A. C. 10. 11. CoiloBbeB,A. T.

CBOPHI4K110 JU4NEPEHLU4AJThHOfl ITEOMETPI4H

H Touonor'1114

MocKoscKoro YHMBepcwreTaMocxBa

A.5.Mishch.nko. Yu.P.and AT. Fominko

Probisms inDifF.r.ntiat G.omstry

andTopology

Translated from the Russianby

Oleg Efimov

Mir Pubtishsrs

Moscow

First published 1985

Revised from the 1981 Russian edition

TO THE READER

Mir Publishers would be grateful for your comments on thecontent, translation and design of this book. We would also bepleased to receive any other suggestions you may wish to make.Our address is: Mir Publishers, 2 Pervy Pizhsky Pereulok, 1-110,

GSP, Moscow, 129820, USSR.

Ha au2JIuuCKoM

© H3naTenbcTao MocKoBcKoro YHHBePCUTeTa, 1981

© English translation, Mir Publishers, 1985

Preface

This book of problems is the result of a course indifferential geometry and topology, given at the mechanics-and-mathematics department of Moscow State University.It contains problems practically for all sections of theseminar course. Although certain textbooks and booksof problems indicated in the bibliography list were usedin preparation of this volume, a considcrable number ofthe problems were prepared for this book expressly.

The material is distributed over the sections as in text-book [3]. Some problems, however, touch upon topicsoutside the lectures. In these cases, the corresponding sec-tions are supplied with additional definitions andexplanations.

In conclusion, the authors express their sincere gratitudeto all those who helped to publish this work.

• Contents

Preface 5

1. Application of Linear Algebra to Geometry 7

2. Systems of Coordinates 9

3. Riemannian Metric 14

4. Theory of Curves 16

5. Surfaces 34

6. Manifolds 53

7. Transformation Groups 60

8. Vector Fields 64

9. Tensor Analysis 70

10. Differential Forms, Integral Formulae, De Rham

Cohomology 75

11. General Topology 81

12. t-Iomotopy Theory 87

13. Covering Maps, Fibre Spaces, Riemaun Surfaces 97

14. Degree of Mapping 105

15. Simplest Variational Problems 108

Answers and Hints 113

Bibliography 208

1

Applicationof Linear Algebra to Geometry

1.1. Prove that a vector set a1 ak in a Euclidean space is linearlyindependent if and only if

det O(ai, 0.

1.2. Find the relation between a complex matrix A and the real matrixrA of the complex linear mapping.

1.3. Find the relations between

det A and det rA, Tr A and Tr rA, det (A — XE) and det (rA XA).

1.4. Find the relation between the invariants of the matrices A, B andA B, A ® B.

Consider the cases of det and Tr.1.5. Prove the formula

det e4 =

1.6. Prove that= + + C' [A, BIC"

for a convenient choice of the matrices C' and C", where [A,B] = AB BA.

1.7. Prove that if A is a skewsymmetric matrix, then is an orthogonalmatrix.

1.8. Prove that if A is a skewhermitian matrix, then is a unitarymatrix.

1.9. Prove that if [A, A*] = 0, then the matrix A is similar to a diagonalone.

1.10. Prove that a unitary matrix is similar to a diagonal one witheigenvalues whose moduli equal unity.

1.11. Prove that a hermitian matrix is similar to a diagonal one withreal eigenvalues.

1.12. Prove that a skewhermitian matrix is similar to a diagonal onewith imaginary eigenvalues.

7

1.13. Let A ((aijO be a matrix of a quadratic form, and Dk= det

Prove that A is positive definite if and only if for all k, I kthe inequalities Dk > 0 are valid.

1.14. With the notation of the previous problem, prove that a matrixA is negative definite if and only if for all k, I k n, the inequality(_1)kDk > 0 holds.

1.15. Put ((A = > Prove the inequalities

((JI —i— B(((( ((

—F (I B

((A (( . ((B

1.16. Prove that if A2 = then the matrix A is similar to the matrix

/'Ek 0 \I,k4-I=n.\0 —E,/

1.17. Prove that if A2 —E, then the order of the matrix Ais (2n x 2n), and it is similar to a matrix of the form

( 0

0

1.18. Prove that if A2 = A, then the matrix A is similar to a matrix

of the form(E 0\0 0

1.19. Prove that varying continuously a quadratic form from the classof non-singular quadratic forms does not alter the signature of the form.

1.20. Prove that varying continuously a quadratic form from the classof quadratic forms with constant rank does not alter its signature.

1.21. Prove that any motion of the Euclidean plane R2 can be resolvedinto a composition of a translation, reflection in a straight line, androtation about a point.

1.22. Prove that any motion of the Euclidean space R3 can be resolvedinto a composition of a translation, reflection in a plane and rotationabout a straight line.

1.23. Generalize Problems 1.21 and 1.22 for the case of the Euclideanspace

8

2Systems of Coordinates

A set of numbers q', q2,..., qfl determining the position of a pointin the space is called its curvilinear coordinates. The relation betweenthe Cartesian coordinates X2 of this point and curvilinearcoordinates is expressed by the equalities

X5 = Xs(q', q2, . ., qfl), (1)

or, in vector form, by

r = r(q', q2 qfl),

where r is a radius vector. Functions (1) are assumed to be continuousin their domain and to have continuous partial derivatives up to the thirdorder inclusive. They must be uniquely solvable with respect toq2 qfl; this condition is equivalent to the requirement that theJacobian

(2)

should not be equal to zero. The numeration of the coordinates isassumed to be chosen so that the Jacobian is positive.

Transformation (I) determines n families of the coordinatehypersurfaces qr = The coordinate hypersurfaces of one and the samefamily do not intersect each other if condition (2) is fulfilled.

Owing to condition (2), any n — I coordinate hyperplanes whichbelong to different families meet in a certain curve. They are called coor-dinate curves or coordinate lines.

The vectors rk = are directed as the tangents to the coordinate

lines. They determine the infinitesimal vector

dr rkdq"

in a neighbourhood of the point M(q', q2, . .. , q"). The square of itslength, if expressed in terms of curvilinear coordinates, can be found fromthe equality

ds2 = (dr, dr)=

r,dq',k1

rkdci) = th15 dqk

where (,) is the scalar product defined in

9

The quantities = = rk) define a metric in the adopted coor-dinate system.

An orthogonal curvilinear coordinate system is one for which(0

s = k

The quantities are called the Lamé coefficients. They are equal to themoduli of the vectors

I /dxi'Y fax2\2=

= + + ... +

The square of the linear element in orthogonal curvilinear coordinatesis given by the expression

ds2 = + + ... +

2.1. Calculate the Jacobian J = of transition from Cartesian

coordinates (x1, . . . , to orthogonal curvilinear coordinates (q',q2 qfl) in the space

2.2. Calculate the gradient grad f of the function f: R3 R in an or-thogonal curvilinear coordinate system.

2.3. Calculate the divergence div a of a vector a E R3 in an orthogonalcurvilinear coordinate system.

2.4. Find the expression for the Laplace operator Lxf of the functionf: R3 R in an orthogonal curvilinear coordinate system.

2.5. Cylindrical coordinates in R3

q1=r, q3=zare related to Cartesian coordinates by the formulae

x = r cosça, y = r Z = Z.

(a) Find the coordinate surfaces of cylindrical coordinates.(b) Compute the Lamé coefficients.(c) Find expression for the Laplace operator in cylindrical coordinates.2.6. Spherical coordinates in R3

q'=r, q2=O, q3 =are related to rectangular coordinates by the formulae

z=rcosO.(a) Find the coordinate surfaces of spherical coordinates.(b) Compute the Lamé coefficients.(c) Find expression for the Laplace operator in spherical coordinates.

10

2.7. Elliptic coordinates in R3

qt = X, q2 = /L, q3 = Z

are defined by the formulae

x y = l)(l —z =

where c is a scale factor.(a) Find the coordinate surfaces of elliptic coordinates.(b) Compute the Lamé coefficients.2.8. Parabolic coordinates in R3

qt = X, q2 = q3 = Z

are related to Cartesian by the formulae

x=

— X2) Ni. Z Z.

(a) Express parabolic coordinates in terms of cylindrical.(b) Find the coordinate surfaces of parabolic coordinates.(c) Compute the Lamé coefficients.2.9. Ellipsoidal coordinates in R3 are introduced by the equations

(a > b > c):X Z 2

2+

2+

21 (X > — c ) (ellipsoid),

a +X b +X c +XX

+ + = 1 (—c2 > > —b2) (hyperboloid ofa b c

one sheet),

2+ +

2

Z= I (— b2 > > a2) (hyperboloid of

a +v b2+i' c +vtwo sheets).Only one set of values X, v corresponds to each point (x, y, z) E R3.

The parameters

= X, q2 = /L, q3 = P

are called ellipsoidal coordinates.(a) Express Cartesian coordinates x, y, z in terms of ellipsoidal

coordinates X, jz, v.

(b) Compute the Lamé coefficients.(c) Find expression for the Laplace operator in terms of ellipsoidal

coordinates.

II

2.10. Degenerate ellipsoidal coordinates (a, j3, in R3 for a prolateellipsoid of revolution are defined by the formulae

x = c sin/3 cosp, y c sinha sinj3 z = c cosha cosl3,

(a) Find the coordinate surfaces in this coordinate system.(b) Compute the Lamé coefficients.(c) Find expression for the Laplace operator.2.11. Degenerate ellipsoidal coordinate system (a, w) in R3 for an

oblate ellipsoid of revolution is defined by the formulaex = c cosha sini3 c cosha singz =

(a) Find the coordinate surfaces for this coordinate system.(b) Compute the Lamé coefficients.(c) Find expression for the Laplace operator.2.12. Toroidal coordinate system (a, in R3 is defined by the

formulaec sinha c sinha c sin/3

cosha — cosha — cos$ cosha — cosfl

where c is a scale factor, 0 a < —ir < j3 —ir < ir.(a) Find the coordinate surfaces in a toroidal coordinate system.(b) Compute the Lamé coefficients.(c) Find expression for the Laplace operator.2.13. Bipolar coordinates in R3

q'=a, q2=j3, q3=zare related to Cartesian coordinates x, y, z by the formulae

a sinha a—, y= z=z,cosha — cosI3 cosha — cos/3

where a is a scale factor.Compute the Lamé coefficients for a bipolar coordinate system.2.14. Bispherical coordinates in R3

q'=a,are defined by the formulae

c sina cosç c sina c sinhflx= y= ,— cosa

c is a constant factor, 0 a < j3, — < — < ir.

12

These formulae can be written shorter:

a+i13z + Q

2=/x +y).

(a) Find the coordinate surfaces in a bispherical coordinate system.(b) Compute the Lamé coefficients.(c) Find expression for the Laplace operator.2.15. Prolate spheroidal coordinates in R3

are defined by the formulae

x = y = c'J(X2 — 1)(1 —

z = — l)(l —

where X 1, — 1 1, 0 2ir, and c is a constant factor.Compute the Lamé coefficients for this coordinate system.2.16. Oblate spheroidal coordinates in R3

are defined by the formulae

x = cX/L y = — z = cX/L cosw,

Compute the Lamé coefficients for an oblate spheroidal coordinatesystem.

2.17. Paraboloidal coordinates in R3

q'=X,are defined by the relations

x = y = z = —

(a) Compute the Lamé coefficients fir a paraboloidal coordinatesystem.

(b) Find the coordinate surfaces.2.18. Let H1, H2, H3 be the Lamé coefficients for a certain curvilinear

coordinate system in R3.Prove the relations

8 1 8H2 3 1 c3H1 1 0H1 8112(1)——----— + ——— = 0dq' 3q1 8q2 H2 8q2 8q3 8q3

13

8 1 81-13 8 1 0112 1 OH2 OH3(2)— + + = 0;i3q2 H2 3q2 0q3 H3 0q3 Hf Oq' Oq'

0 1 OH1 3 1 OH3 I OH3 OH1(3)—-—-----—----- + + = 0;

3q3 H3 0q3 0q1 Oq' 0q2 3q2

— I OH3 1 OHi 0112

0q20q3 — H3 3q2 3q3+

H2 Oq2 3q3'

32H2 1 OH1 OH2 I OH2 OH3

3q3Oq' — H1 Oq3 0q1+

H3 0q3 Oq''

6— I OH2 OH; 1 0113

Oq'Oq2 — H2 Oq' 0q2 + Oq' 3q2

2.19. Prove that if functions q2, q), H2(q', q2, q3), H;(q', q2,q) of class C3 satisfy the relations of the previous problem, then theyare the Lamé coefficients for a certain transformation

Xs = xs(q', q2, q3), s = 1, 2, 3.

3Riemannian Metric

3.1. Prove that the metric ds2 = dx2 + f(x)dy2, 0 < f(x) <co can betransformed to the form ds = g(u, v)(du2 + dv2) (isothermal coor-dinates).

3.2. Prove that local isothermal coordinates can be defined on any realanalytic surface M2. Find the conformal representation of the metric ds2.

3.3. Mercator's projection is defined as follows: rectangular coordinates(x, y) are defined on a map so that a constant bearing line (where thecompass needle remains undeflected) on the earth's surface is put intocorrespondence with a straight line on the map.

(a) Prove that to a point on the surface of the globe with sphericalcoordinates (0, cc) on the map, there corresponds, in Mercator'sprojection, the point with coordinates x = cc, y = In cotO/2.

(b) How can the metric on the terrestrial globe be written in terms ofthe coordinates (x, y)?

3.4. Prove that the metric ds2 on the standard hyperboloid of two sheetswhich is embedded in the pseudo-Euclidean space R coincides with themetric on the Lobachevski plane.

14

3.5. Write the metric on the sphere S2 in complex form.3.6. Find a metric on the two-dimensional space of velocities in

relativity theory.3.7. Change the coordinates in the previous problem so that v tanhx

(where v is the velocity of the moving point).3.8. Write the metric of the previous problem in polar coordinates for

the unit circle.3.9. Calculate the length of a circumference and the area of a circle

on (a) the Euclidean plane, (b) a sphere, (c) the Lobachevski plane.3.10. Let the Lobachevski plane be realized as the upper half-plane of

the Euclidean plane. We call Euclidean semicircumferences with centreson the axis Ox and Euclidean half-lines resting upon the axis Ox andorthogonal to it "straight lines" of the Lobachevski plane. We call a figureformed by three points and the segments of "the straight lines" joiningthem a triangle in the Lobachevski plane.

Prove that the sum of the angles of a triangle in the Lobachevski planeis less than

3.11. (Continuation of Problem 3.10.) Let ABC be an arbitrary trianglein the Lobachevski plane, a, b, c the non-Euclidean lengths of the sidesBC, AC, AB, and cs, j3, -y the values of its angles at the vertices /1, 8,

C. Prove the following relations:

cosa + cosf3 cos-y(I) cosha =

sinf3 siny

cosf3 + cos-y cosa(2) coshh =

SIfly SIfla

cos-y + coscs cosl3(3) coshc =

slncs

3.12. (Continuation of Problem 3.11.) Prove the analogue of the lawof sines for the Lobachevski plane:

sinha sinhb sinhc \JQ

sin$ siny sina sinj3 Siny

where Q = cos2cs + + cos2y + 2coscx cosl3 cosy — I.3.13. (Continuation of Problem 3.12.) Prove the following formulae ex-

pressing the angles of a triangle in the Lobachevski plane in terms ofits sides:

coshb coshc cosha(1) cosa =

sinhb sinhc

15

coshc cosha — coshb=

sinhc sinha

cosha coshb — coshc(3) cOs'y =

sinha sinhb

3.14. (Continuation of Problem 3.13.)Assume that y ir/2, i.e., the triangle ABC is right. Prove the fol-

lowing relations:

(I) sinha = sinhc sina;(2) tanha = tanhc cosL3;

(3) tanha = sinhb tana;(4) coshc = cosha coshb;(5) coshc = cota cotj3;

(6) cosha = cosa/sin/3.

3.15. Let ABC be a spherical triangle on a sphere of radius R, a,y the values of the angles at the vertices A, B, C and a, b, c the lengthsof the sides BC, AC, AB. Prove the following relationship

a b c b.ccos = cos — cos — + sin —sin cos a.

R R R R R

4Theory of Curves

4.1. Let C be a plane curve, M0 a point of the curve C, and XOY arectangular system of coordinates given in the plane of the curve. Denotethe points of intersection of the tangent and the normal to this curvewith the axis OX by T and N, respectively. Let P be the projection ofthe point M0 onto the axis OX.

(a) Find the equation of the curve C if its subnormal PN is constantand equal to a.

(b) Find the equation of the curve C if its subtangent PT is constantand equal to a.

(c) Find the equation of the curve C if the length of its normal M0Nis constant and equal to a (for any point M0 on the curve).

4.2. Find the equation of the curve C whose tangent MT is constantin length and equal to a.

16

4.3. An arbitrary ray OE intersects the circumference

2 / a\2 a2x + - =

and a tangent to it passing through the point C which is diametricallyopposite to 0 at points D and E. Straight lines are drawn through thepoints D and E parallel to the axes Ox and 0y, respectively, to meet eachother at a point M. Set up the equation of the curve formed by suchpoints M (witch of Agnesi).

4.4. A point M moves uniformly along a straight line ON which rotatesuniformly around a point 0. Form the equation of the path of the pointM (Archimedes' spiral).

4.5. A straight line OL rotates around a point 0 with constant angularvelocity w. A point M moves along the straight line OL with a velocitywhich is proportional to the distance IOM1. Form the equation of thepath described by the point M (logarithmic spirai).

4.6. A circle of radius a rolls along a straight without slipping.Set up the equation of the path of a point M to the circlerigidly and placed at a distance d from its centre (when d a, this isa cycloid; when d < a, a curtate cycloid; and when d> a, a prolatecycloid).

4.7. A circumference of radius r rolls without slipping along a circum-ference of radius R and remains outside it. Form the equation of the pathof a point M of the rolling circumference (epicycloid).

4.8. A circumference of radius r rolls without slipping along a circum-ference of radius R and remains inside it. Construct the equation of thepath of a point A'! of the rolling circumference (hypocycloid).

4.9. Find a curve given by the equation r r(t), c < t < d, if it isknown that r'(t) = X(1)a, where X(t) > 0 is a continuous function, anda is a constant nonzero vector.

4.10. Find a curve given by the equation r = r(1), < I < ifr" (t) = a is a constant nonzero vector.

4.11. A vector function r(t) satisfies the differential equationr" = Er' x a], where a is a constant vector. Express (a) [r' X r"]2;(b) (r', r", r' ") in terms of a and r'.

4.12. Let 'y be a closed curve of class C'. Prove that, for any vectora, there is a point x E -y at which the tangent to .y is orthogonal to a.

4.13. Two points move in space so that the distance between themremains constant. Prove that the projections of their velocities onto thedirection of the straight line joining these points are equal.

4.14. Prove that if a vector function r(t) is continuous on a segment[a, bi together with its derivative r', and r II r', but r' 0 and r 0,then the hodograph of the vector function r = r(t) is a straight linesegment.

2—2018 17

4.15. Prove that if a vector function r = r(f) is continuous on a certainsegment [a, bJ together with its two first derivatives r' and r", these deriv-atives are different from zero for all 1 E [a, bi, and collinear, i.e., r' II r"for all 1 E [a, bJ, then the hodograph of the vector function r = r(t) isa straight line segment.

4.16. A plane curve is given by the equation r = lçt(f) Underwhat condition does this equation determine a straight line?

4.17. Find the function r = r(W), given that this equation describes astraight line in polar coordinates on the plane.

4.18. Prove that a point describes a plane path under the action ofa central force F = Fr.

4.19. A plane translational motion is given on the plane by the lawsr r1(1) and r = r2(1) of motion of the ends of a solid rod. Find theequation of the centre surface (a centre surface is the set of all pointsof intersection of straight lines passing through the ends of the rod andperpendicular to the directions of the velocities of its ends).

4.20. The set of instantaneous centres of rotation with respect to amoving rod is called a centrode in a plane translational motion (see theprevious problem). Set up the equation of a centrode.

4.21. Prove that the linear velocity v of a point in any plane transla-tional motion is determined by the relation v = u[r], where r is the radiusvector of the point M(R) under consideration with respect to the instan-taneous centre of rotation (see Problems 4.19, 4.20), and Er] is the vectorobtained from r by rotation through + ir/2. Express w in terms of r1 andrz and find the velocity v of the point M(R).

4.22. The differential equation of the motion of a material particle Mis as follows:

r" = (X>0).

Prove, on the basis of this relation, that the point moves along a curveof the second order.

4.23. A material particle moves under the action of a central forceF = Fr°. It follows from the result of Problem 4.18 that the motion takesplace in a certain plane. Form the equation of the motion and the dif-ferential equation of the path in polar coordinates.

Consider the case

F

4.24. The motion of an electron in a constant magnetic field isdetermined ,by the following differential equation

= [r' x H], H = const.

Prove that the path is a helix.

18

4.25. Find the curves determined by the differential equation

r'

4.26. Find the curves determined by the differential equation

r' = [e x [r Xe]],

where e is a constant unit vector.4.27. Find the curves determined by the differential equation

r' = ae + [e x rJ,

where a = const and e = const.4.28. Find the curves determined by the differential equation

— r(r, e),

where e = const and el = 1.

4.29. Form the equations of the tangent and normal to the followingcurves:

(I) r = [a cost, b sint) (ellipse);

(2) r= +

_!—), — (hyperbola);

(3) r [a cos3 t, a sin3 t) (astroid);

(4) r = {a(t — sint), a(l — cost)[ (cycloid);

(1 1 1 2 3(5) r =2 3

at the point t = 0;

(6) r = a,ø (Archimedes' spiral).

4.30. At what angle do the curves + y2 8 and y2 = 2x intersect?4.31. At what angle do the curves

+ y2 = 8x, y2 = x3/(2 — x)

intersect?4.32. At what angle do the curves

x2=4y, y=8/(x2+4)intersect?

4.33. Prove that the length of the segment of the tangent to the astroid

+ = a213

between the coordinate axes equals a.

19

2*

4.34. Prove that the length of the segment between the axis Oy andthe point of contact of the tangent to the tractrix

a—

equals a.4.35. Prove that the cardioids

r = a(l + cosW), r a(l —

are orthogonal.4.36. Find the envelope of the family of straight lines joining the ends

of pairs of conjugate diameters of an ellipse.4.37. Find the envelope of the family of straight lines cutting a triangle

of constant area off the sides of a right angle.4.38. Find the envelope of the family of straight lines cutting segments

of given area off a given parabola.4.39. Find the envelope of the family of straight lines cutting a triangle

of given perimeter off the sides of a given angle.4.40. Find the envelope of the family of circumferences constructed on

parallel chords of a circumference as on diameters.4.41. Find the envelope of the family of ellipses that have common

principal axes and a given semi-axis sum.4.42. A beam of parallel rays falls on a spherical mirror. Find the enve-

lope of the reflected rays (caustic).4.43. Find the envelope of the family of ellipses that have a given area

and common principal axes.4.44. Find the envelope of the family of circumferences with centres

on an ellipse and passing through one of its foci.4.45. Find the envelope of the family of circumferences of radius a and

centres on a curve r = r(v).4.46. Find the envelope of the normals of a curve r = r(v). The vector

function r(v) is defined, continuous and twice differentiable on a segment[a, b]. The vectors r' and r" are noncollinear at each point of thissegment.

4.47. Find the envelope of the rays reflected from a circumference ifthe luminous point is on the circumference.

4.48. Calculate the curvature of the following curves:

(1) y = sinx at the vertex (sine curve);(2) x = a(1 + m)cosmt — amcos(l + m)t

y = a(1 + m)sinmt — amsin(l + m)t (epicycloid);

(3) y = a cosh(x/a) (catenary curve);(4) x2y2 = (a2 — y2)(b + y)2 (conchoidal curve);

20

(5) r2 = a2cos2ço (lemniscate);

(6) r = a(l + (cardioid);(7) r = acp (Archimedes' spiral);(8) r = a cos3t, a sin3t (astroid).

4.49. Calculate the curvature of the following curves:

(I) y = —In cosx;

(2) x = 3t2, y = 3! — t3 for t = 1;

(3) x = a(cost + t sint), y = a(sint t cost) for I

(4) x = a(2cost cos2l), y = a(2sint — sin2t).

4.50. Find the curvature of the following curves given in polarcoordinates:

(I) r = (2) r = aço"; (3) r = at the point = 0.

4.51. Find the curvature of the curve given by the equation

F(x, y) = 0.

4.52. Curves arc given by their differential equation P(x, y)dx ++ Q(x, y)dy 0. Find their curvature.

4.53. Calculate the length of the following curves:

(I) y = a cosh(x/a);

(2) y = x312

(3) y =(4) y = mx;(5) r = a(1 + cosp);

(6) r = (a(t — sint), a(l — cost) I;(7) r = {a(cost + I sin!), a(sint — t cost));

(8) r=

(2cost + cos2r), (2sint + sin2t)};

(9) r = {acos3t,(10) y = ex;

(11) r = {a (In cot —i-- — cost). a sint].

4.54. Find the arc length of the curve

x = —

y = —

21

The natural equations of a plane curve are equations of the form:

(1) k = k(s), (2) F(k, s) = 0 or (3) k = k(t), s = s(i).

If the natural equations of a curve are given, then the parametrizationof the curve can be given in the form

x = Scosa(s)ds, y =4.55. Form the natural equations of the curves:

(I) x = a cos3l, y a sin3!;

(2) y = x312•

(3) y = x2;

(4) y = lnx;(5) y = a cosh(x/a);

(6) y =

(7) x = a (In tan + y = a sin!;

(8) r = a(1 + cos4,);

(9) x = a(cost + t sin!), y = a(sint — I cost).

4.56. Find the parametric equations of the curves if their naturalequations are given (here R = 1/k):

(I) R = as;

R2(2) —i- + = 1;

(3) Rs = a2;

(4) R = a + s2/a;

(5) + 9R2 = 16a2;(6) S2 + R2 16a2;

(7) R2 = 2as;

(8) R2 + a2 = a2e - 2s - a

4.57. Let p be the distance from the origin of radii vectors to the tangentto a curve 'y at a point M, and r the distance from the point 0 to thepoint M. Prove that

k=rdr

4.58. At a certain point of a curve r = r(s), we have: k 0, 0.Having taken the equation of the osculating circumference in the form

22

— ro — Rono)2 = prove that the osculating circumferenceintersects the given curve in a neighbourhood of the indicated point.

4.59. Given that the following conditions are fulfilled at a certain pointof a curve: k0 0, R0 = 0, ko 0, prove that the osculating circum-ference at this point of the curve does not intersect the curve in asufficiently small neighbourhood of this point.

4.60. Given an equation R = f(a), where R is the curvature radius ofa curve, and a the angle from a constant vector a to the tangent vectorr to the curve, form the parametric equations of the curve.

4.61. Given an equation a = f(R) (see the previous problem), form theparametric equations of the curve.

4.62. Given an equation s = f(a), where s is an arc and a the anglefrom a constant vector a to the tangent vector r to the curve, form theparametric equations of the curve.

4.63. Given an equation a = f(s) (see the previous problem), form theparametric equations of the curve.

4.64. Given that a beam of luminous rays falls on a plane curve r = r(s)from the origin of radii vectors, form the equation of the envelope ofthe reflected rays (caustic).

4.65. What form will the equation of the caustic of a plane curve withrespect to the origin of the radii vectors have if the equation of the curveis given in the form r = r(t)?

4.66. A beam of parallel rays with the direction of a vector e = I)falls on a plane curve given by an equation r = r(s). Form the equationof the envelope of the rays reflected from the given curve (caustic).Consider the cases where the curve is given by an equation r = r(t) andwhere it is given by an equation y = f(x).

4.67. Write the equation of the tangent line and the normal plane ofthe curve

r = (U3 — u2 — 5, 3u2 + 1, 2u3 —

at the point where u = 2.

4.68. Find the tangent line and the normal plane at the pointA(3, —7, 2) of the curve

r=(u4+u2+l,4u3+5u+2,u4—u3J.4.69. Find the tangent line and the normal plane at the point

A(2, 0, —2) of the curve

r=(u2—2u+3,u3—2u2+u,2u3—6u+2).4.70. Write the equation of the osculating plane of the curve

r = (u2, u, U3 — 20)at the point A(9, 3, 7).

23

4.71. Show that the curve

r {au + b, cu + d, u2j

has the same osculating plane at all points.4.72. Form the equations of the osculating plane, principal normal, and

binormal of the curve

y2=x, x2=zat the point (I, 1, 1).

4.73. Given a helix

r lacost, asint, bi),

form the equations of the tangent, normal plane, binormal, osculatingplane, and principal normal.

4.74. Given a curve

r 112, 1 — t

form the equations of the tangent, normal plane, binormal, osculatingplane and principal normal at the point t 1.

4.75. Form the equations of the tangent line and the normal plane ofthe curve given by the intersection of two surfaces

F1(x, y, z) = 0 and F2(x, y, z) = 0.

4.76. The curve in which a sphere meets a circular cylinder, whose baseradius is twice less and which passes through the centre of the sphere,is called a Viviani curve. Make up the equation of a Viviani curve inimplicit and parametric forms. Find the equations of the tangent, normalplane, binormal, principal normal and osculating plane.

4.77. Find the length of the arc of the helix

x = 3a cost, y 3a sint, z = 4at

from the point of intersection with the plane xOy to an arbitrary pointM(t).

4.78. Find the length of one turn between the two points of intersectionwith the plane xOz of the curve

x = a(t — sint), y = a(l — cost), z = 4a cost/2.

4.79. Find the length of the arc of the curve

x3=3a2y, 2xz=a2

between the planes y = a/3 and y = 9a.4.80. Find the length of the closed curve

x = cos3t, y = sin3t, z = cos2t.

24

4.81. Reparametrize the helix

r = {a cost, a sift, bt), b > 0,

by the natural parameter.4.82. Reparametrize the curve

r = letcost, e'sint, e

by the natural parameter.4.83. Reparametrize the curve

r = {cosht, sinht, 1)

by the natural parameter.4.84. Find the vectors r, v, of the Frenet frame for the helix

r = lacost, asint,

Calculate the curvature and torsion of the helix.4.85. Given the curve

r = I — 1, t3),

find the vectors r, v, of the Frenet frame. Calculate the curvature andtorsion of this curve.

4.86. Find the vectors r, v, of the Frenet frame, curvature, and torsionof a Viviani curve (see Problem 4.76).

4.87. Find the curvature and torsion of the following curves:

(1) r = {t — sint, 1 — cost,

(2) r = te', e',(3) r = e'sinf, etcost, e' 1;

(4) r = 12t, int, t2J;

(5) r = (3t — t3, 3t + t3);

(6) r = tcos3t, sin3t, cos2tj.

4.88. At each point of the curve

x = t — sint, y = 1 — cost, z 4sint/2,

a segment equal to four times the curvature at this point is laid off inthe positive direction of the principal normal.

Find the equation of the osculating plane of the curve described bythe end of the segment.

4.89. Calculate the curvature and torsion radii for the curve

x3 = 3a2y, 2xz = a2.

25

4.90. Deduce the formulae for the calculation of the curvature andtorsion of the curve given by equations y = y(x) and z = z(x) and findthe Frenet frame for this curve.

4.91. Find the curves intersecting the rectilinear generators of thehyperbolic paraboloid xy = az at right angles.

4.92. A curve on a sphere that intersects all the meridians of the sphereat a given angle is called a loxodrome. Find the equation of a loxodromeand the vectors r, v, $ of the Frenet frame for this curve at an arbitrarypoint. Calculate its curvature and torsion.

4.93. Given a curve

r = (v cos u, v sin u, kv

where v = v(u), prove that this curve is placed on a cone. Define thefunction v(u) so that this curve intersects the generators of the cone ata constant angle 0.

4.94. The tangent vector T = T(t) 0 is given at each point of a curver = r(t). The function r(t) is defined, continuous, and has a continuousderivative r' (t) on a segment [a, bJ. The function T(t) is continuous onthe segment [a, bJ. Prove that this curve can be parametrized so that

dt

4.95. A curve C is given by an equation r = r(/), the function r(t)is defined on a segment [a, b] and possesses noncoplanar derivatives r',r", r'" at a point M. Prove that the osculating plane of the curve Cat the point M intersects the curve C.

4.96. Prove that if all osculating planes of a curve are concurrent, thenthe curve is plane.

4.97. A curve C is given by an equation r = r(t); the function r(t)is defined on a segment [a, ii] and possesses derivatives r', r", r'" atsome point M(t) with r' r'. Calculate the limit

durn

where d is the distance from the point M(t + itt) to the osculating planeof the curve C at the point M. Consider the special case where the curveis given by an equation r = r(s) (s being the natural parameter).

4.98. Find a necessary and sufficient condition for the given familyof curves

r = Q(u) + Xe(u) (lel = 1)

to have the envelope. Find this envelope.

26

4.99. For what value of b is the torsion of the helixr = (a cost, a sint, bt( (a = coast)

at its maximum?4.100. Prove that if the torsion of a curve C at some of its points M

is other than zero, then the osculating plane of the curve C at the pointM intersects the curve.

4.101. Express t, r, r in terms of t, v, k and x.4.102. Prove that T1313 = x.4.103. Prove that if the principal normals of a curve form a constant

angle with the direction of a vector e, then

d k2+x2ds

k

and conversely, if this relation is fulfilled, then the principal normals ofthe curve form a constant angle with the direction of some vector. Findthis vector.

4.104. Prove that if all normal planes of a line contain a vector e, thenthis line is either straight or plane.

4.105. Prove that if all the osculating planes of a curve which is nota straight line contain the same vector, then this curve is plane.

4.106. Prove that if fi = const, then the curve is plane.4.107. Prove that if the osculating planes of a curve have the same

inclination, then the curve is plane.4.108. A space line is called a generalized helix if all its tangents form

a constant angle with a fixed direction.Prove that a line is a generalized helix if and only if one of the following

conditions is fulfilled:(a) the principal normals are perpendicular to a fixed direction;(b) the binormals make a constant angle with a fixed direction;(c) the ratio of the curvature to the torsion is constant.4.109. Prove that the condition = 0 is necessary and sufficient

for a line to be a generalized helix.4.110. Prove that the line x2 = 3y, 2xy 9z is a generalized helix.Let r = r(s) be a curve parametrized by the natural parameter. Then

the mapping r: (a, b) R3 determines a curve s —p i(s). This curve maybe non-regular. Since r(s)I = 1, the image r(s) lies on the sphere withradius I and the origin at its centre. This curve is called the tangent spheri-cal image of the curve r = r(s). The normal spherical images —p v(s) and the binormal spherical image s fl(s) may be definedsimilarly.

27

4.111 Find the tangent, normal and binormal spherical images of thehelix

r = cost, a sin!, bt).

4.112. Let r r(s) be a curve parametrized by the natural parameter.(a) Prove that the tangent spherical image of the curve r = r(s) degener-

ates into a point if and only if r = r(s) is a straight line.(b) Prove that the binormal spherical image of the curve r = r(s)

degenerates into a point if and only if r = r(s) is a plane curve.(c) Prove that the normal spherical image of the curve r = r(s) cannot

be a point. -4.113. Let s be the length of the tangent spherical image of a curve

r = r(s):

=

(a) Prove that = k.ds

(b) Find necessary and sufficient conditions for the tangent sphericalimage to be a regular curve.

4.114. Les be the length along the normal (resp. binormal) sphericalimage of a curve r = r(s). Prove that

= ,i2 (resp.ds

4.115. Let r = r(s) be a curve parametrized by the natural parameter,kx 0. Prove that the tangent to the tangent spherical image is parallelto the tangent to the binormal spherical image at the correspondingpoints.

4.116. Let r r(s) be a curve parametrized by the natural parameter.Prove that if the tangent spherical image of this curve lies in a planepassing through the origin, then the curve r = r(s) is plane.

4.117. Prove that the curve r = r(s) is a helix if and only if the tangentspherical image is an arc of a circumference.

By definition, a spherical curve is a curve r = r(t) for which there existsa constant vector m such that

< r(t) — m, r(t) — m > = r2.

4.118. Let r = r(t) be a regular curve, and a a point which lies in eachnormal plane to r = r(t). Prove that r = r(t) is a spherical curve.

4.119. Prove that

r = ( —cos2t, —2 cost, sin2tj

28

is a spherical curve by showing that the point (— I, 0, 0) lies in everynormal plane.

4.120. Let r r(s) be a curve which is parametrized by the naturalparameter, k 0, x 0, and 1/k, a = 1/n. Assume that

+ 'a)2 = a2 = coust, a > 0. Prove that the image of the curver r(s) lies on a sphere of radius a.

4.121. Prove that if r = r(s) is a curve which is parametrized by thenatural parameter, k 0, x 0, then r(s) lies on a sphere if and only if

x fk'\' /—=1 I (orxQ= —I—-—k \xk2J \

4.122. Using the results of the previous problems, prove that a curver = r(s) lies on a sphere if and only if there exist constants A and Bsuch that

+ i.

4.123. Two curves r = rj(1) and r = rz(() are said to form a pair ofBertrand curves if for any value of the parameter to, the normal to r,(t)coincides with the normal to r2(t).

(a) Prove that two arbitrary concentric circumferences which lie in thesame plane form a pair of Bertrand curves.

(b) Let

1(1r1(t) = — — t\/1 — I — t2, 0

2 (cost

r2(t) =—

— r, 1 — 12 +2 (cost

Prove that rj(t) and r2(t) form a pair of Bertrand curves.4.124. Prove that the distance between the corresponding points of a

pair of Bertrand curves is constant.4.125. Prove that the angle between the tangents to the two curves of

a Bertrand pair at corresponding points is constant.4.126. Let r = ri(s) be a curve parametrized by the natural parameter,

and kx 0. Prove that the curve r = r2(s) (s is not the natural parameterof F2(S)) which forms a pair of the Bertrand curves with ri(s) exists ifand only if there are constants X and such that

IA = k +

29

4.127. Let r = r(t) be a regular curve of class C3,x 0. Prove thatr(t) is a circular helix if and only if r(t) possesses at least two differentcurves which are related in the sense of Bertrand.

Let m be a constant vector, r r(s) a curve, c(s) = r(s) — mj2, anda a positive number. The curve r(s) is said to possess at a point s so

a spherical contact of order j with the sphere of radius a and centre atthe endpoint of m if

C(So) = a2, C'(So) = C"(So) . . . = (so) = 0,

cU+1)(so) 0

4.128. Given that k 0, calculate the first three derivatives of thefunction c(s) in terms of r, v, k and x.

4.129. Prove that a curve r = r(s) possesses a spherical contact of order2 at a points = if and only if m r(so) + v(so)/k(so) + whereX is an arbitrary number.

4.130. Given that x(SO) 0, prove that a curve r = r(s) possesses aspherical contact of order 3 if and only if

I k'(so)m = r(so) + — v(So) — 2k(so) k (So)x(So)

4.131. Let a curve r = r(s) be of constant curvature. Prove that theosculating sphere and the circumference have the same radius.

Let r r(s), s a], be a plane piecewise regular curve of class C2parametrized by the natural parameter. The number

k is the curvature of the curve, s(O I n — 1) are the singularpoints, r (si) = lim i(s), r4 (se) = urn i(s), and is the angle

s—.si— s__.si+

between the vectors i — (se) and r + (s,), is called the rotation number er(s)of the curve.

4.132. Compute the rotation number of the curve y represented inFig. 1.

4.133. Compute the rotation numbers of the curves given by the fol-lowing equations (the parametrization is not natural):

(1) r = (a + cost, sint), 0 t <

(2) r = (a + Q cost, sint), 0 t 0 < e < lal;

(3) r = cos2t, —Q sin2tj, 0 t > 0;

30

(4) r = 0 t

(5) r (2cosf, — sintj, 0 t

(6) r = El, sin2tJ, 0 t

/ Arcs of cirdurn-I ferencesof rod/vs 2

(-1,0) (1,0)

Fig. 1

4.134. Prove that if r(s) is a simple, closed, regular, and plane curve,then the tangent circular image r: [0, L] S' of this curve is a mapping"onto,,.

An oval is a regular, simple, closed and plane curve for which k > 0.The vertex of a regular plane curve is a point at which the curvature khas a relative maximum or minimum.

Let r(s) be an oval and P a point on r(s). Then there exists a pointP' such that the tangent r to the oval at this point is opposite to thetangent at the point P, i.e., r(P') — r(P). The tangents at the pointsP and P' are parallel. Thus, for a given point P. there exists a uniquepoint P' (said to be opposite of P) on the oval, so that the tangents atP and P' are parallel and distinct.

