The Geometry of Vector Fields
Yu. Aminov
Gordon and Breach Science F'ublishers
The Geometry of Vector Fields
Yu. Aminov Institute for Low Temperature Physics and Engineering Kharkov, Ukraine
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Dedicated to my parents
Contents Preface -
I
xi
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 Normal Curvatures 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 .
1.8
The Second Divergent Representation of Total Curvature of the
.
------
Second Kind 1.9
32 34
The Interrelation of Two Divergent Representations of the Total
Curvatures of the Second Kind ..........................
37
1.10 The Generalization of the Gauss-Bonnet Formula for the Closed Surface
......
....
39
1.11 The Gauss-Bonnet Formula for the Case of a Surface with a 42
Boundary
1.12 The Extremal Values of Geodesic Torsion .................. vii
49
CONTENTS
viii
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 Vectnr Field and the Size of Domain of Definition ... . . .... . . . .. . . .
55
1.15 The Behavior of Vector Field Streamlines in a Neighborhood of a
Closed Streamline ---------------------------------1.16 The Complex Non-Holonomicity .........................
60
66
1.17 The Analogues of Gauss-Weingarten Decompositions and the Bonnet Theorem 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 Ideal Incompressible Liquid .................................
1.21 The Caratheodory-Rashevski Theorem ....................
82 87
1.22 Parallel Transport on the Non-Holonomic Manifold and the Vagner Vector
2
-----------------
---
92
............................
99
Vector Fields and Differential Forms in Many-Dimensional Euclidean and Riemannian Spaces 2.1
The Unit Vector Field in Many-Dimensional Euclidean Space
2.2
The Regular Vector Field Defined in a Whole Space ...........
2.3
The Many-Dimensional Generalization of the Gauss-Bonnet
99 101
Formula 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 .............
III
2.6 On Symmetric Polynomials of Principal Curvatures of a Vector Field on Riemannian Space .............................
114
The System of Pfaff Equations . ..........................
116
2.8 An Example from the Mechanics of Non-Holonomic Constraints ..
123
The Exterior Differential Forms ..........................
124
2.7
2.9
2.10 The Exterior Codifferential
211 Some Formulas for the Exterior Differential
129 -
131
2.12 Simplex, the Simplex Orientation and the Induced Orientation of a Simplex Boundary . ...................................
133
-
-
-
-
-
-
-
-
-
-
-
-
-
ix
CONTENTS
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 of
the 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 --------
----
Subject Index
Author Index ....-
161
169
------
171
Preface For a regular family of surfaces given in a domain of three-dimensional Euclidean space one can discover geometrical properties of this family from the very fact that the surfaces totally fill the domain and lie in space in a good way, i.e. without mutual intersections. These properties cannot be reduced to the geometry of the surfaces themselves but are connected with their spatial configuration. We can study these properties by considering the vector field of unit normals on the surfaces. Many
surface 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 of the distribution of planes orthogonal to the vector field, i.e. non-holonomic geometry, includes the theory of families of surfaces. The set of all vector fields is wider
than the set of normal vector fields and our study of arbitrary vector fields in contrast to the fields of normals of surfaces is necessary, for various reasons. Many results concerning the family of surfaces stay valid, as well, for an arbitrary vector field and this particular case makes these results clearer. It is possible to introduce the concepts of Gaussian and mean curvatures, the two analogues of geodesic lines and lines of curvature, and the analogues of asymptotic lines for the case of a vector field. Also, a vector field has its own geometrical invariant which shows how much a given vector field differs from the field of normals of a family of surfaces. This quantity we call the non-holonomicity value of a vector field. Besides the study of the distribution of planes orthogonal to a given field, it is natural to study the behavior of the streamlines of a vector field from the metric viewpoint. In this context the geometry of a vector field is closely related to the theory of ordinary differential equations. It is useful to provide some background information on the development of the geometry of vector fields and the mathematicians associated with this study. The foundations of the geometry of vector fields were proposed by A. Voss at the end of
the 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 curvature, asymptotic lines, and he also considered some special classes of vector fields and mechanical systems with non-holonomic constraints. Liliental studied the analogue of geodesic lines - the shortest lines orthogonal to the field. Rogers gave the
interpretation 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 and others. D. Sintsov produced the generalization of the Beltrami-Enneper formula, stated the relations between the curvatures of the first and second kind, considered the 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 vector fields from the projective point of view. The papers by Vaguer on non-holonomic geometry were awarded the Lobachevski
prize. Vaguer introduced the notion of parallel transport with respect to a vector field, and generalized the Gauss-Bonnet formula to the case of a strictly nonholonomic vector field. In this monograph we present these results in simple language, avoiding the difficult methods in the Vaguer paper. Another generalization of this formula, based on the concept of spherical image, which works for the case of an arbitrary vector field in n-dimensional Euclidean space, was found by the author. In the preparation of this book, in addition to the papers of other mathematicians, the
author's results on 'divergent' properties of symmetric functions of principal curvatures, on the behavior of streamlines, and on complex non-holonomicity have been included. We also present the author's result on mutual restrictions of the fundamental 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 papers by N. Efimov on the geometry of surfaces of negative curvature. An important area of applications of non-holonomic geometry is the dynamics of mechanical systems with non-holonomic constraints. Usually, these constraints arise
in 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 generalized
the 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 the geometrical viewpoint, mechanical systems with non-holonomic constraints have
been considered by V. Vaguer and G. Vransceanu. Currently, systems with nonholonomic 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, which present necessary and sufficient conditions for the vector field to be the field of velocity directions of the stationary flow of an ideal incompressible fluid. Vector fields occur in a natural manner also in general relativity. Recently it has become clear that vector fields of constant length are used in the description of liquid crystals and ferromagnets. It is interesting to note that the Hopf invariant, involved in 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 is devoted to vector fields in three-dimensional Euclidean space, to triply-orthogonal systems and to applications in mechanics. In Chapter 2 vector fields, Pfaffian forms and systems in n-dimensional space, foliations and their Godbillon-Vey invariant are considered. We also present the connection stated by Whitehead between the integrated non-holonomicity value and the Hopf invariant. The study of geometrical objects in n-dimensional Euclidean space is an exciting problem. The notions introduced here have descriptive interpretations and the theory can be demonstrated easily by concrete examples of vector fields. The methods include no complicated or, more precisely, awkward mathematical tools. Nevertheless, there are some problems where complexity is essential. Note that some important applied investigations lead to vector fields in E".
I would like to convey my appreciation to V. M. Bykov for his useful and encouraging positive remarks. I would also like to express my special thanks to Professor A. Yampol'ski for the translation and to M. Goncharenko for her technical assistance.
1 Vector Fields in Three-Dimensional Euclidean 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-dimensional Euclidean space P. The family of surfaces can be determined in different ways. For instance, we can regard the family surface as a level surface 4 (x,, x2. xz) = const of some function t(x1, x3) which depends on three arguments x1, x2, x3. We suppose that 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 welldefined unit vector a - the normal vector of the surface which passes through that
point. Therefore, the vector function n = a(M) is defined in the domain G. Let xl, x2, x3 be Cartesian coordinates in E' and et, e2, e3 be unit basis vectors. Then we can regard the vector field a as the vector function of the parameters x1. x2, x3. This
means that each component (; of a is a function of three Cartesian coordinates xi, x2. x3 of a point M in G: i = 1,2,3.
f, =t;(xi,x2,x3),
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
curl a = .L
1
e3
IL
THE GEOMETRY OF VECTOR FIELDS
2
We shall need the following formulas in the sequel. If v, a and b are some vector fields and A is a function of x, then curl Av = A curl v + [grad A, v], (1)
curl [ab] = a div b - b div a + Vba - V,b,
where Vba means the derivative of a via b,
the vector of components
i.e.
The following theorem holds. Theorem (Jacobi) If (n, curl n) = 0 at each point then there is a family of surfaces orthogonal to the vector field n and vice versa.
Let (n, curl n) = 0 in a domain G. Show that there is a family of surfaces f (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 directions dr = {dxl, dx2, dx3} each of which is orthogonal to u, i.e. those directions which satisfy the equation CIdx1+e2dx2+C3dx3=0.
(2)
We may assume that the system of coordinates in E3 has been chosen in such a way that at Mo and, as a consequence, in some its neighborhood 3 # 0. We can rewrite equation (2) as follows:
ti
dx3 = - 6 dx1 - = dx2.
(3)
We try to find a function x3 = z(x1, x2) which satisfies equation (3) and such that z(x°, x02) = x°. In this case dx3 is the differential of that function and the coefficients of dx1 and dx2 in (3) are the partial derivatives of the function z(x1 i x2), i.e. the equation (3) is equivalent to the following system: 8z 8x1
_
-
6 S3 '
8z 8x2
_ _ f2
6
( 4) .
The right-hand sides of these equations depend on X1, x2 and z. System (4) is solvable if the compatibility condition (5)
is satisfied, where one ought to find the derivatives taking into account (4). We have
.
CO = (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 13
x = (6x i 3 - 2c, 66 - t3xg3t2 + 1;3xg1 t2) f 3 3. I
The difference of expressions above produces the compatibility condition 1 3Jx, - 1
c)
\
=
3x, - Ix,) + t3(tIx2 - t2x1)k.
_ -(a,curl n)f32 = 0.
2
(6)
Since the latter equation is satisfied by hypothesis, there is a function z = z(xl , x2, X03), where x3 is a fixed parameter, which satisfies the system (4) and such that z(x°, xZ, x3) = x3. This function defines the surface of the position vector r = {xI, x2, z(xI, x2i X03)), where x3 is fixed. Due to the condition z(x°, x2, x3) = x3 this surface passes through the point Mo. Varying x3 we obtain a family of surfaces satisfying equation (2) at each point. That equation means that the tangent plane of the surface is orthogonal to a. Conversely, if the family of surfaces orthogonal to a exists then equation (2) is
satisfied and for the surface which passes through the point Mo it is possible to consider one of parameters, say x3, as a function of x, and x2. Then system (4) will be satisfied and the compatibility condition (5) will be fulfilled. Due to (6) this means that (a, curl a) = 0. So, the vector field a which is orthogonal to the family of surfaces is special and is called holonomic. The field o holonomicity condition can be represented in terms of non-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 orthogonal to n. The vector Vba - V.b is called the Poisson bracket and is denoted by V(a, b). It is natural to expand the class of vector fields under consideration, avoiding the requirement (n, curl n) = 0. We shall call the value p = (a, curl o) the value of nonholonomicity of a vector field a. It is independent of the choice of direction of o. Since a is a unit vector field, the value p has some geometrical sense. We give two geometrical interpretations of the non-holonomicity value p. 1. The Vagner interpretation. Let Mo be some point in a domain G where the vector field n is defined. Draw a plane through Mo which is orthogonal to a(Mo) and a circle L in this plane such that L passes through Mo. Set that positive orientation in
L 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 the direction of n(M). We obtain some ruled surface. Starting from Mo, draw a curve in this surface which is orthogonal to the elements (see Fig. 1). Consider one coil of that curve which is one-to-one projectable with the elements onto the positively oriented L. Denote it by V. The starting point of L' is at Mo. The curve L' meets the straight line element through Mo at another point MI. Denote by l the length of M I Mo and by a the area of disk bounded by L. Then the value of nonholonomicity of a vector field n at Mo is equal to the limit of ratio -1/a when L tends to 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 in the 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 through Mo 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. At the 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 is greater than a and, as a consequence, I cos VI < cos a. The angle a is determined by the vector field n. We can regard a as a function of coordinates .x,y in the plane which 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 = this we get
&x2 + oy2. From
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 are bounded from above in G with some number. The area of ruled surface bounded by L and L" can be estimated from above using (8):
a' =
f L
l u(s)I
< S3C3.
THE GEOMETRY OF VECTOR FIELDS
6
Therefore, the difference of circulations of n along L and L" is of the order S3:
f(n,dr) L
J e
(n,dr) < C4S3,
where C4 = const. Let fi be a disk bounded by L. The area of is a = S1/41r. That is why 3
J(ndr) - J(n,dr) o' <47rC4sumoS=0. s-o lim
Hence
-Oo f (n,dr). L
(9)
L"
Now find the circulation of a along V. Write an expansion of the vector function a along 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 orthogonal to d r along L' this integral has an upper bound
CI lu(s)Ids
That is why its quotient to a tends to zero, when S
J
0. Analogously,
(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.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
7
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 answers the question of why it is impossible to construct the surface orthogonal to n in the non-holonomic case. Construct the vector fields a and b which are orthogonal to each other and a in some neighborhood of a point Mo. To do this we can take, for
instance, a constant vector field f which is not collinear to n(Mo) and set a = [n, £) / I [n, t;) . Then b = [n, a). Draw a curve ! through Mo which has a as a tangent vector field. Then draw a curve m through Mo which has b as a tangent vector field. Starting from the points of m, emit the lines tangentially to the vectors of the field a. They form some surface 4D (see Fig. 2). Denote by n(M) a unit normal of that surface at its arbitrary point M. The n(M) is orthogonal to a(M), hence it is
located in the plane spanned by n(M) and b(M). Denote by ip the angle between n(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 curve tangential 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 the surface 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 at the 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 is
orthogonal to n does not exist. Indeed, let us suppose that at Mo and, as a consequence, in some part of its neighborhood (n, curl n) 0. A surface which is orthogonal to a and passes through Mo must contain the curves 1 and nz because they are orthogonal to n. Hence, it must contain curves tangential to a emitted from the points of m because the field a is orthogonal to n, i.e. that surface must coincide with 4' and its normal field n must coincide with n. But from (12) it follows that -A 0 in some 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 nonzero 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 of the First Kind
The surface behavior about a point Mo on it can be described with the help of the surface 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 the direction of r. Analytically, k can be found as a ratio of the surface second fundamental form 11= (d 2r, u) to the first one I = dr2: k"
_ (d2r,n) (dr,dr)'
where r is a position vector of the surface, n is a unit normal at Mo. Since n is a vector of unit length, it is possible to rewrite the expression above as r, d n) k - - (d(dr,dr)
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
9
n(VJ n#Q 00
dr
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 different from a constant vector field) at Mo. In the holonomic case the curvature will coincide with 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 shift in this plane. Draw another plane through n(Mo) and dr. Let ii(M) be a projection of 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) (or simply kn) of a vector field n in the direction dr is the limit of the ratio -cp/ldrl when M -+ Mo, i.e.
-,p kn(dr) = lim ldrl. M-wo
(1)
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 normal curvature. As a sine of o is equal to the cosine of an angle between ii(M) and dr, we see that sin V
-
(n(M ), d r)
lolldrl
(2)
As the infinitesimals V and sin cp are of the same order, it is possible to replace V with 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)
THE GEOMETRY OF VECTOR FIELDS
10
Note that Iiil -+ 1 when M
Mo. Using (2) and (3) we write lire
-(n(M,dr)
M-Mo
_ - (dd,dr)
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 its extremal values. These extremal normal curvatures are called the principal curvatures of the fast kind. The corresponding directions are called the principal directions of the first kind. In analogy with the theory of surfaces, the curves tangent to these
principal directions are called the lines of curvature of the first kind. Find the equations on the principal directions and lines of curvature. To do this write the expression 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 provided that dr is in the plane orthogonal n(Mo), i.e. dr = (dxl, dx2, 0). In these supposed conditions k takes the form , C t 6,q) (4) 1+(flx2+dx1dx2+f2,.d4].
As dxi + dx2 = 1, we use the Lagrange multiplier method to find the extremal values. Taking dxl and dx2 as independent variables, we want to find the extremal values 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 principal directions: (f lx,
+ \)dx1 +2 &r, dx2 = 0,
flx, +f2n dx 1+(f2,Z+A)dxz=0.
(5)
2
The characteristic equation for A has the form a2 +A1
IX,
rf IX: + 2
The matrix of system (5) is symmetric, so that we have two real roots \1 and A2. Just as in the theory of surfaces, it is possible to prove that these roots are the principal
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
II
curvatures of the field and the principal directions of the first kind are mutually orthogonal. The mean curvature of the field is a half-sum of principal curvatures:
H=
I (y lx, +42X2).
(AI
(6)
The total curvature of the first2 kind is a product of principal curvatures:
6x, +6,, 1 2 2
A point Mo is
called elliptic, parabolic or hyperbolic if at this point KI > 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' Ixz dx2 + A dxl = 0, 2
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-th component 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 a
projection 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 dr and n we obtain the system of the equation which determines the lines of curvature of the 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) as 3
Aij dxf dxj = 0, l,j=1
THE GEOMETRY OF VECTOR FIELDS
12
C1dx1 +6 dx2+{3dx3 =0,
where A;, are some coefficients depending on n and its derivatives. Take one of th coordinates, say X3, as a parameter t of the curve. From the second equation we find dx2
1 dX1
dt
42
dt - 3
Substitute this into the first equation. Solving the latter quadratic equation with respect to dx1 /d:, we find drk dt
k = 1, 2,
Fk(Ci, tt dx3
dt
= 1.
The right-hand sides of the equations above depend on x1, x2, x3. Integrating that system we can find the required lines of curvature. The directions of principal normal curvatures of the first kind are mutually orthogonal. Indeed, returning to system (5), note that the required directions are the eigenvectors of the symmetric matrix
2
From this the analog of the Euler theorem follows. Take the principal directions as
the coordinate axes. Set dx1 Ids = cos a, dx2/ds = sin a. From the assumption dx3 = 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 are the principal curvatures k1 and k2 of the first kind.
and -C ,
Therefore, the Euler formula for the case of a vector field has the same form as in the 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
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
13
FIGURE 4
Cartesian system of coordinates (xi, x2, x3). Then the streamline must satisfy the system of differential equations dxi
T
i= 1,2,3.
ds
We introduce the Hamilton formula for the vector of curvature of the streamline of a given field n. The required vector is directed along the streamline principal normal and its length is equal to the streamline curvature. We denote it by k. The Hamilton formula 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 parameter by d/ds then
the first component of
k
is equal to
r2 2+6r,y3-S2r,S2-S3x,6 1
= bIx,yr - 2
0
t22
+
C32)
= SIx,Si,
which completes the proof. From this it follows that for the case of a field of directions with congruent straight lines we have -[n, curl n] = 0, i.e. curls is collinear with n. Let us take a family of level surfaces U(xl, x2, x3) = coast. Let n be the field of normals. Find the o streamline curvature vector k in this case, i.e. the orthogonal
THE GEOMETRY OF VECTOR FIELDS
14
trajectories of surfaces family. Let n = { } be the unit tangent vector of the streamline. Set H = 1grad 1-1. Then
a
dx;8U d_ H, ax;
i
1,2,3.
The first component of k is equal to
d'x, Note that
8H 8U OUH+ 82U 8UH2
8U1 dxj 8 = ax; CH ax, J
ax, ax, 9X,
ax,axj ax;
1O H3. Therefore, the following is true: d 2x,
_
Hr, + (grad H, n) dx,
H
H
ds'
ds
Analogous calculations with respect to the second and third components of k show that
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: they are the straightest lines (that is, their geodesic curvature is equal to zero) and shortest local fines simultaneously (that is, the geodesic line has the smallest length among all curves joining two sufficiently close points PI and P2). In the case of a non-holonomic vector field we can determine the straightest and the shortest 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 to the field n is the length of the projection of the curvature vector of 1', i.e. of rs,, into
the plane orthogonal to n. We set it positive if the triple (r,, a, r.) is positively oriented and negative in the opposite case, i.e. we set I Pg
=_ (0,
r,,r,)
Set r, = e. Suppose that the unit vector field e is defined in a three-dimensional domain. Then curl e is also defined and rys = -[e, curl ej by Hamilton's formula. The geodesic curvature with respect to the field n of lines tangent to e takes the form I
Pg
= -(n, curl e).
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
15
In the case of the straightest line the geodesic curvature with respect to a is equal to zero. 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 line
curvature 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 the form
This is the differential equation of second order with respect to the vector function r(s) parametrized with some parameter s. If the curve is orthogonal to a at the initial point Po then it stays orthogonal to a for all s. Indeed, the function (r,, a) satisfies the following 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) then take the initial condition for r, such that Cr,I = 1. Then 1r,1 =_ 1 and, as a consequence, s is the are length parameter. It is sufficient to show that r; = const. We have
(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 straightest line 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 the smallest length among all nearly admissible ones. We consider a position vector r of the shortest line L as a function of the arc length parameters which varies from 0 to 1. We vary the curve L in such a way that it will stay admissible and the curve coordinate variations 6x1, 8x2, 6x3 turn into zero at
THE GEOMETRY OF VECTOR FIELDS
16
PO and PI . Take the parameter s as a parameter in the varied curve. Let dr,/ds be a tangent vector of the varied curve. Then the following equation must be satisfied: dr, ds
- 0.
(3)
The shortest line L gives the extremal value of the integral p;
ds =
jjdrj + dr; + dx-3 L
0
among all curves which satisfy (3). Applying the Lagrange method, we find the conditional extremum or, equivalently, the curve which gives the ordinary extremum of the following integral: r
r
f I + P(sk
dx
ds,
0
where W(s) is an unknown function, provided that & = 0 at the points of the required curve L. Consider the variation of , in an arbitrary direction or = (6x1, 6x2i 6x3). We shall consider the admissible curves as lying on the surface r = r(s, a) provided that for any fixed a we get the admissible curve. Then 8
ax,
as '
aO, ax,
6'x,
(4)
as = as as +' asaa'
Set -curl n = {l,}:
11 = 6a, - 6.r
12 = Sax, -
6x,,
13 = 6x, - far,,
or using briefer notation
ft
axj
axjax, =1Q,-;0 where e'j' is the Kronecker symbol. Then we can rewrite (4) as 8
as
'
ax,
ai;, axj OX,
as
axj ba as + asaa
8=x,
_ 1"?j(' + axj) 6Xj ax +
ax)
t\
asaa
6xj ax, +
+
a'x, _ ' asaa -
6a aS
6a as (curl
a''xi ; asaa 6xj
ds 6a
d r br do 6r a, d, T.) + (ds' 6a)
(5)
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
17
Using (3) and (5), we get
T=J
E(t)
L
_ rr L
L
8x; dip(s)
t` as
d r br
(curl n,
ds' ba)
111
(d n br
d r br
aa - (curl
+ f v(s)
dr
b }ds
aa
,
bx; d2x;
`
ba ds2 } ds.
The condition bl/ba = 0 and the expression above imply
d'r
+0
+ cp(s)
[Curl
n,] = 0.
(6)
Our next aim is to exclude the function ip(s). Take the mutually orthogonal unit vectors n, 2, {n, 2] as the basis along L. Decompose curl o and A, with respect to that basis. We have
+ b.
curl n = (n, curl n)n + a l n,d J
(7)
We can represent the curvature vector of the shortest line L as d2r ds2
1
r
= RI n,
dr]
1
+ La.
(8)
Rz
Substituting (7) and (8) into (6), we get
n,-
R2 L J Rz
n,- ,-
n) n,L 2J
ds
L L ZJ ds
=0.
Setting the coefficients of n and [n, !M equal to zero, we obtain RI
+ sp(s) (a, curl n) = 0,
(9)
R2+Jspa=0.
