Solutions to the Exercises in
Elementary Differential Geometry
Chapter 1
1.1.1 It is a parametrization of the part of the parabola with x ≥ 0.
1.1.2 (i) γ (t) = (sec t, tan t) with −π/2 < t < π/2 and π/2 < t < 3π/2. Note that γ is
defined on the union of two disjoint intervals: this corresponds to the fact that
the hyperbola y 2 − x2 = 1 is in two pieces, where y ≥ 1 and where y ≤ −1.
(ii) γ (t) = (2 cos t, 3 sin t).
1.1.3 (i) x + y = 1.
(ii) y = (ln x)2 .
1.1.4 (i) γ̇γ (t) = sin 2t(−1, 1).
(ii) γ̇γ (t) = (et , 2t).
1.1.5 γ̇γ (t) = 3 sin t cos t(− cos t, sin t) vanishes where sin t = 0 or cos t = 0, i.e. t =
nπ/2 where n is any integer. These points correspond to the four cusps of the
astroid (see Exercise 1.3.3).
1.1.6 (i) The squares of the distances from p to the foci are
(p cos t ± ǫp)2 + q 2 sin2 t = (p2 − q 2 ) cos2 t ± 2ǫp2 cos t + p2 = p2 (1 ± ǫ cos t)2 ,
so the sum of the distances is 2p.
(ii) γ̇γ = (−p sin t, q cos t) so if n = (q cos t, p sin t) then n.γ̇γ = 0. Hence the
distances from the foci to the tangent line at γ (t) are
n
(p cos t ∓ ǫp, q sin t).n
pq(1 ∓ ǫ cos t)
= 2 2
knk
(p sin t + q 2 cos2 t)1/2
1
2
and their product is
p2 q 2 (1−ǫ2 cos2 t)
= q2 .
(p2 sin2 t+q 2 cos2 t)
(p−f1 ).n
(iii) It is enough to prove that
both sides are equal to q.
=
kp−f 1 k
(p−f2 ).n
kp−f 2 k
t
0
. Computation shows that
a
at
1.1.7 When the circle has rotated through an angle t, its centre has moved to (at, a),
so the point on the circle initially at the origin is now at the point
(a(t − sin t), a(1 − cos t))
(see the diagram above).
1.1.8 Suppose that a point (x, y, z) lies on the cylinder if x2 + y 2 = 1/4 and on the
sphere if (x + 21 )2 + y 2 + z 2 = 1. From the second equation, −1 ≤ z ≤ 1
so let z = sin t. Subtracting the two equations gives x + 14 + sin2 t = 43 , so
x = 12 − sin2 t = cos2 t − 21 . From either equation we then get y = sin t cos t (or
y = − sin t cos t, but the two solutions are interchanged by t 7→ π − t).
√ √
1.1.9 γ̇γ = (−2 sin t + 2 sin 2t, 2 cos t − 2 cos
2t)
=
√ 2( 2 − 1, 1) at t = π/4. So the
√
1
√
tangent line is y − ( 2 − 1) = (x − 2)/( 2 − 1) and the normal line is
√ √
1
y − ( √ − 1) = −(x − 2)( 2 − 1).
2
1.1.10 (i) Putting x = sec t gives y = ± sec t tan t so γ (t) = (sec t, ± sec t tan t) gives
parametrizations of the two pieces of this curve (x ≥ 1 and x ≤ −1).
3t
3t2
(ii) Putting y = tx gives x = 1+t
3 , y = 1+t3 .
1.1.11 (i) From x = 1 + cos t, y = sin t(1 + cos t) we get y = x sin t so
y 2 = x2 (1 − (x − 1)2 ) = x3 (2 − x).
(ii) y = tx so x4 = t2 x3 + t3 x3 = y 2 x + y 3 = y 2 (x + y).
1.1.12 (i) γ̇γ (t) = (− sin t, cos t + cos 2t) so γ̇γ (t) = 0 if and only if t = nπ (n ∈ Z) and
(−1)n + 1 = 0, so n must be odd.
3
(ii) γ̇γ (t) = (2t+3t2 , 3t2 +4t3 ). This vanishes ⇐⇒ t(2+3t) = 0 and t2 (3+4t) = 0,
i.e. ⇐⇒ t = 0.
1.1.13 (i) Let OP make an angle θ with the positive x-axis. Then R has coordinates
γ (θ) = (2a cot θ, a(1 − cos 2θ)).
(ii) From x = 2a cot θ, y = a(1 − cos 2θ), we get sin2 θ = y/2a, cos2 θ =
cot2 θ sin2 θ = x2 y/8a3 , so the Cartesian equation is y/2a + x2 y/8a3 = 1.
1.1.14 Let the fixed circle have radius a, and the moving circle radius b (so that b < a in
the case of the hypocycloid), and let the point P of the moving circle be initially
in contact with the fixed circle at (a, 0). When the moving circle has rotated
through an angle ϕ, the line joining the origin to the point of contact of the
circles makes an angle θ with the positive x-axis, where aθ = bϕ. The point P
is then at the point
γ (θ) = ((a + b) cos θ − b cos(θ + ϕ), (a + b) sin θ − b sin(θ + ϕ))
a+b
a+b
= (a + b) cos θ − b cos
θ , (a + b) sin
θ
b
b
in the case of the epicycloid,
b
a
'
P
and
γ (θ) = ((a − b) cos θ + b cos(ϕ − θ), (a − b) sin θ − b sin(ϕ − θ))
a−b
a−b
θ , (a − b) sin
θ
= (a − b) cos θ − b cos
b
b
in the case of the hypocycloid.
1.1.15 γ̇γ (t) = (et (cos t − sin t), et (sin t + cos t)) so the angle θ between γ (t) and γ̇γ (t) is
given by
cos θ =
e2t (cos2 t − sin t cos t + sin2 t + sin t cos t)
1
γ .γ̇γ
=
= ,
2t
2
2
k γ kk γ̇γ k
e ((cos t − sin t) + (sin t + cos t) )
2
4
so θ = π/3.
1.1.16 γ̇γ (t) = (t cos t, t sin t) so a unit tangent vector is t = (cos t, sin t). The distance
of the normal line at γ (t) from the origin is
|γγ (t)..t| = | cos2 t + t sin t cos t + sin2 t − t sin t cos t| = 1.
1.2.1 γ̇γ (t) = (1, sinh t) so k γ̇γ k = cosh t and the arc-length is s =
Rt
0
cosh u du = sinh t.
1.2.2 (i) k γ̇γ k2 = 14 (1 + t) + 14 (1 − t) + 12 = 1.
2
2
2
9
2
2
(ii) k γ̇γ k2 = 16
25 sin t + cos t + 25 sin t = cos t + sin t = 1.
1.2.3 Denoting d/dθ by a dot, γ̇γ = (ṙ cos θ − r sin θ, ṙ sin θ + r cos θ) so k γ̇γ k2 = ṙ 2 + r 2 .
Hence, γ is regular unless r = ṙ = 0 for some value of θ. It is unit-speed if and
only if ṙ 2 = 1 − r 2 , which gives r = ± sin(θ + α) for some constant α. To see
that this is the equation of a circle of radius 1/2, see the diagram in the proof
of Theorem 3.2.2.
1.2.4 Since u is a unit vector, |γ̇γ.u| = k γ̇γ k cos θ, where θ is the angle between γ̇γ and
Rb
Rb
q − p)..u = (γγ (b) − γ (a))..u = a γ̇γ .u dt ≤ a k γ̇γ k dt.
u, so γ̇γ.u ≤k γ̇γ k. Then, (q
q − p)/ k q − p k gives the result.
Taking u = (q
√
1.2.5 γ̇γ (t) = (6t, 1 − 9t2 ) so k γ̇γ (t) k = 36t2 + 1 − 18t2 + 81t4 = 1 + 9t2 . So the
arc-length is
Z
t
(1 + 9u2 )du = t + 3t3 .
s=
0
1.2.6 We have γ̇γ (t) = (−2 sin t + 2 sin 2t, 2 cos t − 2 cos 2t) so
k γ̇γ (t) k =
p
8(1 − sin t sin 2t − cos t cos 2t =
So the arc-length is
s=
Z
x
0
p
t
8(1 − cos t) = 4 sin .
2
x
x
t
= 16 sin2 .
4 sin dt = 8 1 − cos
2
2
4
1.2.7 The cycloid is parametrized by γ (t) = a(t − sin t, 1 − cos t), where t is the angle
through which the circle has rotated. So
t
γ̇γ = a(1 − cos t, sin t), k γ̇γ k2 = a2 (2 − 2 cos t) = 4a2 sin2 ,
2
and the arc-length is
Z
0
2π
t
t
2a sin dt = −4a cos
2
2
t=2π
= 8a.
t=0
5
1.2.8 The curve intersects the x-axis where cosh t = 3, say at t = ±a where a > 0. We
have k γ̇γ (t) k2 = 2 cosh2 t − 2 cosh t so the arc-length is
Z ap
Z ap
2
2 cosh t − 2 cosh t dt = 2
2 cosh2 t − 2 cosh t dt.
s=
−a
0
To evaluate the integral put u = cosh t. Then,
√
Z 3r
hp
i3
√ √
√
√
2+ 3
u
√ .
du =
s=2
u(u + 1) − ln( u + u + 1) = 2 3− 2−ln
u+1
1
1+ 2
1
...
b + t2 c, where a, b, c are constant vectors. This implies
1.2.9 γ = 0 =⇒ γ = a + tb
that γ is contained in the plane passing through the point with position vector
a and parallel to the vectors b and c (i.e. perpendicular to b × c). (If b and c
are parallel there are infinitely-many such planes.)
1.3.1 (i) γ̇γ = sin 2t(−1, 1) vanishes when t is an integer multiple of π/2, so γ is not
regular.
(ii) γ is regular since γ̇γ 6= 0 for 0 < t < π/2.
(iii) γ̇γ = (1, sinh t) is obviously never zero, so γ is regular.
1.3.2 x = r cos θ = sin2 θ, y = r sin θ = sin2 θ tan θ, so the parametrization in
terms of θ is θ 7→ (sin2 θ, sin2 θ tan θ). Since θ 7→ sin θ is a bijective smooth
map (−π/2, π/2) → (−1, 1), with smooth inverse t 7→ sin−1 t,√ t = sin θ is a
reparametrization map. Since sin2 θ = t2 , sin2 θ tan θ = t3 / 1 − t2 , so the
reparametrized curve is as stated.
1.3.3 (i) γ̇γ = 0 at t = 0 ⇐⇒ m and n are both ≥ 2. If m > 3 the first components of
...
γ̈γ and γ are both 0 at t = 0 so γ̈γ and γ̈γ are linearly dependent at t = 0; similarly
...
if n > 3. So there are four cases: if (m, n) = (2, 2) or (3, 3) then either γ̈γ or γ
is zero at t = 0, so the only possibilities for an ordinary cusp are (m, n) = (2, 3)
...
and (3, 2) and then γ̈γ and γ are easily seen
to 3be linearly independent at t = 0.
t
(ii) Using the parametrization γ (t) = t2 , √1−t
, we get γ̇γ = 0, γ̈γ = (2, 0),
2
...
γ = (0, 6) at t = 0 so the origin is an ordinary cusp.
(iii) Let γ̃γ (t̃) be a reparametrization of γ (t), and suppose γ has an ordinary
cusp at t = t0 . Then, at t = t0 , dγ̃γ /dt̃ = (dγγ /dt)(dt/dt̃) = 0, d2 γ̃γ /dt̃2 =
(d2γ /dt2 )(dt/dt̃)2 , d3γ̃γ /dt̃3 = (d3γ /dt3 )(dt/dt̃)3 + 3(d2γ /dt2 )(dt/dt̃)(d2 t/dt̃2 ).
Using the fact that dt/dt̃ 6= 0, it is easy to see that d2γ̃γ /dt̃2 and d3γ̃γ /dt̃3 are
linearly independent when t = t0 .
1.3.4 (i) If γ̃γ (t) = γ (ϕ(t)), let ψ be the inverse of the reparametrization map ϕ. Then
γ̃γ (ψ(t)) = γ (ϕ(ψ(t))) = γ (t).
(ii) If γ̃γ (t) = γ (ϕ(t)) and γ̂γ (t) = γ̃γ (ψ(t)), where ϕ and ψ are reparametrization
maps, then γ̂γ (t) = γ ((ϕ ◦ ψ)(t)) and ϕ ◦ ψ is a reparametrization map because it
d
is smooth and dt
(ϕ(ψ(t)) = ϕ̇(ψ(t))ψ̇(t) 6= 0 as ϕ̇ and ψ̇ are both 6= 0.
6
1.3.5 (i) γ̇γ (t) = (2t, 3t3 ) which vanishes ⇐⇒ t = 0. So γ is not regular.
(ii) γ̇γ (t) = (− sin t − sin 2t, cos t + cos 2t) so k γ̇γ (t) k = 2 cos 2t . Hence, γ̇γ (t) =
0 ⇐⇒ t is an odd multiple of π. Thus, when t is retricted to the interval
−π < t < π, γ is regular.
4t
cos t
1.3.6 γ̇γ (t) = 2, − (1+t
is obviously never zero, so γ is regular. Let ϕ(t) = 1+sin
2 )2
t.
1
Then, dϕ
dt = − 1+sin t is never zero so ϕ is a reparametrization map. Setting
u = ϕ(t) we get
! 2
2 cos t 2(1 + sin t)2
2 cos t
γ (u) =
= γ̃γ (t),
,
,
=
cos2 t
1 + sin t 1 + (1+sin
1
+
sin
t
2
+
2
sin
t
2
t)
so γ is a reparametrization of γ̃γ .
1.3.7 γ̇γ (t) = (aω cos ωt, b cos t) = 0 ⇐⇒ cos ωt = cos t = 0 ⇐⇒ ωt = (2k +1) π2 and t =
(2l + 1) π2 , where k, l ∈ Z. This means that ω = 2k+1
2l+1 is the ratio of two odd
integers.
R t̃
Rt
1.3.8 We have s = t0 k dγγ /du k du, s̃ = t̃0 k dγ̃γ /dũ k dũ. By the chain rule,
Rt
dγγ /du = (dγ̃γ /dũ)(dũ/du), so s = ± t0 k dγ̃γ /dũ k (dũ/du) du = ±s̃, the sign
being that of dũ/du.
1.3.9 We can assume that the curve γ is unit-speed and that the tangent lines all pass
through the origin (by applying a translation to γ ). Then, there is a scalar λ(t)
such that γ̇γ (t) = λ(t)γγ (t) for all t. Then, γ̈γ = λ̇γγ + λγ̇γ = (λ̇ + λ2 )γγ . This implies
that γ̈γ is parallel to γ̇γ . But these vectors are perpendicular since γ is unit-speed.
b where a , b are constant vectors, i.e. γ is a straight
Hence, γ̈γ = 0, so γ (t) = taa +b
line.
If all the normal lines are parallel, so are all the tangent lines. If γ is assumed to
be unit-speed, this means that γ̇γ is constant, say equal to a. Then γ (t) = taa + b
as before.
1.4.1 It is closed because γ (t + 2π) = γ (t) for all t. Suppose that γ (t) = γ (u).
Then cos3 t(cos 3t, sin 3t) = cos3 u(cos 3u, sin 3u). Taking lengths gives cos3 t =
± cos3 u so cos t = ± cos u, so u = t, π − t, π + t or 2π − t (up to adding multiples
of 2π). The second possibility forces t = nπ/3 for some integer n and the third
possibility is true for all t. Hence, the period is π and for the self-intersections
we need only consider t = π/3, 2π/3, giving u = 2π/3, π/3, respectively. Hence,
there is a unique self-intersection at γ (π/3) = (−1/8, 0).
1.4.2 The curve γ̃γ (t) = (cos(t3 + t), sin(t3 + t)) is a reparametrization of the circle
γ (t) = (cos t, sin t) but it is not closed.
1.4.3 If γ is T -periodic then it is kT -periodic for all k 6= 0 (this can be proved by
induction on k if k > 0, or on −k if k < 0). If γ is T1 -periodic and T2 -periodic
7
then it is k1 T1 - and k2 T2 -periodic for all non-zero integers k1 , k2 , so
γ (t + k1 T1 + k2 T2 ) = γ (t + k1 T1 )
as γ is k2 T2 -periodic, which = γ (t) as γ is k1 T1 -periodic.
1.4.4 If γ is T -periodic write T = kT0 + T1 where k is an integer and 0 ≤ T1 < T0 . By
Exercise 1.4.3 γ is T1 -periodic; if T1 > 0 this contradicts the definition of T0 .
1.4.5 (i) Choose T1 > 0 such that γ is T1 -periodic; then T1 is not the smallest positive
number with this property, so there is a T2 > 0 such that γ is T2 -periodic.
Iterating this argument gives the desired sequence.
(ii) The sequence {Tr }r≥1 in (i) is decreasing and bounded below, so must converge to some T∞ ≥ 0. Then γ is T∞ -periodic because (using continuity of γ )
γ (t + T∞ ) = limr→∞ γ (t + Tr ) = limr→∞ γ (t) = γ (t). By Exercise 1.4.3, γ is
(Tr − T∞ )-periodic for all r ≥ 1, and this sequence of positive numbers converges
to 0.
(iii) If {Tr } is as in (i) and Tr → 0 as r → ∞, then by the mean value theorem
0 = (f (t + Tr ) − f (t))/Tr = f˙(t + λTr ) for some 0 < λ < 1. Letting r → ∞ gives
f˙(t) = 0 for all t, so f is constant.
1.4.6 Following the hint, since T0 = (ki /k)Ti is an integer multiple of Ti , each γi is T0 periodic. Let T be the union of the finite sets of real numbers {Ti , 2Ti , . . . , ki Ti }
over all i such that γi is not constant, and let P = {T ′ ∈ T | γ is T ′ -periodic}.
Then P is finite (because T is) and non-empty (because T ∈ P). The smallest
element of P is the smallest positive number T0′ such that γ is T0′ -periodic (since
if γ is T ′ -periodic either T ′ > T or T ′ ∈ P). By Exercise 1.4.4, T0 = k ′ T0′ for
some integer k ′ and then there are integers ki′ such that T0′ = ki′ Ti for all i such
that γi is not constant. Then, ki Ti /k = k ′ ki Ti so kk ′ divides each ki . As k is
the largest such divisor, k ′ = 1, so T0 = T0′ .
2
2
u −3
2
2
1.4.7 If γ (t) = γ (u) then tt2 −3
+1 = u2 +1 . This implies t = u . This shows that
γ is not periodic and that for a self-intersection we must have t = −u. The
√
2
−3)
= 0, so t = 0 or ± 3. Hence, the unique
equation γ (t) = γ (−t) implies t(tt2 +1
√
√
self-intersection is at γ ( 3) = γ (− 3) = 0.
