb,c) is invariant under translatory

variation when c 6= kq for all k ∈ Z.

(ii) Say that q is even and q ≥ 2. Then E(γ, µb,c) is invariant under translatory

variation when c 6= k q2

for all k ∈ Z.

Numerical evidence suggests that we could strengthen the conclusion of Propo-

sition 5.35 by replacing “when” in each of Parts (i) and (ii) with “if and only if,” but

this remains a conjecture.

To prove Proposition 5.35, we must first build up a series of other results.

Proposition 5.32 describes a necessary and sufficient condition for translatory invari-

ance.

Proposition 5.32. Say ` = 2π, and let µb,c = b+ cos(cs) for b, c ∈ N. Then for any

γ parametrized by arclength, E(γ, µb,c) will be translatory invariant exactly when the

integrals ∫ `

0

∫ `

0

g(s, t) sin(cs) ds dt∫ `

0

∫ `

0

g(s, t) cos(cs) ds dt∫ `

0

∫ `

0

g(s, t) cos(c(s+ t)) ds dt

180

∫ `

0

∫ `

0

g(s, t) sin(cs) cos(ct) ds dt

all vanish.

Proof. Keeping elementary trigonmetric identities at hand, observe the following:

µb,c(s+ τ)µb,c(t+ τ)

= (b+ cos(c(s+ τ)))(b+ cos(c(t+ τ)))

= b2 + b cos(cs+ cτ) + b cos(ct+ cτ) + cos(cs+ cτ) cos(ct+ cτ)

= b2 + b cos(cs) cos(cτ)− b sin(cs) sin(cτ) + b cos(ct) cos(cτ)− b sin(ct) sin(cτ)

+ (cos(cs) cos(cτ)− sin(cs) sin(cτ))(cos(ct) cos(cτ)− sin(ct) sin(cτ))

= b2 + b cos(cs) cos(cτ) + b cos(ct) cos(cτ)

− b sin(cs) sin(cτ)− b sin(ct) sin(cτ)

+ cos(cs) cos(ct) cos2(cτ) + sin(cs) sin(ct) sin2(cτ)

− cos(cs) sin(ct) cos(cτ) sin(cτ)− sin(cs) cos(ct) cos(cτ) sin(cτ)

So we can use this derivation to expand E(γ, µb,c;τ ) as follows:

E(γ, µb,c;τ )

=

∫ `

0

∫ `

0

g(s, t)µb,c(s+ τ)µb,c(t+ τ) ds dt

= b2

∫ `

0

∫ `

0

g(s, t) ds dt

+ b cos(cτ)

∫ `

0

∫ `

0

g(s, t) cos(cs) ds dt+ b cos(cτ)

∫ `

0

∫ `

0

g(s, t) cos(ct) ds dt

− b sin(cτ)

∫ `

0

∫ `

0

g(s, t) sin(cs) ds dt− b sin(cτ)

∫ `

0

∫ `

0

g(s, t) sin(ct) ds dt

+ cos2(cτ)

∫ `

0

∫ `

0

g(s, t) cos(cs) cos(ct) ds dt

181

+ sin2(cτ)

∫ `

0

∫ `

0

g(s, t) sin(cs) sin(ct) ds dt

− cos(cτ) sin(cτ)

∫ `

0

∫ `

0

g(s, t) cos(cs) sin(ct) ds dt

− cos(cτ) sin(cτ)

∫ `

0

∫ `

0


Using symmetry, we can collapse this sum of integrals into the following:

E(γ, µb,c;τ )

= b2

∫ `

0

∫ `

0

g(s, t) ds dt

+ 2b cos(cτ)

∫ `

0

∫ `

0

g(s, t) cos(cs) ds dt− 2b sin(cτ)

∫ `

0

∫ `

0

g(s, t) sin(cs) ds dt

+ cos2(cτ)

∫ `

0

∫ `

0


+ sin2(cτ)

