[yu. aminov] the geometry of vector fields(bookfi.org)

The Geometryof Vector Fields

Yu. Aminov

Gordon and Breach Science F'ublishers

The Geometry of Vector Fields

Yu. AminovInstitute for Low Temperature Physics and Engineering

Kharkov, Ukraine

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ISBN: 90-5699-201-5

Dedicated to

my parents

Contents

Preface - xi

I Vector Fields in Three-Dimensional Eadldean Space I

1.1 The Non-Holonomicity Value of a Vector Field .............. 1

1.2 Normal Curvature of a Vector Field and Principal NormalCurvatures of the First Kind ............................ 8

1.3 The Streamline of Vector Field 12

1.4 The Straightest and the Shortest Lines . 14

1.5 The Total Curvature of the Second Kind 22

1.6 The Asymptotic Lines . 29

1.7 The First Divergent Form of Total Curvature of the Second Kind . 32

1.8 The Second Divergent Representation of Total Curvature of theSecond Kind ------ 34

1.9 The Interrelation of Two Divergent Representations of the TotalCurvatures of the Second Kind .......................... 37

1.10 The Generalization of the Gauss-Bonnet Formula for the ClosedSurface .... ...... 39

1.11 The Gauss-Bonnet Formula for the Case of a Surface with aBoundary 42

1.12 The Extremal Values of Geodesic Torsion .................. 49

vii

viii CONTENTS

1.13 The Singularities as the Sources of Curvature of a Vector Field 52

1-14 The Mutual Restriction of the Fundamental Invariant-, of a VectnrField and the Size of Domain of Definition ... . . .... . . . .. . . . 55

1.15 The Behavior of Vector Field Streamlines in a Neighborhood of aClosed Streamline ---------------------------------- 60

1.16 The Complex Non-Holonomicity ......................... 66

1.17 The Analogues of Gauss-Weingarten Decompositions and the BonnetTheorem Analogue 69...................................

1.18 Triorthogonal Family of Surfaces ......................... 71

...........................1.19 Triorthogonal Bianchi System 79

1.20 Geometrical Properties of the Velocity Field of an IdealIncompressible Liquid ................................. 82

1.21 The Caratheodory-Rashevski Theorem .................... 87

1.22 Parallel Transport on the Non-Holonomic Manifold and the VagnerVector ----------------- --- 92

2 Vector Fields and Differential Forms in Many-DimensionalEuclidean and Riemannian Spaces 99............................2.1 The Unit Vector Field in Many-Dimensional Euclidean Space 99

2.2 The Regular Vector Field Defined in a Whole Space ........... 101

2.3 The Many-Dimensional Generalization of the Gauss-BonnetFormula to the Case of a Vector Field ..................... 104

2.4 The Family of Parallel Hypersurfaces of Riemannian Space ...... 109

2.5 The Constant Vector Fields and the Killing Fields ............. I I I

2.6 On Symmetric Polynomials of Principal Curvatures of a Vector.............................Field on Riemannian Space 114

..........................2.7 The System of Pfaff Equations . 116

2.8 An Example from the Mechanics of Non-Holonomic Constraints .. 123

2.9 The Exterior Differential Forms .......................... 124

2.10 The Exterior Codifferential 129

211 Some Formulas for the Exterior Differential - - - - - - - - - - - - - - 131

2.12 Simplex, the Simplex Orientation and the Induced Orientation of a...................................Simplex Boundary . 133

CONTENTS ix

2.13 The Simplicial Complex, the Incidence Coefficients ............ 134

2.14 The Integration of Exterior Forms . 135

2.15 Homology and Cohomology Groups ...................... 139

2.16 Foliations on the Manifolds and the Reeb's Example .......... 141

2.17 The Godbillon-Vey Invariant for the Foliation on a Manifold .... 143

2.18 The Expression for the Hopf Invariant in Terms of the Integral ofthe Field Non-Holonomicity Value ....................... 147

2.19 Vector Fields Tangent to Spheres . 150

2.20 On the Family of Surfaces which Fills a Ball ................ 159

References -------- ---- 161

Subject Index 169

Author Index ....- ------ 171

Preface

For a regular family of surfaces given in a domain of three-dimensional Euclideanspace one can discover geometrical properties of this family from the very fact thatthe surfaces totally fill the domain and lie in space in a good way, i.e. without mutualintersections. These properties cannot be reduced to the geometry of the surfacesthemselves but are connected with their spatial configuration. We can study theseproperties by considering the vector field of unit normals on the surfaces. Manysurface metric invariants can be found with the help of normal vector fields.Therefore, the geometry of a vector field or, in other terminology, the geometry ofthe distribution of planes orthogonal to the vector field, i.e. non-holonomic geo-metry, includes the theory of families of surfaces. The set of all vector fields is widerthan the set of normal vector fields and our study of arbitrary vector fields incontrast to the fields of normals of surfaces is necessary, for various reasons. Manyresults concerning the family of surfaces stay valid, as well, for an arbitrary vectorfield and this particular case makes these results clearer. It is possible to introducethe concepts of Gaussian and mean curvatures, the two analogues of geodesic linesand lines of curvature, and the analogues of asymptotic lines for the case of a vectorfield. Also, a vector field has its own geometrical invariant which shows how much agiven vector field differs from the field of normals of a family of surfaces. Thisquantity we call the non-holonomicity value of a vector field.

Besides the study of the distribution of planes orthogonal to a given field, it isnatural to study the behavior of the streamlines of a vector field from the metricviewpoint. In this context the geometry of a vector field is closely related to thetheory of ordinary differential equations.

It is useful to provide some background information on the development of thegeometry of vector fields and the mathematicians associated with this study. Thefoundations of the geometry of vector fields were proposed by A. Voss at the end ofthe nineteenth century and were followed by the publication of papers by Lie,

xi

xii PREFACE

Caratheodory, Liliental and Darboux. A. Voss defined the notions of lines of cur-vature, asymptotic lines, and he also considered some special classes of vector fieldsand mechanical systems with non-holonomic constraints. Liliental studied the ana-logue of geodesic lines - the shortest lines orthogonal to the field. Rogers gave theinterpretation of the non-holonomicity value in terms of mean geodesic torsion.Later on, the study of the geometry of vector fields was developed in papers by D.Sintsov, Schouten, van Kampen, Rogers, J. Blank, V. Vagner, G. Vransceanu andothers. D. Sintsov produced the generalization of the Beltrami-Enneper formula,stated the relations between the curvatures of the first and second kind, consideredthe lines of curvature, and introduced the notion of the indicatrix of geodesic torsion,and his works have been published as a monograph [16]. J. Blank considered vectorfields from the projective point of view.

The papers by Vaguer on non-holonomic geometry were awarded the Lobachevskiprize. Vaguer introduced the notion of parallel transport with respect to a vectorfield, and generalized the Gauss-Bonnet formula to the case of a strictly non-holonomic vector field. In this monograph we present these results in simple lan-guage, avoiding the difficult methods in the Vaguer paper. Another generalization ofthis formula, based on the concept of spherical image, which works for the case of anarbitrary vector field in n-dimensional Euclidean space, was found by the author. Inthe preparation of this book, in addition to the papers of other mathematicians, theauthor's results on 'divergent' properties of symmetric functions of principal cur-vatures, on the behavior of streamlines, and on complex non-holonomicity havebeen included. We also present the author's result on mutual restrictions of thefundamental field invariants and the size of the domain of definition of the field.These results belong to 'geometry in the large'. One can find their origins in papersby N. Efimov on the geometry of surfaces of negative curvature.

An important area of applications of non-holonomic geometry is the dynamics ofmechanical systems with non-holonomic constraints. Usually, these constraints arisein the description of the rolling of a rigid body over the surface of another body,taking friction into account. A. Voss, S. Chaplygin and other scientists generalizedthe Lagrange system of equations to the case of non-holonomic constraints.S. Chaplygin studied the rolling of a ball-like body over the plane in detail. From thegeometrical viewpoint, mechanical systems with non-holonomic constraints havebeen considered by V. Vaguer and G. Vransceanu. Currently, systems with non-holonomic constraints are studied intensively in mechanics ([51], [96], [971).

The second field of applications is the geometry of velocity fields of fluid flow.From the literature on this subject we highlight the papers by S. Bushgens, whichpresent necessary and sufficient conditions for the vector field to be the field ofvelocity directions of the stationary flow of an ideal incompressible fluid.

Vector fields occur in a natural manner also in general relativity. Recently it hasbecome clear that vector fields of constant length are used in the description of liquidcrystals and ferromagnets. It is interesting to note that the Hopf invariant, involvedin the description of vector fields defined in space with the point at infinity included,is also used by physicists, although they use the term 'topological charge' [76].

PREFACE xm

The contents of this book are divided naturally into two chapters. Chapter I isdevoted to vector fields in three-dimensional Euclidean space, to triply-orthogonalsystems and to applications in mechanics. In Chapter 2 vector fields, Pfaffian formsand systems in n-dimensional space, foliations and their Godbillon-Vey invariantare considered. We also present the connection stated by Whitehead between theintegrated non-holonomicity value and the Hopf invariant.

The study of geometrical objects in n-dimensional Euclidean space is an excitingproblem. The notions introduced here have descriptive interpretations and the the-ory can be demonstrated easily by concrete examples of vector fields. The methodsinclude no complicated or, more precisely, awkward mathematical tools. Never-theless, there are some problems where complexity is essential. Note that someimportant applied investigations lead to vector fields in E".

I would like to convey my appreciation to V. M. Bykov for his useful andencouraging positive remarks.

I would also like to express my special thanks to Professor A. Yampol'ski for thetranslation and to M. Goncharenko for her technical assistance.

1 Vector Fields in Three-DimensionalEuclidean Space

1.1 The Noo-Holoaomicity Value of a Vector Field

Let us consider a family of surfaces defined in some domain G of three-dimensionalEuclidean space P. The family of surfaces can be determined in different ways. Forinstance, we can regard the family surface as a level surface 4 (x,, x2. xz) = const ofsome function t(x1, x3) which depends on three arguments x1, x2, x3. We supposethat only one surface of the family passes through each point M in the domain G, i.e.the family under consideration is regular. Then at each point M in G there is a well-defined unit vector a - the normal vector of the surface which passes through thatpoint. Therefore, the vector function n = a(M) is defined in the domain G. Letxl, x2, x3 be Cartesian coordinates in E' and et, e2, e3 be unit basis vectors. Then wecan regard the vector field a as the vector function of the parameters x1. x2, x3. Thismeans that each component (; of a is a function of three Cartesian coordinatesxi, x2. x3 of a point M in G:

f, =t;(xi,x2,x3), i = 1,2,3.

The unit vector field a. being orthogonal to the family of surfaces. is not arbitrary.Construct another vector field curl is by means of the vector field as follows:

curl a=el(43,. -t2r,)+e2(41 -}e3(S2r, - (ir)-We may represent it in symbolic form as

ei e2 e3

I Lcurl a = .L

1

2 THE GEOMETRY OF VECTOR FIELDS

We shall need the following formulas in the sequel. If v, a and b are some vectorfields and A is a function of x, then

curl Av = A curl v + [grad A, v],

curl [ab] = a div b - b div a + Vba - V,b,(1)

where Vba means the derivative of a via b, i.e. the vector of components

The following theorem holds.

Theorem (Jacobi) If (n, curl n) = 0 at each point then there is a family of surfacesorthogonal to the vector field n and vice versa.

Let (n, curl n) = 0 in a domain G. Show that there is a family of surfacesf (xt , x2i X3) = const orthogonal to the vector field n.

Consider at some point Mo E G of coordinates (x°, x2, x3) the set of directionsdr = {dxl, dx2, dx3} each of which is orthogonal to u, i.e. those directions whichsatisfy the equation

CIdx1+e2dx2+C3dx3=0. (2)

We may assume that the system of coordinates in E3 has been chosen in such a waythat at Mo and, as a consequence, in some its neighborhood 3 # 0. We can rewriteequation (2) as follows:

6 tidx3 = - dx1 - = dx2. (3)

We try to find a function x3 = z(x1, x2) which satisfies equation (3) and such thatz(x°, x02) = x°. In this case dx3 is the differential of that function and the coefficientsof dx1 and dx2 in (3) are the partial derivatives of the function z(x1 i x2), i.e. theequation (3) is equivalent to the following system:

8z _ 6 8z _ f2 4

- '

_.

( )

8x1 S3 8x2 6The right-hand sides of these equations depend on X1, x2 and z. System (4) is solvableif the compatibility condition

(5)

is satisfied, where one ought to find the derivatives taking into account (4). Wehave

C O. =

(6x2C3 - fix,2b 3

VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE 3

Here we replaced zx, with -W6 by (4). Interchanging the roles of indices 1 and 2,we get

1 3x I = (6x i 3 - 2c, 66 - t3xg3t2 + 1;3xg1 t2) f 3 3.

The difference of expressions above produces the compatibility condition

1 3Jx, - 1 c) = 3x, - Ix,) + t3(tIx2 - t2x1)k. 2

\ _ -(a,curl n)f32 = 0. (6)

Since the latter equation is satisfied by hypothesis, there is a functionz = z(xl , x2, X03), where x3 is a fixed parameter, which satisfies the system (4) andsuch that z(x°, xZ, x3) = x3. This function defines the surface of the position vectorr = {xI, x2, z(xI, x2i X03)), where x3 is fixed. Due to the condition z(x°, x2, x3) = x3this surface passes through the point Mo. Varying x3 we obtain a family of surfacessatisfying equation (2) at each point. That equation means that the tangent plane ofthe surface is orthogonal to a.

Conversely, if the family of surfaces orthogonal to a exists then equation (2) issatisfied and for the surface which passes through the point Mo it is possible toconsider one of parameters, say x3, as a function of x, and x2. Then system (4) willbe satisfied and the compatibility condition (5) will be fulfilled. Due to (6) this meansthat (a, curl a) = 0.

So, the vector field a which is orthogonal to the family of surfaces is special and iscalled holonomic. The field o holonomicity condition can be represented in terms ofnon-collinear vector fields a and b which are orthogonal to a. Let [a, b] = )n, where

[a, b] 10. Applying formula (1), we can write

(n, curl n) = 0([a, b], curl A-1 [a, b]) = A-2([a, b], curl [a, b])\ - 2 ,adivb-bdiva+Vba-V.b)

= )C' (n, Vba - V.b).

From this it follows that (a, curl a) = 0 if and only if the vector Vba - V.b is or-thogonal to n. The vector Vba - V.b is called the Poisson bracket and is denoted byV(a, b).

It is natural to expand the class of vector fields under consideration, avoiding therequirement (n, curl n) = 0. We shall call the value p = (a, curl o) the value of non-holonomicity of a vector field a. It is independent of the choice of direction of o. Sincea is a unit vector field, the value p has some geometrical sense. We give two geo-metrical interpretations of the non-holonomicity value p.

1. The Vagner interpretation. Let Mo be some point in a domain G where thevector field n is defined. Draw a plane through Mo which is orthogonal to a(Mo) anda circle L in this plane such that L passes through Mo. Set that positive orientation inL to be defined by n(Mo). The field n is defined in some neighborhood of Mo

4 THE GEOMETRY Of VECTOR FIELDS

FIGURE 1

including the points of L. Draw a straight line through each point M in L in thedirection of n(M). We obtain some ruled surface. Starting from Mo, draw a curve inthis surface which is orthogonal to the elements (see Fig. 1).

Consider one coil of that curve which is one-to-one projectable with the elementsonto the positively oriented L. Denote it by V. The starting point of L' is at Mo. Thecurve L' meets the straight line element through Mo at another point MI. Denote by lthe length of M I Mo and by a the area of disk bounded by L. Then the value of non-holonomicity of a vector field n at Mo is equal to the limit of ratio -1/a when L tendsto Mo:

(n, curl n) E,yo = - lim 1.1. MO a

(7)

We find the limit of 1/a under assumption that the radius of circle L tends to zero.Let M be an arbitrary point in L and s be the arc length of L between Mo and MI inthe positive direction (see Fig. 1). Let M' be a point in L' projecting onto M in L.Denote the distance between M and M' by u(s). Let S be the length of the circle L.Show that there is a positive constant C such that for all circles which pass throughMo and lie in some neighborhood of Mo the following inequality holds:

Iu(s)I < S`C. (8)

To prove this, represent a position vector of L' in the form

r(s) = rI (s) + u(s) n(s),

where rI (s) is a position vector of L and n(s) is the field vector at the points of L. Atthe points of L' the equality

(r,n(s)) =0

must be satisfied. We have

r'=r'I+u'n+un'.

VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDCAN SPACE 5

Note that r, is a unit tangent vector of L. Multiplying by n, we get

u'=-(ri,o)=-cos

where V is the angle between n and the tangent to L. So,

u(s) f cos Ip ds.

0

Denote by a the angle between n(M) and the plane where L is located. The angle V isgreater than a and, as a consequence, I cos VI < cos a. The angle a is determined bythe vector field n. We can regard a as a function of coordinates .x,y in the planewhich contains L. Moreover, a(O, 0) = a/2. Then we are able to write an expression

cos a = (cos a), Ax + (cos a),.Ay + o(Ari ),

where o(Arl) is the infinitesimal of a higher order than Orl = &x2 + oy2. Fromthis we get

I cos V1 < l cos al < C, Ar, < C2S.

So,

lu(s)I <_ f Icosal ds < 2252 = CS2,0

where C, C1 and C2 are constants.Let L" be a closed curve formed with L' and the intercept M1 M0. Consider the

circulation of a along the closed curves L and L". They enclose a ruled surface F.Denote its area by o'. By Stokes' theorem we have

I (a, d r) - f(ndr) = Jf(curln,zi)dor,L V F

where v is the normal to F, do is the area element. The following inequality holds:

f f(curl n, v) do < max Icurl nIo'.6

As the vector field n is regular, the derivatives of n and, as a consequence, Icurl al arebounded from above in G with some number. The area of ruled surface bounded byL and L" can be estimated from above using (8):

a' = f l u(s)I < S3C3.

L


Therefore, the difference of circulations of n along L and L" is of the order S3:

f(n,dr) -J

(n,dr)L e

< C4S3,

where C4 = const. Let fi be a disk bounded by L. The area of is a = S1/41r. Thatis why

lims-o

Hence

J(ndr) - J(n,dr)3

o' <47rC4sumoS=0.

-Oo f (n,dr).L L"

(9)

Now find the circulation of a along V. Write an expansion of the vector function aalong each straight line element of the ruled surface

n(M') = n(M) +F(M)u(s) +o(u(s)), (10)

where 8/8r is the derivative of the vector field n at M in the direction of n(M),o(u(s)) is the infinitesimal of a higher order than the infinitesimal u(s). We can write

J(nidr) = / (n(M'),dr)+ r (a,dr).L" L' MI M0

In the first integral on the right we may replace n(M') by (10). As n(M) is ortho-gonal to d r along L' this integral has an upper bound

CI lu(s)Ids<CS3.L

That is why its quotient to a tends to zero, when S 0. Analogously,

J(n,dr) -- -1, (11)

M, Mo

moreover, this integral differs from -1 by an infinitesimal of the order S3.The circulation of n along the circle L can also be evaluated by means of a double

integral over the disk 0. We have

fJ(curlnn(Mo))da = f(ndr);t: a(n,curl n)IM0.


FIGURE 2

Divide both sides by c and pass to the limit, when S - 0. Taking into account (9)and (11) we obtain (7).

2. The second interpretation of the non-holonomicity value geometrically answersthe question of why it is impossible to construct the surface orthogonal to n in thenon-holonomic case. Construct the vector fields a and b which are orthogonal toeach other and a in some neighborhood of a point Mo. To do this we can take, forinstance, a constant vector field f which is not collinear to n(Mo) and seta = [n, £) / I [n, t;) . Then b = [n, a). Draw a curve ! through Mo which has a as atangent vector field. Then draw a curve m through Mo which has b as a tangentvector field. Starting from the points of m, emit the lines tangentially to the vectorsof the field a. They form some surface 4D (see Fig. 2). Denote by n(M) a unit normalof that surface at its arbitrary point M. The n(M) is orthogonal to a(M), hence it islocated in the plane spanned by n(M) and b(M). Denote by ip the angle betweenn(M) and n(M). Then at Mo and the other points of m

dw-(a, curl n), (12)

where d/ds is the derivative with respect to the arc length parameter of the curvetangential to the field a.

To prove (12) consider the decomposition

ii(M) = cos cpn(M) + sin Spb(M ).

Prolongate the vector field ii(M) into some neighborhood of the surface -0. Since thesurface which is orthogonal to n exists by construction, then (n, curl n) = 0.

We can write

curl n = cos W curl n + sin cp curl b + in, grad cp) sin p - [b, grad rp) cos p.

Therefore

(n, curl n) = cos 2 cp(n, curl n) + sin 2 cp(b, curl b)

- (n, b, grad cp) + cos W sin W ((b, curl a) + (n, curl b)) = 0.

S THE GEOMETRY OF VECTOR FIELDS

Since [n, b] = -a, then from the latter equation we find

(n, curl n) + sin'- p { (n, curl n) - (b, curl b) }

- cos cp sin y { (b, curl a) + (n, curl b) }. (13)

At each point of m the normal 11(M) of the surface 4' coincides with n. Therefore atthe points of that curve p = 0. From (13) it follows that at the points of m we have (12).

Formula (12) shows that for a non-holonomic vector field n the surface which isorthogonal to n does not exist. Indeed, let us suppose that at Mo and, as a con-sequence, in some part of its neighborhood (n, curl n) 0. A surface which is or-thogonal to a and passes through Mo must contain the curves 1 and nz because theyare orthogonal to n. Hence, it must contain curves tangential to a emitted from thepoints of m because the field a is orthogonal to n, i.e. that surface must coincide with4' and its normal field n must coincide with n. But from (12) it follows that -A 0 insome neighborhood of Mo. Therefore, in some neighborhood of Mo we have iP 0 0,i.e. the normal field of it does not coincide with the given field n.

At the end of this section we give an example of a vector field with a constant non-zero non-holonomicity value. Set

1 = r cos gyp, {_ = r sin cp, ; = 1 - r'-.

where r is positive constant, tp = V(x,). We have

curla = sin WW ,,0}.

Hence,

(n,curln) _

If p = axz + b with constant a and b then the value of non-holonomicity(n, curl n) = -r=a = const.

1.2 Normal Curvature of a Vector Field and Principal Normal Curvatures ofthe First Kind

The surface behavior about a point Mo on it can be described with the help of thesurface normal curvature k which can be found for any surface tangent direction r.Geometrically, k is interpreted as a curvature of surface normal section at MO in thedirection of r. Analytically, k can be found as a ratio of the surface second fun-damental form 11= (d 2r, u) to the first one I = dr2:

_ (d2r,n)k" (dr,dr)'

where r is a position vector of the surface, n is a unit normal at Mo. Since n is a vectorof unit length, it is possible to rewrite the expression above as

k - - (d r, d n)

(dr,dr)


00

n(VJ

dr

n#Q

FIGURE 3

Suppose now that an arbitrary regular unit vector field n is defined in a domain G.We shall introduce a curvature of the field (i.e. a function which makes n differentfrom a constant vector field) at Mo. In the holonomic case the curvature will coincidewith the normal curvature of the surface which is orthogonal to the field.

Draw a plane through Mo which is orthogonal to n(Mo). Let MOM = dr be a shiftin this plane. Draw another plane through n(Mo) and dr. Let ii(M) be a projectionof n(M) into the latter plane (see Fig. 3).

Denote by p the angle between n(Mo) and n(M ). The normal curvature kn(dr) (orsimply kn) of a vector field n in the direction dr is the limit of the ratio -cp/ldrl whenM -+ Mo, i.e.

kn(dr) = lim-,p

(1)M-wo

ldrl.

If n is a constant vector field then V = 0 at any point M and, as a consequence,kn=0.

Prove that in the holonomic case we come to the definition of the surface normalcurvature. As a sine of o is equal to the cosine of an angle between ii(M) and dr, wesee that

sin V - (n(M ), d r) (2)lolldrl

As the infinitesimals V and sin cp are of the same order, it is possible to replace Vwith sin V in (1). As ii(M) is the projection of n(M) into the plane of vectors n(Mo)and dr,

(ii(M ), d r) = (n(M ), d r).

In the neighborhood of Mo it is possible to produce the following expression:

n(M) = n(Mo) + do + o(dr),

where lo(dr)l is the infinitesimal of a higher order than ldrl. Therefore,

(o(M ), d r) = (dn, dr) + (o(dr), d r). (3)


Note that Iiil -+ 1 when M Mo. Using (2) and (3) we write

lire-(n(M,dr) _ - (dd,dr)

M-Mo

This expression coincides with the expression of the surface normal curvature.The normal curvature of a vector field depends on the direction d r. If we rotate d r

in the plane which is orthogonal to n(Mo) then it varies, in general, and reaches itsextremal values. These extremal normal curvatures are called the principal curvaturesof the fast kind. The corresponding directions are called the principal directions ofthe first kind. In analogy with the theory of surfaces, the curves tangent to theseprincipal directions are called the lines of curvature of the first kind. Find theequations on the principal directions and lines of curvature. To do this write theexpression of k in terms of components oft n:

k,, _ - [Slxidx, +(f1X: +f2,,)dx1 dx2 +f2,_dx;

+ (6x, + t3x,) dxl dx3 + (6 + 6,_) dx2 dx3

(d. Y; +dxz +dx3)-1.

Choose the coordinate axes in E3 in such a way that t j = f2 = 0 at Mo. Since

f3 = J1 - tf-- f21 fax, = 0 at Mo for i = 1, 2, 3. The value of k does not depend on

the length of dr. So, set jdri = 1. We shall find the extremal values of k providedthat dr is in the plane orthogonal n(Mo), i.e. dr = (dxl, dx2, 0). In these supposedconditions k takes the form

t 6,q) C ,1+(flx2+dx1dx2+f2,.d4]. (4)

As dxi + dx2 = 1, we use the Lagrange multiplier method to find the extremalvalues. Taking dxl and dx2 as independent variables, we want to find the extremalvalues of the following function:

,,x,d 1+(flx2+f2,,)dxldx2+f2r,dx2+A(dxi+ z).

This leads us to the following system of equations with respect to the principaldirections:

+ \)dx1 +2 &r,dx2 = 0,(f lx,

flx, +f2n dx1+(f2 +A)dxz=0.2

,Z

The characteristic equation for A has the form

(5)

a2 +A1 IX,rf IX: +

2

The matrix of system (5) is symmetric, so that we have two real roots \1 and A2. Justas in the theory of surfaces, it is possible to prove that these roots are the principal

VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE I I

curvatures of the field and the principal directions of the first kind are mutuallyorthogonal. The mean curvature of the field is a half-sum of principal curvatures:

H =

2

(AI I (y lx, +42X2). (6)

The total curvature of the first kind is a product of principal curvatures:

6x, +6,, 21

2

A point Mo is called elliptic, parabolic or hyperbolic if at this pointKI > 0, KI = 0, KI < 0 respectively.

Let us find the equations of lines of curvature of the first kind. Rewrite system (5)in the form

4 + S2x'

2

Ixz dx2 + A dxl = 0,

dd2 +GIx2

2

br, dx2 + A dx2 = 0.(8)

Note that t;2,, - f Ix, is a third component of curl n. If we set (curl n), to be the i-thcomponent of n then we can write (8) as

2(d 1 +.\ dxI) + (curl n)3 dx2 - (curl n)2 dx3 = 0,

2(dk2 + A dx2) + (curl n) I dx3 - (curl n)3 dxI = 0,

or as

2(d{r+Adx,)+[dr,curln]t=0, a=1,2, (9)

where [ ]r means the i-th component of the vector product. System (9) shows that aprojection of

2(dn+Adr)+[dr,curl a] (10)

into the plane orthogonal to n is zero, i.e. the vector (10) is collinear with o.Therefore, the triple of vectors 2 da + [dr, curl a], dr and o is linearly dependent and,as a consequence, their mixed product is zero:

2(dn, dr, n) + ([dr, curl n], dr, n) = 0.

Using the formula [[a, b], c] = (a, c)b - a(b, c) and the orthogonality property of drand n we obtain the system of the equation which determines the lines of curvature ofthe first kind:

2(d n, d r, a) + (n, curl n)dr2 = 0,(n,dr) = 0.

As d n = EI nx, dxI, we can rewrite system (11) as3

Aij dxf dxj = 0,l,j=1


C1dx1 +6 dx2+{3dx3 =0,

where A;, are some coefficients depending on n and its derivatives. Take one of thecoordinates, say X3, as a parameter t of the curve. From the second equation we find

dx2 1 dX1dt 42

dt - 3

Substitute this into the first equation. Solving the latter quadratic equation withrespect to dx1 /d:, we find

drk ttdt Fk(Ci, k = 1, 2,

dx3= 1.

dt

The right-hand sides of the equations above depend on x1, x2, x3. Integrating thatsystem we can find the required lines of curvature.

The directions of principal normal curvatures of the first kind are mutually or-thogonal. Indeed, returning to system (5), note that the required directions are theeigenvectors of the symmetric matrix

2

From this the analog of the Euler theorem follows. Take the principal directions asthe coordinate axes. Set dx1 Ids = cos a, dx2/ds = sin a. From the assumptiondx3 = 0 and (4) it follows that

k = cos2a+(Clx, +C2x,) cos a sin

From system (5), we find

(Cl., + b2r,)(dx` - drj) + E2x,) dx1 dx2 = 0.

Due to the choice of coordinate axes the solutions of the latter equations are (&x1, 0)and (0, dx2), hence i,, +{2, = 0. Setting a = 0, ?r/2 we find that and -C ,are the principal curvatures k1 and k2 of the first kind.

Therefore, the Euler formula for the case of a vector field has the same form as inthe theory of surfaces:

k =klcos2a+k2sin2a.

1.3 The Streamline of Vector Field

The streamline of vector field n is a line which is tangential to n at each of its points(see Fig. 4). Let be the components of the vector field n with respect to the


FIGURE 4

Cartesian system of coordinates (xi, x2, x3). Then the streamline must satisfy thesystem of differential equations

dxiTds

i= 1,2,3.

We introduce the Hamilton formula for the vector of curvature of the streamline of agiven field n. The required vector is directed along the streamline principal normaland its length is equal to the streamline curvature. We denote it by k. The Hamiltonformula has the following form:

k = -[n,curln].

To prove this it is sufficient to show, for instance, that the first components of k and- [n, curl o] coincide.

If we denote the differentiation with respect to the n streamline arc length para-meter by d/ds then

k the first component ofis equal to

r2 2+6r,y3-S2r,S2-S3x,61

= bIx,yr - 20 t22 + C32)

= SIx,Si,

which completes the proof. From this it follows that for the case of a field ofdirections with congruent straight lines we have -[n, curl n] = 0, i.e. curls is collinearwith n.

Let us take a family of level surfaces U(xl, x2, x3) = coast. Let n be the field ofnormals. Find the o streamline curvature vector k in this case, i.e. the orthogonal


trajectories of surfaces family. Let n = {a} be the unit tangent vector of thestreamline. Set H = 1grad 1-1. Then

dx;8UH, i 1,2,3.

d_ax;

The first component of k is equal to

d'x, 8

CH

8U1 dxj 8H 8U OUH+ 82U 8UH2= ax; ax, J ax, ax, 9X, ax,axj ax;

Note that 1O H3. Therefore, the following is true:

d 2x, _ Hr, + (grad H, n) dx,ds' H H ds

Analogous calculations with respect to the second and third components of k showthat

k = -grad In H + n(grad In H, n).

The expression above was stated by Darboux.

1.4 The Straightest and the Shortest Lines

The geodesic lines in a surface possess two properties of straight lines in a plane: theyare the straightest lines (that is, their geodesic curvature is equal to zero) and shortestlocal fines simultaneously (that is, the geodesic line has the smallest length among allcurves joining two sufficiently close points PI and P2).

In the case of a non-holonomic vector field we can determine the straightest and theshortest lines among the curves which are orthogonal to the field.

Let the field n be orthogonal to the curve I'. Let r = r(s) be a position vector of 1',where s is the arc length parameter. The geodesic curvature I /pg of r with respect tothe field n is the length of the projection of the curvature vector of 1', i.e. of rs,, intothe plane orthogonal to n. We set it positive if the triple (r,, a, r.) is positivelyoriented and negative in the opposite case, i.e. we set

I =_ (0, r,,r,)Pg

Set r, = e. Suppose that the unit vector field e is defined in a three-dimensionaldomain. Then curl e is also defined and rys = -[e, curl ej by Hamilton's formula. Thegeodesic curvature with respect to the field n of lines tangent to e takes the form

I = -(n, curl e).Pg


In the case of the straightest line the geodesic curvature with respect to a is equal tozero. Therefore, that curve satisfies the equation

(a, r.,, ru) = 0. (1)

Equation (1) means that

r...=pr,+An.

As s is the arc length parameter, (r 0, i.e. J' = 0. Hence, the straightest linecurvature vector is directed along a:

rn = An.

The coefficient X is equal to the normal curvature of the field in the direction r i.e..1= (r,,, a) = -(r a). Therefore the vector equation of the straightest line has theform

This is the differential equation of second order with respect to the vector functionr(s) parametrized with some parameter s. If the curve is orthogonal to a at the initialpoint Po then it stays orthogonal to a for all s. Indeed, the function (r,, a) satisfies thefollowing equation:

(r a), = (r,,, a) + (r q,) = 0,

i.e. (r a) = const. As (r n) = 0 at Po then (r a) __ 0. If r(s) is a solution of (2) thentake the initial condition for r, such that Cr,I = 1. Then 1r,1 =_ 1 and, as a con-sequence, s is the are length parameter. It is sufficient to show that r; = const. Wehave

(r,2), = 2(r r1,) = 2(r n).1= 0.

The solution of (2) is determined by the initial conditions r(0), r,(0).Thus, we can draw a unique straightest line in the given direction r,(0) which is

orthogonal to the field through any point of the domain G. We call the straightestline the geodesic.

Turn now to the consideration of shortest lines.Let the points Po and PI be joined with the curves orthogonal to the field a. We call

those curves admissible. Among all admissible curves we shall find a curve with thesmallest length among all nearly admissible ones.

We consider a position vector r of the shortest line L as a function of the arc lengthparameters which varies from 0 to 1. We vary the curve L in such a way that it willstay admissible and the curve coordinate variations 6x1, 8x2, 6x3 turn into zero at


PO and PI . Take the parameter s as a parameter in the varied curve. Let dr,/ds be atangent vector of the varied curve. Then the following equation must be satisfied:

dr,- 0. (3)ds

The shortest line L gives the extremal value of the integral

p;

ds = jjdrj + dr; + dx-3

0 L

among all curves which satisfy (3). Applying the Lagrange method, we find theconditional extremum or, equivalently, the curve which gives the ordinary extremumof the following integral:

rr dx

f I + P(sk ds,

0

where W(s) is an unknown function, provided that & = 0 at the points of therequired curve L.

Consider the variation of , in an arbitrary direction or = (6x1, 6x2i 6x3). Weshall consider the admissible curves as lying on the surface r = r(s, a) provided thatfor any fixed a we get the admissible curve. Then

8 ax, aO, ax, 6'x,as ' as = as as +' asaa'

(4)

Set -curl n = {l,}:

11 = 6a, - 6.r 12 = Sax, - 6x,, 13 = 6x, - far,,

or using briefer notation

ft axj =1Q,-;0axjax,where e'j' is the Kronecker symbol. Then we can rewrite (4) as

8 ax, ai;, axj OX, 8=x,

as ' as axj ba as + asaa

_ 1"?j(' + axj) 6Xj ax + ; a''xit\ ax) 6a aS asaa

+6xj ax, + 6xj

asaa 6a as ds 6a

a'x, d r br do 6r_ ' asaa -

(curla, d, T.) + (ds' 6a) (5)


Using (3) and (5), we get

d r br (d n brT = J dr + f v(s) aa - (curl , aa b }ds

L E(t) L

_ r r 8x; dip(s)(curl

d r br bx; d2x;t`as n,

ds' ba)` ba ds2 } ds.L 111

The condition bl/ba = 0 and the expression above imply

d'r +0 + cp(s) [Curl n,] = 0.

=0.

Our next aim is to exclude the function ip(s). Take the mutually orthogonal unitvectors n, 2, {n, 2] as the basis along L. Decompose curl o and A, with respect tothat basis. We have

curl n = (n, curl n)n + a l n,dJ+ b.

We can represent the curvature vector of the shortest line L as

d2r 1 r dr] 1

ds2= RI n, +

RzLa.

Substituting (7) and (8) into (6), we get

n,- n) n,- n,- ,-R2 L J Rz ds L 2J L L ZJ ds

Setting the coefficients of n and [n, !M equal to zero, we obtain

RI+ sp(s) (a, curl n) = 0,

(6)

(7)

(8)

(9)

R2+Jspa=0. (10)

In the case of a holonomic vector field we have (a, curl a) = 0 and the equation of ageodesic line obtains the form

1 _=_(d2! dr _R, ds2' ds'

n) 0,

i.e. in this case the geodesic curvature along the shortest tine is zero.Suppose that (n, curl a) # 0 along some part of L. Then the function ip(s) can be

excluded from these two equations. Resolving (9) with respect to W(s) and sub-stituting into (10), we obtain

I d I aR2 ds R,(n,-Curl n) RI(n,curln)


From (5) and (6) we obtain the expression for 1/R2 and a. Namely

1 d2r _ A doR2 = (ds2 °) - (ds' ds)'

i.e. I /R2 is the field normal curvature in the direction of L. Denote it by k,,. Also,

a= (curln, n, ) ds) ,

i.e. -a with the opposite sign is equal to the projection of the field streamlinecurvature vector onto the tangent of L. So, the equation of the shortest line has theform

dod + + °) - d r d n (,,, d r)(i0.ds (n, curl n) (ds' ds) + ds (n, curl n)

Substituting W = into (6), we obtain the vector form for the equation ofthe shortest line:

d2r d L+ °)r drds2

+ 1n(n, curl n) + (n, curl n)

I curl n, ds] = 0. (12)

This is the third-order differential equation with respect to the position vector r(s).Write the projections of the right and left sides of the latter equation with respect

to movable axes adjoint to the field a. Let a and b be the unit mutually orthogonalvector fields such that [a, b] = a. Set

drz = cos as + sin ab.

Therefore

2

dd-s2 = (- sin as + cos ab) + cos a + sin a .

We can represent the derivatives of a and b with respect to the arc length parameterin terms of derivatives with respect to the Cartesian coordinates xi:

da dxi d b drci

ds = a'`,

'T'

'is = b, ds .

As a and b are of unit length, then ax, is orthogonal to a and bY, is orthogonal to b.Consider the decompositions

a', = Lib + Tin,k, = Nia + Min.


As a and b are mutually orthogonal, then Li = -Ni. Find ( , a). We have

d2r dr da 2 da \2 db(2, 1, n)

= -+cos a( ,a, a)+sin al ,b,a

da_ -ds

+ Li(cos aa; + sin abi) (13)

where a; and b; are Cartesian components of vector fields a and b respectively.Multiply (12) by n scalarly. We obtain

d--r d(d2r

' j'n) ( ' 'n) dr(dsZ , al + ds (n, curl a) + (a, curl a)

(a' curl a, 1 = 0. (14)

We have

/d2

Z, n) =cosa( , n l +sina(, a\ Ill

= cost aA + cos or sin aB + sin 222 aC, (15)

n, curl n, ds = (a, curl n, cos as + sin ab)

= cos aP + sin aQ, (16)

where A, B, C, P and Q are some functions of x;. Using (13), (15) and (16), we canrepresent equation (14) as

d L;(cos aa; + sin ab1))ds (n, curl n)

+ cost aA + cos a sin aB + sin2 ac

cos aP + sin aQ da(n, curl a) ds + L;(cos as; + sin ab1) = 0.

