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10. 11. CoiloBbeB,
A. T.
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A. C.
10. 11. CoiloBbeB,
A. T.
CBOPHI4K 110 JU4NEPEHLU4AJThHOfl ITEOMETPI4H
H Touonor'1114 MocKoscKoro YHMBepcwreTa MocxBa
A.5.Mishch.nko. Yu.P.
and AT. Fominko
Probisms in
DifF.r.ntiat G.omstry and Topology Translated from the Russian by
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 the content, translation and design of this book. We would also be pleased to receive any other suggestions you may wish to make. Our address is: Mir Publishers, 2 Pervy Pizhsky Pereulok, 1110, 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 in differential geometry and topology, given at the mechanicsandmathematics department of Moscow State University.
contains problems practically for all sections of the seminar course. Although certain textbooks and books of problems indicated in the bibliography list were used in preparation of this volume, a considcrable number of the problems were prepared for this book expressly. The material is distributed over the sections as in textbook [3]. Some problems, however, touch upon topics It
outside the lectures. In these cases, the corresponding sections are supplied with additional definitions and explanations. In conclusion, the authors express their sincere gratitude to 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. tIomotopy 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
Application of Linear Algebra to Geometry 1.1.
Prove that a vector set a1
ak in
a Euclidean space is linearly
independent if and only if det O(ai,
0.
1.2. Find the relation between a complex matrix A and the real matrix rA 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 and A
B, A ® B. Consider the cases of det and Tr. 1.5. Prove the formula det e4 = 1.6.
Prove that =
for B]
+
+ C' [A, BIC"
a convenient choice of the matrices C' and C", where [A, = AB
BA.
1.7. Prove that if A is a skewsymmetric matrix, then matrix.
is an orthogonal
1.8. Prove that if A is a skewhermitian matrix, then
is a unitary
matrix. 1.9. Prove that if [A, A*] = 0, then the matrix A is similar to a diagonal one.
1.10. Prove that a unitary matrix is similar to a diagonal one with eigenvalues whose moduli equal unity. 1.11. Prove that a hermitian matrix is similar to a diagonal one with real eigenvalues.
1.12. Prove that a skewhermitian matrix is similar to a diagonal one with imaginary eigenvalues. 7
Let A
1.13.
be
((aijO
a matrix of a quadratic form, and Dk
= det Prove that A is positive definite if and only if for all k, I k the inequalities Dk > 0 are valid. 1.14. With the notation of the previous problem, prove that a matrix A 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((
((A ((
1.16.
/'Ek
\0
((
((
.
—F
(I
B
((B
Prove that if A2 =
then the matrix A is similar to the matrix
\I,k4I=n.
0
—E,/
Prove that if A2 —E, then the order of the matrix A (2n x 2n), and it is similar to a matrix of the form
1.17.
is
(
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 class of nonsingular quadratic forms does not alter the signature of the form. 1.20. Prove that varying continuously a quadratic form from the class
of quadratic forms with constant rank does not alter its signature. 1.21. Prove that any motion of the Euclidean plane R2 can be resolved
into a composition of a translation, reflection in a straight line, and rotation about a point. 1.22. Prove that any motion of the Euclidean space R3 can be resolved
into a composition of a translation, reflection in a plane and rotation about a straight line. 1.23. Generalize Problems 1.21 and 1.22 for the case of the Euclidean space
8
2
Systems of Coordinates A set of numbers q', q2,..., qfl determining the position of a point
is called its curvilinear coordinates. The relation between in the space the Cartesian coordinates of this point and curvilinear X2 coordinates is expressed by the equalities X5
= Xs(q', q2,
, .
qfl),
(1)
.
or, in vector form, by r
= r(q',
qfl),
q2
where r is a radius vector. Functions (1) are assumed to be continuous in their domain and to have continuous partial derivatives up to the third order inclusive. They must be uniquely solvable with respect to qfl;
q2
this condition is equivalent to the requirement that the
Jacobian (2)
should not be equal to zero. The numeration of the coordinates is assumed to be chosen so that the Jacobian is positive. Transformation (I) determines n families of the coordinate hypersurfaces qr = The coordinate hypersurfaces of one and the same
family do not intersect each other if condition (2) is fulfilled. Owing to condition (2), any n — I coordinate hyperplanes which belong to different families meet in a certain curve. They are called coordinate curves or coordinate lines. The vectors rk =
are
directed as the tangents to the coordinate
lines. They determine the infinitesimal vector dr in
rkdq"
a neighbourhood of the point M(q', q2, . .. , q"). The square of its
length, if expressed in terms of curvilinear coordinates, can be found from the equality
r,dq',
ds2 = (dr, dr) =
where (,)
is
k1
rkdci)
the scalar product defined in 9
=
th15 dqk
The quantities dinate system.
=
=
rk) define a metric in the adopted coor
An orthogonal curvilinear coordinate system is one for which
(0 s= k The quantities are called the Lamé coefficients. They are equal to the moduli of the vectors I
=
/dxi'Y
fax2\2
+ ...
+
=
+
The square of the linear element in orthogonal curvilinear coordinates is given by the expression + ds2 = + ... + of transition from Cartesian
2.1. Calculate the Jacobian J =
to orthogonal curvilinear coordinates (q', coordinates (x1, . , qfl) in the space q2 2.2. Calculate the gradient grad f of the function f: R3 R in an orthogonal curvilinear coordinate system. 2.3. Calculate the divergence div a of a vector a E R3 in an orthogonal curvilinear coordinate system. 2.4. Find the expression for the Laplace operator Lxf of the function f: R3 R in an orthogonal curvilinear coordinate system. 2.5. Cylindrical coordinates in R3 .
.
q1=r,
q3=z
are 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, are
q2=O,
q3 = 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
q2 =
X,
=Z
q3
/L,
defined by the formulae
are
y=
x
l)(l
—
z
=
a scale factor. (a) Find the coordinate surfaces of elliptic coordinates. (b) Compute the Lamé coefficients.
where c is
2.8. Parabolic coordinates in R3
qt =
X,
=
q2
=
q3
Z
are related to Cartesian by the formulae =
x
Ni.
X2)
—
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
a
+X
2
X
+
b +X
+
a
Z
+
2
c +X = 1 (—c2 >
+ b
1
2
(X >
2
—c)
(ellipsoid),
> —b2) (hyperboloid of
c
one sheet),
a2
+v
+
Z
+
b2+i'
c
two sheets). Only one set of values X, The parameters =
X,
q2 =
/L,
2
+v v
q3 =
=
I
(— b2
>
>
a2)
(hyperboloid of
corresponds to each point (x, y, z) E R3. P
called ellipsoidal coordinates. (a) Express Cartesian coordinates x,
are
coordinates X,
jz,
y,
z in terms of ellipsoidal
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 prolate ellipsoid 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 formulae x = c cosha sini3 c cosha sing
z= (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 formulae c sin/3 c sinha c sinha cosha — cos$
cosha —
cosha — cosfl
where c is a scale factor, 0 a< —ir < —ir < j3 (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,
ir.
q3=z
are related to Cartesian coordinates x, y, z by the formulae a sinha a cosha —
—, cosI3
y=
cosha —
z=z,
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
x=
—
c
cosa
y=
is a constant factor, 0
c sina
c sinhfl ,
a < j3, — 12
0).
Prove, on the basis of this relation, that the point moves along a curve of the second order. 4.23. A material particle moves under the action of a central force F = Fr°. It follows from the result of Problem 4.18 that the motion takes place in a certain plane. Form the equation of the motion and the differential equation of the path in polar coordinates. Consider the case F
4.24. The motion of an electron in a constant magnetic field determined ,by the following differential equation = [r' x H], H = const. Prove that the path is a helix. 18
is
4.25. Find the curves determined by the differential equation
r' Find the curves determined by the differential equation
4.26.
r' =
[e
x [r Xe]],
where e is a constant unit vector. 4.27. Find the curves determined by the differential equation
r' =
a=
where
+
ae
x rJ,
[e
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 following curves:
(I) r =
[a cost, b sint) (ellipse); _!—),
(2) r
(hyperbola);
—
+
=
[a cos3 t, a sin3 t) (astroid); — sint), a(l — cost)[ (cycloid);
(3) r
(4) r = {a(t (5) r =
(1
1
1
2
3
2
at the point t = 0; (6) r =
a,ø
3
(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 the point of contact of the tangent to the tractrix a
and
—
equals a. 4.35. Prove
r=
that the cardioids 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 semiaxis sum. 4.42. A beam of parallel rays falls on a spherical mirror. Find the envelope 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 this
segment.
4.47. Find the envelope of the rays reflected from a circumference if the 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= (3) y =
a(1
+ m)sinmt
—
amsin(l + m)t (epicycloid);
cosh(x/a) (catenary curve); (4) x2y2 = (a2 — y2)(b + y)2 (conchoidal curve); a
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 (4)
x=
a(2cost
cos2l),
y=
a(2sint
t
cost) for
—
sin2t).
I
4.50. Find the curvature of the following curves given in polar coordinates:
(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 = (2) y = (3)
(4)
y= y=
a
cosh(x/a);
x312
mx;
(5) r = a(1 + cosp); (6) r = (a(t — sint), a(l — (7)
r=
cost) I; {a(cost + I sin!), a(sint — t cost)); (2cost + cos2r),
(8) r
(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 parametrization
of the curve can be given in the form
x=
Scosa(s)ds,
y=
4.55. Form the natural equations of the curves:
(I) x = (2) y =
a
cos3l, y
a sin3!;
x312•
(3) y = x2;
(4) y = lnx; (5) y = a cosh(x/a); (6) y = (7)
x=
(8) r = (9) x =
a
(In tan
+
y=
a
sin!;
+ cos4,);
a(1
sin!), y = a(sint — I cost). the parametric equations of the curves if their natural
a(cost + t
4.56. Find
equations are given (here R = 1/k):
(I) R =
as;
(2) —i +
R2
= 1;
(3) Rs = a2; (4) R = a + s2/a; (5)
+ 9R2 = 16a2;
(6) S2 + R2
(7) R2 = (8) R2 +
16a2;
2as; a2
= a2e  2s  a
4.57. Let p be the distance from the origin of radii vectors to the tangent
to a curve 'y at a point M, and r the distance from the point 0 to the point M. Prove that
k=
rdr
0. 4.58. At a certain point of a curve r = r(s), we have: k 0, Having taken the equation of the osculating circumference in the form
22
prove that the osculating circumference — ro — Rono)2 = intersects the given curve in a neighbourhood of the indicated point. 4.59. Given that the following conditions are fulfilled at a certain point of 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 a sufficiently small neighbourhood of this point. 4.60. Given an equation R = f(a), where R is the curvature radius of a curve, and a the angle from a constant vector a to the tangent vector r to the curve, form the parametric equations of the curve. 4.61. Given an equation a = f(R) (see the previous problem), form the parametric equations of the curve. 4.62. Given an equation s = f(a), where s is an arc and a the angle from a constant vector a to the tangent vector r to the curve, form the parametric equations of the curve. 4.63. Given an equation a = f(s) (see the previous problem), form the parametric 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 of the reflected rays (caustic). 4.65. What form will the equation of the caustic of a plane curve with respect to the origin of the radii vectors have if the equation of the curve is 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 equation of 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) and where it is given by an equation y = f(x). 4.67. Write the equation of the tangent line and the normal plane of the 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 point A(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
{au + b, cu + d,
r has
u2j
the same osculating plane at all points.
4.72. Form the equations of the osculating plane, principal normal, and
binormal of the curve
x2=z
y2=x, at
the point (I, 1, 1). 4.73. Given a helix r
lacost, asint, bi),
form the equations of the tangent, normal plane, binormal,
plane,
osculating
and principal normal.
4.74. Given a curve
112,
r
1
—
t
form the equations of the tangent, normal plane, binormal, osculating
plane and principal normal at the point t 1. 4.75. Form the equations of the tangent line and the normal plane of the curve given by the intersection of two surfaces F1(x, y, z) = 0 4.76.
and
F2(x, y, z) = 0.
The curve in which a sphere meets a circular cylinder, whose base
radius 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 in implicit and parametric forms. Find the equations of the tangent, normal plane, binormal, principal normal and osculating plane. 4.77. Find the length of the arc of the helix 3a sint, z = 4at x = 3a cost, y from
the point of intersection with the plane xOy to an arbitrary point
M(t). 4.78.
Find the length of one turn between the two points of intersection
with the plane xOz of the curve
x=
a(t
—
sint),
y=
a(l
cost),
—
z=
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
4a
cost/2.
4.81. Reparametrize the helix r by
= {a cost, a sift, bt),
b > 0,
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,
of the Frenet frame for the helix
v,
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 and torsion of this curve. 4.86. Find the vectors r, v, of the Frenet frame, curvature, and torsion of a Viviani curve (see Problem 4.76). 4.87. Find the curvature and torsion of the following curves:
(1) r = {t
—
sint, 1 — cost,
= te', e',
(2) r
(3) r = e'sinf, etcost, e' 1; (4) r = 12t, int, t2J;
3t + t3); (5) r = (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 in the positive direction of the principal normal. Find the equation of the osculating plane of the curve described by the 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 and torsion of the curve given by equations y = y(x) and z = z(x) and find the Frenet frame for this curve.
4.91. Find the curves intersecting the rectilinear generators of the hyperbolic paraboloid xy = az at right angles. 4.92. A curve on a sphere that intersects all the meridians of the sphere at a given angle is called a loxodrome. Find the equation of a loxodrome and the vectors r, v, $ of the Frenet frame for this curve at an arbitrary point. Calculate its curvature and torsion. 4.93. Given a curve
r=
(v cos u, v sin u, kv
v = v(u), prove that this curve is placed on a cone. Define the function v(u) so that this curve intersects the generators of the cone at a constant angle 0. 0 is given at each point of a curve 4.94. The tangent vector T = T(t) r = r(t). The function r(t) is defined, continuous, and has a continuous derivative r' (t) on a segment [a, bJ. The function T(t) is continuous on the segment [a, bJ. Prove that this curve can be parametrized so that where
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 C at the point M intersects the curve C. 4.96. Prove that if all osculating planes of a curve are concurrent, then the 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'" at r'. Calculate the limit some point M(t) with r' urn
d
d is the distance from the point M(t + itt) to the osculating plane of the curve C at the point M. Consider the special case where the curve is given by an equation r = r(s) (s being the natural parameter). 4.98. Find a necessary and sufficient condition for the given family of curves where
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 helix r = (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 point M 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+x2
ds k
and conversely, if this relation is fulfilled, then the principal normals of the curve form a constant angle with the direction of some vector. Find this vector. 4.104. Prove that if all normal planes of a line contain a vector e, then this line is either straight or plane. 4.105. Prove that if all the osculating planes of a curve which is not a 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 may be nonregular. Since r(s)I = 1, the image r(s) lies on the sphere with radius I and the origin at its centre. This curve is called the tangent spheri
cal image of the curve s —p v(s)
r = r(s). The normal spherical image and the binormal spherical image s fl(s) may be defined
similarly. 27
4.111 Find the tangent, normal and binormal spherical images of the helix
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) degenerates 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)
= k.
Prove that ds
(b) Find necessary and sufficient conditions for the tangent spherical image to be a regular curve. 4.114. Les be the length along the normal (resp. binormal) spherical image 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 parallel
to the tangent to the binormal spherical image at the corresponding points. 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 plane passing 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 tangent spherical image is an arc of a circumference. By definition, a spherical curve is a curve r = r(t) for which there exists a 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 each
normal 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 every normal plane. r(s) be a curve which is parametrized by the natural 4.120. Let r 0, x parameter, k 0, and 1/k, a = 1/n. Assume that + 'a)2 = a2 = coust, a > 0. Prove that the image of the curve r r(s) lies on a sphere of radius a. 4.121. Prove that if r = r(s) is a curve which is parametrized by the natural parameter, k 0, x 0, then r(s) lies on a sphere if and only if
fk'\'
/(orxQ= \
x —=1 k \xk2J
I
4.122.
r=
—I——
Using the results of the previous problems, prove that a curve
lies on a sphere if and only if there exist constants A and B such that r(s)
+
i.
4.123. Two curves r = rj(1) and r = rz(() are said to form a pair of Bertrand 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 the same plane form a pair of Bertrand curves. (b) Let
1(1
r1(t) = — 2
(cost
2
(cost
r2(t) =
—
—
t\/1
—
I
—
—
t2, 0
r,
1
—
12
+
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, 0. Prove that the curve r = r2(s) (s is not the natural parameter and kx of F2(S)) which forms a pair of the Bertrand curves with ri(s) exists if and only if there are constants X and such that
IA =
k +
29
x 0. Prove that 4.127. Let r = r(t) be a regular curve of class C3, r(t) is a circular helix if and only if r(t) possesses at least two different curves which are related in the sense of Bertrand. Let m be a constant vector, r r(s) a curve, c(s) = r(s) — mj2, and a 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 at
the endpoint of m if C(So)
= a2, C'(So) = C"(So)
cU+1)(so)
.
. .
(so) = 0,
=
0
0, calculate the first three derivatives of the 4.128. Given that k k and x. function c(s) in terms of r, v, 4.129. Prove that a curve r = r(s) possesses a spherical contact of order where 2 at a points = if and only if m r(so) + v(so)/k(so) + X is an arbitrary number. 0, prove that a curve r = r(s) possesses a 4.130. Given that x(SO) spherical contact of order 3 if and only if I k'(so)
m = r(so) +
— k(so)
v(So)
— k 2(So)x(So)
4.131. Let a curve r = r(s) be of constant curvature. Prove that the osculating sphere and the circumference have the same radius. a], be a plane piecewise regular curve of class C2 Let r r(s), s parametrized by the natural parameter. The number
k is the curvature of the curve, s(O
points, r (si) = lim i(s), r4 (se) = s—.si—
between the vectors i —
(se)
I
urn
s__.si+
n
—
are the singular
1)
i(s), and
is
the angle
and r + (s,), is called the rotation number er(s)
of the curve.
4.132. Compute the rotation number of the curve y represented in Fig. 1.
4.133. Compute the rotation numbers of the curves given by the following equations (the parametrization is not natural): < t (1) r = (a + cost, sint), 0 (2)
(3)
r= r=
(a + Q cost,
sint), 0
cos2t, —Q sin2tj, 0
30
t
t
0 < e < lal;
>
0;
(4) r =
0
(5) r
(2cosf, — sintj, 0
(6) r =
El,
sin2tJ, 0
t
t
t
/
Arcs of cirdurn
I
ferences
of 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 k has a relative maximum or minimum. Let r(s) be an oval and P a point on r(s). Then there exists a point
P' such that the tangent r to the oval at this point is opposite to the tangent at the point P, i.e., r(P') — r(P). The tangents at the points P and P' are parallel. Thus, for a given point P. there exists a unique point P' (said to be opposite of P) on the oval, so that the tangents at P and P' are parallel and distinct. The width w(s) of an oval at the point P = r(s) is the distance between
the 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 is independent of the choice of P 4.135*. Prove that any oval possesses at least four vertices. (This statement is known as the fourvertex theorem.) 4.136*. Prove that if r(s) is an oval of constant width w, then its length equals 7rw. 4.137*. Let r
r(s) be an oval of constant width. Prove that the straight line joining a pair of opposite points P and P' of the oval is orthogonal
to the tangents at the points P and P'. 4.138*. Given that r = r(s) is an oval,prove that r" is parallel to r at least at four points. 31
4.139. Prove that the notion of vertex does not depend on the choice of parametrization. 4.140. Show that the fourvertex theorem (see Problem 4.135) is not valid if the requirement of closedness is omitted. 4.141. Let r1 [0, aJ —÷ R2 be a segment of a curve parametrized by the 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 that the 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 sum of the curvature radii I/k is constant at opposite points and does not depend on the choice of the points. 4.143. (a) Let r(s) be an oval of length L and with natural parametrization. Denote the angle between the horizontal and tangent vector r(s) by 0. Prove that the mapping 0 : [0, LI [0, 2irI is a parametrization of the oval r(s). (b) Let be an oval parametrized by a parameter 8 so that r(s) = = Q(O(s)). Prove = Q(O(s) + ir).
that the point which is opposite to r(s) is R(s) =
(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 the oval 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 = —— + k(O)
do2
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 tangent spherical image cannot lie in any open hemisphere. 32
Prove that the tangent spherical image of a regular closed curve cannot lie in any closed hemisphere except for the case when it is a great circumference bounding the hemisphere. Let 'y be a closed C'curve on the unit sphere S2. Prove that the 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 semicircumferences.
the image C
is
not the union of two great
Using the results of Problems 4.1464.148, prove the following statement: the total curvature of a closed space curve y is not less than 2,r and equal to 2,r if and only if is a plane convex curve (Fenchel theorem).
hR for Let be a space closed curve. Assume that 0 k a certain real number R > 0. Prove that the length 1 of the curve y satisfies 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 into account? Let be an oriented great circumference on the sphere S2. Then there
exists on S2 a unique point w associated with w, viz., the pole of the hemisphere which is on the left when moving along w in the positive direction (Fig. 2).
