DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
Yu Aminov Institute for Low Temperature Physics and Engineering Kharkov, ...
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DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
Yu Aminov Institute for Low Temperature Physics and Engineering Kharkov, Ukraine
CRC P R E S S Aoca Raton London New York Washington, D.C. © 2000 CRC Press
Copyright (Q 2000 Gordon and Breach Science Publishers imprint.
All rights reserved. No part of this book may be reproduced or utilized in any form or by any means. electronic or mechanical, including photocopying and recording, or by any information storage or retrieval system, without permission in writing from the publisher. Printed in Singapore.
Reprinted 2003 by Taylor & Francis I I New Fetter Lane London
EC4E 4EE
Transferred to Digital Printing 2003 Printed in Great Britain by Biddies Short Run Books. King's Lynn
British Library Cataloguing in Publication Data
A catalogue record for this book i available from the British L1brary. ISBN : 9056990918
Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Definition of a Curve
...................................
vii
1
Vectorvalued Functions Depending on Numerical Arguments . . . . . . .
5
The Regular Curve and its Representations . . . . . . . . . . . . . . . . . . . .
9
Straight Line Tangent to a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
Osculating Plane of a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
The Arc Length of a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
The Curvature and Torsion of a Curve . . . . . . . . . . . . . . . . . . . . . . . .
27
Osculating Circle of a Plane Curve . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
Singular Points of Plane Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
Peano's Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
Envelope of the Family of Curves . . . . . . . . . . . . . . . . . . . . . . . . . . .
51
Frenet Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53
Determination of a Curve with Given Curvature and Torsion . . . . . . .
59
Analogies of Curvature and Torsion for Polygonal Lines . . . . . . . . . . .
63
© 2000 CRC Press
CONTENTS
vi
15 Curves with a Constant Ratio of Curvature and Torsion . . . . . . . . . .
65
16 Osculating Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69
17 Special Planar Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
73
18 Curves in Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
19 Curve Filling a Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89
20 Curves with Locally Convex Projection . . . . . . . . . . . . . . . . . . . . . . .
91
21
Integral Inequalities for Closed Curves . . . . . . . . . . . . . . . . . . . . . . .
97
22
Reconstruction of a Closed Curve with Given Spherical Indicatrix of Tangents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
101
23
Conditions for a Curve to be Closed . . . . . . . . . . . . . . . . . . . . . . . . .
103
24
Isoperimetric Property of a Circle . . . . . . . . . . . . . . . . . . . . . . . . . . .
115
25 One Inequality for a Closed Curve . . . . . . . . . . . . . . . . . . . . . . . . . .
119
26 Necessary and Sufficient Condition of the Boundedness of a Curve with Periodic Curvature and Torsion . . . . . . . . . . . . . . . . . . . . . . . .
121
27 Delaunay's Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
125
28 Jordan's Theorem on Closed Plane Curves . . . . . . . . . . . . . . . . . . . .
133
29 Gauss's Integral for Two Linked Curves . . . . . . . . . . . . . . . . . . . . . .
139
30 Knots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
149
3 1 Alexander's Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
157
32 Curves in ndimensional Euclidean Space . . . . . . . . . . . . . . . . . . . . .
169
33 Curves with Constant Curvatures in ndimensional Euclidean Space . .
177
34 Generalization of the Fenchel Inequality . . . . . . . . . . . . . . . . . . . . . .
183
Knots and Links in Biology and One Mystery . . . . . . . . . . . . . . . . . .
187
35
36 Jones' Polynomial, Its Generalization and Some Applications . . . . . . .
191
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
199
© 2000 CRC Press
Preface Differential geometry is a wide domain of modern mathematics, whose significance is incrcasing at present. One of its origins is in the theory of curves. Everybody who wishes to study geometric problems has to begin by studying the theory of curves, where exact definitions, notions, and invariant characteristics are introduced for the first time. Here the initial geometric intuition is formed and then it is developed in the studying of surfaces theory and the geometry of submanifolds. There exist good and extensive monographs devoted to special curves, but the problems of the general theory are not presented. On the other hand, many interesting and important questions on curves are not discussed, in most cases, in the courses on differential geometry in universities. This book is devoted to the general topics of the geometry of curves as well as to some particular results. Presentation begins with important definitions, including definition of a curve. We also introduce basic notions by using sufficiently accessible language. Next, we discuss properties 'in the large' of the curves in Euclidean space, which were presented earlier in scientific articles only. For a plane curve, the conditions on the curvature of a closed curve as a function of the arc length are well known. Therefore Efimov, Fenchel and other geometers state the following problem: what are the necessary and sufficient conditions on the curvature and torsion of a space curve in order for the curve to be closed? Probably effective conditions do not exist. But this question connects with other interesting questions. In this book we investigate problems for special classes of curves, give the working method to obtain the conditions for closed polygonal curves, and give the proof of the BakelmanWerner theorem on necessary and sufficient conditions of the boundedness for curves with periodic curvature and torsion. We investigate the question of the connection between curvature and torsion for curves which we know are closed  curves of the trigonometrical type. An important geometrical characteristic of a curve is its indicatrix of tangents which we construct in the following way. Let P be a point on curve r and T(P)a unit vii © 2000 CRC Press
viii
PREFACE
tangent vector of r at point P. We translate T ( P )in such a manner that the origin of T ( P )coincides with the origin 0 of Cartesian coordinates. The set of end points of translated vectors r ( P ) is called the spherical indicatrix of tangents of the curve l?. For a closed regular curve this set is not arbitrary: it cannot lie on a hemisphere. This circumstance was probably first noticed by Poznjak. It was observed that the mentioned necessary condition is also sufficient. In this book we give the proof of the Vygodsky theorem: if y is a closed curve on the unit sphere such that y does not lie on any hemisphere, then y is the spherical indicatrix of some space curve. The short proof given in the book belongs to the wellknown mathematician M. Krein. The wonderful French mathematician and astronomer Ch. Delaunay proposed a problem to obtain the curve of constant curvature k = 1 which passes through given two points and has the smallest or the greatest length. We give the solution of this problem by K. Weierstrass. Later Schwarz formulated a theorem that the length of such a curve cannot lie in some interval, but did not publish any proof. Then Schur, motivated by Hilbert, proved this theorem by using the twisting of a plane curve. By the twisting of y Schur meant a transformation of y preserving the curvature and the length of y. So, the twisting here is some process. Schur proved that if a plane curve y with end points P and T forms, together with the span PT, a closed convex curve, then as a result of the twisting the length of the span PT only increases. In 1929, Fenchel proved a theorem that the integral of curvature for a closed curve is not less than 27r. Borsuk proposed that for a knotted curve this integral is not less than 47r. Fery and Milnor proved his assumption almost simultaneously in 19491950. Later this theorem was generalized for ndimensional submanifolds by Chern, Lachof, Ferus and others. In 1995, V. Gorkavy suddenly obtained a generalization of the Fenchel inequality in a new direction  for the higher curvatures of a curve in ndimensional Euclidean space. In accordance with the title of the book, we pay much attention to topological questions. First, we discuss Gauss's classical integral for two linked curves. We observe the links between two infinitesimally close curves and prove the formulas of Calagareanu and White, which have application in biology. Here also twisting arose, but as a number characterizing the form of a curve. Later we discuss knots and knot groups, and give proof of the PontryaginFrank1 theorem that every knot is the boundary of some oriented surface. In a geometrical way we construct Alexander's polynomial. It is the knot invariant and the proof is founded on the three kinds of changes in the structure of the knot diagram. A list of the simplest knots and their Alexander polynomials are given. In 1983, Jones constructed a new polynomial using the braid theory and some Markov theorems. This polynomial was a great surprise to topologists. In 1985, almost simultaneously six mathematicians constructed a new polynomial depending on three arguments. It was called the HOMFLY polynomial. It is possible to obtain two previous polynomials from the HOMFLY polynomial. We give the method for calculating the HOMFLY polynomial, moving from simple knots and links to more complicated ones. A very interesting direction for the application of differential geometry and knot theory arose in the biology of DNA molecules. Because biologists obtained closed DNA molecules, the following question began to concern them: how could the © 2000 CRC Press
PREFACE
ix
process of replication of DNA molecules take place, if two chains of nucleotides are very closely linked? The chapter devoted to this question and the list of works give readers the possibility of acquainting themselves with important research in this direction. Three chapters are devoted to the theory of curves in ndimensional Euclidean space. We give the definition and formulas for the calculation of all curvatures of curves, and obtain the canonical form for curves with all constant curvatures. Its behavior is essentially different in spaces of odd and even dimensions. The curves are applied in mathematics and technology, so investigation of them is very relevant at present. I would like to express my thanks to V. Gorkavy for the translation and other assistance. I am very grateful to Dr R. Rennie for his useful indications on the works of Jones and Witten. I am also grateful to Professor Nigel Hitchin who encouraged me to introduce the section on DNA.
© 2000 CRC Press
l Definition of a Curve The notion of a curve is one of the most important notions in differential geometry. In antiquity this notion had no explicit mathematical definition. Euclid, for example, defines a curve as a "length without width". At this time many wonderful and interesting curves were discovered and studied; however, the idea of a general curve remained at a trivial, obvious level. Further technological progress required the development of natural science, especially the evolution of mechanics and mathematics. It was necessary to understand clearly the foundations of mathematics and, in particular, to construct an accurate definition of a curve. The coordinates method proposed by Descartes prepared the way for a general definition of curves; mathematicians contemporary to Descartes defined a plane curve given by an equation @(x,y)= 0 as a set of points such that their Cartesian coordinates satisfy this equation. Another idea arose in mechanics: a curve is imagined as the trace of some moving point, whose coordinates depend on the time t. Jordan proposed the following definition: a space curve is a set of points whose Cartesian coordinates X, y, z are continuous functions
of some parameter t varying inside a real axis segment (a, 6); in other words, a curve is defined as the image of a real axis segment under a continuous map into the space. This definition seemed to be natural, but in 1890 Peano constructed a continuous map of a segment (a, h) into the space such that the image of (a, h) under this map covered the whole square (we will consider Peano's example in one of the following chapters). In 1897 Klein remarked: "What is an arbitrary curve?. . . One may say that at present in mathematics there exists no more dark and more indefinite notion
© 2000 CRC Press
2
DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
than the mentioned one. The object, which we call a mass curve, is a strip, whose length is sufficiently great with respect to another strip's measures. But for a curve to be a subject of strong mathematical consideration, we must idealize a curve in the same way as a point is idealized. And here some difficulties arise. . . Let us turn to a proposition playing an essential role in Riemann's investigations into foundations of geometry: the space can be viewed as a threedimensional continuous manifold.. . We start from a construction of some scale on a mass straight line; then we decompose the scaled line into smaller parts and continue this operation until it is realizable. After that we make the most important step from an experience to an axiom: we postulate that the correspondence between points and real numbers is valid not only empirically, hut also absolutely. . . ." We remark that Veronese considered a geometry with the following assumption: on the real axis there exist numbers different from the rational and irrational numbers; but it seems that this supposition does not lead to essential geometrical statements and there is no natural foundation for it at present. Another extreme way of looking at space is proposed by discrete geometry. Riemann noted: "The question of the validity of the assumptions of geometry 'in the small' is closely connected with the question of inner sources of metric relations in the space. Certainly, this question belongs to the theory of space and we must take into account that in the case of a discrete manifold the principle of metric relations is contained in the notion of this manifold, whereas in the case of a continuous manifold we have to seek it in some other place. From this it follows that either the real space is a discrete manifold, or we must explain the appearance of metric relations by something exterior.. . ." In modern differential geometry a slightly modified definition of Jordan is used. First, we will give the definition of'un elementary curve. Let p be a map of a segment (a,h) of the real axis into the Euclidean space; we denote by y the image of (a,b) under p. The map p is called continuous at a point X r (a, b), if for any positive t there exists a positive S such that the following condition is fulfilled: if a point Y E (a,b) satisfies the inequality IX  YI S, then the distance between the points p(X), p(Y) is less than t. The map p is said to be continuous if it is continuous at each point of the segment (a,h). The map p is called onetoone if the preimage of any point P E 7 consists of one point. If y is onetoone, then one can construct the assigns to a map p' inverse to p. The domain of definition of p' is y;the map point P E y its preimage under the map p; in other words, if P E y and P = p ( X ) , then p P 1 ( P )= X by definition. The inverse map pp' is continuous at P E y if for any positive t there exists a positive S such that the following condition is fulfilled: if Q E y and the distance between P, Q is less than S, then I p p l ( P ) p'(Q)l 5 t .
<
Definition A set of the Euclidean space is called an elementary curve if this set is the image of an interval of the real axis under a onetoone continuous map, whose inverse map is continuous too.
We remark that a onetoone continuous map, whose inverse map is continuous, is called a hnmeomorphism or a topological map. © 2000 CRC Press
DEFINITION OF A CURVE
3
Suppose the considered set y is an elementary curve. A point X E (a, h) can be viewed as a real number t E (a, h); under the map cp some point P = p ( X ) is assigned to the point X. Assume that Cartesian coordinates X, y, z are fixed in the space. Then one can consider the coordinates of the point P as functions of the real parameter t:
These equalities are called a parametric representation of the curve y;sometimes y is said to be parumetrized by the parameter t E (a, 6). If the interval ( a ,h) is topologically mapped onto an interval ( c , d ) , we can view t as a monotone continuous function of a parameter r t ( c , d ) . Since the composition of the topological maps (c,d ) 1 (a,h) and (a,b) + y is a homeomorphism, the curve y can be parametrized by the parmeter 7 . Thus y can be presented as
as well as in form (1.1). We note that an elementary curve can be very complicated. For example, the projection of some elementary curve into a plane can be Peano's curve covering a square. Definition An elementary curve y is called a sinzpl~curve if y is the image either of a .segment of the real axis, or of a circle under a homeomorphism. The image of a circle under u homc~ornorphisrnis called a clo.sed Jordan curve.
Connection property of curves
We will prove the connection property of curves, which is one of the most important properties of curves. First, we say that a point X. of the Euclidean space is a limit point for a set M if in any neighborhood of X. there exists a point of M. The set M is said to be connected if it cannot be decomposed into two disjoint subset M * ,M2 such that each of M , , M2 does not contain limit points of the other. (We note that these definitions are valid not only for the sets of Euclidean space, but also for the subsets of any set of points, where the notion of neighborhoods is defined correctly.) Let us prove that a segment (a, h) of the real axis is a connected set. Assume the converse; then the segment (a, h) is decomposed into two sets M 1 ,M 2 such that limit points of M, do not belong to M, at i # j. We observe that M , is closed. Indeed, from the assumption it follows that if a point X belongs to M ] , then there exists a neighborhood Ux of X consisting of points of the set M ] . A similar discussion demonstrates that the set M 2 is closed. © 2000 CRC Press
4
DlFFERENTlAL GEOMETRY AND TOPOLOGY O F CURVES
Suppose the point h is contained in the set M 2 . We denote by c the least upper bound of the points of M1. The point c is a limit point of M I , hence it belongs to the closed set M, and does not coincide with b. But from the definition of the least upper bound it follows that the points satisfying the inequality X > c belong to M*. Hence c is a limit point of M2; from this fact it follows that c E M 2 . This contradiction proves the connectedness of the segment (a, 6). Now we will demonstrate that the connection property is preserved under any homeomorphism. Let f b e a homeomorphism of a connected set M. Suppose the image of M under f i s not connected. Then there exist two disjoint subsets A, B of f ( M ) such that f ( M ) = A U B and each of A , B does not contain limit points of the other. We consider the sets f''(A), f'(B) situated in M. Since M is connected, there exists a point X. E f  ' ( A )which is a limit point of ,fpl(B).Because of the continuity ofJ; for an arbitrary neighborhood Vyo of the point Yo = f'(Xo) t A there exists a neighborhood Ux, of X. such that f(Uxo) c V,,,. Since X. is a limit point of , f p ' ( ~ ) , the neighborhood Vy,, contains a point of B. Hence Yo is a limit point of B that contradicts the assumption. Thus the image of M under f i s connected as well as M. Now it is easy to see that a ,simple curve is connected. In the next chapters we will consider curves with selfintersection points. So, for the sequel we need to construct a more general notion of a curve. Let an interval ( a , 6) (or a circle) be continuously mapped into the space in such a way that for any point of (a, b) there exists a neighborhood whose image is an elementary curve; then the image y of the whole interval (a, 6) is called a general curve. The general curve y can contain a point corresponding to different points of (a, b); such a point is said to be a selfintersection point of y. When a point of the interval moves from a to h, the corresponding moving point on the curve y passes through any selfintersection point at least twice.
© 2000 CRC Press
2 Vectorvalued Functions Depending on Numerical Arguments Let a real parameter t vary in an interval (a, h). If to each value t E ( a ,h) we assign a vector r(t), then we say that a vectorvaluedfunction r(t) with argument t E (a, h) is given. Assume Cartesian coordinates X , y , z are fixed; then the representation of the vectorvalued function r(t) is equivalent to the representation of three scalar (realvalued) functions s ( t ) , y(t), z(t). We can write r(t) = { x ( t ) ,y(t), z ( t ) } , but the brief notation r(t) is more convenient. One can define many notions connected with vectorvalued functions similarly to notions corresponding to the usual scalar functions. First, we can define the limit of r(t) as t + to. A vector ro is called the limit of the vectorvaluedfunction r(t) as t + to if the length of the vector r(t)  r(to) tends to zero as t + to. Here we write lim r ( t ) = rg.
l+l"
It is clear that the vectorvalued function r(t) has a limit iff each one of the functions x ( t ) , y(t), z ( t ) has a limit as t i to. Limits of vectorvalued functions have the same properties as limits of scalar functions. The vectorvalued function r(t) is said to be coneinuous at tg iff lim r(t) is equal to 1410 the value of r(t) at to: lim r ( t ) = r(to).
!+l,,
Differentiating the components x ( t ) , y(t), z ( t ) of r(t), we obtain a function called the derivative of r(t) and denoted by
© 2000 CRC Press
6
DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
Moreover, one can define the derivative rl(t) in the same way as derivatives of scalar functions. Precisely, the derivative rl(t) of the vectorvalued function r(t) we call the limit: lim r ( t )  r(t0) t  to
[to
If this limit exists, then r(t) is called dzfferentiable. The differentiation of vectorvalued functions has the same properties as the differentiation of scalar functions. The sum of the derivatives of two functions is equal to the derivatives of the sum of these two functions:
If f ( t ) is a scalar function and p(t) is a vectorvalued function, then
i.e. the rule of differentiation in this case is the same as the diffcrentiation rule for products of scalar functions. The differcntiation of inner products, vector products, and mixed products of vectorvalued functions is computed by the consecutive dif1 ) vectorvalued ferentiation of the cofactors. To be precise, if pl(t), p2(t), ~ ~ ( are functions, then
For example, let us prove the differentiation rule for the vector product:
If rl(t) = 0, then r(t) is a constant vector: r(t) = c. By definition, the second derivative rl'(t) of r(t) is the derivative of rl(t). By induction, one can define the nth derivative r(")(t) of r ( t ) as the derivative of r(n')(t). © 2000 CRC Press
VECTORVALUED FUNCTIONS DEPENDING ON NUMERICAL ARGUMENTS
7
Assume that k derivatives of the vectorvalued function r(t)exist and are continuous; then we can write Taylor's expansions for the components x(t), y(t), z(t) of r(t):
+
x ( t ) = x ( t o ) xl(to) ( t  to) ( l )=( l )t
+ ~ "2!( t "()t 
o ( t  to) t

to)
2!
+...+
2
to)
+
0, (
It
+%(It
k

to1 ),

folk).
This system of three equations can be rewritten as r ( f )= r(to)
+ rl(tO)( t

r" (to) to) t  ( t  to)2 2!
+ . . . + o( It  talk ),
where o( It  f o l k ) denotes a vector whose length is an infinitesimal with respect to It  tolk as t , t o We remark that there exists one essential difference between Taylor's expansion of vectorvalued functions and Taylor's expansion of scalar functions. If we consider Taylor's expansion for a scalar function ,f(t), then we have
<
where is a point situated between t and to. For a vectorvalued function we cannot write a similar formula for the corresponding infinitesimal vector, because in general for different components of the vector o ( J t tolk ) the corresponding points E are different. Nevertheless, it is more important that the length of the vector o( It  t o l k ) is an infinitesimal with respect to It  toll'. If we have a continuous vector function r(t), we can define the integral of r(t) as a vector whose components are the integrals of the components x(t), y(t), z(t) of r(t):
The following properties of the defined integral are obvious:
J
Xr(t) dt
=
X
J
r ( t )d t ,
where X
= const.
Moreover, there exist new properties for the integrals of vectorvalued functions. If a is a constant vector, then for integrating the inner product (a,r(t)) or the vector product [U, r(t)] we can apply the formulas J;a, r ( f , , dt
© 2000 CRC Press
=
(a, J r ( o dt) >
/[U,
rill1
= [a, / l ( [ )dt]
8
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
For example, let us prove the second formula. The first component of the vector
is equal to
where a, are the components of the vector a. We see that on the righthand side of the last equality we have the first component of the vector
Considering the second and third components similarly, we prove the desired formula.
© 2000 CRC Press
3 The Regular Curve and its Representations A curve y is called ~ " r e ~ u l a riff, there exists a parametrization of y such that each component of the position vector r(t) is a C'regular function and r: does not vanish. The C'regular curve is called srnootlz . The condition ri # 0 is essential for the definition of regular curves. For example, consider the planar curve formed by two rays and represented by the position vector
where a, h, C, d are constants. If the vectors { a ,h ) , {c, d ) are linearly independent, the considered line has a singularity at the point (0,O). At the same time, each component of the position vector is C1smooth everywhere. But ri(0) = 0. If y is onetoone projected onto a segment [a, h] of the xaxis, then there exists one of the simplest representations:
Indeed, let r(t) = {x(t), y(t), z(t)} be some representation of y. Because y is onetoone projected onto the segment (a,b) of the xaxis, we can assign to each X E (a, 6) a unique value of the parameter t such that the point (X, 0,O) is the image of the point P(t) of y under the projection. Thus t can be viewed as a function of X, i.e. t = t(x). Substituting t(x) into the expressions of the functions y(x), z(x) we obtain:
Thus, the position vector of y parametrized by x has the form (3.1).
© 2000 CRC Press
10
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
When is the curve y regularly projected onto a segment of the xaxis? Assume that y is Clregular and xi # 0. Consider the function x = x(t). Because xi # 0, the function x(t) is monotone on some interval (a, b) of the xaxis, and there exists the inverse function t = t(x). Since t: = l l x i , t ( x ) is Clregular. Therefore to each point x E (a,b) one can assign a unique value of t and a unique point P(t) E y.Thus it is easy to see that y is projected onto the segment (a, b ) of the xaxis. We remark that in this case j ( x ) and Z(x) are C'smooth. Often a planar curve is represented implicitly, i.e. by an equality
that is, this curve consists of the points whose coordinates X , y satisfy equality (3.2). But for some functions @(X, y ) equation (3.2) has no solution or the solution consists of isolated points. So, it is natural to ask: when does equation (3.2) represent a curve (in the sense of chapter l)? Theorem Let @ ( X , y ) be a C'regular function. Assume that a point P ~ ~ i coth ordinates x ~y~, satisfies the equation
and grad cP = {Q,, Q,) # 0 at P. Then the points satisjying (3.2) and situated in some sufficiently small neighborhood of P ,form a C' regular curve. Proof We will apply the theorem on an implicit function: Let a function @ ( Xy, ) be defined and C'smooth in some neighborhood of a point (xo,y o ) Assume that @(Q, yo) = 0 and Q,(xo, yo) # 0. Then there exist S > 0 and a C'regular function y = y ( x ) defined in the interval 6 xo 5 x 5 S + x~ such that y(xo) = yo and @ ( X , y(x)) = 0. By the assumption, grad @(xo,yo) # 0. We suppose without loss of generality that Qy(xO,yO)# 0. The assumptions of the theorem on an implicit function are fulfilled. Then the points with coordinates ( X ,y(x)) in some neighborhood of ( x o ,yo) form a curve y coinciding with the curve represented implicitly by the equation @ ( X , y ) = 0. The coordinate x is chosen as a parameter on y. Because the function y ( x ) is C'regular and r: = {l,y:) # 0, the curve is smooth. Sometimes it is useful and interesting to consider a family of curves. Here it is convenient to assign to each curve one or more numbers called parameters of the family. Then according to the number of parameters the family is called oneparametric, twoparametric and so on. For instance, in the family of straight lines x = c = const on the ( X ,y)plane, every straight line is determined by the value of the constant c, i.e. c is the parameter of this family. A second example: the family of circles on the plane is threeparametric. Let H be a set of planar curves such that for any point P of a domain G on the ( U ,v)plane there exists a unique curve y E H passing through P. We will say that the set H forms a Ckregular oneparametric,family, @for any point (uo,vo) there exist a neighborhood and a smooth map x(u, v),y(u, v) of'the neighborhood onto the circle
+
© 2000 CRC Press
THE REGULAR CURVE AND ITS REPRESENTATIONS
+
11
x2 y2 < 1 with Jacohian J # 0 such that the curves of H are tmnsj'ormed into the straight lines x = c. The level lines a)(u,v) = c of a smooth function @(U,v ) form a smooth family in a neighborhood of a point (uo,vo), if grad cP(uo,vo) f 0. Indeed, let for instance @,(uo, v") # 0 . The map x = @(U,v), y = v has the nonzero Jacobian J = @,(uo, vo) and transforms the level lines @(U,v ) = c into the straight lines x = c. Problems 1. Find the prqjection of the curve
into the ( X , y)plane. Find the projection into the 2. Does the curve
(X,z)plane.
pass through the points ( 1 , l, 1 ) and ( 1 , 0 , 0)? 3. Are the following curves intersecting?
4. Find the point where the curve
intersects the plane z 5. Show that the curve
= 0.
r ( t ) = { a sin2 t , h cos t sin t , c cos t ) is situated on an ellipsoid. 6. Prove that
is the position vector of a circle whose radius is equal to v.'
© 2000 CRC Press
4 Straight Line Tangent to a Curve Let PO be a point on a curve y and U C y some neighborhood of PO such that PO decomposes U into two halfneighborhoods U l and U2 (see Figure 4.1). Take another point Q E y and consider the ray PoQ, whose origin is PO.Assume that Q , which is situated in U;, tends to PO.Then the limit position of the ray PoQ, if it exists, is called the ray tangent to y at PO with respect to O;. If there exist both rays tangent to y at POand they form a straight line, then this line is called the straight line tangent to y at P. or the tangent to y at PO. Let us explain the notion of the limit position of the ray PoQ. Consider the unit sphere with its center at PO. The ray PoQ intersects this sphere at some point M. Suppose that Q tends to PO.If the corresponding points M converge to some point MO,then we say that the ray PoMo is the limit position of the rays PoQ. Assume that y is a smooth curve with the position vector r = r(t). What is the directing vector of the tangent to y? We denote by a prime the derivative with respect to t . Theorem For any point POo f t h e smooth curve y there. cxists the tangent to y at PO, and the directing vector of the tangent is r'.
FIGURE 4.1
© 2000 CRC Press
14
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
Proof Let POcorrespond to the value to of the parameter t , and some close point Q correspond to to A t ; we assume that if A t > 0, then Q is situated in U , , and if A t < 0, then Q lies in U2. The directing vector of the ray PoQ is r(to A t )  r(to). Since y is smooth, we have
+
+
where o ( A t ) is a vector such that
It follows from (4.1) that the vector
r(to
+At) At

r(to)
+4 A t )
= rl(to) A t
is collinear to the ray PoQ at A t > 0 and is opposite to this ray if A t < 0. Because y is smooth and property (4.2) holds, the limit of the expression on the lefthand side of (4.3) as A t 4 0 exists and is equal to rl. Thus there exist both rays tangent to y at PO,they form the tangent to y at PO,and r' is the directing vector of the tangent.. For some nonregular curve, both rays tangent at a point Q" exist and coincide. In this case we say that the curve has the halftangent at Q. and Q. is called the cusp. Let us write the equation of the tangent to y at PO.Again we denote by r(t) the position vector of y and by ?(X)the position vector of the tangent. Since the tangent passes through the point PO with position vector r(to) and rl(to) is the directing vector, the position vector of the tangent is
Rewriting equality (4.1) as
r(to
+ A t ) = r(t0) + rl(to)At+ o ( A t ) ,
and setting X = A t , we see that the difference between the position vectors r(to + A t ) and ?(At) is an infinitesimal o(At):
r(to
+ A t )  ? ( A t )= o ( A t ) .
Therefore if A t is sufficiently small, the curve y is close to the tangent. In other words, the tangent is the first approximation of the curve. When the parameter X varies from 00 to +oo, the corresponding point passes the whole tangent. In terms of Cartesian coordinates the tangent is represented as
© 2000 CRC Press
STRAIGHT LINE TANGENT T O A CURVE
Consider a space curve given by two equations
Assume that the rank of the matrix
is equal to two at a point Po(xo,.yo,zo).If x = x ( t ) , y = y(t), z = z(t) is the position vector of the considered curve, then, substituting these three functions into (4.4), we obtain two equalities 4 ( x ( t ) r y ( t )4, t ) )
=
0,
l i / ( x ( t ) , y ( t )4, t ) ) = 0;
the differentiation of the equalities leads to two additional equalities
Thus, the components { X ' , y', z') of the tangent vector satisfy the system consisting of two equations (4.6), therefore they are proportional to the corresponding minors of matrix (4.5):
Because the rank of matrix (4.5) at point PO is equal to two, some minor of (4.5) is nonzero. If we have a planar curve given by the equations 4 ( x ,y ) = 0 , z = 0 satisfying the condition 4; 4: # 0, then the components of the tangent vector v' = {X', y') is a solution of the linear equation
+
Therefore
and the equation of the tangent is
Let y be a smooth space curve. The plane passing through a point PO t y and orthogonal to the vector tangent to y at PO is called the plane normal to y at PO. © 2000 CRC Press
16
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
Denote by i the position vector of the normal plane. Because this plane is orthogonal to the vector rf(to)and contains the point with position vector i  r(to),the equation of the normal plane is ( i  r ( t O )r,l ( t O ) )= 0.
The vectors orthogonal to the tangent are called the vectors norrnul to y.
© 2000 CRC Press
5 Osculating Plane of a Curve Let PO be a point of a curve y.Take two points Q , , Q2 t y situated on different sides with respect to PO and construct the plane passing through PO,Q , , Q 2 . If the points Q , , Q2 tend to PO,then the limit position of the plane containing PO,Q , , Q2 is called the osculatingplur~eof the curve y at the point PO.By this definition, if the osculating plane exists, then it is unique. Obviously the osculating plane of y at PO passes through P().Let PO correspond to the value to. Theorem Let y he a C2regular curve represented ac r = r(t). Assume tlzut at cr point PO the vectors rl(to)and rl'(to)are not collinear. Then there exists the osculating plurze o f y at PO und it i.s syunned by the vectors rl(tO),rl'(lo) (F~gure5.1) Proof Let points Q , correspond to values to + h, of the parameter t. The position vector of PO is r ( f o ) ,and the position vectors of Q , are r(to +h,). Hence the vectors PoQ, are r(fo + h , )  Y ( ~ o ) .
FIGURE 5.1
17 © 2000 CRC Press
18
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
Since y is regular, one can write a Taylor expansion of the position vector r(t) at a neighborhood of to:
where o(h:) is a sufficiently small vector such that o(h?)/h; + O as hi t 0. Using the Taylor expansion, we find that the vectors PoQl and PoQ2 have the following forms:
The vectors PoQl and PoQz span some plane and this plane contains the following linear combinations of P o Q l , f o e z :
Because the points Q,, Q2 are situated on different sides of y with respect to PO,h1 and h2 have different signs. Therefore I h ,  h2 I > max( I h1 I , ( h2 I ) and