∫ `

0

∫ `

0


− 2 cos(cτ) sin(cτ)

∫ `

0

∫ `

0


Applying the Pythagorean identity to the third line then gives the following:

E(γ, µb,c;τ )

= b2

∫ `

0

∫ `

0

g(s, t) ds dt

+ 2b cos(cτ)

∫ `

0

∫ `

0


∫ `

0

∫ `

0


+ cos2(cτ)

∫ `

0

∫ `

0


+ (1− cos2(cτ))

∫ `

0

∫ `

0



∫ `

0

∫ `

0


= b2

∫ `

0

∫ `

0

g(s, t) ds dt

182

+ 2b cos(cτ)

∫ `

0

∫ `

0


∫ `

0

∫ `

0


+

∫ `

0

∫ `

0

g(s, t) sin(cs) sin(ct) ds dt+ cos2(cτ)

∫ `

0

∫ `

0

g(s, t) cos(cs+ ct) ds dt


∫ `

0

∫ `

0


Now define the following:

U =

∫ `

0

∫ `

0

g(s, t) ds dt

V =

∫ `

0

∫ `

0

g(s, t) cos(cs) ds dt

W =

∫ `

0

∫ `

0


X =

∫ `

0

∫ `

0


Y =

∫ `

0

∫ `

0

g(s, t) cos(c(s+ t)) ds dt

Z =

∫ `

0

∫ `

0


(Note then that the integrals specified the proposition statement are V , W , Y , and

Z.) Using these definitions, we can write

E(γ, µb,c;τ ) = b2U + 2b cos(cτ)V − 2b sin(cτ)W

+X + cos2(cτ)Y − 2 cos(cτ) sin(cτ)Z.

(5.2)

Proof of (⇒): Assume that E(γ, µb,c) is translatory invariant.

Then we know that the expression on the right side of Equation (5.2) is the

same for all choices of τ . We can then acquire the following equations for different

183

choices of τ :

cτ = 0 E(γ, µb,c;τ ) = b2U + 2bV +X + Y

cτ = π E(γ, µb,c;τ ) = b2U − 2bV +X + Y

cτ =π

2E(γ, µb,c;τ ) = b2U − 2bW +X

cτ =3π

2E(γ, µb,c;τ ) = b2U + 2bW +X

This implies that W = V = 0, so then the above equations become the following:

cτ = 0, π E(γ, µb,c;τ ) = b2U +X + Y

cτ =π

2,3π

2E(γ, µb,c;τ ) = b2U +X

But this also tells us that Y = 0. Hence, we may finally observe the following:

cτ =π

4E(γ, µb,c;τ ) = b2U +X − Z

cτ =3π

4E(γ, µb,c;τ ) = b2U +X + Z

And thus Z = 0. Hence, we have shown that V = W = Y = Z = 0, as desired.

Proof of (⇐): Assume that V = W = Y = Z = 0.

Then it immediately follows from Equation (5.2) that

E(γ, µb,c;τ ) = b2U +X = b2

∫ `

0

∫ `

0

g(s, t) ds dt+

∫ `

0

∫ `

0

g(s, t) sin(cs) sin(ct) ds dt,

which is independent of τ . Hence, E(γ, µb,c) is translatory invariant.

We know by definition that g will have certain properties, in particular, g will

be symmetric and `-periodic in both variables. The combination of two additional

properties with the satisfaction of a certain condition can actually ensure translatory

constancy. But first, we must prove a lemma.

184

Lemma 5.33. Let c, q ∈ N with q ≥ 2 and a ∈ R. Then if c 6= kq for all k ∈ Z, we

haveq−1∑j=0

cos

(ca+

2πcj

q

)= 0.