Fig. 2
Conversely, every point of S2 is related to a certain orientable great circumference. Thus, the set of oriented great circumferences is in oneto
one correspondence with the points of The measure of the set of oriented great circumferences is the measure of the corresponding set of points in S2. If w E S2. then w1 denotes the great oriented circumference associated with w. 3—2018
33
For a regular curve
with the spherical image C, we denote the number
of points in Cl)
(which may be infinite) by Note that the does not depend on a parametrization of the curve y.
number 4.152!' Let C be the image on S2 of a regular curve 'y of length 1. Prove
that the measure of the set of oriented great circumferences which intersect C (taking the multiplicities into account) equals 4/. In other words,
41 (the Crofton formula). 4.153!' A closed simple curve y is said to be unknotted if there exists a onetoone 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 total curvature is greater than or equal to 4.154!' Using the Crofton formula, prove that for any closed, regular 2ir. curve, Skds
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) (s being the natural parameter) placed on the sphere S2 equals zero. 4.157!'LetMbe a surface in R3 such that 0 for all closed curves
placed 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.
5
Surfaces 5.1.
Make up a parametric equation of the cylinder for which the curve
Q(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 origin = Q(u) is directing.
of the radius vector for which the curve 34
5.3. Make up a parametric equation of the surface formed by the tangents to a given curve = Q(u). Such a surface is called a developable surface. 5.4. A circumference of radius a moves so that its centre is on a given curve = 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 parametric equation of the surface described by the circumference. 5.5. A plane curve x = revolves about the axis Oz. Make z = up parametric equations of the surface of revolution. Consider the special case 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 of revolution.
5.7. A straight line moves translationally with a constant velocity while intersecting another straight line at right angles and uniformly rotating about it. Make up the equation of the surface which is described by the moving straight line (right helicoid). 5.8. Make up the equation of the surface formed by the principal normals of a helix. 5.9. Make up the equation of the surface formed by the family of normals to a given curve 5.10. A straight line moves so that the point M where it meets a given circumference moves along it, the straight line remaining in the plane normal to the circumference at responding point and rotating through an angle equal to the angle MOM0 through which the point was turned while moving along the circumference. Make up the equation of the sur
face described by the moving straight line assuming that the original position of the moving straight line was the axis Ox and the circumference
is given by two equations x2 + y2 a2, z = 0. 5.11. Given two curves r = r(u) and Q(v). Make up the equation of the surface described by the middle point of the line segment whose extremities lie on the given curves (translation surface). 5.12. Make up the equation of the surface formed by the rotation of the catenary line y = a coshx/a about the axis Ox. This surface is called a catenoid. 5.13. Make up the equation of the surface formed by the rotation of the 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 =
+ 4 = I,
z=
0
0,
z=
h, and the directing curve —j +
a
(i.e., ellipse) are given,
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 which are parallel to a plane. A cylindroid can be determined by two directing curves (lying on it) and a director plane (the generators of the cylindroid being parallel to it). Make up the equation of a cylindroid if its generators are 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
a=
= Q(u) is a vector function determining a certain curve, and
a(u) a vector function determining the distribution of the rectilinear
generators of the surface, is said to be ruled. Make up the equation of a ruled surface whose generators are parallel to the plane y — z = 0 and intersect two parabolas y2 = 2px, z 0 and z2 = — 2px, y = 0. 5.18. Make up the equation of the ruled surface whose generators intersect the axis Oz, arc parallel to the plane xOy, and intersect the line = b2. xyz = a3, x2 + 5.19. Make up the equation of the ruled surface whose generators intersect the straight line r = a + ub, curve = Q(v), and are perpendicular to a vector n. 5.20. Make up the equation of a ruled surface whose generators are parallel to the plane xOy and intersect two ellipses y2
z2
b2
c2
x=a;
y2
z2
c2
b2
x=—a.
5.21. Make up the equation of a ruled formed by the straight lines 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 lines parallel to the plane x + y + z = 0, intersecting the axis Oz, and circumference = I b, a cos u, a sin u j.
5.23. Make up parametric equations of the surface formed by the straight lines intersecting the circumference x2 + = 1, y = 0 and straight lines y = 1, z = and x = 1, z = 0. 5.24. Make up the equation of the surface formed by the tangents to the helix = (acosv, asinv, bvj (developable helicoid). 5.25. Make up the equation of the conic surface with the vertex at the point (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. The curve moves uniformly in the plane so that each of its points travels parallel to AB. The plane ir is, at the same time, in uniform rotation about AB. Make up the equation of the surface described by the curve 1
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 a straight 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 circumference
with centre on the curve r = r(u) and given radius a, a >
0,
ak
= tx(u), Q(u)cosço,
(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) =
j/
x(u) = ± I aln
\
—u 2 —
u
a2
—
U,
u
2
,
a
>
0
(pseudosphere);
(iv) find the mean curvature H at an arbitrary point of the surface of revolution;
(v) select the function Q = 0 on the whole surface.
Q(x)
for the special case x =
H=
u
so that
5.91. Given a curve with the natural parameter u, curvature = k(u) 0, and torsion x = x(u) 0. Let r = i(u) be the unit tangent vector of this curve. Find (a) K, (b) H for the surface of tangents
k=
r(u,
v) = Q(u) + vr(u), V > 0.
5.92. Find the expression for the total curvature of the surface whose first 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 equation F(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 z
y = 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= z=
v3
— 3v — 3u2v,
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 usinv, f(v), x = ucosv, y z where
f(v) is an arbitrary analytic function of variable v, have unlike
signs.
5.102. Find the total and mean curvatures of the surface formed by the binormals of a given curve. 5.103. Find the total and mean curvatures of the surface formed by the principal normals of a given curve. 5.104. Let S be a certain given surface. Mark off segments of the same length and direction on the normals to the surface S. The ends of these segments describe a surface S* "parallel" to the surface S. If the equation of 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 in terms of the coefficients of the first and second funthe surface damental forms of the surface S. 5.105. Express the total curvature K* of the surface S* "parallel" to a surface S in terms of the total and mean curvatures of the surface S. "parallel" to 5.106. Express the mean curvature H* of the surface a 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" to
a surface S if for the surface S the ratio H/K = const. 5.108. Given a surface of constant mean curvature H. Segments of length 1/2 Hare marked off on all its normals. Prove that the total curvature 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 normals of a surface with constant positive total curvature K. Prove that the mean
curvature of the surface so formed is constant. Calculate it. 5.110. Prove that the total and mean curvatures at the corresponding points of two parallel surfaces are related by the formula H2 — 4K
—
K2

H*2
4K*
—
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 the differential equation dv2
—dudv
du2
E
F
G
L
M
N
=0.
5.111. Find the lines of curvature on the surface uv a b x=—(u—v), y=—(u+v), z=—. 5.112.
x=
Find the lines of curvature of the helicoid ucosv,
y=
usinv,
z=
av.
5.113. Prove that, in covering (local isometry) the catenoid
y=
x = Vu2 + a2cosv, z
= aln(u +
a2)
with the helicoid
x=
ucosv.
y=
usinv,
z=
av,
lines of curvature are transformed into asymptotic lines. 5.114. Find the lines of curvature of the surface
the
r(u, v) = Q(u) + f(v)a + g(v) [r(u) x where
r(u) = r'(u),
i(u)I
= 1, (r(u), a) =
a], 0, al
= I, a is a constant
vector.
= 5.115. A plane curve y is given by an equation where u is the natural parameter, k = k(u) its curvature (0 < k < 1/a), p the principal normal unit vector of 'y, e the unit normal vector to the plane in 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 = Q(u) + av(u)cosçe' +
r(u,
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, curvature k(u) < 1/a, and torsion x(u). 5.117. Find the lines of curvature of the surface r(u, v) =
[u(3v2
J_), v(3u2
u2
—1).
— v2
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 L{I(X, X) + I(Y, Yfl,
H
where 1(,) is the second fundamental form of the surface.
5.120. Assume that the first fundamental form of a surface is as follows:
= E du2 + Gdv2.
ds2
Prove that
K
( /3E\\ av
1
=
5.121.
/3G
a!
au
+— 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 angle
between the normals of M1 and M2. Prove that k2sin20 =
+
— 2XlX2cosO.
Two directions in the tangent plane of a surface which are determined by 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,
(ai, a2), b = (b1, b2). A net of lines on a surface is said to be conjugate if the tangent vectors to the lines of different families of this net are conjugate at each point. where a
48
A direction determined by a vector h is said to be asymptotic if 0. A line on a surface is said to be asymptotic if the tangent has the asymptotic direction at each of its points. An asymptotic line is
h) =
= 0 which is held at all its points. Asymptotic
acterized by the equality
lines are determined by the differential equation 2Mdudv + Ndv2 = 0. Ldu2
5.122. Find the asymptotic lines of the surface
/x \y
y
x
5.123. Find the asymptotic lines of the surface z = 5.124. Find the asymptotic lines of the surface
x=u
2
xy2.
2 y=u +uv, z=u +—uv. 2
3
Construct the projections of the asymptotic lines, passing through the point 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 x=—(u+v), y=—(u—v), z=—uv
are straight lines. Find the lines of curvature. 5.130. Show that the coordinate lines on the surface z = fz(u) + P2(V) x = fi(u), y = çoi(v), plane and form a conjugate system. 5.131. Show that the coordinate lines on the surface
are
r= are
M(u) + Q(v)
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
2018
49
5.133. Prove that the product of the normal curvature radii for a pair of conjugate directions attains its minimum for the lines of curvature. 5.134. Prove that the ratio of the principal curvature radii is constant for the surface of revolution obtained by rotating a parabola about its
directrix.
5.135. Prove that if one of the lines of curvature of a developable surface lies on a sphere, then all the remaining nonrectilinear lines of curvature lie on concentric spheres. 5.136. Prove that the normal curvature of an orthogonal trajectory of the 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 plane forms a constant angle with the tangent plane to the surface. We call a line whose principal normal coincides with the normal to a surface at each of its points a geodesic line of the surface. There is a unique geodesic line passing through each point of the surface and having a given direction. The length of the projection of the curvature vector kn onto the tangent
plane of a surface is called the geodesic curvature kg of a line placed on the surface. The geodesic torsion associated with a given direction is the torsion of the geodesic line passing in this direction. 5.138. Prove that a geodesic line on a surface can be fully determined by one of the following properties: (i) The normal to a surface at each point of the line, where its curvature
is other than zero, is a principal normal. (ii) The normal to a surface lies in the osculating plane of the line at each 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, where
N is the normal vector of the surface. 5.142. Prove that geodesic lines of' the plane are straight lines and only they.
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 + 50
= z2.
5.146. Find the geodesic lines of the helicoid
usinv,
r
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 meridian meets 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 if it is plane. 5.153. Prove that a geodesic line is asymptotic if and only if it is straight.
5.154. Prove that the geodesic curvature of a line u = u(s), v = r(u, v) can be calculated by the formula
v(s)
on a surface r = kg
m is the unit normal vector of the surface. 5.155. Find the geodesic curvature of the helical lines of the helicoid
where
r=
(ucosv, usinv,
5.156. Prove that the geodesic torsion of a line u = u(s), v = r(u, v) can be calculated by the formula on the surface r
v(s)
= 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 torsion should equal zero at each of its points. 5.158. Show that on a surface with the first fundamental form (du2 + dv2) ds2 = [ç,(u) + (a Liouville surface) the geodesic lines are determined by the equation du '.Ico(u) + a
±
dv —
a
=0,
where a is an arbitrary constant. 51
5.159. Given a triangle Twhose area is a and the sides are arcs of great
circumferences on a sphere of radius R0, find the sum of the interior angles of the triangle T 5.160. Let T be a triangle whose sides are geodesic lines constructed on a surface with constant Gaussian curvature K = — a2 < 0. Assuming that 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 a 2
X2
portion of the ellipsoid —i +
2
+ —i =
1
determined by the
inequalities x > 0, y > 0, z > 0, find the area 0* of the spherical image of the surface S. 5.162. Given that a surface R R(u, v), u1 < u I
(the endpoints are the same, and the homotopy is fixed endpoint). 12.18. Prove that any connected, cellular complex is homotopy equivalent to a cellular complex with one vertex. ' can be represented as the union 12.19. Prove that the sphere x
1)
x
U
with
Sr x 12.20. Consider the standard sphere and two spheres embedded in it: fXr+i — .
.
.=
the
the Euclidean space R'
= tXi =
=
boundary
common
.
.
=
= ol.
can be joined by Prove that any pair of points y E xE a unique arc on a great circle having no other points of intersection with these spheres. 12.21. Find the topological type of the closed hyperboloid of one sheet = 1) in the projective space Re3. F = 1x2 + — 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 two connected components. 12.23. Prove that the space of polynomials of the third degree without multiple roots is homotopy equivalent to the complement of a trifolium in the sphere Construct an explicit deformation. with pairwise different coordinates. 12.24. Consider the set of points Show that the space obtained has the same type as the EilenbergMacLane complex K(ir, 1). 12.25. Construct an example of two spaces X1, X2, which are not homotopy equivalent, and also of two continuous onetoone mappings X1 such that the spaces themselves may not be X2, g: X2 f: X1 homotopy equivalent. 88
12.26. Let h : X —
be a continuous mapping, and the corre
spondence 4) {X',  fX, YJ be defined by the formula 4)(a) a h. Prove that the correspondence 4) transforms homotopic mappings into homotopic. 12.27. Prove that the following homotopy equivalence holds: 'V sm± 1 x Se') :
12.28. Prove that the finitedimensional sphere is contractible to a point on itself. 12.29. Prove that a connected finite graph is homotopy equivalent to the 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 pathconnected subspace A. Prove that the space X is pathconnected.
12.32. Fix two points Xo and x, in a space X. Let Y be the space of paths starting at Xo and passing through x,. Prove that the space Y is contractible.
12.33. Prove that the space of all paths tX; with X held fixed. X; 12.34. Let a sequence of spaces X,, C with held fixed. be contractible to X
Prove that the space X
U
,
be
given, and let
i
is contractible to Xo with X0 held fixed.
12.35. Prove that any open ndimensional manifold is homotopy equi1)dimensional complex. valent to an (n 12.36. Prove that if a space Xis contractible to a subspace A on itself with 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.
and
Find
and Cat2(RP")
where
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 2dimensional 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 nonhomogeneous coorin Find the homotopy type of dinates X, where
(a)
=
+. .+ .
=
89
= 1,
=. .= .
(b) RP" \
—l=
where
O,Xk+2 '=
...
M
IC
= tx2i + ...
I — •..
I
=
(c) SIC and
12.42. Consider a small ball in the open manifold x Sn_k and glue the projective space in its place, i.e., identify points x and —x = S"
on the boundary of the ball that
Prove
the
space
obtained
is
homotopy
equivalent
to
v whose boundary is a topologibeing contractible to a point
12.43. Given a topological manifold the boundary of cal manifold
in the manifold M". (a) Prove that the manifold is contractible to a point. (b) Prove that if the manifold P" 'is 1connected, then the manifold is homeomorphic to the disk D" (assuming that P" is contractible to a point in Ma). (c) Give an example of a pair (M'1, pn I) such that the manifold p" is contractible to a point in the manifold but is not homeornorAs a corollary, prove that iri(P" 0. phic to the disk I
12.44. Find the homotopy type of the space C"
\
= U
where
U
12.45.
Calculate the number of mappings (up to homotopy): (a') CP" CP"; (b') CF"'' —p CP"; (c')
(a) RP" RP";
(b)
(c) ERP" (d)
l; (d') EcP" +
—*
12.46. Prove that
(a) Cat [join(X, Y)] = min[Cat(X), Cat(Y)1, where Cat is the LusternikSchnirelmann category (the spaces X and Y being connected). (b) Find Cat(S1 x S2). 12.47.
Let
spaces
X,,
I
N,
1
X=XIXX2X...XXN. Prove
that [Cat(X1)]
Cat(X)
1
+
be
pathconnected, [Cat(X1)
and
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 , Vq, where q n, then tliere always exists at least connected) V,, V2, . .
one set V1 such that it contains two diametrically opposite points of the
sphere s", viz., —x and x. 90
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
homotopy
following
equivalence
holds
Reminder. Let X be a topological space, and X x I its direct product by a linesegment. After identifying the "upper base" X x (Ij C X x I
with a point and the "lower base" X x
C X x I with another
point, we obtain a factor space called the suspension of X. 12.50. The relation between the LusternikSchnirelmann category and "cuplong" of a compex (or manifold). Let M' be a smooth, compact, connected, and closed manifold. Consider the ring H*(Mfl; G), where G = Z if M' is orientable, and G = Z2 if M' is nonorientable. Denote by l(M'; G) the greatest integer for which there exists a sequence of x, of the ring H*(M"; G) elements XL, X2 > 0, 1 a I) A X2 A . . A XI(M";Q) such that 0 in the ring G). The number G) is denoted by cuplong (Ma). Prove that l(M'; G). Cat(M') .
12.51. Prove that for any pathconnected topological space X and any of its points x0, the group iri((IX, xo) is Abelian. 12.52. Prove that any contractible space is 1connected. 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 isomorphisms hold: iri(X) iri(Y) and lrk(X) 2. lrk(Y), k 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 no homeomorphism of Euclidean space (or sphere) onto itself which would transform the trefoil knot into the standardly embedded, unknotted circumference, 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 twodimensional standardly embedded torus T2 C R3, on which it traverses its parallel p 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 knot
I'; where p = 2, q = 3.) Make out the role of the condition for the numbers p and q to be prime to one another. 12.58. Let X = Y where Y, Z, W are finite CWcomplexes, W = V fl Z, W is pathconnected, and X = Y the common subset W. Calculate the group iri(X), given the groups ?rL(Y), and 7r1(W). Consider the case where W is disconnected separately.
91
12.59. Given an arbitrary group G with a finite number of generators and relations, prove that there exists a finite complex X whose fundamen
tal group is isomorphic to G. Can a finitedimensional manifold, e.g., fourdimensional. be selected as such a complex X? 12.60. Calculate
the group iri(X), where X is the wedge of three
circumferences.
12.61. Construct a twodimensional complex X whose fundamental
group equals Z/pZ. For which values of p can a twodimensional, smooth, closed, and compact manifold be selected as such a complex? 12.62. Calculate the fundamental group of the twodimensional sphere
with three handles. Check this group on commutativity and find its commutator subgroup. Calculate the fundamental group of the twodimensional torus. 12.63. Let a simplicial complex X have N onedimensional simplexes.
Prove that its fundamental group has no mare than N generators. 12.64. Prove that iri(X) = 7r1(X2), where X is a CWcomplex and X2 its twodimensional skeleton, i.e., the union of all cells of dimensions 1 and 2. where X = S' V S2. Is this group finitely generated? 12.66. Find the knot group of the figureofeight (i.e., wedge of two 12.65. Find
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 pathconnected space. Prove that the group 7r1(X, y) for any two points x, y E X. xo) is isomorphic to the group V S". where X is the wedge and 12.69. Calculate 12.70. Prove that if X is a onedimensional CWcomplex, then 7ri(X) is a free group. Z 12.71. Prove that the group C Z Z Z cannot be the fundamental group of any threedimensional manifold. 12.72. Calculate iri(Pg), where Pg is a twodimensional, compact, closed, and orientable surface of genus g. 12.73. Calculate iri(TPg), where TPg is the manifold of linear elements of a surface of genus g. 12.74.
Calculate the fundamental group of the Klein bottle by
constructing the covering space with the action of a discrete group. 12.75. Let P be a twodimensional surface with nonempty boundary (i.e., open surface). Prove that irt(P) is a free group. 12.76. Prove that if X is a CWcomplex, then is a group whose generators are onedimensional cells, and the whole set of relations is determined by the boundaries of twodimensional cells. 92
12.77. Let G be a topological groupoid with identity. Prove that G is homotopically 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 C a subgroup. Prove that (a) it is possible to introduce the operation of multiplication in X (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 times
k tunes
12.80. Prove that the groups ir(X) are commutative when i > 1 for any CWcomplex X. 12.81. Demonstrate by way of example that the excision axiom does not 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 pathconnected space Yand any point xo E Y, the isomorphism holds Xo) 1(tL0, V. where is the constant 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 CWcomplexes) 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; 93
(b) I: X V
Y
—'
X x Y is an embedding, then the exact sequence is
given rise: 7rg(X
V Y)
12.85.