Consider the projection of equation (12) onto a. We have

(dSZ, a) = -sinaia+sina a)

da dx;_ -sinads

+L;ds

Comparing with (13), we get

dzr \\ dzr drCdsZ,al=sina(,

(17)

(18)


Next we have

(a, curl a, ) = (a curl n, cos as + sin ab)

_ -sin a(n, curl n). (19)

Taking into account (18) and (19), we conclude that the equation obtained as aprojection of (12) onto a is satisfied identically. The same is true for the projection of(12) onto b. Thus, only equation (17) is left which can be rewritten in general form

d da/ds ad-= F x1.x2 x; a (20)ds (n, curl n)

i , ,ds

To this equation we associate the equation

dr= cos as + sin ab. (21)

ds

The equations (20), (21) form a system of ordinary differential equations with respectto the functions x-(s), i = 1, 2, 3, a = a(s). Let Po be a point of the coordinates x,(0)such that (n, curl a) 0 0 at this point. Suppose that this point corresponds tothe value s = 0. Give the initial values to the unknown functions at s = 0: x1(0),i = 1, 2,3 a(0), a (0). The choice of xi(0) means the choice of the starting point of theshortest line, the choice of a(0) gives the direction of the shortest line at Po. Thus, thefollowing theorem holds.

Theorem If (a, curl n) # 0 at some point PO then a bundle of shortest lines withrespect to the given direction passes through this point.

Give an example of a vector field n with a simple structure but non-constant withrespect to which the equation (17) is integrable in quadratures. Suppose the field n isparallel to the plane of xI, x2, i.e. it has the following components:

t1 = cos cp, 6 = sin cp, ; = 0.

For the sake of simplicity, we suppose that cp = cx3 + d, where c, d are constants.Then we may set

a = e;, b = - sin (pet + cos;pe2.

We see that

curl a = { -c cos gyp, -c sin gyp, 0} = -c n.

Therefore, (n, curl n) _ -c, [n, curl n[ = 0. Since a is a constant vector, Li = T, = 0,

d2r dr da(ds2' ds' a) ds.

Let us find ( d-r,n) for this case. Differentiating b, we find

db dip dYzds= -n =-ncd.


From the equation

dr= cos as + sin ab

ds

we find dx /ds = cos a. Hence, d b/ds = -nc cos a. We rewrite equation (15) as

d2r, s - sin a cos ac.

dS` ) -Therefore, equation(14) of the shortest line has the form

d2a_ -c- sin a cos a.

Suppose that da/ds 34 0. Multiplying by 2da/ds and integrating, we find

dal2Cds/

=c2cos2a+c1,

where cl is a constant. We find the function a(s) from the following equation:r das=J

c2 cos2 a + cl -

To find a position vector of the shortest line we use the representation of d r/ds via aand b, which in coordinate form is

_ - sin a sin gyp, d-2 = sin a cos gyp,- = cos a, (22)

where the dependence of W from s and a is given by

c cos adadcp= -ccosads= -CcoS2a+cl

Therefore sin (cp + c2) = -c sin a/ c'- -c1. One of the solutions of (21) is evident.This is a family of straight lines parallel to the x3 axis.

FIGURE 5


As in each plane of X3 = const the field n is constant, the other solution is a familyof straight lines in these planes which are orthogonal to a (see Fig. 5).

1.5 The Total Curvature of the Second Kind

In this section we introduce another notion of the Gaussian curvature of a vectorfield analogous to the Gaussian curvature of a surface. Consider at some point Mo asmall piece of plane orthogonal to n (Mo). Take some neighborhood of Mo in thisplane and map it into the unit sphere S2 by means of a vector field n (see Fig. 6).

The total curvature K of the second kind is the limit of the quotient of thespherical image area and the area of the original when the original tends to Mo. If wedenote the areas of the original and of the spherical image by AS and Eo respect-ively then

K=lim SQ

Let

.

us find some analytic expressions for K. Suppose, at first, that o is holonomic.Let (u, v) be the curvilinear coordinates in the surface orthogonal to n and x(u, v) bea position vector of that surface, such that x., x,., n form the positively orientedtriple. Denote by W du dv the area element of the surface. Then

_ (n,,, n,., n) _ I (0X1 axe ax1 ax2K

W W (n I , n, n) au av av au(7X2 0X3 ax2 aX3

+ (nYj, nt,, n) au av aV au

aX3 ax1 aX3 aXI+ (n.,,, nY,, n) au av - av au

Note that n can be represented as

a=W

[y-U, xV}

FIGURE 6


Introduce a vector P setting

P= {(nx2,ax,.n),(nx31 nx,r0),(ari,nx2+0}

It is easy to show that the definition of P is independent of the choice of Cartesiancoordinates, i.e. in a coordinate change the components of P behave as vectorcomponents. So, it is possible to write

K = (P, n). (1)

Now represent K by means of components of the vector which is collinear with n butis not in general of unit length. Suppose that such a vector is A = {A;} = An, whereA = (A2 + A2 + A3)1I2. Then evidently

P = \-3{(Ax2A13A), (AX,AX,A), (Ax,A.,A)}.

Therefore, from (1) we see that

K = J1-4 { (Ax2 Ax3A)A l + (Ax, Ax, A)A2 + (Ax, Ax2A)A3 }

Al,,, Al., Al-v, Al

Azx, A2x2 A,.3 A2

A3x, A3x2 A3x3 A3

Al A2 A3 0

(A2 + A2 + A3)2(2)

Let us be given a family of surfaces fi(xl, x2, x3) = const, which are orthogonal to n.Then we may take grad 4' as A and obtain the Neumann formula

K=-

4'xix, 4'xix2 'XIX3 4'xi

4'x1 X2 4'x2-V2 4'X2-V3 4x21x,x3 4x2x3 4'x3X3 4'X3

4,x, 4'x2 4x3 0

(4' + tz2 + 423 )2 (3)

Turn now to the general case, when the field n may be non-holonomic. We shallshow that (1) holds in this case, too. At first, clarify the geometrical meaning of P.Let F be an arbitrary surface located in the domain of definition of the field n. Letthe parameters (a, ,B) in the surface induce the normal vector field v. We can map thesurface F into the unit sphere S 2 by putting each point M E F into correspondencewith the end-point of n(M) in S2. Denote this mapping by ,0. The area element da ofunit sphere can be represented as

do = (na, nf, n) da d,Q. (4)


Let xi = xi(a, 0) be the components of the surface F position vector with respect toCartesian coordinates. Then we have

ax2 aX3 aX2 ax;(°x °) as ap as aa

ax349TI ax3OXI

+ °) as as - as asaxI Ox-, axI ax2)}

+ 5c as - as aa' (5)

Let vi be the components of surface F unit normal, dS be the area element of F. From(4) and (5) we obtain

do = (P, v) dS. (6)

Consider a neighborhood of some point M in F. Let AS and oa be the areas of thatneighborhood and its spherical image respectively. Then (6) shows that

lim 03 = (P, v), (7)

where v is the normal to F at M. Thus, there is a vector field P depending on n suchthat for any unit vector v the transversal Jacobian of the mapping t' of the planeorthogonal to v into the unit sphere S2 is equal to (P, v). We call P the vector ofcurvature of the field n.

The total curvature K can be introduced in another way. Consider the Rodriguesequation do = -A dx. In general, this equation is insolvable for arbitrary field n,excluding the trivial case A = 0. To find A we have the equation

det Abiill = 0.

Since det II is the Jacobian of the mapping of a three-dimensional domain in E3into the unit two-dimensional sphere S2, det IIExfll = 0. Hence A = 0 is a root. Theother two roots satisfy the characteristic equation

A2 + 52x2 + 6xi) +2 I

S/Y, 'tY' I

0.i.!- I

We call the roots of the above equation the principal curvatures of the second kind. Inspite of the fact that the roots can be complex when the equation do = -A dx isinsolvable, the half-sum of roots is called, as usual, the mean curvature H and theproduct of roots is called the total curvature K:

1 1H=-2divn=2(AI+A2),

l 3 Six, tifY,K = AIA2 = -I I. (8)

2i,/=1 1x Yi


Indeed, since the expression for A1A2 is invariant with respect to the choice ofCartesian coordinates, we shall prove (8) with respect to the special system of co-ordinates. Directing the axis x3 along n(M), we get

Slxl SIx2AIAz = t

b2Y1 t2Y2 .

(9)

On the other hand, we have (1). Taking into account that

GV3 Cl 'r3 6x,P - {I C2Y2 2x1 I ' I 2x) 2Y1 ('I blxl bl.Y' I i f

&,,. SLY_ 1(

and the fact that the third component of P is its projection onto n, we find thatK = AIA2 for any vector field n.

Comparing (6), (7) of Section 1.2 with (8) and (9), we see that the mean curvaturesof the first and second kinds coincide, while the total curvatures of the first and secondkinds are different in the non-holonomic case: K - KI = 4

5)2

In the theory of surfaces it is well known that H2 > K. The generalization of thisinequality to the case of an arbitrary vector field is the following:

(10)

We may state (10) in the special system of coordinates constructed above withrespect to which {I = C2 = fax, = 0. In this case we have

H = - 2(Slxl + 6X02 (a, curl a) = 2Y1 - t, x2,

K I Ixl 6X2

e2Y1 tbq

H2 + (a, curl a)2 - K = r{I YI c2"212+ ( I 12 2 2Y2>0.

We call the curvatures H, K, the value of non-holonomicity (o, curl o) and the cur-vature of streamlines Ikl the basic invariants of the field. Since Ai = H ± H2 - K,from (10) we obtain the estimate of the imaginary part of the root of the char-acteristic equation

I ImA; I < < I (n, curl a) I

2

Let AI,A2 be the roots of equation do = -Adx, where the shift dx is orthogonalto n. The following theorem holds.

Theorem If at each point A 1 = A2 # 0 then the field n is holonomic and the surfacesorthogonal to a are spheres.


Denote by A the common value of A . It is real. Let -r, ,v be the fields orthogonal ton. As Ai = A2, an arbitrary direction which is orthogonal to n, satisfies the equation

do = -Adx.

We have

nx,rf = -Ar, nxyvj = -AV.

Differentiating the first equation via v and the second via r and then subtracting,we get

(nxiri)x,vj - (nxJVj)xfri = (-A1r),vj +

The tatter expression can be rewritten as

nx,(Tix1vj - vi,T) = V ,-v) - [[v, TI, grad 11].

If we multiply both sides scalarly by n then we obtain zero on the left-hand side dueto the fact that n is of unit length and as a consequence (n, nr,) = 0. Therefore

(n,V,r-Viv)=0.

In Section 1.1 we stated that this means holonomicity of the vector field n. Since thesurface which is orthogonal to n is totally umbilic, it is a sphere by the well-knowntheorem from the theory of surfaces.

As an exercise, find the total curvature of the second kind for the special field nwhose streamlines are the family of small circles in concentric spheres. Let 0 be acommon center of the spheres. Suppose that the centers of the circles - thestreamlines in the fixed sphere - are on the same diameter I. Also, we suppose thatthe circles are in the parallel planes orthogonal to I. At two opposite points of thesphere the circles degenerate into points - the singular points of the vector field n.The diameter I is a function of the sphere radius. We consider the simplest case, whenthe diameter is in the x, y-plane. Denote by a the angle which / makes with the axisOx. If x,y, z are the coordinates of some point on a sphere then a = a(x2 +y2 +z2).In each sphere of the family under consideration the circles make no intersectionswith each other. Moreover, since the spheres are concentric, the circles of differentspheres have no intersections. The tangent vector field a of the circles is regular at allpoints excluding the points of some curve which lies in the x, y-plane and is sym-metric with respect to O. Let us find the components of n = {Q as functions of thecoordinates. Let the diameter I coincide with the axis Ox. Then at a point withcoordinates x, y, z we have

6 = 0, 6 = -z y

z2 -+y2'6 = z2 + y2 .

Let 1 make an angle a with Ox. If we rotate the coordinate axes by angle a in sucha way that the axis Ox coincides with I then we obtain the previous situation.


FIGURE 7

The new coordinates u, v, w are related to the old coordinates by the followingformulas:

u= cos ax+sinay,v= -sin ax+cos ay,

Let 77j, 7r, 713 be the new components of the vector field n on the sphere whichcorresponds to the diameter I. They are of the form stated above:

w V

'lI=0, 172=- 173=W.2+1,2 K,2+V2

The old components of n have the following expression in terms of the new ones:

I =ri1 cosa-772 sin a,= 77I sina+7h cos a,

3=113.

Therefore, the representative of the vector field n at a point with coordinates x, y,has components

I = z sin a/A,z = -z cos a/A,3 = (- sin ax + cos ay)/A,


where a = a(x2 + y2 + z2) is a given function, A = z- +(- sin ax + cos ay)2. Tofind the total curvature K we use formula (2). Set A = (z sin a, -z cos a, - sin ax+ cos ay). Write the numerator of formula (2)

Z cos aaX Z cos aa), sina + Z cos aa- z sin az sin aax z sin aa). - cos a + z sin aa_ -z cos a

- sin a - ua.r cos a - uay -ua: v

z sin a -z cos a v 0

We multiply the elements of the first row by x/z, the second one by y/z and then addto the third one. We multiply the elements of the fourth row by 1 /z and add to thethird one. As a result we obtain a row of zeros. Therefore, K = 0. To find the meancurvature of the field we use the formulas

A div A (A, grad A)-2H = div n = div , =

A- A2

It is easy to see that div A = 0. Write the expression for the components of grad A:v

Ax = j (- sin a - uar),

At. _ v (cos a - ua,.),

A:(z-uva:).The direct calculation gives (A, grad A) = 0. Therefore, H =- 0. It is easy to calculatethe non-holonomicity of the vector field just constructed. We have

A2 (n, curl a) = (A, curl A) = -2z(x2 + y2 + z2)a'.

Therefore, if a' 0 and z 96 0 then n is non-holonomic. Thus, for the vector fieldunder consideration we have K = H =- 0 and (n, curl n) 96 0.

Now we give the geometric meaning of the field P. If P # 0 at some point M in thedomain of definition of the field n then a unique line n = const, which is tangent to P ateach of its points, passes through that point.

Indeed, suppose that dr = (dxl,dx2,dx3) is tangent to the line n = const.Choosing the special system of coordinates as above, we find that dr satisfies thefollowing equations at M:

Slx, dxI + Slx= dx2 + SIx, dX3 = 0,

6c, dxI + 2 2 dx2 + dx3 = 0.

From this we find that dr = AP, where A is some number. Therefore, the linen = const is tangent to P.

From the formula K = (n, P) it follows a simple way to construct the field n withzero total curvature: K = 0. Indeed, that condition means that n and P are mutuallyorthogonal. Suppose that P 54 0. Since n is constant along the streamline of P, these


streamlines are planar if K = 0. Therefore, to construct the required field it is suf-ficient to define an arbitrary congruence of planar curves and set n = const on eachcurve of the family. To do this, take a surface F2 of the position vector r(u, v). Thenat each point x E F2 take a unit vector n(u, v) and a vector a(x) in a plane which isorthogonal to n(x). We realize that selection with the help of a single functioncp(u, v). We emit a curve in the direction of a(x) which lies in the plane orthogonal ton(x). Those curves are defined by curvature k(u, v,s) as a function of the arc lengthparameter. So, the arbitrariness in the definition of n is in six functions of twoarguments and one function of three arguments.

In the case of P = 0, the image of any surface generated on the unit sphere by thevector field n is a line. Conversely, the inverse image of any point of that line is asurface.

To construct the field n satisfying P = 0 define a regular family of surfaces in E3and then define some constant value of n on each of the surfaces which variescontinuously on the family.

1.6 The Asymptotic Lines

We call a direction d x in a plane orthogonal to the field n asymptotic if the normalcurvature of the field in the direction dx is zero. Since the normal curvature of thefield is equal to the quotient of -(d n, d x) and d x2, the asymptotic direction isorthogonal to d n. Therefore, d x is proportional to [n, dn]. Thus, the asymptoticdirection can be found from the following vector equation:

In, dnj = p d x,

where µ is some numeric coefficient. Find the condition when the asymptoticdirection exists. To do this we use a special system of coordinates such thatI = 6 = 0, 6 = I at some fixed point Pa . Then at Po the system on the asymptoticdirection takes the form

(6xi + µ) dCXI - 52x2 dx2 = 0,

S I r, dx I + (S I r2 - µ) dx2 = 0,

dx3 = 0.

Therefore, the coefficient p satisfies the quadratic equation

µ - µ(f I r: - 2ri ) + VIx,{2rz - GIx2Gx,) = 0.

Rewrite this equation in terms of the field invariants:

µ2 - µ(n, curl n) + K = 0.

We have

µI2=(n, curl n) ± (n, curl n)2 - 4K

2


Note that -4K, = (n, curl n)2 - 4K. Therefore, the asymptotic direction exists if andonly if the total curvature of the first kind satisfies K, < 0. If K, < 0 then there aretwo asymptotic directions. If K, = 0 then either only one direction is asymptotic or anydirection orthogonal to n is asymptotic.

The line which is tangent to the asymptotic direction at each of its points is calledthe asymptotic line. Let us clarify the geometrical meaning of p. Let s be the arclength parameter of an asymptotic line. Then along the asymptotic line we have

dp d r d2r d2n

Z ds + pds2_ rn'

, n) = 0, provided that p 96 0. Therefore, the vector nMultiplying by n we have (`tTyr

is a binormal of the asymptotic line or differs from it by sign. Denote a torsion of theasymptotic line by c and the principal normal and binormal vectors by v and Qrespectively. If n =13 then d n/ds = -isv. We have

drIn

dnl drpds= ,dsj =-'[n,vj=-'c2.

Therefore, the value of p differs from is only in its sign. The same result occurs whenn = -,Q. Thus, the torsion of the asymptotic line can be expressed in terms of the fieldbasic invariants as

is=- ((ne curl a) ± (n, curl n)2 - 4KK

2

Denote the torsion of the first and the second asymptotic lines by is,, i = 1, 2, re-spectively. The following generalization of the Beltrami-Enneper theorem holds:

Theorem The sum of the torsions of the asymptotic lines is equal to the field non-holonomicity with a negative sign and the product of the torsions of the asymptotic linesis equal to the total curvature of the second kind.-

K (+ K2 = - (n, curl n), !6, K2 = K.

If K = 0 then the torsion of one of the asymptotic lines is zero and the torsion of theother is equal to the field non-holonomicity value up to a sign.

As an example, find the asymptotic lines of the helical field, i.e. the field of helicalstreamlines. Consider the regular family of helices on the cylinders with a commonaxis. Suppose that the helices on the same cylinder have the same pitch of screw. Wesuppose, also, that the pitch of a screw depends on the cylinder radius. Consider theequation of some of these helices:

x=pros ,y = p sin gyp,

z=bcp,


where p = x2 + y2, b = b(p2) is some function of p2. The helix tangent vector is

fx'O -yA Y: = x

Z:p b

Also, consider the unit vector field

-y x bn

p2 + b2' p2 + b2' p2 + b2

The field does not depend on :. If b(0) 0 0 then the field is well defined for x = 0,y = 0, also. Find the mean curvature of the field

2H=-divn= 8Y - a

x_ =0.

ax p2 + b2 aY p2 + b2

We find the total curvature of the second kind by (2) of Section 1.5:

0 -1 0 -y

K1 1 0 0 x -b(2p2b' - h)

p2 + b2 b'2x b'2y 0 b (p2 + b2)2

-y x b 0-

It is easy to see that (A, curl A) = 2(b'p2 - b). Therefore, the field non-holonomicityvalue is

(n,curln) =2(b p- - b)

(p2 + b--)2

From this we see that n is holonomic if b = cp2, where c is a constant. Write theequation of asymptotic lines in coordinate form:

f243 -6 d6 =µdY,

b dd1 - 1 d 3 =A dY,

1db -S241 =µd:.Represent the field components in terms of p and gyp:

-p sin cp p cos w b1 = p2 + b2 ' p2 + b2 ' 3 =

p2 _+b_1

The equations of asymptotic lines can then be represented as

(p cos Vdb - bd(p cos v)) = (p2 + b2)pd(p cos p),

(p sinOb-bd(p sin V)) = (p2+b2)pd(p sin gyp), (1)

p2 dip = (p2 + b-)µ rL-.


For the subradical expression in the formula for evaluating 'a at the beginning of thissection we have

(n, curl n)2 - 4K = 2(p'- + b222)

Therefore, the helical field has two, maybe coinciding, asymptotic directions. Thecoefficient p in the equation of asymptotic lines takes one of two values:

-b 2b'p'- - bpIp2+b2' p2= p`+b2

If p = p, = -b/(p'- + b2) then from the first and second equations of (1) it followsthat db = 0, while from the third we see that p2 dcp = -b dz. Therefore, in this casethe asymptotic line is on the cylinder p = const and represents a helix with torsion ofthe sign opposite to the sign of torsion of the field streamline. The family of thesehelices together with the original family of helices form a regular orthogonal net oneach cylinder. Consider the asymptotic lines corresponding to p = 142. The equationsof those asymptotic lines have the form

p cos W db = 2b'p2d(p cos ap),

p sin cc db = 2b' p2d(p sin cp),

p'dcp=(2b'p'-b)d.

Provided that b' 0 0, it follows from the first and the second equations that dip = 0.From the third equation we see that dz = 0, provided that K # 0. Therefore, in thiscase asymptotic lines are the rays emitted parallel to the x, y-plane from the points ofthe common axis of the cylinders. This solution is evident from a descriptive view-point. In this case we ought to regard the torsion h2 as a value which characterizesthe rotation of the vector field n in moving along that ray.

If b' = 0 then K, = 0 and p, = p2. In this case the first and second equations in (I)turn into identities. The third equation p2 dcp = -b d: expresses the orthogonalitycondition of the required asymptotic direction and given field n. Therefore, in thecase of N = 0 any direction which is orthogonal to n is asymptotic.

1.7 The First Divergent Form of Total Curvature of the Second Kind

Now we are going to state an integral formula which gives the expression for theintegral of total curvature K in a domain G in terms of some integral along theboundary 8G or in terms of an integral over some domain on the unit sphere S2. Letx be a position vector of M E G. Suppose that (a, f3) are local curvilinear coordinatesin 8G with the usual orientation, i) is a mapping of 8G into the unit sphere S2 bymeans of the vector field n , do = (n,,, n,,, n) da df3 is a signed area element of unit


sphere. Then supposing no singular points of the vector field n in G. the followingformula holds:

JKdV= J(xn)do,G c'(8G)

where dV is the volume element of a three-dimensional domain G.

Proof We can represent the integrand on the right as

(x, n) da = (x, n)(n0 ny, n) da d$

aX I axe OX 18X2

(act OQ OQ as

0X2 0X3 0X2 0x3\+(n,,,n 3,n)(act as aQ as

aX30XI OX30XI

as aQaa) dadQ= (x,n)(P,v)dS,

(1)

where v is the normal of 8G.By the Gauss-Ostrogradski formula we have

f (x, n) (P, v) dS= J div ((x, n)P) dV = f(x,n)divPdV+ f (P, grad (x, a)) dVaG G

= f(Pn)dV+G G

+ (x, nY3)(n Y,, nx2, n)} dV.

The first integral is the integral of total curvature K because of (1) Section 1.5. Sincen is of unit length, (nx,, nx3, qr,) = 0. Consider the coefficient of xl in the integrand ofthe second integral:

SI r, (nr,, n.r2, n) + SIx, (nr;, nx3, n) + G r' (nx,, nx,, n)

SI.ri l[:l Y; `;,x3

1;2Y, 1;2YZ 1;2Y3 2bx, 6x2 b3.x3 C3

IY, 6x'2 1;1x3 0

_ (n.rl,nx:,nx,) =0.

In an analogous way we see that the coefficients of x2 and x3 are both zero. Thus, thesecond integral turns into zero and, as a consequence, (1) is proved. The calculationsabove shows that K, as well as the H, is a divergence of some vector. Namely,

K = div(x,n)P. (2)


1.8 The Second Divergent Representation of Total Curvature of the Second Kind

Consider (8) of Section 1.5 in detail:

3

K = 2 E,.j-I

t2a C f

(Sl= rz - e2SI.c:) + aY,I

i",

where the dots mean the terms having an analogous structure with respect to re-placement of index I by 2 or 3. Consider the expression under derivation -1, It hasthe form

j16 (C2,, + 6c:) - S2CIc; Skx7}

Add and substract Then the previous expression obtains the form

2 16 (S Ix, + 6- v , + S3x%) - S1SI.c, - 2SI.c. "2

{l;I div n - k, },

where k, is the first component of the field n streamline curvature vector. Transformin an analogous manner the expressions under the derivations ;,'f . . We get

2K = -div (2Hn + k), (1)

where k is the field n streamline curvature vector.Now consider a closed surface F containing no interior singular points of the field

a. Let v be the normal of F. Then

2 / KdV = - J(2Hn + k, v) dS. (2)

JG G

Consider some particular cases.

(1) Suppose that the boundary of G is formed by connected surfaces F, and F2, whereF2 is inside Fl. Suppose that n coincides with the normal vector field on F; (seeFig. 8). In this case (k, v) = 0 at the points of surfaces. Since F; are orthogonal ton, the curvatures of F, coincide with the curvatures of the field n at the points ofF;. From (2) it follows that for any vector field n inside G with F, and F2 as theboundary the volume integral of field curvature K in G can he expressed as adifference of integrated mean curvatures of boundary surfaces:

JKdV= - jHdS+ / HdS. (3)

F,fil

Note that the right-hand side is independent of a.


FIGURE 8

(2) Suppose that at each point of a surface F the field n is tangent to F. Then thestreamline of n starting at F belongs to F. In this case we say that F is theinvariant submanifold of the field n. The value of (k, v) is the normal curvature ofthe streamline in F, where v is directed outside F. If n is regular at each point of Fthen F is necessarily homeomorphic to the torus because only the torus admitsthe regular unit tangent vector field (see Fig. 9). Formula (2) for this case has thefollowing form:

2 J KdV = - r k dS, (4)G F

where G is the interior of F, k is the streamline normal curvature in F of thefield n.

Apply formula (4) to the field defined inside the tubular surface. Denote it by F asabove. Let r be a closed curve of length 1, k and ti be its curvature and torsionrespectively. The tubular surface F meets any plane normal to F by the circle ofconstant radius R with its center in r. If p(u) is a position vector of r and (fi, t2, £3)is a natural frame along I', then we can represent a position vector of F as

r(u, v) = p(u) + R(cos vv2 + sin vv3).

FIGURE 9


We suppose that u is the arc length parameter of r. Set

a = cos sin v6,

b = - sin vi;2 + cos ve;3.

Then

r,, (I -kRcosv)+RKb, r,=Rb,[r,,, r,.) = -aR(1 - kR cos v).

Hence, EI and b are tangent to F. The first fundamental form of F has the form

dsz = { (1 - kR cos v)'" + R2,c2 } due + 2R 2x du dv + R2 dv`.

The unit output normal vector v of F coincides with a.Let us find the second fundamental form. Since the coefficients of the second

fundamental form are equal to the scalar products of the second derivatives of r(u, v)and the unit normal to F, in second derivatives of r(u, v) we present only those termswhich contain no I or b. We have

r,,,, =kt2(1

r,, =-Ra.Hence, the second fundamental form of F with respect to the normal a has the form

11= -{Ric2 - (l - kR cos v)k cos v} due - 2Rn, dudv - Rdv2.

Since we are interested in the normal curvature of Fin the direction of a streamlineof n, the differentials du and dv in the expression for I and II must correspond to thedirections of the field n streamline. Introduce the following notations for them:du dv = rl. The area element of F is

dS = R(1 - kR cos v) dudv.

The following relation holds:

11 dS = [kR cos v1- R2(ng + s))'`] du dv.

Hence, the integral of streamline normal curvature over F has the form

Jr 21T

12i !

jdS=JJ Rkcosvdudv-rJR2(`et

+'1)2dudv. (5)

F F 0 0 0 0

The first integral on the right is zero because R and k are independent of v. Using (4)and (5) we can write

l 2x I

J KdV = ff R2( 21 , )2 dude.

G 0 0


FIGURE 10

As the first fundamental form may be transformed to

I= (1 -kRcosv)2 2+R2(s +rt)'

the integrand on the right is less then 1/2. So, we have already proved the followingstatement.

Theorem If the tubular surface F with the axial line of length I is an invariantsurface of the vector field a then the integral of curvature K of the field n over a domainbounded by F is non-negative and does not exceed irl.

0< fKdV<Trl.G

The lowest value of the integral occurs when n is such that the streamlines in Fsatisfy the equation

r4 +n=0(in the particular case, when the torus F axial curve is planar, these curves arev = const).

The largest value corresponds to the field n whose streamlines are the circles inplanar sections of F perpendicular to the axial curve (see Fig. 10). Note that theupper boundary of the integral does not depend on the tube radius.

1.9 The Interrelation of Two Divergent Representations of theTotal Curvatures of the Second Kind

In previous section we stated two divergent representations for K.

K = div (x, n)P, 2K = -div (2Hn + k).


These relations imply formulas (1) of Section 1.7 and (2) of Section 1.8. Comparingthem we conclude that if the interior of a domain G contains no singular points ofthe field n then

J (x,n)do=-I (Hn+2,v)dS.R If') F

(1)

We are going to show that (1) is also valid if the singular points are strictly interior toG. Define on F the following external forms:

a = 2(x, n) do, 0 = -{2H(n, v) + (k, v)} dS, y = (dn, n, x),

where x is a position vector of F, v is a unit normal of F, dS is an area element of Fand do is an area element of the spherical image of F generated by the vector field n.The definitions of exterior form and exterior differentiation are given in Section 2.9.

Theorem The difference of Q and a is equal to the exterior differential of ry on F.

Q-a=d1' (2)

Let u, v be the curvilinear coordinates in F which induce the normal v. Then Cr canbe represented as

a = 2I(x, n) do = 2(x, n)(n,,, n,, n) du dv = 2(nd, nr, x) du dv

= 15.(n, n,., x) + a (n,,, n, x) + (n, nu, X0 - (n, or, xu)I du dv

= -dy + (n, [n, x,.] - [n,., XU I) dudv. (3)

Show that the second term on the right is 8. Indeed,

3 = ((n, v) div n - (k, v)) [[x,,, xr] I du dv

_ { (n, x,,, x,.) div n - (k, x,,, x,)} du dv.

The forms (n, [n,, x,.] - [n x, j) du dv and /3 do not depend on the coordinate systemsin E3 or F. Introduce the coordinates in E3 and the coordinates u, v in F in such away that at Mo the tangent plane to F coincides with the xl,x2-coordinate plane andthe basis el = x,,, e2 = xr, e3 = v. Then at Mo

nu = fix,, n,. = nx2.

Moreover, at Mo we haveff t C ii `` `

(n,

([nw,xr]

- [nr,x,]) = blx,e3 - S'_r213 - (4)

(n,x,,,xr)divn-(k,x.,x,)=CIx,b +J2xb -S3x,SI (5)

Comparing (4) and (5), we conclude

(n, In., xr] - [nr, xu]) = (n, x,,, xr) div n - (k, x,,, xr).


From (3) it follows that

So, (2) is proved.Since F is a closed surface,

a=-dy+f3.

Jd-7=0.

Hence, (2) implies (1), while the singular points of the field n possibly exist strictlyinside of the domain bounded by F.

Observe that if n is the field of normals of F the equation (1) implies the Minkowskitheorem on the integrated support function of a closed surface and on its integratedmean curvature, i.e. (1) turns into the following equality:

f(xn)KdS= - f HdS.F F

Thus, (1) generalizes the Minkowski formula.

1.10 The Generalization of the Gauss-Bonnet Formula for the Closed Surface

In the theory of surfaces the remarkable relation between the integrated Gaussiancurvature over some domain G in a surface F and the integrated geodesic curvatureof the boundary of G is well known. It is the so-called Gauss-Bonnet formula:

fKdS=f-!-ds+2r. (1)P9

G XHere 8G is a smooth closed curve on a surface F. It is also supposed that theboundary normal v, with respect to which one evaluates l 1p., is directed outwardfrom G. If the surface F is closed then

fKdS=41rrx, (2)

F

where X is an integer called the Euler characteristic of F. We are going to generalizethese formulas to vector fields.

Suppose that a vector field n is defined, maybe with singularities, in a domain Q ofthree-dimensional Euclidean space. Let F be a closed surface in Q not passingthrough the singular points.

Theorem For any closed surface F C Q not passing through the singular points ofthe vector field n

J(nK + 2Hk + Vkn, v) dS = 4irO, (3)

F


where B is an integer called the degree of mapping of F into S2 generated by the vectorfield n.

Here Vin means the derivative of n along the curvature vector k of the n fieldstreamline. In the case when a is holonomic and F is a surface orthogonal to n, i.e.n = v, then (k, v) = 0, (Vrn, v) = 0 and the integral in (3) coincides with the integralof the Gaussian curvature of F. To prove (3) we consider the mapping of F into theunit sphere S2 generated by the vector field n. In Section 1.5 we denoted thatmapping by t/' and found that the area elements of image and original are related toeach other by the following formula:

do = (P, v) dS,

where P is some invariant vector defined by n. In the mapping V) the image covers S2an integer number times. We recall the notion from topology of the mapping degreeof a closed surface F onto the unit sphere. Denote it by 0. The mapping degree isconnected with 1() image area by the following formula:

J do = 42rO.

u(F)

Hence

J(Pv)dS=47r9. (5)

F

Consider the vector P in detail. Choose the coordinate axes in such a way that thex3-axis at M E F is directed along n(M). Then the mean and the total curvatures atthat point obtain the following form:

1/ [ C2-,,!),H= _+K- \SI Y. S2Y, -SI C_SZYI )'

The vector P = {P;} which was introduced in Section 1.5 has the following com-ponents at M:

6)P = J I `Ix: CIx3 l I tI.C SI CI I I ti, 1 C2 I (

S2r; S2.r, S2.r, e2 r. S2r S2r:O

Thus, the third component of P, i.e. the projection of P onto n(M), is equal to K.Transform the first and the second components of P as follows. Add and subtractIr,Slx3 to P1, to P2. We obtain

`` ttPI = SI.xiblxa +SIS2rt + 2H£I

P2 = 2Ht;2ri

The vector with components {Slx,. S2r 0} is the vector of curvature of the n fieldstreamline (see Section 1.3). Denote it[ by k[. Now consider the following vector:

I = {SI.r1 SIx) + Slx2 S2r11 S2r. S I.rt 0} -


FIGURE I I

If a and b are vector fields of components aj and b1 respectively with respect toCartesian coordinates then V1b means, as usual, the vector of components(V1b); _ )Y, We call V1b the derivative of b along a.

It is easy to see that I is the derivative of n along k, i.e. Vkn = I.Thus, P can be represented as

P=nK+2Hk+Vkn.

Substituting the latter expression into (5) we obtain (3). If there are no singularpoints of n inside F then 0 = 0. Indeed, contract the surface F with a continuousdeformation to a sufficiently small neighborhood of some intrinsic point x0 where adiffers slightly from n(xo). The image of F after deformation will be located in somesmall neighborhood in S2. Therefore, the degree of t/i is zero. This fact also followsfrom the equation div P = 0.

Suppose that there is a finite number of singular points F1,. .. , Pt inside Q atwhich the field o loses regularity. Then we can define the index I of each singularpoint P1. To do this take a sphere SE, of radius s containing only one singular point(see Fig. 11). The index 1, of a singular point P, with respect to the field a is thedegree of mapping of SE onto S2 by means of the field a.

Suppose that Fcontains the field o singular points PI,..., Pt. Then Ftogether withSE, bounds a domain QE. Since this domain does not contain any singular point,

J(PzI)dS=0,J(Pv)dS+>F

where the normal v on S£, is directed inside SE,. Therefore, each integral over SE, isequal to the degree of mapping SE, onto S2 with the negative sign. Hence, if thesurface F encloses the singular points of the vector field n then

r r

/ (P, v) dS = > !i.F ,_1


1.11 The Gauss-Bonnet Formula for the Case of a Surface with a Boundary

Now consider a regular surface F having a regular boundary 1'. The image of thissurface in the unit sphere is some domain g with boundary y, where -y is the image of1' under the mapping 0 (see Fig. 12). The domain g we shall consider as a many-valued Riemannian surface. The surface F is partitioned into connected domains G,in which (P, v) preserves the sign. If (P, v) > 0 in some domain G; then we take itsspherical image with a + sign . If (P, v) < 0 in some domain G; then we take itsspherical image area with a - sign. We call the algebraic sum of areas of all g; thespherical image area of the surface F. Denote it by o(g).

FIGURE 12

Applying the Gauss-Bonnet formula to the -y bounded domain gin the unit sphereS22, represent the o(g) in terms of the contour integral of y geodesic curvature

a(g) = Jkg(Y)dr+21rm, (1)

I

where kg(y) means the geodesic curvature of -y in S2, r means the arc length para-meter of -y.

The geodesic curvature of -y can be expressed in terms of the vector field n if werecall its definition. Let s be the arc length parameter of r. Then

kg(y) = -(n,nr,nn) = -(n,n.,,nss)1 ds 3

Since n7 = I and ns = )2= Hence, we can represent o(g) in terms of theintegral along -y of some value which depends on n. Applying (4) of Section 1.10, weobtain

J(PP)dS= n2 nu) ds + 21rm. (2)

F r

We may consider this relation as the analogue of formula (1) of Section 1.10. Nowconsider the cases when the integrand in (2) can be expressed in terms of the usualgeometric values.


Remember that by definition from Section 1.4 the geodesic curvature of 1' withrespect to the vector field a is the value . = -(a, x5, x.,), where x = x(s) is a positionvector of r with respect to the arc length parameter s.

The following theorem holds.

Theorem Let t be a regular boundary of a surface F. If the field a is orthogonal tor at the points of t then

J(P,v)dS= JPg

F r

where k is an integer.

To prove the theorem we use (2) and the following lemma [251.

Lemma Let C(t), q(t) be C2 regular vector functions defined in the segment0 < t < T such that C(t) and 11(t) are three-dimensional unit vectors mutually ortho-gonal for each t. Let p be the angle between & and ri. Then

T T

Jdt.& ,Crr)C2 di J(C,1,,rh)dt+0

` 0 0

Set v = r)]. Let us take C, n, v as a basis in E3. For each t the following holds:

f, = A(cos ni + sin pv),

= A, cosd- ( + sin pv) + A(- sin M + cos pv) dt + A(cos sin pv,).

We have

A2 { d`f + cost ftr) + sin2 rP(F,, v, u,)

sinW (({,17,v,) + (C, v,rt,))

From the definition of v we see [C, n] = v, [C, v] = -ri. Therefore,

v, v,) _ >>!r), vr) _ (v, vr) _ O, z , vr) _ -(n, q,) = 0.

Hence

The lemma is proved.Now consider the curve I' and the vector field a. Suppose that r is orthogonal to a

at each of its points. Apply the lemma, taking as a and n as xs, i.e. as a unit tangent


to F. By the hypothesis of the theorem, (n, x,) = 0.1f cp is the angle between n., and x,then by the lemma we have

f(")i..n,(a,x.,,x.«)ds+fdsds=-

J-ds

+27rn,lS P.9r r r r

where n is an integer, l /pr is the geodesic curvature of r with respect to n. So, theintegral on the right in (2) is expressed in terms of the integral of the geodesiccurvature of r. The theorem is proved.

The second interesting case occurs when r is a closed streamline of n.