Proof. Recalling that cos(z) = 12(eiz + e−iz) and using the geometric sum formula, we

may observe the following:

q−1∑j=0

cos

(ca+

2πcj

q

)=

q−1∑j=0

1

2

(eica+ 2πicj

q + e−ica−2πicjq

)=eica

2

q−1∑j=0

e2πicjq +

e−ica

2

q−1∑j=0

e−2πicjq

=eica

2

(1− e

2πicqq

1− e2πicq

)+e−ica

2

(1− e

−2πicqq

1− e−2πicq

)

=eica

2

(1− e2πic

1− e2πicq

)+e−ica

2

(1− e−2πic

1− e−2πicq

)=eica

2

(1− 1

1− e2πicq

)+e−ica

2

(1− 1

1− e−2πicq

)=eica

2

(0

1− e2πicq

)+e−ica

2

(0

1− e−2πicq

)= 0

Notice that the condition c 6= kq for all k ∈ Z is indeed required, as otherwise the

denominators in the geometric sum formulas would become 0.

Corollary 5.34. Let c, q ∈ N with q ≥ 3 and a ∈ R. Then we have two cases:

(i) Say that q is odd. If c 6= kq for all k ∈ Z, we have

q−1∑j=0

cos

(ca+

4πcj

q

)= 0.

(ii) Say that q is even. If c 6= k q2

for all k ∈ Z, we have

q−1∑j=0

cos

(ca+

4πcj

q

)= 0.

185

Proof. Case 1: q is odd.

We first show that 2c 6= kq for all k ∈ Z. By way of contradiction, say that there

exists k′ ∈ Z such that 2c = k′q. Then k′q is even, but since q is odd, it follows that

k′ is even, so then k′

2∈ Z. But then c = k′

2q and k′ ∈ Z, which is a contradiction.

Hence, we have that 2c 6= kq for all k ∈ Z.

Define c′ = 2c. Then we can immediately write

q−1∑j=0

cos

(ca+

4πcj

q

)=

q−1∑j=0

cos

(c′a

2+

2πc′j

q

).

Since q ≥ 3 and c′ = 2c 6= kq for all k ∈ Z by the above work, we can apply

Lemma 5.33 to the right-hand expression to obtain the desired result.

Case 2: q is even.

We can immediately write

q−1∑j=0

cos

(ca+

4πcj

q

)=

q−1∑j=0

cos

(ca+

2πcj

q/2

).

Since q is even and q ≥ 3, we know that q ≥ 4, hence q2≥ 2 and q ∈ N. Thus, we

can then immediately apply Lemma 5.33 to the right-hand expression to obtain the

desired result.

And now to the proof of the main result:

Proposition 5.35. Say ` = 2π, and let µb,c(s) = b + cos(cs) for b, c ∈ N. Then let

q ∈ N, and choose γ parametrized by arclength so that g satisfies the following:

(a) g(s, t) = g(−s,−t)

(b) g(s, t) = g(s+ 2πq, t+ 2π

q)

186

Then we have two cases:

(i) Say that q is odd and q ≥ 1. Then E(γ, µb,c) is invariant under translatory

variation when c 6= kq for all k ∈ Z.

(ii) Say that q is even and q ≥ 2. Then E(γ, µb,c) is invariant under translatory

variation when c 6= k q2

for all k ∈ Z.

Proof. Notice that in the cases q = 1 and q = 2 the proposition statement tells us

nothing about the effects of the translatory variation. Henceforth, we may assume

that q ≥ 3.

Note that the combination of Property (a) with the periodicity of g tells us

that

g(s, t) = g(2π − s, 2π − t).

Furthermore, note for all a ∈ [0, 2π] that

sin(c(2π − a)) = sin(−ca) = − sin(ca).