Tg(X X 1')
Prove that
7r1(CP'1)
0.
=
0;
=
1r2
(CP")
=
Z,
n > 0;
Irk = (CP") = Irk(S2" + '), k 2. 12.86. Prove that if a CWcomplex X has no cells of dimensions from to k inclusive, then iri(X) = 0 when i k. 1
12.87. Let X, Y be two CWcomplexes. 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 = Z, where is a sphere.
(0
12.89. Prove that IrE(S3) = ir'(S2) when I
q
n) and prove that
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 = is the division ring of a + '), where B nontrivial O(n + 1)fibration of disks on S'. 12.93. If a mapping f(X, A) —' (} B) sets up the isomorphisms Irg(Y) and lrg(A) Ir8(B) for all g, then it sets up the isomorphisms irg(X, A) wg(X, B) for all g. 12.94. Calculate the groups Ir,, where is the real Stiefel manifold. Irg(X)
12.95. Prove that the groups cannot become trivial, as k increases, 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 onedimensional relative homotopy group of the pair (CP2, S2), where S2 CP' C CP2 standardly. 12.99. Prove that if: (a) a threedimensional, compact, and closed manifold M3 is 1connected, then M is homotopy equivalent to the sphere (i.e., M3 is a homotopy sphere); 94
(b) M" is a smooth compact and closed manifold such that
=
0
then W is homotopy equivalent to the sphere
when i
12.100. Construct an example of a threedimensional, closed, and compact manifold M3 such that M3 is a homology sphere (i.e., it has the same integral homology as S3), but iri(M3) 0. Construct an example of a finitely generated group G which coincides with its first commutator subgroup. 12.101. Prove that the set of homotopy classes of mappings ISa, Xi is 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 figureofeight embedded into the twodimensional plane. 2n + I. 12.103. Calculate ir1(CP") when i = 0, and a finite group G act on X and Y without 12.104. Let fixed points. Prove that there exists, and is unique up to homotopy, a mapping f: Y X which commutes with the action of the group G. where [X, Y] is the set of 12.105. Prove that [CP2, S2l = homotopy classes of mappings of X into Y.
12.106. Let (X, A), X D A, be a pair of topological spaces, and X pathconnected. Let A be the set of paths in the space X starting at a certain 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 to nconnectedness:
(a) iro(S'1, X) consists of one element when q ving maps); 5q
any continuous mapping continuous mapping of the disk (h)
12.108. Prove that iro(X,
n (basepoint preser
X
can be extended to any X, q n. an Abelian group, where X, Z are
is
two topological spaces, and fIX is the loop space. Prove that fIX is an Hspace.
I, for any point 12.109. Let A be a retract of X. Prove that when n the homomorphism induced by the embedding
xo E A,
Xo)
Xo)
is
a monomorphism, and when
ii
2,
decomposition into the direct sum irn(X, xo)
xo)
A, xo). 95
determines the following
X) is an Abelian group. Establish a relation 12.110. Prove that with iro(Z, [lOX). S" be the standard tangent bundle over the sphere 12.111. Let Calculate 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 homomorphism. 12.113. Let Y
xo)
yo)
is
a
group
F0 3
i
X3 be
yo)
xo
a fibration witn iixed points Xo,
yo
and fibre F. Prove that
F0;
ir,,(X, xo).
12.114. Let E, X be two topological spaces, X pathconnected, and p: E —' X a continuous mapping such that for any points x E X, yE
'(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 the group structure). Prove that for any points x1 and the topological spaces p '(Xi) and 1(X2) are weakly homotopy equivalent. 12.115. Prove that for the homotopy groups of a pair (X, A) the exact sequence is valid
ir(A)
ir(X)
A)
i(A) *
12.116. Prove that if X is a smooth, compact and closed submanifold of codimension one in Euclidean space, then X is orientable. 12.117. Prove that if the fundamental group of a compact closed manifold is trivial, then the manifold is orientable. Prove that if a manifold X is nonorientable, then there is a subgroup of index two in 12.118. Prove that if X is a nonorientable space, then the suspension is not a manifold.
12.119. Prove that the Euler characteristic X(X) of any compact, closed manifold is trivial. 12.120. Give examples of
(a) a nonorientable manifold doubly embedded in another manifold (whose dimension is one greater); (b) an orientable manifold singly embedded in another manifold. 96
12.12 1. Let X1 and X2 be two tori (meaning the solids), f: 0X1
a diffeomorphism, and
oX2
= 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 the mapping iri(0Xi)
Z
i.e.,
Z
which is induced by the diffeomorphism between the solid tort X1 and
is given by the integral matrix
X2. It is obvious that the
(a
b
\c
d
that this matrix is unimodular and calculate the fundamental group of the manifold in terms of the matrix Prove
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 CWcomplex is homotopy equivalent to a manifold with boundary.
13
Covering Maps, Fibre Spaces, Riemann Surfaces 13.1. Let p: X—' V be a covering map such that [In (X, xo)1 is a normal subgroup of the group ini(Y, yo), p(xo) yo. Prove that each element of the covering, i.e., a E ( Y, Yo) generates a homeomorphism
pcc(x)
p(x).
Let p X
yo. Prove that V be a covering map, 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 13.2.
irt(X, xo)
each element a E ini(X) is determined by a homeomorphism of the space
Y onto itself, a: V
Y,
and the
7—20l8
97
is
commutative.
13.5. Letp : X
Y be a covering map where the space Xis connected, E F. E 1',
and let F = p'(yo) be the inverse image of a point Prove that F and Xo)
Yo) are in onetoone correspondence if irt(X,
= 0.
13.6. Let p : X function, where j2
pf(l) = F(t,
be a covering map, F: j2 —+ Y a continuous a square, and f: P —÷ X also continuous, with
Y is
0).
Prove that f can be extended to the mapping G: j2
G(t, 0) = f(t). 13.7. Let p : X g(0). Let pf(1) that f(l) g(l).
f(O)
X, with pG
Y be a covering map, j, g two paths on X, and pg(1), and the paths pf and pg homotopic. Prove
13.8. Let p : X Y be a covering map, f, g two paths on X, and g(l)? g(0). Does it follow from pf(l) = pg(1) that f(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 path g in X such that pg f. 13.11. Prove that a covering map is a Serre fibre map. 13.12. Prove that any twosheeted covering is regular. What purely algebraic fact corresponds to this statement?
f(0) =
13.13. Prove that a threesheeted covering of a pretzel (i.e., sphere with two handles) is nonregular. 13.14. Let M2 be a nonorientable, compact, smooth and closed manifold. Prove the existence of a twosheeted covering map p : M2+ —p M2, where M2+ is an orientable manifold, and find M2÷ in explicit form. What is the property of the fundamental group of a nonorientable manifold? 13.15. Construct the covering map S'1 RP'1 and prove that:
(a) RP" is orientable when n = n = 2k; (b) irt(RP'1)
2k— 1,
ir(S'1) when
Z2,
and nonorientable when >
,>
13.16. Prove that a covering space is regular if and only if its paths lying over one path in a basis are either all closed or all nonclosed. 13.17. Let p : X . X be a covering map. Prove that any path in X can be covered in X in a unique way up to the choice of the origin of the path in the inverse image, and the multiplicity of the projection p is the same at all points of the base space. 13.18. Construct all coverings over the circumference and prove that in(S1)
Z,
i
2.
98
Zk
P2, where Construct the regular covering map p: Pk 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'). 13.19.
where P2 is a pretzel, 13.21. Construct a covering map çø : X —p 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 twodimensional closed manifold and ,ri(M2) an infinite group, 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 and find them. Construct an example of two nonequivalent, but homeomorphic 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 with one handle) and Klein bottle (i.e., sphere with two crosscaps) and calculate the homotopy groups of P1 and N2. Can the torus P1 be a twosheeted
and regular covering of the Klein bottle? If so, find the covering and calculate the image of the group 7r1(P1) in 7r1(N2) under the covering monomorphism. 13.26. Prove that if = 0 or is a simple or finite group of order p 2 (where p is prime), then the manifold LW is orientable. 13.27. Construct the explicit form of seven smooth linearly independent
vector fields on the sphere S7. Use the algebra of octaves (Cayley numbers). Construct the integral curves of these vector fields. 13.28. Prove that if k linear operators A1, . are given in R" such , that = —E and + = 0 (for all i, j), then k linearly independent vector fields can be specified on the sphere C R". 13.29. If the homotopy groups of the base space and fibre of a fibre space have finite rank, then the homotopy groups of the total space also have finite rank, the rank of the qdimensional group of the total space .
.
—
not exceeding the sum of the ranks of the qdimensional homotopy groups of the base space and fibre. 13.30. Let the fibre map p E B admit an image set of a section x : B —* E, and eo = x(bo). Prove that when n I, the mapping is an epimorphism, and when n 2, it determines a decomposition into the direct sum b0) eo) = 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 finite and their orders dO not exceed the product of the orders of the homotopy
groups of the base space and fibre of the same dimension. 13.32. Prove that the mapping p EX topy axiom (Serre fibre map). 13.33. Prove that Xi,2 = lm(p2),. then (X1,
Xsatisfies the covering homo
are covering maps and 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 any connected complex X such that iri(X) 0 (i.e., existence of a universal covering).
13.35. Prove that the set of vector bundles with a structural group G over the sphere is isomorphic to ir,,_ j(G), and that the group G is pathconnected. 13.36. Show by way of example that there exists no exact homology sequence of a fibration.
13.37. Let p E plexes. Then 13.38.
(
B be a locally trivial fibre map, and B, F finite com
x(fl.
Given that a material particle moves with constant (in modulus)
velocity (a) on the torus 7", (b) sphere S", find phase space for this system. B be a fibre map with pathconnected B and F. 13.39. Let p E Let Cat = Cat — be a reduced LusternikSchnirelmann i.e., + Cat(B) + Cat (of a point) = 0. Prove that Cat(E) 1
+ CatE(F), where CatE(F) is a relative category of the fibre F respective
to E. 13.40. Prove that if p: X + Yis a Serre fibre map, then p is a mapping "onto". and 13;41. Prove that if p X Y is a Serre fibre map, then p are homotopy equivalent for any y2 E Y. 13.42. Prove that the manifold of linear elements of a manifold M is a fibration with the base space M.
13.43. Prove that a locally trivial fibration (twisted product) is a Serre fibration. 13.44. Prove that the space of paths EX whose starting point is fixed
in 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 its total and unimodular tangent bundles (fibrations) are orientable. 13.46. Prove that the direct product of topological spaces X x Y with the projection onto one of the factors is a Serre fibration. 100
X Y be a covering map, p(xo) = yo, and f g two paths thatf(O) = g(O) = yo, f(1) g(1). LetJg E (irj(X, Xo)) and let be two coverings of these paths. Prove that JO) =
13.47. Let p
J
13.48. Represent
the torus T2 as T2 = (g where g
Consider the following equivalence relations R:
\0
(a) (e"l, (b) (etwl, (c)
Find the space X T2/R and calculate the image C where 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
(b) T"
where T3 is a threedimensional torus; S', where 7" is an ndimensional torus (fibrations are
considered up to homotopy equivalence)? 13.50. Let C = A * B be the free product of arbitrary groups A and B. Prove that for any subgroup M C C, the equality M A1 * B1 * F
holds, where A1 C A, B1 C B and F is a free group. Give a topological proof using covering spaces. 13.51. Let be a 1connected compact Lie group, and a: ® an arbitrary involutive automorphism (i.e., a2 = I). Put = (g E = Prove that = = i.e., any element = gl; V = fg e admits a representation in the form gE
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 Lie group. Put V = fg E Then a(g) = E ® = V and V C is a totally geodesic submanifold. Therefore, V
is a symmetric space. The submanifold V is called Cartan's model of the symmetric space Any symmetric space admits Cartan's model (which is almost always uniquely determined). (a) Prove that the projection p: determines the principal fibration 0 —' 101
V, —'
where p(g) = —4
0.
ga(g
= 0, e E V, and e the identity (b) Let V be Cartan's model, Prove that if a point x E V is conjugate of e along a then the point x is conjugate of e geodesic C V in the group itself. along y in the manifold V C 1333. Prove that a compact, closed manifold M2 with Euler characteristic N can be represented as a (2N + 4)gon such that some of its sides are glued together to yield the word
element in
aIa2..
—1
. aN÷2a1
—I
—1
a2
. .
. aN÷2
2 are the designations of the sides) in traversing , az, . the sides one after another. Prove that the last factor is if and only if M2 is orientable. 1354. Classify compact, closed smooth and connected manifolds M and calculate their fundamental groups in terms of the generators and (where
.
.
relations.
13.55. Prove that any orientable, twodimensional, and compact manifold is determined by a unique invariant, viz., the genus of the manifold.
1336. Prove that any nonorientable, twodimensional and compact manifold can be represented as the connected sum of an orientable manifold and several replicas of projective planes. 13.57. Describe the semigroup of twodimensional manifolds under the connected 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 twodimensional
manifold 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
0
\—1 13.62. Given two odd functions on the sphere S2, prove that they have a common zero. 13.63. Let ir be the fundamental group of a twodimensional surface,
and f:
an epimorphism. Prove that f is an isomorphism. 13.64. Prove by three totally different techniques that there exists no 102
continuous vector field without singular points (i.e., different from zero
at each of its points) on the sphere S2. 13.65. Let in C2(z, w) the Riemann surface of the algebraic function w= be given, where the polynomial has no multiple roots. Prove that this Riemann surface turns, after completing it with a point at infinity, into a twodimensional, smooth, compact, and orientable manifold. 13.66. Find the genus of a twodimensional manifold described in the
previous problem as a function in the degree n of the polynomial 13.67. Can the twodimensional projective plane RI'2 be the Riemann w(z) in C2(z, w) in the sense surface of a certain algebraic function w of Problem 13.65, i.e., after completing the Riemann surface with a point at infinity? 13.68. Prove that the Riemann surface of an algebraic function in C2 is always an orientable manifold. 13.69. Investigate what happens to the Riemann surface of the function when some roots of the polynomial w= merge to yield a multiple root. For example,_what is the structure of the Riemann surface
of the function w = — 13.70. Describe the topological structure of the Riemann surfaces of the following analytic functions:
z=w+1/w, 13.71. Prove that the Riemann surface of the function w = lnz is homeomorphic to a finite part of the complex plane. 13.72. Let p X V be a twosheeted covering. Prove that any path in Y can then be covered by precisely two paths. 13.73. Construct a universal covering space for the orthogonal group SO(n).
13.74. Prove that any twodimensional, closed, oriented and smooth manifold can be locally isometrically covered by the Lobachevski plane (which is supplied with the standard metric of constant negative curvature). In other words, prove that the fundamental group of a surface of the indicated form can be represented as a discrete subgroup (operating effectively) of the Lobachevski plane isometry group. Corollary. A twodimensional, compact, closed, orientable, and smooth manifold can be supplied with the Riemannian metric of constant negative curvature. 13.75. What spaces can cover the Klein bottle? 13.76. Let Sg be a sphere with g handles. What Sh can cover 103
13.77. Prove that for any compact, nonorientable, and twodimensional manifold, there is precisely one compact, twodimensional, and orientable manifold which serves as its twosheeted covering. 13.78. Prove that the Beltrami surface (i.e., surface of constant negative
curvature standardly embedded in R3) can be infinitelysheeted and locally isometrically covered by a certain region lying in the Lobachevski plane.
Find this region. Prove that it is homeomorphic to the twodimensional disk. Find the corresponding group of this covering (it is the group Z). 13.79. Can a twodimensional torus be a twosheeted covering of the Klein bottle? 13.80. Calculate the permutation group of the sheets of the Riemann surface of the algebraic function w = arising in traversing around the branch point of this function (point 0). 13.81. Let M2 be an ellipsoid, and p one of its vertices. Consider all geodesics emanating from the point p. Find the locus of the first conjugate points (i.e., mark the first conjugate point of p on each geodesic and describe this set). 13.82. Prove that the fundamental group of a complete Riemannian manifold of nonpositive curvature contains no elements of finite order. Prove that (M) (where Mis a complete Riemannian manifold of strictly negative curvature) possesses the following property: if two elements commute (ab = ba, a, b E irj(M)), then a and b belong to the same cyclic subgroup.
13.83. Prove that a closed, orientable Riemannian manifold of strictly positive curvature and even dimension is 1connected. 13.84. (a) Prove that any compact, closed Riemannian manifold of constant curvature X is isometric either to the sphere S" or (of radius Use Problem 13.83. (b) Let be a compact, closed, 1connected, complete Riemannian manifold, and C(I) the set of the first conjugate points of a certain point 1 E
Prove that if is a symmetric space, then the complement Mn/C(1) homeomorphic to the open disk. 13.85. Prove that a complete, noncompact Riemannian manifold of positive curvature and dimension m, where either m 2 or m 5, is diffeomorphic to Rm. 13.86. Let x, y be two near points on the standard sphere S2, and a function f(z) the area of the geodesic triangle with vertices at the points x, y, z. (a) Is the function f(z) harmonic on the sphere S2? is
104
(b) Investigate the case of the ndimensional sphere (where f(z) is the volume of the geodesic simplex whose one face is fixed, and z is a free vertex).
(c) Investigate the same problem for the Lo 'achevski plane. 13.87. Prove that if M" is a complete, lccinnected Riemannian manifold such that n is odd and there exists a point p on the set of the first conjugate points of p being regular and each point of constant order k, then k n — I, M" is homeomorphic to the sphere (order of a point is understood to be its multiplicity). 13.88. Let y C R2 be a simple closed curve of length I bounding a region G of area S (on the plane). Prove that ,2 4irS and that the equality holds if and only if 'y is a circumference.
13.89. Let 'y C R2 be a closed curve (not necessarily simple, i.e., in contrast with the previous problem, selfintersecting is possible). Prove that ,2
w(x)dS, where the function
47r
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, transparent, isotropic, but nonhomogeneous substance filling the twodimensional 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) f(z) = z", = I.
14.2. Calculate the degree of the mappingf:
S' —
where
S2, wheref(z) =
zE CU Lool.
be an orientable, smooth, and compact manifold. Prove 14.3. Let 5" is fully determined by that the homotopy class of a mapping the degree of the mapping. 52n be a continuous mapping. Prove that there is 14.4. Let f: a 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 that there 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 number
of points (meaning the difference between the numbers of positively oriented and negatively oriented points); (b) if g are homotopic, then the difference between the number of positively oriented and negatively oriented points of the inverse image is the same for the two mappings; (c) if the difference between the number of positively oriented and negatively 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 surface in R3. Prove that f(w) = K(w') where and are elements of area and K the Gaussian curvature. Prove that 2 degf = the Euler characteristic of the surface.
and also equals
14.S. Prove that any continuous mapping of the ball D" into itself always 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 ndimensional (n < p) manifold in R". Let v(f) be the normal bundle of this immersion, and Sv(j) the associated fibre is the boundary of a certain bundle of spheres, i.e., Sv(J) = sufficiently small tubular neighbourhood of the submanifold be a usual Gauss (spherical) f(M") C R". Let T: Sv(J)
mapping.
Find deg T (dimSv(J) = p — 1) if the Euler characteristic of the manifold M" is known. Does deg T depend on the method of immersing M" What happens if M" is nonorientable? Separately consider the into case where p = ii + 1. 14.11. Given that a twodimensional, orientable, closed, and compact manifold M2 of genus g is embeddable in the Euclidean space R3, find the 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 the height function. 14.12. Prove that nondegenerate critical points of a smooth function on a smooth manifold are isolated. 14.13. Letf(x) be a function on a twodimensional, compact, orientable 106
surface of genus g (i.e., sphere with g handles) having a finite number of critical points, all of them being nondegenerate. Prove that the number of minima minus the number of saddle points plus the number of maxima equals 2g — 2. —i R be a smooth function on a smooth manifold. 14.14. Let f: Prove that almost every value of the function f is regular. 14.15. Prove that the alternating sum of singular (critical) points of a smooth function f(x) (assuming that all its singularities are nondegenerate) given on a smooth compact manifold does not depend on the function (by the alternating sum, we understand
where
n=
dimM,
X the index of a critical point, and mx the number of the critical points of index X). 14.16. Let f(x) be a complex analytic function of one variable x. Prove that the set of critical values of the function f(x) S2 has measure zero.