Hence the infinitesimal summed to rl'(to) in the second vector of (5.1) tends to O as Q, PO. Therefore the limit positions of the vectors presented in (5.1) are rl(to), r1I(to).If these vectors are not collinear, they determine uniquely the limit position of the plane passing through Q , , PO, Q 2 , i.e. the osculating plane of y at PO. If the vectors rl(to),r1I(to)are collinear, the limit position of the considering plane is not determined. For instance, take a straight line r(t) = a
+ bt,
where a, b are constant vectors. Then rl(tO= ) b,
r M ( t o= ) 0,
so the osculating plane of the straight line is not determined uniquely. In this case, one can think that any plane containing the straight line is its osculating plane. © 2000 CRC Press
OSCULATING PLANE OF A CURVE
If r r ( t ) , rl'(t) are collinear, then the straighteningpoint of y.We will view any
19
corresponding point of y is called the plane passing through the tangent to y at
this point as the osculating plane of y. We remark that the notion of an osculating plane does not depend on the choice of the parameter on y. Instead, if r = r ( ~ is) another parametrization of y and t = t ( ~ ) , then, differentiating r(t(7)) as a composite function, we obtain
Calculating the vector [ r : , r : ] , we see that it is collinear to the vector [ r : , r i ] :
Because the osculating plane is passing through PO and [ r i ( t o ) , r i ( t o ) ] is its normal vector, the obtained equality means the independence of the osculating plane on the parametrization of y. The osculating plane of a planar curve coincides with the plane containing this curve. Let us consider the Taylor expansion of the position vector r ( t ) at a neighborhood of PO: r(to
The curve
+ A t ) = r(to) + r r ( t o ) A t+ rl'(to)A2t 2 + o ( A t 2 )
r:
determined by this expansion is situated in the osculating plane of y at PO; the difference between the position vectors of y and 7 is a sufficiently small vector: r(tn
+ A t ) ?(At)= o(at2).
Hence a sufficiently small neighborhood of PO on the space curve y is near to the planar curve 7 situated in the osculating plane of y at PO. Now let us write the equation of the osculating plane of y at PO.Denote by i the position vector of the osculating plane. The vector product [ r l ( t o ) ,r " ( t o ) ] is orthogonal to the osculating plane, and the vector I:  r(to) belongs to the osculating plane, hence the inner product of these vectors is equal to zero: ( I :  to), + ( t o ) , i l ( t O ) )= 0.
© 2000 CRC Press
20
DIFFERENTIAL GEOMETRY A N D TOPOLOGY OF CUKVES
This is just the desired equation. With respect to Cartesian coordinates it has the following form:
With the help of the osculating plane one can distinguish two special straight lines normal to y at PO.The straight line normal to y and situated in the osculating plane of y at PO is called the principal normal straight line to y at PO.It is the intersection of the plane normal to y and the osculating plane at PO. We have proved that the vector [ r l ( t o ) , r " ( t o ) ]is orthogonal to the osculating plane. Because the principal normal straight line is orthogonal to the normal of the osculating plane and to the tangent vector of y,it is collinear up to a sign to the vector
Problems
1. Write the equations of the tangent to the curve x = 3 t  t  1,
y=3t2,
z=3t+t3
at the point (0,0, 0). 2. Write the equations of the plane normal to the curve considered in Problem 1 at the point (0,0,0). 3. Find the tangent vector and the normal vector to the planar curve
4. Write the equation of the osculating plane of the curve X = P'' COS U ,
y
= P"
sin U , z = e"
5. Prove that the curve having the tangent straight line at each of its points is smooth.
© 2000 CRC Press
6 The Arc Length of a Curve Now we will define the notion of the arc length of a curve. Let y be a space curve given by its position vector u = r(t), t E [U, h]. The length of an arbitrary polygonal line consisting of a finite number of segments is equal to the sum of the lengths of all these segments. Inscribe a polygonal line into the curve y in the following way. Take some points a = to < t l < .  . < t, = h belonging to the segment [a, h]. Some points PO,P I , . . . , P, on y correspond to the chosen values of the parameter t. If we connect
F: FIGURE 6.1
21 © 2000 CRC Press
22
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
POwith P , , P I with P2 and so on by segments, we obtain a polygonal line, which is inscribed into y and whose vertices are P;. Consider the length of the inscribed polygonal line. If we take some new points in [a, h] and the corresponding new points on y, we obtain a new polygonal line inscribed into y, and its length is greater than the length of the preceding inscribed polygonal line. It is clear that if we have chosen a new point Q E y situated between points Pi, Pi+,, the sum of the lengths of the segments PiQ and QPi+l is greater than the length of the segment Pip,+'; hence the length of the new polygonal line P O . .. P;QPi+l . . . P, inscribed into y is greater than the length of the preceding one P O ...P;P;+l . . . P,. The curve y is called rectifiable, if the lengths of all polygonal lines inscribed into y as described is bounded from above. If y is rectifiable, then the supremum of the lengths of inscribed polygonal lines exists by the Weierstrass theorem, and it is called the length of y. We will denote it by s(y). Theorem If y issmooth, then it is rectifiable, and the length of y is equal to the integral
Proof First, we will prove that the smooth curve y is rectifiable. It is necessary to find some estimate of the length of an arbitrary polygonal line y, inscribed into y. Let r(t,) be the position vector of the vertices of 7,. Then the length of y, is equal to the sum of the lengths of vectors A,r. Because y is smooth,
In addition, from the smoothness of y it follows that r: is a continuous vectorvalued function defined on the segment [a,h], hence it is bounded. There exists a constant M such that
Therefore we have
thus the length of an arbitrary polygonal line inscribed into y is bounded from above by the constant M(h  a) depending only on the curve y. So, y is rectifiable. Let us prove that the length of y is computed with the help of formula (6.1). Because y is smooth, i.e. the components x(t), y(t), z(t) of the position vector r(t) are C'smooth, we have © 2000 CRC Press
THE ARC LENGTH OF A CURVE
6
where r: are some points of the segment [ t i P lti]. , In general, the points are different. We will replace r;, r:, r; by one point ~i E [til,ri]and consider the mistake which appears. We have
where the vector a; has the following form:
Since every component of r' is a continuous function defined on the segment [a, h],it is uniformly continuous. Hence for any positive E there exists a positive 6 such that I a; I < E whenever I t ,  t,l 1 5 S. We denote c = rl(r;)A,t.From (6.2) it follows that
Air  c
= a,A,t.
Therefore It is easy to obtain the following estimates:
and
Thus the difference between / Air I and Ir1(ri)lAlt is less than &Ai(. We apply the obtained estimate in order to estimate the difference between the length of y, and the sum C:=,Ir1(ri)/Ait:
4%)

The sum
C:=,(rl(ri)lAitis a Riemann sum for the integral
so the difference between this sum and integral is sufficiently small for a suitable choice of the points ti decomposing the interval [a,h]. On the other hand, one can © 2000 CRC Press
24
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
inscribe a polygonal line y, into the curve y such that the difference between the lengths of y and y, is sufficiently small. Because E is arbitrary, s(y) coincides with integral (6.7). Thus formula (6.1) is proved. From the definition it follows that the length of any curve does not depend on the parametrization. Hence one can use any parameter T in formula (6.1) to compute the length of y. With the help of the notion of the length, one can introduce some special parameter a , which is most naturally connected with y. Fix a point PO t y. Assume that y is parametrized by a parameter t , some value to corresponds to POand t corresponds to P. The parametrisation t gives us some "positive" direction (orientation) on y. To any point P E y we assign a equal to .(POP) in the case when the arc POPhas the positive direction; if the direction of P O Pis negative, we assign a = s(PoP) to P. Then the parameter a is computed by the formula
1 r i ( t )1 dt.
~ ( t=) 10
Since
the parameter a is a monotone function of t , hence we can consider a as a new parameter on y. This parameter is called the arc length and for sin~plicitywe use the previous notation S. The parametrization of y by the arc length is called natural. Using (6.8) we obtain
Thus the absolute value of the differential of the arc length is equal to the absolute value of the differential of the position vector of y. Therefore
Thus we have the following characterizing property of the natural parametrization: iJ r(.r) is the position vector of'y with respect to the arc length S, then the length of the vector r(,(s) is equal to l :
4f:Jorsome parameter s the length of the tangent vector ri is equal to l identically, then s is the arc length. © 2000 CRC Press
THE ARC LENGTH OF A CURVE
Problems l . Find the arc length of the curve X =
In sin t,
t
y=JZ'
z=.
t
JZ
2. Find the length of the arc Q 5 t 5 1 of the curve
3. Find the length of the arc 0 5 t X
© 2000 CRC Press
= h(t

< 27r of the cycloid
sin t),
y = h(1

cost).
7 The Curvature and Torsion of a Curve In this chapter we will define and study the curvature and torsion of curves. Let P be a point of a curve y. For some point Q E y sufficiently near to P we denote by A 0 the angle between the tangents to y at P, Q and by As the length of the arc PQ. The limit
k
=
lim
Q+P
A0
as

is called the curvature of the curve y at the point P. We remark that by supposition here both A$ and A s are positive numbers. Denoting by r(s) the position vector of y with respect to the arc length s we will prove a formula for the curvature k. l a r Then the curvature k o f y exists and it can be Theorem Let y be a ~ ~  r e ~ ucurve. calculated with the help of the jormulu
The vector ry>is called the curvature vector of y. Its length is equal to the curvature. The vector 11 = r:,/k is called the principal normal vector of y;it is defined at points where k # 0 (compare with the definition of the principal normal straight line given in chapter 5).
+
Proof Let points P, Q correspond to values S , s A s of the arc length. We denote the vectors tangent to y at P, Q by r(s), r ( s As). The angle A 0 is the angle between T ( S ) and r ( s + As). Consider the triangle P M N such that the vectors P M , PN are equal to r ( s ) and ~ (+sAs) respectively (see Figure 7.1). This triangle is isosceles, because the vectors r(s), r(s AS)are unit. The altitude P L is the bisectrix
+
+
© 2000 CRC Press
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
FIGURE 7.1
of the angle / M P N that
=
AQ. It is easy to show, considering the right triangle PLM,
I T(S+ a s )  T(S) I 2
.
= sin
AQ . 2

(7.1)
Now we divide both parts of equality (7.1) by As/2 and set As t 0. Because y is c2regular and we have denoted the vector r:.(s) by ~ ( s ) we , obtain
I r;(s + As)  Y:(s) / I riy(s)( = AS+O lim AS
2 sin(AO/2) ASo As sin(AO/2) A812 AB = lim 2 = lim  = k. ASo A012 AS ASo As =
lim
The theorem is proved. Now let us define the torsion of y. Since T = rt is a unit vector, the curvature vector v,: and the principal normal vector U are orthogonal to the tangent vector T. The vector product [T, U ] is called the hinormal of y and is denoted by P. It is obvious that the binormal P is well defined at points where the curvature k is nonzero. Here we suppose that As can be positive and negative.
FIGURE 7.2
© 2000 CRC Press
29
THE CURVATURE AND TORSlON OF A CURVE
Let Ad denote the angle with sign between the binormals P(P), P(Q) at sufficiently near points P, Q E y;this angle is equal to the angle between the osculating planes of y at P, Q (see Figure 7.2). We define the sign of AB as follows. The normal plane N p to y at P is spanned and oriented by the basis vectors u(P),P(P). We translate the vector P(Q) in such a way that its origin is the point P and then we consider the orthogonal projection of the translated vector P(Q) into the normal plane N p . If the length As of the arc PQ is sufficiently small, the vector P(Q) does not vanish. Assume that As > 0; then the sign of AH coincides with the sign of the angle counted from P(P) to with respect to the given orientation of NI.. If As < 0, then the sign of At4 is opposite to the sign of the considered angle between P(P), P(Q). The limit
P(Q)
B(Q)
K =
AH lim 
QP
as
is called the torsion of y at P. Theorem Suppose y is C'regular and the curvature k qf y is not equal to 0 everywhere. Then ,for any point P E y there exists the torsion K of y at P and the fillo wing forrnula holds: ,,l ,.l1 Ill ( .S 2 s s , r.s.7.71 K = k2 '
Proof Consider a triangle PMN C N p such that the vectors P M , PN are equal to the vector P ( P ) and the translated vector P(Q) respectively. This triangle is isosceles. Similarly to the proof of the previous theorem, consider the altitude PL, which is the median and the bisectrix of the triangle PMN. Using relations between elements of the right triangle PLM we get
We divide both parts of this equality by lAs1/2 and set As
thus I /3:(s)
I = I K 1. Consider
the vector
P:.
+
0. Then we obtain
We have
Since r(s) is smooth of class c3and k # 0 by the assumptions, the curvature k(s) is a C'smooth function and @(S)is a C' vectorvalued function. Then we have
© 2000 CRC Press
30
DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
Both summands situated on the righthand side of the last equality are vector products; each one of these products contains the vector T = ri as a factor. Hence f l is orthogonal to T. Since P is a unit vector, is orthogonal to P. Therefore P', is collinear to the principal normal vector U . Because the length of 0;.is equal to I K ~ , we get Pi. = f1. 1 ~. To determine the sign, we consider Taylor's expansion of the binormal ,B at a neighborhood of P:
The projection ,&Q) of the vector @(Q)is equal to the vector P(P) + ,Bi(P)As up to an infinitesimal o(As). Take Q such that A s > 0. It is easy to see that the considered angle between @(P)and ,&Q) is positive iff pi(P)As is collinear to v. Hence
In order to prove the desired formula presenting the torsion we multiply equality (7.2) by v and obtain:
Note that the sign of the torsion does not depend on thc choice of orientation of y; from formula (7.3) it follows that the sign of K does not change if we replace s by S. It is useful to know the formulas of k and K in the cases when y is parametrized by a parameter t different from the arc length. Viewing the arc length s as a function of t , we differentiate the position vector r of y as a composite vectorvalued function and get:
Since r: is unit, from the first equality it follows that Idt/ds( = l / l r i ( . The vectors r: and rtS are orthogonal, hence we have
=
;l3
I [rl r"] l l'
It
in.c11
=  .
i r :13
Thus the curvature k of y with respect to an arbitrary parameter t is given by the formula
© 2000 CRC Press
THE CURVATURE AND TORSION OF A CURVE
The formula just proved can be written in the following coordinate form:
Let us now find a formula for the torsion. We will denote the rnixedproduct ([a,h],c) by (a, h, C). It is known that the mixed product is linear with respect to each of its arguments; for example, (a,h c, d ) = (a,6, d ) + (a,c, d ) . Also, the mixed product is equal to zero iff at least two factors are collinear. Using these facts we get
+
Substituting the found expression of k in formula (7.4) and applying the equality Idtldsl = I/lril, we obtain that the torsion of y with respect to an arbitrary parameter t is computed with help of the formula (r'/ > r"I t > r"') l//
/E.=
l E >r312
.
Curvature of planar curves Planar curves are a particular case of space curves, hence formula (7.4) can be used to find the curvature of planar curves; in this case this formula has a simpler form. Let y be a planar curve. If y is situated in the plane z = 0, then z' = z" = 0. Therefore,
k
=
(x'y"  y'x" (xt2+ y
This expression for k is the simplest if x" = 0, hence
X
l
'2 3i2
'
is taken as the parameter on y;then
X' =
1,
For the planar curve y one can define the notion of the curvature with sign. We will denote the curvature with sign by the same notation k. Let r denote the tangent © 2000 CRC Press
DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
FIGURE 7 . 3
vector to y,e be a unit vector orthogonal to the plane n containing y.The vector 7 = [e,T ] is unit and orthogonal to T ; also it lies in the plane T . It is clear that 7 coincides with the principal normal vector v of y or with U (Figure 7.3). When a point P is moving along y the vector v(P) is varying continuously. The coefficient k given in the formula
is called the curvuture with sign of y.It is different from the curvature defined earlier maybe only in the sign. We remark that using the curvature with sign k one can describe the changing of the convexity direction of y (see Figure 7.3); for k one of the two following formulas holds:
Now let us find a formula for k in terms of the angle between the tangent vector T to y and a fixed vector a. Without loss of generality we can assume that a is collinear to the positive direction of the xaxis. Suppose the arc length s of y is the parameter given in formulas (7.7). Since = T is a unit vector, the components X',, y: of r: have the following form: X:. ( S ) = COS a ( S ) ,
y:. (S) = sin oi ( S ) .
Substituting the derivatives
into formulas (7.5) we obtain
As an example, we find the curvature of a circle of radius R. Denote by p the angle between the xaxis and the position vector of a point of the considered circle. Obviously, ds = R d p . If the circle is oriented counterclockwise, then a = p n/2.
+
© 2000 CRC Press
THE CURVATURE AND TORSION OF A CURVE
33
Applying the formula k = dalds, we obtain that the curvature of the circle is equal to k = 1 / R . Assume that y is given by an equation @ ( x , y )= 0. What is the formula for the curvature k? The vector {Q,, Q,,,) is normal to y at the corresponding point. For the unit normal vector T,J we have
where W = \/(B,):
+ (a,):. T =
Hence the unit tangent vector to y is
dx dy
{;i;.5} =
Q,, {&}'
@
where d/ds is the differentiation with respect to the arc length of y. From the formula T: = krl it follows that /c =  (Tf .\ > 7 ) . We can find the derivative of q with respect to s by differentiating function:
as a composite
Moreover, it is easy to see that
Thus, we can write
+
Since @,dx QYdy = d@ = 0 at the points of y,the last term on the righthand side of equality (7.9) is equal to zero. Substituting expressions (7.8) for the components of the vector T into equality (7.9) we get
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DlFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
It is easy to see that this formula can be rewritten in the following divergent form:
Problems 1. Find the curvature and torsion of the curve
x=acosu,
y=asinu,
z=bu.
2. Find the curvature and torsion of the curve x=t,
y = a t 2,
z=bt2+ct+d.
3. Find the binormal of the curve x=cost,
y=sint,
z=cos2t.
4. Find the curvature of the ellipse
at the vertices. 5. Prove that the curve
is planar. Find the plane containing this curve. 6 . Find the unit tangent vector T and the principal normal vector v to the curve
7. Find the curvature of the conical spiral
8. Prove that any curve with constant torsion
K
# 0 can be presented in the form
where [ = [(t) is a unit vectorvalued function. 9. Let some family of regular planar curves satisfy two conditions: (i) the curvature k of any curve of the family is bounded, k > ko > 0, ko = const; (ii) all curves of the family are situated inside a circle of radius R. Prove that R 5 2/ko. © 2000 CRC Press
THE CURVATURE AND TORSION OF A CURVE
35
10. A closed convex planar curve y is called an oval. The distance between two straight lines, which are collinear to a fixed direction T and tangent to y,is said to be the width of y with respect to the vector 7 . Prove Barbier's theorem: if the width d of an oval does not depend on the direction 7 (one can say that this oval has the constant width d), then the length of the considered oval is equal to nd.
© 2000 CRC Press
8 Osculating Circle of a Plane Curve Consider a smooth planar curve y with curvature k # 0. For any point PO E y the difference between the behavior of y and the behavior of the tangent to y at P" is negligible at a sufficiently small neighborhood U of PO;we say that the tangent to y at PO is the first approximation of y at U. The second approximation of y at the sufficiently small neighborhood U of PO is a circle called the osculating circle of y at PO.In order to define the osculating circle we fix Cartesian coordinates X, y in such a way that PO is the origin and the xaxis coincides with the tangent to y at PO. Let y = y(x) be the equation for the curve y and y = j ( x ) be the equation of some circle C passing through PO. This circle C is called the osculating circle of y at POif the values of the function y(x), ? ( X ) are equal at x = 0 and the values of their first and second derivatives coincide at x = 0:
From these equalities it follows that Taylor's expansions
are equal up to an infinitesimal 5 ( A x 2 ) . Hence if we consider all circles passing through PO,the osculating circle is the nearest to the curve y. Since the tangent to y at PO is completely determined by the values y(O), yl(0),the tangent to y at PO and the tangent to the osculating circle at PO coincide, so we can say that the osculating circle is tangent to y. Also, the center 0 of the osculating circle is situated on the straight line that contains P,) and is collinear to the principal normal vector v(Po) (see Figure 8.1).
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DIFFERENTIAL GEOMETRY A N D TOPOLOGY OF CURVES
FIGURE 8.1
The curvature of a planar curve y = , f ( x ) is computed by the formula
Because the first and second derivatives of y , j are equal at PO, we see that the curvature k of y at PO is equal to the curvature of the osculating circle of y at PO. Thus the radius R of the osculating circle is equal to l / k ( P o ) .The center of the osculating circle is called the curvature center of y at PO;the radius of the osculating circle is called the curvature radius of y at PO. Generally, the curvature center depends on the point PO E y. All curvature centers of y form a curve y called the evolute. Let us find the position vector of the evolute. Let r = r(s) denote the position vector of y with respect to the arc length S . If r(so)is the position vector of a point P, then the position vector p(so) of the curvature center 0 of y at P is the sum of the vector r(so) and the vector PO. This last vector is collinear to the principal normal vector v(so)and its length is equal to the curvature radius R(so) = l/k(so).Thus we obtain the position vector of the evolute:
From the formula evolute is
7: =
kv it follows that v', = kr. Hence the vector tangent to the
Therefore if (ilk): # 0, then the vector tangent to the evolute of y is collinear to the principal normal vector to y (Figure 8.2). Assume that y is the evolute of some planar curve 7; then 7is called the evolvent of y. Let us find the position vector T; = r(s) of the evolvent with respect to the arc length s of y. Since any point of the evolvent is situated on the corresponding straight line tangent to y,we have © 2000 CRC Press
OSCULATING CIRCLE OF A PLANE CURVE
FIGURE 8.2
The curve y is the evolute of 7,hence its tangent vector r is collinear to the principal normal vector to the evolvent. Therefore T is orthogonal to the vector tangent to 7at the corresponding point. We have
This implies that X(S) = c
 S,
C
= const.
Thus the position vector of the evolvent of y is
where c is a constant. It can be demonstrated that for any arbitrary constant c the curve given by the position vector (8.2) is the evolvent of y.Hence there exist many evolvents of the given curve y. Indeed, from formula (8.1) we get
Let S denote the arc length and k be the curvature of the evolvent 7.Since 7; is a unit vector,we have
Suppose that c  s > 0. Then
r
1
"
© 2000 CRC Press
l ds "d3
I/  =
kT
1 = (cs)k
1 cs
7.
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
FIGURE 8.3
Therefore the curvature k of 7 is equal to I/(c  S ) and v normal vector of 7.Equation (8.2) can be rewritten as
=
r is the principal
hence y is really the evolute of 7. It is very easy to demonstrate the form of the evolvent. Fix a point Q E y and consider the arc of y with length c; denote by P the end of this arc. Now imagine that a thread covers the arc QP and the origin of the thread is fixed at Q. If we take the end of the thread and reel the thread off the arc QP viewed as a pattern, then the end of the moving thread forms the evolvent of the arc QP of y.Indeed, some part Q A of the thread coincides with Q A C QP; if the length of Q A is equal to S, then the rest of the thread is a segment of the tangent to y and its length is equal to c  ,F. Therefore from (8.2) it follows that the end of the moving thread forms the desired evolvent. The evolutes and evolvents are frequently used in technology.
© 2000 CRC Press
9 Singular Points of Plane Curves Let a plane curve y be given by an equation cp(x, y) = 0 and cp E ck,k > 1. A point M = (xo, yo) of the curve y is called a singularity of y,if the following conditions are fulfilled at this point:
There exist different types of singularities. Assume that M = (xo,yo) is a singularity of y and some second derivatives of the function p do not vanish at M. Let us introduce some notations:
There are three cases with respect to D: (a) if D > 0, then M is called an isolated point of y (Figure 9. la); (h) if D < 0, then M is called a point of selfintersection of y (Figure 9.lb); (c) if D = 0, then M is either an isolatedpoint, or a cusp, which can be of two types, or an osculate point of y (Figure 9. lc).
a
6
C FIGURE 9.1
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DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
The first case in Figure 9. l c shows a curve such that its singularity M is a cusp of the first type (the corresponding branches of the curve are situated on different sides with respect to the straight line containing the tangent ray to this curve at M); in the second case, M is a cusp of the second type (the tangent ray does not decompose the branches); in the third case, M is an isolated point; and in the fourth case, M is a point where two branches are tangent (in particular, these branches can coincide). If we assume that the first, second and third derivatives of the function p(x, y) are continuous, we can write Taylor's formula:
+
where o(Ar2) is an infinitesimal with respect to Ar2 = (x  x ~ )0,~ y0)2. Let us denote Ax = X  xo, Ay = y  y". Since the conditions (9.1) are fulfilled at the singularity (xo,yo), we obtain:
Denoting Ax Ar2, we get
= Ar
cos a, Ay
= Ar
sin a and dividing the equation p(x, y) = 0 by
Consider case (a), D > 0. Suppose that there exists an angle a" such that A cos2a g
+ 2Bcos no sin a0 + csin2a" = 0.
Assume without loss of generality that A > 0 (we can make a rotation in the (X,y)plane in order that A # 0); then dividing the last equation by sin2 ao, we see that tan a,)is a solution of an algebraic equation of the second degree; on the other hand, the discriminant of this equation is equal to D and by our assumption it is negative, so this equation has no solution. Thus we obtain a contradiction. Therefore, for any value of a the expression A cos2 Q + 2Bcos a sin a
+ csin2a
(9.3)
is greater than some positive number. Suppose now that there exists a sequence of points of y converging to the singularity M; then o(Ar2)/Ar2 + 0. Since expression (9.3) does not converge to 0, we see that equation (9.2) cannot be fulfilled as Ar2 + 0. Hence M is an isolated point of y. Consider case (b), shown in Figure 9.1b. Assume that A # 0. We will demonstrate that the value of lAy/Axl cannot be infinitesimally small at points of y near to M. Suppose that there exists a sequence of points M, + M such that © 2000 CRC Press
SINGULAR POINTS OF PLANE CURVES
at these points. If we divide equation (9.2) by cos2a , we get
At the points M, we have the inequality Ar2 5 ( l + &Ax2. Hence the sequence of points M, converging to M satisfies the condition
Since the coefficients corresponding to B and C tend to zero as M, + M and A # 0, equation (9.4) cannot be fulfilled for the sequence of points M,. Thus there exists a positive ko such that (AylAxj 2 ko for all points situated in a sufficiently small neighborhood of M. We rewrite this inequality as
Dividing equation (9.2) by sin2cu, we obtain
We solve this equation as an algebraic equation of the second degree with respect to Axl Ay:
Because the points of y satisfy
we have
as A y + 0. From this conclusion and from (9.5) it follows that the values of A x l d y at points of y are infinitesimally near to the following two numbers:
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DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
This means that the points of y,which are sufficiently near to M, are sufficiently near to two straight lines
intersecting at M. Hence the singular points M are points where two branches of y intersect. That is why M is called a seljintersection point. Now let us consider the most intricate case (Figure 9. lc). Using a suitable rotation of Cartesian coordinates in the plane, we can obtain B = 0. Since AC = 0 in this case, AC = 0. We assume that A = 0. Because some second derivatives of p do not vanish, we have C # 0. The behavior of y at the point M depends on the sign of C and on the properties of the infinitesimal o(Ar2).IJ'the signs of C and o(Ar2)are equal, then M is an isolatedpoint o f M . For our next consideration it is necessary to consider Taylor's formula including the third derivatives of p:
where k , are constants. We will prove that
as points of y converge to M. Suppose that there exists a sequence of points M, E y converging to M such that
for some arbitrary fixed positive number k. If we divide equation (9.6) by ( A y ) * ,we obtain
Since
cm.Then from (9.8) it follows that C = 0, at the points M,, o ( A r 3 ) / A y 2+0 as n contradicting the assumption that C # 0. Therefore property (9.7) holds, i.e. the curve y is tangent to a straight line collinear to the xaxis. Let us rewrite the equation of y:
© 2000 CRC Press
SINGULAR POINTS OF PLANE CURVES
Then from this equation we obtain:
If k l # 0, then the expression under the square root is equivalent to 4CklAx3 as Ar 4 0. In this case the curve is defined either for A x 2 0 or for A x 5 0, depending on the sign of 4Ckl. It is easy to see that the denominator of the expression for A y is equivalent to f The curve y at a small neigborhood of M is similar to the curve
Js.
consisting of two branches situated on different sides with respect to the tangent ray at M. Tlze singularity M is a cusp of the,fir.rt type. If kl = 0, then the expression under the square root can be equivalent to the infinitesimal o(Ar3).So we must consider the fourth derivatives of the function cp. Here the equation of y can be rewritten as
where li are constants defined by the fourth derivatives of cp at the point M = (xo,yo). We write the lefthand side of this equation as an algebraic equation of the second degree with respect to Ay;
where by cu we denote the infinitesimal
The discriminant of equation (9.9) is
If k:  4Cll # 0, then the last expression is equivalent to the first summed. In the case k i  4C11 < 0, the determinant is negative for small A x ; therefore equation (9.9) does not have a solution different from A x = A y = 0. Hence M is an isolutedpoint of y. If k i  4Cll 2 0, the curve y is defined for small values A x of different signs; here the equation of y is
© 2000 CRC Press
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
46
The curve y in a small neighborhood of M consists of two mutually tangent branches. In this case M is called an osculate point. If k:  4Cll = 0, then the discriminant is equivalent to o(Ax4). Assume o ( A x 4 ) > 0 for all sufficiently small values of A x . Then for A y we have two solutions; hence I' consists of two tangent branches and M is an osculate point. If o ( A x 4 ) 0 for sufficiently small values of A x of sorne.fi'xed sign, then M is a cusp of y.In the case k2 # 0, both branches of y are situated in one side with respect to the tangent ray at M; the singularity M is called a cusp ofthe second type. If k2 = 0, then M is either a cusp of the,first type (for example, in the case 12 = 0), or a cusp of the second type (in the case when l2 # 0 and Jo(ax4) = o(Ax")). The considered investigation is valid iff the function p(x, y) is sufficiently smooth at M. If cp is not differentiable at M, then the behavior of y can be very complicated. For example, let p be the function given by the following equality:
>
The curve presented by the equation cp(x,y) = 0 consists of the point 0 = (0,O) and of a spiral going to 0. The point 0 is a singularity of this curve and cp is not differentiable at 0 .
Problems Find the singularities of the curve "y aax2 + X ' . What are the types of these singularities? What are the types of singularities of the curve
called a Descartes leaf? How does the type of the singularity of the curve
vary as the parameters a, h, c vary? Find and investigate the singularities of the curve y2 = ux3 is called the divergent parabola. Find and investigate the singularities of the cissoid
which can be represented in the parametric form
This curve was discovered by Diocles (second century © 2000 CRC Press
BC).
+ bx2 + cx + d, which
SINGlJLAK POINTS OF PLANE CURVES
6. Find the singularity of Maclaurin's curve
presented in the parametric form asin 3t sin t '
= 
y
a sin 3t
= COS
7. Find the singularities of the curve y2 = x4. 8. Draw the tractrix x = a(1og tan p/2 + cos p) + c,
© 2000 CRC Press
t
y = a sin p.
10 Peano's Curve In 1890, Peano constructed the beautiful example of a continuous map of an interval whose image is a square. The construction is the following. Decompose an interval A into four equal intervals A i l ,i l = 1,2,3,4 indexed from left to right. Then decompose a square A into four equal squares A i l ,il = 1,2,3,4, enumerating these small squares in such a manner that consecutive squares have a mutual side. We call the constructed decomposition a first step. Let us assign to each interval Ail the square A,, and denote this correspondence by f i , i.e. A,, = , f l ( A i , ) .Decompose every interval A,, in a similar way into four equal intervals AiIi2.We also decompose every square A,, into four equal squares here we choose the index i2 in such a way that the Ail,, have a mutual side with AiIi,,I and AiI4have a mutual side with A i l + l l We . call this decomposition a second step. Let us assign to each interval A,,,, the square Ai,;* and denote this correspondence by f 2 , i.e. A,,;, =,f2(AiIi2). We continue the decomposition in this way and denote intervals of the nth step by A,, ..., and squares by A,, ...i,,. If two intervals of the nth step have a mutual point, then the corresponding
1
1
1
1
1
4, 1
FIGIJRE 10.1
49 © 2000 CRC Press
1
1
1
50
DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
squares of the nth step have a mutual side. The correspondencef, between intervals and squares has the following property: if Ail...;,,c Ail...;,,_l,then fn(Ai,...,,) c j;,,(Ail. Let us assign to every point of A a point of the square A in the following manner. Each point t E A belongs to an infinite sequence of enclosed intervals A i l ,A;,,,, . . . , A ;,..., , , . . . . The squares corresponding to these intervals form an infinite sequence of enclosed squares:
Because the lengths of the sides of the squares converge to 0, there exists a unique point P E A, belonging to all squares of the sequence. Thus we define the map f : A 4 A, assigning to every point t E A the corresponding point P = f ( t ) . It follows from the definition that if t E Ail...i,, then f ( t ) E ,fn(Ai,...i,,). Let us prove that every point P E A is an image of some point t E A . For any point PO E A there exists at least one infinite sequence of enclosed squares (A;, > A;,,2 > . . . 3 Ai, ...,,,) > . . . such that Po belongs to all squares of the sequence. This sequence corresponds to an infinite sequence of enclosed intervals A,, A;,;,> . . . 3 A ;,...i,,) > . . . such that j;,(Ai,...;,,) = A ,,...i,,, n = l,m, and the intervals of this sequence have a unique mutual point to, hence we obtain by the definition off: .f(to)= .P Therefore the image of the interval A under f is the square A! The mapf is continuous. In order to prove this fact let us take a positive t. When n tends to infinity, the length of the sides of the nth step's squares converges to zero. Therefore there exists a number n, such that all squares of the n,th step , which contain the point f ( t o ) , belong to the Fneighborhood of j'(to). Let us take the intervals of the nth step containing to. To every point of these intervals the map f' assigns a point of the squares of the nth step belonging to the tneighborhood of f(to). This means that the map f is continuous. An analog of Peano's curve was constructed in [57] with the help of a complex power series F(z) = C z ocnzr'.A series F(z) was found, with the following properties: inside the circle / z /< 1 it was convergent ; it was continuous inside the closed circle / z I< l; the set of values of the function ~ ( e " )t ,E [O, 2n]contained an open set of the zplane.
© 2000 CRC Press
11 Envelope of the Family of Curves Above, we considered a regular family of plane curves given by an equation @(x,y) = 0; different curves of the family were mutually disjoint. Now we will consider families of curves such that different curves of a family can be mutually intersecting. Let us take , for example, the family of circles of radius R whose centers are situated on the Xaxis.The equation of any such circle is
For different a we have equations of different circles of the family. Two arbitrary sufficiently near circles are intersecting. In general we will assume that an equation
defines a family of plane curves. This means that for every fixed a the equation ( l 1.2) represents some curve y,,of the family.
FIGURE 1 1.1
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52
DIFFERENTIAL GEOMETRY A N D TOPOLOGY OF CURVES
On examination of a family of plane curves, it can be seen that some plane curves are singular with respect to the family. For example, the lines y = R, y = R are the singular curves concerning the family given by (1 1.1); at each point these lines are tangent to a circle of the family, i.e. they are the enveloping curves of the family.
Definition A plane curve I' is called an envelope of a family of plane curves, if I' is tangent at each point P to a curve ofthe,family passing through P. We will find the equation of an envelope I' of the family given by (1 1.2). Let P be a point of F with coordinates X, y. Suppose that there exists a unique curve y,,of the family, passing through P. Then we can assign to the point P a unique value of the parameter a ; thus we can consider the coordinates X,y of points of the envelope as functions of a: x = x(a), y = y(a)  this is a parametrization of the envelope by the parameter of the family. Because P lies on the curve y,, we have
Differentiating with respect to
ai
we get
The envelope I' and the curve y,,have a common tangent at the point P. Since y, is given by equation (1 1.2), the normal of y, at P has the coordinates (f,(P), f;(P)). The tangent vector of F at P is (dxlda, dylda), therefore
Hence the points of the envelope must satisfy the two following equations:
Eliminating (if possible) the parameter a from these two equations, we obtain an equation
which represents the envelope.
© 2000 CRC Press
12 Frenet Formulas Let a space curve y be given by its position vector r = r(s) viewed as a vectorvalued function of the arc length S. We will denote by T the unit tangent vector r l ( s ) , by v the principal normal, by 0 the binormal of y. Three vectors T , v, P depend on the parameter S, hence we consider these vectors to be vectorvalued functions of S . At any point of y the vectors T , v, p are mutually orthogonal and form a basis of Euclidean space. We say that T , v, P.form the natural frame. The derivatives of these vectors are decomposed into linear combinations with respect to the natural frame at the corresponding point:
Let us find the coefficients of the decomposition. It follows from the definition of the principal normal that d2r dr = = kv. ds2 ds Therefore a1 = 0, a2 = k, a3 = 0. When we proved the theorem about torsion (see chapter 7), we proved the formula
hence cl
= 0, c2 =
K,
c3 = 0. Because v is a unit vector, 62 = (v,VI, ) = '(V, V): = 0. 2
© 2000 CRC Press
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DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
Using the decomposition of v,: and the found values of al, c2 we obtain
Thus
These decompositions of the vector ri, v.:, /3: are called Frenetformulas. They have a very important significance in the differential geometry of curves. Let us apply the Frenet formulas in order to investigate the behavior of y at a neighborhood U of some point P E y, where s = so, k(so)# 0 and &(so)# 0. We write a Taylor expansion of the vectorvalued function r(s) at U:
According to our notations we have
It is easy to see, with the help of the Frenet frame, that
Using the obtained expressions we can write
We will describe the behavior of y by considering the projections of y into the coordinate planes of the natural frame at P. Assume that Cartesian coordinates x,y, z are fixed in such a way that P is the origin and the xaxis, yaxis, zaxis are collinear to the vectors T , v, p respectively. The projection of y onto the plane spanned by r and v is given by the coordinates x(s), y(s), which are the coefficients at T and v of the Taylor expansion (12.1):
© 2000 CRC Press
FRENET FORMULAS
FIGURE 12.1
So, the projection of y into the the parabola
( 7 ,U)plane (Figure
12.1) is approximately equal to
y =  kX .2
2
The projection into the (r,O)plane is a curve given by the coordinates
+ A S )= as+ +S), as3+ o ( A s ~ ) , Z ( S ~+ as)= kK 6 
hence it is similar to the cubic parabola
Such a curve with K > 0 is shown in Figure 12.2. The projection into the (v, P)plane is a curve given by the coordinates
y(so
as2+ o ( A s 2 ) , + As) = k 2
z(so
As3 + As) = k~ + o(As". 6
FIGURE 12.2
© 2000 CRC Press
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
FIGURE 12.3
It is clear that if As is sufficiently small, the coordinate y is positive. Eliminating As, we find that the projection behaves similarly to the curve
Therefore the point P is a cusp of the first type, and both branches of the projection are tangent to the yaxis (Figure 12.3). Using three constructed projections one can find the form of y at a sufficiently small neighborhood U. Assume that the torsion K. is positive. Because the projection
FIGURE 12.4
© 2000 CRC Press
FRENET FORMULAS
57
into the (X,y)plane is similar to the parabola y = 5x2, z = 0, the curve y is approximately situated on the cylinder y = 5x2 with the generator collinear to the zaxis. It follows from the behavior of the projection into the (x,z)plane that the branch P A corresponding to As > 0 of y lies above the (X,y)plane, and the branch PB corresponding to As < 0 is under the (X,y)plane (Figure 12.4). For some curves y one can define the curvature with a sign. Suppose that there exists a continuous and differentiable field of orthonormal frames e,, e2, e3 satisfying two conditions: (1) el is the tangent vector; (2) a t points where the principal normal is defined, one of the vectors e2, e2 is equal to the principal normal. Writing the derivatives of e, with respect to the arc length S , one can obtain an analog of the Frenet formula:
where k, are some functions of S . The function k l is called the curvature with sign. It is equal to the curvature k up to the sign; it may be useful when k = 0 at some points and the principal normal passing through such points changes its direction.
© 2000 CRC Press
13 Determination of a Curve with Given Curvature and Torsion Earlier we defined two geometric notions  the curvature and torsion of space curves. The curvature depends on the first and second derivatives of the position vector; the torsion is calculated with the help of the first, second and third derivatives. Using higher derivatives one can construct other geometrical concepts characterizing the behavior of curves. But it is unnecessary, because any curve is completely determined by its curvature and torsion. More accurately, we will prove Theorem Let k(s) and ~ ( sbe) given continuous functions qf'a parameter s E [0,I ] , k(s) is positive everywhere. Assume that a point PO and three orthogonal unit vectors 7 0 , V O , PO = 170, uO]arefixed in Euclidean space. Then there exists a unique C'regular curve y having the.following properties: ( I ) y passes through PO and s is its arc length countedfrom PO; ( 2 ) 7 0 , U O , PO is the natural frame of y at PO; ( 3 ) k(s) is the curvature and K ( S ) is the torsion of y. ,
Proof Let us consider the system of linear differential equations similar to Frenet formulas:
, are unknown functions. The coefficients k(s), ~ ( sof) this system where [ ( S ) , ~ ( s )<(S) are given. ODE theory guarantees that (13.1) has a unique solution [(S),~ ( s )( ,( S ) with given initial values [(O) = TO, q(0) = uo, ((0) = 0";the solution exists on the whole segment [0,l ] . At any point s the vectors [(S),v(s), <(S) are unit and mutually orthogonal. In order to prove this fact, consider the functions
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DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
Their values at s = 0 are 1, 1, l , 0 , 0 , 0. One can demonstrate, using (13.1),that these functions are a solution of the system
It is a system of homogenous linear differential equations with respect to six considered functions. We observe that this system has the constant solution 1, 1, 1, 0 , 0, 0, whose initial value coincides with the initial value of the solution E2, r12,i2, (E,v), (E, C), (v,c). But by ODE theory there exists a unique solution of the corresponding initial value problem, therefore
i.e. at any points the vectors I , rl, (' are unit and mutually orthogonal. Let us define a curve y with the position vector
where r(0) is the position vector of PO.If k(s), &(S) are continuous, then r(s) E C 2 ;if k(s) E C' and &(S) is continuous, then r(s) E C 3 .Because r: = [(S)is the unit tangent vector, the parameter s is the arc length of y. At the initial point PO we have r(,(O)= ( ( 0 ) = 70,hence ro is tangent to y at PO.It follows from the Frenet formulas and system (13.1) that
therefore k(s) is the curvature of y and 7 is the principal normal. Hence the vector = [E, v] is the binormal of y, and ~ ( sis) the torsion due to the third equation of system (1). Thus, any curve y is determined by its curvature k(s) and torsion & ( S ) uniquely up to a motion of Euclidean space. That is why we represent y in the form of two equations
This representation is called the natural equations of y. It does not depend on Cartesian coordinates. If the natural equations of y are given, the reconstruction of y leads to finding a solution of the Frenet formulas. We note that it can be reduced to © 2000 CRC Press
DETERMINATION OF A CURVE WITH GIVEN CURVATURE AND TORSION
61
one Ricatti equation for a complex vector. Let us denote by a, h, c the first components of the vectors r, V, p and set
+
a ih lc
m = 
Using the Frenet formulas we can obtain:
Because the unit vectors r, V, P are mutually orthogonal, a2 + h2 + c2 = 1. Taking this equality into account, we can rewrite the sum of the two last summed of (13.3):