(10)
In the case of a holonomic vector field we have (a, curl a) = 0 and the equation of a geodesic line obtains the form 1
_=_(d2! dr ds2' ds'
R,
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 substituting into (10), we obtain I
R2
d
I
ds R,(n,-Curl n)
a RI(n,curln)
THE GEOMETRY OF VECTOR FIELDS
18
From (5) and (6) we obtain the expression for 1/R2 and a. Namely
_
d2r
1
R2 =
(ds2 °)
A do
- (ds'
ds)'
i.e. I /R2 is the field normal curvature in the direction of L. Denote it by k,,. Also,
a=
(curl n, n,
)
ds) ,
i.e. -a with the opposite sign is equal to the projection of the field streamline curvature vector onto the tangent of L. So, the equation of the shortest line has the form d do °) ds (n, curl n) +
+
-
drdn
(,,, d r)
(ds' ds) +
(i0. (n, curl n)
ds
into (6), we obtain the vector form for the equation of
Substituting W = the shortest line: d2r ds2
+n
L+ °)r
d
1
(n, curl n) + (n, curl n)
dr 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 orthogonal vector fields such that [a, b] = a. Set
dr
z = 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 parameter in terms of derivatives with respect to the Cartesian coordinates xi: dxi
da
'T'
ds = a'`,
db
drci
'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.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
As a and b are mutually orthogonal, then Li = -Ni. Find (
d2r dr (2, 1, n)
da
=-
_-
+cos2
da ds
,
19
a). We have
\
a(da ,a, a)+sin2 al db,b,a
+ 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 d2r
d--r
(dsZ
,
al
d( ' j'n) ( ' 'n) (a' curl a, dr 1 = 0. + ds (n, curl a) + (a, curl a)
(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 can represent equation (14) as
d ds
L;(cos aa; + sin ab1)) (n, curl n)
+ cost aA + cos a sin aB + sin2 ac
cos aP + sin aQ da (n, curl a)
L;(cos as; + sin ab1) ds +
= 0.
(17)
Consider the projection of equation (12) onto a. We have (dSZ, a) =
-sinaia+sina
a)
dx; _ -sina da +L; ds ds
Comparing with (13), we get dzr \\ dzr dr CdsZ,al=sina(,
(18)
20
THE GEOMETRY OF VECTOR FIELDS
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 a projection 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 ds (n, curl n)
= F x1.x2 i x; , a ,
d-a ds
(20)
To this equation we associate the equation
dr ds
= cos as + sin ab.
(21)
The equations (20), (21) form a system of ordinary differential equations with respect to 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 to the 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 the shortest line, the choice of a(0) gives the direction of the shortest line at Po. Thus, the following theorem holds. Theorem If (a, curl n) # 0 at some point PO then a bundle of shortest lines with respect to the given direction passes through this point.
Give an example of a vector field n with a simple structure but non-constant with respect to which the equation (17) is integrable in quadratures. Suppose the field n is parallel to the plane of xI, x2, i.e. it has the following components: t1 = cos cp,
; = 0.
6 = sin cp,
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
ds= -n
dip
dYz =-ncd.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
21
From the equation dr ds
= cos as + sin ab
we find dx /ds = cos a. Hence, d b/ds = -nc cos a. We rewrite equation (15) as d2r dS`
,s
)
- - sin a cos ac.
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
dal2 Cds/
=c2cos2a+c1,
where cl is a constant. We find the function a(s) from the following equation:
r
s=J
da c2 cos2 a + cl -
To find a position vector of the shortest line we use the representation of d r/ds via a and 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
dcp= -ccosads= -
c cos ada
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
THE GEOMETRY OF VECTOR FIELDS
22
As in each plane of X3 = const the field n is constant, the other solution is a family of 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 vector field analogous to the Gaussian curvature of a surface. Consider at some point Mo a small piece of plane orthogonal to n (Mo). Take some neighborhood of Mo in this plane 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 the spherical image area and the area of the original when the original tends to Mo. If we denote the areas of the original and of the spherical image by AS and Eo respectively 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) be
a position vector of that surface, such that x., x,., n form the positively oriented triple. Denote by W du dv the area element of the surface. Then K
_
(n,,, n,., n)
W
_
I
(0X1 axe
W (n
I,
n, n) au av
+ (nYj, nt,, n) + (n.,,, nY,, n)
Note that n can be represented as
a=
[y-U, xV}
W
FIGURE 6
(7X2 0X3
au av aX3 ax1
ax1 ax2
av au ax2 aX3 aV
au
aX3 aXI
au av - av au
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
23
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 Cartesian
coordinates, i.e. in a coordinate change the components of P behave as vector components. 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 but is not in general of unit length. Suppose that such a vector is A = {A;} = An, where A = (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
(2)
(A2 + A2 + A3)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
4x2
1x,x3 4,x,
4x2x3
4'x3X3
4'X3
4'x2
4x3
(4' + tz2 + 423
0 )2
(3)
Turn now to the general case, when the field n may be non-holonomic. We shall show 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. Let the parameters (a, ,B) in the surface induce the normal vector field v. We can map the surface F into the unit sphere S 2 by putting each point M E F into correspondence with the end-point of n(M) in S2. Denote this mapping by ,0. The area element da of unit sphere can be represented as do = (na, nf, n) da d,Q.
(4)
THE GEOMETRY OF VECTOR FIELDS
24
Let xi = xi(a, 0) be the components of the surface F position vector with respect to Cartesian coordinates. Then we have
(°x
ax2 aX3
aX2 ax;
°) as ap
as aa
ax349TI
ax3OXI
°) as as - as as
+
axI 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 that neighborhood 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 such that for any unit vector v the transversal Jacobian of the mapping t' of the plane orthogonal to v into the unit sphere S2 is equal to (P, v). We call P the vector of curvature of the field n. The total curvature K can be introduced in another way. Consider the Rodrigues equation 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 Abiill = 0.
det
II is the Jacobian of the mapping of a three-dimensional domain in E3 Since det into the unit two-dimensional sphere S2, det IIExfll = 0. Hence A = 0 is a root. The
other two roots satisfy the characteristic equation A2 +
52x2 + 6xi) +
2
I
i.!- I
0. 'tY'
S/Y,
I
We call the roots of the above equation the principal curvatures of the second kind. In
spite of the fact that the roots can be complex when the equation do = -A dx is insolvable, the half-sum of roots is called, as usual, the mean curvature H and the product of roots is called the total curvature K: 1
1
H=-2divn=2(AI+A2),
I 2i,/=1 1x Yi l
K = AIA2 = -
3
Six,
tifY,
I.
(8)
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
25
Indeed, since the expression for A1A2 is invariant with respect to the choice of Cartesian coordinates, we shall prove (8) with respect to the special system of coordinates. Directing the axis x3 along n(M), we get Slxl
SIx2
b2Y1
t2Y2 .
AIAz = t
(9)
On the other hand, we have (1). Taking into account that P
- {I
GV3 C2Y2
2x1
I
'
I
Cl 'r3
6x,
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 that K = AIA2 for any vector field n. Comparing (6), (7) of Section 1.2 with (8) and (9), we see that the mean curvatures of the first and second kinds coincide, while the total curvatures of the first and second 5)2
kinds are different in the non-holonomic case: K - KI = 4 In the theory of surfaces it is well known that H2 > K. The generalization of this inequality to the case of an arbitrary vector field is the following: (10)
We may state (10) in the special system of coordinates constructed above with respect to which {I = C2 = fax, = 0. In this case we have (a, curl a) = 2Y1 - t, x2,
H = - 2(Slxl + 6X02 K
H2 + (a, curl a)2
I Ixl
6X2
e2Y1
tbq
- 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 curvature 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 characteristic 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 orthogonal to n. The following theorem holds. Theorem If at each point A 1 = A2 # 0 then the field n is holonomic and the surfaces orthogonal to a are spheres.
26
THE GEOMETRY OF VECTOR FIELDS
Denote by A the common value of A . It is real. Let -r, ,v be the fields orthogonal to n. 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 due to 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 the surface which is orthogonal to n is totally umbilic, it is a sphere by the well-known theorem from the theory of surfaces. As an exercise, find the total curvature of the second kind for the special field n whose streamlines are the family of small circles in concentric spheres. Let 0 be a
common center of the spheres. Suppose that the centers of the circles - the streamlines in the fixed sphere - are on the same diameter I. Also, we suppose that the circles are in the parallel planes orthogonal to I. At two opposite points of the sphere 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, when the diameter is in the x, y-plane. Denote by a the angle which / makes with the axis Ox. 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 intersections with each other. Moreover, since the spheres are concentric, the circles of different spheres have no intersections. The tangent vector field a of the circles is regular at all points excluding the points of some curve which lies in the x, y-plane and is symmetric with respect to O. Let us find the components of n = {Q as functions of the
coordinates. Let the diameter I coincide with the axis Ox. Then at a point with coordinates x, y, z we have
6 = 0,
6=-
z z2 -+y2'
6=
y
z2 + y2 .
Let 1 make an angle a with Ox. If we rotate the coordinate axes by angle a in such
a way that the axis Ox coincides with I then we obtain the previous situation.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
27
FIGURE 7
The new coordinates u, v, w are related to the old coordinates by the following formulas:
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 which corresponds to the diameter I. They are of the form stated above:
'lI=0,
172=-
w W.2+1,2
V
173=
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,
THE GEOMETRY OF VECTOR FIELDS
28
where a = a(x2 + y2 + z2) is a given function, A = z- +(- sin ax + cos ay)2. To find 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),
z sin aax
z sin aa).
sina + Z cos aa- cos a + z sin aa_
-ua:
- sin a - ua.r cos a - uay -z cos a z sin a
z sin a
-z cos a v
0
v
We multiply the elements of the first row by x/z, the second one by y/z and then add to the third one. We multiply the elements of the fourth row by 1 /z and add to the third one. As a result we obtain a row of zeros. Therefore, K = 0. To find the mean curvature of the field we use the formulas A
-2H = div n = div , =
div A A
- (A, grad 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 calculate the non-holonomicity of the vector field just constructed. We have A2 (n, curl a) = (A, curl A) = -2z(x2 + y2 + z2)a'.
0 and z 96 0 then n is non-holonomic. Thus, for the vector field under 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 the domain of definition of the field n then a unique line n = const, which is tangent to P at each of its points, passes through that point.
Therefore, if a'
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 the following 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 line n = const is tangent to P. From the formula K = (n, P) it follows a simple way to construct the field n with zero total curvature: K = 0. Indeed, that condition means that n and P are mutually orthogonal. Suppose that P 54 0. Since n is constant along the streamline of P, these
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
29
streamlines are planar if K = 0. Therefore, to construct the required field it is sufficient to define an arbitrary congruence of planar curves and set n = const on each curve of the family. To do this, take a surface F2 of the position vector r(u, v). Then at each point x E F2 take a unit vector n(u, v) and a vector a(x) in a plane which is orthogonal to n(x). We realize that selection with the help of a single function cp(u, v). We emit a curve in the direction of a(x) which lies in the plane orthogonal to n(x). Those curves are defined by curvature k(u, v,s) as a function of the arc length
parameter. So, the arbitrariness in the definition of n is in six functions of two arguments and one function of three arguments. In the case of P = 0, the image of any surface generated on the unit sphere by the vector field n is a line. Conversely, the inverse image of any point of that line is a surface.
To construct the field n satisfying P = 0 define a regular family of surfaces in E3
and then define some constant value of n on each of the surfaces which varies continuously 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 normal curvature of the field in the direction dx is zero. Since the normal curvature of the
field is equal to the quotient of -(d n, d x) and d x2, the asymptotic direction is orthogonal to d n. Therefore, d x is proportional to [n, dn]. Thus, the asymptotic direction can be found from the following vector equation: In, dnj = p d x,
where µ is some numeric coefficient. Find the condition when the asymptotic direction exists. To do this we use a special system of coordinates such that I = 6 = 0, 6 = I at some fixed point Pa . Then at Po the system on the asymptotic direction 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
30
THE GEOMETRY OF VECTOR FIELDS
Note that -4K, = (n, curl n)2 - 4K. Therefore, the asymptotic direction exists if and only if the total curvature of the first kind satisfies K, < 0. If K, < 0 then there are two asymptotic directions. If K, = 0 then either only one direction is asymptotic or any direction orthogonal to n is asymptotic. The line which is tangent to the asymptotic direction at each of its points is called the asymptotic line. Let us clarify the geometrical meaning of p. Let s be the arc length parameter of an asymptotic line. Then along the asymptotic line we have dp d r d2r Z ds + pds2
_ rn' d2n
Multiplying by n we have (`tTyr , n) = 0, provided that p 96 0. Therefore, the vector n is a binormal of the asymptotic line or differs from it by sign. Denote a torsion of the
asymptotic line by c and the principal normal and binormal vectors by v and Q respectively. If n =13 then d n/ds = -isv. We have
dr pds=
In
dnl dr ,dsj =-'[n,vj=-'c2.
Therefore, the value of p differs from is only in its sign. The same result occurs when n = -,Q. Thus, the torsion of the asymptotic line can be expressed in terms of the field basic invariants as
- ((ne curl a) ± is=
(n, curl n)2 - 4KK 2
Denote the torsion of the first and the second asymptotic lines by is,, i = 1, 2, respectively. The following generalization of the Beltrami-Enneper theorem holds: Theorem The sum of the torsions of the asymptotic lines is equal to the field nonholonomicity with a negative sign and the product of the torsions of the asymptotic lines is 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 the other 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 helical streamlines. Consider the regular family of helices on the cylinders with a common axis. Suppose that the helices on the same cylinder have the same pitch of screw. We suppose, also, that the pitch of a screw depends on the cylinder radius. Consider the equation of some of these helices:
x=pros , y = p sin gyp,
z=bcp,
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
where p =
31
x2 + y2, b = b(p2) is some function of p2. The helix tangent vector is
-y
fx'O A
x
=
Y:
b
Z:p
Also, consider the unit vector field -y
x
b
n 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
8 2H=-divn= ax
Y
p2 + b2
- aYa
x
_ =0.
p2 + b2
We find the total curvature of the second kind by (2) of Section 1.5: 0 1
K
1
p2 + b2 b'2x
-y
-1 0
0 -y 0 x
b'2y
0
x
b
-b(2p2b' - h)
b 0
(p2 + b2)2
-
It is easy to see that (A, curl A) = 2(b'p2 - b). Therefore, the field non-holonomicity value is 2(b p- - b)
(n,curln) =
(p2 + b--)2
From this we see that n is holonomic if b = cp2, where c is a constant. Write the equation 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: 1=
-p sin cp
p2 + b2 '
p cos w p2 + b2 '
3=
b 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), p2 dip = (p2 + b-)µ rL-.
(1)
32
THE GEOMETRY OF VECTOR FIELDS
For the subradical expression in the formula for evaluating 'a at the beginning of this section we have (n, curl n)2 - 4K =
2 (p'- + b222)
Therefore, the helical field has two, maybe coinciding, asymptotic directions. The coefficient p in the equation of asymptotic lines takes one of two values:
-b
pIp2+b2'
2b'p'- - b
p2= p`+b2
If p = p, = -b/(p'- + b2) then from the first and second equations of (1) it follows that db = 0, while from the third we see that p2 dcp = -b dz. Therefore, in this case the asymptotic line is on the cylinder p = const and represents a helix with torsion of the sign opposite to the sign of torsion of the field streamline. The family of these helices together with the original family of helices form a regular orthogonal net on each cylinder. Consider the asymptotic lines corresponding to p = 142. The equations of 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 this case asymptotic lines are the rays emitted parallel to the x, y-plane from the points of the common axis of the cylinders. This solution is evident from a descriptive viewpoint. In this case we ought to regard the torsion h2 as a value which characterizes the 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 orthogonality condition of the required asymptotic direction and given field n. Therefore, in the case 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 the integral of total curvature K in a domain G in terms of some integral along the boundary 8G or in terms of an integral over some domain on the unit sphere S2. Let x be a position vector of M E G. Suppose that (a, f3) are local curvilinear coordinates
in 8G with the usual orientation, i) is a mapping of 8G into the unit sphere S2 by means of the vector field n , do = (n,,, n,,, n) da df3 is a signed area element of unit
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
33
sphere. Then supposing no singular points of the vector field n in G. the following formula holds:
JKdV= J(xn)do, G
(1)
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
(act
OX 18X2
OQ as
OQ
0X2 0x3\
0X2 0X3
+(n,,,n 3,n)(act as
aQ as
aX30XI
OX30XI
as
aQ
aa) dadQ= (x,n)(P,v)dS,
where v is the normal of 8G. By the Gauss-Ostrogradski formula we have
f
(x, n) (P, v) dS=
aG
=
JG
f(x,n)divPdV+ f (P, grad (x, a)) dV
div ((x, n)P) dV =
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. Since n is of unit length, (nx,, nx3, qr,) = 0. Consider the coefficient of xl in the integrand of the 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
bx,
6x2
b3.x3
2 _ (n.rl,nx:,nx,) =0. C3
IY,
6x'2
1;1x3
0
In an analogous way we see that the coefficients of x2 and x3 are both zero. Thus, the second integral turns into zero and, as a consequence, (1) is proved. The calculations above shows that K, as well as the H, is a divergence of some vector. Namely, K = div(x,n)P.
(2)
THE GEOMETRY OF VECTOR FIELDS
34
1.8 The Second Divergent Representation of Total Curvature of the Second Kind
Consider (8) of Section 1.5 in detail: 3
K=2E ,.j-I
=
a
Y,I
t
C
rz - e2SI.c:) + i",a
2(Sl
f
where the dots mean the terms having an analogous structure with respect to replacement of index I by 2 or 3. Consider the expression under derivation -1, It has the form j16 (C2,, + 6c:) - S2CIc;
Add and substract
Skx7}
Then the previous expression obtains the form {l;I div n - k, },
2 16 (S Ix, + 6- v , + S3x%) - S1SI.c, - 2SI.c. "
2
where k, is the first component of the field n streamline curvature vector. Transform in 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)
G
JG
Consider some particular cases. (1) Suppose that the boundary of G is formed by connected surfaces F, and F2, where F2 is inside Fl. Suppose that n coincides with the normal vector field on F; (see Fig. 8). In this case (k, v) = 0 at the points of surfaces. Since F; are orthogonal to n, the curvatures of F, coincide with the curvatures of the field n at the points of F;. From (2) it follows that for any vector field n inside G with F, and F2 as the
boundary the volume integral of field curvature K in G can he expressed as a difference of integrated mean curvatures of boundary surfaces:
JKdV=
-
jHdS+ / HdS. F,
fil
Note that the right-hand side is independent of a.
(3)
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
35
FIGURE 8
(2) Suppose that at each point of a surface F the field n is tangent to F. Then the streamline of n starting at F belongs to F. In this case we say that F is the invariant submanifold of the field n. The value of (k, v) is the normal curvature of the streamline in F, where v is directed outside F. If n is regular at each point of F then F is necessarily homeomorphic to the torus because only the torus admits the regular unit tangent vector field (see Fig. 9). Formula (2) for this case has the following form:
2 J KdV = -
r k dS,
(4)
F
G
where G is the interior of F, k is the streamline normal curvature in F of the field n.
Apply formula (4) to the field defined inside the tubular surface. Denote it by F as
above. Let r be a closed curve of length 1, k and ti be its curvature and torsion respectively. The tubular surface F meets any plane normal to F by the circle of constant 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
THE GEOMETRY OF VECTOR FIELDS
36
We suppose that u is the arc length parameter of r. Set
a = cos
sin v6,
b = - sin vi;2 + cos ve;3.
Then
(I -kRcosv)+RKb,
r,,
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 terms which 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 streamline of n, the differentials du and dv in the expression for I and II must correspond to the directions 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
r
J F
2i
21T
!
1
rJR2(`et
jdS=JJ Rkcosvdudv-
F
+'1)2dudv.
(5)
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
J KdV = G
2x I
ff R2(
0 0
21 , )2 dude.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
37
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 following statement. Theorem If the tubular surface F with the axial line of length I is an invariant surface of the vector field a then the integral of curvature K of the field n over a domain bounded by F is non-negative and does not exceed irl.
0< fKdV
The lowest value of the integral occurs when n is such that the streamlines in F satisfy the equation
r4 +n=0 (in the particular case, when the torus F axial curve is planar, these curves are v = const). The largest value corresponds to the field n whose streamlines are the circles in
planar sections of F perpendicular to the axial curve (see Fig. 10). Note that the upper boundary of the integral does not depend on the tube radius. 1.9 The Interrelation of Two Divergent Representations of the Total Curvatures of the Second Kind
In previous section we stated two divergent representations for K. K = div (x, n)P,
2K = -div (2Hn + k).
THE GEOMETRY OF VECTOR FIELDS
38
These relations imply formulas (1) of Section 1.7 and (2) of Section 1.8. Comparing them we conclude that if the interior of a domain G contains no singular points of the field n then
J (x,n)do=-I (Hn+2,v)dS.
(1)
F
R If')
We are going to show that (1) is also valid if the singular points are strictly interior to G. Define on F the following external forms:
0 = -{2H(n, v) + (k, v)} dS,
a = 2(x, n) do,
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 F and 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 can be 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 (3)
= -dy + (n, [n, x,.] - [n,., XU I) dudv.
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 systems in E3 or F. Introduce the coordinates in E3 and the coordinates u, v in F in such a way that at Mo the tangent plane to F coincides with the xl,x2-coordinate plane and the basis el = x,,, e2 = xr, e3 = v. Then at Mo n,. = nx2.
nu = fix,,
Moreover, at Mo we ([nw,xr] have ff
(n,
t
- [nr,x,]) = blx,e3 -
C
ii
`` S'_r213 -
`
(n,x,,,xr)divn-(k,x.,x,)=CIx,b +J2xb -S3x,SI Comparing (4) and (5), we conclude (n, In., xr] - [nr, xu]) = (n, x,,, xr) div n - (k, x,,, xr).
(4)
(5)
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
39
From (3) it follows that
a=-dy+f3. So, (2) is proved. Since F is a closed surface,
Jd-7=0. Hence, (2) implies (1), while the singular points of the field n possibly exist strictly inside of the domain bounded by F. Observe that if n is the field of normals of F the equation (1) implies the Minkowski theorem on the integrated support function of a closed surface and on its integrated mean 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 Gaussian curvature over some domain G in a surface F and the integrated geodesic curvature of the boundary of G is well known. It is the so-called Gauss-Bonnet formula:
fKdS=f-!-ds+2r.
X
G
(1)
P9
Here 8G is a smooth closed curve on a surface F. It is also supposed that the boundary normal v, with respect to which one evaluates l 1p., is directed outward from 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 generalize these formulas to vector fields. Suppose that a vector field n is defined, maybe with singularities, in a domain Q of
three-dimensional Euclidean space. Let F be a closed surface in Q not passing through the singular points. Theorem For any closed surface F C Q not passing through the singular points of the vector field n J(nK + 2Hk + Vkn, v) dS = 4irO, F
(3)
THE GEOMETRY OF VECTOR FIELDS
40
where B is an integer called the degree of mapping of F into S2 generated by the vector field n.
Here Vin means the derivative of n along the curvature vector k of the n field streamline. 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 integral of the Gaussian curvature of F. To prove (3) we consider the mapping of F into the
unit sphere S2 generated by the vector field n. In Section 1.5 we denoted that mapping by t/' and found that the area elements of image and original are related to each 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 S2 an integer number times. We recall the notion from topology of the mapping degree
of a closed surface F onto the unit sphere. Denote it by 0. The mapping degree is connected with 1() image area by the following formula: do = 42rO.
J u(F) Hence
J(Pv)dS=47r9.