The width w(s) of an oval at the point P = r(s) is the distance betweenthe tangent lines to the oval at the points P and P'.

An oval is said to be of constant width if its width at a point P isindependent of the choice of P

4.135*. Prove that any oval possesses at least four vertices. (Thisstatement is known as the four-vertex theorem.)

4.136*. Prove that if r(s) is an oval of constant width w, then its lengthequals 7rw.

4.137*. Let r r(s) be an oval of constant width. Prove that the straightline joining a pair of opposite points P and P' of the oval is orthogonalto the tangents at the points P and P'.

4.138*. Given that r = r(s) is an oval,prove that r" is parallel to rat least at four points.

31

4.139. Prove that the notion of vertex does not depend on the choiceof parametrization.

4.140. Show that the four-vertex theorem (see Problem 4.135) is notvalid if the requirement of closedness is omitted.

4.141. Let r1 [0, aJ —÷ R2 be a segment of a curve parametrized bythe natural parameter, and t2(S) a curve

r2(s) = r1(s) + (ao — S)r(s),

where i(s) is a tangent vector to ri(s) and ao > a a constant. Show thatthe unit tangent to r2(s) is orthogonal to i(s) at every point.

4.142. Let r(s) be a plane curve of constant width. Show that the sumof the curvature radii I/k is constant at opposite points and does notdepend on the choice of the points.

4.143. (a) Let r(s) be an oval of length L and with natural parametri-zation. Denote the angle between the horizontal and tangent vector r(s)by 0. Prove that the mapping 0 : [0, LI [0, 2irI is a parametrizationof the oval r(s).

(b) Let be an oval parametrized by a parameter 8 so that r(s) == Q(O(s)). Prove that the point which is opposite to r(s) is R(s) == Q(O(s) + ir).

(c) Prove that the curve R(s) is regular.4.144. Let be an oval parametrized by an angle 8 in a manner

similar to that of the previous problem. Let w(0) be the width of theoval at a point Q(O). Prove that

wdO = 2L,

where L is the length of the oval.4.145. Let be an oval parametrized by an angle 0, k(O) and w(0)

its curvature and width, respectively. Prove that

d2w 1+ w = —— +do2 k(O) k(O + 7r)

The total curvature of a regular space curve r = r(s) parametrized by

the natural parameter is the number Since k = ir'(s)I, the total cur-

vature is the length of the tangent image

r : [0, LI —p

Prove that if r = r(s) is a regular closed curve, then its tangentspherical image cannot lie in any open hemisphere.

32

Prove that the tangent spherical image of a regular closed curvecannot lie in any closed hemisphere except for the case when it is a greatcircumference bounding the hemisphere.

Let 'y be a closed C'-curve on the unit sphere S2. Prove thatthe image C of the curve is contained in an open hemisphere if

(a) the length I of the curve 'y is less than(b) I = 2ir, but the image C is not the union of two great

semi-circumferences.Using the results of Problems 4.146-4.148, prove the following

statement: the total curvature of a closed space curve y is not less than2,r and equal to 2,r if and only if is a plane convex curve (Fencheltheorem).

Let be a space closed curve. Assume that 0 k hR fora certain real number R > 0. Prove that the length 1 of the curve y satis-fies the inequality I 2irR.

4.151. Calculate the tangent spherical image for the ellipse

r t2cost, sint, Oj, 0 t

What can be said about the image taking the Fenchel theorem intoaccount?

Let be an oriented great circumference on the sphere S2. Then thereexists on S2 a unique point w associated with w, viz., the pole of thehemisphere which is on the left when moving along w in the positivedirection (Fig. 2).

Conversely, every point of S2 is related to a certain orientable greatcircumference. Thus, the set of oriented great circumferences is in one-to-one correspondence with the points of

The measure of the set of oriented great circumferences is the measureof the corresponding set of points in S2.

If w E S2. then w1 denotes the great oriented circumference assoc-iated with w.

3—2018 33

Fig. 2

For a regular curve with the spherical image C, we denote the numberof points in Cl) (which may be infinite) by Note that thenumber does not depend on a parametrization of the curve y.

4.152!' Let C be the image on S2 of a regular curve 'y of length 1. Provethat the measure of the set of oriented great circumferences whichintersect C (taking the multiplicities into account) equals 4/. In otherwords,

41 (the Crofton formula).

4.153!' A closed simple curve -y is said to be unknotted if there existsa one-to-one continuous function g : D2 R3 (D2 being the unit disk)which maps the boundary S' of the disk D2 onto the image of the curve-y. Otherwise, the curve is said to be knotted.

Prove that if 'y is a simple, knotted, and regular curve, then its totalcurvature is greater than or equal to

4.154!' Using the Crofton formula, prove that for any closed, regularcurve, Skds 2ir.

We call the number Sxds the total torsion of a regular space curve

r = r(s) parametrized by the natural parameter.4.155!' Prove that for any real number r, there exists a closed curve

y such that its total torsion r.

4.156!' Prove that the total torsion of a closed curve r = r(s) (sbeing the natural parameter) placed on the sphere S2 equals zero.

4.157!'LetMbe a surface in R3 such that 0 for all closed curvesplaced on M. Prove that M is a part of a plane or sphere.

4.158*. Prove that ds = 0 for any closed, spherical curve param-

etrized by the natural parameter.

5Surfaces

5.1. Make up a parametric equation of the cylinder for which the curveQ(u) is directing and whose generators are parallel to a vector e.

5.2. Make up a parametric equation of the cone with vertex at the originof the radius vector for which the curve = Q(u) is directing.

34

5.3. Make up a parametric equation of the surface formed by thetangents to a given curve = Q(u). Such a surface is called a developablesurface.

5.4. A circumference of radius a moves so that its centre is on a givencurve = a(s) and the plane in which the circumference is placed is,at each particular moment, a normal plane to the curve. Make up a pa-rametric equation of the surface described by the circumference.

5.5. A plane curve x = z = revolves about the axis Oz. Makeup parametric equations of the surface of revolution. Consider the specialcase where the meridian is given by an equation x = /(z).

5.6. The circumference x = a + bcosv, z bsinv (0 < b < a) re-volves about the axis Oz. Make up the equation of the surface ofrevolution.

5.7. A straight line moves translationally with a constant velocity whileintersecting another straight line at right angles and uniformly rotatingabout it. Make up the equation of the surface which is described by themoving straight line (right helicoid).

5.8. Make up the equation of the surface formed by the principal nor-mals of a helix.

5.9. Make up the equation of the surface formed by the family of nor-mals to a given curve

5.10. A straight line moves so that the point M where it meets a givencircumference moves along it, the straight line remaining in the plane nor-mal to the circumference at responding point and rotating throughan angle equal to the angle MOM0 through which the point was turnedwhile moving along the circumference. Make up the equation of the sur-face described by the moving straight line assuming that the originalposition of the moving straight line was the axis Ox and the circumferenceis given by two equations x2 + y2 a2, z = 0.

5.11. Given two curves r = r(u) and Q(v). Make up the equationof the surface described by the middle point of the line segment whoseextremities lie on the given curves (translation surface).

5.12. Make up the equation of the surface formed by the rotation ofthe catenary line y = a coshx/a about the axis Ox. This surface is calleda catenoid.

5.13. Make up the equation of the surface formed by the rotation ofthe tractrix

= Ia In tan(ir/4 + t/2) — a sint, a cost I

about its asymptote (pseudosphere).5.14. The surface formed by a straight line moving parallel to a given

plane (director plane) so that its generator intersects a given curve(directing curve) is called a conoid. A conoid is determined by a directing

3*35

line, director plane, and curve which the moving straight line intersects(i.e., the directing curve). Make up the equation of a conoid if the director

plane yOz, directing line y = 0, z = h, and the directing curve —j-- +

+ 4 = I, z = 0 (i.e., ellipse) are given,

a

5.15. Make up the equation of the conoid for which the directing line,director plane and directing curve are given by the following equations,respectively:

(a) x = a, y = 0;

(b)z=0;(c) y2 = 2pz, x = 0.

5.16. We call a cylindroid the surface formed by straight lines whichare parallel to a plane. A cylindroid can be determined by two directingcurves (lying on it) and a director plane (the generators of the cylindroidbeing parallel to it). Make up the equation of a cylindroid if its generatorsare two circumferences x2 + — 2ax = O,y = Oandy2 + z2 — 2ay = 0,

x = 0, and the director plane is the plane xOy.5.17. A surface given by the parametric equation

r = r(u, v) Q(u) + va(u),

where = Q(u) is a vector function determining a certain curve, anda = a(u) a vector function determining the distribution of the rectilineargenerators of the surface, is said to be ruled. Make up the equation ofa ruled surface whose generators are parallel to the plane y — z = 0 andintersect two parabolas y2 = 2px, z 0 and z2 = — 2px, y = 0.

5.18. Make up the equation of the ruled surface whose generatorsintersect the axis Oz, arc parallel to the plane xOy, and intersect the linexyz = a3, x2 + = b2.

5.19. Make up the equation of the ruled surface whose generatorsintersect the straight line r = a + ub, curve = Q(v), and are per-pendicular to a vector n.

5.20. Make up the equation of a ruled surface whose generators areparallel to the plane xOy and intersect two ellipses

y2 z2 y2 z2x=a; x=—a.b2 c2 c2 b2

5.21. Make up the equation of a ruled formed by the straightlines intersecting the curve = (u, u2, u3), parallel to the plane xOy,and intersecting the axis Oz.

36

5.22. Make up the equation of the surface formed by the straight linesparallel to the plane x + y + z = 0, intersecting the axis Oz, and circum-ference = I b, a cos u, a sin u j.

5.23. Make up parametric equations of the surface formed by thestraight lines intersecting the circumference x2 + = 1, y = 0 andstraight lines y = 1, z = 1 and x = 1, z = 0.

5.24. Make up the equation of the surface formed by the tangents tothe helix = (acosv, asinv, bvj (developable helicoid).

5.25. Make up the equation of the conic surface with the vertex at thepoint (0, 0, — c) and the directing line (x2 + y2)2 = a2(x2 — y2).

5.26. Given a straight line AB and a curve in a plane ir. Thecurve moves uniformly in the plane so that each of its points travelsparallel to AB. The plane ir is, at the same time, in uniform rotationabout AB. Make up the equation of the surface described by the curve

This surface is called a helical surface. A special case of a helical sur-face is a right helicoid (see Problem 5.7); in this case, = Q(u) is astraight line orthogonal to AB.

5.27. Let r r(u) be a curve whose curvature k is other than zero.Normal planes are drawn through each of its points, and a circumferencewith centre on the curve r = r(u) and given radius a, a > 0, ak < I,is constructed in every such plane. The locus of these circumferences isa tubular surface S.

(a) Make up the equation of the surface S.(b) Prove that any normal to the surface S intersects the curve r = r(u)

and is a normal to this curve.5.28. Find the surface S, given that all its normals meet at one point.5.29. Show that the volume of the tetrahedron formed by the

intersection of the coordinate planes and the tangent plane to the surface

x=u, y=v, z=a3/uvdoes not depend on the choice of the point of tangency on the surface.

5.30. Show that the sum of the squares of the coordinate axis interceptsof a tangent plane to the surface

x = u3sin3v, y u3cos3v, z = (a2 — u2)312

is constant.5.31. Show that the tangent plane meets the conoid

x ucosv, y = usinv, z = asin2vin an ellipse.

5.32. Prove that the planes which are tangent to the surface z = xf(y/x)are concurrent.

5.33. Make up the equation of the tangent plane and normal to thehelicoid

r = lvcosu, vsinu, kul.

37

5.34. Make up the equation of the tangent plane to the surface

xyz = a3.

5.35. Given that a surface is formed by the tangents to a curve C, provethat this surface possesses the same tangent plane at all points of atangent to the curve C.

5.36. Given that a surface is formed by the principal normals of a curveC, make up the equation of the tangent pJane and the normal at anarbitrary point of the surface.

5.37. Make up the equation of the tangent plane and the normal tothe surface formed by the binormals of a curve C.

5.38. Prove that the normal of a surface of revolution coincides withthe principal normal to the meridian and intersects the axis of rotation.

5.39. Prove that if all normals of a surface intersect one and the samestraight line, then the surface is a surface of revolution.

5.40. A ruled surface (see the definition in Problem 5.17) is said to bedevelopable if the tangent plane to the surface is the same at all pointsof an arbitrary generator.

Prove that the ruled surface

R r(u) + va(u)

is developable if and only if

r'aa' = 0.

5.41. Prove that any developable surface may be partitioned into thefollowing parts:

(i) a part of the plane;(ii) a part of a cylinder;(iii) a part of a cone;(iv) a part of a figure consisting of the tangents to a certain non-plane

line. In the last case, the indicated line is called an edge of regression.5.42. Find the envelope and the edge of regression of the family of

ellipsoids

v2\ :2

+ —I -f = I,b2J

where cs is the parameter of the family.5.43. Find the envelope of the family of spheres constructed on the

chords parallel to the major axis of the ellipsexl

y2-f = I, z = 0

a2 b2

as on diameters.

38

5.44. Find the envelope and the edge of regression of the family ofspheres whose diameters are the chords of the circumference

x2+y2—2x=O, z=O,that pass through the origin.

5.45. Two parabolas are placed in perpendicular planes, possessing thecommon vertex and the common tangent to the vertex. Find the envelopeof the family of planes which are tangent to both parabolas.

5.46. Find the envelope of the family of spheres with constant radius,whose centres are placed on a given curve = a(s) (canal surface).

5.47. Find the edge of regression of the family of spheres with constantradius a, whose centres are placed on a curve =

5.48. Find the envelope and the edge of regression of the family ofspheres with radius a, whose centres are placed on the circumference

x2+y2=b2, z=O.5.49. Find the envelope and the edge of regression of the family of

spheres passing through the origin and whose centres are placed on thecurve

r = u2, uJ.

5.50. Find the envelope of the family of ellipsoids

x2 y2 z2—+-—+—=la2 b2 c2

whose semi-axis sum

a+b+c=1is given.

5.51. Find the surface whose tangent planes cut off on the coordinateaxes line segments such that the sum of their squares equals a2.

552. Find the surface whose tangent planes cut a tetrahedron ofconstant volume off the coordinate angle.

5.53. Find the envelope and the edge of regression of the family ofplanes

xa2 + ya + z = 0,

where a is the parameter of the family.5.54. Find the envelope and edge of regression of the family of planes

xsina — ycoscr + z = aa,

where a is the parameter of the family.

39

5.55. Find the envelope, the characteristics, and the edge of regressionof the family of osculating planes of a given curve.

5.56. Find the envelope, the characteristics, and the edge of regressionof the family of normal planes to a given curve.

5.57. Find the characteristics, the envelope, and the edge of regressionof the family of planes

rn + D = 0, ii = n(u), D = D(u), 1,

where u is the parameter of the family.5.58. Find the developable surface through the two parabolas

(1) y2 = 4ax, z = 0;

(2) x2 = 4ay, z = b.

5.59. Show that the surface x cosv — (u + v)sinv, y sinv ++ (u -1- v)cosv, z = u + 2v is developable.

5.60. Show that the surface x = u2 + 1/3 v, y = 2u3 + uv,

z = + 2/3 u2v is developable.5.61. Given a paraboloid

x = 2aucosv, y = 2businv, z = 2u2(acos2v + bsin2v),

where a and b are constants, make up the equation of a curve on thesurface so that the tangent planes to the surface may form a constantangle with the plane xOy along the curve.

Show that the characteristics of this family of tangent planes form aconstant angle with the axis z. Find the edge of regression of the envelope.

5.62. Find the edge of regression of the developable surface whichtouches the surface az = xy at the points where it meets the cylinder

= by.5.63. Show that the developable surface passing through two circum-

ferences x2 + y2 = a2, z 0 and x2 = b2, y = 0 intersects theplane x 0 in an equilateral hyperbola.

5.64. Calculate the first fundamental form of the following surfaces:

(1) r = asinucosv, (sphere);

(2) r Lacosucosv, bsinucosv, (ellipsoid);

(a/ i\ b/' '\. ci(3) r = ( v + — I cosu, — I v + — I sinu, — I V + —(2\ vJ 2\ v/ 2\ V

(hyperboloid of one sheet);

(4)r = +b

V —_U, UV — 1](hyperboloid of one sheet);

v + u v + u v + u

40

(af 1\ bf I'\. ci(5) r = ( V — — J cosu, — V Sinu, — V —(2\ vi 2\ vJ 2\ v

(hyperboloid of two sheets);

(6) r = (elliptic paraboloid);

(7) r = + (U — 2uv) (hyperbolic paraboloid);

(8) r = by sinu, (cone);

(9) r = {a cosu, b sinu, vJ (elliptic cylinder);

(af l\ b/ l\(10) r = + —(U — —I, V(2\ uJ 2\ u/(hyperbolic cylinder).

5.65. Calculate the first fundamental form of the following surfaces:

(1) r = Q(s) + Xe, e = const (cylindrical surface);(2) r = vQ(s) (conical surface);

(3) r = Q(s) + Xe(s) = 1) (ruled surface);(4) r = Q(s) + + (canal surface);(5) r = çc(v)cosu, (surface of revolution);(6) r = t(a + bcosv)cosu, (a ÷ bcosv)sinu, bsinv( (torus);

(7) r = vcosu, vsinu, ku (minimal helicoid);(8) r = a(s) + Xv(s) (surface of principal normals);(9) r = Q(s) + X/3(s) (surface of binormals).

5.66. The first fundamental form of a surface is the following:

ds2 = du2 + (U2 + a2)dv2.

(i) Find the perimeter of the curvilinear triangle formed by the intersect-ing curves

u = ± 1/2 av2, v = 1.

(ii) Find the angles of this curvilinear triangle.(iii) Find the area of the triangle formed by the intersecting curves

v=l.5.67. The first fundamental form of a surface is the following:

ds2 = du2 + (U2 + a2)dv2.

41

Calculate the angle at which the curves

u+v=O, u—v=Ointersect each other.

5.68. Find the equations of the curves which bisect the angles betweenthe coordinate lines of the paraboloid of revolution

x = ucosv, y = usinv, z = 1/2 U2.

5.69. Find the curves intersecting the curve v = const at a constantangle 0 (loxodromes) on the surface

x = ucosv, y = usinv, z = aln(u + ,,/u2 a2)

5.70. Given a surface

r = (usinv, ucosv,

Find:(i) the area of the curvilinear triangle 0 u sinhv, 0 v v0;

(ii) the lengths of the sides of this triangle;(iii) the angles of this triangle.5.71. Let £ be a curve whose equation is r = r(u), the curvature k(u),

and the torsion where u is natural parameter of the curve £. LetS be the surface

r(u, = r(u) + +

where p, are the unit vectors of the principal normal and the binormalof the curve £, respectively, a = const > 0, ak(u) < 1. One and thesame point on S is assumed to correspond to the coordinates (u, and(u, + 22r).

(i) Find the first fundamental form for the surface S.(ii) Find the curves on S which are orthogonal to the circumferences

u = const.(iii) Calculate the area of the region on the surface S bounded by the

circumferences u = u1, u = u2.

(iv) Using the result of (iii), find the area of the torus obtained byrotating the circumference (x — b)2 + = a2, b > a > 0, about theaxis z.

(v) Find the area of the surface S in that special case where £ is anarc of the helix x = rcost, y = rsint, z = bt, 0 f ir, r > a, b 0.

5.72. Given a surface

r = {Q(u)cosv, Q(u)sinv, z(u)J,

where Q'(u)2 + z'(u)2 = 1, u1 < U < U2, 0 < v < v0, v0 < 27r, provethat it can be placed inside a cylinder x2 + = with arbitrarily small

42

positive radius e by bending itself. In bending, a self-covering of the sur-face is possible.

5.73. Find the surface of revolution which is locally isometric to thehelicoid

r = (usinv, ucosv, v).

5.74. Show that the following helical surface (conoid)

X = QCOSV, 3' Qsinv, Z = Q + V

covers (i.e., is locally isometric to) the surface of revolution (hyperboloidof revolution)

x = y = z =

the correspondence of the covering points being given by the equations

= V + r2 = + 1.

5.75. Show that the helical surface

covers (is locally isometric to) the surface of revolution

x = y = z = aV2In(r + ,1r2 — a2).

5.76. Show that any helical surface

x = ucosv, y = usinv, z = F(u) + av

covers (i.e., is locally isometric to) a surface of revolution so that the heli-cal lines are transformed into parallels.

5.77. Prove that with a convenient choice of curvilinear coordinateson a surface of revolution, its first fundamental form can be transformedto the following:

ds2 = du2 + G(u)dv2.

5.78. Transform the first fundamental form of the sphere, torus,catenoid, and pseudosphere to the following:

ds2 = du2 + G(u)dv2.

5.79. A curvilinear coordinate system on a surface is said to beisometric if the first fundamental form of the surface is, with respect tothese coordinates, as follows:

ds2 = A(u, v)(du2 + dv2).

Find isometric coordinates on the pseudosphere.

43

5.80. A right-angled triangle whose sides are arcs of great circum-ferences of a sphere is given on the sphere. Find (a) the relations betweenthe sides of the triangle, (b) its area.

5.8L A spherical tune is a figure formed by two great semi-circumferences with common ends. Calculate the area S of a June withan angle a at the vertex.

5.82. Prove that any cylindrical surface is locally isometric to the plane.5.83. Prove that any conical surface is locally isometric to the plane.5.84. A Liouville surface is one whose first fundamental form can be

transformed to the following:

ds2 = (1(u) + g(v))(du2 + dv2).

Prove that a surface locally isometric to a surface of revolution is aLiouville surface.

5.85. Prove that any surface of revolution can be locally conformallymapped onto the plane.

5.86. Calculate the second fundamental form of the following surfacesof revolution:

(I) r = (Rcosucosv, Rcosusinv, Rsinu (sphere);

(2) r = (acosucosv, acosusinv, csinu( (ellipsoid of revolution);

(3) r = (acoshucosv, acoshusinv, csinhu( (hyperboloid of revolutionof one sheet);

(4) r = (asinhucosv, asinhusinv, ccoshu) (hyperboloid of revolutionof two sheets);

(5) r = (ucosv, usinv, u2 J (paraboloid of revolution);

(6) r = (Rcosv, Rsinv, u} (circular cylinder);

(7) r = (ucosv, usinv, kuj (circular cone);

(8) r = ((a + bcosu)cosv, (a + bcosu)sinv, bsinu( (torus);

(9) r = {acosh acosh (catenoid);

(10) r = 1asinucosv, asinusinv, a (Intan -f-- + cosu2

(pseudosphere).

5.87. Calculate the second fundamental forth of the right helicoidx = ucosv, y = usinv, z = av.

44

5.88. Calculate the second fundamental form of the catenoidx = + a2cosv,

= a2sinv,

z = aln(u +

5.89. Calculate the second fundamental form of the surface xyz a3.5.90. Given the surface of revolution

r(u, = tx(u), Q(u)cosço, Q(u) >

(i) find the second fundamental form;(ii) find the total curvature K at an arbitrary point of the surface and

the dependence of the sign of K on the sense of convexity of the meridian;(iii) calculate K for the special case Q(u) = U,

/ 2j —u 2 2x(u) = ± I aln — a — u , a > 0\ u

(pseudosphere);(iv) find the mean curvature H at an arbitrary point of the surface

of revolution;(v) select the function Q = Q(x) for the special case x = u so that

H = 0 on the whole surface.5.91. Given a curve = with the natural parameter u, curvature

k = k(u) 0, and torsion x = x(u) 0. Let r = i(u) be the unittangent vector of this curve. Find (a) K, (b) H for the surface of tangents

r(u, v) = Q(u) + vr(u), V > 0.

5.92. Find the expression for the total curvature of the surface whosefirst fundamental form with respect to these coordinates is

ds2 = du2 + G(u, v)dv2.

5.93. Find the total curvature of a surface whose first fundamental form

ds2 = du2 + eZudv2.

5.94. Find the total curvature of the surface given by the equationF(x, y, z) = 0.

5.95. Find the total and mean curvatures of a surface z = f(x, y).5.96. Find the principal curvature radii of the surface

zy = xtan—.

a

45

5.97. Find the principal curvature radii of the surface

x = cosv — (u + v)sinv,

y = sinv + (U + v)cosv,

z = U + 2v.

5.98. Calculate the total and mean curvatures of the helical surface

x=ucosv, y—usiflv, z=u+v.5.99. Calculate the total and mean curvatures of the surface

x = 3u + 3uv2 — u3,

y = v3 — 3v — 3u2v,

z = 3(u2 — v2).

5.100. Show that the mean curvature of the helicoid (see Problem 5.7)equals zero.

5.101. Show that the principal curvature radii of the right helicoid

x = ucosv, y usinv, z f(v),

where f(v) is an arbitrary analytic function of variable v, have unlikesigns.

5.102. Find the total and mean curvatures of the surface formed bythe binormals of a given curve.

5.103. Find the total and mean curvatures of the surface formed bythe principal normals of a given curve.

5.104. Let S be a certain given surface. Mark off segments of the samelength and direction on the normals to the surface S. The ends of thesesegments describe a surface S* "parallel" to the surface S. If the equationof the surface S is r = r(u, v), then the equation of S* is

= r(u, v) + an(u, v),

where n(u, v) is a unit normal vector of S.Express the coefficients of the first and second fundamental forms of

the surface in terms of the coefficients of the first and second fun-damental forms of the surface S.

5.105. Express the total curvature K* of the surface S* "parallel" toa surface S in terms of the total and mean curvatures of the surface S.

5.106. Express the mean curvature H* of the surface "parallel" toa surface S in terms of the total and mean curvatures of the surface S.

5.107. Make up the equation of the minimal surface S* "parallel" toa surface S if for the surface S the ratio H/K = const.

5.108. Given a surface of constant mean curvature H. Segments oflength 1/2 Hare marked off on all its normals. Prove that the total curva-ture of the surface so formed and "parallel" to the given one is constant.

46

5.109. Segments of length I/s/k are marked off on all the normalsof a surface with constant positive total curvature K. Prove that the meancurvature of the surface so formed is constant. Calculate it.

5.110. Prove that the total and mean curvatures at the correspondingpoints of two parallel surfaces are related by the formula

H2 — 4K — H*2 — 4K*

K2 -A line on a surface is called a line of curvature if it has the principal

direction at each of its points. Lines of curvature are determined by thedifferential equation

dv2 —dudv du2

E F G =0.L M N

5.111. Find the lines of curvature on the surface

a b uvx=—(u—v), y=—(u+v), z=—.

5.112. Find the lines of curvature of the helicoid

x = ucosv, y = usinv, z = av.

5.113. Prove that, in covering (local isometry) the catenoid

x = Vu2 + a2cosv, y =

z = aln(u + a2)

with the helicoid

x = ucosv. y = usinv, z = av,

the lines of curvature are transformed into asymptotic lines.5.114. Find the lines of curvature of the surface

r(u, v) = Q(u) + f(v)a + g(v) [r(u) x a],

where r(u) = r'(u), i-(u)I = 1, (r(u), a) = 0, al = I, a is a constantvector.

5.115. A plane curve y is given by an equation = where uis the natural parameter, k = k(u) its curvature (0 < k < 1/a), p theprincipal normal unit vector of 'y, e the unit normal vector to the planein which the curve -y lies. A surface S is given by the equation

r(u, = + +

47

(i) Find the Gaussian curvature of the surface S.(ii) Find the mean curvature of the surface S.(iii) Find the lines of curvature of the surface S.5.116. Find the lines of curvature of the surface

r(u, = Q(u) + av(u)cosçe' +

where p and are the unit vectors of the principal normal and the binorm-al to the curve = Q(u) having the natural parameter u, curvaturek(u) < 1/a, and torsion x(u).

5.117. Find the lines of curvature of the surface

r(u, v) = [u(3v2 u2—

J_), v(3u2 — v2—1--).

2uv}.

Find its total and mean curvatures at each point.5.118. Prove that H2 K. When does the equality hold?5.119. Let X and V be orthogonal tangent vectors at some point of

a surface. Prove that

H -L{I(X, X) + I(Y, Yfl,

where 1(,) is the second fundamental form of the surface.5.120. Assume that the first fundamental form of a surface is as

follows:

ds2 = E du2 + Gdv2.Prove that

( /3E\ /3G1 av \ a! au

K=

+ —5.121. Assume that two surfaces M1 and M2 meet in a curve C. Let

k be the curvature of C, X1 normal curvatures of C in and 0 the anglebetween the normals of M1 and M2. Prove that

k2sin20 = + — 2XlX2cosO.

Two directions in the tangent plane of a surface which are determinedby two vectors a and b are said to be conjugate if v'z(a, b) = 0, i.e., if

La1bi + M(a1bz + a2b1) + Na2b2 = 0,

where a (ai, a2), b = (b1, b2). A net of lines on a surface is saidto be conjugate if the tangent vectors to the lines of different familiesof this net are conjugate at each point.

48

A direction determined by a vector h is said to be asymptotic ifh) = 0. A line on a surface is said to be asymptotic if the tangent hasthe asymptotic direction at each of its points. An asymptotic line isacterized by the equality = 0 which is held at all its points. Asymptoticlines are determined by the differential equation

Ldu2 2Mdudv + Ndv2 = 0.

5.122. Find the asymptotic lines of the surface

/x y\y x

5.123. Find the asymptotic lines of the surface z = xy2.5.124. Find the asymptotic lines of the surface

2 32 2x=u y=u +uv, z=u +—uv.

Construct the projections of the asymptotic lines, passing through thepoint u = 1, v = 1/2, onto the plane xy.

5.125. Find the asymptotic lines of the surface

x = a(l + cosu)cotv, y = a(1 + cosu), z =5.126. Prove that for an asymptotic line on the surface, x2 = —K

(where x is the torsion and K the total curvature).5.127. Find the torsion of the asymptotic lines of the surface formed

by the binormals to a given curve.5.128. Find the torsion of the asymptotic lines of the surface formed

by the principal normals to a given curve.5.129. Show that the coordinate lines of the surface

a b uvx=—(u+v), y=—(u—v), z=—

are straight lines. Find the lines of curvature.5.130. Show that the coordinate lines on the surface

x = fi(u), y = çoi(v), z = fz(u) + P2(V)

are plane and form a conjugate system.5.131. Show that the coordinate lines on the surface

r = M(u) + Q(v)

are conjugate.5.132. Prove that the sum of the normal curvature radii for each pair

of conjugate directions is the same at an arbitrary point of a surface.

4 201849

5.133. Prove that the product of the normal curvature radii for a pairof conjugate directions attains its minimum for the lines of curvature.

5.134. Prove that the ratio of the principal curvature radii is constantfor the surface of revolution obtained by rotating a parabola about itsdirectrix.

5.135. Prove that if one of the lines of curvature of a developable sur-face lies on a sphere, then all the remaining non-rectilinear lines of curva-ture lie on concentric spheres.

5.136. Prove that the normal curvature of an orthogonal trajectory ofthe asymptotic lines of a surface equals the mean curvature of the surface.

5.137. Prove that a line of curvature is plane if its osculating planeforms a constant angle with the tangent plane to the surface.

We call a line whose principal normal coincides with the normal toa surface at each of its points a geodesic line of the surface. There isa unique geodesic line passing through each point of the surface and hav-ing a given direction.

The length of the projection of the curvature vector kn onto the tangentplane of a surface is called the geodesic curvature kg of a line placedon the surface.

The geodesic torsion associated with a given direction is the torsionof the geodesic line passing in this direction.

5.138. Prove that a geodesic line on a surface can be fully determinedby one of the following properties:

(i) The normal to a surface at each point of the line, where its curvatureis other than zero, is a principal normal.

(ii) The normal to a surface lies in the osculating plane of the line ateach of its points where its curvature is other than zero.

(iii) The geodesic curvature equals zero at each point of the line.(iv) The curvature equals the absolute value of the normal curvature

at each point of the line.(v) The rectifying plane coincides with the tangent plane to the surface

at each point of the line where its curvature is other than zero.5.139. Prove that any straight line on a surface is a geodesic line.5.140. Given that two surfaces touch each other along a line 1, prove

that if / is a geodesic line on one surface, then it is geodesic on the other.5.141. Prove that the differential equation of the geodesic lines of a

surface r = r(u, v) can be represented in the form Ndrd2r 0, whereN is the normal vector of the surface.

5.142. Prove that geodesic lines of' the plane are straight lines and onlythey.

5.143. Find the geodesic lines of a cylindrical surface.5.144. Find the geodesic lines of a developable surface.5.145. Find the geodesic lines of the circular cone x2 + = z2.

50

5.146. Find the geodesic lines of the helicoid

r usinv,

5.147. Find the geodesic lines of an arbitrary conical surface.5.148. Prove that the meridians of a surface of revolution are geodesic

lines.5.149. Prove that a parallel of a surface of revolution is geodesic if

and only if the tangent to a meridian at the points where the meridianmeets the parallel is parallel to the axis of rotation.

5.150. Find the geodesic lines on the sphere.5.151. Show that the geodesic lines of a surface whose first fundamental

form is

ds2 = v(du2 + dv2)

are parabolas on the plane u, v.

5.152. Prove that a geodesic line is a line of curvature if and only ifit is plane.

5.153. Prove that a geodesic line is asymptotic if and only if it isstraight.

5.154. Prove that the geodesic curvature of a line u = u(s), v = v(s)on a surface r = r(u, v) can be calculated by the formula

kg

where m is the unit normal vector of the surface.5.155. Find the geodesic curvature of the helical lines of the helicoid

r = (ucosv, usinv,

5.156. Prove that the geodesic torsion of a line u = u(s), v = v(s)on the surface r r(u, v) can be calculated by the formula

= tthm,

where m is the unit normal vector of the surface.5.157. Prove the following statement: for a line on a surface to be a

line of curvature, it is necessary and sufficient that the geodesic torsionshould equal zero at each of its points.

5.158. Show that on a surface with the first fundamental form

ds2 = [ç,(u) + (du2 + dv2)

(a Liouville surface) the geodesic lines are determined by the equation

du dv± =0,

'.Ico(u) + a — a

where a is an arbitrary constant.

51

5.159. Given a triangle Twhose area is a and the sides are arcs of greatcircumferences on a sphere of radius R0, find the sum of the interiorangles of the triangle T

5.160. Let T be a triangle whose sides are geodesic lines constructedon a surface with constant Gaussian curvature K = — a2 < 0. Assumingthat the area a of T is given, find the sum of its interior angles.

5.161. Given that a surface S is obtained by a certain bending of aX2 2 2

portion of the ellipsoid —i- + + —i-- = 1 determined by the

inequalities x > 0, y > 0, z > 0, find the area 0* of the spherical imageof the surface S.

5.162. Given that a surface R R(u, v), u1 < u <U2, v1 < v << V2, has the first fundamental form ds2 = du2 + B2(u, v)dv2, findthe area 0* of the spherical image of this surface.

5.163. Let -y be a closed geodesic line without self-intersections on aclosed convex surface S. Prove that thespherical image of the curve -ydivides the Gaussian sphere into two parts equal in area.

5.164. Given that ds2 = dQ2 + sinh2Q in the geodesic polar coor-dinates co) on the non-Euclidean plane, find the length 5(Q), geodesiccurvature kg(Q), and rotation H(Q) of the geodesic circumference

= const. Calculate urn kg(Q), lim H(Q). Compare the results obtained

with the similar quantities for the Euclidean plane.5.165. Given a plane P1 with the metric ds2 = du2 + cosh2udv2,

— < u < co, —00 < v < 00, and a plane P2 on which ds2 = dQ2 ++ sinh2Qdço2 with respect to the geodesic polar coordinates ço), prove

that the planes and P2 (with the first fundamental forms given onthem) are isometric.

5.166. Given that a surface S is defined by a vector function of theform R = R(u, v) of class C2, verify that the quantity dn2 = (dn, dn),where n is the normal unit vector of the surface 5, is a quadratic formwith respect to the differentials du, dv (the so-called third fundamentalform of the surface 5). Express dn2 in terms of the first and second fun-damental forms of the surface S.

5.167. Prove that the sum of the squares of the curvature and torsionof a geodesic. line is equal to —K on a minimal surface.

5.168. Prove that the plane and catenoid are the unique minimal sur-faces of revolution.

5.169. Prove that among ruled surfaces, the minimal .are the plane andthe right helicoid.

5.170. Prove that for the mean curvature of a surface 5, the followingformula is valid:

do — da*H= urn

a-0 2ada52

where d*i and da* are the corresponding elements of the area of the paral-lel surfaces S and

5.171. Prove that the area of any portion of a minimal surface cannotbe less than the area of the corresponding portion of a parallel surface.

5.172. Prove that the limit of the ratio of the area of spherical represen-tation of a surface S to the area of the corresponding region of the surfaceS equals the total curvature of the surface in magnitude and sign.

5.173. Prove that if one of the principal curvature radii of a surfaceis constant, then the surface is the envelope of a family of spheres withconstant radius whose centres lie on a certain curve.

5.174. Given that a circular cylinder is intersected by a plane not parallelto the axis of the cylinder, what line will the line of intersection be trans-formed into in covering (locally isometric mapping) the plane with thecylinder?

5.175. Prove that if a material point moving across some surface is notacted upon by external forces, then it is moving along a geodesic line.

5.176. Prove that in a locally isometric mapping of surfaces, geodesiclines are transformed into geodesic.

5.177. Prove that two surfaces of the same constant Gaussian curvatureare locally isometric.

5.178. Prove that any surface of constant positive Gaussian curvatureis locally isometric to the sphere.

5.179. Prove that any surface of constant negative Gaussian curvatureis locally isometric to the pseudosphere.

5.180*. Prove that all geodesic lines which are different from meridiansare closed on the surface S given by the equations

1 1

x = — cosu y = — cosu

z = Jl — ±-sin2u du,

- -f-- u -f-, 0 'p 2ir.

5.181. Given the differential equation of motion of a point electricalcharge in the field of a magnetic pole, viz.,

r"(t) = c Ir(t)13 [r(t), r'(t)], c = const,

prove that the path of the charge is a geodesic line of a circular cone.

53

5.182. Prove that the Gaussian curvature of the metric ds2 =v)(du2 + dv2) can be represented in the form

K =24

a2 a2where = + —i- is the Laplace operator.

0u2 av

5.183*. Prove that there are no closed geodesic lines on 1-connectedsurfaces such that the Gaussian curvature is non-positive at all of theirpoints.

6Manifolds

6.1. Prove that an n-dimensional sphere determined in 11' by theequation + + . . . + = 1 is a smooth manifold. Constructthe atlas of charts for

6.2. Prove that the two-dimensional torus T2 obtained by rotating aboutthe axis Oz of a circumference lying in the plane Oxz and not intersectingthe axis Oz is a smooth manifold. Construct the atlas of charts.

6.3. Prove that the union of two coordinate axes in R' + is not amanifold.

6.4. Show that an atlas consisting of only one chart cannot beintroduced on the sphere S" C

6.5. Determine whether the following plane curves are smooth man-ifolds: (a) a triangle, (b) two triangles with only one common point, viz.,a vertex.

6.6. Prove that the n-dimensional projective space is a smooth (andreal-analytic) manifold.

6.7. Prove that the n-dimensional complex projective space CP" is asmooth (and complex-analytic) manifold.

6.8. Prove that the graph of the smooth function = f(xi,is a smooth manifold.

6.9. Prove that the group SO (2) is homeomorphic to the circumference.What manifold is the group 0(2) homeomorphic to?

6.10. Prove that the group SO(3) is homeomorphic to the projectivespace RP3.

6.11. Prove that the groups GL(n, R), GL(n, C) are smooth manifolds.6.12. What manifold is the set of all straightlines on the plane R2

homeomorphic to?Form the equations of the following manifolds in R3:

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6.13. The cylinder with a directing curve Q = Q(u) and a generatorparallel to a vector e.

6.14. The cone with the vertex at the origin and directing curve =

6.15. The surface made up of the tangents to a curve = Q(u).6.16. The surface formed by a circumference moving so that its centre

is on a curve = (u ) and its plane normal to the curve at each ofits points.

6.17. Prove that the Jacobian matrix of the composite of smooth map-pings is the product of the Jacobian matrices of the factors.

6.18. Prove that the rank of a Jacobian matrix does not depend onthe choice of a local coordinate system.

6.19. Calculate the rank of the Jacobian matrix of the mapping

6.20. Let f: U —÷ R" be a smooth manifold of an open domainU C and the Jacobian 811 0 at a point p E U

Prove that there exists an open domain V C p E V such that f( V) == W is an open set, 11 v a homeomorphism, and the inverse mapping(fvY' smooth.