Theorem Let t be a closed streamline of the field n. If K is a torsion of I' and F is asurface with F as a boundary then

J(P7L')dS=-JKds+2irm. (4)

F r

Consider (2). In our case n = x,., where x, is tangent to P. So we have

(a, ns, n") ds = J(x,,xx)ds = JK ds,

oY x2. rwhich settles (4).

Finally, for the third case we suppose that r is a closed curve with n being constantalong it.

If P 0 at some point from the domain of definition of n then one and only onecurve tangential to P with the required property passes through this point. Thecurves n = const are the streamlines of P.

Let F be some surface based on I'. The image in S2 of F under the mapping -0 has aboundary consisting of a single point because o is constant along F. Hence, the >!i(F)covers the sphere an integer number times. The following theorem holds.

Theorem if 1 is a closed curve such that n = const along it then for any surface Fbased on I'

J(Pv)ds=47roi (5)

F

where 0 is an integer.

Let us give the interpretation of 0 in the case when the field is defined at all pointsof E3 and there is a limit of n(M) at infinity. In this case we can put the continuousmapping V of the sphere S3 onto the sphere S2 into correspondence with the field a.We consider S3 as the usual sphere in E°. At the north pole of it we take a tangentspace E3 where the field n is defined. We denote the stereographic projection of S3onto E3 from the south pole by a. This projection is formed with the rays from thesouth pole. We put the point P of the ray intersection with S3 into correspondencewith the point Q of intersection of the same ray with P. The south pole of S3


FIGURE 13

corresponds to a point at infinity in E3. We can put each Q E E3 into correspon-dence with a point A in the unit sphere S2 C E3 of the center at some point O. To dothis, we put n(Q) at 0 and denote the end-point by A. Denote the latter mapping ofE3 onto S2 by ip. By hypothesis, the vector field n is defined at all points of E3 andalso at infinity. The composite mapping W = via defines the continuous mapping ofS3 onto S2 (see Fig. 13). So, each continuous vector field n, defined in E3 includinginfinity, defines a continuous mapping of S3 onto S2. The inverse statement is alsotrue: each continuous mapping of S3 onto S2 we can put into correspondence with aunit vector field in E3 having a limit value at infinity. To do this, if A E S2 corres-ponds to P E S3 then define the vector of the field at Q E E3 setting n = OA.

The continuous vector fields n, (x) and 02(x) defined in E3 + oo are said to behomotopic if there is a continuous family of vector fields n(x, t), 0 < t < I such thatn(x,O) = n, (x), n(x, 1) = n2 (x). The vector fields n, (x) and n2(x) are homotopic ifand only if the corresponding mappings i,, and t'2 of S3 onto S2 are homotopic.Various continuous mappings of S3 onto S22 can be separated into classes ofhomotopic mappings. We know from topology the following theorem (see [78], p. 98).

Theorem The homotopy classes of mappings V: S3 -+ S2 are in one-to-one cor-respondence to integers. For any integer y the corresponding homotopy class is a classof mappings W = p4: S2 S3 where p: S3 S2 is the Hopf mapping and &:S' -+ S' is a mapping of degree y.

The number y is called the Hopf invariant of cp: S3--+S2 . For any point P E S3 alinear mapping V,, of tangent space TP(S3) into the tangent space TA(S2) is defined.We say that W is normal at P if the image of the mapping above coincides with T4S2.We say that W is normal over A E S2 if cp is normal at any point P E cp-I (A). Fromtopology (see [51], p. 206) we know that for any point A E S2 there is however closeto W a smooth mapping 0 which is homotopic to W and normal over A. Further onwe suppose that W is normal over A. It can be proved that if cp is normal over A thenthe inverse image r4 = W_ I (A) is a set of smooth curves in S3. To find the Hopfinvariant we can apply the Whitehead formula. Orient the curve in I'4 as follows.Choose in the tangent space of S3 at P a basis e,, e2, e3 which defines a positiveorientation of S3 and such that V,,e, = 0 while V,,e2, V,e3 defines a positiveorientation in S2; then e, will define a positive orientation of I'4. Let F be a smooth


surface in S3 based on rA such that its orientation is coordinated with the or-ientation of rA. Let w be an area form of S2 and p*w be an induced form on S3.Then for the Hopf invariant -y we have

7=4tr J' .,.'=4a fw.

F pF

Since V maps the boundary of F, i.e. the I'A, into a single point, the image of Funderthe mapping V is a two-dimensional cycle on S2. The degree of mapping W restrictedto F is defined and is equal to the Hopf invariant.

Turning back to (5), we note that 0 will be the Hopf invariant if the boundary of Fis a pre-image of some point A E S2, i.e. can be constituted from some number ofclosed curves with an orientation described as above.

Consider now in three-dimensional Euclidean space E3, with a point at infinityincluded, the unit vector field n generated by the Hopf mapping. The Hopf mappingis constructed as follows. Represent the three-dimensional sphere S3 as a unit spherein complex space C'- of two complex variables z I , z2, i.e. as a set of points(:1, z2) E C2 such that

ZIZI +Z2 f2 = 1,

where z; means the complex conjugate to z;. At the same time, the two-dimensionalsphere S2 can be represented as a complex projective straight line, i.e. as a set ofpairs (zi,z2) not being zero simultaneously complex numbers and such that (zl,z2)and (z zz) are identical to each other if there is a complex J 0 such that zi = Az'Z2 = A:. We denote the class of identical pairs (zl,z2) with respect to the equival-1ence above by The Hopf mapping p: S3 -+ S2 is defined by the followingformula:

p(zI, z2) = [zI, z2],

where (zi, z2) E S2. In this mapping the inverse image of each point of S2 is a greatcircle SI in S3. The three-dimensional sphere S3 is decomposed in a family of greatcircles; moreover, as the quotient space of that decomposition we have the two-dimensional sphere S2. Now construct the vector field in E' which is generated by theHopf mapping. Since we can put any point of S3 into correspondence with a point inE3, excluding a point at infinity, and we can put a point of S2 into correspondencewith a vector of constant length, the mapping p: S3 -+ S2 generates some vector fieldin E3. At infinity we are also able to obtain a definite vector. Proceed to the vectorfield construction. Let S3 be a sphere of unit radius in El defined by the equationx, + +.v42 = 1, where xf are Cartesian coordinates in E4. Let E3 be the tangentspace to the sphere at a point (0, 0, 0, 1) and y, be Cartesian coordinates in that E3.By means of a stereographic projection from (0, 0, 0, -1) we put the point in E3 intocorrespondence with a point (x1, x2, x3, x4). The coordinates of the correspondingpoint in E3 are as follows:

2x,yr x4+l, i=1,2,3.


Set zI = xl + ix2i z2 = X3 + ix4. Suppose that z2 96 0. Next we put the point in S2into correspondence with the point (xl, ... , x4) E S3 : [ ,1] . The complex number

u + iv , provided that x3 + ix4 0 0, has the corresponding point in somex3+ix4

plane E`2 tangent to S2. We put the point u + iv in E2 into correspondence with thevector t having components

_ 2u _ 2v u2+v2-1l+u2+v2' £2 l+u2+v2' t3 l+u2+v2

If we putt into the origin then the t end-point defines the point in S2. Represent thecomponents of { as a functions of y,. At first, represent {; in terms of xi. We have

XIX3 + x2x4 + i(x2x3 - x1x4)u+iv=x3+x4

From this it follows that u2 + v2 =. Therefore,

2u 2(xIx3 + x2x4)SI ]+u2+V2

_XI+x2+xj+JC4.

Taking into account that ° x = 1, we obtain

t1 = 2(xlx3 + x2x4)

In the same way we find

2 = 2(x2x3 - x1x4)

Substituting the expression for u2 + v2 above into the expression for (3, we have

3=xI+X -x3-x4.Now express xi in terms of yi. It is evident that

3

1-x2=fix?.

Set

i=1

3

3 4ExYi2 j=1 4(1 - x4)a= -

=1 (1 + x4)2 (1 + x4)

From this we find the expressions of x4 and x4 + 1 in terms of a:

8X4 4+a' x4+1 =4+a

Then we have

x, = Yi(X4 2+ 1) _ 4+

aii = 1, 2, 3.


Substituting the expressions above into the ones for ti, we obtain

8(4YIY3 + y2(4 - a)) 8(4y2y3 - YI(4 - a))

(4 + a)2, C2 =

(4 + a)2

1;3 =16(y; +.y22 - y3) - (4 - a)2

(4 + a)2

Consider the behavior of this vector field. If y3 + y2 +.y23 0o then CI -+ 0, C2 - 0,C3 - 1. Thus, when the point tends to infinity, the vector field C tends to a definitelimit. Along the y3-axis the vector field C is constant, namely (0, 0, -1). It is easy tosee that along the circle y, +y22 = const the component C3 stays constant. From theexpressions for C, and C2 we see that £ goes into itself in rotating around the y3-axis,i.e. C is invariant with respect to rotation around the y3-axis. Therefore, if we knowthe field behavior in the y, , y2-plane then we are able to reconstruct the field be-havior in space. Consider the field behavior along the y,-axis. If y2 = Y3 = 0 then

_ _ z

Ci = 0, 2 =(4f+

Y), C3 -16y2

(4

_

+ a)2Yz)I)

Thus, along the y,-axis the vectors of the field f lie in the plane which is orthogonalto this axis. In moving along this axis C is rotating in that plane clockwise (from the+oo viewpoint). If 0 < y, < 2 then C2 < 0 and if y, > 2 then C2 > 0. Therefore, invarying y, from 0 to +oo the C turns around in an angle 2ir. At the points P, and P2in a half-axis [0, oo) with y; = 4(3 - 2f) and y, = 4(3 + 2f) respectively we haveb = 0. Also, at P, the field vector C is directed opposite to the direction of the y2-axis,while at P2 it is directed along the y2-axis. The circles in the yIy2-plane centered atthe origin which are the vector field i; streamlines pass through these points. How-ever, they have the opposite bypass to each other in moving along the C direction.

The total rotation of the vector field C vectors in varying yl from -oo to +oo isequal to 4a (see Fig. 14).

Consider the streamlines of that vector field. They satisfy the following system ofdifferential equations:

dyi -i = 1, 2, 3.

Suppose that a point moves along the t streamline in the t direction. From the fasttwo equations we find

d arctan - a-4 dQ + y2) _ 64(+ y2)y3_L1 -ds (a + 4)2' ds (4 + a)2

From this it follows that if a point on the streamline lies inside the bally; + y2 + y3 < 4 then its projection into the yIy2-plane moves clockwise; if a pointlies outside that ball then its projection moves in the opposite direction. Wheny3 oo, the streamlines turn around the y3-axis infinitely many times. When y3 > 0


FIGURE 14

the distance between the projection and the yy3-axis increases, while for y3 < 0 thatdistance decreases. Consider the behavior of y3 along the streamline. We havedy3/ds = 0 if y i = -(4 + p) + 4. The latter equation represents the convex curvewith the end-points in the p-axis in a plane of parameters y;, p. This curve corre-sponds to some closed curve -y in the y3, yI-plane which generates a torus T inrotating around the Y3-axis. A streamline emitted from a point in the annulus.bounded by the circles p = 4(3 - 2v) and p = 4(3 + 2f) in the plane y3 = 0, risesat first and then descends.

1.12 The Extremal Values of Geodesic Torsion

The torsion of a geodesic (straightest) fine having a given direction is what we callthe geodesic torsion of the field in that direction. Let dr/ds be the tangent vector of


a geodesic line. This principal normal of this curve coincides with the field a. Denotethe binormal by b. Then by the Frenet formulas we get

where k is the curvature and c is the torsion of the straightest line. From this we getthe expression for the geodesic torsion:

(dr,a,dn)ds2

(1)

Thus, the geodesic torsion with respect to the principal direction of the second kind iszero. Let i, and n2 be the extremal values of (1) in rotating dr/ds in a plane which isorthogonal to n. Then, as Rogers stated, rcl + rC2 = (n, curl a). Prove this. Choose thebasis in such a way that I = 2 = 0, 1. The expression (1) obtains the followingform

II d, dd2 d3K _ , drl dx2 0 = dsl ds2 ds do

0 0 11

drl1' dc2 dxl dx2

ds ds +(ds)To find the extremal values of re, form the symmetric matrix and the characteristicequation in the following manner:

S2Yi - t

A

In expanded form we have

=0.

2 /CC C` tt iS2x, - SIY,} - (S2Y_

4S,_q}_

+6c2 S2xi = 0.

From this we find that the extremal values of the torsion satisfy the equation

A2-2Ap-(H2-K)=0,

KI2=Pf p2+H2-K.So, we get

IdI + K2 = 2p = (n,curln).

It is possible to consider (1) with respect to arbitrary directions of dr/ds, i.e. not onlythose which are orthogonal to a. Let -r1, i = 1, 2, 3 be the extremal values of (1). We


call them the principal geodesic torsions. Introduce the total, mixed and mean geodesictorsion as

T = TIT2T3, M = TI T2 + riT3 + r2T3i S = rI +T2 + r3.

respectively. It happens that S = (n, curl n). As above, we consider (1) with respect tothe special choice of basis:

Al 42 43dxI dx2 dx3

d{I dx2 - dd2 dxI

ds20 0 11

dxI 2 dX2 dXI [ dX3 dxI- S2.YI s) Js ds - S2x; ds dS

dxI dx2 rdx212 dx3 dx2

+ ds d ds ds

We can find the extremal values from the following characteristic equation:

-6_ "' .1fz., fz.3

2 2

[ fl,-2

fly,2 2

-A

=0.

In expanded form we have

-t2rz)2+tIxl+t22.,

ll 4J

+

2

-fall2

fl .1

2

0

= 0.

Thus, for the mean and the mixed torsions of the field we have

S = (n,curln), M=K-H2 - J[n,curln]j4

The value T represents some field invariant. With respect to the chosen coordinateswe have

l 2 z 1T 4 1 -S2X,6xj + (62X, - fIx1 )SIx,C2.YI +SI,, 6x11. (2)


1.13 The Singularities as the Sources of Curvature of a Vector Field

Suppose that there are singular points of the field n in a domain G. We denote the setof singular points by M. Introduce a geometrical characteristic of the singular points.Let T, be the e-neighborhood of M and F be the boundary of T,. Formula (2) ofSection 1.8, which gives the expression for the integrated field curvature K over thevolume, has been stated on the supposition that there are no singular points insidethe surface F. Let v be the normal of F5 directed inward to T. We say that thesingularity is the curvature source of power Q if there is the following limit:

Q=1im 1 / {(n,v)diva-(k,v)}dS. (1)e-o 2.11

F.

If there is the curvature source in G with F as the boundary then we shall consider theintegral of K over G as an improper integral, i.e. set

JKdV= lim JKdV.:-oG 6A T,

It is possible to rewrite formula (2) in Section 1.8 with regard for singular pointsinside G as

JKdV = - f {(n, v)H + (k. v)} dS + Q.G 9G

Thus, if the vector field n is fixed on F then the higher the field source power themore the integrated curvature K is enclosed inside F.

There may be an arbitrarily powerful source enclosed in an arbitrarily smallvolume. For instance, take some closed curve -y of length / in a domain G. Define thevector field u in a neighborhood of -y as a tangent vector field for the family ofconcentrated circles in the planes normal to -y with their centers in -y. As we havestated in Section 1.8, the integral on the right of (2) is equal to -7r/, i.e. the curve isthe curvature source of power -7rl. Note that v is directed inside the tube.

Consider the isolated singular point Mo. If the singularity is such that for any ehowever small the modulus of the area of the image of the sphere F. of radius a centeredat Mo is bounded, i.e.

J Idal < C = const,

1JF )

then the curvature source power is zero.Indeed, use the relation, stated in Section 1.8, between the integral on the right of

(1) and the integral of (x,n) over V'(F). It is also possible to write

Q = li m J (x, n) da.

VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE

Put the origin into Mo. Since I(x,n)I < E over OF, then

V,(F,)

1 (x,n)dQI

<EC.

53

Passing to the limit when a 0, we obtain Q = 0. From this it follows, for instance,that the isolated singular points of algebraic fields have zero power.

Let the vector field n be singular at the points of some closed curve y; moreover,we suppose that at each point P E -y the field n is in the normal plane of -y and isdirected along the radius of the circle centered at P in that plane.

Suppose that the field n is directed outward at the points of a tubular surface F ofsufficiently small radius R with y as an axial curve. In this case (n, v) = -1. Let usfind the curvature source power of such a singularity. It is equal to the integratedmean curvature of F with respect to o directed outward.

Thus, we need to find the mean curvature of the tubular surface with respect to anextrinsic normal. We know from differential geometry that

H - EN - 2MF+ GL2(EG - F2)

Here E, F, G are the coefficients of the first fundamental form and L, M, N are thecoefficients of the second fundamental form of the tubular surface. Use the formulasfrom Section 1.8. We have

EG - F2 = R2(l - kR cos v)2,

EN - 2MF+ GL = R(1 - kR cos v)(2kR cos v - 1).

Hence, the mean curvature is

H 2R(1 - kR cos v)

So, the curvature source power is

r 2,r I

Q=JHdS= rr 2k Rcosv-1 EG-F2dudvJ J 2R(1 - kR cos v)

F 0 0

2,-fI

= 2 J J (2kR cos v - 1) dudv = -Trl,

00

where I is the length of the axial curvature y. If n is directed inside the tube then Qgets the opposite sign.

Consider the following problem: in which case the vector field singularity is topo-logically removable? Let M be a connected set of the singular points of the field

2kR cos v - I


having the --neighborhood where the vector field n is defined. Set M, to be thee-neighborhood. We say that a singularity is topologically removable if for any suf-ficiently small e > 0 the field n can be smoothly prolongated from 8M_. inside M.

Theorem Suppose that in some e-neighborhood of the set M of singular points ofthe field n the field invariants are bounded as follows:

IKI < Co, IHI < CO, Ikl <- Co, 14(n,curln)TI < Co,

where T is the total geodesic torsion, Co = const. Suppose that the surface measure Sof the set M is sufficiently small, namely S < 4

71n. Then the singularity is topolo-

gically removable.

Observe that in the case of a holonomic vector field it is sufficient to require theboundness of IK 1, IHI and Ikl. To prove the theorem we need to estimate the lengthof vector P = nK + 2Hk + Okn. Set I = Vka. Consider the scalar product of I and Pin a special choice of coordinate system:

(1,SI.I7

(SL.Y)SIYj + I Y11I Ya)+SI'

S2Y2 Sir,

SLIVITST, +

C22,:

ttxI

VI Yz - S2YI) (6-11, - t2x?)l;I YtS2 Yi -

(S x, I.YIS2.Y. - I Y_S2.Yi/

Now use the expression for the total geodesic torsion from Section 1.12. We get

(1, P) = -4(n, curl a) T - KIk12.

On the other hand, from the expression for P above we see that

(1, P) = +2H(k, I) + 12.

Hence, to find I I I we have the equality

12 + 211(k, 1) + 4(n, curl n) T+ K Ik12 = 0.

Let cp be the angle between k and 1. Then

I I I = H IkI cos w f H22k22 cost P - 4(n, curl n) - K Ik12.

From this we find that 111 < X02. Next we consider the square of vector P length:

P2 = K2 + 12 + 2H(k, 1) + 4Hk2 + 2H(k, I).

So, the modulus of P satisfies

IPI < 13Co.


Apply formula (5) of Section 1.10 to the closed surface LIME. The area of this surfacediffers from S arbitrarily small. Since

e

S <1-3 C."

J (P,v) dSl <4-FM:

Therefore, the degree of mapping of 8ME onto the unit sphere S2 by means of thefield n is equal to zero. As we know (see, for instance, [52], p. 125), in this case thevector field n can be prolongated without singular points inside 8ME.

1.14 The Mutual Restriction of the Fundamental Invariants of aVector Field and the Size of Domain of Definition

Let us ask the following question: for a given fixed domain of definition, is it possibleto find a vector field of an arbitrarily large curvature? The following example gives theanswer. Let us consider the cube: 1 < x, < 2, i = 1, 2, 3. Define the vector field asicn = (A,, A2i A3), where AI = x1 cos Ax2, A2 = xI sin axe, A3 = 1, A > 0 is someconstant. Then by (2) of Section 1.5 we have

K_Aix,A2x,--A2x,Aix, - XI A

(1 +A2 +A2)2 (1 +x;)2 25'

Thus, we are able to construct in the fixed domain the vector field of arbitrarily largecurvature. Observe that in this example the value of non-holonomicity increasestogether with K increasing. The following theorem holds.

Theorem If a regular vector field n of curvature K> Ko > 0 is defined in a ball Dof radius r and the value of the non-holonomicity satisfies I(n, curl n)I < po then

3r<

2 Ko - pu

To prove the theorem we use the inequality stated in Section 1.5:

H2 + (n, curl n)2 > K,4

which implies that div n > 2V4 - p.2 in the ball D. Integrating the latter inequalityinside the ball D and applying the Gauss-Ostrogradski formula, we find

2 rJKo - po 7rr3 <

J div n dV = / (v, n) dS < 7rr2,

D an

where v is the outward normal of the boundary LID of the ball D. Thereforer <

i(& - P02) -1/2.


FIGURE 15

Now consider the vector field of large negative total curvature K < 0. We shallobtain in this case the restriction on the size of domain of definition provided thatthe curvature of vector field streamline is bounded from above. The followingtheorem holds.

Theorem Let a regular vector field n is defined in a cube of the edge length a inEuclidean space E3 such that the total field curvature satisfies the inequalityK < -K0 < 0, where K0 = const, and the field streamline curvature is bounded fromabove Iki < µa. Then the cube size is limited, namely

a<K+F(4M1'+K.

(1)

To prove the theorem we shall state some integral formula which relates the totalcurvature and the streamline curvature of the field n. Let V(t) be a cube with thecenter at the origin and with the edge 2t {x, = ft} (see Fig. 15). Represent the totalcurvature K in the divergent form

K = I C. I G 'i-"I=

2 1((If2i2 2tIx: +66x., - C3SIr,) +

By the Gauss--Ostrogradski theorem, it is possible to reduce the integral of K in thecube V(t) to the integral over the surface of the cube. Consider separately the ex-pression which corresponds to the cube face x1 = t. It has the form

'ff( I tb,2 - t2' 1x: + 6bx, - bf1x,) dv2 dx3

=2 f f 8x2(f1(2)+8x3(U3)-2fI.ru-2fIx,b)dx2dX3. (2)

X1 =1


We turn the integrals of the first two terms in this expression into integrals along theedges. To transform the third and the fourth terms we note that we can represent thethird and the fourth components of curl n as

3r, - ICI _ 92P + Ik3 - 3k1,

I.C. -b2r, = 2

where p = i (n, curl n), k = (k1i k2, k3) is the streamline curvature vector. From thiswe substitute ,.r, and 1 r, into (2). Then (2) turns into the sum of integrals along thecube edges and faces:

1

2 f U2 dX3 - f2dx3 + f fib b dX2 - Ji3dc2

- fJ{2(.z; + 93P + 2k, - 1k2) + Wbc, - 222P + 3k1 - SIk3)} dX2 dX3,

x, =/

where 1r are the edges of the face x; = t; for instance, 1, : x1 = 1, x2 = t. We addand subtract i k, to the expression integrated in the face x1 = t. Since k I a, theintegrand in the face x, = t turns into

2 x k1.

Take the derivative with respect to x, in the first term out of the integral and replaceit with the derivative with respect to t. Since the face depends on t, we ought to addthe integrals of (t;2 + t;3)/2 along the edges which bound the face x, = t. Presentingnot the integrals along the edges 12,13 and 14, we obtain the following expression:

2

\f 3)dX3+...-f kidx2dx3)

Zat,l ,l (C2+Cs)dX2dx3 (3)x,=1

Transforming the integrals in the other faces in the same manner, we find that alongthe edge 11, for instance, we integrate the following expression:

3 =2l(f1+l;z)z+ 3}=Ia3I2,

where we denoted by a,, i = 1, ... , 6 the projection of a into the plane containing theedge 1; and the center of the cube. Let v be a normal vector of the cube V(t), dS acube surface area element. So, we have

3

f f 2 2 dx; dx1-if (n, curl n, v) dS + F f Ja,12 dx;. (4)fK d V = -dt

V(I) k-I t 8V ,_II,tfit,


Integrating (4) in t from 0 to t = a/2 and taking into account the inequalities

J (a, curl n, v) dS

at,

< po6(2t)3, - JKdV > K0(2r)`.

2 dr;dv, < 12t2,Jf+

k- -I

3a2 > Ko

8- paa3.

From this the required inequality follows.We say that the vector field A and the family of orthogonal planes are algebraic

of order in if the components A&x1, x2: x3) are the polynomials of order not higher thanin. (Here, the vector A is not necessarily unit; if we have the algebraic family ofsurfaces then we get the algebraic family of planes.) It is evident that the algebraiccharacter of a vector field and its order do not depend on the choice of Cartesiancoordinates. With the help of formula (1) of Section 1.7 we can prove the followingtheorem.

Theorem Ler us be given a regular algebraic vector field of order min a cube V withedge length a. Suppose that the total curvature K satisfies inside V the inequality

IKl > Ko > 0,

where Ko is constant. Then the edge length is bounded from above, namely

cina< , C=2 ;;V3.

Here the regularity of the vector field means that JAI 0. We represent the mappingt+ of the boundary 0V onto a unit sphere by means of the vector field n as acomposition of the mapping +p of OV into three-dimensional Euclidean space bymeans of a vector field A and the subsequent projection X onto unit sphere by meansof the rays from the origin: t' = Xcp.

Consider the image of the face xi = const under the mapping V. Denote it by D.A common point of D and the ray from the origin directed as (a 1. cx_, a3) hascoordinates satisfying

Al AZ A3

a, a2 a3

We may assume that a2 36 0, for instance. Set

P(x2,x3) = a2AI - a1A2, Q(x2,x3) = a3A2 - a2A3.


Then at the common point of D and the ray we have P = Q = 0 due to the equationsabove. Since the orders of polynomials P and Q do not exceed m, the number ofpoints in the face xI = const such that the Jacobian I () 0 and P = Q = 0 doesnot exceeds m2. Consider the points at which I ( ) = 0. At those points we havethe following equation:

a201I IA3x2Ar2 AZY, I + a21 A3x: A1x, I

+ a2a31Atx, A2x2 I = 0. (5)

A3x, 2 A3x, AIx, Alx, A2x3

Evidently, the normal to D is collinear to the vector

_ A2r3A3x; AIx2 Alx, A2x,( A3x2 A3x, ' A3x3 A113 I'

I

Aix, A2x, ()-Therefore, by virtue of (5) either (a1, a2, a3) is tangent to D or I = 0. For the firstcase the Jacobian of X is zero, for the second case the Jacobian of W is zero. By theSard theorem the set of those points in the image, in our case in the unit sphere, is ofzero measure. So, each of the rays from the origin, excluding the set of zero measure,intersect D no more than m2 times. Now put the origin into the center of the cube V.Then I(x, n)) < -232A Hence, using (1) of Section 1.7 , we have

Koa3<I KdVI < J (x,o)do <12fa,rm2.xv(av)

The theorem is proved.The problems considered in this section was brought into being by the ideas of the

well-known Efimov theorem on the regular projection of a surface Z = Z(x, y) ofnegative curvature K < -1 onto the square in the xy-plane (see [62]). That theoremstates that the edge length a of the square is bounded from above by the independentconstant a < 14.

That theorem founded Efimov's remarkable investigations of negatively curvedsurfaces which were awarded Lenin's prize. The central result of those investigationsrepresents the following statement.

Theorem For any C2 regular complete surface of negative curvature in Euclideanspace E3 holds inf IK I = 0 (see [631-(65]).

Thus, the generalization of Hilbert's theorem from the surfaces of constantnegative curvature to the surfaces of variable negative curvature was given. Thatproblem interested many famous geometers and was not solved for a long time.

The proof is based on the following lemma, which we cite because of its applica-tions not only to the theory of surfaces but, as Efimov suggested, to the theory ofvector fields.

Efimov's Lemma Let us be given a mapping of the whole xy-plane into a pq-plane:

p = p(x,y), q = q(x,y),


FIGURE 16

where p, q are of class C'. Suppose that the Jacobian 0 satisfies 0 = I pv q` 0and the rotation I = p, - qx satisfies the inequality p`. q).

IAI>aIII+a'-where a > 0 is some constant. Then the image of the whole xy-plane is either thewhole pq-plane or a half-plane or a zone between parallel straight lines; the mappingof the xy-plane onto these domains is a homeomorphism.

In the theory of surfaces the lemma is used only when I = 0. The value I (therotation) is analogous to the non-holonomicity value of the vector field. If n isindependent of x3 and {3 # 0 then setting p = q = 6/6 we obtain the fol-lowing expression for the non-holonomicity value and the curvature K:

(n, curl n) = py - q'`0

1+p2+q2K

(1+p2+q2)2

Therefore, if the invariants of the field n satisfy the inequality K < 0.IKI > aI(n, curl n)I + a2 then the mapping p(x, y). q(x, y) satisfies the hypothesis ofEfunov's lemma.

Examples of mappings satisfying the hypothesis of the lemma and which map aplane onto a plane or a half-plane are known. In [92] the mapping onto a half-plane:p = ln(x + x + e- Y') + y, q = x +e- 7 has been constructed. The problem ofthe existence of such mapping onto a zone is still open.

1.15 The Behavior of Vector Field Streamlines in the Neighborhood of a ClosedStreamline

Let us be given a unit vector field n defined in some domain of three-dimensionalEuclidean space E3. Let L be a closed streamline of that vector field. Consider thefollowing aspects of streamline behavior near L: (1) the rotation of bands formed bythe field a streamlines with respect to the band of principal normals of L; (2) thestability of L as a closed trajectory of the solution, differential equation d x/dt = n(x)


where z E E3. In some cases we can gain information about streamline behavior nearL by the geometrical invariants of the field only at L. In considering the problemswhich are posed one of those invariants arises naturally, namely

A = (div n)Z - 4K + (n, curl n)' (1)

It happens that if A = 0 then the streamlines infinitely close to L behave uniformly.Take a point Po E L and draw a plane F orthogonal to L at Po. In that plane we

take the infinitely small intercept a which passes through Po and then draw the fielda streamlines through the points of a (see Fig. 17). The band intersects the normalplane at an arbitrary point P in L by some intercept a(P) which makes some angle Vwith the principal normal of L. We suppose that V is a continuous function of the arclength parameter of L. We also suppose that the field of principal normals of L iscontinuously and uniquely prolongable. After the first bypass along L the bandintersects F at an by infinitely small intercept a, which will be twisted in some angleDip with respect to a.

Theorem Let L be a closed streamline of the field n. Suppose that A = 0 at L.Then the twist angle of all bands formed with the streamlines is the same and isequal to

f{(ncurln) - rc }ds(2)JJJJJJ

where rc is the torsion of L.

If the integral (2) is a rational number p/q after being divided by 2ir then the bandwill take its starting position after q turns with respect to L and p turns with respectto the band of principal normals; if that number is irrational then the band will windaround L everywhere dense.

FIGURE 17


Consider the invariant A. In the case when n is holonomic, i.e. it is orthogonal tothe family of surfaces with 1 /R; as the principal curvature, then

A = 1 - (3)RI R2I

In the sequel we shall need another expression for A:

A = [(a, curl a) - (b, curl b)]2+[(a, curl b) + (b, curl a)]2, (4)

where a, b are unit mutually orthogonal vectors such that [a, b] = n. To prove this,we use the following formula which is valid for any two vector fields a and b:

curl [a, b] = a div b - bdiv a + Vba - V.b, (5)

where, for instance, Vba means the derivative of a along b. It is possible to write

curl a = curl [b, a] = b div n - n div b + V.b - Vbn,

curl b = curl [n, a] = n div a - a div a + V.n - V.a.

From this we find

(a, curl a) = (a, V.b - Vbn),

(b, curl b) = (b, V,n - V.a).

Subtracting the second from the first and taking into account that (a, V.b)+(b, V.a) = 0, we obtain

(a, curl a) - (b, curl b) = -(a, Vbn) - (b, Van). (6)

Multiplying the expression for curl a and curl b by b and a respectively and thenadding, we get

(a, curl b) + (b, curl a) = (a, Van) - (b, Vbn). (7)

From (6) and (7) follows:

[(a, curl a) - (b, curl b)]2+[(a, curl b) + (b, curl a)]2

= [(a, Vbn) + (b, V.n)]2+[(a, V.n) - (b, Vbn)]2.

Now choose the Cartesian coordinates x1,x2ix3 in such a way that at some fixedpoint M the x3-axis would be directed along n(M), the xI-axis along a and the x2-axis along b. Suppose that 1;1, 2,1;3 are the components of the field a with respect tothat system of coordinates. Note that at M we have 0, i = 1, 2, 3.Then we can write

(a, Vbn) + (b, V.n) = Ixs +

(a, V,n) - (b, Vbn) = ylx, - Zx2.


From this we find

[(a, Vbo) + (b, V1n)]2+[(a, V1n) - (b, Vbn)]2

= (SI``Y, + c l,"),

+ (S Cl.cl S2Y2)2-

= (SIxi + 52x,)2 - 4(CIx,'2Y2 - Ix,e2Y1) + (Slx,

= (divn)'- - 4K + (n, curl n)2 = A2,

which proves (4).Below we shall need the following relation:

divn=Slx, +f2Y, = (a,V.n)+(b,Vbn). (8)

Lemma 1 Let q, v be the fields of principal normals and binormals of the field nstreamlines respectively. Then the torsion is of the field n streamline is

K = 2 ((n, curl n) - (q, curl (v, curl v)). (9)

Use formula (6). We apply it to the unit vector fields n, q and v. We have

curl v = curl [n, n] = n div q - q div n + Onn - V.ir. (10)

Hence, the non-holonomicity value of the binormal field is

(v, curl v) = -K + (v, V,In). (i 1)

Next, by means of (5), we find

curl n = curl [v, nJ = v div n - n div v + V.v - (12)

Therefore, the non-holonomicity value of the principal normal field is

(q, curl 77) = -/L - (,7, (13)

Finally, we find the non-holonomicity value of the basic vector field n.

(n, curl n) = (n, V ,,q) - (n, 0,,v) = (v, Vqn) - (17, (14)

Adding (11) and (13) and then subtracting (14), we obtain the required formula (9).

Lemma 2 Let a, b be the vector fields such that

a = cos nj + sin ,pv, b = - sin <pq + cos,pv.

Then

2k = (n, curl n) - (a, curl a) - (b, curl b) - 2d

, (15)

where dip/ds is a derivative along the streamline of the field o.


By direct calculation we find

(a, curl a) = cos 2 W(rl, curl rt) + sin2 V(v, curl v)

+ cos P sin p{ (v, curl il) + (ii, curl v)) -dids

We optain the corresponding expression for (b, curl b) from the previous expressionreplacing W with W + it/2. We have

(b, curl b) = sin` cp(rl, curl q) + cos2 W(v, curl v)

- cos p sin gyp{ (v, curl r!) + (n, curl v)) -dip

ds

Adding the latter expression we find

(a, curl a) + (b, curl b) = (q, curl r)) + (v, curl v) - 2 d `p

From this and lemma 1, (15) follows.Now consider a band formed by the streamlines. Evidently, it may be included in a

regular family of such bands. Define a vector field b as a field of normals to thosebands. By definition, the field b is holonomic, i.e. (b, curl b) = 0. Take a vector field atangent to the band and orthogonal to a. From (15) we find that the turn of such aband with respect to the band of principal normals q is equal to

r

J 11(n, curl n) - n - I (a, curl a) } ds.

LJJJ

For an arbitrary field n the value (a, curl a) depends on the band choice. If at L wehave A = 0 then by virtue of (4) (a, curl a) = (b, curl b) = 0. Therefore, all the bandsturn by the same angle. This proves the theorem.

Turn now to the stability problem of the closed streamline. Choose a positivedirection in L. We are going to find a condition, in terms of the geometrical in-variants along that curve of the field n, under which L will be the orbitally asymp-totically stable limit cycle of the differential equation dx/dt = n(x).

The stability condition below is a generalization of the Poincare condition for thefield in a plane and expresses the Vagewski stability criterion in geometrical terms.

Theorem Let L be a closed trajectory of the differential equation dx/dt = n(x). If

J divnds < -JAds,L L

then L is an orbitally stable limit cycle.The integral is assumed in the positive direction of L which corresponds to the

positive direction of t. Observe that if A = 0 at L then all the nearby trajectoriesbehave uniformly with respect to the stability property.

Let r°(s) be a position vector of L, s its arc length parameter, I the length of thecurve. Suppose that a and b are the two mutually orthogonal unit vector fields which


are orthogonal to n and periodic at L. Let P(s) be a point in L and F(s) be a normalto the L plane at this point. In the plane F(O) we take the intercept a passing throughPo = P(O) and emit the field n streamlines from the points of a. Let Po be a point ina and co = IPoPPI. Denote by L' a streamline emitted from Po and infinitely close toL. Let P(s) be a common point of L' and F(s). Then we set E(s) = PP'. Werepresent a position vector infinitely close to the L trajectory, namely L', as

r(s) = r°(s) + E(s, Eo){cos W(s,,-o)a(s) + sin V(s, £o)b(s)}.

Denote the arc length parameter of L' by u. We have

r Z = r° + e, (cos pa + sin rp b) + e(cos Wa + sin cp b)3 .

The vector r coincides with n(P'). Multiply it by (cos W a + sin cp b). Then

dE(u,co)= (n(P'),cosW a+sinW b). (17)

du

Consider the expansion

n(P') = n(P) + £(u, Eo)(cos cp V,n + sin W Vbn) + o(E).

Set a° = f. Substitute n(P') into (17), divide both sides by E° and then pass to thelimit with respect to co. We consider only one loop of V. It is possible to show thatthere exists lim cp(s, co) when co 0. Denote this limit by cp(s). Then at the points ofL we obtain

d in If(s) I= (V,n, a) cos 2 V + (Vbn, b) sin2 cp + cos V sin V [(V,n, b) + TO, a)]

= 2 [(oan, a) + (Vbn, b)) +2

cos 2,p [(Van, a) - (Vbn, b)]

+ 2 sin 2tp[(Van, b) + (Vbn, a)]

Use formulas (8), (7) and (6):

d Ink If(s) I= 2 div n -

2

sin 2W[(a, curl a) - (b, curl b)]

+ 2 cos 2cp[(a, curl b) + (b, curl a)]. (18)

Note that by virtue of (4) the sum of the two last terms is less than A(s). Integrating

(18) along L in the positive direction and using condition (16), we see that moll < 1

uniformly for any starting intercept a in the plane F(0). This inequality is sufficientfor stability.

It is easy to produce an example of a field with a closed stable trajectory Lsatisfying the hypothesis of the theorem. Take any closed spatial curve L and drawthrough any point P E L a piece of surface of strictly positive curvature turned with


its concavity to the positive direction of L and perpendicular to L. Let R, and R2 bethe principal curvatures of that surface. In moving P along L we obtain the regularfamily of surfaces in some neighborhood of L. Define n as a field of normals of that

family. In this case A = IR - -I, div n = - (R- + R) < -A. Hence, Lisa stable limitcycle. If at least one of the principal curvatures is zero then the closed trajectoriesoccur, maybe, near L, i.e. L cannot be the limit cycle.

1.16 The Complex Non-Hooonomicity

We shall introduce a new geometrical object connected with the behavior of thestreamlines for the given field n in a domain G C E3. Let a, b be the unit mutuallyorthogonal vectors such that [a, b] = n. Consider the field of complex vectors in G:

,Q=a+ib.

It is natural to define curl ,3 as follows:

curl /3=curl a+icurl b.