Then we can observe that

W =

∫ 2π

0

∫ 2π

0


=

∫ 2π

0

∫ π

0

g(s, t) sin(cs) ds dt+

∫ 2π

0

∫ 2π

π


Now performing a change of variables on the second integral:

W =

∫ 2π

0

∫ π

0


+

∫ 0

2π

∫ 0

π

g(2π − s, 2π − t) sin(c(2π − s)) (−ds) (−dt)

187

=

∫ 2π

0

∫ π

0


∫ 2π

0

∫ π

0

g(2π − s, 2π − t) sin(c(2π − s)) ds dt

=

∫ 2π

0

∫ π

0


∫ 2π

0

∫ π

0

g(s, t)(− sin(cs)) ds dt

=

∫ 2π

0

∫ π

0

g(s, t)(sin(cs)− sin(cs)) ds dt = 0

Hence, we have that W = 0. In preparation for Z, notice that

sin(c(2π − s)) cos(c(2π − t)) = sin(−cs) cos(−ct) = − sin(cs) cos(ct).

It then follows from a similar argument to the one for W = 0 (we simply replace

sin(cs) with sin(cs) cos(cs)) that

Z =

∫ 2π

0

∫ 2π

0

g(s, t) sin(cs) cos(ct) ds dt = 0.

Now note that the second condition on g generalizes to

g(s, t) = g

(s+

2πk

q, t+

2πk

q

)for all k ∈ Z.

This tells us that g will take on the same values over the set

{(s+

2πk

q, t+

2πk

q

)| 0 ≤ k < q − 1

}.

Regardless of the parity of q, we know that q ≥ 2 and c is not a multiple of q. Then

we have for all s ∈ [0, 2π] that

q−1∑i=0

cos

(c

(s+

2πj

q

))= 0

as a result of Lemma 5.33. Set α = 2πq

; we can expand the integral representing V as

188

follows:

V =

∫ 2π

0

∫ 2π

0

g(s, t) cos(cs) dt ds

=

q−1∑j=0

q−1∑i=0

∫ (i+1)α

iα

∫ (i+j+1)α

(i+j)α

g(s, t) cos(cs) dt ds

=

q−1∑j=0

q−1∑i=0

∫ α

0

∫ α

0

g(s+ iα, t+ (i+ j)α) cos(c(s+ iα)) dt ds

=

q−1∑j=0

q−1∑i=0

∫ α

0

∫ α

0

g(s, t+ jα) cos(c(s+ iα)) dt ds

=

q−1∑j=0

∫ α

0

∫ α

0

g(s, t+ jα)

q−1∑i=0

[cos(cs+ αci))

]︸︷︷︸

= 0, shown above

dt ds = 0

The fourth line follows from Property (b), and the last line follows from the fact that

cos(c(s+ iα)) is independent of t and g(s, t+ jα) is independent of i. Hence, we have

that V = 0 in both cases. Following the approach used with V , we can expand the

definition of Y using sums, perform a change of variables, and use Property (b) to

acquire the following:

Y =

∫ 2π

0

∫ 2π

0

g(s, t) cos(c(s+ t)) dt ds

=

q−1∑i=0

q−1∑j=0

∫ (i+1)α

iα

∫ (i+j+1)α

(i+j)α

g(s, t) cos(c(s+ t)) dt ds

=

q−1∑i=0

q−1∑j=0

∫ α

0

∫ α

0

g(s+ iα, t+ (i+ j)α) cos(c(s+ iα + t+ (i+ j)α)) dt ds

=

q−1∑i=0

q−1∑j=0

∫ α

0

∫ α

0

g(s, t+ jα) cos(c(s+ t+ 2iα + jα))

=

q−1∑j=0

∫ α

0

∫ α

0

g(s, t+ jα)

q−1∑i=0

[cos(c(s+ t+ jα) + 2αci)

]︸︷︷︸

= 0 by Corollary 5.34

dt ds

189

The fourth line follows from Property (b), and the last line follows from the fact that

g(s, t + jα) is independent of i. Note that Corollary 5.34 does indeed apply here, as

q ≥ 3 and c is not a multiple of q (if q is odd) or q2

(if q is even). Thus, Y = 0 in for

appropriate c in both cases of q.