1.4.8 If γ (t) = γ (u) then sin t = sin u. Also, k γ (t) k2 = k γ (u) k2 , which gives
5 + 4 cos t = 5 + 4 cos u, hence cos t = cos u. So t = u + 2nπ for some integer n.
We must have cos ω(t + 2nπ) = cos ωt so ω(t + 2nπ) = ωt + 2mπ for some integer
m. Hence, ω = m/n is rational. The period T of γ is the smallest positive
number such that γ (t + T ) = γ (t) for all t. From the first part we know that
T = 2N π for some integer N . Then, N is the smallest positive integer N such
that ω(t+2N π) = ωt+2M π for some integer M , i.e. the smallest positive integer
such that mN/n is an integer. If m and n have no common factor, N = n.
8
1.5.1 x(1 − x2 ) ≥ 0 ⇐⇒ x ≤ −1 or 0 ≤ x ≤ 1 so the curve is in (at least) two pieces.
The parametrization is defined for t ≤ −1 and 0 ≤ t ≤ 1 and it covers the part
of the curve with y ≥ 0.
1.5.2 If γ (t) = (x(t), y(t), z(t)) is a curve in the surface f (x, y, z) = 0, differentiating
f (x(t), y(t), z(t)) = 0 with respect to t gives ẋfx + ẏfy + żfz = 0, so γ̇γ is perpendicular to ∇ f = (fx , fy , fz ). Since this holds for every curve in the surface, ∇f
is perpendicular to the surface. The surfaces f = 0 and g = 0 should intersect in
a curve if the vectors ∇ f and ∇g are not parallel at any point of the intersection.
1.5.3 Let γ (t) = (u(t), v(t), w(t)) be a regular curve in R3 . At least one of u̇, v̇, ẇ
is non-zero at each value of t. Suppose that u̇(t0 ) 6= 0 and x0 = u(t0 ). As in
the ‘proof’ of Theorem 1.5.2, there is a smooth function h(x) defined for x near
x0 such that t = h(x) is the unique solution of x = u(t) for each t near t0 .
Then, for t near t0 , γ (t) is contained in the level curve f (x, y, z) = g(x, y, z) = 0,
where f (x, y, z) = y − v(h(x)) and g(x, y, z) = z − w(h(x)). The functions f
and g satisfy the conditions in the previous exercise, since ∇f = (−v̇h′ , 1, 0),
∇ g = (−ẇh′ , 0, 1), a dash denoting d/dx.
1.5.4 The equation cannot be satisfied if x = 1, so the line x = 1 separates the curve
into at least two connected pieces. The curve γ is a parametrization because
(x − 1)2 (x2 + y 2 ) = cos2 t((1 + cos t)2 + (sin t + tan t)2 )
= cos2 t(1 + cos t)2 (1 + tan2 t) = (1 + cos t)2 = x2 .
But since tan t is undefined when t = ±π/2, ±3π/2, ..., γ must be defined on one
of the intervals (−π/2, π/2), (π/2, 3π/2), etc. The restriction of γ to the first of
these intervals parametrizes the part of the curve with x > 1, the second gives
the part with x < 1. This shows that the two parts of the curve with x < 1 and
with x > 1 are connected.
1.5.5 Letting f (x, y, z) = x2 + y 2 = 1/4, g(x, y, z) = x2 + y 2 + z 2 + x − 3/4, we find
∂f ∂f ∂f
∂g ∂g ∂g
that ( ∂x
, ∂y , ∂z ) = (2x, 2y, 0), ( ∂x
, ∂y , ∂z ) = (2x + 1, 2y, 2z). If these vectors
are parallel, z = 0; if y 6= 0 the vectors must then be equal and this is impossible;
so y = 0 and then we must have x = 1/2 to satisfy both equations. Hence, the
vectors are parallel only at the point (1/2, 0, 0).
For the parametrization γ in Exercise 1.1.8, γ̇γ (t) = (− sin 2t, cos 2t, cos t). This
is never zero as k γ̇γ (t) k2 = 1 + cos2 t, so γ is regular.
1.5.6 Define Θ(t) = tan πθ(t)/2, where θ is the function defined in Exercise 9.4.3.
Then Θ is smooth, Θ(t) = 0 if t ≤ 0, and Θ : (0, ∞) → (0, ∞) is a bijection. The
curve
(Θ(t), Θ(t))
if t ≥ 0,
γ (t) =
(−Θ(−t), Θ(−t)) if t ≤ 0,
is a smooth parametrization of y = |x|.
9
There is no regular parametrization of y = |x|. For if there were, there would
be a unit-speed parametrization γ̃γ (t), say, and we can assume that γ̃γ (0) = (0, 0).
The unit tangent vector γ̃γ˙ would have to be either √12 (1, 1) or − √12 (1, 1) when
x > 0, so by continuity we would have γ̃γ˙ (0) = ± √1 (1, 1). But, by considering the
2
part x < 0 in the same way, we see that γ̃γ˙ (0) = ± √12 (1, −1). These statements
are contradictory.
Chapter 2
2.1.1 (i) γ is unit-speed (Exercise 1.2.2(i)) so
1
1
1
κ = k γ̈γ k = k ( (1 + t)−1/2 , (1 − t)−1/2 , 0) k = p
.
4
4
8(1 − t2 )
(ii) γ is unit-speed (Exercise 1.2.2(ii)) so κ = k γ̈γ k = k (− 45 cos t, sin t, 53 cos t) k
= 1.
t,0)×(0,cosh t,0)k
2
cosh t
(iii) κ = k(1,sinh
= cosh
3 t = sech t using Proposition 2.1.2.
k(1,sinh t,0)k3
(iv) (−3 cos2 t sin t, 3 sin2 t cos t, 0)×(−3 cos3 t+6 cos t sin2 t, 6 sin t cos2 t−3 sin3 t, 0)
sin2 t cos2 t)k
= (0, 0, −9 sin2 t cos2 t), so κ = k(−3 k(0,0,−9
= 3| sin 1t cos t| . This
cos2 t sin t,3 sin2 t cos t,0)k3
becomes infinite when t is an integer multiple of π/2, i.e. at the four cusps
(±1, 0) and (0, ±1) of the astroid.
2.1.2 The proof of Proposition 1.3.5 shows that, if v(t) is a smooth (vector) function
of t, then k v(t) k is a smooth (scalar) function of t provided v(t) is non-zero
for all t. The result now follows from the formula in Proposition 2.1.2. The
curvature of the regular curve γ (t) = (t, t3 ) is κ(t) = 6|t|/(1 + 9t4 )3/2 , which is
not differentiable at t = 0.
2.1.3 Working in R3 , γ̇γ (t) = (1 − cosh 2t, 2 sinh t, 0) = 2 sinh t(− sinh t, 1, 0), γ̈γ (t) =
γ ×γ̈γ k
kγ̇
(−2 sinh 2t, 2 cosh t, 0) which gives γ̇γ ×γ̈γ = (0, 0, 4 sinh2 t cosh t) and κ = kγ̇γ k3 =
1/2 sinh t cosh2 t. This is non-zero if t > 0, but → 0 as t → ∞.
10
2.1.4 We have
γ̇γ (t) = (sec t tan t, sec t tan2 t + sec3 t, 0),
γ̈γ (t) = (2 sec3 t − sec t, 6 sec3 t tan t − sec t tan t),
which gives γ̇γ × γ̈γ = sec4 t(0, 0, 2 sec2 t − 3). Hence, the curvature vanishes where
sec2 t = 3/2. If −π/2 < t < π/2, cos t > 0 so the curvature vanishes at the
two values
of t between −π/2 and π/2 at which cos t = 2/3, i.e. at the point
√
(3/2, 5/2).
ns = 0 and use ṫt = κsns .
2.2.1 Differentiate t.n
2.2.2 If γ is smooth, t = γ̇γ is smooth and hence so are ṫt and ns (since ns is obtained
ns is smooth.
by applying a rotation to t). So κs = ṫt.n
2.2.3 For the first part, from the results in Appendix 1 it suffices to show that κ̃s = −κs
if M is the reflection in a straight line l. But this is clear: if we take the fixed
angle ϕ0 in Proposition 2.2.1 to be the angle between l and the positive x-axis,
then (in the obvious notation) ϕ̃ = −ϕ. Conversely, if γ and γ̃γ have the same
non-zero curvature, their signed curvatures are either the same or differ in sign.
In the first case the curves differ by a direct isometry by Theorem 2.2.5; in
the latter case, applying a reflection to one curve gives two curves with the same
signed curvature, and these curves then differ by a direct isometry, so the original
curves differ by an opposite isometry.
2.2.4 The first part is obvious as the effect of the dilation is to multiply s by a and
leave ϕ unchanged.
For the second part, consider the small piece of the chain between the points with
arc-length s and s + δs. The net horizontal force on this piece is (in the obvious
notation) δ(T cos ϕ), and as this must vanish T cos ϕ must be a constant, say λ.
The net vertical force is δ(T sin ϕ), and this must balance the weight of the piece
of chain, which is a constant multiple of δs. This shows that T sin ϕ = µs + ν
for some constants µ, ν, and ν must be zero because ϕ = s = 0 at the lowest
point of C. From T cos ϕ = λ, T sin ϕ = µs, we get tan ϕ = s/a where a = λ/µ.
= 1/a, so the signed curvature is
Hence, sec2 ϕ dϕ
ds
κs = dϕ/ds = 1/a sec2 ϕ = 1/a(1 + tan2 ϕ) =
s2
a
.
+ a2
Using the first part and Example 2.2.4 gives the result.
ns /dt = (1 − λκs )ds/dt t, so the arc-length sλ
2.2.5 We have dγγ λ /dt = dγγ /dt + λdn
of γ λ satisfies dsλ /dt = |1 − λκs |ds/dt. The unit tangent vector of γ λ is tλ =
ns . Then,
(dγγ λ /dt)/(dsλ/dt) = ǫtt, hence the signed unit normal of γ λ is nλs = ǫn
11
the signed curvature κλs of γ λ is given by
dttλ /dt
ǫ
dtt
dttλ
=
=
λ
ds
|1 − λκs |(ds/dt)
|1 − λκs | ds
κs
ǫκs
κs
=
ns =
nλs =
nλ .
1 − λκs
1 − λκs
|1 − λκs | s
κλsnλs =
2.2.6 ǫ(s0 ) lies on the perpendicular bisector of the line joining γ (s0 ) and γ (s0 + δs).
This gives (ǫǫ(s0 ) − 21 (γγ (s0 ) +γγ (s0 + δs)))..(γγ (s0 + δs) −γγ (s0 )) = 0. Using Taylor’s
theorem, and discarding terms involving powers of δs higher than the second, this
gives (with all quantities evaluated at s0 ) (ǫǫ − γ )..γ̇γ δs + 21 (ǫǫ.γ̈γ − 1 − γ .γ̈γ )(δs)2 = 0.
This must also hold when δs is replaced by −δs; adding and subtracting the
two equations gives (ǫǫ − γ )..γ̇γ = 0 and (ǫǫ − γ )..γ̈γ = 1. The first equation gives
ns for some scalar λ, and since γ̈γ = κsns the second gives λ = 1/κs .
ǫ = γ + λn
2.2.7 The tangent vector of ǫ is t + κ1s (−κs t) − κκ̇2s ns = − κ̇κs2 ns so its arc-length is
s
s
R
R
u = k ǫ̇ǫ k ds = κκ̇s2 ds = u0 − κ1s , where u0 is a constant. Hence, the unit
s
ns and its signed unit normal is t. Since −dn
ns /du =
tangent vector of ǫ is −n
κ3s
3
κst/(du/ds) = κ̇s t, the signed curvature of ǫ is κs /κ̇s .
ns (s) for some λ. Hence,
Any point on the normal line to γ at γ (s) is γ (s) + λn
the normal line intersects ǫ at the point ǫ(s), where λ = 1/κs (s), and since the
tangent vector of ǫ there is parallel to ns (s) by the first part, the normal line is
tangent to ǫ at ǫ(s).
Denoting d/dt by a dash, γ ′ = a(1 − cos t, sin t) so the arc-length s of γ is
given by ds/dt = 2a sin(t/2) and t = dγγ /ds = (sin(t/2), cos(t/2)). So ns =
1
(− cos(t/2), sin(t/2)) and ṫt = (dtt/dt)/(ds/dt) = 4a sin(t/2)
(cos(t/2), − sin(t/2)) =
ns , so the signed curvature of γ is −1/4a sin(t/2) and its evolute
−1/4a sin(t/2)n
ǫ(t) = a(t − sin t, 1 − cos t) − 4a sin(t/2)(− cos(t/2), sin(t/2))
= a(t + sin t, −1 + cos t).
Reparametrizing ǫ by t̃ = π + t, we get a(t̃ − sin t̃, 1 − cos t̃) + a(−π, −2), so ǫ is
obtained from a reparametrization of γ by translating by the vector a(−π, −2).
2.2.8 The free part of the string is tangent to γ at γ (s) and has length ℓ − s, hence the
ns
stated formula for ι(s). The tangent vector of ι is γ̇γ − γ̇γ + (ℓ − s)γ̈γ = κs (ℓ − s)n
ι
(a dot denotes d/ds). The arc-length v of is given by dv/ds = κs (ℓ − s) so
ns /dv =
its unit tangent vector is ns and its signed unit normal is −tt. Now dn
1
−1
n
κs (ℓ−s) ṅ s = ℓ−s t , so the signed curvature of ι is 1/(ℓ − s).
√
2.2.9 The arc-length parametrization of the catenary is γ̃γ (s) = (sinh−1 s, 1 + s2 ).
The involute
1
s
−1
˙
,√
= (u − tanhu, sechu)
ι(s) = γ̃γ (s) − sγ̃γ (s) = sinh s − √
1 + s2
1 + s2
12
if u = sinh−1 s. Thus,
if (x, y) is a point on the involute ι, u = cosh−1 (1/y) and
p
x = cosh−1 (1/y) − 1 − y 2 .
2.2.10 The rotation ρ−θ(s) takes the tangent line of γ at γ (s) to l and the line joining
q and γ (s) to a line parallel to that joining Γ(s) to p + saa. Hence,
p + saa) = ρ−θ(s) (q
q − γ (s)),
Γ(s) − (p
which gives the stated equation. Now,
d
q − γ (s)) − ρ−θ(s)γ̇γ (s).
Γ(s) = a +
ρ−θ(s) (q
Γ̇
ds
The last term is clearly parallel to a and as they are both unit vectors they are
equal. So we want to prove that
d
q − γ (s))..ρ−θ(s) (q
q − γ (s)) = 0.
ρ−θ(s) (q
ds
v)t dA
If A = ρ−θ(s) , v = q − γ (s), we have to show (in matrix notation) (Av
ds v = 0,
t t dA
i.e. v A ds v = 0. Since A is orthogonal, this follows from parts (i) and (ii) of
the hint.
P
P
P
To prove (i), use components: vt Svv = i,j vi vj Sij = i,j vj vi Sji = − i,j vi vj Sij .
For (ii), differentiate At A = I.
2.2.11 If two unit-speed curves have the same non-zero curvature, their signed curvatures are either the same or differ in sign. In the first case the curves differ by a
direct isometry by Theorem 2.2.6; in the latter case, applying a reflection to one
curve gives two curves with the same signed curvature by Exercise 2.2.3, and
these curves then differ by a direct isometry.
2.2.12 By applying a translation we can assume that all the normal lines pass through
ns (t) for all t.
the origin. Then, there is a scalar λ(t) such that γ (t) = λ(t)n
ns − λκs t using Exercise 2.2.1. It follows that λ̇ = 0,
Differentiating gives t = λ̇n
so λ is constant, and that λκs = −1, so κs = −1/λ is constant. Hence, γ is a
circle by Example 2.2.7.
2.2.13 The arc-length
Z
Z
s=
k γ̇γ k dt =
k ekt (k cos t − sin t, k sin t + cos t) k dt
√
Z p
k 2 + 1 kt
kt
k 2 + 1e dt =
e + c,
=
k
where c is a constant. Taking c = 0 makes s → 0 as t → ∓∞ if ±k > 0.
13
Since t = γ̇γ / k γ̇γ k = √k12 +1 (k cos t − sin t, k sin t + cos t), the signed unit normal
is ns = √k12 +1 (−k sin t − cos t, k cos t − sin t). So
dtt/dt
e−kt
e−kt
dtt
=
= 2
(−k sin t − cos t, k cos t − sin t) = √
ns ,
ds
ds/dt
k +1
k2 + 1
hence κs = 1/ks. By Theorem 2.2.6, any other curve with the same signed
curvature is obtained from the logarithmic spiral by applying a direct isometry.
2.2.14 Let δ (t) be the foot of the perpendicular from p to the tangent line of γ at γ (t).
ns for some scalar λ = (δδ − p)..ns . But γ − δ is parallel to the
Then δ = p + λn
tangent line, so is perpendicular to ns : (γγ − δ )..ns = 0. Hence, δ .ns = γ .ns and
we get λ = (γγ − p)..ns .
p)..t )n
ns +((γγ −p
p)..ns )tt). So δ̇δ = 0
Using Exercise 2.1.1, we find that δ̇δ = −κs (((γγ −p
ns + ((γγ − p)..ns )tt) = 0. Since ns and t
if and only if either κs = 0 or (((γγ − p)..t)n
are perpendicular, the last equation can hold only if (γγ − p)..t = (γγ − p)..ns = 0,
which implies that γ = p. Hence, δ is regular if and only if γ has nowhere
vanishing curvature and does not pass through p.
For the circle γ (t) = (cos t, sin t), ns = −γγ and so
δ (t) = (−2, 0) + (2 + cos t, sin t)..(cos t, sin t)(cos t, sin t)
= ((1 + 2 cos t) cos t − 2, (1 + 2 cos t) sin t),
which is obtained by translating the limaçon in Example 1.1.7 by (−2, 0).
ns . Since t and ns are perpendicular
2.2.15 (i) Differentiating γ = rtt gives t = ṙtt + κs rn
vectors, it follows that κs = 0 and γ is part of a straight line.
ns gives t = ṙn
ns + rṅ
ns = ṙn
ns − κs rtt (Exercise 2.1.1).
(ii) Differentiating γ = rn
Hence, ṙ = 0, so r is constant, and κs = −1/r, hence κs is constant. So γ is part
of a circle.
(iii) Write γ = r(tt cos θ + ns sin θ). Differentiating and equating coefficients of t
and ns gives ṙ cos θ − κs r sin θ = 1, ṙ sin θ + κs r cos θ = 0, from which ṙ = cos θ
and κs r = − sin θ. From the first equation, r = s cos θ (we can assume the
arbitrary constant is zero by adding a suitable constant to s) so κs = −1/s cot θ.
By Exercise 2.2.13, γ is obtained by applying a direct isometry to the logarithmic
spiral defined there with k = − cot θ.