The complex non-holonomicity value is a complex number:

((3, curl #) = (a, curl a) - (b, curl b) + i{(b, curl a) + (a, curl b)}.

The modulus of this number does not depend on the choice of a and b (see Section1.14), i.e. it is the invariant of the field n and is equal to A:

curl ,0)12 = (diva)' -4K+(n,curl n)2 = A2.

In rotating a and b in the plane orthogonal to n the value (/3,curl,3) changes in aspecial manner. Let a', b' be the vector fields obtained from a and b as a result ofrotation by the angle cp = W(xi,.r2, x.1):

a'b' = - sinWa+cosipb.

i.e. We have

J).curl,' = e''"' (curl,0 - i [grad V, 6

Hence, the complex non-holonomicity of the field ,l3' has the form

(!3', curl D') = e -24' { (i3, curl /3) - i(13 grad gyp, 0))

= e-Up (/3,curl,3),

because of (Q, grad9, [3) - 0.We shall say that a complex vector field .0 is holonomic if there are complex

functions A(x1, x2 i x3) and ii(x 1, x2, x2) of real variables x1 , x_2, x3 in a domainG C E3 such that

0 = A grad tb. (1)


Theorem A complex vector field 13 is holonomic if and only if the complex non-holonomicity value is zero:

(/3, curl /3) = 0. (2)

If (1) is satisfied then

curl Q) = A(grad'i, A curl grad vP + [grad A, grad tfl)

= A(grad 0, grad A, grad ii) = 0,

i.e. (2) is satisfied, too. Conversely, suppose that (/3, curl /3) = 0. Show that everyvector field /3 = a + i b generated by the vector fields a and b and orthogonal to n isholonomic.

Find, first, a single holonomic vector field /3. To do this we rewrite the holo-nomicity condition in terms of real functions. Let A = A, + i A2, 0 = ', + i ii2. Then(1) can be expanded to the system of two differential equations:

a = A, grad 4b, - A2

grad grad V52.

We draw a surface F through the point Po E G such that in a sufficiently smallneighborhood of P0 the field n restricted on F is not tangent to F. We define a regularfamily of curves in that neighborhood restricted on F. Denote it by -y,. Then we drawa streamline of the field n through the points of each curve. We obtain a familyof surfaces 01(x,, x2, x3) = const made up from the field a streamlines. Let a be aunit normal field of that family of surfaces, i.e. a = A, gradW,. Since this field isholonomic by construction, (a, curl a) = 0. Set b = [n, a]. By the hypothesis(b, curl b) = (a, curl a) = 0. Hence, the field b is also holonomic, i.e. there exists afamily of surfaces p2(x,,x2,x3) = const such that

a = Al grad gyp, ,

b = A2 grad W2.

Since n is orthogonal to b, n is tangent to the surfaces W2 = const. Hence, thosesurfaces are formed with the field n streamlines and the surfaces from differentfamilies, namely gyp, = const and cp2 = const, intersect each other by a streamline.From the condition (a, curl b) + (b, curl a) = 0 the relation for A, and A2 follows.Since

(a, curl b) = A, (grad gyp,, grad A2, grad ;p2),

(b, curl a) = A2 (grad V2, grad Al, grad ;p,),

it follows that

(a, curl b) + (b, curl a) = (A2 + A2) (grad gyp, , grad W2, grad arctan Az = 0.

The gradient of A, /A2 is a linear combination of grad gyp, and grad (P2 and hence, isorthogonal to the streamlines of the field n. Therefore, the surfaces A,/A2 = coast are


formed by the streamlines of the field n, i.e. those surfaces have the same structure ascp, = const and W2 = const. Namely, those surfaces intersect the surface F by somefamilies of curves ry, and y2 and the corresponding field streamlines are drawnthrough the points of the curves of the families. By means of the families yI and y2,we introduce curvilinear coordinates cp, and <p2 such that the curves 'pi = constcoincide with the curves 'yi. Assume that the surfaces A, /A, = const intersect F by thefamily of curves '(1, W2) = const. Then Al /A2 is a function of V1, (p2:

A,

A, = 4 (`pl,'P2)

Therefore, for the vector fields a, b we have

a = Az'P(So,,<PZ)gradipl,b = A, grad ip2.

Choose the functions t/' (,p, , ,p,) and 1/ 2 (,p, , cp,) such that a and b will have the form (3):

a = grad i('1 +'p1,h grad tP2),

b = A2(W2v., grad 01 +'p grad 02)

The following system must be satisfied:

4'Pii, = p2c

Supposing that the Jacobian of change 'pi by 4i is non-zero, we can represent thelatter system as a system with respect to iii('p,, cp2):

aO2 001

(4)

Excluding one of the functions, say we come to a single equality of elliptic type:

a (a>G)/atP, +-L (.6 " = 0.8,p, 4 ) a,Pz 0W2

2 2

Find a local solution of this equation such that(00.

) +(` 96 0 in the neigh-borhood of a given point. Then the Jacobian of change ,pi by 01 will be non-zero. So,the fields a, b just constructed, being orthogonal to n, are represented in the form (3).The field 0 = a + i b is holonomic: 6 = A grad 4/,. Any other field /3' = e-4,8 _e-"' A grad 0 is also holonomic. The theorem is proved.

If the field n is holonomic then the complex value of the non-holonomicity value(J3, curl l3) is zero if and only if the field n is afield of normals to the family of spheresor planes. Indeed, from the expression for A it follows that H2 = 4K, i.e. the surfacesorthogonal to n are totally umbilical, i.e. they are spheres or planes.

VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE

FIGURE 18

69

The geometrical property of streamlines of the field n with the complex value ofthe non-holonomicity (Q, curl Q) = 0 is as follows. Consider a point Po and draw aplane having n(Po) as the normal. Take the intercepts ao and bo in that plane havingPo as their common end-point. Draw the streamlines of the field n through eachpoint of the intercepts. So, we obtain two bands of streamlines. The angle betweenthose bands stays constant along their common streamline (see Fig. 18). Indeed, aswe have stated in Section 1.15, if A = 0 then all the streamline bands turn in the sameangle with respect to the band of principal normals. Hence, the angle between themstays constant.

1.17 The Analogues of Gauss-Weingarten Decompositions and theBonnet Theorem Analogue

Suppose that besides a, the unit mutually orthogonal and orthogonal to a vectorfields S,, 22 in a domain G are given. Then a,, 22 and n form the moving basis inspace. We can describe the behavior of the vector field a in terms of coefficients inderivation formulas which are analogous to the Gauss-Weingarten decompositionsof the surface. Set the indices i, j to be distinct. Consider the decompositions ofthe field a; and n derivatives in Cartesian coordinates xk with respect to the basisa, , a2, n:

aar__

axkI'akaj + N;kn,

= Mikal + M2ka2.f7xk


From the conditions (21, 22) = 0, (a;, n) = 0 it follows that rlk = -r2k, M;k = -N;k.Therefore, the system of derivative formulas have the form

Oa' 1

xk= r;kaj + N;kn, ( )a

an_axk

-Nikal - N2ka2 (2)

We call rik the connection coefficients and Na the second fundamental form coeffi-cients.

At each point of G three r/k and six Na are defined. Let us find the differentialequations that these values satisfy. Differentiating (1) in xi and using (1), (2), weobtain

-2a; ark-

BN,kN;kNji of + (r;kr l - N;kN;i)a; + + r;kNJI n.

axkaxi - (axi ) axi )

Interchanging the roles of k and 1, we obtain the expression for As xk mean theCartesian coordinates in E3, the second derivatives of a, do not depend on the orderof differentiation. Equating the expressions for corresponding second derivatives, weobtain the equalities on coefficients of aj and n:

ar;k _ or,,axt axk

+ N;;N)k - N;kNj; = 0,

ON;k ON,,

ax. - axk+ r;kNji - r;iNjk = 0.

(3)

(4)

By virtue of rlk = -r2k the equality for the coefficients of a; holds automatically. Ifwe write the equality condition on mixed second derivatives of n then we arrive atequation (4). The values r2k can be completely excluded from those equations.Therefore, the system (3), (4) consists of three differential equations with derivativesof r,k and of six differential equations with derivatives N;A.

The following is the analogue to the Bonnet theorem.Let us be given for G C E3 the functions A;k and B;k of Cartesian coordinates, where

Alk = -A2k. Suppose that the functions satisfy the system (3), (4) when we replaceA;k instead of r;k and B& instead of Na. Then there are the unit vector fields al, a2, n inG with Ak as the connection coefficients and B;k as the second fundamental formcoefficients. The vector fields a1, a2, n are defined uniquely up to their choice at onepoint.

Indeed, the system (3), (4) is a compatibility condition for the system (1), (2)having the fields a1, a2, n as a solution. If (3), (4) are satisfied then the solution of (1),(2) exists. Moreover, if at the initial point the triple a1, a2, n is orthonormal then it iseasy to verify that the solution is orthonormal in G because of the nature of thesystem.


The set values r ik and Na characterize not only the field a but the fields a,, a2 also.We would like to obtain some invariants of the field n itself. Consider the vectorfields b1, b2 obtained from a1, a2 by rotation by the angle W as follows:

b, = cos cp a, + sin v a2,

b2 = - sin spa, +cos ;p a2.

Let us have

b;., = rikbj + Nun,

n.« = -N,kb, - N2kb2.

The new coefficients of that decomposition can be expressed in terms of the old ones:

Tik = r1k +8az (5)

,R

91k = N1k cos <p + N2k sin <p,

N2k = -N,k(6)

Hence, the values F1k are defined by the field n up to the derivatives of some functionV in a fixed Cartesian coordinate system. We can regard the set of r1k as some vectorfield r in G. From (5) it follows that

curl r = curl F, (7)

Conversely, if (7) holds then f and 1: differ by the gradient of some function.Thus, the vector field F is the invariant of the field n. Equations (3) produce the

expressions of curl r components in terms of N;k. If we introduce two more vectorfields N, and N2 of components {Na}, i = 1,2 then (3), (4) may be rewritten as

curb _ -[N1,N2],cur1N1 = -[N2,I'],curl N2 = -[I', N,].

Denote the complex vector N, + i N, by N. From (6) it follows that the modulus of

the field N, i.e. Ni + N;, is the invariant of the field a.

1.18 The Triortbogonal Family of Surfaces

Suppose that three surfaces each of which belongs to one of the three families ofsurfaces defined as u1 = const, u2 = const, u3 = const, where u = u;(x1, x2, x3) passthrough any point of a domain G c E3. If they meet each other at right angles thenwe say that a triorthogonal family of surfaces is defined in G (see Fig. 19). Varioustriorthogonal families are used in the construction of coordinate systems in E3. Weare going to state the famous Dupin theorem on those families.


FIGURE 19

Theorem The surfaces of a triorthogonal family intersect each other by the lines ofcurvature.

Let r;, i = 1, 2, 3 be the unit fields tangent to the lines of intersections. Then wehave

(TI,T'-) = 0, (T1_, 73) = 0, (T3, T1) = 0.

Differentiate the first, the second and the third equation in the directions of ri. T,and r2 respectively. Then we get

+ (TI, Qr,T2) = 0,

V,, TI) + (T2,Vr,Ti) = 0. (1)

(Or.ri,TI) + (ri, Or.rI) = 0.

From the first equation of (1) we subtract the second and group together the terms

(V117_1 - V ,r3,T2) + (r1,Vr,T2) - (r3,VT,T2) = 0. (2)

Since T2 is holonomic and rl, T3 are orthogonal to r2, VT,TI - V,ri is orthogonal tor2. Hence, (2) implies

(TI, V ,r2) = (T3, V-, T2).

In an analogous way we obtain two more equations

(3)

(T2,Or,T1) = (r3,V ,,T1), (4)

(TI,Vr.T3) = (T2,Vr,T3) (5)

The left-hand sides of (3)-(5) are equal to each other. Indeed, consider the fieldholonomicity condition

Vr,T2 = Vr1T3 + QT2 + /ITi.


Hence,

(TI,Vr,r2) = (r1,Vr;r3). (6)

From (5) and (6) we see that the left-hand sides of (3) and (5) are equal to each other.Both sides of (3)-(5) are the scalar products of some rt: and the derivative of anotherfield r, with respect to the third field rm, where kit 10 m # k. So, such scalarproducts are equal to each other. From (1) we see that all of them are equal to zero.We have, for instance,

(T2, V71 73) = 0. (7)

Since r3 is a unit field,

v,,-r3 = \T1,

i.e. r1 is principal on the surface which is orthogonal to r3. In the same way we canprove that r1 is principal on the surface which is orthogonal to r,. Thus, the line ofintersection of those surfaces is the line of curvature with respect to both of them.The theorem is proved.

If we have two families which meet each other at right angles at the lines ofcurvature then we can complete the triorthogonal family.

Let u1 = const and u2 = const be given surface families; r1 i r2 be their normalvector fields respectively; r3 be the field of tangents to their lines of intersection. Weassert that r3 is holonomic.

Since r3 is a principal direction on the surface u2 = const, then TI being ortho-gonal to r3 but tangent to the surface is principal, too. Hence,

Vr,T2 = µI rI.

At the same time, T, is principal on the surface u1 = const and hence,

V r.T l = a,T2.

So, we have

Vr,772 - 0r2r1 = ILIT1 - -\2T2.

Therefore, r3 is holonomic: there is a family of orthogonal T3 surfaces with u3 = const.The direction 72 is tangent to the surface u1 = const and to the surface u3 = const

also, because (T2i Ti) = 0. Therefore, r2 is tangent to the line of intersection ofuI = const and u3 = const. Analogously, TI is tangent to the line of intersection ofu2 = const and u3 = const. Hence, the new family completes the given two to thetriorthogonal family.

If we have only one family given, say u(XI,x2i X3) = 0, then it is natural to ask thequestion: whether or not to complete this family to the triorthogonal one? If this ispossible then both of the fields of principal directions are necessarily holonomic,which does not hold for an arbitrary family of surfaces. In order that the familyu(xl,X2ix3) = const be complemented to the triorthogonal family it is necessary andsufficient that the function u satisfies some differential equation. Such a family is calledthe Lame family.


Let r1 = r3 = {(}. Then from (7)

(I fl (l,k0 = (r371r3,06 = (2 6 (24

(3 6 (3,iyi

_ kimC (,(m,ij _ >Bitf1ri

where B,, = rdu"(,c(Consider the system of homogeneous equations with respectto (,:

EBi,(,(i=0,i.i

0.

From this system we find two solutions for r1:

yi = AF,,

(8)

(9)

where the functions F, = F,((k, (i,;) depend on 7-3, A is some multiplier. Since r1 isholonomic, F, satisfy the equation

F 8Fz _ 8F3 8F3 - 8F1+ F3

8F1 _ 8F20.

8x38x2) + Fz

8x1 8x3 8x2 8xI(10)

Substituting here the expression for F, in terms of (j and (ii, we obtain the equationwhich must be satisfied for components of the normal field of a Lame family.

Consider the system (8H 10). We can represent equation (8) as

(2 (21 -IG(1 (I1 I(1 (1JI

(3 (3.iC (2

(3.i6 (3

(2.ii! =

0.

Rotate the coordinate system axes in such a way that at a fixed point Po the fol-lowing would be satisfied: (1 = (2 = 0, (3 = 1. We can regard equation (11) as theequation of some surface in the space of coordinates ((1,(2i(3). At Po the equation(11) has the form

-(j (2.1 + (1 (2((1.1 -- (2.2) - (1 6 (23 + (i (1,2 + (2 6 (1.3 = 0. (12)

Form the matrix B with coefficients of the homogeneous polynomial in the left-handside of (12):

-(z I '.' (22 -c ,, 2

, -C2G 2 ,B= . ,

2 (1.2 2

" 02 2


The characteristic polynomial of this matrix coincides with the characteristic poly-nomial on the extremal geodesic torsion (see Section 1.12) and has the followingform:

-A3 + Az(r3, curl r3) - AM + T = 0,

where M and T are the mixed and the total geodesic torsions of r3. As r3 isholonomic, then (r3, curl r3) = 0. Therefore, the sum of roots of this equation is zeroAi + Az + A3 = 0. Equation (9) gives {3 = 0 at Po. Hence, (12) at Po has the form ofthe equation on principal directions

-(1 (z.l + (I 2 ((] , I - (2.2) + (i (1.2 = 0.

Both of the principal directions r1, T2 are the solutions of the latter equation.Find now another form of equations of the Lame family. Let ul denote the

derivative of u in x;, Du the derivative along grad u, i.e.

aDu = E u# .i ax;

Suppose that the level surfaces of functions u, v, w generate the triorthogonal system.Differentiating the equality >, u; v, = 0 in xm, we get

Duvet+Dvum=0. (13)

In the same way we find

DuWm + D. um = 0. (14)

Differentiating the equality E. wmvn, = 0 along grad u, we obtain E.(w,. D. vm+vet Du um) = 0. Using (13) and (14) we see that

E(Wr D, um + vet D. um) = 0.m

Changing the notation of the indices, we can write the last equation in the followingform:

E vet)VnUmn=0. (15)m.n

Apply to (15) the operator D which performs differentiation and use (13) and (14).Appropriately changing the summation indices, we obtain

1: vmwn Du umn - 2 umk ukn = 0.M." k

(16)

Introduce the values Amn which depend on derivatives of u up to the third order:

Amn = Du umn - 2 E umk U.k


Evidently, A,,,,, = A,,,,,. We can rewrite equation (16) briefly as:

E vmtiv,,Amn = 0. (17)m.n

For the following six values

al = vlw1,

a3 = V2W2,

v, w2 + v2 W 1,

vIw3+v3W1ia2 =

a4 =

a5 = v2w3 + V3W2, a6 =

we have three linear homogeneous equations

al+a2+a3=0,ul I al + u1202 + 112203 + 111304 + 112305 + 113306 = 0, (18)

A1Ia1 + A12a2 + A22a3 + A13a4 + A23a5 + A30t6 = 0.

Also, if we involve the equations

11, V1 + U2 V2 + U3 V3 = 0,

ulwl +u2W2+u3W3 =0,

then, multiplying the first by wl and the second by v, and adding them, we obtain

2ulvlwl + u2(v,w2 + v2w,) + u3(v,w3 + v31v1) = 0. (19)

In an analogous manner, multiplying by w2 and V2 and then by w3 and v3, we obtaintwo more equations:

ul(v1w2 + v2w,) + 2112v2w2 + u3(v2w3 + V3w2) = 0,

ul(vlw3 + v3w,) + u2(v2w3 + v3w2) + 2u3v3w3 = 0-(20)

Thus, ai satisfy the system of six equations (18), (19), (20). Since ai are not equal tozero simultaneously the determinant of the system is zero:

All A22 A33 A23 A31 A,2uI, U22 1133 U23 U31 11,2

1 1 1 0 0 0= 0. (21)

2u, 0 0 0 U3 U2

0 2ul 0 113 0 u1

0 0 2113 112 u, 0

So, we obtain the equation with respect to u which contains the derivatives of u up to thethird order. From the general theory of differential equations it follows that we cantake three analytic functions u(xl, x2i 0), u.Y, (x,, x2, 0) and ur, x, (xl , x2, 0) in somedomain of the coordinate plane x3 = 0 as initial conditions. By the Cauchy-Kova-levskaya theorem, the solution of (21) is defined uniquely after that.

Thus, the arbitrariness in the construction of a triorthogonal system of surfaces in E3consists of three functions of two arguments.


Now we give some examples of triorthogonal systems. Laplace used the tri-orthogonal net of confocal surfaces of second order in studying celestial mechanics.All of these net surfaces can be represented with a single equation

I X2 X22 3

l a, + t a2 + t a3 l'

(22)

where a; are distinct constants. The value oft defines the surface. For any fixed pointP(x,, x2i x2) except the origin we can find three values oft for which (22) is satisfied.Each value oft corresponds to the surface from the family (22) which passes throughP. Indeed, to find t when x; and a, are fixed we have the cubic equation. Suppose thata, > a2 > a3 > 0. Denote by f (t) the left-hand side of (22). If t = a; then f (t) turnsinto infinity. Hence, the equation f (t) - I = 0 has three real roots tl, t2, t3, where

t,>a,>12>a2>13>a3.Thus, the point P(x,,x2,x3) determines the three values of the parameter t, namely11, t2,13. Substituting one of these values into (22), we obtain the equation of thecorresponding surface. If t = t, then the surface is an ellipsoid; if t = t2 then thesurface is a hyperboloid of one sheet; if t = t3 the surface is a two-sheeted hyper-boloid.

Check that the family under consideration is triorthogonal. The normal to thesurface from the family 4'(x,, x2, x3) = const is grad'. Hence the vector

XI X2 X311,-al ' tI-a2' t,-a3

is normal to the surface corresponding to t;. Consider the scalar product of thenormals

L.r xk C ( xkt,ak

(f(') -f(1j)) = 0.tj - t;

Therefore, the normals to the family surfaces which pass through the fixed point Pare orthogonal to each other, i.e. the family is triorthogonal.

Consider the curvilinear coordinates in a domain G which can be introduced withthe help of a triorthogonal family of surfaces u; = const. For each point P E G weput into correspondence the numbers (u1,u2iu3), namely, the values of the para-meters u; which determine three surfaces from the system each of which passesthrough P. We represent a position vector r of P as a vector function of u,:

r = r(ul, u2, u3).

If one of the parameters is fixed, i.e. if we set u; = const, then in varying the otherparameters the position vector end-point moves in the surface from a given tri-orthogonal system. If we fix two of the parameters, i.e. if we set u; = const and


uj = const, then in varying the left parameter the point P moves along the line ofintersection of those surfaces. Thus, the surfaces u; = const are the coordinate onesand their intersections are the coordinate curves. The tangent vector of the co-ordinate curve of parameter u; is r4,. Since the system is triorthogonal, (r,,,, 0 ifi # j. Therefore, the first fundamental form of E3 with respect to the curvilinearcoordinates u; has the form

ds2 = (r,,, du1 +

r dui + i du2 + rrj du3.

Introduce the Lame coefficients r2, = H. Then ds2 has the form

ds2=H, du,+H;du2+H3du3.

The coefficients H; are functions of u;. As we introduce the curvilinear coordinates ina domain G of flat space, the components of the Riemann tensor R;jk, are zero withrespect to u;. This gives six Lame equations for functions H;.

To write them in a brief form, we introduce the Darboux symbols

aui, i5j.

Then the Lame equations have /the form

d aj + daj+ f3k,akj = 0,

du, duj(23)

000= QaQkj.

ddk

where Q, k are distinct and of the range 1, 2, 3. The first of these equations corres-ponds to the equality R;j;, = 0, the second one corresponds to the equality Rj;k = 0.

Let us find the expression of principal curvatures of the surface x3 = const interms of Lame coefficients and Darboux symbols. The unit vector = r 3/H3 is theunit normal to that surface. Therefore, the second fundamental form of the surface is

2 (ru,.,, r,3)= E

H3du; duj.1113

q= 1

We denote the normal curvature of the u1 curve in the surface u3 = const by k31. Thisis equal to the ratio of the second and first fundamental forms, i.e.

k. (ru,,rJ, ,) _ - I (r,". )-J31

H3H2 H3H, 2 H3H1

I OH1 031=-HIH30u3H,


In an analogous manner we can find that

kj1 = - (24)

where kjj means the normal curvature of the curve uj in the surface uj = const. Weshall use this formula in the next section.

1.19 The Triorthogonal Bianchi System

We shall apply the Lame equations to study the triorthogonal system, consideredby Bianchi, which contains the family of surfaces of constant negative Gaussiancurvature. Assume that the family is u3 = const and the Gaussian curvature of thefamily surfaces is Ko. Denote by ds3 the first fundamental form of the surfaceu3 = const:

ds, = H2 th + H2 du.

The product of principal curvatures k31 and k32 is equal to K0. Therefore

131032 = Ko.H1 H2

Suppose that Ko < 0. Set

131 = -tanw Ko. (1)HI

Then

132 = cot (2)H,

' 'Using the Lame equations = 032121 ands 131112, we findato au

1 aHltan

8w 1 a112 _ cotw aw- .

HI 8u2w

au2 ' H2 (9141

Solving them, we get

HI = cos w7G(uI, u3), H2 = sin wtP(u2, u3),

where a, does not depend on u2 while p does not depend on ul.We assert that thesefunctions do not depend on U3. Indeed, from (1) and (2) we can find two expressionsfor H3:

raw 81n 7yl IH3 = l - cot w J

au3

.9u3

-Ko ,

H 3 =f 8 w

+ t a n wa In p l I

X33 J V K.


Hence, the following equation holds:

tan w a V+cotw auk = 0.

If 34 0 then the latter equation implies

tan w =M(u,, 143)

N(u2i 143)

Consider some fixed surface u, = const and, preserving the same coordinate net,parametrize it as follows:

P1 = f t1(ui, u3) du,, p,-= f co(u2, 143) du2.

Then, the first fundamental form of that surface becomes very simple

ds = COS2 w dpi + sin` w dpi.

The angle w is defined by the relation

Atanw= 8 (3)

where A does not depend on p2 while B does not depend on p,. Since the curvature ofu3 = const is equal to K0, the Gauss equation for this surface has the form

82w 82w- sin w cos w ICo. (4)

Opt 8pz

We are going to show that (3) and (4) imply the degeneration of A or B intoconstants. Indeed, they imply the equation

A + B (A2 + B2) _ -Ko (A2 + B2) + 2A i2 + 2Bi2, (5)

where the prime over A means the derivative in p, while the prime over B means thederivative in p2. Differentiating (5) in p, /first and then the result in p2, we obtain

C AA"

BB' + I B AA' = 0.

The latter equation has separable variables. Thus, there is a constant C such that

(-)A'l= CAA',

(-A-)"/_ -CBB'.

Integrating these relations, we get

2Ai2=CZa

+ 2C, A2 +C3,A"u =CZ

2

+ C, ,


4 Bu 2

2Bi2 = - C B + 2C2B2 + C4,B

= - C2 + C2,

where C; are constants. Substituting these expressions into (5), we find

(C2-CI +Ko)A2+(CI - C2 + Ko)B 2 = C3 + C4.

The latter equation can be satisfied if and only if either A or B is constant on thesurface u3 = const and one of the coefficients of A2 and B2 is zero. In the latter casethe surface u3 = const is a surface of revolution.

So, in general, when the surfaces of a Lame family are not surfaces of revolution,the following equations are satisfied:

0 = o,app

=0.0U3 0U3

Then for the Lame coefficient H3 we have H3 = -Ko With respect to newparametrization the first fundamental form of space can be represented as

2

ds2 = cos' w 42, + sine w dp2+ C l l`"' dp23-

5P3

So, the Lame coefficients depend on a single function w and the derivative Thefunction w satisfies the system of equations which can easily be obtained from theLame system. The solution of that system, as Bianchi proved (see [13]), depends onfive arbitrary functions of one argument. From the expression just obtained for thefirst fundamental form of E3 it follows that the Bianchi system formed with regularsurfaces with Ko j4 0 cannot exist in a domain of large size and uniform expansion in alldirections. Suppose, for instance, that the domain is a ball of radius r. Show that r isbounded from above. To do this, take a curve p3 which passes through the center ofthe ball. Then the length of that piece of the curve which is located inside the ballwhich the boundary included is not less then 2r. Since for the regular family ofsurfaces we have H3 0 0, by means of the choice of positive direction in the p3 curvewe can fulfill the inequality > 0. Let Pi be the end-points of the curve under

003consideration which are located in the boundary of the ball. We have

(w(Pz) - w(PI )) = f of a dpi >_ 2r.

The angle w satisfies the inequalities 0 < w < 7r/2. Otherwise, i.e. if w = 0 or w = 7r/2at some point, then at this point one of the first fundamental form coefficientsbecomes zero. But this means that in the surface p3 = const there is a singular point.By hypothesis the Bianchi family is formed with the regular surfaces. Hence,r < 7r/4vf--Ko.


1.20 Geometrical Properties of the Velocity Field of an idealIncompressible Liquid

Vector fields occur in various problems of mathematics and mechanics. We shallconsider geometrical properties of the motion of the velocity field of a liquid. Themotion of a continuous medium can be described, by the Euler method, in terms ofthe velocity V of particles of the medium as a function of time t and coordinatesx, y, z of points in space where the movement takes place, i.e. by the vector field

V = V(x,y,:, t).

The values x, y, z, t are called the Euler variables. The velocity field is called sta-tionary if V does not depend on t, otherwise it is called non-.stationary. At fixed t, thecurves tangent at each point (x, y, z) to the velocity vector V (x, y, z, t) are calledstreamlines. The individual medium particle describes the motion of some trajectorywhich is tangent to V(x, y, z, t) at a point (x, v, z) for each t. If the velocity field isstationary then the streamlines coincide with trajectories of the motion of particles ofthe medium, otherwise they do not since the streamlines are different for differentvalues of time.

For an ideal incompressible liquid the velocity field V satisfies the following funda-mental system of equations which were found by Euler:

divV = 0 (the equation ofincompressibility).

ate + VvV = F - - gradp (the equation of tension.the Euler equation).

(1)

Here V is a derivative along V, F is a density of solid forces, i.e. F = lim J,,,_.0where AR is the resultant force vector affixed to the points in a small volume Orr,Am = pir is a mass in that small volume, where p is the liquid density beingconstant for given liquid. The function p - the hydrodynamic pressure -- depends oncoordinates and time; it characterizes the action of the forces onto every infinitesimalplane area passing through the given point at the given instant.

System (1) is a system of four equations with respect to four functions, namely,three components of V and the p. To find a specific solution we must impose theboundary conditions and the initial conditions, in addition, for the non-stationarycase. It is supposed that the solid force F is a given vector function of x, y, Z.

Generally, the solid forces assumed the potential

F = -grad U.

Then the second equation in (1) can be reduced to the following form

at,+grad u) +[curl V,V] =0.


The function H + E+ U is called the Bernoulli junction. We shall consider astationary velocity field. For the stationary velocity field in the motion of an idealincompressible liquid the following system of equations holds:

divV = 0,grad H = [V, curl V].

From this it follows that V and curl V are tangent to the surface H = const. Letn(x, y, z) be a unit vector field along V, i.e. V = vn, where v is the value of thevelocity of liquid motion. Settle the necessary geometrical properties of the field n inthe case of a stationary velocity field. The first equation in (1) gives

div V = v divn + (n, grad v) = 0.

If we denote by A. the derivative along the field n streamlines then

div ndlnv= - . (2)

We see that if the velocity value is constant along the field n streamlines then themean curvature of the field n is zero.

Now transform the second equation. Using the formula

curl vn = v curl n + [grad v, n],

we get

grad H = v- [n, curl n] + v[n, [grad v, n]]

= r2 [n curl n] - v((n, grad r)n - grad v).

Taking into account (2), we obtain

grad H = v2 [n, curl n] + vv'n divn + v grad v.

Since the field n streamline curvature vector k is equal to -[n, curl n], due to theHamilton formula, which we stated in Section 1.3, we can write

grad H = tie (n div n - k) + 2 grad v`.

So, we have

n div n - k = v= grad (H - I v`) = v grad (p + U). (3)

Relation (3) shows that the field ndivn - k is holonomic. The vector fieldI = n div n - k is called the field of vectors adjoins to n. Observe that the field ofadjoint vectors occurred in Section 1.8. It was proved there that the divergence ofthis field is equal to double the total curvature of the second kind, i.e. div 1= 2K.


Now settle the second necessary geometrical condition for the field n. To do this inthe case div n 96 0, we define the secondary adjoint field I' as

[curl I, n]

2divn

Then the second necessary condition will have the form

curl I' = 0;

more precisely, the field I' is a field of gradient vectors of the function - In v:

I' = -grad Inv.

Since

we can write

Next we get

I = y2 (grad H - gradv= )

curl l = - 2 [grad v, grad H].

[curl I, n] [[grad v, grad H], n]

2divn v3 div n

_ (grad v, n) grad H - (grad H, n) grad vv3 div n

(4)

By virtue of the second equation in (1), we have (grad H, n) = 0. Using (2), we obtain

[curl I, n] grad H2divn v2

Then

I' = I - [curl I, n] = 1 ad H - grad - grad H2 divn v2 ( v2

= -grad In v. (5)

Hence, the vector field I' is a gradient and therefore curl I' = 0.Let P and Q be nearby points in a domain of a stream and ry be a curve joining P

and Q. If d r is the differential of the position vector of that curve then

In v(P) = In v(Q) - J(J*,dr).

7

Thus, the function In v is defined by its value at one point and the geometry of flow,i.e. by the geometry of the vector field n.


Condition (2) follows from equation (5). Indeed, (I', n) = (I, n) = diva.Conditions (3) and (4) were stated by S. Bushgens. So, the following theorem

holds.

Theorem If n is a unit vector field of the velocity directions of an ideal in-compressible liquid in the field of potential forces then the vector field

I=ndivn - k

is holonomic, and the vector field

I' = I _ [curl I, n]

2 div n

is a gradient.

Bushgens assumed that these conditions are also sufficient. But that is not true.Besides these necessary conditions, we need one more condition which is generatedby the relation between functions expressing the holonomicity of I and the possibilityof I' to be expressed as a gradient. Let

I = A grad gyp, I` = -grad

The functions A, W and ip are not arbitrary. They are connected with each other dueto (3) and (4). From (5) we see that v = et''. If n is a vector field of liquid flow velocitythen by (3) we get

Agrad o = e-20 grad p

Hence, the vector field e't'' Agrad W is a gradient. From this it follows that[grad A e"i , grad cp] = 0, i.e. A e2'`' is some function of

A e2t' =f (V). (6)

Conditions (3), (4) and (6) are necessary and sufficient for the vector field n to be thefield of liquid flow velocity. Indeed, if (6) holds then

A grad cp =f (V) grad p - grad 0(,p)

e2 'e2V,

where 0(cp) is some function. Set + U = O(cp). We define the value of flow velocityby (2). Then (1) will be satisfied.

One of the central problems in liquid theory is the problem of the existence of flowwith a given boundary and initial conditions. Note that for three-dimensional flowsof an ideal incompressible liquid the solution of system (1) which is global in time isnot known. Therefore, for the detailed study of liquids only some classes of fields areacceptable.

One of the important classes of streams widely used in applications is a streamwithout rotations, i.e. streams for which

curl V = 0.


Such an assumption is justified by the following facts:

(1) The Lagrange theorem holds: if at some initial instant we have curl V = 0 at allpoints of an ideal liquid moving under the action of solid forces with a potential thencurl V = 0 at any subsequent instant.

(2) In describing the motion of the body within a liquid it is supposed that the liquidis initially immovable. Since the immovable liquid contains no rotations, themotion remains irrotational.

For the irrotational motion the field n is holonomic. Indeed, curl V = 0 means that ateach instant

V = grad gyp,

i.e. the field n is orthogonal to the family of surfaces V = cont. The functiondepends on x, y, z and t; also, (1) implies that cp satisfies the Laplace equation

(Ave + T__ = 0.

This equation is solvable in a given boundary condition. Next, from (1) we find

grad(aw

+ H) = 0.at

Hence

8V+pY2 +W2E

IR

(7)

where f (t) is some function of time but independent of space coordinates. Then from(7) we find the pressure

p = p{f(t) - 8t - 1grad ol21.

Another class, considered by Gromeka and Beltrami, forms the screw streams, i.e.the streams satisfying

curl V = W.

where A is a scalar function. This class is also interesting from the practical view-point. Those streams arise in descriptions of liquid motion generated by the water-screw of a ship, in the theory of wings, propellers etc.

A necessary and sufficient condition for the vector field n to have a stationaryscrew flow is the following requirement: the adjoint field I is a gradient.

Indeed, for the screw flow we have an equation curl V = aV or

vcurl n + [grad v, n) = \v n.

Find the vector product of both sides of this equation with n:

v[curl n, n] + (grad v, n)n - grad v = 0.


Earlier we obtained (grad In v, n) = -div n. Hence

-I = [curl n, n] - n div n = grad In v.

Conversely, if [curl n, n] - n div n = grad tb, where 0 is some function, then(grad t', n) = -div n. Set the velocity of flow to be v = er'. We have

divV =e' ((gradtfi,n)+div n) =0.

Next,

curl V = e" ((grad tI', n) + curl n) = (n, curl n)V.

This means that V is a screw flow. Both equations for liquid flow are fulfilled and thefunction H =_ const.

Note the following interesting property of screw flow: the non-holonomicity(n curl n) is constant along the streamlines of the field n.

1.21 The Caratheodory-Rashevski Theorem

In the case of (n, curl n) - 0 the curves orthogonal to the field and passing throughthe fixed point lie in the same surface, i.e. in this case the space is a foliated one. Theset of points which can be joined to a fixed point with the curves orthogonal to thefield forms a surface. In the case of a non-holonomic field we have essentiallyanother picture. The following theorem holds.

Theorem Let us be given a regular unit vector field n in a domain G C E3 with thenon-holonomicity value (n, curl n) 0 0. Then any two points in G can be joined by thecurve orthogonal to the field n.

A more general statement was proved in 1909 by C. Caratheodory in thermo-dynamics [12]. He proved the following.

Let us be given the Pfaff equation dxo + XI dx, + + X" dx" = 0, where X, arecontinuously differentiable functions of x,. If in each neighborhood of an arbitrary pointP in the space of xi there is a point inaccessible by the curve satisfying the equation thenthere is a multiplier which turns the equation into a total differential.

The proof is short but contains some vague arguments. A more general theoremwas proved by P. Rashevski [29]. Let us be given m vector fields A'1,. .. , X," in adomain G of Euclidean space E". Let us take all various Poisson brackets[Xi, XJ] = V X,Xi - V X,Xi

Theorem (Rashevski) Suppose that among the vector fields X1,..., X,,, and all theirconsecutive Poisson brackets it is possible to select n fields which are linearly in-dependent at each point of G. Then it is possible to pass from every point of G to anyother point of G by a finite number of moves along trajectories of the vector fieldsX,,..., X,".


We use the Rashevski proof and, for the sake of simplicity, consider the vectorfields in E3.

At first, we state an elementary lemma from calculus. Let (xi,x2ix3) be a pointin G.

Lemma] Let us be given a function F(x) of class C. Suppose that xo, xi are thepoints which can be joined by the intercept in G. Then

F(x) - F(xo) = E(xi - xio)Ai,i-i

where Ai are the functions of xi and xio of class C".

Introduce the function of the parameter t:

fi(t) = F(xo + t(x - xo)).

Its derivative is

dfi3

T Fx,(xi - xio)

Then

ri

F(x) - F(xo) - dt

0

3 '

_ E(xi - xio) I Ft,(xo + t(x - xo)) dt.i-1 o

Denote the integral coefficients of (xi - xio) by A,, A2i A3. Note that A, = Fx, (xo) forxi = xio. The lemma is proved.

Suppose that there are given in G the vector fields depending on the parameters tiand 12, namely

1(x, t1) and m(x, t2).

We can regard the field l(x, t) as a mapping which maps x into the I end-point. Wesuppose that the following conditions are fulfilled:

(I) There are continuous derivatives of those fields up to the k-th order with respectto each argument.

(1I) There is an inverse vector field. The inverse vector field 1-' (x, t,) is that with thefollowing property: if til(x, t) maps x into x' then 111-'(x', t) maps x'-x.Denote by to the field l(x,0). By means of the fields I and m we are able toconstruct one more vector field V as follows: shift the point x into y byt11-' (x, ti), then shifty into z by 12M-'(Y, 12), next shift z into u by tiI(z, ti),finally shift u into v by 12m(u, t2). The difference between v and x is not equal to


zero. As a matter of fact, this difference depends on x, tl and t2. Denote it byD(x, tl, t2):

V(x, tl, t2) = tl(I(z, tl) - I(}', ti)) + t2(m(u, t2) - m(z, t2))

By the lemma, we can write

s

1(z,tl) -1(y,11)=

E(zi-yi)L3,i=1

where the L; are the vector functions of y, z. Also, if z coincides with y then L;coincides with alAnalogously

3

m(u, 12) - m(z, t2) = F,(ui - zi)M;.i=1

Ifu=zthen

M; =OM(z,12)

az,.