6.21. Letf: U Vbe a smooth mapping of open domains in whichhas a smooth inverse mapping.

Prove that the Jacobian 18J1 0 at each point p E U6.22. Set up explicit formulae for a smooth homeomorphism of the

open disk = E < onto the Euclidean space

6.23. Prove that any smooth manifold has an atlas such that each chartis homeomorphic to a Euclidean space.

6.24. Give an example of a smooth one-to-one mapping which is nota diffeomorphism.

6.25. Construct a smooth function x,) (of class C°) equalto unity on a ball of unit radius, vanishing outside a ball of radius 2,and such that 0 f 1.

6.26. Let M be a manifold, p E U C M a neighbourhood of a pointp. Prove that there exists a smooth function f such that 0 f 1,

f(p)=l,f(x)=OonM\U6.27. Let M be a manifold, A = A a closed set, and U J A an open

domain. Prove that there exists a smooth function! such that 0 f 1,

j1A = 1, JiM/V 0.6.28. Prove that any continuous function in R" can be uniformly ap-

proximated, as close as we please, by a smooth function.6.29. Prove that any continuous mapping of smooth manifolds can be

approximated, as close as we please, by a smooth mapping.

5.5

6.30. Let a torus T2 C R3 be formed by rotating a circumference aboutsome axis (standard embedding). Prove that coordinates x, y, z are smoothfunctions on the torus

6.31. Let a torus T2 C R3 be standardly embedded in R3, and thefunction f: T2 S2 associate each point p E T2 with a vector of unitlength normal to the torus T2 at the point p. Prove that f is a smoothmapping.

6.32. Prove that a mapping f: S2 RP2 associating a point p on thesphere S2 with the straight line which passes through the origin and thepoint p is a smooth mapping.

6.33. Prove that two smooth structures on a manifold coincide if andonly if the spaces of smooth functions (with respect to these structures)coincide.

6.34. Let be the solution set of the equations

gi(xi XN) = 0, (i = 1 N — n),

and the equality for the rankfi

= N — n be held.Prove that M" is a submanifold.6.35. Let M" C RN be a submanifold. Prove that for any point p E

there exists a coordinate set . . . , such that the projection ofonto the subspace = tx,1, . . . , is a local diffeomorphism of aneighbourhood of the point p of the manifold onto an open domainin

6.36. Let C RN be a submanifold. Prove that the manifold W isspecified locally by a system of equations gi(xi, . . . , XN) = 0(i = 1,. . . ,N—n), the equality for the rank = N—n beingfulfilled.

6.37. Let C RN be a compact submanifold. Prove that there existsa set of smooth functions Ii, . . . , 1k on R such that the solution setof the system of equations fi = 12 = . . . = 1k = 0 coincides with M'7and the rank of the Jacobian matrix equals N — n (k N— n).

6.38. Show that the stereographic projection of a sphere onto a tangentplane from the pole placed opposite the point of contact is a diffeomor-phism everywhere except the projection pole.

6.39. Prove that the spaces and Rm are not diffeomorphic whenn m.

6.40. Prove that the groups SL(n, R), SL(n, C) are smooth subman-ifolds in spaces of real (or complex) square matrices of order n.

Prove that the group S0(n) is a smooth submanifold of the spaceof all square matrices of order n.

6.42. Prove2 that the groups U(n), SU(n) are smooth submanifolds inthe space of complex square matrices of order n.

56

6.43. Show that the matrix mapping A —f exp(A) is a smooth homeo-morphism in a neighbourhood of the null matrix from the inverse image,and a neighbourhood of the unit matrix from the image. Show that theinverse mapping can be specified by the corres )ondence B ln(B).

6.44. Prove that some of the Cartesian coordinates of the matrixln(A - 'X) can be taken as local coordinate systems in a neighbourhood(IA of a matrix A on each of the groups listed in Problems 6.40-6.42.Show that coordinate changes are smooth functions of class C°° respectiveto the coordinate systems indicated.

6.45. The Riemann surface of the algebraic function w = whereP(z) is a polynomial, is given by the equation w" — P(z) = 0. Find acondition for the roots of the polynomial under which the Riemann sur-face is a two-dimensional submanifold in C2.

6.46. Show that the projection of the direct product X x Y of twomanifolds X and Y onto the factor X is a smooth mapping.

6.47. Prove that a compact smooth manifold W can be embedded inthe Euclidean space RN for a convenient dimension N <

6.48. Prove that a smooth function on a compact smooth manifoldM can be represented as a coordinate under a certain embeddingM C

6.49. Prove that the product of spheres can be embedded in RN ofcodimension 1.

6.50. Prove that if dim X < dim Y and f: X Y is a smooth map-ping, then the image of the mapping f does not coincide with Y.

6.51. Prove that a two-dimensional, compact, smooth and closed man-ifold can be immersed into R3.

6.52. (Whitney lemma.) Prove that a compact, smooth and closed man-ifold M" can be embedded in the Euclidean space and immersedinto

6.53. Let f: X Y be a smooth mapping of a compact and closedmanifold into a manifold r. Let yo E Y be a regular point of themapping f Prove that the inverse image '(yo) consists of a finitenumber of points.

Letf: X Ybe a smooth mapping of smooth manifolds, and M C Va smooth submanifold. The mappingf is said to be transverse along the

M if for every point x E '(M), the tangent space Tf(x)(Y)to the manifold V is the sum (generally speaking, not direct) of thetangent space to the manifold M and the image ofthe tangent space to the manifold X. iWo submanifolds M1 and M2 ofthe manifold Xare said to intersect transversally if an embedding of oneof them is transverse along the other.

6.54. Prove that if y E Y is a regular point of a mapping f: X Y,then f is a mapping transverse along y.

57

6.55. Prove that the definition of a transversal intersection does notdepend on the choice of order in the pair M,, M2.

6.56. Prove that if f: X Y is a mapping transverse along a subman-ifold M C Y, then the inverse imagef '(M) is a submanifold of the man-ifold X. Calculate the dimension of (M).

6.57. Investigate whether the following submanifolds intersecttransversally: (a) the plane xy and the axis z in R3; (b) the plane xy andplane spanned by the vectors 1(3, 2,0), (0,4, —1)1 in R3; (c)thesubspaceV x and the diagonal of the product V x V; (d) the spaces ofsymmetric and skewsymmetric matrices in the space of all matrices.

6.58. For what values of a will the surface x2 + y2 — z2 = 1 intersectthe sphere x2 + + z2 = a transversally?

6.59. Let all the points of a mapping f: X V be regular, and X,Y compact manifolds. Prove that f is a locally trivial fibre map (or fib-ration), i.e., the inverse image f '(U) of a sufficiently small neigh-bourhood of each point y E Y is homeomorphic to the direct productU x In particular, if Yis a connected manifold, then all subman-ifolds 1(y), y E Y are pairwise homeomorphic.

6.60. Let f: S" — be a mapping associating a point x E withthe straight line passing through the point x and the origin in R" Provethat all points of the mapping f are regular.

6.61. Let f: S0(n) -' associate every orthogonal matrix with itsfirst column. Prove that all points of the mapping f are regular. Findthe inverse image f 1(y)

6.62. Letf: U(n) S2" -' associate every unitary matrix with its firstcolumn. Prove that all points of the mapping fare regular. Find the in-verse image f '(y)

6.63. Show that the set of all orthonormal systems consisting ofk vectors from the Euclidean space R' admits a smooth manifoldstructure. Find its dimension. Show that = S" - = 0(n).

6.64. Show that the set Gn,k of all k-dimensional subspaces in theEuclidean space Ri admits a smooth manifold structure. Find itsdimension. Show that Ga,, =

6.65. Vn,ic s k be a mapping associating an orthonor-mal system consisting of k vectors with its first s vectors. Prove that everypoint for the mapping f is regular. Show that the inverse image f '(y)is homeomorphic to the manifold V,5, k-s.

6.66. Let f: 0(n) be a mapping associating every orthogonalmatrix with the subspace generated by the first k columns. Show thatall points for the mapping f are regular. Prove that the inverse imagef '(y) is homeomorphic to the manifold 0(n — k) x 0(k).

6.67. Let f .- X x V M be a smooth mapping, and m0 E Ma regularpoint. Consider the family of mappings X M, = f(x, y).

58

Prove that the point m0 is regular for the almost for all valuesof the parameter y, i.e., when y ranges over an open, everywhere densesubset of Y.

6.68. Solve Problem 6.67 if the point mo is replaced by a submanifoldN C M, and its regularity by transversality of the mappings along thesubmanifold N.

6.69. Verify whether the following manifolds are orientable: (a) asphere (b) a torus 7"; (c) a projective space RP"; (d) a complexprojective space CP"; (e) groups GL(n, R), U(n), SO(n).

6.70. Prove that the Klein bottle is a non-orientable, two-dimensionalmanifold.

6.71. Prove that an arbitrary complex analytic manifold is orientable.6.72. Let M be a manifold with boundary 3M. Prove that the manifold

M can be embedded in the half-space i 0) of the Euclidean spaceRN+ so that 3M lies in the subspace (XN+ = 0).

6.73. Let a boundary 3M consist of two connected components3M = M1 U M2, fl M2 = 0. Prove that the manifold M can beembedded in RN x [0, 1] so that M1 lies in RN x (0), and M2 inRNx (1).

6.74. Prove that an orientable two-dimensional surface possesses acomplex structure.

6.75. Prove that the manifolds S' x S2" -', S2"' x S2" -' possess acomplex structure.

6.76. Prove that a compact closed odd-dimensional Riemannian man-ifold of positive curvature is orientable.

A function w = f(z1,..., z"), = x1' + is said to be holomor-phic if it is continuously differentiable and its differential is a complexlinear form at each point (z', . . . , z").

6.77. Show that if f is a holomorphic function, then

3Ref 3Imf—

öImf 3Refôxk —

6.78. Let WI = . . , z") be a holomorphic vector function mappingC" into Ctm. Find the relation between the real Jacobian matrix of thismapping and its complex Jacobian matrix.

6.79. Prove that a holomorphic vector function f: C" C" producesa local coordinate system if and only if its complex Jacobian is otherthan zero.

59

6.80. Show that S2 admits a complex analytic structure. Describe thesimplest atlas of charts explicitly.

6.81. Show that complex projective spaces CP2 admit a complexanalytic structure. Describe the simplest atlas of charts explicitly.

6.82. Identify S2 with CP'.

7Transformation Groups

7.1 Prove that all one-parameter smooth homeomorphism groups ona compact manifold are in one-to-one correspondence with smooth vectorfields of point trajectory velocities.

7.2. Let X be a smooth connected manifold, and x0, x1 two arbitrarypoints. Find a one-parameter group of smooth transformations suchthat 9'l(Xo) x1. Show that we can assume, without loss of generality,all the transformations to be identity outside a certain compactum.

7.3. Give an example of a vector field on a lion-compact manifoldwhose trajectories are not generated by the action of any one-parametertransformation group.

7.4. Let be a constant vector field respective to angular coordinateson the two-dimensional torus T2. Investigate under which conditions forthe coordinates of the field the integral curves are closed.

7.5. Generalize the previous problem to the case of the torus viz.,let =

s") be a constant vector field respective to angular coor-dinates on the torus Prove that the closure of any trajectory is honieo-morphic to the torus Tk, where k is the number of linearly independentnumbers over the field of rational numbers.

7.6. Let a finite group G act smoothly on a smooth manifold X. Provethat if the action of the group G is free (i.e., each point x E X is trans-formed into itself only under the action of the unit element of the group0), then the factor space X/G is a manifold.

7.7. Show that the projective space RP' is a factor space undera certain action of the group Z2 on the sphere 5".

7.8. Show that the complex projective space CP" is the factor spaceunder the action of the group S1 on the sphere

7.9. Let a finite group G act smoothly on a manifold X, and Xo E Xbe a fixed point under the action of any element of the group 0. Provethat in a neighbourhood of the point xo, there is a local coordinate systemwith respect to which the action of the group G is linear.

60

7.10. Generalize the previous problem to the case of an arbitrary com-pact Lie group.

7.11. Prove that the set of all fixed points under the action of a finitegroup G on a smooth manifold is the union of smooth submanifolds(generally speaking, of different dimensions).

7.12. Let G be a Lie group. Show that the action of the group G onitself via left (or right) translations is smooth.

7.13. Let a Lie group G act on itself via inner automorphisms. Provethat the set of fixed points coincides with the centre of the group G.

7.14. Prove that the group of isometry of a Riemannian space is asmooth manifold.

7.15. List all finite-dimensional Lie groups of transformations of thestraight line R1.

7.16. Find the group of all linear fractional transformations preservingthe disk Izi I in the complex plane. Prove that this group is isomorphicto the group SL(2, R)/Z2 and also to the group of all transformationspreserving the form dx2 + dy2 — dt2 in y, t). Establish a relationto Lobachevskian geometry.

7.17. Prove that the connected component of the Unit element of theisometry group on the Lobachevski plane (under the standard metric ofconstant curvature) is isomorphic to SL(2, Find the total numberof components in the group of motions of the Lobachevski plane.

7.18. A solid ball is pressed in between two parallel planes (which aretangent to it). With the planes moving (so that they remain parallel andat the same distance from each other), the ball rotates without slippingat the points of contact. Consider all motions of' the ball induced by themotion of the upper plane such that the lower point of contact of theball describes a closed trajectory on the lower plane, i.e., the point ofcontact returns to the original position. What part of the group SO(3)can be obtained by such .ball rotations (rotations of the ball are consideredafter its centre returns to the original point)?

7.19. Prove that the isometry group of Euclidean space is generatedby orthogonal transformations and parallel displacements.

7.20. Prove that the isometry group of the standard n-dimensionalsphere is isomorphic to the group of orthogonal transformations of the(n + 1)-dimensional Euclidean space.

7.21. Prove that the groups Sp(l) and SU(2) are isomorphic (as Liegroups). Prove that they are diffeomorphic to the sphere S3. Establishthe relation to quaternions.

7.22. Prove that in the algebra of quaternions, multiplication by a qua-ternion A : x Ax generates the transformation group SU(2). Prove thatthe transformations of the form x —* AxB, where A, B are quaternions,generate the group SO(4). Prove that SO(4) is isomorphic to the factor

61

group S3 x S3/Z2, where S3 is supplied with the structure of the groupSU(2) Sp(l). Find the fundamental group S0(4), and also S0(n) forany n.

7.23. Prove that the Lie groups S0(n), SU(n), U(n), Sp(n) areconnected. Prove that there are two connected components in the group0(n). Find the number of connected components in the group of motionsof the pseudo-Euclidean plane of index 1. Prove that the groupSL(2, R)/Z2 is connected.

7.24. Let us realize the group U(n) and its Lie algebra u(n) as subman-ifolds in the Euclidean space of all square complex matrices of ordern x n and consider the natural embedding of unitary and skewhermitianmatrices in this space.

2,,2—L 2n2—L(a) Prove that U(n) C S , where the sphere S is standardlyembedded in R2n2 = C"2 and has radius

(b) Prove that the Riemannian metric on the group SU(n),which is considered as a submanifold of S2" - coincides with the Car-tan-Killing metric invariant on the group SU(n).

(c) Find the intersection U(n) fl u(n) by considering these sets assubmanifolds in the space C"2.

Solve similar problems for the groups 0(n) and Sp(n).'7.25. Find the factor group where is the group of motions

of the 1.obachevski plane (under the standard metric), the connectedcomponent of the unit element, Indicate all conformal transformationsof the standard metric.

7.26. Find all discrete subgroups of the group of affine transforma-tions of the straight line R'.

7.27. Describe all discrete normal subgroups of the following compactLie groups: 0(n), S0(n), SU(n), U(n), Sp(n).

7.28. Find all symmetry groups of all regular polygons. Find allsymmetry groups (groups of motions) of all regular convex polyhedra inR3. Indicate the non-commutative groups among them.

7.29. Prove that left-invariant vector fields on a Lie group G are inone-to-one correspondence with the vectors of the tangent spaceto the group G at the unit element.

7.30. Prove that the Poisson bracket of two left-invariant vector fieldsis also a left-invariant vector field, i.e., the commutator operation trans-forms the space Te(G) into a Lie algebra.

7.31. Let be a left-invariant vector field, and a one-parameter trans-formation group associated with it. Prove that is a right translationfor any t, i.e., tr E G.

Let G be a Lie group, and x' x" a local coordinate system ina neighbourhood of the unit element (we will assume its coordinates tobe zeroes). Then the operation of multiplication induces the vector-valued

62

function q = q(x, y) = x = (x',. . . , y = (j'1 -. ,

If the function q = q(x, y) is expanded into Taylor's series, then it willassume the following form:

= + 4,j,k

where is an infinitesimal of the third order with respect to the coor-dinates x', y1.

The bilinear expression

= = EE

j,k

determines a certain operation (called the Poisson bracket of the vectorsE and over the tangent vectors in the unit element of the group G.Thus, the tangent space Te(G) has been transformed into an algebra calledthe Lie algebra of the Lie group G. Usually, it is denoted by the smallletter g.

7.32. Show that the following properties are fulfilled in a Lie algebra:(a) [E, 'ii = — En, E];

(b) + tEn, Eb El + El, 'ii = 0.

7.33. Verify that an operation in a Lie algebra g is transformed into

the Poisson bracket of vector fields if a vector E is associated with a(right-) left-invariant vector field.

7.34. Let x(t), y(t) be two curves passing through the unit element ofthe group G,

dx dyE = '7 =

Show that

EE = (x( fi)y' = 0.

7.35. Let y(t) be a one-parameter subgroup of a Lie group. Assumethat y intersects itself. Show that there exists a number L > 0 such thaty(t + L) = 7(t) for all I E R.

7.36. Let G be a compact, connected Lie group. Show that each pointx E G belongs to a certain one-parameter subgroup.

7.37. Let G be a compact group acting smoothly on a manifold M.Show that there is a Riemannian metric on M such that G is the isometrygroup.

7.38. Show that a commutative, connected Lie group is locally isomor-phic to a finite-dimensional vector space.

63

7.39. Show that a compact, commutative, connected Lie group isisomorphic to the torus.

7.40. Show that a commutative, connected Lie group is isomorphic tothe product of the torus and a vector space.

7.41. Let a Lie group G be a subgroup of the matrix group GL(n,C) C C"2 = End(n, C). Show that the commutator operation in the Liealgebra g of the group G, which is understood to be a subspace of End(n,C), coincides with the usual commutator of matrices, i.e.,

= — E g.7.42. Describe the Lie algebras of the following matrix Lie groups:

SL(n, C), SL(n, R), U(n), 0(n), 0(n, m), Sp(n).

7.43. Prove that the operator Y: I IJ is determined by a skew-symmetric matrix. Find the relation between the coefficients of this mat-rix and the coordinates of the vector j1.

7.44. Let Y, Z be two matrices of vector multiplication operators bytwo vectors y, z. Prove that the matrix of the vector multiplication operat-or by [y, z] equals Z} YZ — ZY.

7.45. Prove that a finite group cannot operate effectively on R7.

8Vector Fields

8.1. Prove the equivalence of the three definitions of a tangent vectorto a manifold at a point P:

(a) tensor of rank (1, 0);(b) differentiation operator of smooth functions at the point P;(c) a class of osculating curves at the point P.8.2. Find the derived function f at a point P in the direction of the

vector E:

(a)f = + z2; P = (1, 1, 1), = (2, 1,0);

(b)f = x2y + xz2 — 2; P = (1, 1, —1), = (1, —2, 4);

(c) f xc! + yex — z2; P = (3, 0, 2), = (1, 1, 1);

(d) f = - -s-; P = (1, 1), E = (4, 5).y x

8.3. Find the derivative of the function f = ln(x2 + y2) at the pointP = (1, 2) along the curve = 4x.

8.4. Find the derivative of the function f = tan '(y/x) at the pointP = (2, —2) along the curve x2 + y2 — 4x = 0.

64

8.5. Find the derivative of the function fat the point P along the curve

(a) f = ÷ y2, P = (1, 2), 'y : + y2 5;

(c) f = — y2, P = (5, 4), y : — 9;

(d)f = ln(xy + yz + xz), P (0, 1, 1), -y : x = cOs4 y siflt, Z = 1;

(e)f = x2 + + z2, P = (0, R, ira/2), y = Rsint,

z = at.2

8.6. Find the derivative of the function f = — + --- + at ana2 b c

arbitrary point P = (x, y, z) in the direction of the radius vector of thispoint.

8.7. Find the derivative of the function f = 1/r, rin the direction of its gradient.

8.8. Find the derivative of the function f = yzex in the direction ofits gradient.

8.9. Find the derivative of the function f = f(x, y, z) in the directionof the gradient of the function.

8.10. Let v be a vector differential operator in 93 whose components

are as follows: V = (-f-, -f-, . Show that\8x öy öz/

(a) gradF = V F;(b) divX = (V. X);(c) rotX = [V x

8.11. Prove the formula

div(uX) = u div X + (X, gradu),

where X is a vector field, and u a function in R3.8.12. Prove the formula

rot(uX) = u rot X — [X, gradu].

8.13. Calculate divX[X x A'].8.14. Prove that the vector x = u grad v is orthogonal to rot X.8.15. Show that

(a) div (rot A') = 0;

(b) rot rot X = grad divX —

5-—2018

65

82 82 82where — + — +

3x2 8y2 0z2

8.16. Let X = (x, y, z). Show that

(a) div X = 3;

(I,) rot X 0;

(c) div = 0;

(d) rot = 0;

(e) grad — = —

Find a function such that X =8.17. Let y, z) be the field of velocities of a solid rotating about

some axis. Show that

(a) div(v) = 0;

(b) rot(v) 2w,

where w is an angular velocity vector.8.18. Let X = (x, y, z), and Y a constant vector field. Show that

rot[Y x X] = 2Y.8.19. Show that rot gradF = 0.

8.20. Prove the formula

FMJ + + gradG).

8.21. Solve the equation rotX = Y if

(a) Y = (1, 1, 1);

(b) V = (2y, 2z, 0);

(c) = (0 0 gX —

(d) Y = (6y2, 6z, 6x);(e) Y = (3y2, — 3x2, — (y2 + 2x));(I) V = (0, 2cosxz, 0);(g) V = (—y/(x2 + y2), x/(x2 + y2), 0);(h) Y = (yex2, 2yz, — (2xyzexZ + z2)).

8.22. Prove that to each smooth vector field on a manifold, there corre-sponds a one-parameter group of diffeomorphisms whose trajectoriesare tangent to the given vector field.

66

8.23. Show that the Poisson bracket of vector fields (as differentiationoperators) is a vector field.

8.24. Let be a one-parameter group of diffeomorphisms associatedwith a vector field Show that

d[a?, — - —

8.25. Let E, be two vector fields, and g two smooth functions.Prove the formula

LIE, = + —

8.26. Let be two vector fields, and the one-parametertransformation groups associated with them. Sfiow that if = 0, thenthe transformations and commute.

8.27. Let V be a linear finite-dimensional space of vector fields whichis closed under the Poisson bracket operation, i.e., [E, E V when

E V Show that V is a Lie algebra.8.28. (See the previous problem.) Show that the Lie group G corre-

sponding to the algebra V acts on a manifold, each field E E V specifyinga one-dimensional subgroup of the group G whose orbits under thisaction are tangent to the vector field

8.29. Let Q be two arbitrary points of the disk C Find adiffeomorphism on such that ço(P) = Q, = x, when x E

8.30. Let be a vector field on a manifold X, P E X. Show that ifEjp 0, then there exists a local coordinate system (x', . . . , x") in aneighbourhood of the point P such that E =

8.31. Construct three linearly independent smooth vector fields at eachpoint of the standard sphere S3. Find the explicit forms of the integralcurves on these fields.

8.32. Construct the integral curves of the following vector fields onthe plane:

a a(a) = x— + y—;

ax ay

a a(b) E = y— — X—;

äx

(c) E = —ox ay

(d) E = (x + +Ox Oy

67

a a(e) = (x — y)— + x—;

ax ay

(f) = X2 +ax

8.33. Prove that the singular points (zeroes) of the vector fieldsgradRe(f(z)), grad Imf(z) coincide with the zeroes of the derivative

8.34. Find the integral curves of a flow vi(x) orthogonal to a flow Vz(X),where v2(x) = gradf(x), x R2, f(x) is the magnitude of the angle AxB(A, B being certain two points of the plane R2 and x a variable point).

8.35. Specify the quality characteristics for the integral curvedistribution of the flows v1 = grad Re(f(z)), v2 = grad Im(f(z)) of thecomplex-analytic functions f(z) listed below. Find the singular points ofthe flows v1, vz. Investigate the stability of the singular points. Specifythe quality characteristics of the behaviour of the trajectories of the flowsv1, V2 on the sphere S2 (extended plane R2: S2 = R2 U co). Specify theresolution process of the singularity z = 0 of these vector fields for asmall perturbation of the original function 1(z) leading to a function g(z)with all the singular points of the flows v1, vz non-degenerate:

(a) 1(z) = z" (where n is an integer);(b) 1(z) = z + l/z (Zhukovski function);(c) 1(z) z + I/z2b (where b is an integer);(d) 1(z) z + l/(z — 2);

(e) 1(z) z4(2(z — 5)2 + 12z6) (investigate in a neighbourhood ofthe point z = 0);

(f) 1(z) = z3(z — 1)100(z —

(g) 1(z) 2z — lnz;(h) 1(z) = 1 + z4(z4 — — (investigate in a neigh-

bourhood of the point z = 0);

(i) 1(z) = ln[(z — 2i)/(z —4)]3;

(j) 1(z) l/(z2 + 2z — 1);

(k)f(z) = + 2llnz2;

(l)f(z) z5 + 2Inz;

(m)f(z) = 21n(z — 1)2 — 4/3 In(z +JQJ)3;

68

(n) 1(z) = l/z3 — l/(z — j)3;

(0)1(z) = (2 + 5i/2)ln[(4z — 2)/(64z + i)];

/18z — i\2(p) 1(z) = (I — i/2)41n

\1Oz + 1

8.36. Prove that the irrotational flow v = (P. Q), where P, Q are thecomponents of the flow on the plane y), is potential andv = gradf(x, y) for a certain smooth function 1. What can be said aboutthe potential of f, given additionally that the flow is incompressible, i.e.,div(v) = 0?

8.37. Let a vector field satisfy the condition div(s) = 0. Show thatthe displacement operator along the integral curves is unitary.

8.38. Find all homotopy classes of the vector fields on the torus T2.8.39. Prove that if a vector field X on the two-dimensional torus is

homotopic to then it possesses a periodic trajectory.8.40. Find the greatest number of linearly independent tangent vector

fields on a smooth closed surface M2.8.41. Prove that the indices of two vector fields on an arbitrary two-

dimensional and closed surface are equal. Does the statement hold fora manifold of any dimension?

8.42. Let m, n be the rotation numbers of the vector field on the torusX = (m, n). Prove that this field has X periodic solutions (closed

trajectories).8.43. (The Poincaré-Bendixson theorem.) Prove that if an arbitrary

integral curve of some vector field on the plane is compact and containsno singular points, then it is periodic.

8.44. Prove that if P on the plane is a limit point for some trajectoryof a vector field, then the trajectory passing through P is limiting forthe original trajectory.

8.45. Prove that the set of vector fields possessing only isolated singu-larities is connected.

8.46. Prove that the sum of the indices at singularities of a vector fieldon a compact and closed manifold is unaltered in smooth deformations.

8.47. Prove that the set of all integral curves of the vector field v(x)= (x', —x°, x3, —x2), where x = (x°, x', x2, x3) E S3 = 1) C R4,is homeomorphic to the sphere S2. Find the relation to the Hopf map

S3 S2. How is this vector field related to quaternions?8.48. Let v(x) be a smooth vector field on the plane R2, L a smooth

self-intersecting contour on the plane R2, JL the index of the contour Lin the vector field v(x), J the number of points where the field v andcontour L touch internally, and E the number of points touchingexternally.

69

Prove that if the number of all points of contact of the field with the

contour is finite, then IL + J — E). Prove that the index of any

isolated singular point of a smooth irrotational vector field v(x)= gradf(x) is always not greater than unity (note that this singular pointmay, certainly, be degenerate).

8.49. How many solutions can the equation sinz = z have over thefield of complex numbers?

a a .8 0 0 .88.5O.Put— =— — — + i—.Showthatafunctionliz Ox dy 3z Ox dy

f is holomorphic if and only if (f) 0 for all k.Oz

8.51. Show that a vector field is holomorphic if and only if it has the

form E = a' where a' are holomorphic functions, with respect toliz

a local coordinate system (z', . . . , zn).

9Tensor Analysis

9.1. Determine the type of the following tensors:

(a)

821(b) = at those points where the gradient of the function f

vanishes;(c) i.e., the components of the matrix of a linear operator on a

vector space;(d) i.e., the components of the matrix of a bilinear form on a vector

space.

9.2. Let

when

(1, when i=jShow that yields a tensor of type (1, 1).

70

9.3. Let (EU) be a tensor of type (2, 0). Show that the numberssatisfying the condition = M yield a tensor of type (0, 2).

9.4. Show that if f: is a linear mapping of tensor spaces,then the mapping components yield a tensor of type (m, n).

9.5. Determine the dimension of the tensor space9.6. Show that any tensor of type (2, 0) can be decomposed uniquely

into the sum of a symmetric and a skewsymmetric addend.

9.7. Determine the dimension of the space A" of skew-symmetrictensors.

9.8. Determine the dimension of the space of symmetric tensors.9.9. Prove the formula

Ak(VI ® V2) ®a + iS = R

9.10. Calculate the components of the fundamental tensor of the planewith respect to a system of polar coordinates.

9.11. Calculate the components of the fundamental tensor of R3:(a) with respect to a system of cylindrical coordinates;(b) with respect to a system of spherical coordinates.9.12. Calculate the components of the fundamental tensor of the sphere

(a) with respect to a system of spherical coordinates;(b) with respect to Cartesian coordinates on the stereographic

projection.9.13. Assuming that the gradient of a function f is the composite of

two operations, viz., that of partial differentiation and that of raisingthe indices, write the gradient of the function with respect to:

(a) a system of polar coordinates;(b) a system of cylindrical coordinates;(c) a system of spherical coordinates.

______________

9.14. Find the gradient of the function f = lnVx2 + + z2.9.15. Derive the following formulae for the functions f and g with

respect to an arbitrary system of coordinates:

(a)

grad (f g) = gradf ± gradg;

(c) grad(fg) = fgradg + g gradf;

g gradf — fgradg(d) grad(f/g) = 2

, g 0;g

(e) grad(f(g)) = gradg.dg

71

9.16. Let! = f(u, v), where u, v are two functions. Show that grad! =

= grad(u) — grad(v).dv

9.17. Write the formula for the derivative of a function f with respectto an arbitrary system of coordinates:

(a) in the direction of its gradient;(b) in the direction of the gradient of the function.

9.18. Derive a formula enabling us to determine the greatest changeof a function f at a given point with respect to an arbitrary system ofcoordinates.

9.19. Write a solid medium deformation tensor

•k / du' duk du'U' = —I--——— + +

2 \ dx"

with respect to an arbitrary system of coordinates using the fundamentaltensor. Write out separately similar formulae for the terms which arelinear respective to u'.

9.20. Prove that the Christoffel symbols of two connections differ byaddends which are the components of a tensor.

9.21. Show that the covariant derivative along a curve depends on thevalue of the Christoffel symbols of this curve.

9.22. Show that if two submanifolds osculate to some curve thenthe parallel displacement operation does not depend on the choice of asubmanifold.

9.23. Show that the parallel displacement operation can be obtainedon a submanifold by passage to the limit of the composite of a paralleldisplacement in the ambient manifold and the orthogonal projection ontothe tangent space to the submanifold.

9.24. Calculate the angle through which the tangent vector to a rightcircular cone turns after parallel displacement along a closed curve.Establish the dependence on the kind of the curve.

9.25. Calculate the angle through which the tangent vector of a sphereturns after parallel displacement along a curve -y if:

(a) -y is a parallel;(b) -y is made up of two meridians and a part of the equator which

is included in between them;

72

(c) is made up of two meridians and a part of the parallel whichis included in between them.

9.26. Establish a dependence between the angle of rotation of thetangent vector to a sphere after parallel displacement along a closed curve-y and the area of the region bounded by the curve y.

9.27. Generalize Problem 9.26 to the case of a surface with constantGaussian curvature.

9.28. Prove that if the curvature tensor of a Riemannian manifold isidentically equal to zero, then the operation of parallel displacementalong a curve -y does not depend on a homotopy of the path -y.

9.29. Calculate the scalar curvature of the following Riemannianmanifolds:

(a) S2;(b) the torus T2 embedded in R3;(c) the Lobachevski plane;(d) the right circular cone;(e) the cylinder;(f) the group SO(n) with a bi-invariant metric.

9.30. Show that any two sufficiently near points on any compact,Riemannian manifold can be joined by a geodesic line, the geodesic ofthe least length being unique.

9.31. Describe the geodesics in the following Riemannian manifolds:

(a) R2;(b) the torus T2 under the flat metric;(c) S2;(d) the Lobachevski plane.

9.32. Let = + + + + - = 0 fl

n{

= 1] be the Brieskorn spheres (k 28). Prove

that the Riemannian metric induced on these spheres (in embedding

—. S" C C5 = R'°) is not a metric with positive curvature.

9.33. Prove that there always exist a pair of conjugate points on a com-pact, I-connected manifold What happens if is not I-connected?Is there a conjugate point on every geodesic y(1) (the geodesics emanatingfrom one point)?

9.34. The following kinds of manifolds of strictly positive curvatureare known, being the classical symmetric spaces of rank I, viz., S", RPM,

73

CPA, K'6. Calculate the curvature (find the limits within which thecurvature varies) of R(a) on CPA, K'6.

9.35. Let ® be a Lie group, and a bi-invariant metric on ®, whereg E ® is the variable point on ®.

Recall that a metric is said to be bi-invariant if it is preserved underleft and right translations, viz.,

Lg0 : g g0g; Rg0 : g ggo.

Prove that it follows from the bi-invariance of the metric on ®that the form (,)e is invariant under all transformations of the formAdg : X gXg -'.

9.36. Give examples of matrix Lie algebras G such that the quadraticform <XY>e = Tr(XYT) is non-singular, where X, Y E G, the bar denotescomplex conjugacy, and Tr transposing.

9.37. Let (,)g be a bi-invariant Riemannian metric on a Lie group ®.Let V be a symmetric Riemannian connection on ®, compatible withthe metric Prove that the geodesics of the connection V are thefollowing trajectories only: one-parameter subgroups of the group ® andtheir translations (left and right).

9.38. Let X be a left-invariant vector field on a group ®. Prove thatthe integral curves -y(t) of this field are left translations of a one-parame-ter (i.e., one-dimensional) subgroup which passes through the unit elementof the group in the direction of the vector X(e), where X(e) is the valueof the field X at a point e E ®.

9.39. Let X, Y be two invariant vector fields on a group ®, and Va symmetric connection compatible with a bi-invariant Riemannianmetric on®. Prove that Vx(Y) = 1/2 [X, Yl, where [X, 1'] = XY — YX(the Lie group being a matrix).

9.40. Let <,) be a bi-invariant metric on the Lie algebra G of a group®, and [X, fl = XY — YX the commutator in G. Prove that ([X, 11,Z) = (X, Z]).

Note. The operation X —÷ [X, Y} is sometimes denoted asady : X [X, fl; then the required relation is written as

(adyX, Z) = —<X, adyZ>.

9.41. Let ® be a compact Lie group with a bi-invariant metric, andX, Y, Z left-invariant vector fields on ®. Prove that R(X, Y)Z = 1/4yI,z'.

74

9.42. Let be a compact Lie group with the bi-invariant metric<,>, and X, Z, W left-invariant vector fields on

Prove that

<R(X, Y)Z, W) = 1/4 ([X, YI, [Z, W]).

Let be a compact Lie group, and X, V two orthogonal unit vectors(bi-invariant metric (,) is given on tb). We call the number o(X,Y) = (R(X, Y)X; Y> the sectional curvature determined by thevectors X, Y.

Prove that a(X, Y) 0 and a(X, Y) = 0 if and only if (X, 1'] = 0.

Hint: a(X, Y) 1/4 ([X, YJ, [X, Y]) = 1/4 hEX, Y]112

10Differential Forms, Integral Formulae,

De Rham Cohomology

10.1. Prove that if vectors Vi, . . . , vi,, E V are linearly dependent, thenT(vi = 0 for any form TE AI)(V*).

10.2. Prove that if forms E are linearly dependent, then

10.3. Let E and v1,. . . , v,, E V. Prove that

(WiA . . ., =

10.4. Prove that the element of volume equals '[det(g11)dx'A .. . Adf,where gij is the Riemannian metric with respect to the coordinates(x', . . . ,

10.5. Show that the exterior differentiation operation of a differentialform can be represented as the composite of the gradient covariantcomponent operation and that of alteration for an arbitrary symmetricconnection on a manifold.

10.6. Calculate the exterior differential of the following differentialforms:

(a) z2dx A dy + (z2 + 2y)dx A dz;

(b) l3xdx + y2dy + xyzdz;

(c) (x + 2y3)(dz A dx + 1/2 dyAdx);

(d) (xdx + ydy)/(x2 +(e) (ydx — xdy)I(x2 +(f)f(x2 + y2)(xdx + ydy);

(g) fdg g being two smooth functions);(h) f(g(x',..., x"))dg(x' x").

10.7. Prove the validity of the formula

2(dw)(X, Y) = X(w(Y)) — Y(w(X)) — w(IX, Y]),

where w is a differential form of degree 1 and V two vector fields.10.8. Generalize the formula in 10.7 to the case of differential forms

of an arbitrary degree.Given a scalar product on the vector space R", there are two isomorphic

operations. One of them associates each vector X with a linear formw V(X) such that (X, Y) = V(X)(Y). The other associates eachmultilinear skewsymmetric form w of degree p with a form * (w) ofdegree n — p as follows: let w1 w,, be the orthonormal basisconsisting of linear forms, and w = f W1A . . Then*(w) = (— 1)°fw11A . . . where o is parity of the permutation

. . . . Jfl_p).

10.9. Show that the following formulae hold for the space R3:

(a) gradF = V '(dfl;(b) divX = *d* V 1(X);

(c) rotX = —

10.10. Show that Green's, Stokes' and Ostrogradsky's formulae are spe-cial cases of general Stokes' theorem for differential forms.

10.11. Derive the formula of integration with respect to the volumeV bounded by a closed surface E:

(a) + gradi,1'))dv=

da;

(b) — = (p — da,

where ô/t9n denotes the derivative in the direction of the normal to thesurface

76

10.12. Calculate the surface integral dc with respect to a

closed surface

(a) for = z2, = x2 + — z2 if bounds the region x2 + ++ 1 and y 0;

(b) for ç 2x2, = x2 + z2 if bounds the region x2 + I

and 0 z 1;

= = (x + y + if E is the spherex2 + + = r2;

(d) for = 1, = exsiny + + z if is the tn-axial ellipsoid

x2 y2 z2+

b2+ 1.

10.13. Find the gradients of the functions with respect to cylindricalcoordinates:

(a) u + —

(b) u = + —

10.14. Find the gradients of the functions with respect to sphericalcoordinates:

(a) u r2cosO;

(b) u 3r2sinO ercosço — r;(c) u cosO/r2.

10.15. Find divX with respect to cylindrical coordinates:

(a) X(b) X = 2, Z2CZ)

10.16. Find the divergence of the vector field X = (r2, —+ I)) with respect to spherical coordinates.

10.17. Find the rotors of the vector fields with respect to sphericalcoordinates:

(a) X = (2r + —asinO, rcosO), a = const;(b) X = (r2, 2cosO, —

10.18. Verify that the following vector fields are potential with respectto spherical coordinates (r, 0,

(a) X = (2cos0/r3, sinO/r3, 0);(b) X = (f(r), 0, 0).

77

10.19. Find the potentials on the following vector fields with respectto cylindrical coordinates z):

(a) )( = (1, l/Q, 1);(b) X = 'p/Q, z);(c) X = z,

(d) X 2z);

(e) X = cosz, — QçOSinZ).