1
1
(1, 2t), ns = √1+4t
2.2.16 For the parabola γ (t) = (t, t2 ), t = √1+4t
(−2t, 1), so
2
2
λ
2λ
2λ
2
√ λ
γ
(1, 2t).
,
t
+
.
This
gives
γ̇
=
1
−
γ λ (t) = t − √1+4t
2
(1+4t2 )3/2
1+4t2
Hence, γ̇γ λ (t) = 0 ⇐⇒ 2λ = (1 + 4t2 )3/2 . If λ < 1/2 this has no solution
for t, so γ λ is regular. If λ = 1/2, the only solution is t = 0 so (0, 1/2) is the
unique singular point of γ . If λ > 1/2 there are two solutions for t, hence γ λ has
two singular points.
14
2.2.17 Ḟ = 2γ̇γ.(γγ − c). So F (t0 ) = Ḟ (t0 ) = 0 ⇐⇒ k γ (t0 ) − c k = R and γ̇γ (t0 ) is
perpendicular to γ (t0 ) − c. This means that γ (t0 ) is on the circle C and that
γ̇γ (t0 ) = t(t0 ) is parallel to the tangent to C at that point (since this tangent is
perpendicular to the radius γ (t0 ) − c).
Now F̈ = 2ṫt. (γγ −cc)+2 (since γ is unit-speed), so F̈ (t0 ) = 0 ⇐⇒ ṫt(t0 ).(γγ (t0 )−cc) =
ns (t0 ) is perpendicular to t(t0 ) (since γ is unit-speed),
−1. Now ṫt(t0 ) = κs (t0 )n
hence parallel to γ (t0 )−cc from the condition Ḟ (t0 ) = 0. So the condition F̈ (t0 ) =
ns (t0 ) for some scalar λ and we have κs (t0 )n
ns (t0 )..λn
ns (t0 ) =
0 gives γ (t0 ) − c = λn
−1 which gives λ = −1/κs (t0 ). So the conditions F (t0 ) = Ḟ (t0 ) = F̈ (t0 ) = 0 are
1
satisfied if and only if the centre of C is c = γ (t0 ) + κs (t
ns (t0 ) and its radius is
0)
k γ (t0 ) − c k = |λ| = 1/|κs (t0 )|. Comparing with Exercise 2.2.6 shows that C is
the osculating circle of γ at γ (t0 ).
2.2.18 From the solution of Exercise 2.2.16, t =
1
√ 1
(1, 2t), ns = √1+4t
(−2t, 1),
2
1+4t
√2
k = 1 + 4t2 (a dot denoting d/dt).
and if s is the arc-length of γ , ds/dt = k γ̇γ
Hence,
1
2
4t
dtt
,
=√
,
ṫt = −
ds
(1 + 4t2 )2 (1 + 4t2 )2
1 + 4t2
hence κs = 2(1 + 4t2 )−3/2 . The evolute is therefore
1
1
ǫ = γ + ns = (t, t2 ) + (1 + 4t2 )(−2t, 1) =
κs
2
1
−4t , 3t +
2
3
2
.
p
a2 sin2 t + b2 cos2 t = 1/λ, say. Then,
2.2.19 If s is the arc-length of γ , ds/dt =
denoting d/ds by a dot, t = γ̇γ = λ(−a sin t, b cos t), ns = λ(−b cos t, −a sin t),
ns
ṫt = λ[λ(−a cos t, −b sin t) − λ3 (a2 − b2 ) sin t cos t(−a sin t, b cos t)] = λ3 abn
after some simplification. Hence, κs = λ3 ab =
λ
ǫ = (a cos t, b sin t) + 3 (−b cos t, −a sin t) =
λ ab
ab
(a2 sin2 t+b2 cos2 t)3/2
and so
a2 − b2
b2 − a2
3
3
cos t,
sin t .
a
b
ns . Then, γ̇γ λ = γ̇γ +λṅ
ns = (1−λκs )n
ns . Assuming that λκs 6= 1 (so
2.2.20 Let γ λ = γ +λn
λ
λ
t
that γ λ is regular), we have (in an obvious notation) ds
ds = |1 − λκs |, t = ±t ,
nλs = ±n
ns , where ± is the sign of 1 − λκs . Hence,
±ṫt
κsns
κsnλs
dttλ
=
=
±
=
.
dsλ
|1 − λκs |
|1 − λκs |
|1 − λκs |
15
So κλs =
κs
|1−λκs |
and
ns +
ǫλ = γ + λn
|1 − λκs |
1 − λκs
ns ) = γ + λn
ns +
(±n
ns = ǫ.
κs
κs
2.2.21 (i) With the notation in Exercise 2.2.7, the involute of ǫ is
ι(u) = ǫ + (ℓ − u)
1
dǫǫ
ns = γ − (ℓ − u0 )n
ns ,
= γ + ns − (ℓ − u)n
du
κs
since u = u0 − κ1s , so ι is the parallel curve γ −(ℓ−u0 ) .
(ii) Using the results of Exercise 2.2.8, the evolute of ι is
ι + (ℓ − s)(−tt) = γ + (ℓ − s)tt − (ℓ − s)tt = γ .
2.2.22 (i) If γ (θ) = (x(θ), y(θ)), we have
x cos θ + y sin θ = p.
Denoting d/dθ by a dot, we also have
ẏ
dy
cos θ
=
=−
.
ẋ
dx
sin θ
Differentiating the first equation and using the second gives
−x sin θ + y cos θ = ṗ.
Solving the first and third equations for x and y gives the stated formula for γ .
(ii) From (i), γ̇γ = (p + p̈)(− sin θ, cos θ), so k γ̇γ k = |p + p̈|. Hence, γ is regular
if and only if p + p̈ is never zero. In that case p + p̈ must be either always > 0
or always < 0. But the latter
R 2π case is impossible since it implies that p̈ < 0,
contradicting the fact that 0 p̈ dθ = 0 since p is smooth and 2π-periodic.
1
d
θ + π2 = dθ
(iii) If s is the arc-length of γ , the signed curvature is ds
ds = ds/dθ =
1
1
γ k = p+p̈ .
kγ̇
(iv) Since
ds
dθ
= p + p̈, the length of γ is
Z
0
2π
(p + p̈)dθ =
Z
2π
p dθ.
0
(v) The first part is obvious. The foot of the perpendicular from the origin to
the tangent line at γ (θ) is p(θ)(cos θ, sin θ), so w(θ) is the distance between this
16
point and p(θ + π)(cos(θ + π), sin(θ + π)) = p(θ + π)(− cos θ, − sin θ) which is
p(θ) + p(θ + π).
(vi) The function q(θ) = w(θ) − w(θ + π) is smooth and q(0) = −q(π/2) because
w(0) = w(π). Hence, for some angle θ0 with 0 ≤ θ0 ≤ π/2, we have q(θ0 ) = 0,
i.e. w(θ0 ) = w(θ0 +π/2). This means that the tangent lines at θ0 , θ0 +π/2, θ0 +π
and θ0 + 3π/2 form a square.
(vii) From (iv) and (v), the length is
Z
π
(p(θ) + p(θ + π))dθ =
0
Z
π
w(θ) dθ,
0
which is equal to πD if w(θ) is equal to the constant D. kπ
2 kθ
(viii) If p(θ) is as stated, w(θ) = a cos2 kθ
+ 2b = a + 2b.
2 + cos
2 + 2
(ix) If |k| = 1 we find that
γ (θ) =
1
1
a, 0 +
a + b (cos θ, sin θ),
2
2
which is the circle with centre ( 21 a, 0) and radius
− 21 ka sin kθ, p̈ = − 12 k 2 a cos kθ, so
κs =
1
a cos2 kθ
2
−
1 2
k a cos kθ
2
+b
=
1
2a
+b+
1
2a
+ b. In general, ṗ =
1
,
− k 2 )a cos kθ
1
2 (1
which is constant ⇐⇒ k = 0 or ± 1 (and k = 0 does not give a regular curve).
2.2.23 Parametrize the parabola by x = sinh t, y =
at the origin) is
s=
Z
t
1
2
sinh2 t. The arc-length (starting
(cosh2 u + cosh2 u sinh2 u)1/2 du =
0
1
(t + sinh t cosh t).
2
In the notation of Exercise 2.2.10, θ is the angle between (cosh t, cosh t sinh t)
and the x-axis, so cos θ = secht, sin θ = tanht, p = (0, 0), a = (1, 0) and we get
Γ = (s, 0) +
cos θ
− sin θ
sin θ
cos θ
− sinh t
1 − 12 sinh2 t
.
Using the formulas for s and θ from above gives the stated formula for Γ after
simplification.
2.2.24 Differentiating γ (t+a) = γ (t) gives t (t+a) = t(a); rotating anti-clockwise by π/2
ns (t + a) = κs (t)n
ns (t),
gives ns (t + a) = ns (t); differentiating again gives κs (t + a)n
so κs (t + a) = κs (t).
17
2.3.1 (i) t = ( 21 (1 + t)1/2 , − 12 (1 − t)1/2 , √12 ) is a unit vector so γ is unit-speed;
p
ṫt = ( 14 (1 + t)−1/2 , 14 (1 − t)−1/2 , 0), so κ = k ṫt k = 1/ 8(1 − t2 );
n = κ1 ṫt = √12 ((1 − t)1/2 , (1 + t)1/2 , 0); b = t × n = (− 12 (1 + t)1/2 , 21 (1 − t)1/2 , √12 );
p
b = (− 14 (1 + t)−1/2 , − 14 (1 − t)−1/2 , 0) so the torsion τ = 1/ 8(1 − t2 ). The
ḃ
n = −κtt + τn
n is easily checked.
equation ṅ
(ii) t = (− 54 sin t, − cos t, 53 sin t) is a unit vector so γ is unit-speed;
ṫt = (− 54 cos t, sin t, 35 cos t), so κ = k ṫt k = 1; n = κ1 ṫt = (− 54 cos t, sin t, 53 cos t);
b = 0 and τ = 0. By the proof of Proposition 2.3.5,
b = t × n = (− 35 , 0, − 54 ), so ḃ
γ is a circle of radius 1/κ = 1 with centre γ + κ1 n = (0, 1, 0) in the plane passing
through (0, 1, 0) perpendicular to b = (− 35 , 0, − 54 ), i.e. the plane 3x + 4z = 0.
2.3.2 Let a = κ/(κ2 + τ 2 ), b = τ /(κ2 + τ 2 ). By Examples 2.1.3 and 2.3.2, the circular
b
a
helix with parameters a and b has curvature a2 +b
2 = κ and torsion a2 +b2 = τ .
By Theorem 2.3.6, every curve with curvature κ and torsion τ is obtained by
applying a direct isometry to this helix.
2.3.3 Differentiating t .aa (= constant) gives n.aa = 0; since t, n, b are an orthonormal
b for some scalar µ; since a is a unit vector, µ = ± sin θ;
basis of R3 , a = t cos θ+µb
differentiating a = t cos θ ± b sin θ gives τ = κ cot θ. Conversely, if τ = λκ, there
b sin θ is a constant
exists θ with λ = cot θ; differentiating shows that a = t cos θ+b
t
a
t
a
vector and .a = cos θ so θ is the angle between and . For the circular helix in
Example 2.1.3, the angle between the tangent
vector dγγ /dθ = (−a sin θ, a cos θ, b)
√
−1
2
and the z-axis is the constant cos (b/ a + b2 ).
n.(γγ −aa) =
2.3.4 Differentiating (γγ −aa).(γγ −aa) = r 2 repeatedly gives t.(γγ −aa) = 0; t.tt +κn
b).(γγ −aa) = κ̇/κ2 , and so b.(γγ −aa) = κ̇/τ κ2 ;
0, so n.(γγ −aa) = −1/κ; n.tt +(−κtt +τb
n.(γγ − a ) = (κ̇/τ κ2 )˙, and so τ /κ = (κ̇/τ κ2 )˙. Conversely, if Eq.
and finally b.tt − τn
(2.22) holds, then ρ = −σ(ρ̇σ)˙, so (ρ2 + (ρ̇σ)2 )˙ = 2ρρ̇ + 2(ρ̇σ)(ρ̇σ)˙ = 0, hence
n + ρ̇σb
b; then
ρ2 + (ρ̇σ)2 is a constant, say r 2 (where r > 0). Let a = γ + ρn
a
t
n
t
b
b
n
ȧ = + ρ̇n + ρ(−κt + τb) + (ρ̇σ)˙b + (ρ̇σ)(−τn) = 0 using Eq. (2.22); so a is a
constant vector and k γ − a k2 = ρ2 + (ρ̇σ)2 = r 2 , hence γ is contained in the
sphere with centre a and radius r.
Γ = P γ̇γ so T = P t and k Γ̇
Γ k2 = (P γ̇γ )..(P γ̇γ ) = γ̇γ.γ̇γ since P is orthogonal. Then,
2.3.5 Γ̇
Γ = P γ̈γ , taking lengths shows that γ and Γ have the same curvature κ, and then
Γ̈
dividing by κ gives N = P n. Then B = P t × P n. If P corresponds to a direct
isometry (i.e. a rotation), this is equal to P (tt × n) = P b, but if P corresponds
to an opposite isometry, P t × P n = −P b (Proposition A.1.6).
vj . The vectors v1 , v2 , v3 are orthonormal if and only if λij = δij
2.3.6 Let λij = vi .v
(= 1 if i = j and = 0 if i 6= j). So it is enough to prove that λij = δij
vj gives
for all values of s given that it holds for s = s0 . Differentiating vi .v
P3
λ̇ij =
k=1 (aik λkj + ajk λik ). Now λij = δij is a solution of this system of
differential equations because aij +aji = 0. But the theory of ordinary differential
18
equations tells us that there is a unique solution with given values when s = s0 .
√
√
√
√
2.3.7 (i) t = γ̇γ = ( √12 sin t 2 cos t 2, √12 ), ṫt = (cos t 2, − sin t 2, 0) so κ = k ṫt k = 1,
√
√
b = n so τ = 1.
n = ṫt, b = t × n = ( √12 sin t 2, √12 cos t 2, − √12 ). Then, ḃ
(ii) t = γ̇γ = (− √13 sin t + √12 cos t, − √13 sin t, − √13 sin t − √12 cos t), so κ = k ṫt k = 1
and n = ṫt. Then b = t × n = (− √1 , √2 , − √1 ) is constant so τ = 0.
6
6
6
1
√ (sinh t, cosh t, 1)
2
2.3.8 Denoting d/dt by a dot, γ̇γ =
so k γ̇γ k = cosh t. Hence, the
arc-length is s = sinh t, up to adding a constant. The unit tangent vector is
1
t = ds/dt
γ̇γ = √12 (tanht, 1, secht) so
dtt/dt
secht
dtt
=
= √ (sech2 t, 0, −secht tanht)
ds
ds/dt
2
and κ = k dtt/ds k = √12 sech2 t. Then, n =
t × n = √12 (−tanht, 1, −secht). Finally,
1 dt
κ ds
= (secht, 0, −tanht) and b =
b
secht
db
sech2 t
2
√
=
(−sech t, 0, secht tanht) = − √ n,
ds
2
2
√1 sech2 t.
2
...
1
γ̇γ = (− t2 + 1, 1, − t12 ),...γ̈γ = ( t23 , 0, t23 ), γ = (− t64 , 0, − t64 ). Hence γ̈γ
...
γ ×γ̈γ ). γ
1+t2
γ , and so τ = (γ̇ γ
=
0.
Alternatively,
note
that
− (1−t)
t
t
γk
kγ̇ ×γ̈
so τ =
2.3.9
is parallel to
= t + 1 so γ
is contained in the plane x − z = y.
2.3.10 We have
γ̇γ = (cos(ln t) − sin(ln t), sin(ln t) + cos(ln t), 1),
1
γ̈γ = (− sin(ln t) − cos(ln t), cos(ln t) − sin(ln t), 0),
t
1
∴
γ̇γ × γ̈γ = (sin(ln t) − cos(ln t), − sin(ln t) − cos(ln t), 2),
t
q
γ ×γ̈γ k
kγ̇
so κ = kγ̇γ k3 = 1t 29 .
2.3.11 The torsion is defined whenever the curvature is non-zero. In that case k γ̇γ × γ̈γ k
> 0 for all t and so is a smooth
function of t (compare the proof of Proposition
...
γ ×γ̈γ ). γ
(γ̇
1.3.5). Hence, τ = kγ̇γ ×γ̈γ k is smooth.
n, so the unit tangent vector of δ
2.3.12 Let a dot denote d/dt. Then, δ̇δ = γ̈γ = κn
T/ds = ṅ
n/(ds/dt) =
is T = n and its arc-length s satisfies ds/dt = κ. Now dT
2
τ
τ
−1
b) = −tt + κ b. Hence, the curvature of δ is k −tt + κ b k = (1+ κτ 2 )1/2 =
κ (−κtt +τb
19
µ, say. The principal normal of δ is N = µ−1 (−tt + κτ b) and its binormal is
b + κτ t). The torsion T of δ is given by dB
B/ds = −T N , i.e.
B = T × N == µ−1 (b
−1 B
κ Ḃ = −T N . Computing the derivatives and equating coefficients of b gives
T = (κτ̇ − τ κ̇)/κ(κ2 + τ 2 ).
2.3.13 The curve is γ (t) = (eλt cos t, eλt sin t, eλt ). The angle between the tangent
vectorγ̇γ = eλt (λ cos t − sin t, λ sin t + cos t, λ) and the vector k = (0, 0, 1) is
γ̇γ.k
cos−1 kγ̇γ k = cos−1 √2λλ2 +1 , which is constant.
2.3.14 We find that
2(9b2 c2 t4 + 9a2 c2 t2 + a2 b2 )1/2
3abc
κ=
.
,
τ
=
9b2 c2 t4 + 9a2 c2 t2 + a2 b2
(a2 + 4b2 t2 + 9c2 t4 )3/2
Now γ is a generalized helix if and only if there is a constant λ such that τ = λκ,
i.e.
3abc(a2 + 4b2 t2 + 9c2 t4 )3/2 = 2λ(9b2 c2 t4 + 9a2 c2 t2 + a2 b2 )3/2 ,
which is equivalent to
(*)
9b2 c2 t4 + 9a2 c2 t2 + a2 b2 = µ(a2 + 4b2 t2 + 9c2 t4 )
where µ = (3abc/2λ)2/3. The last equation holds if and only if
9b2 c2 = 9µc2 , 9a2 c2 = 4µb2 , a2 b2 = µa2 .
The first and third equations give µ = b2 and the second gives µ = 9a2 c2 /4b2 .