(5)
F
Consider the vector P in detail. Choose the coordinate axes in such a way that the x3-axis at M E F is directed along n(M). Then the mean and the total curvatures at that point obtain the following form:
H= _
1/ [
C2-,,!),
+K- \SI Y. S2Y, -SI C_SZYI )'
The vector P = {P;} which was introduced in Section 1.5 has the following components at M: P=
JI
`Ix:
S2r;
CIx3 l
S2.r,
I
tI.C
SI CI
S2.r,
e2 r.
I
I ti, S2r
1 C2 I
( 6) O
S2r:
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 subtract Ir,Slx3 to P1, to P2. We obtain `` tt PI = SI.xiblxa +SIS2rt + 2H£I
P2 =
2Ht;2ri
The vector with components {Slx,. S2r 0} is the vector of curvature of the n field streamline (see Section 1.3). Denote it[ by k[. Now consider the following vector: I = {SI.r1 SIx) + Slx2 S2r11 S2r. S I.rt
0} -
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
41
FIGURE I I
If a and b are vector fields of components aj and b1 respectively with respect to
Cartesian coordinates then V1b means, as usual, the vector of components 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
(V1b); _ )Y,
P=nK+2Hk+Vkn. Substituting the latter expression into (5) we obtain (3). If there are no singular points of n inside F then 0 = 0. Indeed, contract the surface F with a continuous deformation to a sufficiently small neighborhood of some intrinsic point x0 where a differs slightly from n(xo). The image of F after deformation will be located in some small neighborhood in S2. Therefore, the degree of t/i is zero. This fact also follows from the equation div P = 0.
Suppose that there is a finite number of singular points F1,. .. , Pt inside Q at which the field o loses regularity. Then we can define the index I of each singular point 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 the degree of mapping of SE onto S2 by means of the field a. Suppose that Fcontains the field o singular points PI,..., Pt. Then Ftogether with SE, bounds a domain QE. Since this domain does not contain any singular point,
J(Pv)dS+> J(PzI)dS=0, F
where the normal v on S£, is directed inside SE,. Therefore, each integral over SE, is
equal to the degree of mapping SE, onto S2 with the negative sign. Hence, if the surface F encloses the singular points of the vector field n then
r
r
/ (P, v) dS = > !i. F
,_1
THE GEOMETRY OF VECTOR FIELDS
42
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 this surface in the unit sphere is some domain g with boundary y, where -y is the image of 1' under the mapping 0 (see Fig. 12). The domain g we shall consider as a manyvalued 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 its spherical image with a + sign . If (P, v) < 0 in some domain G; then we take its spherical image area with a - sign. We call the algebraic sum of areas of all g; the spherical 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 sphere S22, 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 parameter of -y.
The geodesic curvature of -y can be expressed in terms of the vector field n if we recall 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 the integral along -y of some value which depends on n. Applying (4) of Section 1.10, we obtain
J(PP)dS= F
n2 nu) ds + 21rm.
(2)
r
We may consider this relation as the analogue of formula (1) of Section 1.10. Now consider the cases when the integrand in (2) can be expressed in terms of the usual geometric values.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
43
Remember that by definition from Section 1.4 the geodesic curvature of 1' with respect to the vector field a is the value . = -(a, x5, x.,), where x = x(s) is a position vector 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 to
r at the points of t then
J(P,v)dS=
J Pg
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 segment
0 < t < T such that C(t) and 11(t) are three-dimensional unit vectors mutually orthogonal for each t. Let p be the angle between & and ri. Then T
T
& ,Crr) C2 `
di
Jdt.
J(C,1,,rh)dt+0
0
0
r)]. Let us take C, n, v as a basis in E3. For each t the following holds:
Set v =
f, = A(cos ni + sin pv),
-= A, (cos
d + sin pv) + A(- sin M + cos pv) dt + A(cos
sin pv,).
We have
A2 { d`f + cost
ftr) + sin 2 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
THE GEOMETRY OF VECTOR FIELDS
44
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=l r r r
S
J-ds
+27rn, P.9
r where n is an integer, l /pr is the geodesic curvature of r with respect to n. So, the integral on the right in (2) is expressed in terms of the integral of the geodesic curvature 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 a surface with F as a boundary then
J(P7L')dS=-JKds+2irm. F r
(4)
Consider (2). In our case n = x,., where x, is tangent to P. So we have (a, ns, n") ds = oY
J(x,,xx) ds = JK ds, x2.
r
which settles (4).
Finally, for the third case we suppose that r is a closed curve with n being constant along it. If P 0 at some point from the domain of definition of n then one and only one
curve tangential to P with the required property passes through this point. The curves 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 a boundary 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 F based 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 points of E3 and there is a limit of n(M) at infinity. In this case we can put the continuous mapping 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 tangent space E3 where the field n is defined. We denote the stereographic projection of S3 onto E3 from the south pole by a. This projection is formed with the rays from the south pole. We put the point P of the ray intersection with S3 into correspondence
with the point Q of intersection of the same ray with P. The south pole of S3
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
45
FIGURE 13
corresponds to a point at infinity in E3. We can put each Q E E3 into correspondence with a point A in the unit sphere S2 C E3 of the center at some point O. To do this, we put n(Q) at 0 and denote the end-point by A. Denote the latter mapping of E3 onto S2 by ip. By hypothesis, the vector field n is defined at all points of E3 and also at infinity. The composite mapping W = via defines the continuous mapping of S3 onto S2 (see Fig. 13). So, each continuous vector field n, defined in E3 including infinity, defines a continuous mapping of S3 onto S2. The inverse statement is also true: each continuous mapping of S3 onto S2 we can put into correspondence with a unit vector field in E3 having a limit value at infinity. To do this, if A E S2 corresponds 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 be homotopic if there is a continuous family of vector fields n(x, t), 0 < t < I such that n(x,O) = n, (x), n(x, 1) = n2 (x). The vector fields n, (x) and n2(x) are homotopic if and 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 of homotopic 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 correspondence to integers. For any integer y the corresponding homotopy class is a class
of mappings W = p4: S2 S3 where p: S3 S' -+ S' is a mapping of degree y.
S2 is the Hopf mapping and &:
The number y is called the Hopf invariant of cp: S3--+S2 . For any point P E S3 a linear 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). From topology (see [51], p. 206) we know that for any point A E S2 there is however close to W a smooth mapping 0 which is homotopic to W and normal over A. Further on we suppose that W is normal over A. It can be proved that if cp is normal over A then
the inverse image r4 = W_ I (A) is a set of smooth curves in S3. To find the Hopf invariant 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 positive
orientation of S3 and such that V,,e, = 0 while V,,e2, V,e3 defines a positive orientation in S2; then e, will define a positive orientation of I'4. Let F be a smooth
46
THE GEOMETRY OF VECTOR FIELDS
surface in S3 based on rA such that its orientation is coordinated with the orientation 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 f w. F
pF
Since V maps the boundary of F, i.e. the I'A, into a single point, the image of Funder the mapping V is a two-dimensional cycle on S2. The degree of mapping W restricted to 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 F is a pre-image of some point A E S2, i.e. can be constituted from some number of closed curves with an orientation described as above. Consider now in three-dimensional Euclidean space E3, with a point at infinity included, the unit vector field n generated by the Hopf mapping. The Hopf mapping is constructed as follows. Represent the three-dimensional sphere S3 as a unit sphere
in 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-dimensional sphere S2 can be represented as a complex projective straight line, i.e. as a set of pairs (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 equivalence above by The Hopf mapping p: S3 -+ S2 is defined by the following formula: 1
p(zI, z2) = [zI, z2],
where (zi, z2) E S2. In this mapping the inverse image of each point of S2 is a great circle SI in S3. The three-dimensional sphere S3 is decomposed in a family of great
circles; moreover, as the quotient space of that decomposition we have the twodimensional sphere S2. Now construct the vector field in E' which is generated by the Hopf mapping. Since we can put any point of S3 into correspondence with a point in E3, excluding a point at infinity, and we can put a point of S2 into correspondence with a vector of constant length, the mapping p: S3 -+ S2 generates some vector field in E3. At infinity we are also able to obtain a definite vector. Proceed to the vector
field construction. Let S3 be a sphere of unit radius in El defined by the equation +.v42 = 1, where xf are Cartesian coordinates in E4. Let E3 be the tangent space 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 into correspondence with a point (x1, x2, x3, x4). The coordinates of the corresponding point in E3 are as follows:
x, +
2x,
yr
x4+l,
i=1,2,3.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
47
Set zI = xl + ix2i z2 = X3 + ix4. Suppose that z2 96 0. Next we put the point in S2 into 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 some x3+ix4 plane E`2 tangent to S2. We put the point u + iv in E2 into correspondence with the vector t having components
_
_
2u
l+u2+v2'
2v
t3
l+u2+v2'
£2
u2+v2-1 l+u2+v2
If we putt into the origin then the t end-point defines the point in S2. Represent the components of { as a functions of y,. At first, represent {; in terms of xi. We have
u+iv= XIX3 + x2x4 + i(x2x3 - x1x4) x3+x4
From this it follows that u2 + v2 =. Therefore,
_
2u SI
Taking into account that
]+u2+V2
2(xIx3 + x2x4)
XI+x2+xj+JC4.
° 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?. i=1
Set 3
a= 3
=1
4Ex
Yi2
-
j=1
(1 + x4)2
4(1 - x4) (1 + x4)
From this we find the expressions of x4 and x4 + 1 in terms of a: X4
4+a'
8 x4+1 =4+a
Then we have
x, = Yi(X4 2+ 1) _
4
ia
+
i = 1, 2, 3.
THE GEOMETRY OF VECTOR FIELDS
48
Substituting the expressions above into the ones for ti, we obtain 8(4YIY3 + y2(4 - a)) ,
(4 + a)2
C2 =
8(4y2y3 - YI(4 - a)) (4 + a)2
16(y; +.y22 - y3) - (4 - a)2 1;3 =
(4 + a)2
Consider the behavior of this vector field. If y3 + y2 +.y23 0o then CI -+ 0, C2 - 0, - 1. Thus, when the point tends to infinity, the vector field C tends to a definite C3 limit. Along the y3-axis the vector field C is constant, namely (0, 0, -1). It is easy to see that along the circle y, +y22 = const the component C3 stays constant. From the expressions 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 know
the field behavior in the y, , y2-plane then we are able to reconstruct the field behavior in space. Consider the field behavior along the y,-axis. If y2 = Y3 = 0 then
_
_
_
z
16y2
Y), C3 -
2 =
Ci = 0,
(4f
+
(4
I)
+ a)2Yz)
Thus, along the y,-axis the vectors of the field f lie in the plane which is orthogonal to 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, in varying y, from 0 to +oo the C turns around in an angle 2ir. At the points P, and P2
in a half-axis [0, oo) with y; = 4(3 - 2f) and y, = 4(3 + 2f) respectively we have b = 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 at the origin which are the vector field i; streamlines pass through these points. However, 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 is equal to 4a (see Fig. 14). Consider the streamlines of that vector field. They satisfy the following system of differential equations: dyi
-
i = 1, 2, 3.
Suppose that a point moves along the t streamline in the t direction. From the fast two equations we find
- a-4 _L1 -
d arctan ds
(a + 4)2'
dQ + y2) ds
_ 64(+ y2)y3 (4 + a)2
From this it follows that if a point on the streamline lies inside the ball y; + y2 + y3 < 4 then its projection into the yIy2-plane moves clockwise; if a point
lies outside that ball then its projection moves in the opposite direction. When y3
oo, the streamlines turn around the y3-axis infinitely many times. When y3 > 0
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
49
FIGURE 14
the distance between the projection and the yy3-axis increases, while for y3 < 0 that
distance decreases. Consider the behavior of y3 along the streamline. We have dy3/ds = 0 if y i = -(4 + p) + 4. The latter equation represents the convex curve with the end-points in the p-axis in a plane of parameters y;, p. This curve corresponds to some closed curve -y in the y3, yI-plane which generates a torus T in rotating 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, rises at 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 call the geodesic torsion of the field in that direction. Let dr/ds be the tangent vector of
THE GEOMETRY OF VECTOR FIELDS
50
a geodesic line. This principal normal of this curve coincides with the field a. Denote the 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 get the expression for the geodesic torsion:
(dr,a,dn) (1)
ds2
Thus, the geodesic torsion with respect to the principal direction of the second kind is zero. Let i, and n2 be the extremal values of (1) in rotating dr/ds in a plane which is orthogonal to n. Then, as Rogers stated, rcl + rC2 = (n, curl a). Prove this. Choose the basis in such a way that I = 2 = 0, 1. The expression (1) obtains the following form K_
II d,
dd2
d3
, drl
dx2
0
0
0
= dsl
ds do
ds2
11
drl1'
+(ds) ds ds
dc2 dxl
dx2
To find the extremal values of re, form the symmetric matrix and the characteristic equation in the following manner: t
S2Yi -
=0. A
In expanded form we have
`
2
/CC
C
S2x, - SIY,} -
(S2Y_
tt
4S,_q}_
i
+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 only those which are orthogonal to a. Let -r1, i = 1, 2, 3 be the extremal values of (1). We
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
51
call them the principal geodesic torsions. Introduce the total, mixed and mean geodesic
torsion 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 to the special choice of basis:
Al
42 43
dxI
dx2
0
0
d{I dx2 - dd2 dxI
dx3 ds2
- S2.YI
11
dxI
s)
dX2 dXI
2
[
dX3 dxI
Js ds - S2x; rdx212
dxI dx2
dx3 dx2
ds d
+
ds dS
ds ds
We can find the extremal values from the following characteristic equation: -6_ "'
fz.,
.1
2
[
-
fz.3 2
fl, 2
fly, 2
2
=0.
-A
In expanded form we have -t2rz)2+tIxl+t22.,
4
ll
J
-fall 2
fl .1
+
2
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]j 4
The value T represents some field invariant. With respect to the chosen coordinates we have T
l 41
2
z
1 -S2X,6xj + (62X, - fIx1 )SIx,C2.YI +SI,, 6x11.
(2)
THE GEOMETRY OF VECTOR FIELDS
52
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 set of 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) of Section 1.8, which gives the expression for the integrated field curvature K over the volume, has been stated on the supposition that there are no singular points inside
the surface F. Let v be the normal of F5 directed inward to T. We say that the singularity is the curvature source of power Q if there is the following limit:
Q=1im 1 / {(n,v)diva-(k,v)}dS. e-o 2.11
(1)
F.
If there is the curvature source in G with F as the boundary then we shall consider the integral of K over G as an improper integral, i.e. set
JKdV= lim JKdV. :-o G
6A T,
It is possible to rewrite formula (2) in Section 1.8 with regard for singular points inside G as
JKdV = G
f {(n, v)H + (k. v)} dS + Q. 9G
Thus, if the vector field n is fixed on F then the higher the field source power the more the integrated curvature K is enclosed inside F. There may be an arbitrarily powerful source enclosed in an arbitrarily small volume. For instance, take some closed curve -y of length / in a domain G. Define the
vector field u in a neighborhood of -y as a tangent vector field for the family of concentrated circles in the planes normal to -y with their centers in -y. As we have stated in Section 1.8, the integral on the right of (2) is equal to -7r/, i.e. the curve is the 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 e however small the modulus of the area of the image of the sphere F. of radius a centered at 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
53
Put the origin into Mo. Since I(x,n)I < E over OF, then
1 (x,n)dQ <EC. V,(F,)
I
0, we obtain Q = 0. From this it follows, for instance, Passing to the limit when a 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 is directed 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 of sufficiently small radius R with y as an axial curve. In this case (n, v) = -1. Let us find the curvature source power of such a singularity. It is equal to the integrated mean curvature of F with respect to o directed outward. Thus, we need to find the mean curvature of the tubular surface with respect to an extrinsic normal. We know from differential geometry that
H-
EN - 2MF+ GL 2(EG - F2)
Here E, F, G are the coefficients of the first fundamental form and L, M, N are the coefficients of the second fundamental form of the tubular surface. Use the formulas from 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 2kR cos v - I 2R(1 - kR cos v)
H So, the curvature source power is
r
2,r I
2k Rcosv-1 EG-F2dudv Q=JHdS= JrrJ 2R(1 - kR cos v) F
00
2,fI
=2
J00J
(2kR cos v - 1) dudv = -Trl,
where I is the length of the axial curvature y. If n is directed inside the tube then Q gets the opposite sign. Consider the following problem: in which case the vector field singularity is topologically removable? Let M be a connected set of the singular points of the field
THE GEOMETRY OF VECTOR FIELDS
54
having the --neighborhood where the vector field n is defined. Set M, to be the e-neighborhood. We say that a singularity is topologically removable if for any sufficiently 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 of the 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 S of the set M is sufficiently small, namely S71< 4n. Then the singularity is topologically removable.
Observe that in the case of a holonomic vector field it is sufficient to require the boundness of IK 1, IHI and Ikl. To prove the theorem we need to estimate the length of vector P = nK + 2Hk + Okn. Set I = Vka. Consider the scalar product of I and P in a special choice of coordinate system: SI.I7
(1,
Sir,
S2Y2
+SI' SLI VITST, + ttxI
(SL.Y)SIYj + I Y11I Ya)
C22,:
VI Yz - S2YI)
(6-11, - t2x?)l;I YtS2 Yi I.YIS2.Y. - I Y_S2.Yi/
(S x,
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.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
55
Apply formula (5) of Section 1.10 to the closed surface LIME. The area of this surface
differs from S arbitrarily small. Since S <
1-3 C."
e
4-F
J (P,v) dSl < M:
Therefore, the degree of mapping of 8ME onto the unit sphere S2 by means of the field n is equal to zero. As we know (see, for instance, [52], p. 125), in this case the vector field n can be prolongated without singular points inside 8ME. 1.14 The Mutual Restriction of the Fundamental Invariants of a Vector Field and the Size of Domain of Definition Let us ask the following question: for a given fixed domain of definition, is it possible to find a vector field of an arbitrarily large curvature? The following example gives the
answer. Let us consider the cube: 1 < x, < 2, i = 1, 2, 3. Define the vector field as icn = (A,, A2i A3), where AI = x1 cos Ax2, A2 = xI sin axe, A3 = 1, A > 0 is some constant. Then by (2) of Section 1.5 we have
K_Aix,A2x,--A2x,Aix, (1 +A2 +A2)2
- XI (1
A
+x;)2 25'
Thus, we are able to construct in the fixed domain the vector field of arbitrarily large
curvature. Observe that in this example the value of non-holonomicity increases together with K increasing. The following theorem holds. Theorem If a regular vector field n of curvature K> Ko > 0 is defined in a ball D of radius r and the value of the non-holonomicity satisfies I(n, curl n)I < po then 3
r< 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 inequality inside the ball D and applying the Gauss-Ostrogradski formula, we find 2
JKo - po 7rr3 <
r
JD
div n dV = / (v, n) dS < 7rr2, an
where v is the outward normal of the boundary LID of the ball D. Therefore r < (& - P02) -1/2.
i
THE GEOMETRY OF VECTOR FIELDS
56
FIGURE 15
Now consider the vector field of large negative total curvature K < 0. We shall obtain in this case the restriction on the size of domain of definition provided that the curvature of vector field streamline is bounded from above. The following theorem holds. Let a regular vector field n is defined in a cube of the edge length a in
Theorem
Euclidean space E3 such that the total field curvature satisfies the inequality K < -K0 < 0, where K0 = const, and the field streamline curvature is bounded from above Iki < µa. Then the cube size is limited, namely
a
F(4M1'+K.
(1)
To prove the theorem we shall state some integral formula which relates the total curvature and the streamline curvature of the field n. Let V(t) be a cube with the center at the origin and with the edge 2t {x, = ft} (see Fig. 15). Represent the total curvature 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 the cube V(t) to the integral over the surface of the cube. Consider separately the expression 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
X1 =1
8x2(f1(2)+8x3(U3)-2fI.ru-2fIx,b)dx2dX3.
(2)
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
57
We turn the integrals of the first two terms in this expression into integrals along the edges. To transform the third and the fourth terms we note that we can represent the third 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 this we substitute ,.r, and 1 r, into (2). Then (2) turns into the sum of integrals along the cube edges and faces:
f
1
2
U2 dX3 -
- fJ{2(.z;
Ji3dc2
f2dx3 + f fib b dX2 -
+ 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 add and subtract i k, to the expression integrated in the face x1 = t. Since k I a, the integrand in the face x, = t turns into k1.
2x
Take the derivative with respect to x, in the first term out of the integral and replace it with the derivative with respect to t. Since the face depends on t, we ought to add the integrals of (t;2 + t;3)/2 along the edges which bound the face x, = t. Presenting not 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 along the 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 the edge 1; and the center of the cube. Let v be a normal vector of the cube V(t), dS a cube surface area element. So, we have
f V(I)
3
KdV=-
dt
ff22
k-I tfit, t
dx;
dx1-if (n, curl n, v) dS + F f Ja,12 dx;. (4) 8V
,_I
I,
THE GEOMETRY OF VECTOR FIELDS
58
Integrating (4) in t from 0 to t = a/2 and taking into account the inequalities
J
(a, curl n, v) dS < po6(2t)3,
- JKdV >
K0(2r)`.
at,
k- -I
+
Jf
dr;dv, < 12t2,
2
- paa3.
3a2 > Ko
8
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 than
in. (Here, the vector A is not necessarily unit; if we have the algebraic family of surfaces then we get the algebraic family of planes.) It is evident that the algebraic character of a vector field and its order do not depend on the choice of Cartesian coordinates. With the help of formula (1) of Section 1.7 we can prove the following theorem. Theorem Ler us be given a regular algebraic vector field of order min a cube V with edge 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
a< cin
C=2 ;;V3.
,
Here the regularity of the vector field means that JAI
0. We represent the mapping
t+ of the boundary 0V onto a unit sphere by means of the vector field n as a composition of the mapping +p of OV into three-dimensional Euclidean space by means of a vector field A and the subsequent projection X onto unit sphere by means of 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) has coordinates 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.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
59
Then at the common point of D and the ray we have P = Q = 0 due to the equations above. Since the orders of polynomials P and Q do not exceed m, the number of
points in the face xI = const such that the Jacobian I () 0 and P = Q = 0 does not exceeds m2. Consider the points at which I ( ) = 0. At those points we have the following equation: Ar2
AZY, I + a21 A3x:
a201I IA3x2
Atx, AIx, + a2a31 Alx, A1x, I
2 A3x,
A3x,
A2x2 I A2x3
= 0.
(5)
Evidently, the normal to D is collinear to the vector
_
(
A3x2
A2r3A3x;
AIx2
Alx,
A2x,
A3x, ' A3x3
A113
I' Aix,
A2x,
()-
I Therefore, by virtue of (5) either (a1, a2, a3) is tangent to D or I = 0. For the first case the Jacobian of X is zero, for the second case the Jacobian of W is zero. By the Sard theorem the set of those points in the image, in our case in the unit sphere, is of zero 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
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) of negative curvature K < -1 onto the square in the xy-plane (see [62]). That theorem states that the edge length a of the square is bounded from above by the independent constant a < 14. That theorem founded Efimov's remarkable investigations of negatively curved surfaces which were awarded Lenin's prize. The central result of those investigations represents the following statement. Theorem For any C2 regular complete surface of negative curvature in Euclidean space E3 holds inf IK I = 0 (see [631-(65]).