10.20. Find the potentials on the following vector fields with respectto spherical coordinates:

(a) X = (0, 1, 0);

(b) X = (2r, l/rsinO, l/r);(c) X = (çô2/2 0/r);(d) X = — sinç);

(e) X = (ersin0, etcos0/r, +

10.2!. Calculate the circulation of the vector field X = (r, 0,(R + r)sin0) with respect to spherical coordinates along the circum-ference = R, 0 =

10.22. Calculate the line integral along the line L of the vector fieldX, both given with respect to cylindrical coordinates:

(a) X = (z. L is the line-segment = a, = 0,

o z Ii:

(b) X = z), L is the semi-circumference Le 1, z = 0,

(c) X = L is the helix = R, z =o 2ir};

(d) X = (z. z

L is the circumference z =

X = L is

the XL in

X L = {r 1, 0, 0 71);

(b) X = Oço, L = = 71/2, 0

0 r 1);

78

(c) X = (sin28, sinO, r = l/sinO,

0

(d) X = (rO, 0, rsinO), L = [r = 1, 9 = ir/4(e) X = (rsinO, Oe°, 0), L = (r = sinw, 0 = 0

(fl X = (0, 0, L is the contour bounding the half-disk R,

=

10.24. Find the flow of the vector field X on the surface S given incylindrical coordinates:

(a) X = z), S bounds the region fQ 2, 0 z 2j;

(b) X = — 2z), S bounds the region 1, 0 ir/2,—l z ( U.

10.25. Find the flow of the vector field X on the surface S given inspherical coordinates:

(a) X = (l/r2, 0, 0), S encloses the origin;(b) X = (r, rsin0, — S bounds the region r R, 0 ir/2

(c) X = (r2, 0, R };

(d) X = (r, 0, 0), S bounds the region Ir R, 0 ( 7r/2J;

(e) X (r2, 0, S bounds the region tr R,

0 ir/2, 0

10.26. Let : X x [0, 1] —* Ybe a smooth mapping, and w a differen-tial form on dw = 0. Prove that ft(w) = d(l for aconvenient form on X.

10.27. Prove that if a manifold X is contractible, then, for any formw(dw = 0), the equation = w is solvable.

10.28. Let F be a vector field in a three-dimensional region W witha smooth boundary 3W, and n a vector normal to 0W.

Prove that

= (nF)dA,

where dA is the element of area on OW.

79

10.29. (See the previous problem). Prove that

n)dA = (ftdxi + f2dx2 + f3dx3),

where S is a smooth surface with a smooth boundary 3S.10.30. Calculate the de Rham cohomology groups of the following

manifolds: (a)S', (b)S2, (c)RP2, (d)T2, (e)T", (I) the plane with exclusionof a finite number of ppints.

10.31. Describe the differentials of left-invariant forms on a Lie groupin terms of the commutator in the Lie algebra.

10.32. Prove that the bi-invariant forms are closed on a compact Liegroup.

Prove that bi-invariant forms are not homologous to zero on a compactLie group.

10.33. Prove that there is a bi-invariant metric on a compact Lie group.10.34. Show that the de Rham cohomology on a compact Lie group

is isomorphic to the space of bi-invariant forms.10.35. Prove that each differential form on a complex manifold with

respect to complex coordinates z', . . , z" = + is as follows:

w = EWk,dzk A dz',

where dZk A .. . A dz11, dz' = dz11 A . .. A

10.36. Let X be a complex-analytic manifold of dimension n, and wa holomorphic form of degree n. Show that the integral of the form walong the boundary of an (n + 1)-dimensional real submanifold in Xequals zero.

10.37. Derive the Cauchy theorem = 0 from Stokes' formula

for a holomorphic function in a region bounded by a curve 'y.10.38. Derive the Cauchy residue theorem from Stokes' formula.10.39. Let f: M N be a smooth mapping of orientable, closed, and

compact manifolds of dimension n, and w an n-dimensional differentialform on the manifold N.

Prove that f*(w) = degf w.

10.40. Let p and q be two arbitrary polynomials in variables (z'zn), and a, k real numbers. Let there exist a differential form w such that

80

dp A w pdz, dw = adz, dq A w kdz. Prove that 0(where dz dZt A . . . A

10.41. Let C S', a curvature form, and w a connection form.PrOve that

(a) dcc = f*(w), dw = 0;

(b) 4w are integers for any closed cycle

10.42. Prove that if w are integers for any closed cycle, then there exists

a connection such that w is its curvature form.10.43. Construct a connection in the fibration G C/H (where C is

a Lie group and H its subgroup) so that the form is invariant with respectto all the motions.

10.44. Let M2 be a smooth, closed and compact manifold, g,j thecomponents of its fundamental tensor, and K(x) the Gaussian curvature.

1(g; M2) = K(x)do(g).

Given that M2) = 0, derive the classical Gauss-Bonnet formula

K(x)thi(g) K(x)da = x(M2).2irj2M M

10.45. Prove that one-dimensional de Rham cohomology groups areisomorphic to the group Hom(iri(X), R').

10.46. Let a smooth triangulation of a manifold M be given. Provethat the simplicial cohomology groups with real coefficients are isomor-phic to the de Rharn cohomology groups.

-11

Genera) Topo)ogy

11.1. Prove that any finite CW-complex can be embedded in a finite-dimensional Euclidean space (of sufficiently large dimension).

11.2. If a compact, smooth and closed manifold is talen as a CW-com-

6—2018 81

plex, then the result formulated in the previous problem can be mademore precise, viz.,

(a) prove that M" can be embedded in the Euclidean space R2flk, wherek is the number of open balls forming a covering of M";

(b) prove that M" can be embedded in the Euclidean space wherethe number has been defined in item (a).

11.3. Prove that any compact, smooth and closed manifold M"(a) can be embedded in the Euclidean space 1;

(b) can be immersed into the Euclidean space g2n•

11.4. Construct the immersion of the projective plane RP2 into theEuclidean space

11.5. Describe the set of nodes of the immersion of KP2 into R3constructed in Problem 11.4. Indicate the of the nodes, i.e.,how many sheets of the surface intersect at each node of the surface.

11.6. Consider the immersion of RP2 into R3 described in Problem11.4. Denote the image of RP2 in R3 by i(RP2). Consider a line-segmentof length 2€ orthogonal to with the centre at each point x E i(RP2)which is not a node of the surface, where is sufficiently small. Sincei is a smooth mapping, the pencil of orthogonal line-segments obtainedcan be additionally defined at each node. In doing so, we shall obtainas many line-segments at each node as the multiplicity of the node is.Consider in R3 the set consisting of the ends of all the orthogonal line-segments. Prove that it is the image of a two-dimensional sphere undera certain smooth immersion into R3.

11.7. Construct an example of a topological space not satisfying thefirst countability axiom (resp. the second countability axiom).

11.8. Given two continuous functionsf(x), g(x) on the two-dimensionalsphere S2 such that f(x) = —f(rx), g(x) = —g(rx), where i- is thereflection through the centre of the sphere. Prove that these functionshave a common zero.

11.9. Construct an example of a topological space X such that a certainof its subsets Y C X (indicate Y) is closed and bounded, but notcompact.

11.10. Prove that a one-dimensional cellular complex is a space of typeK(7r, I), where is a free group.

11.11. Prove that any finite siniplicial complex is a subcomplex of asimplex of sufficiently large dimension. In particular, it can be embeddedin the Euclidean space so that the embedding is linear on each simplex.

11.12. Prove that a contractible space is homotopy equivalent to apoint_

11.13. Prove that a universal covering space of X is also a coveringspace for any other covering space.

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11.14. Prove that any two spaces of type K n) are weakly homotopyequivalent.

11.15. Prove that

(a) A=

(b) S"/S" is homotopy equivalent to V Sk + 1, S" \ Sk homotopyequivalent to and S?k diffeomorphic to x ifSk c is the standard embedding.

11.16. Prove that a function is continuous on a CW-complex if andonly if it is continuous on each finite subcomplex.

11.17. Let M = X x Y, where X, Y are two topological spaces. Weshall consider a set from M to be open if it is the product of open setsfrom X and Y or the union of any number of such sets.

Prove that such a system satisfies all the axioms determining a topologyon the set M.

11.18. Prove that if a space X is both Hausdorff and locally compact,while a space Y Hausdorff, then, for any space T, the spaces !-f(X x7) and H(Y, H(X, 7)) are homeomorphic, where H(X, Y) =

11.19. Prove that the standard fibre map EX X (Serre fibre map),where X is a manifold, is a locally trivial fibre map.

11.20. Prove that there exists a homeomorphism of the Cantordiscontinuum onto itself commuting two given points.

11.21. Let a mappingf: E Fbe a continuous mapping "onto", andlet E be compact. Prove that F is compact.

11.22. Prove that the n-dimensional sphere (n < co) is compact. Is ittrue for n = 00?

11.23. Let A, B be two connected spaces and A fl B 0. Prove thatA U B is connected.

11.24. Prove that if F, F are two connected spaces, then E x F isconnected.

11.25. Let f: E —* F be a continuous mapping "onto", and Econnected. Prove that F is a connected space.

11.26. Prove that:

(a) the intervals 0 < x < 1, 0 x 1, 0 x < 1 are connected;(b) if A C R' is connected, then A is of the form a < x < b,

a x b, a < x b, a x < b, where a, b may assume the values± 00.

11.27. Letf: E —' E = A U B, A = A, B = B. Thenfis continuousif and only if and are continuous. If A A, then, generallyspeaking, this does not hold. Give an example.

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11.28. Prove thatf: E —' Fis continuous if and only if '(U) is openfor any open subset U C F.

11.29. Letf: X Y. Prove that f is continuous if and only if the in-verse image of every closed set is closed.

11.30. Prove that A U B = AU Band A nB = An B.

11.31. Let lnt(A) U LG CA: G is open). Then p E lnt(A) if andonly if there exists a neighbourhood U of the point p such thatU C Int(A), p E A = fl (F A :F is closed) if and only if we haveU fl A 0 for any neighbourhood U ? p.

11.32. Prove that the open disk (lxi < 1) in Euclidean space is an openset.

11.33. Let X be a locally path-connected metric space. Prove that ifX is connected, then X is path-connected.

11.34. Let X be a metric, compact and connected space. Can any twoof its points be connected with a continuous path?

11.35. Prove that the cube and sphere S" are connected.

11.36. Let G C I' be an open set on a closed line-segment. Prove thatG is the union of disjoint open intervals.

11.37. Let X be a metric space. Prove that each of its one-point subsetsis closed.

11.38. Prove that if the product of two topological spaces is homeomor-phic to the suspension of some topological space, then either both factorsor one of them is contractible to a point.

11.39. Let X be a compact space, Y a metric space, and f: X Y acontinuous map. Prove that f is a uniformly continuous map.

11.40. Prove that if f: X Y is a sequence of continuous mappingsand In uniformly converges to f( Y being a metric space), then f iscontinuous.

11.41. Let X C Y, and V a compact space. Prove that X is a compactspace if and only if X is a closed subspace.

11.42. Let A fl B 0, and X = A U B. Prove that if A and B areconnected spaces, then X is a connected space.

11.43. Prove that the cube is a compact space.11.44. Prove that the sphere and ball are homeomorphic to

cellular spaces.+

11.45. Prove that the ellipsoid { -4 i} is homeoinorphic

to the sphere

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11.46. Prove that the ball { i} and the upper hemisphere

{i, o} are

11.47. Prove that the cube 1, 1 1, 2,..., nl and the ball

{i} are homeomorphic. Prove that an open cube and an

open ball are diffeomorphic.11.48. Are the line-segment 0 x 1 and the letter T

homeomorphic?11.49. Prove that the interval (— 1, 1) is homeomorphic to the straight

line (— co). Prove that any two intervals are homeomorphic.11.50. Are a ball and a sphere homeomorphic?11.51. Prove that the Hamming metric on the n-dimensional cube

cannot be embedded in any R", i.e., there exists no embedding such thatthe Hamming metric is induced by the standard Euclidean metric (thecube being considered only as the set of its vertices, i.e., as a discreteset, and then the distance Q(a, b), where a and b are the vertices of thecube, equals the number of different coordinates of these two vertices).

11.52. Letf.• M2 S2 be a mapping of class C2 of a closed, smooth,and compact manifold M2 onto S2, f being open (i.e., the image of anyopen set is open) and finitely multiple (i.e., the inverse image of each pointx E S2 is a finite number of points).

Prove that M2 is diffeomorphic to the sphere S2. What can be saidabout a similar mapping f:

11.53. Prove that a metric topological space satisfies Hausdorff separ-ation axiom.

11.54. Is it true that the distance between two disjoint, closed sets onthe plane (straight line) is always greater than zero?

Recall that the distance between two subsets X and V of a metric spaceZ is the number Q(X, Y) = sup inf r(x, y) + sup inf r(x, y), where

x€X yE)' yE)' xEX

r(x, y) is the distance between two points x and y in the space Z. Thisis not the only way of defining the distance between two subsets of ametric space. Give other examples of the metric (X, Y).

11.55. Prove that a set whose elements are closed subsets of a metricspace can itself be transformed into a metric space in a natural mannerby introducing a metric described, e.g., in Problem 11.32.

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11.56. Prove that any contracting mapping of a metric space iscontinuous.

A mappingf : X —* X of a metric space X into itself is said to be con-tracting if there exists a real constant X < 1 such that Q(f(X), f(y))

y) for any two points x, y E X.11.57. Prove that any contracting mapping of a complete metric space

into itself always has a fixed point (which is unique).11.58. Give an example showing that a condition for a metric space

to be complete (see Problem 11.57) cannot be discarded.11.59. Show that a group of orthogonal matrices of order 3 x 3 is

a compact topological space.11.60. Prove that the group of orthogonal transformations of the

n-dimensional Euclidean space is a compact topological space.11.61. Prove that the group S0(3) considered as a topological space

(in embedding SO(3) in the linear space of all real matrices of order3 x 3) is homeomorphic to RP3.

11.62. Prove that S0(n) is a connected topological space, and that 0(n)consists of two connected components. Prove that U(n) and SU(n) areconnected topological spaces.

11.63. Prove that the open disk x2 ÷ < I and the plane R2(x, y)are homeomorphic. Prove that the open square x) < 1, < 1) andplane R2(x, y) are homeomorphic. Prove that the interval 0 < x < Iand the open square < 1, < 11 are not homeomorphic.

11.64. Prove that the group of unitary matrices U(n) considered as atopological space is homeomorphic to the direct product of S' and SU(n)(as topological spaces).

11.65. Prove that the group GL(n, G) considered as a subset in thespace of all complex matrices of order n x n is an open and connectedsubset.

11.66. Prove that the group GL + (n, R) consisting of real matrices oforder a x n with positive determinants is a connected topological space.

11.67. Prove that the group GL(n, R) of real non-singular matrices oforder n x a is a topological space consisting of two connectedcomponents.

11.68. Prove that a Mdblius strip is not homeomorphic to the directproduct of a line-segment by a circumference.

11.69. Construct an immersion of a Mäbius strip into Euclidean three-dimensional space so that the boundary circumference of the former maybe standardly embedded in Euclidean two-dimensional plane.

11.70. Prove that the set of all straight lines on the Euclidean planeR2 is homeomorphic to a MObius strip.

11.71. Prove that for any compact set K C there exists a smoothreal-valued function f such that K f '(0)

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11.72. The group SO(3) is, naturally, embeddable, in the Euclideanspace R9 (each element of SO(3) is a real matrix of order 3 x 3). Provethat, virtually, SO(3) C S8 C R9, where S8 is the sphere of radiusstandardly embedded in R9.

12Homotopy Theory

12.1. Represent (a) the torus, (b) Klein bottle and (c) suspension ofa complex K as cellular complexes.

12.2. Prove that the topology of a CW-complex is the weakest of alltopologies respective to which all the characteristic mappings arecontinuous.

12.3. Prove that a torus with a disk generated by a meridian arehomotopy equivalent to the wedge S' V S2.

12.4. Prove that a torus with a disk generated by a meridian and aparallel are homotopy equivalent to the sphere S2.

12.5. Generalize Problems 12.3 and 12.4 to the case of the products4 x

12.6. Prove the homotopy equivalence of the spaces (X x S")/(X V S")and 1"X.

12.7. Prove the homotopy equivalence

(a) E(XV Y)

(b) Y) x Y).

12.8. Let IX; A, B) be the space of paths starting at A and endingat B, and A C R Prove that (X; A, B J contains a subspace homeo-morphic to A.

12.9. Let f: X be a continuous mapping of simplicial complexes,and Y C X a subcomplex such that the mapping f is simplicial on it.Prove that there exists a subdivision of the complex X such that it isidentity on Yand the mappingf is homotopic to a certain simplicial map-ping q, the homotopy being constant on Y.

12.10. Let X be a simplicial complex, and the star of a vertex x E XProve that any two simplexes of the star Sx meet in a certain face.

12.11. Prove that a simplicial mapping of simplicial complexes iscontinuous.

12.12. Let X be a simplicial complex, and > 0. Prove that there existsa subdivision of the complex X such that the diameter of each new sim-plex is less than

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12.13. Let f be a mapping of the unit line-segment [0, 1] into itself,and f(0) 0, f(l) 1. Prove that there exists a homotopy which leavesthe endpoints of the line-segment fixed and deforms the mappingf intothe identity.

12.14. Is the vector space contractible to a point on itself?12.15. Let a space X be contractible to a point on itself. Prove that

any two paths with the same endpoints are homotopic to each other (afixed endpoint homotopy).

12.16. Let a space X be contractible to a subspace Y, with the homotopyleaving V fixed (constant). Prove that any path in X with the endpointsin Y is homotopic to a path wholly lying in Y (a fixed endpointhomotopy).

12J7. Prove that any two paths are homotopic on the sphere S", n > I(the endpoints are the same, and the homotopy is fixed endpoint).

12.18. Prove that any connected, cellular complex is homotopy equi-valent to a cellular complex with one vertex.

12.19. Prove that the sphere -' can be represented as the unionx U x 1) with the common boundary

Sr x12.20. Consider the standard sphere the Euclidean space R'

and two spheres embedded in it:

fXr+i — . . . = = = tXi = . . = = ol.

Prove that any pair of points y E x E can be joined bya unique arc on a great circle having no other points of intersection withthese spheres.

12.21. Find the topological type of the closed hyperboloid of one sheetF = 1x2 + — = 1) in the projective space Re3.

12.22. Cut a Möbius strip (embedded in R3) along its midline.Is the manifold obtained orientable?Repeat the cutting process several times. Describe the manifold

obtained (it is disconnected) and find the linking number of any twoconnected components.

12.23. Prove that the space of polynomials of the third degree withoutmultiple roots is homotopy equivalent to the complement of a trifoliumin the sphere Construct an explicit deformation.

12.24. Consider the set of points with pairwise different coordinates.Show that the space obtained has the same type as the Eilenberg-MacLanecomplex K(ir, 1).

12.25. Construct an example of two spaces X1, X2, which are nothomotopy equivalent, and also of two continuous one-to-one mappingsf: X1 X2, g: X2 X1 such that the spaces themselves may not behomotopy equivalent.

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12.26. Let h : X — be a continuous mapping, and the corre-spondence 4) : {X', - fX, YJ be defined by the formula4)(a) a h.

Prove that the correspondence 4) transforms homotopic mappings intohomotopic.

12.27. Prove that the following homotopy equivalence holds:x Se') 'V sm± 1

12.28. Prove that the finite-dimensional sphere is contractible to apoint on itself.

12.29. Prove that a connected finite graph is homotopy equivalent tothe wedge of circumferences VS1.

12.30. Let a mapping p X Y satisfy the covering homotopy axiom.Prove that the inverse images of the points are homotopy equivalent.

12.31. Let a space X be contractible to its path-connected subspace A.Prove that the space X is path-connected.

12.32. Fix two points Xo and x, in a space X. Let Y be the space ofpaths starting at Xo and passing through x,. Prove that the space Y iscontractible.

12.33. Prove that the space of all paths tX;X X; with X held fixed.

12.34. Let a sequence of spaces X,, C , be given, and let i

be contractible to with held fixed.Prove that the space X U is contractible to Xo with X0 held fixed.

12.35. Prove that any open n-dimensional manifold is homotopy equi-valent to an (n 1)-dimensional complex.

12.36. Prove that if a space Xis contractible to a subspace A on itselfwith A held fixed, then A is homotopy equivalent to X.

12.37. Calculate the sets ir(S' x S'. S2) and ir(S" x Sn).12.38. Find and where and Cat2(RP")

are the minimal numbers of closed subsets X, such that X = U

X are homotopic to constant mappings.12.39. Calculate Cat,(K) and Cat2(K) for the case of a sphere with

three identified points.12.40. Let M2 be a compact, closed, oriented, and 2-dimensional man-

ifold of genus h, i.e., M2 is the sphere S2 with h handles. Find E2M2(i.e., double suspension) up to homotopy type.

12.41. Consider some standard chart with non-homogeneous coor-dinates X, in Find the homotopy type of

(a) where = + . . . + = 1, = . . . ==

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(b) RP" \ where M IC = tx2i + ... I — •.. I

—l= O,Xk+2 '= ... =(c) SIC and

12.42. Consider a small ball in the open manifold x Sn_k andglue the projective space in its place, i.e., identify points x and —xon the boundary of the ball = S"

Prove that the space obtained is homotopy equivalent tov

12.43. Given a topological manifold whose boundary is a topologi-cal manifold the boundary of being contractible to a pointin the manifold M".

(a) Prove that the manifold is contractible to a point.(b) Prove that if the manifold P" 'is 1-connected, then the manifold

is homeomorphic to the disk D" (assuming that P" is contractibleto a point in Ma).

(c) Give an example of a pair (M'1, pn I) such that the manifold p" I

is contractible to a point in the manifold but is not homeornor-phic to the disk As a corollary, prove that iri(P" 0.

12.44. Find the homotopy type of the space C" \ where = UU

12.45. Calculate the number of mappings (up to homotopy):

(a) RP" (a') CP" CP";(b) RP"; (b') CF"'' —p CP";

(c) ERP" (c')(d) —* l; (d') EcP" -+

12.46. Prove that

(a) Cat [join(X, Y)] = min[Cat(X), Cat(Y)1, where Cat is the Luster-nik-Schnirelmann category (the spaces X and Y being connected).

(b) Find Cat(S1 x S2).

12.47. Let spaces X,, I 1 N, be path-connected, andX=XIXX2X...XXN.

Prove that [Cat(X1)] Cat(X) 1 + [Cat(X1) 11.

12.48. (a) Calculate Cat(RPn); Cat(T"); Cat(Sm x 5").(b) Prove that if the sphere S" is covered by q closed sets (not necessarily

connected) V,, V2, . . , Vq, where q n, then tliere always exists at leastone set V1 such that it contains two diametrically opposite points of thesphere s", viz., —x and x.

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12.49. Let M C k' be an arbitrary subset of Euclidean space (e.g.,smooth submanifold), and let R' C R' be the standard embedding.Prove that the following homotopy equivalence holds

Reminder. Let X be a topological space, and X x I its direct productby a line-segment. After identifying the "upper base" X x (Ij C X x Iwith a point and the "lower base" X x C X x I with anotherpoint, we obtain a factor space called the suspension of X.

12.50. The relation between the Lusternik-Schnirelmann category and"cuplong" of a compex (or manifold). Let M' be a smooth, compact,connected, and closed manifold. Consider the ring H*(Mfl; G), whereG = Z if M' is orientable, and G = Z2 if M' is non-orientable. Denoteby l(M'; G) the greatest integer for which there exists a sequence ofelements XL, X2 x, of the ring H*(M"; G) > 0, 1 a I)such that A X2 A . . . A XI(M";Q) 0 in the ring G). Thenumber G) is denoted by cuplong (Ma). Prove that

Cat(M') l(M'; G).

12.51. Prove that for any path-connected topological space X and anyof its points x0, the group iri((IX, xo) is Abelian.

12.52. Prove that any contractible space is 1-connected.12.53. Prove that the group lrL(VAS') is a free group with A generators.12.54. Prove that if X and Y are homotopy equivalent, then the isomor-

phisms hold: iri(X) iri(Y) and lrk(X) lrk(Y), k 2.12.55. Prove that IrL(XV Y) = irj(X) * iri(Y), where lrL(X) * iri(Y)

is the free product of the groups and iri(Y).12.56. Find the knot group of the trefoil in R3 (and also in the sphere

S3) and prove that one cannot "untie" the trefoil, i.e., there exists nohomeomorphism of Euclidean space (or sphere) onto itself which wouldtransform the trefoil knot into the standardly embedded, unknotted cir-cumference, i.e., trivial knot.

12.57. Find the knot group of a knot I' in R3 given thus: the circum-ference which represents the knot is placed on the two-dimensionalstandardly embedded torus T2 C R3, on which it traverses its parallelp times, and its meridian q times, p and q being prime to one another.(The trefoil knot from Problem 12.56 can be represented as such a knotI'; where p = 2, q = 3.) Make out the role of the condition for thenumbers p and q to be prime to one another.

12.58. Let X = Y where Y, Z, W are finite CW-complexes,W = V fl Z, W is path-connected, and X =

Y the common subset W. Calculatethe group iri(X), given the groups ?rL(Y), and 7r1(W). Consider thecase where W is disconnected separately.

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12.59. Given an arbitrary group G with a finite number of generatorsand relations, prove that there exists a finite complex X whose fundamen-tal group is isomorphic to G. Can a finite-dimensional manifold, e.g.,four-dimensional. be selected as such a complex X?

12.60. Calculate the group iri(X), where X is the wedge of threecircumferences.

12.61. Construct a two-dimensional complex X whose fundamentalgroup equals Z/pZ. For which values of p can a two-dimensional,smooth, closed, and compact manifold be selected as such a complex?

12.62. Calculate the fundamental group of the two-dimensional spherewith three handles. Check this group on commutativity and find itscommutator subgroup. Calculate the fundamental group of the two-dimensional torus.

12.63. Let a simplicial complex X have N one-dimensional simplexes.Prove that its fundamental group has no mare than N generators.

12.64. Prove that iri(X) = 7r1(X2), where X is a CW-complex and X2its two-dimensional skeleton, i.e., the union of all cells of dimensions 1and 2.

12.65. Find where X = S' V S2. Is this group finitely generated?12.66. Find the knot group of the figure-of-eight (i.e., wedge of two

circumferences).12.67. Let f be a path in X, a E 7r1(X, xo), and f(O) = Xo. Prove that

there exists a path g such that g(O) = xo, g(l) = f(1) and fg' E a.12.68. Let X be a path-connected space. Prove that the group 7r1(X,

xo) is isomorphic to the group y) for any two points x, y E X.12.69. Calculate and where X is the wedge V S".

12.70. Prove that if X is a one-dimensional CW-complex, then 7ri(X)is a free group.

12.71. Prove that the group C Z Z Z Z cannot be the fun-damental group of any three-dimensional manifold.

12.72. Calculate iri(Pg), where Pg is a two-dimensional, compact,closed, and orientable surface of genus g.

12.73. Calculate iri(TPg), where TPg is the manifold of linear elementsof a surface of genus g.

12.74. Calculate the fundamental group of the Klein bottle byconstructing the covering space with the action of a discrete group.

12.75. Let P be a two-dimensional surface with non-empty boundary(i.e., open surface). Prove that irt(P) is a free group.

12.76. Prove that if X is a CW-complex, then is a group whosegenerators are one-dimensional cells, and the whole set of relations isdetermined by the boundaries of two-dimensional cells.

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12.77. Let G be a topological groupoid with identity. Prove that G ishomotopically simple in all dimensions and, as a corollary, that iri(G)is an Abelian group.

12.78. Let X be a topological groupoid with identity, and G Ca subgroup. Prove that

(a) it is possible to introduce the operation of multiplication inX (where Pa is the projection of the covering space

onto X) becomes a homomorphism;(b) if X is a group, then Xc (i.e., covering space relative to the subgroup

G) is also a group. Consider the example Z2 Spin(n)— SO(n), n > 2.

12.79. Prove that the following isomorphism holds

k tunes k times

12.80. Prove that the groups ir(X) are commutative when i > 1 forany CW-complex X.

12.81. Demonstrate by way of example that the excision axiom doesnot hold for the group Y) (the axiom being held for the usual (co)homology theories), i.e., there exist pairs (X, Y) such that

Y) irj(X/ Y).

12.82. Prove that for any path-connected space Yand any point xo E Y,the isomorphism holds Xo) 1(tL0, V. where is theconstant 1oop at the point Xo.

12.83. Prove that = Z2, n > I, and ?rk(RP") = Irk(S"),a I, k > 1, where is the real projective space.

12.84. Prove that if

(a) A is a contractible subspace in a space X (X and A being CW-com-plexes) to a point x0 E A, then the homomorphism : xo)Xo) is trivial when a 1, and when n 3, the decomposition

A, Xo) x0) 7r,,i(A, xo)

holds;

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(b) I: X V Y —' X x Y is an embedding, then the exact sequence isgiven rise:

7rg(X V Y) Tg(X X 1') 0.

12.85. Prove that 7r1(CP'1) = 0; 1r2 = (CP") = Z, n > 0;Irk = (CP") = Irk(S2" + '), k 2.

12.86. Prove that if a CW-complex X has no cells of dimensions from1 to k inclusive, then iri(X) = 0 when i k.

12.87. Let X, Y be two CW-complexes. Prove that ir(X x 1') == lr(X) Calculate the action of iri(X x F) on iri(X x Y).Construct a universal covering of X x Y.

12.88. Find the homotopy groups (0 q n) and prove that= Z, where is a sphere.

12.89. Prove that IrE(S3) = ir'(S2) when I 3 and, as a corollary, that= Z.

12.90. Prove that

(a) 7ri(SO(3)) = Z2, = 1r2(SO) = 0, where SO = limSO(n);

(b) 7r3(SO(4)) = Z, iri(U) = Z, 1r2(U) 0, where U = limU(n) (em-

beddings U(n) C U(n + 1) and SO(n) C SO(n + 1) being standard);(c) = Z.

12.91. Find the groups irg(S' V S'), g 0.12.92. Calculate the groups iri(X), Irn(X) and action of the group iri(X)

on the group irn(X) for the following cases: (a) X = RP"; (b) X= S' V (c) X = + '), where B is the division ring of anon-trivial O(n + 1)-fibration of disks on S'.

12.93. If a mapping f(X, A) —' (} B) sets up the isomorphismsIrg(X) Irg(Y) and lrg(A) Ir8(B) for all g, then it sets up the isomor-phisms irg(X, A) wg(X, B) for all g.

12.94. Calculate the groups Ir,, where is the real Stiefelmanifold.

12.95. Prove that the groups cannot become trivial, as k in-creases, beginning with a certain number k.

12.96. Prove that 7r3(S2) Z, and 1(S") = Z2, when n 3.12.97. Find 1r3(32 V S2), 7r3(S' V and 3r3(S2 V S2 V S2).12.98. Calculate the one-dimensional relative homotopy group of the

pair (CP2, S2), where S2 CP' C CP2 standardly.12.99. Prove that if:(a) a three-dimensional, compact, and closed manifold M3 is

1-connected, then M is homotopy equivalent to the sphere (i.e., M3 is ahomotopy sphere);

94

(b) M" is a smooth compact and closed manifold such that = 0

when i then W is homotopy equivalent to the sphere

12.100. Construct an example of a three-dimensional, closed, andcompact manifold M3 such that M3 is a homology sphere (i.e., it has thesame integral homology as S3), but iri(M3) 0. Construct an exampleof a finitely generated group G which coincides with its first commutatorsubgroup.

12.101. Prove that the set of homotopy classes of mappings ISa, Xiis isomorphic to the set of classes of conjugate elements of the group

xo) under the action of iri(X, Xo) (X being a connected complex).12.102. Calculate ir2(R2, X), where R2 is a plane and X a figure-of-eight

embedded into the two-dimensional plane.12.103. Calculate ir1(CP") when i 2n + I.12.104. Let = 0, and a finite group G act on X and Y without

fixed points. Prove that there exists, and is unique up to homotopy, amapping f: Y X which commutes with the action of the group G.

12.105. Prove that [CP2, S2l = where [X, Y] is the set ofhomotopy classes of mappings of X into Y.

12.106. Let (X, A), X D A, be a pair of topological spaces, and Xpath-connected. Let A be the set of paths in the space X starting at acertain point Xq and ending at points of the subspace A. Prove that irg(X,A, a) irg_j(A, Xa), where Xa is an arbitrary path from Xo to a E A.

12.107. Prove that the following conditions are equivalent ton-connectedness:

(a) iro(S'1, X) consists of one element when q n (base-point preser-ving maps);

(h) any continuous mapping 5q X can be extended to anycontinuous mapping of the disk X, q n.

12.108. Prove that iro(X, is an Abelian group, where X, Z aretwo topological spaces, and fIX is the loop space. Prove that fIX is anH-space.

12.109. Let A be a retract of X. Prove that when n I, for any pointxo E A, the homomorphism induced by the embedding

Xo) Xo)

is a monomorphism, and when ii 2, determines the followingdecomposition into the direct sum

irn(X, xo) xo) A, xo).

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12.110. Prove that X) is an Abelian group. Establish a relationwith iro(Z, [lOX).

12.111. Let S" be the standard tangent bundle over the sphereCalculate the homomorphism : —hr,, (S" I) from the

exact homotopy sequence of this bundle.12.112. Letf: X— V be a continuous mapping (f(xo) = Yo). Prove that

the induced mapping xo) yo) is a grouphomomorphism.

12.113. Let

Y F0 3

iX 3 xo

be a fibration witn iixed points Xo, yo and fibre F. Prove that F0;

yo) ir,,(X, xo).12.114. Let E, X be two topological spaces, X path-connected, and

p: E —' X a continuous mapping such that for any points x E X,y E '(x), the isomorphism holds true

p1(x), y) ir1(X, x), I 0

(for I = 0, 1, a set isomorphism is valid, the sets being stripped of thegroup structure). Prove that for any points x1 and the topologicalspaces p '(Xi) and 1(X2) are weakly homotopy equivalent.

12.115. Prove that for the homotopy groups of a pair (X, A) the exactsequence is valid

ir(A) ir(X) A) i(A) -*

12.116. Prove that if X is a smooth, compact and closed submanifoldof codimension one in Euclidean space, then X is orientable.

12.117. Prove that if the fundamental group of a compact closed man-ifold is trivial, then the manifold is orientable. Prove that if a manifoldX is non-orientable, then there is a subgroup of index two in

12.118. Prove that if X is a non-orientable space, then the suspensionis not a manifold.

12.119. Prove that the Euler characteristic X(X) of any compact, closedmanifold is trivial.

12.120. Give examples of

(a) a non-orientable manifold doubly embedded in another manifold(whose dimension is one greater);

(b) an orientable manifold singly embedded in another manifold.

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12.12 1. Let X1 and X2 be two tori (meaning the solids), f: 0X1 oX2a diffeomorphism, and = X1U1X2. Give examples of diffeomor-phisms f such that the manifold M) is diffeomorphic to: (a) S3,(b) S2 x S1, (c) RP3.

12.122. With the notation of the previous problem, consider themapping

iri(0Xi) i.e., Z Z

which is induced by the diffeomorphism between the solid tort X1 andX2. It is obvious that the is given by the integral matrix

(a b

\c d

Prove that this matrix is unimodular and calculate the fundamental groupof the manifold in terms of the matrix

12.123. Let X,, be the space of polynomials (of one complex var-iable) without multiple roots. Find the groups

12.124. Prove that a finite CW-complex is homotopy equivalent to amanifold with boundary.

13Covering Maps, Fibre Spaces,

Riemann Surfaces

13.1. Let p: X—' V be a covering map such that [In (X, xo)1 is a norm-al subgroup of the group ini(Y, yo), p(xo) yo. Prove that each elementa E ( Y, Yo) generates a homeomorphism of the covering, i.e.,

pcc(x) p(x).13.2. Let p X V be a covering map, yo. Prove that

irt(X, xo) yo) is a homomorphism.13.3. Let p X —* V be a covering map, and p(xo) = Yo. Prove that

the induced mapping : irt(X, xo) If .yo) is a monomorphism.13.4. Let p: X V be a covering map, and irj(Y) = 0. Prove that

each element a E ini(X) is determined by a homeomorphism of the space

Y onto itself, a: V Y, and the is commutative.

7—-20l8

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13.5. Letp : X Y be a covering map where the space Xis connected,and let F = p'(yo) be the inverse image of a point E 1', E F.

Prove that F and Yo) are in one-to-one correspondence if irt(X,Xo) = 0.

13.6. Let p : X Y be a covering map, F: j2 —+ Y a continuousfunction, where j2 is a square, and f: P —÷ X also continuous, withpf(l) = F(t, 0).

Prove that f can be extended to the mapping G: j2 X, with pGG(t, 0) = f(t).

13.7. Let p : X Y be a covering map, j, g two paths on X, andf(O) g(0). Let pf(1) pg(1), and the paths pf and pg homotopic. Provethat f(l) g(l).

13.8. Let p : X Y be a covering map, f, g two paths on X, andf(0) = g(0). Does it follow from pf(l) = pg(1) that f(l) g(l)?- 13.9. Let p : X Y be a covering map,

j two paths on X such that pf = pg g, f(0) = g(0).Prove that if f and g are homotopic, then f and g are also homotopic.13.10. Let p : X —+ Y be a covering map,! a path in Y, and Xo a ooint

in X such that p(xo) = f(0). Prove that there exists, and is unique, a pathg in X such that pg f.

13.11. Prove that a covering map is a Serre fibre map.13.12. Prove that any two-sheeted covering is regular. What purely

algebraic fact corresponds to this statement?13.13. Prove that a three-sheeted covering of a pretzel (i.e., sphere with

two handles) is non-regular.13.14. Let M2 be a non-orientable, compact, smooth and closed man-

ifold. Prove the existence of a two-sheeted covering map p : M2+ —p M2,where M2+ is an orientable manifold, and find M2÷ in explicit form. Whatis the property of the fundamental group of a non-orientable manifold?

13.15. Construct the covering map S'1 RP'1 and prove that:

(a) RP" is orientable when n = 2k— 1, and non-orientable whenn = 2k;

(b) irt(RP'1) Z2, ir(S'1) when > , >

13.16. Prove that a covering space is regular if and only if its pathslying over one path in a basis are either all closed or all non-closed.

13.17. Let p : X -. X be a covering map. Prove that any path in Xcan be covered in X in a unique way up to the choice of the origin ofthe path in the inverse image, and the multiplicity of the projection pis the same at all points of the base space.

13.18. Construct all coverings over the circumference and prove thatin(S1) Z, i 2.

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Zk13.19. Construct the regular covering map p: Pk P2, where

k > 2 and Pk is a sphere with k handles.13.20. Construct a universal covering bf v4S1 and prove that

= 0 when i > 1. Find lrl(VAS').13.21. Construct a covering map çø : X —p where P2 is a pretzel,

such that X is contractible to the graph. Prove, as a corollary, that

(a) a universal covering space P2 is contractible, P2 — K(ir, 1);

(b) if M2 is a two-dimensional closed manifold and ,ri(M2) an infinitegroup, then M2 — K(ir, 1) (i.e., homotopy equivalent).

13.22. Establish the relation between universal coverings over Pk (i.e.,sphere with k handles) and the Lobachevski plane.

13.23. Prove that all covering maps of the torus T2 are regular andfind them. Construct an example of two non-equivalent, but homeomor-phic covering maps of the torus T2.

13.24. Let X be a finite complex. Find the relation between G C iri(X)(arbitrary subgroup), x(X) (the Euler characteristic) and X(XG) — (XG)(covering map constructed relative to the subgroup G C iri(X)).

13.25. Construct a universal covering of the torus P1 (i.e., sphere withone handle) and Klein bottle (i.e., sphere with two cross-caps) and calcu-late the homotopy groups of P1 and N2. Can the torus P1 be a two-sheetedand regular covering of the Klein bottle? If so, find the covering andcalculate the image of the group 7r1(P1) in 7r1(N2) under the coveringmonomorphism.

13.26. Prove that if = 0 or is a simple or finite groupof order p 2 (where p is prime), then the manifold LW is orientable.

13.27. Construct the explicit form of seven smooth linearly independentvector fields on the sphere S7. Use the algebra of octaves (Cayleynumbers). Construct the integral curves of these vector fields.

13.28. Prove that if k linear operators A1, . . . , are given in R" suchthat = —E and + = 0 (for all i, j), then k linearlyindependent vector fields can be specified on the sphere — C R".

13.29. If the homotopy groups of the base space and fibre of a fibrespace have finite rank, then the homotopy groups of the total space alsohave finite rank, the rank of the q-dimensional group of the total spacenot exceeding the sum of the ranks of the q-dimensional homotopygroups of the base space and fibre.

13.30. Let the fibre map p E B admit an image set of a sectionx : B —* E, and eo = x(bo). Prove that when n I, the mapping isan epimorphism, and when n 2, it determines a decomposition intothe direct sum eo) = b0) eo).

13.31. Prove that if all the homotopy groups of the base space and

99

fibre are finite, then the homotopy groups of the total space are also finiteand their orders dO not exceed the product of the orders of the homotopygroups of -the base space and fibre of the same dimension.