The above work then shows that V = W = Y = Z = 0 for appropriate c in

both cases of q. It then follows immediately from Proposition 5.32 that E(γ, µb,c) will

be translatory invariant in these circumstances.

As alluded to in the exposition above, (p, q) torus knots are examples of γ for

which g satisfies Properties (a) and (b) of Proposition 5.35.

Example 5.2. In our calculations we calculated the energy with respect to the ar-

clength parametrization for curves of length 2π, and we used µb,c : [0, 2π] → R+

defined by µb,c = b+ cos(cs) as our weight function.

Given relatively prime p, q ∈ N, (p, q)-torus knots in R3 are parametrized as

follows:

γp,q(t) = φ(pt, qt) for 0 ≤ t ≤ 2π,

where

φ(u, v) = ((R + r cos v) cosu, (R + r cos v) sinu, r sin v) for fixed 0 < r < R.

It then follows that |γp,q(t)| is periodic with period 2πq

and γp,q

(t+ 2π

q

)is a rotation

of γp,q(t) about the z-axis by 2πpq

radians. Define `p,q = length(γp,q), and let γp,q :

[0, `p,q] → R3 be a reparametrization of γp,q with respect to arclength, then γp,q is

invariant under rotations of 2πpq

radians about the z-axis.

190

We can then rescale γp,q via a dilation to obtain γp,q : [0, 2π] → R3, an ar-

clength parametrization of our (p, q)-torus knot, now with length 2π. We conclude

by recognizing that for γp,q we have

g(s, t) = g

(s+

2π

q, t+

2π

q

)and g(s, t) = g(−s,−t),

which tells us that Proposition 5.35 does indeed apply to γp,q.

Remark. There are examples of non-torus knots that satisfy Properties (a) and (b)

of Proposition 5.35, such as

γ(t) =

(2 + cos 3t)(cos t+ cos 7t2

)(2 + cos 3t)(sin t+ sin 7t

2)

32

sin 3t

.This knot is illustrated by Figure 5.6.

Figure 5.6: A non-torus knot satisfying Properties (a) and (b) of Proposition 5.35.

Our original experiments with torus knots led to Conjecture 5.36. Experimen-

tal evidence supports this conjecture, but we have yet to find a formal proof.

191

Conjecture 5.36. There are two cases:

(i) q is odd. Then Eµb,c(γp,q) only fails to be invariant under translatory variation

when c = kq for some k ∈ N.

(ii) q is even. Then Eµb,c(γp,q) only fails to be invariant under translatory variation

when c = k q2

for some k ∈ N.

We usually refer to Conjecture 5.36 as the “resonance conjecture.” We have

shown experimentally that the conjecture holds for all c, p, q ≤ 10, but have yet to

prove it in the general case.

Given the reference to torus knots above, one may wish to inquire about the

behavior of torus curves on the flat torus (Clifford torus) in R4. We are intentional

here about the use of the word “curve”, as such curves cannot be knotted, since any

closed curve in R4 will be isotopic to the unknot. However, this does not discourage us

from discussing their nonuniform energy. In particular, we claim that the nonuniform

energy of every torus curve on the flat torus is invariant under translatory variation,

regardless of the weight distribution. It is helpful here to recall the definitions of the

flat torus and a (p, q)-curve upon it.

Definition 5.7. The flat torus T in R4 is given by the parametrization

T (s, t) = (cos s, sin s, cos t, sin t).

Definition 5.8. A (p, q)-curve γp,q on the flat torus is given by the parametrization

γp,q(s) = (cos ps, sin ps, cos qs, sin qs).

192

These definitions equip us with the tools necessary to prove the translatory

invariance of any (p, q)-curve on the flat torus, regardless of the weight function used.

Proposition 5.37. Let γp,q be a (p, q)-curve on the flat torus T in R4. Then we

claim that E(γp,q, µ) is translatory invariant for any choice of µ.