Thus, the generalization of Hilbert's theorem from the surfaces of constant negative curvature to the surfaces of variable negative curvature was given. That problem 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 applications not only to the theory of surfaces but, as Efimov suggested, to the theory of vector fields.
Efimov's Lemma Let us be given a mapping of the whole xy-plane into a pqplane: p = p(x,y),
q = q(x,y),
60
THE GEOMETRY OF VECTOR FIELDS
FIGURE 16
where p, q are of class C'. Suppose that the Jacobian 0 satisfies 0 = and the rotation I = p, - qx satisfies the inequality
I pv
q`
p`.
q).
0
IAI>aIII+a'where a > 0 is some constant. Then the image of the whole xy-plane is either the whole pq-plane or a half-plane or a zone between parallel straight lines; the mapping of 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 (the rotation) is analogous to the non-holonomicity value of the vector field. If n is independent of x3 and {3 # 0 then setting p = q = 6/6 we obtain the following expression for the non-holonomicity value and the curvature K:
(n, curl n) = py - q'`
1+p2+q2
K
0 (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 of Efunov's lemma. Examples of mappings satisfying the hypothesis of the lemma and which map a plane 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 of the existence of such mapping onto a zone is still open.
1.15 The Behavior of Vector Field Streamlines in the Neighborhood of a Closed Streamline
Let us be given a unit vector field n defined in some domain of three-dimensional Euclidean space E3. Let L be a closed streamline of that vector field. Consider the following aspects of streamline behavior near L: (1) the rotation of bands formed by the field a streamlines with respect to the band of principal normals of L; (2) the stability of L as a closed trajectory of the solution, differential equation d x/dt = n(x)
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
61
where z E E3. In some cases we can gain information about streamline behavior near L by the geometrical invariants of the field only at L. In considering the problems which 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 field a streamlines through the points of a (see Fig. 17). The band intersects the normal plane at an arbitrary point P in L by some intercept a(P) which makes some angle V with the principal normal of L. We suppose that V is a continuous function of the arc length parameter of L. We also suppose that the field of principal normals of L is
continuously and uniquely prolongable. After the first bypass along L the band intersects F at an by infinitely small intercept a, which will be twisted in some angle Dip 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 is equal to
f{(ncurln)
- rc }ds(2) JJJ JJJ
where rc is the torsion of L.
If the integral (2) is a rational number p/q after being divided by 2ir then the band will take its starting position after q turns with respect to L and p turns with respect to the band of principal normals; if that number is irrational then the band will wind around L everywhere dense.
FIGURE 17
THE GEOMETRY OF VECTOR FIELDS
62
Consider the invariant A. In the case when n is holonomic, i.e. it is orthogonal to the family of surfaces with 1 /R; as the principal curvature, then
A=
1RI -
(3)
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 then adding, 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 fixed point M the x3-axis would be directed along n(M), the xI-axis along a and the x2axis along b. Suppose that 1;1, 2,1;3 are the components of the field a with respect to 0, i = 1, 2, 3. that system of coordinates. Note that at M we have
Then we can write (a, Vbn) + (b, V.n) = Ixs + (a, V,n) - (b, Vbn) = ylx, - Zx2.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
63
From this we find [(a, Vbo) + (b, V1n)]2+[(a, V1n) - (b, Vbn)]2 ``Y,
= (SI
+ c l,"),
+ (S Cl .cl
-
S2Y2)2
= (SIxi + 52x,)2 - 4(CIx,'2Y2 - Ix,e2Y1) + (Slx,
= (div n)'- - 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 n streamlines respectively. Then the torsion is of the field n streamline is K = 2 ((n, curl n) - (q, curl
(9)
(v, curl v)).
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 (13)
(q, curl 77) = -/L - (,7,
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
Then
2k = (n, curl n) - (a, curl a) - (b, curl b) - 2
d
where dip/ds is a derivative along the streamline of the field o.
,
(15)
THE GEOMETRY OF VECTOR FIELDS
64
By direct calculation we find (a, curl a) = cos 2 W(rl, curl rt) + sin 2 V(v, curl v)
+ cos P sin p{ (v, curl il) + (ii, curl v))
-di ds
We optain the corresponding expression for (b, curl b) from the previous expression replacing 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 those bands. By definition, the field b is holonomic, i.e. (b, curl b) = 0. Take a vector field a tangent to the band and orthogonal to a. From (15) we find that the turn of such a band with respect to the band of principal normals q is equal to
r
J
11(n, curl n) - n - I (a, curl a) } ds. JJJ
L
For an arbitrary field n the value (a, curl a) depends on the band choice. If at L we have A = 0 then by virtue of (4) (a, curl a) = (b, curl b) = 0. Therefore, all the bands turn by the same angle. This proves the theorem. Turn now to the stability problem of the closed streamline. Choose a positive direction in L. We are going to find a condition, in terms of the geometrical invariants along that curve of the field n, under which L will be the orbitally asymptotically stable limit cycle of the differential equation dx/dt = n(x). The stability condition below is a generalization of the Poincare condition for the field 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
JL
divnds < -JAds, 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 trajectories behave 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 the curve. Suppose that a and b are the two mutually orthogonal unit vector fields which
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
65
are orthogonal to n and periodic at L. Let P(s) be a point in L and F(s) be a normal to the L plane at this point. In the plane F(O) we take the intercept a passing through Po = P(O) and emit the field n streamlines from the points of a. Let Po be a point in a and co = IPoPPI. Denote by L' a streamline emitted from Po and infinitely close to
L. Let P(s) be a common point of L' and F(s). Then we set E(s) = PP'. We represent 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). du
(17)
Consider the expansion n(P') = n(P) + £(u, Eo)(cos cp V,n + sin W Vbn) + o(E).
a° = f. Substitute n(P') into (17), divide both sides by E° and then pass to the limit with respect to co. We consider only one loop of V. It is possible to show that 0. Denote this limit by cp(s). Then at the points of there exists lim cp(s, co) when co L we obtain Set
d in If(s) I
= (V,n, a) cos 2 V + (Vbn, b) sin2 cp + cos V sin V [(V,n, b) + TO, a)] cos 2,p [(Van, a) - (Vbn, b)]
= 2 [(oan, a) + (Vbn, b)) + 2
+ 2 sin 2tp[(Van, b) + (Vbn, a)]
Use formulas (8), (7) and (6): d Ink If(s) I
= 2 div n
-
sin 2W[(a, curl a) - (b, curl b)] 2
+ 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 sufficient for stability.
It is easy to produce an example of a field with a closed stable trajectory L satisfying the hypothesis of the theorem. Take any closed spatial curve L and draw through any point P E L a piece of surface of strictly positive curvature turned with
THE GEOMETRY OF VECTOR FIELDS
66
its concavity to the positive direction of L and perpendicular to L. Let R, and R2 be the principal curvatures of that surface. In moving P along L we obtain the regular family 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 limit cycle. If at least one of the principal curvatures is zero then the closed trajectories occur, 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 the streamlines for the given field n in a domain G C E3. Let a, b be the unit mutually orthogonal 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 Section 1.14), 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 a special manner. Let a', b' be the vector fields obtained from a and b as a result of rotation by the angle cp = W(xi,.r2, x.1):
a'
b' = - sinWa+cosipb. i.e.
We have
curl,' = e''"' (curl,0 - i [grad V, 6 J). 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 domain G C E3 such that 0 = A grad tb.
(1)
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
67
Theorem A complex vector field 13 is holonomic if and only if the complex nonholonomicity 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 every vector field /3 = a + i b generated by the vector fields a and b and orthogonal to n is holonomic. Find, first, a single holonomic vector field /3. To do this we rewrite the holonomicity 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 small neighborhood of P0 the field n restricted on F is not tangent to F. We define a regular family of curves in that neighborhood restricted on F. Denote it by -y,. Then we draw
a streamline of the field n through the points of each curve. We obtain a family of surfaces 01(x,, x2, x3) = const made up from the field a streamlines. Let a be a unit normal field of that family of surfaces, i.e. a = A, gradW,. Since this field is holonomic 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 a family 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, those surfaces are formed with the field n streamlines and the surfaces from different families, 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, is orthogonal to the streamlines of the field n. Therefore, the surfaces A,/A2 = coast are
68
THE GEOMETRY OF VECTOR FIELDS
formed by the streamlines of the field n, i.e. those surfaces have the same structure as cp, = const and W2 = const. Namely, those surfaces intersect the surface F by some
families of curves ry, and y2 and the corresponding field streamlines are drawn through the points of the curves of the families. By means of the families yI and y2, we introduce curvilinear coordinates cp, and
A, = 4
(`pl,'P2)
Therefore, for the vector fields a, b we have a = Az'P(So,,
a=
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 the latter 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. 4 ) a,Pz 0W2
8,p,
Find a local solution of this equation such that
(00.
2
) +(`
2
96 0 in the neighborhood 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 spheres or planes. Indeed, from the expression for A it follows that H2 = 4K, i.e. the surfaces
orthogonal to n are totally umbilical, i.e. they are spheres or planes.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
69
FIGURE 18
The geometrical property of streamlines of the field n with the complex value of the non-holonomicity (Q, curl Q) = 0 is as follows. Consider a point Po and draw a plane having n(Po) as the normal. Take the intercepts ao and bo in that plane having Po as their common end-point. Draw the streamlines of the field n through each point of the intercepts. So, we obtain two bands of streamlines. The angle between those bands stays constant along their common streamline (see Fig. 18). Indeed, as we have stated in Section 1.15, if A = 0 then all the streamline bands turn in the same angle with respect to the band of principal normals. Hence, the angle between them stays constant. 1.17 The Analogues of Gauss-Weingarten Decompositions and the Bonnet Theorem Analogue
Suppose that besides a, the unit mutually orthogonal and orthogonal to a vector fields S,, 22 in a domain G are given. Then a,, 22 and n form the moving basis in space. We can describe the behavior of the vector field a in terms of coefficients in derivation formulas which are analogous to the Gauss-Weingarten decompositions of the surface. Set the indices i, j to be distinct. Consider the decompositions of the field a; and n derivatives in Cartesian coordinates xk with respect to the basis a, , a2, n: aar__ I'akaj + N;kn, axk f7xk
= Mikal + M2ka2.
THE GEOMETRY OF VECTOR FIELDS
70
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'
axk
= r;kaj + N;kn,
an_ axk
(1)
-Nikal - N2ka2
(2)
We call rik the connection coefficients and Na the second fundamental form coefficients.
At each point of G three r/k and six Na are defined. Let us find the differential equations that these values satisfy. Differentiating (1) in xi and using (1), (2), we obtain -2a;
ark
axkaxi - (axi
- N;kNji) of + (r;kr l - N;kN;i)a; +
BN,k
axi
+ r;kNJI n.
)
Interchanging the roles of k and 1, we obtain the expression for As xk mean the Cartesian coordinates in E3, the second derivatives of a, do not depend on the order of differentiation. Equating the expressions for corresponding second derivatives, we obtain the equalities on coefficients of aj and n:
ar;k _ or,, + N;;N)k - N;kNj; = 0, axt axk 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. If we write the equality condition on mixed second derivatives of n then we arrive at equation (4). The values r2k can be completely excluded from those equations. Therefore, the system (3), (4) consists of three differential equations with derivatives of 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 replace A;k instead of r;k and B& instead of Na. Then there are the unit vector fields al, a2, n in
G with Ak as the connection coefficients and B;k as the second fundamental form coefficients. The vector fields a1, a2, n are defined uniquely up to their choice at one point.
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 is
easy to verify that the solution is orthonormal in G because of the nature of the system.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
71
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 vector fields 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 +
(5)
8az , R
91k = N1k cos
N2k = -N,k
(6)
Hence, the values F1k are defined by the field n up to the derivatives of some function V in a fixed Cartesian coordinate system. We can regard the set of r1k as some vector
field 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 vector fields 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 of surfaces defined as u1 = const, u2 = const, u3 = const, where u = u;(x1, x2, x3) pass through any point of a domain G c E3. If they meet each other at right angles then we say that a triorthogonal family of surfaces is defined in G (see Fig. 19). Various triorthogonal families are used in the construction of coordinate systems in E3. We are going to state the famous Dupin theorem on those families.
THE GEOMETRY OF VECTOR FIELDS
72
FIGURE 19
Theorem curvature.
The surfaces of a triorthogonal family intersect each other by the lines of
Let r;, i = 1, 2, 3 be the unit fields tangent to the lines of intersections. Then we have (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 to r2. Hence, (2) implies (TI, V ,r2) = (T3, V-, T2).
(3)
In an analogous way we obtain two more equations (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 field holonomicity condition Vr,T2 = Vr1T3 + QT2 + /ITi.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
73
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 another
field r, with respect to the third field rm, where kit 10 m # k. So, such scalar products 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 can prove that r1 is principal on the surface which is orthogonal to r,. Thus, the line of intersection 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 of curvature then we can complete the triorthogonal family. Let u1 = const and u2 = const be given surface families; r1 i r2 be their normal vector fields respectively; r3 be the field of tangents to their lines of intersection. We assert that r3 is holonomic. Since r3 is a principal direction on the surface u2 = const, then TI being orthogonal 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 of uI = const and u3 = const. Analogously, TI is tangent to the line of intersection of u2 = const and u3 = const. Hence, the new family completes the given two to the triorthogonal family. If we have only one family given, say u(XI,x2i X3) = 0, then it is natural to ask the question: whether or not to complete this family to the triorthogonal one? If this is possible 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 family u(xl,X2ix3) = const be complemented to the triorthogonal family it is necessary and sufficient that the function u satisfies some differential equation. Such a family is called the Lame family.
THE GEOMETRY OF VECTOR FIELDS
74
Let r1 =
r3 = {(}. Then from (7)
fl (l,k
(I
0 = (r371r3,06 = (2 6 (24
(3 6 _
kimC
(3,iyi
(,(m,ij _ >Bitf1 ri
where B,, = rdu"(,c(Consider the system of homogeneous equations with respect to (,:
EBi,(,(i=0,
(8)
i.i
0.
(9)
From this system we find two solutions for r1: yi = AF,,
where the functions F, = F,((k, (i,;) depend on 7-3, A is some multiplier. Since r1 is holonomic, F, satisfy the equation
F 8Fz _ 8F3
8x2) + Fz
8x3
8F3
- 8F1
8x1
8x3
+ F3
8F1
_ 8F2
8x2
8xI
0.
(10)
Substituting here the expression for F, in terms of (j and (ii, we obtain the equation which 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 (3
(21
(3.i
C
- (2
(1
(I1
IG
(3.i
6
(3
I(1
(1JI
(2.i
i! = 0.
Rotate the coordinate system axes in such a way that at a fixed point Po the following would be satisfied: (1 = (2 = 0, (3 = 1. We can regard equation (11) as the equation 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-hand side of (12):
-(z I
B=
G ., -C2 ,2
'.' ,
2
(1.2
2
" 2
(22
-c , 2 ,
2
0
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
75
The characteristic polynomial of this matrix coincides with the characteristic polynomial on the extremal geodesic torsion (see Section 1.12) and has the following form: -A3 + Az(r3, curl r3) - AM + T = 0,
where M and T are the mixed and the total geodesic torsions of r3. As r3 is holonomic, then (r3, curl r3) = 0. Therefore, the sum of roots of this equation is zero Ai + Az + A3 = 0. Equation (9) gives {3 = 0 at Po. Hence, (12) at Po has the form of the 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.
Du = E i
a u# 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)
DuWm + D. um = 0.
(14)
In the same way we find
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 following form: (15)
E vet)VnUmn=0. 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 M."
umk ukn
= 0.
(16)
k
Introduce the values Amn which depend on derivatives of u up to the third order: Amn = Du umn - 2 E umk U. k
76
THE GEOMETRY OF VECTOR FIELDS
Evidently, A,,,,, = A,,,,,. We can rewrite equation (16) briefly as: (17)
E vmtiv,,Amn = 0. m.n
For the following six values al = vlw1, a3 = V2W2,
a2 =
v, w2 + v2 W 1,
a4 = vIw3+v3W1i
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 obtain two more equations: ul(v1w2 + v2w,) + 2112v2w2 + u3(v2w3 + V3w2) = 0, (20)
ul(vlw3 + v3w,) + u2(v2w3 + v3w2) + 2u3v3w3 = 0-
Thus, ai satisfy the system of six equations (18), (19), (20). Since ai are not equal to zero simultaneously the determinant of the system is zero: All uI,
A22
A33
A23
A31
A,2
U22
1133
U23
U31
11,2
1
1
1
0
0
2u,
0
0
0 0
U3
U2
0 0
2ul
0
113
0
u1
0
2113
112
u,
0
= 0.
(21)
So, we obtain the equation with respect to u which contains the derivatives of u up to the
third order. From the general theory of differential equations it follows that we can take three analytic functions u(xl, x2i 0), u.Y, (x,, x2, 0) and ur, x, (xl , x2, 0) in some domain of the coordinate plane x3 = 0 as initial conditions. By the Cauchy-Kovalevskaya theorem, the solution of (21) is defined uniquely after that. Thus, the arbitrariness in the construction of a triorthogonal system of surfaces in E3 consists of three functions of two arguments.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
77
Now we give some examples of triorthogonal systems. Laplace used the triorthogonal net of confocal surfaces of second order in studying celestial mechanics. All of these net surfaces can be represented with a single equation I
l a,
X2
X2
2
3
+ t a2 + t a3
(22) l'
where a; are distinct constants. The value oft defines the surface. For any fixed point P(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 through P. Indeed, to find t when x; and a, are fixed we have the cubic equation. Suppose that a, > a2 > a3 > 0. Denote by f (t) the left-hand side of (22). If t = a; then f (t) turns into 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, namely 11, t2,13. Substituting one of these values into (22), we obtain the equation of the corresponding surface. If t = t, then the surface is an ellipsoid; if t = t2 then the surface is a hyperboloid of one sheet; if t = t3 the surface is a two-sheeted hyperboloid.
Check that the family under consideration is triorthogonal. The normal to the surface from the family 4'(x,, x2, x3) = const is grad'. Hence the vector XI 11,-al '
X2
X3
tI-a2' t,-a3
is normal to the surface corresponding to t;. Consider the scalar product of the normals
L.r
xk
C ( t,ak xk tj - t; (f(') -f(1j)) = 0.
Therefore, the normals to the family surfaces which pass through the fixed point P are orthogonal to each other, i.e. the family is triorthogonal. Consider the curvilinear coordinates in a domain G which can be introduced with the help of a triorthogonal family of surfaces u; = const. For each point P E G we put into correspondence the numbers (u1,u2iu3), namely, the values of the parameters u; which determine three surfaces from the system each of which passes through 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 other parameters the position vector end-point moves in the surface from a given triorthogonal system. If we fix two of the parameters, i.e. if we set u; = const and
THE GEOMETRY OF VECTOR FIELDS
78
uj = const, then in varying the left parameter the point P moves along the line of intersection of those surfaces. Thus, the surfaces u; = const are the coordinate ones and their intersections are the coordinate curves. The tangent vector of the coordinate curve of parameter u; is r4,. Since the system is triorthogonal, (r,,,, 0 if i # j. Therefore, the first fundamental form of E3 with respect to the curvilinear coordinates 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 in a domain G of flat space, the components of the Riemann tensor R;jk, are zero with respect 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
du,
duj
+ f3k,akj = 0, (23)
000 = QaQkj. ddk
where Q, k are distinct and of the range 1, 2, 3. The first of these equations corresponds 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 in terms of Lame coefficients and Darboux symbols. The unit vector = r 3/H3 is the unit normal to that surface. Therefore, the second fundamental form of the surface is 2
1113
=
E q= 1
(ru,.,, r,3)
H3
du; duj.
We denote the normal curvature of the u1 curve in the surface u3 = const by k31. This
is equal to the ratio of the second and first fundamental forms, i.e.
k. 31
(ru,,rJ, ,) H3H2
H3H,
I OH1 =-HIH30u3
031
H,
_ - I (r,". )-J 2 H3H1
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
79
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. We shall use this formula in the next section. 1.19 The Triorthogonal Bianchi System
We shall apply the Lame equations to study the triorthogonal system, considered by Bianchi, which contains the family of surfaces of constant negative Gaussian curvature. Assume that the family is u3 = const and the Gaussian curvature of the family surfaces is Ko. Denote by ds3 the first fundamental form of the surface u3 = const:
ds, = H2 th + H2 du. The product of principal curvatures k31 and k32 is equal to K0. Therefore 131032
H1 H2
= Ko.
Suppose that Ko < 0. Set
131 = -tanw
Ko.
HI
(1)
Then 132
H,
Using the Lame equations
ato
'
1 aHl HI 8u2
= cot
(2)
' = 032121 ands au tan w
8w au2 '
1
131112, we find
a112
H2 (9141
_ cot
-
w aw
.
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 these functions do not depend on U3. Indeed, from (1) and (2) we can find two expressions for H3:
raw
H3 = l H3=
au3
f8w
- cot w
81n .9u37yl J
a In p l
I
-Ko I
X33 J V K. +tanw
,
THE GEOMETRY OF VECTOR FIELDS
80
Hence, the following equation holds: tan w
If
a V+cotw auk = 0.
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 A
tanw= 8
(3)
where A does not depend on p2 while B does not depend on p,. Since the curvature of u3 = const is equal to K0, the Gauss equation for this surface has the form 82w
82w
Opt
8pz
(4)
- sin w cos w ICo.
We are going to show that (3) and (4) imply the degeneration of A or B into constants. 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 the derivative 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',
"/_ -CBB'. (-A-)
Integrating these relations, we get =CZa
2Ai2
+ 2C, A2 +C3,
A"u
2
=CZ
+ C, ,
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE 4
2Bi2
Bu B
= - C B + 2C2B2 + C4,
= - C2
81
2
+ 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 the surface u3 = const and one of the coefficients of A2 and B2 is zero. In the latter case the 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,
0U3
app
0U3
=0. -Ko With respect to new
Then for the Lame coefficient H3 we have H3 =
parametrization the first fundamental form of space can be represented as 2
ds2 = cos' w 42, + sine w dp2
+C
l
l`"'
dp23-
5P3
The So, the Lame coefficients depend on a single function w and the derivative function w satisfies the system of equations which can easily be obtained from the Lame system. The solution of that system, as Bianchi proved (see [13]), depends on five arbitrary functions of one argument. From the expression just obtained for the first fundamental form of E3 it follows that the Bianchi system formed with regular surfaces with Ko j4 0 cannot exist in a domain of large size and uniform expansion in all
directions. Suppose, for instance, that the domain is a ball of radius r. Show that r is bounded from above. To do this, take a curve p3 which passes through the center of the ball. Then the length of that piece of the curve which is located inside the ball
which the boundary included is not less then 2r. Since for the regular family of surfaces we have H3 0 0, by means of the choice of positive direction in the p3 curve we can fulfill the inequality 003 > 0. Let Pi be the end-points of the curve under consideration 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/2
at some point, then at this point one of the first fundamental form coefficients becomes 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.