13.32. Prove that the mapping p EX Xsatisfies the covering homo-topy axiom (Serre fibre map).

13.33. Prove that Xi,2 are covering maps and= lm(p2),. then (X1, X) and (X2, p2, X) are fibre homeomor-

phic, where lm(p1), C iri(X).

13.34. Prove that there exists a covering map p —p X over anyconnected complex X such that iri(X) 0 (i.e., existence of a universalcovering).

13.35. Prove that the set of vector bundles with a structural group Gover the sphere is isomorphic to ir,,_ j(G), and that the group G ispath-connected.

13.36. Show by way of example that there exists no exact homologysequence of a fibration.

13.37. Let p E B be a locally trivial fibre map, and B, F finite com-plexes. Then ( x(fl.

13.38. Given that a material particle moves with constant (in modulus)velocity (a) on the torus 7", (b) sphere S", find phase space for thissystem.

13.39. Let p E B be a fibre map with path-connected B and F.Let Cat = Cat — 1 be a reduced Lusternik-Schnirelmann i.e.,Cat (of a point) = 0. Prove that Cat(E) + Cat(B) ++ CatE(F), where CatE(F) is a relative category of the fibre F respectiveto E.

13.40. Prove that if p: X -+ Yis a Serre fibre map, then p is a mapping"onto".

13;41. Prove that if p X Y is a Serre fibre map, then p andare homotopy equivalent for any y2 E Y.

13.42. Prove that the manifold of linear elements of a manifold M isa fibration with the base space M.

13.43. Prove that a locally trivial fibration (twisted product) is a Serrefibration.

13.44. Prove that the space of paths EX whose starting point is fixedin the space X is a Serre fibre space with the base space X.

13.45. Prove that_if M" is a smooth manifold, then the spaces of itstotal and unimodular tangent bundles (fibrations) are orientable.

13.46. Prove that the direct product of topological spaces X x Y withthe projection onto one of the factors is a Serre fibration.

100

13.47. Let p X Y be a covering map, p(xo) = yo, and f g two pathsthatf(O) = g(O) = yo, f(1) g(1). LetJg E (irj(X, Xo)) and let

J be two coverings of these paths. Prove that JO) =

2 213.48. Represent the torus T as T = (g where g \0Consider the following equivalence relations R:

(a) (e"l,

(b) (etwl,

(c)

Find the space X T2/R and calculate the image Cwhere f: T2 —' X = T2/X is the projection induced by the relation R.Is f a covering map?

13.49. How many fibrations are there of the following form:

(a) T3 where T3 is a three-dimensional torus;

(b) T" S', where 7" is an n-dimensional torus (fibrations areconsidered up to homotopy equivalence)?

13.50. Let C = A * B be the free product of arbitrary groups A andB. Prove that for any subgroup M C C, the equality M A1 * B1 * Fholds, where A1 C A, B1 C B and F is a free group. Give a topologicalproof using covering spaces.

13.51. Let be a 1-connected compact Lie group, and a: ® anarbitrary involutive automorphism (i.e., a2 = I). Put = (g E == gl; V = fg e = Prove that = i.e., any elementg E admits a representation in the form

g = vhv, yE

V (homogeneous space).13.52. The following construction (by Cartan) is known. Let a:

be an arbitrary involutive automorphism of a compact connected Liegroup. Put = E ® a(g) V = fg E = ThenV and V C is a totally geodesic submanifold. Therefore, Vis a symmetric space. The submanifold V is called Cartan's model of thesymmetric space Any symmetric space admits Cartan's model(which is almost always uniquely determined).

(a) Prove that the projection p: V, where p(g) = ga(gdetermines the principal fibration 0 —' —' —4 0.

101

(b) Let V be Cartan's model, = 0, e E V, and e the identityelement in Prove that if a point x E V is conjugate of e along ageodesic C V in the group then the point x is conjugate of ealong -y in the manifold V C itself.

1333. Prove that a compact, closed manifold M2 with Euler charac-teristic N can be represented as a (2N + 4)-gon such that some of itssides are glued together to yield the word

—1 —1 —IaIa2.. . aN÷2a1 a2 . . . aN÷2

(where az, . . . , 2 are the designations of the sides) in traversingthe sides one after another. Prove that the last factor is if and onlyif M2 is orientable.

1354. Classify compact, closed smooth and connected manifolds Mand calculate their fundamental groups in terms of the generators andrelations.

13.55. Prove that any orientable, two-dimensional, and compact man-ifold is determined by a unique invariant, viz., the genus of the manifold.

1336. Prove that any non-orientable, two-dimensional and compactmanifold can be represented as the connected sum of an orientable man-ifold and several replicas of projective planes.

13.57. Describe the semigroup of two-dimensional manifolds under theconnected sum operation.

1338. Calculate the homotopy groups ir(Tg) (i I) of a two-dimensional manifold T5 of genus g.

13.59. Let M2 be a compact, closed, oriented and two-dimensionalmanifold of genus g. Find the homotopy type of

13.60. (a) Prove that RP2 \ D2 is diffeomorphic to the Mdbius strip.(b) What spaces is the sphere Sz, with a Möbius strip glued into,

homeomorphic to? with two Möbius strips?13.6!. Let S' x S' c R3 be the standard embedding of the torus in

Euclidean space. Prove that there exists no homeomorphism of the pair(R3, S' x S') onto itself whose restriction to the torus is determined by

the matrix ( 0 1

\—1 0

13.62. Given two odd functions on the sphere S2, prove that they havea common zero.

13.63. Let ir be the fundamental group of a two-dimensional surface,and f: an epimorphism. Prove that f is an isomorphism.

13.64. Prove by three totally different techniques that there exists no

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continuous vector field without singular points (i.e., different from zeroat each of its points) on the sphere S2.

13.65. Let in C2(z, w) the Riemann surface of the algebraic functionw = be given, where the polynomial has no multiple roots.Prove that this Riemann surface turns, after completing it with a pointat infinity, into a two-dimensional, smooth, compact, and orientablemanifold.

13.66. Find the genus of a two-dimensional manifold described in theprevious problem as a function in the degree n of the polynomial

13.67. Can the two-dimensional projective plane RI'2 be the Riemannsurface of a certain algebraic function w w(z) in C2(z, w) in the senseof Problem 13.65, i.e., after completing the Riemann surface with a pointat infinity?

13.68. Prove that the Riemann surface of an algebraic function in C2is always an orientable manifold.

13.69. Investigate what happens to the Riemann surface of the functionw = when some roots of the polynomial merge to yielda multiple root. For example,_what is the structure of the Riemann surfaceof the function w = —

13.70. Describe the topological structure of the Riemann surfaces ofthe following analytic functions:

z=w+1/w,

13.71. Prove that the Riemann surface of the function w = lnz ishomeomorphic to a finite part of the complex plane.

13.72. Let p X V be a two-sheeted covering. Prove that any pathin Y can then be covered by precisely two paths.

13.73. Construct a universal covering space for the orthogonal groupSO(n).

13.74. Prove that any two-dimensional, closed, oriented and smoothmanifold can be locally isometrically covered by the Lobachevski plane(which is supplied with the standard metric of constant negative curva-ture). In other words, prove that the fundamental group of a surface ofthe indicated form can be represented as a discrete subgroup (operatingeffectively) of the Lobachevski plane isometry group.

Corollary. A two-dimensional, compact, closed, orientable, andsmooth manifold can be supplied with the Riemannian metric of constantnegative curvature.

13.75. What spaces can cover the Klein bottle?13.76. Let Sg be a sphere with g handles. What Sh can cover

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13.77. Prove that for any compact, non-orientable, and two-dimensional manifold, there is precisely one compact, two-dimensional,and orientable manifold which serves as its two-sheeted covering.

13.78. Prove that the Beltrami surface (i.e., surface of constant negativecurvature standardly embedded in R3) can be infinitely-sheeted andlocally isometrically covered by a certain region lying in the Lobachevskiplane.

Find this region. Prove that it is homeomorphic to the two-dimensionaldisk. Find the corresponding group of this covering (it is the group Z).

13.79. Can a two-dimensional torus be a two-sheeted covering of theKlein bottle?

13.80. Calculate the permutation group of the sheets of the Riemannsurface of the algebraic function w = arising in traversing around thebranch point of this function (point 0).

13.81. Let M2 be an ellipsoid, and p one of its vertices. Consider allgeodesics emanating from the point p. Find the locus of the first con-jugate points (i.e., mark the first conjugate point of p on each geodesicand describe this set).

13.82. Prove that the fundamental group of a complete Riemannianmanifold of non-positive curvature contains no elements of finite order.Prove that (M) (where Mis a complete Riemannian manifold of strictlynegative curvature) possesses the following property: if two elementscommute (ab = ba, a, b E irj(M)), then a and b belong to the same cyclicsubgroup.

13.83. Prove that a closed, orientable Riemannian manifold ofstrictly positive curvature and even dimension is 1-connected.

13.84. (a) Prove that any compact, closed Riemannian manifold ofconstant curvature X is isometric either to the sphere S" or (of radius

Use Problem 13.83.

(b) Let be a compact, closed, 1-connected, complete Riemannianmanifold, and C(I) the set of the first conjugate points of a certain point1 E

Prove that if is a symmetric space, then the complement Mn/C(1)is homeomorphic to the open disk.

13.85. Prove that a complete, non-compact Riemannian manifold ofpositive curvature and dimension m, where either m 2 or m 5, isdiffeomorphic to Rm.

13.86. Let x, y be two near points on the standard sphere S2, and afunction f(z) the area of the geodesic triangle with vertices at the pointsx, y, z.

(a) Is the function f(z) harmonic on the sphere S2?

104

(b) Investigate the case of the n-dimensional sphere (where f(z) is thevolume of the geodesic simplex whose one face is fixed, and z is a freevertex).

(c) Investigate the same problem for the Lo 'achevski plane.

13.87. Prove that if M" is a complete, l-ccinnected Riemannian man-ifold such that n is odd and there exists a point p on the set of thefirst conjugate points of p being regular and each point of constant orderk, then k n — I, M" is homeomorphic to the sphere (order of apoint is understood to be its multiplicity).

13.88. Let -y C R2 be a simple closed curve of length I bounding aregion G of area S (on the plane).

Prove that ,2 4irS and that the equality holds if and only if 'y isa circumference.

13.89. Let 'y C R2 be a closed curve (not necessarily simple, i.e., incontrast with the previous problem, self-intersecting is possible). Provethat ,2 47r w(x)dS, where the function is the number of rotations

R2

of the curve about a point x E R2.

13.90. Is it true that if n(x, y) is the refractive index of a planar, transpa-rent, isotropic, but non-homogeneous substance filling the two-dimensional plane, then the integral curves of the vector field grad n(x,y) (n = clv) are the trajectories of light rays? (Certainly, not only they.)

14.Degree of Mapping

14.1. Calculate the degree of the mapping f(z) S' — wheref(z) = z", = I.

14.2. Calculate the degree of the mappingf: S2, wheref(z) =zE CU Lool.

14.3. Let be an orientable, smooth, and compact manifold. Provethat the homotopy class of a mapping 5" is fully determined bythe degree of the mapping.

14.4. Let f: 52n be a continuous mapping. Prove that there isa point for which either f(x) = x or f(x) = —x.

105

14.5. Let the degree of the mapping f: S" —+ be equal to 2k + 1.

Prove that there exists a point x such that f(x) = —f( — x). Prove thatthere exists no even tangent vector field v(x) without singularities (i.e.,v( — x) = v(x) has no zeroes) on the sphere —

14.6. Let g: 5" 5" be two simplicial mappings. Prove that:

(a) the inverse image of each interior point Consists of the same numberof points (meaning the difference between the numbers of positivelyoriented and negatively oriented points);

(b) if g are homotopic, then the difference between the number ofpositively oriented and negatively oriented points of the inverse imageis the same for the two mappings;

(c) if the difference between the number of positively oriented andnegatively oriented points of the inverse image coincides for the two map-pings, then they are homotopic.

14.7. Let f: M S2 be a mapping of the normals of a closed surfacein R3. Prove that f(w) = K(w') where and are elements of areaand K the Gaussian curvature. Prove that 2 degf = and also equalsthe Euler characteristic of the surface.

14.S. Prove that any continuous mapping of the ball D" into itselfalways has a fixed point.

14.9. Let f: SU(n) SU(n) be a smooth mapping, and f(g) = g3.

Find deg f.14.10. Let!: M" R" be a smooth mapping of a connected, compact,

orientable, and closed n-dimensional (n < p) manifold in R". Let v(f)be the normal bundle of this immersion, and Sv(j) the associated fibrebundle of spheres, i.e., Sv(J) = is the boundary of a certainsufficiently small tubular neighbourhood of the submanifoldf(M") C R". Let T: Sv(J) be a usual Gauss (spherical)mapping.

Find deg T (dimSv(J) = p — 1) if the Euler characteristic of the man-ifold M" is known. Does deg T depend on the method of immersing M"into What happens if M" is non-orientable? Separately consider thecase where p = ii + 1.

14.11. Given that a two-dimensional, orientable, closed, and compactmanifold M2 of genus g is embeddable in the Euclidean space R3, findthe minimal number of the saddle points (generally speaking, degenerate)of the function f(p) = z, p E i(M2), where i is an embedding and f theheight function.

14.12. Prove that non-degenerate critical points of a smooth functionon a smooth manifold are isolated.

14.13. Letf(x) be a function on a two-dimensional, compact, orientable

106

surface of genus g (i.e., sphere with g handles) having a finite numberof critical points, all of them being non-degenerate. Prove that the numberof minima minus the number of saddle points plus the number of maximaequals 2g — 2.

14.14. Let f: —i R be a smooth function on a smooth manifold.Prove that almost every value of the function f is regular.

14.15. Prove that the alternating sum of singular (critical) points ofa smooth function f(x) (assuming that all its singularities are non-degenerate) given on a smooth compact manifold does not depend onthe function (by the alternating sum, we understand

where n = dimM,

X the index of a critical point, and mx the number of the critical pointsof index X).

14.16. Let f(x) be a complex analytic function of one variable x. Provethat the set of critical values of the function f(x) S2 has measurezero.

14.17. Let = lxf(x) = c}. Prove that if contains no criticalpoints of the function f, then is a submanifold in and

= 1.

14.18. Prove that the notion of non-degenerate critical point of asmooth function does not depend on the choice of the local chartcontaining this point.

14.19. Show that for the standard embedding of the torus T2 C R3(i.e., surface of revolution about the axis Oz), the coordinate x orthogonalto the axis of rotation of T2 has only non-degenerate critical points.

14.20. (a) Construct functions with only non-degenerate critical pointson RI" and CI"' so that their values at all critical points may be different.

(b) Construct functions on RP" and CP" such that f(xx) = X =where are non-degenerate critical points of index X.

14.21. Let F(x, y) be a non-degenerate bilinear form on R". Considera smooth functionf(x) = F(x, x), where x = 1, i.e., F(x, x) is a functionon the sphere S" C R". Let Xo . . . 1 be all the eigen-values of the form F (recall that all 0 i n — 1, are real).

Prove that X are the critical values of the function F(x, x) on the sphereFind all the critical points of the function F(x, x). Prove that

= inf maxf(x) where 5' are the standard i-dimensional equators ofS' xES'

the sphere

107

14.22. Prove that if a point p is a non-degenerate critical point for asmooth function f(x) on a smooth manifold, then there exists a localcoordinate system in which the function f(x) in a neighbourhood of thepoint p can be represented as a non-singular quadratic form.

14.23. Prove that if is a non-critical level for a function f(x) ona manifold M(i.e., the level hypersurfacef(x) c const not containingcritical points for f(x)), then the neighbourhood is diffeomorphic to

x I.14.24. If and are consecutive critical levels, then the interval

between them is diffeomorphic to >< I, where Ci < C < C2.14.25. If there are no critical levels between and (i.e., level

hypersurfacesf(x) = const with critical points) and and are non-critical either, then they are diffeomorphic.

14.26. (a) Construct a smooth function f(x) having one point ofmaximum, one point of minimum (both being non-degenerate), and ano-ther critical point, perhaps, degenerate, on every compact, orientable,two-dimensional, and smooth manifold M2. Find the relation betweensuch a function and the representation of M2 as the Riemann surfaceof a certain many-valued analytic function. Investigate the case of a non-orientable two-dimensional manifold M2 (e.g., case of the projective planeRP2).

(b) Construct a smooth function f(x) having only non-degeneratecritical points, precisely one point of maximum, precisely one point ofminimum and s saddle points (find the number s) on every compact man-ifold M2. Construct the function so that it takes the same value at allthe saddle points. Investigate the non-orientable case. Indicate the relationto the problem of point (a) and Construct the confluence of all the saddlepoints into one degenerate critical point.

15Simplest Variational Problems

15.1. Prove that the extremals of the action functional E[y]

on a smooth Riemannian manifold M'1 (where -y(t) are smooth trajectorieson M", 0 t 1, and is the velocity vector of the curve 'y(t)) aregeodesics.

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15.2. Establish the relation between the extremals of the length func-

tional L[y] = and action functional E['y] = Prove that any

extremal 'yo(t) of E[yJ is that of LFy]. Prove that if so(t) is an extremalof then by replacing the parameter t = t(r) by so(t), this trajectorycan be transformed into an extremal of E[-y].

15.3. Let S(f) = be a functional associating each

smooth function z = f(x, y) which is defined on a bounded regionD = D(x, y) C R2(x, y) (where x, y, z are Cartesian coordinates in R3)with the area of the graph of the function z = f(x, y). Prove that theextremality of a function fo relative to a functional S is equivalent to thecondition H = 0, where H is the mean curvature of the graph of z =y) considered as a two-dimensional, smooth submanifold in R3 -

15.4. Prove the statement formulated in the previous problem for thecase of the (n — 1)-dimensional graphs off =f(x1, . . . , in

15.5. Prove that the action functional E[-yl and length functional L ['ylare related by the formula E[-yJ, the equality being held if andonly if -y(t) is a geodesic.

15.6. Prove that the areal functional

SV] = — F2dudv

(where ,' 1'(u, v) is a radius vector in R3 depending smoothly on (u, v))

and Dirichlet's functional D[?[

D

E ± Gdudv are related by the

formula S[,1 D[f'].15.7. Remember that the radius vector f'(u, v) determining a two-

dimensional surface M2 in Euclidean three-dimensional space is said tobe harmonic if v) is an extremal of Dirichlet's functional DVI =

= +D

+ G)dudv. Prove that if the mean curvature H of a

surface M2 given by a radius vector v) equals zero, then local coor-dinates (p. q) can be introduced in a neighbourhood of each point onthe surface so that the radius vector q) in these coordinates becomesharmonic.

109

15.8. Construct an example of a harmonic radius vector r-..(u, v) suchthat the surface M2 C R3 described by it may not be minimal (i.e., sothat H 0).

15.9. The Wirtinger Inequality. Let H be a hermitian symmetric positivedefinite form in and C" R2" a realification of C". Then

H HR = ( Swhere H = S + IA, S and A are real\-A SJ

matrices and 5T = 5, = —A, If T = H.

The form S defines the Euclidean scalar product in R2" and the formA an exterior 2-form in B2". For simplicity, we may assume that

A dzk. Consider the form 0(2r) = I. . . Aw, r n.

r!

(a) If WI W2r is an arbitrary orthonormal basis in R2" C" rela-tive to the scalar product S ReH, then

1 and W2,-)l =

if and only if the plane L(wi W2r) generated by the vectorsW2r is a complex subspace in R2" C".

Hint: Let r = 1, and Wi, be an orthonormal pair of vectors. It isrequired to prove that W(Wl, W2) I, where = A(Wl, W2).

Consider

H = (WI, W2) = (S + 1A)(wi, W2) =

= S(wi, W2) + 1A(wi, W2) = W2).

Hence, H(Wi, w2) = A(wi, W2) W2 = I. Now, let A(wi, W2) = 1.

Then H(Wi, W2) = 1, i.e., 5(Iwi, W2) = 1. Since = lW = 1, it followsthat i.e., the two-dimensional plane spanned by WI, W2 is thecomplex straight line. For r > I, the relation — W2) =should be used, where gjj is the skewsymmetric scalar product definedby the 2-form

(b) Let Wt C C", r < n (r being complex dimension) be a complexsubmanifold in C" (if W' is an algebraic submanifold, then singular

110

points on are possible). Let V2' be a real submanifold in suchthat VU W = where is a real (2r + 1)-dimensionalsubmanifold in whose boundary is V U W. Let K = V fl W. ThenV012r(V \ K) V012r(W/K).

Note. This statement means that complex submanifolds W in thecomplex space C" are minimal submanifolds, i.e., after any "pertur-bation" of V, the 2r-dimensional volume (V0l2r) does not decrease.

Hint: The statement follows from the Wirtinger inequality (see above)and Stokes' formula. In fact, consider the exterior 2-form

k=lA dzk,

and let

= wA . . . (r times).r!

Since = 0, = 0. It follows from Stokes' formula that

=

While integrating the form with respect to a 2r-dimensional subman-ifold, the expression of the sort U2r)dX A . . . A

where is an orthonormal basis in the tangent plane to thesubmanifold (with respect to the Riemannian metric induced by theunderlying Euclidean metric in C" = R2") should be considered (in localcoordinates x' x2'). If the submanifold W is complex, then

W2r) = I and vol2r(W') =

If the submanifold V is of general form (i.e., real), then1, i.e., vol(I/), which proves the statement.

111

(c) Prove that the statement of problem (b) remains valid if C" is re-placed by any Kähler manifold, i.e., complex manifold supplied with anexterior 2-form (non-degenerate and closed).

15.10. Consider functions of the form F(x' f) on R"(x'

x") and the functional J [F] = where D is the domain of

the functions F. Let F0 be an extremal of the functional J. Prove thatthe level surfaces F0 (x' x") = const considered as hypersurfacesin R"(x' x") are locally minimal.

Answers and Hints

2Systems of Coordinates

2.1. J = J-11H2 . . .

(1 of 1 Of 1 Of2.2. grad! = — —,

(H1 Oq' H2 aq2 H3 0q3

2.3.diva = iP__(H2H3ai) + +

H1H2H3 3q2

+ , where a1, a2, a3 are the coordinates of the vector a.Oq j

2.4. 4f = LH2H3

+H3H1 +

1-11H2H3 LOq'\ Oq'J 0q2\ H2 9q2J

+ 0 (HIH2 Of

H3 0q3

2.5. (a) The coordinate surfaces are: cylinders r = const, planes= const and planes z = const.

= 1,1-12 = r,H3 1.

1 0 / Ou\ 1 02u 02u= —--—(r—J -I + —.

r Or\ OrJ r 0z2

2.6. (a) The coordinate surfaces are: concentric spheres r = const,

planes = const and cones 0 = const.

(b) H1 = 1, 1-12 = r, H3 = rsin0.

8—2018 113

I O/23u\ 1 0/. Ou\———--tr —) + —lsin0-----—-I +r2 Or \ Or / r2 sin0 30 \ 30 J

I I 02u+

r2 sin20

2.7. (a) The coordinate surfaces are: cylinders of elliptic section andfoci at the points x y = 0 when X = const, the family of confocalhyperbolic cylinders = const and planes z = const.

(b)H1=

=

2.8. (a) q' sin —p--, q2 = cos = z.

(b) The coordinate surfaces are: parabolic cylinders with generators par-alLel to the axis z when X = const, = const.

= I.

2.9.

a2)

a2)

+ + b2)(v + b2)4..j (c2 — b2)(a2 — b2)

—c2)Q1 + c2)(v + c2)— ±

(a2 — c2)(b2 — c2)

(b) H1 = —= X)

2 R20..)

2 R2(v)

where R(s) = a2)(s s = v.

114

(c)=

_________________

— v)R(X) +(X — — v)(fL — v) L ax \ 3X /

+ (p + (X —]

2.10. (a) The coordinate surfaces are: prolate ellipsoids of revolutiona = const, hyperboloids of revolution of two sheets j3 = const and planes

= const.

(b) H1 = 112 c sinha sinl3.

i r 1

(c) = I ( sinha +c2(sinh2a + sin2f3) Lsinha &x \ /

1 3 7. äu\ / 1 1 \ 32u]+———— Isin13—I + 1 + —I —I = 0.

sinj3 313 \. 313/ \sinh2a sin213J

2.11. (a) The coordinate surfaces are: oblate ellipsoids of revolutiona = const, hyperboloids of revolution of one sheet 13 const and planes

= const passing through the axis z.

(b) H1 = H2 — sin, H3 = C cosha sin 13.

1 [1 3/(c) = i cosha +

c2(cosh2a — sin2f3) [cosha 3a äa /

I 3u\ / 1 1 \ 32u+ ——--tsin13—1 + I— — I

313 \ c913J \sin2j3 cosh2a/

2.12. (a) The coordinate surfaces are: the tori a const,

— c + =

2

=\sinh a,'

the sphere 13 = const,

I 2

2 2(z ccotf3) + =in 13

and the plane = const;

115

13 / sinha 0u\(c) = —J +

ôa \Cosha cos(3 3aJ

13 / sinha äu\ I t32u+—( —1+13/3 \coshcr — 13/3 / (cosha — cos$)sinha

a2.13. H1 = H2 = , H3

cosha — cosfl

2.14. (a) The coordinate surfaces are: spindle-shaped surfaces ofrevolution a = cOflst

(Q — ccota)2 4- z2 =

spheres í3 = const,

+ (z — c cot$)2 =

/ 2(C'\Sifla

(si:hfi)2'

C(b) =

= cosha —

and planes = const.

(b) H1 H2cosh/3 — cosa'

c sinha1-4=cosha — cosf3

C

coshf3 — C05a

13 / sina(c)

= 3at\coshl3 — cosa —) +

/ sina i3u\

+

— /32

2.15. H1 =— 1'

+sina(cosh/3 cosa)

H2 —fL2

2'—/3

H3

116

2.16. H1 =—

H2=

H3 = cXv.

2.17. (a) H1 = H2 H3 =

X = const, = const ofrevolution about the axis of symmetry Oz.

3

Riemannian Metric

3.2. Let the surface M2 C R3 be given by equations x, = Xj(p, q), I = 1,

2, 3, and the variables p and q have a plane region as their domain. Letthe functions x = x(p, q) be real-analytic. The pair (p. q) can be regar-ded as the coordinates of a point on the surface M2. A curve C on M2is given by the equations

p = p(t), q = q(t), a t b.

An element of arc length is expressed in terms of the vector x = (xi,X2, thus:

ds2 = dxdx = + xqdq) -i- xqdq),

or

ds2 (xv, + + (xq, xq)dq2

= Edp2 + 2Fdpdq + Gdq2,

where E = (xv, xe), F = (xe, Xq), G = (Xq, Xq).Since the element of length ds2 is always positive, W2 = EG — F2 is

also positive. Let us find the coordinate system (u, v) with the elementof arc ds2 = X(u, v)(du2 + dv2). We have

ds2+ (F + (F -1W) do).

Assume that we can find an integrating factor = + 1a2 such that

117

(F+iW) \+

Then

ti+ (F

= du — idv,

and, finally, = du2 + dv2. Assuming 012 l/X, we obtain therequired isothermal coordinates (ii, v). Thus, we have obtained isothermalcoordinates by having found the integrating factor which transforms theexpression

+ (Fdq

into a total differential. The differential dii + idv can be written in thefollowing form:

du + idv = + i dp + + I dq.\Op OPJ \Oq OqJ

Further,

t3v — Ou Ov (F + 1W)—+I—=aOp Op Oq Oq

Eliminating c, we obtain

fOu Ov\ /Ou •Ov

\Oq OqJ \Op Op

or

c3u Ov OvE— = F— - W—, E— = W— + F—.rIq Op Op lIq Op Op

Solving this system for the unknowns Ov/Op and Ov/Oq, we obtain

OuF— - F—Ov Op Oq

OP -

Ov Op — Oq

Oq =(1)

118

Similarly,

av avE— - F—

ap

avF— - G—a1,

(2)3q

Therefore, u satisfies the equation

IF - E IF G

—p--- (3q

+ ( =W J ap W J

which is called the Beltrami-Laplace equation. Given a second family ofisothermal coordinates (x, y) in a neighbourhood of a point, we haveds2 = + dy2). Using the coordinates (x, y) instead of the coordinates(p, q), we obtain E = G = F = 0 and

— 3u av 3u—

— ay —

Thus, the Cauchy-Riemann equations have been obtained, and hencethe functions u and v are conjugate harmonic functions, whereas thefunction f = u + iv is analytic in z = x + iy. The Beltrami equationassumes the form of the well-known Laplace equation 82u/äx2 ++ a2u/ay2 = 0. A complex-valued function lip, q) defined on M2 is saidto be a complex potential on M2 if its real and imaginary parts satisfyequations (1). Thus, the real and imaginary parts of a complex potentialon the manifold M2 determine isothermal coordinates in a neigh-bourhood of every point on M2 (the coordinates being local and not serv-ing, generally speaking, the whole of the two-dimensional manifold;while transferring from one point to another, the complex potential willvary).

3.3 (a) Consider some curve ço = on the surface of the sphere.In moving along this curve, the compass needle forms an angle withthe direction of motion determined by the relations

= (I)

119

(The angle is measured from the y axis clockwise.) We obtain on themap:

dy / ir\(2)

dx \ 2/

It follows from relations (1) and (2) that

dx äx 8x dip

dip dO 80 dOsinO = — =

dO dy ôy3y dip'dO 80 8ço dO

/ 8y 0y dip \ dip 8x 8x dip(3)

\ 80 aip dO I dO 80 aip do

Since relation (3) must be fulfilled at the point in question for any valueof dip/dO, we obtain, by equalizing the coefficients of the same powersof the derivative dip/dO on the right-hand and left-hand sides, that

0, y =y(0), (4)aip

(5)

8y 3x—sinO— =

80 äip

It follows from (4) and (5) that the left-hand side of relation (6) dependsonly on 0, whereas the right-hand side only on ip; therefore, both sidesof this relation should be constant. We put this constant equal to unity.Thus, in Mercator's projection, the mapping is given by the formulae

rdO 0x = y = — i— = Incot—.

,jsinO 2

(b)ds2 = dO2 + sin2Odip2 = sin2O(dx2 -4- dy2) = (dx2 + dy2)/cosh2y.

3.5. ds2=

(1 + zzl2

120

2 dv2 v2dça23.6.ds = +

2(i-p) (1-v)3.7. ds2 = +

3.8. ds2= (42 + Q2d.p2)

(1 —

3.9. (a) In polar coordinates,

ds2=dr2+r2dso2,

(b) If the sphere has radius a, then

ds2 = a2(d02 + 0 0 ir, 0 2ir.

(c) a's2 + sinh2xdc,2), 0 x < 0

In case (a),= (1

, and the equation of the circumference is

r(t) = R const, = t, 0 t

whereas the length of the circumference is

L = 2irR.

The circle is given by the relations 0 r R, 0 2ir, and

its area equals S = rdrdçô = irR2.

(b)(a2

2and the equation of the circumference is

\0 a sinOaO(t) R const, = t, 0 I 2ira, 0(1) R/a, = (Ia,whereas the length of the circumference is

(' flRl RL = a J sm2 dt = 2wa sin —.

J \J aa a0

121

The circle is given by the relations 0 aG R, 0 2ir, i.e.,

'.Ig=a2sino,

and the area of the circle is

=

R/a

= —

(c)= (a 02 and the equation of the circumference is

a

ax(t) = R const, aW(t) = t, 0 t 2ira,

and

x(l) = R/a, = t/a,

whereas the length of the circumference is

2,ro

1' 1 RI RL = a lsinh2 dt = 2ira sinh —.aa a

The circle is given by the relations 0 ax R, 0 2ir, ie.,

and the area of the circle is

s = a2 (cosh — i).

122

4.3. y = a cot!, y = a

4.4. r =4.5. r where wI.

4.6. x = at — d sin!, y = a — d cost.

r R+r4.7. x = (R + r)cos— I — rcos I,

R R

r •R+ry = (R + r)sin — I — r sin I.

R R

4.8. x = (R — mR)costnt + mR cos(t —

y = (R ,nR)sinmt - mR sin(t — mo, m = r/R.

4.9. The equation of the required curve is

r(t) = + b,

where b is a constant vector, and /L(t) the antiderivative of the functionX(!), c < 0 < d. Geometrically, the following cases are possible: a

d

straight line collinear with a if X(i)dt diverges when I = c and 0 = d;

d

a ray with the direction of the vector a if\

X(t)dt converges when I = C,

but diverges when I = a ray with the direction of the vector —a if

X (t)dt diverges when I c, but converges when r = d; an open line-

segment collinear with a if Mod! converges.

4.10. The equation of the required curve is

r(t) = —02a+ lb + c,

where b, c are arbitrary constant vectors.If b 0, then this equation determines (with b and c fixed) a parabola

with the axis whose direction coincides with that of the vector a. If b = 0,

then we obtain two coincident rays parallel to a.

4.11. (a) (r')2 Er' x a]2;

(b) —(r', a) [r' x aJ2.

4.12. Apply the Rolle theorem to the function (a, r(t) r(to)).

124

4.13. Use the equality (r2(t) — ri(r))2 = const, where r1, r2 are radiivectors of the moving points and t is time.

4.14. Put = X, X(t) being a function continuous on the segment

La, b] and having the same sign on it. We have — Xr 0, whencer = Since the derivative of the function eMt equals it doesnot change sign on the segment [a, b], i.e., eXdJt is a monotonic andcontinuous function of 1.

4.15. Applying the method of solution of the previous problem, wehave r' = whence

r = a + b.

The derivative of equals > 0; therefore, is amonotonic increasing function of I E [a, b].

4.16. r' = [ço', + r" = + j, [r' x r"] == 2 — pç,". The given equation determines a straight line if and onlyif 2sc'2 — = 0. Solving this equation, we find cc = 1/(at + b),where a and b are constants.

4.i7. r = rr°, + Since r° = sinccl,dcc dco

cosccl, i.e., is obtained from r° by rotating it throughdcc

-- in2. Denote the vector obtained from r° by rotating it through + ir/2by [r°l. Therefore,

dr= r'r + r[r°J.

dcc

Furthermore,

= r"r° + — rr° = (r" — r)r° + 2r'[r°],dcc

Idr d2rl r' r =2r'2—rr"+r2=zO.Ldcc dçc2j r" — r 2r'

Putting r' = w, we find

dwr = = —r' =dcc dr dr

2c/u, 2 2w2 wdw

2w =0, ——--—--+ 1=0.dr r2 rdr

125

Put w2 = p. r2 = q, then dp/dq = 2p/q + 1. Solving_this equation, wefind that p = aq2 — q, or = ar4 — r2, r' = r'iar2 — 1. Substitutinghr = we easily obtain

= C1 + c2),

where C1 and C2 are arbitrary numbers.4.18. F = Fr = mr". Differentiating, we obtain X'r + Xr' = mr";

therefore, the vectors r', r", r " are coplanar.4.19. The radius vector of an arbitrary point of the centre surface

can be determined by one of the relations:

= r1 + X[rI] = r2 +

r1 — r2 + X[rf] =

(ri r2)r2 + = 0,

= —

r1 + , ,[r1 )< r2]

and in coordinates,

(X2x — yl,

—

(X2 — Xi)Xj + (y2 —n=yi— x1.x(y2' - xiyi'

4.20. Consider the vector X[r(], where= (r2 — ri)rI

. If this vectorx

is marked off from the end M1 of the rod, then its end will fall on theinstantaneous centre of rotation. The projections of the vector X[r(1 ontothe vectors r2 — r1 and [r2 r1] are equal, respectively, to

r1)and

X[r(][r2 — nj— ri r2 — rI!

Therefore, the equations of the centrode are:

= (r2 — r1)r( lr( x (r2 — rj)][r( x ri]: r2 r11

126

= (rz — ri)rI rf(r2 — ri:

[r, x r2 — r1j

or

((X2 —yl

X2—Xi Y2—Y'Xi

— x1)2X2 Y2

= [(xi — x1)xI+ (Y2 — yi)yfl[(xi — x)x( + (yi —

YIH — xj)2+Xi Yi

4.21. R = r1 + p ri + Ea + where a = — r1, = const,

= const (point M being rigidly connected with the rod) andR' rI + a = ri — = const, a' I a. Therefore,

a' = sEa], rj — ri = s[r2 — ri],

ri')[r(] = s[r2 —

= s(r2 — ri)ri,[ri x rfl 1

s=(ri — r1)r( X

Thus:

a' = — [a], [a'] = — — a,>'

K' = ri + [a] — a=

+ E[a] — na).

On the other hand,

r= R— r1 + — X[rfl = Ea+ ,1[a]

[r] = — E[a] —

therefore,

I IR' = —(r], w =

x x

127

4.22. r" II r, Er, r"] = 0, Err']' = Err"]. Therefore, Err'] = a = const,and

Ear"] — (r' r2 — r(rr'))

— rr'=

Thus,

Ear'] +

X = b = const.

Multiplying both sides of this equality by r and noticing that [ar']r= a[r'r] —a2, we have: —a2 + Xr = br. The motion is in the sameplane perpendicular to the vector a (since it follows from the relationErr'] = a that ar = 0). Introducing a polar coordinate system on thisplane and making the pole coincident with the origin of the radii vectors,while directing the polar axis along the vector b, we obtain — a2 + Xr == br whence r = a2/(X b is a curve of the second order.

/d2u \ F4.23. u2 I + u I = — , u = l/r c = const.

\dçc2 / mc2

En the case of the Newtonian force, F = — km/r2 = kmu2, whence

d2u+ u = a (a = k/c2).

4.25. The circumferences whose centres are placed on the straight linepassing through the origin of the radii vectors and collinear with the vec-tor w, whereas the planes of these circumferences are perpendicular tothe indicated straight line.

4.26. The straight lines along which the planes perpendicular to thevector e intersect with those passing through the straight line drawn thro-ugh the pole 0 and collinear with the vector e.

4.27. Introducing Cartesian rectangular coordinates with the axis Ozcollinear with the vector e, we have ae + [cr1 = —yi + xi + ae, andthe given differential equation assumes the following form: x' = —y,

= x, z' = a. We find from the relations x' = —y, y' = xthat x2 + y2 = C1 is the family of circular cylinders whose axes coincide

128

with the straight line passing through the origin of the radii vectors andcollinear with the vector e. Furthermore,

dx — y dy = x

dz a' dz a'whence

xdy — ydx x2 + y2 xdy ydx 7 y2 \= , a = I + — idzdz a x2 \

x

is the family of right helicoids whose axis is the axis of the cylindersmentioned above. The integral curves are helical. rinally, z = at + C3.

Now, to express x, y, z in terms of t is easy from relations obtained.4.28. Semi-circumferences touching the axis Oz (which is collinear with

the vector e) at the origin.4.31. 7r/4 and ir/2.4.32. tan13.

4.36. 2 + 2 1, where a and b are the semi-axes of thefa\ fb

given ellipse.

4.37. xy = ±s/2, where s is the given area.

4.38. y ax2 + where the parabola is given by the equation

= and s is the area of a segment.

4.39. (x — /)2

+ y2 = (i tan where is the given

cos2

angle, and I the semi-perimeter of the triangle.

x2 y24.40. + = 1, where a is the radius of the given circumference.

a2 2a2

4.41. r cos3v, / sin3vl, where / is the given semi-axis sum.9—2018

129

4.42. x = (3 cosv — cos3v), y = (3 sinv — sin3v) is a hypo-

cycloid.

4.43. xy = ± 1 where c is the given area.

4.44. (x — c)2 + = 4a2, where a is the major semi-axis of the ellipse,

c = Va2 — b2.

[r']4.45. ± r ± a—.

r'2

lEr' x r"]Iand in coordinates,

, x'2 + x'2 +x'y" — x"y' x'y" — x"y'

4.47. A cardioid.

4.48. (1) 1; (2)4am(1 + m)

(3)1+2m 2 a

(4)a(y — 2by3 + a2b2)312

— —' 3r'

4 (1 +(6) a cos —; (7) a

3 2

(8) 3a1sint cost1.

4.49. (1) (2) 1/6; (3) -- ir; (4)a

4.50. (1)2 +

2k(k+1)+ço2

(aço"'(1c2 +

(3)1

'11 + ln2a

130

4.51.k= F,, 0

' x yl

(Q - ±Q) + - Q4.52. k L ax ay ax

(P2 +

4.53. (1) s = + y'2dx = —

(2) s = y'2dx = [(4 + 9x)312 — 8];

(4)s=

(5)s = + =

(6)s = + y'2dt = 4a(1 — cosf);

(7)s = y'2dt =

(8)s y'2dt =J 3 20

131

t = — sin2t;(9) s = + y'2d3a

20

x

(10) s=

0

= Vi + e2x +—

2In

+ 1— — — 1);

(Ii) s=

a In sint.