14.17. Let contains no critical = lxf(x) = c}. Prove that if points of the function f, then is a submanifold in and =
1.
14.18. Prove that the notion of nondegenerate critical point of a smooth function does not depend on the choice of the local chart containing 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 orthogonal to the axis of rotation of T2 has only nondegenerate critical points. 14.20. (a) Construct functions with only nondegenerate critical points on 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 nondegenerate critical points of index X. 14.21. Let F(x, y) be a nondegenerate bilinear form on R". Consider a smooth functionf(x) = F(x, x), where x = 1, i.e., F(x, x) is a function C R". Let Xo on the sphere S" 1 be all the eigenvalues 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 sphere
Find all the critical points of the function F(x, x). Prove that = inf maxf(x) S'
where
5' are the standard idimensional equators of
xES'
the sphere
107
14.22. Prove that if a point p is a nondegenerate critical point for a smooth function f(x) on a smooth manifold, then there exists a local coordinate system in which the function f(x) in a neighbourhood of the point p can be represented as a nonsingular quadratic form. is a noncritical level for a function f(x) on 14.23. Prove that if a manifold M(i.e., the level hypersurfacef(x) c critical points for f(x)), then the neighbourhood
const not containing is diffeomorphic to
x I. 14.24. If and are consecutive critical levels, then the interval >< I, where Ci < C < C2. between them is diffeomorphic to 14.25. If there are no critical levels between (i.e., level and hypersurfacesf(x) = const with critical points) and and are noncritical either, then they are diffeomorphic.
14.26. (a) Construct a smooth function f(x) having one point of maximum, one point of minimum (both being nondegenerate), and ano
ther critical point, perhaps, degenerate, on every compact, orientable, twodimensional, and smooth manifold M2. Find the relation between such a function and the representation of M2 as the Riemann surface of a certain manyvalued analytic function. Investigate the case of a nonorientable twodimensional manifold M2 (e.g., case of the projective plane RP2).
(b) Construct a smooth function f(x) having only nondegenerate critical points, precisely one point of maximum, precisely one point of minimum and s saddle points (find the number s) on every compact manifold M2. Construct the function so that it takes the same value at all the saddle points. Investigate the nonorientable case. Indicate the relation to the problem of point (a) and Construct the confluence of all the saddle
points into one degenerate critical point.
15
Simplest 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 trajectories on M", 0 1, and is the velocity vector of the curve 'y(t)) are t geodesics. 108
15.2. Establish the relation between the extremals of the length functional L[y] =
Prove that any
and action functional E['y] =
extremal 'yo(t) of E[yJ is that of LFy]. Prove that if so(t) is an extremal of then by replacing the parameter t = t(r) by so(t), this trajectory can be transformed into an extremal of E[y]. 15.3.
be a functional associating each
Let S(f) =
smooth function z = f(x, y) which is defined on a bounded region D = 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 the extremality of a function fo relative to a functional S is equivalent to the condition H = 0, where H is the mean curvature of the graph of z = y) considered as a twodimensional, smooth submanifold in R3 15.4. Prove the statement formulated in the previous problem for the in case of the (n — 1)dimensional graphs off =f(x1, . , 15.5. Prove that the action functional E[yl and length functional L ['yl E[yJ, the equality being held if and are related by the formula only if y(t) is a geodesic. .
.
15.6. Prove that the areal functional SV] =
(where ,'
—
F2dudv
1'(u, v) is a radius vector in R3 depending smoothly on (u, v))
E± G
and Dirichlet's functional D[?[
dudv are related by the
D
D[f']. formula S[,1 15.7. Remember that the radius vector f'(u, v) determining a twodimensional surface M2 in Euclidean threedimensional space is said to v) is an extremal of Dirichlet's functional DVI = be harmonic if
=+
+ G)dudv.
Prove that if the mean curvature H of a
D
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 on the surface so that the radius vector harmonic. 109
q) in these coordinates becomes
15.8. Construct an example of a harmonic radius vector r..(u, v) such that the surface M2 C R3 described by it may not be minimal (i.e., so
that H
0).
15.9. The Wirtinger Inequality. Let H be a hermitian symmetric positive definite form in C" R2" a realification of C". Then and HR =
H
S (\A
where
SJ
H=
S
S and A are
+ IA,
real
= —A, If T = H. matrices and 5T = 5, The form S defines the Euclidean scalar product in R2" and the form A an exterior 2form in B2". For simplicity, we may assume that
dzk. Consider the form 0(2r) =
A
I
.
r!
.
. Aw, r
W2r is an arbitrary orthonormal basis in R2" (a) If WI tive to the scalar product S ReH, then
1
and
W2,)l
if and only if the plane L(wi W2r is
1,
and Wi,
= S(wi,
W2)
W2)
=
+ 1A(wi, W2) =
Hence, H(Wi, w2) =
A(wi,
Then H(Wi, W2) =
i.e., 5(Iwi, W2) =
1,
=
an orthonormal pair of vectors. It is I, where = A(Wl, W2).
be
required to prove that W(Wl, W2) Consider
H = (WI, W2) = (S + 1A)(wi,
C" rela
W2r) generated by the vectors C".
a complex subspace in R2"
Hint: Let r =
n.
W2).
W2) 1.
W2 = I. Now, let A(wi, W2) = 1. Since = lW = 1, it follows
i.e., the twodimensional plane spanned by WI, W2 is the complex straight line. For r > I, the relation — W2) = should be used, where gjj is the skewsymmetric scalar product defined by the 2form that
(b) Let Wt C C", r < n (r being complex dimension) be a complex submanifold in C" (if W' is an algebraic submanifold, then singular 110
are possible). Let V2' be a real submanifold in
points on
that VU W =
where
a real (2r
such
+ 1)dimensional
whose boundary is V U W. Let K =
submanifold in V012r(V \ K)
is
V
fl W. Then
V012r(W/K).
Note. This statement means that complex submanifolds W in the complex space C" are minimal submanifolds, i.e., after any "perturbation" of V, the 2rdimensional volume (V0l2r) does not decrease. Hint: The statement follows from the Wirtinger inequality (see above) and Stokes' formula. In fact, consider the exterior 2form A dzk,
k=l and let
=
wA .
.
(r times).
.
r!
=
= 0,
Since
0.
It follows from Stokes' formula that
=
While integrating the form
with respect to a 2rdimensional subman
ifold, the expression of the sort
U2r)dX A
.
.
.A
an orthonormal basis in the tangent plane to the submanifold (with respect to the Riemannian metric induced by the underlying Euclidean metric in C" = R2") should be considered (in local coordinates x' x2'). If the submanifold W is complex, then where
is
W2r)
=
I
and
vol2r(W') =
If the submanifold V is of general form (i.e., real), then 1,
i.e.,
vol(I/), which proves the statement.
111
(c) Prove that the statement of problem (b) remains valid if C" is replaced by any Kähler manifold, i.e., complex manifold supplied with an exterior 2form (nondegenerate and closed). 15.10. Consider functions of the form F(x'
x") and the functional J [F] =
f) on R"(x'
where D is the domain of
the functions F. Let F0 be an extremal of the functional J. Prove that the level surfaces F0 (x' x") = const considered as hypersurfaces x") are locally minimal. in R"(x'
Answers and Hints 2
Systems of Coordinates
2.1. J =
J11H2
.
.
(1 of (H1 Oq'
2.2. grad! =
2.3.
.
diva =
Of
1
— —, H2 aq2
Of
1
H3 0q3
iP__(H2H3ai) + H1H2H3
+
j
Oq
2.4. 4f
= L
+
,
where
a1, a2, a3 are the coordinates of the vector a. H2H3
0 (HIH2
Oq'J
= 1,112 =
2.6.
H3H1
0q2\ H2
+ 9q2J
Of 0q3
2.5. (a) The coordinate surfaces = const and planes z = const.
planes
+
111H2H3 LOq'\
H3
=
+ 3q2
r,H3
0 / Ou\ ——(r—J I r Or\ OrJ 1
are: cylinders r
= const, planes
1.
1
r
02u
02u
+ —. 0z2
(a) The coordinate surfaces are: concentric spheres r = = const and cones 0 = const.
(b) H1 = 8—2018
1,
112 =
r,
H3 =
rsin0. 113
const,
O/23u\
I
———tr —) + Or \
r2
r2 sin0 30
\
30 J
02u
I
I
r2
sin20
+
Or /
0/.
Ou\ —lsin0—I +
1
2.7. (a) The coordinate surfaces are: cylinders of elliptic section and y = 0 when X = const, the family of confocal foci at the points x hyperbolic 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= const.
alLel to the axis z when X = 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
=— 2
2
where R(s) =
=
X)
R20..)
R2(v)
s=
a2)(s 114
v.
_________________ =
(c)
— v)R(X)
— v)(fL — v)
(X —
+ (p
ax \
L
3X
/
+ (X —
+
]
2.10. (a) The coordinate surfaces are: prolate ellipsoids of revolution const, hyperboloids of revolution of two sheets j3 = const and planes = const.
a=
(b) H1 =
c
112
r
i
=
(c)
1
(
I
c2(sinh2a + sin2f3) Lsinha &x \
7.
3
1
sinha sinl3.
/
äu\
1
+———— Isin13—I + sinj3 313 \. \sinh2a 313/ 1
sinha
+
/
+ sin213J —I\ 32u] —I 1
= 0.
2.11. (a) The coordinate surfaces are: oblate ellipsoids of revolution const, hyperboloids of revolution of one sheet 13 const and planes = const passing through the axis z.
a=
(b) H1 =
H2
—
1
(c)
=
c2(cosh2a — sin2f3)
sin, i
[cosha 3a
/
2.12.
c913J
+
c
=
=
\sinh a,'
the sphere 13 = const, I
(z
ccotf3)2 +
and the plane
2
=
2
in 13
= const; 115
äa /
\
32u
I
cosh2a/
(a) The coordinate surfaces are: the tori a 2
—
1
1
+ ——tsin13—1 + \sin2j3 I— — 313 \
C cosha sin 13.
[1 3/cosha
3u\
I
H3 =
const,
+
/
13
(c)
=
ôa
\Cosha
/ +—( 13/3 \coshcr — 2.13. H1 =
—J + 3aJ
cos(3
äu\ —1+ 13/3 /
sinha
13
0u\
sinha
(cosha
a
H2 =
t32u
I
cosha — cosfl
cos$)sinha
—
, H3
2.14. (a) The coordinate surfaces are: spindleshaped surfaces of revolution a = cOflst
(Q — ccota)2
4
z2 =
/ (C
2
'\Sifla
spheres í3 = const,
+ (z —
c
= cosha
and planes (b) H1
(si:hfi)2' C
=
(b)
cot$)2 =
cosha —
—
cosf3
= const. C
H2
cosh/3 — cosa'
/
sina
i3u\
+
+
—)
— 1'
H2
+
sina(cosh/3
— /32
2.15. H1 =
coshf3 — C05a
/
sina = 3at\coshl3 — cosa 13
(c)
c sinha
14=
cosa)
fL2 —
H3 116
—/3 2'
=
—
2.16. H1 = 2.17. (a) H1
H3 = cXv.
H2
= H2
H3 = X
= const,
= const of
revolution 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. Let the functions x = x(p, q) be realanalytic. The pair (p. q) can be regarded as the coordinates of a point on the surface M2. A curve C on M2 is given by the equations
p= An
p(t),
q=
q(t),
a
t
b.
element of arc length is expressed in terms of the vector x = (xi, thus:
X2,
ds2 = dxdx =
+ xqdq)
i
xqdq),
or
(xv,
ds2
+ (xq, xq)dq2
+
= Edp2 + 2Fdpdq + Gdq2,
E = (xv, xe), F = (xe, Xq), G = (Xq, Xq). the element of length ds2 is always positive, W2 = EG — F2 is also positive. Let us find the coordinate system (u, v) with the element of arc ds2 = X(u, v)(du2 + dv2). We have where
Since
ds2
+ (F
+ (F
Assume that we can find an integrating factor 117
=
1W) do). + 1a2
such that
+
\
(F+iW)
Then
+ (F
ti
= du
idv,
—
= du2 + dv2. Assuming 012 l/X, we obtain the required isothermal coordinates (ii, v). Thus, we have obtained isothermal coordinates by having found the integrating factor which transforms the expression and, finally,
+ (F
dq
into a total differential. The differential dii + idv can be written in the following form:
du +
idv =
\Op
+i
OPJ
dp +
\Oq
+ I
OqJ
dq.
Further, t3v
Op
—+I—=a (F +
—
Op
Ou
Ov
Oq
Oq
1W)
Eliminating c, we obtain
fOu \Oq
/Ou \Op
Ov\ OqJ
•Ov Op
or Ov c3u  W—, E— = F—
rIq
Op
Op
E— = W— lIq
Op
+
Ov F—.
Op
Solving this system for the unknowns Ov/Op and Ov/Oq, we obtain Ov
Ou F—  F— Oq Op
OP 
Ov
Op —
Oq
(1)
Oq = 118
Similarly, av av E—  F— ap
av F—  G— a1,
(2)
3q
Therefore, u satisfies the equation

IF —p (
E
IF
3q
W
J
+
ap
(
G
W
=
J
which is called the BeltramiLaplace equation. Given a second family of
isothermal coordinates (x, y) in a neighbourhood of a point, we have ds2 =
+ dy2). Using the coordinates (x, y) instead of the coordinates F = 0 and
(p, q), we obtain E = G = 3u
— —
—
av
3u
ay —
Thus, the CauchyRiemann equations have been obtained, and hence the functions u and v are conjugate harmonic functions, whereas the function f = u + iv is analytic in z = x + iy. The Beltrami equation assumes the form of the wellknown Laplace equation 82u/äx2 + + a2u/ay2 = 0. A complexvalued function lip, q) defined on M2 is said to be a complex potential on M2 if its real and imaginary parts satisfy equations (1). Thus, the real and imaginary parts of a complex potential on 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 twodimensional manifold; while transferring from one point to another, the complex potential will vary).
on the surface of the sphere. 3.3 (a) Consider some curve ço = In moving along this curve, the compass needle forms an angle with the direction of motion determined by the relations (I)
=
119
is measured from the y axis clockwise.) We obtain on the
(The angle map:
ir\ 2/
/ \
dy
dx
(2)
It follows from relations (1) and (2) that
sinO
dip
=—
dO
dx
äx
dO
80
dy
=
8x
dip dO
ôy3y dip' 80
dO
8ço
dO
/ 8y
0y
dip \ dip
8x
8x
dip
\
aip
dO I dO
80
aip
do
Since
80
(3)
relation (3) must be fulfilled at the point in question for any value
of dip/dO, we obtain, by equalizing the coefficients of the same powers
of the derivative dip/dO on the righthand and lefthand sides, that 0,
y =y(0),
(4)
aip
(5)
8y
3x
80
äip
—sinO— =
It follows from (4) and (5) that the lefthand side of relation (6) depends
only on 0, whereas the righthand side only on ip; therefore, both sides of 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
0
y = —,jsinO i— = Incot—. 2
x =
(b)ds2 = dO2 + sin2Odip2 = 3.5. ds2
sin2O(dx2 4 dy2) = (dx2
= (1
+ zzl2 120
+ dy2)/cosh2y.
dv2
3.6.ds 2 =
3.7. ds2 =
3.8. ds2
v2dça2
+
(ip) (1v) 2
+
(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
(c) a's2
ir,
0
x
0; therefore,
The
is
a
E [a, b].
= [ço', + + r" = j, [r' x r"] = — pç,". The given equation determines a straight line if and only = 0. Solving this equation, we find cc = 1/(at + b), if 2sc'2 —
r'
4.16.
2
=
a and b
where
are
constants.
4.i7. r = rr°,
Since r° =
+
sinccl,
dco
dcc
is obtained from r°
cosccl, i.e.,
by
rotating it through
dcc
 in2. Denote the vector obtained from r° by rotating it through + ir/2 by [r°l. Therefore,
dr
r'r
=
dcc
+ r[r°J.
Furthermore,
=
r"r°
+
—
rr° = (r"
—
r)r° + 2r'[r°],
dcc
Idr
d2rl
r'
Ldcc
dçc2j
r" — r
Putting
r
=
r' = dw
w,
2w
2
dr
c/u,
dr
=2r'2—rr"+r2=zO.
we find
= —r' =
dcc
r 2r'
2
dr
2w2 wdw =0, ———+ 1=0. r2
125
rdr
Put w2 = p. r2 = q, then dp/dq = 2p/q + 1. Solving_this equation, we = ar4 — r2, r' = r'iar2 — 1. Substituting find that p = aq2 — q, or hr = we easily obtain = where
+ c2),
C1
C1 and C2 are arbitrary numbers.
4.18. F = Fr = mr". Differentiating, we obtain X'r + Xr' = therefore, the vectors r', r", r " are coplanar.
mr";
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
= 0,
r2)r2 +
=
—
r1
+
,
[r1
and
=
,
)
'
ri +
K' =
—
[a]
a
=
+
E[a]
—
na).
On the other hand,
r= R— [r]
=
r1 +
—
— E[a] —
therefore,
R'
=
I
I
x
x
—(r], w =
= const,
rigidly connected with the rod) and a = ri — = const, a' I a. Therefore,
127
X[rfl = Ea+
,1[a]
4.22. r" and
r,
II
Ear"]
Er,
r"] =
0,
Err']' = Err"]. Therefore, Err'] =
(r'
—
rr'
—
r2
—
a
= const,
r(rr'))
=
Thus, Ear'] +
= b = const.
X
Multiplying both sides of this equality by r and noticing that [ar']r —a2, we have: —a2 + Xr = br. The motion is in the same
= a[r'r]
plane perpendicular to the vector a (since it follows from the relation Err'] = a that ar = 0). Introducing a polar coordinate system on this plane 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 = whence r = a2/(X is a curve of the second order. = br b
/d2u
4.23. u2 I
\dçc2
\ /
+ u
I
=
F —
mc2
,u =
l/r c =
En the case of the Newtonian force, F = — km/r2 = d2u
+u
= a
(a =
const.
kmu2, whence
k/c2).
4.25. The circumferences whose centres are placed on the straight line passing through the origin of the radii vectors and collinear with the vector w, whereas the planes of these circumferences are perpendicular to the indicated straight line.
4.26. The straight lines along which the planes perpendicular to the vector 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 Oz collinear with the vector e, we have ae + [cr1 = —yi + xi + ae, and the given differential equation assumes the following form: x' = —y, = x,
that x2 +
z' = a. We find from the relations x' = —y, y2 = C1 is the family of circular cylinders whose axes 128
y' =
x coincide
with the straight line passing through the origin of the radii vectors and
collinear with the vector e. Furthermore, dx
dy
y
—
=
x
a'
a' dz
dz whence
xdy — ydx
x2 + y2
=
dz
,a
ydx
xdy
=
x2
a
7I
\
y2 \ +— idz
x
the family of right helicoids whose axis is the axis of the cylinders mentioned above. The integral curves are helical. rinally, z = at + C3. relations obtained. Now, to express x, y, z in terms of t is easy from is
4.28. Semicircumferences touching the axis Oz (which is collinear with
the vector e) at the origin. 4.31. 7r/4 and ir/2.
4.32. tan13.
fa\
4.36.
2
+
fb
1,
2
where a and b are the semiaxes of the
given ellipse.
4.37. xy = ±s/2, where s
=
and s is
(x
4.39.
the given area.
where
ax2 +
4.38. y
is
the parabola is given by the equation
the area of a segment.
)2
/
—
+ y2 =
(i tan
where
is the given
cos 2
angle, and I the semiperimeter of the triangle. 4.40.
x2
a2
4.41. r
+
y2
= 1, where a
is
the radius of the given circumference.
2a2
cos3v,
/ sin3vl, where / is the given semiaxis 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. = 4a2, where a is the major semiaxis of the ellipse,
4.44. (x — c)2 + c = Va2 — b2.
[r'] ± r ± a—.
4.45.
r'2 lEr'
x r"]I
and in coordinates,
x'2 +
x'2 +
,
x'y" — x"y'
x'y" — x"y' 4.47. A cardioid. 4.48. (1) 1; (2)
(4)
a(y
—
4am(1 + m)
'
4
a
(3)
2
2by3 + a2b2)312 —
(6)
1+2m
3r'
—
cos —; (7) a
(1 +
2
3
(8) 3a1sint cost1.