82
THE GEOMETRY OF VECTOR FIELDS
1.20 Geometrical Properties of the Velocity Field of an ideal Incompressible Liquid
Vector fields occur in various problems of mathematics and mechanics. We shall consider geometrical properties of the motion of the velocity field of a liquid. The motion of a continuous medium can be described, by the Euler method, in terms of the velocity V of particles of the medium as a function of time t and coordinates x, 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 stationary if V does not depend on t, otherwise it is called non-.stationary. At fixed t, the curves tangent at each point (x, y, z) to the velocity vector V (x, y, z, t) are called streamlines. The individual medium particle describes the motion of some trajectory which is tangent to V(x, y, z, t) at a point (x, v, z) for each t. If the velocity field is stationary then the streamlines coincide with trajectories of the motion of particles of the medium, otherwise they do not since the streamlines are different for different values of time. For an ideal incompressible liquid the velocity field V satisfies the following fundamental system of equations which were found by Euler:
divV = 0
(the equation of incompressibility). (1)
ate + VvV = F - - gradp (the equation of tension. the Euler equation). Here V is a derivative along V, F is a density of solid forces, i.e. F = lim J,,,_.0 where 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 being constant for given liquid. The function p - the hydrodynamic pressure -- depends on coordinates and time; it characterizes the action of the forces onto every infinitesimal plane 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 the boundary conditions and the initial conditions, in addition, for the non-stationary case. 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.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
83
The function H + E+ U is called the Bernoulli junction. We shall consider a stationary velocity field. For the stationary velocity field in the motion of an ideal incompressible 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. Let n(x, y, z) be a unit vector field along V, i.e. V = vn, where v is the value of the velocity of liquid motion. Settle the necessary geometrical properties of the field n in the 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 n
= - dlnv .
(2)
We see that if the velocity value is constant along the field n streamlines then the mean 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 the Hamilton 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 field I = n div n - k is called the field of vectors adjoins to n. Observe that the field of adjoint vectors occurred in Section 1.8. It was proved there that the divergence of this field is equal to double the total curvature of the second kind, i.e. div 1= 2K.
THE GEOMETRY OF VECTOR FIELDS
84
Now settle the second necessary geometrical condition for the field n. To do this in the 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;
(4)
more precisely, the field I' is a field of gradient vectors of the function - In v:
I' = -grad Inv. Since
I = y2 (grad H - grad
v= )
we can write
curl l = - 2 [grad v, grad H]. Next we get [curl I, n]
[[grad v, grad H], n]
2divn
v3 div n
_
(grad v, n) grad H - (grad H, n) grad v v3 div n
By virtue of the second equation in (1), we have (grad H, n) = 0. Using (2), we obtain [curl I, n]
grad H
2divn
v2
Then
I' = I -
[curl I, n]
2 divn
=
1
v2
(
ad H - grad
= -grad In v.
-
grad H v2
(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.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
85
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 incompressible 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 generated by the relation between functions expressing the holonomicity of I and the possibility of 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 due to (3) and (4). From (5) we see that v = et''. If n is a vector field of liquid flow velocity then 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 the field of liquid flow velocity. Indeed, if (6) holds then A grad cp =
f (V) grad p e2 '
- grad 0(,p) e2V,
where 0(cp) is some function. Set + U = O(cp). We define the value of flow velocity by (2). Then (1) will be satisfied. One of the central problems in liquid theory is the problem of the existence of flow with a given boundary and initial conditions. Note that for three-dimensional flows of an ideal incompressible liquid the solution of system (1) which is global in time is not known. Therefore, for the detailed study of liquids only some classes of fields are acceptable. One of the important classes of streams widely used in applications is a stream without rotations, i.e. streams for which curl V = 0.
THE GEOMETRY OF VECTOR FIELDS
86
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 all points of an ideal liquid moving under the action of solid forces with a potential then
curl V = 0 at any subsequent instant. (2) In describing the motion of the body within a liquid it is supposed that the liquid is initially immovable. Since the immovable liquid contains no rotations, the motion remains irrotational. For the irrotational motion the field n is holonomic. Indeed, curl V = 0 means that at each instant V = grad gyp,
i.e. the field n is orthogonal to the family of surfaces V = cont. The function
IR
depends 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 +W2
E
(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 viewpoint. Those streams arise in descriptions of liquid motion generated by the waterscrew of a ship, in the theory of wings, propellers etc. A necessary and sufficient condition for the vector field n to have a stationary screw 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.
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
87
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 the function 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 through the fixed point lie in the same surface, i.e. in this case the space is a foliated one. The set of points which can be joined to a fixed point with the curves orthogonal to the field forms a surface. In the case of a non-holonomic field we have essentially another picture. The following theorem holds. Theorem Let us be given a regular unit vector field n in a domain G C E3 with the non-holonomicity value (n, curl n) 0 0. Then any two points in G can be joined by the curve orthogonal to the field n.
A more general statement was proved in 1909 by C. Caratheodory in thermodynamics [12]. He proved the following. Let us be given the Pfaff equation dxo + XI dx, + + X" dx" = 0, where X, are continuously differentiable functions of x,. If in each neighborhood of an arbitrary point P in the space of xi there is a point inaccessible by the curve satisfying the equation then there is a multiplier which turns the equation into a total differential. The proof is short but contains some vague arguments. A more general theorem was proved by P. Rashevski [29]. Let us be given m vector fields A'1,. .. , X," in a
domain 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 their
consecutive Poisson brackets it is possible to select n fields which are linearly independent at each point of G. Then it is possible to pass from every point of G to any other point of G by a finite number of moves along trajectories of the vector fields
X,,..., X,".
THE GEOMETRY OF VECTOR FIELDS
88
We use the Rashevski proof and, for the sake of simplicity, consider the vector fields in E3.
At first, we state an elementary lemma from calculus. Let (xi,x2ix3) be a point in G.
Lemma] Let us be given a function F(x) of class C. Suppose that xo, xi are the points 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 3
T dfi
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) for xi = xio. The lemma is proved. Suppose that there are given in G the vector fields depending on the parameters ti and 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. We suppose that the following conditions are fulfilled: (I) There are continuous derivatives of those fields up to the k-th order with respect to each argument. (1I) There is an inverse vector field. The inverse vector field 1-' (x, t,) is that with the
following 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 to construct one more vector field V as follows: shift the point x into y by t11-' (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
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
89
zero. As a matter of fact, this difference depends on x, tl and t2. Denote it by D(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
= 1(z,tl) -1(y,11)
s
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 components {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 by means of I and m. The field tl(x,11, t2) is invertible. To find the ??-I (X, tl, 12) we shall
z' by ill-1(y , ti), z' u' by y' by 121111-1 (x, t2), y' 12m(z, 0, u'- v' by tll(u, t1). Now find the %, i.e. the function 1(x, tl, t2) for
make the chain of shifts: x
FIGURE 20
THE GEOMETRY OF VECTOR FIELDS
90
11 = 12 = 0. We have I(z, 0) =10. But z(x) = x for it = 0. Also M1 turns into a partial x.0l. Analogously, m1(z, 0) = mi0, Li0 = ` t0 . Hence derivative of m: M10 = a oxf 71o(x) = l1oMo - mioL1o =11o(x)
x, 0)
exi
- min(x)
x, 0) 8x
If we interpret the vector fields l0i m0, 710 as the fields of operators
10=E4o8x1a 1
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 the following kind: a shift from x into x' by tll(x, it), then a shift from x' into x" by t2m(X', t2) and so on; or use the vectors it]--' (x, it), t2m- I (x, 12) in the procedure above. 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 means of 1 and m. Lemma 2 Let us be given the vector functions 1(x, it), m(x,12), q(x, it, t2) in G C E3. Suppose that the corresponding 10, mo, 710 are linearly independent at each point 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, v be 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).
(1)
Fix r2 - 0 and find Ov/8tl for II = 12 = TI = 0. Evidently,
8 (0, 0, 0, r2) =1(x, 0) =10(x)Itlt
Analogously 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) is equal to 72 (l0, mo, 710) 0 0-
(2)
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
91
Therefore, for any point v we can find ti, t2i r1 such that (1) will be satisfied. This implies that for some sufficiently small neighborhood of x for the values of v and of 0 for the values of 11, t2, r1 there exists a one-to-one correspondence between the neighborhoods. Let us be given in G the non-holonomic vector field o, i.e. (a, curl n) # 0. Take the vector fields X, and X2 orthogonal ton. Each of the vector fields XI, X2 generates the streamlines
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 the initial point x. Now define the shift vector i(x, ti) as a vector from x to the point in the trajectory which corresponds to the parameter value ti. Restricting the values of t, to some sufficiently small interval about zero, we are able to achieve the situation when for any x in a domain G' c G the corresponding trajectory segment exists and does not go out of the main domain G. So, A(x, ti) is already defined. Evidently,
0(x,11) = fXi(ui,u2u3)drii 0
where u1,u2iu3 are the coordinates in the i-th trajectory being functions of ri and initial coordinates Xi, X2, x3. The parameter r; varies from 0 to ti. Introduce a new parameter t = ri/ti, so ri = tti. Then rI
0(x, ti) = ti
J0
X;(u1, U2, u3) dt,
where UI, 142, U3 are continuously differentiable up to the k-th order functions of the initial coordinates x1 , x2i X3 and t, t;. Denote the integral on the right of the expression 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 we are 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 of 110 and 120. Also, i0(x, 0, 0) is a Poisson bracket of the vector fields 110 = XI and 110 = X2. By the previous lemma, any point is accessible by means of 11, 12 and n. But accessibility by means of 11,12, 1 implies accessibility by means of 11,12 . The theorem is proved.
92
THE GEOMETRY OF VECTOR FIELDS
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 orthogonal to n. V. Vaguer introduced the notion of parallel transport of tangent vectors with respect 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 the infinitesimal motion along the curve orthogonal to n. We call those curves admissible. As usual, we denote the field differential in E3 by dv. Denote by by the projection of dv into the plane orthogonal to n: by = dv - n(n, dv).
(1)
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-transportable along the admissible curve if its absolute differential along the curve is zero. For that field
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 representation with 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 7 is 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 vector field defined over the whole domain of definition of the non-holonomic manifold such that it will be parallel-transportable along every admissible curve? Evidently, along a single admissible curve such a field exists: this is a solution of an ordinary differential equation with the right-hand side linear in v. However, the two points in space can be joined to each other with the different admissible curves and the transport along each of them gives, maybe, different results at the same point. Therefore, our question may be reformulated as follows: in which case does not the parallel transport depend on 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 v,
Y3
4 q
v3
Sl
S2
S3
vi
01
ti
t3
vj
a
2
6
VLC FOR I ILLDS IN THRLI -DIMENSIONAL LUCLIDLAN SPACI
where s;,' =
.
93
The differentials do and dv can he represented in terms of
%0
those operators as
dn=
dn
dt =drdr.
Note that since n is a unit vector, "x is always orthogonal n . where x is an arbitrary vector. Let a be any vector field tangent to the non-holonomic manifold. For any
parallel-transportable vector field v along any admissible curve the following equation holds: do 1
dv dra-n
v,clra
(3)
Let x and y be constant vector fields in E'. Denote by a and b the projections of x and y respectively into the plane orthogonal to n. We want to extend the concept of parallel transport to arbitrary curves. To do so. Find the result of the action of the operator d v/dr over a. Set A = do/dr. Lenwia Parallel transport along admissible curves induces the parallel transport along the field n streamlines in the case of (a, curl a) ¢ 0 and this transport is defined by the following equation-
dv Sr
n = -n(v, An)
K In, vj
(a, curl n)
Differentiating both sides of (3) in the direction of the vector b. we obtain S-v (a,
dr=
b) +
dv A
d,
- b=
!dv - .96{v, Aa) - a (
7r
F {a,b)}
b, Aa A
b),
(4)
where and d! are the linear vector functions of two arguments. They are formed from the second derivatives of components of the vector fields v and a respectively. For instance, the ca-th component of ch- has the form
-dbl
02,04
where a', b' are components of a and b respectively. The equality of mixed derivatives of 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 second
term on the left of (4) into the right-hand side. Since a = x - a(n, x) and x is a constant vector field,
db = -Ab(n, x) - n(x, Ab).
(5)
THE GEOMETRY OF VECTOR FIELDS
94
Using (5), transform the next two terms in the right-hand side of (4):
-drdab-nA (v,
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 into account all the transformations, rewrite (4) in the following form: z
dr2 (a, b) = (x, Ab)
dv
n + n(x, Ab)(v, An) 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 a and b. Subtract the equation obtained from the original. Using the symmetry of the vector 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
= -(a, b, curl n) = -(a, curl n) (n, a, b).
f; )
(9)
We can represent the difference of the two last terms in (8) as the following vector product [[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 second symmetric function of principal curvatures of the second kind) can be found by the following 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) is invariant with respect to rotation of coordinate axes in a space, choose the co-
95
VECTOR FIELDS IN THREE-DIMENSIONAL EUCLIDEAN SPACE
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' (Aa,Ab,n) _ tjbi
{?a'
0
jb1 0 =
0
0
-b'a2)
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), the equation (8) can be represented as
dv
n = -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 is proved. Decomposing any vector x into components collinear and orthogonal to n, we are able to give the equation of parallel transport with respect to each of them. Adding those equations, we get
dvx dr
= -n(v,Ax)
-K[n,v](a,x)
(13)
(n, curl n)
In the case when x is orthogonal to n we shall come to the equation of a paralleltransportable field along the admissible curve introduced above.
We shall call the parallel transport defined by (13) the prolongated parallel transport. It is easy to check that if the scalar product (n, v) is zero at some initial point on the curve of parallel transport then by virtue of (13) this product is zero
identically. This means that the parallel transport takes a tangent to the nonholonomic 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 it contains 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-transportable vector 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 of equation (13). Let x and y be constant vector fields in E3. Differentiating both sides of (3) in the direction of the vector y we obtain d1v dr2
(x, y)
Ay(v, Ax) - n (dr - In, v]
(x,
d
dd1n
Y'
Ax I -0 (VI ///
Kn
dr ((n, curl n)) Y)
r'-
(x, Y)
_ K(n,x) (n, curl n) ([AY' v]
dv
+
n' dr Y] )
14
(
)
THE GEOMETRY OF VECTOR FIELDS
96
Make a substitution of d v/dr by means of (13). Also, using the orthogonality of v and Ay to the field n, we can write [Ay, v] = n(n. Ay. v).
Equation (14) can be rewritten as d-W (x,
dr-
y) _ - Ay(v, Ax) - [n, vJ x,
d Kn y dr (n, curl n)
V,
-a(
d n (x dr-
Y)
{(n.x)(a,v,Ay)+(n,y)(n,v.Ax)1
+(n, unln)
+ I(,curln))-(° x)(n.y)[n.[n.vJ]. The last three terms on the right of the equation are symmetric with respect to x and y. Interchanging the roles of x and y and subtracting the equation obtained from the original, we get In, vJ
X. Y, curl (n,
[v. [Ax, Ay]J = 0. curl n)) +
(15)
Consider the second term. Transform the vector product in it as [Ax, Ay] = n(n, Ax. Ay).
(16)
With respect to the special system of coordinates defined above, we obtain ;vi
(n, Ax. Ay) =
,Z,r
Mfr
'}'
0
0
0J 0 1
YI
L
V1
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).
Turn back to (15). Substitute it into (16) and (17). We get Kn
In, vJ
(x, y, curl (\\\(n, curl n)
) - P) = 0-
(17)
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 t
Kn
`\{n, curl n)
-P=0.
We shall call the vector on the left the Vaguer vector. So. the following theorem holds. Theorem In order that the rector Iirld parallel-transportable on the Itolono,nu manifold exhits it is Fircesxarr and sufficient that the Vaguer rector
curl
Kn
l
{n,curln)J
lion-
-(h°-kdivn+Dkn}
is zero.
Give an example of a vector field having a zero Vagncr vector. Define the components 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 the
non-holonomicity value of that field is (n. curl n) = -c. As the field n streamlines are the 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 surface into the arc of the same great circle in the unit sphere. Hence, K = 0. We see that all the terms in the expression for the Vagner vector become zero.
2 Vector Fields and Differential Forms in ManyDimensional 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 Euclidean 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 principal curvature of the first kind. We shall call the directions d r for which these values are achieved the principal directions of the first kind. Let us find the system of equations to determine them from. Introduce the Cartesian system of coordinates x' in El" We denote the vector components with respect to this system with superscripts; for instance, the components of n will be denoted by '. To simplify notations, we denote the derivative 8{'/8xi by E . Consider the field n in the neighborhood of M. Choose the Cartesian coordinates in such a way that the basis vector a",., will be directed along 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:
(J
+ A ) dx '
+
2
+
x
dv" = 0 ,
t1
t2
+2
2ei
dx'
+
dx1
+
(f2 + A) dx-
+
+ 99
...
+
+
(
...
= 0
r'+A) dx'"
=0 .
,
100
THE GEOMETRY OF VECTOR FIELDS
The principal curvatures of the first kind are the roots of the characteristic polynomial (1)
1i+
As the matrix +ek)/2II is symmetric, all the roots A, are real. Denote by al, the k-th symmetric polynomial of the principal curvatures of the first kind. The ok can be expressed in terms of the main minors of the matrix II e, )/2II as follows:
Qk =
(_I)k
(2)
I
We define the principal curvatures and principal directions of the second kind by means of the Rodrigues equation
do = -p dr.
(3)
We call the direction d r satisfying (3) and orthogonal to n the principal direction of the second kind.
We can find the number ls, which satisfies the Rodrigues equation, from the following equation:
tl + p
C;
l;z + l +1
...
...
C,"+i + m+
p
Here we do not suppose a special choice of space coordinate system. One of the roots um+, 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-dimensional domain G in the unit sphere S'. Therefore, the Jacobian of 0, which is is equal to zero. Hence, one of the roots of (4) is equal to the determinant of zero. For definiteness, set µm,., = 0. The other roots are called, by definition, the principal curvatures of the second kind. Since the matrix IIc;II is not symmetric in general, the p- can be complex. Therefore, the real principal directions do not exist for these pi. But in all cases the symmetric polynomials of the p, are real ones. They can be expressed in terms of coefficients of the polynomial (4). Denote the symmetric polynomial of the principal curvatures of the second kind by Sk. It is easy to express them in terms of derivatives of components of the vector field n: i
4
Sipk Observe 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'll including a point at infinity. We can put every such field into correspondence with a regular 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 means of 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 the corresponding point in E" in stereographic projection. Now construct a mapping
ip of Euclidean space E"' onto Sm. To do this, we put n(x) into the origin O E E"' I. The end-point of n(x) then marks a point q in S'. Put into correspondence 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 of S"' including the south pole. Conversely, by a given mapping of Sm+1 onto Sm we are 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) the following theorem has been proved. For m > 3 there are exactly two homotopic classes of mappings S"' -+ S". Whether or not the mapping f : Sm+I --+ S" belongs to one of those classes depends on some invariant 5(f) which is the residue modulo 2 and 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-dimensional submanifold in S". Denote by M' the image of this submanifold in a stereographic projection of Sm+I onto E" ". In terms of a vector field the submanifold M' consists of one or several closed curves where the field n is constant. If the set f- I (q) does not contain the south pole then MI is a smooth closed submanifold of Euclidean space Em. By means of the mapping IIi: E" I S" we are able to frame MI with the fields
orthogonal to MI. To do this we consider the tangent space Tq at q E Sm. The mapping iti induces the linear mapping ti'. of E"+' onto Tq at x E MI. Choose the basis e,, . . . , e" in Tq. We take the vectors u; = (i.-1e; as a framing of MI at x. By means of a continuous deformation, this framing can be transformed into both an orthonormal and normal to an MI framing. We denote the vector fields of the latter framing 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 framed one-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 those matrices along M' defines a continuous mapping h of the curve M' into the manifold
Hm+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 residue 3(h) = 0 if h is homotopy equivalent in Hm+I to a point and (3(h) = I otherwise. For
THE GEOMETRY OF VECTOR FIELDS
102
the case of a many-component manifold the residue /3(h) is equal to the sum of each of the component residues. If m = 1 then i(h) is a degree modulo 2 mapping of MI onto the circle H2. Next, let r(MI) be the number of components of MI. Then the
invariant b(f) of the mapping f and, consequently, the field n invariant under continuous 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 infinity Eo+1. By means of a stereographic projection we are which is parallel to some space
able 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 correspond to those deformations of n in E"+' which preserve the property of n to be parallel to Eo+1
The classes of continuous mappings f form the homotopy group an+,(S"). Some element of this homotopy group corresponds to the field n. The problem of finding all the homotopy groups of spheres is an unsolved one. The solution was given for r not very large. Pontryagin stated that 7rn+2(S") = Z2 and pointed out the method of evaluating the invariant which classifies the mappings S"+2 __+ S" up to homotopies. We apply that method to the vector field defined in Euclidean space En+2. Let Q E S" be a regular point. Then M2 - the preimage of Q - is the two-dimensional surface in E`2 on which the field n is constant and the end-point of n coincides with Q when the initial point of n is put to the center of S". If M2 consists of a single component and is a
surface of genus p then it is possible to select a set of smooth closed curves A1,. .. , AP, B1,. .. , BP such that A, intersects B; at a single point avoiding tangency while any other curves have no intersections. For any curve C of this set we are able to determine the invariant b(C ). To do this we shall construct a framing of C. Set un+1 to be the vector which is normal to C and tangent to M2. The surface M2 is
framed 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. Decomposing the vectors u1, . . . , u,,. , r with respect to the fixed orthonormal basis in En,'`' we obtain as above the mapping of C into the group Hn+2 of orthogonal
matrices of order n + 2. If this mapping is homotopic to the constant one then b(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 each component. As an example, evaluate the b invariant for the vector field n defined in E4 and parallel to E. The field n has the following components: 2a 2b t a2 +b2 - 1 l;l =1+a2+b2' 2=1+a2+b2' 3=1+a2+b2' y4=0,
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
103
where a = x1 +72, b = %(v23 + x4. Set eI, e2i e3, e4 to be a fixed basis in E4. Evidently, 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 defined by Q = const, while B2 is a curve defined by a = coast. Construct the framing of the
curves generated by the field n. The vectors pa and ft (the derivatives of n in parameters a and b respectively) are tangent to the sphere S2. Consider the expansion of the field n nearby xo E M2:
n(x) = n(xo + ix) = n(xo) + (n0-+ ub
Ob
= 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
8x, - cos a, 8b OXI =
0,
_=0 8a
8a _
8a sin a, 8x2 =
0,
ax3
ab =
ab
ax2
ax3
8x4
ab
=cosh,
8x4
=sing.
Under the induced mapping of the tangent space of E4 at xo the framing vectors u1 (x) and u2 (x) must be turned into the vectors na and ob respectively. Consider the torus 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, namely do = oo da. If Ox is directed along n2 then do = ab A. Hence, the framing vectors u1 and u2 coincide with n1 and 02 respectively. Next, the vector r, = - sin ae1 + cos ae2 is tangent to A,, while r2 = - sin /3e3 + cos (3e4 is tangent to M2 and orthogonal to
A,. The matrix of transformation of the fixed basis e, , e2, e3, e4 into the basis ui, n2i r,, 72 has the form cos a
sin a
0
0
0
0
cosQ 0
sin(3 0
0
0
- sing
cos Q
-sins cosy
On the curve A, we have fl = coast, while a varies from 0 to 2r. The theorem has been stated (see [72], p. 132) which allows a check to be made as to whether or not a
given mapping h of the circle S' into the group H. of orthogonal matrices is homotopic to the constant one. Every mapping h: S1 - H. is continuously homotopic to a mapping g: S1 - H2 = S1. It happens that g is homotopic to the constant mapping in H. (n > 3) if and only if the degree of g is even.