Hence, if γ is a generalized helix 9a2 c2 /4b2 = b2 , which is equivalent to the
stated condition. Conversely, if this condition holds, taking µ = b2 (and hence
λ = 3ac/2b2 ) satisfies (*) so τ = 3ac
2b2 κ and γ is a generalized helix.
n(s), where a is a scalar possibly
2.3.15 (i) From the definition, γ̃γ (s̃) = γ (s) + a(s)n
depending on s. Then,
n
da
dn
da
dγγ
ds̃
b.
+ n+a
= (1 − κa)tt + n + τ ab
t̃t =
ds
ds
ds
ds
ds
n = ±n
n, so t̃t is perpendicular to n. This implies that
But we are given that ñ
da/ds = 0 and hence a is constant.
(ii) We have
d
ds̃
n = 0,
n.t̃t + κ̃tt.ñ
(tt.t̃t) = κn
ds
ds
n and hence to n, and similarly t is perpendicular to ñ
n.
since t̃t is perpendicular to ñ
Hence, t.t̃t is a constant, say cos α (it must lie between −1 and 1 as t and t̃t are unit
20
b,
vectors). Then, since t is perpendicular to n, we must have t̃t = cos αtt ± sin αb
and we can assume the sign is − by changing α to −α if necessary. Then,
b = t̃t × ñ
n = ±(cos αtt − sin αb
b) × n = ±(sin αtt + cos αb
b).
b̃
(iii) From (i) and (ii),
ds̃
b) = (1 − κa)tt + τ ab
b.
(cos αtt − sin αb
ds
Equating coefficients of t and b gives the first pair of equations. The second pair
follows by interchanging the roles of γ and γ̃γ , which changes α to −α and a to
−a.
(iv) This follows from the first pair of equations in (iii).
(v) The first equation follows from the two equations in (iii) involving the torsion,
ds̃ ds
= 1. The first equation follows similarly from the other two
noting that ds
ds̃
equations in (iii).
n for any constant a 6= 0. If
2.3.16 Let γ (s) be a unit-speed plane curve, and let γ̃γ = γ +an
t̃t = (1 − κa)tt.
s̃ is the arc-length of γ̃γ , the unit tangent vector t̃t of γ̃γ is given by ds̃
ds
n = ±n
n.
Hence t̃t = ±tt and so (since γ is planar) ñ
2.3.17 The ‘only if’ part was proved in Exercise 2.3.15(iv). Conversely, if aκ + bτ = 1,
n. Then,
let γ̃γ = γ + an
ds̃
b = τ (btt + ab
b).
t̃t = (1 − κa)tt + τ ab
ds
b). Differentiating,
Hence, t̃t = ± √a21+b2 (btt + ab
n = ±√
κ̃ñ
b
db
κb − τ a ds
1
ds dtt
(b + a ) = ± √
n.
2
ds
+ b ds̃ ds
a2 + b2 ds̃
a2
n = ±n
n.
It follows that ñ
n is a Bertrand
2.3.18 The solutions of Exercises 2.3.15 and 2.3.17 show that γ̃γ = γ + an
mate of γ (a being a constant) if and only if aκ + bτ = 1 for some constant b, and
that such curves γ̃γ are all the Bertrand mates of γ . Thus, γ has more than one
Bertrand mate if and only if the equation aκ + bτ = 1 is satisfied by more than
one value of a (and some corresponding values of b). In that case κ and τ must be
constant, so γ is a circular helix (Exercise 2.3.2). To see that γ̃γ is then a circular
helix with the same axis and pitch as γ , take γ to have axis the z-axis (this can
be achieved by applying an isometry of R3 ). Then, γ (t) = (c cos λt, c sin λt, dλt),
where c is the radius of γ , 2πd is its pitch and λ = (c2 + d2 )−1/2 . Hence,
n = (− cos λt, − sin λt, 0) so the most general Bertrand mate of γ is
γ̃γ = γ a + n = ((c − a) cos λt, (c − a) sin λt, dλt) ,
21
which is a circular helix with axis the z-axis, pitch 2πd and radius |c − a|.
2.3.19 If κ is constant, Eq. (2.22) in Exercise 2.3.4 gives τ = 0, so the curve is a circle
(constant curvature and zero torsion).
2.3.20 Let γ (t) be a unit-speed parametrization of C. If all the normal planes of C pass
through a point p, there are scalars λ, µ (possibly depending on t) such that
n + µb
b=p
γ + λn
for all t. Differentiating and using the Frenet-Serret equations,
n + λ(−κtt + τb
b) + µ̇b
b − µτn
n = 0.
t + λ̇n
Hence,
1 − λκ = 0, λ̇ − µτ = 0, λτ + µ̇ = 0.
Using the first and third equations,
d 2
(λ + µ2 ) = 2λλ̇ + 2µµ̇ = 2λ(µτ ) + 2µ(−λτ ) = 0,
ds
p
so k γ − p k = λ2 + µ2 is constant. Hence, γ lies on a sphere with centre p.
(An alternative is to make use of Exercise 2.3.4.)
2.3.21 (i) is obvious.
(ii) Ḟ = γ̇γ .N so Ḟ (t0 ) = 0 if and only if the tangent vector of γ at γ (t0 ) is parallel
to Π. Using (i) this gives (ii).
n.N, so F̈ (t0 ) = 0 if and only if n(t0 ) is parallel to Π. Thus,
(iii) F̈ = γ̈γ.N = κn
the unique plane Π such that F (t0 ) = Ḟ (t0 ) = F̈ (t0 ) = 0 is that passing through
γ (t0 ) and parallel to t(t0 ) and n(t0 ), i.e. perpendicular to b(t0 ).
(iv) If γ is contained in the plane v.N′ = d′ , then γ̇γ .N ′ = γ̈γ .N′ = 0 so by (iii)
the osculating plane Π of γ at any point γ (t0 ) is perpendicular to N′ , and so is
parallel to Π′ . But both Π and Π′ pass through γ (t0 ), so they must coincide.
(v) The (signed) distance from γ (t) to Π is d(t) = (γγ (t) − γ (t0 ))..b(t0 ). Then,
d(t0 ) = 0 and
˙ 0 ) = t(t0 )..b(t0 ) = 0,
d(t
¨ 0 ) = κ(t0 )n
n(t0 )..b(t0 ) = 0,
d(t
...
n + κ(−κtt + τb
b))(t0 )..b(t0 ) = κ(t0 )τ (t0 ).
d (t0 ) = (κ̇n
If τ (t0 ) 6= 0, Taylor’s theorem gives
d(t) =
1
κ(t0 )τ (t0 )(t − t0 )3 + higher order terms.
6
22
As t passes from values < t0 to values > t0 , the leading term changes sign, so γ
crosses Π.
2.3.22 For the circular helix γ (t) = (a cos t, a sin t, bt), b = √a21+b2 (b sin t, −b cos t, a), so
by Exercise 2.3.21 the osculating plane at γ (t0 ) is (γγ (t) − γ (t0 ))..b(t0 ) = 0, i.e.
xb sin t0 − yb cos t0 + az = abt0 .
2.3.23 Proceeding as in the preceding exercise, the osculating plane at γ (t) is found to
be
3t2 x − 3ty + z = t3 .
If P1 , P2 , P3 correspond to t = t1 , t2 , t3 , the three osculating planes intersect at
a single point if and only if the determinant
3t21
3t22
3t23
−3t1
−3t2
−3t3
1
1 6= 0.
1
But the determinant equals 9(t1 −t2 )(t2 −t3 )(t3 −t1 ), and this is indeed non-zero
as t1 , t2 , t3 are distinct.
If Q is the point (X, Y, Z), then t1 , t2 , t3 are the roots of
t3 − 3t2 X + 3tY − Z = 0.
On the other hand, if the plane Π passing through the points γ (t1 ), γ (t2 ) and
γ (t3 ) has equation
ax + by + cz + d = 0,
then t1 , t2 , t3 are also the roots of
at + bt2 + ct3 + d = 0.
The two cubic equations must be the same, so Π must be the plane
3Y x − 3Xy + z = Z,
and the point (X, Y, Z) obviously lies on this plane.
2.3.24 We can assume that γ is unit-speed. If all the osculating planes pass through a
point p, there are scalars λ, µ such that
n = p.
γ + λtt + µn
Proceeding as in Exercise 2.3.20 we find that
µτ = 0, λ̇ − µκ = −1, λκ + µ̇ = 0.
23
If µ = 0 the third equation gives λ = 0, which contradicts the second equation.
So µ 6= 0 and the first equation gives τ = 0, i.e. γ is planar.
2.3.25 Let γ be a unit-speed curve. Its orthogonal projection onto the normal plane at
γ (t0 ) is
Γ = γ − (γγ .t(t0 ))tt(t0 ).
Then,
Γ = t − (tt.t (t0 ))tt(t0 ) = 0
Γ̇
if t = t0 .
Hence, Γ has a singular point at t = t0 . Similarly, we find that
...
Γ(t0 ) = κ(t0 )n
n(t0 ), Γ (t0 ) = κ̇(t0 )n
n(t0 ) + κ(t0 )τ (t0 )tt(t0 ).
Γ̈
These vectors are linearly independent if κ(t0 ) and τ (t0 ) are non-zero, so in that
case Γ has an ordinary cusp at t = t0 .
The orthogonal projection of γ onto its osculating plane at γ (t0 ) is
Γ = γ − (γγ.b(t0 ))b
b(t0 ),
Γ̃
so
˙
Γ(t0 ) = t(t0 ) − (tt(t0 )..b(t0 ))b
b(t0 ) = t(t0 ) 6= 0,
Γ̃
Γ.
so γ (t0 ) is a regular point of Γ̃
2.3.26 (i) is obvious.
(ii) Ḟ = 4tt.(γγ − c) so Ḟ (t0 ) = 0 if and only if t(t0 ) is perpendicular to γ (t0 ) − c,
i.e. γ is tangent to S at γ (t0 ). Next,
...
n.(γγ − c) + 2, F = 2κ̇n
n.(γγ − c) + 2κ(−κtt + τb
b)..(γγ − c).
F̈ = 2κn
...
Hence, F (t0 ) = Ḟ (t0 ) = F̈ (t0 ) = F (t0 ) = 0 if and only if γ is tangent to S at
γ (t0 ) and
κ̇
n. (γγ − c) = −1, κτb
b.(γγ − c) =
κn
κ
(all quantities evaluated at t = t0 ). These equations and the fact that γ (t0 ) − c
is perpendicular to t(t0 ) are equivalent to
1
κ̇
γ − c = − n + 2 b,
κ
κ τ
so the centre c of S is uniquely determined and its radius is
R =k γ −c k=
r
1
κ̇2
+
κ2
κ4 τ 2
24
(again, all quantities evaluated at t = t0 ). The centre c is independent of t0 if
and only if
d
1
κ̇
γ+ n− 2 b =0
dt
κ
κ τ
when t = t0 . Using Frenet-Serret, the derivative is found to be
κ̇
d
τ
b.
−
κ dt κ2 τ
By Exercise 2.3.4, c is independent of t0 if and only if γ is spherical, in which
case the solution of Exercise 2.3.4 shows that γ lies on the sphere with centre c
and radius R, i.e. the osculating sphere.
2.3.27 Since b(t0 ) is perpendicular to the osculating plane at γ (t0 ), the centre of the
b(t0 ) for some scalar λ. As this
osculating circle will be of the form c′ = c(t0 ) + λb
n(t0 )
point lies on the osculating plane, it must be of the form γ (t0 ) + µtt(t0 ) + νn
for some µ, ν. Hence,
γ+
κ̇
1
b = γ + µtt + νn
n,
n − 2 b + λb
κ
κ τ
all quantities being evaluated at t = t0 . So λ =
osculating circle is
b(t0 ) = γ (t0 ) +
c(t0 ) + λb
κ̇
κ2 τ
and the centre of the
1
n(t0 ).
κ(t0 )
If R is the radius of the osculating sphere, the radius r of the osculating circle
is given by
κ̇2
1
1
2
2
′
2
c
c
r = R − k − k = 2 + 4 2 − λ2 = 2 ,
κ
κ τ
κ
so r = 1/κ(t0 ).
Chapter 3
3.1.1 γ̇γ = (− sin t − a sin 2t, cos t + a cos 2t), k γ̇γ k2 = 1 + a2 + 2a cos t ≥ 1 + a2 − 2|a| =
(1 − |a|)2 so γ is regular if |a| =
6 1. If |a| = 1 then k γ̇γ k = 2(1 + a cos t) so
the origin is a singular point of γ . If a = 0 then γ is a circle. If 0 < |a| < 1,
then γ (t1 ) = γ (t2 ) =⇒ 1 + a cos t1 = 1 + a cos t2 =⇒ cos t1 = cos t2 =⇒ t2 = t1
or 2π − t1 . In the latter case, γ (t2 ) = ((1 + a cos t1 ) cos t1 , −(1 + a cos t1 ) sin t1 )
so γ (t1 ) = γ (t2 ) =⇒ sin t1 = 0 =⇒ t1 = 0 or π. In all cases, t2 − t1 is a
multiple of 2π, so γ is a closed curve with period 2π without self-intersections.
If |a| > 1, γ passes through the origin when cos t = −1/a, which has two roots
25
with 0 ≤ t < 2π, say t1 < t2 , so the origin is a self-intersection. The picture is
qualitatively similar to that in Example 1.1.7 (which is the case a = 2), so the
complement of the image of γ is the union of two bounded regions enclosed by
the part of the curve with t1 ≤ t ≤ t2 , and an unbounded region.
3.1.2 γ̇γ = (− sin t − 2a sin t cos t, cos t + a(cos2 t − sin2 t)) so
cos t + a cos 2t
,
sin t + a sin 2t
dϕ
1 + 3a cos t + 2a2
∴ sec2 t
=
,
dt
(sin t + a sin 2t)2
dϕ
1 + 3a cos t + 2a2
a(cos t + a)
=
=1+
.
2
dt
1 + 2a cos t + a
1 + 2a cos t + a2
tan ϕ = −
∴
If |a| < 1, γ is a simple closed curve of period 2π (Exercise 3.1.1), so by the
Umlaufsatz,
Z 2π
dϕ
dt = 2π,
dt
0
giving
Z
2π
0
a(cos t + a)
dt = 0.
1 + 2a cos t + a2
If |a| > 1, it is clear geometrically that the tangent vector of γ rotates by 4π on
going once around γ (see the diagram in Example 1.1.7 for the case a = 2), so
Z
giving
Z
0
2π
2π
0
dϕ
dt = 4π,
dt
a(cos t + a)
dt = 2π.
1 + 2a cos t + a2
v) = P v +b
b, where P is a
3.2.1 By Appendix 1, any isometry M of R2 is of the form M (v
3×3 orthogonal matrix and b is a constant vector. If γ̃γ = M (γγ ), then γ̃γ˙ = P γ̇γ , so
k γ̃γ˙ k = k γ̇γ k, which implies that γ and γ̃γ have the same length. If we think of γ
RT
as a curve in the xy-plane in R3 , Eq. (3.2) can be written A(γγ ) = 0 (γ̇γ × γ̈γ )..k dt,
where k = (0, 0, 1). It now follows from Proposition A.1.6 that γ and γ̃γ have the
same area (note that if M is opposite, the area appears to change sign, but it does
not because in that case γ̃γ is negatively-oriented when γ is positively-oriented).
3.2.2 Parametrizing
the ellipse by γ (t) = (p cos t, q sin t), with 0 ≤ t ≤ 2π, its area is
R 2π p
√
2
pq sin t + pq cos2 t dt = 2π pq. By the isoperimetric inequality, the length
0
√
√
ℓ of the ellipse satisfies ℓ ≥ 4π × πpq = 2π pq, with equality if and only
26
if the ellipse is a circle, i.e.
R 2π p
p2 sin2 t + q 2 cos2 t dt.
0
p = q.
But its length is
R 2π
0
k γ̇γ k dt =
3.2.3 γ (t′ ) = γ (t) ⇐⇒ cos t′ = cos t and sin t′ = sin t ⇐⇒ t′ − t is a multiple of 2π, so
γ is simple closed
with period 2π. Taking x = p cos t, y = q sin t in Eq. 3.2 gives
R
1 2π
the area as 2 0 pq dt = πpq.
3.3.1 Let (x1 , y1 ) and (x2 , y2 ) be points in the interior of the ellipse, so that
for i = 1, 2. A point of the line segment joining the two points is
x2i
p2
y2
+ qi2 < 1
(tx1 + (1 − t)x2 , ty1 + (1 − t)y2 )
for some 0 ≤ t ≤ 1. This is in the interior of the ellipse because
(ty1 + (1 − t)y2 )2
(tx1 + (1 − t)x2 )2
+
p2
q2
2
x1
x22
y12
y22
y1 y2
x1 x2
2
2
=t
+ 2 + (1 − t)
+ 2 + 2t(1 − t)
+ 2
p2
q
p2
q
p2
q
2
2
2
2
y
x
y
x1
+ 12 + 22 + 22 ≤ t2 + (1 − t)2 + 2t(1 − t) = 1.
< t2 + (1 − t)2 + t(1 − t)
2
p
q
p
q
√
3.3.2 γ̇γ = (− sin t − 2 sin 2t, cos t + 2 cos 2t) and k γ̇γ k = 5 + 4 cos t, so the angle ϕ
sin t−2 sin 2t
√ t+2 cos 2t .
between γ̇γ and the x-axis is given by cos ϕ = − √
, sin ϕ = cos
5+4 cos t
5+4 cos t
Differentiating the second equation gives ϕ̇ cos ϕ =
ϕ̇ =
2
sin t(24 cos t+42 cos t+9)
(5+4 cos t)(sin t+2 sin 2t)
=
9+6 cos t
.
5+4 cos t
− sin t(24 cos2 t+42 cos t+9)
,
(5+4 cos t)3/2
so
Hence, if s is the arc-length of γ , κs =
dϕ/ds = (dϕ/dt)/(ds/dt) = (9 + 6 cos t)/(5 + 4 cos t)3/2 , so
κ̇s =
12 sin t(2 + cos t)
.
(5 + 4 cos t)5/2
This vanishes at only two points of the curve, where t = 0 and t = π.
3.3.3 From ǫ(s) = γ (s) + κ1s ns we get ǫ̇ǫ = −κ̇s ns /κ2s , so ǫ has a singular point where
κ̇s = 0, i.e. where γ has a vertex.
3.3.4 Parametrize by q
γ (x) = (x, f (x)) and denote d/dx by a dot. The arc-length s is
ds
given by dx
= 1 + f˙2 , so the unit tangent vector is t = √ 1 ˙2 (1, f˙) and we
1+f
find that
1
1
dtt
(−f˙f¨, f¨).
=
ṫt =
ds
ds/dx
(1 + f˙2 )2
27
The signed unit normal ns = √
f¨
.
(1+f˙2 )3/2
Then,
1
(−f˙, 1),
1+f˙2
so the signed curvature is κs =
...
f (1 + f˙2 ) − 3f˙f¨2
.
κ̇s =
(1 + f˙2 )5/2
...