4.54. f(co) + f" (a).

4.55. (1) R2 + — 6as = 0;

(2) (27s + 8)2 [4 + 936R2

2];(27s + 8)

(3)s=I 3 1

E

3—1 + V2R + —in — 1 + 12R].

4 N(4) The parametric natural equations are

,,Ji+x2_1 xs=Vi+x2+in andk=x (1 + X2)3/2'

(5) R = a + s2/a;(6) The parametric natural equations are

Is VTTr?x + —in and k =

2 + 1 (1 +

(7) 1?2 + a2 = a2e — 2S/a;

(8) s2 + 9R2 16a2;

(9) R2 = 2as.

4.56. (1) r = a logarithmic spiral;

a b . (a+b) b a—b \(2) x sin t + sin t )2(a+b b a—b b

b a+b b a—b"— cos b cos

2 a+b a—b b

132

S S

(C s2 C s2(3) r = lcos—ds, % a clothoid;(j 2a2 J 2a2 )

0 0

(4)x = aintan + = line;4 2 cost

(5) r = + 2 sin2t), — + 2 cos2t)};

(6) r = ta(2t + sin2f), a(2 — cos2t)), a cycloid;

(7) r = [a(cost + I sint), a(sint — I cost)), the evolute of thecircumference;

( 71 t(8) r = a cost, a In tan + — — a sint a tractrix.

4 2 )

4.57. p = ml. Assume that rn > 0. Then p = mn. Hence

= rn + rñ = —rrk = —rtk = =ds ds

whence the required relation.

4.58. Rewrite the equation — ro — Rono)2 = R02 in the form— ro) = 0 and consider the function p(s) = (r —

— ro)2 — 2Rono(r — r0). We have (s) = 2(r — ro)r — 2Ronor,= 0, çc"(s) = 2 + 2kn(r — ro) — 2R0nokn, = 0,

(s) = 2thi(r — ro) — 2k2r(u — no) — + 2Ronokr,(So) = — 2R0ko 0; therefore p(s) changes sign when s crosses thro-

ugh and since 97(5) is the index of the point on the circumference,the proposition has been proved.

4.59. See the solution to the previous problem. We have

97'(So) = 97"(So) = c°"(So) = 0,

= 2kn(r — ro) — 2kkr(r — ro) — 4k/cr(r — ro) —

— 2k3n(r — ro) — 2k2 — 2Ron0kn + + 4Rnokkr+

+ 2R0u0k3n,

= — — 2R0k0 + = 2R0k!0 o.

Therefore, the index of a point on the osculating plane does not changesign in crossing through c0.

133

4.60. = k , ds =f(a)

x = cosaf(a)da, y = sinaf(a)da.

4.61. dcx/ds = 1IR,f'(R)dR/ds = hR. ds = Rf'(R)dR,

x = cos[f(R)]Rf'(R)dR,

y = sin[f(R)jRf'(R)dR.

4.62. x= f

cos y = sin [f(s)]ds.

4.64. R = r +rn

— (rr)'r).2kr + rn

4.65. R = r +2

Er r] ([r'r][r'J —[2r r', r"] + [r'rlr'2

4.66. R = r + (n(en) — r(erfl.

If the curve is given by an equation r = r(t), then,

R = r +Er e]

(Er', e] [r'] — (r', e)r').2[r', r"]

If the curve is given by an equation y = f(x), then

[m — , (m — If' (x))(1 + mf' (x))f(x)—2f"(x) 2f"(x)

Y = f(x) + Em if' (x)]2 — (m — If' (x))(1 + mf' (x)) f (x),

2f" (x) 2f" (x)

where 1 = Li, ml.4.67. (x + 1)/2 = (y — 13)/3 = z/6, 2x + 3y + 6z — 37 = 0.

4.68. u = — I at the point A. The tangent is (x — 3)/6 = (y + 7)!(— 17) = (z — 2)17, and the normal plane 6x i7y + 7z — 151 = 0.

4.69. u = 1 at the point A. Since r'(l) = 0 and r"(l) = (2, 2,

121 0, the direction of the tangent is determined by this vector, or (1,1, 6 collinear with it. The tangent is (x — 2)/i = y/1 = (z + 2)/6, andthe normal plane x + y + 6z + 10 = 0.

4.70. 9x — — z + 7 0.

134

4.71. For the osculating plane we find the equation cx — ay = bc —

ad not containing the parameter u. Substituting the expression for x, yin terms of u in this equation, we obtain an identity, whence the curve,in fact, lies in its osculating plane.

4.72. The osculating plane is 6x — 8y — z + 3 = 0, the principal nor-mal x=l—31X, y=l—26X, z=l+22X, and the binormalx=l+6X,y=l — 8X,z=l —X.

4.73. The tangent is

r = cost — Xa sin!, a sin! + aX cost, b(X + t)l,

The normal plane

axsint — aycost — bz + b2t = 0,

The binormal

r = {acost + Xbsin!, asin! — Xbcost, bt + Xal,

The osculating plane

bxsint by cost + az — abt = 0,

The principal normal

r = (a + X)cost, (a + X)sint, bt I.

4.74. The tangent is x = I + 2X, y = —X, z = 1 + 3X,

The normal plane 2x — y + 3z — 5 0,The binormal x = 1 — 3X, y = —3X, z 1 + X,

The osculating plane 3x + 3y — z — 2 0,The principal normal x = 1 — 8X, y = 11X, z = 1 + 9X.

4.75. The tangent is

3F1 t3Fi

3y azX=x+X0F2 oF2

Oy Oz

OF1 OF1

Oz OxY=y+X

OF2 OF2

Oz Ox

135

OF1 OF1

Oxz + X

OF2 OF2

Ox 0,'

The equation of the normal plane is

X-x Y-y Z-zOF, OF1 OF,

Ox 3z —

OF2 OF2 OF2

Ox 0,'

4.76. Having chosen a convenient coordinate system, we shall write theequations of the Viviani curve in the form

orx2 +y2 + = a2,x2 — aX= 0.

To make up the parametric equations, we put

a a a.x — — cost, y = — slnt.

2 2 2

Then

a2 2 a2• 2 2— (1 + cost) + — sin2t + z a, z = a sin —4 4 2

(sign can be omitted, since if is added to t, then x andy are unalteredand z changes sign). Thus,

(a a . . Ir = + cost), —sint, asin—(2 2 2

The equation of the tangent is

r= + cost)— + +(2 2 2 2

that of the normal plane

xsint — ycost — = 0,

136

of the binormal

(ar = (1 + cost) + X sin — (2 + cost).(2 2

sint — X cos (1 + cost), a sin +

of the principal normal

(ar 1— (1 + cost) + X[_cos2 —f-- (1 + cost) — 2(2

a. t t-- —sint(6 + cost), asin— Xsin—1,

2 2 2 2J

and that of the osculating plane

1 tsin — (2 + cost)x — cos — (I + cost)y + 2z

2 2

a L(s + cost) = 0.sin2 2

4.77. s = 5at.

4.78, s =4.79. s = 9a.

4.80. s = 10. The curve has four cusps with ds/dt changing sign atthe points t = 0, 3ir/2.

S s bs4.81. r [acos asin

+ b2' + +

4.82. r cos In

sinin ,—j.4 83 r

{,',J

2 +

2 2

137

4.84. T+ b2

— a sin I, a cost, b),

p = [—cost, —sint, 0),

= 1

{bsint, —bcost, a),Va2 -4- b2

a bK=a2+b2•

[2t,4.85. i =

Vi + + 9,4'

— 2t + 9f3, 3t + 6t3I'

= + 9?)2± (3t + 6,3)2'

— (—3t, I)-

k2(1 + + 9j4)1/2

3

= (1 + 4t2 + ' = + +

cost,

J 2

2 cost, —! sint(6 + cost),2

— sin—J

2

[— cos2(1 + cost) — 2 cost] + sin2 ((6 + cost)2 +

4 2

j3 = sin—(2 + cost), —cos—(1 +2

2

t

2k = 13

+\3' X =

a(13 + 3 cost)a 2(1+cos_12/138

—5)

4.87. (1) k = — Ii + sin2 —, ,c =2 4(3 — cost)

(2)k= ,

(e' + e - 1)2 (e' + e — 1)2

2t —2t(3)k= ,

(1 + 2t2)2 (1 + 212)2

(4)k = —-, = — —i-;3e 3e

(5)k=x=3(t2 + 1)2

3 4(6)k— , x=

25 sint cost 25 sint cost

4.88. y = L

xIx2 a2\z xfx2 a2\2r=

+ + (y'z" —4.90. k

— +

"Z" — Y" Z"

+ + (y'z" —

— 11, y', z'}+ y'2 +

=[—z'z" —y'y", y" —z'(y'z" —y'z"),y'(y'z" —z'y")+z"

pV(z'z"—y'y")2+ — —y'z")]2+[z" +y'Vz" —z'y")12

— — y"z', —z",y"J

139

4.91. The two families of curves are:

(a) y2 + = const, xy az;

and(b) x2 + = const, xy = az.

4.92. Let the equation of the sphere be of the form:

r tacosvcosu, acosvsinu, asinvj.

Then the equation of the loxodrome is

U tanO +

where 0 is the given angle.

cos0( —sinv cosu — sinu tan0, —sinv sinu + cosu tanO,cosv);

cosO ( CO5U 2— cosu cosv + tanv sinu tanO — tan 0,tan2o ( CO5V

I +

COSV sjnu 2— — tan 0 — tanv tan0cosu, —sinvcosV COSV

(. tan0—cosu,

( cosv

I tan2oIl +cos2v

cos0 I tan20 tanO

a cos2v a(cos2v + tan20)

u cotO

JI+k24.93.v=Ce

r r r, and in the special case, — kjx'.

6 [r' x r"]( 6

4.98. The necessary and sufficient condition is e' 0, 'ee' = 0,

while the equation of the envelope

Q'e'e.

e

140

4.99. When a =4.101. r = r, r = kv, 1 = —k2r + kv * kx/3.

4.103. ev = C, e(—kr + = 0, ei- — efl = 0,

k2 +evk=k k

Differentiating once again, we obtain the required relation. Note thatin view of the above relations, we may assume

x k2+x2 k2+x2r+v+k

\kJ \kIf the relation is held

k2 + +x=0,

then this vector is constant. This constant vector e forms with the vectorp an angle whose cosine equals = const.

4.104. er = 0, key = 0; hence either k = 0 (straight line) orev = 0; if ev = 0, then e(—kr + x/3) = 0, whence x = 0 (plane line).

4.105. = 0; xev = 0; hence x = 0, since, if the inequality x 0were held, we would have ep = 0, e(—ki + x13) = 0,ker = 0, er 0.

Therefore, k = 0, and the line is straight.4.106. fi = — xv = 0, x = 0.

4.108. (a) Let a be a unit vector with a fixed direction. Then

ar = cosv (v = const). (1)

We have (an = = 0. Therefore, kay = 0. Excluding the case wherek = 0 (i.e., of straight lines), we obtain

av=0. (2)Therefore, the normals are perpendicular to the fixed direction.

Conversely, if v is perpendicular to the fixed direction, then equality(1) holds.

(b) Let x 0. It follows from (2), with the use of the third Frenetformula, that

aj3 = 0,

whence = const.

141

Conversely, differentiating this formula, we obtain (2).(c) Differentiating (2), we obtain

kar =whence

k = = const.x ar

Conversely, it follows from the first and third Frenet formulae that

+ =k x

whence

+ = 0, ----r + f3 = const = a.k k

Multiplying scalarly by v, we obtain av = 0. Therefore, condition (2) hasbeen fulfilled.

4.109. Thke into account that

= k5 (—f-- ,

and use the previous problem.

5Surfaces

5.1. r = + ye.5.2. r =5.3. r = + VQ'.5.4. r Q(s) + + j3(s)sinct,.5.5. r = (W(v)cosu, p(v)sinu, i/i(v)).In the special case, r = (f(v)cosu, f(v)sinu, vi.5.6. r = t(a + bcosv)cosu, (a + bcosv)sinu, b sinv).5.7. Let the moving straight line coincide with the axis Ox at the initial

moment, and the second line in question with the axis Oz. Then theequation of the right helicoid is of the form

r = vsinu, kuj,

where v is the distance of a point of the helicoid from its axis (i.e., theaxis Oz), and u the longitude of the point.

142

5.8. If the equation of the helix is given in the form

= (a cosu, a sinu, but,

then the vector n = { — cosu, — sinu, 0 is the principal normal vector.Hence, the required equation

r = — Xn = ((a + X) cosu, (a + X)sinu, bu) =

= (vcosu, vsinu, bu)

is that of a right helicoid.5.9. r = a(s) + Xtn(s)cosço(s) + where cc(s) is an

arbitrary function of the variable s.5.10. The normal plane to the circumference = (a cosu, a sinu, 0)

is determined by the vectors n = (cosu, sinu, 0) and (0, 0, ii. Thevector lying in the normal plane and inclined at the angle u to the vectorn is a = n cosu + k sinu. Therefore, the equation of the required surface

r = {acosu, asinu, 0) + va

= Ia cosu + v cos2u, a sinu + v sinu cosu, v sinu).

Eliminating the parameters u and v, we find

cos2ux = acosu + z cotu(asinu + zcosu),sin u

y = a sinu + z cosu, cotu, = (a + z cotu)2,y sinu

y2 (i + 4-) = (a +,

or y2(x2 + y2) = (ay + xz)2,

a surface of the fourth order.

5.l1.R = (r(u) + e(v))

5.12. r = a cosu cosh —f--, a sinu cosh where u is thea a)

longitude and v the oriented distance from a point of the surface to thegorge section of the catenoid.

5.13. r = ía In tan —f- + — a sint, a cost cosu, a cost sinu\4 2/

143

5.14. The equation of the given straight line is r1 = u, 0, h andthat of the ellipse

r2 = {acosv, bsinv,

Furthermore,

r1—r2=(u—acosv,—bsinv,h),u--acosv=0,r1 — r2 = [0, —bsinv, hi,

and the required equation of the conoid is

r = {acosv, bsinv, + X[0, —bsinv, h

= [a cosv, b(I X)sinv, Xh).

Eliminating the parameters X and v, we obtain the implicit equation ofthe conoid

\ \h / b2

I (5.15.ri = = = —v,u —

( 2p) ( 2p

u————--=0, u=—,

r = v, v, 0) = laX, v(l X),

or

a2y2 = 2pz(x a)2.

5.16. The parametric equations of the given circumferences are

r1 = [a(I + cosu), 0, a sinu r2 = (0, a(l + cosv), a sinv J.

We find

r1 — r2 = (a(1 + cosu), —a(l + cosv), a(sinu — sinv)j.

We have sinu — sinv = 0, whence

(1) v u + 2kw,

(2) v = ir u + 2kw.

In the first case, we have

r2 = (a(1 + cosu), —a(I + cosu), 0) II (1, —1, 0),

144

and thus obtain the elliptic cylinder

= {a(i + cosu), 0, a sinul + X(!, —1, =

= La (1 + cosu) + X, —X, asinul.

in the second case,

r1 — r2 {a(i + cosu), —a(l cosu), 01,

and the second surface making up the given cylindroid is determined bythe equation

R = ta(i + cost,), 0, a sinul ++ A{a(l + cosu), —a(i — cosu), 01 == La(1 + X)(1 + cosu), —aX(i — cosu), a sinu).

Eliminating the parameters X and u, we obtain

+ — 2a(x + y)] + 4a2xy = 0.

(u2 I v25.17. rj = u, 0, v

) ( 2p

(u2 + v2r1 —r2= ,u, —v( 2p

The condition for this vector to be collinear with the plane y — z = 0is given by the relations

u+v=0, v=—u,ri—r2=1u2/p,u,u),and the required equation is the following:

r = + v u} = + 2v),u(l +

Eliminating the parameters u and v, we obtain y2 — z2 = 2px, ahyperbolic paraboloid.

5.18. The equation of the axis Oz is of the form: r1 = [0, 0, uJ andthe equation of the given curve

I a3r2 = b cosv, b sinv,

cosv sinv

hence

rz — r1 = cosv, b sinv,a

— u( b2 cosv sinv

10- 2018145

u = a, — r1 fbcosv, bsinv, 0),

b cosvsinv

r = 0, 0,a

+ X(b cosv, b sinv,b2cosvsinv)

= Xb cos v, Xb sin v,a3

b2cosvsinvEliminating the parameters X and v, we obtain

b2xyz = a3(x2 + y2).

5.19. (a + ub — = o, u =n(Q — a)

nb

— a)a + ub — = b — a),

nb

R = + x["(Q— a)b — + a].

5.20. Take the equations of the given ellipses in the form:

= a, b cos u, c sin u), r2 a, c cos v, b sin v),

ri — r2 = L2a, b cosu — c cosv, c sinu b sinv),

c sinu — b sinv = 0,

= = ± —

b b

— r2 = 12a, b cosu ± —s-- sin2u, 0( b

The required equation is

R = La, b cosu, c sinu) + v 2a, b cosu ± 0b

or

R = + 2av,bcosu) + vlbcosu ± — c2sin2u,csinu( b

5.21. The equation of the axis Oz is p 0, and we find

u3—v=0, v=u3,p = 0, u3).

146

The required equation is

r = p + — p) = 10, 0, U3) + v{u, U2, 0) = {uv, u2v, u3J.

5.22. r = (by, av cosu, (b + a cosu)(l — v) + a sinu).5.23. The equations of the given straight lines are = (u, 1, 1 and

p (1, v, 0). The equation of the straight line passing through twoarbitrary points of these straight lines is the following:

r = (1, v, 0) + X(u, 1, 1).

For the point where this straight line meets the plane xOz, we have:

v+X=0, X= —v,r= (l,v,0) — v(u,l,l) = [1 —uv,0, —v(.

This point must lie on the circumference

x = y = 0, z =Therefore,

1 — uv = V =

whence

u= =—tan-—-.2

It remains to make up the equations of the straight line passing through

the points (_tan —f--, i, i) and (1, —sinw, 0). Finally, we obtain:

r = (1, —sinw, o( + — 1,

= {i + + +

5.24. r = (a(cosv — u sinv), a(sinv + u cosv), b(u + vfl.5.25. c2(x2 + = a2(x2 — y2)(z + c)2.5.26. We will assume that rectangular Cartesian coordinates (E, are

given on the plane ir. Then the equation of the curve = Q(u) can bewritten in coordinate form thus: E = 'j = In addition, weassume that the straight line AB is the axis z in space and that the axis

of the moving plane ir slips along it. For the appropriate choice of theaxes x, y and positive directions on the coordinate axes, we have:

R(u, v) = (E(u)cosv, + av).

5.27. R(u, v) = r(u) + ap(u)cosv + afl(u)sinv, where v and i3 arethe principal normal and binormal unit vectors to the curve r = r(u),and the points (u, v) and (u, v + 2ir) regarded as identical.

10*147

5.28. Thke the point of intersection of the normals to be the originof the radii vectors. Then

r r 0,

whence r2 = const. Therefore, the given surface is either a sphere or apart of a sphere.

5.29. The volume of the tetrahedron is 9a3/2.5.30. The tangent plane is determined by the equation

X Z 2+ +2 2

=a.usinv ucosv

The required sum equals a6.5.31. The equations of the line of intersection in curvilinear coordinates

are u = u1cos(v + v1)/cos2vi (except for the generator v = vj), whereu1, v1 are the coordinates of the point of contact. The parametricequations of the same line in Cartesian coordinates are

cos(v + v1) cos(v + Vt)X = Ut COSV, y = U1 5IflV,

cos2v1 cos2vi

z = a sin2v.

The equation of its projection on the plane xy is

x2 += Uj

(x cosvi — y sinv1).

Since the projection is a circumference, the line itself (being a plane line)is an ellipse.

5.32. The equation of the tangent plane is

Z - xf = (i !_f')(X - x) + (Y or

z = (f - + Yf',

and all the tangent planes pass through the same point, viz., the origin.Besides, it is also clear from the fact that the given equation determinesa cone with vertex at the origin (z being a homogeneous function in xand y).

5.33. The tangent plane has the equation

kxsinu — kycosu + vz — kuv = 0,

and the normal

r = cosu + Xk 5inU, V SiflU — Xk cosu, ku + XvJ.

148

x Y Z5.34. + — ÷ = 3.

x y z

5.35. Let the equation of the curve C be = Q(s). The equation ofthe surface is as follows: r = + Xr, where r is the unit vector of thetangent to the curve C. We find that

8r Far an= r + Xkv, = r, I—,---—I = Xkfl;as ax L3X ôsj

when s = const (i.e., at the points of the same tangent), this vector hasthe same direction (for then i3 = const), from which it follows also thatthe tangent plane to such a surface at all points of the curve C is theosculating plane to this curve.

5.36. The equation of the surface is

8rr=Q+Xv, —=,-+X(—kr-i-xfl), —=v,as ax

= (I — Xk)/3 —

Las axjThe equation of the tangent plane is

(R — — Xv)(f3 — Xk13 — Xx'r) = 0,

or

Xxr) — Xxr) + X2x = 0,

and that of the normal

R=

5.37. r = + Xfl, = r — Xxv, = 13,as ax

Far an= —v — Xxr.

Las 3XJ

The equation of the tangent plane is

(R— —X/3)(z'+Xxr)=O,

(R — Q)(v + Xxr) = 0.

The equation of the normal is

+ X13 + E(v + Xxi).

149

5.39. If a is the direction vector of the given straight line, and the originof the radii vectors is taken on it, then the vectors r, a,

and lie in the same plane, whileLau 3v]

I Far är]]r Ia, —i I = 0.

L Lau avJJHence,

(r . (a . - (r . (a . = o.\ auJ \ avJ \ ôvJ \ 3uJ

But this equality can be written as the vanishing of the functional deter-minant, viz.,

8r2 a ar2 a——(aS r) — ——(a • r) = 0,öu av äv

from which it follows that the entities r2 and a r are in the functionaldependence

r2 = f(a r).

Choosing the axis Oz along the vector a, we obtain x2 + y2 = f(z), asurface of revolution.

5.42. 4z2

:2

= I, the edge of regression being imaginary.

5.43. x2 + + = a2.a2 + b2

5.44. The envelope has the equation (x2 + + — x)2 =+ y2, and the edge of regression degenerates into the point (0, 0, 0).

5.45. Taking the equation of the parabolas in the form y2 = 2px, z = 0

and y2 = 2qz, x = 0, we obtain the equation of the envelope in theform y2 = 2px + 2qz, i.e., a parabolic cylinder with + q2 as aparameter.

5.46. (R — = a2. Differentiating with respect to s, we obtain(R — = 0. Hence R — = Xb + Since (R —

= a2, X + = a2, and we can putX = a cosço, = a sinw

so that the equation of the envelope isR = + a(b + v

150

5.47. The edge of regression is a curve whose points are obtained byintersecting the curvature axes of the curve Q = a(s) with the corre-sponding spheres of the given family.

5.48. The equation of the family is the following:

(x — b cosco)2 + y b sin2w + :2 — a2 = 0,

and the envelope is a torus whose equation may be obtained by2

eliminating from the equations F = 0 and = 0; (x2 + + z +

+ b2 — a2)2 — 4b2(x2 + y2) = 0 is a surface of the fourth order. Theedge of regression when a > b is reduced to the two points

(0, 0, ±

or one point (0, 0, 0) if a = b.

5.49. The equation of the family is as follows

+ + :2 — 2u3x 2u2y — 2uz 0.

We shall find the envelope by eliminating u from this equation and from

3u2x + 2uy + z = 0.Thus, the equation of the envelope is:

3x[9x(x2 + + z2) — 2zy]2 + 2y[9x(x2 + + :2) —

— (l2xz — 4y2) + z(l2xz — 4y2)2 = 0.

The edge of regression is found by adjoining another equation to thetwo indicated above, viz., 6ux + 2y = 0, or 3ux + y = 0.

Hence, u = —y/3x, and the equation of the edge of regression is thefollowing

27x2(x2+y2+z2)—4y3+l8xyz=0,y2—3XZ=0.The edge of regression can also be obtained in parametric form:

( 2u3 —6u4 6u5r

= L 9u4 + 9u2 + 1 ' + 9u2 + I ' + 9u2 + 1

5.50. x2"3 + + z213 12/3.

5.51. x2"3 + p2/3 + =5.52. xyz = 2/9 a3.5.53. The envelope is y2 = 4xz, and the edge of regression is degener-

ated into a point, viz., the origin.5.54. The characteristic is x = a(cosa + a sina) — z sincx, y =

= a(sina — a coscr) + z cosa, and the edge of regression is a helix

x = acosa, y = asina, z = aa.

151

5.55. Let = Q(s) be the equation of the given curve. The equationof the family of the osculating planes is

(r 0.

Differentiating with respect to s, we obtain (r = 0. The charac-teristic is the tangent

(r—Q)b=0, (r—Q)v=0.The envelope is r = + Xr, i.e., the surface formed by the tangentsto the given curve. Differentiating the relation (r — 0 once more,we obtain (r = 0. Hence, taking into account the relations

(r — = 0, (r — = 0

we have

r =

i.e., the edge of regression is the given curve.5.56. The characteristics are the curvature axes of the given curve, and

the envelope is the surface formed by the curvature axes. The edge ofregression is the curve described by the centres of the osculating spheresof the given curve.

5.57. rn' + D' = 0, r = an + fin' + X[nn'],

rn' D'a = rn = —D, fi = —i- = —n' n

The equation of the envelope is

D' n'r = —Dn— 2

+ X[nn']

(with the parameters u and X). The characteristics are straight linesu = const. The edge of regression is found by solving the equations

rn+D=0, rn'+D'=O, rn"+D"=Ofor r, viz.,

r= (rn) [n'n"] + (rn') + (rn") [nn']

= D[n'n"] + D'[n"nJ + D"[nn']nfl 'n"

152

5.58. The envelope of the family of planes tangent to both parabolas.The equation of the family is

Ia2 4a3\ 4a3Xa — 2aY — I— + —lZ + = 0,\2b ba) a

where a is the parameter of the family.5.59. The vector of normal is n = (u + v)(i sinv — j cosv + k) is

parallel to the vector i sinv — cosv + k, which is unaltered if the par-ameter v remains constant. Hence, the lines v = const are rectilineargenerators of the surface, and u + v = 0 is the edge of regression, sincethe modulus of the vector n vanishes at each of the points of the surface.

5.61. The equation of the curve is u = const and the edge of regression

x = 2(a — b)u cos2v, y = 2(a — b)u sin3v,

z = 2u2[(a — 2b)cos2v + (b — 2a)sin2v].

5.62. x = 3t, y = — 3t2/b, z = — t3/ab.5.63. The required developable surface envelops the family of planes

Xx + + Z f-f-- - x2 = a2,

where x is the parameter of the family.5.64. (1) r2(cos2v du2 + dv2);

(2) (a2 sin2u + b2 cos2u)cos2v du2 +

+ 2(a2 — b2)sinu cosu sinv cosv du dv

+ cos2u + j,2 sin2u)sin2v + c2 cos2vjdv2;

/(3) — ( v + —) (a2 sin2u + b2 cos2u)du2 +4\ v/

+ (b2 — a2)sinu cosu (v dudv

+ [-1- (i—

(a2 cos2u + b2 sin2 U) + (i +1)2]

(4) 1)2 + 4b2v2 + c2(v2 + +(u+v) -

+ 2(a2(u2 l)(v2 — 1) — 4b2uv + c2(u2 + 1)(v2 + +

+ (a2(u2 — 1)2 + 4b2u2 + c2(u2 + 1)2 Jdv2];

153

/ \2(5) —'-- ( v — —') (a2 sin2u + b2 cos2u)du2 +4\ VI

+ — a2)sinu cosu—

dudv +

+ + cos2u + b2 sin2u) + —

(6) (p sin2u + q cos2u)v2du2 + 2(q — p)sinu cosu dudv

+ (p cos2u + q sin2u + v2)dv2;

(7) (p + q + 4v2)du2 + 2(p — q + 4uv)dudv + (p + q + 4u2)dv2;

(8) v2(a2 sin2u + b2 cos2u)du2 + 2(b2 a2)sinu cosu dudv

+ (a2 cos2u + b2 sin2u + c2)dv2;

(9) (a2 sin2u + b2 cos2u)du2 + dv2;

(10) (i—

2 +(i

+

2]du2 + dv2.

5.65. (1) ds2 + 2redsdX + dX2;

(2) v2ds2 + 2vrQdsdV + Q2dv2;

(3) + X ds2 + 2erdsdX + c1X2

(4) ((1 kcosço)2 + x2lds2 + +

(5) + +

(6) (a + b cosv)2du2 + b2dv2;

(7) (v2 + k2)du2 + dv2;

(8) 1(1 — Xk)2 + x2X2)ds2 + dX2;

(9) (1 + X2x2)ds2 +

5.66. (1) The curves u ± 1/2 av2, V I intersect at the points

A(u—0,v=0); B(u=l/2a,v=1); C(u=—1/2a,v=l);the differentials of curvilinear coordinates on these curves being relatedby the formulae:

du = avdV for the curve AB with the equation u = av2;

154

du —avdv for the curve AC with the equation u =

dv 0 for the curve BC with the equation v 1.

Substituting these values in the first fundamental form, we obtain

ds2 = a2(1/4 V4 + v2 + 1)dv2, ds (1/2 V2 + 1)dv for the curve

ds2 a2(1/4 v4 + v2 + l)dv2, ds (1/2 v2 + 1)dv for the curveAc;

ds2 = du2, ds = du for the curve BC.

It remains to evaluate the integral between the limits determined by thecoordinates of the points A, B, C, viz.,

AB = AC = a v2 + 1)dv = 7a/6, CD=

du = a.

u = — 1/2a

10Thus, the perimeter of the triangle equals — a.

(2) cosA = 1, cosB = 2/3, cosC = 2/3, i.e.,

A 0, B = C = cos' 2/3.

(3)S + in(l +

5.67. cosO =I +

r5.68. v =

±+ 1 + In

+ 1]+ const.

5.69. v = tanO in [U + Vu2 — a2J + const.

5.70. (1) 1/4 + sinh2vo);(2) sinhvo,(3) in2, in!4, in/4.5.71. (1) ds2 = UI — + [ax(u)12)du2 +

+ +(2) = —

(3) 2iraluz — uji;

(4) 4rr2ab;

(5)

155

5.72. Consider the family of the surfaces

R(u, v, t) = t)cos t)sinv/t, I)),

where

t) = tQ(u), t) = t2Q '(u)2du, I t >0.

5.73. R(u, = Vi + ln(u + '.11 +or

R(z, = zi.

This is a catenoid, a surface of revolution of the catenary curvex = coshz.

5.75. Hint: The first equation determining the correspondence betweenpoints is as follows:

r2 = + a2.

5.78. For the sphere ds2 = du2 + R2 cos2(u/R)dv2,

for torus ds2 = du2 + (a + b cos dv2,bJfor catenoid ds2 = du2 + (a2 + u2)dv2,for pseudosphere ds2 = du2 +

Hint: u is the natural parameter of the meridian.5.79. = dIP +

Putting u = g v = we obtain

ds2 = (du2 + dv2).

5.80. (a) If a and b are the sides containing the right angle of a right-angled spherical triangle, c its hypotenuse, and R the radius of the sphere,then the following relation is held

cosc/R = cosa/R cosb/R.

(b) Let A, B be the angles opposite to the sides a and b. Then

S = R2(A + B — ir/2).

5.81. S = where R is the radius of the sphere.5.83. Hint: Take the equation of a conic surface in the form r = ve(u),

where = I, and compare its first fundamental form with the quad-ratic form of the plane in polar coordinates.

156

5.86. (1) R(du2 + cos2udv2);

ac(2)

2 2(du2 + cos2udv2);

C COS U

—ac(du2 — cosh2udv2);(3)

+ c2 cosh2u

ac(4) — (du2 + sinh2udv2);

V a2 cosh2u + C2 sinh2u

(5)2

(du2 + u2dv2);Vi + 4u2

(6) Rdv2;

(7)ku

dv2;Vi + k2

(8) bdu2 + cosu(a + b cosu)dv2;

(9) — — (du2 — a2dv2);a

(10) —a cotu(du2 — sin2u dv2).

5.87. — 2adudv/Vu2 + a2.

a5.88. — du2 + adv2.

u2 + a2

5.89.2a3

dx2 + dxdy + —f- dY2]Vx4y4 + a6(x2 + y2) x y

(Q'x" Q"x')du2 +5.90. (1)

(x')2 +

(2) K— x'(Q'x" — Q"X')— Q[(x')2 +

K > 0 if the convexity of the meridian is directed from the axis ofrotation; K < 0 if the convexity of the meridian is directed towards theaxis of rotation; K = 0 if the meridian has a point of inflexion or ifit is orthogonal to the axis of rotation (when 0).

(3) K = — 1, when x 0; K is undetermined, when x = 0;

— + x'[(x')2 + (e')2](4) H =

2Q[(x')2 +

157

(5) Q(x) = Xo),a

where xo and a > 0 are arbitrary constants (catenoid).5.91. (a) K = 0;

'C

(b) H = —

2kv

ouu5.92. K = —

5.93.K —1.

5.94.dXXF

1K=- I

(âfl2 + (öyfl2 +

0

rt — S25.95. K =

(1 + p2 + q2)2

H = (1 + p2)t + (1 + q2)r — 2pqs

2(1 + p2 + q2)312

where

p = ôxz, q = äyZ, r = = 8xyZ, t = t9yyZ.

5.96.X2 + + a2

a

1 1 15.97.—=0, —=R1 R2 (u +

1 2(1+u2)5.98.K= H= —

(2u2 + 1)2' (2u2 + 1)3/2

45.99.K= — 11=0.9(u2 + v2 +

5.102. r = Q(s) + ufi(s),

k + kx2u2 — udsK= ———______

(1 + u2x2)2 ' (1 +

158

5.103.K= — 2'1(1 — ku)2 + u2x2]

— + uxH=21(1 — ku)2 + u2x2]3"2

5.104. r* = r + am, = + ama, = + amp,

E* = = + + E — 2aL + a2(2HL EK).

Similarly,

F* = (1 — a2K)F + 2a(aH — l)M,

G* = (1 — a2K)G + 2a(aH 1)N,

L* = aEK + (1 — 2aH)L,

M*=M_a(2MH_FK),N* = N - a(2NH — OK).

5.105. K* =1 — 2aH + a2K

5.106. H*H — aK

1 — 2aH + a2K

5.107. r* r + —m.K

5.108. K* 4H2 = const.

5.109. H* = —

2

du2 dv25.111. — = 0.a2+b2+u2 a2+b2+v2

5.112. v = ± ln[u + '.Ju2 + a2] = const.

5.114. u = const, v = const.

k5.115. (1) K = —

_______________

a(1 — ak cosçc')

(2) H = —(1—ak

(3) u = const, = const.

5.116. u = const, = —

159

5.117. The lines of curvature are u ± v = const,

H=O.(3u2 + 3v2 +

5.122. (1) The rectilinear generators are y/x = const;

(2) — — —i- = const.x2

5.123. (1) The rectilinear generators are y = const;

(2) x2y = const.

5.124. The equation of the asymptotic lines is

(1) u const;

(2) u5 = v2(C —

5.125. v = const,

cos2u cosv= const.

(1 + cosu)2

5.126. = ±n on the asymptotic line. Therefore, x2 = We

select a coordinate system (u, v) in a special way so that the followingconditions may be fulfilled at the point u = u0, v = v0 under consider-ation: (1) the lines u = const and v = const have principal directions;(2) E(uo, vo) = G(uo, vo) = 1. Then F(uo, VO) = 0 due to theorthogonality of the principal directions and — = —

by the Rodrigues theorem. Therefore,

1 dn'\2 / du dv\2 += (flu— + flv—I =\dsj \ ds dsJ ds2

Let be the angle between the line v = v0 and asymptotic direction.Then with respect to this direction du/ds = cos dv/ds = sin çô becauseE = G = 1, F = 0. On the other hand, we find from the Euler formulathat

k1 cos2ç, + k2 = 0.

Finally, we have= (d)2

= — = —K.

5.127. + u2x2).

160

5.128.(1 — ku)2 + u2x2

u + a2 + b25.129. = const;

v + Vv2 + a2 + b2

[u + + a2 + b2] [v + ,,/v2 + a2 + b2J = const.

5.143. Assume that the rectilinear generators are parallel to the axisOz. Then the equation of the surface can be written in the form

r = f(u)ei + + ve3,

where u is the natural parameter of the directing line. We will seek theequation of the geodesic in the form

v = v(u). (*)

Then

N = [rn, ço'ei — f'e2,

dr = (f'ei + cc"ez + v'e3)du,

d2r = (f"ej + + v"e3)du2,

and the equation for determining the geodesic lines is

—f, 0

f' v' = 0,

jflF v

or

(w'2 + f'2)v" + f'f")v' = 0.

Since w'2 + = 1, we have

+= ÷ + = 0

Thus, v" = 0 and v = CIU + C2. The vector equation of the familyof geodesics is

r = f(u)ei + + (c1u + c2)e3,

whence

C'cosO = OZ) =

2+ Ci

11—2018161

Therefore, the geodesics found are generalized helices. Besides, therectilinear generators are also geodesics which were not included in thegeneral solution, since their equation cannot be represented in the form(*)

5.144. Let the equation of a developable surface be in the form

r = Q(s) + uT(s),

and let 0 be the angle at which a geodesic intersects a rectilinear generators = const. If k is the curvature of the curve R a(s), then the differentialequation of the geodesic is

— ukcot0 + 1 0.ds

It is a linear differential equation of thequadratures.

5.145. The equations of the geodesics are

first order integrable by

5.147. Consider the equation of a cone in the form r = up(v) andassume that I pI = I, p I = I. Then the equations of the geodesics areof the form

r = p(v).sin(C v)

5.150. Great circumferences of the sphere.5.154. By the Meusnier theorem, the curvature radius R of the curve -y

at a certain point equals the projection of the geodesic curvature radius Rg.(= i/kg) onto the osculating plane of the curve y, i.e, R = I Rg cos 01;the vector e = [t, m] is the unit vector lying in the tangent plane to thesurface, and orthogonal to -y; n is the unit vector of the principal normalto the curve -y;

IcosOl = leni, kg= klcos0l = klenl = lell = lItmI = Imirl.

5.155. kg = u/(u2 + a2).

162

Ccosv Csinv

Cdu5.146. v = C1 ±

+ h2)(u2+h2

5.158. Considering v as a function of u along a geodesic, we obtain thefollowing differential equation for the geodesics:

du2 du J du du du du'Or

+ = (du2 + dv2)(d1'du2 — d'pdv2),

whence

d =\. + dv2 J

Integrating this relation, we obtain the required equations.5.159. +5.160. ii — a2a.5.161. a* = ir/2.

V2 U2

5.162.G* dv du.

UI

5.164. s(p) = 2ir sinhp; kg(P) = cothp — I as p — +00;11(p) = 271 coshp; 11(p)— asp— +oo. On the Euclidean plane,s(p) 2lrP;kg(p) = 1/p — Oasp — +oo;Il(p) = 271.

5.165. First, we establish that the metrics defined onP1 and P2 have thesame curvature K = — I. Then we introduce semi-geodesic coordinates

on the plane P2 so that:(1) the geodesics are the lines = const and = const;(2) is the natural parameter of the line = 0;(3) is the natural parameter of the line = 0.Then ds2 = + 77)th72, with = B(E,

B(0, 77) = 1, Bt(0, = 0. It follows from these equalities that= cosh

5.166. dn2 = 2H(n, d2R) — Kds2.5.167. Apply the Frenet formula ñ = — kt + xb. For the geodesic line

(m = n), we have

= (dm= k2 + x2.\dsj \ds/

On the other hand,

= II — K 1,

\ ds/

163

where I and II are the first and second fundamental forms of the surface.1dm \ 2

When H = 0, we have ( ) = —K, and therefore, k2 + = —K.\ ds/5.168. If the equations of a surface of revolution are written in the form

x = cos v, y = sin v, z = u,

then the vanishing of the mean curvature implies that

=0.Putting p = and considering as a new variable, we obtain

1 + 2—

dp= 0,

= 1

d(ln (1 +dço 2

whence

= i + p2.

With respect to the original variables,

= du,

(1 + u2)f'(u) = a, f'(u) = a/(1 + u2).

Integrating this equation, we get

f(u) + b = z + b = u.