Proof. If we substitute the definition of γp,q into E, simplify, and apply some trigono-

metric identities, we obtain the following:∫ 2π

0

∫ t+π

t−π

[p2 + q2

4− 2 cos(p(s− t))− 2 cos(q(s− t))− 1

(s− t)2

]µ(s)µ(t) ds dt.

(It is worth noting that γp,q has a constant speed of√p2 + q2, which is what gives us

our numerator here.) We now perform the change of variable s = x + τ , t = y + τ .

Note that the τ will cancel out in each of the (s− t), which gives

=

∫ 2π−τ

−τ

∫ y+π−τ

y−π−τ

[p2 + q2

4− 2 cos(p(x− y))− 2 cos(q(x− y))− 1

(x− y)2

]· µ(x+ τ)µ(y + τ) dx dy.

But the periodicity of the integrand allows us to adjust the limits of integration, which

gives

=

∫ 2π

0

∫ y+π

y−π

[p2 + q2

4− 2 cos(p(x− y))− 2 cos(q(x− y))− 1

(x− y)2

]· µ(x+ τ)µ(y + τ) dx dy.

Note that each of these steps maintains the value of the original integral. Furthermore,

recognize that the final expression is the translatory variation of E(γp,q, µ) for an

arbitrary µ. Since the energy is the same in the first and last expressions, our claim

is proved.

193

5.2.3 The Compromisal Variation

Let us consider the variation of E(γ, µ) in which the curve γ is fixed and

parametrized by arclength, but the weight is described by a variable balance two

weight functions.

Definition 5.9. Let η1, η2 : [0, `]→ R+ be weight functions, and define

µε(s) =1 + ε

2η1(s) +

1− ε2

η2(s).

We call µε(s) the compromise of η1 and η2, and we call E(γ, µε) the compromisal

variation of E.

Proposition 5.38. The compromisal energy can be written as

E(γ, µε) =1

4

(E(γ, η1) + E(γ, η2) + 2

∫ `

0

∫ `

0

g(s, t)η1(s)η2(t) ds dt

)+ε

2(E(γ, η1)− E(γ, η2))

+ε2

4

(E(γ, η1) + E(γ, η2)− 2

∫ `

0

∫ `

0


).


E(γ, µε) =

∫ `

0

∫ `

0

g(s, t)µε(s)µε(t) ds dt

=

∫ `

0

∫ `

0

g(s, t)

(1 + ε

2η1(s) +

1− ε2

η2(s)

)·(

1 + ε

2η1(t) +

1− ε2

η2(t)

)ds dt

=

∫ `

0

∫ `

0

g(s, t)

[1

4η1(s)η1(t)

]ds dt+

∫ `

0

∫ `

0

g(s, t)

[1

4η1(s)η2(t)

]ds dt

+

∫ `

0

∫ `

0

g(s, t)[ ε

4η1(s)η1(t)

]ds dt−

∫ `

0

∫ `

0

g(s, t)[ ε

4η1(s)η2(t)

]ds dt

194

+

∫ `

0

∫ `

0

g(s, t)

[1

4η2(s)η1(t)

]ds dt+

∫ `

0

∫ `

0

g(s, t)

[1

4η2(s)η2(t)

]ds dt

+

∫ `

0

∫ `

0

g(s, t)[ ε

4η2(s)η1(t)

]ds dt−

∫ `

0

∫ `

0

g(s, t)[ ε

4η2(s)η2(t)

]ds dt

+

∫ `

0

∫ `

0

g(s, t)[ ε

4η1(s)η1(t)

]ds dt+

∫ `

0

∫ `

0

g(s, t)[ ε

4η1(s)η2(t)

]ds dt

+

∫ `

0

∫ `

0

g(s, t)

[ε2

4η1(s)η1(t)

]ds dt−

∫ `

0

∫ `

0

g(s, t)

[ε2

4η1(s)η2(t)

]ds dt

−∫ `

0

∫ `

0

g(s, t)[ ε

4η2(s)η1(t)

]ds dt−

∫ `

0

∫ `

0

g(s, t)[ ε

4η2(s)η2(t)

]ds dt

−∫ `

0

∫ `

0

g(s, t)

[ε2

4η2(s)η1(t)

]ds dt+

∫ `

0

∫ `

0

g(s, t)

[ε2

4η2(s)η2(t)

]ds dt.