104
THE GEONMETRN 01' VECTOR I IELDS
In our case the mapping li by itself has the form of g: S' -. H,. This mapping defines 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 to the 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 the other hand, their stabilization in growth of n became clear. Here we cite some results on homotopy groups [78]: 0 for it > 6;
7r".j., = Z_4 for it > 5;
= 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 Formula to the Case of a Vector Field Consider the unit vector field defined in some domain G of E'11-". We suppose that n is regular of class C' except for some points. maybe. Let F"' be a closed hypersurface
in E"''' which does not pass through the singular points of the field n. Denote the curvilinear coordinates in some domain of the hypersurface F"' by a .... ; o,,,. By means of the vector field n we can construct the mapping 1' of this domain onto the unit sphere S"' with the center at the origin 0. To do this we put into correspondence the point M E F" and the point in S"' which is the end-point of n(M) being taken to the origin. The field n depends on along F"' and hence they can be considered as the coordinates in S'. as well. Denote by da the in-volume element of S". 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'. then by the composite function differentiation rule we have (2)
,,
For the sake of convenience. denote the derivative n,, by n,. If the set of indices differs from the set ii, ... , % only in their order then (a,,) ...,a,.,,n) =E,', 1" (n,,
-,aa», a)
(3)
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
105
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 set 1 , 2, ... , m + 1 . Let ( i i , . . . , j,,,) E A. On the right of (2) we rearrange the terms of the sum in such a way that each string of indices (ii,... , i,,,) would be different from the fixed 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 determinant ax',
ar+-
act,
on.
x
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. If r(01, ... , a) is a position vector of F" then the unit normal v is the normalized vector product of r,,,, ... , r I [r.,
, .... r..11
Let e ' , . .. , em+I be a positively oriented orthonormal basis in Em+1. We assume the
vector product such that [el,... , em] = em+,. Then the vector product can be found in terms of the signed determinant, namely eI
...
m ant
art.
em+l
IF'" 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)
THE GEOMETRY OF VECTOR FIELDS
106
Note that the m-volume element dV of Fm has the following expression dV = I[r,,,, ... , r..] I do1 ... dan,.
(7)
Introduce the vector field P as follows: P=(-1)m{(n2,03,...,Onr+1,
0),-(01,133....,Din ,1,0),...
(-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 hypersurface then the 0 image of Fm covers the unit sphere some integer number of times. More precisely, 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 found by the formula
0=I fdo, wm
y'-(Fl)
where the integral expresses the volume of the image of the mapping I' if both the sign 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 the hypersurface 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 defined with respect to any Cartesian system of coordinates. Now we state the invariant representation of P which is analogous to one in Section 1.10. Define the vector sequence k, inductively:
kp = 0,...,k;+1 = Q);,n,..., where i = 0, 1, ... m. Remember that we denoted by Si the symmetric polynomials of the principal curvatures of the second kind. The following theorem holds. Theorem The curvature vector P can be expressed in terms of k; and symmetric polynomials 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
C+I Sl
...
II. +
S,n+1
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
107
', where the summation over p is assumed.
The vector k1 has components
Therefore, the components of k2 are as follows: t,' of k1 are
I
p
P. In general, the components
where the summation over j, r, s, ... 1,p is assumed and the number of indices under summation is i. The index a shows that the expression stands for the a-th component of 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 = AAn has the form
1I
1
"nr+I
A2n = m-fI
SI
...
M+fI
Comparing components, we again conclude that k2 and A2n coincide. In an analogous manner we find that k1= A'n. Thus, the vectors k1 are the results of consequent 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 of aj1 in the matrix A. Show that P = Bn. To do this we compare, for instance, the first components of those vectors. The first component of Bn is c2
...
Sm+I _ t2
b111 = cI
Cni, I
I'
fin. I
m
rI
r2
m+l
t2
Q2
+1
tm h 1
...
1
C+4
...
m+11
Sm+ C.
THE GEOMETRY OF VECTOR FIELDS
108
The first component of P has the form
m()
;
e2
el
2 Stt1m+
i
SI c2 SCm+I
...
em+I
...
V+ 1
t.2'+1
ttm f+ 11
Sm
...
S21
Sm+l
C2
`mti l
m+2I
Sm+ l
Thus, the first component of P coincides with the first one of Bn. The analogous result 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. The element 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) in terms of the coefficients of the characteristic polynomial 0(µ) = Io,j - Cijl and the powers 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, (-1)m-I Am-1
B(0) = (- l )mAm - PI
- ... +Pm- I A - pmE.
Let us find pi in terms of symmetric polynomials of principal curvatures of the They are the roots of the equation second kind µI, ... IAbij
lµb,j - C,i1= 0,
i.e. µl, ... ,.a are the eigenvalues of matrix C. Thus, we have the following expression 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 case of 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 the generalization 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 to state the geometrical nnarung of the liele! P streamlines the field a is constant along them. In the previous section we denoted these curves by M1. From (9) it evidently follows that if P - 0 then the volume of the n/' image in S' of any hypersurface F"' is
zero. i.e. the dimension of the image is lower than in. Therefore, the vector P characterizes the curvature of n in dimension in. However, if P s 0 then the field n is not 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 regular
field of normals offly) defined in some domain G c t We draw a geodesic of ambient space in the direction of n(x) through each point x E F. We shall call the field of tangent vectors of geodesic lines the geodesic field. Denote it with the same letter n. Let M he a point in a distance s on the geodesic from y E Fin the direction of n(t ). If r ranges over all points in the domain G then the set of marked points M form, in general. some hypersurface. That hypersurface is called geodexicalli parallel
or 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 geodesically parallel hypersurfaces filling some domain in the Riemannian space. The geodesic lines are orthogonal to those hypersurfaces at each point. If we takes to be large then it 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 submanifolds of lower dimension, moreover into points. We consider the following question: how do the principal curvatures of geodesically parallel htpersurfaces change in moving along the geodesic which is orthogonal to the surfaces of the famil r. A tangent direction r on a hypersurface of a Riemannian space
is called principal if the covariant derivative of a in the direction of r is collinear with 7-
V'0 = -Ar. The coefficient A is called the principal curvature pf the hypersurface. At each point of
the hypersurface we have n - I principal directions. Theorem Let r he a principal direction an a htpersurface from the famllr c1 geodesically parallel hrpersurfaces and ,\ flit, a corrc'sprntdnrg principal curvature. Then
THE GCOMFTRY OF VLCTOR FIFLDS
110
the derivative of A nith respect to the arc length parameter of the geodesic orthogonal to the hlpersurface has the following form (1)
d+K(r,a),
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. As
usual, we denote the vector contravariant components by superscripts and the covariant derivative by "comma". In these notations the equation for principal directions 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 the right-hand side reduces itself to -dA/ds. We have r n',,Ar' n A + r, n', , r', A ll = -
(/A
(4)
ds
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 tt
A
(5)
= lm' +
where T. stands for the principal direction different from r and having A as the corresponding principal curvature, lc and v are some numbers. Let us find it. To do this 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 coThen we get variant derivative in k and as a compensation subtract
(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 consequence. 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), where R't,A is a curvature tensor of R.
(7)
AND RIEMANNIAN SPACES III
VECTOR FIELDS IN MAN)-1)IMI NSIONAL
The first term on the right-hand side of (7) is zero because n is a geodesic field. The second term is
Ar r, = -112.
-n', AW,,r-'T, =
(8)
So, after applying (6)-(8). equality (4) gives (1).
If we set A _ -'
then (1) has the form of the Jacobi equation
`k(r.a)q.
dt'
This equation allows us to estimate the change in principal curvatures of geodesically parallel hypersurfaces along the geodesics orthogonal to the family under the given restrictions 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 derivative
of this field with respect to any direction is equal to zero: V,EA = 0
Since n= 0. Therefore, the length of such a field is conV,n' = stant. If two constant vector fields are given then they make a constant angle. The
curvature vector of the constant vector field streamline is tr'V,t;A = 0. i.e. the streamlines of the constant vector field are geodesic lines. Let a and b be orthogonal to 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 ' ,
such
that the surfaces xt = const are orthogonal to n. As xt we take the distance from some fixed surface of v' = const along the streamline. Then the first fundamental form 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. With respect to the coordinate system introduced the field a has the following compon0. i > 2. Consider the derivative ents:
V, I =
v4- rz, r:,
THE GEOMETRY OF VECTOR FIELDS
112
From this it follows that rj = 0. Since gIr = 0 for a 0 1,
r' - g -
11
1
091i + 1991k _ ag,k axe axl
i axk
=
-g 111 ag,k = 0. 2 axl
Thus, if a constant vector field exists in a Riemannian space then the metric can be reduced 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, by definition, 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 in R". 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 the metric 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
air di' d 2 J = d x ' dxJ +
8xk
d x k dx bt.
d x ` d x ' bt + axk
Hence b ds'- _
(
" dx' dxj + g y 6OXk 'e dxk dxj + g;; 8 dxk dx' 1 bt. k
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 third We get
bdi
a
laxa
+ gkj
ae +a aavj)
dYj br.
By the definition of a Killing field, b ds2 must be the infinitesimal of higher order than bt in all choices of d_xi or dxj. Therefore, the Killing field satisfies the following system of equations: ag;;
Since
aaxj
aaxi
= 0. + g.k axt ` + gkj = 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 a metric 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 4kj =
-gki
j
Further, 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 the Killing 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 of tensors, we get
(RU + Ri + R°kjAn + 2G& = 0.
114
THE GEOMETRY OF VECTOR FIELDS
We can replace the sum of the two first components of the Riemannian tensor in parentheses with R°kii. Hence CjJk = -'R°kJI u.
(2)
Also, there are the relations between the first-order derivatives of Killing field components and the original components: C. ( o.k - Rpkji.!) + R /j, E,°.k - R Lit F,,j -
jklO.1 - Rikl
F
Sf- = 0.
These relations arise as the compatibility conditions of (2). 2.6 On Symmetric Polynomials of Principal Curvatures of a Vector Field on Riemannian Space
The symmetric polynomials of principal curvatures of the second kind of a vector field with a constant length in Riemannian space have the same expressions as in the case of a vector field in Euclidean space if one replaces the usual derivatives with covariant derivatives:
Sk=(-1)k E II <..
71
...
Sff4
We are going to find the analogs of the divergent representations for S2 and S3. Let k = {t'jtJ} be the field C streamline curvature vector. Consider the divergence of the field -(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 Ricci curvature for the direction C. It is equal to the sum of sectional curvatures of Riemannian space along the planes spanned on C and n - I vectors which are mutually orthogonal 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
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
115
We shall denote the determinant of a matrix formed with this column and its derivatives by brackets. Then 6S3 = [a,i,a,i2a,i3J. Take out the sign of covariant differentiation 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 a
tensor t. In detail the first of the determinants above is ,,t, I
Sri,
[aj,i2 - a,,21,aaj,] = I
t", I
I R°i, i21;.
_
+R;;.3itoele - 2R0 to 0.
It is easy to check that the other two determinants containing second-order derivatives of a are equal to the first one. Now consider the determinant [aa,i2a,3] which stands under covariant differentiation in zi, . Expanding it with respect to the first column, we have tth ti,
t1
[aaj2a,i,] = St;,(SJ2 Sj, - Si2,i,) -
SI2
C;, c;, el tI2 (et;, ttl, S,;, - Sj2,i,) + S (S ,i2,
°, ;1
Evidently, the expression in the first parentheses is equal to 2S2. Now gather the
terms containing t; and 22. The other two terms are equal to each other after summation 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 for the case of a vector field in Euclidean space. Note that the expressions for the other two determinants, namely [a,1, as j3 ] and [a,,, a j2 a], can be obtained from the expressions presented by the substitution of i1 by i2 and i3 respectively. Therefore, we obtain the following formula for the third symmetric polynomial S3 in the case of a vector field in Riemannian space:
3S3 = divP+R°0t"40SI +R,,,3t°k'3 - R0
1
tft
Consider the expression for the sum of terms with Riemannian and Ricci tensors in the case of space of constant curvature K0. We have
-R°;irCp = (n- I)Ko,
R.,3f°V = ng°uCa k" =0.
THE GEOMETRY OF VECTOR FIELDS
116
Introduce a coordinate system in such a way that at some fixed point g,,,j = 6,,;j and
the 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 the principal 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 be represented 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". This expression 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 the transformation law of covariant tensor components. Namely, if dk are its components 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' =aA
aax'dv=dkdvk.
Consider the two symbols of differentiation, namely 6 and d. We shall adopt that d bx' = 6 dx'. The differentiation b produces the differentials 6x'. We denote the Pfaff form with respect to bx' by w(b). To each Pfaffian form w we are able to put into correspondence the form of second order, 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;b
ax-'
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 =
axi
ax/
I Qak _ aaj (dri 2
Xj
axk
bxJ dxk ).
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES 117
a ) form a tensor. The equation w = 0
It is easy to check that the coefficients (a 'TO
is called the Pfaff equation. Let us also define the bilinear form w(d, b) which depends
on both shifts dr and br: w(d, b) = a;1(dx' bx' - dxj bx'),
where a; form a tensor of second order, i.e. a;; = dk1
i
03A IV 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 some shifts dr and Or then it is zero with respect to any other coordinate system for these shifts.
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 equal
to m. This definition can be interpreted geometrically. Consider the system of m vectors A, = {a11}, ... , A. = {am;} in a Euclidean space. The condition of nondegeneracy means that A1,.. . , Am are linearly independent. We denote the space they span by Tm (see Fig. 21). The surface Vn_m of dimension n - m in Euclidean space E" is called the integral surface if any shift {dx'} in tangent space of V"_m satisfies 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
118
THE GEOMETRY OF VECTOR FIELDS
The condition, under which the system (2) is totally integrable, gives the following theorem. Frobenius theorem The Pfaff system of equations w1 = 0, j = 1.....m is totally integrable in a domain G C E" if and only if for all w, the bilinear Frobenius forms are zero by virtue of the system wj = 0. We cite the proof from [26]. The conversion to zero of bilinear Frobenius forms by
virtue 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. This mapping maps all the vectors of T"-" into zero and is identical on T'. By the hypothesis, 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 fixed point P is determined by the matrix of order n. The elements of this matrix are the continuously differentiable functions of P, i.e. of parameters u,. The derivative of M means, 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, du, = M. dr',
M,,, but = Mr bx', 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 we set
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 given Pfaffian system ,v, = 0. To do this. we find the matrix of Al. Suppose that M takes the 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 the
Pfaff system (6) to zero, EA nil, dt1 is a linear combination of systems of Pfaff equations nI,A dr'
j = 1, ... , it,
dxA ,
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,,, } into themselves. A,,a,Aa/A =a!,.
Here the first subscript in al, is a number of Al and the second is the number of these vector components with respect to E". The matrix 11a,,11 is not square, but it may be completed 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 basis in E". We denote the complemented matrix by A. Then we may take the components of 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,=III s' i a.vok
The components of MA, d r - Mjt br have the form am,A
ax"
6x" d,k -
am,A
Ox"
dr" 6x' =
aa,, a,k
av,
bx" dxk
aa,A
+
a
_ 8a,,,
a,A
&e'
dv" ax"
d.%A
bx"(aA dxk) - O ; dx"(a,A dk).
By virtue of (6), the latter expression is zero.
THE GEOMETRY OF VECTOR FIELDS
120
The shifts dr, br E T"-"'. Therefore, the last two terms are zero. Since the determinant 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 zero bw,(d) - dw.,(6) = 0 by virtue of the system w,(d) = w,(b) = 0. Thus, the necessary condition is proved.
Now prove that this condition is sufficient, also, for the system wf = 0 to be holonomic. We shall construct some surface and prove that this surface is tangent to
T` at each of its points. Consider at Po E G a unit vector T E T"-m(Po) with the initial point at P0. Its end-point lies in a unit sphere S which is centered at Po and located in T"-"'(Po). The dimension of this sphere is n - m - 1. Introduce the coordinates u1, ... , u"-m-1 in this sphere. Then r can be considered as a vector function of these coordinates: Span the (m + 1)-dimensional plane T"'+I (Po) on r and Tm(Po). Let P be an arbitrary 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 projection 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 field orthogonal to Tm(P). Draw the streamline of this field through Po in the direction of T. This curve is orthogonal to Tm(P) because it is orthogonal to the projection of Tm(P) into T" (Po). If we take the various r at Po and draw the curves -y as above then 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 is orthogonal to T"'(P). Hence,
Mr,=O.
(7)
i=I,...,n-m-1.
(8)
Show that
Mr,,,=0,
This means that all tangent vectors of the constructed surface lie in T"-m(P) for every P, i.e. the surface is the integral one of the Pfaff system (2) and, as a consequence of the definition, the system is holonomic. Introduce a vector a = Mr,, and find a linear differential equation this vector satisfies. We have
a, = Mr,,, +
8M 80
r,,, = Mr,,,, +
8M 8xJr,,,
8x) 80
= Mr,,,, + Mrsrw
(9)
If we differentiate (7) in u,, we get
M r,,,; + M,,r, = M r,,,, +
8M 8x!
axe au
r, =
M,., r,, = 0.
(10)
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
121
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 the system (2). Therefore, the differentials d r, br E T" satisfy (6). If it were known that r", 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, we have r,,, = r,,, (ul , ... u"_,"_ 1, 0) because the initial point Po is the same for all 7 starting from Po. Therefore, a(0) = Mr,,,(ul,... un_rn_1,0) = 0. Hence, (12) implies
a(a) = 0, i.e. at an arbitrary point P of the constructed surface we have r,,, 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. The field a is holonomic if the family of hypersurfaces having this field as a normal one exists. By the theorem just proved, this occurs if and only if the bilinear Frobenius form 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 be expressed as
" Cep a'o
al; p)
axa - ax°
(dx° bx0 - dxa bx°) = 0.
(13)
THE GEOMETRY OF VECTOR FIELDS
123
Choose the system of coordinates in E" in such a way that at a fixed point P the components of the vector n are (0,0,..., 1). Then for every admissable shifts dr and 6r one must have de" = 0 and 6x" = 0. Hence, in (13) the summation over a and 0 goes from 1 to it - 1. If we take the shifts dr = (0, ... , . j,0,-..,0) and r = (0, ... ,1, ... , 0), where the unit takes the a-th place in d r and the 13-th place in or. then we obtain the system of (n - 1)(n - 2)/2 equations: 6r=(0,-..,0
art+a, = 0,
1 < a,
aF1'-kK
8:5n-1.
(14)
Denote the derivative 8/8x" by 8,,. Consider the set of the following expressions with respect to an arbitrary system of coordinates in E":
where the brackets which include the indices a, /3, 7 mean the sum of various terms which can be obtained from g (, g, with permutations of indices a, Q, y; also, the corresponding expression is included in this sum with a *'+ " sign if the permutation
is even and a "-" sign if the permutation is odd. Consider the way Sl,,i, changes under the coordinate change in space. Let x" = x"(,ti''). Then the components , of n with respect to the system y' are related to its components with respect to the system x" by the tensor rule 8x"
Let us form the expression analogous to Sl a, with respect to the system aaM
,
,(6ayk
)ev,
,1
' 8x"' 80 ex" + Oaf' ay" a eayll yla)A TV-, The last term on the right is zero due to the equality to each other of the mixed derivatives of the functions xri = xs(y) and the convention given above for the a2
_ &001 Sri
brackets. The first term can be transformed by relabelling the summation indices as sl;
ax" 8.0 ax's
ay, aye ay, =
ax" ax' axtf
ay 8y, ayk .
So, the quantities Sl h, satisfy the tensor rule of transformation. If the field is holonomic then all components 0 at the point P with respect to the system of coordinates chosen above. Observe that if a tensor fl",s, and bilinear Frobenius form are zero with respect to a given system of coordinates then they are zero with respect to any other system of coordinates. Therefore, the condition Slap, = 0 is necessary and sufficient for the field n to be holonomic. In the particular case, when the field a is in a domain G C E3, that condition gives (n, cunl a) = 0. The tensor fl"p, is called the object of non-holonomicity of the field a in E".
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES 123
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 expressed in the form of a system of differential equations. This system may either be integrable or not. As a typical example, we consider the equations of constraints which arise in a disk D rolling over the xy-plane. Suppose that the disk is always perpendicular to the xy-plane and the disk plane intersects the xy-plane in a straight line 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 xyplane, B is a fixed point on the boundary circle of D which was the point of contact at the initial instant. Denote the angle between OA and OB by t/, (see Fig. 23). As the rolling of the disk is assumed to be without sliding, the length of the curve -y in the xy-plane is equal to the 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 system of Pfaff equations
w, = dx - R cos Wd4& = 0,
w2=dy-RsinVdb=0.
(I)
The bilinear Frobenius forms are dwl (6) - 6w1 (d) = R sin (p(dW 6o - bcpd,b), d w 2 (6) - b w 2 (d)
(2)
= -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
THE GEOMETRY OI VECTOR 1IELDS
I24
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 is non-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 coordinate 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. they
possess 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 , h if
=
I
if the permutation is even,
-1 if the permutation is odd.
The exterior differential form of order p with respect to local coordinates is the following sum:
w E a,, It<
,,d"A...Adx''
<1,
or
w='I E a,, P ,I
,I,dx" A...Adx+r,
1r
where the summation is taken over the indices it ... ip which takes values from I to n independently and the form coefficients a,, 1,, satisfy the skew-symmetry condition a,, 1, = ±a,,
,,
('.is even and "-" if the perwhere "+" is assigned if the permutation mutation is odd. t such that Suppose that in D we have other curvilinear coordinates
'
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES 125
Consider the coefficients of w with respect to the system of coordinates u' . We have r du; A ...
w = li >2 al,...i,
Therefore, the coefficients have the form
A du°'.
of w with respect to a new system of coordinates
an,...°, = al,...i,
Oxi"
Oxi,
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, ry=Sdx'Adr2Adr3,
(1)
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, (2) 1
>2 bl,..i. dxf' A ... A dxi'. w2 = q'i,. .ip The exterior product wi A w2 of the forms w, and w2 is the following form of order
p+q: w, Awl = p
- 2 ai, ...,, bi, ...i dx" A ... A dX" Adze' A ... A dx
.
(3)
i, #
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 only changes the sign. Indeed, w2 A w, can be obtained from w, A w2 by transpositions of the 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 the linear form by dal,._.j,, namely the differential of the function a,,..;,. Then the
THE GEOMETRY OF VECTOR FIELDS
126
exterior differential of the form w of order p is the following exterior form of order
p+ 1: dw=
I
P1
A...A
dais...,, A
(4)
it ... i,,
The following rule of differentiation holds for the exterior product the forms w, and w2:
d(w1 Awe) =dw1 Awe+(-1)"WI Adw2.
(5)
Indeed, taking into account (3), we have dw1 Awl =
E{(dais...i,,)bj,...j,, dx" A... Adxj' +ai,...i dbj,...j, Adix'' A... AdxJ,
PQI
The first term is dw1 A w2. In the second term we make p transpositions of the differential dbi,,,,i, with the differentials dxi', ... , dx". Then this term gives the form for 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 identically zero.
ddv=0. Consider the expression for the exterior differential
E
CAW
aaxa1
W A ... A dxj'.
The coefficients of this form are the functions e1°. Hence daai,..."
dew I
=
1
(p +
0dx"t A dxa A &0 A ... A t:&'.
1)I
.