c 1c
K  + K
ia(a + ih) (1cc)2
If we find the halfsum of the right and left sides of the last equality, we obtain that the sum of the two last summed of (13.3) is equal to:
Substituting the found sum into (13.3), we see that m is a solution of the Ricatti equation dm  i m 2 K ds 2
irc 2
ikm  .
If a solution m is known, then the coordinates a, h, c can be computed by the formulas
where E is the conjugate of m. © 2000 CRC Press
DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
Problems
1. Write the equations of the principal normal and binormal of the curve
at the point t = 0. 2. Find the natural frame of the curve X
© 2000 CRC Press
= 1  cos 1,
y = sin t ,
z
=
t.
14 Analogies of Curvature and Torsion for Polygonal Lines Let POPI. . .P, be a polygonal line formed by segments a,+] = PiP;,l, i = 0 , . . . ,n  1. Instead of the curvature and torsion of regular curves we will define two sequences of angles: p;, i = 1,. . . ,n  1 and $,, j = l , . . . ,n  2. It is convenient to consider the segments a, as vectors directed from P I P Ito P,. We take two segments a,, a;+].Assume that a, and a,, are linearly independent. Then we can consider an oriented plane E;, which is spanned and oriented by the basis of vectors a,, a;+l.Now we define an angle p, as an angle between the vectors a, and a,+l computed with respect to the orientation. Obviously 0 < p; < 7r. The oriented plane E; has a unique unit normal vector v,. If q, and qi+l are linearly independent, we can construct a plane ET spanned by the vectors q,, q;+.l;because a;+l is orthogonal to E;" we can assume that E;" has an orientation determined by Now we define an angle as an angle between v; and v;+l computed with 5 27r. The angles p ] , . . . ,p,_l respect to the orientation. It is clear that 0 5 .. are called the torsion are called the curvature angles, and the angles angles. There exists a statement on polygonal lines similar to the theorem on unique determination of a regular curve by the curvature and torsion.


Theorem Let S1,.. . ,S,, g l , . . . ,$,l, $ 1 , . . . ,$12 be number sequences satisfying 5 LT. Let P be an arbitrary point, E* an oriented plane S; > 0, 0 < pi < T , 0 passing through P and ii an arbitrary vector parallel to E". Then there exists a unique polygonal line g = POPl. . .P, satisfying the following conditions:
<
g; ( i ) S ] , . . . ,S, are the lengths of segments a ] ,. ._,a, forming (ii) $ 1 , . . . ,$,_l are the curvature angles and + I , .. . , are the torsion angles of g; (iii) PO is the same as P, a1 is collinear to ii and El coincides with E.
© 2000 CRC Press
64
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
Proof Let us construct the desired polygonal line g. We set PO= P and form a vector al, which is collinear to ij and whose length is equal to il;the end point of a l we denote by P I .The points PO,P1 belong to the oriented plane El = E*. One can construct a vector a2 such that: the length of a2 is equal to i2, the plane El is spanned and oriented by the basis of vectors al, az, the angle between a1 and a2 computed with respect to the orientation of El is equal to pl, P , is the origin point of a2. Let us denote the end point of a2 by PZ The unit vector 71normal to El is orthogonal to the vector a2. In the oriented plane with normal vector a2 we can construct a unit vector 772 in such a manner that the angle between 71and 72 computed with respect to the orientation of E; is equal to $1 . We denote by E? a plane passing through P,, P2 orthogonally to 712. One can continue this process. Let us remark that, denoting by a, the unit vector collinear to a,, we have
ET
The resulting polygonal line g satisfies conditions (i), (ii), (iii), and it is easy to see that by the construction this polygonal line is unique.
© 2000 CRC Press
15 Curves with a Constant Ratio of Curvature and Torsion Let y be a curve with a constant ratio of the curvature k and torsion Because k / =~ const, we can write
p=
Jm.
Iz
= p cos p ,
K, =
K.
Set
p sin p,
where p = arctan ~ / =k const. The Frenet formulas have the following form: 1d7

P ds
= cospu
We introduce a new parameter o = Spds. Using the Frenet formulas we obtain the equation
d2v

do2
d~ + sin do
 COS p 
do
 = v.
do
It follows from the theory of differential equations that the general solution of the last equation is
+
where a and b are constant vectors. Since v is a unit, a2 b2 = 1 and (a, b) = 0. Hence the principal normal vector of y is situated in a fixed plane. Integrating the
© 2000 CRC Press
66
DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
first equation of the Frenet formulas, we obtain the vector tangent to y:
where c is a constant vector. If c = 0, y is planar. Let c # 0. We have
Because this equality is fulfilled identically with respect to a,
( a ,c) = (b, c)
= 0,
c2
2
= sin
p.
Thus the vector c is orthogonal to the plane spanned by a and b, and the length of c is equal to sin cp. Hence an angle between the vector tangent to y and the fixed vector c is constant. Curves whose tangent vectors have such properties are called slope curves. Since drlds = T , by integrating this relation we find the position vector of 7
Thus, the position vector of the curve y with a constant ratio of the curvature and torsion has form (1 5. 1). One may show that the ratio of the curvature and torsion of a curve with the position vector of form (15.1) is constant. A partial case of curves with a constant ratio of the curvature and torsion is a curve y whose curvature and torsion are constant. Then p = p0 = const and the parameter a is proportional to S, i.e. a = pos. The position vector of such curves has the following form: r=ro+cosp ~ ( ~ i n ~ ~ a  c o s p ~ s b ) d s + c s 0
cos cp = ro  q (cosposu+sinposb) P O
+CS.
Construct a plane E2 spanned by a, b and passing through a point 0 with the position vector ro. The projection of y into is the circle C with its center at 0 and with radius equal to cos J t . Hence the curve is situated on a cylinder whose axis 1is parallel to c. When s varies from m to +m, the projection passes the circle C infinitely many times; at the same times, the projection of y onto the axis of the cylinder passes the whole Zaxis once and the motion of a point of y in the direction of the laxis is proportional to the angle of rotation around the axis. Such a curve is called a helix. The value of translation of point P E y, when a corresponding © 2000 CRC Press
CURVES WITH A CONSTANT RATIO O F CURVATURE AND TORSION
67
projection point P* E E2 passes the whole circle C, is called the step of one helix; it is ' . equal to 27r sin cp J The representation of a helix in the case when c is parallel to the zaxis is
© 2000 CRC Press
16 Osculating Sphere Above, we defined the osculating circle of a planar curve. Let y be a space curve. Take four points P, Q , , Q2, Q3 situated on y. If they do not lie on a plane, then there exists a unique sphere S passing through these points. The limit sphere Sp at Q, + P, i = 1 , 2, 3, if it exists, is called the osculating sphere of y at P. The center and radius of Sp are the limits of centers and radii of spheres S as Q; P. Let r(s) be the position vector of y and s be the natural parameter. The point P corresponds to a value s = so and points Q; correspond to values s = so + h;. To find the center C of the sphere S we take two points P and Q;: because P, Q; E S, the point C lies in the plane E,? passing through the center of the segment PQ, orthogonally to this segment. Thus, C is a point of intersection of the planes E;, E:. Let us write the equation of some plane E,?. The position vector of the center of the segment PQ; is (r(so)+ r(so lzi))/2,and the vector r(so h;)  r(so)is orthogonal to E:; thus, if Yi is the position vector of the points of E:, then

+

r(so
+
+ h,) + r(so) r(so + h,) 2

r(so)
So, the position vector ? of the center C is a solution of the system of three equations of form (16.1) (i = 1,2,3). Let r(to) = 0. Using formula (12.1) of chapter 12 we obtain (Y,
© 2000 CRC Press
SO + h;) 
1 2
+
SO))   r2(s0 h;) = (Y, 7)
70
where
DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES T,
v , /3 is the natural frame of y.Next we have (F, r)h;
h! +I{(r,P)kn 6
h? +{ ( r ,v ) k  1) 2
+ k1(?,U )

k2(?,7 ) ) + o(h;) = 0.
(16.2)
Let us denote the coefficients of h? and h? by A and B respectively. If we divide (16.2) by h; and tend hi to 0, it follows from (16.2) that lim (F, T ) = 0 as h, t 0. Thus the center of the osculating sphere Sp is situated in the normalplane to y at P. Let us take two equations (16.2) corresponding to i = 1, 2 and subtract one from the other. We obtain
(h1  h2)A + (h:

G:) hi)^ = 4 h : )  hl
h2
We divide (16.3) by h1  h2 and set h, t 0. Suppose that the points Q; converge to P uniformly and set h , = h2. Then
Hence lim { ( U , v ) k  1)
Q,+p
= 0.
If we write an equation similar to (16.3) for h , and h3, then with the help of these two equations we obtain
Set, for example, hl = h2, h1 = 2hl. Then it is easy to see that the expression on the righthand side of the previous equality has the form o(h:)/h: 4 0 as h , t 0. Therefore
Thus one can write
and hence
© 2000 CRC Press
OSCULATING SPHERE
Obviously the radius of the osculating sphere S p is equal to
Let us consider a spherical curve y,i.e. a curve situated on a sphere S. Because the osculating sphere of y is constant and coincides with S, by differentiating equality (16.4) and using the Frenet formulas we obtain
Thus we see that the curvature and torsion qf a spherical curve satisfy the equation
It is more convenient to write this equation using the curvature radius p = Ilk instead of the curvature k of y. Let us replace s by the parameter a = J k d s . Then equation (16.6) is equivalent to the following equation:
The general solution of (16.7) is p = cl cos a
+ c2 sin a,
where c; are constant. Assume now that the curvature k and torsion K of some space curve y satisfy equation (16.7). Then for radius R of the osculating sphere of y we have
= const. Moreover, it follows from (16.5) that the center of the osculating Hence sphere is constant. From (16.4) we find
(Y  r ( s ) )2
=
R 2 = const.
Thus y is situated on the sphere with its center at Y and with radius equal to R. So we conclude that (1 6.6) is the equation characterizing spherical curves.
© 2000 CRC Press
17 Special Planar Curves There exist some remarkable planar curves known to everyone who begins to study geometry: they are the ellipse, hyperbola and parabola. In this chapter we will consider some other special planar curves which have beautiful forms and wonderful properties. The lemniscate of'Bevnoulli is one of the most splendid curves of the fourth degree. To construct this curve we fix points F l , F2; a curve formed by points M such that the product of the lengths of segments M F I and MF2 is a constant p, is called the lemniscate. The points F,, F2 are called the foci of the lemniscate. What is the equation representing the lemniscate? We choose Cartesian coordinates X, y on the plane in such a way that the origin 0 is situated at the center of the segment F , F2 and the xaxis contains F 1 ,F2. Denote the distance between F , and F2 by 2a. Then F1 and F2 have the coordinates (a,O) and ( a,O) respectively. The distances between a point M with coordinates (X,y) and the points F, are
Hence the equation
represents the desired lemniscate. It follows from the definition that the lemniscate is symmetric with respect to the xaxis and yaxis. Setting y = 0, we can find from (17.1) that the coordinate X of the points where the lemniscate intersects the xaxis is a solution of the equation
© 2000 CRC Press
DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
FICURE 17.1
FICURE 17.2
If a2 = p, the intersection points are (0,O) and (f1,0,'). In this case the lemniscate has the form of a figure of eight with one selfintersection point situated at 0 (Figure 17.1). Let p < a2. Then there exist four points where the lemniscate intersects the xaxis: 0). The lemniscate consists of two convex curves (Figure 17.2). they are (fJG, If p > a*, then. there exist two points of intersection : (fJG, 0). In this case the lemniscate is a closed curve homeomorphic to a circle (Figure 17.3). For every fixed p, equation (17.1) determines some curve. When p changes we have a family of curves (see Figure 17.4). Note that the lemniscate can be drown by a hinge mechanism.
© 2000 CRC Press
SPECIAL PLANAR CURVES
FIGURE 17.4
One can define lemniscates with an arbitrary number of foci. Let F ] ,. . . ,F,, be different points of the plane. Assume that a curve y has the following properties: for any point M of y the product of the distances between M and F, is MFI . . . MF, = p = constant.
Then y is called a lemniscate with njoci. An Archimedean spiral is a curve represented with respect to the polar coordinates p, cp by the equation p = cop + cl, where co and cl are constants. This curve consists of an infinite number of twists; if cl = 0, it starts from the pole 0, and when cp increases, the corresponding point of the spiral is receding (see Figure 17.5). With the help of the Archimedean spiral, any angle can be divided into any number of equal parts using compasses and ruler. This is carried out in the following manner. Assume that an angle AOB with the vertex at O must be divided into three equal angles. We place the angle AOB in such a way that the ray O A is tangent to the Archimedean spiral at 0 . We denote by C some point where O A intersects the spiral. One can divide the segment OC into three equal parts using a ruler and compasses. Construct a circle centered at 0 with the radius equal to lOC//3.This circle intersects the spiral at a point D. It is clear that
FIGURE 17.5
© 2000 CRC Press
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
FIGURE 17.6
hence
Thus the angle AOD is equal to 113 of the angle AOB. The logarithmic spiral was discovered by Descartes and Toricelli independently. This curve may occur both in nature and in artificial life. The representation of the logarithmic spiral with respect to the polar system of coordinates is
where p0 and c are constant. Let c > 0, for example. As cp t m the curve, making an infinite number of twists, comes arbitrarily close to the pole 0 ;as cp 4 +m the curve goes away from the pole (Figure 17.6). The logarithmic spiral has a remarkable property: the curve intersects all rays whose origin is O at a constant angle. To prove this fact let us find an angle formed by a vector tangent to the spiral and by the polar axis cp = 0. Denote by (.x(p), y(cp)) the position vector of the spiral in Cartesian coordinates. It is clear that
Therefore
Hence
y ($I=
(f
+
e 2 y 2+ I).
The unit vector tangent to the spiral has the following form: © 2000 CRC Press
SPECIAL PLANAR CURVES
where C
cos p() =  sln p0
Jm'
Therefore the angle between
1
Jm'
= 
T
and the position vector r is equal to
p
+ po