= 0 ⇐⇒ f (1 + f˙2 ) = 3f˙f¨2 .
dκs
s
Hence, dκ
ds = 0 ⇐⇒ dx
3.3.5 γ̇γ = (a − b cos t, b sin t), so if ϕ is a turning angle for γ ,
b sin t
,
a − b cos t
b(a cos t − b)
∴ ϕ̇ sec2 ϕ =
,
(a − b cos t)2
(a − b cos t)2
b(a cos t − b)
b(a cos t − b)
.
ϕ̇ =
=
2
2
2
(a − b cos t) (a − b cos t)2 + b2 sin t
a − 2ab cos t + b2
tan ϕ =
∴
If s is the arc-length of γ , the signed curvature is
κs =
Hence,
dϕ
dϕ/dt
ϕ̇
b(a cos t − b)
=
=
= 2
.
ds
ds/dt
k γ̇γ k
(a − 2ab cos t + b2 )3/2
ab sin t(2b2 − a2 − ab sin t cos t)
dκs
.
=
dt
(a2 − 2ab cos t + b2 )5/2
Thus, γ (t) is a vertex if and only if sin t = 0 or cos t =
2b2 −a2
ab .
Hence, there are
vertices when t is an integer multiple of π, and no other vertices if
On the other hand, if
2b2 −a2
ab
≤ 1, i.e. if
many values of t such that cos t =
points on the curve.
2
2
2b −a
ab
a−b
b
≤
2b
a
≤
a+b
,
b
2b2 −a2
ab
> 1.
there are infinitely-
, and these all correspond to different
Chapter 4
4.1.1 Let U be an open disc in R2 and S = {(x, y, z) ∈ R3 | (x, y) ∈ U, z = 0}. If
W = {(x, y, z) ∈ R3 | (x, y) ∈ U }, then W is an open subset of R3 , and S ∩ W
is homeomorphic to U by (x, y, 0) 7→ (x, y). So S is a surface.
4.1.2 The image of σ x± is the intersection of the sphere with the open set ±x > 0 in
R3 , and its inverse is the projection (x, y, z) 7→ (y, z). Similarly for σ y± and σ z± .
A point of the sphere not in the image of any of the six patches would have to
have x, y and z all zero, which is impossible.
28
4.1.3 Multiplying the two equations gives (x2 − z 2 ) sin θ cos θ = (1 − y 2 ) sin θ cos θ, so
x2 + y 2 − z 2 = 1 unless cos θ = 0 or sin θ = 0; if cos θ = 0, then x = −z and
y = 1 and if sin θ = 0 then x = z and y = −1, and both of these lines are also
contained in the surface. The given line Lθ passes through (sin 2θ, − cos 2θ, 0)
and is parallel to the vector (cos 2θ, sin 2θ, 1); it follows that we get all of the
lines by taking 0 ≤ θ < π.
Let (x, y, z) be a point of the surface; if x 6= z, let θ be such that cot θ =
(1 − y)/(x − z); then (x, y, z) is on Lθ ; similarly if x 6= −z. The only remaining
cases are the points (0, 0, ±1), which lie on the lines Lπ/2 and L0 . To get a
surface patch covering S, define σ : U → R3 by
σ (u, v) = (sin 2θ, − cos 2θ, 0) + t(cos 2θ, sin 2θ, 1).
By the preceding paragraph, this patch covers the whole surface.
Let Mϕ be the line (x−z) cos ϕ = (1+y) sin ϕ, (x+z) sin ϕ = (1−y) cos ϕ. By the
same argument as above, Mϕ is contained in the surface and every point of the
surface lies on some Mϕ with 0 ≤ ϕ < π. If θ + ϕ is not a multiple of π, the lines
sin(θ−ϕ) cos(θ+ϕ)
Lθ and Mϕ intersect in the point ( cos(θ−ϕ)
sin(θ+ϕ) , sin(θ+ϕ) , sin(θ+ϕ) ); for each θ with
0 ≤ θ < π, there is exactly one ϕ with 0 ≤ ϕ < π such that θ + ϕ is a multiple of
π, and the lines Lθ and Mϕ do not intersect. If (x, y, z) lies on both Lθ and Lϕ ,
with θ 6= ϕ, then (1 − y) tan θ = (1 − y) tan ϕ and (1 + y) cot θ = (1 − y) cot ϕ,
which gives both y = 1 and y = −1 (the case in which θ = 0 and ϕ = π/2, or
vice versa, has to be treated separately, but the conclusion is the same). This
shows that Lθ and Lϕ do not intersect; similarly, Mθ and Mϕ do not intersect.
√
4.1.4 For the first part, let U = {(u, v) ∈ R2 | 0 < u2 + v 2 < π 2 }, let r = u2 + v 2 ,
and define σ : U → R3 by σ (u, v) = ( ur , vr , tan(r − π2 )).
If S 2 could be covered by a single surface patch σ : U → R3 , then S 2 would
be homeomorphic to the open subset U of R2 . As S 2 is a closed and bounded
subset of R3 , it is compact. Hence, U would be compact, and hence closed. But,
since R2 is connected, the only non-empty subset of R2 that is both open and
closed is R2 itself, and this is not compact as it is not bounded.
σ α : Uα → R3 } is an atlas for a surface S, and if W is an open subset of R3 ,
4.1.5 If {σ
σ |Uα ∩σ −1
} form an atlas of S ∩ W (one should discard
then the restrictions {σ
α (W )
−1
the restrictions for which Uα ∩ σ α (W ) is empty).
4.1.6 Parametrize the curve by γ (t) = (cos θ, sin θ, z) where θ and z are smooth functions of t, and assume that γ is unit-speed. Then, γ̇γ = (−θ̇ sin θ, θ̇ cos θ, ż) so
γ interescts the rulings of the cylinder at a constant angle if and only if ż is
constant, i.e. z = at + b for some constant a, b. Since γ is unit-speed, θ̇ 2 + ż 2 = 1
so θ̇ is a constant c such that a2 + c2 = 1 and θ = ct + d where d is a constant.
Then, γ (t) = (cos(ct + d), sin(ct + d), at + b). This is a circle if a = 0, a straight
line if c = 0 and a circular helix in all other cases.
29
4.1.7 Noting that (x, y, z) lies on the ellipsoid if and only if ( xp , yq , zq ) ∈ S 2 , we can use
the latitude-longitude parametrization of S 2 to get the parametrization
σ (θ, ϕ) = (p cos θ cos ϕ, q cos θ sin ϕ, r sin θ)
of the ellipsoid. σ is continuous and is a homeomorphism onto its image if (θ, ϕ)
is restricted in the same way as for S 2 (Example 4.1.4).
4.1.8 If x = sin u, y = sin v and z = sin(u + v), then
x2 − y 2 + z 2 = sin2 u − sin2 v + sin2 (u + v)
= sin2 u − sin2 v + sin2 u cos2 v + cos2 u sin2 v + 2 sin u sin v cos u cos v
= sin2 u + sin2 u cos2 v − sin2 v(1 − cos2 u) + 2 sin u sin v cos u cos v
= (1 − sin2 v) sin2 u + sin2 u cos2 v + 2 sin u sin v cos u cos v
= 2 sin2 u cos2 v + 2 sin u sin v cos u cos v
= 2 sin u cos v(sin u cos v + cos u sin v)
= 2 sin u cos v sin(u + v).
Hence, (x2 − y 2 + z 2 )2 = 4 sin2 u(1 − sin2 v) sin2 (u + v) = 4x2 (1 − y 2 )z 2 . Now, σ
is clearly continuous, as is its inverse map (x, y, z) 7→ (sin−1 x, sin−1 y), so σ is a
homeomorphism.
4.2.1 σ is obviously smooth and σ u × σ v = (−fu , −fv , 1) is nowhere zero, so σ is
regular.
√
√
4.2.2 σ z± is a special case of Exercise 4.2.1, with f = ± 1 − u2 − v 2 ( 1 − u2 − v 2 is
smooth because 1 − u2 − v 2 > 0 if (u, v) ∈ U ); similarly for the other patches.
The transition map from σ x+ to σ y+ , for example, is Φ(ũ, ṽ) = (u, v), where
√
σ y+ (ũ, ṽ) = σ x+ (u, v); so u = 1 − ũ2 − ṽ 2 , v = ṽ, and this is smooth since
1 − ũ2 − ṽ 2 > 0 if (ũ, ṽ) ∈ U .
σ v = (−v, −u, 1)
4.2.3 (i) is clearly injective and is regular because σ is smooth and σ u ×σ
is never zero.
(ii) is injective but is not regular since σ u × σ v = (0, −3v 2 , 2v) vanishes when
v = 0.
σ (−u − 1, v) and is also not regular since
(iii) is not injective because σ (u, v) = (σ
σ u × σ v = (0, 2v(1 + 2u), 1 + 2u) vanishes when u = −1/2.
4.2.4 This is similar to Example 4.1.4, but using the ‘latitude-longitude’ patch σ (θ, ϕ) =
(p cos θ cos ϕ, q cos θ sin ϕ, r sin θ).
4.2.5 A typical point on the circle C has coordinates (a+b cos θ, 0, b sin θ); rotating this
about the z-axis through an angle ϕ gives the point σ (θ, ϕ); the torus is covered
30
by the four patches obtained by taking (θ, ϕ) to lie in one of the following open
sets:
(i) 0 < θ < 2π, 0 < ϕ < 2π;
(ii) 0 < θ < 2π, −π < ϕ < π;
(iii) −π < θ < π, 0 < ϕ < 2π;
(iv) −π < θ < π, −π < ϕ < π.
Each patch is regular because
σ θ × σ ϕ = −b(a + b cos θ)(cos θ cos ϕ, cos θ sin ϕ, sin θ)
is never zero (since k σ u × σ v k = b(a + b cos θ) ≥ b(a − b) > 0).
4.2.6 Suppose the centre of the propeller is initially at the origin. At time t, the
centre is at (0, 0, αt) where α is the speed of the aeroplane. If the propeller is
initially along the x-axis, the point initially at (v, 0, 0) is therefore at the point
(v cos ωt, v sin ωt, αt) at time t, where ω is the angular velocity of the propeller.
Let u = ωt, λ = α/ω. Next, σ u = (−v sin u, v cos u, λ), σ v = (cos u, sin u, 0), so
the standard unit normal is N = (λ2 + v 2 )−1/2 (−λ sin u, λ cos u, −v). If θ is the
angle between N and the z-axis, cos θ = −v/(λ2 +v 2 )1/2 and hence cot θ = ±v/λ,
while the distance from the z-axis is v.
4.2.7 σ is the tube swept out by a circle of radius a in a plane perpendicular to
n + τ a cos θb
b,
γ as its centre moves along γ . σ s = (1 − κa cos θ)tt − τ a sin θn
n + a cos θb
b, giving σ s × σ θ = −a(1 − κa cos θ)(cos θn
n + sin θb
b);
σ θ = −a sin θn
this is never zero since κa < 1 implies that 1 − κa cos θ > 0 for all θ. The first
fundamental form is ((1 − κa cos θ)2 + τ 2 a2 ) ds2 + 2τ a2 dsdθ + a2 dθ 2 , so the area
R s R 2π
is s01 0 a(1 − κa cos θ)dsdθ = 2πa(s1 − s0 ).
σ = σ + a , where a is a constant vector, then σ̃
σ is smooth if σ is smooth, and
4.2.8 If σ̃
σ u = σ u , σ̃
σ v = σ v , so σ̃
σ is regular if σ is regular. If A is an invertible 3 × 3
σ̃
σ
σ
σ is smooth if σ is smooth and σ̃
σ u = Aσ
σ u , σ̃
σ v = Aσ
σv ,
matrix and σ̃ = Aσ , then σ̃
σ u and σ̃
σ v are linearly independent if σ u and σ v are
so since A is invertible σ̃
linearly independent.
4.2.9 See Exercise 4.1.5. The restriction of a smooth map U → R2 , where U is an
open subset of R2 , to an open subset of U is smooth.
4.2.10 The map σ is a diffeomorphism, since it is smooth, obviously bijective, and has
smooth inverse (x, y, z) 7→ (x, y).
4.2.11 We find that σ u × σ v = (− cos v cos(u + v), − cos u cos(u + v), cos u cos v). This
is non-zero when −π/2 < u, v < π/2 since cos u and cos v are non-zero in this
range.
4.2.12 The torus is covered by the patches σ : Un → R3 , n = 1, 2, 3, where σ is as in
31
the question and
U1 = {(θ, ϕ) | 0 < θ < 2π, 0 < ϕ < 2π},
2π
4π
4π
U2 = (θ, ϕ) | −
<θ<
,−
<ϕ<
3
3
3
2π
2π
4π
<θ<
,−
<ϕ<
U3 = (θ, ϕ) | −
3
3
3
2π
,
3
4π
.
3
Suppose for a contradiction that the torus is covered by two patches σ : V → R3
and σ : V ′ → R3 , where σ is as before and V, V ′ are open subsets of R2 . We
might as well assume that
V = {(θ, ϕ) | α < θ < α + 2π, β < ϕ < β + 2π},
V ′ = {(θ, ϕ) | α′ < θ < α′ + 2π, β ′ < ϕ < β ′ + 2π},
for some α, α′ , β, β ′ ∈ R. But neither of the patches σ : V → R3 or σ : V ′ → R3
contains the point σ (α′ , β) (for example).
4.2.13 Taking f = 3xy − z(z + 4) + c, we have ∇ f = (3y, 3x, −2z − 4) which is zero
only at the point (0, 0, −2). But f (0, 0, −2) = 0 only if c = −4. So if c 6= −4,
f = 0 is a smooth surface.
4.2.14 Taking f = x3 + 3(y 2 + z 2 )2 − 2, we have ∇f = (3x2 , 12y(y 2 + z 2 ), 12z(y 2 + z 2 ))
which is zero only at the origin. But f (0, 0, 0) 6= 0, so f = 0 is a smooth surface.
4.2.15 Take f = x2/3 + y 2/3 + z 2/3 = 1. Not that f is smooth provided x, y and z
are never zero, i.e. if we exclude the points on the coordinate planes. Then
∇ f = ( 23 x−1/3 , 32 y −1/3 , 23 z −1/3 ) which is non-zero.
4.2.16 If xyz = 1 then x, y and z are all non-zero. By the intermediate value theorem,
a smooth (or even just continuous) curve on the surface cannot connect a point
with x > 0 (for example) to a point with x < 0 (as it would have to pass
through a point with x = 0). Thus, two points in the same connected piece of
the surface must have x-coordinates of the same sign (and similarly for their yand z-coordinates). Note that the sign of any two of the coordinates determines
that of the third, since xyz = 1.
Consider the piece with x, y and z all > 0. This can be parametrized using the
single patch σ (u, v) = (u, v, 1/uv), defined on the open set {(u, v) | u > 0, v > 0}.
It is clear that σ is smooth and injective.
4.2.17 Consider the circular helix γ (θ) = (a cos θ, a sin θ, bθ). The mid-point of the
chord joining γ (θ) and γ (ϕ) is the point
1
1
1
a(cos θ + cos ϕ), a(sin θ + sin ϕ), b(θ + ϕ) = (v cos u, v sin u, bu),
2
2
2
32
where u = 12 (θ + ϕ), v = a cos 12 (θ − ϕ). Hence, the set of mid-points of the
chords of the helix is the subset of the helicoid in Exercise 4.2.6 with λ = b given
by |v| ≤ |a|.
4.3.1 If S is covered by a single surface patch σ : U → R3 , then f : S → R is smooth
σ : Ũ → R3 is
if and only if f ◦ σ : U → R is smooth. We must check that, if σ̃
σ is smooth if and only if f ◦ σ is smooth.
another patch covering S, then f ◦ σ̃
σ = (f ◦ σ ) ◦ Φ, where Φ is the transition map from σ to
This is true because f ◦ σ̃
σ , and both Φ and Φ−1 are smooth. The last part is true because if σ : U → R3
σ̃
is a smooth map, where U is an open subset of R2 , then each component of σ
(which is a map U → R) is smooth (this is because a vector function such as σ
is differentiated ‘componentwise’).
4.3.2 f is not a diffeomorphism as it is not injective: f (0, y, z) = f (0, y, z + 2π).
Take an atlas for the cone consisting of the patches σ (u, v) = (u cos v, u sin v, u),
defined on the open sets
U1 = {(u, v)|u > 0, 0 < v < 2π} and U2 = {(u, v)|u > 0, −π < v < π}
(call these σ 1 and σ 2 ), and parametrize the half-plane by π (u, v) = (0, u, v)
with u > 0. If (0, a, b) is any point in the plane, assume first that b is not
an even multiple of π, say 2nπ < b < 2(n + 1)π for some integer n. Then,
π(u, v)) = σ 1 (u, v − 2nπ) if 2nπ < v < 2(n + 1)π. So f is a diffeomorphism
f (π
from the open subset {(0, y, z)|2nπ < z < 2(n + 1)π} of the half-plane to the
cone with the half-line y = 0, x = z > 0 removed. Similarly if b is not an odd
multiple of π. This proves that f is a local diffeomorphism.
4.4.1 (i) At (1, 1, 0), σ u = (1, 0, 2), σ v = (0, 1, −2) so σ u × σ v = (−2, 2, 1) and the
tangent plane is −2x + 2y + z = 0.
(ii) At (1, 0, 1), where r = 1, θ = 0, σ r = (1, 0, 2), σ θ = (0, 1, 2) so σ r × σ θ =
(−2, −2, 1) and the equation of the tangent plane is −2x − 2y + z = 0.
σ (ũ, ṽ) be a reparametrization of σ . Then,
4.4.2 Let σ̃
σu =
∂ṽ
∂ṽ
∂ ũ
∂ ũ
σ ũ +
σ ṽ , σ v =
σ ũ +
σ ṽ ,
σ̃
σ̃
σ̃
σ̃
∂u
∂u
∂v
∂v
σ ũ and σ̃
σ ṽ . Hence, any linear combination
so σ u and σ v are linear combinations of σ̃
σ ũ and σ̃
σ ṽ . The converse is also true
of σ u and σ v is a linear combination of σ̃
σ
σ
since is a reparametrization of σ̃ .
d
F (γγ (t)) = Fx ẋ + Fy ẏ + Fz ż = ∇F .γ̇γ . Since
4.4.3 If γ (t) = (x(t), y(t), z(t)) then dt
∇ S F − ∇F is perpendicular to Tp S, it is perpendicular to γ̇γ (t0 ) for every curve
γ on S passing through p when t = t0 . It follows that ∇S F .γ̇γ = ∇F .γ̇γ at p. If
the restriction of F to S has a local maximum or a local minimum at p, so does
33
d
F (γγ (t)) = 0 at p, which
F (γγ (t)) for all curves γ on S passing through p, hence dt
implies that ∇F is perpendicular to γ̇γ , and hence perpendicular to the tangent
plane of S at p. This means that ∇S F = 0.