Therefore,

u = tan (z + b)/a, y/x + tan (z + b)/a,

which is an implicit equation of the right helicoid

z=mi—b.5.170. Let the coordinate lines coincide on the surface S with the lines

of curvature. Then

r = (I — r = (1 —

Therefore, the coefficients of the first fundamental forms of the surfacesS and S * are related by the formulae

E* = (I — ak1)2E, G* = (1— ak2)2G, F* = F= 0.

Hence,

dcr* = (1 — ak1)(l — ak2)dcr,

anddo_da*

. 7k1+k2 I \lim = lim I + ak1k2 I = = H.a-Q 2ada u-.O\ 2 2 / 2

164

5.171. Let S be a minimal surface, and a surface parallel to it, thedistance between them along the normal being equal to a. As it followsfrom the previous problem, the corresponding elements of the areas ofthe surfaces S * and S are related by the formula

da* = (I + a2K)da,

where K is the Gaussian curvaturc of the surface S. Therefore,

= do + a2 Kdo.

Since K 0 on the minimal surface,

do.

5.174. Take the axis of the cylinder to be the axis Oz and place the axisOx in the sectional plane. Then the equations of the cylinder assume theform

x=acosl, y=asint, z=u,and the equation of the sectional plane is

z = Ày.

Cut the cylinder along a generator intersecting the axis Ox, and place it onthe plane xOz. Since after the superposition, the part of the abscissa isplayed by the length of an arc of the perpendicular section of the cylinders = at, the equation of the required line is

z = aAa

i.e., a sine curve.5.175. The general equation of the motion of a point across the surface

is of the form

d2rm- =F+Rm—pIRIt,

dt2

where F is an external force, R the normal reaction of the surface, ji thecoefficient of friction, t the unit tangent vector to the trajectory, and mthe unit vector of the normal to the surface. Since

d2r d2s /ds'\2dt-••=-1+I- Idt2 dt2 \dt/ ds

165

when F = 0, the equation assumes the form

fd2s fds\2dt'\m(—2t+l- 1- J=Rm—piRIt.\dt \d(/ ds/

Multiplying it scalarly by [t, m}, we obtain

dt dr d2rtm- = m- =0,

ds ds ds2

i.e., the point moves along a geodesic.5.177. Take a semi-geodesic coordinate system on the surface. Then

ds2 = du2 + G(u, v)dv2.

On the line u = 0, we have TG I = = 1. Besides, we obtain from theequation of geodesic lines that

=0.u=O

In the semi-geodesic coordinate system,

IfK = 0, then

0

and the solution of this equation satisfying the initial conditions indicatedabove is './G = 1. Therefore, for all surfaces of zero Gaussian curvaturethe first fundamental form can be reduced to the form

ds2 = du2 + dv2;

hence, all of them are locally isometric to each other.

JfKa2

(a = const), then

'[G=cos", ds2=du2+cos2Udv2.a a

If K =—

(a = const), then

ds2 = du2 + cash2U

dv2.a

166

5.180. The surface S can be obtained by bending the hemisphere so thatthe two halves of its boundary circumference may overlap each other, andthen glue the surface along these semi-circumferences, from which itfollows that the geodesics on the surface S not passing through its singularpoints (ends of meridians) become closed after traversing around the sur-face twice (i.e., after increasing by 47r).

5.182. It follows from the formula

H Kdii + kgdS =

when kg = 0 that

H Kda =

And this equality cannot be valid if K 0 at all points of the surface.

6Manifolds

6.1. As the atlas of charts, the sets determined by the inequality= {Xk > = Xk < should be taken. As the coordinate

functions, all Cartesian coordinates exceptxk should be taken in the chart

6.2. Notice that T2 is homeomorphic to the Cartesian product51 x S1, and reduce the problem to the previous when n = I.

6.3. Any neighbourhood U of the origin 0 can be split into at least 4connected components, while discarding the point 0, which is impossibleon a manifold.

6.4. The sphere 5" is a compact space.6.5. (a) Yes. (b) No.6.6. The space RP" is the set of collections (x0:x1: . . . :x,,), where

x, eR, * 0, with the equivalence relation (x0:x1: . . . :x,,)(\x0: Xxi: . . Xx,1). Introduce a real analytic structure on RP". To

this end, cover RP" by a set of n + 1 charts. Consider the collections(x0:x1: . . . :x,,) such that x, * 0. The set of such collections can benaturally considered identical with R", viz.,

(x0:x1: . .

x, x, x,

167

It is easy to see that the definition of this correspondence is correct. It re-mains to consider the functions of transition from the /-th chart to thej-th. Let be the k-th coordinate of the collection (X0: X1: . . inthe i-th chart, and the 1-th coordinate in the j-th, respectively (let, forsimplicity, i < j). Then

(j) (I)(i) — X1 (j) — 1

—— x/1

X1

—' — 4f

Thus, the transition functions are not only smooth, but also real andanalytic.

6.7. See Problem 6.6.6.8. The atlas consists of one chart with coordinate functions

(x1

6.9. Represent the elements of the group SO (2) as rotations of the planethrough a certain angle about the origin. The group 0(2) is homeomor-phic to the union of two replicas of S'.

6.10. Represent the elements of the group SO(3) as rotations of thespace about a certain axis through a certain angle.

6.11. The groups GL(n, R), GL(n, C) are open sets of the space of allmatrices.

6.12. A cylinder.6.17. Use the rule for differentiating a function of a function.6.19. 1.6.20. Apply the implicit function theorem.6.22.

U1 = Rx,

6.23. Use Problem 6.22.6.24.y = x3.6.25. Use the functiony = e

6.26. Use the function from the previous problem.6.27. Let U21 be an atlas of charts sufficiently fine for the following

conditions to be fulfilled: if A fl 0, then C U. Let be apartition of unity, subordinate to the covering

6.28. Use the average operation

= f(x)dx.Ix-• yl <

Put f = where the summation is over all indices a for which

A fl * 0.6.30. The composite of smooth mappings is a smooth mapping.

168

6.31. Write formulae explicitly expressing the coordinates of the nor-mal in terms of local coordinates on the torus.

6.32. Homogeneous coordinates on the straight line depend smoothlyon local coordinates on the sphere and local coordinates on RP2 are ex-pressed in terms of homogeneous coordinates.

6.33. Coordinate functions are a special case of a smooth function in amanifold.

6.34, 6.35, 6.36. Use the implicit function theorem.6.37. Use Problem 6.36 and partition of unity.6.38. Using local coordinates, calculate the rank of the Jacobian matrix

of the mapping.6.39. Use the properties of the rank of the product of two matrices.6.40, 6.41, 6.42. Consider the group GL (n, R) of all square matrices of

order n with non-zero determinants. Each matrix from GL (n, R) can beassociated with a vector from the space while the mappingdet: R is a continuous function. Therefore, the group GL (n, R) isan open subset in Any open subset is a smooth manifold.

Definition. A linear group is said to be algebraic if it can be singled outof the group GL (n, R) by some set of algebraic, i.e., polynomial, rela-tions among matrix elements.

Theorem. Any algebraic linear group G is a Lie group.Proof. Consider the space M(n, R) of all square matrices of order n

with elements in R. Let J be the ideal, formed by all polynomialsvanishing on G, of the ring S of polynomials on M(n, R). By the Huberttheorem [5], there exist polynomialsf1 e J (forming the idealf)such that any polynomial f e J can be represented in the form f == where €8 (a = 1, . . . , a). Letp be the rank of the Jaco-bian matrix (having a rows and n2 columns) when (x,1) = E.

Lemma. The rank of the matrix equals p at all points of thegroup.

Let A E G. For any polynomialf e 5, we put (Af) (X) = f(A —

Xe M(n, R). The transformationf — Af is an automorphism of the ringS, transforming the ideal J into itself. Therefore, the polynomialsAl1 Af0 are generators of the ideal J as well as off1 ,...,f0. Wehave

= fa =

where hap e S. At the points of the group G, we have

a(Afa) — 9fa — 'ç'h

ax,1 — ar,; ax,:, — — ax

169

from which it follows that the ranks of the matrices (3(Afa)/ôxu) and(t9fa/ôxij) are equal at all the points of the group G. On the other hand,the rank of the matrix at the point A equals the rank of thematrix at the point E, i.e., equals p. Therefore, the rank of thematrix lox,1) at the point A equals p. and thus the lemma has beenproved.

Now, let be a minor of order p, not vanishing atE, of the functionalmatrix (of1 Assume, for definiteness, that it is contained in the firstp rows. It has been proved that all the minors bordering it are identicallyequal to zero on the group G, i.e., belong to the ideal J. Similarly to theproof of the classical matrix rank theorem, we obtain

(1)OX,j Ox,j

I

where S, a = 1, . . . , a. Consider the set O of matrices satisfyingthe equationsf1 = . . . = f,, = 0. It is obvious that G C C. By the im-plicit function theorem, there exists a neighbourhood U of the unit matrixin M(n, R) such that the intersection C fl U can be given parametrically,the number of parameters being equal to d = n2 p. We can see to itthat does not vanish on U, and the range of parameters is connected.Let be a curve in C fl U, with (x,1(0)) = E. When a = 1, .

p, dfaldt = 0 along this curve. We obtain from (1)

where a is any index. The unique solution of this system, with the initialconditionsfa(0) = 0, is the zero solution. Therefore,fa = 0 for all a,i.e., C fl U = G fl U, which shows that G is a Lie group.

The group SO (n) in question is a Lie group (consequently, a smoothmanifold), since it is an algebraic group. Its relations are

= '5ik' det (a1) = 1.

The dimension of the group SO(n) equals a(n — 1)/2. Consider the caseof the groups U(n) and SO(n). Every linear transformation over C withthe matrix A can be treated as a linear transformation over R. Thistransformation will have the following matrix with respect to the basis

• . . , e,,, ie1 (where e1 are basis vectors in C"):

( ReA ImA

ReA

170

The group GL (n, C) is a subgroup of the group GL (2n., R). Therefore,the theorem proved can be applied to the groups of linear transformationsof C'1. Note that U(n) is the group of linear transformations of C'7, pre-serving the hermitian metric. The group SU(n) C U(n), det A = 1 ifA SU(n). The condition for unitarity is written with respect to the or-thonormal basis thus:

=

In the passage to real coordinates, we obtain symmetric relations for thematrix elements. It follows from what has been considered previously thatthe group U(n) is an algebraic group. The group SU(n) is also algebraicand singled out of the group U(n) by the additional equation det A = 1.

6.43. Calculate the rank of the Jacobian matrix of the mapping.6.45. All the roots are of multiplicity one.6.47. Since M" is compact, it can be covered with a finite number of

charts each of which is homeomorphic to the open ball D". Let there be acovering of charts D'7 (where are local coordinates).Thereby, each pointx M'1 is put into correspondence with the collectionof its coordinates in D". This is a smooth mapping f. Extend! to thewhole manifold M'7 by constructing a new covering with the charts

C and also constructing the = 0 whenx(x) = 1 on Vt,, 0 1. Suppose there were k charts in the cover-

ing. Then, to each point of the manifold, we assign the nk-dimensionalvector

x — .

Under such a mapping, one-to-one correspondence is not achieved atthose points x, y which belong to \ V,,, since here we smoothen thefunctions in an arbitrary manner. To eliminate this defect, constructanother covering C Perform the same constructions for thecoverings and Ua as for and to obtain the collection of func-tions Then the correspondence

x — . . . e

is one-to-one.6.49. Verify that the neighbourhood of S'1 C R" + is diffeomorphic to

S'7 x K', and apply the method of induction.6.50. Apply the Sard lemma.6.51. Use the two-dimensional surface classification.6.52. Prove, at first, that the manifold M'7 can be immersed into It

is known that any compact, smooth manifold M'1 can be embedded inRN, where N is a sufficiently large number. We assume then that

171

ç. We shall decrease the dimension of N by projectingM" alonga certain vector E onto its orthogonal complement. Under these projec-tions there may appear points at which the smoothness of isdisturbed, i.e., such points at which the tangent space contains avector parallel to the vector

Thus, for the to be a smooth mapping at every point, thevector should be selected so as not to belong to the tangent space (at anypoint x The tangent space is diffeomorphic to R", butsince we are interested in the directions of the vectors only, the manifoldof directions is — The dimension of all the tangent directions is notmore than n + (n — 1) = 2n — I. lfN — 1 > 2n — 1, then a direc-tion independent of all the directions of the tangent spaces to themanifold M (i.e., not contained in them) can be chosen in the spaceRPN_ Select one of such directions and project along it. This pro-cedure is repeated while the inequality 2n — 1 < N — 1 is fulfilled.Finally, when 2n — 1 = N — I, this reasoning will not hold for the firsttime, since the existence of a convenient direction no longer follows fromthe dimension inequalities. Thereby, it has been proved that N can belowered to 2n (at any rate), i.e., the manifold M" is immersed into thespace Let us now prove the existence of the embedding

+ For the immersion, it is sufficient to ban all the directions(as projected) in the tangent spaces to the manifold. Now, we have to banpossible self-intersections which can arise under a projection, i.e., ban allthe chords of the form (x, y) parallel to the projecting direction PE. Thespace of chords is the space of pairs (x, y), wherex M",y e Con-sider the mappingf: (x, y) — RPN_ The mappingf is not smooth,since singular points appear on the diagonal in M" x M'2. Restrictthe mapping f to x where = {(x,

x = 2n. Nowf is a smooth mapping. Consider theimage f(M'7 x C Close this image in RPN_ 1;dim Im f is unaltered under this operation (note that under the closureoperation, the directions from the tangent spaces are necessarily taken in-to account). Furthermore, if N — I > 2n, then we project, similarly tothe investigation of immersion, along any direction not belonging toIm f. Thereby, we decrease the dimension of the space by unity and con-tinue the process whileN — 1 > 2n. WhenN — I = 2n, a "free" direc-tion may not exist.

6.53. The zero-dimensional compact manifold consists of a finitenumber of points.

6.54. Since the pointy is a zero-dimensional manifold, its tangent spacevanishes.

6.55. In this case, the condition for transversality is equivalent to thesubspaces TM1 and TM2 generating the whole tangent space of the am-bient manifold.

172

6.56. Use the implicit function theorem.6.57. Transversally in all the cases.6.58.a * 1.

6.59. Construct linearly independent vector fields which are normal tothe fibres under a certain metric, and then construct the requiredhomeomorphisms by means of motions along the integral curves.

6.60-6.62. Calculate the rank of the Jacobian matrices of the mappingsin local coordinates.

6.69. (a), (b), (d), (e): orientable;(c) orientable when n is odd; non-orientable when even.6.70. Represent the Klein bottle as a square whose opposite sides are

identified, and transfer the basis consisting of the tangent vectors alongthe midline.

6.71. A manifold is said to be orientable if there exists a collection ofcharts such that the Jacobians of all the transition functions are positive(i.e., there exists at least one such collection of charts for the manifold).Let be a transition function of variables zt z", and

/9za 0. Let A be the Jacobian of the transition function andA = (au). The mapping A can be considered as a linear operatorC" — C". The realification of the mapping!: GL (n, C) GL (2n, R)

will assume the formf(A) = RA=

where A = B + ID.

We take the basis e1, . . . , e,,, ie1, . . . , in Let us prove the for-mula det RA = I det A 12 by induction on n. When n = 1, forA = a + bi, we obtain

IdetA 12 = a2 + b2, RA = (adetRA = a2 + b2.

Let the statement be proved for k n. We now prove it for k = n + 1.

Reduce the operator A to the Jordan normal form (the determinant re-mains unaltered):

fa1+ib1 0\A =

),where e, =

0

0 00 0 IdetAI2A b1 0 a1 0

0

= +

173

where

7a2+ib1 0 \ 2D = deti

Let us calculate detR A by expanding it along the last row:

detR A = (1)3fl+ 1det

(aiei _bi)b1 a1e1

=(_1)3fl+3bn+i(_bn+i)(_1)3fl+3Dn

+ = +

a collection of charts such that in changing thecoordinates (the change being smooth)

(z1 ,...,Zn, z1, . . . ,z,1) (x1, . . . ,xhl,yl yfl)

(realification) the Jacobians of all the transition functions are positive.6.74. We obtain from the existing classification of two-dimensional,

closed, differentiable manifolds (which are orientable), that all of themare spheres with g handles, i.e., surfaces of genus g. Each of suchmanifolds is the Riemann surface of a certain polynomial without

multiple roots, where g = [n—

]. The function w = is

complex and analytic; therefore, by taking z and w as coordinate patches,we obtain an atlas with a complex and analytic transition function.

6.75. Obviously, a complex structure can be introduced only on even-dimensional manifolds. Let F be a group operating on \ (0) andgenerated by the transformation z — 2z. Consider the factor spacerelative to z — 2z. It carries a complex structure induced by the structureof the space \ (0) and is homeomorphic to — x S'. Therefore,

x S1 also has a complex structure. There exists a fibration' x —

—— 1 X CP'1 with the fibre

F = S' x S' = T2. The fibre and base space have a complex structure(proved). A complex structure on can be defined by means of a formwhich is the restriction of the hermitian form on C'1 to the sphere S2'1 — 1:

dS2 = —

This form is obtained from the form on CP'1 — since the former is in-

variant with respect to the action of S1, where S2" — CP'1 — Wedefine the action of thus:

(z°, . . . , z'1— )e"1 — (e"1z°, . .

174

— = = + = —

= + — +

= — hxdcx, + kxda.

Therefore,— = —

Let zk be the coordinates of the first factor — and z of the secondfactor Let

VkJ = z') E — x — 1 : zkz 0).

The sets VkJ form an open covering of the space x In-troduce complex coordinates

r 'S

tkj = ± (lnzk + y2iri

on each set VkJ, where y is a vector from and are determinedmodulo I. Therefore, tkj is a point of the torus T(1, y) obtained by gluingtogether the opposite sides of the parallelogram constructed on the vec-tors I and -y. Thus, we have 2n + 1 coordinates in which determinethe mappingf: VkJ — x T(l, 'y). The mappingf is a homeomor-phism. The quantities kwr, 'S,

tkj uniquely determine the coordinates zkand In fact,

r—r rr rrZ Z = Z Z — 1 = —

r

The quantities I z are determined similarly (uniquely). Besides,

= lnIzkI + i argzk,

whence

2lrtkJ = + 'y + argzk + y argz'-1.

If I I and z '-' I are known, then arg z"< and arg z can be found hasbeen chosen so that Im -y * 0). Consequently, zk and z are also deter-mined uniquely. The transition function in VkJ fl is complex andanalytic because it is determined by the formulae:

uwr == L°'.

175

tuv 'kj + (in kW + y2iri

Thereby, a complex structure has been introduced on S" x j.

This construction introduces a complex structure on any productS2" ' x S2'1 I, wherep, q > 0 and can be different.

6.76. If two piecewise smooth paths c0, c1 : [cr, 3] — M'1 are freelyhomotopic, then, in traversing them, the orientations are either bothaltered or both unaltered. There is the shortest periodic geodesic in eachfree homotopy class of paths on M". We can now prove the requiredstatement. It is clear that it suffices to give the proof for a connectedmanifold M". Let p eM". It suffices to prove that in traversing anysmooth loop with the origin and end at the pointp, the orientation in thespace is unaltered.

Assume the contrary, i.e., that there is a smooth, closed path with theorigin and end at the point p, and, in traversing it, the orientation in

is altered. Then there exists a non-trivial periodic geodesicc [0, 11 —. M" which is freely homotopic to this path and shortest in itsfree homotopy class. Let c(0) = c(I) = q eM". Then the paralleldisplacement along the geodesic c induces an automorphism r reversingthe orientation on the subspace Mq' C TqM orthogonal to the vectorc(0). Since c is a geodesic, r is an orthogonal automorphism, and sincedim Mq' = n — 1 = 2k, there exist two-dimensional Euclidean spacesE1 Ek invariant with respect to r such that

= E1 . . . It is clear that

det r = fi det ri E. = — 1,

and then the relation det nE, = — 1 is fulfilled for a certain i so that rreverses the orientation onE1. But then r has a non-zero fixed vector u,i.e., ru = u * 0. Now let Y be a parallel vector field along c, for whichY(0) Y(l) = u. Then there exists an open interval Ye R containingzero such that & Y(t) lies in the domain of the exponential mapping exp onM" foralLe El,! e [0,1]. Wedefine V:[O, 1] x I—. M" bytheequalityV(t, e) = exp (e Y(t)). Let L (z) be the length of the curve V(t, Then,since c is a geodesic, L '(0) = 0. It follows from Y being parallel thatY' = 0 and <Y, c> = 0. Since — V(t, e) is a geodesic for any

e [0, 1],L "(0) = — <R(Y, è)è, Y>dt, whereR is the Riemann ten-

sor of the manifold M". If follows from the curvature along the geodesicc being positive that L "(0) < 0, and, therefore, L has a relative max-imum on c, i.e., c is not shortest, which is a contradiction.

176

6.78. If A is a complex Jacobian matrix, then the real Jacobian matrix

( ReA lmA

\—lmA ReA

6.79. Use Problem 6.78.6.80. Changing the coordinates z2 = l/z1, we obtain that the atlas

consists of two charts.

7Transformation Groups

7.1. Use the theorem on the existence and uniqrieness of a solution to asystem of ordinary differential equations of the irst order.

7.2. Construct a vector field such that one of the trajectories may jointhe points x0 and x1 and that it may be trivial outside a certainneighbourhood of this trajectory.

7.4. The ratio of the coordinates of the field should be a rationalnumber.

7.6. Select an atlas of charts so fine that each orbit may intersect an ar-bitrary chart at no more than one point.

7.7. The action of the group Z2 on the sphere should be given by theformulax — —x.

7.8. The action of S' on C C is given by the formula

(X,x)—Xx, X€S'CC'.7.9. Use the differential of the mapping determining the action of an

element of the group C. — -7.18. Fix an orthonormal coordinate 3-frame e1, e2, in R3. An ar-

bitrary state of the described system is uniquely determined by a pointx S2 and the velocity vector v(x) E _whereIv(x)I = C = const = 0. It is obvious that the mapping x — x,v(x) — v(x)/c is a homeomorphism, xis the unit vector in R3 emanatingfrom the point 0, and v (x) is a unit vector in R3. Shift the origin of v (x) tothe point 0. This transformation is the identity on the vectorsx and v(x),andx and v are orthogonal. Lety be a vector mR3 such that IyI = 1, itis orthogonal tox and and the system teL, e3} is oriented in the samesense as {x, v, y }. Obviously, the mapping x, v — v, y is a homeomor-phism. All systems v, y} are in one-to-one and continuous cor-respondence with the matrices associated with the linear transformations

177

in which map the orthonormal coordinate 3-frame e2, e3} into theorthonormal coordinate 3-frame Ix, v, y I. These matrices form the groupSO(3) : A SO(3) AA1 = E, and det A = +1. Thus, the space ofthe states of the system under consideration is homeomorphic to themanifold SO(3). Any orthogonal transformation of R3 preserving theorientation is a rotation about a certain axis in a plane perpendicular to itthrough an angle where — ir < < ir.

Therefore, all elements of the group SO(3) are in one-to-one and con-tinuous correspondence with the points of a ball of radius ir in R3 whosediametrically opposite boundary points are considered to be identical, Itremains to show that the ball glued in this manner is homeomorphic toRP3. In fact, RP3 = S3/Z2 where S3 is standardly embedded in R4.Therefore, RP3 can be considered as a hemisphere of S3 placed in theregion with xt 0 and with the diametrically opposite boundary pointsconsidered to be identical:

fl = 01 = I(0,x2,x3,x4)€R4 : (x2)2 + (x3)2 + (x4)2 = II

which is homeomorphic to the sphere S2 of radius7.21. Let us prove that if A SU(2), then

A= ( 11312= l,a,13eC.\—13 a/

ía 13\Let E SU(2). Then

1al2+ 11312= 1 (1)

(2)

1y12+ lô12= 1 (3)

detA = ab — 13'y = 1. (4)

Substituting a = —13ö/-y from (2) in (4), we obtain — +

+ I.y12)= l,ory = —f3,whencea =

Sp(1)= (q1,q2),q1,q2eQj,

where (q1, q2) = Re q1q2. It is easy to see that IqI = I, since, for= q2 = 1, we have Ii' q 12 = I, i.e., Iqi = 1. Conversely, if

Iqi = 1, then (q1q, q2q) = (q1, q2). Thus, Sp(1) consists of quater-nions of length 1, i.e., S3 C Q = R4. Further,

q = a + lb +jc + kd= (a + ib)+j(c — id)= z1 +jz2,

178

wherez1andz2eCCQandlz1l2+1z212=1q12.lfIqI=1,then1z112 + 1z212 = 1. Let us arrange for ofSp(J)into

= +jz2)= (

zi1z112 + 1z212 =

—z2 z1

It is obvious that the element 40(g) belongs to SU(2). It is easy to verifythat is an isomorphism.

7.23. It suffices to show for the group G = SL(2, that anyelement of the group G can be joined to ±E. Let be the eigen-values of the matrix A. Then either (a) X1, X2€R, X2 = since

det A = ± 1, or (b) X2 = = X2 = Consider case (a).With respect to the basis consisting of eigenvectors, the matrix A has the

form A' = CAC1=

We can assume that X > 0. Con-

struct a path 'y : I — G

0=

0 (X(l — t) +

Consider case (b). There exists a basis on the plane with respect to whichA is of the form

/cosço —sin40I\sin cos

Construct a path I G

— (cos(1 — —sin(1 — 1)40

\,sin(i — cos(1 —

7.25. Consider a model of the Lobachevski plane L2 in the upper half-plane (Im z > 0 in the complex notation). The metric is of the formds2 = (dx2 + dy2)/y2, or, in complex terms, ds2 = dzdz/(z — z)2.Consider the linear fractional transformations of C' into C', keeping theupper half-plane fixed (i.e., transform it into itself). These are transfor-mations of the form

( az+b( cz+d

This transformation class preserves the metric, but there are othertransformations preserving it. E.g., the transformation w = —z which is,evidently, a motion, but does not belong to the group G, at least becauseit is not an analytic function. Similarly, it is easy to verify that the wholeclass of transformations of the form

H = = (az + b)/(cz + d); a, b, c, d e R, ad — bc = — l}

179

preserves the metric. The group of motions of the Lobachevski plane con-sists of transformations of forms G and H only. In fact, G U H is agroup, G U H = ® acts on the Lobachevski plane L2 transitively. Con-sider a subgroup S of the group ® (i.e., subgroup of transformationskeeping a point i fixed) and a certain motion h : L 2 L 2, h (i) = i. Weprove that h a S. We shall show that the motion keeping i fixed is fullydetermined by its action in the tangent plane at the point I. Let h, g be twomotions, and = g: — T1L2. Then the transformations h andgact on the geodesics passing through i in the same way; therefore, coincideon them, and since any point of L2 can be joined to i with a geodesic,Ii = g on L2. It remains to establish that, for any a a 0(2), there existsan element g a such that g,, = a. Let g(z) = (az + b)/(cz + d) a

: R2 —. R2. The differentialg '(i), whereg '(z) denotes the derivative with respect to complex variablez, viz.,g'(t) = 1/(ci + d)2. Let

a = c = d =

cos + I sin i.e., it is a rotation of C' through theangle In the case of symmetry, consider the transformation w = —zwhose differential is a symmetry, and then apply a linear fractionaltransformation which carries out a rotation. Now, let h be an arbitrarymotion and h (i) = z0. Due to the transitivity of the group there existsan element g a ® such that g h

h (i) = I. The subgroup G is a connected subgroup containing theidentity element. The transformation w = —z does not belong to G.Therefore, the group G is a subgroup of index 2 of the group ®.

7.27. Consider the cases n = 2k + 1 and n = 2k. The group 0(n) isdisconnected and is the disjoint union of two path-connected com-ponents, viz., 0+ (n), i.e., the collection of matrices with det = + 1, and0 (n), i.e.; the collection of matrices with det = — 1. Whenn = 2k + 1, the unit matrixE a and the matrix —E a 0(n).Consider a discrete normal subgroup H of 0(n). The element ghg a Hfor anyh a H, and anyg a 0(n). Ifg a (n), theng can be joined toEwith a continuous path p(r) so that = g, = E. If g = 0 (n),then the elements g and —E can be joined with a continuous path sothat = g, = —E. It is possible to construct two mappingsM(t)and JV(t) such thatM(t) = andN(t) =g a 0 + (n) and g a 0 (n), respectively. Then

= ghg' = R, = ghg' =

LM(1 = h, tN(l) = h.

180

The elements fi and h are joined with a continuous path which lies in Hwholly. Since H is discrete, we obtain that h = i.e., ghg = h forany h H, g n 0(n).

(The Schur Lemma.) Let p1 : G — GL(V1), p2 : G —. GL(V2) be twoirreducible representations of a group G, and letf be a linear mapping ofthe space V1 into the space V2 such that p2(S)f = fp' (S) for each S G.Then (a) if pt and p2 are not isomorphic, thenf = 0, (b) if V1 = V2,

= thenf is a homothety (i.e., multiplication by a certain number).The mapping p: 0(n) — 0(n) C GL (Re) is an irreducible representa-

tion, since reflections with respect to each of the axes can be considered.These are matrices of the form

1 is placed at (I, 1). All such transformations are contained in thegroup 0(n). Collectively, they keep fixed only the point (0, 0 0).Besides R'7 and 0, there are no other invariant subspaces in R'1. Applyingthe Schur lemma and using ghg -' = h for any h e H we obtain that h isa scalar matrix. But there are only two scalar matrices in 0(n), viz., E and—E. It is they that make up the discrete normal subgroup of 0(n).

Consider the case n = 2k. The matrices E and —En (n) are asubgroup of a discrete normal subgroup H of 0(n). We show that H con-tains no other elements. 0 + (n) contains no other elements from H, sinceH fl (n) is a discrete normal subgroup of 0 + (n), but only the group± E can be that in (n). The reasoning is similar to the previous. Weprove that (n) contains no elements from H. Assume thath 0 (n) fl H. Then ghg — = h for any g (n). The matrix h canbe reduced to block triangular form with an odd number of eigenvalues— 1 by a certain orthogonal transformation of the basis with determinant+ 1. If dim W1 > 2, then, by an even number of transpositions of thebasis vectors, the diagonal elements can be interchanged. Generallyspeaking, we will obtain a new matrix then, i.e., ghg * h. E.g., we in-terchange — I and the block; if there is no block, then we interchange —and the block formed by + 1, + 1. In the case where n = 2, we have onlytwo kinds of matrices to which any matrix from (n) can be reduced byan orthogonal transformation of the variable, viz.,

/1 0\ 1—1 0I and

\0 —1, \ 0 1

Then

/ cos sin /1 0\ /1 0\ / cos sinII 1*1 II

— sin cos / \0 — I / \0 — I) \ — sin cos

181

Therefore, there are no elements from the discrete normal subgroup of0(n).

7.28. Theorem. Any group of motions of finite order N in R3 is isomor-phic (assuming that the action has no kernel) to one of the followinggroups: CN, a cyclic group; DN, a dihedral group; T, the tetrahedralgroup; W, the hexahedral (octahedral) group; and P, the dodecahedral(icosahedral) group.

Proof. Let F be a finite rotation group of order N. Consider the fixedpoints (poles) of all transformations from F different from the identitytransformation. Let the multiplicity of a pole p (number of transforma-tions from F, leavingp fixed) be equal to v. The number of operations dif-ferent from the identity transformation I and leaving the pole p fixedequals ii — 1. Let g be the set of points into which the pole p istransformed under the action of elements from the group F. Then I is

an orbit consisting of points equivalent to each other. The number ofpoints g equivalent to p equals N/v. In fact, the multiplicity of g alsoequals The transformation L e 1' reduces p into g (i = I n).Let . . . , be transformations leaving the point p fixed, and

All these transformations are different, each element g e F is contained inthis set, and IF I = N, i.e., N nv for any orbit N = where cis a certain orbit. Consider all pairs (S, p), where Sc F are fixed on p,and S * I. The number of such pairs equals, on the one hand, 2(N — 1),

and, on the other hand, — viz.,

2(N— l),2— 2/N= (1— 2

(because if N = 1, we shall have a trivial group). Therefore, 2, and,from the evident relations, we get that 2 ( c 3. The following casesare possible:

= 2. Then 2/N = 1/v1 + l/v2, 2 = N/v1 + N/v2 = n1 + n2,=

Each of the two classes of equivalent poles consists of one pole ofmultiplicity N, i.e., we have obtained a cyclic group of order N of rota-tions about one axis.

2. c = 3. Then 1/v1 + 1/v2 + I/v3 = 1 + 2/N, v1 v2 At leastone v = 2. Let v1 = 2. Then 1/v2 + 1/v3 = 1/2 + 2/N. The numbers v1,

cannot be greater than or equal to 4, i.e., v2 = 2 or 3. (a) v1 = P2 =

= 2, N = 21'3, P3 n, a dihedral group DN; (b) = 2, v2 = 3,

1/v3 = 1/6 + 2/N. Then the following cases are possible: P3 = 3,

182

N = 12, the tetrahedral group T; v3 = 4, N = 24, the hexahedral groupW; = 5, N = 60, the dodecahedral group P.

The dodecahedral group T contains two classes of poles with four polesof multiplicity 3, I TI = I + 4 2 + I 3 = 12. The generators andrelations of the group T are abc = adb = acd = bdc == 1 (a, b, c, d being rotations about all four vertices through the angle

a3 = b3 = c3 = d3 = 1. Let e,f, g be rotations about the axesand ef = fe = g. To T, reflections may be added. Let h be an im-

proper rotation, and he = eli, a'h = (i = 0, I, 2). With all theimproper motions added, we obtain I TI = 24. The cube and octahedronpossess the same group of motions I WI = I + 3 3 ± 1 6 ++ 4 2 = 24. We have one class of six poles of order 4 (the vertices of thecube), eight poles of order 3 (the centres of the faces), and twelve poles oforder 2 (the midpoints of the edges) for W. T is a subgroup of W. This isobvious from geometry (the tetrahedron can be inscribed in the cube).The relations are a4 = b3 = c2 = = I (where d is a reflection);a1d = da4',b'd = = dc,ac = b, IPI = 60 are only pro-per motions. With the reflections added, we obtain IPI = 120. Thesubgroup of proper motions inP is isomorphic to A5. It has twelve polesof order 5 (the vertices of the icosahedron), twenty poles of order 3 (thecentres of the faces) and thirty poles of order 2 (the midpoints of theedges). This group is commutative only partly. The relations in thedodecahedral group are

abcde = 1, = 1, = 1,

= 1, = 1, 1,

bce = l,bkei_1 = l,i = k,ci'g'e = 1,bg and k, we get

bce = I, = I, citbe = 1, bict = 1.

It follows from the relations bce = 1 and = 1 that i = cb, andfrom biei = 1, tbe = 1 and i = cb that 'b2c = I and

= I. From all the relations obtained, we deduce thestatement of the complete non-commutativity of the group P, i.e.,P = [p. p1. Another variant of the corepresentation is

bc = cb = a5 = b3 = abab 1,

where a is the rotation about the axis of order 5, a = (12345), b the rota-tion about the axis of order 3, b = (452) and P = A5.

183

7.45. Let G be a finite group operating effectively on R'7, i.e., ifxg = x, x e R'7, g e G, then g = e. The group Zk generated by the ele-ment g also operates effectively on Consider the space x = G /Zk,where x y if y = g'x, g Zk; 7r1(X) = Zk, ir(X) = = 0

when i > 1, since R'1 — X is a covering map. Therefore, X is homotopyequivalent to K (Zk, 1), i.e., to a lens space. But the homology K(Zk, 1) isdifferent from 0 in an infinite number of dimensions, whereas X has nocells of dimensions greater than n. Thus, substantiating the statements:(a) if a discrete group G acts on R'1 without fixed points, and

= /G is the set of orbits, then the natural mapping p R'7 — XG is

a covering map; (b) (XG) = 0 (the proofs of statements (a) and (b) areleft to the reader), we complete the proof.

8Vector Fields

8.2. (a) (b) 3v'21 /7; (c) /3; (d) — 2/5.8.3.8.4. 1/4. V38.5. (a) 0; (b) 2 -- + 3); (c) 0; (d) —2; (e)

8.7. hr2.8.8. 1.

8.9. (gradf, grad g).

8.21. (a) (0, x, y —(b) (0, 0,y2 — 2xz);

(c) (0, — xe-p', 0);

(d) (0, 3x2, 2y3 — 6xz);(e) (0, —x(x + y2), x3 +

— 2xyz);

(g) (sin xz/x, 0, —sin xz/y);(h) (xz/(x2 + y2), yz/(x2 + z2), — 1).

8.22. Use the existence and uniqueness theorem for a solution of asystem of ordinary differential equations.

8.23. Investigate the action of the Poisson bracket on the product oftwo smooth functions.

184

8.26. Consider the case, where = 3/3x1.8.33. Let z0 = x0 + iy0 be a singular point, and

f(z) = u(z) = iv(z),3u

=cIV (x0,y0) (x0,y0)

Since

au = = = -=

3x (x0,y0) ôy (x0,Yo) 8y (x0,y0) aX (x0,y0)

f(Zo) = 0. Let = 0, i.e.,

•av(z0)+ 1—- (z0)= 0,

3x 3x

av au(z0) = (zn) = — —- = 0.

ax ay

Then grad Ref(z0) = 0.8.31. Represent the sphere S3 as the group of quaternions of unit

length.8.34. We shall seek the integral curves only in the half-plane lying over

the straight lineAB. The level curves for the functionf(x) are the arcs ofthe circumferences for which the line-segment AB is a chord. The vectorgradf(x) is orthogonal to the level curve. Therefore, the vector or-thogonal to it is tangent to the level curve, i.e., a circumference, and allthe arcs of the circumferences described are the integral curves of the flowv1(x).

8.35. (a) The integral curves of the vector field grad (Re are thelevel curves of the conjugate function Im = sin The uniquesingular point of the field v = grad(Re is: = 0, since! = 0 onlyat this point. The point z = 0 is the case of a degenerate saddle

point. Let us give a small perturbation to the function fl (z —

Then the singular point splits into n — I non-degenerate saddle points ofthe second order. Consider the behaviour of the integral curves near toone of the singular points. Expanding the function in Taylor's series, viz.,

f(z) = f(a1) + f'(a1) (z — a,) + . . . , wheref'(a,) = 0,

we see that the expansion starts from a term of the second order, since* 0 (non-degenerate critical point);f"(a1) * 0 if and only if all a

185

are multiples of but this is not true, since all c1 are different.Therefore, we have, near to the point a1, that f(z) = k(z — a,)2 ++ o(z2), i.e., the saddle point is non-degenerate. If the equality

f"(a1) = 0 were valid, then we would have the case of a degeneratesingular point (i.e., saddle point of the third or higher order).

(b)f(z) = z + l/z;

Re (1(z)) = p cos + cos = (p + l/p) cos

Im (1(z)) = P sin sin (p — 1/p) sin

(z =

A singular point is at the origin, since the function liz is discontinuous.The derivative of the function f(z) equals I l/z2, i.e., the singularpoints are z = 1, z = I. Both points are non-degenerate. Consider theintegral curves for Im f(z), emanating from and returning to the singularpoints, i.e., the separatrices (p — l/p) sin = c (c = 0 at the point(1, 0)). Thus, (p — lip) sin = 0, whence = Kir, or p = 1. Hence,we find the separatrices viz., the unit circumference consisting of twoseparatrices, and the real axis consisting of four separatrices. In the caseof grad[Re(f(z))], the separatrices are given similarly, by the equation(p + lip) cos 2, and have the shape of two loops tangent to eachother.

(c)f(z) = z + l/z2. Consider grad[Re(,f(z))]. The integral curves ofthis flow are the level curves of the function lm(f(z))

2xy .

= — 2 2 2 = r sin—(x+y) r

(in polar coordinates on R2). Similarly, we seek the level curves of thefunction Re(f(z)):

cos2r

(d) 1(z) z + l/(z 2). The singular points are z = I, z = 2,

z = 3. In a neighbourhood of the point z = 1, f'(z) = — 2(z — I) (thefirst term in Taylor's expansion). This is a singular point, a saddle.Similarly, the singular point z = 3 is also a saddle point. In aneighbourhood of the point z = 2, the expansion in Laurent's series off'isf'(z) = — l/(z — 2)2 + . . . . Therefore, the integral curves of thisflow in a neighbourhood of the point z = 2 have the form of the integralcurves of the flow grad[Re(l/(z 2))1.

186

(e)f(z) = z3(z — l)'00(z 2)900. The singular points, i.e., zeroes ofthe derivative,f'(z), are the following: z1 = 0, a saddle point of the sec-ond order; z2 = 1, a saddle point of the 100-th order; z3 = 3, a saddlepoint of the 900-th order; z4 (1109 ± l093)/2006, non-degeneratesingular points.