p = p() = const.
A curve, represented by the equation
is called a spirul winding onto a circle. As p + m the function p(p) tends to po, i.e. the curve winds onto the circle with its center at the pole 0 and with radius equal to p(). A spiral winding onto two rircles. The equation of such a curve is
where a, and b, are positive constants. Assume that alh2  u2bl < 0. Then p is the monotone function of p. As p i m the curve comes arbitrarily close to the circle C l of radius u l / h l with its center at the pole; as cp + +m the spiral winds onto the circle C2 of radius u2/h2with the center at the pole. By assumption, C , is situated inside C2, and the spiral is situated between Cl and C2. The witch of Agnesi (Figure 17.7) is a curve represented with respect to Cartesian coordinates by the equation u3 y=a2 + x2 '
FIGURE 17.7
© 2000 CRC Press
DIFFERENTIAL GEOMETRY A N D TOPOLOGY OF CURVES
FIGURE 17.8
where a = const. It is named in honor of a woman who investigated the solvability of equations of the third degree. The cissoid of Diocles (second century B C ) was discovered during discussions of the problem of duplication of the cube. This curve is constructed with the help of a circle. Let us fix a circle of radius a and with diameter O A (Figure 17.8). We draw the straight line I tangent to the circle at point A . Consider a ray iwith origin at 0 and denote by B a point where I intersects we denote by C a point of the intersection of lI with the circle. On the segment OB there exists a point M such that the length of the segment O M is equal to the length of BC. The set of points Mconstructed in the described manner form a cissoid. It is a curve of the third degree given by the equation
The parametric representation of the cissoid is
The word "cissoid" originated from a Greek name for ivy, whose leaves have a form similar to a cissoid. An interesting class of curves having a wide application in technology consists of throchoids and roulets. Any such curve is formed by some point of a curve moving on another fixed curve. A more specific class consists of cycloids, epicycloids and hypocycloids. These are formed by a point of a circle moving on another circle without sliding. © 2000 CRC Press
SPECIAL PLANAR CURVES
79
Let a circle of radius a move on some straight line I; without loss of generality, we think that I is an xaxis with respect to Cartesian coordinates. When we fix a point P on the circle, this point moves together with the circle and forms a curve called the cycloid. The equation representing the cycloid is
Points of the cycloid, which are situated on the xaxis, are singular points (Figure 17.9). Consider a circle C of radius Xa moving on a fixed circle of radius a. Let P be a fixed point of C. If the first circle is outside the second one, the moving point P forms a curve called the epzcycloid (Figure 17.10); if the first circle is inside the second one, a curve formed by the moving point P is called the huvpocycloid (Figure 17.11). The epicycloid corresponding to X = l is called the curdioid; the hypocycloid corresponding to X = 113 is called the astroid. Conchoids are characterized by the following property: a curve y is a conchoid, if there exists a point 0 and a straight line g such that for any straight line 1 passing through 0 a segment AB C I contained between y and g has a constant length L. The equation defining a conchoid can be written in the form
FIGURE 17.9
FIGURE 17.10
FIGURE 17.11
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80
DIFFERENTlAL GEOMETRY AND TOPOLOGY O F CURVES
where d is equal to the length L and a is a distance between 0 and l. Thus conchoids are curves of the fourth degree. They are sometimes called the conchoids of Nicomede. T o draw a conchoid, one may use special instruments called after the Greek mathematician Nicomede. Two examples of conchoids corresponding to different values of the parameters a and d are shown in Figure 17.12.
FIGURE 17.12
FIGURE 17.13 © 2000 CRC Press
SPECIAL PLANAR CURVES
81
The snail of Pascal is a curve named after the French mathematician Etien Pascal. First, we fix a circle C and a point 0 E C; then a snail is the locus of points M such that for a segment OM the length of its part CM C OM contained between M and the circle C is constant (Figure 17.13). Curves given with respect to the polar coordinates by the equation p
= a sin kp,
where u and k are constants, are called roses. If k is rational, then the corresponding roses are algebraic curves of an even degree.
Problem Prove that the representations of an epicycloid and a hypocycloid are
+ X) cos Xp = (1 + X) sin Xp a
X

a
= (1


+ X)p, X sin ( l + X)p, X cos ( l
and X
a

a
respectively.
© 2000 CRC Press
+ X cos (X X) sin Xp + X sin ( X
= (1  X) cosXq = (l



1)q
1)p
18 Curves in Mechanics
Let a mass point move on a space curve y with position vector r. Denote the time by t and the length of arc of y by S. We can consider the length of arc as a function of time. Then dsldt is the value of the velocity of the moving mass point. The vector of velocity is tangent to y: v = drldt = drlds. dsldt, where r = r(s) is the position vector of y. The vector of acceleration dvldt lies on the osculating plane of y:
By Newton's law the value of acceleration is proportional to a force F having effect on the mass point, so the trace y of a mass point is more curved if the value of F is greater. When the mass point moves on y with a constant value of velocity Ivj = dsldt = const, the vector of acceleration is collinear to the principal normal of y and its value is equal to kv2, where k is the curvature of y. For example, a particle with electric charge moves in a similar way in a magnetic field. Let us consider the motion of a mass point in a force field. Suppose that a force F is a vectorfunction depending on the position vector of the mass point, the velocity and the time, i.e. F = F(r, drldt, t ) . In some cases F depends only on r (similarly, for example, to the gravitation of fixed center); in others F depends only on the velocity (like the free fall of a mass point in a resistant medium). In the general case we have:
© 2000 CRC Press
84
DIFFERENTIAL GEOMETRY A N D TOPOLOGY O F CURVES
Motion in a Parallel Force Field Let F be a force parallel to a fixed direction a, i.e. F = X(r, v)a, where X(r, v ) is a scalar function. Along the trace y of a mass point with position vector r(t), the function X(r, v) is a function of the time 1. Integrating equation (18.1) we obtain
dr m(4 dt
dr dt
= m(to)
+ad)([),
where
Thus a mass point in a paralle1,force field moves on u plane spanned by a constant vector and an initial value of velocity vector passing through r(to).
Motion in a Central Force Field Let F be a force represented by the formula F = A(r(t),v(t))r, where r is a position vector of a mass point and X(r(t), v(t)) is a scalar function; this means that F is a central force with its center at the origin of Cartesian coordinates. The equation of motion is
It follows from equation (18.2) that
Hence [v, drldt] is equal to a fixed vector c. Because the inner product
vanishes identically, we obtain: (c,r) = 0. Thus a trace of a mass point in a central .force field is situated on a j x e d plane, which is uniquely defined by an initial position vector and an initial velocity of the mass point. The converse statement is true also: © 2000 CRC Press
CURVES IN MECHANICS
85
Theorem of Halphen If'the,field of,forces does not depend on time and if all traces of some mass point are plane, then the field o f forces, which have un qffect on the mass point, is a central force field or a paralle1,force field. Motion of a Mass Point in the Gravitation Field Set r3 = lr13 and F = p r / r 3 , where p is a positive constant. The equation of motion of a mass point in the gravitation force field F is
Using this equation we obtain
Let us choose Cartesian coordinates in such a manner that the of motion of the mass point. Setting we obtain
x=pcos4,
(X,
y)plane is a plane
y=psin&,
Integrating equation (1 8.3) we have
where A is a constant. Above we have proved that [r, $1 only for the coordinate z we obtain
= c. If
we take this equation
The last equation can be rewritten as
Substituting the expression of d4/dt in terms of p, c0 into (18.4) we have
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86
DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
Now we find the expression of dpldt from the last equation and multiply it by dtld4. With the help of (18.5) we obtain
where c, are constants. Integration of this equation results in the equation of the trace of mass point:
where 40 is an initial value of 4. The obtained equation represents a conic section with focus at the pole.
Motion of a Charged Particle in an Electromagnetic Field Assume that a particle of mass m and electric charge e is moving in an electromagnetic field. Let E be the electric component and H the magnetic component of the electromagnetic field. The force of the electric field having effect on the particle is equal to eE; by the law of Lorentz the force of a magnetic field is equal to : [ H , $1, where c is the velocity of light. Then we can write the equation of motion: d2r m=eE+dz2
:[
H.
If we multiply equation (1 8.6) by drldt we obtain m 2
d(v2) = e(E, dr),
where Ivl is the velocity of the particle. Thus we have the following statement: ifthe electric component of an electromagnetic field is vanishing, then a charged particle moves with a constant velocity. Suppose that E vanishes and the magnetic field H is generated by a unique magnetic pole situated at the origin of the Cartesian coordinates; then H has the form 'k,o3 where k is a constant and p = We can introduce the length of arc s of a particle's trace in place of the time t: s = vo(t  t o ) Then the equation of motion is
,/m.
where p is a constant. Multiplying this cquatiorr by r we >tax{ © 2000 CRC Press
CURVES IN MECHANICS
The integration of this equation leads to the following:
where a is a constant vector. From the last equation it follows that p
+ ( a ,r ) = 0.
This means that the trace of a charged particle is situated on a cone with its center at the magnetic pole. Moreover a trace o f a chargedparticle moving in a magnetic field is a geodesic. curve of a cone, i.e. a curve which transforms in a straight line when tlze cone develops onto the plane. Motion of a Charged Particle in a Constant Electromagnetic Field Let E and H be electric and magnetic components of a constant in the space and time electromagnetic field. The equation of motion (18.6) can be rewritten:
where
P
and p are constants. From this equation it follows that
where c is a constant vector. Let us choose Cartesian coordinates as follows: the zaxis is collinear to H and the yaxis is orthogonal to the plane spanned by E and H. Then
and we can write
© 2000 CRC Press
88
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
From the first and second equations we obtain
where a = % p and p is a constant. The general solution of this equation is X=
p
+ sintat + a ) A ,
tu = cons!,
A
= corzst.
Substituting the obtained expression of x into the second equation of system (18.7) we can find y=qt+hcos(at+or)A,
q=const,
h=consl.
The general solution of the third equation of system (18.8) is
Thus we can represent the motion of a charged particle in the form of a sum of two motions: a uniform motion along a circle X = Asin(at
+ a),
J, =
Acos(at
+
tu),
and a motion of this circle whose center is moving along a plane curve
Note that the moving circle is parallel to the ( X , y)plane. In the case of vanishing XI and q the particle moves along a spiral.
© 2000 CRC Press
19 Curve Filling a Surface Some simple curves have a very complicated behavior. In this chapter we will construct a space curve situated on a torus, which fills this torus densely everywhere. First, let us write the equation of a torus. This surface is formed by points of a circle (1, which is moving in such a way that the center of a is moving along a circle P and the planes containing a and are mutually orthogonal. We will write the position vector r of the points of the torus in the form of a vectorfunction of two variables 4 and H. Let us place the circle P into the (X, y)plane with the center of P at the origin 0 of Cartesian coordinates. Denote the radius of B by R and the radius of a by p. Let e, be unit coordinate vectors, P an arbitrary point of P and 4 an angle between el and OP. Then O P = R(cos 4 e l
+ sin 4ez).
+
We will denote the unit vector cos 4 el sin 4 e2 by ~ ( 4 )The . circle a is situated in a plane y, which passes through O and is spanned by e3 and OP; the center of cw coincides with P. When the plane y rotates, the points of the circle a form the torus (Figure 19.1). Let Q be a point of a. Then we can represent OQ in the form of the sum OP PQ. The vector PQ is a linear combination of U and e3. Denote an angle between v and PQ by 0; then we have
+
PQ
= p(cos B I /
+ sin H e3).
Thus the position vector of the points of the torus has the following form: r(4,H) = R(cos 4 el + sin 4 e2) p(cos B(cos 4 el sin 4 e2)
+
© 2000 CRC Press
+
+ sin Q e3).
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF ClJRVES
FIGURE 19.1
Now let 0 be a linear function of 4:
where X is an irrational number. When 4 is varying from m to +m, the points with coordinates 4, 0 = A4 on the torus form some curve r. If X is a rational number p/q then the curve F will be closed after p rotation with respect to 0 and q rotation with respect to 4. Let us fix the circle a0 given on the torus by the formula 4 = 0. If X is irrational then the number of points where F intersects a0 is infinite. We will show that these points (i.e the points with coordinates 4 = 27rn, 0 = A27rn, n E Z ) form a set on cu0 which is dense everywhere. Let us consider the Poincarl map g: a0 , a0 defined by the formula g(0) = 0 X27r. Denote a composition of k maps g by $. The map g preserves lengths of arcs of tro. Let us take an arbitrary point Q E a0 with angle 00, a positive t , and an arc U c a0 containing Q and whose length is less than 4 2 . Because the arcs U, g ( U ) , . . . ,g k ( u ) ,. . . have an equal length, there exists k > l such that & U ) n g'(U) is not empty. Every gh has an inverse map which we denote by g p k . We can consider the set M = g''(U); it is easy to see that g'(A4) = g k ( u ) . Let P E $ ( U ) n g ' ( ~ ) Then . g  ' ( ~ )C M and g  ' ( ~ )C U. Therefore the intersection g k p ' ( u ) n U is not empty and 100  g"(OO)l< t, where m = k  l. Hence the points 00, g)ll(OO), g2m(00),. . . ,gnm(O()), . . . decompose a0 into equal arcs whose length is less than t. Thus for any fixed arc of a0 we can find a positive t and an integer m such that points gn"'(Oo) are situated inside this arc. This means that the set of points gN(Oo),n E N is dense everywhere on ao. Now it is easy to show that the curve F filling the torus is dense everywhere. Note that the projection of T onto the ( X , y)plane is represented by the formulas
+
X
= cos4(R+pcosX4),
y
= sin 4 (R
+ p cos X4).
If R > p then this projection is a regular curve filling an annulus densely everywhere between two concentric circles.
© 2000 CRC Press
20 Curves with Locally Convex Projection Let us translate a straight line l parallel to a vector e along a space curve y. A surface formed by the moving line l is called a cylinder. The curve y is called the d k c t r i x of the cylinder and the line 1 is called the generutor of the cylinder. We devote this chapter to curves on the cylinder. Without loss of generality we can suppose that y is a plane curve situated on the ( X , y)plane and 1 is parallel to the zaxis. Denote by ro(t)the position vector of y,by t the length of arc of y and by e the unit vector spanning l. Let be a regular curve lying on the cylinder. Its position vector has the form r(t) = ro(t) z(t)e, where z(t) is the third coordinate of the points of r. The curve y can be viewed as the projection of r into the ( X y)plane. , For the torsion r; of l? we have the following expression:
+
K
=
(r', r", r"') zl'lp + z'lpl + z l / p l [ r y 1 2 = I +z12 +z/12p2
where we denote by I/p the curvature of y and by a prime the derivative with respect to t. If p(t) is the angle between the vector tangent to y at point ro(t) and a fixed vector in the ( X , y)plane, then dtldp = p. Suppose that p > 0, i.e. y is locally convex. By this assumption y is parametrized regularly by p and z' = $(p). Because
we obtain from (20.1):
It is easy to see that ds = p ~ m d is pthe differential of the length of arc s of We will use formula (20.2) to investigate the behavior of l?.
© 2000 CRC Press
r.
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CUJIVES
02
Lemma Assume that the curvature k and torsion projection y of is locally convex, then
K
of'r are not vanishing. lj the
Proof Rewrite expression (20.2) as follows:
Multiplying both parts of (20.4) by J1+1/,2and integrating with respect to cp we obtain
for any points PO, P I of I? and corresponding points P ~ , of y. Inequality (20.3) B follows from (20.5) immediately.
Corollary Assumr that the curvature k of r is not vanishing, the torsion r; is bounded K > KO = const > 0 and the total length of is infinite. Ifthe curvature I / p of y is bounded l / p 5 const, then the total length o j ' y is infinite. Proof Because 1 S,. K dsJ 2 1 S,. KO dsl = ~ o L ( ris) infinite, we obtain from (20.3) that dt/p is infinite, and it follows from the boundness of l / p that S*, dt is infinite. Now let us consider the following problem: assume that some curve satisfies three conditions: Jy
(a) its total length is infinite; (h) its curvature is positive and its torsion is bounded from below by a positive constant; (c) it has a locally convex projection. When is this curve bounded? This question is not trivial. For example, the curve x = cos p, y = sin p, z = cp  2 sin cp has infinitely many stationary points with respect to the zcoordinate.
Theorem [5J Suppose that a regular curve r of infinite tolal length has a positive curvature k > 0, a torsion K bounded from below by a positive constant KO > 0 and a closed strongly convex projection y. Then I' is unbounded. Proof First, the curve r is not tangent to the zaxis, because of the convexity of y and k # 0, K # 0. Therefore the function zi is regular. © 2000 CRC Press
CURVES WITH LOCALLY CONVEX PROJECTlON
Rewrite (20.2) in the following way:
+
where Q(p) = & ( l $2 i(d$/dp)'). It follows from the differential equations theory that li, can be represented by the formula
/
011
$(h)=
sin($

i.)Q(p)pdp
+A
COS
Cl + Bsin 80,
0
where A and B are constants. Assume that the origin of Cartesian coordinates is a point of y. The coordinates of the position vector rO(s) are X($) =
S
COS
p(7) dr,
y(s) =
Denote by I the total length of y and set ~ ( s=) Q(p(s)). When a point of the closed curve y goes along y once, a corresponding point of l? goes along l? and a variation of its zcoordinate is
We observed that x d y  y dx = r i do, where cu is the angle between vector ro and the xaxis. Since y is strongly convex, dolds > 0. Thus
where S is the area of the plane domain bounded by y. From the corollary it follows that if a point P goes along the whole curve F, the corresponding point P goes along the whole curve y infinitely many times. Hence the variation of the zcoordinate of P is infinite. Thus, l? is unbounded. It must be pointed out that the condition of strong convexity is essential. One can construst a closed curve r with k > 0 , K KO > 0 and with a locally convex projection. For this purpose let us set
>
+('P) = 1  4 cos X'P)
© 2000 CRC Press
DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
p(p)
=
l
+ 21 cos Xp,
00
< p < +cm,
where X is a rational number such that X2 = 1 + t, 0 < t < 114. We define the curve l? in the following manner: 'P
'P
X =
c
o
s
y =Ipsinpdp, 0
0
z= /+pdp 0
It is easy to obtain that 112 < p 5 312 and
If X2  1 < 114, then K is greater than some positive constant Q. On the other hand, because X is rational and X2 # 1, the curve r is closed. But if the constant KO is sufficiently great, we cannot construct such an example. Theorem Assume that a curve r of injinite length is regularly projected into U locully convex curvcj y with curvature Ilp. Let 0 < l/p < const und the torsion K c,f r satisjy the inequu1it.v K sup&. Then l? is unbounded.
>
Proof Let us show that there exists cpl such that the corresponding function $(p) is monotone on the ray p pl. Assume that there exist two points PO= P(po) and PI = P ( p l ) such that W l d p is vanishing at these points. Using (20.5) we have
>
>
Because K sup ,'2~ 4)j,r 1 on the segment [po,p,]. Therefore 4)E 1 on the whole ray [po, cm), or there exists p1 such that is monotone on the ray cp 2 p,. Then y', does not change the sign on some ray p 2 ( ~ 2Let . us take an arbitrary segment A belonging to the ray [p2, m). From the inequality
+
+
+
we have
where KO = inftc, AZ is a variation of z at A. Because one can take A to be an arbitrarily great length, lAz/can be sufficiently great too. Therefore r is unbounded. © 2000 CRC Press
CURVES WITH LOCALLY CONVEX PROJECTION
95
We remark that the value 112 is essential for lcp. If lcp > 112, then z,, does not change sign when s is sufficiently great; but if i n f ~ p< 112  t, where F is a small positive constant, the function z,, can have the form of an oscillating function.
© 2000 CRC Press
21 Integral Inequalities for Closed Curves Let r' be a closed regular curve in Euclidean space E ~Denote . the length of arc of F by s and the curvature by k(s). We will prove Fenchel's inequality [12]:
and we have the equality in (21.1) ij'and only if F is a convex plane curve. Suppose that the curvature of is positive. We will define a spherical indicatrix of tangents of the curve .'I Let P be a point on r and T ( P )a unit tangent vector of r at point P. We translate r ( P ) in such a manner that the origin of T ( P )coincides with the origin 0 of Cartesian coordinates. Then the end point P* of the translated vector r ( P ) is situated on the unit sphere S'. When P moves along l?, the point P* moves forming a curve 7 on S 2 . This curve is called U spherical indicatrix of tangents of the curve l?. The position vector of 7 is represented by a vectorvalued function ~ ( s ) . Using the Frenet formulas we have
Hence (dr/ds,dr/ds) = k2. From the supposition of positiveness of k it follows that y is parametrized regularly by S . Let rr be a length of arc of y.Then the length of arc s of F can be viewed as an increasing function s = s(a) of m. The derivative of T with respect to is a unit vector, therefore from
d r ( s ( a ) ) &(S) ds da ds drr
© 2000 CRC Press
98
DIFFERENTIAL GEOMETRY AND TOPOLOGY 01: ClJRVES
we obtain drr
= kds.
Thus the integral in formula (21.1) is equal to the length l of y:
So, in order to prove inequality (21.1) for some closed curve we must evaluate the length of its spherical indicatrix of tangents. Assume that r is closed and nonplanar. Because is closed, for any plane a there exist points on which are most (or least) distant from a. The tangent vector T of r at each of these points is parallel to cr. Therefore the indicatrix y meets any plane a passing through 0. Thus the spherical indicatrix of tangents of a closed nonplanar curve has the following property: this curve intersects any great circle at least in two points and therefore it does not belong inside any hemisphere. Now we will prove that any closed curve, which is situated on S' and whose length is less than 2n, belongs to some open hemisphere. For this purpose we will use the following property of the great circles on S': for two arbitrary points on S2 the shortest curve on connecting these points is an arc of great circle and this arc is defined uniquely if the points are not antipodal; we will call the length of the shortest curve an inner distance between two points on S*. We will expound the proof of the following more general statement (see [ l l]): Theorem Let y be a closed curve on S* and suppose that length L qf y is less than 2n. Then there exists a point Q E S' such that for any point P E y the inner distance between Q and P is not greater than L/4. Proof We will denote by p(*, *) the inner distance on the unit sphere. It is easy to observe that if the inner distance between two points A and B is less than n, then there exists a unique point C on the arc AB such that p(A, C ) = p(C, B) = p(A, B)/2; the point C is called the center of the pair A, B. Lemma Let p(A, B) < n , C be the center of the pair A, B and X a point on the sphere such that p(X, C ) < 71.12. Then 2p(C,X ) = p(A, X ) p(B, X ) .
+
Proof In order to prove the lemma let us consider the great circle connecting C and X, and fix a point X on this circle in such a way that p(C, X ) = p(C, X ) , i.e. X is symmetric to X with respect to C. Because A is symmetric to B with respect to C, we have p ( ~A) , = p(X, B). Since p(C, X ) < n/2, p(X, X ) = p ( ~ C, ) + p(C, X ) = 2p(X, C ) . On the other hand, by the triangle inequality: p(X, X ) 5 p(X, A) + p(A, X ) = p(X, A) p(B, X ) . Thus 2p(C,X ) = p(A, X ) p(B, X ) . Now let us fix two points A, B E y; these points decompose y into two arcs AB and BA of equal lengths L(y) = 2L(AB) = 2L(BA). Because p(A, B) < L(AB) = L(y)/2 < n,there exists a unique center C of the pair A, B. Let X be an arbitrary point of y; we assume without loss of generality that X belongs to the arc AB. Suppose that p(C, X ) < 71.12.Because p(A, X ) is less than the length of the arc AX and p(B, X ) is less than the length of the arc XB, using the proved lemma we obtain 2p(C, X ) < p(A, X)+p(B, X ) < L ( A X ) L(XB) = L(AB) = L(y)/2. Therefore p(C, X ) < L(y)/4.
+
+
+
© 2000 CRC Press
INTEGRAL INEQUALITIES FOR CLOSED CURVES
99
Thus we have proved that if p(C, X ) < L(y)/2, then p(C, X ) 5 L(y)/4. So, for any point X of y two alternatives exist, p(C, X) 5 L(y)/4 and p(C, X ) > 7r/2. Since p(C, X ) , when viewed as a function of X E y, is continuous and p(C, A)= p(A, R)/2 5 L(AB)/2 = L(y)/4, for any point X E y we have p(C, X ) L(y)/4. Because the spherical indicatrix of any closed nonplanar curve does not belong to any hemisphere, we can conclude, using the proved theorem, that the length of the indicatrix is greater than 27r; it is easy to complete the proof of Fenchel's inequality. Let us consider knotted curves without selfintersections. By a knotted curve we understand a closed curve which cannot be deformed continuously without selfintersections into the unit circle. Fenchel's inequality is generalized by the FaryMilnor inequality:
<
for an arbitrary knorted curve 1'. One can carry out the demonstration with the help of one formula of integral geometry. Let a closed curve y with length l be situated on the unit sphere S; Cc C S denotes a great circle lying in a subspace E2 with a unit normal vector (. We denote by N i t ) the number of points where y intersects CF. Considering N ( [ ) as a function of we have (see [15]);
<
Let y be the spherical indicatrix of a knotted curve r. We will prove that y intersects any great circle of S at least at four points. Suppose that y meets some great circle C of S at exactly two points, which means that the distance Iz(P) from a point P E l? to the plane containing C has exactly two stationary points PI and P2. Then one can decompose the curve r into two arcs I'[,r2in such a manner that h(P) is strictly increasing on rl and strictly decreasing on r 2 . Therefore any plane 2, which is parallel to and situated between PI and PZ,intersects I'at two points Q l , Q2. Let us connect Q [ , (22 by the segment Ql Q2. The family of all such mutually disjoint segments forms a surface containing r. A domain bounded by on the surface is homeomorphic to a disk, therefore is not knotted. Thus our assumption is false, so the function h(P) has at least three stationary points on l?. But the number of stationary points must be even. Hence h(P) has at least four stationary points, and y intersects any great circle at least at four points, i.e. N(<) > 4. Applying this fact we obtain
and this completes the proof of the FaryMilnor inequality.
© 2000 CRC Press
22 Reconstruction of a Closed Curve with Given Spherical Indicatrix of Tangents In the previous chapter we proved that any great circle of the unit sphere meets the spherical indicatrix of a closed space curve l?, it means that the spherical indicatrix does not belong to any hemisphere. This necessary condition was first discovered by E. Poznjak when he was a postgraduate student of Vygodsky. We will discuss the following problem: Let y he u closed curve on the unit sphere. Does there exist a closed curve l? such that y is the spherical indicatrix of r?It was observed that the mentioned necessary condition is sufficient.
Theorem Let y he a closed curve on the unit sphere such that y does not belong to uny hemisphere. Then y is the spherical inu'icatrix oj' some space curve r. This statement was proved by Vygodsky [l01 and we will sketch the proof proposed in [g]. Let [(U)be the position vector of y,u E [ G . h]. Then for any constant vector c the inner product (c,[(U)) is a function of u with alternating sign. Denote by r(s) the position vector and by s the length of arc of the desired curve r. Then drlds = [(U). Assume that s is a strongly increasing function of U ,i.e. ds/du = =(U) > 0. We have r(u) = r(a) ( ( U )d~ = r(a) [ ( u ) a ( ~du. i ) Let us set r(a) = 0. Considering all possible positive functions a(u) with a(a) = n(h), we can define a set M formed by points with position vectors r(h) = ~;j'[(u)a(u)du. This set is convex. In fact, if r l ( b ) , rz(b) belong to M , i.e. r,(b) = [(u)a,(u)du, then for any positive constants X,, X?:
+ Jr
Xlrl( h )
+ 1;
+ X2r2(h) = X I h
© 2000 CRC Press
I ( U ) (~U I) d~ + X 2
102
DlFFERENTlAL GEOMETRY A N D TOPOLOGY OF ClJllVES
hence the point with position vector Xlrl(h)+ X2r2(h) belongs to M , therefore M is convex. Assume that M does not contain the origin 0 of Cartesian coordinates. Then there exists a plane E', which contains 0 and does not intersect M. Denoting by e the unit vector normal to E', we see that the inner product (r. e), when viewed as a function of the points of M, has a constant sign; we can suppose that it is positive. Therefore ([, e)a(u)du > 0 for any positive function n(u) satisfying a(n) = o(h). On the other hand, ( [ ( U )e) , is a function of 24 with alternating sign, and we can choose the function o*(u)in such a manner that a*(u) is sufficiently small when ( [ ( U )e) , > 0 and sufficiently great when ( < ( U ) , c ) < 0. Then J ' ([, e)o*(u)du< 0 and we obtain a contradiction. Thus the set M contains 0,i.e. there exists a positive function @ ( U ) satisfying @(U) = @(h)and i ( h ) = (@(U) du = 0. This means that a curve with the position vector ?(U)= h f [ @ ( u ) r his closed, and it is obvious that this curve is regularly parametrized by u and that y is its spherical indicatrix of tangents.
h:
© 2000 CRC Press
23 Conditions for a Curve to be Closed Let us consider a planar curve y with position vector r = r(s). Assume that the curvature k(s) of y is a continuous periodic function and its period is equal to T. We have
The vector tangent to y has the following form: r:, = { cos a , sin a ),
where a denotes the angle between rt. and the xaxis. The curvature of the planar curve y,when viewed as the curvature with sign, can be expressed in terms of a:
Therefore u(s) = a0
+
1 S
k(s) ds,
where
a0 = consr.
0
Assume that y is closed and that its length is equal to L. If a point goes around y, then obviously the variation of a is divided by 27r. Thus
© 2000 CRC Press
104
DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
where m is an integer. We obtain, after integration of the expression of rr(s):
S
cos ads = 0,
S
sin ads = 0.
0
0
Substituting (23.1) into (23.2) we conclude that satisfies the equalities
if
y is closed, then its curvature k(s)
One can demonstrate that if the curvature of some planar curve satisfies equality (23.3), then this curve is closed. Efimov, Fenchel and other geometers independently studied the following problem: what are the necessary and sufficient conditions on the curvature and torsion o f a space curve in order for the curve to be closed? It is assumed that the curvature k(s) and torsion IF.(S)are functions of the arc length S. Fenchel thought that this question is the most important one in the theory of closed curves. There exists an ineffective way to solve the problem. The reconstruction of a space curve with given curvature and torsion is reduced to one integral Volterra equation. By the Neumann formula a solution of such an equation can be written in the form of a series, any sum of which is determined by the curvature and torsion. Effective conditions may be found in some special cases. We will consider one such case. Let y be a closed space curve with curvature k(s) and torsion IF.(s). Naturally we assume "k tc2 > 0. We replace the arc length s by the parameter
If L denotes the length of y, we set T = t(L) and define a function cp(t) in the following way: cos cp
=
k
sin cp =
,/W'
IF.
JFT2
It is clear that if the functions cp(t) and t(s) are given, then the curvature k(s) and torsion ~ ( sare ) reconstructed uniquely, and if the curvature k(s) and torsion ~ ( sare ) given, they determine the functions p(t) and t(s) uniquely. (We assume that K is defined with a sign.)
Theorem Let the curvature k(s) and torsion ~ ( sof) a space curve y be such that p = a1 t + az, ai = const. The curve y is closed and its length is equal to L qj'there exist two integers p, q such that a1 = (g  p)/2&Gj and © 2000 CRC Press
CONDITIONS FOR A CURVE 7'0 BE CLOSED
I
sin p ds = 0,
(23.5)
Proof Denote by t l , G, t3the natural frame of the curve y.Frenet equations have the following form:
3=