This expression appears somewhat overwhelming at first, we can collapse it pretty

quickly. First, every integral with matching weight functions can be rewritten us-

ing the definition of the nonuniform energy. Second, any integral containing mixed

weight functions can be rewritten using symmetry to make η1 dependent on s and η2

dependent on t. This particular manipulation gives

E(γ, µε) =1

4E(γ, η1) +

1

4

∫ `

0

∫ `

0

g(s, t)η1(s)η2(t) ds dt+ε

4E(γ, η1)

− ε

4

∫ `

0

∫ `

0

g(s, t)η1(s)η2(t) ds dt+1

4

∫ `

0

∫ `

0


+1

4E(γ, η2) +

ε

4

∫ `

0

∫ `

0

g(s, t)η1(s)η2(t) ds dt− ε

4E(γ, η2)

+ε

4E(γ, η1) +

ε

4

∫ `

0

∫ `

0

g(s, t)η1(s)η2(t) ds dt+ε2

4E(γ, η1)

− ε2

4

∫ `

0

∫ `

0

g(s, t)η1(s)η2(t) ds dt− ε

4

∫ `

0

∫ `

0


− ε

4E(γ, η2)− ε2

4

∫ `

0

∫ `

0

g(s, t)η1(s)η2(t) ds dt+ε2

4E(γ, η2).

And then some quick factoring causes this to become

E(γ, µε) =

(1

4+ε

2+ε2

4

)E(γ, η1) +

(1

4− ε

2+ε2

4

)E(γ, η2)

195

+

(1

2− ε2

2

)∫ `

0

∫ `

0

g(s, t)η1(s)η2(t) ds dt.

Sorting these terms by powers of ε then gives us the desired result.

Corollary 5.39. The first and second derivatives of the compromisal energy with

respect to ε are given as

dE

dε(γ, µε) =

1

2(E(γ, η1)− E(γ, η2))

+ε

2

(E(γ, η1) + E(γ, η2)− 2

∫ `

0

∫ `

0


),

and

d2E

dε2(γ, µε) =

1

2

(E(γ, η1) + E(γ, η2)− 2

∫ `

0

∫ `

0


).

Proof. This follows directly from taking derivatives of the result of Proposition 5.38.

196

REFERENCES

[1] Ahlfors, L. (1966). Complex analysis. New York: McGraw-Hill.

[2] Cantarella, J., Kusner, R., & Sullivan, J. (2002). On the minimum ropelength ofknots and links. Inventiones Mathematicae, 150(2):257-286.

[3] do Carmo, M. (1992). Riemannian geometry. Boston: Birkhauser.

[4] Freedman, M., He, Z., & Wang, Z. (1994). Mobius energy of knots and unknots.Annals of Mathematics, 139:1-50.

[5] Litherland, R., Simon, J., Durumeric, O., & Rawdon, E. (1999). Thickness ofknots. Topology and Its Applications, 91:233-244.

[6] Navanlinna, R., & Paatero, V. (1969). Introduction to complex analysis. London:Addison-Wesley.

[7] O’Hara, J. (2000). Energy of knots. Preprint retrieved fromhttp://www.comp.tmu.ac.jp/knotNRG/texfiles/esurvey.pdf.

[8] Rudin, W. (1976). Principles of Mathematical Analysis (3rd Ed.). New York:McGraw-Hill.

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