(2) 1/6; (3)  ir;
4.49. (1)
a
2+
4.50. (1)
(2
(3)
(4)
k(k+1)+ço2 aço"'(1c2 + 1
'11 + ln2a 130
a
4.51.k=
F,,
'
(Q
ax
L
4.52. k
0
yl
x
 ±Q) (P2
ay
+
+ y'2dx =
4.53. (1) s =
s=
+
—
y'2dx =
[(4 + 9x)312
(5)s =
+
=
(6)s =
+ y'2dt = 4a(1
(2)
—
(4)s=
(7)s =
cosf);
—
y'2dt =
(8)s
y'2dt = J
3
0
131
2
8];
Q
ax
3a
+ y'2dt = — sin2t;
s=
(9)
2
0
x
(10)
s =
0
= Vi
+ e2x +
—
In
—
—
+1
2
(Ii) s
a
—
1);
In sint.
=
4.54. f(co) + f" (a). 4.55. (1)
R2 +
(2) (27s +
— 6as = 0;
[4
8)2
+ 9
(3)s=I
36R2
(27s + 8)2]; 3
+ V2R +
1
3—
1
—in 4
—
+ 12R].
1
EN
(4) The parametric natural equations are
s=Vi+x2+in
x
x
andk=
(1 +
X2)3/2'
= a + s2/a;
(5) R (6)
,,Ji+x2_1
The parametric natural equations are
VTTr?x
s
+ a2 =
(7)
1?2
(8)
s2 +
9R2
I
—in
+
+
2
and k = 1
(1 +
a2e — 2S/a;
16a2;
(9) R2 = 2as.
4.56. (1) r = (2) x
a logarithmic spiral;
a
b
.
2(a+b
sin
(a+b) t + b
b 2
—
a+b
cos
b
a—b
a+b
b
b
a—b
132
sin
a—b t\
)
b cos
a—b" b
S
(3)
S
(C
r=
s2
(j
2a2
2a2
)
0
aintan
=
a clothoid;
%
J
0
(4)x
s2
C
lcos—ds, +
4
(5) r =
+
(6) r = ta(2t +
2
=
2
line;
cost
sin2t), —
+
2
cos2t)};
sin2f), a(2 — cos2t)), a cycloid;
(7) r = [a(cost + I sint), a(sint — I cost)), the evolute of the circumference; (8)
r = ( a cost, a In tan
4.57.
ds
p=
t
71
+ — — a sint
4
2
ml. Assume that rn > 0. Then p =
—rrk = —rtk =
= rn + rñ =
)
mn.
a
tractrix.
Hence
=
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 — — (s) = 2(r — ro)r — 2Ronor, — ro)2 — 2Rono(r — r0). We have = 0, çc"(s) = 2 + 2kn(r — ro) — 2R0nokn, = 0, (s) = 2thi(r — ro) — 2k2r(u — no) — + 2Ronokr, 0; therefore p(s) changes sign when s crosses thro(So) = — 2R0ko 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 —
— 2k3n(r — ro) — 2k2
—
ro) —
+ 4Rnokkr+
2Ron0kn +
+ 2R0u0k3n, =
—
—
=
2R0k0 +
2R0k!0
o.
Therefore, the index of a point on the osculating plane does not change sign in crossing through c0. 133
4.60.
=k
x=
cosaf(a)da,
, ds =
f(a)
y=
sinaf(a)da.
4.61. dcx/ds = 1IR,f'(R)dR/ds = hR.
x =
cos[f(R)]Rf'(R)dR,
y=
sin[f(R)jRf'(R)dR.
4.62. x
=
ds = Rf'(R)dR,
f
y=
cos
rn
4.64. R = r +
R=r+
4.66.
R=r+
—
+ rn
2kr
4.65.
sin [f(s)]ds.
Er r]
r"]
[2r 2 r',
[r'rlr'2
+
(rr)'r).
([r'r][r'J
—
r(erfl.
(n(en) —
If the curve is given by an equation r = r(t), then,
R=
e]
Er
r +
(Er', e] [r'] — (r', e)r').
2[r', r"]
If the curve is given by an equation y = f(x), then [m
,
—
f(x)—
2f"(x) Y = f(x)
+
Em
(m
If' (x))(1 + mf' (x)) 2f"(x)
if' (x)]2
—
(m
—
2f" (x)
= Li, ml. 4.67. (x + 1)/2 = (y
—
If' (x))(1 + mf' (x)) 2f" (x)
f
(x),
where 1 4.68.
(— 17) 4.69.
u= = (z
u=
13)/3 = z/6, 2x + 3y + 6z — 37 = 0.
—
at the point A. The tangent is (x — 3)/6 = (y + 7)! i7y + 7z — 151 = 0. 2)17, and the normal plane 6x
—I — 1
at the point A. Since r'(l) =
0
and r"(l) =
(2, 2,
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, and 121
the normal plane x + y + 4.70. 9x —
—z +
6z + 10 = 0. 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, y
in 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 northe binormal
mal x=l—31X, y=l—26X, z=l+22X, and x=l+6X,y=l — 8X,z=l —X. 4.73. The tangent is
r=
cost —
sin!, a sin! +
Xa
The normal plane axsint — aycost
—
t)l,
aX cost, b(X +
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= 4.74.
(a +
(a + X)cost,
The tangent is x =
X)sint, bt I.
+ 2X,
I
y=
—X,
z=
+ 3X,
1
The normal plane 2x — y + 3z — 5 0, The binormal x = — 3X, y = —3X, z 1 + X, The osculating plane 3x + 3y — z — 2 0, The principal normal x = — 8X, y = 11X, z = + 4.75. The tangent is 1
1
X=x+X
Y=y+X
3F1
t3Fi
3y
az
0F2
oF2
Oy
Oz
OF1
OF1
Oz
Ox
OF2
OF2
Oz
Ox
1
135
9X.
OF1
OF1
Ox z
+X OF2
OF2
Ox
0,'
The equation of the normal plane is
Xx Yy Zz OF,
OF,
OF1
3z
Ox OF2
OF2
Ox
0,'
—
OF2
4.76. Having chosen a convenient coordinate system, we shall write the
equations of the Viviani curve in the form
orx2 +y2 +
— aX=
= a2,x2
0.
To make up the parametric equations, we put x—
a 2
a — cost,
y=
2
a.
— slnt. 2
Then a2
— (1 + cost) 2 4
a2•sin2t +— +z
a, z =
2
2
4
a
sin — 2
is added to t, then x andy are unaltered
(sign can be omitted, since if
and z changes sign). Thus,
r=
(a (2
a
.
.
I
+ cost), —sint, asin— 2
2
The equation of the tangent is
r=
(2
+ cost)—
+
that of the normal plane xsint — ycost
—
+ 2
2
= 0, 136
2
of the binormal
(a
r=
+ cost) + X sin —
(1
(2
(2
+ cost).
2
sint —
(1 + cost), a sin
cos
X
+
of the principal normal
(a 1—
r
(2
+ cost) +
(1
a.
X[_cos2 —f (1 + cost) —
2
t
—sint(6 + cost), asin—

2
2
2
t
Xsin—1,
2J
and that of the osculating plane 1
sin — (2 2
+ cost)x —
t
cos — (I 2
a
sin
2
+ cost)y + 2z
L(s
+ cost) =
0.
2
4.77. s = 5at.
4.78, s = 4.79. s =
9a.
4.80. s =
10.
the points t = 4.81. r
The curve has four cusps with ds/dt changing sign at 3ir/2.
0, S
[acos
+ b2'
asin
s
bs +
+
cos In
4.82. r
sinin
4
2
83 r
{,',J
+ 2
2
137
,—j.
— a sin I, a cost, b), + b2 = [—cost, —sint, 0),
4.84. T p
=
1
Va2
{bsint, —bcost, a), b2
4
b
a
4.85. i =
K=a2+b2• [2t,
Vi
+ 9,4'
+
2t + 9f3, 3t + 6t3 + 9?)2± (3t + 6,3)2'
—
I'
= —
I)
(—3t,
k
+ 9j4)1/2
2(1 + =
(1
+ 4t2 +
3
'
=
+
+
cost,
J
2
2
cost, —! sint(6 + cost),
—
sin
2 —J
2
[— cos2
(1 + cost) —
cost] +
sin2 ((6 + cost)2 + 4
2
j3 =
2
2
sin—(2 + cost), —cos—(1 + 2
t
k=
13
a
\3' X =
+ 2(1+cos_1 2/
138
2
a(13
+ 3 cost)
—5)
Ii +
4.87. (1) k = —
(2)k=
sin2 —, ,c =
,
(e' + e  1)2
(1 +
—2t ,
2t2)2
(4)k = —,
3e
3(t2 + 1)2 3
(6)k—
+ 212)2
(1
= — —i;
3e
(5)k=x=
e — 1)2
(e' +
2t
(3)k=
4(3 — cost)
2
25 sint cost
x=
,
4
25 sint cost
4.88. y = L
xIx2
4.90. k
r=
xfx2
a2\2
+ (y'z" —
+ —
+
"Z" + —
a2\z
—
Y" Z"
+ (y'z"
—
11, y', z'} + y'2 +
p
=
[—z'z" —y'y", y" —z'(y'z" —y'z"),y'(y'z" —z'y")+z" V(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;
(b) x2 +
= const, xy =
az.
4.92. Let the
equation of the sphere be of the form:
and
r
tacosvcosu, acosvsinu, asinvj.
Then the equation of the loxodrome is U
+
tanO
the given angle.
where 0 is
cos0( —sinv cosu — sinu tan0, —sinv sinu + cosu tanO,cosv);
(
cosO
— cosu
cosv + tanv sinu tanO —
tan2o (
I+
—
cosV
(.
—cosu,
( I
Il + cos0
tan2 0,
CO5V
COSV
—
CO5U
sjnu
tan2 0 — tanv tan0cosu, —sinv
COSV
tan0 cosv
tan2o
cos2v tan20
I
a
tanO
a(cos2v +
cos2v
tan20)
u cotO
4.93.v=Ce
JI+k2
r r r 6
[r' x r"](
,
and
in the special case, — kjx'. 6
4.98. The necessary and sufficient condition is e' while the equation of the envelope
Q'e'
e.
e
140
0,
'ee' =
0,
4.99. When a = 4.101. r = r, 4.103. ev =
r=
C,
kv,
1
= —k2r + kv
e(—kr +
= 0,
* kx/3.
ei —
efl = 0,
k2 +
evk=
k
k
Differentiating once again, we obtain the required relation. Note that in view of the above relations, we may assume
x k2+x2 k
r+v+
\kJ
k2+x2
\k
If the relation is held
k2 +
+x=0,
then this vector is constant. This constant vector e forms with the vector = const. p an angle whose cosine equals
key = 0; hence either k = 0 (straight line) or then e(—kr + x/3) = 0, whence x = 0 (plane line). 0 4.105. = 0; xev = 0; hence x = 0, since, if the inequality x 0. were held, we would have ep = 0, e(—ki + x13) = 0,ker = 0, er Therefore, k = 0, and the line is straight. 4.104. er = 0,
ev =
0;
if ev =
0,
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 = k = 0 (i.e., of straight lines), we obtain
av=0.
0.
Excluding the case where (2)
Therefore, the normals are perpendicular to the fixed direction. Conversely, if v is perpendicular to the fixed direction, then equality (1) holds.
0. It follows from (2), with the use of the third Frenet (b) Let x formula, that aj3 = 0, whence
= const. 141
Conversely, differentiating this formula, we obtain (2). (c) Differentiating (2), we obtain
kar = whence k
=
= const.
ar
x
Conversely, it follows from the first and third Frenet formulae that + k x whence +
k
=
= 0,
r +
= const = a.
f3
k
Multiplying scalarly by v, we obtain av = been fulfilled. 4.109. Thke into account that = k5 (—f
0.
Therefore, condition (2) has
,
and use the previous problem.
5
Surfaces 5.1.
r=
5.2. r = 5.3. r = 5.4.
r
+ ye.
+ VQ'. 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 the equation 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., the axis 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, Hence, the required equation
r=
((a +
— Xn =
X)
—
is the principal normal vector.
sinu, 0
cosu, (a +
X)sinu,
bu) =
= (vcosu, vsinu, bu) is that of a right helicoid.
where cc(s) is an 5.9. r = a(s) + Xtn(s)cosço(s) + 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. The vector lying in the normal plane and inclined at the angle u to the vector n is a = n cosu + k sinu. Therefore, the equation of the required surface
asinu, 0) +
r = {acosu, = Ia
cosu +
v
va
cos2u, a sinu +
v
sinu cosu, v sinu).
Eliminating the parameters u and v, we find
x=
acosu +
y=
a
sinu +
cos2u sin u
z
cotu(asinu + zcosu),
z cosu,
cotu,
sinu
y
= (a + z cotu)2,
,
y2
(i
+
4)
=
(a
or
+
y2(x2 + y2) = (ay + xz)2,
a surface of the fourth order.
5.l1.R = 5.12. r =
(r(u) + e(v)) a
cosu cosh
—f,
a
a
sinu cosh
a)
where
u is the
longitude and v the oriented distance from a point of the surface to the gorge section of the catenoid. 5.13. r = ía In tan
\4—f + 2/
—a
143
sint, a cost cosu, a cost sinu
5.14. The equation of the given straight line is r1 = that of the ellipse r2 = {acosv, bsinv,
u,
0, h
and
Furthermore,
r1—r2=(u—acosv,—bsinv,h),uacosv=0, r1 —
= [0,
r2
—bsinv, hi,
and the required equation of the conoid is + X[0, —bsinv, h r = {acosv, bsinv, = [a cosv, b(I
X)sinv, Xh).
Eliminating the parameters X and v, we obtain the implicit equation of the conoid
\h
\
/
5.15.ri =
=
b2
I 2p)
(
( (
=
—v,u —
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 We
+ cosu),
0, a sinu
r2
= (0,
a(l +
cosv), a sinv J.
find
r1 —
r2
= (a(1 + cosu),
We have sinu — sinv = (1) v (2)
v=
0,
u+
2kw,
ir
u + 2kw.
—a(l + cosv), a(sinu — sinv)j. whence
In the first case, we have r2
= (a(1 + cosu), —a(I + cosu), 0) 144
II
(1,
—1, 0),
and thus obtain the elliptic cylinder
= {a(i + = La (1 +
0, a sinul + X(!,
cosu),
=
—1,
—X, asinul.
cosu) + X,
in the second case, r1 —
{a(i +
r2
cosu), —a(l
cosu), 01,
and the second surface making up the given cylindroid is determined by the equation
R = ta(i +
0, a sinul +
cost,),
+ A{a(l + cosu), —a(i — cosu), 01 =
= La(1 +
+ cosu), —aX(i — cosu), a sinu).
X)(1
Eliminating the parameters X and u, we obtain +
2a(x + y)] + 4a2xy
—
5.17. rj =
r1 —r2=
(u2 u,
(u2 + (
) v2
2p
I
v2
(
2p
= 0.
0, v
,u, —v
The condition for this vector to be collinear with the plane y — is given by the relations
z
=0
u+v=0, v=—u,ri—r2=1u2/p,u,u), and the required equation is the following:
r=
+ v
+ 2v),u(l +
u} =
Eliminating the parameters u and v, we obtain y2 —
z2
= 2px, a
hyperbolic paraboloid.
5.18. The equation of the axis Oz is of the form: r1 = [0, 0, uJ and the equation of the given curve
r2 =
I b cosv, b sinv,
a3
cosv sinv
hence rz —
r1
10 2018
=
(
cosv, b sinv,
a b2 cosv sinv 145
—
u
u=
a
b cosvsinv
r=
a
0, 0,
=
fbcosv, bsinv, 0),
r1
—
,
+ X(b cosv, b sinv,
b2cosvsinv)
a3
Xb cos v, Xb sin v,
b2cosvsinv Eliminating the parameters X and v, we obtain b2xyz = a3(x2 + y2).
5.19. (a +
a+
ub
ub
—
=
—
R=
= o, u =
—
a)
b
a),
+
—
a)
—
nb
b—
nb
x["(Q—
+
a)
n(Q
a].
5.20. Take the equations of the given ellipses in the form: = ri
—
a,
r2
b cos u, c sin u), r2 =
L2a,
b cosu —
c sinu — b sinv =
c
= ±
cosv, c sinu
b
sinv),
—
b
b r2
c cos v, b sin v),
0,
=
—
a,
sin2u, 0
= 12a, b cosu ± —s
(
b
The required equation is
R=
La,
b cosu, c sinu) +
v
2a, b cosu ±
0
b
or R =
+ 2av,bcosu) + vlbcosu ±
(
—
5.21. The equation of the axis Oz is p
u3—v=0, p=
0, u3). 146
c2sin2u,csinu
b
and we find
0,
v=u3,
The required equation is r
=p
+
r=
5.22.
(by,
— p) = 10, 0, U3) + v{u,
U2,
av cosu, (b + a cosu)(l —
v)
0) =
{uv, u2v, u3J.
+ a sinu).
5.23. The equations of the given straight lines are
= (u,
1,
1
and
(1, v, 0). The equation of the straight line passing through two arbitrary points of these straight lines is the following: p
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=
z=
0,
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,
r=
(1,
—sinw, o( +
=
{i
+
i)
and (1, —sinw, 0). Finally, we obtain: 1,
—
+
+
5.24. r = (a(cosv
—
5.25. c2(x2 +
= a2(x2 — y2)(z + c)2.
u
sinv), a(sinv + u cosv), b(u + vfl.
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 be written in coordinate form thus: E = In addition, we 'j = assume 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 the axes x, y and positive directions on the coordinate axes, we have:
R(u, v) =
+ av).
(E(u)cosv,
5.27. R(u, v) = r(u) + ap(u)cosv + afl(u)sinv, where v and i3 are the 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 origin of the radii vectors. Then r
r
0,
whence r2 = const. Therefore, the given surface is either a sphere or a part of a sphere. 5.29. The volume of the tetrahedron is 9a3/2. 5.30. The tangent plane is determined by the equation X
usinv
+
ucosv
=a.
Z
+
2
2
2
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), where u1, v1 are the coordinates of the point of contact. The parametric equations of the same line in Cartesian coordinates are cos(v + Vt) cos(v + v1) X = Ut
y=
COSV,
cos2v1
5IflV,
U1
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 z=
=
(f 
(i !_f')(X +

+ (Y
x)
or
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 determines a cone with vertex at the origin (z being a homogeneous function in x and y). 5.33. The tangent plane has the equation kxsinu — kycosu + vz — kuv =
and the normal r = cosu + Xk 5inU, V
SiflU
—
148
0,
Xk cosu, ku
+ XvJ.
x +— Y ÷
5.34.
x
Z
y
= 3.
z
= Q(s). The equation of 5.35. Let the equation of the curve C be the surface is as follows: r = + Xr, where r is the unit vector of the tangent to the curve C. We find that
as
=r
Far an = r, I—,—I = Xkfl; ôsj L3X
8r
+ Xkv,
ax
when s = const (i.e., at the points of the same tangent), this vector has the same direction (for then i3 = const), from which it follows also that the tangent plane to such a surface at all points of the curve C is the osculating plane to this curve. 5.36. The equation of the surface is 8r —=v, r=Q+Xv, —=,+X(—krixfl), ax as
Las
= (I
axj
— Xk)/3
—
The equation of the tangent plane is — Xv)(f3
(R —
Xk13 —
—
Xx'r) = 0,
or Xxr)
—
Xxr) +
X2x
= 0,
and that of the normal
R= 5.37. r =
Far
+ Xfl,
an
=
—v
=r
as —
—
Xxv,
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
ax
=
13,
5.39. If a is the direction vector of the given straight line, and the origin
of the radii vectors is taken on it, then the vectors r, a, and
3v]
Lau
I
Far
L
Lau
Ia,
r
lie
in the same plane, while
är]]
—i avJJI
= 0.
Hence,
(r
\
.
(a
\
auJ
.
avJ
 (r ôvJ \ .
(a
\
.
3uJ
= o.
But this equality can be written as the vanishing of the functional determinant, viz., 8r2 a ——(aS
öu
r) —
av
ar2 a ——(a
•
r) = 0,
äv
from which it follows that the entities r2 and a
r are in the functional
dependence
r2 = f(a
r).
Choosing the axis Oz along the vector a, we obtain x2 + surface of revolution.
y2 =
f(z), a
= I, the edge of regression being imaginary.