cos p<, + sin &,
=

sin pc2.
* dt
dt
Introducing the vector obtain
/L =
sin cpEl
By the proposition dpldt =
+ cos pt3 and using the Frenet equations one can
It is easy to show that
It follows from (23.8) and (23.9) that d&/dt satisfies the linear differential equation
We denote 41 + o:
=r.
Hence ~dt= ~ c o s r t + B s i n r t ,
© 2000 CRC Press
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DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
where A and B are constant vectors. Since d&/dt is a unit vector, the vectors A, B are unit and mutually orthogonal. Integrating (23.10) we get: 1
E2 = ( A s i n r t  B c o s r t 7) +
(23.11)
C,
c2
where C is a constant vector; because is unit, the vector C is unit and orthogonal to A and B. The second derivative of & is
Using expressions (23.1 1),(23.12) and equality (23.8) one can find the vector p. Then it follows from the Frenet formulas that the vectors JI, t3 are a solution of the following system: 
coscptl +sincpG
=Acosrt+Bsinrt,
Therefore €1
=A[
2 ' 1
s i n r t sincocosrt coscol
Since d&/ds is a periodical function, its values at t (23.8) we get T T = 2nn, where n is an integer. On the other hand, k and
=0
and t
=
T coincide. From (23.14)
K.
are periodic functions, hence
i.e. a1 T = 27rm, where m is an integer. So (23.14) and (23.15) lead to the following equalities:
Set p = n  m, g = n + m. Then equality (23.4) follows from (23.14). Because the vectors A, B, C are mutually orthogonal, the integrations of the coefficients of (23.13) and the condition © 2000 CRC Press
CONDITIONS FOR A CURVE TO BE CLOSED
are leading to equalities (23.5)(23.7). Now we will sketch an investigation of the considered problem in the case of polygonal lines (see [49]).The solution of this problem may be useful in various technologies. Let a polygonal line y = POPl. . .P,, be formed by segments a;+, = PiP,+l.Analogues of the curvature and torsion of regular curves are two sequences of angles p;, i = fi and Q,, i = G defined in chapter 14. Denote by ei a vector collinear to the segment Pipi+, with the origin at Pi and the end point at Pi+l. Then cos pi = (e;,e;,,). The plane spanned and oriented by the basis of vectors ei, ei+1 determines uniquely the unit normal vector Pi= [ci,ei+l]/sin pi, which is an analog of the binormal vector of regular curves. By definition the angle Hi is equal to the angle between the vectors o,, p;, I situated in an oriented plane with the unit normal orienting vector e,+l. Suppose y is closed. The last segment a, of y is determined uniquely by another n  I segments a,. Therefore the closed polygonal line y is determined uniquely up to a space motion by 3n  6 parameters
where by si we denote the length of a;. Thus the problem can be formulated in the following manner: what conditions on the length si and angles p,, Qk are necessary and sufficient in order that the polygonal line y is closed? There are two methods: (1) one can obtain the expressions of si, p;, Ok in terms of some other parameters satisfying some equations; (2) one can find a system of six equations 
with respect to 3n parameters si, p;, H;, i = 1 , n. Let us consider the first method. Because y is closed, the lengths s; satisfy the equality Denote by ( X , , y;, z;) the components of e; with respect to Cartesian coordinates X , y, z in E ~Then . equality (23.16) is equivalent to the system of three equations for X , , yi,2;:
Since all e; are unit, we have n equations:
© 2000 CRC Press
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
108
The angles pi are computed by the formula
The angles
P, also can be expressed in terms of Cartesian coordinates:
Assume that X I , .. . ,z, are given in such a way that the three lines of the matrix
are independent; let for instance
Then we can express S,, Q , s3 in terms of s4,. . . , S , and X I , . . ,z,. Because X,, y,, z, satisfy n equations, the number of independent parameters is equal to 3n  3. The positions of the vectors e l , are not fixed. We set xi = 1, yl = zl = z2 = 0. Then the number of independent parameters is equal to 3n  6. Thus, the system
can be viewed as necessary and sujjicient conditions in order for y to be closed. These conditions are in terms of the parameters S;,xj,. y k , ZI,but it is of more interest to find similar conditions in terms of si, pj,Bk. Assume that y is nonplanar and three vectors e t , ez, e3 are mutually independent. Set aq = (ei,e,). If we multiply equality (23.16) by e l , ez, e3, we obtain the system
Consider the ndimensional symmetric matrix A the chapter a proof of
= Ila,ll.
We will sketrk at the end of
Lemma All elements of the matrix A are expressed in terms of ttit ungles © 2000 CRC Press
v,,
8,.
CONDITIONS FOR A CUKVE TO BE CLOSED
So equalities (23.17) are conditions with respect to
cos Q,
=
S,,
109
R,Bk. It is easy to see that
.,+l], k , + l , e,+21) (P,, P]+l) = (l.,? sin cpi sin R+I
By these formulas the elements 012, 4 3 , . . . , a,l, and Q,, are expressed in terms of pi, and the elements u13, ~ 2 4 ,... , an2, and al,l, a2, have expressions in terms of pi, Q,. Consider the case when y has four segments, n = 4. Then the system consisting of equalities (23.18) has the following form:
From the last four equations one can find a13 and angles p,, Q, satisfy two equations
U24
It is possible to make iff the
cos 81 sin p1 sin p2  cos pi cos p2 sin p4 sin p3  cos p4 cos p?,
= cos B3
cos Q2 sin p 3 sin p2  cos p3 cos p2 = cos Q4 sin p1 sin p 4  cos pl cos p4. Four vectors el ,e2, e3, e4 belong to threedimensional Euclidean space, therefore the corresponding determinant is vanishing:
To prove equality (23.20) we observe that system (23.17), consisting of four linear equations, has a nontrivial solution, hence its determinant is equal to zero, i.e. (23.20) is satisfied. Condition (23.20) is equivalent to the equality
© 2000 CRC Press
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DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
Since all elements ak, are expressed in terms of the angles p,, B;, equality (23.21) is a condition with respect to pi, 8;. Thus, the polygonal line y with four segments is closed if and only if the lengths sl, s2, ~ 3 s4, and the angles pl, p2, p3, p4,81, 82, &,84 satisfy the six equalities (23.17), (23.19), (23.20). Consider the case when y has five segments, i.e. n = 5. Similarly to the case of four segments, elements of the matrix
have expressions in terms of the angles p,, 0;:
a24 = cos p2 cos p3  cos H2 a35 = cos p3 cos p4  COS O3 a14= COS p4 COS p5  COS O4 a25 = cos p5 cos p1  cos 05
sin p2 sin p3 sin p4 sin p5
sin p3, sin ( ~ 4 , sin p5, sin p1
Thus ten elements ak, of the matrix A are expressed in terms of ten angles pi, 8,. Because c l , . . . ,es belong to threedimensional Euclidean space, by taking any four vectors e; and writing the corresponding equality similar to (23.20) we obtain three independent equalities
These equalities are independent because a12 is included only in the first equation, a15 only in the second and 0 2 5 only in the third. Substituting the obtained expressions of akl into system (23.22) we get three equalities similar to (23.21) in terms of p,, 8,. Thus, the polygonal line with five segments is closed if and only if the lengths sl, . . . ,$5 and the angles qq, . . . ,ps, 0 1 ,. . . ,O5 satisfy six equalities (23.17), (23.22). If y is formed by six st  m n t s , with the help of formulas (23.18) one caii express all elements akr of the corresponding matrix A except al4, 0 2 5 , U36 in terms of the angle

© 2000 CRC Press
CONDITIONS FOR A CURVE TO BE CLOSED
pi, el,
Hi.
11 1
In order to find these elements, we observe that any four vectors among
. . . ,e6 are contained in threedimensional Euclidean space, therefore the follow
ing conditions similar to (23.20) must be satisfied:
These equalities are quadratic equations with respect to 014, 025, a16; if some inequalities are fulfilled, they have real solutions. Thus all elements of A can be expressed in terms of p;, C),. Note that al5, a16,~2~ are not included in the three equations (23.23). Then we can write another three equations similar to (23.20):
which can be viewed now as equations with respect to pi, Hi. They are independent because als is given only in the first equation, a16 only in the second and a26 only in the third. The proposed method is applicable in the case of polygonal lines with n segments. Let us prove the lemma formulated above. We have observed that all elements
are expressed in terms of p,, H,. Then we can find any elements aii+3 using an equation similar to (23.20). For example, in order to obtain a14we have the equation
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112
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
where all other a? are found. After all aii+3 are obtained, we seek the expression for aii+4 and so on. Continuing in the same way, one can express in terms of the angles pi, 0, all elements akl of the matrix A . We observe that three equations
are not used to find the elements akl. They are independent because al,l is included only in the first equation, a,, only in the second and a2, only in the third. Thus we have proved Theorem . The polygonal lirze y with n segments is closed f u n d only if the lengths sl , . . . ,S , and the angles pl, . . . ,p,, H I , . . . ,Q, satisfy six qualities (23.17 ) , (23.24), where all coefficients akl are expressed in terms of p,,8., Let us return to smooth curves. It seems that there does not exist an effective way to solve Fenchel's problem; this opinion was supported by Nikolaevsky [61]. He considers the following problem leading up to Fenchel: given a curve y on the unit sphere S, find necessary and sufficient conditions depending on the geodesic curvature of y C S for this curve to be closed; it is assumed that the geodesic curvature is given as a function of the arc length of y. (The length of the projection of the curvature vector of a curve situated on a surface into the tangent plane to the surface at the corresponding point is called the geodesic curvature of the curve.) The theorems following below show that the necessary conditions in the form of simple natural integral equalities are either empty or far off the sufficient conditions. Let l > 0 be a fixed positive number, c 2 ( 1 )denote the class of c2regular closed curves of length l that are situated on the unit sphere S. Assume s denotes the arc length, and k(s) stands for the geodesic curvature of curves from C2(1). Theorem Let F(x, y) be a continuous .function. Then either ( l ) the set of' values of the integral
c :imputed for all cuvves 1, W I C2(l) contazi 7 int~rval, or ( 2 i F(x,y ) p b ) , whei e p(y) u a perio~ fun^. )n M 111 od

© 2000 CRC Press
j
~
h
L
CONDITIONS FOR A CURVE TO BE CLOSED
113
Theorem Let F(x, y, z ) and G ( x ,y) be analytical functions defined in the space and in the ( X , y)plane respectively. Then either ( l ) the set cf values of' the integral
contains an interval, or computed,for all curves.from c2(1) &), where p(y) is a periodic ,function with prriod I, or (2a) F(x,y, z ) (2h) G ( x , y ) cp(y), where PO/) is a periodic filnction with period l, and F ( x , y , z ) = Y'i(x,y). Remark Cases (2) of the stated theorems mean that the considered integral equalities do not depend on the geodesic curvature, hence they are empty. Necessary conditions for space closed curves were investigated in [62] for socalled trigonometric curves. By definition, a space curve y is called trigonometric if each component of its position vector r(s) with respect to the arc length s is a trigonometric polynomial:
where ck are constant complex vectors satisfying the equations ck = c,:. Of course, any trigonometric curve is closed. So, we can interpret the EfimovFenchel problem in this case as follows: what are the properties of the curvature k(s) and the torsion r;(s) of the trigonometic curve y with position vector (23.25)? It is more convenient to (s) of k(s) and r;(s). One can demonstrate consider the functions /c2 and ~ ( s ) k ~ instead easily that if r(s) is a trigonometric polynomial of degree n, then k 2 and r;(s)k2(s)are trigonometric polynomials of degree 2(n  1) and 3(n  1 ) respectively:
It is observed that the coefficients of these two expansions are mutually connected.
Theorem I f r(s) is a trigonometric polynomial of degree n with respect to the arc  1) are connected by three algebraic rrlations length, then the coefficients ao, . . . ,
involving the complex conjugates 5,. © 2000 CRC Press
1 14
DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
Each coefft'cient
pp is connected with
the coefficients a, by an algebraic relation
The last two coeflicients ,L$ are
Here F,, Q iand 0, denote algebraic polynomials with integer coefficients As we have mentioned, the degrees of k2 and &k2 are not arbitrary. Theorem Let k(s)and ~ ( sbe) the curvature and torsion ofa trigonometric curve. Then the degree v of the trigonometric polynomial k2 is even, and the degree p of the trigonometric polynomial &k2is divided by 3. Also, ij'p is equal to 31, then v is equal to 21. It is natural to consider a trigonometric curve whose position vector is a trigonometric polynomial of degree n such that the degree of the corresponding polynomial k 2 is less than 2(n  l) or the degree of the corresponding polynomial &k2is less than 3(n  1). One can show that if the last 2k coefficients of the polynomial k 2 are equal to 0, then the last 2k vectors cp have a very specific form. First, we say that two complex vectors c, d are collinear if there exists a complex number X such that c = X4 next, any complex vector c = a ib with nonzero real part a and nonzero imaginary part b determine a unique plane in E3 spanned by the vectors a and h; we say that this plane is the plane of the complex vector c. It can be proved that if ~ 2 ( ,  1 ) = . . . = Gt2(,k) = 0, then the vectors c,, . . . ,c,k, c]  (c,, e)e, j = n  k  1 , n  2k are collinear, where e is a unit real vector orthogonal to the plane of the vector c,.
+
Theorem Let the curvature k of a trigonometric curve y C is an arc qf a circle. Theorem Let the torsion an arc of a circle.
K
he constant. Then y
of a trigonometric curve y C E3 be constant. Then y is
We remark that the condition y C is essential. In Euclidean spaces, whose dimensions are even and greater than three, there exist trigonometric curves with constant curvature and torsion, which are different from an arc of a circle (see chapter 32).
© 2000 CRC Press
24 Isoperimetric Property of a Circle Let r be a closed rectifiable plane curve, L denote the length of T,F stand for the area of a plane domain bounded by F. The length L and the area F are mutually connected by the following isoperimetric inequality:
= 47rF is true ij" is a circle. One can interpret this fact in the The equality following two ways: among closed rectifiable plane curves with the same length, the circle bounds a plane domain of greatest area; among closed rectifiable plane curves bounding plane domains with the same area, the circle has the smallest length. We will sketch the proof of the isoperimetric inequality in the case of smooth curves. Let %(S),j(s) be Cartesian coordinates of the position vector and s the length of arc of r. It is more convenient to use a parameter t = 2 m l L in place of S. Functions x ( t ) = i ( t L / 2 7 ~ y(t) ) , = j(tL/27~)are periodic functions with the period 27~. Since
we have
therefore
© 2000 CRC Press
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
1 16
The function x(t), y(t) can be represented by Fourier series
+ C ak cos kt + bk sin ktl m
x(t)= an/2
k= l
y ( t ) = co/2 +
C 02
ci, cos kt
+ 4 sin kt
k= l
as well as the derivatives
"
dx dt
x (  a h sin kt
dy dt
" C(ck sin kt + dk cos kt)k.
=
=
+ bk cos kt)k,
k= l
k= l
It is easy to show, using the orthogonality of functions sin kt, cos kt, sin nzt, cos mt, k # m, that
The area F is computed with the help of a standard formula of analysis:
+ C ( u kCOS kt + bh sin k t ) W
0
(Xi
sin kt
k= l
Therefore
0
© 2000 CRC Press
0
l
+ dk COS kt)k
dt
ISOPERIMETRIC PROPERTY O F A CIRCLE
 47rF > 0. The Because the righthand side of this expression is not less than 0, equality appears iff ak = bk = ck = dk= 0, k 2 2 and a1 = dl, 61 = cl. In this case the position vector of has the following form:
+
x(t) = a012 ul cost y(t) = col2  bl cost
+ hl sin t , + a1 sin t.
Obviously it is the representation of a circle. The isoperimetric inequality was proved by Steiner; the proof expounded above was given by Hurwitz.
Problems 1 . Let y be a closed convex planar curve and 0 E y a fixed point. Consider a function h(P) assigning to each point P E y the distance between 0 and the straight line tangent to y at P. This function is called the support function of y. Assume that y is parametrized by an angle a formed by the straight line tangent to y and a fixed straight line. Prove the formula: p(a) = h(o)
+ hrr(cu),
where p(@)denotes the curvature radius of y. 2. Assume that two planar closed curves yl and y2 are mutually tangent at some point P in such a way that they are situated in the same halfplane with respect to the mutual tangent straight line at the point P. Suppose that for any pair of points P I E yl, P2 E 7 2 such that the straight lines 11, 12 tangent to y l , 7 2 at P I , P:, respectively are parallel, the curvatures k l of yl and k2 of y2 satisfy the inequality k l ( P I )2 k2(P2).Demonstrate that yl is situated inside 7 2 . 3. Prove that the area of the domain bounded by a closed planar curve y is equal to
/"
I h($)ds. 2 4. Prove that there does not exist a regular function fik) satisfying the following condition: for any closed planar curve y the integral
where k(s) denotes the curvature of y, is equal to the area of the domain bounded by 7.
© 2000 CRC Press
25 One Inequality for a Closed Curve Let F be a closed curve situated inside a ball of radius R. We will expound the proof of the following inequality demonstrated in [31]:
where s denotes the length of arc, k is the curvature, K is the torsion and L is the length of S. Let r(s) be the position vector and ( S ) , &(S), &(S) the natural frame of r. We define functions xi(s) = (ti(s),r(s)). Using the Frenet formulas, one can see that the functions XI,x2,x3 satisfy the differential equations:
cl
Because S is closed, we have
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DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
for arbitrary constant p. Since r is situated inside a ball of radius R, 1x21 5 R. Let us set
Then we can obtain
Substituting (25.3) into (25.2) we obtain the desired inequality (25.1). Since R 5 L/4, a n inequality involving the length, curvature and torsion of l? follows from (25.1):
The integration of the identity I r (r, r')'

(v, r") leads to the following inequality:
In some cases inequality (25.1) is more optimal than (25.4), in other cases inequality (25.4) is more optimal.
© 2000 CRC Press
26 Necessary and Sufficient Condition of the Boundedness of a Curve with Periodic Curvature and Torsion Let the curvature k(s) and torsion &(S)of a curve r, when viewed as functions of the length of arc S, be periodic functions with a period T. Denote by r(s) the position vector and suppose that r(0) = 0; by cp we denote a space rotation transforming the natural frame @(O) of at point r(0) into the natural frame @(gat the point with the position vector r(T). We will prove the following: Theorem (Bakel'man and Werner, [16].) A curve r of infinite length is bounded iff the vector r(T) is orthogonal to any eigenvector of cp corresponding to the eigenvalue l. One can reformulate this statement in the following way. Let l be an axis of the rotation cp; this means that the vector a spanning l is a stationary vector of the rotation cp. If @(O)# @(T),then a is unique. The theorem says that r is bounded iff (r(T), a) = 0. If @(O)= @(T), then l? is bounded iff r(T) = 0, i.e. iff is closed. Proof Because k(s) and &(S) are periodic functions, the curve r is formed by congruent parts. The part rocorresponding to 0 5 s 5 T is congruent to the part r, corresponding to T s I 2T. Since @(O)is transformed into @ ( T )by the rotation cp and F is defined uniquely by the curvature and torsion, the composition of cp and the space translation with respect to r(T) transform rointo rl.Consider two vectors r(T) and r(2T)  r(T). We have
<
Because cp(a) = a,
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122
DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
By induction one can demonstrate the following equality:
Thus if I' is bounded, then ( r ( T ) ,a) = 0. On the other hand, assume (r(T),a) = 0; then the points P, with the position vector r(nT) belong to the plane E ~which , passes through POand is normal to the axis l of rotation cp. We construct a polygonal line L formed by the segments P O P IP, , P2,. . . connecting the points P I ,P2,. . . . The lengths of the segments coincide and the angles formed by adjacent segments also coincide. If cP(0) # @(g, these angles do not vanish and L is inscribed into a circle. Therefore I' is bounded. If @(O) = @(T), then L is a point or a ray. By assumption r(T) = 0, hence L is degenerated into a point and I' is bounded. There exist necessary and sufficient conditions for unboundedness of a curve I' with periodic curvature k(s) and torsion ~ ( s )One . statement using the properties of the spherical indicatrix y of r is of interest to us. We denote by y,!, the part so<s<so+Tofy.
Theorem [l71 If
r is nonplanar and y,,, is convex for
any so, then
is unbounded.
We say that a domain G on the sphere is convex, if for any two points P, Q of G the smaller arc of a great circle connecting P and Q belongs to G; a closed curve on the sphere is convex, if it is the boundary of a convex domain; a curve is convex, if it is a part of a closed convex curve.
Proof Preserving the notations from above, we denote by ~ ( s )the unit tangent vector to l?, i.e. r(s) is the position vector of y. One can write
Let G,, be the convex hull of y,s,; the cone formed by rays which have their origin at 0 and pass through points of G,sois convex too. Because of the convexity of this set and (26.1), the point with the position vector r(so T )  r(so) belongs to G,,. Let q(s) be a vector tangent to y at the point Q, with the position vector T(s). The rotation cp induces a rotation of the unit sphere; this rotation transforms Q,, into and ~ ( $ 0 into ) q(s0 T). The rotation cp has the following form. Because orbits of points on the sphere under G are circles, the points Q,$,,Q,,+T are situated on a circle C and 7)(sO), q(sO T ) are tangent to C . Let W be the spherical center of C; this means that the inner distance on the sphere between W and an arbitrary point of C is constant. The straight line passing through W and the center 0 of the sphere is parallel to the axis I of rotation cp, hence the position vector W of W is an eigenvector of cp. If for arbitrary s and arbitrary point Q of y, the inner distance on the sphere between Q and W is not equal to ~ / 2then , any y, belongs to a hemisphere with the
+
+
+
© 2000 CRC Press
NECESSARY AND SUFFICIENT CONDITION OF THE BOUNDEDNESS OF A CURVE
123
spherical center at the point W or at the point antipodal to W. Therefore any G, is situated inside this hemisphere. Thus (r(s T )  r(s),W ) # 0, and it follows from the theorem proved above that l? is unbounded. Assume that there exists a part T,~,of y and a point Q, E y,, such that the inner distance between W and Q, is equal to 7r/2. Consider the corresponding part y,,. The points Q,,, are situated on the great circle C with the spherical center at W. Because r is nonplanar, yYYO does not belong to C. It follows from the convexity of y that C is a unique great circle passing through Q,, and any inner point of yy, belongs to a hemisphere bounded by C. Therefore any inner point of G,, belongs to this hemisphere, so that (r(so T )  r(so),W ) # 0. Using the theorem just proved above we can easily conclude that r is unbounded. With the help of this theorem the following statement was proved.
+
+
K*
Theorem [l71 Assume that the total curvature k* = Jr k(s)ds and torsion = Sr K ( S ) ~ Sof a nonplanar curve l? sarisfy one of the following two conditions:
( i ) k* < ~ / 2 , K* i i 2 k* < X , Suppose that
K
© 2000 CRC Press
5 n/2 + 1 / 2 J m .
2 0. Then r is unbounded.
27 Delaunay's Problem The wonderful French mathematician and astronomer Ch. Delaunay proposed the following problem of calculus of variations: given two points PIand P2,find a curve of constant curvature k = 1 , which passes through P I , P2 and has the smallest or the greatest length. Karl Weierstrass solved this question under an additional supposition: given vectors rp, and rp, are tangent to the desired curve at P I and P2 respectively. We will sketch his investigation. Let r(t), t t [ t l ,t z ] denote the position vector of y and r(tl),r(t2)the position vector of P , , PI! respectively. Denoting by a prime the derivative with respect to t we have
this equality can be rewritten
where s stands for the length of arc of y. The length
of y has an extremal value iff the integral
I=
11
[ r'
1I
© 2000 CRC Press
1  ( t )( 1
[ rr
1  I ds/dt 13)]
dl
126
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
has an extremal value, here ~ ( tis) an unknown function. Let y be transformed into a curve with the position vector r(t) + Sr(t). The variation of the integrand in (27.3)has the following form:
'"" (6, €((m 6r')

"" , [Sr',r"] + [r',Sr''])