4.4.4 d(f ◦ γ )/dt) = Dγ (t) f (γ̇γ (t)) is non-zero because γ̇γ is non-zero (γγ is regular) and
Dγ (t) f is invertible (Proposition 4.4.6).
√
√
4.4.5 We find that, at θ = ϕ = π/4, σ θ = 21 b(−1, −1, 2), σ ϕ = 21 (a 2 + b)(−1, 1, 0),
so the unit normal of the torus at this point is parallel to
√
√
√
(−1, −1, 2) × (−1, 1, 0) = − 2(1, 1, 2).
Hence, the tangent plane is x + y +
√
2z = 0.
4.5.1 The transition map Φ(t, θ) = (t̃, θ̃) is defined on the union of the rectangles given
by 0 < θ < π and π < θ < 2π (and −1/2 < t < 1/2). Obviously Φ(t, θ) = (t, θ)
if 0 < θ < π. If π < θ < 2π, we must have θ̃ = θ − 2π. Since sin θ̃2 = − sin θ2 ,
σ (t̃, θ̃) forces t̃ = −t. So Φ(t, θ) = (t, θ) if 0 < θ < π,
cos 2θ̃ = − cos θ2 , σ (t, θ) = σ̃
and = (−t, θ − 2π) if π < θ < 2π. The Jacobian determinant is +1 on the first
rectangle, −1 on the second.
σ α : Uα → R3 } be an atlas for S such that the transition map Φαβ between
4.5.2 Let {σ
σ α and σ β satisfies det(J(Φαβ )) > 0 for all α, β (Definition 4.5.1). By Proposition 4.3.1, {f ◦ σ α } is an atlas for S̃, and the transition maps for this atlas are
the same as those for the atlas of S, because (f ◦ σ β )−1 ◦ (f ◦ σ α ) = σ −1
β ◦ σα
(where this composite is defined). So the atlas {f ◦ σ α } gives S̃ the structure of
an oriented surface.
σ (θ, ϕ).
4.5.3 If σ (θ, ϕ) = (cos θ cos ϕ, cos θ sin ϕ, sin θ), we find that σ θ × σ ϕ = − cos θσ
Hence, the standard unit normal points inwards if cos θ > 0 (if −π/2 < θ < π/2,
for example) and outwards of cos θ < 0. Note that the points at which cos θ = 0
(i.e. the north and south pole) must be excluded for σ to be regular.
σ x± )u ×(σ
σ x± )v = ± √1−u12 −v 2 σ x± , so
For the patches σ x± , for example, we find that (σ
the standard unit normal of σ x+ points outwards and that of σ x− points inwards.
v.
4.5.4 Apply the proof of Proposition 2.2.1 with f = γ̇γ .v, g = γ̇γ.ṽ
4.5.5 From σ ∗ = σσ.σ , we get
σ ∗u =
σ.σ )σ
σ u − 2(σ
σ u.σ )σ
σ
(σ
,
4
kσ k
34
with a similar formula for σ ∗v . This gives
1
2
σ
σ
σ
σ
σ
σ
k
σ
k
(σ
×
σ
)
−
2σ
×
((σ
.
σ
)σ
−
(σ
.
σ
)σ
)
u
v
u
v
v
u
k σ k6
1
2
σ
σ
σ
σ
k
σ
k
(σ
×
σ
)
+
2σ
×
(σ
×
(σ
×
σ
))
=
u
v
u
v
k σ k6
1
σ.(σ
σ u × σ v ))σ
σ − k σ k2 (σ
σu × σ v )
2(σ
=
6
kσ k
σ .N)σ
σ
k σ u × σ v k 2(σ
−N .
=
k σ k4
k σ k2
σ ∗u × σ ∗v =
.N )σ − N is a unit vector. But its squared
Hence, it suffices to show that 2(σ
kσ k2
length is
σ. N ) 2
σ.N )2
4(σ
4(σ
−
+ k N k2 = 1.
k σ k2
k σ k2
Chapter 5
5.1.1 (i) fx = 2x, fy = 2y and fz = 4z 3 vanish simultaneously only when x = y =
z = 0, but this does not satisfy x2 + y 2 + z 4 = 1. So by Theorem 5.1.1 this is a
smooth surface.
(ii) Let f (x, y, z) be the left-hand side minus the right-hand side; then,
fx = 4x(x2 + y 2 + z 2 − a2 − b2 ),
fy = 4y(x2 + y 2 + z 2 − a2 − b2 ),
fz = 4z(x2 + y 2 + z 2 + a2 − b2 );
if fz = 0 then z = 0 since x2 + y 2 + z 2 + a2 − b2 > 0 everywhere on the torus;
if fx = fy = 0 too, then since the origin is not on the torus, we must have
x2 + y 2 = a2 + b2 , but then substituting into the equation of the torus gives
(2a2 )2 = 4a2 (a2 + b2 ), a contradiction.
For the last part, let σ (θ, ϕ) = (x, y, z) be the parametrization in Exercise 4.2.5.
Then, x2 + y 2 + z 2 + a2 − b2 = 2a(a + b cos θ), so
(x2 + y 2 + z 2 + a2 − b2 )2 = 4a2 (a + b cos θ)2 = 4a2 (x2 + y 2 ).
p
Conversely, if (x, y, z) satisfies the given equation, let r = x2 + y 2 . A little
algebra gives (r 2 + z 2 − a2 − b2 )2 = 4a2 (b2 − z 2 ). Hence, |z| ≤ b so z = b sin θ
for some θ ∈ R. Then we find r 2 = a2 + b2 cos2 θ ± 2ab cos θ, so (since r ≥ 0)
r = a ± b cos θ. With the plus sign (x, y, z) = σ (θ, ϕ) for some ϕ ∈ R; with the
35
minus sign, (x, y, z) = σ (π − θ, ϕ) for some ϕ. Thus, the image of σ : R2 → R3
coincides with the set of solutions to the given equation.
5.1.2 See the solution of Exercise 4.4.3 for the first part. Since S has a (smooth)
choice of unit normal ∇ f k ∇f k at each point, it is orientable. The solution of
Exercise 4.4.3 also shows that if the restriction of F to S has a local maximum
or a local minimum at p, then ∇ F is perpendicular to the tangent plane of S at
p. But ∇ f is also perpendicular to the tangent plane. Hence, if the restriction
of F to S has a local maximum or a local minimum at p, then ∇F is parallel to
∇f for some scalar λ.
∇ f at p, i.e. ∇ F = λ∇
5.1.3 Let f (x, y, z) = xyz − 1, F (x, y, z) = x2 + y 2 + z 2 . Then, f = 0 is a smooth
surface S by Theorem 5.1.1 and F defines a smooth function on S. To see that
F has a smallest value on S, let B be the closed ball given by x2 + y 2 + z 2 ≤ 3.
Then, B ∩ S is compact as it is closed and bounded and it is non-empty because
it contains the point (1, 1, 1). Hence, the continuous positive function F must
attain its lower bound, say ℓ, on B ∩ S, and ℓ ≤ 3 since F (1, 1, 1) = 3. Obviously
F (x, y, z) > 3 if (x, y, z) ∈
/ B, so ℓ is the smallest value of F on S.
By Exercise 5.1.2, the local maxima or minima of F on S occur where (2x, 2y, 2z)
= λ(yz, xz, xy) for some λ. Since xyz = 1 on S this gives x2 = y 2 = z 2 = λ/2, so
x, y, z are equal up to sign. Since their product is 1, there are four possibilities:
(x, y, z) = (1, 1, 1), (1, −1, −1), (−1, 1, −1) or (−1, −1, 1). The value of F is 3
at each of these points, which is the smallest value of F on S√from above. The
distance between any two of these points of R3 is the same (2 2), so they form
the vertices of a regular tetrahedron.
5.2.1 (i) (p cos u cos
q v, q cos u sin v, r sin u) (cf. Exercise 4.2.4); (ii) see Exercise 4.1.3;
(iii) (u, v, ± 1 +
u2
p2
+
v2
);
q2
(iv), (v), (vi) see Exercise 4.2.1;
(vii) (p cos u, q cos u, v); (viii) (±p cosh u, q sinh u, v); (ix) (u, u2 /p2 , v);
(x) (0, u, v); (xi) (±p, u, v).


1
−1/3 0
5.2.2 In the notation of Theorem 5.2.2, A =  −1/3
1
0 . The eigenvalues
0
0
−2
are
2/3,
4/3,
−2
and
the
corresponding
unit
eigenvectors
are
the columns of P =
√
 

 ′
 √
x
x
1/√2 −1/√ 2 0
 1/ 2 1/ 2 0 . If  y ′  = P  y , then z ′ = z and the quadric
z′
z
0
0
1
2
2
4 ′2
2 ′2
′2
′
becomes 3 x + 3 y − 2z + 4z = c, i.e. 32 x′ + 43 y ′ − 2(z ′ − 1)2 = c − 2.
Comparing with the standard forms in Theorem 5.2.2 gives the stated results
when c > 2 and c < 2. If c = 2 we have a cone with axis the z-axis (which is
the same as the z ′ -axis), vertex at x′ = y ′ = 0, z ′ = 1, i.e. x = y = 0, z = 1, and
2
2
cross-section perpendicular to the z-axis an ellipse 23 x′ + 43 y ′ = constant.
36
b into the equation of the
5.2.3 Substituting the components (x, y, z) of γ (t) = a + tb
quadric gives a quadratic equation for t; if the quadric contains three points on
the line, this quadratic equation has three roots, hence is identically zero, so the
quadric contains the whole line.
For the second part, take three points on each of the given lines; substituting
the coordinates of these nine points into the equation of the quadric gives a
system of nine homogeneous linear equations for the ten coefficients a1 , . . . , c of
the quadric; such a system always has a non-trivial solution. By the first part,
the resulting quadric contains all three lines.
5.2.4 Let L1 , L2 , L3 be three lines from the first family; by the preceding exercise,
there is a quadric Q containing all three lines; all but finitely-many lines of the
second family intersect each of the three lines; if L′ is such a line, Q contains
three points of L′ , and hence the whole of L′ by the preceding exercise; so Q
contains all but finitely-many lines of the second family; since any quadric is a
closed subset of R3 , Q must contain all the lines of the second family, and hence
must contain
S.
5.2.5 z =
x
p
−
y
y
1
x
q
p + q = uv, x = 2 p(u+v),
σv
( 21 p(u+v), 12 q(v −u), uv); σ u ×σ
y = 12 q(v−u), so a parametrisation is
σ (u, v) =
= (− 21 q(u+v), 12 p(v −u), pq) is never
zero so σ is regular. For a fixed value of u, σ (u, v) = ( 21 pu, − 12 qu, 0)+v( 21 p, 12 q, u)
is a straight line; similarly for a fixed value of v; hence the hyperbolic paraboloid
is the union of each of two families of straight lines.
a c
5.2.6 Let A =
. Since A is a real symmetric matrix, there is an orthogonal
c b
′
a 0
t
′
′
matrix P such that P AP = A is a diagonal matrix, say A =
(see
0 b′
Appendix 0). We can assume that det(P ) = 1 by changing the sign of one of the
rows of P if necessary. Then the equation of the conic becomes
2
2
a′ x′ + b′ y ′ + d′ x′ + e′ y ′ + f = 0,
′
d
x
d
x′
= P
. In other words, we might as well
= P
,
where
e
y
e′
y′
assume that c = 0 from the beginning.
Assume then that c = 0. If a 6= 0, we can assume that d = 0 by replacing x by
x + ad (a translation). Similarly, if b 6= 0 we can assume that e = 0.
If a and b are both 6= 0, we can therefore assume that d = e = 0 and we are in
case (i) or (iii).
If a = 0 and b 6= 0, the equation takes the form by 2 + dx + f = 0. If d 6= 0
replacing x by x + fd we can assume that f = 0 and then we are in case (ii). If
d = 0 we are in case (v). The case a 6= 0, b = 0 is similar.
37
If a = b = 0 we have a straight line.
5.2.7 (i) From the classification in Theorem 5.2.2, the connected quadric surfaces are
those of types (i), (ii), (iv), (v), (vii), (ix) and (x) (remember that for type
(vi) to be a smooth surface, we have to remove the vertex which results in a
disconnected surface). Type (i) is diffeomorphic to S 2 by (x, y, z) 7→ ( xp , yq , zr );


type (ii) to the unit cylinder by (x, y, z) 7→  q
x
p
x2
p2
,q
2
+ y2
q
y
q
x2
p2
2
+ y2
q
, z ; types (iv)
and (v) to a plane by (x, y, z) 7→ (x, y, 0); type (vii) to the unit cylinder by
(x, y, z) 7→ ( xp , yq , z); type (ix) to a plane by (x, y, z) 7→ (x, 0, z); and type (x) is
a plane.
(ii) For types (iii) and (vi) the connected pieces are the parts of the quadric with
z > 0 and with z < 0. Each part of type (iii) is diffeomorphic to the plane by
(x, y, z) 7→ (x, y,
0). Each part of type (vi)is diffeomorphic to the unit cylinder
by (x, y, z) 7→  q
x
p
x2
p2
,q
2
+ y2
q
y
q
x2
p2
2
+ y2
q
, ln |z|. The two connected pieces of type
(viii) are given by x > 0 and x < 0; each piece is diffeomorphic to the plane by
(x, y, z) 7→ (0, y, z).
5.3.1 From Example 5.3.2, the surface can be parametrized by
σ (u, v) = (cosh u cos v, cosh u sin v, u),
with u ∈ R and −π < v < π or 0 < v < 2π.
5.3.2 k σ (u, v) k2 = sech2 u(cos2 v + sin2 v) + tanh2 u = sech2 u + tanh2 u = 1, so
σ parametrizes an open subset of S 2 ; σ is clearly smooth; and σ u × σ v =
−sech2 u σ (u, v) is never zero, so σ is regular. Meridians correspond to the parameter curves v = constant, and parallels to the curves u = constant.
5.3.3 (i) γ̃γ .aa = 0 so γ̃γ is contained in the plane perpendicular to a and passing through
the origin;
(ii) simple algebra;
(iii) ṽ is clearly a smooth function
of (u, v) and the Jacobian matrix of the map
1 0
, where a dot denotes d/du; this matrix is invertible
(u, v) 7→ (u, ṽ) is
γ
γ̇ .aa 1
so γ̃γ is a reparametrization of γ .
5.3.4 σ u = γ̇γ +vδ̇δ , σ v = δ (a dot denotes d/du) so δ̇δ (u) is perpendicular to the surface at
σ (u, v) ⇐⇒ δ̇δ. (γ̇γ + vδ̇δ ) = 0, δ̇δ.δ = 0. The second equation follows from k δ k = 1
so the two conditions are satisfied ⇐⇒ v = −(γ̇γ.δ̇δ )/ k δ̇δ k2 . Hence, Γ(u) =
Γ.δ̇δ = γ̇γ.δ̇δ − (γ̇γ .δ̇δ )δ̇δ.δ̇δ / k δ̇δ k2 = 0.
γ − (γ̇γ .δ̇δ )δδ / k δ̇δ k2 . Using δ̇δ.δ = 0 again, Γ̇
38
5.3.5 The vector σ u is tangent to the meridians, so a unit-speed curve γ is a loxodrome
σ u / k σ u k = cos α, which gives u̇ = cosh u cos α; since γ is unit-speed, γ̇γ =
if γ̇γ .σ
(−u̇ sech u tanh u cos v−v̇ sech u sin v, −u̇ sech u tanh u sin v+v̇ sech u cos v, u̇ sech2 u)
is a unit vector; this gives u̇2 + v̇ 2 = cosh2 u, so v̇ = ± cosh u sin α. The correponding curve in the uv-plane is given by dv/du = v̇/u̇ = ± tan α, and so is a
straight line v = ±u tan α + c, where c is a constant.
5.3.6 The ruling that makes an angle θ with the positive x-axis is given by y =
x tan θ, z = f (θ), so a point at a distance u from the z-axis has coordinates
df
df
sin θ, − dθ
cos θ, u) and this is
x = u cos θ, y = u sin θ, z = f (θ). σ u × σ θ = ( dθ
non-zero if u 6= 0 (i.e. if we exclude the line L itself from the surface).
5.3.7 (i) The assumption means that N is constant. Hence, σ .N is constant, since
σ.N)u = σ u.N = 0, etc. So σ is contained in a plane v.N = constant.
(σ
N for some scalar λ.
(ii) If all the normal lines pass through a point p, σ = p + λN
Nu so 0 = σ u.N = λu since N u.N = 0 as N is a unit vector.
Then, σ u = λuN + λN
Similarly λv = 0. So λ is a constant and σ is part of the sphere k v − p k = |λ|.
(iii) By applying an isometry we can assume that the given straight line is
the z-axis. Reparametrize the surface using polar coordinates in the xy-plane:
N is the poσ (r, θ) = (r cos θ, r sin θ, f (r, θ)), say. For some scalar µ, σ + µN
sition vector of a point on the z-axis. If N = (a, b, c), setting the x- and
N equal to zero gives µa = −r cos θ, µb = −r sin θ,
y-components of σ + µN
and hence either µ = 0 (i.e., r = 0) or a sin θ = b cos θ. But σ r × σ θ =
(fθ sin θ − rfr cos θ, −fθ cos θ − rfr sin θ, r), so if µ 6= 0 then
sin θ(rfr cos θ − fθ sin θ) = cos θ(rfr sin θ + fθ cos θ),
i.e. fθ = 0. We have now proved that fθ = 0 except possibly when r = 0; but
then fθ = 0 for all (r, θ) as f is smooth (and hence continuous). Thus, f depends
only on r, and hence σ is the surface of revolution obtained by rotating the curve
r 7→ (r, 0, f (r)) in the xz-plane around the z-axis.
5.3.8 Use the parametrization γ (u) = (cos u, sin u, 0) of the ‘waist’ of the hyperboloid.
From Exercise 4.1.3, there is a straight line on the hyperboloid passing through
γ (u), and it takes the form
(x − z) cos θ = (1 − y) sin θ, (x + z) sin θ = (1 + y) cos θ.
Substituting x = cos u, y = sin u, z = 0 gives θ = u2 + π4 . The vector δ (u) =
(cos 2θ, sin 2θ, 1) = (− sin u, cos u, 1) is parallel to this line, so a parametrization
of the hyperboloid as a ruled surface is
σ (u, v) = (cos u, sin u, 0) + v(− sin u, cos u, 1) = (cos u − v sin u, sin u + v cos u, v).
Since γ̇γ.δ̇δ = 0, the line of striction is γ (see the solution to Exercise 5.3.4).
39
5.3.9 We use the notation in Theorem 5.2.2.
(a) Types (vii) through (xi).
(b) Types (vi) and (x).
(c) Types (ii), (v) and (vi) through (xi).