5.42. 4z2
:2
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 the form y2 = 2px + 2qz, i.e., a parabolic cylinder with + q2 as a parameter. 5.46. (R — = a2. Differentiating with respect to s, we obtain = 0. Hence R — = Xb + Since (R — (R — = a2, X + = a2, and we can put X = a cosço, = a sinw
so that the equation of the envelope is
R=
+ a(b
+v 150
5.47. The edge of regression is a curve whose points are obtained by intersecting the curvature axes of the curve Q = a(s) with the corresponding spheres of the given family. 5.48. The equation of the family is the following: b sin2w + :2 — a2 = 0, (x — b cosco)2 + y
and the envelope eliminating + b2
—
is
a torus whose equation may be obtained by
from the equations F =
0
+ z2 +
= 0; (x2 +
and
is a surface of the fourth order. The is reduced to the two points
a2)2 — 4b2(x2 + y2) = 0
edge of regression when a >
b
(0, 0, ± one point (0, 0, 0) if a =
or
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 +
—
— (l2xz
—
4y2) + z(l2xz
+ :2) —
4y2)2
—
=
0.
The edge of regression is found by adjoining another equation to the two indicated above, viz., 6ux + 2y = 0, or 3ux + y = 0. Hence, u = —y/3x, and the equation of the edge of regression is the following
27x2(x2+y2+z2)—4y3+l8xyz=0,y2—3XZ=0. The edge of regression can also be obtained in parametric form: r
(
6u5
—6u4
2u3
= L 9u4 + 9u2 +
1
'
+ 9u2 +
I '
+ 9u2 +
1
12/3. + z213 5.50. x2"3 + p2/3 + = 5.51. x2"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 equation of the family of the osculating planes is (r
0.
Differentiating with respect to s, teristic is the tangent
(r—Q)b=0,
we
obtain (r
= 0. The charac
(r—Q)v=0.
The envelope is r = + Xr, i.e., the surface formed by the tangents to the given curve. Differentiating the relation (r — 0 once more, we obtain (r = 0. Hence, taking into account the relations = 0,
(r —
=0
(r —
we have
r= the edge of regression is the given curve. 5.56. The characteristics are the curvature axes of the given curve, and
i.e.,
the envelope is the surface formed by the curvature axes. The edge of regression is the curve described by the centres of the osculating spheres of the given curve. 5.57. rn' + D' = 0, r = an + fin' + X[nn'],
D'
rn'
fi = —in'
a = rn = —D,
=
—
n
The equation of the envelope is
r=
—Dn
D' n' —
2
+ X[nn']
(with the parameters u and X). The characteristics are straight lines u = const. The edge of regression is found by solving the equations
rn+D=0, rn'+D'=O, rn"+D"=O for r, viz., r
[n'n"] + (rn')
=
(rn)
=
D[n'n"]
+ D'[n"nJ +
+ (rn") [nn']
D"[nn']
nfl 'n" 152
5.58. The envelope of the family of planes tangent to both parabolas. The equation of the family is Xa — 2aY
—
4a3\ Ia2 + —lZ + I— \2b ba)
4a3
a
=
0,
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 parameter v remains constant. Hence, the lines v = const are rectilinear generators of the surface, and u + v = 0 is the edge of regression, since the 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
z=
2u2[(a
b)u cos2v, y = 2(a
—
—
2b)cos2v
—
b)u sin3v,
+ (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 ff  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
4\
+
+
+ —)
v/
(a2
sin2u + b2 cos2u)du2 +
(b2 — a2)sinu cosu
[1
(i
(a2
(v
dudv
cos2u + b2 sin2 U) +
(i
+
1)2]
—
(4)
1)2
(u+v)
+ 2(a2(u2
+ (a2(u2 —
+ 4b2v2 + c2(v2 +
+

l)(v2 — 1) — 4b2uv + c2(u2 + 1)(v2 + 1)2
+ 4b2u2 + c2(u2 153
+ 1)2 Jdv2];
+
\2
/
4\
(5) —'
(v
—') (a2 sin2u +
—
cos2u)du2 +
b2
VI
dudv +
— a2)sinu cosu
+
—
+
cos2u +
+
sin2u) +
b2
(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 + v2(a2 sin2u +
(8)
cos2u)du2 +
b2
2(b2
q + 4u2)dv2;
a2)sinu cosu dudv
+ (a2 cos2u + b2 sin2u + c2)dv2;
(9) (a2 sin2u + b2 cos2u)du2 + 2
(i
(10)
+
dv2;
(i
—
5.65. (1) ds2 + 2redsdX +
(2) v2ds2 + 2vrQdsdV +
+X
(3)
(4) ((1
(6) (a + (7)
b
du2 +
dv2.
+
dX2;
Q2dv2;
2erdsdX +
kcosço)2 + x2lds2
+
(5)
ds2 +
2]
c1X2
+
+
+
cosv)2du2 + b2dv2;
(v2 + k2)du2 +
dv2;
(8) 1(1 — Xk)2 + x2X2)ds2 +
dX2;
(9) (1 + X2x2)ds2 + 5.66.
(1) The curves u
A(u—0,v=0);
± 1/2 av2, V
B(u=l/2a,v=1);
I
intersect at the points
C(u=—1/2a,v=l);
the differentials of curvilinear coordinates on these curves being related by the formulae:
du =
avdV for the curve AB with the equation u = 154
av2;
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 (1/2 V2 + 1)dv for the curve ds2 = a2(1/4 V4 + v2 + 1)dv2, ds ds2
a2(1/4 v4 + v2 + l)dv2,
ds
(1/2
v2
+ 1)dv for the curve
Ac;
ds =
ds2 = du2,
du
for the curve BC.
It remains to evaluate the integral between the limits determined by the coordinates of the points A, B, C, viz., =
AB = AC =
v2 + 1)dv = 7a/6,
a
du =
CD u = — 1/2a
10
Thus, the perimeter of the triangle equals —
(2) cosA = A
1,
0, B =
cosB = 2/3, cosC = 2/3, C =
cos'
(3)S
I
5.68.
in(l +
+
r
v=
i.e.,
2/3.
+
5.67. cosO =
a.
+
1
+
In
+
±
5.69. v =
tanO in [U + Vu2 — a2J
sinhvo,
(3) in2, in!4, in/4. 5.71. (1) ds2 = UI
+ [ax(u)12)du2 +
—
+
+ (2)
+ const.
+ sinh2vo);
5.70. (1) 1/4 (2)
1]
=
—
(3) 2iraluz — uji; (4) 4rr2ab;
(5) 155
+ const.
a.
5.72. Consider the family of the surfaces R(u, v, t) =
I)),
t)sinv/t,
t)cos
where
t) =
t) =
tQ(u),
5.73. R(u,
t2Q
'(u)2du,
Vi +
=
I
t >0.
ln(u + '.11 +
or R(z,
This is
x=
zi. a catenoid, a surface of revolution of the catenary curve =
coshz.
5.75. Hint: The first equation determining the correspondence between points is as follows:
r2 =
+ a2.
5.78. For the sphere ds2 =
for torus ds2 =
du2 +
du2
(a
+ R2 cos2(u/R)dv2,
+ b cos
bJ
dv2,
for catenoid ds2 = du2 + (a2 + u2)dv2, for pseudosphere ds2 = du2 +
Hint: u is the natural parameter of the meridian. 5.79.
Putting u = ds2 =
= dIP + g
v=
we obtain
(du2 + dv2).
5.80. (a) If a and b are the sides containing the right angle of a rightangled 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)
C
(5) (6)
2
(du2 + cos2udv2); U
+ c2 cosh2u ac —
V a2 cosh2u + C2 sinh2u 2
Vi +
(du2 — cosh2udv2); (du2
+ sinh2udv2);
(du2 + u2dv2); 4u2
Rdv2;
ku (7)
COS
—ac
(3)
(4)
2
Vi
+ k2
(8) bdu2
dv2;
+ cosu(a + b cosu)dv2;
(9) — — (du2
—
a
a2dv2);
(10) —a cotu(du2 — sin2u dv2). 5.87. — 2adudv/Vu2 5.88.
5.89.
—
2a3
x
Vx4y4 + a6(x2 + y2)
5.90. (1)
(2) K
+ a2.
a du2 + adv2. u2 + a2
— —
(Q'x"
x'(Q'x"
dx2 + dxdy + —f dY2] y
Q"x')du2 + (x')2 + —
Q"X')
Q[(x')2 +
K > 0 if the convexity of the meridian is directed from the axis of rotation; K < 0 if the convexity of the meridian is directed towards the axis of rotation; K = 0 if the meridian has a point of inflexion or if it is orthogonal to the axis of rotation (when 0). 0; K is undetermined, when x = 0; (3) K = — 1, when x (4)
H=
—
+
x'[(x')2
2Q[(x')2 + 157
+ (e')2]
(5) Q(x) =
Xo), a
where xo and a > 0 are arbitrary constants (catenoid). 5.91. (a) K = 0; (b)
'C
H=
5.92.
—
2kv
K=
5.93.K
ouu —
—1.
5.94. dXXF
K=
1 I
(âfl2
(öyfl2 +
+
0 5.95.
rt —
K=
H=
+ p2)t +
(1
S2
+ p2 + q2)2
(1
+ q2)r
(1
2(1 +
2pqs
—
p2 + q2)312
where
p=
q=
ôxz, X2
5.96.
r =
äyZ,
t =
= 8xyZ,
+ a2
+
a
5.97.—=0, —= R1 R2 1
1
1
5.98.K=
(2u2 + 1)2'
5.99.K= — 5.102. r =
1
(u +
H=
4
(2u2 +
+ ufi(s),
k + kx2u2 —
u
ds
K= ———______ (1
1)3/2
11=0.
9(u2 + v2 +
Q(s)
2(1+u2) —
+ u2x2)2 '
(1
158
+
t9yyZ.
_______________
5.103.K=
—
H=
ku)2 + u2x2]
—
1(1
2'
+ ux
—
21(1 — ku)2 + u2x2]3"2
5.104. r* = r + am, E* = = +
+ ama,
=
=
E — 2aL + a2(2HL
+
Similarly,
F* =
(1
a2K)F + 2a(aH
—
l)M,
—
G* = (1 — a2K)G + 2a(aH L* = aEK + (1 — 2aH)L,
1)N,
M*=M_a(2MH_FK), N* = N  a(2NH OK). —
5.105. K* = 1
—
2aH + a2K
H — aK
5.106. H*
1 — 2aH + a2K
r + —m.
5.107. r*
K
4H2 =
5.108. K*
5.109. H* =
const.
—
2
5.111.
du2
dv2
a2+b2+u2
—
a2+b2+v2
5.112. v = ± ln[u + '.Ju2 + 5.114. u = const, v = const. 5.115. (1) K =
a2] = const.
k —
a(1
—
(2) H = —(1—ak (3) u = const, =
const.
5.116. u =
=
const,
+ amp,
ak cosçc')
—
159
= 0.
EK).
5.117. The lines of curvature are u
= const,
H=O.
+ 3v2 +
(3u2
±v
(1) The rectilinear generators are y/x = const;
5.122.
(2) — x2
— —i = const.
(1) The rectilinear generators are y = const; (2) x2y = const. 5.123.
5.124. The equation of the asymptotic lines is (1) u
const;
(2) u5 = v2(C — 5.125. v = const,
cos2u cosv
(1 +
cosu)2
= const.
= ±n on the asymptotic line. Therefore, x2 =
5.126.
We
select a coordinate system (u, v) in a special way so that the following conditions may be fulfilled at the point u = u0, v = v0 under consideration: (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 the orthogonality of the principal directions and = — — by the Rodrigues theorem. Therefore,
/ du + flv—I dv\2 = = (flu— \dsj \ ds dsJ 1 dn'\2
+ ds2
be the angle between the line v = v0 and asymptotic direction. Then with respect to this direction du/ds = cos dv/ds = sin çô because E = G = 1, F = 0. On the other hand, we find from the Euler formula that Let
k1 cos2ç, +
= 0.
k2
Finally, we have =
5.127.
(d)2
=
—
+ u2x2). 160
= —K.
5.128.
(1 — ku)2
u+ v+
5.129.
+ u2x2 a2 + b2
Vv2 +
[u +
+
a2
a2
= const;
+ b2
[v +
+ b2]
,,/v2 + a2 + b2J = const.
5.143. Assume that the rectilinear generators are parallel to the axis Oz. Then the equation of the surface can be written in the form r
= f(u)ei +
+
ve3,
u is the natural parameter of the directing line. We will seek the equation of the geodesic in the form where
v=
(*)
v(u).
Then
N =
ço'ei — f'e2,
[rn,
(f'ei + cc"ez + v'e3)du, = (f"ej + + v"e3)du2,
dr = d2r
and the equation for determining the geodesic lines is
0
—f,
v'
f' jflF
= 0,
v
or
Since w'2
= 1,
+
+ Thus,
+ f'f")v'
f'2)v"
+
(w'2
v" =
=
we have
÷
+
and v =
0
= 0.
CIU
= 0
+ C2. The vector equation of the family
of geodesics is
r = f(u)ei
+
+ (c1u + c2)e3,
whence cosO =
OZ) =
C'
+ Ci2
11—2018
161
Therefore, the geodesics found are generalized helices. Besides, the rectilinear generators are also geodesics which were not included in the general 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 generator s = const. If k is the curvature of the curve R a(s), then the differential equation of the geodesic is —
ds It is
ukcot0 +
0.
1
a linear differential equation of the first order integrable by
quadratures.
5.145. The equations of the geodesics are
Csinv
Ccosv
5.146.
Cdu
v = C1 ± +
h2)(u2+h2
5.147. Consider the equation of a cone in the form r = up(v) and assume that I pI = I, p of the form
r=
sin(C
v)
I
= I. Then the equations of the geodesics are
p(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 the surface, and orthogonal to y; n is the unit vector of the principal normal to the curve y; IcosOl = leni,
kg= klcos0l
5.155. kg = u/(u2 +
=
a2).
162
klenl = lell = lItmI = Imirl.
5.158. Considering v as a function of u along a geodesic, we obtain the following differential equation for the geodesics: du2
J
du
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. U2
V2
5.162.G*
dv
du. UI
5.164. 11(p)
s(p) =
2ir
sinhp;
kg(P) = cothp —
asp— +oo.
= 271 coshp; 11(p)—
On
I
as
p — +00;
the Euclidean plane,
2lrP;kg(p) = 1/p — Oasp — +oo;Il(p) = 271. 5.165. First, we establish that the metrics defined onP1 and P2 have the same curvature K = — I. Then we introduce semigeodesic 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.
s(p)
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 = (m = n), we have = (dm
\dsj On
\ds/
= k2 + x2.
the other hand,
\ ds/
=
II — K
1,
163
— kt
+ xb. For the geodesic line
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/
If the equations of a surface of revolution are written in the form cos v, sin v, z = u, y=
5.168.
x =
then the vanishing of the mean curvature implies that
=0. Putting p = dp
2
+
1
as a new variable, we obtain
and considering =
= 0,
—
1
d(ln
(1
+
2
dço
whence
=
+ p2.
i
With respect to the original variables, = du,
(1 + u2)f'(u) =
f'(u) = a/(1 + u2).
a,
Integrating this equation, we get
f(u) + b =
z
+b=
u.
Therefore,
u=
tan
y/x +
(z + b)/a,
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
= (I
r
r
—
= (1
—
Therefore, the coefficients of the first fundamental forms of the surfaces S and S * are related by the formulae E* = (I — ak1)2E, G* = (1— ak2)2G, F* = F= 0. Hence,
dcr* = and
lim
aQ
(1
— ak1)(l — ak2)dcr,
do_da* 2ada
7k1+k2 + u.O\ 2 .
= lim I
164
I
2
ak1k2
\= / I
2
= H.
a surface parallel to it, the 5.171. Let S be a minimal surface, and distance between them along the normal being equal to a. As it follows from the previous problem, the corresponding elements of the areas of the 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.
0 on the minimal surface,
Since K
do. 5.174. Take the axis of the cylinder to be the axis Oz and place the axis Ox in the sectional plane. Then the equations of the cylinder assume the
form
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 on
the plane xOz. Since after the superposition, the part of the abscissa is played by the length of an arc of the perpendicular section of the cylinder at, the equation of the required line is
s=
z=
aA
a
i.e., a sine curve. 5.175. The general equation of the motion of a point across the surface is of the form d2r dt2
m
=F+Rm—pIRIt,
F is an external force, R the normal reaction of the surface, ji the coefficient of friction, t the unit tangent vector to the trajectory, and m the unit vector of the normal to the surface. Since where
••=1+I\dt/ ds dt2 dt2 d2r
d2s
/ds'\2dt I
165
when F =
the equation assumes the form
0,
fds\2dt'\ fd2s m(—2t+l1 ds/ J=Rm—piRIt. \d(/ \dt Multiplying it scalarly by [t, m},
dt
tm
=
ds
dr
ds
d2r
m
we
obtain
=0,
ds2
the point moves along a geodesic. 5.177. Take a semigeodesic coordinate system on the surface. Then
i.e.,
ds2 = du2 + G(u, v)dv2.
On the line u = 0, we have TG I equation of geodesic lines that u=O
=
= 1. Besides, we obtain from the
=0.
In the semigeodesic coordinate system,
IfK
=
0,
then 0
and the solution of this equation satisfying the initial conditions indicated
above is './G = 1. Therefore, for all surfaces of zero Gaussian curvature the first fundamental form can be reduced to the form ds2 = du2 + dv2; hence,
all of them are locally isometric to each other.
JfK
a2
(a =
const), then
ds2=du2+cos2Udv2.
'[G=cos", a
If
K=
—
a
(a
= const), then
ds2 = du2 + cash2
U
a
dv2.
166
5.180. The surface S can be obtained by bending the hemisphere so that the two halves of its boundary circumference may overlap each other, and
then glue the surface along these semicircumferences, from which it follows that the geodesics on the surface S not passing through its singular points (ends of meridians) become closed after traversing around the surface twice (i.e., after increasing by 47r).
5.182. It follows from the formula
H Kdii + when
kg =
0
kgdS
=
that
H Kda = And this equality cannot be valid if K
0 at all points of the surface.
6
Manifolds As the atlas of charts, the sets determined by the inequality should be taken. As the coordinate = {Xk > = Xk < functions, all Cartesian coordinates exceptxk should be taken in the chart 6.1.
6.2. Notice that T2 is homeomorphic to the Cartesian product
51 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 4 connected components, while discarding the point 0, which is impossible on 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,,)
* 0, with the equivalence relation (x0:x1: .
x, eR, (\x0: Xxi: .
.
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 be naturally considered identical with R", viz., (x0:x1: .
.
Xx,1).
.
x,
167
x,
x,
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 the be the kth coordinate of the collection (X0: X1: . jth. Let in .
the ith chart, and the 1th coordinate in the jth, respectively (let, for simplicity, i < j). Then (i) — X1(j)
(j) —
X1
— x/1
(I) 1
'
—
—
— 4f
Thus, the transition functions are not only smooth, but also real an analytic.
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 plane
through a certain angle about the origin. The group 0(2) is homeomorphic to the union of two replicas of S'. 6.10. Represent the elements of the group SO(3) as rotations of the space 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 all matrices. 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 a partition of unity, subordinate to the covering 6.28. Use the average operation
f(x)dx.
= Ix• yl
2n, then we project, similarly to the investigation of immersion, along any direction not belonging to Im f. Thereby, we decrease the dimension of the space by unity and continue the process whileN — > 2n. WhenN — I = 2n, a "free" direction may not exist. 1
6.53. The zerodimensional compact manifold consists of a finite number of points. 6.54. Since the pointy is a zerodimensional manifold, its tangent space vanishes.
6.55. In this case, the condition for transversality is equivalent to the subspaces TM1 and TM2 generating the whole tangent space of the ambient 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 to
the fibres under a certain metric, and then construct the required homeomorphisms by means of motions along the integral curves. 6.606.62. Calculate the rank of the Jacobian matrices of the mappings in local coordinates. 6.69. (a), (b), (d), (e): orientable; (c) orientable when n is odd; nonorientable 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 along the midline. 6.71. A manifold is said to be orientable if there exists a collection of charts 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 and
A=
(au).
The mapping A can be considered as a linear operator
C" — C". The realification of the mapping!: GL (n, C)
will assume the formf(A) = We take the basis e1, .
.
.
RA
, e,,, ie1,
where
= .
.
.
A = B + ID.