3
2)
.
We apply the following formula:
and obtain ( [r', r"], [Srl,r"]) = ( [[r", rl],rl'],6r'), ([r',r"],[r',6r1'])= ([[r',rl'],r'],6r").
Also
Thus, the variation of I is equal to the following integral:
+
[ ( A ,6r1) ( B ,Sr")]d t , 11
where we have denoted
Because the end points P I ,P2 and the tangent vectors r p , , r p 2 of y are fixed when this curve is varied, 6 r ( P I )= 6r(P2)= 0, 6r1(P1)= 6r1(P2)= 0. Hence we can obtain, using integration by parts:
Thus, since the variation 6r of the position vector is arbitrary, the position vector of the desired curve y necessarily satisfies the following equation:
A' BN = 0, © 2000 CRC Press
DELAUNAY'S PROBLEM
127
i.e. A  B' is a constant vector C. Assume that the length of arc s of y is taken in place of t . Then Ir'l = 1, I[[rU,r'], #']I = r', Irl'I = 1 and the desired curve has to satisfy the equation
It follows from this equation that
[r',r'(1
+ 2c) + (tr")'] = [r',C ] ,
and one can obtain by integration:
t[rl',r'] = [C,r] + D ,
(27.5)
where D = ( d l ,d 2 ,d 3 ) is a constant vector. Now it will be convenient to use Cartesian coordinates X , y , z chosen in such a way that C = ( c 1 ,0,O). We denote by ([, 7, C ) the coordinates of r'. Equation (27.4) is equivalent to the system of equations
Since (r", r") = 1 and (r', r') = 1 , (r',r") = 0 and (r', v"') =  1 . Therefore we obtain from (27.4), that (C, r') = l + t , i.e.
where I = l / c l . Let us denote m equations of system (27.6) we get
= d 2 / c 1 , n = d 3 / c 1 . From
Using the first equation of (27.6) we write
© 2000 CRC Press
the second and third
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
128
Therefore equation (27.8) is equivalent to the following:
where h
= d ' / c l . Hence
Because ([rl,rl'](= 1, using system (27.6) we obtain
Thus
I satisfies the following equation:
The general solution of this equation is an elliptic function. We will consider partial where is a constant satisfying the equation cases. The first is the case of J = 6,
The function we find
t = (Eo 
/ ) c 1 is constant. From the second equation of system (27.6)
d< d2z tolo  = C ds = toto ds2
,
(Z
 m).
Let us denote p = ~ ' / ( t ~ [ ~Then ) . the solution of equation (27.10) is
Similarly one can obtain y The coordinate
X
=
qcosfis+psinfis+n
is obtained from the equation dxlds
= CO:
Thus the desired curve is a helix. The second case is dl = 0. From equation (27.9) we find
© 2000 CRC Press
DELAUNAY'S PROBLEM
129
c
It is easy to see that 7 = p sin ( S  so), = q sin ( s  so), where p, q are constants satisfying p2 + q2 = l . Therefore the desired curve is an arc of the circle
x = sin ( S  so) + xo, y = p cos ( s  so) yo, z = qcos(sso) f z o .
+
Later the problem of Delaunay was investigated by many authors (see 1201, [33][35]).Schwartz observed the following f x t : the problem of Delaunay has no solution in some special cases. For example, if the distance d between two points PI and P2 is greater than 2, then for any l > d there exists a curve connecting P I and P2, which has a constant curvature k = 1 and length equal to l; this curve is a suitable helix. This means that Delaunay's problem has no solution. If d < 2 there exist two solutions; they are arcs of the circle with radius equal to 1. If we rotate these arcs around the segment P IP2, we obtain two families of solutions. Schwartz proved the following statement in 1884. Theorem 1 Let the distance d between P I and P2 be less than 2. Let cu < n and
p = 27r  0 be the lengths of' arcs of'the unit circle, which connect P I and P2. Then any other curve, which has curvature k one of two inequalities:
=
1 and connects P I and P*, has length l satisfying
Schwartz did not publish any proof. In 1921 A. Schur, motivated by David Hilbert, proved Theorem 1 with the help of Theorem 2 Let a plane curve y with end points P, T and the span PTjhrm a closed convex curve. I f we twist the curve y , the length oj'the span P T increases. (By the twisting of y Schur meant a transformation of y preserving the curvature and the length of y). We will demonstrate the proof of Theorem 2 in the case of polygonal lines. An analog of the curvature for a polygonal line L is the set of angles between adjacent segments of L. An analog of the torsion is the set of angles between the planes spanned by the pairs of adjacent segments. So, let y be a plane polygonal line PQl . . . QnT which forms together with the span P T a closed convex curve y*.Let p be a straight line containing PT, qi a straight line passing through Qi, Qi+1; by Mi we denote an intersection point of p and qi, if this point exists. Since y* is convex, the points Mi do not belong to the segment PT. A twisting p of y can be represented by a composition of suitable rotations around q,. More accurately: at the first step the segment PQl is rotated around ql and the part Q l Q 2 . .. Q,T is fixed, so we obtain a polygonal line P ' Q I Q2.. . Q,T; at the second step P 1 Q 1 Q 2is rotated around q2 and Q2Q3 . . . Q, T is fixed, so we obtain a polygonal line P'Q; Q2Q3. . . Q, T , and so on; in the same way we consider n  3 further rotations leading to a polygonal line Q ~  ' Q ;  ~. . . Q,1 Q, T , and this line is an image of PQl Q2 . . . Q, T under cp. © 2000 CRC Press
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
FIGURE 27.1
Let S be the sphere centered at T of radius R equal to the length of the segment PT, S; the sphere centered at M; of radius R;equal to the length of the segment M;P (see Figure 27.1). Assume that the points M 1 ,. . . , Mk are situated on the ray PM1 of p and M ~ + I ., ..,M,I belong to the ray PM,I. Then S1 C . . . C Sk and 3 ... 3 S,I 3 S. Obviously the sphere Si is transformed into itself under the rotation around q;. The point P', which is an image of P under the rotation around g;, belongs to S 1 , therefore lies inside S2. The point p2 is the image of P' under the rotation around 92, hence p2 is situated inside S2. Continuing in the same way, we see that P' is inside Si for any i 5 k, therefore the length of P'T is greater than the length of PT. Because pk is situated outside the image P"+' of Pk under the rotation around qk+l lies © 2000 CRC Press
DELAUNAY'S PROBLEM
FIGURE 27.2
outside Sktl.Continuing, we obtain that the last point P"' is situated outside S,,, , hence it is outside S . Thus the length of the segment P"' T is greater than the length of PT. Let us apply Theorem 2 in order to prove Theorem 1. Assume that there exists a curve y such that its curvature is equal to 1, it connects P and T, and its length l satisfies the inequality a l 5 P. Suppose that the distance between P and T is equal to d. We take a unit circle passing through P, T; the circle is decomposed into two arcs P T and TP, whose lengths are equal to a and 0respectively. Fix a point N such that the length of the arc PTN is equal to l, and a point T* on the circle in such a way that the length of the segment PT* is equal to d. The point N belongs to the arc TT* and it is easy to see that the length d of the segment PN is not less than d (see Figure 27.2). On the other hand, the curve y is the image of the arc PTN under some twisting, hence d > d by Theorem 2. Thus we have obtained a contradiction. So the assumption a 15 p is false.
<
<
© 2000 CRC Press
28 Jordan's Theorem on Closed Plane Curves In this chapter we will consider topological properties of closed curves situated on the plane. The image of a circle under a homeomorphism will be called a closed Jordan curve; the image of an interval under a homeomorphism will be called a simple arc.
Theorem A plane closed Jordan curve y decomposes the plane into two connected components. Moreover, y is the mutual boundary of these components. Similarly, a closed curve on the sphere decomposes the sphere into two connected components, but some closed curves on closed surfaces different from the sphere do not have such properties. The formulated theorem is called Jordan's Theorem. The brief elementary proof, which we are going to sketch, was proposed by Phylippov [51]. Let us fix a ray R. with the origin at a point 0. This ray is called a vertical ray. Given a polygonal line g and a straight line (or ray) I, a point P where g is intersected by l is called a proper intersection point if P is not a vertex of g or if P is a vertex of g such that two edges adjoining P are situated on a different halfplane with respect to 1. We construct some special function N(Q,g) in the following manner. Consider a ray RQ collinear to R. with the origin at a point Q; if the number of points where RQ intersects g properly is even, we set N(Q,g) = 0. If not, we set N Q , g ) = 1. Lemma l Let g and h be disjoint polygonal lines satisfying one of the two following conditions: ( i ) h is closed, (ii) g lies between two vertical lines passing through the end points of h. Then .for all points Q E g the value N(Q, h) is constant.
Proof Construct a vertical straight line passing through the vertices of h. Then on any part of g situated between two adjacent vertical lines the value N(Q, h) is constant. Passing from one part of g to another, we see that N(Q, h) is constant on the whole polygonal line g. H
© 2000 CRC Press
DIFFERENTIAL GEOMETRY A N D TOPOLOGY OF CURVES
FIGURE 28.1
Lemma 2 Any closed Jordan curve y decomposes the plane into at least two connected components. Proof Let Q be the most left point and P the most right point of y. These points decompose y into two arcs yl and 72. We fix a vertical straight line 1 passing between the points Q, P and denote by H the highest point where l intersects y. We can assume that H belongs to the arc yl. Let L be the lowest point where l intersects yl. We take a point C E l such that C is lower than L and the distance d(L, C) between L and C is less than the distance d(L, y2) (see Figure 28.1). Without loss of generality, we can fix the ray R. in such a way that y does not intersect Ro. We will prove that the point 0 cannot be connected with C by a polygonal line disjoint with y. Assume that there exists a polygonal line h, which does not intersect y and connects 0 with C. Then d(h,y)=a>O, d(yl,R,)=h>O, d ( C L U L H U R,,, 7 2 ) = C > 0 , where R, is the vertical ray opposite to R. with the origin at C, C L is the vertical segment with end points C and L, L H is the part of y , between L and H. Let © 2000 CRC Press
JORDAN'S THEOREM ON CLOSED PLANE CURVES
135
E = min (a,h, c). We inscribe a closed polygonal line g in y such that the lengths of its edges are less than and the points H , L, P, Q are vertices of g. Let us denote by g ) , g2, g* parts of g whose vertices belong to yl , y;?,LH respectively. The distance between any point of g and the curve y is less than (12. One may state:
<
( 1 ) because
Let us compute N(C,gl) and N(C,g2). The end points Q, P of yl are situated on different halfplanes with respect to l. Therefore the number of points where l intersects g1 properly is odd, i.e. N(C,gl) = 1. The polygonal line C L Ug* does not intersect g2 and is situated between the end points Q, P of g2. It follows from Lemma 1 that N(C,g2) = N(H,g2). Since the vertical ray RH does not intersect g2, we have N(H, g2) = 0. Thus N(C, g2) = 0. Because g is formed by g1 and g2,
Hence N(C,g) = l . On the other hand, N ( 0 ,g) = 0 and g does not intersect h. By Lemma 1 N(C, g ) = N ( 0 , g) = 0, contrary to the obtained result N(C,g) = 1 . Thus our assumption leads to contradicting conclusions, so 0 and C cannot be connected by a polygonal line disjoint with y.
Lemma 3 A simple arc does not decompose the plane.
Proof Let A B be a simple arc with end points A and B. Assume that A B decomposes the plane, i.e. there exist points P, Q such that any polygonal line connecting P and Q intersects AB. Let C be a point of A B such that the two following conditions are fulfilled: (i) for any inner point C of the arc AC, there exists a polygonal line which connects P, Q and does not intersect the arc AC; (ii) for any point C+ outside of AC, any polygonal line connecting P and Q intersects the arc AC+. We will prove that the arc A C does not separate the points P and Q, i.e. that there exists a polygonal line which connects P, Q and does not intersect AC. Let ql be a square such that C is its center and P, Q do not belong to ql. We denote by A I the last point where the arc AC intersects the boundary dql of the square ql, and construct a small square q2 inside ql in such a way that C is its center and the arc A A I is situated outside q2. We denote by A2 the last point where the arc A C intersects the boundary dqz. Because AA2 C AC, the arc AA2 does not separate the points P and Q, i.e. some polygonal line g connects P and Q and does not intersect AA2. If g does not intersect AB, the lemma is proved. So we assume that g intersects AB. Let P*Q* be a part of g, which is contained in ql and whose end points P*,Q* belong to dql. The points P", Q* decompose dql into two parts r l , r2;we suppose that A I E l?,. The polygonal line P*Q* decomposes ql into connected components; the © 2000 CRC Press
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
FIGURE 28.2
boundary dG of one of these components G contains r2.Because A I does not belong to the closure G of the component G and the arc A I A 2 does not intersect dG = P Q * U r2,the arc A I A 2does not have mutual points with G (see Figure 28.2). The set G \ g2 is formed by polygons; r2belongs to the boundary i3R of one of these polygons R. The points P*, Q* are situated on 3R. Let us replace P'Q* by the polygonal line h* = dR \ r2.The line h* connecting P* and Q* does not intersect the arc AC. Thus any part P'Q* of g can be replaced by some polygonal line h*, which does not intersect AC. The resulting polygonal line h, which is constructed starting from g by replacing some parts of g, connects P, Q and does not intersect the arc AC. Now let us prove that C coincides with B. We take a point Cl inside the arc CB in such a way that the arc CCI lies inside a circle with radius r < d(C,h). Then it is easy to see that the arc AC1 does not separate P and Q , contrary to condition (ii) defining C. Thus C coincides with B.
Lemma 4 Let a closed curve y decompose the plane into connected components G1,G2,. . . . Then y is the boundary of each component Gi. Proof Let us remember that a set dG is called the boundary of a domain G, if for any point P of dG and for any neighborhood U of P the following conditions are satixred: (i) Inside U there exists an inner point of G, (ii) inside U there exists a point which ioes not belong G. © 2000 CRC Press
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Assume that a point A belongs to y and A does not belong to the boundary dGi of some component Gi. Let us take a point PiE Gi and a point P; in some component G, different from Gi. Denote by X C y an arc situated in the circle whose center is at A and whose radius is equal to d(A, i3Gi). Then dGi lies inside the simple arc y \ X. Because dGi separates the points P and Q, the arc y \ X separates these points, contrary to Lemma 3. Now it is easy to complete the proof. Let us prove Jordan's Theorem. The curve y decomposes the plane into connected components G I , (32,. . . . Let A be an inner point of G1, B and C points outside the closure c l . We construct two rays h l ,h2 with the origin at B, which intersect y; the points nearest to B, where h l , h2 intersect y, we denote by B1,B2. In the same way we construct rays cl, c2 with origin at C and points C l , C2. The segments BBi and CC, do not intersect y. There exist only two possibilities: (i) two points B I , B2 on y are separated by C l ,C2, (ii) the pairs B I , B2 and C l , C2 do not separate each other. Without loss of generality we suppose that these points are arranged on y as follows: (i) B1C1B2C2,(ii) B I B 2 C I C 2 . Let us consider the closed Jordan curve h = Bl BB2 U B2C2U C2CCIU C lB I , where BIBB2, C2CCI are the corresponding polygonal lines and C I B I ,B2C2 are the corresponding arcs of y. The line h decomposes the plane into connected components H I , H2,. . . . Let A E H I . By Lemma 4, in neighborhoods of the points B, C there exist inner points B3, C3 of H2. There exists a polygonal line BB3C3C which connects B, C , and all points of this line excluding B and C are inner points of H2. Assume that the considered polygonal line intersects y at some point D. By Lemma 4, in a neighborhood of D there exists an inner point Dl of C l . We connect D1 with A by a polygonal line H* situated inside C l . The line h* does not intersect h, because h is formed by arcs of y and segments BBi, CC, disjoint with G,. Therefore A E H2, contrary to the supposition A E H I . Thus B and C are situated in one connected component. Let us consider a closed smooth curve y on the oriented plane. Assume that y has a finite number m of selfintersection points; for each point of selfintersection Q, only two branches of y pass and these branches intersect transversely. Denote by T(P)the unit vector tangent to y at a point P and by P*the point with position vector T(P). When P passes the whole curve y once, the corresponding point P* passes the unit circle n times. Gauss observed a connection between m and n. The integer n is equal to the integral
where k(s) is the curvature with sign of y. Under a smooth deformation of y the integer n does not vary. But m varies, because new selfintersection points may appear. Clearly the difference between the old and ihe new values of m is an odd integer. Trivial examples demonstrate that m  n is an even integer. To see a closer connection between m and n, we classify the selfintersection points in the following manner. Let A be a point of y different from selfintersection points. Let us fix an orientation of y. When A moves on y in the positive direction, it passes through each © 2000 CRC Press
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selfintersection point P twice. When A passes through P the first time, we denote by q ( P ) the unit vector tangent to the corresponding branch of y at P; the unit vector tangent to the second branch of y at P we denote by 72(P).By assumption, 71( P ) and T ~ ( Pform ) a basis of the plane. If the orientation given by this basis in the plane coincides with the initial orientation, we assign to P the number cp(P) = I , otherwise we set cp(P) =  I . The sum J = C cp(P), where P runs over all selfintersection points, is called the Whitney number for y. It has been proved that J = n  l or J = n + 1 depending on the choice of A (see [60, p. 5141).
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29 Gauss's Integral for Two Linked Curves We will consider some topological properties conected with links of closed space curves. Let yl, y2 be disjoint closed space curves. We say that yl and 7 2 are not linked, if they can be deformed continuously into curves y;, y; situated inside two disjoint balls; otherwise yl, y2 are said to be linked. Let us denote by r = rj(ti)the position vectors of the curves y;. If the position vectors are given, then one can construct some number I ( y l ,y2) called the link coefficient of 7 172. , This notion was discovered by Gauss in 1833 (see[55]); using the link coefficient, one can distinguish linked curves. First, we will demonstrate a construction of the link coefficient I ( y l , y 2 ) .Let us take a plane of parameters t1, t2 and some rectangle ll = {al L tl 5 az, b l i t2 I b2 ) in this plane. Since the curves yi are closed, rl(a1) = rl(a2) and r 2 ( b l )= r2(b2). We denote the vertices ( a l ,bl), (a2,hl), (a2,b2), ( U I ,b2)of rI by A I , A2, A3, A4 respectively. To any point ( a l , t2)of the side A1A4 we assign the point (a2,t2)of the side A2A3;in this way, we identify side A1A4with side &A3. Similarly one can identify side A IA2 with side A 4 A 3 Next we obtain a set T; it is easy to demonstrate that T is homeomorhic to a torus. Besides, a pair of points Q 1 E yl, Q2 E y2 correspond to a point of T and this correspondence is a homeomorphism of the set of pairs Q l E yl, Q2 E 7 2 onto T. Consider the vectorvalued function rl( t l ) r2(t2)defined on the set T. Since yl, 7 2 are mutually disjoint, this function is not vanishing everywhere. Hence we can define the unit vector
for each point P E T with coordinates tl, t2. Thus, we get a map q5 : T + S 2 transforming a point P = ( t l ,t2)into a point with position vector n ( t l ,t 2 ) Let us find the area of the domain $ ( T ) on the unit sphere S 2 .
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DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
We remark without detailed explications that the area S of a surface F with position vector r(u, v) is computed by the formula
where h denotes the domain of definition of F. The vectors v,, r , are tangent to F, therefore the vector [ v , , v,,] is normal to F. Let us return to the set $ ( T ) situated in the unit sphere S 2 . The vectorvalued function n ( t l ,t2) is the position vector of the points of $ ( T ) . At the same time, n(tl,t Z )is a unit normal vector to s2for any tl, t2. Hence, the area element dA of s2 at the points of $ ( T ) is
where (*, *, *) denotes the mixed product. Next, we will compute the "oriented" area of $ ( T ) . For this purpose, we decompose T into domains T, where (n,n,, ,n,,) are of constant signs. Then we sum the areas of domains T, where (n,n,,,n,,) is positive, and subtract the areas of domains T, where (n,n,,,n,,) is negative; we call the obtained number the "oriented" area of $ ( T ) . Denoting the element of the "oriented" area of $ ( T ) by the same notation d A , we get
We have
Therefore,
To every continuous map p of a closed surface into the sphere one can assign some integer called the degree of cp; the preimage of any point of the sphere consists of n points and the degree of p is equal to n by definition. It is known that the degree of cp is equal to the oriented area divided by 4~ of the image under cp. The degree of the map $, which we have constructed starting from the curves y,, we denote by I(yl,y2). From (29.1) it follows that
Thc integral on the righthand side is called the Gauss integral. Since I(yl,y2) is an integer, it does not vary under continuous deformations of disjoint y,. © 2000 CRC Press
GAUSS'S INTEGRAL FOR TWO LINKED CURVES
141
If y l , 7 2 are not linked, then we deform these curves onto some curves y;, y; situated in disjoint arbitrarily small balls. The function n * ( t l ,t 2 ) corresponding to y ; , y,* is almost constant and the image $ * ( T ) is situated in an arbitrarily small domain on the unit sphere. Hence the degree of Ijl* is equal to 0 as well as the degree of ,$. Thus, f y l , 7 2 are not linked, then Z(yl,y2) = 0. However, there exist linked curves, whose link coefficient is equal to 0. Now, let us consider the links of sufficiently near curves. Let y be a closed space curve without selfintersections. We take a curve sufficiently close and parallel to y. More exactly, if r = r(s), s E [0,L] is the position vector of y with respect to the arc length S , then we take a curve y, given by the position vector
where a ( t ) is a unit vector field orthogonal to y and E is a sufficiently small positive number such that y and y, are disjoint. We connect every point r(t) of y with the corresponding point i ( t ) by a straight line segment. All such connecting segments form a closed strip R; if we view this strip as a surface, then its position vector is r = r(t, v) = r(t) + va(t), where s E [0, L], v E [0, E]. The components y , y, of the strip's boundary are linked iff this strip cannot be deformed continuously without singularities onto a closed cylindrical strip. If E is sufficiently small, then the linking coefficient of the curves y , y, does not depend on E . We have
='/I L
I(%%)
47r
0
L ( ~ ( 3)
+
~ ( t) & ~ ( tr)l (, s ) ,r l ( t ) Eal(t)) Ir(s)  r ( t )  &a(t)l3 dt ds,
(29.2)
0
where we have denoted the length of the closed curve y by L. But if we put E = 0, then the value of the integral on the righthand side of formula (29.2) is not equal to values of I(?,?,) at sufficiently small positive E. We introduce the function
defined in the square Q = (0 5 s < L, 0 < t 5 L ) excluding the diagonal s = t. Setting A(s,s) = 0, we complete the definition of A. To demonstrate that this operation is well defined, we apply the Taylor formula for A in a neighborhood of the diagonal. For fixed S , t we have
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DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
where ci are bounded vectors determined by values of the third derivative of r(t) at points of the segment [ S , t]. Denoting A = s  t, we find an expression for A at points near to the diagonal:
where o + 0 as A + 0. Therefore, A + 0 as A + 0. Thus, if we complete the definition of A in the proposed way, we get a continuous function defined on the whole square; also, one can demonstrate that A is differentiable. The integral
is well defined, but its value is not an integer. We are interested in the difference between J ( y ) and I(y,yE). We denote the function integrated on the righthand side of (29.2) by A(s, t, E). Let us decompose the square Q into two parts: a domain U* = { I  S 5 S 5 t + S}, where S > 0 is a sufficiently small number, and the domain Wg = Q \ Uh. It is easy to see that
60
A(s, t , E ) dt r l s + lim lirn
471.
E0
h0
60
W6
47r
uh
Consider the integral
If E is sufficiently small, the function r(s)  r(t)  m ( t ) does not vanish everywhere in Wg; hence the limit of the considered integral is equal to J(y): 60
E0
47r
.
A(s, t , E ) dt ds = lim 60
'/

47r
.
A(s, t , 0 ) dt ds = J ( y ) .
It is more complicated to find the integral of A(s, t, E ) with respect to the domain U&.To estimate the limit of this integral as S, E + 0, we apply formula (29.3) and obtain:
A(s, t , E )
= A{
+
 ~ ~ ( a ( rl(t), t ) , a l ( t ) ) E ARI ( t ,E )
+ a4&(t, s) + & a 2 ~ 3 ( t E, ) ) , S,
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GAUSS'S INTEGRAL FOR TWO LINKED CURVES
where Riare functions bounded in Ub:
R, = ( a , r u , r ' + € a ' ) , 1 2
+ (cl,r",rl), l R3 = ( r ' , v", a')  (a, Q , r' + + A 2 ( q r", + c 2 A ,a ) 2 + A(c1 rl,a') R 2 = A ( c l , c 2 , r 1 ) +(r11,c2,r1)
&U')