(d) Type (i) if p2 , q 2 , r 2 are not distinct; types (ii), (iii), (iv), (vi) and (vii) if
p2 = q 2 ; and types (x) and (xi).
5.3.10 Take the ruled surface to be σ (u, v) = γ (u) + vδδ (u), where γ is unit-speed and
δ is a unit vector. Consider the ruling corresponding to u = u0 , say. Since
σ u × σ v = (γ̇γ + vδ̇δ ) × δ (a dot denoting d/du, evaluated at u = u0 ), the union of
the normal lines along the ruling u = u0 is the surface
b + wcc + vwd
d,
Σ (v, w) = σ (u0 , v) + w(γ̇γ + vδ̇δ ) × δ = a + vb
where a = γ (u0 ), b = δ (u0 ), c = γ̇γ × δ and d = δ̇δ × δ are constant vectors.
By applying a translation we can assume that a = 0. Then, by applying a
rotation we can assume that b = (1, 0, 0) and that δ̇δ is parallel to (0, 1, 0) (note
that δ̇δ is perpendicular to δ ). Then, c is parallel to the yz-plane and d is parallel
to the z-axis.
If d = 0, i.e. if δ̇δ (u0 ) = 0, Σ is a plane. If d = (0, 0, λ) 6= 0 and c = (0, µ, ν),
then Σ(v, w) = (v, µw, νw + λvw). If µ = 0 we have the xz-plane. Otherwise,
the translation x 7→ x − λν followed by a rotation by π/2 around the z-axis takes
Σ to a parametrization of the hyperbolic paraboloid z = µλ (x2 − y 2 ).
5.4.1 Both surfaces are closed subsets of R3 , as they are of the form f (x, y, z) = 0,
where f : R3 → R is a continuous function (equal to x2 − y 2 + z 4 − 1 and
x2 + y 2 + z 4 − 1 in the two cases). The surface in (i) is not bounded, and hence
not compact, since it contains the point (1, a2 , a) for all real numbers a; that in
(ii) is bounded, and hence compact, since x2 + y 2 + z 4 = 1 =⇒ −1 ≤ x, y, z ≤ 1.
The surface in (ii) is obtained by rotating the curve x2 + z 4 = 1 in the xz-plane
around the z-axis:
5.4.2 A closed curve γ with period T can be identified with the unit circle by γ (t) 7→
40
(cos(2πt/T ), sin(2πt/T )). This gives rise to a diffeomorphism from the tube
around γ to a tube around the circle, i.e. a torus. We have to make the tube
have a sufficiently small radius as otherwise it might intersect itself and then it
would not be a surface.
5.5.1 Both parts are geometrically obvious.
5.5.2 Let (a, b, c) ∈ R3 with a and b non-zero. Then Ft (a, b, c) → ∞ as t → ∞ and as
t approaches p2 and q 2 from the left; and Ft (a, b, c) → −∞ as t → −∞ and as t
approaches p2 and q 2 from the right. From this and the fact that Ft (a, b, c) = 0
is equivalent to a cubic equation for t, it follows that there exist unique numbers
u, v, w with u < p2 , p2 < v < q 2 and q 2 < w such that Ft (a, b, c) = 0 when
t = u, v or w. The surfaces Fu (x, y, z) = 0 and Fw (x, y, z) = 0 are elliptic
paraboloids and Fv (x, y, z) = 0 is a hyperbolic paraboloid, and we have shown
that there is one surface of each type passing through each point (a, b, c).
To show that the system is orthogonal, note that ∇ Ft = p22x−t , q 22y−t , −2 so
if (a, b, c) is a point at the intersection of the surfaces Fu (x, y, z) = 1 and
Fv (x, y, z) =, we have
4a2
4b2
+
+4
(p2 − u)(p2 − v) (q 2 − u)(q 2 − v)
4(Fu (a, b, c) − Fv (a, b, c))
4−4
=
=
= 0.
u−v
u−v
∇Fu.∇Fv =
Thus, the first two families intersect orthogonally. Similarly for the other two
pairs of families.
To parametrize these surfaces, write Ft (x, y, z) = 0 as the cubic equation
x2 (q 2 − t) + y 2 (p2 − t) − 2z(p2 − t)(q 2 − t) + t(p2 − t)(q 2 − t) = 0,
and note that the left-hand side must be equal to (t−u)(t−v)(t−w);
putting t =
q
p2 , q 2 and then equating coefficients of t2 (say) gives x = ±
q 2
2 −v)(q 2 −w)
y = ± (q −u)(qp2 −q
, z = 21 (u + v + w − p2 − q 2 ).
2
(p2 −u)(p2 −v)(p2 −w)
,
q 2 −p2
5.5.3 (i) Let U = xy − uz 2 , V = x2 + y 2 + z 2 − v, W = x2 + y 2 + z 2 − w(x2 − y 2 ).
Then, ∇ U = (y, x, −2uz), ∇V = (2x, 2y, 2z), ∇ W = (2(1 − w)x, 2(1 + w)y, 2z).
Hence, at a point of intersection of the surfaces U = 0, V = 0, W = 0, we have
∇ U..∇V = 4xy − 4uz 2 = 0 by the equation U = 0;
∇V .∇ W = 4((1 − w)x2 + (1 + w)y 2 + z 2 ) = 4(x2 + y 2 + z 2 − w(x2 + y 2 )) = 0
by the equation V = 0; and
∇W .∇ U = 2(1 − w)xy + 2(1 + w)xy − 4uz 2 = 4(xy − uz 2 ) = 0
41
by the equation U = 0.
(ii) Let
p
p
p
p
U = yz − ux, V = x2 + y 2 + x2 + z 2 − v, W = x2 + y 2 − x2 + z 2 − w.
y
x
x
z
√
√
√
√
Then, ∇U = (−u, z, y), ∇V =
+ x2 +z2 ,
, x2 +z2 , ∇ W =
x2 +y 2
x2 +y 2
√ 2x 2 − √ 2x 2 , √ 2y 2 , − √ 2z 2 . Hence, at a point of intersection of the
x +z
x +z
x +y
x +y
surfaces U = 0, V = 0 and W = 0, ∇ U..∇V = (yz − ux) √ 21 2 + √x21+z2 =
x +y
2
2
2
2
0 by the equation U = 0; ∇V .∇W = x2x+y2 − x2x+z2 + x2y+y2 − x2z+z2 = 0;
1
1
∇ W .∇U = (yz − ux) √ 2 2 − √x2 +z2 = 0 by the equation U = 0.
x +y
5.5.4 Two families of plane curves U (x, y) = u and V (x, y) = v form an orthogonal
system if ∇ U..∇ V = 0 at each point of intersection
of the first family
of a curve
∂U ∂U
with a curve of the second family. Here, ∇U = ∂x , ∂y , etc.
(i) U (x, y) = x, V (x, y) = y, i.e. the two families of straight lines parallel to the
y- and x-axis, respectively. The orthogonality is obvious.
(ii) U (x, y) = y cos u − x sin u, V (x, y) = x2 + y 2 − v 2 , i.e. the family of straight
lines passing through the origin and the family of circles with centre the origin.
The orthogonality is obvious.
5.5.5 The function Ft (a, b) is a continuous function of t (for fixed non-zero real numbers
a, b) in each of the open intervals (−∞, p2 ), (p2 , q 2 ) and (q 2 , ∞). Also, Ft (a, b) →
0 as t → ±∞, and Ft (a, b) → ±∞ as t approaches p2 or q 2 from the left (+ sign)
or right (− sign). It follows that there is at least one value of t in each of the
open intervals (−∞, p2 ) and (p2 , q 2 ) such that Ft (a, b) = 1. On the other hand,
Ft (a, b) = 1 is equivalent to a quadratic equation for t which has at most two real
roots. It follows that there are unique numbers u ∈ (−∞, p2 ) and v ∈ (p2 , q 2 )
such that Fu (a, b) = Fv (a, b) = 1. The conics Fu (x, y) = 1 and Fv (x, y) = 1
are ellipses and hyperbolas, respectively, and we have shown that there is one
of each passing through each point on the plane that does not lie on one of the
coordinate axes.
To show that this is an orthogonal system of curves, we calculate ∇Fu .∇Fv
at
b) that lies on the conics Fu = 1 and Fv = 1. Since ∇ Fu =
a point (a,
2y
2x
p2 −u , q 2 −u
, we get
4a2
4b2
+
(p2 − u)(p2 − v) (q 2 − u)(q 2 − v)
4(Fu (a, b) − Fv (a, b)
4−4
=
=
= 0.
u−v
u−v
∇Fu.∇Fv =
42
2
For the last part, let Ft (x, y) = p2x−t − 2y + t. Arguing as in Exercise 5.5.2,
for any point (a, b) ∈ R2 not on one of the coordinate axes, there are unique
numbers u ∈ (−∞, p2 ) and v ∈ (p2 , ∞) such that Fu (a, b) = Fv (a, b) = 1. Both
systems of curves Fu (x, y) = 1 and Fv (x, y) = 1 are parabolas. At (a, b) we find
that
∇Fu.∇ Fv =
4(Fu (a, b) − Fv (a, b))
4−4
4a2
+4=
=
= 0,
2
2
(p − u)(p − v)
u−v
u−v
showing that the parabolas form an orthogonal system.
5.5.6 Let U (x, y) = u and V (x, y) = v be an orthogonal system of plane curves (see
Exercise 5.5.4). We will show that the system of surfaces U (x, y) = u, V (x, y) =
v, z = w is orthogonal. The first two families of surfaces consist of generalized
cylinders, the third consists of parallel planes. Suppose that (a, b, c) is a point
of
intersection
of a surface
from each family. We have to prove that the vectors
∂U ∂U
∂V ∂V
and (0, 0, 1) are mutually perpendicular at (a, b, c).
∂x , ∂y , 0 ,
∂x , ∂y , 0
But it is obvious that the third vector is perpendicular to the other two, and
the first two are perpendicular because the curves U (x, y) = u and V (x, y) = v
intersect orthogonally at (a, b).
5.6.1 Let F : W → V be the smooth bijective map with smooth inverse F −1 : V → W
constructed in the proof of Proposition 4.2.6. Then, (u(t), v(t)) = F −1 (γγ (t)) is
smooth.
5.6.2 Suppose, for example, that fy 6= 0 at
(x0 , y0). Let F (x, y) = (x, f (x, y)); then F
1 fx
is smooth and its Jacobian matrix
is invertible at (x0 , y0 ). By the in0 fy
verse function theorem, F has a smooth inverse G defined on an open subset of R2
containing F (x0 , y0 ) = (x0 , 0), and G must be of the form G(x, z) = (x, g(x, z))
for some smooth function g. Then γ (t) = (t, g(t, 0)) is a parametrization of the
level curve f (x, y) = 0 containing (x0 , y0 ).
fx fy fz
The matrix
has rank 2 everywhere; suppose that, at some point
gx gy gz
fy fz
(x0 , y0 , z0 ) on the level curve, the 2 × 2 submatrix
is invertible.
gy gz
Then, the function
 F (x, y, z)= (x, f (x, y, z), g(x, y, z)) is smooth and its Ja1 0 0

fx fy fz  is invertible at (x0 , y0 , z0 ). Let G(x, u, v) =
cobian matrix
gx gy gz
(x, ϕ(x, u, v), ψ(x, u, v)) be the smooth inverse of F defined near (x0 , 0, 0). Then
γ (t) = (t, ϕ(t, 0, 0), ψ(t, 0, 0)) is a parametrization of the level curve f (x, y, z) =
g(x, y, z) = 0 containing (x0 , y0 , z0 ).
43
5.6.3 Let σ (u, v) = (f (u, v), g(u,
 v), h(u,
v)). The condition σ u × σ v 6= 0 at (u0 , v0 )
fu fv
means that the matrix  gu gv  has rank 2 at (u0 , v0 ), so at least one 2 × 2
hu hv
fu fv
submatrix is invertible, say
. If F (u, v) = (f (u, v), g(u, v)), then as
gu gv
in the proof of Proposition 4.2.6 there is an open subset V of R2 containing
F (u0 , v0 ) and an open subset W of U containing (u0 , v0 ) such that F : W → V
is bijective, in particular injective. Then the restriction of σ to W is injective.
5.6.4 Let σ (u, v) = (f (u, v), g(u, v), h(u, v)). The condition
that N (u0 , v0 ) is not par
fu fv
allel to the xy-plane means that the matrix
is invertible at (u0 , v0 ).
gu gv
If F (u, v) = (f (u, v), g(u, v)), then as in the proof of Proposition 4.2.6 there
is an open subset V of R2 containing F (u0 , v0 ) and an open subset W of U
containing (u0 , v0 ) such that F : W → V is bijective with smooth inverse. If
F −1 (u, v) = (α(u, v), β(u, v)), then near (x0 , y0 , z0 ) the surface coincides with
the graph z = h(α(x, y), β(x, y)). If N (u0 , v0 ) is parallel to the xy-plane, then
at least one of the other two 2 × 2 submatrices of the Jacobian matrix of σ (u, v)
is invertible, and then the surface coincides near (x0 , y0 , z0 ) with a graph of the
form x = ϕ(y, z) or y = ϕ(x, z).
5.6.5 Let γ (t) = (γ1 (t), . . . , γn (t)). As γ is regular, γ̇γ (t0 ) 6= 0, so at least one component of γ̇γ (t0 ) is non-zero, say γ̇1 (t0 ) (the proof is the same in the other cases).
By the inverse function theorem for real-valued function of one variable, there
is a smooth function α, say, defined on an open interval containing γ1 (t0 ), and
some ǫ > 0 such that α(γ1 (t)) = t for all t ∈ (t0 − ǫ, t0 + ǫ). This implies that the
restriction of γ1 to the interval (t0 − ǫ, t0 + ǫ) is injective. Hence, the restriction
of γ to the same interval is injective.
Chapter 6
6.1.1 (i) Quadric cone x2 + z 2 = y 2 ; we have
σ u = (cosh u sinh v, cosh u cosh v, cosh u),
σ v = (sinh u cosh v, sinh u sinh v, 0),
k σ u k2 = 2 cosh2 u cosh2 v,
σ v = 2 sinh u cosh u sinh v cosh v,
σ u .σ
k σ v k2 = sinh2 u cosh 2v,
and the first fundamental form is
2 cosh2 u cosh2 v du2 + sinh 2u sinh 2v dudv + sinh2 u cosh 2v dv 2 .
44
(ii) Paraboloid of revolution; (2 + 4u2 ) du2 + 8uv dudv + (2 + 4v 2 ) dv 2 .
(iii) Hyperbolic cylinder; (cosh2 u + sinh2 u) du2 + dv 2 .
(iv) Paraboloid of revolution; (1 + 4u2 ) du2 + 8uv dudv + (1 + 4v 2 ) dv 2 .
6.1.2 Applying a translation to a surface patch σ does not change σ u or σ v . If P is
σ u ), (P σ )v = P (σ
σ v ), and P preserves
a 3 × 3 orthogonal matrix, (P σ )u = P (σ
p).P (q
q) = p.q
q for all vectors p, q ∈ R3 ). Applying the dilation
dot products (P (p
(x, y, z) 7→ a(x, y, z), where a is a non-zero constant, multiplies σ by a and hence
the first fundamental form by a2 .
6.1.3 Since both sides define bilinear forms on the tangent plane, it suffices to prove
σ u , σ v }. This is easily
that the two sides agree when v, w belong to the basis {σ
σ u ) = dv(σ
σ v ) = 1, du(σ
σ v ) = dv(σ
σ u ) = 0.
checked using du(σ
∂v
σ ũ = σ u ∂∂uũ + σ v ∂∂vũ , σ̃
σ ṽ = σ u ∂u
6.1.4 By the chain rule, σ̃
∂ ṽ + σ v ∂ ṽ , which gives Ẽ =
2
2
σ ũ .σ̃
σ ũ = E ∂u
+ 2F ∂∂uũ ∂∂vũ + G ∂∂vũ . Similar expressions for F̃ and G̃ can be
σ̃
∂ ũ
found; multiplying out the matrices shows that these formulas are equivalent to
the matrix equation in the question.
Following the procedure given, Edu2 + 2F dudv + Gdv 2 =
2
∂v
2
∂u
∂v
∂v
∂v
∂u
dṽ
dũ
+
dṽ
dũ
+
dṽ
+G
dũ
+
dṽ
+2F
. The
E ∂∂uũ dũ + ∂u
∂ ṽ
∂ ũ
∂ ṽ
∂ ũ
∂ ũ
∂ ṽ
∂ ṽ
∂u 2
∂u ∂v
∂v 2
2
coefficient of dũ is E ∂ ũ + 2F ∂ ũ ∂ ũ + G ∂ ũ , which agrees with the expression for Ẽ found above. Similarly for F̃ and G̃.
6.1.5 (i) ⇐⇒ (ii): Ev = Gu = 0 ⇐⇒ σ u.σ uv = σ v .σ uv = 0 ⇐⇒ σ uv is parallel to N.
Consider the quadrilateral bounded by the parameter curves
R v u = u0 , u = u1 , v =
σ v (u0 , v) k dv =
v0 , v = v1 . The length of the side given by u = u0 is v01 k (σ
R v1 p
G(u0 , v)dv.
v0
(i) =⇒ (iii): If Gu = 0, G depends only on v so the last integral is unchanged
when u0 is replaced by u1 . So the two sides u = u0 and u = u1 have the same
length; and similarly for the other two sides.
Rv p
(iii) =⇒ (i): If the lengths are equal then v01 G(u, v)dv is independent of u;
Rv
√u dv = 0 for all v0 , v1 , so Gu = 0;
differentiating with respect to u gives v01 2G
G
and similarly Ev = 0.
Rp
Rp
E(u) du, ṽ =
G(v) dv.
Assuming conditions (i) - (iii) are satisfied, set ũ =
Then,
(u,
v)
→
7
(ũ,
ṽ)
is
a
reparametrization
map
because
its
Jacobian
matrix
√
√
E √0
has non-zero determinant EG. The first fundamental form of
0
G
σ (ũ, ṽ) of σ (u, v) is dũ2 + √2F dũdṽ +dṽ 2 . Now
the reparametrization σ̃
EG
√2F
EG
<1
since EG−F 2 > 0, so there is a smooth function θ(ũ, ṽ) with 0 < θ < π such that
cos θ = √2F
. This gives the first fundamental form as dũ2 + 2 cos θdũdṽ + dṽ 2 .
EG
45
Since ũ = 12 (û + v̂), ṽ = 21 (û − v̂), the first fundamental form becomes
1
1
1
(dû + dv̂)2 + cos θ(dû2 − dv̂ 2 ) + (dû − dv̂)2
4
2
4
1
1
θ
θ
= (1 + cos θ)dû2 + (1 − cos θ)dv̂ 2 = cos2 dû2 + sin2 dv̂ 2 .