The components {zi - y,} are the components of the shift from y to z, i.e. of-t2m(z,12). Let mi(z, t2) be the components of m(z, t2). Analogously, the compon-ents {u, - zi} are the components of the shift from z to u, i.e. of the vector 111(z, 11).Set I(z, tl) = {1,(z, tl)}. So, V can be expressed as

V = -11 t2mi(z, t2)Li + it t21i(z, t1)Mi

= tl t2(li(z, tl)M, - mi(z, t2)Li}.

We denote the expression in braces by ii(x,11, t2). Note that q is constructed bymeans of I and m. The field tl(x,11, t2) is invertible. To find the ??-I (X, tl, 12) we shallmake the chain of shifts: x y' by 121111-1 (x, t2), y' z' by ill-1(y , ti), z' u' by12m(z, 0, u'- v' by tll(u, t1). Now find the %, i.e. the function 1(x, tl, t2) for

FIGURE 20


11 = 12 = 0. We have I(z, 0) =10. But z(x) = x for it = 0. Also M1 turns into a partial

derivative of m: M10 = a x.0l. Analogously, m1(z, 0) = mi0, Li0 = ` t0 . Henceoxf

x, 0) x, 0)71o(x) = l1oMo - mioL1o =11o(x)

exi - min(x) 8x

If we interpret the vector fields l0i m0, 710 as the fields of operators

10=E4o a1

8x1

then 180(x) can be interpreted as the Poisson bracket of operators to and m0.We say that a pointy is accessible from x by means of some set of vector functions

l(x, ft), m(x, 12).... if we are able to pass from x to y as a result of shifts of thefollowing kind: a shift from x into x' by tll(x, it), then a shift from x' into x" byt2m(X', t2) and so on; or use the vectors it]--' (x, it), t2m- I (x, 12) in the procedureabove. Since the vector field TO, it, t2) has been constructed by means of 1 and m,then the point which is accessible by means of 1, m and 71 will be accessible by meansof 1 and m.

Lemma 2 Let us be given the vector functions 1(x, it), m(x,12), q(x, it, t2) inG C E3. Suppose that the corresponding 10, mo, 710 are linearly independent at eachpoint of G. Then any two points are mutually accessible by means of 1, m, 17.

We shall prove that any point in some neighborhood of x is accessible. Let x, y, z, vbe a sequence of points formed as follows:

y is a result of a shift from x by t1l(x, it),z is a result of a shift from y by t2m(y, t2),v is a result of a shift from z by riT271(z,TI, r2).

We obtain

v(tl,12,TI,T2) = x+111(x,11)+t2m(y,t2)

+T1T27J(z,TI,T2).

Fix r2 - 0 and find Ov/8tl for II = 12 = TI = 0. Evidently,

8Itlt

(0, 0, 0, r2) =1(x, 0) =10(x)-

Analogously

(1)

iit2(0, 0, 0,,r2) = m(x, 0) = mo(x),

(TTI= T2710,

and we see that the Jacobian of transformation from (it, 12, TO to (VI, V2, V3) isequal to

72 (l0, mo, 710) 0 0- (2)


Therefore, for any point v we can find ti, t2i r1 such that (1) will be satisfied. Thisimplies that for some sufficiently small neighborhood of x for the values of v and of0 for the values of 11, t2, r1 there exists a one-to-one correspondence between theneighborhoods.

Let us be given in G the non-holonomic vector field o, i.e. (a, curl n) # 0. Take thevector fields X, and X2 orthogonal ton. Each of the vector fields XI, X2 generates thestreamlines

dI = X I (x), dt X2(x) (3)

Now introduce the parametric vector fields 11 (x, ti) and 12(x, t2) as follows.Draw the integral curve of (3) through each point x in G, i.e. the trajectory of the

operator Xi, where t; is the parameter of that trajectory. We assume that t; = 0 at theinitial point x. Now define the shift vector i(x, ti) as a vector from x to the point inthe trajectory which corresponds to the parameter value ti. Restricting the values oft, to some sufficiently small interval about zero, we are able to achieve the situationwhen for any x in a domain G' c G the corresponding trajectory segment exists anddoes not go out of the main domain G. So, A(x, ti) is already defined. Evidently,

0(x,11) = fXi(ui,u2u3)drii0

where u1,u2iu3 are the coordinates in the i-th trajectory being functions of ri andinitial coordinates Xi, X2, x3. The parameter r; varies from 0 to ti. Introduce a newparameter t = ri/ti, so ri = tti. Then

rI

0(x, ti) = tiJ X;(u1, U2, u3) dt,

0

where UI, 142, U3 are continuously differentiable up to the k-th order functions ofthe initial coordinates x1 , x2i X3 and t, t;. Denote the integral on the right of theexpression above by li(x, ti). Then

,&(x, ti) = t; Ii(x, ti).

Thus, for each operator X; we introduced the corresponding vector function li(x, y).This function is invertible: 1; 1(x, ti) = -11(x, -ti).

If ti = 0 then Iio = Xi(x). To prove the theorem it is now sufficient to show that weare able to reach any pointy starting from x by means of 11(x, t,) and 12(x,12). Since(n, curl n) # 0, q, being constructed by means of lto and 120, is linearly independent of110 and 120. Also, i0(x, 0, 0) is a Poisson bracket of the vector fields 110 = XI and110 = X2. By the previous lemma, any point is accessible by means of 11, 12 and n. Butaccessibility by means of 11,12, 1 implies accessibility by means of 11,12 . The theoremis proved.


1.22 Parallel Transport on the Non-Hooonomic Manifold and the Vagner Vector

We shall call the set of curves orthogonal to the field n the non-holonomic manifold.We say that a vector field is tangent to the non-holonomic manifold if it is orthogonalto n. V. Vaguer introduced the notion of parallel transport of tangent vectors withrespect to the field n, i.e. parallel transport on the non-holonomic manifold [15].

Let us be given a vector field v orthogonal to n at each point. Consider theinfinitesimal motion along the curve orthogonal to n. We call those curves ad-missible. As usual, we denote the field differential in E3 by dv. Denote by by theprojection of dv into the plane orthogonal to n:

(1)by = dv - n(n, dv).

We call by the absolute differential of v on the non-holonomic manifold. Since(n, v) = 0 then we can rewrite (1) as

by = dv + n(v, do). (2)

The tangent vector on the non-holonomic manifold is called parallel-transportablealong the admissible curve if its absolute differential along the curve is zero. For thatfield

dv = -u(v, dn),

where the differentials are assumed to be along the curve under consideration.Let us answer the question of what is the vector which is tangent to the admissible

curve y and parallel-transportable along it. If r = r(s) is the curve ry representationwith respect to the arc length parameter s and v = r., is parallel-transportable, then

r' = -n(r.,, n,).

According to Section 1.4, this is the equation of the straightest lines. Conversely, if 7is a straightest line then its tangent vector is parallel-transportable.

It is natural to ask the question: in which case are we able to construct the vectorfield defined over the whole domain of definition of the non-holonomic manifold suchthat it will be parallel-transportable along every admissible curve? Evidently, along asingle admissible curve such a field exists: this is a solution of an ordinary differentialequation with the right-hand side linear in v. However, the two points in space can bejoined to each other with the different admissible curves and the transport along eachof them gives, maybe, different results at the same point. Therefore, our questionmay be reformulated as follows: in which case does not the parallel transport dependon the curve of the transport?

Denote by do/dr and dv/ds the operators generated by derivatives of these fields.The corresponding matrices have the form

Sl S2 S3 vi v, Y3

01 ti t3 vj

a 26 4 q v3

VLC FOR I ILLDS IN THRLI -DIMENSIONAL LUCLIDLAN SPACI 93

where s;,' = . %0 The differentials do and dv can he represented in terms ofthose operators as

dn= dn dt =drdr.

Note that since n is a unit vector, "x is always orthogonal n . where x is an arbitraryvector. Let a be any vector field tangent to the non-holonomic manifold. For anyparallel-transportable vector field v along any admissible curve the followingequation holds:

dv do 1dra-n v,clra (3)

Let x and y be constant vector fields in E'. Denote by a and b the projections of xand y respectively into the plane orthogonal to n. We want to extend the concept ofparallel transport to arbitrary curves. To do so. Find the result of the action of theoperator d v/dr over a. Set A = do/dr.

Lenwia Parallel transport along admissible curves induces the parallel transportalong the field n streamlines in the case of (a, curl a) ¢ 0 and this transport is definedby the following equation-

dvn = -n(v, An)

K In, vj

Sr (a, curl n)

Differentiating both sides of (3) in the direction of the vector b. we obtain

S-vb)

dv Ab=!dv

b, Aadr=

(a, + d, - - .96{v, Aa) - a (7r

F {a,b)} A b), (4)

where and d! are the linear vector functions of two arguments. They are formedfrom the second derivatives of components of the vector fields v and a respectively.For instance, the ca-th component of has the form

ch-

-dblwhere a', b' are components of a and b respectively. The equality of mixed derivativesof V' implies that the vector function JQ is symmetric in the arguments a and b.

Since v is parallel-transportable, then dF b is collinear to a, AS is orthogonal to n.Therefore, the second term on the right-hand side of (4) is zero. Transfer the secondterm on the left of (4) into the right-hand side. Since a = x - a(n, x) and x is aconstant vector field,

02,04

db = -Ab(n, x) - n(x, Ab). (5)


Using (5), transform the

(v,

next two terms in the right-hand side of (4):

-drdab-nA b l = (n, x)+n(v,Azb) )

+ (x, Ab) _d n + n(x, Ab)(v, An). (6)

Since Ab is orthogonal to n, by virtue of (3) when a is replaced with Ab we obtain

dr Ab = -n(v, A2b).

Therefore, on the right-hand side of (6) the multiplier of (n, x) is zero. Taking intoaccount all the transformations, rewrite (4) in the following form:

z

n + n(x, Ab)(v, An)dr2 (a, b) = (x, Ab)dv

z

- Ab(v, Ae) - n (VI (a, b)) . (7)

Since Ab is orthogonal to n, x may be replaced with a. Now interchange the roles of aand b. Subtract the equation obtained from the original. Using the symmetry of thevector functions dzv/dr'- and dzn/drz with respect to the arguments, we find

{(a, Ab)- (b, As) } ( n + n(v, An) } + Aa(v, Ab) - Ab(v, Aa) = 0. (8)

Transform the expression in the first braces:

(a, Ab) - (b, An) = 4J'a'b' - OP = ? (a'bJ - a f; )

= -(a, b, curl n)= -(a, curl n) (n, a, b). (9)

We can represent the difference of the two last terms in (8) as the following vectorproduct [[Ab, Aa], v]. But Ab and Aa are orthogonal to n. Therefore

[[Ab, Aa], v] = -[n, v](Aa, Ab, n).

Now show that the total curvature K of the field n (remember that K is the secondsymmetric function of principal curvatures of the second kind) can be found by thefollowing formula:

K= (Aa,Ab,n)(n, a, b)

(10)

where a, b are the vectors orthogonal to a. Since the right-hand side of (10) isinvariant with respect to rotation of coordinate axes in a space, choose the co-


ordinate axes in a special way, namely direct e3 along n(M), where M is some point.At this point £3 = 1, tj = 0, a3 = b3 = 0. We have

.ja' {?a' 0

(Aa,Ab,n) _ tjbi jb1 0 = -b'a2)0 0 1

= K(n,a,b),

which completes the proof of (10). On account of (10) we get

[[Ab, Aa], v] = -K [n, v](n, a, b).

By the hypothesis of the lemma, (a, curl a) 3A 0. Taking into account (9) and (11), theequation (8) can be represented as

dvn = -n(v, An) - K In, v]

(12)(n, curl a)

Thus, we obtain the value of the action of the operator dv/dr over n. The Lemma isproved.

Decomposing any vector x into components collinear and orthogonal to n, we areable to give the equation of parallel transport with respect to each of them. Addingthose equations, we get

dvx = -n(v,Ax) -K[n,v](a,x)(13)

dr (n, curl n)

In the case when x is orthogonal to n we shall come to the equation of a parallel-transportable field along the admissible curve introduced above.

We shall call the parallel transport defined by (13) the prolongated paralleltransport. It is easy to check that if the scalar product (n, v) is zero at some initialpoint on the curve of parallel transport then by virtue of (13) this product is zeroidentically. This means that the parallel transport takes a tangent to the non-holonomic manifold vector into the tangent, as well.

Equation (13) is a consequence of equation (3). Conversely, equation (13) implies(3). Therefore, (13) is equivalent to (3). But (13) is preferable to (3) because itcontains the operator d v/dr which acts over an arbitrary vector field x.

The equivalence of (3) and (13) means that if there is a parallel-transportablevector field in the sense of (3) then it will be parallel-transportable in sense of (13),i.e. with respect to prolongated parallel transport. Find the integrability condition ofequation (13). Let x and y be constant vector fields in E3. Differentiating both sidesof (3) in the direction of the vector y we obtain

d1v

dr2(x, y) Ay(v, Ax) - n (dr

Y'Ax I -0 (VI

dd1nr'- (x, Y)

/// 14d Kn _ K(n,x) dv ( )

- In, v] (x,dr ((n, curln)) Y) (n, curl n) ([AY' v] + n' dr Y] )


Make a substitution of d v/dr by means of (13). Also, using the orthogonality of vand Ay to the field n, we can write

[Ay, v] = n(n. Ay. v).

Equation (14) can be rewritten as

V, d n (xY)d-W

d Kn(x, y) _ - Ay(v, Ax) - [n, vJ x, y

-a(drdr (n, curl n) dr-

+(n,unln) {(n.x)(a,v,Ay)+(n,y)(n,v.Ax)1

+ I(,curln))-(° x)(n.y)[n.[n.vJ].

The last three terms on the right of the equation are symmetric with respect to x andy. Interchanging the roles of x and y and subtracting the equation obtained from theoriginal, we get

In, vJ X. Y, curl (n,curln)) + [v. [Ax, Ay]J = 0.

Consider the second term. Transform the vector product in it as

[Ax, Ay] = n(n, Ax. Ay).

With respect to the special system of coordinates defined above, we obtain

;vi Mfr 0J

(n, Ax. Ay) = ,Z,r '}' 0

0 0 1

YI

V1

L

(15)

(16)

The latter expression is the scalar product of the curvature vector P (see Section 1.5)and [x, yJ. In Section 1.9 we stated the invariant form of the curvature vector

P=Kn+Hk+Vkn.So,

(n, Ax, Ay) = (P, x, y). (17)

Turn back to (15). Substitute it into (16) and (17). We get

In, vJ (x, y, curl (K n ) - P) = 0-

\\\(n, curl n)

G1"(_TOIt I1111)5IN 111RI I:I)IMLNS1(INAI_ I UCI1t)I-AN'PACti 97

But [n, v] # 0 and %, y are arbitrary vectors. Therefore.

curt tKn -P=0.

`\{n, curl n)

We shall call the vector on the left the Vaguer vector. So. the following theoremholds.

Theorem In order that the rector Iirld parallel-transportable on the lion-Itolono,nu manifold exhits it is Fircesxarr and sufficient that the Vaguer rector

curlK n l

{n,curln)J -(h°-kdivn+Dkn}

is zero.

Give an example of a vector field having a zero Vagncr vector. Define the com-ponents of that field as

1=cosip, ==sinp, a=0.where c p = c x' + d. c # 0 and d are the constants. In Section 1 .1 we found that thenon-holonomicity value of that field is (n. curl n) = -c. As the field n streamlines arethe straight lines, the vector of the curvature of the streamlines k = 0. Hence Vk = 0.The field n is parallel to the x'x=-Mane: Therefore, the mapping maps any surfaceinto the arc of the same great circle in the unit sphere. Hence, K = 0. We see that allthe terms in the expression for the Vagner vector become zero.

2 Vector Fields and Differential Forms in Many-Dimensional Euclidean and Riemannian Spaces

2.1 The Unit Vector Field in Many-Dimensional Euclidean Space

Let us given a unit vector field n in some domain G of (m + 1)-dimensional Euc-lidean space El ` ' . For any shift d r from M E G in the direction orthogonal to n(M)we define the normal curvature of the first kind k setting k = -(do,dr)/dr'.

We shall call the extremal values of the first kind of normal curvature the principalcurvature of the first kind. We shall call the directions d r for which these values areachieved the principal directions of the first kind. Let us find the system of equationsto determine them from. Introduce the Cartesian system of coordinates x' in El"We denote the vector components with respect to this system with superscripts; forinstance, the components of n will be denoted by '. To simplify notations, we denotethe derivative 8{'/8xi by E . Consider the field n in the neighborhood of M. Choosethe Cartesian coordinates in such a way that the basis vector a",., will be directedalong n(M ). The fact that dr is orthogonal to n(M) implies d r = {dx' , ... , dx, 0}.In solving the extremal problem we come to the system of principal directions:

d 'AJ 2 dv" = 0x+ ) +( x + ,

t2 +t1

dx' + (f2 + A) dx- + ... + ... = 02

,

2ei dx1 + + + '+A) dx'"( =0r .

99


The principal curvatures of the first kind are the roots of the characteristic poly-nomial

1i+ (1)

As the matrix +ek)/2II is symmetric, all the roots A, are real. Denote by al, thek-th symmetric polynomial of the principal curvatures of the first kind. The ok canbe expressed in terms of the main minors of the matrix II e, )/2II as follows:

Qk =(_I)k

I <il <i: < <;i(2)

We define the principal curvatures and principal directions of the second kind by meansof the Rodrigues equation

do = -p dr. (3)

We call the direction d r satisfying (3) and orthogonal to n the principal direction ofthe second kind.

We can find the number ls, which satisfies the Rodrigues equation, from thefollowing equation:

tl + p C;

l;z + l

+1 ... ... C,"+i + pm+

Here we do not suppose a special choice of space coordinate system. One of the rootsum+, of equation (4) is zero. Indeed, by means of the vector field n we are able

to construct the mapping i of the (m + 1)-dimensional domain G onto the m-di-mensional domain G in the unit sphere S'. Therefore, the Jacobian of 0, which isequal to the determinant of is equal to zero. Hence, one of the roots of (4) iszero. For definiteness, set µm,., = 0. The other roots are called, by definition, theprincipal curvatures of the second kind. Since the matrix IIc;II is not symmetric ingeneral, the p- can be complex. Therefore, the real principal directions do not existfor these pi. But in all cases the symmetric polynomials of the p, are real ones. Theycan be expressed in terms of coefficients of the polynomial (4).

Denote the symmetric polynomial of the principal curvatures of the second kind bySk. It is easy to express them in terms of derivatives of components of the vector field n:

i 4

SipkObserve that a1 = S1 and Sm+1 = 0 because of µm I = 0.

VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES 101

2.2 The Regular Vector Field Defined in a Whole Space

We shall consider the unit continuous vector fields defined at all points of E'llincluding a point at infinity. We can put every such field into correspondence with aregular mapping of the unit sphere Sm+ I onto the unit sphere S" and vice versa.Indeed, if we regard E" I as a tangent space of Sm+I at the north pole then by meansof stereographic projection from the south pole we are able to map Sm+I onto Ell'.The south pole corresponds to a point at infinity in Ell'. Let p E Sm+I and x be thecorresponding point in E" in stereographic projection. Now construct a mappingip of Euclidean space E"' onto Sm. To do this, we put n(x) into the originO E E"' I. The end-point of n(x) then marks a point q in S'. Put into correspon-dence the points p E Sm" and q E Sm. Since n(x) is defined at all points of Ell 'including a point at infinity, the mapping of S"f I onto Sm is defined at all points ofS"' including the south pole. Conversely, by a given mapping of Sm+1 onto Sm weare able to construct a vector field in Em+I

The Freudenthal theorem on the classification of mappings S"+' - S" is known:the homotopy group ltm+I(Sm) = Z2. In Pontryagin's book (see [72], p. 154) thefollowing theorem has been proved. For m > 3 there are exactly two homotopicclasses of mappings S"' -+ S". Whether or not the mapping f : Sm+I --+ S" belongsto one of those classes depends on some invariant 5(f) which is the residue modulo 2and therefore takes only values 0, 1. The construction of this invariant is as follows.Let q E S' be a regular point of the mapping f. Then f-'(q) is a one-dimensionalsubmanifold in S". Denote by M' the image of this submanifold in a stereographicprojection of Sm+I onto E" ". In terms of a vector field the submanifold M' consistsof one or several closed curves where the field n is constant. If the set f- I (q) does notcontain the south pole then MI is a smooth closed submanifold of Euclidean spaceEm. By means of the mapping IIi: E" I S" we are able to frame MI with the fieldsorthogonal to MI. To do this we consider the tangent space Tq at q E Sm. Themapping iti induces the linear mapping ti'. of E"+' onto Tq at x E MI. Choose thebasis e,, . . . , e" in Tq. We take the vectors u; = (i.-1e; as a framing of MI at x. Bymeans of a continuous deformation, this framing can be transformed into both anorthonormal and normal to an MI framing. We denote the vector fields of the latterframing along M' by u1,. .. , um. We denote the vector field tangent to M' by 0M,1.Thus, for the mapping f: Sm + 1 Sm we can put into correspondence some framedone-dimensional closed submanifold M'.

Let r, , ... , T,,,+ I be some positively oriented orthonormal basis in Ell'. Then

M+ I

u;(x) = Eh;,(x)Tj, i = l,...,m+ 1,j=I

where jJh;1(x)JJ is an orthogonal matrix with a positive determinant. The field of thosematrices along M' defines a continuous mapping h of the curve M' into the manifoldHm+I of all rotations of Euclidean space Ell 1. Now we define the invariant 13(h) -the residue modulo 2. If m > 2 then for the unicomponent manifold M' the residue3(h) = 0 if h is homotopy equivalent in Hm+I to a point and (3(h) = I otherwise. For


the case of a many-component manifold the residue /3(h) is equal to the sum of eachof the component residues. If m = 1 then i(h) is a degree modulo 2 mapping of MIonto the circle H2. Next, let r(MI) be the number of components of MI. Then theinvariant b(f) of the mapping f and, consequently, the field n invariant undercontinuous deformations is the number

b(f) _ /3(h) + r(MI) (mod 2).

Consider the unit field n defined in Euclidean space E"+' including a point at infinitywhich is parallel to some space Eo+1. By means of a stereographic projection we areable to state the one-to-one correspondence between the points of S"+' and E"+'Therefore, the field n generates a mapping f: S"+' -' S". The homotopies off cor-respond to those deformations of n in E"+' which preserve the property of n to beparallel to Eo+1

The classes of continuous mappings f form the homotopy group an+,(S"). Someelement of this homotopy group corresponds to the field n. The problem of findingall the homotopy groups of spheres is an unsolved one. The solution was given for rnot very large.

Pontryagin stated that 7rn+2(S") = Z2 and pointed out the method of evaluating theinvariant which classifies the mappings S"+2 __+ S" up to homotopies. We apply thatmethod to the vector field defined in Euclidean space En+2. Let Q E S" be a regularpoint. Then M2 - the preimage of Q - is the two-dimensional surface in E`2 onwhich the field n is constant and the end-point of n coincides with Q when the initialpoint of n is put to the center of S". If M2 consists of a single component and is asurface of genus p then it is possible to select a set of smooth closed curvesA1,. .. , AP, B1,. .. , BP such that A, intersects B; at a single point avoiding tangencywhile any other curves have no intersections. For any curve C of this set we are ableto determine the invariant b(C ). To do this we shall construct a framing of C. Setun+1 to be the vector which is normal to C and tangent to M2. The surface M2 isframed with normal fields uI,... , un which are induced by mapping onto S".Therefore, C is framed with u1,.. . , un, un+,. Let r be the tangent vector of C. De-composing the vectors u1, . . . , u,,. , r with respect to the fixed orthonormal basis inEn,'`' we obtain as above the mapping of C into the group Hn+2 of orthogonalmatrices of order n + 2. If this mapping is homotopic to the constant one thenb(C) = 1, otherwise b(C) = 0. Then the Pontryagin invariant is a residue modulo 2:

P

b(M2) = > b(A;)b(B;).1=1

If M2 consists of several components then b(M2) is the sum of its values on eachcomponent.

As an example, evaluate the b invariant for the vector field n defined in E4 andparallel to E. The field n has the following components:

2a 2b a2 +b2 - 1 tl;l =1+a2+b2' 2=1+a2+b2' 3=1+a2+b2' y4=0,


where a = x1 +72, b = %(v23 + x4. Set eI, e2i e3, e4 to be a fixed basis in E4. Evid-

ently, we can represent the Cartesian coordinates in E4 in the following form:

x, = a cos a, x2 = a sin a, x3 = b cos 9, x4 = b sin P.

We get the surface M2 on which the field n is constant taking a and b fixed. Hence,all of the surfaces M2 for a 0 0, b 0 0 are the Clifford tori. Let A 1 be a curve definedby Q = const, while B2 is a curve defined by a = coast. Construct the framing of thecurves generated by the field n. The vectors pa and ft (the derivatives of n inparameters a and b respectively) are tangent to the sphere S2. Consider the expan-sion of the field n nearby xo E M2:

n(x) = n(xo + ix) = n(xo) + (n0-+ ubOb

= n(x0) + n.(cos a Ox, + sin a 0x2) + nb(COS Q ax3 + sin Q A x4).

Here we have used the following expressions:

Oa 8a 8a _ _=08x, - cos a,

8x2 =sin a,

ax3 0, 8x4

8b ab ab ab0,

= =cosh, =sing.OXI = ax2 ax3 8x4

Under the induced mapping of the tangent space of E4 at xo the framing vectorsu1 (x) and u2 (x) must be turned into the vectors na and ob respectively. Consider thetorus normals

n I = cos ae 1 + sin ae2, 02 = cos /3e3 + sin j3e4.

Let the shift Ax be directed along nI. Then du is proportional to a, namelydo = oo da. If Ox is directed along n2 then do = ab A. Hence, the framing vectors u1and u2 coincide with n1 and 02 respectively. Next, the vector r, = - sin ae1 + cos ae2is tangent to A,, while r2 = - sin /3e3 + cos (3e4 is tangent to M2 and orthogonal toA,. The matrix of transformation of the fixed basis e, , e2, e3, e4 into the basisui, n2i r,, 72 has the form

cos a sin a 0 00 0 cosQ sin(3

-sins cosy 0 00 0 - sing cos Q

On the curve A, we have fl = coast, while a varies from 0 to 2r. The theorem hasbeen stated (see [72], p. 132) which allows a check to be made as to whether or not agiven mapping h of the circle S' into the group H. of orthogonal matrices ishomotopic to the constant one. Every mapping h: S1 - H. is continuously homo-topic to a mapping g: S1 - H2 = S1. It happens that g is homotopic to the constantmapping in H. (n > 3) if and only if the degree of g is even.

8a

104 THE GEONMETRN 01' VECTOR I IELDS

In our case the mapping li by itself has the form of g: S' -. H,. This mappingdefines a rotation of 01, ri in the e1, e2-plane. The degree of g is 1. Therefore,g: A I -+ H4 is not homotopic to the constant mapping and by definition b(A1) = 0.So. 6(M2) = 0. It is easy to check directly that the field n(M) is homotopic tothe constant one. To do this, replace a with ta, It with tb in the expressions for ,.Then t = I produces the given field n. while t = 0 produces the constant field.

The groups ir,,.,., (S") for r > 15 have been found by V. Rohlin, J.-P. Serre. A.Cartan. Toda and others. The variety of the structure of these groups and, on theother hand, their stabilization in growth of n became clear. Here we cite some resultson homotopy groups [78]:

7r".j., = Z_4 for it > 5; 0 for it > 6;= 0 for it > 7; Z_, for ii > 5;

2r,,, 7 = Z,.,,, for it > 9:

The latter isomorphism is a most impressive result.

2.3 The Many-Dimensional Generalization of the Gauss-Bonnet Formulato the Case of a Vector Field

Consider the unit vector field defined in some domain G of E'11-". We suppose that nis regular of class C' except for some points. maybe. Let F"' be a closed hypersurfacein E"''' which does not pass through the singular points of the field n. Denote thecurvilinear coordinates in some domain of the hypersurface F"' by a .... ; o,,,. Bymeans of the vector field n we can construct the mapping 1' of this domain onto theunit sphere S"' with the center at the origin 0. To do this we put into correspondencethe point M E F" and the point in S"' which is the end-point of n(M) being taken tothe origin. The field n depends on along F"' and hence they can beconsidered as the coordinates in S'. as well. Denote by da the in-volume element ofS". Then

da =n) do1 ...dc(,,,, (1)

where the parentheses mean the mixed product of the (in + 1) vector set in E"*'.Since the field n components are regular functions of Cartesian coordinates x'. thenby the composite function differentiation rule we have

,,(2)

For the sake of convenience. denote the derivative n,, by n,. If the set of indicesdiffers from the set ii, ... , % only in their order then

(a,,) ...,a,.,,n) =E,', 1" (n,, -,aa», a) (3)


where e;:.' ;m is the Kronecker symbol. Consider the following set of index strings

(2,3,...,m+1),(1,3,...,m+ 1),

(1,2,...,m).Denote this set by A. Each of these strings consists of m distinct integers from the set1 , 2, ... , m + 1 . Let ( i i , . . . , j,,,) E A. On the right of (2) we rearrange the terms of thesum in such a way that each string of indices (ii,... , i,,,) would be different from thefixed string (jl, ... , j,,,) only in their order. Then this sum obtains the form

m+1 ax'-(111,,...,nj.,n)..

,Ei,..r

The multiplier of (n,...... n,., n) is the determinantax', ar+-

xact, on.

Using that notation, we can write

(n0,... , n) _ (112, 113, ... , nm+ I, 11)123.. m+I

+ (n1,113,...,nm+1,n)113 . m+) +...+(n1,n2,...,nm,n)112..m (4)

The values I,,...,. can be expressed in terms of the Fm normal vector components. Ifr(01, ... , a) is a position vector of F" then the unit normal v is the normalizedvector product of r,,,, ... , r

I [r., , .... r..11

Let e ' , . .. , em+I be a positively oriented orthonormal basis in Em+1. We assume thevector product such that [el,... , em] = em+,. Then the vector product can be foundin terms of the signed determinant, namely

m

eI ... em+l

ant IF'"

art. 00.

From this we find the components of the Fm unit normal. By means of (4) and (5),we obtain

(nQ,..... nn) = (-l)ml(n2,n3,...,nm+I,n)vl- (111,133,...11m+h n)V2 + ...

... + (_l)m(111,112,...,nmen)vm+i}I[ra,,...,ro.] (6)


Note that the m-volume element dV of Fm has the following expression

dV = I[r,,,, ... , r..] I do1 ... dan,.

Introduce the vector field P as follows:P=(-1)m{(n2,03,...,Onr+1,

0),-(01,133....,Din ,1,0),...

(7)

(-W'(111, 112, ... , llm, p))- (8)

From (1), (6)-(8) the formula for the volume element of the unit sphere S"' follows,namely

do = (P, v) dV. (9)

Here the parentheses mean the scalar product in Em' 1. If Fm is a closed hypersurfacethen the 0 image of Fm covers the unit sphere some integer number of times. Moreprecisely, we are able to put the mapping i' iinto correspondence with the integer,namely, the degree of mapping >' (see for definitions [78] ). This number can be foundby the formula

0=I fdo,wm

y'-(Fl)

where the integral expresses the volume of the image of the mapping I' if both thesign of da and the covering multiplicity of the domains in S' are taken into account.The w", stands for the volume of the unit sphere S'. Integrating (8) over thehypersurface Fm, we obtain

J(P, v) dV = wm8.

F'

We call the vector P the curvature vector of the field n. This vector P is definedwith respect to any Cartesian system of coordinates. Now we state the invariantrepresentation of P which is analogous to one in Section 1.10. Define the vectorsequence k, inductively:

kp = 0,...,k;+1 = Q);,n,...,

where i = 0, 1, ... m. Remember that we denoted by Si the symmetric polynomials ofthe principal curvatures of the second kind. The following theorem holds.

Theorem The curvature vector P can be expressed in terms of k; and symmetricpolynomials Si with the following formula:

P=(-1)m{SO+Sn,_1k1+ +km}. (10)

There is another way to represent k;. Let A = IIai;II be the matrix of order (m + 1):

A = IIi... II.

C+I +Sl S,n+1


The vector k1 has components ', where the summation over p is assumed.Therefore, the components of k2 are as follows: t,' Ip

P. In general, the componentsof k1 are

where the summation over j, r, s, ... 1,p is assumed and the number of indices undersummation is i. The index a shows that the expression stands for the a-th componentof k1. On the other hand, the vector An has the form

Ci ... Cm+I

An =mf1 tm+I p

Comparing the components of kl and An, we see that k1 = An. Further, A2n = AAnhas the form

A2n =

I 11 "nr+I

m-fI ... fIS M+I

Comparing components, we again conclude that k2 and A2n coincide. In an ana-logous manner we find that k1= A'n. Thus, the vectors k1 are the results of con-sequent actions over n of a linear operator with matrix A.

Let B = Ilbuli be a matrix formed with the elements buj which are the cofactors ofaj1 in the matrix A. Show that P = Bn. To do this we compare, for instance, the firstcomponents of those vectors. The first component of Bn is

c2 ... Sm+I

b111 = cI _ t2

Cni, I ...C.

m+11Sm+

I ' fin. Im

rI

r2 t2 tm h 1

m+lQ2

+1 ...C+4

1


The first component of P has the form

m() el e2 ... em+I

2 ; ... V+ 1

Stt1m+i

t.2'+1Smttm f+ 11

SI S21...

Sm+lc2 C2

m+2I `mti lSCm+ISm+ l

Thus, the first component of P coincides with the first one of Bn. The analogousresult is true for all other components. Denote by C = IIC,iII the matrix -A. In [75],p. 93 the definition of the adjoint matrix, denoted B(µ), of the matrix C is given. Theelement bij(µ) of the matrix B(µ) is a cofactor of µbj - Cij in the matrix II;`bij - CijII.In a particular case, the matrix B(O) is formed with cofactors in the matrix -C, i.e.with cofactors in A. Therefore, B(O) = B. In [75], p. 94 the expression for B(µ) interms of the coefficients of the characteristic polynomial 0(µ) = Io,j - Cijl and thepowers of the matrix is presented. If we write the characteristic polynomial of C as

001) _Am+1 -pill, -...-pmfl,

then for u = 0 we obtain

B(0) = C"' -PICm-I pE,where E is a unit matrix.

Since A = -C,

B(0) = (- l )mAm - PI(-1)m-I Am-1

- ... +Pm- I A - pmE.

Let us find pi in terms of symmetric polynomials of principal curvatures of thesecond kind µI, ... They are the roots of the equation

IAbij lµb,j - C,i1= 0,

i.e. µl, ... ,.a are the eigenvalues of matrix C. Thus, we have the following ex-pression for pi:

P1 = SI9 P2 = -S2, ..., P,

Hence

B = B(0) _ (-1)m(SmE+ Sm_IA + + Am).

As P = Bn,

P (-l)m{Smn+Sm-IAn+...+A"'n).

Replacing the Ain here with Iri, we come to formula (10).

VL(TOR I I L ! 1)S IN M ANl-I)IMl N' IONA1. rt.(LII)I: %N AND RII.MANNIAN SPACI.S 110)

Using (9), we obtain the following generah:e(l Gauss-Bonnet formula for the caseof a rector field in (in + I)-clrmrnsional Euclidean .spare:

1)", r (S,,.n + 5,,,-Ik1 + . + k,,,, ii) th, = u)..0.

In the case when the field n coincides on F" with the field of normals we come to thegeneralization of the Gaus'-Bonnet formula to the hypersurface

Observe some properties of the vector field P. Just as in Section 1.5, it is easy tostate the geometrical nnarung of the liele! P streamlines the field a is constant alongthem. In the previous section we denoted these curves by M1. From (9) it evidentlyfollows that if P - 0 then the volume of the n/' image in S' of any hypersurface F"' iszero. i.e. the dimension of the image is lower than in. Therefore, the vector Pcharacterizes the curvature of n in dimension in. However, if P s 0 then the field n isnot necessarily constant.

Since (a1. -..a,,,;1) = 0. it is easy to check that div P = 0.

2.4 The Family of Parallel Hypersurfaces in Riemannian Space

Let us be given a hypersurface F in a Riemannian space R". Let n(x) he a regularfield of normals offly) defined in some domain G c t We draw a geodesic ofambient space in the direction of n(x) through each point x E F. We shall call thefield of tangent vectors of geodesic lines the geodesic field. Denote it with the sameletter n. Let M he a point in a distance s on the geodesic from y E Fin the directionof n(t ). If r ranges over all points in the domain G then the set of marked points Mform, in general. some hypersurface. That hypersurface is called geodexicalli parallelor parallel simply with respect to the hypcrtiurface F. Denote this surface by F,Considering various small fixed values of s, we obtain a family of geodesicallyparallel hypersurfaces filling some domain in the Riemannian space. The geodesiclines are orthogonal to those hypersurfaces at each point. If we takes to be large thenit may happen that the geodesics from different points of F intersect each other and,as a consequence. either we have singular points in F, or F, degenerates into sub-manifolds of lower dimension, moreover into points.

We consider the following question: how do the principal curvatures of geodesicallyparallel htpersurfaces change in moving along the geodesic which is orthogonal to thesurfaces of the famil r. A tangent direction r on a hypersurface of a Riemannian spaceis called principal if the covariant derivative of a in the direction of r is collinearwith 7-

V'0 = -Ar.

The coefficient A is called the principal curvature pf the hypersurface. At each point ofthe hypersurface we have n - I principal directions.

Theorem Let r he a principal direction an a htpersurface from the famllr c1geodesically parallel hrpersurfaces and ,\ flit, a corrc'sprntdnrg principal curvature. Then

110 THE GCOMFTRY OF VLCTOR FIFLDS

the derivative of A nith respect to the arc length parameter of the geodesic orthogonalto the hlpersurface has the following form

d+K(r,a), (1)

where K(r, n) is the curvature nf'space with respect to the plane generated by r and n.

Let us be given a coordinate system in the space domain under consideration. Asusual, we denote the vector contravariant components by superscripts and thecovariant derivative by "comma". In these notations the equation for principaldirections has the form

n', rr' = -Ar'. (2)

Differentiating both sides of (2) in the direction of n, we obtain

n',,Ar'nA+n',,r',An' _-4

r'-Ar,4nA. (3)

Multiply both sides by r. Since r is a unit vector field. (r, A nA, r) = 0 and hence theright-hand side reduces itself to -dA/ds. We have

r n',,Ar' n A + r, n', , r', A ll = -(/A

ds (4)

Consider the second term on the left-hand side of (4). The r,k nA is orthonormal to r,so that we have the following decomposition

r', A ttA = lm' + (5)

where T. stands for the principal direction different from r and having A as thecorresponding principal curvature, lc and v are some numbers. Let us find it. To dothis we multiply both sides of (5) by a scalarly:

n, r', At` = la

Transform the expression on the left-hand side. Bring it, under the sign of the co-variant derivative in k and as a compensation subtract Then we get

(nir'), AW - r' n At = lr.

Since n is orthogonal to r and the field a is a geodesic one: n AnA = 0. as a con-sequence. is = 0. Using (5) with It = 0. write

r, it', , ,r'. A11A =,r, n',, V,, rii 0. (6)

Consider the first term on the left-hand side of (4):

r,n', , rift = r,(n', A, -(nl, An') IT'r, - n', Ant, ,r'r, - K(r, n), (7)

where R't,A is a curvature tensor of R.