 C*,
and A = r1(t)A
A2 + r"(t) +clA3 2

m(t)
Next, one can write
where for the function li, WC have the estimate
Integrating the first sum of (29.4) with respect to Ub and letting
From the estimate for the function Il, and from the equality
it follows that limit (29.5) is equal to 2
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1 o
( a ( t ) ,r 1 ( t ) ,a t ( t ) ) d t .
E
tend to 0, we get
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Let us consider the integral of the second sum of (29.4) with respect to Uh:
Again, from the estimate for 1+4 it follows that this integral tends to 0 as E + 0. The integral of AA4R2 tends to 0 as 6 ,0. The integral of the last sum of (29.4) with respect to Ug also tends to 0 as E 4 0. Thus, we have:
+'I/ L
47r
Cl
o
L
( ~ ( s) r ( t ) ,rl(&s), ~ ' ( t) ) dt ds.
I ~ ( s) r(t)13
The first sum in formula (29.6) is called the total twisting of the strip. Suppose the principal normal vector v and the binormal P to y are continuous periodic vector fields defined on y. We put
Then one can obtain the equality
If the strip is formed by the moving principal normal vector, then the total twisting of the strip is equal to the integral divided by 2 n of the twisting of y;in this case formula (29.6) was proved in [21], and the general case was investigated by White in [23].A simpler proof of formula (29.6) was given in [23]. We have remarked that, generally, the integral J ( y ) is not integer. Let us consider one simple case when J(?; is an integer. Cut out a narrow paper strip of rectangular form, whose length ~x much greater than its width. Bend and twist © 2000 CRC Press
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145
the strip and then paste together its ends. We can say that such a strip is a plane. More precisely, let a be a straight line segment orthogonal to the curve y. Let a move along y being always orthogonal and so form a closed strip. This strip is called a plane if any small piece can be developed into a Euclidean plane. The position vector of this strip is given below by (29.8). It turns out that the boundary curve of a closed plane strip is not arbitrary. Namely the following statement by Fenchel holds. Theorem A regulm curve y ofclass c3is tlze boundary of aplane strip if and only if the integral
is an integer. Proof To prove this theorem, write the position vector r(u, v) of the strip viewed as a surface in Euclidean space. The position vector r(u, v) of a point M on the strip is the sum of the position vector p(u) of a corresponding point on the boundary and of the vector va(u) that is orthogonal to y,lies on the strip and has the length v:
where vector a(u) is assumed to be a unit. Suppose that u is the length of an arc of y. In the theory of surfaces, to describe the behavior of surfaces the Gauss curvature K which is one of the most important characteristics of surfaces is used. The Gauss curvature of the Euclidean plane is vanishing: K = 0. Gauss proved that K does not change under bending. By this theorem the curvature of any plane strip is vanishing, since it is equal to the curvature of the Euclidean plane. Gauss found an expression of K in terms of the coefficients of the first fundamental form of surface ds2 = E ~ U 2Fdudv ~ ~ d vwhere ~ , the coefficients E, F, G are given by the formulae
+
+
If F = 0 then the Gauss formula for curvature is as follows:
We can differentiate the position vector in (29.8) to obtain
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146
where 7, V, /3 is the Frenet frame of y. Therefore,
Find the Gauss curvature of the strip by applying (29.9):
Since the Gauss curvature of the plane strip is vanishing, the following condition must be fulfilled:
Our plane strip is closed, hence
where m is integer. So
On the other hand, if y is closed and the equality (29.7) holds, then the vector field a = ocos p @sinp with p satisfying (29.10) defines a plane strip which is closed because of (29.7), q.e.d. It follows from the formula (29.6) and from the condition (29.7) on the boundary y of a plane strip, that the integral J ( y ) is an integer equal to the link coefficient of the boundary curves. W Now, we will demonstrate a geometric way to find the link coefficient of two arbitrary closed space curves yl, 7 2 . We construct an oriented surface F such that yl is its boundary. (The existence of such a surface will be proved in the next chapter.) Suppose there exist a finite number of points where y2 intersects F; also, we assume that the curve 7 2 is not tangent to F at these intersection points. We fix an orientation of F and an orientation of y2; the orientation of F naturally induces an orientation of 7 1 To . any point Pi, where 7 2 intersects F, we assign a number ~i = jc1 in the following way. If the inner product of the vector normal to F at P; and of the vector tangent to 7 2 at Piis positive, we put E ; = 1; otherwise, E ; =  1. By definition, the intersection index {F,;,2} is the sum of all E;. It is proved in topology that the intersection index (F,y 2 ) is equal to the link cnefficient of the curves yl,7 2 :
+
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GAUSS'S INTEGRAL FOR TWO LINKED CURVES
147
Applying this equality, one can find that the link coefficients of curves shown in Figures 29.129.4 are 0,1,2,3.
© 2000 CRC Press
FIGURE 29.1
FIGURE 29.2
FIGURE 29.3
FIGURE 29.4
30 Knots A closed space curve without selfintersections is called a knot. A polygonal knot is the union of a finite number of straight line segments called edges. A knot y is tame, if there exists a homeomorphism of the space onto itself, which maps y onto a polygonal knot; otherwise this knot is called wild. The following theorem is known (we do not demonstrate its proof): if a knot is C'regular, then it is tame. We will consider only tame knots. A knot is said to be trivial if there exists a homeomorphism of the space E3 onto itself, which maps this knot onto a circle. It is natural to ask: how can one distinguish trivial and nontrivial knots? How does one classify knots? To solve the first problem, we use the notion of the knot group. For any knot y we consider the fundamental group irl of the set \ y;this group is called the group of the knot y. It was discovered by Dehn as a particular case of the Poincar'e group. To construct the group of the knot y,we take a point 0 outside y and consider a set R of all closed oriented space curves I passing through 0 and disjoint with y. We view 0 as the start point and the end point of these curves. Any curve I E R is the image of the segment [0, l] under a map cpl such that cpl(0) = cpl(l) = 0. For any two curves 11,l2 r S1 we define the product I = 11 o 12 as the image of [0, l] under the map
It is evident that l E R. We say that two curves 11, 12 E S2 are equivalent, if 11 cannot be continuously deformed onto 12 without intersecting y.One can verify that this equivalence is a true equivalence relation; the set S1 is partitioned into equivalence classes. We will denote equivalence class of a curve l E R by I*. Moreover, one can define a product of the equivalence classes in the following way: if 1; and 1; are the equivalence classes of curves l , , 12, then we define the product I* = 1; o 1; of the classes as the equivalence
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class of the product l = 1, o I2 of the curves. It is trivial to show that this construction is well defined. If l* is the equivalence class of an oriented curve l, then the inverse element l*'is the equivalence class of the curve l' different from l only by orientation. The set of equivalence classes equipped with the constructed product operation is the group of the knot y. We will denote it by G(y). In general, the group of a knot is noncommutative. To determine a group it is necessary and sufficient to give the generators of this group and the relations between the generators. We will consider a geometric way to construct generators and generator relations for the group G(y). Let us project y into some plane, which we will call the horizontal plane. Without loss of generality we assume that the projection 9 has only double selfintersection points; this means that for any selfintersection point P E j. exactly two branches of y pass over P; the arc of 9 corresponding to the overpassing arc of y is shown without a break at P and the arc of j. corresponding to the underpassing arc of y is shown with a break at P (Figure 30.1). Using this agreement, we can identify y with its projection j. and think that y is situated in the horizontal plane. Next, we decompose the oriented knot y into oriented connected arcs cri such that any arc a, appears under some arc a k and finishes under some arc Q,. For any arc a; we construct an oriented trivial knot a, passing through 0 and satisfying two conditions: (1) a; cannot be continuously deformed into the point 0 without intersecting y; (2) the orientations of y and a; are connected by the screw rule (Figure 30.2).
FIGURE 30.1
U FIGURE 30.2
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KNOTS
FIGURE 30.3
We apply the following statement: Any curve 1 E f2 is equivalent to a product of some curves a,. (We will not demonstrate a proof of this fact.) Thus, the equivalence classes of the curves a; can be viewed as generators of the knot group G(?); we will denote these classes by the same notation ai. The constructed generators are connected by some relations. In order to find these relations, we consider a selfintersection point P E One branch of ? passes through P without a break and another branch is drawn with a break at P. Take a small circle S with center P; it is evident that one arc ai intersects S twice, one arc a k passes inside S, and one arc a, passes from P to outside S. So, we have an image shown in Figure 30.3 or 30.4. We decompose S by four points A , , A2, A3, A4 into four arcs in such a way that any arc AiAi+{ intersects only one arc a, of ;j.. Next, let us lower S vertically under the horizontal plane (we remember that 0 lies over the horizontal plane) and connect the points A; with 0 by some arcs disjoint with ?. Then we can write:
+.
O A IA20 = U ; , O A 2 A 3 0= u k ,
0 A 3 A 4 0= U ; ' , 0 A 4 A 1 0= a;'.
Hence, we have
Because the circle S lies under the horizontal plane, it can be continuously deformed into the point A I without intersecting ;j., therefore curve (30.1) is trivial:
FIGURE 30.4
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DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
Similar relations exist for each selfintersection point of G. The formal rule for writing such relations is the following: we go around S in a clockwise manner and write the arcs ol,,which we meet, in order from left to right; then we replace every ap by the corresponding generator a,, taken with degree 1 in the case when ap passes from outside to inside S and with degree 1 otherwise. The obtained generator relations demonstrate the following fact: if G(?) is commutative, then there exists a unique generator of G(y). If y is trivial, then the knot group G(?) is isomorphic to the additive group of integers (2,+); any closed curve I E R is equivalent to a curve winding around the knot y n times. The converse statement is also true; we will not demonstrate its proof based on Dehn's lemma ([25], 1291). Finally, we have:
Theorem A knot is trivial iJf its knot group is isomorphic to the additive group (2, +). Thus, the knot group is a good tool for distinguishing trivial knots. However, it is not sufficient to classify nontrivial knots. There exist knots which are not equivalent and whose knot groups are isomorphic. One such example constructed by Fox is demonstrated in Figures 30.5 and 30.6. Let us find the knot groups of the drawn knots yl, 72. We decompose the knots by oriented arcs q,as above and consider the corresponding generators a,, which are the same for both knots y,. For knot yl we can write six generator relations corresponding to six selfintersection points on Figure 30.5:
From the first and third relations of system (30.2) we get:
If we substitute the found expression of a1 into the first relation of (30.2), we obtain the equality U?' Oa2 Oa3 Oa2 003' O U T ' = 1, which can be rewritten as al o a,' = 1 , i.e. a1 = a4. Then, using relations (30.2) and (30.3), it is easy to find expressions of a3 and of a5 in terms of a ' , u2, a6:
FIGURE 30.5
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KNOTS
FIGURE 30.6
Thus, the knot group G(yl) is determined by three generators p = a ] , q = a2, r = a6 connected by two relations r o p o r = p o r o p , q o p o q = p o q o p . Now consider the knot group G(y2) of knot y2 shown in Figure 30.6. The right side of this knot coincides with the right side of knot yl, hence the first three generator relations are the same. Another three relations are
From (30.2) it follows that a1 terms of a1 and as:
Let us denote a1 = p ,
a2 =
= ae;
using (30.4), we obtain an expression of a6 in
g, a5 = v . Then we get
Thus, knot group G(72) is isomorphic to knot group G(yl), because their generators p, q, r are connected by the same generator relations. One classification of nontrivial knots was proposed in [26], [27]. It seems surprising, but to distinguish two nontrivial knots yl, 7 2 it is necessary to complicate these knots by attaching some standard knot R to yi and constructing some windings of the obtained knots. For the standard knot R one can take Listing's knot shown in Figure 3 1.7. To explain these constructions, we will give some definitions. Without loss of generality, we assume that both knots are situated in the sphere S3. We say that two arbitrary knots in S3 are equivalent, if there exists a homeomorphism of S3 onto itself, which maps one knot onto another one. The attachment of the standard knot R to a knot y c S' is constructed as follows. First, let us place y and R in s3in different sides with respect to some sphere s2C s3; © 2000 CRC Press
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DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
next y and R we replace and deform them continuously without selfintersections in such a way that R and y (being situated in different sides with respect to S 2 ) have a common arc e situated in S 2 . We suppose that the orientations of R and y are fixed in such a manner that the orientations of e induced by the orientations of R and y are opposite. By definition, the knot y#R obtained by attachment of R to y is the knot (7 \ 4 U (R\ e). Now, what is a winding of a knot y? Consider a small tubular neighborhood U of y and its boundary dU. Let p, X be two closed curves such that the linking coefficient for the pair p, y is equal to l and X can be deformed continuously without selfintersections onto y. Then a linear composition u p + bX with some integer coefficients U , b is called a winding of y. Let us return to the knots yl,y2. Construct the knots yl#R, y2#Rand consider two windings l: = p 2X, 1; = p  2X for each knot ?,#R. As a classifying group we take the group
+
where * denotes the free product of the groups. (We remember that the free product of groups A and B is a group whose generators consist of the generators of A, B and whose generator relations are the union of generator relations of A, B.) The classifying statement is Theorem Let yl,y2 be two knots situated in the sphere S'. The groups C G ( y l ) , CG(y2) are isomorphic iff yl,y2 are equivalent. Next we will consider one property of knots discovered by Pontryagin and Frank1 [30]: Theorem A knot y is the boundury of some oriented suvJuce F. We will sketch the proof proposed by Seifert. He was applying surfaces spanned by knots to investigate the behavior of knots. Following Seifert's method, we project knot y into the horizontal plane. Suppose the projection 9 has q double selfintersection points. Then there exist 2q points P; E y projected into the selfintersection points of 9. The points P;decompose y into connected arcs. For any pair of points Pi, P, projected into a selfintersection point of i., we construct a vertical segment connecting these points. Let us fix an orientation of y. Starting from some point P; we move along y in a positive direction to the point P, which follows Pi, after which we go along the vertical segment connecting Pi with some point P,, continue the movement along y in a positive direction and so on. Finally, we return to the point Pi. The oriented closed curve we have described is called Seifert's circle. This curve, excluding its vertical segments, is onetoone projected into the horizontal plane. If some point PI does not belong to the constructed Seifert circle, we repeat the described process starting from PI. In such a manner, we obtain f Seifert circles. Any vertical segment contained in a Seifert circle is passed by a moving point of this Seifert circle only once. Besides, any vertical segment P;Pj belongs to two Seifert circles and the orientations of Pip, induced by the orientations of the Seifert circles are opposite. For each Seifert circle C we construct an oriented surface Fc homeo© 2000 CRC Press
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155
morphic to a disk and spanned by C (it means that C is the boundary of the surface Fc and the orientations of Fc and C are related in the natural way). It is easy to construct the described surfaces in such a manner that any two surfaces are disjoint. Next, we join these surfaces along the vertical segments Pipj and obtain a surface F with boundary y. Because two surfaces Fc, whose boundaries contain a common vertical segment Pip,, induce opposite orientations of Pipj, the surface F is oriented. From topology it is known that any oriented surface H, whose boundary has a unique connected component, is homeomorpic to the sphere with h handles and without one disk. The number of handles is called the genus of H. There exist surfaces of different values of genus, which are spanned by y. The minimum of these values is called the genus of the knot y.
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31 Alexander's Polynomial The knot group is an extremely powerful invariant. Unfortunately, the problem of determining when two arbitrary groups are isomorphic appears to involve most of the difficulties of the knot problem itself. However, in many cases, it is possible to distinguish one type of knot from another by means of several numeral invariants. In particular, one can use Alexander's polynomial constructed in [36]. This polynomial is intimately related to the knot group, but we are going to describe it in a geometrical way. As in the previous chapter, we think of a knot as a simple, closed, directed polygon in the Euclidean threedimensional space; a knot is composed of a finite number of vertices and directed edges. We allow ourselves to operate on a knot in the following three ways: (i) To subdivide an edge into two subedges by creating a new vertex at a point of the edge. (ii) To reverse the last operation: that is to say, to amalgamate a pair of consecutive collinear edges, along with their common vertex, into a single edge. (iii) To change the shape of the knot by continuously displacing a vertex (along with the two edges meeting at the vertex) in such a manner that the knot never acquires a singularity during the process. More precisely the operation (iii) is following. Construct on an edge PQ some triangle PQR that the domain bounded by PQR does not contain points of the knot. Change the edge PQ on the two edges PR and RQ. Two knots are said to be of the same type, iff one of them is transformable into the other by a finite succession of elementary operations of the three kinds just described. An oriented knot y is represented schematically by a diagram.We can think of the diagram as a plane oriented curve traced picturing the knot as viewed from a point of space sufficiently removed so that the entire knot comes within the field of vision.
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DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
FIGURE 31.1
We assume that the point of observation is in a general position so that the selfintersection points of the diagram are all double points with distinct tangents; the selfintersection points are called vertices of the diagram and the regions into which the diagram subdivides the plane regions of the diagram. At each selfintersection point, two of the four corners will be dotted to indicate which of the two branches through the selfintersection point is to be thought of as the one passing under, or behind, the other. The convenient way is to place the dots in such a manner that a point moving in the positive direction along the "lower" branch through a selfintersection point has the two dotted corners on its left (Figure 31.1). To each region of a diagram a certain integer, called the index of the region, is assigned in the following way. We allow ourselves to choose the index of any one region at random, but after that we fix the indices of all the remaining regions by imposing the requirement that whenever we cross the curve from right to left (with reference to our point moving along the curve in the positive direction) we must pass from a region of index p to a region of the next higher index p + 1 (Figure 31.2). Evidently, this condition determines the indices of all the remaining regions fully and without contradiction. For brevicy, we shall say that a corner of a region of index p is itself of index p. It is easy to verify that at any selfintersection point A there are always two opposite corners of the same index p and two opposite corners of indices p  1 and p + 1 respectively. The index p associated with the first pair of corners is referred as the index of the selfintersection point A . The same diagram represents an infinite number of different knots, but this lack of determination is an advantage, as the knots so represented are all of the same type. To tell the type of knot determined by a diagram, it is not necessary to know the exact shapes of the various elements of the diagram, but only the relations of in
FIGURE 3 1.2
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ALEXANDER'S POLYNOMIAL
159
cidence between the elements. Because of this fact, the essential features of a diagram may all be displayed by a properly chosen system of linear equations, as we shall now demonstrate. If the diagram has v selfintersection points C l , . . . ,C,,, we find, by a simple application of Euler's theorem on polyhedra, that it must have v + 2 regions rg, . . . ,r,+l. Suppose the four corners at a selfintersection point C1 belong respectively to the regions rj, rk, r,, and r,,, that we pass through these regions in the cyclical order just named as we go around the point Ciin the counterclockwise direction, and that the two dotted corners are the ones belonging to the regions r, and rk respectively. Then, we write the linear equation
c;(r)= xr,

xrk
+ r,  Y ,
=0
(31.1)
corresponding to the selfintersection point C,. The v equations (3 1.1) determined by v points C , are called the equations of the diagram. By way of illustration we write out the equations of the diagram of the trefoil knot (Figure 31.3). They are as follows:
c1 ( r ) = xr2  xro + r3  rq = 0 , c2(r)= X Y ~ xrg + rl  r4 = 0, c3 ( r ) = X Y I  xrg + r2  r4 = 0.
(31.2)
We remark that the equations of a diagram tell us the incidence relations between the edges and selfintersection points, and the relative position of the four edges at a selfintersection point; the distribution of the coefficients X tells us how the corners must be dotted. Therefore, we have all the information needed to reconstruct the diagram. Let us now treat the equations of the diagram as a set of ordinary linear equations E in which the ordering of the terms in the various lefthand members is immaterial. Then, the matrix of the coefficients of equations E is a certain rectangular array M of v rows and U + 2 columns, one row corresponding to each selfintersection point and one column to each region of the diagram. We observe that M has the following property:
Theorem 1 If the matrix M is reduced to a square matrix MO by striking out two of its columns corresponding to regions with consecutive indices p and p + l , the
FIGURE 3 1.3
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DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
determinant of the residual matrix MO will be independent of the two columns struck out, to within a factor qf the form fX". Proof To prove Theorem 1, let us introduce the symbol RP to denote the sum of all the columns corresponding to the regions of index p and the symbol 0 to denote a column made up exclusively of zero elements. Then, we obviously have the relation
for in each row of the matrix there are only four nonvanishing elements, namely X, X, 1 , 1, and the sum of these elements iz zero. We also have the relation
for if we multiply the elements of each column by a factor xP, wherep is the index of the region corresponding to the column, the four nonvanishing elements in a row of index q become X'4, X'q, Xq, and  x O respectively, so that their sum is again zero. By combining reletions (3 1.3) and (3 1.4) we obtain the relation
in which the term R. disappears. Now, let
be the determinant of any one of the matrices Mpq obtained by striking out from the matrix M a pair of columns of indices p and q respectively. Then we clearly have
Indeed, let Mopbe obtained by striking out from the matrix M a column a of index 0 and a column b of index p, MOqbe obtained by striking out from the matrix M a column c of index 0 and a column d of index g. If we denote the determinants of Mop,MO, by brackets, we have
where V" denotes that column a is absent. By using (31.3) we can find column a in the form of a linear combination of column c and the columns of the matrix MOq;then
AOq= [. . . ,C  d , . . . V'. . . b . . . v d . ..].
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161
Let us multiply A,, by XP  1. From (31.5) one can obtain an expression for (xpP  1)h in the form of a linear combination of column d and all the remaining columns with nonzero indices of MO,. We get (xpP  l)&,
=
(X
p

= +(Yq 
l)[. . . ,C  d,. . . V'. . . d . . . v d . ..] l)[. . . C . . . V'. . . vh. ". d . ".]
= f(xpq 
l)AOp,
that is, just formula (3 1.6). Moreover, since indices are determined up to an additive constant only, relation (31.6) g'Ives us
whence
But, as a special case of (31.7) p
=r
+ l; we have the relation
which proves the theorem. Let us divide the determinant A,,+l by a factor of the form fX" chosen in such a manner as to make the term of lowest degree in the resulting expression A(x) a positive constant. Then, Theorem 2
The polynomial A(x) is a knot invariant.
First, let us evaluate the invariant A(x) in a simple concrete case. From the equations of the diagram of the trefoil knot (Figure 3 1.3), we obtain the matrix
If we assign indices in such a way that the first row of M is of index 2, the next three rows will be of index 1 and the last row of index 0. The determinant Aol obtained after striking out the last two rows of matrix (31.8) is
then we have A=1x+x2 To prove the formulated theorem, it will be necessary to obtain a somewhat more precise statement about the matrix M than Theorem 1. Two matrices M' and M" will © 2000 CRC Press
DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
162
be said to be equivalent if it is possible to transform one of them into the other by means of the ordinary operations allowed in the theory of matrices with integer coefficients:
( a ) Multiplication of a row (column) by  1. (p) Interchange of two rows (columns). (y) Addition of one row (column) to another. (S) Bordering the matrix with one new row and one new column, where the element common to the new row and column is l and the remaining elements of the new row and column are 0's; or the inverse operation of striking out a row and a column of the type just described. Two matrices M' and M" will be said to be Eequivalent if it is possible to transform one of them into the other by means of the operations ( a ) ,(P), (y), (S) along with the further operation (E)
Multiplication or division of a row (column) by X.
Two polynomials will be said to be &equivalentif they differ, at most, by a factor of the form &X". The following theorem, which we state without a proof, holds. Theorem 3 Ij' two diagrams represent knots of the same type, the corresponding matrices M are &equivalent. When a knot is deformed, the equations of its diagram remain invariant so long as the topological structure of the diagram does not change. A change in the structure of the diagram may come about in one or another of the following ways (Reidemeister transformations): (A) The diagram may acquire a loop and selfintersection point (Figure 31.4) or it may lose a loop and selfintersection point by a deformation of the inverse sort. (B) One branch of the curve may pass under another with the creation of two new selfintersection points (Figure 3 1.5); or by a deformation of the inverse sort, one branch may slide out from under another with the loss of two selfintersection points. (C) If there is a three cornered region in the diagram, bounded by three arcs and three selfintersection points, and if the branch corresponding to one of the three arcs passes beneath the branches corresponding to the other two, then any one of the three branches may be deformed past the selfintersection point formed by the intersection of the other two (Figure 31.6).
FIGURE 3 1.4
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ALEXANDER'S POLYNOMIAL
FIGURE 31.5
FIGURE 31.6
It is a simple matter to verify that any allowable variation in the structure of the diagram may be compounded out of variations of the three simple types indicated above. Now, to prove Theorem 3, it is sufficient to show that under each of the transformations (A), (B), (C) the matrix of the diagram is carried into an &equivalentone. First, consider case (A), where a branch of the curve acquires a new loop and selfintersection point (Figure 31.4). The equation of the diagram with respect to the new selfintersection point is
None of the remaining equations of the diagram change. If MO is the matrix of the original diagram, then the matrix M of the deformed diagram is obtained from MO by bordering with a new row and column:
1
r;
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(l
+X)
r
;
0 MO 0 0
... 0
X
v l'"
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Using operations (a), (y), (E), and (S), we can transform M into MO;this means that the matrices M and MO are €equivalent. Under case (B), three new regions v',, rb, rp and two new selfintersection points arise instead of one region r2 (Figure 3 1.5). Hence, we have two new equations of the diagram:
The equations of the original diagram are transformed as follows. If a vertex C belongs to the boundary of the region r2, then after the deformation it is situated either in the boundary of rk, or in the boundary of r',. Therefore, the transformation of the equation corresponding to C is the replacement of r2 by r', or by r;'. If MO is a matrix of the original diagram, then the matrix M of the deformed diagram is obtained from MO in the following way. We replace the columns corresponding to the region r2 by two columns corresponding to the regions 6,r; and obtain the matrix M; then we borcjer the matrix M by two rows and one column so that we have
M=
Applying the operations (y), (a), (E), and (S) consecutively, we see that M is Eequivalent to M O . Finally, let us consider the deformation C of the diagram and the corresponding transformation of the equations of the diagram (Figure 3 1.6). We have the following equations related to the points Ci, C::
+ r3 c,(r) = x r ~ + r3 ( r ) = xrg  x r ~+ rg c , ( r ) = xr2
C,
 XYO

r4 = 0,
 XYS

rg = 0,

r2 = 0,
+ r; Y;) = 0, c ; ( r ) = xr; xr;) + r: r; = 0, cT(r) = xr;)  xr; + r;  r: = 0; c ; ( r ) = xr;,  xr; 