Adx" Adx" A... Adxi'
aloll,....,ip
axax
(6)
Since dx-6 A dxa = -dxa A dx' and the second mixed derivatives of ai,,..1, are equal to
each other, the sum on the right of (6) contains the term of opposite sign for each term. 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 exterior differential of some other form. In other words, if dw = 0 then in a sufficiently small neighborhood there is a form wl such that w = dw,.
VI CI OR
IN
AND RIFMANNIAN SPACES
LUC
127
To prove this, consider firstly the form w of order 1, say w = a, dY. Then dw = 0 implies all, c7v
Arc,
car'
_ 0.
This condition, evidently, is sufficient (and necessary) for the existence of such a function / that a, = /;,, i.e. w = dl. For the case of a second-order form a dr' A dr' the condition dw = 0 implies the system
Ou
0t4, (7)
As w, we take the first-order form w, = h, dr'. The equation w = &,I can be represented as i)h,
ah,
ox'
Ox'
- a,
We rewrite this system in a definite order. First we write the equation containing the derivatives of h,. then the remaining equations with derivatives of h, and so on. We get Oh,
_ 8h nzl,
car-
(8) I
_ Oh
Jxn
= a, _ l
We shalt solve this system step by step from the bottom. We take an arbitrary C2 function as h,,. From the last equation we can find hi_1 by integrating with respect to x". Substitute the functions h,, and into the previous equation. Suppose that we already found the functions Let us find b,-,. This function is included in a group of equations of (8). namely Oh,_ 1
_
8h,
_ (9)
Oh,-1
Oh
t'i?1"
(fit'-1
aln- I
This system is integrable if the compatibility conditions are satisfied: for any o, /3 > i the difference of mixed derivatives of b,_1 is zero:
0=
O-h,_ I
8 "'ar'
8h
0--b,_ 1
Ox'
6x" +OJ.Vl-I
0x days)
(10)
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 +
a. =
su.*, = 0
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 the case 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 is equivalent to the system ah,, t 9.V
,
ab,,, ,
- .. - U,,,
,P
(12)
OY1"
Introduce a multl-index i = (i1 ... i,,). where i1 ... i,,. It is possible to compare multiindices.
We say that i<jifi1 <j1 In the case ofil =j1 we set i< /1(/: <j2 and so on.
, 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 of equations containing the derivative of h1. These derivatives are with respect to from the .r"i 1, ..,x". Then. select the group containing the derivatives of h1 remaining equations. They are the derivatives with respect to x0:1, ... , x". The derivative of this function with respect to xP is already included in the first group of equations. We continue this process for the next functions h,. For every function h ,, the corresponding group of equations contains the derivatives of this function with respect to x" only for o > Indeed, if a < i,, then the equation We put the coefficients h,,
is already included into the previous group corresponding to the function b,, ,,, since
... 0) < (i1 ... ip). Therefore, we may assume that a > i,,. This implies that the group of equations corresponding to h,, , includes the functions of greater multi-index. The groups of equations contain. in general, a different number of equations. The function h,, ,,_,,, has no corresponding group. Therefore, we can take the function (i1
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
129
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 every group satisfies the compatibility condition. Suppose that we have already found the function 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
axn Obi,
r=1
abi,...,r...i, axi,
;,
abi,
P
-
E -Tx"-
where a, a > iP. The compatibility condition has the form A
a'-bi 0x°ax;3
a_bi
- aviax° aa:r;,
i,
P
.. ;,
+
aX3
a
abi, ... ,r... i,
aX (
ab;, ... a ... ip (13)
axe
where the r which are superscripts to a and Q mean that the corresponding index takes the r-th position. In parenthesis we have the derivatives of functions of multiindices
(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,
.
aX'
-
...
aX`
ab;,
P
A=1.k#r
... aA
.. ;r
ax'
are satisfied. On the right of (13) we obtain aa.i, ax;l
aa3i,
.
-
..
axi
aas;,. r>'...
axe-
+ v a2bi, ..,, 9 i..4
axi'axik
The last sum in the equation above is zero due to the skew-symmetry of b;,...;, with respect 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 we find the form w, . This completes the proof of the Poincare lemma. Observe that w, is not determined uniquely. For instance, it is always possible to add 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 which
puts the exterior form w of order p and the exterior form bw of order p - I into
THE GEOMETRY OF VECTOR FIELDS
130
correspondence. Let V denote the covariant derivative in Riemannian space and Vka;,..;, be the covariant derivative of the tensor a;, Set vka;,..i,.
Vi ai,...i, =
Since V'a;;,...;,_, is a skew-symmetric tensor of order p - 1, we are able to define the following 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 scalar function. 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 differentiation. If we take into account that the difference of second-order covariant derivatives of a tensor has an expression in terms of the tensor contractions with a Riemannian 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,
$1 '
Interchanging the roles of a and i, I and
= - R3jk/ sai
e,l gJk gl '
j and k we obtain
A I = Rlkja aieg'I gk'g°w = - RJ, ,k a0, g" '$g'r
gk,
= -A,.
Therefore A, = 0. In an analogous way we find that the second contraction with a Riemannian 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 -g,kOy/vk bi,... i, = g'kev,Ok aij;,...;, =
ajii,... i,
_ - gJk$'l vk v/ ajii,... i, = -gjkg'1(vl vk ajii, + R jkl aoii, ... i,
. iv
+ R11 ajn;, ... i, + R'i, kl ajin i, + ... + R`i,kl aji -J' .
(3)
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
131
Just in the same way as for the second-order form we can see that the contraction of gfkgd with the first two terms on the right of (3), where the Riemannian tensor involved, is zero. Consider the following expression: A3
=8
jk !
R°,kl olio. ip = RBi,k/ aji...ip 8
_ -R+,akr ajinii ..io e
egd
Let us make the index change as Q - k
jk
g
I
Q and
I-
a. Since
ainji, .rp = aij,. _...rp,
A3 = -1ti,kraajrn...ip g'9ngjkgI
Making the same index change and adding the obtained expressions, we find 3n e 3A3 = -(Ri,:9kr + Ri,kl3 + Rr,l9k) ajrni:.. ip
By virtue of the Bianchi identity we see that A3 = 0. In an analogous way we obtain zero in all consequent terms of (3). Therefore,
k
-g gjlVjV aij...ip
bi, ip =
Comparing this and the initial expression for bi,.. j', we see that the coefficients bi,...,p are zero. So, 6=w = 0. One can get a simpler proof of this fact by means of the Hodge
star operator. 2.11 Some Formulas for the Exterior Differential It is easy to represent the integrability condition of the Pfaff equation in terms of the exterior differential and exterior product. The following theorem holds. Cartan Lemma In order for the second-order exterior form F = aij dx' A dxj to be
equal to zero by virtue of the equation f = a; dx' = 0 it is necessary and sufficient that 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 linear form f. Suppose that f, ... fn is the required basis, wheref, =f Consider n
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
= Eblif j=I
THE GEOMETRY OF VECTOR FIELDS
132
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 a linear form w such that
dw=wAcp.
(2)
This is the necessary and sufficient condition for the Pfaff equation w = 0 to be integrable. In other words, in the integrable case the form dw is divisible by w. This condition 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}. Let N = (Ni) be a unit vector field of principal normals of the streamlines of the field r and k be the curvature of these streamlines. We may also adjoin the linear form
17 = Ni dx' for this field. Let r be the unit vector field. Let us show that if r is holonomic or, equivalently, the equation w = 0 is totally integrable then it is possible to set cp = ki7 in equation (2), i.e. dw=kwA17.
This equation can be regarded as the formula analogous to the Hamilton formula from Section 1.3. The form ki is
kq =
8a,
axi
;
aj dx .
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 the
coordinate curve x" to be tangent to r at some fixed point Po while the other coordinate 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
`' =
2
e
49an
4W) dx° A dx' +n-14 (8x°
Bart
- 8x") dr" A dac".
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
133
Since the conditionsexi a"" -OX" "" = 0 a,,6 < n - I holds (see (14) of Section 2.7) and n-I
dw
- E 8" 0=1
de A dx".
(3)
Next we consider w A kt. Since at P0 ak = ai = 0 for i, k < n, n-1
aaj wAki1=E8x"deAdx'.
(4)
!= i
Comparing (3) and (4) we conclude that w A ki1 and dw coincide with each other at P0. The coincidence of the forms does not depend on the choice of coordinate system because 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 correspondence its value on k vectors YI,... , Yk. namely, the contraction of the tensor al,..,;,, with the components of a multi-vector generated by Y1,. .. , Yk. Denote this value by w(YI, ... , Yk). Then YI''
w(YI,...,Yk) = a, ...,
.. Yk
...
Y1'k
.
...
Yk
Let us state a useful formula. Let w = a, dx' be a linear form, r the corresponding unit vector field, r, and rp the fields orthogonal to T. Then
V rn)
(r0, r,) = (r,
(5)
Indeed we have
dw(rQ,ry)
-
1
8a;
2 (8xJ
8a
rj
8x')
=21-aim
ral
+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 Simplex
Bounds" 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,
134
THE GEOMETRY OF VECTOR FIELDS
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 of vertices if necessary: t"(Mo, MI,_, Mn). Define the simplex orientation as the orientation in E" generated by the basis MOM,,..., MoM". The odd permutation of vertices 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 face obtained by deleting Mi, where V' means the omission of Mi. The orientation of a simplex (Mo,... , V',... Mn) is called induced by the simplex t"(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 simplicial complex in El of order n is a finite or countable set of simplices satisfying the following 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 say that the smooth manifold M" is triangulated if it is mapped onto some polyhedron homeomorphically. We shall call the image of a simplex in M" the curvilinear simplex. The fundamental hypothesis of combinatorial topology ("Hauptvermutung")
asserts that any two triangulations of homeomorphic polyhedra have isomorphic partitions. In recent years this hypothesis has been refuted. Suppose that every simplex in a simplicial complex is oriented. Introduce the incidence coefficient of the two simplices in and t"'. We say that the incidence
coefficient (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 the given one and (t"t"-1) = -1 for the opposite case. Let t"-2 be some subsimplex in t"
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
135
and be the (n - 1)-simplices in t" which include in-'. Then for any choice of 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 will not change if we change the orientations in to" and in-'. Suppose that to-' does not contain the vertex Mo. while t," -I does not contain the vertex MI. Orient to" and in-1 in such a way that
(to-I t" `) = 1,
(tn-I to-2) = I.
Then the simplices t" and in-' have the forms n-I
= (MIM2...Mn), tj"-' = (M0M2 ... M"). to
Under the conditions posed on the incidence coefficients the formula we need to prove 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) is reduced 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
THE GEOMETRY OF VECTOR FIELDS
136
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 multiplier which is the Jacobian of the coordinate transformation 1(" ,,). It is possible to define the integral of the form w over D C R" in the usual way as the integral of a, .
1w= R,
f D
The value of the form integral does not depend on the choice of the coordinate system. 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 is induced. We shall call this form the induced one and denote it by w: Of
yi ...
1
/,
Of
A...Ady"
'
Thus, form coefficients are, evidently, skew-symmetric with respect to indices j,,. .. , j,,,. The integral of w over the m-dimensional surface Fm is, by definition, the integral of w over F"'
F'
F"
Consider, as a particular case, the form w of order n - I in some n-dimensional oriented 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 the following 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 differential and boundary. We prove this formula in the case when R" is the atfine space of coordinates x1, . , " 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 orientation. We shall represent the simplex as (An... ..4,,). We are able to represent any subsimplex 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 equation x' = 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' formula for 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
f do
On the face D = (A,..
. A,;)
wi =
Jii + fwi 1),
rn,
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
THE GEOMETRY OF VECTOR FIELDS
138
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 its orientation as part of the boundary of D. So, we have
Jwl=_Ja2...n(Ox...x")dxA...Adx" DD -...,X2,...,x")ar?
a2...n(1-17
+J
A...Adf
Do
',...,xn)dx'Adx2A...Adx", a,...n(x',x,
87
_ no 0
where xa = 1 - x2 -
- - x". The latter expression is the integral over D. On the
other hand, consider the exterior differential of w1:
dwI =da,...n_IAdX2A...Adr"= _
a
a2...n-1dx'Adx2A...Adx".
Hence Jwi DD
=Jdthi. D
In an analogous manner one can prove this formula for any other wl. For the case of a curvilinear simplex in some manifold R" the proof can be reduced to the case of a simplex in affine n-dimensional space. Further on we assume that D is triangulated, and also that the boundary of D is
represented as (n - I)-simplices of triangulation. The theorem that any smooth manifold 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 consistently then for any (it - 1)-simplex in a common boundary of two distinct ii-simplices the orientations induced on it from those two simplices are opposite. Therefore, in the sum under consideration the integrals over (n - I )-simplices interior to D cancel each other out. The remaining integrals are ones over the (n - I )-simplices on the boundary 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 the
partition is already endowed with an orientation. Denote the i-th simplex of dimension 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 of simplices 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 the boundary 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 is sufficient to prove this for the boundary of a single simplex. We have ddt,' r
-)tR
i
`=ED,; I.
r
In Section 2.13 we proved the equality ' t t'-I
Dtr
)(t/'-1141-2)
= 0.
The chain x' with zero boundary is called closed or rite cycle, i.e. for the case of a
closed 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;,
140
THE GEOMETRY OF VECTOR FIELDS
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 form the Abelian group Z, with respect to the adding operation as above. The quotient group of Z, with respect to the subgroup of cycles homologous to zero is called the rdimensional Betti group. If T, is a subgroup of Zr formed by the cycles homological
to zero, then the Betti group H, is
H, = Zr/T,. In an analogous way the cohomology group can be defined. For each cell jr the coboundary is the (r + 1)-dimensional chain of the form
bt, = (t; t +1)t +1
i 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 called cohomologous to zero. If a coboundary of the chain x' is zero then this chain is called a cocycle: bx' = 0. The set of all cocycles, being endowed with the natural operation of adding, forms the additive group denoted by Zr. Let T' be a subgroup of Zr formed by the cocycles homologous to zero. Then the quotient 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 the circle 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 the orientation of the arc. The coboundary of A is formed by two arcs, namely y, and y3 giving
bA =Y3 - Yl Analogously, bB = y, - y2, bC = y2 - y3. Hence, the arcs y3 and Y2 are cohomologous 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 following de Rham theorem holds:
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
141
The quotient space of closed forms of order p with respect to the subspace of exact
forms of order p is isomorphic to the p-dimensional cohomology group with real coefficients 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 Riemannian manifold of class C' without a boundary. Suppose that M" is decomposed into connected subsets which we shall call the fibers of foliation or the leaves and denote them by L. Let Mn be covered by the set of neighborhoods UQ each endowed
with a coordinate system x1,. .. , x" (i.e. there is the C' regular homeomorphism v,, : U V C E" of U. onto some domain V. in E") such that the intersection of each 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 is called the codimension of the foliation. Each fiber is a submanifold of M". Thus, the foliation is structured locally like a family of parallel planes of dimension p in Euclidean
space E". If M" is an analytic manifold and V,, are analytic functions then we say that the foliation is analytic. Among the closed surfaces M2 only the torus and the Klein bottle admit foliations without singular points. For n = 3 the foliations of class C' exist on any manifold. However, the manifold containing an analytic foliation must satisfy the condition with respect to its fundamental group. Haefliger stated that a compact manifold with a finite fundamental group does not admit analytic foliation of codimension 1. The analytic requirement in the theorem is essential. Reeb constructed the foliation of class Cx on the three-dimensional sphere S3. We describe this foliation here after Lawson. At first, construct the foliation of the toroidal body K3. The boundary of this body is a two-dimensional torus which we denote by T2. It will be one of the fibers, moreover the limit fiber for the other fibers. As Lawson said figuratively, each fiber within K3 is like a snake that eats its tail. While the phrase above expresses everything you need to understand, we continue the explanation. In the xy-plane we take a zone between two parallel straight lines x = -1 and x = 1. Construct the line foliation in it. Let us be given a curve y = f (x) defined in -1 < x < 1. The function f (x) is supposed to be of class C-, infinitely large when IxI 1 and for all derivatives f(k) (X)
lim f ck 1(x) = oc,
IYI-1
k = I , ... , M.
We define each line of zone foliation with the equation y(x) =f(x) + c, where c is an arbitrary real number (see Fig. 25). Revolving this foliation around the y-axis, we obtain the foliation of a cylindrical body. Let us map the cylindrical body onto the toroidal body K3 in the following standard way. Consider the axial curve (the circle S') 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
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THE GEOMETRY OF VECTOR FIELDS
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 body K3 as a product S' x D2. Introduce the polar coordinates (p, 0) in D2 taking P as a 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 the
cylindrical body onto K3 setting s = y, p = r, 0 = V. Every torus point will be covered infinitely many times under this mapping; also, if the points a, and a2 of a cylindrical body go into the same point of K3 then the difference of corresponding ordinates satisfies y1 - y2 = 27rk, where k is an integer, while the corresponding parameters (r, lp) coincide. Let us show that under this mapping the foliation of the
cylindrical body goes into the of K3 (see Fig. 26). Locally, the mapping t' is a homeomorphism. Suppose that the fibers 11 and 12 of the cylindrical body went into the surfaces rb(11) and ap(12) which have a common point b. Then the inverse images
of this point, say a, and a,, differ in ordinates as y, - y2 = 27rk, while the polar parameters are the same, r1 = r2, W1 = V2. The fibers 11 and 1, have been obtained in
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
143
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 a common boundary, namely the torus T2. Within each toroidal body we construct the 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 the fibers, namely the axial curves of the toroidal bodies. Later, the theory of foliations was developed in topological and geometrical directions. Novikov [73] proved that every codimension-one foliation on S3 has a compact leaf. Thurston [100] proved the existence of p-dimensional foliations on any opened manifold. These foliations exist on closed manifolds if there exists a field of p-dimensional tangent spaces. Reviews of topological results from the theory of foliations 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 spaces where the leaves were totally geodesic, minimal, umbilical and so on. It was proved that the codimension of a totally geodesic foliation on a complete manifold of positive curvature is more than 1. Moreover, the Frankel theorem is known: in a compact manifold with positive curvature, two compact totally geodesic submanifolds whose summed dimensions are not less than the dimension of the manifold do not have an empty intersection. So the dimension of a totally geodesic foliation with compact leaves on a compact manifold with a positive curvature is less than half the manifold's dimension. Ferus [103] proved the following interesting theorem in which the completeness of the manifold is not necessary: if in some domain of the manifold Mn` there exists a totally geodesic foliation with dimension v of class C2 and with complete leaves and a mixed curvature of M"+" tangential to a leaf of vector X and orthogonal to a leaf of vector Y, K(X, Y) = const > 0,
so v < p(n) , where (p(n) - 1) is the maximal number of linearly independent vector
fields on S".. Rovenskii [104] showed that the condition K(X, Y) = const is essential and he generalized 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 manifold M" of class C°°. The foliation is called transversally orientable if there is a smooth
144
THE GEOMETRY OF VECTOR FIELDS
vector field on M" transversal (i.e. not tangent) to the foliation leaf through the initial point. Consider the transversally orientable foliation. Then in every neighborhood of a point the foliation leaves can be defined by the Pfaff equation w, = 0. The form w1 must satisfy the integrability condition w, A dw1 = 0. This means that dw1 is divisible by 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.
(3)
If we differentiate this new relation then we obtain
W1 A(w2Aw3-d,,3)=0. Consider the third-order form 1 = w1 A w2 A w3 which was introduced by Godbillon and 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 nonholonomicity of some vector field. Since is linear, it can be written as w2 = P dv I + Q d.2 + R dr3. If we introduce the vector field v = {P, Q, R} then 11= -(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 only on 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 by some differential form. We have
d i=w1Aw2=w1Aw2. From this we conclude that w, A (w; - w2) = 0, which means that w2 - w2 is proportional 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
to the foliation. w1 A dw1 = 0. Also, w1 A (ko, = 0,
Since w1 corresponds 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 of H3(Af, R). This class is called the Gadh(llon-Vev charucieristic cuss. Denote it by 10)(F). Now we are going to establish the independence of [fl](F) of the concrete choice of w1. i.e. we show that it depends only on the foliation F itself. The form w1 is definite
to 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 A dw, = 0,
w1 Ad/'Aw1 =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 the differential of some second-order form. Thus, the cohomology class [ft](F) depends on 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 a manifold V onto M. Then the inverse images of fibers of the foliation F define a foliation in t : Denote this foliation by f' (F). The mapping / induces a mapping
THE GEOMETRY OF VECTOR FIELDS
146
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 GodbillonVey 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 of C2-regular foliations and Cl-diffeomorphisms. (2) If F and F' are the foliation of codimension 1 in the same manifold M which are differentiable and homotopic to each other then their Godbillon- Vey classes coincide.
(3) If the foliation F is given in a three-dimensional manifold M3 then the integral of St gives some number -y, namely
M;
M3
This number is uniquely defined by the foliation and is called the Godbillon- Vey number. This number is not necessarily an integer. (4) Let us settle the relation of ft to both the curvatures of lines which are orthogonal to the fibers and the second fundamental forms of the fibers. Denote the lines orthogonal 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 be represented 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. Let a, $ > 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 codimension 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 any vector 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
-VrTi).
(r2.-k,.-1tn-1
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 ,
ct
where r is some unimportant coefficient. So. we find the following cxprerslrm for the Godhillon- /er form:
it = -l, ,.:2 A d(2 = k (A'2(z A <1 A 4i -F 1
,
!(T2. T,.
A (I n
n=3
This expression was established in [691.
2.18 The Expression for the Hopf Invariant in Terms of the Integral of the Field Non-Holonomicity Value
Let f: S' -' S-' be a C2-regular mapping of spheres, We are going to find the expression for the Hopf invariant ry of a mapping, in terms of the integrated nonholonomicity value of some vector field given on S. Let ul, u2 be the local coordinates in S- and ds- = g,, du' du' be the first fundamental form of S- with respect to these coordinates. g = det 1Ig 11. Let x1,x=, x3 be the local coordinates in V. The mapping / can be represented with the functions t1 =1e(x1,x_1,x3),
i=1,23.
THE GEOMETRY OF VECTOR FIELDS
148
Set
uI
uI
u'
u!'.
Let us introduce in S' a second-order skew-symmetric tensor a",3 = 4!, tensor generates the second-order exterior form, namely
This
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 8x3
8xI
+ X21 dr1 A dx2 A dr3. 8x2
I
Since f is a function of u' which are the functions of x" on the other hand, Oat
8x"
_ 1 (8f 8ukl +81;11 47r
8uk 8x" 1
81"
It is easy to find the following equalities: 8123
0131
Uxi
u0
8u' 811i 112 = 141 8x1 123 + axe I31 + C7r;
u111
8u'
8112
8xi + 8x2 + 8x' = 0'
Iu,
u?
u'rt
urs = 0. '4 1
These equalities imply dw = 0, i.e. the form w is closed. Since the cohomology group
H2(S', R) is trivial, then by the de Rham theorem there is a global linear form w, = 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 integrated non-holonomicity value of the field b = {b;) -y =
fwi
A dw, =
S3
J(b, curl b) dV,
(1)
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 mapping f can be approximated with the normal mapping. Further on we assume that f is already normal within each cell E2. Let Mq = f - I (q). As f is normal, the set M. consists of one or several closed curves. Let us orient each curve of Mq as follows. Let p be a point in M9. Select the basis e,, e2, e3 at p which produces a positive orientation of S' and such that e3 is tangent to Mq, while the directions of el and e2 go into the directions of e, and e2 under 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 orientation in Mq.