Let us prove the for1, for
in
,
GL (2n, R)
mula det RA = det A 12 by induction on n. When n = A = a + bi, we obtain I
IdetA 12 =
a2
(a + b2, RA =
detRA = a2 + b2.
n. We now prove it for k = n + 1. Let the statement be proved for k Reduce the operator A to the Jordan normal form (the determinant remains unaltered):
A=
0\ ),
fa1+ib1 0 0 b1
0
e, =
0 0
0 A
where
IdetAI2
0
a1
0
= 173
+
where
D = deti7a2+ib1
\2
0
Let us calculate detR A by expanding it along the last row: 1)3fl+
detR A = (
1det
_bi)
(aiei a1e1
b1
=(_1)3fl+3bn+i(_bn+i)(_1)3fl+3Dn +
=
+
a collection of charts such that in changing the coordinates (the change being smooth)
(z1 ,...,Zn, z1,
.
.
.
(x1,
,z,1)
.
.
.
yfl)
,xhl,yl
(realification) the Jacobians of all the transition functions are positive. 6.74. We obtain from the existing classification of twodimensional, closed, differentiable manifolds (which are orientable), that all of them
are spheres with g handles, i.e., surfaces of genus g. Each of such manifolds is the Riemann surface of a certain polynomial
multiple roots, where g = [n
—
without
]. 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) and generated by the transformation z — 2z. Consider the factor space relative to z — 2z. It carries a complex structure induced by the structure — x S'. Therefore, of the space \ (0) and is homeomorphic to
x S1 also has a complex structure. There exists a fibration
' x
—
—
F = S' x S' =
—
1
X CP'1
with
fibre
the
T2. The fibre and base space have a complex structure
(proved). A complex structure on can be defined by means of a form which 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
variant with respect to the action of S1, where S2" — define the action of thus: (z°,
.
.
.
, z'1
—
)e"1 — (e"1z°,
.
.
174
the former is inCP'1
—
We
—
=
=
+
= +
= =
—
—
+
—
+ kxda.
hxdcx,
Therefore, Let zk be
—
=
—
the coordinates of the first factor
factor
—
and z
of the second
Let
z') E
VkJ =
x
—
—
: zkz
1
0).
The sets VkJ form an open covering of the space troduce complex coordinates r
x
In
'S
(lnzk + y tkj = ± 2iri and on each set VkJ, where y is a vector from are determined modulo I. Therefore, tkj is a point of the torus T(1, y) obtained by gluing together the opposite sides of the parallelogram constructed on the vectors I and y. Thus, we have 2n + 1 coordinates in which determine x T(l, 'y). The mappingf is a homeomorthe mappingf: VkJ — phism. The quantities kwr, 'S, tkj uniquely determine the coordinates zk In fact, and
Zrr Z=
r—r
rr
ZZ—1=
—
r
The quantities
Iz
are
determined similarly (uniquely). Besides,
= lnIzkI + i argzk, whence
+ argzk + y argz'1.
+ 'y
2lrtkJ =
and z '' are known, then arg z"< and arg z can be found has been chosen so that Im y * 0). Consequently, zk and z are also determined uniquely. The transition function in VkJ fl is complex and analytic because it is determined by the formulae: If
I
uwr
I
=
I
= L°'.
175
'kj + 2iri (in kW + y
tuv
Thereby, a complex structure has been introduced on S" This S2"
x
j.
construction introduces a complex structure on any product
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 freely homotopic, then, in traversing them, the orientations are either both altered or both unaltered. There is the shortest periodic geodesic in each free homotopy class of paths on M". We can now prove the required statement. It is clear that it suffices to give the proof for a connected '
manifold M". Let p eM". It suffices to prove that in traversing any smooth loop with the origin and end at the pointp, the orientation in the space is unaltered. Assume the contrary, i.e., that there is a smooth, closed path with the
origin and end at the point p, and, in traversing it, the orientation in is altered. Then there exists a nontrivial periodic geodesic c [0, 11 —. M" which is freely homotopic to this path and shortest in its free homotopy class. Let c(0) = c(I) = q eM". Then the parallel displacement along the geodesic c induces an automorphism r reversing the orientation on the subspace Mq' C TqM orthogonal to the vector c(0). Since c is a geodesic, r is an orthogonal automorphism, and since dim Mq' = n — = 2k, there exist twodimensional Euclidean spaces E1 with respect to r such that Ek invariant = E1 It is clear that 1
.
det r =
.
.
fi det ri E. =
— 1,
and then the relation det nE, =
— 1 is fulfilled for a certain i so that r reverses the orientation onE1. But then r has a nonzero fixed vector u, i.e., ru = u * 0. Now let Y be a parallel vector field along c, for which Y(0) Y(l) = u. Then there exists an open interval Ye R containing
zero such that & Y(t) lies in the domain of the exponential mapping exp on
M" foralLe El,! e [0,1]. Wedefine V:[O, 1] x I—. M" bytheequality V(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 that Y' = 0 and = 0. Since — V(t, e) is a geodesic for any e [0, 1],L "(0) =
—
dt, whereR is the Riemann ten
sor of the manifold M". If follows from the curvature along the geodesic
c being positive that L "(0) < 0, and, therefore, L has a relative maximum 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.
7
Transformation Groups 7.1. Use the theorem on the existence and uniqrieness of a solution to a system of ordinary differential equations of the irst order. 7.2. Construct a vector field such that one of the trajectories may join
the points x0 and x1 and that it may be trivial outside a certain neighbourhood of this trajectory.
7.4. The ratio of the coordinates of the field
should be a rational
number. 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 the formulax — —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 3frame e1, e2, in R3. An arbitrary state of the described system is uniquely determined by a point and
vector v(x) E _where obvious that the mapping x — x, v(x) — v(x)/c is a homeomorphism, xis the unit vector in R3 emanating from the point 0, and v (x) is a unit vector in R3. Shift the origin of v (x) to
x
S2
Iv(x)I =
the
velocity
C = const = 0.
It is
the 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, it is orthogonal tox and and the system teL, e3} is oriented in the same v, y is a homeomorsense as {x, v, y }. Obviously, the mapping x, v — phism. All systems v, y} are in onetoone and continuous correspondence with the matrices associated with the linear transformations 177
which map the orthonormal coordinate 3frame e2, e3} into the orthonormal coordinate 3frame Ix, v, y I. These matrices form the group SO(3) : A SO(3) AA1 = E, and det A = +1. Thus, the space of in
the states of the system under consideration is homeomorphic to the manifold SO(3). Any orthogonal transformation of R3 preserving the orientation is a rotation about a certain axis in a plane perpendicular to it through an angle < ir. where — ir < Therefore, all elements of the group SO(3) are in onetoone and continuous correspondence with the points of a ball of radius ir in R3 whose diametrically opposite boundary points are considered to be identical, It remains to show that the ball glued in this manner is homeomorphic to RP3. In fact, RP3 = S3/Z2 where S3 is standardly embedded in R4. Therefore, RP3 can be considered as a hemisphere of S3 placed in the region with xt 0 and with the diametrically opposite boundary points considered 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 radius 7.21. Let us prove that if A SU(2), then
A= ( \—13 ía
Let
13\ E
1al2+
11312= l,a,13eC.
a/ SU(2). Then
11312= 1
(1)
(2)
1y12+ lô12= 1
detA = ab
(3)
— 13'y
Substituting a =
=
—13ö/y
+ I.y12)= l,ory =
(4)
1.
from (2) in (4), we obtain —
+
—f3,whencea =
Sp(1)=
(q1,q2),q1,q2eQj,
(q1, q2) = Re q1q2. It is easy to see that IqI = I, since, for = 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 quaternions of length 1, i.e., S3 C Q = R4. Further, where
=
q2
q=
a
+ lb +jc + kd=
(a + ib)+j(c —
178
id)=
z1
+jz2,
wherez1andz2eCCQandlz1l2+1z212=1q12.lfIqI=1,then ofSp(J)into 1z112 + 1z212 = 1. Let us arrange for =
zi
+jz2)
(
=
—z2
1z112 + 1z212 = z1
It is obvious that the element 40(g) belongs to SU(2). It is easy to verify that is an isomorphism. that any 7.23. It suffices to show for the group G = SL(2, be the eigenelement of the group G can be joined to ±E. Let
values of the matrix A. Then either (a) X1, X2€R, X2 = since X2 = Consider case (a). det A = ± 1, or (b) X2 = = 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
= (X(l —
0
t) +
Consider case (b). There exists a basis on the plane with respect to which
A is of the form /cosço I
\sin
I
Construct a path —
—sin40
cos
G
(cos(1
—
—sin(1 —
\,sin(i
—
cos(1
1)40
—
7.25. Consider a model of the Lobachevski plane L2 in the upper halfplane (Im z > 0 in the complex notation). The metric is of the form ds2 = (dx2 + dy2)/y2, or, in complex terms, ds2 = dzdz/(z — z)2. Consider the linear fractional transformations of C' into C', keeping the upper halfplane fixed (i.e., transform it into itself). These are transformations of the form
( ( This
az+b cz+d
transformation class preserves the metric, but there are other
transformations preserving it. E.g., the transformation w = —z which is, evidently, a motion, but does not belong to the group G, at least because it is not an analytic function. Similarly, it is easy to verify that the whole class of transformations of the form
H=
= (az + b)/(cz + d); a, b, c, d e R, ad — bc = 179
— l}
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 a group, G U H = ® acts on the Lobachevski plane L2 transitively. Consider a subgroup S of the group ® (i.e., subgroup of transformations keeping a point i fixed) and a certain motion h : L 2 L 2, h (i) = i. We prove that h a S. We shall show that the motion keeping i fixed is fully determined by its action in the tangent plane at the point I. Let h, g be two = g: motions, and — T1L2. Then the transformations h andg act on the geodesics passing through i in the same way; therefore, coincide
on 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 exists an element g a such that g,, = a. Let g(z) = (az + b)/(cz + d) a R2 —. R2. The differential g '(i), whereg '(z) denotes the derivative with respect to complex variable z, viz.,g'(t) = 1/(ci + d)2. Let :
a= angle
d=
c=
cos i.e., it is a rotation of C' through the + I sin In the case of symmetry, consider the transformation w = —z
whose differential is a symmetry, and then apply a linear fractional transformation which carries out a rotation. Now, let h be an arbitrary motion and h (i) = z0. Due to the transitivity of the group there exists
an element g a ® such that gh h (i) = I. The subgroup G is a connected subgroup containing the
identity 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) is disconnected and is the disjoint union of two pathconnected components, viz., 0+ (n), i.e., the collection of matrices with det = + 1, and 0 (n), i.e.; the collection of matrices with det = — 1. When n = 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 H for anyh a H, and anyg a 0(n). Ifg a (n), theng can be joined toE with 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 so that = —E. It is possible to construct two mappingsM(t) = g,
and JV(t) such thatM(t) = andN(t) = g a 0 + (n) and g a 0 (n), respectively. Then =
ghg'
LM(1 = h,
=
R,
=
tN(l) = 180
ghg' h.
=
The elements fi and h are joined with a continuous path which lies in H wholly. Since H is discrete, we obtain that h = i.e., ghg = h for any h H, g n 0(n). (The Schur Lemma.) Let p1 G — GL(V1), p2 : G —. GL(V2) be two irreducible representations of a group G, and letf be a linear mapping of the 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 representation, since reflections with respect to each of the axes can be considered. These are matrices of the form
=
is placed at (I, 1). All such transformations are contained in the group 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. Applying the Schur lemma and using ghg ' = h for any h e H we obtain that h is a 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 a subgroup of a discrete normal subgroup H of 0(n). We show that H contains no other elements. 0 + (n) contains no other elements from H, since H 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. We prove that (n) contains no elements from H. Assume that (n). The matrix h can h 0 (n) fl H. Then ghg — = h for any g be reduced to block triangular form with an odd number of eigenvalues — by a certain orthogonal transformation of the basis with determinant + 1. If dim W1 > 2, then, by an even number of transpositions of the basis vectors, the diagonal elements can be interchanged. Generally speaking, we will obtain a new matrix then, i.e., ghg * h. E.g., we in1
1
and the block; if there is no block, then we interchange — 1. In the case where n = 2, we have only two kinds of matrices to which any matrix from (n) can be reduced by terchange —
I
and the block formed by + 1, +
an orthogonal transformation of the variable, viz.,
0\
/1
I
\0 —1,
and
1—1
0
\
1
0
Then
/
cos
sin
— sin
cos
/
/1 0\ / cos 1*1 II \0 I) \ — sin I/
0\
/1
II \0
—
—
181
sin
cos
Therefore, there are no elements from the discrete normal subgroup of
0(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 following groups: CN, a cyclic group; DN, a dihedral group; T, the tetrahedral group; W, the hexahedral (octahedral) group; and P, the dodecahedral (icosahedral) group. Proof. Let F be a finite rotation group of order N. Consider the fixed
points (poles) of all transformations from F different from the identity transformation. Let the multiplicity of a pole p (number of transformations from F, leavingp fixed) be equal to v. The number of operations dif
ferent from the identity transformation I and leaving the pole p fixed equals
ii
— 1. Let
g
be the set of points into which the pole p is
transformed under the action of elements from the group F. Then is an orbit consisting of points equivalent to each other. The number of points g equivalent to p equals N/v. In fact, the multiplicity of g also equals The transformation L e 1' reduces p into g (i = I n). Let , be transformations leaving the point p fixed, and I
.
.
.
these transformations are different, each element g e F is contained in nv for any orbit N this set, and IF I = N, i.e., N where c = All
is 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), viz., and, on the other hand, —
l),2— 2/N=
2(N—
(because if N =
1,
(1—
2
2, and, we shall have a trivial group). Therefore, 3. The following cases
from the evident relations, we get that 2 ( c are 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 of multiplicity N, i.e., we have obtained a cyclic group of order N of rotations about one axis. v2 At least 2. c = 3. Then 1/v1 + 1/v2 + I/v3 = + 2/N, v1 one 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, v2 = 3, n, a dihedral group DN; (b) = 2, N = 21'3, P3 1/v3 = 1/6 + 2/N. Then the following cases are possible: P3 = 3, 1
182
N= W;
the tetrahedral group T; v3 = 4, N = 24, the hexahedral group = 5, N = 60, the dodecahedral group P.
12,
The dodecahedral group T contains two classes of poles with four poles I + 4 2 + 3 = 12. The generators and relations 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 axes and ef = fe = g. To T, reflections may be added. Let h be an improper rotation, and he = eli, a'h = (i = 0, I, 2). With all the improper motions added, we obtain I TI = 24. The cube and octahedron
of multiplicity 3, I TI =
same group of motions I WI =
possess the
+42=
I
+3 3±
I
6+
1
We have one class of six poles of order 4 (the vertices of the cube), eight poles of order 3 (the centres of the faces), and twelve poles of order 2 (the midpoints of the edges) for W. T is a subgroup of W. This is 24.
obvious 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 =
The
120.
subgroup of proper motions inP is isomorphic to A5. It has twelve poles of order 5 (the vertices of the icosahedron), twenty poles of order 3 (the
centres of the faces) and thirty poles of order 2 (the midpoints of the edges). This group is commutative only partly. The relations in the dodecahedral group are =
abcde = 1, = bce
=
1,
=
1,
1,
1,
1,
= l,bkei_1 = l,i = k,ci'g'e = 1,b g and k, we get
bce = I,
= I,
citbe =
1,
bict =
1.
It follows from the relations bce = and = that i = cb, and tbe = and i = cb that 'b2c = I and from biei = 1, = I. From all the relations obtained, we deduce the 1
1
1
statement of the complete noncommutativity of the group P, i.e.,
P=
[p. p1. Another variant of the corepresentation is
bc =
a5 =
cb =
b3
= abab
where a is the rotation about the axis of order 5, a = (12345), b tion about the axis of order 3, b = (452) and P = A5. 183
1, the
rota
7.45. Let G be a finite group operating effectively on R'7, i.e., if xg = x, x e R'7, g e G, then g = e. The group Zk generated by the eleConsider the space x = G /Zk, ment g also operates effectively on = 0 where x y if y = g'x, g Zk; 7r1(X) = Zk, ir(X) = when i > 1, since R'1 — X is a covering map. Therefore, X is homotopy equivalent to K (Zk, 1), i.e., to a lens space. But the homology K(Zk, 1) is
different from 0 in an infinite number of dimensions, whereas X has no cells of dimensions greater than n. Thus, substantiating the statements:
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) are left to the reader), we complete the proof. (a) if
8
Vector Fields /3; (d)
(b) 3v'21 /7; (c)
8.2. (a)
—
2/5.
8.3.
8.4. 1/4. V3 8.5. (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,
—
xep', 0);
(d) (0, 3x2, 2y3 — 6xz);
(e) (0, —x(x + y2), x3 + —
2xyz);
(sin xz/x, 0, —sin xz/y); (h) (xz/(x2 + y2), yz/(x2 + z2), — 1). (g)
8.22. Use the existence and uniqueness theorem for a solution of a system of ordinary differential equations. 8.23. Investigate the action of the Poisson bracket on the product of two smooth functions. 184
8.26. Consider the case, where = 3/3x1. 8.33. Let z0 = x0 + iy0 be a singular point, and 3u
f(z) = u(z) = iv(z), cIV
(x0,y0)
= (x0,y0)
Since
=
au
f(Zo) =
0.
=
=
3x (x0,y0)
ôy
Let
8y
(x0,Yo)
= aX
(x0,y0)
(x0,y0)
= 0, i.e.,
•av
3x
(z0)+ 1— (z0)= 0, 3x
(z0) =
av
ax
au (zn) = — —
=
ay
0.
grad Ref(z0) = 0. 8.31. Represent the sphere S3 as the group of quaternions of unit
Then
length. 8.34. We shall seek the integral curves only in the halfplane lying over
the straight lineAB. The level curves for the functionf(x) are the arcs of the circumferences for which the linesegment AB is a chord. The vector
gradf(x) is orthogonal to the level curve. Therefore, the vector orthogonal to it is tangent to the level curve, i.e., a circumference, and all the arcs of the circumferences described are the integral curves of the flow v1(x).
are the 8.35. (a) The integral curves of the vector field grad (Re The unique = sin level curves of the conjugate function Im = 0 only is: = 0, since! singular point of the field v = grad(Re
at this point. The point z =
is the case of a degenerate saddle
0
point. Let us give a small perturbation to the function
fl (z —
I nondegenerate saddle points of the second order. Consider the behaviour of the integral curves near to one of the singular points. Expanding the function in Taylor's series, viz.,
Then the singular point splits into n —
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 (nondegenerate critical point);f"(a1) * 0 if and only if all a 185
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 nondegenerate. If the equality f"(a1) = 0 were valid, then we would have the case of a degenerate are multiples of
singular point (i.e., saddle point of the third or higher order).
(b)f(z) = Re
z
(1(z)) =
Im (1(z)) =
+
l/z;
cos
+ cos
P sin
sin
p
=
(p
+ l/p) cos 1/p) sin
(p —
(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 singular points are z = 1, z = I. Both points are nondegenerate. Consider the integral curves for Im f(z), emanating from and returning to the singular points, 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 two separatrices, and the real axis consisting of four separatrices. In the case of grad[Re(f(z))], the separatrices are given similarly, by the equation 2, and have the shape of two loops tangent to each (p + lip) cos other.
(c)f(z) = z + l/z2. Consider grad[Re(,f(z))]. The integral curves of this flow are the level curves of the function lm(f(z)) =
2xy —
(x+y)2 2 2
=r
.
sin
—
r
(in polar coordinates on R2). Similarly, we seek the level curves of the function Re(f(z)): cos r2 2). The singular points are z = I, z = 2, (d) 1(z) z + l/(z z = 3. In a neighbourhood of the point z = 1, f'(z) = — 2(z — I) (the first term in Taylor's expansion). This is a singular point, a saddle. Similarly, the singular point z = 3 is also a saddle point. In a
neighbourhood of the point z =
isf'(z) =
— l/(z
—
2)2
+
.
.