VECTOR FIELDS IN MAN)-1)IMI NSIONAL AND RIEMANNIAN SPACES III

The first term on the right-hand side of (7) is zero because n is a geodesic field. Thesecond term is

-n', AW,,r-'T, = Ar r, = -112. (8)

So, after applying (6)-(8). equality (4) gives (1).

If we set A _ -' then (1) has the form of the Jacobi equation

dt' `k(r.a)q.

This equation allows us to estimate the change in principal curvatures of geodesicallyparallel hypersurfaces along the geodesics orthogonal to the family under the givenrestrictions on the curvature of space.

2.5 The Constant Vector Fields and the Killing Fields

The field n = {e) in Riemannian space R" is called ewistanr if the covariant derivativeof this field with respect to any direction is equal to zero:

V,EA = 0

Since n= V,n' = 0. Therefore, the length of such a field is con-stant. If two constant vector fields are given then they make a constant angle. Thecurvature vector of the constant vector field streamline is tr'V,t;A = 0. i.e. thestreamlines of the constant vector field are geodesic lines. Let a and b be orthogonalto n, i.e. (a, n) = 0. (b, a) = 0 Then

To - ©ab. n) = (Vba, n) - (V8b. 0)= Vb(a, o) - \7.(b. o) - (a, L7bn) + (b. V,n) = 0.

Hence, n is holonomic. Introduce the special system of coordinates _ Y ' , suchthat the surfaces xt = const are orthogonal to n. As xt we take the distance fromsome fixed surface of v' = const along the streamline. Then the first fundamentalform of the Riemannian space has the following expression:

it

,IV-' = (dxt + g,dx" dr".

Since n is supposed to be constant, the functions g,,;, do not depend on .rt. Withrespect to the coordinate system introduced the field a has the following compon-ents: 0. i > 2. Consider the derivative

V, I = v4- rz, r:,


From this it follows that rj = 0. Since gIr = 0 for a 0 1,

r' - 111 091i + 1991k _ ag,k 111 ag,k = 0.- g i axk axe axl =

-g2 axl

Thus, if a constant vector field exists in a Riemannian space then the metric can bereduced to the following form:

n

ds2 _ (dCl)2+ > gnj(x2,...,x")dx"dxd",d=2

and vice versa.From this it is easy to see that the Riemannian space R" which is not locally

Euclidean admits no more then n - 2 constant vector fields.Now we turn to the consideration of Killing vector fields. Let x' be the local

coordinates in Riemannian space R". The infinitesimal transformation of R" is, bydefinition, the correspondence between the points of coordinates x' and x' given by

z'=x'+bt,where are components of some vector field on R", bt is some fixed infinitesimal inR". The field is, in general, of variable length. The infinitesimal shift bx' = z' - x'has the form

bx'=e bt.

We shall call t; the Killing field if corresponding to infinitesimal transformation themetric of space does not change up to the infinitesimals (6t)2, i.e.

b ds2 = 0

up to (6t)2. In more detail, if £ is the Killing field then

bds2 =gri(x")dz'd. 1-g, (x")dY' dxj

is the infinitesimal of a higher order than bt. We have

dx=d_Y'+ dxkbt,

gi(X") = gr1(f) + x bx" = gr1(x") + ax"bt.

Up to infinitesimals (61)2, we have

airdi' d 2 J = d x ' dxJ +

8xkd x ` d x ' bt +

a x kd x k dx bt.

Hence

b ds'- _ ( " dx' dxj + g y 6OXk'e dxk dxj + g;; 8k

dxk dx' 1 bt.

VECTOR t'ILLDS IN MANY-DIMENSIONAL LUCLIDtAN AND RIEMANNIAN SPACES 113

In the second term on the right we make the substitution k -' i - k, and in the thirdWe get

bdi +a ae

laxa gkj +a aavj) dYj br.

By the definition of a Killing field, b ds2 must be the infinitesimal of higher orderthan bt in all choices of d_xi or dxj. Therefore, the Killing field satisfies the followingsystem of equations:

ag;;

axt ` + gkjaaxi

+ g.k

aaxj= 0.

Since = 1';,,; + f,,,, we are able to represent the latter equations as

gkj1 +I'n +ga xj +r =0.

We have the covariant derivatives in parentheses. Since the covariant derivative of ametric tensor is zero, the equations for a Killing field can be expressed as

(l)

These equations are called the Killing equations. The divergence of Killing field is zero.Indeed, contracting both sides of the Killing equations with gki, we obtain

4k = -gki jjFurther, setting k = i and contracting in this index, we obtain

divC = -dive = 0.

If the length of a Killing field is constant then its streamlines are geodesics.Show that the second covariant derivatives of Killing field components e; can be

expressed in terms of these components and the Riemannian tensor. Consider theKilling equations (l) for three pairs of indices

&.j + Gi = 0,CI,k + k.J = 0,Ck,i + 6k = 0.

Covariantly differentiate the first equation in xk, the second in xi, the third in xJ.Add the first and second equations and subtract the third one. We get

Silk - &.kj + Ck ji - Sk.iU + f jki + t j.ik = 0.

Using the expression for the difference of second-order covariant derivatives oftensors, we get

(RU + Ri + R°kjAn + 2G& = 0.


We can replace the sum of the two first components of the Riemannian tensor inparentheses with R°kii. Hence

CjJk = -'R°kJI u. (2)

Also, there are the relations between the first-order derivatives of Killing field com-ponents and the original components:

Sf- = 0.C. ( o.k - Rpkji.!) + R /j, E,°.k - R Lit F,,j - jklO.1 - RiklF

These relations arise as the compatibility conditions of (2).

2.6 On Symmetric Polynomials of Principal Curvatures of a Vector Fieldon Riemannian Space

The symmetric polynomials of principal curvatures of the second kind of a vectorfield with a constant length in Riemannian space have the same expressions as in thecase of a vector field in Euclidean space if one replaces the usual derivatives withcovariant derivatives:

Sk=(-1)k EII <.. <I4 71 ... Sff4

We are going to find the analogs of the divergent representations for S2 and S3. Letk = {t'jtJ} be the field C streamline curvature vector. Consider the divergence of thefield -(SIC + k):

div (-S1C - k) = (Sii,`'2 - el,C" ).r,

+ 2S2 + Rit C° {s,

where R,,.r are the Ricci tensor components. The expression -R°j t° tj is the Riccicurvature for the direction C. It is equal to the sum of sectional curvatures of Rie-mannian space along the planes spanned on C and n - I vectors which are mutuallyorthogonal and orthogonal to C. Therefore, we have

2S2 = div (-Sit - k) + Ric (e).

In the case of a space of constant curvature K0

2S2 =div(-S1 -k)+(n- I)K9.

It is convenient further on to introduce a column


We shall denote the determinant of a matrix formed with this column and its deriv-atives by brackets. Then 6S3 = [a,i,a,i2a,i3J. Take out the sign of covariant differ-entiation and subtract the extra terms. We have

18S3 = [aa,i2a.;,];, + [a,i,aaj,].i2 + [a,;,a,i2a].i,

- [aa.i2i, a,ry] - [aaj2a.i,i,] - [aj,;2 aa.;,]

- [a,i, aa,, ,2 ] - [a.;,;, a,i2 a] - [a.1 , a.;2;, a]

Consider the sum of terms containing covariant derivatives of the second order.Gathering them in pairs, we get

-[a,i,,2 - a,i2i,aa,i3] - [aa.i2a,i3i, - a,1113] - 1a.;,aa,i3i2 - a,,24 ).

Now we use the formula for the difference of second-order covariant derivatives of atensor t. In detail the first of the determinants above is

I

,,t, Sri,[aj,i2 - a,,21,aaj,] = I

I R°i, i21;. t", I

_ +R;;.3itoele - 2R0 to 0.

It is easy to check that the other two determinants containing second-order deriv-atives of a are equal to the first one. Now consider the determinant [aa,i2a,3] whichstands under covariant differentiation in zi, . Expanding it with respect to the firstcolumn, we have

t;, tth ti, t1 t;, ttl, C;, c;, el (S , tI2 ;1[aaj2a,i,] = S (SJ2 Sj, - Si2,i,) - SI2 (e S,;, - Sj2,i,) + S ,i2 °,

Evidently, the expression in the first parentheses is equal to 2S2. Now gather theterms containing t; and 22. The other two terms are equal to each other aftersummation over the index change. We obtain

[aaj,aj,] = 2(t" S2 + k" SI + ey 1;1i).

The expression above is the i-th component of the vector P which we introduced forthe case of a vector field in Euclidean space. Note that the expressions for the othertwo determinants, namely [a,1, as j3 ] and [a,,, a j2 a], can be obtained from the expres-sions presented by the substitution of i1 by i2 and i3 respectively. Therefore, we obtainthe following formula for the third symmetric polynomial S3 in the case of a vector fieldin Riemannian space:

3S3 = divP+R°0t"40SI +R,,,3t°k'3 - R01tft

Consider the expression for the sum of terms with Riemannian and Ricci tensors inthe case of space of constant curvature K0. We have

-R°;irCp = (n- I)Ko, R.,3f°V = ng°uCa k" =0.


Introduce a coordinate system in such a way that at some fixed point g,,,j = 6,,;j andthe coordinate curve x' is tangent to t. Then

R".,jfrEjei = -R1111C; = K0S1

So, in the case of a space of constant curvature the third symmetric polynomial of theprincipal curvatures of the second kind of a vector field has the following expression:

3S3 = div P - (n - 2)KOS,.

Since S, is the divergence of a vector field -t, then S3, in contrast to S2, can berepresented as a divergence of some vector field, namely of 3 (P + (n - 2)KoC).

2.7 The System of Pfaff Equations

Consider the expression of the form

(1)

where a, = ae(x' , ... , x") are differentiable functions of coordinates x',... , x". Thisexpression is linear with respect to dx' and is called the Pfaff form.

We shall suppose that in a coordinate change the set of functions at satisfies thetransformation law of covariant tensor components. Namely, if dk are its compon-ents with respect to a new system of coordinates y', ... , y" then

at=dkaYk

ax'

In this case the form expression is invariant with respect to the coordinate change:

w=a1dx' =aAaax'dv=dkdvk.

Consider the two symbols of differentiation, namely 6 and d. We shall adopt thatd bx' = 6 dx'. The differentiation b produces the differentials 6x'. We denote the Pfaffform with respect to bx' by w(b).

To each Pfaffian form w we are able to put into correspondence the form of secondorder, namely dw(b) - bw(d) which is called the bilinear Frobenius form.

In differentiating w we use the functional character of at. We have

bw(d) = b(aj dx') = ba; dx' + a;bax-'

bxi d.4' + a;b dx',

dw(b) = a " dxjbxk + akdb.%'.

From this we find the expression for the bilinear Frobenius form:

dw(li) - bw(d) = auk drj bxk -auk bxjdak = I Qak _ aaj (dri bxJ dxk ).axi ax/ 2 Xj axk


It is easy to check that the coefficients (a 'TO a ) form a tensor. The equation w = 0is called the Pfaff equation. Let us also define the bilinear form w(d, b) which dependson both shifts dr and br:

w(d, b) = a;1(dx' bx' - dxj bx'),

where a; form a tensor of second order, i.e.

03A IVa;; = dk1i iix-i

The form of this expression is invariant in coordinate change:

w(d, b) = dk, a a (dx' bx' - dx' bx')

= ak,(dyk b)' - dy'byk)-

Therefore, if the form w is zero with respect to some coordinate system for someshifts dr and Or then it is zero with respect to any other coordinate system for theseshifts.

Let us be given the system of Pfaff equations in a domain G C E"

wI =alidx'=0, i= 1,...,n,(2)

wrn = ami dx' = 0, m< n .

We say that (2) is not degenerate at each point a if the rank of the matrix IJa,311 is equalto m. This definition can be interpreted geometrically. Consider the system of mvectors A, = {a11}, ... , A. = {am;} in a Euclidean space. The condition of non-degeneracy means that A1,.. . , Am are linearly independent. We denote the spacethey span by Tm (see Fig. 21). The surface Vn_m of dimension n - m in Euclideanspace E" is called the integral surface if any shift {dx'} in tangent space of V"_msatisfies system (2). This surface is orthogonal to Tm at each point. In this case,i.e. when Vn_m exists, we say that the Pfaff system is totally integrable or holonomic.

FIGURE 21


The condition, under which the system (2) is totally integrable, gives the followingtheorem.

Frobenius theorem The Pfaff system of equations w1 = 0, j = 1.....m is totallyintegrable in a domain G C E" if and only if for all w, the bilinear Frobenius forms arezero by virtue of the system wj = 0.

We cite the proof from [26]. The conversion to zero of bilinear Frobenius forms byvirtue of the system w. = 0 means that these bilinear forms are zero for all shifts{dr'} and {bx'} which satisfy the system (2).

Prove, firstly, the necessity. Suppose that the system wj = 0 is totally integrable.Let T"' be a space perpendicular to T'. Consider the linear mapping M of E"

onto T"' which is the orthogonal projection onto T' at each point P E G. Thismapping maps all the vectors of T"-" into zero and is identical on T'. By thehypothesis, the integral surface V"_", of (2) exists. Let

be its parametric representation.The tangent vectors r,,, of this surface are in T"-'. Hence,

Mr,, =0, (3)

The linear mapping M with respect to the fixed coordinate system in E". at a fixedpoint P is determined by the matrix of order n. The elements of this matrix are thecontinuously differentiable functions of P, i.e. of parameters u,. The derivative of Mmeans, in our considerations, the derivative of the corresponding matrix.

Differentiate (3) in u;. We get

Mr,,, + M,,,r, = 0. (4)

Interchanging the roles of i and j and subtracting the result from (4), we find

M,,,r,,, - M,,,r", = 0.

Multiply this by bu, and dud. Then sum over i and j from I to n - m. Then we obtain

(M., bui)(r+,, du,) - (M,,, du;) (r,,, buy) = 0. (5)

We can express the first differentials of M and r in terms of the coordinate x, in E":

M,,, but = Mr bx', M, du, = M. dr',

r,,, du, = rr, dx-' = dr, r", bud = r,,k bxk,

where dr and Or are tangent to V"_" i.e. they are in T"`"'. To simplify notations weset

MG bx' = M,,r, M,. dr' = Mar.

VECTOR FIhLDS IN MANY-DIMENSIONAL I UCLIDFAN AND RIEMANNIAN SPACFS 119

We rewrite the equation (5) in a brief form

Me, d r - M,,, fir = 0, (6)

where dr, Fr E T"-'. Express condition (6) in terms of the coefficients of a givenPfaffian system ,v, = 0. To do this. we find the matrix of Al. Suppose that M takesthe vector {d.'} into the vector {di'} by

dr' = E nn,A dt ,

A

where IIm,A II is the matrix of M. Since M takes any vector {de'} which satisfies thePfaff system (6) to zero, EA nil, dt1 is a linear combination of systems of Pfaffequations

nI,A dr' dxA , j = 1, ... , it,

where A,, are some indefinite multipliers. The latter expression holds for all shifts{dxA }. Therefore, ni,A = A,, u,A. Since M takes all of the A, _ {a,1, - - , a,,,} intothemselves.

A,,a,Aa/A =a!,.

Here the first subscript in al, is a number of Al and the second is the number of thesevector components with respect to E". The matrix 11a,,11 is not square, but it may becompleted to a square matrix in such a way that the determinant will be non-zero.This corresponds to the possibility of completing m independent vectors A, to a basisin E". We denote the complemented matrix by A. Then we may take the componentsof the inverse matrix A-I as the A,,. Consider (6). We have

M,,, =8M S.r".

The matrix aM/a " consists of elements am,A /Ox". Therefore, we have

Mk,=IIIs'a.vok i

The components of MA, d r - Mjt br have the form

am,A6x" d,k -

am,Adr" 6x' =

aa,, a,kbx" dxk

a,Adv" ax"

ax" Ox" av, &e'aa,A _ 8a,,,

d.%Aa+ bx"(aA dxk) -

O; dx"(a,A dk).

By virtue of (6), the latter expression is zero.


The shifts dr, br E T"-"'. Therefore, the last two terms are zero. Since the de-terminant JA,,J -A 0, the system

a/.s{bws(d) - &r(b)) = 0

has trivial solutions only, i.e. for all s the linear Frobenious forms are zerobw,(d) - dw.,(6) = 0 by virtue of the system w,(d) = w,(b) = 0. Thus, the necessarycondition is proved.

Now prove that this condition is sufficient, also, for the system wf = 0 to beholonomic. We shall construct some surface and prove that this surface is tangent toT` at each of its points. Consider at Po E G a unit vector T E T"-m(Po) with theinitial point at P0. Its end-point lies in a unit sphere S which is centered at Po andlocated in T"-"'(Po). The dimension of this sphere is n - m - 1. Introduce the co-ordinates u1, ... , u"-m-1 in this sphere. Then r can be considered as a vector functionof these coordinates:

Span the (m + 1)-dimensional plane T"'+I (Po) on r and Tm(Po). Let P be an arbit-rary point in Tm. I (P0). At each point P the corresponding plane T(P) is defined.Project Tm(P) into Tm"(Po). In a sufficiently small neighborhood of Po the pro-jection is a one-to-one mapping and the result of projection is of dimension m.Denote it by Tm(P) (see Fig. 22). At each point P E Tm+I (Po) we have a vector fieldorthogonal to Tm(P). Draw the streamline of this field through Po in the direction ofT. This curve is orthogonal to Tm(P) because it is orthogonal to the projection ofTm(P) into T" (Po). If we take the various r at Po and draw the curves -y as abovethen we obtain some (n - m)-dimensional surface of the position vector

r = r(ul,...,u"_m_t,o),

where a is the arc length parameter in ry. By construction, r, being tangent to ry isorthogonal to T"'(P). Hence,

Mr,=O. (7)

Show that

Mr,,,=0, i=I,...,n-m-1. (8)

This means that all tangent vectors of the constructed surface lie in T"-m(P) forevery P, i.e. the surface is the integral one of the Pfaff system (2) and, as a con-sequence of the definition, the system is holonomic. Introduce a vector a = Mr,, andfind a linear differential equation this vector satisfies. We have

a, = Mr,,, +8M

r,,, = Mr,,,, +8M 8xJr,,,

= Mr,,,, + Mrsrw (9)80 8x) 80If we differentiate (7) in u,, we get

8M 8x!M r,,,; + M,,r, = M r,,,, + axe au r, = M,., r,, = 0. (10)


FIGURE 22

Subtracting (10) from (9), we obtain

aQ = Mr,r., - (11)

By the hypothesis, all the bilinear Frobenius forms of wj are zeroes by virtue of thesystem (2). Therefore, the differentials d r, br E T" satisfy (6). If it were known thatr", E T"-` then we could apply (6). Consider the decomposition of r.,,

r,,, = a + b,

where a = Mr,,, E Tm(P) and b E T"-"(P). Also, we can write

Mr., = M. + Mb.

Then (11) can be rewritten by virtue of (6) as

a Mr,(a+b)-(M.+Mb)r., =Mr,a-M.r". (12)

So, the vector a satisfies some linear differential equation. At Po, where or = 0, wehave r,,, = r,,, (ul , ... u"_,"_ 1, 0) because the initial point Po is the same for all 7starting from Po. Therefore, a(0) = Mr,,,(ul,... un_rn_1,0) = 0. Hence, (12) impliesa(a) = 0, i.e. at an arbitrary point P of the constructed surface we haver,,, E T"-'"(P). The theorem is proved.

Consider the particular case, when the system (2) consists of a single equation

EIIdx'=o.1=1

We may assume that in a domain G C E" a unit vector field a = {f,} is defined. Thefield a is holonomic if the family of hypersurfaces having this field as a normal oneexists. By the theorem just proved, this occurs if and only if the bilinear Frobeniusform dw(b) - 6w(d) of the form w is zero for the shifts dr and or which are ortho-gonal to n. In terms of the components of n, dr and Or this condition can beexpressed as

" Cep al; p)(dx° bx0 - dxa bx°) = 0. (13)

axa - ax°a'o


Choose the system of coordinates in E" in such a way that at a fixed point P thecomponents of the vector n are (0,0,..., 1). Then for every admissable shifts dr and6r one must have de" = 0 and 6x" = 0. Hence, in (13) the summation over a and 0goes from 1 to it - 1. If we take the shifts dr = (0, ... , . j,0,-..,0) andr = (0, ... ,1, ... , 0), where the unit takes the a-th place in d r and the 13-th place in

6r=(0,-..,0

or. then we obtain the system of (n - 1)(n - 2)/2 equations:

aF1'-kK = 0, 1 < a, 8:5n-1. (14)art+a,Denote the derivative 8/8x" by 8,,. Consider the set of the following expressions withrespect to an arbitrary system of coordinates in E":

where the brackets which include the indices a, /3, 7 mean the sum of various termswhich can be obtained from g (, g, with permutations of indices a, Q, y; also, thecorresponding expression is included in this sum with a *'+ " sign if the permutationis even and a "-" sign if the permutation is odd. Consider the way Sl,,i, changesunder the coordinate change in space. Let x" = x"(,ti''). Then the components , of nwith respect to the system y' are related to its components with respect to the systemx" by the tensor rule

8x"

Let us form the expression analogous to Sl a, with respect to the system

aaM )ev,

, ,(6ayk ,1

_ &001 Sri8x"' 80 ex" + Oaf' a2 '

yla)A TV-, ay" a eayll

The last term on the right is zero due to the equality to each other of the mixedderivatives of the functions xri = xs(y) and the convention given above for thebrackets. The first term can be transformed by relabelling the summation indices as

ax" 8.0 ax's ax" ax' axtfsl; ay, aye ay, = ay 8y, ayk .

So, the quantities Sl h, satisfy the tensor rule of transformation. If the field is holo-nomic then all components 0 at the point P with respect to the system ofcoordinates chosen above. Observe that if a tensor fl",s, and bilinear Frobenius formare zero with respect to a given system of coordinates then they are zero with respectto any other system of coordinates. Therefore, the condition Slap, = 0 is necessaryand sufficient for the field n to be holonomic. In the particular case, when the field ais in a domain G C E3, that condition gives (n, cunl a) = 0. The tensor fl"p, is calledthe object of non-holonomicity of the field a in E".


2.8 An Example from the Mechanics of Non-Holonomic Constraints

In mechanics, the restrictions posed on the motion of the body can often be ex-pressed in the form of a system of differential equations. This system may either beintegrable or not. As a typical example, we consider the equations of constraintswhich arise in a disk D rolling over the xy-plane. Suppose that the disk is alwaysperpendicular to the xy-plane and the disk plane intersects the xy-plane in a straightline which makes an angle cp with the x-axis.

Set 0 to be a center of D, R is a radius of D, A is a point of contact of D to the xy-plane, B is a fixed point on the boundary circle of D which was the point of contactat the initial instant.

Denote the angle between OA and OB by t/, (see Fig. 23). As the rolling of the diskis assumed to be without sliding, the length of the curve -y in the xy-plane is equal tothe length of the circle arc AB. In an infinitesimal motion of A this arc of length Rd:l,is the infinitesimal length of -y, namely, ds = dx2 + dye. Hence, we have

dx = R cos w d O,

dy = R sin cp dO.

Consider the four-dimensional space E° of the parameters x, y, p, r/, and the systemof Pfaff equations

w, = dx - R cos Wd4& = 0,

w2=dy-RsinVdb=0.

The bilinear Frobenius forms are

(I)

dwl (6) - 6w1 (d) = R sin (p(dW 6o - bcpd,b), (2)

d w 2 (6) - b w 2 (d) = -R cos cp(dW 60 - bp dpi).

The admissible shifts (dx, dy, dco, dpi) and (bx, by, bhp, bpi) satisfy the system (1).Therefore, they can be written as (R cos cp dr/,, R sin p d pi, dW, dl/,) and (R cos cp 60,R sin cp 6i1', bcp, 60). Hence, the differentials dip, dpi, 6cp, 60 are arbitrary in admissible

FIGURE 23

I24 THE GEOMETRY OI VECTOR 1IELDS

shifts and can be taken in such a way that dtib+p - depbtp 96 0. It follows from (2)that

2

>[b'a+(d) - dca,(b)j = R2(dpft - bpdLi)2 # 0.1=t

Therefore, the system of equations of constraints for the disk rolling over the plane isnon-holonomic.

2.9 The Exterior Differential Forins

Let x....... " be curvilinear coordinates in a domain D of a Riemannian space V".The structure of exterior differential forms can be expressed in terms of the co-ordinate differentials &e. Introduce terms of the form

a "Aa A...Adt1P.

It is possible to multiply them by numbers and to add them to each other. Also. theypossess a skew-symmetry property. i.e. if ft,... , jl, is a permutation of numbers

then

d.0Adt1A...AdSp=E' Fdx''Adt"A...Adx'r,

where the generalized Kronecker symbol is

11 , = I if the permutation is even,h if -1 if the permutation is odd.

The exterior differential form of order p with respect to local coordinates is thefollowing sum:

w E a,, ,,d"A...Adx''It< <1,

or

w='I E a,, ,I,dx" A...Adx+r,P ,I 1r

where the summation is taken over the indices it ... ip which takes values from I to nindependently and the form coefficients a,, 1,, satisfy the skew-symmetry condition

a,, 1, = ±a,, ,,

where "+" is assigned if the permutation ('.is even and "-" if the per-mutation is odd. t '

Suppose that in D we have other curvilinear coordinates such that


Consider the coefficients of w with respect to the system of coordinates u' . We have

w = li >2 al,...i, r du; A ... A du°'.

Therefore, the coefficients of w with respect to a new system of coordinateshave the form

Oxi" Oxi,an,...°, = al,...i, ale,

i.e. the quantities a1,,.,,, form the covariant components of a tensor of order p.As an example, we present the forms w; of order i in three-dimensional space

a=Adr'+Bdx2+Cdx3,0 =Pdx2Adx3+Qdx3Adx'+Rdx'Adx2, (1)

ry=Sdx'Adr2Adr3,

where A.... S are functions of x', x2, x3.For two given exterior forms their exterior product is defined. Let w, and w2 be the

exterior forms of order p and q respectively:

W1 = l1 >2 a!, ...i, dx" A ... A dxip,

w2 =1

>2 bl,..i. dxf' A ... A dxi'.q'i,. .ip

(2)

The exterior product wi A w2 of the forms w, and w2 is the following form of orderp+q:

w, Awl = p - 2 ai, ...,, bi, ...i dx" A ... A dX" Adze' A ... A dx .

i, #

(3)

For example, the exterior product of a and Q from (1) is the form of order three,namely

aA,3 = (AP+BQ+CR)dx' Adze Adr-;.

If we transpose the cofactors then the exterior product either stays the same or onlychanges the sign. Indeed, w2 A w, can be obtained from w, A w2 by transpositions ofthe differentials dxj^ and dxi-,. The number of those transpositions is pq. Hence,

w, Awe = (-1)Rw2 Aw,.

Let us define the exterior differential do of the exterior form w. Denote thelinear form by dal,._.j,, namely the differential of the function a,,..;,. Then the


exterior differential of the form w of order p is the following exterior form of orderp+ 1:

d w = I dais...,, A A . . . AP1

it ... i,,

(4)

The following rule of differentiation holds for the exterior product the forms w, andw2:

d(w1 Awe) =dw1 Awe+(-1)"WI Adw2. (5)

Indeed, taking into account (3), we have

dw1 Awl =PQI

E{(dais...i,,)bj,...j,, dx" A... Adxj' +ai,...i dbj,...j, Adix'' A... AdxJ,

The first term is dw1 A w2. In the second term we make p transpositions of thedifferential dbi,,,,i, with the differentials dxi', ... , dx". Then this term gives the formfor the exterior product w, A (-1)"dw2. Formula (5) is proved.

The following lemma plays an important role.

Poincare lemma The second exterior differential of a differential form is identi-cally zero.

ddv=0.

Consider the expression for the exterior differential

aaxa1 W A ... A dxj'.CAW EThe coefficients of this form are the functions e1°. Hence

dew daai,..." Adx" Adx" A... Adxi'I

= 10dx"t

A dxa A &0 A ... A t:&'. (6)(p +

1)I

. axaxaloll,....,ip

Since dx-6 A dxa = -dxa A dx' and the second mixed derivatives of ai,,..1, are equal toeach other, the sum on the right of (6) contains the term of opposite sign for eachterm. Therefore, this sum is zero.

The inverse statement is also true: if the exterior differential of some form w is zero,i.e. dw = 0, then locally, in a sufficiently small neighborhood, the form w is the exteriordifferential of some other form.

In other words, if dw = 0 then in a sufficiently small neighborhood there is a form wlsuch that w = dw,.

VI CI OR IN LUC AND RIFMANNIAN SPACES 127

To prove this, consider firstly the form w of order 1, say w = a, dY. Then dw = 0implies

all, Arc, _ 0.c7v car'

This condition, evidently, is sufficient (and necessary) for the existence of such afunction / that a, = /;,, i.e. w = dl.

For the case of a second-order form a dr' A dr' the condition dw = 0 impliesthe system

Ou 0t4,(7)

As w, we take the first-order form w, = h, dr'. The equation w = &,I can be re-presented as

i)h, ah,- a,

ox' Ox'

We rewrite this system in a definite order. First we write the equation containingthe derivatives of h,. then the remaining equations with derivatives of h, and so on.We get

Oh, _ 8hcar- nzl,

(8)

I _ Oha, _ l=Jxn

We shalt solve this system step by step from the bottom. We take an arbitrary C2function as h,,. From the last equation we can find hi_1 by integrating with respect tox". Substitute the functions h,, and into the previous equation. Suppose that wealready found the functions Let us find b,-,. This function is included in agroup of equations of (8). namely

Oh,_ 1 _ 8h, _

(9)

Oh,-1 Ohaln- I

t'i?1" (fit'-1

This system is integrable if the compatibility conditions are satisfied: for any o, /3 > ithe difference of mixed derivatives of b,_1 is zero:

0 = O-h,_ I 0--b,_ 1 8h(10)8 "'ar' Ox' 6x" +OJ.Vl-I 0x days)

128 THEGEOMI:TRI OF VECrOR FIELDS

Since a, 13 > i, the h and b,1 are already known and satisfy the equation

Oh,, 8b;,0.0 ar'

Therefore, on the right of (10) we obtain

aa,,,-1 + Oa,-i;i + su.*, = 0a. =

which is zero by virtue of system (7). Therefore, there is a function hr_1 satisfying (9).The sequence of functions h,,, ... , b ... , h1 provides the solution of (8). So, for thecase of the second-order form w the statement is proved.

The same method is applicable to the form

w - a,, , dx" A . A dx''".

The condition dw = 0 implies the system

Consider the form w1 = h ,, dY" A.. A dV' of order p. The equation w = a0w1 isequivalent to the system

ah,, , ab,,, ,t 9.V OY1"

- .. - U,,, ,P (12)

Introduce a multl-index i = (i1 ... i,,). where i1 ... i,,. It is possible to compare multi-indices.

We say that i<jifi1 <j1 In the case ofil =j1 we set i< /1(/: <j2 and so on.We put the coefficients h,, , into the sequence taking b, earlier then b, if i <J.

We also rewrite system (12) in a definite order. At first, consider the group ofequations containing the derivative of h1. These derivatives are with respect to.r"i 1, ..,x". Then. select the group containing the derivatives of h1 from theremaining equations. They are the derivatives with respect to x0:1, ... , x". The de-rivative of this function with respect to xP is already included in the first group ofequations. We continue this process for the next functions h,. For every functionh ,, the corresponding group of equations contains the derivatives of this functionwith respect to x" only for o > Indeed, if a < i,, then the equation

is already included into the previous group corresponding to the function b,, ,,, since(i1 ... 0) < (i1 ... ip).

Therefore, we may assume that a > i,,. This implies that the group of equationscorresponding to h,, , includes the functions of greater multi-index.

The groups of equations contain. in general, a different number of equations. Thefunction h,, ,,_,,, has no corresponding group. Therefore, we can take the function


bi,...i,_,,, to be arbitrary requiring only class C2 regularity. Then we solve the system(12) ordered as above group by group starting from the bottom. Show that everygroup satisfies the compatibility condition. Suppose that we have already found thefunction bj of the multi-index j, where j > i. Let us show that the function b; exists,too. In the group which corresponds to b;,_ we consider the equations

8bi,.. P abi,...,r...i, -axn - axi,r=1

abi, ;,Obi, P

E-Tx"-where a, a > iP. The compatibility condition has the form

A a'-bi a_bi-0x°ax;3 aviax°

i, aa:r;, .. ;, P a abi, ... ,r... i, ab;, ... a ... ip

aX3 + aX ( axe(13)

where the r which are superscripts to a and Q mean that the corresponding indextakes the r-th position. In parenthesis we have the derivatives of functions of multi-indices (i, ... V'. .. i,,a) and (i, ... V' ... i,'6) which are greater than (ii ... ip).Therefore, by virtue of the inductive hypothesis the equations

abil .. . Obi, ... - P ab;, ... aA .. ;raX' aX` A=1.k#r ax'

are satisfied.On the right of (13) we obtain

aa.i, . aa3i, .. aas;,. r>'... + v a2bi, ..,, 9ax;l - axi axe- axi'axiki..4

The last sum in the equation above is zero due to the skew-symmetry of b;,...;, withrespect to indices. The sum of the remaining terms is zero because of (11). Therefore,the group of equations on bi is compatible. If we consequently find all the bi then wefind the form w, . This completes the proof of the Poincare lemma.

Observe that w, is not determined uniquely. For instance, it is always possible toadd the form dw2.

The form w having a zero exterior differential is called an exact form.

2.10 The Exterior Codifferential

In a Riemannian space the operator b of exterior codifferentiation is defined whichputs the exterior form w of order p and the exterior form bw of order p - I into


correspondence. Let V denote the covariant derivative in Riemannian space andVka;,..;, be the covariant derivative of the tensor a;, Set

Vi ai,...i, = vka;,..i,.

Since V'a;;,...;,_, is a skew-symmetric tensor of order p - 1, we are able to define thefollowing exterior form:

bwV'a;i,...,,_,dx; A...Adx'"

The codifferential satisfies the Poincare lemma b2w = 0.The proof of the latter statement is more complicated than for the case of the

differential d. It is based on the properties of the Riemannian tensor. We produce it,first, in the case of the second-order form w = aj dx n dx. Then b'-w is a scalarfunction. We have

62w = Ov1(g'kvkalj) =g'Ig1k 10ka;j = -gjlg'k01Vkaj;.

(1)

Make the index change 1- k -i 1 and then change the order of covariant differ-entiation. If we take into account that the difference of second-order covariantderivatives of a tensor has an expression in terms of the tensor contractions with aRiemannian tensor then we see that

bZw = -gjk grl V k v/aj; _ -gik g'I (V,Vkaji + R jkl aa; + R`;k, aj., ). (2)

Consider the following expression (we denote it by A I):=A i = R jk; aR, $1 ' - R3jk/ sai e,l gJk gl'

Interchanging the roles of a and i, I and j and k we obtain

A I = Rlkja aieg'I gk'g°w = - RJ, ,k a0, g" '$g'rgk, = -A,.

Therefore A, = 0. In an analogous way we find that the second contraction with aRiemannian tensor is zero:

A, = g!k gd R°;k1 aj0 = R,;k, a., g°i gJk g'I = R:ijkl a,. g' g'k gj1

= Rpj,k aia eg'I gf k = Rijk, ani gJ* gr1 gJk = A. = 0.

So, we have

6zw = -gikg'IV/Vk aji = -g'kgJlvlvkaii,

which differs from the initial expression only by the sign. Therefore, 62w = 0.Now, consider the form w = a;jl,...;,dx' A ... dx'o of order p + 2, where p > 1. Let

;, be the coefficients of 62w. We have

bi,... i, = g'kev,Ok aij;,...;, =-g,kOy/vk

ajii,... i,

_ - gJk$'l vk v/ ajii,... i, = -gjkg'1(vl vk ajii, . iv

+ R jkl aoii, ... i, + R11 ajn;, ... i, + R'i, kl ajin . i, + ... + R`i,kl aji -J' (3)


Just in the same way as for the second-order form we can see that the contraction ofgfkgd with the first two terms on the right of (3), where the Riemannian tensorinvolved, is zero. Consider the following expression:

jk ! jk IA3 = 8 R°,kl olio. ip = RBi,k/ aji...ip 8 g

_ -R+,akr ajinii ..io eegd

Let us make the index change as Q - k I - Q andainji, .rp = aij,. _...rp,

a. Since

A3 = -1ti,kraajrn...ip 9ng' gjkgI

Making the same index change and adding the obtained expressions, we find

3A3 = -(Ri,:9kr + Ri,kl3 + Rr,l9k) ajrni:.. ip3n e

By virtue of the Bianchi identity we see that A3 = 0. In an analogous way we obtainzero in all consequent terms of (3). Therefore,

kbi, ip = -g gjlVjV aij...ip

Comparing this and the initial expression for bi,.. j', we see that the coefficients bi,...,pare zero. So, 6=w = 0. One can get a simpler proof of this fact by means of the Hodgestar operator.

2.11 Some Formulas for the Exterior Differential

It is easy to represent the integrability condition of the Pfaff equation in terms of theexterior differential and exterior product. The following theorem holds.

Cartan Lemma In order for the second-order exterior form F = aij dx' A dxj to beequal to zero by virtue of the equation f = a; dx' = 0 it is necessary and sufficientthat there is a linear form W such that

F=fAW.

To prove this result, we go from the basis dxi to the basis which includes the linearform f. Suppose that f, ... fn is the required basis, wheref, =f

Considern n

F=bijfi (1)j=1

Since F = 0 for f, = 0 and since f, A f,, are linearly independent second-order forms,b,,t, = 0. Therefore, F = f, A E I b, jf . Introduce the linear form

n

= Eblifj=I


Then (1) can be written as F =.f A cp.Let us be given the Pfaff equation w = 0. We can represent the bilinear Frobenius

form dw(b) - 6w(d) as the exterior differential of w if we set

dx'bx1 -br'd.Yi =dx' Adr,

dw(b) - bw(d) = dw.

The Pfaff equation w = 0 is integrable if and only if the bilinear Frobenius form. i.e. do,is zero by virtue of the equation w = 0. The lemma just proved implies that there is alinear form w such that

dw=wAcp. (2)

This is the necessary and sufficient condition for the Pfaff equation w = 0 to beintegrable. In other words, in the integrable case the form dw is divisible by w. Thiscondition can also be represented as w A do = 0.

For every linear Pfaff form w = a; dx' we may adjoin the vector field r = {a1}. LetN = (Ni) be a unit vector field of principal normals of the streamlines of the field rand k be the curvature of these streamlines. We may also adjoin the linear form17 = Ni dx' for this field. Let r be the unit vector field. Let us show that if r isholonomic or, equivalently, the equation w = 0 is totally integrable then it is possibleto set cp = ki7 in equation (2), i.e.

dw=kwA17.

This equation can be regarded as the formula analogous to the Hamilton formulafrom Section 1.3. The form ki is

8a, ;dx .kq =

axiaj

Find the exterior product

w A k17 = ak dxk A aX-t aj dx' = 2 I ax 8 f - ai aj d k A dx'.

Prove the equality of dw and w A kq in a special choice of coordinate system. Set thecoordinate curve x" to be tangent to r at some fixed point Po while the othercoordinate curves are orthogonal to r. Then at Po we have a1 = = an-I = 0,a,, = 1, 8a,,/8x'=0. The form w at P0 is

n-1 49an Bart

`' = 2 e 4W) dx° A dx' + 4 (8x° - 8x") dr" A dac".