Locally, in a neighbourhood of each singular point, the integral curvesplay the role of saddle points (degenerate or non-degenerate at the pointsz4 0.008, z5 1.09) of the corresponding order.

(f) 1(z) = 1 + z4(z4 — In a neighbourhood ofz = 0, 1(z) can be replaced by 1(z) = 1 + The qualitativebehaviour of the curves in a neighbourhood of the point z = 0 is never-theless unaltered. But adding a constant does not change the form of thetrajectories. Therefore, the function 11(z) = cz4, c = wherez 0, can be considered. The equations of the trajectories are of theform cp4 cos = k. The point z = 0 is a degenerate singular point,which, after a slight disruption, splits into four non-degenerate points.More precisely, g(z) = c(z — c1)(z E2)(z — e4) is a slightdisruption of the function 1(z).

(g) 1(z) = 1/100 ln[(z — 2i)/(z At the points z = 21 andz = 4, we have logarithmic singularities. Apart from them, there are noother singular points.

(h)f(z) = 1/(z2 + 2z 1). To simplify the notation, we perform atranslation w = z + 1. Then 1(w) = l/(w2 — 2). The singular pointsare w = w = The singular points of grad[Re(f(z))] coincidewith the zeroes of f'(w), i.e., w = 0, a singular point (it is non-degenerate, sincef"(w) 0).

(i)f(z) = 2/z + 21 In z2. The singular points are (0,0), (1/21, 0). Theseparatrices are the curves 21 = sin and the axis x from 0 to + oo.

(j)f(z) = zt + 2 ln z. The singular points arez = 0, z5 = — 2/5, thevertices of a pentagon. At these points,f(z) kz2, k * 0.

(k)f(z) = 2 ln(z — 1)2 4/3 ln(z + lOi)3. The singular points arez = l,z = —lOi,f'(z) * Oforallz.

(1) 1(z) = 1/z3 — l/(z — The singular points are z = 0, z =(the poles of the third order). Differentiating,

f'(z) =+ (z

=

We obtain four other points:

z = — i), z = + I),z + Th/(1 + z = +

The integral curves behave at infinity as the integral curves of the flowgrad[Re(l/z4)], i.e., as the integral curves of a pole of order four.

187

8.36. Let the flow v = (P, Q) be irrotational, i.e.,

3P aQ 8P c9Qrotv= — E0, or8)/ ay äx

Let us find a functionf such that P = 3f/Ox, Q = 3f/ôy. To this end,we integrate the first relation with respect to x between 0 and x, viz.,

f(x, y) Pdx + g(y). To find g(y), we integrate the latter relation

with respect toy,

x

a_fQ(x,y) = - (x,y) = dx + g'(y)8y jay

0

= + = Q(x,y) Q(0,y) +

rhus, Q(x, y) = Q(x, y) — Q(0, y) + g'(y). Therefore, g'(y) == Q(O,y), i.e.,

g(y) = çQ(o,y)dy + C.

Consequently,

f(x,y) = P(x,y)dy + Q(0,y)dy + C.

Let and be two paths from (0, 0) to the point (x, y) in the plane(x, y). Consequently, if rot v = 0, then

Pdx + Qdy + C,

where is an arbitrary path from (0,0) to (x,y).

188

Besides, let the flow be incompressible, i.e.,

f(x,y)=

Consider the flow

v' = (Q,P), rotv' = 0.

Therefore, the field is potential. Thus, there exist functions a(x, y)and b(x, y) such that v = grad a(x, y), v = grad b(x, y). Sincediv V = div v' = 0, we obtain that a(x, y) and b(x, y) are harmonicfunctions, i.e., Lsa 0. Consider the functionf = a + ib. It is

complex and analytic, since the Cauchy-Riemann equations are valid,viz.,

ôa c9b 3a c3b-. = = P(x,y), = — -• = Q(x,y).0x 3y iix

Such a functionf is called a complex potential of the flow.

8.39. Hint: is homotopic to dçc1.

8.47. Consider the differential equation in

= Ax, where A =0

,

The required set consists of the integral curves of this equation, whichbelong to the sphere. It is clear that x(t) = If R4 is regarded as

C2, then the integral curve passing through a point (z1, z2) e S3 is of theform e'1z2), since eAt may be written in the complex notation as

fehbo\\ .

follows: Let us assign to this trajectory the point (z1 : z2)

belonging to CP1. The definition of this correspondence is correctbecause any other pair (z3: z4) lying on the same trajectory differs from(z1 : z2) only by the factore't, and therefore, determines the same point ofCP'. it remains to note that the mapping is one-to-one and continuous.

189

9Tensor Analysis

9.1. (a) (0, 1); (b) (0, 2); (c) (1, 1); (d) (0, 2).9,5. If k = dim V, then dim 1/,'," =

9.14. gradf= (x,y,z).Vx2 + +

9.25. (a) Use Problem 9.22 while replacing the sphere by the cone.(b) The meridians and the equator are geodesic lines.(c) Apply (a) and (b).9.28. Perform covariant differentiation with respect to the parameter

which parametrizes the family of curves.9.39. Hint: The integral curves of the left-invariant vector field X are

left translations of a one-parameter subgroup, i.e., geodesics. Therefore,we may assume that X = where is the vector field of the velocities ofthe geodesic -y(). Since, by the definition of a geodesic, 71(y) = 0, forany left-invariant field X, we have Vx(X) = 0. In particular,

+ Y) = 0, i.e., + = 0. On the other hand,— = [X, Y]. in fact,

= YIVXk = + — +

k 3yk ,3Xk= x' Y' - + X1 — rk ) = x' .

ax' ax' ax'

since = (the connection is symmetric). The required statementfollows from the system: + = 0 and V.1X == [X, Y].

9.41. Hint: The invariant definition of a curvature tensor is of the formR(X, Y)Z = — Since VxY = 1/2[X, Y], we obtain

R(X, Y)Z = ([X, [Y, Zfl — [Y, [X, Z]]) Y], Z].

Taking into account the Jacobi identity.

[X, [Y, Zfl + [Z, [X, Y]] + [Y, [Z, X]] = 0,

we get

R(X, Y)Z = Y], Z].

190

10Differential Forms, Integral Formulae,

De Rham Cohomology

10.6. (a)—2(z + 1)dxAdyAdz;(b)yzdxAdz + xzdyAdz;(c)6ydxAA dy A dz; (d) 0; (e) 0; (f) 0; (g)df A dg; (h) 0.

10.7. Reduce the problem to the case of constant coefficients.10.12. (a) — 471/15; (b) — w/3; (c) 4irR3/3; (d) 0.

10.13. (a) (2 cos sin

cos sin 0, 0);/ 2cos0 sinO

3,— 3,0.

r r

10.15. (a) 2 + zip cos sin z;

(b) tan ' p +2 — (z2 + 2z)ez.

p l+p10.17. 4r — cotO +

r r(r2+ 1)sin0

10.19. (a) p + + z + C; (b) + + z2) + c; (c) + c;

(d) sin + + c; (e) cos z +

10.20. (a) rO + c; (b) r2 + + 0 + C; (c)l + 02) + c;

(d) r cos sin 8 + c; (e) er sin 0 + In (1 + + c.10.21. 471R2.

10.22. (a) 1; (b) 712; (c) 2irR; (d) 0; (e) —271R4; (f) ir.

10.23. (a) 712; (b) 1; (c) ± — 1; (d) w; (e) 0; (f) 0.

10.24. (a) 2471; (b) 71/2.

10.25. (a) 471; (b) R3; (c) 471R4; (d) 271R3; (e) R4R

10.30. (a) 1-I '(S1) = R1; (b) H2(S2) = RI; (c) H"(RP2) = 0, k 1;

(d)H'(T2)= R2;H2(T2)= = (11);(f)dim H'

= k, where k is the number of points excluded.

191

11

General Topology

11.1. Note that the open ball B" and punctured sphere S" arehomeornorphic. We prove the statement by induction on the dimensionof the complex. If the dimension n = 0, then the statement is evident. Letthe statement be held for all numbers less than n. Then, by the inductivehypothesis, the (n — 1)-dimensional skeleton K" — of the complex inquestion is embeddable in the Euclidean space W'. This means that con-tinuous real functions f1(x) are given on K'1 such that(f)(x) fN(x)) (f1(y), . . Let e7 (I = 1,

k) be all n-dimensional cells of our complex. Then the functions.f, (x) are defined on the boundary of each cell e"' (we denote it by Lete"' be homeomorphic to the interior B" of the closed ball D". We mayassume then that the functionsf,(x) are given on D"\B". Their con-tinuity is preserved, but they may not be one-to-one now. Let us extendthese functions from D" \ B" to B" (i.e., from to e7) as follows. Letz B", and z 0. We putf1(z) If z = 0, then we setf1(z) = 0. Thus, we have extended thetions on the whole complex K. Now, we define

1(x).

We put 0 (s 1 ii + 1) outside e7and

= sin . . .

,sin cos + one".\Ix! xl /

We define F K —. k(n ± by the equality

l(x)= (J1(x)

The mapping F is thus one-to-one, and the statement proved.11.4. A section of the Klein bottle by a plane should be considered so

that there may be two MObius strips. Then this plane should be lifted(while discarding one MObius strip), thereby carrying out the boundarycircumference deformation represented. When this circum ferencebecomes free of self-intersections and turns into the standardly embedded

192

circumference, it should be glued to the two-dimensional disk. Consider-ing the surface obtained in lifting the plane and which is the trace of thecircumference deformed, we get an embedding of RP2 in H3.

11.5. The set of the points of self-intersection is homeomorphic to thewedge of three circumferences S1 V S V S'. The vertex of this wedge is atriple self-intersection point, and any of its points different from thevertex is double.

11.6. The boundary M2 of the normal tubular neighbourhood of radiusc constructed is, obviously, projected onto HP2 (two endpoints of thenormal line-segment are sent to its centre lying on RP2). Thus, M2 is asmooth, two-dimensional, compact, and closed manifold and a two-sheeted covering of the projective plane. If we prove that this manifold isconnected, then we shall prove thereby that it is a two-dimensionalsphere, since S2 is the unique two-sheeted connected covering of HP2.

To establish the connectedness, it suffices to consider two points on M2which are the endpoints of the same normal line-segment, and find thepath on M2 joining these two points. To construct such a path, it sufficesto consider a point Ton RP2 which is the centre of the line-segment underconsideration, and take on RP2 a closed path starting and ending at thepoint Tand such that the orientation of the two-frame slipping along thepath and always tangent to HP2, changes in moving along it. Then, byadding to this frame a third vector orthogonal to HP2 and considering thetrace of this vector which is cut out by the frame in moving continuouslyalong the closed path, we obtain the continuous path on M2 joining thetwo selected points.

Note. The embedding of the two-dimensional sphere in Euclideanthree-dimensional space helps to prove a remarkable topological fact,viz., the possibility of "turning the two-dimensional sphere in R3 insideout". This task is outside the scope of our course, and we confineourselves to a short sketch only. The embedding of S2 indicated is suchthat admits interchanging the exterior and interior of the two-dimensionalsphere while remaining in the regular embedding class. In fact, it sufficesto consider a smooth deformation of the two-dimensional sphere alongthe normal vector field determined by the normal line-segments describedabove. In doing so, the interior and exterior surfaces of the sphere inter-change.

11.7. Consider a vector space over R with a basis of the power of thecontinuum. We introduce the following topology on it. Consider the"cube" B = — I < < I for all are the coordinates ofthe vector x, and the cross-section B is of finite codimension, viz.,

B fl = O,x13 = 0, . . . =

13— 20

193

We call the sets the neighbourhoods of the point 0. It is obviousthat the point 0 has no countable base for the neighbourhoods in such atopology, i.e., the space constructed does not satisfy the first countabilityaxiom, nor does it satisfy the second, since the first axiom is a corollary tothe first.

11.8. Consider the mapping F : S2 —. R2, x — (J(x), g(x)), whereF(S2) C R2 is the image of the sphere. The image F(S2) is a set sym-metric about the point (0, 0), since if (a, b) E F(x), then(—a, —b) = F(rx). Assume that (0, 0) F(S2) and project the planewith the exclusion of the origin onto the unit circumference. In polarcoordinates, where this projection can be written in the formh = Then h (F(S2)) is a certain centrally symmetric set on theunit circumferenceS1, where h(F(S2)) is the image of the connected setS2 under a continuous mapping h F. Therefore, it is also connected. It isobvious that a connected, centrally symmetric set on S must coincidewith Further, h(F(S2)) must be 1-connected as the image of the1-connected set which is contrary to the equality h(F(S2)) =

11.9. As the space X, take a space /2 whose elements are sequences ofreal numbers x = (x1 , x2 x,,, . . .) satisfying the condition 11x211 =

= < Asthespace Y C X, takeasphereinX, i.c.,theset

of x such that 11x112 = 1. Consider a sequence of pointsx1 in Y, so thatunity is at place i, and the other coordinates are zeroes. This infiniteseguence has no limit point, since — xIII = V2 for all i,j. Therefore,Y is not a compactum.

11.11. Let e1 be the vertices of the complex K. Take the pointse in general position in i.e., anyj points are linearly in-

dependent whenj 2n + 2. To each skeleton

T= le. ...e.IEK,'0 r

we assign the simplex

T'= le'1r

This simplex exists, since due to the points e e being in generalposition in K2" + and inequality r n, the points e,' arelinearly independent. The simplexes form a complex isomorphic to thecomplex T', since to each vertex, there corresponds one and only onevertex from K.

The complex K is a triangulation. To prove this; it suffices to show thatno two simplexes 7', c K - intersect. Let be the vertices ofT/, and eJ0 eJ0 those of T,' (some of the vertices may be common).

194

Let be all the points which are the vertices of at least one ofthe simplexes and T. The number of these points r + I satisfies theinequality

r + I (p + I) + (q + 1) (ii + I) + (n + I) = 2n -i- 2.

Since the points are in general position in the pointsare the vertices of a certain degenerate simplex T0 of dimeti-

sion not higher than 2n + I. The simplexes and TJ are two faces ofT0, and therefore, do not intersect each other if different.

11.19. The main fact to prove is to show that any two fibres fl(x0, x)and II (x0,y) are homeomorphic for any pointsx andy from the space X.We assume here that X is a connected manifold, x andy two points fromX, and the space of continuous paths from the base pointx0tox.We should be able to assign to each path from x0 to x, a path from x0toy so that this correspondence may define a fibre homeomorphism. Onjoiningx andy with a path S we consider a tubular neighbourhood U(s)of the path S and define a diffeomorphism : U(s) — U(s) which isidentity outside U(s) and sends the point x to a point 5(1) eS, whereo i I, s(O) = x, s(1) = y. As to the rest, the diffeomorphism 5c1 isarbitrary. The family (0 1 I) determines a homotopy in themanifold X. In constructing the homotopy, we have used the fact that Xis a manifold. Now, we define the homotopy

f : x) — lI(x0, y), f(—y(1 )) =

It is easy to verify that this mapping establishes a homeomorphism be-tween the spaces cl(x0, x) and tl(x0, y).

11.22. The existence of a convergent sequence in any infinite sequenceof points on a finite-dimensional sphere follows, e.g., at least from thisfact being valid for infinite-dimensional sequences on the unit line-segment. Each vector is specified by a set of n + I real numbers (coor-dinates of the vector). It is clear that we use here the finiteness of thenumber of coordinates. An infinite sphere is non-compact: it is easy tofind a sequence from which a convergent one cannot be singled out. E. g.,the endpoints of the unit vectors of an orthonormal frame can be taken assuch a sequence. Since the distance between any two of such (non-coincident) points equals a convergent sequence cannot be singledout.

11.34. No. If a topological, metric and compact space is connected,then it is not necessarily path-connected, the well-known example beingthe set of points on the plane (x, y) specified as follows:

195

12Homotopy Theory

12.2. Axiom (W). IfK is aCW-complex, then the setF C K is closed ifand only if, for all cells e?, the full inverse image (fq)— 1(F) C is

closed in B". Assume that there are two topologies in the space X:and . We will say that (stronger) if for any point x Xand any x, there is x such that C

Assume apart from the topology determined by axiom (W), thereis another topology in the CW-complex. Take an arbitrary pointx€K, i.e., a point belonging to the CW-complex, and A

neighbourhood is the union of mutually disjoint open intersectionsfl Consider the full inverse image (J9 (Ua0). It is open in

(this follows from the continuity of the Therefore, for thecomplement (K\ the full inverse image 1(K\ is closedin B" for all It follows from axiom (W) that (K\ is closed in K.Therefore, is open in the topology determined by axiom (W), i.e.,Ucy0 belongs to the system of open sets determined by (W).

12.21. The Klein bottle.12.26. Let cxi, H(X', Y), a

homotopyF: X' x / — YsuchthatF(x,O) = a1(x),F(x, I) = a2(x).Put F' = Then F' : Xx I — Y, F'(x, 0) = F(h(x), 0)

= a1(h(x)), F'(x, I) = F(h(x), I) = a2(h(x)). Therefore,a1h a2h.

12.28. Let S = urn s", where S" + is the suspension of

S so defined is a CW-complex. Consider a andf af: 5' — sending the base point inS' to the base point of S. Letf: K — L be a continuous mapping of a complex K into a complex L, themap being cellular on the subcomplex K1 C K. Then there exists a mapg:K — L such that (a) f is hornotopic to g; (b) g is cellular on K;(c)flk1 g k1; (d) the homotopy connecting! and g is the identity onK1. Therefore, there exists a mapping homotopic tof which transforms S'into the i-dimensional skeleton of S, i.e., into S', but S' C 5' + CConsequently, g:S' — Since ir,(S") = 0 when i < n, any map-pingf€ a ir1(5) is homotopic to the mapping sending the whole of S'to the base point of S" (i.e., constant mapping). This means that the map-ping f : S' — S is homotopic to constant. The mapping f has beenchosen arbitrarily. Therefore, ir1(S) = 0.

If X and Y are cell complexes and a mappingf:X —Y induces theisomorphism of all the homotopy groups, then f is a homotopyequivalence. We take the mapping — * as!. The isomorphism of the

196

homotopy groups is induced, since all of them are zero. Therefore, thesphere is homotopy equivalent to a point, and contractible to apoint.

12.30. Letp'(x0) = F0,p'(x1)= Xan embed-ding. Then p F0 x0 e V. Join x0 to x1 with a path, i.e., arrange fora homotopy between the mappings of. the fibre F0 into x0 and x1, viz.,

F0 — Y, 1'1(F0) = y(t), where y is our path. Then it follows from thecovering homotopy axiom that there exists a covering homotopy (familyof mappings :F0 — X) such that (p• = i.e., (p= y(l) = x1, from which it follows that C F1. Thus, we haveconstructed the F0 — F1 by means of the path y. We nowprove that depends only on the homotopy class of the path -y, i.e., ify1 is homotopic to then is homotopic to Note that the con-structed mapping F0 —. F1 does not depend on the choice of a coveringhomotopy in the sense that any two such mappings are homotopic. Infact, let ta1 and cover Then the mapping : F0 — F1 is homotopic to

F0 — F0, = whereas the latter is homotopic to : F0 — F1.Now, let the family of paths y, be given. We shall show that is

homotopic to We have the mapping

x I— X; X 1) =

In Y, there exists a homotopy of 'y0 into which can be covered by amapping 4):(F0 x 1) x / — X such that 4"(F0 x o = and

nx = f1 with (p f)(F0 x I) = Therefore, the mappingf1can be taken as a covering map for all 'Yl = Then 4) '(F0 x 1) x Ja homotopy between and Note furthermore that the mapping

(— y)X : F1 — F0 can be constructed similarly by means of the path (—It remains to prove that the mapping F0 — F0 is homotopicto the identity. But this mapping can be considered as the one induced bythe path -y + (—y) which is, evidently, homotopic to a mapping into apoint.

12.37. Let S" x — k be a cell complex with only four cells, viz., e0,— k e". Consider the mapping f S" — k

— wheref(e0) = * is a certain point in 5t)• We take the latter as the zero-dimensional cell in S".

By the cellular approximation theorem, there exists a mapg: x — k — which is cellular and homotopic to f, the equalityf(e0) = g(e0) being held one0 and the whole homotopy connectingf andg coinciding one0 C f. Since 5" consists of only two cells, viz., the zero-dimensional (*) and n-dimensional, then, under the mapping g, the cellsek and e" — k are transformed into a point on 5". We obtain that the map-ping g may not be equal to a constant only on the n-dimensional cell.

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Therefore, all the mappings x k differ only by a certainmapping of the n-dimensional cell into Sk x S" — k and then into S",which transforms the whole boundary into a point on (due to the path-connectedness of the choice of a point is immaterial). But they aremappings of type — (more exactly, a one-to-one correspondencecan be established between 7r(Sk x s" — k S") and ir(S", Sn))

12.49. Let (x1 x'1) — (x' 0) be the standard embed-ding R'7 — + (in the form of a hyperplane). Consider two pointsA = (0 0, 1) and B(0 0, 1) in Rn + and construct conesCAM and CBM with vertices at the points A and B, respectively, and acommon base H C Then any deformation of the subset R' \ H in Rncan be extended to a deformation of the subset E(R'1 \ F!) in R'1 +

12.50. Assume the contrary, viz., let < I(M'1; G), i.e., thatthere exists a covering of M" by closed sets Xk, k < I (M'1; G)each of which contracts on to a point. Due to Poincaré duality,

G) JJI7 — G), to the cocycles x1 x1 there corre-spond cycles y1 y1 and to the product h = x1 A . . . A x1 (of thecocycles x1 x1) there corresponds the cycle a = y1 fl . . . flwhich is the intersection of all the cyclesy1 y1. Since the Poincaréduality operatorD is an isomorphism, the intersectiony1 fl . . . ñ == a is different from zero (i.e., the cycle a is not homologous to zero).Since each subset X, (1 i k) is contractible on M'1 to a point,H* (Mn; X,) = H* (Mn) (where * > 0). Therefore, the cycle y ce (Ma) can be assumed to be homologous to the cycle ee (M'1; i.e., the carrier of the lies in Mn \ X, (1 / k).Hence, it follows that the intersectiony1 fl . . . fl (homologous to theintersection y1 fl . . . fl Yk) lies in the complement of (the union)

U . .. U Xk; the more so, y1 1) . . fl fl .. flfl y1 C \ (X1 U . . . U Xk) = 0, since X1 Xk forms acovering of Mn. Since the intersection of the carriers of the cycles y1 flfl . . . fl = 0, the corresponding product of the cocyclesx1 A . . . AA x1 = 0, which contradicts the condition that x1 A . . . A x1 0, andthe theorem is thus proved.

12.51. Consider the fibration (E,p, x), whereE is the space of all pathsof the spaceX starting at the pointx0, andp the mapping associating eachpath with its endpoint. The total space E is considered here in thecompact-open topology. The fibre of this fibration is the

space X the point x0. It is easy to see that the space Eis contractible on itself to a point (each path is contractible on itself to thepoint x0). Therefore, lrn(E) = 0, and the homotopy sequence of thisfibration

1(E) L(X) lrn(tlx) . .

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generates the isomorphism + 1(X). In particular,The group is Abelian whenn 2.

12.52. Definition. A space X is said to be contractible if the identitymapping X — X is homotopic to the mapping X X sending all X to apoint.

Definition. A spaceX is said to be 1-connected if ir1(X) = 0.

Since X is contractible, there exist X X, being the identitymapping X — X, and the mapping X — x0 e X. Since the definitionof a fundamental group does not depend on a base point (up to isonior-phism), let I — X be an arbitrary path on X, y(O) = y(l) = x0,b(r) x0, c5 : I —. X. The same homotopy : X — X stipulates that theloops 'y and ô are homotopic. Thus, any two paths on X are homotopic,i.e., ir1(X) = 0.

12.53. We prove that (a) any element from (where is thewedge of circumferences) is representable as the finite product ofelements and where is the class of the mapping(which is the standard embedding); (b) such a representation is unique upto cancelling the factors and 7 placed in a row.

(a) This is equivalent to 7r1 being a free group with the generatorsa A. Consider the mapping f: S1 — Represent each cir-

cumference S' and e as the sum of three one-dimensionalsimplexes P. Q, R and By the simplicial approximationtheorem, the mappingf is homotopic to a simplicial mapping F of a cer-tain subdivision of the complex S1 into Multiply the mapping F onthe right by a homotopy where is the identity mapping,transforms into a base point and stretches to the whole ofWe obtain a mapping P honiotopic to the original. The mapping P eithertransforms each of the equal parts, into which S1 is divided, into a pointor winds it round one of a A. The class of such a mapping in

is the product of elements of the form and e (identityelement of the fundamental group), i.e., the constant mapping class.

(b) The product . = ± I, k I), which has no andtin a row, is not equal to the unit element in i.e., there exist

no relations in Under the covering map lr: T — X, the inverseimage of each point ir(x) = D is found to be in one-to-one cor-respondence with the cosets of the group T1(X) relative to the subgroup

(r1(T)). In particular, ifx1 ,x2 T,x E X, ir(x1) = ir(x2) = x, and Sany path from x1 to x2, then the loop with vertex at the point x is nothomotopic to zero: otherwise, = x2. Let = . where

is a loop traversed in the direction of a circumference of the wedge ac-cording to the sign of Take k + 1 replicas of the wedge, and placethem one over another. We take in the first and second wedges, cutout a line-segment in both replicas, and join their ends "crosswise", while

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extending the projection 71 to them. Similarly, we join the second wedgeto the third by using , etc. If there are two identical letters in the word

one after the other, two line-segments of the same circumferenceshould be cut out. In doing so, the second operation precedes the first if

= 1, and follows it otherwise. We obtain a (k + 1)-sheeted coveringof the path being covered by a path starting at the lower point, andending at the upper. This loop is not homotopic to zero.

12.54. Letf: Y1 — Y2 andg: Y2 — Y1 be two homotopy equivalences,i.e., gf Idy1;fg — Idy2. We define the —

Y1, aCaElr1(Y1),the class of the loopf a S1 — Y2). g,, = (f• g),

7r1(Y2) and :ij-1(Y1) — 711(Y1) are isomor-

phisms, whence ir1(Y1) =12.55. Let 711(X) * 711(Y) be the free product of 711(X) and 711(Y).

Let be two universal coverings of X and Y, respectively. Let x0 be abase point of X, Y and the wedge X V Y. We construct the followingcovering: taking X, we consider p 1(x0), where p : — X is a coveringmap, and glue ? at each point E p — 1(x0) We identify with xb,where x'0 is a certain point from p j 1(x0) andp1 Y a covering map.At each remaining point frompj '(x0), and to each replica of "glued",we glue X in this manner, etc. The projection p ": (X V Y) — X V Y isdefined in a natural manner, viz., each replica of r is mapped into Y viap , and each replica It is obvious that the space obtain-ed is a covering of X V Y. Consider the fundamental group X V Y, pointst1 andt2inXV Ysuchthatt1,t2E andapath&. Undertheprojection p ", this path will be transformed into a certain loop arepresenting the class of a in (X V Y). Note that it follows from theconstruction of the covering map and X and Y being 1-connected that thepath from t1 to t2 is unique up to homotopy.

Let 7r1(X V Y) be decomposed in terms of the generators a€ir1(X) and i.e., a = . . .

BJn. Then thisrepresentation is unique up to the relations in ir1(X) and 711(Y). In fact,let = . . . 67" — I, where 1 is the constant loop at thepointx0 and not all and a5 equal zero (we take a reduced word). Thencan be realized as a path in X V Y which, as it is obvious from the form ofthe covering map, is not closed; therefore, 1. Thus, we have obtain-ed that V Y) = 7r1(X) * 711(Y). The same result follows from thevan Kampen theorem on expressing the fundamental group of a complexin terms of the fundamental groups of its subcomplexes and their intersec-tions.

12.56. Definition. If K is a knot, then the fundamental group711(R3\K) is called the knot group.

Let us find the corepresentation of this group. Consider the upper (or

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lower) corepresentation of the trefoil knot. Let PK be its projection. Thepoints K1 (i = 1, . . . , 6) divide the knot into two alternating classes ofclosed, connected arcs, viz., the class of overpasses and the class of under-passes. Let A1 , A2, A3 be the overpasses, B1 , B2 B3 the underpasses, andF3 the free group with the generators x, y, z. We call a path v in R2 simpleif it is the union of a finite number of closed straight line-segments, itsorigin and endpoint do not belong to PK, and it meets PK in a finitenumber of points which are not the vertices either of PK or v. Weassociate each path v with v * F3: v * = . . . where Xik are thegenerators of the free group, tk = 1 or = — I depending on how vpasses under A,k. The upper corepresentation of the group 1r1 (R3 \ K) isof the form

(x,y,z;r1,r2,r3), (1)

where r1 = v7 are the relations. The upper corepresentation determinedby formula (1) is known to be the corepresentation of ir1(R3\K). Theloops v1, v2, v3 around the overpasses (x, y, z are the generators) satisfythe equalities

v1* =x_IyZy_l, v2* =y_lZxZ_l, v3* =Z_lxyX_I.

We have obtained the corepresentation (x,y, z; x = 1,y = zxzz = 1)• Substitute z = then

ir1(R3\K)= (x,y;x = xyx). (2)

Thus, 'r1(R3\K) = (x, y; xyx = yxy). It is impossible to untie thetrefoil knot, since its type is different from the trivial knot type. If twoknots K' and K" have the same type, then their complementary spacespossess coincident fundamental groups. The group G = (x, y;xyx = yxy) is not the infinite, cyclic group Z. In fact, a homomorphism0 : G — S3 can be constructed, where S3 is generated by the cycles (12) and(23).

Let K' and K" be two connected subcomplexes of a connectedn-dimensional, and simplicial complex K, each simplex from K belongingto at least one of these subcomplexes. Their intersection D = K' fl K"is neither empty nor connected. Let F, F', F", FD be the fundamentalgroups of the complexes K, K', K", D. We take 0 D as the startingpoint of the closed paths. Then each closed path of the complex D is, atthe same time, a path of the complexes K' and K ". We refer here to thewell-known van Kampen theorem. The group F is obtained from the freeproduct F' x F" if each pair of elements of F' and F" corresponding to

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the same element ofF0 are identified, i.e., assuming these elements to beequal, we thereby add relations to the generators of the groups F' and

We now find the fundamental group of the "helical" knot defined asfollows. Draw generators on the lateral surface of a circular cylinder atthe distance of 2ir/m from each other, and then rotate the upper basethrough 2irn/m. Then, identifying the bases, add one point at infinity(oo) and thereby turn R3 into S3. Remove from S3 all the points belongingto the tubular neighbourhood of the knot. We obtain a polyhedron K, thecomplement of the knot. Divide S3 into two parts by the torus which con-tains the "helical" knot. The complex K is then divided into two solid torieach of which has been stripped of the knot tubular neighbourhood onthe surface. We take one solid torus as K', and the other (with the pointat infinity) as K". The fundamental group F' (resp. F") of thepolyhedron K' (resp. K ") is a free group with one generator A (resp. B).The generator A can be represented as the midline of the solid torus of thepolyhedron K (the same be done with B). The intersection D of both thesolid tori is a twisted annulus. The fundamental group D is also free withone generator which we take to be the midline of the annulus. The groupF' ® F" is a free group with the generators A and B. For an appropriateorientation of the paths A and B, the path C considered as an element ofthe group F' equals Am, and as an element of the group F", it equals B".We obtain the relation Am B". Thus, the corepresentation of thegroup \ 'y) is A 2 = B3}, where 'y is the trefoil knot.

The two corepresentations of the fundamental group of the trefoil knotobtained are equivalent. We leave the verification of this proposition tothe reader.

12.58. We choose the point 0 belonging to Was the starting point of theclosed paths. Then each closed path of the complex W is, at the sametime, a path of the complexes Z, Y, i.e., to each element of the group

( W) there correspond an element of the group it1 (X) and an elementof the group ir1(Y). We represent Z, Y, Was simplicial complexes. Joineach vertex of X to 0 with a path. If the vertex lies in W, then the pathmay be drawn in W wholly (because of the connectedness). A simplex ofan arbitrary dimension of the complex X belongs either to Z (but not toY) or Y\ Z or Y fl Z. The set of all sitnplexes is thus divided into threedisjoint subsets Z, Y, W. The generators a, of the group it (X) can be putinto one-to-one correspondence with the edges of the complex X. In ac-cordance with that simplicial complex to which this edge belongs (Z, YorW), we redesignate a into z1 ,y, or respectively. Thus, it1 (X) has as itsgenerators those of the fundamental groups ir1(Y) and ir1(Z) (thegenerators of ir1(W) being included in those of it1(Z) and ir1(Y)). Therelations in the group it1 (X) are in one-to-one correspondence with the

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edges and triangles of the complex X. Since the complex X has been sub-divided into three subsets, these relations also get into three classes. Let uswrite out the relations:

= 1 (inZ), (1)

y,) = I (in y) (2)

= 1 (in W). (3)

Relations (3) are defining relations for the group ( W), relations (2) and(3) for the groups ir 1(W), relations (1) and (3) for the groups711(Z) and ir1(W), and relations (1), (2), (3) for the group 711(X). Theserelations can be rewritten in the following manner:

= 1, = 1, (1')

= 1, = 1, (2')

W. = Wi,-. (3')

Relations (1') and (2') determine the free product of the groups 711(Z)and 7r1 ( Y), and (3') implies that these elements of the groups 71 1(Z) and711(Y) corresponding to the same element w- of the group 7ri ( W) must beidentified. In proof, we have used the fact that W is connected, sinceotherwise the statement derived for the group ir1(X) is incorrect. E.g.,Z = Y = I (a line-segment), W = S0, X = S'. ir1(X) = Z,ir1(Z) = ir1(Y) = e.

12.78. It is known that for any subgroup G C 7r1(X), there exists acovering map p : XG — X such that Im 71* (XG)) = G. Introducemultiplication on XQ. Let ê ep 1(e), where e is the unit element in X,and Joinë andfl with = =

e,91 = 9. = x,andp(j3) = y.Thenxandyarejoinedtoe with the pathscan be multiplied together in X, i.e., we can consider the path

X1 X y1 which joins e to the pointz1 = z = xy. Let; be liftable toin XG, and x 9 = It remains to verify the correctness of the

definition. The following statement can be proved. Let X be a groupoidwith identity, and a, e 7r1(X, e). Then ai3 = a X meaningmultiplication in ir1(X, e) on the left-hand side, and multiplication inXon the right-hand.

We omit the proof leaving it to the reader. The correctness of thedefinition follows from this statement immediately.

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12.79. Whenp > 0 and q > 0, for any n < p + q — 1, the isomor-phism holds: V S4') lrn(SP) + Since the pair

x V is a relative (p + q)-dimensional cell, it followsfrom proposition 1 (see below) that x v S") = 0 whenm <p + q. Therefore,

V = +

if for the triple (X, A , x0), the pair (X, A) is a relative n-dimensional cell,then 7rm(X,A,xQ) = OwhenO < m < n.Theproofislefttothereader.

12.81. It follows from the Freudenthal theorem that thehomomorphism lrm(U, — V) is an isomorphism whenm < 2n, and an epimorphism when m = 2n, where U and V are thenorth and south hemispheres of + We find ir3(D2, ôD2):

— ir 1(3D2). . . —.

When n = 3, we have ir3(aD2) = = 0, = 0, == 0. Hence, ir3(D2, dD2) = 0. From the exact sequence,ir3(S2) = ir3(S2,D2) = Z.

12.82. The proof follows from the exact homotopy sequence of theSerre fibration.

12.83. The proof follows from the cellular representation of the projec-tive space, and from the investigation of the standard covering map.

12.85. Let us prove that ir1 (CPI?) = 0, where is a cell complexhaving one cell in each even dimension, i.e., having no one-dimensionalcells. By the theorem on the fundamental group of a cell complex withone zero-dimensional cell, we obtain that = 0. Further, thesphere + fibres over CP" with the fibre S1. In fact, let + 1

C C" embedding). The point (z1 + i) a + jfand only if = 1. Further, CP" = kz1, . . . , up tomultiplication by X, i.e., X(z1 z,, + i) = (z1, . . . ,z,, + Set themapping — CP", p(z1, . . .) = (z1, . . .). it is continuous,and its image is the whole space CP". Over the point from CP", thefollowing set of points from + "hangs": let (z1 + j)then

— C, — e Zfl÷1;

where is the full inverse image, and 0 In fact,eIW1(ZI z,, + and (z1, . . . ,z,, + i) are the same point in CP",but if then these are distinct points in S2" t Therefore, themapping p Si" + I CP" is a fibre map. It now remains to apply theexact honiotopy sequence of the fibre map.

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12.86. The statements follow from the cellular approximation theorem.12.87. Let : X x Y — Y, Py : X x Y — Y be two projections.

Set the homomorphism : ir,(X x Y) — ir( Y); viz.,

p(a) =

+ i3) = + + 13)) ==

(b) is a monomorphism. Let = 0, i.e., a = 0, a = 0.

Therefore, = : — X is homotopic to a constant mapping,i.e., there exists X such that = = * . Then we

set the homotopy thus: = when t = 0,

'I,(a) = a when t = 1. 4',(a) is a mapping of into(*) x YC Xx x Y) =is a contractible spheroid, i.e., a is contractible.

(c) is an epimorphism. Let 13 e y Then a = (13, y) is

transformed into i3 y under

Let there be two universal coverings: X, Y. Consider themappingp1 xp2:E1 x E2—Xx Y,(p1 X P2) (e1 x == (p1e1 x p2e2). We assert that this is a covering. The proof is left tothe reader.

Lety1 e ii-1(X,x0), Y2 7r1(Y,y0), a1 e x0), a2 y0),and a homotopy along the path 'y1 of the spheroid a1 be given so thatF0(a1) = a1,F1(a1) = y1[a1], Similarly,4(t): c10(a2) = a2,4'1(a2) = C Define ahomotopy along the ioop = (v1 Y2) (t) into X x V of the spheroida (a1 x Then F1(a1) X cF1(a2) == y1[a1] y2][a1 =

y any spheroid from X x Y are of the forms

for certainy1 e ir1(X), E 7r1(Y), a1 e

a2 e then the action of x Y)on lIn(X X Y)has been fullydetermined.

12.89. Use the 1-lopf map — S2. It is arranged as follows:

= 1z112 + 1z212 = 1}eC2,S2 = CP',

S2 = : (Xz1, (Zi,Z2)}.

We obtain the fibre map S3 — S2. The exact sequence may be written forthis fibre map, viz.,

— 7r1(S1) — ir1(S3) — — ir1 —

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From the properties of the sequence, -ic(S3) = r(S2) when i 3. Bythe Freudenthal theorem, the homomorphism 1) ir.(S'1) isan epimorphism when i 2n — 2, and an isomorphism when1 < 2n — 2, i.e., the homomorphisms

— —

are isomorphisms. Therefore, -ir3(S3) = Z. Since ic1(S3) =(1 3), r3(S2) = Z.

15Simplest Variational Problems

15.1. It suffices to consider the Euler equations for the action func-tional and write these equations in local coordinates. In doing so, use ex-plicit formulae for the Christoffel symbols.

15.2. The Euler equations for both functionals should be written out,and then the behaviour of the functionals when changing the parameter(i.e., time) on the extremal solutions considered. The required statementfollows from the length functional being invariant under parameterchanges and the action functional being not invariant.

15.3. The proof is reduced to the direct computation: the Euler equa-tion in Cartesian coordinates should be written out, and the explicit for-mulae for the mean curvature used (the mean curvature is calculated forthe graph of a smooth function).

15.4. The proof is similar to that in Problem 15.3. This analogy is basedon the codimension of the graph of the function being equal to unity inboth problems. Therefore, the mean curvature tensor is given by onefunction only (by the mean curvature itself).

15.5. Use the classical inequality

dr)

15.6. Square the integrands and compare them.15.7. Use local Beltrami-Laplace equation theory and write the equa-

tion in a curvilinear coordinate system. It follows from the theory thatconformal coordinates can be always locally introduced on a two-dimensional surface (given by analytic functions), adding the condi-tion that the mean curvature equals zero transforms these coordinates in-to harmonic.

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15.8. For example, the function i(u, v) = (u, v, — v2).15.9. (c) The calculations performed in point (b) are of local character,

which enables us to carry them out in a neighbourhood of each point onthe Kahler manifold. On the contrary, Stokes' formula is valid for anysmooth manifold (recall that the Kähler exterior 2-form is closed).

15.10. By the implicit function theorem, one of the coordinates, e.g.,can be expressed (on the level surface F0 = const) as a smooth func-

tion of the other coordinates. Substitute this function in the Euler equa-tion for the extremal of the functional J.

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