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
149
Let gl,g2 be points in E, 2. Connect them by a smooth curve I in Ei. Then Mr = f --I (I) is a surface with the oriented boundary LIM, = M, - My, . The whole surface is formed by My, where q E I. Show that
J w= Jwi. M,
M92
Indeed, by Stokes' theorem we have
Jdwi= tth
Jwi_ Mat
Jwi.
(2)
M
Let v', v2 be local coordinates in Mi. Set
T°'; =
i x' W 1 The coordinates ul, u2 of a point in S2 are the functions of v1, v2 along M,. Then
I
u1, U2) VI
_
1 ul x7,
Ju2.
VZ
Y
.,r
x
xir
u'x 1 Xs
1
r
2
u2, r JC
+
7-1 I.
Since the surface Mi goes into the curve under the mapping f, T("; ) = 0. Consider the integral on the left of (2):
J dwl = M, Mi
Jw =
= 4rr
1
- MiJ
f IgFi Lv A
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 q is the same number - some invariant of the mapping f. Let us show that this number coincides with the Hopf invariant ry. Let M. be a base of a piecewise regular surface F and t, 77 the local coordinates in F. Then by Stokes' theorem we have
f
wl
rdwl = T7r_ j
1
(\
,7l
/ dS A
M
As the boundary of F goes into the single point q, then f(F) covers the sphere S2 some integer number of times (more precisely, the total oriented area of the mapping
THE GEOMETRY OF VECTOR FIELDS
150
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 a topological product of M. and some surface for which the mapping f maps onto E,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 curvilinear coordinates u1, u2, u3 and the mapping f translates the point p into the point q with coordinates 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 the domain T as any iterated one as follows: /' T
Adwl = J
-rte
du1 Adu2 / c3(u',u-,u3)du3 = J''-TJwi.
e?
t_
0
st,
Since the integral over Mq of w1r does not depend on q, r wI Adw1
J
r
=-yS(E'') 42r
where S(E?) stands for the area of E! in S2. If we sum over all domains E.' then we obtain (1). The theorem is proved. S. Novikov [741 obtained some many-dimensional generalizations of the Whitehead formula. 2.19 Vector Fields Tangent to Spheres
Consider the vector fields in E" which are tangent to a sphere S"' at each point x E S"-I from the families of spheres with a common center. It is well known that an
even-dimensional sphere admits no regular tangent vector fields. Therefore, we consider the vector fields on the odd-dimensional spheres. The vector fields nI (x), ... , nk(x) on a sphere S"-' are called linearly independent if they are linearly independent at each point x E S"-1. The Hurwitz-Radon-Eckmann
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
151
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 a product of an odd number and a power of 2, i.e. n = (2a + 1)2". Divide b by 4 with
the remainder, i.e. represent it in the form b = 4d + c, where 0 < c < 3. Set p(n) = 2C + 8d. Then there are p(n) - I linearly independent vector fields on the sphere S". Adams [94] proved that this number is maximally possible. We denote the number of linearly independent vector fields on S"- 1 by k. Below we 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 explicit form, avoiding heavy theory. This also gives us a rich set of non-holonomic fields in E". It is easy to find a single vector field on the odd-dimensional sphere. Suppose that S"-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 consequent order. We define the vector field a = (-x2, xt , -x4, x3, ... , -x", x"_ 1) at the point of a sphere having coordinates x1,... , x". Let us construct three mutually orthogonal vector fields on S3. Denote a position
vector 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
x
xl I over each pair. X;
The splitting under consideration gives the vector field a = (1X2i x), -x3, x4). Next
we take the splitting into pairs (x1,X3), (x2,x4) and construct the vector field b = (-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 procedure above, but it is not orthogonal to a. Taking the pairs (x1, x4), (x2, X3), we find the vector field c = (-x4, -x3i .r2, XI). Write the matrix formed by the components of r, 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 centered at the origin. They are mutually orthogonal and of unit length. We shall say that each of the vectors a, b, c defines some permutations over coordinates, say A, B, C.
THE Cii:OMCTRN OF VECTOR FIELDS
152
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 groups of four coordinates: (xI, X2, X3, xa), (x5, x6, .Y7i .Yx). For each of these groups we can
construct the fields analogous to those on S3. We obtain four of the vector fields on S3 by joining three-dimensional sphere fields. i.e. we define them as followsa = (al, a'-) _ (-x2, Xl, -.Y4, x3, -x6,.Yi, -xx, X7), b = (bi, -k,) _ (- x3, x-1;X1, -x2,Y7. -xx, -X5e lb),
(-Xa,
C = (Cl , CA
-X2, X1, -Xx, -X71 x6, X5),
where a,, b e, are the vectors on S3. constructed as above involving the coordinates of 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 the
groups (xi,x2,.Y7,xx) and (x3,X4,Perform the permutations B. C over each of 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
a
-x_
b
-xi
Y4
C
-X4
-X=
x2
.r
xz
Y4
Xc
-YX
X7
Xx
-X4
Y3
-X,,
.xS
-xx
X7
X1
-X2
X7
-V$
--Vs
X6
N.,
X1
-XK
X6
X5
-X7
XK
X1
-x7 -xZ
X7
X.
X
d
-xc
Yh
e
-Y,,
-TS
f
-.Y7
XK
Xc
g
-X
-X7
-x6
X
-X -XS
xz
-X4
-X4
-X,
-x, X1
-X=
X4
X1
xA
X1
X,
The components of this matrix satisfy a,, = -a for i 70 j. It is easy to check that all constructed vector fields on S7 are mutually orthogonal.
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
153
As we know from the table at the beginning of this section, the sphere SII only admits three linearly independent vector fields. To find them, we split twelve coordinates xi into three groups of four and perform the permutations A, B, C over each of them. Write the corresponding matrix of components of vector fields on SI 1: r
X1
X2
X3
X4
X5
X6
X7
a
-X2
XI
-X4
X3
-X6
X5
-X8
X7 -XI0
h
-X3
X4
X1
-X2
X7
-Xg
-X5
C
-X4
-X3
X2
X1
-X8
-X7
X6
X6 -XII X12 X5 -X12 -XII
X9
X8
XII
X12
X9 -X12
X11
X10
X9 -X10 X10
X9
Evidently, a more general statement holds: if the sphere Sn-1 admits k linearly independent vector fields then the sphere S"9-I admits not less than k vector fields. As 23 = 8.3 - 1, S23 admits seven vector fields which are the three times repetition of the seven vector fields on S7.
Let us construct eight vector fields on S15. Since we can split the sixteen coordinates into two groups by eight coordinates, namely (xl, ... , xg) and (X9i ... , X16), and we have already constructed seven vector fields on S7, by double repetition of these fields we obtain seven vector fields on S 15. If a;, bi, . . . , g; i = 1, 2 are the corresponding vector fields on S7 generated by the splitting mentioned above then 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 other fields. 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 respect
to the groups of coordinates given, i.e. the vector obtained by means of the D-type permutation (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 that
h is not orthogonal to b, d, f. To satisfy the condition of orthogonality we ought to change the signs of the last eight components of these vectors, i.e. we ought to set b = (b,, -b2), d = (dl, -d2), f = (fl, -f2). Present the table formed with components of mutually orthogonal vector fields on S I3 and r: r
X1
a
b
-X2 -X3
C
-X4 -X3
d
-X5
e
-X6 -X5
X8
f
-X7
g
-X8 -X7 -X6 -X5
h
-X9
X2 X1
X4
X3
X4
XS
X6
X7
X3 -X6 X5 -X8 X1 -X2 X7 -X8 -X5
-X4 X2
X1
-X8 -X7
X6 -X7
Xg
X1 -X2
X7
X2
XI
X5 -X6 -X3
X4
X1
X3
X2
X8
X10 -XII
X4
X12 -X13
X6
X8
X9
X7 -X10 X6
X10
X11 -X12 -X9
X5 -X12 -XII
X3 -X4 X13 -X14 -X4 -X3 -X14 -X13
X14 -XIS
-X2
X11
X9 -X12
X10
X15
X16
X11 -X14
X13 -X16
XIS
X10 -X15
X16
X12
XIS -X16 -X9 X16
X15
X15 -X16-X13
X14
X1 -X2
X14
X9 -X16 -X15
X1 -X16 -X15 -X14 -X13 X16
X13
X3 -X4
X13 -X14 X14
X13
X10 -XII
X12
X9 -X12 -XII X11 -X12 -X9 X10 X10
X12
X11
X5 -X6
X10
X9
X7 -X8
THE GEOMETRY OF VECTOR FIELDS
154
We can construct the vector field i by means of a permutation of E types over the second 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 we can take i. Consider the geometrical properties of the fields just constructed. Each of these fields is non-holonomic. Let us show, for instance, that a is a field in E" which is non-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.
(1)
The exterior differential of this form is
dw=2{dX,
(2)
Ad.Xn}.
Exclude one of the differentials xi from (1), say dx,, and substitute into (2); we obtain
I
dw= 2{
AdXn i=3
JJJ
where ,pc are some functions of x2, ... , xn. The exterior differential dw is not zero identically 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 the position vector of the streamline satisfy dv1
ds
=
2,
dx2
ds
= x, ,
dx3
ds = -x4,
d.X4
ds = X3,
i.e. d r/ds = a. Hence, d2x, = -X1 i
j2
, d2X4 ;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 the
second kind of the streamlines of one of those p=
x+
fields,
say n = a/Ial. Set
+ xn. The following theorem holds:
Theorem Odd symmetric polynomials of principal curvatures of the second kind are zero, while even symmetric polynomials are Sk = -L Ci ,11,.
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
155
Let us show that all the principal curvatures of the second kind are purely imaginary except one of them which is zero. Denote by & the components of n. The matrix IIaE II has the form xixj
I
P
A= x x,_i
I
P3
P
P1
-i P'
X.X.
:!Lc,, P,
P
P,
P,
To obtain a more convenient form of the characteristic equation IA - AE] = 0 we introduce the following vectors: A
P
-A
P
O
rl
,
0
72 =
,
xn
0
O
0
,
Tn =
0
A
We shall denote the determinant by brackets. Then the equation JA - AE I = 0 can be expressed as
=0. Using the rule of determinant decomposition, we obtain [TI,T2...... r
-XI[TI,l,T3,...,Tn]+... =0,
]+x2[I,T2,...,T,,]
where dots stand for the sum of determinants each of which contains only one column I taking the position of the column Ti. Those determinants are similar to the
two 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:
+II
(n-2)/2
(XI
x2\ -3 Ip , P
ln-zl/2
(A2
[TI,I,T3,...,T] _
+
)
x1/
(_x2A+_)P-3.
156
THE GEOMETRY OF VECTOR FIELDS
The characteristic equation with respect to A has the form
(A2+_) 1
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 the imaginary roots are of multiplicity "-. 2The theorem is proved. Let us find the curvature vector P which has been introduced in Section 2.3 for the case 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. Since every streamline of the field a is a great circle on a sphere, k1 = -r/p2. The field is
constant along the sphere radius, so that k2 = -V.n = 0, k1 = 0, i > 2. Take into account that the odd symmetric polynomials are zero. So, P = S"-2k1 = -r/p". By means of the vector P we can find the degree of mapping generated by the field a/p of the sphere S"-I onto the unit sphere in E":
0=
1 wn-I
J(P,v)dV=-1.
Consider now the question of the holonomicity of pairs of fields. We shall say that the distribution of the pair field a, b is holonomic if there are (n - 2)-dimensional submanifolds orthogonal to both a and b. Otherwise we say that the distribution is non-holonomic. In papers we also meet another definition: the distribution of planes spanned on a and b is called integrable if there are two-dimensional submanifolds tangent to those planes. 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 can prove this fact both analytically in terms of exterior forms and geometrically. The fields c and r are orthogonal to a and b. Therefore, the planes of great circles in the sphere S3 which are tangent to the field c are orthogonal to a and b. This means that the 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,
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES 157
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 distribution
is non-holonomic. Thus, we have an example of two distributions of two-dimensional mutually orthogonal planes in E4 such that one of them is holonomic, while the other is not. The case considered above is exceptional. In a space of larger dimension the distribution of vector pairs is non-holonomic. Consider, for instance, the distribution of 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 find 8
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, the latter will contain the term (a3b5 - b3a5)dx3 A dx5. In collecting similar terms this term does not vanish. Therefore, dw1 0 by virtue of w, = 0, w2 = 0, i.e. the distribution 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 from the Pfaff system and substitute into dw1. In the Pfaff system w; = 0, i = 1, 2,3 a3 denotes the three-element column formed with the coefficients of dx1. The determinant of the matrix of the system with respect to dx1, dx2, dx3 has the following form: -X2 A = [a1a2a3] = -x3 -X4
X,
-X4
X4
X,
-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 dx4 and dx5 in the expressions for dx1, dx2, dx3 Write
dx1 =Q(Adx4+Bdx5)+---,
THE GEOMETRY OF VECTOR FIELDS
158
I
dx2 =(Cdx4+Ddxs)+ , dx3 =
where A,-, F have the form A = -[a4a2a3], C = -[a1a4a3],
B = -[asa2a3],
E_ -[ala2a4],
F= -[ala2asj.
D = -[alasa3j,
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
A=-
C=-
X3
XI
-X4
-X2
X4
X1
X1
-X3
X2
-X2
X3
-X4
-x3 -x2
X1
-X4
XI
X2
-X6
X1
-X4
X4
X1
-Xg
-X3
X2
-X2 -X3
-X6
-X4
X7
X1
-X4
-Xg
X2
-X2
XI
-X6
-X3
X4
X7
-X4
-X3
-Xg
B=- X7
D=F= -
= -X1U,
= -x2 U,
= X2X4X6 + XjXg - X3X4X7 + X4X8 + X1X3X6 + XIX2X7,
= X22X7 - XIX4X6 + X3X4X8 + X42X7 + XIX2X8 + X2X3X6,
_ -X2X4Xg + XIX4X7 + X3X6 + X4X6 + XIx3Xg + X2X3X7.
After the substitution, we obtain
AD-BC-FA =x4x6U2. Therefore at all points where x4, x6 and U are different from zero, the coefficient of dx4 A dx5 in dw1 is not equal to zero. If we set dx6 = dx7 = 6x6 = 6x7 = 0 then all the forms dxj A dx6 and dxj A dx7 are zero. So, dvI $ 0 by virtue of the system, i.e. the Pfaff system is not integrable.
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
159
Each of the vector fields constructed is a Killing field on a sphere. We get the proof from the definition. Such a field defines an infinitesimal mapping r -+ r of a
Riemannian manifold onto itself under which d rz differs from d r-' by the infinitesimal of a higher order than (bt)2. Put the point of the position vector r into correspondence 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 can take 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 fields constructed as the field . It is sufficient to check that (dr, 4) = 0. Each field f at a point with the position vector r is obtained from r by component transpositions and sign 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 a negative sign and once with a positive sign. Therefore, (dr, ek) = 0 and, as a consequence, { 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 of surfaces (i.e. through each point of the ball one and only one surface passes). We shall suppose that each surface of the family corresponds to a unique value of the parameter t from (a, b] and vice versa, and also that this correspondence is continuous. Moreover, suppose that each surface is connected and divides the ball into two bodies such that in varying t from a to b the volume of one of those bodies varies from zero to the volume of the whole ball D". We shall say that such a family fills the ball regularly. Nevertheless, the surfaces themselves may have singularities. For instance, we may take as such a family the set of circular cones with a common axis and the same angle about the vertex if this vertex is within the ball and completed with the family of frustums of the cones if the vertex is out of the ball. Two examples are: the family of coaxial cylinders truncated by the ball; the family of level surfaces
of some differentiable function p(x1 i ... , x") with a non-zero gradient within the ball. The surfaces which correspond to extremal values oft can be degenerated into a point or a curve. We can pose the following simple question: is there a surface among those surfaces
having an (n - I)-dimensional area greater than or equal to the area the diame rral section 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 of the 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 R
regularly. Then there is a surface of the family with an area greater than or equal to
THE GEOMETRY OF VECTOR FIELDS
160
R"-1 C(n), where C(n) = 1(21 " - 1)/2 is constant and (n - 1)-dimensional sphere.
I
is the area of a unit
Let VI and VZ be the volumes of the bodies DI and D2 which are the parts into which the surface divides the ball D". The boundary of each Di consists of the domain in a boundary sphere having an (n - 1)-dimensional area F; and the surface
under 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 Vi"-I)/"
2F>
+ V2"-1)/") - (F1 + F2).
Since the family of surfaces fills the ball regularly, the volume of DI varies continuously 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 the whole boundary sphere. Hence,
2F>
2"a"R"-12(n-1)1"
-w"-IR"-1.
I = w"_ 1. Therefore, 2F > w"_ I R"- I (21/n - 1). The theorem is proved. Note that 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 each surface 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 >
above as div n < -
2(n ,1 ) K° I
.
2;." ;) Ko
(or from
if we integrate this inequality over the ball D', we
obtain
FHence, 2(n - 1)
-1
Ko.
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
VIKo
(n - 2)
2(n
This assertion can be extended to the case of Riemannian space, but note that in Riemannian space it is impossible, in general, to get the upper bound on the radius of a ball for an arbitrary K° where the regular family of surfaces with a second symmetric function S2 > K° > 0 exists. Indeed, in Lobachevski space L3 with curvature
VECTOR FIELDS IN MANY-DIMENSIONAL EUCLIDEAN AND RIEMANNIAN SPACES
161
-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 pieces of geodesically parallel spheres of S2 > 1. But if Ko > I + E, e > 0 then it is possible to find the upper bound on the radius of a ball D3 C L3. However, we fix the ball radius to be I and watch the influence of Ko on the curvature of the space. Integrating the inequality diva > CIK-o over D3 in Riemannian space, we obtain the estimate < 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 in terms 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 we interpret the curvature of the family of surfaces as some loading on the domain of space then we may say that if this loading is greater than some critical value then the metric of ambient space will change, i.e. the Euclidean space will change into the curved one; the value of curvature will be determined to some extent by the value of loading. This influence is similar to the deformation of a flat membrane under the action of interior tension. The membrane loses its flatness, swells into space and gains curvature. The estimate obtained means that if Ko increases, then after some critical point, the space curvature Q also increases. Consideration of the families of the surface in a ball leads to some problems. Let us formulate two of them: (1) Suppose that in a ball D3 C E3 of radius R there is a regular family of C2-regular
surfaces having negative Gaussian curvature K < -1. Does the estimate from above 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 bounded from above. (2) Suppose that in a ball D" the regular vector field a is given. Is there a regular homotopy which preserves the vector field on the boundary of the ball and turns it into the integrable one?
A more general formulation involves a domain G in Riemannian space and a given distribution 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 the totally integrable one? Wood [99] proved that every field of 2-planes which is transversally orientable is homotopic to the totally integrable one. The first example of 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 the totally integrable one. Note the following result obtained by Thurston [100]: every (n -1)-plane field on a closed manifold M" is homotopic to the tangent plane field of a C'°, codimensionone foliation.
162
THE GEOMETR% OF VECTOR FIELDS
2.21 Summary
We have studied the local and global geometrical characteristics of vector fields in n-dimensional Riemannian space. We have also found some relations between these characteristics and topology, the theory of ordinary differential equations and fluid mechanics. In general, the vector field was assumed to be regular and to belong to classical class C2. However, in certain cases the field under consideration could have singular points, which were interpreted as sources of curvature of definite power. In recent decades, the ideas of non-holonomic geometry combined with the theory of generalized functions has allowed a theory of currents on manifolds and varifolds to be constructed. Nevertheless, further development of the classical theory of vector
fields is also very important. I hope that this book will help readers to prepare themselves for new discoveries in this area of mathematics.
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258-286.
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Subject Index A Adjoint vector 83 Algebraic vector field asymptotic lines 22
H Haefliger theorem L41 Hamilton formula 13 holonomic vector field 3 homotopy classes AS Hopf invariant 45 Hurwitz-Radon-Eckmann theorem 1551
58
B
Basic invariants 25 Bernoulli function 83 Bianchi system 22 Bushgens theorem 85 C Caratheodory-Rashevski theorem 82 Closed stremline 64 complex non-holonomicity
D Degree of mapping Dupin theorem 22
J Jacobi theorem 2 K Killing field 66
111, 112, 113
L
Lame equations 28 family 23 Laplace net 22
451
M
Mean curvature, torsion 42
E
Efmov's theorem 60
lemma 60 N
Euler equations 82
Normal curvature 9 24 G Gauss-Bonnt formula 32 geodesic torsion 42 Godbuon-Wey invariant 51
P Principal curvature 10 Poisson bracket 3 169
SUBJECT INDEX
170
R
Reeb's example
L41
V
S
Singularity sources
total torsion 51 triorthogonal family 71
52
52
stable limit cycle 64
T Thurston result total curvature
lb_1
11
Vagewski stability criterion Vagner interpretation 3 vector 92, 97 W Whitehead, theorem Wood theorems
L61
148
64
Author Index Bernoulli, L 82
G Gauss, K.
Bianchi, L. 79 Blank, Ya. xii
H
B
Bonnet, 0. 39 70 Bykov, V. M.
Haefliger. A. 141 Hamilton, W. R. 13 Hodge, W. W. D. 131
xii
C
Caratheodory, C. xii, Cartan, E. 131 Chaplygin, S. xii
J Jacobi, K. G. J.
i87
2
K
van Kampen, E. R. Killing, G. 23
D Darboux, G. xii, 14 Dupin, F. P. Ch. 21
xii
L
Lame, G. 23 Laplace. P. S. 22 Lawson. H. B. L43
E
Efimov, N. V. Euler, L. 82
39. 69
xii, 59
Lie, S.
xi
Lilienthal, R. xii Lobachevski, N. xii
F Ferus, D. L43 Frankel, T. T. 143 Freudenthall, IL 1W Frobenius, F. G. L. 116 Fuks, D. 143
M
Minkowski, H. 34 N Novikov, S. P.
171
143. 154
AUTHOR INDEX
172
P Pfaff, J. F. 116 Poincare, A. 126 Poisson, S. D. 3 Pontryagin, L. S.
T Tamura, F. 143 Thurston, W. P. 143, 164 Tondeur. P. 143 101
V Vagner. V. xii. 3, 92, 97 Voss. A. xi Vransceanu, G. xii
R
Rogers, G. xii Rohlin, V. 104 Rashevski, P. K.
87, 143
S
Schouten, J. A. xii Sintsov, D. xii Serre, J.-R. 104
W Weingarten, J.
69
Whitehead, J. M. C. 45, 148 Wood, J. 161 Y
Yampolski, A. xiii
THE GEOMETRY OF VECTOR FIELDS Yu. 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 (V t 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 (1
sysk n» In n-dimensional ,pace, Ioliations and their Godbillun-Vev inUariant are also ( onsidered. there is much interest in the study of geometrical ohlects in n-dnnensional [u( iI k'ITl spat o and this Volume provides d useful and fill prehensiye presentation.
About the author I'rt ii Yu. Am nt ty is d leading resear( h fellow at the Institute for Low TemperatUrt Phvsit s and 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 interest Ihr (,rometr) of Sub(ttandol(h Yu. Aminnv
I inear Algehrd and Geometry A.I. hostrikin and 1'u.i. Manin Tensorand L'et Ale( hdnit and l'hysh edited by AT. Fumenko, O.V. Manturov and V.V. Troiimus
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