Since the conditions a"" - "" = 0 a,,6 < n - I holds (see (14) of Section 2.7) andexi OX"

n-I

dw - E 8"de A dx".

0=1

Next we consider w A kt. Since at P0 ak = ai = 0 for i, k < n,

(3)

n-1 aajwAki1=E8x"deAdx'. (4)!= i

Comparing (3) and (4) we conclude that w A ki1 and dw coincide with each other atP0. The coincidence of the forms does not depend on the choice of coordinate systembecause of their invariant meaning. Therefore, the equality dw = kw A q is proved.

For every k-th order form w = a;,...;kdx'I A ... A dx'k it is possible to put into cor-respondence its value on k vectors YI,... , Yk. namely, the contraction of the tensoral,..,;,, with the components of a multi-vector generated by Y1,. .. , Yk. Denote this valueby w(YI, ... , Yk). Then

YI'' ... Y1'k

w(YI,...,Yk) = a, ..., ..Yk

.

... Yk

Let us state a useful formula. Let w = a, dx' be a linear form, r the correspondingunit vector field, r, and rp the fields orthogonal to T. Then

(r0, r,) = (r, V rn) (5)

Indeed we have

1 8a; 8adw(rQ,ry) -

2 (8xJ 8x')

rj

ral

=21-aim +a; ro+aiL3r.J -a, 74

= (r,V ,r,3 - V,, ra),

where ra stands for the components of r,,. The formula (5) is established.

2.12 Simplex, the Simplex Orientation and the Induced Orientation of a SimplexBounds"

Let us be given a set of (n + 1) points in Euclidean space E": Mo, Ml'...' M. Let r,be the position vectors. The set of end-points of all vectors

r =x°ro+x'ri +---+x"rn,


where the numbers x' satisfy the condition

Xo+X'+...+X"= 1, 0<X'<

is called the n-simplex. We shall denote the simplex by in pointing out the order ofvertices if necessary: t"(Mo, MI,_, Mn). Define the simplex orientation as the ori-entation in E" generated by the basis MOM,,..., MoM". The odd permutation ofvertices make the orientation opposite, while the even one preserves the orientation.We shall denote the simplex of orientation opposite to In by -t".

Select in a given simplex any p vertices and form a simplex tP(Mio, Mi,,... , M,,).This subsimplex is called the face.

Consider the (n- 1)-faces. Denote by t, -' or T,"-' (Mo, ... , v', ... M") the faceobtained by deleting Mi, where V' means the omission of Mi.

The orientation of a simplex (Mo,... , V',... Mn) is called induced by the simplext"(Mo,... , M") if the simplex (M,, M0, ... , M") has the same orientation as In.The simplex t"-' of orientation induced by In has the form t"-' _(-1)'(Mo, ... , V , ... M").

2.13 The Simplicial Complex, the Incidence Coefficients

First, we consider the simplicial complex in Euclidean space El. The simplicialcomplex in El of order n is a finite or countable set of simplices satisfying thefollowing conditions

(1) Together with the simplex to, this set contains all its faces including the vertices.(2) Every p-simplex tP of a simplicial complex is a face of some n-simplex in (the

homogenity property).(3) The intersection of any two simplices tP n t9 either is empty or is a simplex of a

given complex.(4) Every two simplices can be joined with the chain of simplices having link-wise

non-empty intersections (the connectivity property).(5) Any bounded domain in space intersects a finite number of simplices.

The set of points in El which belongs to the complex is called the polyhedron. We saythat the smooth manifold M" is triangulated if it is mapped onto some polyhedronhomeomorphically. We shall call the image of a simplex in M" the curvilinear sim-plex. The fundamental hypothesis of combinatorial topology ("Hauptvermutung")asserts that any two triangulations of homeomorphic polyhedra have isomorphicpartitions. In recent years this hypothesis has been refuted.

Suppose that every simplex in a simplicial complex is oriented. Introduce theincidence coefficient of the two simplices in and t"'. We say that the incidencecoefficient (t"t"-') is zero if t'-' is not a face of P. If t"' is a face of t" then(t"tn-') = I for the case when the orientation in t"-' induced by In coincides with thegiven one and (t"t"-1) = -1 for the opposite case. Let t"-2 be some subsimplex in t"


and be the (n - 1)-simplices in t" which include in-'. Then for any choiceof orientation of simplices the following holds:

(tr,t0,-1)(ton-I to 2) + Itn 2) _ (1)

Set, for example, that ti-2 is (M2 ... Me). The value of each term of the above willnot change if we change the orientations in to" and in-'. Suppose that to-' does notcontain the vertex Mo. while t," -I does not contain the vertex MI. Orient to" andin-1 in such a way that

(to-I t" `) = 1, (tn-I to-2) = I.

Then the simplices t" and in-' have the formsn-I

to = (MIM2...Mn),

tj"-' = (M0M2 ... M").

Under the conditions posed on the incidence coefficients the formula we need toprove has the form

(tntA-1)+(tr,tl-I) =0.

This relation stays the same if we change the orientation of t" into the opposite one.Therefore, we choose the orientation by setting, for instance,

t" = (M1MIM2 ... Al,).

Then

Hence, formula (1) is true.Consider now the simplicial complex. Let t" and to-2 be some selected simplices.

Then

(tntn-1)(tj'-Itn-2) = 0, (2)

where the summation is assumed over all (n - I)-simplices.If t11-22 is not a subsimplex of t" then every simplex M--' which includes tn-2 is not a

face of t". In this case (t"N`-1) = 0 and (2) is true. Therefore, suppose that t"-2 C in.This simplex contains precisely two simplices to-1 which include t"-2. Hence, (2) isreduced to the formula (1) which has been proved already.

2.14 The Integration of Exterior Forms

Suppose that the n-dimensional manifold R" endowed with curvilinear coordinates x'and the form of order n is given:

w=a1,, dr'A...A1f


or

w= 1 E a;,...;. dx" A ... A dr°.n! i,.. i

In a coordinate change the coefficients of the n-th order form gives the multiplierwhich is the Jacobian of the coordinate transformation 1(" ,,). It is possible todefine the integral of the form w over D C R" in the usual way as the integral of a, .

1w= fR, D

The value of the form integral does not depend on the choice of the coordinatesystem. Indeed, with respect to a u' coordinate system the form w is

w = a,._ , I u ) du' A ... A du".

Let G be a domain in a space of u' . .... u" which corresponds to D. Since

dx' ...d1" a, ."Il )du' ...du",

the integral of the form depends on the form and the domain only.Suppose now that the form w of order p is given in a domain D. Let Fm be an

in-dimensional surface represented as

x' =f'(v'',.....v"), i = 1,...,n.

On the surface F"' the differentials dx' are some linear forms depending on dy',j = 1, ... , m. Therefore, on the surface F" some new exterior form of order m isinduced. We shall call this form the induced one and denote it by w:

1 Of Ofyi ... A...Ady"/, '

Thus, form coefficients are, evidently, skew-symmetric with respect to indicesj,,. .. , j,,,. The integral of w over the m-dimensional surface Fm is, by definition, theintegral of w over F"'

F' F"

Consider, as a particular case, the form w of order n - I in some n-dimensionaloriented domain D. Let r be a smooth boundary of D endowed with the orientation

VI (TOR I I1.U')S IN MANN -I)IMENSIONAI I (It 111)1 AN AND RI1M' NNi NN SPA(.'I S 117

induced from D. In D the form dw of order is is also defined Hence, both of thefollowing integrals are defined:

;.; and 40

Denote the boundary of D by OD.The generalized Siukex' Joiiuula holds

j w= jdw,,i n

This formula establishes the close relation between the notions of exterior differentialand boundary.

We prove this formula in the case when R" is the atfine space of coordinatesx1, . , " and D is an u-simplex defined by the inequalities

b<x'< 1.

Let Ac0, A,,.. , .4 be the simplex vertices. Their ordering gives the simplex ori-entation. We shall represent the simplex as (An... ..4,,). We are able to represent anysubsimplex in an analogous manner. The simplex boundary is the set of all its signed(ii - I)-suhsimphces. namely

r)D= (-I) ...A,,)

At any point of a boundary simplex one of Its coordinates satisfies the equationx' = 0. Let w be the form of order (u - I) in D :

w_ a,, d"A...Adr'°

=cr -idt=A...Ad1"+ f r1,

Each term is itself some exterior form of order a - 1. Let us prove Stokes' formulafor each of them. Take, for instance, the form

w, =a2 n-idx=A... Ad.N".

On the simplex face D,. which is given by x' = 0. the differential dx' = 0. Therefore.on the simplex D, the form w, is zero if i 7 2. Hence, w, is different from zero on D,and D only when

fwi = Jii + fwido 1), rn,

On the face D = (A,.. . A,;) we have x1 = I -.V-' - -.vn. Settle the corres-pondence between the faces Dn and DI by the equality of coordinates. i.e. by means


FIGURE 24

of projection (see Fig. 24). The simplex DI is given as D, = (-l)(AoA2...An).Therefore, the orientation of D1 defined by coordinates -v2,. . ., x" is opposite to itsorientation as part of the boundary of D. So, we have

Jwl=_Ja2...n(Ox...x")dxA...Adx"DD

+ Ja2...n(1-17 -...,X2,...,x")ar? A...Adf

Do

',...,xn)dx'Adx2A...Adx",_ 87a,...n(x',x,

no 0

where xa = 1 - x2 - - - x". The latter expression is the integral over D. On theother hand, consider the exterior differential of w1:

dwI =da,...n_IAdX2A...Adr"= a_ a2...n-1dx'Adx2A...Adx".

Hence

Jwi =Jdthi.DD D

In an analogous manner one can prove this formula for any other wl. For the case ofa curvilinear simplex in some manifold R" the proof can be reduced to the case of asimplex in affine n-dimensional space.

Further on we assume that D is triangulated, and also that the boundary of D isrepresented as (n - I)-simplices of triangulation. The theorem that any smoothmanifold can be triangulated is known. The integral of do over each simplex of

VLCIOR 111 LUS IN MANY-tIMi.NSIONAL FUCI IM AN AND Kit MANNIAN SPACES 139

triangulation is equal to the integral of w over the (n - I)-simplices of the boundary.Let us consider the sum of those integrals. If the n-simplices are oriented consistentlythen for any (it - 1)-simplex in a common boundary of two distinct ii-simplices theorientations induced on it from those two simplices are opposite. Therefore, in thesum under consideration the integrals over (n - I )-simplices interior to D canceleach other out. The remaining integrals are ones over the (n - I )-simplices on theboundary of D. This completes the proof of Stokes* formula.

2.15 Homology and Cohomology Groups

Let us be given a simplicial partition of a topological space. Each simplex of thepartition is already endowed with an orientation. Denote the i-th simplex ofdimension r hr ti l. Let us form a chain of simpliccs as

where a, are some integers A houndarr of the chair: r is a sum of the boundaries ofsimplices t, if we take them with coefficients a,:

clt' _ cr, (it".

To find the boundary of a single simplex we use the incidence coefficients. Then theboundary of t, is the following (r - 1)-dimensional chain:

The boundary taken twice is zero, i.e. the boundary of a boundary is zero. It issufficient to prove this for the boundary of a single simplex. We have

ddt,' -)tR`=ED,;

r i I. r

In Section 2.13 we proved the equality

Dtr' t t'-I)(t/'-1141-2)

= 0.

The chain x' with zero boundary is called closed or rite cycle, i.e. for the case of aclosed chain eft' = 0. The chain x' of dimension r which is the boundary of an(r + 1)-dimensional chain r' is called homologous to zero: x1 = did

The sum of two chednm of the sane dimension

x'=grit;,


is the chain

x'+y'=E(ai+bi)t;.i

The sum of two cycles is a cycle. Therefore, the set of all cycles of dimension r formthe Abelian group Z, with respect to the adding operation as above. The quotientgroup of Z, with respect to the subgroup of cycles homologous to zero is called the r-dimensional Betti group. If T, is a subgroup of Zr formed by the cycles homologicalto zero, then the Betti group H, is

H, = Zr/T,.

In an analogous way the cohomology group can be defined. For each cell jr thecoboundary is the (r + 1)-dimensional chain of the form

bt, = (t; t +1)t +1i

The chain coboundary is the additive sum of coboundaries of chain cells

5x'=Eaibt;.

The chain x' which is a coboundary of some (r - 1)-dimensional chain y'-' is calledcohomologous to zero. If a coboundary of the chain x' is zero then this chain is calleda cocycle: bx' = 0. The set of all cocycles, being endowed with the natural operationof adding, forms the additive group denoted by Zr.

Let T' be a subgroup of Zr formed by the cocycles homologous to zero. Then thequotient group of Z' with respect to T' is called the cohomology group:

H' = Z'/T'

Consider, for instance, the cohomology group H, (S') with integer coefficients of thecircle S'. Divide S' with the points A, B, C into three arcs, namely AB = y,, BC = y2,CA = y3. We assume that the given sequence of end-points of arcs yi produces theorientation of the arc. The coboundary of A is formed by two arcs, namely y, and y3giving

bA =Y3 - Yl

Analogously, bB = y, - y2, bC = y2 - y3. Hence, the arcs y3 and Y2 are cohomolog-ous to y,:

y3 = yl + bA, Y2 = yl + bA + bC.

Therefore, the group H' (S') is isomorphic to the group of integers Z.There is a close relation between the space of different forms defined over the

whole manifold and the cohomology groups of the manifold, namely the followingde Rham theorem holds:


The quotient space of closed forms of order p with respect to the subspace of exactforms of order p is isomorphic to the p-dimensional cohomology group with realcoefficients HP(M, R) of the manifold M.

2.16 Foliation on the Manifolds and the Reeb's Example

Let us define the notion of a foliation on the manifold. Let M" be a regular Rie-mannian manifold of class C' without a boundary. Suppose that M" is decomposedinto connected subsets which we shall call the fibers of foliation or the leaves anddenote them by L. Let Mn be covered by the set of neighborhoods UQ each endowedwith a coordinate system x1,. .. , x" (i.e. there is the C' regular homeomorphismv,, : U V C E" of U. onto some domain V. in E") such that the intersection ofeach leaf L with U can be determined by the system of equations.rP+1 = const,... , x" = const. In this case we say that on the manifold M" the C'-differentiable foliation with the leaves of dimension p is given. The number n - p iscalled the codimension of the foliation. Each fiber is a submanifold of M". Thus, thefoliation is structured locally like a family of parallel planes of dimension p in Euclideanspace E". If M" is an analytic manifold and V,, are analytic functions then we saythat the foliation is analytic.

Among the closed surfaces M2 only the torus and the Klein bottle admit foliationswithout singular points. For n = 3 the foliations of class C' exist on any manifold.However, the manifold containing an analytic foliation must satisfy the conditionwith respect to its fundamental group. Haefliger stated that a compact manifold witha finite fundamental group does not admit analytic foliation of codimension 1.

The analytic requirement in the theorem is essential. Reeb constructed the foliationof class Cx on the three-dimensional sphere S3. We describe this foliation here afterLawson. At first, construct the foliation of the toroidal body K3. The boundary ofthis body is a two-dimensional torus which we denote by T2. It will be one of thefibers, moreover the limit fiber for the other fibers. As Lawson said figuratively, eachfiber within K3 is like a snake that eats its tail. While the phrase above expresseseverything you need to understand, we continue the explanation. In the xy-plane wetake a zone between two parallel straight lines x = -1 and x = 1. Construct the linefoliation in it. Let us be given a curve y = f (x) defined in -1 < x < 1. The functionf (x) is supposed to be of class C-, infinitely large when IxI 1 and for all deriv-atives f(k) (X)

lim f ck 1(x) = oc, k = I , ... , M.IYI-1

We define each line of zone foliation with the equation y(x) =f(x) + c, where c is anarbitrary real number (see Fig. 25). Revolving this foliation around the y-axis, weobtain the foliation of a cylindrical body. Let us map the cylindrical body onto thetoroidal body K3 in the following standard way. Consider the axial curve (the circleS') in the usual torus. We assume that circle radius is 1. Lets be the arc length of S'with respect to some fixed point P0 as the initial one. Make the section of the


FIGURE 25

FIGURE 26

toroidal body K3 through the point P in S' with a plane orthogonal to S' at P.Denote a disk obtained with that section by D2. We can represent the toroidal bodyK3 as a product S' x D2. Introduce the polar coordinates (p, 0) in D2 taking P asa pole. In a section y = const we have a disk. Introduce the polar coordinates (r, gyp)in this disk with a pole at a point in the y-axis. Now define a mapping -r/, of thecylindrical body onto K3 setting s = y, p = r, 0 = V. Every torus point will becovered infinitely many times under this mapping; also, if the points a, and a2 of acylindrical body go into the same point of K3 then the difference of correspondingordinates satisfies y1 - y2 = 27rk, where k is an integer, while the correspondingparameters (r, lp) coincide. Let us show that under this mapping the foliation of thecylindrical body goes into the of K3 (see Fig. 26). Locally, the mapping t' is ahomeomorphism. Suppose that the fibers 11 and 12 of the cylindrical body went intothe surfaces rb(11) and ap(12) which have a common point b. Then the inverse imagesof this point, say a, and a,, differ in ordinates as y, - y2 = 27rk, while the polarparameters are the same, r1 = r2, W1 = V2. The fibers 11 and 1, have been obtained in


revolving the curves y = y(r) + cl and y = y(r) + c2 respectively. As y, = y(r,) + c,,y2 = y(r2) + C2, then cl - c2 = 2itk. Hence, y(r) - y(r) = 27rk for any r. Therefore,the surface t&(l,) coincides with the surface ,p(l2), i.e. they form the same fiber.Evidently, every point within K3 has a neighborhood covered with the fibers '(l ).The boundary torus is also the fiber of foliation just obtained.

The sphere S3 can be represented as a union of two toroidal bodies with acommon boundary, namely the torus T2. Within each toroidal body we constructthe foliation as above. Gluing the toroidal bodies along their common boundary T2,we get S3 endowed with the Reeb foliation.

Observe that in this case we have two closed curves which are orthogonal to thefibers, namely the axial curves of the toroidal bodies.

Later, the theory of foliations was developed in topological and geometricaldirections. Novikov [73] proved that every codimension-one foliation on S3 has acompact leaf. Thurston [100] proved the existence of p-dimensional foliations on anyopened manifold. These foliations exist on closed manifolds if there exists a field ofp-dimensional tangent spaces. Reviews of topological results from the theory offoliations may be found in the work of Fuks [101], Tamura [71] and Lawson [102].

A series of interesting results was obtained for foliations in Riemannian spaceswhere the leaves were totally geodesic, minimal, umbilical and so on. It was provedthat the codimension of a totally geodesic foliation on a complete manifold ofpositive curvature is more than 1. Moreover, the Frankel theorem is known: in acompact manifold with positive curvature, two compact totally geodesic submani-folds whose summed dimensions are not less than the dimension of the manifold donot have an empty intersection. So the dimension of a totally geodesic foliation withcompact leaves on a compact manifold with a positive curvature is less than half themanifold's dimension.

Ferus [103] proved the following interesting theorem in which the completeness ofthe manifold is not necessary: if in some domain of the manifold Mn` there exists atotally geodesic foliation with dimension v of class C2 and with complete leaves and amixed curvature of M"+" tangential to a leaf of vector X and orthogonal to a leaf ofvector Y,

K(X, Y) = const > 0,

so v < p(n) , where (p(n) - 1) is the maximal number of linearly independent vectorfields on S"..

Rovenskii [104] showed that the condition K(X, Y) = const is essential and hegeneralized this theorem for some class of foliations, which he called conformal.Papers [105H113] give various results on foliations with geometrical conditions.

There is a review of geometrical results on foliations in the work of Tondeur [114].

2.17 The Godblllon-Vey Invariant for the Foliation on a Manifold

Let us be given a C2-regular foliation of codimension 1 in a differentiable manifoldM" of class C°°. The foliation is called transversally orientable if there is a smooth


vector field on M" transversal (i.e. not tangent) to the foliation leaf through the initialpoint. Consider the transversally orientable foliation. Then in every neighborhood ofa point the foliation leaves can be defined by the Pfaff equation w, = 0. The form w1must satisfy the integrability condition w, A dw1 = 0. This means that dw1 is divisibleby w1, i.e. there is a form w2 such that

dw1 =w1 Aw2. (1)

The form w2 is not uniquely defined. If we differentiate (1) externally then we obtain

w, Adw2 =0. (2)

Therefore, dsi2 is also divisible by wl. There is a form w3 such that

dW2 = w, A W3.

If we differentiate this new relation then we obtain

W1 A(w2Aw3-d,,3)=0.

(3)

Consider the third-order form 1 = w1 A w2 A w3 which was introduced by Godbillonand Vey in [68]. Using (3), the form 0 can be expressed as

ft=-W2Adw2. (4)

For the case of n = 3 the form Q can be expressed in terms of the value of the non-holonomicity of some vector field. Since is linear, it can be written asw2 = P dv I + Q d.2 + R dr3. If we introduce the vector field v = {P, Q, R} then11= -(v,curl v)dx1 Adx2 Adx3.

The following theorem holds:

Theorem The form Q is closed and its cohomology class in H3(M, R) depends onlyon the foliation F

Consider the exterior differential of SZ:

dfl=-dw2Adw2=-w1AW3Aw,Aw3.

Since w, is a form of odd order, w1 A w1 = 0. Hence dfl = 0.Suppose that w2 is any other form satisfying dot = w, A w2. Let us show that the

Godbillon-Vey form constructed by means of wZ differs from the previous form bysome differential form. We have

d i=w1Aw2=w1Aw2.

From this we conclude that w, A (w; - w2) = 0, which means that w2 - w2 is pro-portional to w1i i.e. there is a function f such that w; - w2 = fw1.

We have

W2 Ad' = W2 Ad,,2 +fw, Add +W2 AdfAw,

+fw1 A df A w1 +fw2 A dw, +f 2w1 A dw1.

VICTOR I ICLnS IN MAN) -DIMI NSUONAL I U(L11)I AN AND RIFM-%NN! \N SPAC PS 145

Since w1 corresponds to the foliation. w1 A dw1 = 0. Also, w1 A (ko, = 0,w1 A (1j Awl =O. Then

-d(/w+Aw1)=-d1Aw,Aw1 -ldw,Aw,+ /w,A(k1r1=w,Ad/Aw1 +/W, Adw1

Hence

w; Adw; = w, A(1o:+ - d(/w, Awl )

So. the forms -w, A dw, and -w; A dw; belong to the same cohomology class ofH3(Af, R). This class is called the Gadh(llon-Vev charucieristic cuss. Denote it by10)(F).

Now we are going to establish the independence of [fl](F) of the concrete choice ofw1. i.e. we show that it depends only on the foliation F itself. The form w1 is definiteto the scalar multiplier Letwi = Awl be the form which gives the same foliation F.where A is it regular function. Set (/wl' = wl A W. We have

i1wi=Ouel=dAAWI+Ad01=dAAw1+\w1Aw'=Aw1 A(w,-dlnA)=w; Aw_.

Hence, there is a function / such that

w:=w,-dInA+1w,

We have

w;A(:;=w:A(!w2-dlnAA(ku,+/w1 Adw?+w,Ad/Aw1 -dlnAAd/Aw1 +/w1 Ad/Awl+/w,Adwi -/dlnAA(tw1 +/'w1 Adw1.

Many of the terms in the expression above vanish by virtue of the equalities

w,A(1fr1 =0, w1 Adw1 =0, w1 Ad/'Aw1 =0,w1 A dw, = 0, w, A dw1 = 0.

Hence, we finally obtain

W3 Adw; = w,Adw,-d(lnAdw,+fw,Awl -d lnAAw1).

This equality means that -w; A dw; and -w± Adw: differ from each other by thedifferential of some second-order form. Thus, the cohomology class [ft](F) dependson the specific foliation F.

We present some properties of [0)(F):

(1) Let us be given a foliation Fin M. Let f : V M be a differentiable mapping of amanifold V onto M. Then the inverse images of fibers of the foliation F define afoliation in t : Denote this foliation by f' (F). The mapping / induces a mapping


f' of the cohomology group H3(M,R) into H3(V,R):

f' : H3(M, R) H3(V, R).

Then the Godbillon-Vey class [11) (F) of the foliation F goes into the Godbillon-Vey class of induced foliation f' (F) under the mapping f'.

f*([Q](F)) = [Q] (f (F))

This property of Godbillon-Vey classes has been proved in [98] for the case ofC2-regular foliations and Cl-diffeomorphisms.

(2) If F and F' are the foliation of codimension 1 in the same manifold M which aredifferentiable and homotopic to each other then their Godbillon- Vey classes coin-cide.

(3) If the foliation F is given in a three-dimensional manifold M3 then the integral of Stgives some number -y, namely

M; M3

This number is uniquely defined by the foliation and is called the Godbillon- Veynumber. This number is not necessarily an integer.

(4) Let us settle the relation of ft to both the curvatures of lines which are orthogonalto the fibers and the second fundamental forms of the fibers. Denote the linesorthogonal to the fibers by 1. Let k,,. .. , k,,_, be the curvatures of those lines.Suppose that r1 i ... , r is the natural orthonormal frame of 1, where r; = {rii}.Let (i be a linear form generated by Ti. The foliation F is given by the equation(1 = 0. In Section 2.11 we proved that the exterior differential of (, can berepresented as the exterior product of (, and k1(2, i.e.

d(, =(I A k l (2

Therefore, the Godbillon-Vey class for the foliation F is represented by the form

Il=-k;(2Ad(2.

Let us express the form d(2 in terms of an exterior product of the forms (i. Leta, $ > 3. Find the value of d(2 on r,,, ,r,3. Using (5) from Section 2.11, we have

d(2(Tca, ra) = (r2, V,.T3 - V r0)

Consider the system of two Pfaff equations: (I = 0, (2 = 0. By virtue of (1) and (3),the differentials of (, and (2 can be expressed as

d(, = k, (, A (2, d(2 = (1 A ,

where W is some linear form. Since d(, = 0 and d(2 = 0 by virtue of (, = 0, the system(1 = 0, (2 = 0 is totally integrable and, as a consequence, defines a foliation of co-dimension 2 such that each fiber of the new foliation is in some fiber of the given

VECTOR 111-LDS IN MANN-DIMENSIONAL LOCI IDFAN AND RILMANNIAN SPACES 147

foliutlon F. Therefore, the vector field r is orthogonal to the Poisson bracket of anyvector fields T,,,T,f for a. fl ? 3. Hence

t,r1 = 0.

Next we consider the value of this form on r1, r,,. We have the Frenet equations

Vr,r1 =klr2,Vr,T_ = -klr1 +k_ri,

0r, r, = 1r,_1 +k,r,: 1.

Taking them into account, we are able to write

(T_.VT,T,. -V ,,T11

(r2.-k,.-1tn-1 -VrTi).Denote the second fundamental form of the Fiber of the foliation F by !(x, r1. Then

dC2(r1 r } = -k,o' - !(r2, r ). cr ? 2.

where hi is the Kronecker symbol. Thus. ik_ can be expressed in terms of Z;, as

d{_ _ -4_(1 A1(r,,r,.)(1 Ac ,ctwhere r is some unimportant coefficient. So. we find the following cxprerslrm for theGodhillon- /er form:

it = -l, ,.:2 A d(2 = k 1 (A'2(z A <1 A 4i -F , !(T2. T,. A (I nn=3

This expression was established in [691.

2.18 The Expression for the Hopf Invariant in Terms of the Integralof the Field Non-Holonomicity Value

Let f: S' -' S-' be a C2-regular mapping of spheres, We are going to find theexpression for the Hopf invariant ry of a mapping, in terms of the integrated non-holonomicity value of some vector field given on S. Let ul, u2 be the local co-ordinates in S- and ds- = g,, du' du' be the first fundamental form of S- with respectto these coordinates. g = det 1Ig 11. Let x1,x=, x3 be the local coordinates in V. Themapping / can be represented with the functions

t1 =1e(x1,x_1,x3), i=1,23.


Set

uI uI

u' u!'.

Let us introduce in S' a second-order skew-symmetric tensor a",3 = 4!, Thistensor generates the second-order exterior form, namely

w = a,,.3 d .e' A dr".

Show that the exterior differential of this form is zero: dw = 0. Indeed, we have

dw =J8a,2 + 8a23 + X21

Idr1 A dx2 A dr3.

8x3 8xI 8x2

Since f is a function of u' which are the functions of x" on the other hand,

Oat _ 1 (8f 8ukl +81;118x" 47r 8uk 8x" 1 81"

It is easy to find the following equalities:

Uxi u0 u'rt8123 0131 8112 8u' 8u' 811i

1

8xi + 8x2 + 8x' = 0' 8x1 123 + axe I31 + C7r;112 = 141 u111 urs = 0.

Iu, u? '4

These equalities imply dw = 0, i.e. the form w is closed. Since the cohomology groupH2(S', R) is trivial, then by the de Rham theorem there is a global linear formw, = b; dx' on S3 such that w = dol. Whitehead [66] stated the following assertion.

Theorem The Hopf invariant -y of a mapping!: S' - S2 is equal to the integratednon-holonomicity value of the field b = {b;)

-y = fwi A dw, = J(b, curl b) dV, (1)

S3 S3

where dV is a volume element of S3.

Let us divide S2 into two-dimensional cells E,l which are so small that the map-ping f can be approximated with the normal mapping. Further on we assume that fis already normal within each cell E2.

Let Mq = f - I (q). As f is normal, the set M. consists of one or several closedcurves. Let us orient each curve of Mq as follows. Let p be a point in M9. Select thebasis e,, e2, e3 at p which produces a positive orientation of S' and such that e3 istangent to Mq, while the directions of el and e2 go into the directions of e, and e2under the induced mapping of tangent spaces f': Tp(S3) - Tq(S2), such that (e*,,e2)produce a positive orientation of S2. Then we say that e3 defines a positive ori-entation in Mq.


Let gl,g2 be points in E, 2. Connect them by a smooth curve I in Ei. ThenMr = f --I (I) is a surface with the oriented boundary LIM, = M, - My, . The wholesurface is formed by My, where q E I. Show that

J w= Jwi.M, M92

Indeed, by Stokes' theorem we have

Jdwi= Jwi_ Jwi.tth Mat M

Let v', v2 be local coordinates in Mi. Set

T°'; =

(2)

i x' W 1

The coordinates ul, u2 of a point in S2 are the functions of v1, v2 along M,. Then

u1, U2) 1 ul x7, u'Y

1 Xs 1I _ .,r x 7-1 I.VI

VZ Ju2. xi u2, JC + 2x r r rSince the surface Mi goes into the curve under the mapping f, T("; ) = 0. Considerthe integral on the left of (2):

J dwl = Jw =- J

f IgFi Lv A

Mi M, Mi

= 4rr 1 2 In;iT"3 dv' A dv2

M,

I A dv2 = 0.

Equality (1) is proved. Thus, the integral of w, over the inverse image of any point qis the same number - some invariant of the mapping f. Let us show that this numbercoincides with the Hopf invariant ry. Let M. be a base of a piecewise regular surfaceF and t, 77 the local coordinates in F. Then by Stokes' theorem we have

f r (1

wl dwl = T7r_ j \ ,7l / dS AM

As the boundary of F goes into the single point q, then f(F) covers the sphere S2some integer number of times (more precisely, the total oriented area of the mapping


f image is a multiple of 41r). This number is called the Hopf invariant of mapping f.So,

M,

Consider a neighborhood T of the curve Mq in S3. It possible to represent T as atopological product of M. and some surface for which the mapping f maps ontoE,2, i.e. T = Mq x fi. Let u3 be the parameter in M. ranging from 0 to 21r and u' , u2

be the coordinates in 0 induced from E,2. Then every point p in T has the curvilinearcoordinates u1, u2, u3 and the mapping f translates the point p into the point q withcoordinates ul, u2. With respect to coordinates III, u'-, u3 the form v can be written as

du" A du" = f du' A due.

Let wI = c; du', where c; are some functions of u', u2, u3. Consider the exterior product

w1 Adw1 =wI nw=`4 du' AdugAdu'.

Note that fg does not depend on u3. Represent the integral of w1 A dwl over thedomain T as any iterated one as follows:

/' -rteAdwl = J du1 Adu2 / c3(u',u-,u3)du3 = J''-TJwi.T e? 0 t_ st,

Since the integral over Mq of w1r does not depend on q,

r wI Adw1=-yS(E'')

J 42rrwhere S(E?) stands for the area of E! in S2. If we sum over all domains E.' then weobtain (1). The theorem is proved.

S. Novikov [741 obtained some many-dimensional generalizations of the White-head formula.

2.19 Vector Fields Tangent to Spheres

Consider the vector fields in E" which are tangent to a sphere S"' at each pointx E S"-I from the families of spheres with a common center. It is well known that aneven-dimensional sphere admits no regular tangent vector fields. Therefore, weconsider the vector fields on the odd-dimensional spheres.

The vector fields nI (x), ... , nk(x) on a sphere S"-' are called linearly independent ifthey are linearly independent at each point x E S"-1. The Hurwitz-Radon-Eckmann


theorem [93] on the possible number of linearly independent vector fields on S"is known. James formulated this result as follows: represent n in the form of aproduct of an odd number and a power of 2, i.e. n = (2a + 1)2". Divide b by 4 withthe remainder, i.e. represent it in the form b = 4d + c, where 0 < c < 3. Setp(n) = 2C + 8d. Then there are p(n) - I linearly independent vector fields on thesphere S". Adams [94] proved that this number is maximally possible.

We denote the number of linearly independent vector fields on S"- 1 by k. Belowwe present a table of k for low-dimensional spheres:

n - I 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29

k 1 3 1 7 1 3 1 8 1 3 1 7 1 3 1

We shall construct the regular vector fields on low-dimensional spheres in explicitform, avoiding heavy theory. This also gives us a rich set of non-holonomic fieldsin E".

It is easy to find a single vector field on the odd-dimensional sphere. Suppose thatS"-1 is given as a set of points in E" whose coordinates satisfy the equation

where p is a radius of S"-1.Since n is odd, the coordinates can be gathered into pairs, for instance, in con-

sequent order. We define the vector field a = (-x2, xt , -x4, x3, ... , -x", x"_ 1) at thepoint of a sphere having coordinates x1,... , x".

Let us construct three mutually orthogonal vector fields on S3. Denote a positionvector of a point in S3 by r = (x1,...,x4). Split the coordinates into two pairs:

(XI, X2), (X3, X4). Perform the permutation of the form(xi

xl I over each pair.x X;

The splitting under consideration gives the vector field a = (1X2i x), -x3, x4). Nextwe take the splitting into pairs (x1,X3), (x2,x4) and construct the vector fieldb = (-x3i x4, x1, -x2) in an analogous way to the construction of the field a. Note,however, that the field (-x3 i -x4, x1, x2) can also be constructed by the procedureabove, but it is not orthogonal to a. Taking the pairs (x1, x4), (x2, X3), we find thevector field c = (-x4, -x3i .r2, XI). Write the matrix formed by the components ofr, a, b, c:

r X1 x2 x3 .x4

a -X2 X1 -X4 X3

b -X3 X4 X1 -X2

C -X4 -X3 X2 X1

It is easy to see that the constructed fields are tangent to the sphere S3 centeredat the origin. They are mutually orthogonal and of unit length. We shall say thateach of the vectors a, b, c defines some permutations over coordinates, say A, B, C.

152 THE Cii:OMCTRN OF VECTOR FIELDS

For instance, the permutation A applied to the quadruple (fi, 2. 3i 1) gives(-6,Cl,

Let us construct the vector fields on S7. We split eight coordinates into two groupsof four coordinates: (xI, X2, X3, xa), (x5, x6, .Y7i .Yx). For each of these groups we canconstruct the fields analogous to those on S3. We obtain four of the vector fields onS3 by joining three-dimensional sphere fields. i.e. we define them as follows-

a = (al, a'-) _ (-x2, Xl, -.Y4, x3, -x6,.Yi, -xx, X7),

b = (bi, -k,) _ (- x3, x-1;X1, -x2,Y7. -xx, -X5e lb),C = (Cl , CA (-Xa, -X2, X1, -Xx, -X71 x6, X5),

where a,, b e, are the vectors on S3. constructed as above involving the coordinatesof the i-th group. Evidently, these fields are mutually orthogonal.

Consider the splitting into the following quadruples: (10, x4, x7, -TX)-Perform the B permutation over each of them:

xi,x,,xt.t6/

.Y3rxa.x7,xs`-.xi,-Y(,xl, -x, -.Y7: Xx:.Y3, -xa

These permutations define the vector field

d = (-x=, xe: -x7: Xx, x,, -X2, X.1- --V4)-

In an analogous way. the permutation C over the same groups defines

e = (-xh, -x, xx, X7, x2, X1, -x4, -x3 )

One can check that the constructed fields are mutually orthogonal. Now consider thegroups (xi,x2,.Y7,xx) and (x3,X4,Perform the permutations B. C over eachof them. This gives two more vector fields:

f = (-x7, -Y$, x5. -x,,, -.Y3,xa, xi, -X_),

g = (-X$, -X7, -x6, -x5, -y4, X3. X2, xl ).

Write the matrix formed with the components of a,... . g on S7 and a position vector r:

r x1 x2 xz Y4 Xc -YX X7 Xx

a -x_ .r -X4 Y3 -X,, .xS -xx X7

b -xi Y4 X1 -X2 X7 -V$ --Vs X6

C -X4 -X= N., X1 -XK -x7 X6 X5

d -xc Yh -X7 XK X1 -xZ xz -X4

e -Y,, -TS X X7 X. X -X4 -X,f -.Y7 XK Xc -X -X= X4 X1 -x,g -X -X7 -x6 -XS xA X1 X, X1

The components of this matrix satisfy a,, = -a for i 70 j. It is easy to check that allconstructed vector fields on S7 are mutually orthogonal.


As we know from the table at the beginning of this section, the sphere SII onlyadmits three linearly independent vector fields. To find them, we split twelve co-ordinates xi into three groups of four and perform the permutations A, B, C overeach of them. Write the corresponding matrix of components of vector fields on SI 1:

r X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 XII X12

a -X2 XI -X4 X3 -X6 X5 -X8 X7 -XI0 X9 -X12 X11

h -X3 X4 X1 -X2 X7 -Xg -X5 X6 -XII X12 X9 -X10C -X4 -X3 X2 X1 -X8 -X7 X6 X5 -X12 -XII X10 X9

Evidently, a more general statement holds: if the sphere Sn-1 admits k linearlyindependent vector fields then the sphere S"9-I admits not less than k vector fields. As23 = 8.3 - 1, S23 admits seven vector fields which are the three times repetition ofthe seven vector fields on S7.