2, .
the expansion in Laurent's series off'
. Therefore, the integral curves of this
flow in a neighbourhood of the point z = curves of the flow grad[Re(l/(z 2))1. 186
2
have the form of the integral
(e)f(z) = z3(z — l)'00(z 2)900. The singular points, i.e., zeroes of the derivative,f'(z), are the following: z1 = 0, a saddle point of the second order; z2 = 1, a saddle point of the 100th order; z3 = 3, a saddle point of the 900th order; z4 (1109 ± l093)/2006, nondegenerate singular points. Locally, in a neighbourhood of each singular point, the integral curves play the role of saddle points (degenerate or nondegenerate at the points 1.09) of the corresponding order. 0.008, z5 z4 (f) 1(z) = + z4(z4 — z = 0, 1(z) can be replaced by 1(z) =
In a neighbourhood of
1
The qualitative behaviour of the curves in a neighbourhood of the point z = 0 is nevertheless unaltered. But adding a constant does not change the form of the trajectories. Therefore, the function 11(z) = cz4, c = where 0, can be considered. The equations of the trajectories are of the z form cp4 cos = k. The point z = 0 is a degenerate singular point, which, after a slight disruption, splits into four nondegenerate points. More precisely, g(z) = c(z — c1)(z E2)(z — e4) is a slight disruption of the function 1(z). (g) 1(z) = 1/100 ln[(z — 2i)/(z At the points z = 21 and z = 4, we have logarithmic singularities. Apart from them, there are no other singular points. (h)f(z) = 1/(z2 + 2z 1). To simplify the notation, we perform a translation w = z + 1. Then 1(w) = l/(w2 — 2). The singular points are w = w= The singular points of grad[Re(f(z))] coincide with the zeroes of f'(w), i.e., w = 0, a singular point (it is nondegenerate, sincef"(w) 0). (i)f(z) = 2/z + 21 In z2. The singular points are (0,0), (1/21, 0). The separatrices 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, the vertices 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 are 1
+
z = l,z = —lOi,f'(z) * Oforallz. (1) 1(z) = 1/z3 — l/(z — The singular points are z = (the
0,
z=
poles of the third order). Differentiating,
f'(z) =
= + (z
We obtain four other points:
z= z
—
i), z =
+ Th/(1 +
+ I), z=
+
The integral curves behave at infinity as the integral curves of the flow grad[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.,
rotv=
3P
aQ —
E0,
or
8)/
8P
c9Q
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_f
Q(x,y) =  (x,y) = 8y
jay
dx + g'(y)
0
=
= Q(x,y)
+
rhus, Q(x, y) =
Q(x, y) — Q(0,
Q(0,y) +
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.
and be two paths from (0, 0) to the point (x, y) in the plane (x, y). Consequently, if rot v = 0, then Let
Pdx
where
+ Qdy + C,
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
rotv' = 0.
v' = (Q,P),
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). Since div V = div v' = 0, we obtain that a(x, y) and b(x, y) are harmonic functions, i.e., Lsa 0. Consider the functionf = a + ib. It is complex and analytic, since the CauchyRiemann equations are valid, viz., ôa
. =
c9b
0x
3y
= P(x,y),
3a
=
c3b —
•
iix
= Q(x,y).
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, which If R4 is regarded as belong to the sphere. It is clear that x(t) = C2, then the integral curve passing through a point (z1, z2) e S3 is of the form e'1z2), since eAt may be written in the complex notation as follows:
fehbo\\
.
Let us assign to this trajectory the point (z1 : z2)
belonging to CP1. The definition of this correspondence is correct because 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 of CP'. it remains to note that the mapping is onetoone and continuous.
189
9
Tensor 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 leftinvariant vector field X are left translations of a oneparameter subgroup, i.e., geodesics. Therefore, where is the vector field of the velocities of we may assume that X = the geodesic y(). Since, by the definition of a geodesic, 71(y) = 0, for any leftinvariant field X, we have Vx(X) = 0. In particular, + = 0. On the other hand, + Y) = 0, i.e., — = [X, Y]. in fact, = YIVXk =
+
+
—
k
= x' since
ax'
=
Y' 
ax'
x'
3yk
,3Xk
.
ax'
(the connection is symmetric). The required statement
follows from the system:
= [X,
— rk ) =
+ X1
+
= 0 and
V.1X =
Y].
9.41. Hint: The invariant definition of a curvature tensor is of the form Since VxY = 1/2[X, Y], we obtain —
R(X, Y)Z =
R(X, Y)Z =
Y], Z].
([X, [Y, Zfl — [Y, [X, Z]])
Taking into account the Jacobi identity.
[X, [Y, Zfl + [Z, [X, Y]] + [Y, [Z, X]] = we get
R(X, Y)Z =
Y], Z]. 190
0,
10
Differential Forms, Integral Formulae, De Rham Cohomology 10.6. (a)—2(z + 1)dxAdyAdz;(b)yzdxAdz + xzdyAdz;(c)6ydxA A 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 sin 0, 0);
cos
2cos0
,—
r3
3,0.
sinO
r
10.15. (a) 2 + zip cos tan
(b)
'
p+
l+p
p
10.17. 4r —
(d)
sin
+
2
(z2
—
+ 2z)ez.
r(r2+ 1)sin0
+ z + C; (b) + c; (e)
10.20. (a) rO + (d) r cos
sin z;
cotO +
r
10.19. (a) p +
sin
cos
c;
cos
(b) r2 +
+
+ z2) + c; (c)
+ c;
z+
+ 0 + C;
sin 8 + c; (e) er sin 0 + In (1 +
(c)
l
+ 02) + c;
+ 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)
R4
R
10.30. (a) 1I '(S1) = R1; (b) H2(S2) = RI; (c) H"(RP2) = 0, k = (11); (d)H'(T2)= R2;H2(T2)= (f)dim H' = k, where k is the number of points excluded. 191
1;
11
General Topology 11.1.
Note that the open ball B" and punctured sphere S" are
homeornorphic. We prove the statement by induction on the dimension of the complex. If the dimension n = 0, then the statement is evident. Let the statement be held for all numbers less than n. Then, by the inductive hypothesis, the (n — 1)dimensional skeleton K" — of the complex in question is embeddable in the Euclidean space W'. This means that consuch that tinuous real functions f1(x) are given on K'1 Let e7 (I = 1, (f)(x) (f1(y), fN(x)) k) be all ndimensional cells of our complex. Then the functions Let .f, (x) are defined on the boundary of each cell e"' (we denote it by e"' be homeomorphic to the interior B" of the closed ball D". We may .
.
assume then that the functionsf,(x) are given on D"\B". Their continuity is preserved, but they may not be onetoone now. Let us extend these functions from D" \ B" to B" (i.e., from to e7) as follows. Let B", and z 0. We putf1(z) If z = 0, then we set z f1(z) = 0. Thus, we have extended the tions on the whole complex K. Now, we define 1(x).
We put
0 (s
=
\Ix!
sin
We define F K —.
ii
1
, .
.
.
xl k(n ±
+ 1) outside e7and
sin
cos
+
/
one".
by the equality
l(x)= (J1(x)
The mapping F is thus onetoone, 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 boundary circumference deformation represented. When this circum ference becomes free of selfintersections and turns into the standardly embedded 192
circumference, it should be glued to the twodimensional disk. Considering the surface obtained in lifting the plane and which is the trace of the circumference deformed, we get an embedding of RP2 in H3. 11.5. The set of the points of selfintersection is homeomorphic to the wedge of three circumferences S1 V S V S'. The vertex of this wedge is a
triple selfintersection point, and any of its points different from the vertex is double. 11.6. The boundary M2 of the normal tubular neighbourhood of radius
c constructed is, obviously, projected onto HP2 (two endpoints of the normal linesegment are sent to its centre lying on RP2). Thus, M2 is a smooth, twodimensional, compact, and closed manifold and a twosheeted covering of the projective plane. If we prove that this manifold is connected, then we shall prove thereby that it is a twodimensional sphere, since S2 is the unique twosheeted connected covering of HP2. To establish the connectedness, it suffices to consider two points on M2
which are the endpoints of the same normal linesegment, and find the path on M2 joining these two points. To construct such a path, it suffices to consider a point Ton RP2 which is the centre of the linesegment under consideration, and take on RP2 a closed path starting and ending at the point Tand such that the orientation of the twoframe slipping along the path and always tangent to HP2, changes in moving along it. Then, by adding to this frame a third vector orthogonal to HP2 and considering the trace of this vector which is cut out by the frame in moving continuously along the closed path, we obtain the continuous path on M2 joining the two selected points. Note. The embedding of the twodimensional sphere in Euclidean threedimensional space helps to prove a remarkable topological fact, viz., the possibility of "turning the twodimensional sphere in R3 inside
out". This task is outside the scope of our course, and we confine ourselves to a short sketch only. The embedding of S2 indicated is such that admits interchanging the exterior and interior of the twodimensional sphere while remaining in the regular embedding class. In fact, it suffices to consider a smooth deformation of the twodimensional sphere along the normal vector field determined by the normal linesegments described above. In doing so, the interior and exterior surfaces of the sphere interchange.
11.7. Consider a vector space over R with a basis of the power of the continuum. We introduce the following topology on it. Consider the "cube" B = —I < < I for all are the coordinates of the vector x, and the crosssection B is of finite codimension, viz., B fl 13—
= O,x13
=
0,
.
.
20
193
.
=
We call the sets the neighbourhoods of the point 0. It is obvious that the point 0 has no countable base for the neighbourhoods in such a topology, i.e., the space constructed does not satisfy the first countability axiom, nor does it satisfy the second, since the first axiom is a corollary to the first. 11.8. Consider the mapping F : S2 —. R2, x — (J(x), g(x)), where
F(S2) C R2 is the image of the sphere. The image F(S2) is a set symmetric 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 plane with the exclusion of the origin onto the unit circumference. In polar coordinates, where this projection can be written in the form = Then h (F(S2)) is a certain centrally symmetric set on the unit circumferenceS1, where h(F(S2)) is the image of the connected set S2 under a continuous mapping h F. Therefore, it is also connected. It is obvious that a connected, centrally symmetric set on S must coincide h
with Further, h(F(S2)) must be 1connected as the image of the which is contrary to the equality h(F(S2)) = 1connected set 11.9. As the space X, take a space /2 whose elements are sequences of real numbers x = (x1 , x2 x,,, .) satisfying the condition 11x211 = .
=
Asthespace
0). Therefore, the cycle y c
(Ma) can be assumed to be homologous to the cycle
e
i.e., the carrier of the (M'1; Hence, it follows that the intersectiony1 fl e
.
e / lies in Mn \ X, (1 k). fl (homologous to the .
.
fl Yk) lies in the complement of (the union) intersection y1 fl . fl fl .. fl U . .. U Xk; the more so, y1 1) . fl y1 C Xk forms a \ (X1 U . . U Xk) = 0, since X1 covering of Mn. Since the intersection of the carriers of the cycles y1 fl fl = 0, the corresponding product of the cocyclesx1 A . A fl . . A x1 0, and A x1 = 0, which contradicts the condition that x1 A the theorem is thus proved. 12.51. Consider the fibration (E,p, x), whereE is the space of all paths of the spaceX starting at the pointx0, andp the mapping associating each .
.
.
.
.
.
.
.
.
.
path with its endpoint. The total space E is considered here in the compactopen topology. The fibre of this fibration is the space X the point x0. It is easy to see that the space E is contractible on itself to a point (each path is contractible on itself to the
point x0). Therefore, lrn(E) = 0, and the homotopy sequence of this fibration 1(E)
L(X)
lrn(tlx) 198
.
.
generates
the
isomorphism The group
+
1(X).
particular,
In
is Abelian whenn
2.
12.52. Definition. A space X is said to be contractible if the identity X is homotopic to the mapping X X sending all X to a
mapping X — point.
Definition. A spaceX is said to be 1connected if ir1(X) = 0. being the identity X, X Since X is contractible, there exist the mapping X — x0 e X. Since the definition mapping X — X, and of a fundamental group does not depend on a base point (up to isoniorphism), let I — X be an arbitrary path on X, y(O) = y(l) = x0, b(r) : X — X stipulates that the x0, c5 : I —. X. The same homotopy loops 'y and ô are homotopic. Thus, any two paths on X are homotopic, i.e., ir1(X) = 0. (where is the 12.53. We prove that (a) any element from wedge of circumferences) is representable as the finite product of is the class of the mapping elements and where (which is the standard embedding); (b) such a representation is unique up and placed in a row. to cancelling the factors being a free group with the generators (a) This is equivalent to 7r1 Represent each cira A. Consider the mapping f: S1 — 7
cumference S' and
e
as the sum of three onedimensional
By the simplicial approximation simplexes P. Q, R and theorem, the mappingf is homotopic to a simplicial mapping F of a cerMultiply the mapping F on tain subdivision of the complex S1 into the right by a homotopy is the identity mapping, where into a base point and stretches transforms to the whole of We obtain a mapping P honiotopic to the original. The mapping P either transforms each of the equal parts, into which S1 is divided, into a point or winds it round one of a A. The class of such a mapping in and e (identity is the product of elements of the form element of the fundamental group), i.e., the constant mapping class. I), which has no and (b) The product = ± I, k i.e., there exist tin a row, is not equal to the unit element in no relations in Under the covering map lr: T — X, the inverse .
image of each point ir(x) = D
found to be in onetoone cor
is
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 S with vertex at the point x is not any path from x1 to x2, then the loop
homotopic to zero: otherwise,
= x2. Let
=
.
where
is a loop traversed in the direction of a circumference of the wedge according to the sign of Take k + 1 replicas of the wedge, and place them one over another. We take in the first and second wedges, cut out a linesegment in both replicas, and join their ends "crosswise", while 199
extending the projection 71 to them. Similarly, we join the second wedge to the third by using , etc. If there are two identical letters in the word one after the other, two linesegments of the same circumference
should be cut out. In doing so, the second operation precedes the first if = 1, and follows it otherwise. We obtain a (k + 1)sheeted covering of the path being covered by a path starting at the lower point, and ending 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, the class
phisms, whence
of the
loopf a
7r1(Y2) and ir1(Y1) =
aCaElr1(Y1),
g,, = (f• g),
S1 — Y2).
:ij1(Y1) — 711(Y1)
are
isomor
12.55. Let 711(X) * 711(Y) be the free product of 711(X) and 711(Y).
be two universal coverings of X and Y, respectively. Let x0 be a base point of X, Y and the wedge X V Y. We construct the following Let
covering: taking X, we consider p 1(x0), where p : — X is a covering map, 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 is defined in a natural manner, viz., each replica of r is mapped into Y via It is obvious that the space obtainp , and each replica ed is a covering of X V Y. Consider the fundamental group X V Y, points
t1 andt2inXV Ysuchthatt1,t2E andapath&. Underthe projection p ", this path will be transformed into a certain loop a representing the class of a in (X V Y). Note that it follows from the construction of the covering map and X and Y being 1connected that the path from t1 to t2 is unique up to homotopy. Let
7r1(X
V Y) be decomposed in terms of the generators
i.e., a =
€ir1(X) and
a
BJn.
Then this representation is unique up to the relations in ir1(X) and 711(Y). In fact, let = . . 67" — I, where 1 is the constant loop at the pointx0 and not all and a5 equal zero (we take a reduced word). Then can be realized as a path in X V Y which, as it is obvious from the form of 1. Thus, we have obtainthe covering map, is not closed; therefore, ed that V Y) = 7r1(X) * 711(Y). The same result follows from the van Kampen theorem on expressing the fundamental group of a complex in terms of the fundamental groups of its subcomplexes and their intersections. .
.
.
.
12.56. Definition. If K is a knot, then the fundamental group 711(R3\K) is called the knot group. Let us find the corepresentation of this group. Consider the upper (or 200
lower) corepresentation of the trefoil knot. Let PK be its projection. The points K1 (i = 1, . . , 6) divide the knot into two alternating classes of closed, connected arcs, viz., the class of overpasses and the class of underpasses. Let A1 , A2, A3 be the overpasses, B1 , B2 B3 the underpasses, and F3 the free group with the generators x, y, z. We call a path v in R2 simple if it is the union of a finite number of closed straight linesegments, its .
origin and endpoint do not belong to PK, and it meets PK in a finite number of points which are not the vertices either of PK or v. We . . associate each path v with v * F3: v * = where Xik are the .
generators of the free group, tk = 1 or
= — I depending on how v under A,k. The upper corepresentation of the group 1r1 (R3 \ K) is
passes
of the form
(x,y,z;r1,r2,r3),
(1)
where r1 = v7 are the relations. The upper corepresentation determined
by formula (1) is known to be the corepresentation of ir1(R3\K). The loops v1, v2, v3 around the overpasses (x, y, z are the generators) satisfy the equalities v1*
=x_IyZy_l,
v2*
=y_lZxZ_l,
v3*
=Z_lxyX_I.
We have obtained the corepresentation (x,y, z; x = 1)•
z=
Substitute z
=
1,y = zxz
then
ir1(R3\K)= (x,y;x
= xyx).
(2)
Thus, 'r1(R3\K) =
(x, y; xyx = yxy). It is impossible to untie the trefoil knot, since its type is different from the trivial knot type. If two knots K' and K" have the same type, then their complementary spaces possess coincident fundamental groups. The group G = (x, y; xyx = yxy) is not the infinite, cyclic group Z. In fact, a homomorphism : G — (23).
0
S3
can be constructed, where S3 is generated by the cycles (12) and
Let K' and K" be two connected subcomplexes of a connected ndimensional, and simplicial complex K, each simplex from K belonging to at least one of these subcomplexes. Their intersection D = K' fl K"
is neither empty nor connected. Let F, F', F", FD be the fundamental groups of the complexes K, K', K", D. We take 0 D as the starting point of the closed paths. Then each closed path of the complex D is, at the same time, a path of the complexes K' and K ". We refer here to the wellknown van Kampen theorem. The group F is obtained from the free
product F' x F" if each pair of elements of F' and F" corresponding to 201
the same element ofF0 are identified, i.e., assuming these elements to be
equal, we thereby add relations to the generators of the groups F' and We now find the fundamental group of the "helical" knot defined as follows. Draw generators on the lateral surface of a circular cylinder at the distance of 2ir/m from each other, and then rotate the upper base through 2irn/m. Then, identifying the bases, add one point at infinity (oo) and thereby turn R3 into S3. Remove from S3 all the points belonging
to the tubular neighbourhood of the knot. We obtain a polyhedron K, the complement of the knot. Divide S3 into two parts by the torus which contains the "helical" knot. The complex K is then divided into two solid tori each of which has been stripped of the knot tubular neighbourhood on the surface. We take one solid torus as K', and the other (with the point at
infinity) as K". The fundamental group F' (resp. F") of the
polyhedron 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 the polyhedron K (the same be done with B). The intersection D of both the solid tori is a twisted annulus. The fundamental group D is also free with one generator which we take to be the midline of the annulus. The group F' ® F" is a free group with the generators A and B. For an appropriate orientation of the paths A and B, the path C considered as an element of the 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 the group A 2 = B3}, where 'y is the trefoil knot. \ 'y) is The two corepresentations of the fundamental group of the trefoil knot obtained are equivalent. We leave the verification of this proposition to the reader. 12.58. We choose the point 0 belonging to Was the starting point of the closed paths. Then each closed path of the complex W is, at the same time, 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 element of the group ir1(Y). We represent Z, Y, Was simplicial complexes. Join each vertex of X to 0 with a path. If the vertex lies in W, then the path may be drawn in W wholly (because of the connectedness). A simplex of an arbitrary dimension of the complex X belongs either to Z (but not to Y) or Y\ Z or Y fl Z. The set of all sitnplexes is thus divided into three disjoint subsets Z, Y, W. The generators a, of the group it (X) can be put into onetoone correspondence with the edges of the complex X. In accordance with that simplicial complex to which this edge belongs (Z, Yor W), we redesignate a into z1 ,y, or respectively. Thus, it1 (X) has as its generators those of the fundamental groups ir1(Y) and ir1(Z) (the
generators of ir1(W) being included in those of it1(Z) and ir1(Y)). The relations in the group it1 (X) are in onetoone correspondence with the 202
edges and triangles of the complex X. Since the complex X has been subdivided into three subsets, these relations also get into three classes. Let us write 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 ir 1(W), relations (1) and (3) for the groups (3) for the groups 711(Z) and ir1(W), and relations (1), (2), (3) for the group 711(X). These relations can be rewritten in the following manner: = = W.
1,
1,
=
1,
(1')
=
1,
(2')
= 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) and 711(Y) corresponding to the same element w of the group 7ri ( W) must be
identified. In proof, we have used the fact that W is connected, since otherwise the statement derived for the group ir1(X) is incorrect. E.g., Z = Y = I (a linesegment), 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 a covering map p : XG — X such that Im 71* (XG)) = G. Introduce multiplication 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.Thenxandyarejoinedto e with the
paths
can 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 to in XG, and x 9 = It remains to verify the correctness of the definition. The following statement can be proved. Let X be a groupoid with
identity, and a,
e 7r1(X,
e). Then ai3 =
a X
meaning
multiplication in ir1(X, e) on the lefthand side, and multiplication inX on the righthand.
We omit the proof leaving it to the reader. The correctness of the definition follows from this statement immediately. 203
12.79. Whenp > 0 and q > 0, for any n < p + q —
1, the isomorSince the pair lrn(SP) + V is a relative (p + q)dimensional cell, it follows x v S") = 0 when from proposition 1 (see below) that
phism x
holds:
V S4')
m