none of the remaining equations of the original diagram change under the deformation. Let M, M* be the matrices of the original diagram and of the deformed one respectively after the columns corresponding to the coinciding equations of the diagrams have been struck out. Then,
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ALEXANDER'S POLYNOMIAL
By the operations ( a t ( & the ) , matrices M, M* can be transformed into the matrices
v;,r f by respectively. If we denote the columns of M, M* corresponding to the regionsd;, 4 and the columns of A by ai, the described transformation of M, M*into M , M' is
and
The regions ro, r; have only three vertices, hence column a0 of matrix A consists of     zero elements. Columns d l , d4, d5, d6 are equal to columns d;,G, 4, 4 respectively.  Only columns do, d2,d3 are different from 4, 4, 4; we write two matrices formed by these last columns:   a
It is easy to see that these matrices are transformed one into another by elementary transformations. Thus, the matrices M and M* are &equivalent, that is, proving Theorem 3. © 2000 CRC Press
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Now, we shall prove that the matrix MO obtained by deleting from the diagram's matrix M two columns with consecutive indices p and p + 1 is &equivalent to the matrix M. To demonstrate this property, it is sufficient to show that any two columns with consecutive indices p, p + 1 can be expressed in the form of linear combinations of the remaining columns in such a way that the coefficients of the linear combinations are polynomials of X, X' with integer coefficients. Since the indices are defined up to an additive constant, we suppose p = 0. The term with R" in equality (31.5) is absent. Therefore, we can divide (31.5) by X'  1 and obtain an equality such that the coeficient at RI is equal to 1 and the coefficients at the remaining Ri are integer polynomials of X , X  ' . Thus, the column with index p = 1 is a linear combination of the columns of indices different from p = 0 , l and the coefficients of this linear combination are polynomials of X , X' with integer coefficients. If we multiply (31.4) by X and subtract equality (31.3), we get the equation
which does not contain R I . From this equality it follows trivially that R. is also a linear combination with polynomial coefficients of the columns of indices different from p = 0, 1. W Applying the theory of elementary transformations of socalled Xmatrices, which is developed in [37], we see that the determinants of Xmatrices are invariant up to a factor fxn under elementary transformations. Thus, we obtain that Alexander's polynomial is an invariant of knots. Some examples, which were studied by Alexander, are shown in Figure 31.7. We remark that the degree of Alexander's polynomial of a knot is equal to or less than the doubling genus bf the knot.
FIGURE 31.7
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ALEXANDER'S POLY NOMlAL
Problems
1. Prove Gauss's theorem on diagrams of knots: Assume a point is moving along the diagram of a knot in the positive direction. If we write and enumerate the diagram's vertices passed consecutively by the moving point, then each vertex will be placed once on an odd place and once on an even place. 2. Prove that ~ l e x i n d e r ' spolynomial of the product 7,#7;!of two knots y1,72 is equal to the product of Alexander's polynomial of these knots. 3. Prove that Alexander's polynomial of the knot shown in Figure 30.5 is (l  X + x212.
© 2000 CRC Press
32 Curves in ndimensional Euclidean Space Let E" be ndimensional Euclidean space provided with Cartesian coordinates x i , . . . , X " . Thus, every point P in E" is defined by n real numbers ( X ' ,. . . ,X"). If we denote by el, . . . ,e, the basic coordinate unit vectors, then the position vector r of any point x E E" can be expressed as a linear combination of the basic vectors:
Let a < t 5 b be a segment on the auxiliary axis of parameters. The elementary curve y in E" is the image of the segment [a,h] under a homeomorphism into Euclidean space. This means that there exists a onetoone correspondence between points of y and [a,h],which has a continuous inverse one. In this case the coordinates ( X ' , . . . ,X " ) of a point of y are continuous functions of parameter t :
This is the socalled parametric representation of y. We shall write the parametric representation of the curve in a brief form using vector notations: r =f( 4 , where f ( t ) is a vectorvalued function of t. To simplify notations we shall often write r(t) instead off (t). If each coordinate function f i ( t ) is differentiable, the vectorvalued function f ( t ) is called diferentiable. By definition
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DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
The curve is called regular of class C k , if there exists its parametric representation with coordinate functions f i ( t ) of class cksuch that f ' # 0. In the case of k = I the curve is called snzooth. For an arbitrary curve y and for any point P E y,similarly to the definitions given in chapter 4 one can define two rays tangent to y at P and the straight line tangent to y at P. If y is smooth, then for any point P there exists a unique straight line tangent to y at P. Modifying slightly the proof of the corresponding statement from chapter 4, one can demonstrate that if r = r(t) is the position vector of y,then rl(t) is the directing vector of the tangent straight line. The length s of a curve is a supremum of lengths of polygonal lines inscribed in this curve. If the length of the curve exists, then the curve is called rectifiable. Just like in threedimensional space, one can prove that a smooth curve is rectifiable. The length of the smooth curve y can be evaluated with the following formula:
Choose the orientation on the given smooth curve y and fix an arbitrary point P E y We can parametrize the curve with the help of the arc length in the following way. For a point Q E y we consider the oriented arc PQ with the origin at P; if the orientations of the arc PQ and y coincide, we assign to Q the length s = s(PQ) of the arc; if the orientations are different, we assign s = s(PQ) . This parameter s we will call the natural parameter of y. It follows from (32.1) that JriJ= 1. Now we can define the concept of curvatures of y E En. Assume that y is parametrized by the natural parameter S. Denote by the unit tangent vector r', . The length of the vector d t l / d . ~is called the first curvature k l of y, i.e. k l = Id<~/dsl. If k l # 0, we denote by 5'2 the unit vector collinear to dtl/ds. Then it is easy to demonstrate that
Let us consider dc2/ds. Obviously this vector is orthogonal to J2. Denote by k z the length of the projection of dt2/dsinto the space orthogonal to [ I and and by C3 the unit vector collinear to this projection. If k2 # 0, the vector I3 is well defined and we can write:
where a is unknown coefficient. But it is very easy to find it. Indeed, multiplying (32.2) by c l we obtain
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CURVES IN nDIMENSIONAL EUCLIDEAN SPACE
Thus
Now we can define curvatures k s , .. . ,kn_l and vectors &, . . . ,tnby induction. We always choose (, in such a way that the derivative d(,lids decomposes into a linear and a new vector I,,which is orthogonal to [ l , . . . combination of ( 1 , . . . , i.e.
The vector E, is well defined in the case of k,l # 0. It is easy to see that in (32.3) and a,l = 0. Indeed, let us multiply (32.3) by t,,
a, = . . . = a,3 = 0, a12 = k,2 a < i  2. Then we obtain
By construction, d<,/ds is a linear combination of the vectors a + l < i  l, hence
Because [;I is a unit vector, U;l = 0. On the other hand, by inductive hypothesis
Hence
Let us now multiply (32.3) by
&2.
So, we get the following formula:
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Then
[l,.
. . ,[,+l
and
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The process of construction of the vectors t;, is interrupted in the case when either one of the curvatures ki is equal to zero, or the number of vectors orthogonal to each other is equal to the dimension of the space, that is to n. In the latter case the derivative dE,/ds is collinear to (,I So. we obtain the following formulas for the curve y E E": k1E2, kl(l
ki
+ k213,
I Ei I
+ ki&+I ,
4l<"l.
They are called Frenet formulas. If we introduce a symbolic vector [with components of the forms:
Ei and the skewsymmetric matrix A
then Frenet formulas can be written in a brief form as
Observe that by the construction of 11,. . . , 1, all the curvatures k, are positive. Assuming E" is oriented one can define the curvature k,_l with the sign: if (1,. .. , &  l are constructed, we denote by a unit vector such that the vectors (1,. ... &, form an orthonormal basis of E" and the orientation induced by this basis coincides with the initial orientation of E". Then define the last curvature k,_l as (%,Fn). The expression of Frenet formulas does not change. But the curvature k,l can be positive, negative or equal to zero. The curvatures k l , ..., k n P lare important due to the fact that they determine the curve in E" uniquely up to a rigid motion. More precisely, the following theorem is true.
Theorem Let k l ( s ) ,..., k n P l ( s )be continuous functions of a parameter s E [0,l ] . Assume that k l > 0,. .., k,2 > i There exists a unique up to a rigid motion c2regular curve y C E" having the functions k , as its curvatures and s as the length of arc. © 2000 CRC Press
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Proof Let us consider Frenet formulas as a system of ordinary differential equations with respect to unknown vectorfunctions E l , . . . ,En. If initial values t l ( 0 ) ,. . . ,&(0) are given, then by the wellknown theorem from ODE theory there exists a unique solution E,,. . . ,E, of the corresponding initial values problem, and the position vector of y can be expressed in the following form:
Let us choose the initial conditions in such a way that (&(O),<,(O)) = S;,. It is easy to verify that, due to the structure of Frenet equations, the vectors [ ~ ( s ). ,. . ,&(S) are orthonormal for each S . To do this, introduce the functions Uii = ([,,E,)  S,. Then
It is trivial to demonstrate that the sum containing in the righthand side of (32.4) is vanishing and the equations can be rewritten in the following form:
Thus U , satisfy a homogeneous system of ODE. Since UV(O)= 0, it follows from the uniqueness theorem that the unique solution of (32.4) is Uii = 0. Therefore, for any s the vectors El(s),. . . ,c&) form an orthonormal basis of E", El(s) is the unit tangent vector to y and s is the arc length. Since the vectorvalued functions [;(S)satisfy Frenet equations, the vectors [ I , .. . ,E, form the natural frame along y and functions k;(s) are the curvatures of y. The abovementioned definitions of curvatures do not give a way to find their value immediately. So it is useful to obtain formulas to compute the curvatures in the case of an arbitrary parametrization of y. First of all, let us recall the formulas for the curvature and torsion of curves in E ~ :
We will define the notion of the simple multivector in E" similar to the vector product in E'. For given k vectors a l , . . . , a k in E", k 5 n with components a; (i = l , . . . ,n; a = 1 , . . . ,k ) , the values
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DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
are called components of a simple multivector. The set {pi''k, i l <. . . < ik)is called a simple multivector and denoted by p = [a,,. . . ,an]. For given multivectors p, q we define their inner product by the following formula:
the length of p is defined as
Let us note the following properties of a multivector: (l) If among a,,. . .,a,, there exist two collinear ones, then the multivector [ U , , . . .,ak] = 0. (2) If a; = Xb pc, then [a,, . . . ,ai,. . . ,ak] = X[ul,. . . ,b, . . . ,ak] p[al,. . . ,c, . . . ,ak],where X and p are numbers. (3) If two vectors ai, ai+l are transposed, then a multivector changes its sign: [ a t , . . . ,a;,ai+r,. . . ,alil =  [ U , , . . . ,ai+l, ai,. . . ,ak]. (4) The inner product of two multivectors p = [ a l , .. . ,ak], q = [b,,. . . ,bk] can be expressed in terms of the inner products of ai, b, c E" as
+
+
First, we will find the formulas expressing the curvatures ki with respect to the natural parametrization. Using the Frenet formulas, we get
where in the expression for v!) the dots mean a linear combination of Then the twovector [ri, r$] can be expressed as
Applying formula (32.5), we can find the length of
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[ E l , <2]:
tl,. . . ,c;,.
CURVES IN nDIMENSIONAL EUCLIDEAN SPACE
Therefore, k l
= I[r:, v:s]l. If
the parametrization is not natural, then evidently
So we obtain the formula for k l :
Moreover, we have k:k2 = I k(,,
r;is1 l l
hence
Let us now find a multivector generated by the vectors rl,, . . . ,r?):
Using (32.5) we obtain have
IIEl
. . .till = 1. If we denote the length of [ r i , . . . ,r!)] by S;, we
S; = kl' ki2 . . . k?,k;, . Let i  2
2 l . Because
the curvature ki1 has an expression in terms of S,:
It is easy to see that for an arbitrary parametrization,
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176
Hence for arbitrary parametrization v(/):
The last curvature k n P l can be defined with a sign. Note that a multivector of n vectors in E" has only one component  it is equal to the determinant of the matrix formed by components of these vectors. Therefore, the curvature kn1 with a sign can be computed by the formula
Problems 1. Find the curvature of the curve r(t) = (t, t2, t 3 , t4) at t = U. 2. Prove that if the curvatures of a curve y satisfy k l # 0,. . . ,kll # 0 and k, = 0, then y belong to some E'. 3. Let y be a curve in E". A straight line is called the principal normal line of y at a point P E y,if it passes through P and is collinear to the principal normal vector ofy at P. Prove that if there exists another curve y* and a mapping y + y* such that the principal normal lines of y and y* at corresponding points coincide, then y belongs to threedimensional Euclidean space, and its curvature kl and torsion k2 satisfy the condition akl bk2 = 1, where a and b are constants. (Such curves are called Bertrand's curves.) 4. Prove that a curve y E E" is situated on a (n  l)dimensional sphere S"' iff the curvature of y satisfies a differential equation obtained from the system
+
by excluding the coefficients ai.Here a prime denotes the derivative with respect to the arc length.
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33 Curves with Constant Curvatures in ndimensional Euclidean Space Now we use Frenet formulas to describe curves with constant curvatures k l , . . . ,k n + ] in E". In threedimensional Euclidean space, any curve with constant positive curvature k and constant twisting K is a circle or a helix. Such curves in E" were studied in [3941]. We will suppose k, # 0. The Frenet system has the form
It is a system of linear homogeneous ordinary differential equations of first order with constant coefficients. Denote by p the differentiating symbol d/ds and rewrite the system (33.1) as
where L:@) are polynomials with respect to p, which have constant coefficients. Similar systems are well known in the theory of differential equations. System (33.2)
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is reduced to one ordinary differential equation of nth order with constant coefficients. If one denotes by D@) the determinant of the matrix ((Li,(p)ll,then the equation corresponding to (33.2) has a simple form [38]:
In particular, the vectorvalued function (1 satisfies this equation. The polynomial D@) is a characteristic one for the skew symmetric matrix A (see the previous chapter); it has the form
The coefficient A, is a sum of the principal minors of order r:
A;
=
C
I , <  <i ,  ~
0 k , , O . . . . . . O k, 2 . . . . . . 0 kj, 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 o . . ,.
O 0 0 ... kiT 1 0
Since the determinant of a skew symmetric matrix of odd order is equal to zero, A , = 0 for odd r. When T is even, it is easy to show that
For example:
k:k:. .. k:_, 0,
.
when n is even, when n is odd.
Thus, one can rewrite (33.2) in the form
+ A2p"2 + . . . + An)J1= 0 , when n is even, p(pnP1+ + ... + = 0 , when n is odd. (pn
For example, when n = 4, we have the following equation:
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CURVES WITH CONSTANT CURVATURES IN nDIMENSIONAL EUCLIDEAN SPACE
This equation can be obtained easily from (33.1), if we find four derivatives of The characteristic equation has the following form:
X4
+ (k: + k; + k i p 2 + k;k:
179
El.
= 0.
Then
Thus the solutions of the characteristic equation are distinct imaginary numbers, which we denote by ial, h l , ia2,icez. Therefore 2
cl = C(A~ cos ais+
sin ais),
i= l
where A , , B, are constant vectors in E ~Integrating . the obtained expression for we find the position vector of the desired curve y.This curve is situated in a bounded part of E4. It is closed iff the trigonometric functions in (33.7) have a common period, i.e. iff the ratio al/a2 = k / l is rational, where k, I E 2. We remark that y is closed iff its constant curvatures satisfy
for some integers k, l. Now let us return to the general case. It is easy to verify that all eigenvalues of a skewsymmetric matrix are imaginary. Indeed, assume that for some eigenvector (complex in general) the following equation holds: Ax
= Xx.
Denote by X the vector complex conjugate to left by X, we get
(33.8) X.
Then, multiplying (33.8) from the
But each number xixj xixjis either imaginary or zero. Therefore the eigenvalue X is either imaginary or zero. As A, > 0 for even r , (33.5) has only imaginary roots for even n. In the case when n is odd, equation (33.6) has one root equal to zero and the others imaginary. Let us prove that if ki # 0, i = l , . . . ,n  1, then all roots are distinct. Matrix A belongs to the Jacobi class. Denote by D,(X) the determinant of a corner minor of order T for the matrix IjXE  AI/. It is easy to derive the recurrence relation for the sequence of polynomials D,(X), r = l , . . . ,n:
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DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
In particular,
If we define a new variable T = X2, then D2/ = F/(T), D 2 / += ~ X@,(r), where F/(r), @/(T)are polynomials of degree l with respect to 7 . Consider the sequence of polynomials:
We have two recurrence relations:
The polynomial sequence mentioned above is similar to a sequence of Sturm polynomials. F/ and @/ have exactly I negative roots (maybe with multiplicities). Using recurrence relations, one can prove by induction: (1) between two neighboring roots of any polynomial in the sequence there is exactly one root of previous and one root of subsequent polynomials; (2) values of any polynomial at any neighboring roots of previous or subsequent polynomials are distinct. This implies that the roots of a characteristic polynomial for A are distinct. Set n = 2m.Denote these roots XI = i a l , X2 = iol,. . . , X2,l = ia,, X = i a , . When n = 2m + 1 there is one more root a,+I = 0. Only in some particular cases exact values of Xk are known. For instance, if kl = k2 = . . . = k,l = 1, then all distinct eigenvalues are in the sequence:
7lk Xk = 2i cos n+l'
k = l, ...
We can write the general solution of (33.5) and (33.6) in the form:
where A; B, are constant vectors in E". Differentiating El we produce h,.. . ,E,. We obtain the position vector of the desired curve y by integrating El(s). Because = Sii and (a, are distinct), the vectors A;, B, are mutually orthogonal and of the same length. If we choose Cartesian coordinates X', . . .,X" by fixing an orthonormal basis of vectors el, . . . ,e, in such a way that A iare collinear to ezil, B, are collinear to ezj, then the position vector of y is
(&,c,)
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CURVES WITH CONSTANT CURVATURES IN nDIMENSIONAL EUCLIDEAN SPACE X'
= a , cos cqs,
x2 = a1 sin als, n = 2m;
181
. . . , X"'  a, cos ams, . . . , xn = a, sin ams,
 a, cos ams, . . . , x "  ~. a,s,  a, sin x2 = a, sinals, . . . ,
X'
= a1 cos als,
xn=bs,
n=2m+1.
Since ri is a unit vector, the constants a,, d; and h must satisfy the relations
CL,aTaf = 1 CL,a f a ? + h2 = 1
for even n; for odd n.
So, the representation of a curve with constant curvatures is quite different when n is even or odd. If n is even, then the curve is bounded. If n is odd, then the curve is unbounded in one direction. If n is even and all ai/cujare rational numbers, then the curve is closed. One class consists of the trigonometric curves: we say that a curve y E E n is trigonometrical, if the components of its position vector with respect to some Cartesian coordinates are trigonometric polynomial depending on some parameter t. The position vector of such a curve has the following form:
where ck are constant complex vectors satisfying ck = c k . An arbitrary trigonometric curve y is closed. It is natural to find some relations between the curvatures of y.One can obtain the following equality:
[ d , . . . , r ( l ) ]=
m1
C dp&P',
where
Using formula (32.7) of chapter 32, we obtain that k:_*is a ratio of two polynomials with respect to e". There is one case of interest: t is the arc length. Then the vector ck must satisfy the following system:
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One can view this system as a set of algebraic equations with respect to the components c: of the vectors c k . Then ( 3 3 . 9 ) represents an algebraic manifold M in the space of the components c:. It follows from formula (33.6) of chapter 32 that
4.
We can where constant coefficients ag are rational functions with respect to consider a@as rational functions on the manifold M. But a theorem of algebraic geometry states that any n 1 rational functions given on an algebraic manifold of the dimension U are mutually dependent. Thus, the coefficients alp are mutually connected by some system of algebraic equations. In conclusion, we make some remarks on the topology of curves in the ndimensional Euclidean space, n > 3 . First, all the closed curves are homotopic to a circle; this means that there exist no nontrivial knots, whenever n > 3 . Also, two arbitrary disjoint closed curves can be continuously deformed without intersection onto two disjoint circles on a plane; hence the linked curves do not exist. For example, let us consider the threedimensional space E' given in the fourdimensional space E4 by the equation x4 = 0 with respect to Cartesian coordinates X I , . . . , x 4 ; next, in E 3 we take two linked circles given by equations x: + xi = 1, (x2 + X : = 1 . Using a translation with respect to x4, we map the second circle xi = 1, x4 = t. Then we translate onto the circle given by the equations (x2 the obtained circle onto the circle (x2  312 x i = 1, x4 = t, and then we again apply a translation with respect to x4 and get the circle (x2  312 + x: = 1, x4 = 0 situated in the initial space E3. Thus, we have deformed the circle (x2 + X: = 1 onto the circle (x2  312 +X: = 1 , which is not linked with the circle X: + xi = 1. Similarly, one can demonstrate that all the knots are trivial. However, metric properties of curves in E", n > 3 are more complicated and they are not determined completely by homotopical theory. Maybe, to find connections of the metric properties of a curve y with topology, we must consider a threedimensional closed simplyconnected manifold M containing y such that this curve can be viewed as a knot in M.
+
+
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34 Generalization of the Fenchel Inequality Let y be a closed C'regular curve in the ndimensional Euclidean space E" with length of arc s and position vector p($). We denote the curvatures of y by k ~ ,. . , k , ,  ~ . Fenchel's inequality ( considered in chapter 21) holds for curves in the ndimensional Euclidean space [63]; if y is a closed C2regular curve with length of arc s and the first curvature k l ( s )in the ndimensional Euclidean space E", then
and Jy kl ( S ) ds = 27r if and only if y is flat and convex. In this chapter we demonstrate one generalization of Fenchel's inequality obtained by Gorkaviy [64]. Theorem 1 Let y be a C'regular closed curve in E", n > 4 . Let for some integer 3 5 m 5 n  1 , r 2 m 3 the curvatures kl(s),. . . ,km(s)of y be positive; s denotes the length of arc. Then
+
Proof Let y be a C'regular closed curve in E", n 2 4, and suppose the curvatures kl(sj,. . . ,km($ of y are positive functions for some m such that 3 5 m _< n  1 , r 2 m + 3. From the assumptions it follows that the first m + 1 vectors El(sj,. . . ,Em+,(s)of the Frenet frame for y are uniquely defined as well as the curvature k m ,,(S), which may vanish somewhere on y. Consider the map G, which assigns to a point s the mdimensional osculating space P ( s ) c E" of y (i.e. spanned and oriented by the basis of vectors El(s),. . . ,Em(s)). G is a map of y into the
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DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES
Grassmannian G(m, n) of oriented mdimensional subspaces of E". The image S of y under G is called the indicatrix of mdimensional osculating spaces of y; it is a generalization of the indicatrix of tangents (see chapter 21). We use the Plucker embedding p1 of G(m,n) into the Euclidean space EN, N = C,: (see [65]): if the subspace 7f" c E", which corresponds to the point P E G(m,n), is spanned and oriented by the orthononnal basis 'rll,. . . ,vm, then the position vector of $(P) is the . is known that p1 is multivector [rll,. . . ,rim] considered as a unit vector in E ~ (It analytic, isometric, and maps G(m, n) into the unit sphere sNpl C EN.) We consider the image S*c ENof l? under p2; its position vector is of the form
Using the Frenet formula we obtain :
and it follows from the assumptions that S*is a C'regular closed curve in EN.Let us select the length of arc a(s) on F* as an increasing function. Then from (34.2) we obtain
and further
+
where w = d
Using the Fenchel inequality for the curve l?*:
and taking into account (34.3), (34.5), we get
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GENERALIZATION OF THE FENCHEL INEQUALITY
9
8,
It is easy to verify by a simple computation that the vector (& $)+ KI whose length is equal to the second curvature of I'* C E ~does , not vanish, so F* is not a flat curve. Hence the inequalities (34.6), (34.7) are strict. Remark Theorem l gives a necessary condition ,for the curve y C E" with curvatures kl (S), . . . , k,_l (S) to he closed. Using properties of curves with constant curvatures in the ndimensional Euclidean space, Gorkaviy [64] investigated the question of optimality of the inequality (*) and proved: Theorem 2 In evendimensional Euclidean space E ~ "n, 2 2, for any 1 < p < n  1 andfiw any positive E there exists a closed curve y with positive constant curvatures k l , . . . , k2n1 such that
Theorem 3 In odddinzensionul Euclidean spuce E~"", n 2 2, Jbr any 1 5 p 5 n  1 and ji)~ any positive E there exists a closed analytical curve y with positive curvatures kl(s), . . . , kz,,+l(s)and with curvature kZn(s)vanishing at least at two points such that
Remark The method used to prove Theorems 2 and 3 does not permit the solution of the question of optimality of (*) in the case of even m. Probably this depends on the different behavior of curves with constant positive curvatures in odd and in evendimensional Euclidean spaces, and maybe the inequality (*) is not optimal for even m, but this remains an open question.
© 2000 CRC Press
35 Knots and Links in Biology and One Mystery In the past twenty years some interesting applications of differential geometry and the topology of curves to the biology of DNA molecules have arisen. In accordance with the Watson and Crick theory, these molecules have the form of a double helix  two chains of nucleotides (1953). Two helices from a double helix are connected to each other by the rule of complementary base pairing and wound about a common imaginary axis. It is interesting that under common conditions they are righthanded helices: the socalled Bform of DNA molecules. Nature prefers asymmetry in the structure of DNA. This wonderful general asymmetry at the foundation of life was a surprise to biologists, as then a new Zform was found, which is a left helix. This form is uncommon. Under some special conditions the right Bform can transform to a Zform. In the process of replication of a DNA molecule, some special enzyme cuts the connection between two helices. After separation every helix builds its complement. The result is two new double helices which from two new cells. This theory of the reproduction of cells is well known. For this result J. Watson and F.H.C. Crick won the Nobel Prize in 1962. But one fact severely complicates this simple and clear picture: in the double helix every strand links to each other. At first, the biologists did not pay attention to this linking, as the known molecules were short, and besides, their axes were intercepts. But then very long molecules were discovered with a length of a few centimeters. Moreover, in 1963 biologists suddenly obtained closed DNA molecules, see Figure 35.1. The following question arose: how could the two strands of the double helix st?. 2 ..' %eafter cutting by an enzyme, if they are twisted about one another? According to one hypothesis, some unknown enzyme cuts one helix and makes separation possible. In reality, such enzymes have been discovered. They are called topoisomerases. They are like a tailor  they can cut and sew. The number of linkages between two h e ~ l ~ is e slarge; for example, for some DNA it is errual to 300 000. So there me problem of explaining the unwinding of one
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GEOMETRY AND TOPOLOGY OF CURVES
helix for separation. For the solution of this problem, it is necessary to study the dynamics of a DNA molecule, using its inner structure. Research in this direction is ongoing [68; 701. We note that in order to describe the mechanical properties of the DNA molecule and its movements during its change, of form it is necessary to use ndimensional space (five or sixdimensional) [66]. Coordinates in this space are the parameters of helices and various angles between planes of complementary pairs of nucleotide bases and sugarphosphate backbones. To describe the linking of two helices, biologists apply the theory of knots and links. As the distance between two chains of a double helix is small and constant along the DNA molecule, we can use the theory of ribbon rotation. Biologists consider three ribbons: two ribbons modelling two chains of nucleotides and a third ribbon with an axis of the double helix which has been built by moving the intercepts a of the line halfway between the first two ribbons, see Figure 35.2. The axis of a double helix can be on a straight line for a linear molecule or can be some space curve as in the case of a closed molecule. Let us consider the third ribbon. Let y be the axis of the double helix. It is natural to use the formula which expresses the number of links of two infinitesimally close curves Lk in terms of two integrals; see (29.6), chapter 29. The first integral is called Twist and is denoted by Tw. It depends on the curve y and on the ribbon. The second integral, which depends only on the curve y,is called the Writhing number and is denoted by Wr. So formula (29.6) has the following form: Lk = Tw Wr. In biological experiments it is possible to observe the number Wr, for example by measuring the velocity of sedimentation in a solvent. There is a group of scientists  W. Pohl, G. Roberts, J. White, F. Crick, W. Bauer, M. FrankKamenezkii and others  who are working in this direction [66691. If Wr # 0, biologists can tell that the DNA molecule has the supercoiling form. The concept of supercoiling of DNA was introduced by the biologist J. Vinograd in 1963. It means that the double helix winds in a new helix, see Figure 35.3.
+
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KNOTS AND LINKS IN BIOLOGY AND ONE MYSTERY
FIGURE 35.2
FIGURE 35.3 (redrawn from [69])
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Therefore the axis of the double helix does not lie on a straight line, but is some space closed curve. The supercoiling form is a widespread phenomenon, which is important from medical and biological points of view. An entire class of DNA tumor viruses, including the polyomas and human papillomas viruses, contains such DNA. It i s interesting to note that in all cases the supercoiling is negative. In the chromatin of higher organisms, the DNA is wound around a core of protein to form a lefthanded superhelix. The length of DNA can be a few centimeters but in chromatin it has an exterior diameter equal to only a few micrometers. Therefore the supercoiling matches the molecular dimensions. The author would like to thank the referee for assistance with the discussed application of Geometry to Biology.
© 2000 CRC Press
36 Jones' Polynomial, Its Generalization and Some Applications In 1970 Conway proposed an inductive way to construct Alexander's polynomials for knots and links which is simpler from a computation viewpoint. He observed that for three oriented links whose diagrams differ only in a neighborhood of a point and such that inside that neighborhood the links have views as in Figure 36.1, the polynomials satisfy a simple relation. Denote by L+, L P ,L. the links corresponding to the first, second and third pictures of Figure 36.1. Let el, t.2 be a positively oriented basis on the diagram plane. For the link L+ the vector el goes through the point of selfintersection above e2 and for the link L_ the vector el goes below e2. Then the Alexander polynomial, when suitably normalized, satisfies the following equation: AL,( t )  & ( t )
+ (t1I2

t   ' / 2 ) ~ L o (= t )0
L0
FIGURE 36.1
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In 1983 Jones constructed a new polynomial V L ( t )using the braid theory and the representation of braid groups in the Hecke algebras [71]. It was a great surprise to topologists. This polynomial opened the way to solving some old problems in knot theory. For example, this new polynomial distinguishes, for some cases, the knot and its mirror image. This polynomial satisfies the condition
In 1985, almost at the same time, four groups of mathematicians, encouraged by Jones' success, obtained one and the same result. They proved the existence and studied the properties of a new polynomial invariant for knots and links. Their results were gathered by the editors of Bull. Amer. Math. Soc. into a joint paper by six authors, P. Freyd, D. Yetter, J. Hoste, W.B.R. Lickorich, K. Millett and A. Ocneanu, and published [74]. The paper defined a homogeneous polynomial of three variables PL(x,y, Z ) with positive and negative degrees of these variables X, y, z. The polynomial was called the HOMFLY polynomial. This polynomial PL(x,y, Z ) has to satisfy the two following conditions:
if the link consists of only one unknotted component, so
Polynomial PL(x,y, Z ) is unique and invariant with respect to the isotopy class of the link L. The Pl,(x, y, z ) polynominal generalizes the Alexander and Jones polynomials in the following sense:
But the polynomial PL cannot be expressed as a function of the Alexander and Jones polynomials, because there is a knot K having a mirror image K such that AK = A, , V K= V K and PK # PE. Let us consider how one can find the polynomial PL for some simple knots and links. If L consists of n unlinked and unknotted components, then
The formula above is true for n = 1. Set n = 2. Let us consider the transformation of the circle as in Figure 36.2. Since L+ and L are unknotted circles, PL+= P L = 1. Hence
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193
FIGURE 36.2
FIGURE 36.3
We have PL, = (X + y ) z r l . Now take two unlinked and unknotted circles and transform one of them as in Figure 36.2. In this case, L" consists of three and L+ and L consist both of two unknotted and unlinked components. Using (36.1) we obtain PLO= (X+ y)2z2. Continuing the process, we come to (36.3). Consider now the polynomial for two linked circles at link number 1. The transformation in a neighborhood of the upper point of the selfintersection gives us the following pictures (see Figure 36.3). We see that L consists of two unknotted and unlinked circles; L. is an ordinary circle. Using (36.3) with n = 2 we get
Thus, the polynomial for two linked circles of link number 1 is
It is easy to obtain the polynomial for two linked circles of link number  1:
Let us consider a trefoil  the link as at the first place in Figure 36.4. © 2000 CRC Press
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FIGURE 36.4
FIGURE 36.5
We see that L is a trivial knot, and L. consists of two linked circles of link number l . Using (36.4) we obtain the polynomial for the knot L = L+ in Figure 36.4: PL = 2yx1
+ X2z2
 xZy2.
For the knot in Figure 36.5 we have
For the knot in Figure 36.6 we have the following pictures. Here L is a trivial knot, and L. consists of two linked circles of link number 1. Using (36.4) we get
Transformations of Whitehead's link in Figure 36.7 in the neighborhood of point a give us the following links: is a trivial knot and L+ consists of two linked circles of the link number  1. So, for Whitehead's link we have
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195
FIGURE 36.6
L FIGURE 36.7
New interesting ideas were proposed by Vassiliev [75].He considered a knot as a point in some space of continuous mappings which is divided into cells. Knots of the same topological type belong to the same cell. They have the same values of Vassiliev invariants. Different cells are separated by walls. The walls are made up of knots having selfintersection points. For every integer m Vassiliev defined the invariant by analogy to the Conway method. Let L, be a knot with a selfintersection point a; L+ and L are knots having, in some neighborhood of this point, views as in Figure 36.8. Let us suppose that L, has m selfintersection points. If Vis a numerical invariant of knots without selfintersection points, one defines the invariant of L, by the inductive hypothesis
This means that for two neighbouring cells, the difference between their Vassiliev invariants is equal to the invariant of the wall. © 2000 CRC Press
DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES
FIGURE 36.8
The Vassiliev invariant is said to be of type m if for all knots V("'+')= 0. It was found that the mth coefficient of the Jones polynomial is the Vassiliev invariant of type m (Birman, Lin) [76]. Now these invariants have many applications in knot theory. Let us consider some applications of knot theory in mathematics and physics. Ordinary threedimensional closed manifolds M' can be constructed with the help of Hegor's diagram. Take two threedimensional bodies T1 and T2 with boundary surfaces MI and M2 homeomorphic to each other. If f : M1 + M? is a homeomorphism, xi E M; and ,f(xl)= xz, then one can identify these points. In other words, two bodies T1and T2 become glued along the surfaces M , and M2. Thus, we obtain a closed threedimensional manifold, every point of which has a neighborhood homeomorphic to a threedimensional ball. But the construction of M~ can also be produced with the help of links. Wallach proved the following theorem, the socalled Bing hypothesis.
Theorem Z f M 3 is U closed oriented ?manifold, there is a collection cf disjoint toroidal bodies T I ,. . . , TAin M M and a collection of disjoint toroidal bodies T,', . . . , T , in the sphere S 3 such that the closure of \ ( T IU . . . T k ) is homeomorphic to the closure of s3\ (T,' U . . . TL). Hempel proved that it is possible to take From this theorem follows
7;'unknotted.
Theorem Every closed oriented 3manifbld M 3 is homeomorphic to S 3 j r o m which some number of bodies of toruses are dragged out and glued buck in various ways. The glueing can be done in one special way. On the surface of each torus Ti and q.' we take the canonical pairs (ai,pi)and (a:,p:) consisting of families of parallels and meridians. Map the boundary of T; onto the boundary of T,' so that ai + P,( and ,h'; t a:. Then identify the corresponding points. How the topology of the link is connected with the topology of the 3iiianifold is an interesting question. W( can meet knots and links in ph7,sics pipers too. In 1989, Witten proposed a topoiogical rnodel i . ' ' elementary parucles whicil has some relation to the Jones © 2000 CRC Press
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197
polynomials [77]. He considers threedimensional (2 + 1) "spacetime" with a moving particle along a trajectory in the form of a knot. In the space there is no notion of metric and only topological properties have physical meaning. But his model contains some integrals with metric terms. To avoid the dependence on metrics and obtain only topological properties, Witten integrates over all the space of metrics Weierstrass could find an occasion for criticism here. Some specialists believe that it is a beautiful paper. The works of Jones and Witten were rewarded with the Fields medal in 1990. One can find the exposition of that paper in the book by Atiyah [78]. But a more rigorous mathematical foundation is needed. The abovementioned paper stimulated the search for further connections between the knot and link theory and the theory of elementary particles. The BarNatan paper [79] contains a definition of a chord diagram related to Feynman diagrams in the particle interaction theory. (See also Sossinsky [SO].) By definition, the chord diagram is a disk with some number of marked chords. If one maps the boundary circle into 3space in such a way that only the ends of a chord have common images, then for every chord diagram there is a corresponding knot with selfintersection points. Hence, one can associate some number of Vassiliev invariants with every chord diagram. Interesting applications of the knot and link theory to physics are given in the article by Kauffman [81].

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Index Agnesi's witch 77 Alexander's polynomial Archimedian sptral 75 Astrotd 79 Atiyah 197
!57
Bakel'man 121 Barbier's theorem 35 BarNatan 197 Bauer 188 Bertrand's curves 176 Bing hypothesis 196 Birman 196
Definition of elementary curve of envelope 52 Delaunay's problem 125 diagram of knot 157 DNA 187
2
Etimov 104 Electromagnetic field 86 elementary curve 2, 169 equatwn characterizmg spherical curves 71 Euclid I evolute 38 evolvent 38
Cardioid 79 center of curvature 38 of osculating sphere 70 c kregular curve 9 cissoid of Diocles 78 classifying group of knots 154 closed Jordan curve 3 conchoids 79 Conway method 195 Crick 187 curvature of curve 27, 171 of planar curve 31 with sign 32, 57, 65 angles 63 curve witb constant torsion 34 cusp 41 , 45, 46 cycloid 79
FaryMilnor theorem 99 Fenche1 97, 104, 183 Feynman diagram 197 Field 197 first approximation 14 curvature 170 Fourier series 116 FrankKamenezkii 188 Frenet formulas 54, 172 Freyd 192 Gauss's integral 140 theorem 167 general curve 4 generator relations of knots genus of a knot 155 203
153
204
Gorkaviy 183 gravitation field group of a knot
INDEX
Parallel force 84 Pea no curve I, 49 Phylippov proof 133 polygonal lines 6:1 Pohl 188 Pontryagin 154 Poznjak 101 pnncipal normal 27
85 I SO
Hegor"s dtagram 196 helix 66 Hilbert 129 homeomorphism 2 Homily polynomial 192 Hoste 192 Hurwitz 117 hypocycloid 79
Rectifiable curve 22 regions of diagram 158 Reidemiester transfo rmatiOns
162
Index of a region 158 intersection mdex 146 isolated point 41 , 45 isoperimetnc properly II 5
Riemann 2 Roberts 188 roses 81 roulets 78
Jones' polynomial 192 Jordan's theorem 133
Second approxtmation 37 Schur 129 Schwartz 129 Seifert circle 154 s1mple curve 3 singular point 41 slope curve 66 smooth curve 9 snail of Pascal 81 spherical indicatrix 97 Steiner 1I 7 step of helix 67 support function 117
Kauffman 197 Klein I knot 149 knotted curve 99 Lemniscate of Bernoulli with II fOCI 75 length of a curve 22 Lickorich 192 linked curve~ 139 Lin 196 listmg's knot 153 logarithmic sp1ral 76 Maclaunn 's curve Millet 192 motion 84 multivector 174
47
Natural parametrization Nicomcde 80 Nikolaevsky I I 2 normal 16 Ocneanu 192 osculating plane point 46 sphere 69 ova l 35
73
17
24, 170
Tangent 13 Taylor's expansions 7 throcboids 78 topological map 2 torsion angles 63 of a curve 29 tractrix 47 trigonometrical curve 181 twist 188 twisting total 144 Vassiliev invanants 195 vectorvalued function 5 Veronese 2 vertices of diagram I 58 Vinograd 188 Vygodsky 101
I
Wallach 196 Watson 187 We1erstra s 1:!5. 197 Werner 121 Wh1tehead's hnk 194
DEX Whitney number 188 White 144, 188 Witten 196 Writhmg number 188
205