Let us construct eight vector fields on S15. Since we can split the sixteen co-ordinates into two groups by eight coordinates, namely (xl, ... , xg) and(X9i ... , X16), and we have already constructed seven vector fields on S7, by doublerepetition of these fields we obtain seven vector fields on S 15. If a;, bi, . . . , g; i = 1, 2are the corresponding vector fields on S7 generated by the splitting mentioned abovethen we represent the seven vector fields on SIS as a = (a,,a2), b = (b1ib2), ... ,g = (91,92)-

Now construct one more vector field on S 15 which is orthogonal to all the otherfields. Consider one more splitting, namely (XI, x2i X3, x4, x9, x1o, XI, , x I2) and(X5i x6, x7, Xg, X13, x14, X 15i x16). Construct the vector h formed as (dl , d2) with respectto the groups of coordinates given, i.e. the vector obtained by means of the D-typepermutation (see the structure of the field d on S7) It = (-x9, xIo, -XI 1, X12, -x13,X14, -X15i X16i X1, -X2, X3, -X4, X5, -X6, X7, -xg). However, a direct check shows thath is not orthogonal to b, d, f. To satisfy the condition of orthogonality we ought tochange the signs of the last eight components of these vectors, i.e. we ought to setb = (b,, -b2), d = (dl, -d2), f = (fl, -f2). Present the table formed with componentsof mutually orthogonal vector fields on S I3 and r:

r X1 X2 X3 X4 XS X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16

a -X2 X1 -X4 X3 -X6 X5 -X8 X7 -X10 X9 -X12 X11 -X14 X13 -X16 XIS

b -X3 X4 X1 -X2 X7 -X8 -X5 X6 X11 -X12 -X9 X10 -X15 X16 X13 -X14

C -X4 -X3 X2 X1 -X8 -X7 X6 X5 -X12 -XII X10 X9 -X16 -X15 X14 X13

d -X5 X6 -X7 Xg X1 -X2 X3 -X4 X13 -X14 XIS -X16 -X9 X10 -XII X12

e -X6 -X5 X8 X7 X2 XI -X4 -X3 -X14 -X13 X16 X15 X10 X9 -X12 -XIIf -X7 X8 X5 -X6 -X3 X4 X1 -X2 X15 -X16-X13 X14 X11 -X12 -X9 X10

g -X8 -X7 -X6 -X5 X4 X3 X2 X1 -X16 -X15 -X14 -X13 X12 X11 X10 X9

h -X9 X10 -XII X12 -X13 X14 -XIS X16 X1 -X2 X3 -X4 X5 -X6 X7 -X8


We can construct the vector field i by means of a permutation of E types over thesecond splitting (see the structure of the field e on S7):

1 = (-X10, --X9, X, 2)X 11, -X14, -X 13, X 16, X15, X2, X I , -X4, -X3, x6, X5, -X8, -X7) -

A direct check shows that i is orthogonal to all the fields except d: (1, d) 54 0. Thus,the set of mutually orthogonal vector fields is not uniquely defined. Instead of d wecan take i.

Consider the geometrical properties of the fields just constructed. Each of thesefields is non-holonomic. Let us show, for instance, that a is a field in E" which isnon-holonomic, i.e. there is no family of the field a orthogonal to hypersurfaces.Consider the Pfaff equation which corresponds to the field a:

W = -X2dX1 +X, dX2 +... - XndXn-1 +Xn-1 dx, = 0.

The exterior differential of this form is

(1)

dw=2{dX, Ad.Xn}. (2)

Exclude one of the differentials xi from (1), say dx,, and substitute into (2); weobtain

dw= 2{ AdXnI

i=3 JJJ

where ,pc are some functions of x2, ... , xn. The exterior differential dw is not zeroidentically by virtue of w = 0. So, the field a is non-holonomic.

The streamlines of every field under consideration are great circles of spheres.Consider, for instance, the streamlines of the field a on S3. The derivatives of theposition vector of the streamline satisfy

dv1 dx2 dx3 d.X4

ds= 2,

ds = x, , ds = -x4, ds = X3,

i.e. d r/ds = a. Hence,

d2x, d2X4j2 = -X1 i , ;2 = -X4,

i.e. d2r/ds2 = -r, which completes the proof.Scaling, we can get the unit vector fields defined at all points of space except the

origin. Let us find symmetric polynomials Sk of the principal curvatures of thesecond kind of the streamlines of one of those fields, say n = a/Ial. Set

p = x + + xn. The following theorem holds:

Theorem Odd symmetric polynomials of principal curvatures of the second kind arezero, while even symmetric polynomials are Sk = -L Ci ,11,.


Let us show that all the principal curvatures of the second kind are purely ima-ginary except one of them which is zero. Denote by & the components of n. The

matrix IIaE II has the formI xixj

P

To obtain a more convenient form of the characteristic equation IA - AE] = 0 weintroduce the following vectors:

A=

xn

P1

:!Lc,,P,

x x,_i I -iP3 P P'

X.X.P, P P,

P

P-A

O

A

, 72 = 0 ,rl

0 0

O , Tn = 0

A

We shall denote the determinant by brackets. Then the equation JA - AE I = 0 can beexpressed as

=0.

Using the rule of determinant decomposition, we obtain

[TI,T2...... r ]+x2[I,T2,...,T,,] -XI[TI,l,T3,...,Tn]+... =0,

where dots stand for the sum of determinants each of which contains only onecolumn I taking the position of the column Ti. Those determinants are similar to thetwo which were presented. The matrix [TI, ... , is formed with second order

matrix boxes. Therefore, its determinant is easy to find [TI, ... , Tn] _ (A2 +Next we find:

I (n-2)/2 x2\ -3+I

(XIP

Ip ,

ln-zl/2 x1/[TI,I,T3,...,T] _

(A2

+ ) (_x2A+_)P-3.


The characteristic equation with respect to A has the form

1(A2+_) A2 = 0.

The first root A = 0 arises from the condition that the field n is a unit vector field.The other roots, namely A = 0 and A = f P, are the principal curvatures, where theimaginary roots are of multiplicity "-. 2The theorem is proved.

Let us find the curvature vector P which has been introduced in Section 2.3 for thecase of the unit vector field n in E":

P = S"_1n + Sn_2k1 + Sn_3k2 + ... +

where k; are defined inductively: k1 = Van, ki+1 = Vk,a, .... As n we take a/p. Sinceevery streamline of the field a is a great circle on a sphere, k1 = -r/p2. The field isconstant along the sphere radius, so that k2 = -V.n = 0, k1 = 0, i > 2. Take intoaccount that the odd symmetric polynomials are zero. So, P = S"-2k1 = -r/p". Bymeans of the vector P we can find the degree of mapping generated by the field a/pof the sphere S"-I onto the unit sphere in E":

0=1 J(P,v)dV=-1.wn-I

Consider now the question of the holonomicity of pairs of fields. We shall say thatthe distribution of the pair field a, b is holonomic if there are (n - 2)-dimensionalsubmanifolds orthogonal to both a and b. Otherwise we say that the distribution isnon-holonomic.

In papers we also meet another definition: the distribution of planes spanned on aand b is called integrable if there are two-dimensional submanifolds tangent to thoseplanes. The context will show in which sense this notion is used.

If we take a pair of fields a, b in E4 then their distribution is holonomic. We canprove this fact both analytically in terms of exterior forms and geometrically. Thefields c and r are orthogonal to a and b. Therefore, the planes of great circles in thesphere S3 which are tangent to the field c are orthogonal to a and b. This means thatthe distribution of the fields a and b is holonomic in E4.

On the other hand, the distribution of fields c and r is not holonomic in E4.Indeed, write the corresponding Pfaff system:

W1 =x1dxl +x2dx2+x3dx3+x4dx4 = 0,w2 = -x4 dxt - x3 dx2 + x2 dx3 + x1 dx4 = 0.

The exterior differentials are

dw1 = 0, dw2 = 2(dx, A dx4 + dx2A dx3).

Set A = x1x3 - x2x4. Exclude the differentials dx1 and dx2 from the Pfaff system:

dx1 = -{(x2 + x3) dx3 + (x3x4 + xix2) dx4}/0,


dx2 =I (xlx2+x3x4)dx3+(xi +X4)dx4}/A.

Substitute them into dw2:

2 , ,dw2 = -0(x; +x; +x3+x4)dx3 ndx4.

Hence, dw2 is not zero identically by virtue of the system. Therefore, the distributionis non-holonomic. Thus, we have an example of two distributions of two-dimen-sional mutually orthogonal planes in E4 such that one of them is holonomic, whilethe other is not.

The case considered above is exceptional. In a space of larger dimension thedistribution of vector pairs is non-holonomic. Consider, for instance, the distributionof the vector pair a, b in E8. The corresponding Pfaff system is

w, = (a, d r) = - x2 dx1 + x1 dx2 - X4 dx3 + X3 dx4

- x6 dx5 + x5 dx6 - x8 dx7 + x7 dxs = 0,

w2 = (b, dr) = - x3 dYl + x4 dx2 + XI dx3 - x2 44

+X7dx5 -x8dx6-X5dx7+x6dx8 =0.

Let us find, for instance, the exterior form dw,:

do, = 2{dx1 Adx2 +dx3 Adx4+dxs Adx6+dx7 Adx8}.

From the system w, = 0, w2 = 0, we find8 8

dx1 = Eas dx,,a dT2 = 1: b3dx0,a=3 3=3

where a, b,7 are some coefficients. After the substitution of dx1 A dx2 into dal, thelatter will contain the term (a3b5 - b3a5)dx3 A dx5. In collecting similar terms thisterm does not vanish. Therefore, dw1 0 by virtue of w, = 0, w2 = 0, i.e. the dis-tribution is non-holonomic.

Let us consider a more complicated system of three Pfaff equations in E8:

w, = (a, d r) = 0, W2 = (b, dr) = 0, w3 = (c, dr) = 0.

We already found the exterior differential of dw, above. Exclude dx1i dx2, dx3 fromthe Pfaff system and substitute into dw1. In the Pfaff system w; = 0, i = 1, 2,3 a3denotes the three-element column formed with the coefficients of dx1. The determin-ant of the matrix of the system with respect to dx1, dx2, dx3 has the following form:

A = [a1a2a3] =-X2 X, -X4-x3 X4 X,-X4 -X3 X2

_ -x4 (x; +X2 + X3 + X4).

Find the coefficient of dx4 A dx5 in dw1. To do this we only need the terms with dx4and dx5 in the expressions for dx1, dx2, dx3 Write

dx1 =Q(Adx4+Bdx5)+---,


Idx2 =(Cdx4+Ddxs)+ ,

dx3 =

where A,-, F have the form

A = -[a4a2a3], B = -[asa2a3],

C = -[a1a4a3], D = -[alasa3j,E_ -[ala2a4], F= -[ala2asj.

Substituting dx1 i dx2, dx3 into dv1, we obtain

dwl = 2(AD-BC-F0)dx4Adx5

where the dots stand for terms without dx4 A dxs. Set U = (x2, + x2 + _x3 + xi). Then

X3 XI -X4

A=- -X2 X4 X1 = -X1U,

X1 -X3 X2

-X2 X3 -X4

C=- -x3 -x2 X1 = -x2 U,

-X4 XI X2

-X6 X1 -X4

B=- X7 X4 X1 = X2X4X6 + XjXg - X3X4X7 + X4X8 + X1X3X6 + XIX2X7,

-Xg -X3 X2

-X2 -X6 -X4

D=- -X3 X7 X1 = X22X7 - XIX4X6 + X3X4X8 + X42X7 + XIX2X8 + X2X3X6,

-X4 -Xg X2

-X2 XI -X6F= - -X3 X4 X7 _ -X2X4Xg + XIX4X7 + X3X6 + X4X6 + XIx3Xg + X2X3X7.

-X4 -X3 -Xg

After the substitution, we obtain

AD-BC-FA =x4x6U2.

Therefore at all points where x4, x6 and U are different from zero, the coefficient ofdx4 A dx5 in dw1 is not equal to zero. If we set dx6 = dx7 = 6x6 = 6x7 = 0 then all theforms dxj A dx6 and dxj A dx7 are zero. So, dvI $ 0 by virtue of the system, i.e. thePfaff system is not integrable.


Each of the vector fields constructed is a Killing field on a sphere. We get theproof from the definition. Such a field defines an infinitesimal mapping r -+ r of aRiemannian manifold onto itself under which d rz differs from d r-' by the in-finitesimal of a higher order than (bt)2. Put the point of the position vector r intocorrespondence with the point of the position vector r = in a unit sphere. As

the metric of the sphere is an induced one from the ambient Euclidean space, we cantake the differential of the square of r in Euclidean space to find this metric:

dr = {dr2 + 2(d r, d.c) bt + dd2(bt)2}/(1 + (bt)2),

where (d r, ik) means the scalar product in Euclidean space. Take one of the fieldsconstructed as the field . It is sufficient to check that (dr, 4) = 0. Each field f at apoint with the position vector r is obtained from r by component transpositions andsign changes. More precisely, if xi takes the j-th place then -xj takes the i-th place.Therefore, the products dx; dxj meet twice in the expression of (d r, dd) - once with anegative sign and once with a positive sign. Therefore, (dr, ek) = 0 and, as a con-sequence, { is a Killing vector field.

2.20 On the Family of Surfaces Which Fills a Ball

Let D" be a ball in Euclidean space E" covered with a one-parametric family ofsurfaces (i.e. through each point of the ball one and only one surface passes). Weshall suppose that each surface of the family corresponds to a unique value of theparameter t from (a, b] and vice versa, and also that this correspondence is con-tinuous. Moreover, suppose that each surface is connected and divides the ball intotwo bodies such that in varying t from a to b the volume of one of those bodies variesfrom zero to the volume of the whole ball D". We shall say that such a family fills theball regularly. Nevertheless, the surfaces themselves may have singularities. For in-stance, we may take as such a family the set of circular cones with a common axisand the same angle about the vertex if this vertex is within the ball and completedwith the family of frustums of the cones if the vertex is out of the ball. Two examplesare: the family of coaxial cylinders truncated by the ball; the family of level surfacesof some differentiable function p(x1 i ... , x") with a non-zero gradient within theball. The surfaces which correspond to extremal values oft can be degenerated into apoint or a curve.

We can pose the following simple question: is there a surface among those surfaceshaving an (n - I)-dimensional area greater than or equal to the area the diame rralsection of D"?

For n = 2 this question asks the lengths of curves which cover a disk of radius R.Among the curves of this family there is the curve which passes through the center ofthe disk. Since this curve comes to the boundary circle, the length of this curve > 2R.For the n-dimensional case we prove the following assertion.

Theorem Let us be given a family of surfaces which fills a ball D' of radius Rregularly. Then there is a surface of the family with an area greater than or equal to


R"-1 C(n), where C(n) = 1(21 " - 1)/2 is constant and I is the area of a unit(n - 1)-dimensional sphere.

Let VI and VZ be the volumes of the bodies DI and D2 which are the parts intowhich the surface divides the ball D". The boundary of each Di consists of thedomain in a boundary sphere having an (n - 1)-dimensional area F; and the surfaceunder consideration has area F. Therefore, the boundary area of Df is F+F,.Consider the isoperimetric inequality with respect to each D;:

F;+F> na,I,1"V;(n-,)'", -=1,2,

where an is the volume of a unit ball in E. Adding these inequalities, we get

2F> Vi"-I)/" + V2"-1)/") - (F1 + F2).

Since the family of surfaces fills the ball regularly, the volume of DI varies con-tinuously from zero to the volume of D" when the parameter t varies from a to b.Therefore, there is a surface which splits the ball into two bodies of equal volume, i.e.in this case VI = V2 = R"a"/2. Also, note that F, + F2 = 1 R"-1 is the area of thewhole boundary sphere. Hence,

2F> 2"a"R"-12(n-1)1" -w"-IR"-1.

Note that I = w"_ 1. Therefore, 2F > w"_ I R"- I (21/n - 1). The theorem is proved.Consider now the impact of the curvature of family surfaces on the size of the

domain of definition. Suppose that the field a of normals of surfaces is given in D"and is differentiable. Suppose that the second symmetric polynomial S2 of eachsurface is greater than some fixed positive number Ko. Since Si > li! 2) S2, the

divergence of a normal field is bounded from below as div o > 2;." ;) Ko (or from

above as div n < - 2(n ,1 ) K° I . if we integrate this inequality over the ball D', we

obtain

2(n - 1)-1 Ko.FHence,the radius of a ball, where a regular family of surfaces with a symmetric

function 52 > K > 0 can be given, is bounded from above as

R <n (n - 2)

VIKo 2(n

This assertion can be extended to the case of Riemannian space, but note that inRiemannian space it is impossible, in general, to get the upper bound on the radius ofa ball for an arbitrary K° where the regular family of surfaces with a second sym-metric function S2 > K° > 0 exists. Indeed, in Lobachevski space L3 with curvature


-1 every geodesic sphere has the normal curvatures > 1. Hence, we can take a ball,however large the radius, filled with a regular family of surfaces, namely with piecesof geodesically parallel spheres of S2 > 1. But if Ko > I + E, e > 0 then it is possibleto find the upper bound on the radius of a ball D3 C L3. However, we fix the ballradius to be I and watch the influence of Ko on the curvature of the space. Inte-grating the inequality diva > CIK-o over D3 in Riemannian space, we obtain theestimate < CS/ V, where S is the area of boundary sphere, V the volume of D3,C = const. The right-hand side of this inequality can be estimated from above interms of the maximum Q of sectional curvatures of Riemannian space (see [54]).Therefore, .,/R o < f(Q), wheref is some monotone increasing function of Q. If weinterpret the curvature of the family of surfaces as some loading on the domain ofspace then we may say that if this loading is greater than some critical value then themetric of ambient space will change, i.e. the Euclidean space will change into thecurved one; the value of curvature will be determined to some extent by the value ofloading. This influence is similar to the deformation of a flat membrane under theaction of interior tension. The membrane loses its flatness, swells into space andgains curvature. The estimate obtained means that if Ko increases, then after somecritical point, the space curvature Q also increases.

Consideration of the families of the surface in a ball leads to some problems. Letus formulate two of them:

(1) Suppose that in a ball D3 C E3 of radius R there is a regular family of C2-regularsurfaces having negative Gaussian curvature K < -1. Does the estimate fromabove on R hold?Observe that in Section 1.14 the existence of such an estimate has been proved,provided that the curvature of the normal vector field streamlines is boundedfrom above.

(2) Suppose that in a ball D" the regular vector field a is given. Is there a regularhomotopy which preserves the vector field on the boundary of the ball and turns itinto the integrable one?

A more general formulation involves a domain G in Riemannian space and a givendistribution of planes.

The latter problem is connected with one posed by Reeb, Chern and others:suppose that on a differential manifold of dimension n the field of k-dimensional(k < n) tangent planes is given. Under which conditions is this field homotopic to thetotally integrable one? Wood [99] proved that every field of 2-planes which istransversally orientable is homotopic to the totally integrable one. The first exampleof a field which is not homotopic to the totally integrable one was given by Bott [90].He found some necessary conditions for the field of planes to be homotopic to thetotally integrable one.

Note the following result obtained by Thurston [100]: every (n -1)-plane field ona closed manifold M" is homotopic to the tangent plane field of a C'°, codimension-one foliation.

162 THE GEOMETR% OF VECTOR FIELDS

2.21 Summary

We have studied the local and global geometrical characteristics of vector fields inn-dimensional Riemannian space. We have also found some relations between thesecharacteristics and topology, the theory of ordinary differential equations and fluidmechanics. In general, the vector field was assumed to be regular and to belong toclassical class C2. However, in certain cases the field under consideration could havesingular points, which were interpreted as sources of curvature of definite power. Inrecent decades, the ideas of non-holonomic geometry combined with the theory ofgeneralized functions has allowed a theory of currents on manifolds and varifolds tobe constructed. Nevertheless, further development of the classical theory of vectorfields is also very important. I hope that this book will help readers to preparethemselves for new discoveries in this area of mathematics.

References

[I]

[2]

Voss, A. Geometrische Interpretation Differetialgleichung Pdx + Q dy + R dz= 0. Math. Ann. 16 (1880) 556-559.Voss, A. Zur Theorie der algemeinen Punkt-ebensysteme. Math. Ann. 23 (1884)43-81.Voss, A. Zur Theorie der algebraischen Differetialgleichungen erster Ordungerster Grades. Math. Ann. 23 (1884) 157-180.Voss, A. Theorie der rationalen algebraischen Punkt-Ebenen-Systeme. Math.Ann. 23 (1884) 359-411.Voss, A. Ober die Differentialgleichungen der Mechanik. Math. Ann. 25 (1885)258-286.Lilienthal, R. Uber kurzsten Integralkurven einer Pfaffschen Gleichungen.Math. Ann. 52 (1899) 417-432.Lie, S. Geometric der Berahrungstransformationen. Bd. I, Leipzig: B. G. Teub-ner, 1896.Rogers, G. Some Differential properties of the orthogonal trajectories of acongruence of curves. Proc. Irish Academy 29 A, No. 6 (1912).Dauthevill, S. Sur les systemes non holonomes. Bull. Soc. Math. France 37(1909).Darboux, G. Sur differentes proprietes des trajectories orthogonales d'unecongruence des courbes. Bull. Soc. Math. France 36 (1912) 217-232.Darboux, G. Lecons sur les systems orthogonaux et les coordones curviignes.Paris: Gauthier-Villars, 1910.Caratheodory, C. Untersuchungen uber die Grundlagen der Thermodynamik.Math. Ann. 67 (1909) 355-386.Bianchi, L. Vorlesungen uber Differentialgeometrie. Leipzig, Berlin: Druck andVerlag von B. G. Teubner, 1910.Chaplygin S. A. Investigation on the dynamics of non-holonomic systems. Mos-cow: GITTL, 1949. [Russian]

163

164 REFERENCES

[15] Egorov, D. F. Proceedings in differential geometry. Moscow: Nauka, 1970.[Russian]

[16] Sintsov D. M. Proceedings in non-holonomic geometry. Kiev: Vistshaya Shkola,1972. [Russian]

[17] Blank, Ya. P. Uber eine geometrische Deutung der Integrabilitats bedingungder Pfaffschen Differntialgleichung. Notes of Kharkov Math. Soc. 13 (1936)75-81.

[18] Schouten, J. A. Ober nicht holonome Uebertragungen in einer L". Math. Zeit.30 (1929) 149-172.

[19] Schouten, J. A. and von Kampen E.R. Zur Einbettungs and Krum-mungstheorie nicht holonomie Gebilde. Math. Ann. 103 (1930) 752-783.

[20] Vranceanu, G. Sur les espases non holonomes. Comp. Rend. 183 (1926) 835-854.

[21] Vranceanu, G. Etude des espaces non holonomes. J. Math. 13 (1934) 131-132.[22] Vranceanu G. La theorie unitaire des champs et les hypersurfaces non holo-

nomes. Comp. Rend. 200 (1935) 2056.[23] Vranceanu, G. Sur les systemes integres d'un systeme de Pfaff. Bull. Math. 35

(1934) 262.[24] Vranceanu, G. Lectii de geometrie differentiald. Bucharest, II 1951.[25] Krein, M. Uber den Satz von "Curvature integra". Izv. Kazan. Phys.-Math.

Soc. 3 (1928) Ser. 3.[26] Lopshits, A. M. The nonholonomic relations system in many-dimensional

Euclidean space. Proc. Semin. Vector Tensor analysis, 4 (1937) 302-332. [Rus-sian]

[27] Goursat, E. Lecons sur le problemm de Pfaff. Paris: Hermann (1922).[28] Griffin, M. Invariants of Pfaffian systems. Trans. Ann. Math. Soc. 35 (1935)

936.[29] Rashevski, P. K. On the connectability of any two points of a totally non-

holonomic space with admissible curve. Sci. notes of Ped. Inst., Ser. phys.-math.2 (1938) 83-94. [Russian]

[30] Vagner, V. V. On the geometrical interpretation of the curvature vector of anon-holonomic V3 in three-dimensional Euclidean space. Math. Sb. 4 (1938)339-356.

[31] Vagner, V. V. Differential geometry of non-holonomic manifolds. Proc. Semin.Vector Tensor analysis, H-HI (1925) 269-314. [Russian]

[32] Vagner, V. V. Differential geometry of non-holonomic manifolds. VIII Int.competition on Lobachevski prize, Kazan (1940) 195-262. [Russian]

[33] Vagner, V. V. Geometry of (n - 1)-dimensional nonholonomic manifold in n-dimensional space. Proc. Semin. Vector Tensor analysis, 5 (1941) 173-225.[Russian]

[34] Vagner, V. V. Geometrical interpretation of moving nonholonomic dynamicalsystems. Proc. Semin. vector tensor analysis, 5 (1941) 301-327. [Russian]

[35] Vaguer, V. V. The theory on circle congruences and the geometry of non-holonomic V. Proc. Semin. Vector Tensor analysis, 5 (1941) 271-283. [Rus-sian]

REFERENCES 165

[36] Bushgens, S. Geometry of vector fields. Rep. Acad. Sci. USSR, Ser. Mat. 10(1946) 73-96. [Russian]

[37] Bushgens, S. Critical surface of Adiabatic flow. Sov. Math. Dokl. LVIII (1947)365-368. [Russian]

[38] Bushgens, S. Geometry of stationary flow of an ideal incompressible liquid.Rep. Acad. Sci. USSR, Ser. Mat. 12 (1948) 481-512. [Russian]

[39] Bushgens, S. On a parallel displacement of vector fields Vestnik MSU No. 2(1950) 3-6. [Russian]

[40] Bushgens, S. On ideal incompressible liquid streamlines. Sov. Mat. DokI. 78(1951) 837. [Russian]

[41] Bushgens, S. Geometry of adiabatic flow. Sci. notes NSU, Ser. Mat. 148 (1951)50-52. [Russian]

[42] Bushgens, S. On the streamlines II. Sci. Math. Dokl. LXXXIV (1952) 861.[Russian]

[43] Pan, T.K. Normal curvature of a vector field. Amer. J. Math. 74 (1952)955-966.

[44] Coburn, N. Intrisic relations satisfied by vorticity and velocity vectors in fluidtheory. Michigan Math. J. 1 (1954) 113-130.

[45] Coburn, N. Note on my paper "Intrisic relations ...... Michigan Math. J. 2(1954) 41-49.

[46] Cheorghiev, Ch. Studii si cercetdri. Acad. R.P.Romania, Stiin matem. Filiala JasiMath. Sec. 1 8 No. 2 (1957).

[47] Sluhaev, V. Non-holonomic manifolds V,,' of zero external curvature. Ukr.Geom. Sb. 4 (1967) 78-84. [Russian]

[48] Sluhaev, V. Towards the geometrical theory of liquid stationary motion. Sov.Mat. Dokl. 196 No. 3 (1971) 549-552. [Russian]

[49] Sluhaev, V. Geometrical theory of ideal liquid stationary motion. Novosibirsk,Numer. methods of continuous medium mechanics. 4 No. 1 (1973) 131-145.[Russian]

[50] Sluhaev, V. Geometry of vector fields. Tomsk: Tomsk University Press, 1982.[Russian]

[51] Neimark, Yu. and Fufaev, N. Dynamics of non-holonomic systems. Moscow:Nauka, 1967. [Russian]

[52] Krasnosel'ski, M. Topological methods for nonlinear integral equations theory.Moscow: Gostehizdat, 1956. [Russian]

[53] Krasnosel'ski, M., Perov, A., Zabrejko, P. and Povolotski, A. Vector fields on aplane. Moscow: Fizmatgiz, 1963. [Russian]

[54] Aminov, Yu. Vector field curvature sources. Math. Sb. 80 (1969) 210-224.[Russian]

[55] Aminov, Yu. The divergent properties of vector field curvatures and the familiesof surfaces. Math. Zametki. 3 (1968) 103-111. [Russian]

[56] Aminov, Yu. Some global aspects in the geometry of vector fields. Ukr. Geom.Sb. 8 (1970) 3-15. [Russian]

[57] Aminov, Yu. On the behavior of vector field streamlines near the cycle. Ukr.Geom. Sb. 12 (1972) 3-12. [Russian]

166 REFERENCES

[58] Aminov, Yu. On an energetic condition for the existence of rotation. Math. Sb.86 (1971) 325-334. [Russian]

[59] Aminov, Yu. On estimates of the diameter and volume of a submanifold ofEuclidean space. Ukr. Geom. Sb. 18 (1975) 3-15. [Russian]

(60] Aminov, Yu. The holonomicity condition for the principal curvatures of asubmanifold. Math. Zametki. 41 (1987) 543-548. [Russian]

[61] Aminov, Yu. The multidimensional generalization of the Gauss-Bonnet formulato vector fields in Euclidean space. Math. Sb. 134 (1987) 135-140. [Russian)

[62] Eftmov, N. The investigation of the unique projection of a negatively curvedsurface. Soy. Mat. Dokl. 93 (1953) 609-611. [Russian]

[63] Eftmov, N. Non-existence of a complete regular surface of negative upperbound for Gaussian curvature in 3-dimensional Euclidean space. Soy. Mat.Dokl. 150 (1963) 1206-1209. [Russian]

[64] Efimov, N. The occurrence of singular points on negatively curved surfaces.Math. Sb. 64 (1964) 286-320. [Russian]

[65] Efimov, N. Differential criteria for some mappings to be homeomorphisms withapplication to the theory of surfaces. Math. Sb. 76 (1968) 489-512. [Russian]

[66] Whitehead, J. H. C. An expression of Hopf's invariant as an integral. Proc.Nat. Acad. Sci. USA 33 (1947) 117-123.

[67] Haefliger, A. Sur les feuilletages analytiques. Comp. Rend. 242 (1956) 2908-2910.

[68] Godbillon, C. and Vey J. Un invariant des feulletages des codimension 1. Comp.Rend. 273 (1971) 92-95.

[69] Reinhart, B. L. and Wood, J. W. A metric formula for the Godbillon-Veyinvariant for foliations. Proc. Amer. Math. Soc. 38 (1973) 427-430.

[70] Tamura, I. Every odd dimensional sphere has a foliation of codimension one.Comm. math. Helv. 47 (1972) 164-170.

[71] Tamura, I. Topology of foliations. Moscow: Mir, 1979. [Russian][72] Pontryagin, L. Smooth manifolds and their applications to homotopy theory.

Moscow: Nauka, 1976. [Russian][73] Novikov, S. Topology of foliations. Proc. Moscow Math. Soc. 14 (1965) 248-

278. [Russian][74] Novikov, S. The generalized analytic Hopf invariant. Multivalued functionals.

Uspehi Mat. Nauk, 36 (1984) 97-106. [Russian][75] Gantmacher, F. Matrix Theory. Moscow: Nauka, 1988. [Russian][76] Kosevich, A., Ivanov, B. and Kovalev A. Nonlinear waves in magnetism. Di-

namical and topological solitons. Kiev: Naukova dumka, 1983. [Russian][77] Whitney, H. Geometrical theory of integration. Moscow: IL, 1960. [Russian)[78] Hu, S. Homotopy Theory. Moscow: Mir, 1964. [Russian][79] Rogovoj, M. Towards the metric theory of non-holonomic hypersurfaces of

n-dimensional space. Ukr. Geom. Sb. 5-6 (1968) 126-138. [Russian][80] Rogovoj, M. Towards the differential geometry of a non-holonomic hyper-

surface. Ukr. Geom. Sb. 7 (1970) 98-108. [Russian][81] Rogovoj, M. On the osculating hypersurfaces of a non-holonomic manifold

VR-1 in P,,. Ukr. Geom. Sb. 18 (1975) 116-125. [Russian]

REFERENCES 167

[82] Rogovoj, M. On the curvature and torsion of a non-holonomic hypersurface.Ukr. Geom. Sb. 7 (1970) 109-113. [Russian]

[83] Rogovoj, M. Towards the projective-differential geometry of non-holonomicsurface. Ukr. Geom. Sb. 8 (1970) 112-119. [Russian]

[84] Rogovoj, M. Projective characteristics of the non-holonomic manifold V.Ukr. Geom. Sb. 11 (1971) 83-86. [Russian]

[85] Lumiste, U. Towards the theory of manifold of planes in Euclidean space. Sci.Notes Tartu Univ. 192 No. 6 (1966) 12-46. [Russian]

[86] Lumiste, U. and Gejdel'man, R. Geometry of family of m-dimensional spacesin n-dimensional spaces. Proc. fourth All-Union Congr. of Math.. Leningrad, 2(1964) 201-206. [Russian]

[87] Laptev, G. and Ostianu, N. The distribution of m-dimensional elements inspaces of projective connection. Proc. Geom. Semin., Viniti 3 (1971) 49-94.[Russian]

[88] Ostianu, N. The distribution of hyperplane elements in projective space. Proc.Geom. Semin., Viniti, 4 (1973) 71-120. [Russian]

[89] Bott, R. Vectorfields and characteristic numbers. Michigan Math. J. 14 (1967)231-244.

[90] Bott, R. A residue formula for holomorphic vector fields. J. Dif. Geom. 1(1967) 231-244.

[91] Baum, B. F., Cheeger J. Infinitesimal isometrics and Pontryagin numbers.Topology 8 (1969) 173-193.

[92] Kantor, B. Towards the problem of the normal image of a complete surface ofnegative curvature. Math. Sb. 82 (1970) 220-223. [Russian]

(93] Eckmann, B. Gruppentheoretischer Beweis des Satzes von Hurwitz-Radon.Comm. Math. Help. 15 (1942/43) 358-366.

[94] Adams, J. F. Vector fields on spheres. Ann. Math. 75 (1962) 603-632.[95] Stepanov, V. A course in differential equations. Moscow, Gostehizdat, sixth

edition 1953. (Russian][96] Rumjantsev, V. On a integral principal for non-holonomic systems. PMM 46

(1982) 3-12. [Russian][97] Rumjantsev, V. and Karapetjan, A. Stability in a moving non-holonomic sys-

tem. Itogi Nauki i Tehn., General Mech. M. VINITI 3 (1976) 5-42. [Russian][98] Raby, G. Invariance des classes de Godbillon-Vey C'-diffeomorphismes. Ann.

Inst. Fourier. 38 (1988) 205-213.[99] Wood, J. Foliations on 3-manifolds. Ann. Math. 89 (1969) 336-358.

[100] Thurston, W.P. Existence of codimension-one foliations. Ann. Math. 104(1976) 249-268.

[101] Fuks, D. Foliations. Itogi Nauki, Seriya: Algebra,Topologiya, Geometriya. 18(1981) 151-213 [Russian]. Translation: J. Soviet Math. 18 (1982) 255-291.

[102) Lawson, H.B. Foliations. Bull. A.M.S.(3) 80 (1974) 369-418.[103] Ferus, D. Totally geodesic foliations. Math. Ann. 188 (1970) 313-316.[104] Rovenskii, V. Totally geodesic foliations. Sibir. Math. J. 23 (1982) 217-219.[105] Tanno, S. Totally geodesic foliations with compact leaves. Hokkaido Math. J.

1 (1972) 7-11.

168 REFERENCES

[1061 Abe, K. Application of a Riccatti type differetial equation to Riemannianmanifold with totally geodesic distribution. Tohoku Math. J. 25 (1973) 425-444.

[107] Johnson, D. and Whitt, L. Totally geodesic foliations. J. Diff. Geom. 15 (1980)225-235.

[108] Johnson, D. and Naveira, A. A topological obstruction of the geodesibility ofa foliation of odd dimension. Geom. Dedicaza. 11 (1981) 347-357.

[109] Gromov, M. Stable mappings of foliations into manifolds. lzv. Acad. Nauk.SSSR. Ser. Mat. 33 (1969). 707-734.

[110] Haefliger, A. Some remarks on foliations with minimal leaves. J. Dif. Geom. 15(1980). 269-284.

[111] Hass, J. Minimal surfaces in foliated manifolds. Comm. Math. Hely. 61(1986).1-32.

[112] Solomon, B. On foliations of R"' 1 by minimal hypersurfaces. Comm. Math.Hely. 61 (1986) 7-83.

[113) Cairns, G. Feuilletages totalement geodesiques de dimension 1, 2 or 3.C. R. Acad. Sci. Paris. 298 (1984) 341-344.

[114] Tondeur, P. Foliations on Riemannian manifolds. Universitext, Springer-Verlag, New York, 1988.

Subject Index

AAdjoint vector 83

HHaefliger theorem L41

Algebraic vector field 58 Hamilton formula 13

asymptotic lines 22

BBasic invariants 25

holonomic vector field 3homotopy classes AS

Hopf invariant 45Hurwitz-Radon-Eckmann

Bernoulli function 83 theorem 1551

Bianchi system 22Bushgens theorem 85 J

Jacobi theorem 2CCaratheodory-Rashevski

theorem 82Closed stremline 64complex non-holonomicity 66

DDegree of mapping 451

Dupin theorem 22

EEfmov's theorem 60lemma 60

Euler equations 82

KKilling field 111, 112, 113

LLame equations 28

family 23Laplace net 22

MMean curvature,torsion 42

NNormal curvature 9 24

GGauss-Bonnt formula 32geodesic torsion 42Godbuon-Wey invariant 51

PPrincipal curvature 10Poisson bracket 3

169

170 SUBJECT INDEX

RReeb's example L41

total torsion 51triorthogonal family 71

SSingularity 52

VVagewski stability criterion 64

sources 52 Vagner interpretation 3stable limit cycle

TThurston result

64

lb_1

vector 92, 97

WWhitehead, theorem 148

total curvature 11 Wood theorems L61

Author Index

B

Bernoulli, L 82Bianchi, L. 79Blank, Ya. xiiBonnet, 0. 39 70Bykov, V. M. xii

CCaratheodory, C. xii, i87

Cartan, E. 131

Chaplygin, S. xii

DDarboux, G. xii, 14Dupin, F. P. Ch. 21

EEfimov, N. V. xii, 59Euler, L. 82

FFerus, D. L43

Frankel, T. T. 143Freudenthall, IL 1W

Frobenius, F. G. L. 116Fuks, D. 143

GGauss, K. 39. 69

HHaefliger. A. 141

Hamilton, W. R. 13Hodge, W. W. D. 131

JJacobi, K. G. J. 2

Kvan Kampen, E. R. xiiKilling, G. 23

LLame, G. 23Laplace. P. S. 22Lawson. H. B. L43

Lie, S. xiLilienthal, R. xiiLobachevski, N. xii

MMinkowski, H. 34

NNovikov, S. P. 143. 154

171

172

PPfaff, J. F. 116Poincare, A. 126Poisson, S. D. 3

Pontryagin, L. S. 101

RRogers, G. xiiRohlin, V. 104Rashevski, P. K. 87, 143

SSchouten, J. A. xiiSintsov, D. xiiSerre, J.-R. 104

AUTHOR INDEX

TTamura, F. 143Thurston, W. P. 143, 164Tondeur. P. 143

VVagner. V. xii. 3, 92, 97Voss. A. xiVransceanu, G. xii

WWeingarten, J. 69Whitehead, J. M. C. 45, 148Wood, J. 161

YYampolski, A. xiii

THE GEOMETRY OF VECTOR FIELDSYu. Aminov. In.IIttlle lol Lots I'hv,its,intl Lngineering, KImrkoU, LIkraine

This volume prosenls a I LissiraI ,ipproat11111 Ia' found.)l inns anti of the genlllel (Vt d Ve( a tr fields and (It s( r fields in 111,11 [urlidvan spa( e, triply-t rlI I()gt n,II,Vstems and applications in met hani( s. Oliler Ittpics in( It I fields, I'iaiiian I( if-Ills .111 (1sysk n» In n-dimensional ,pace, Ioliations and their Godbillun-Vev inUariant are also ( onsid-ered. there is much interest in the study of geometrical ohlects in n-dnnensional [u( iI k'ITlspat o and this Volume provides d useful and fill prehensiye presentation.

About the authorI'rt ii Yu. Am nt ty is d leading resear( h fellow at the Institute for Low TemperatUrt Phvsit sand Lngineering in Ukrainian Aca(lern of Sciences, hharkov, Ukraine. Iie works in classical

geometry and has puhhshed in this aria Of ntathemattcs.

Titles of related interestIhr (,rometr) of Sub(ttandol(hYu. Aminnv

I inear Algehrd and GeometryA.I. hostrikin and 1'u.i. Manin

Tensorand L'et Ale( hdnit and l'hyshedited by AT. Fumenko, O.V. Manturov and V.V. Troiimus

Itilt\ tut-5(114-201-5

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Gordon and Breach Science Publishers Australia I ,tndd.i Pram e India

I,ilmn Lus.enthuurg Malaysia The NIIht'rlancs Russia Singapore Switzerland

[yu. aminov] the geometry of vector fields(bookfi.org)

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