2
2
2
2
6.1.6 (i) σ u = (cos v, sin v, 1/u), σ v = (−u sin v, u cos v, 0) so E = 1 + u−2 , F = 0,
G = u2 , The surface is obtained by rotating the curve z = ln x in the xz-plane
around the z-axis.
(ii) σ u = (cos v, sin v, 0), σ v = (−u sin v, u cos v, 1) so E = 1, F = 0, G = 1 + u2 .
This is a helicoid (Exercise 4.2.6).
(iii) σ u = (sinh u cos v, sinh u sin v, 1), σ v = (− cosh u sin v, cosh u cos v, 0) so E =
1 + sinh2 u = cosh2 u, F = 0, G = cosh2 u. This is a catenoid (Exercise 5.3.1).
6.1.7 The first fundamental form is 2du2 + u2 dv 2 . Hence, the length is
√
Z π
2λ2 + 1 λπ
(e − 1).
(2(λeλt )2 + (eλt )2 )1/2 dt =
λ
0
A ruling can be parametrized by u = t, v = constant, so the angle is
!
r
λt
2
2λ2
2(λe
+
u
(0))
= cos−1
cos−1 p
,
√
2λ2 + 1
(2λ2 + 1)e2λt 2
which is constant and independent of the ruling.
b = t − τ vn
n, σ v = b, so since t, n and b are
6.1.8 Denoting d/du by a dot, σ u = γ̇γ + vḃ
2 2
perpendicular unit vectors, E = 1 + τ v , F = 0, G = 1.
σ u.σ uu . Hence, σ u.σ uu =
6.1.9 From E = σ u.σ u we get Eu = σ uu.σ u + σ u.σ uu = 2σ
1
2 Eu . Similarly, differentiating E with respect to v and G with respect to u
and v gives σ u.σ uv = 12 Gu , σ u.σ uv = 12 Ev , σ v .σ vv = 12 Gv . Differentiating
F = σ u.σ v with respect to u gives Fu = σ uu.σ v + σ u.σ uv and using the formula
σ u.σ uv = 21 Ev gives σ uu.σ v = Fu − 21 Ev . The formula for σ u.σ vv is proved
similarly.
√
√
σ (u, v), say. The im6.2.1 The map is σ (u, v) 7→ u 2 cos √v2 , u 2 sin √v2 , 0 = σ̃
age of this map is √
the sector of the xy-plane whose polar coordinates (r, θ)
satisfy 0 < θ < π 2. The first fundamental
form of σ is 2 du2 + u2 dv 2 ;
√
√
σu =
σ v = −u sin √v , u cos √v , 0 , so k σ̃
σ u k2 =
2 cos √v2 , 2 sin √v2 , 0 , σ̃
σ̃
2
2
σ v = 0, k σ v k2 = u2 and the first fundamental form of σ̃
σ is 2 du2 + u2 dv 2 .
2, σ u .σ
√
6.2.2 No: the part of the ruling (t, 0, t) with 1 ≤ t ≤ 2 (say) has length 2 and is
mapped to the straight line segment (t, 0, 0) with 1 ≤ t ≤ 2, which has length 1.
46
6.2.3 A straightforward calculation shows that the first fundamental form of σ t is
cosh2 u(du2 +dv 2 ); in particular, it is independent of t. Hence, σ (u, v) 7→ σ t (u, v)
is an isometry for all t. Taking t = π/2 gives the isometry from the catenoid to
the helicoid; under this map, the parallels u = constant on the catenoid go to
circular helices on the helicoid, and the meridians v = constant go to the rulings
of the helicoid.
6.2.4 The line of striction is given by v = −(γ̇γ .δ̇δ )δδ / k δ̇δ k2 (Exercise 5.3.4), where in this
case δ = γ̇γ . Since γ is unit-speed, γ̇γ .γ̈γ = 0 so v = 0 and we get the curve γ itself.
For the second part, we can assume that u0 = 0 and by applying an isometry of
R3 that γ (0) = 0, t(0) = i, n(0) = j, b(0) = k (in the usual notation). Then, using
...
Frenet-Serret, γ̈γ (0) = κ(0)jj, γ (0) = (−κ(0)2 , κ̇(0), κ(0)τ (0)) so, neglecting higher
...
powers of u in each component, γ (u) = γ (0) +uγ̇γ (0) + 12 γ̈γ (0)u2 + 61 γ (0)u3 +. . . =
(u, 12 κ(0)u2 , 61 κ(0)τ (0)u3 ). The intersection of the surface with the plane perpendicular to t(0) = i is given by setting the x-component of σ (u, v) equal to zero.
This gives v = −u+ higher terms, so neglecting such terms, u = −v. Then the
intersection is Γ(v) = σ (−v, v) = γ (−v) + vγ̇γ (−v) = (0, − 21 κ(0)v 2 , 31 κ(0)τ (0)v 3 ).
6.2.5 A generalized cylinder can be parametrized by σ (u, v) = γ (u) + vaa, where we can
assume (see Exercise 5.3.3) that γ is unit-speed, a is a unit vector, and γ̇γ .a = 0,
where the dot denotes d/du. Then, σ u = γ̇γ , σ v = a so the first fundamental
form is du2 + dv 2 . This is the same as the first fundamental form of the plane
(u, v, 0), so by Corollary 6.2.3 the map σ (u, v) 7→ (u, v, 0) is a local isometry.
By applying an isometry of R3 (a translation), we can assume that the vertex
of the cone is the origin. It can then be parametrized by σ (u, v) = vγγ (u), where
γ is a unit-speed curve on S 2 . Then, σ u = vγ̇γ , σ v = γ so the first fundamental
form is v 2 du2 + dv 2 . This is the same as the first fundamental form of the plane
in polar coordinates (v cos u, v sin u, 0).
6.2.6 Assume that
f satisfies the given condition and denote d/du by a dot. Define
Rq
g(u) =
1 − f˙2 du; then g is smooth since the term inside the square root
is always > 0. The curve γ (u) = (f (u), 0, g(u)) in the xz-plane is unit-speed,
and by Example 6.1.3 if we rotate γ around the z-axis, the resulting surface of
revolution has first fundamental form du2 + f (u)2 dv 2 . By Corollary 6.2.3, the
given surface is locally isometric to this surface of revolution.
Rp
U (u) du. The given condition on U implies that U (u) > 0 for all
6.2.7 Define ũ =
values of u, so ũ is a smooth function of u and dũ/du is never zero. It follows
that (u, v) 7→ (ũ, v) is a reparametrization map. The first fundamental form in
terms of (ũ, v) is dũ2 + Ũ (ũ)dv 2 , where Ũ (ũ) = U (u). By Exercise 6.2.6, this
df
< 1 for all
surface patch is locally isometric to a surface of revolution if dũ
47
values of ũ, where f =
p
Ũ . But,
df
1 dŨ
1 dU/du
1 dU
= p
= √
=
.
dũ
2U du
2 U dũ/du
2 Ũ dũ
6.3.1 If the first fundamental forms of two surfaces are equal, they are certainly proportional, so any isometry is a conformal map. Stereographic projection is a
conformal map from S 2 to the plane, but it is not an isometry since λ 6= 1 (see
Example 6.3.5).
6.3.2 The first fundamental form of the given surface patch is (1+u2 +v 2 )2 (du2 +dv 2 );
this is a multiple of du2 + dv 2 so the patch is conformal.
2
σ (u, v) is dψ
σ is
6.3.3 The first fundamental form of σ̃
du2 + cos2 ψ(u)dv 2 . So σ̃
du
R
conformal ⇐⇒ dψ/du = ± cos ψ. Taking the plus sign, we get u = sec ψdψ =
ψ
cos ψ
ψ
u
u
−u
= 1+sin
ln(sec ψ + tan ψ), so 1+sin
cos ψ = e . Then 2 cosh u = e + e
cos ψ + 1+sin ψ =
σ (u, v) is the patch in Exercise
2 sec ψ. Hence, cos ψ = sechu, sin ψ = tanhu and σ̃
5.3.2.
6.3.4 Φ is conformal if and only if fu2 + gu2 = fv2 + gv2 and fu fv + gu gv = 0. Let
z = fu + igu , w = fv + igv ; then Φ is conformal if and only if zz̄ = ww̄ and
z w̄ + z̄w = 0, where the bar denotes complex conjugate; if z = 0, then w = 0 and
all four equations are certainly satisfied; if z 6= 0, the equations give z 2 = −w2 ,
so z = ±iw; these are easily seen to be equivalent to the first pair of equations
in the statement of the exercise if the sign is +, and to the second pair if the
f
gu
= ±(fu2 + fv2 ), with a plus sign if the
sign is −. We have det(J(Φ)) = u
fv g v
first pair of equations hold and a minus sign if the second pair of equations hold.
6.3.5 Let S be an orientable surface. Fix a smooth choice of unit normal at each point
of S, and let A be the atlas for S consisting of all the surface patches for S
whose standard unit normal agrees with the chosen normal. On the other hand,
by Theorem 6.3.6 S has an atlas consisting of conformal parametrizations; let
à be the maximal such atlas (i.e. the set of all conformal parametrizations of
S). Then, A ∩ Ã is an atlas for S. Indeed, if p ∈ S, let σ be any conformal
parametrization of S containing p. If σ has the wrong orientation (so that
σ (u, v) = σ (−u, v) is a conformal parametrization containing p
σ ∈
/ Ã), then σ̃
that has the correct orientation. Thus, in any case there is a surface patch of
S containing p that is both conformal and correctly oriented. Let Φ be the
transition map between two of the patches in the atlas A ∩ Ã. Then, Φ is a
conformal diffeomorphism between open subsets of R2 . By Exercise 6.3.4, Φ is
either holomorphic or anti-holomorphic, and in the latter case det(J(Φ)) < 0,
contradicting the fact that Φ is the transition map between two correctly oriented
surface patches. Hence, Φ must be holomorphic.
48
y
x
, z+1
, 0). Identifying (u, v) ∈
6.3.6 Following Example 6.3.5, we find Π̃(x, y, z) = ( z+1
2
1−|w|
σ 1 (w) = ( |w|2w
R2 with w = u + iv ∈ C, we find σ̃
Then σ 1 (w) =
2 +1 , 1+|w|2 ).
σ 1 (1/w̄), so the transition map is w 7→ 1/w̄. This is not holomorphic, so the atlas
σ̃
σ 1 , σ̃
σ 1 } does not give S 2 the structure of a Riemann surface. If σ̂
σ 1 (w) = σ̃
σ 1 (w̄),
{σ
σ 1 is w 7→ 1/w. This is holomorphic (when
the transition map between σ 1 and σ̂
σ 1 , σ̂
σ 1 } gives
w 6= 0, which holds on the overlap of the two patches), so the atlas {σ
2
S the structure of a Riemann surface.
6.3.7 Any circle on S 2 is the intersection of S 2 with a plane, and so (see Appendix 2)
has equation of the form aw + āw̄ + bz = c, where a ∈ C, b, c ∈ R are constants
|ξ|2 −1
(and a and b are not both zero). Substituting w = |ξ|2ξ
2 +1 , z = |ξ|2 +1 gives
(b − c)|ξ|2 + 2aξ + 2āξ̄ = b + c, which is the equation of a Circle in C∞ (a line if
b = c, a circle otherwise). The converse is proved similarly.
σ 1 , σ̃
σ 1 } of S 2 in Ex6.3.8 The expression of the map Π−1 ◦ M ◦ Π in terms of the atlas {σ
ercise 6.3.6, which consists of conformal patches, is of the form w 7→ M (w), w 7→
−1
−1
M (1/w̄), w 7→ M (w) , or w 7→ M (1/w̄) , i.e. a Möbius or conjugate-Möbius
transformation. Since such transformations are conformal (Appendix 2), the
result follows.
6.3.9 The first fundamental form is (1 + f˙2 )du2√+ u2 dv 2 , where a dot denotes d/du.
So σ √
is conformal if and only if f˙ = ± u2 − 1, i.e. if and only if f (u) =
±( 21 u u2 − 1 − 12 cosh−1 u) + c, where c is a constant.
6.3.10 The first fundamental form is (1 + 2vγ̇γ .δ̇δ + v 2δ̇δ .δ̇δ)du2 + 2γ̇γ .δδ dudv + dv 2 . So σ is
conformal if and only if 1 + 2vγ̇γ .δ̇δ + v 2δ̇δ .δ̇δ = 1 and γ̇γ .δδ = 0 for all u, v; the first
condition gives δ̇δ = 0, so δ is constant, and the second condition then says that
γ .δδ is constant, say equal to d. Thus, σ is conformal if and only if δ is constant
and γ is contained in a plane v.δδ = d. In this case, σ is a generalized cylinder.
6.3.11 From σ ∗u = σ1.σ σ u − 2((σσ..σσ)u2) σ , we get (in the obvious notation)
E ∗ = σ ∗u.σ ∗u =
σ.σ u )2
σ. σ u ) 2
4(σ
4(σ
E
E
−
+
=
.
2
3
3
σ. σ )
σ .σ )
σ. σ )
(σ
(σ
(σ
k σ k4
Similarly, we find that F ∗ = F/ k σ k4 , G∗ = G/ k σ k4 . Hence, the first
fundamental form of σ ∗ is a multiple of that of σ .
6.3.12 By Example 6.1.3, any surface of revolution has an atlas consisting of surface
2
2
2
patches σ (u, v) whose first fundamental
R duform is du + f (u) dv for some positive
smooth function f (u). Define ũ = f (u) . Then, the first fundamental form of
σ (ũ, v) of σ (u, v) is f 2 (dũ2 + dv 2 ). This is a conformal
the reparametrization σ̃
surface patch.
6.4.1 Parametrize the paraboloid by σ (u, v) = (u, v, u2 + v 2 ); its first fundamental
form is (1 + 4u2 )du2 + 8uv dudv + (1 + 4v 2 )dv 2 . Hence, the required area is
49
RR p
1 + 4(u2 + v 2 ) dudv, taken over the disc u2 + v 2 < 1. Let
the integral
R1√
u = r sin θ, v = r cos θ; then the area is 2π 0 1 + 4r 2 r dr = π6 (53/2 − 1). This
is less than the area 2π of the hemisphere.
6.4.2 If S is a sphere with centre the origin and radius R, the map S 2 → S given
p multiplies the first fundamental form by R2 , and so is conformal
by p 7→ Rp
but multiplies areas by R2 . It follows from Theorem 6.4.7 that the sum of the
angles of a spherical triangle of area A on S is π + A/R2 . In this case, R is the
radius of the earth and A is ≥ the area of Australia, so the sum of the angles is
30
≥ π + (7 500 000)/(6 500)2 = π + 169
radians. Hence, at least one angle of the
10
radians.
triangle must be at least one third of this, i.e. π + 169
6.4.3 Take a point p inside the polygon and join it to each vertex of the polygon by an
arc of a great circle. This gives n triangles whose sides are arcs of great circles.
The sum of their angles is the sum of their areas (i.e. the area of the polygon)
minus nπ by Theorem 6.4.7, and is also the sum of the angles of the polygon
plus 2π (the angle around p).
6.4.4 The sum of the angles around any vertex is 2π, so the sum of the angles of all the
polygons is 2πV . By the preceding exercise, the sum of the angles of a polygon
with P
n sides is (n − 2)π plus its area. Summing over all polygons gives 2πV =
4π + polygons (n − 2)π, since the sum of the areas of all the polygons is the area
P
4π of the sphere. Since two polygons meet along each edge, polygons n = 2E,
and since there are F polygons altogether we get 2πV = 2πE − 2πF + 4π, which
is equivalent to V − E + F = 2.
√
6.4.5 (i) is obvious as a local isometry preserves E, F, G and hence EG − F 2 .
If Ẽ = λE, F̃ = λF and G̃ = λG, and if Ẽ G̃ − F̃ 2 = EG − F 2 , then λ2 = 1 and
so λ = 1 (as E, Ẽ are > 0). This proves (ii).
The map from S 2 to the unit cylinder in the proof of Theorem 6.4.6 is an equiareal
map that is not a local isometry.
6.4.6 Let σ : U → R3 ; f is equiareal ⇐⇒
ZZ
ZZ
2 1/2
(E1 G1 − F1 ) dudv =
(E2 G2 − F22 )1/2 dudv for all regions R ⊆ U .
R
R
This holds ⇐⇒ the two integrands are equal everywhere, i.e. ⇐⇒ E1 G1 − F12 =
E2 G2 − F22 .
6.4.7 Since N is perpendicular to the tangent plane, N × σ u is parallel to the tangent
σ u + βσ
σ v for some α, √
N × σ u ).σ
σ u = 0, (N
N × σ u )..σ v =
plane, and so = ασ
β. Now (N
2
σ u × σ v )..N = k σ u × σ v k N.N = EG − F by √
(σ
Proposition 6.4.2. This
gives the √
two equations αE + √
βF = 0, αF + βG = EG − F 2 , which imply
2
α = −F/ EG − F , β = E/ EG − F 2 . The formula for N × σ v is proved
similarly.
50
6.4.8 By Exercise 6.1.6(ii), the first fundamental form is du2 + (1 + u2 )dv 2 . So the
area is
Z 2π Z 1 p
√
√
1 + u2 dudv = π( 2 + ln(1 + 2)).
0
0
(The integral can be evaluated by putting u = tan θ.)
6.4.9 If we count the m edges that meet at each vertex, and then sum over all vertices,
we will have counted each edge twice, once from each end. Hence, mV = 2E.
Similarly, counting the n edges of each polygon, and then summing over all the
polygons, gives twice the total number of edges as each edge is an edge of two
polygons. This gives nF = 2E.
−1
1
+ n1 − 12
,
From V − E + F = 2 and V = 2E/m, F = 2E/n we get E = m
and hence values for V and F . The desired inequality follows from the fact that
V, E and F are all > 0.
The inequality can be written (m − 2)(n − 2) < 4. Thus, m − 2 and n − 2 are
two positive integers whose product is less than 4, giving the 5 possibilities 1 × 1,
1 × 2, 2 × 1, 3 × 1, 1 × 3.
6.4.10 We can assume that the sphere is the unit sphere S 2 . If such curves exist they
would give a triangulation of S 2 with 5 vertices and 5 × 4/2 = 10 edges, hence
2 + 10 − 5 = 7 polygons. Since each edge is an edge of two polygons and each
polygon has at least 3 edges, 3F ≤ 2E; but 3 × 7 > 2 × 10. If curves satisfying
the same conditions exist in the plane, applying the inverse of the stereographic
projection map (Example 6.3.5) would give curves satisfying the conditions on
S 2 , which we have shown is impossible.
6.4.11 Such a collection of curves would give a triangulation of the sphere with V = 6,
E = 9, and hence F = 5. The total number of edges of all the polygons in
the triangulation is 2E = 18. Since exactly 3 edges meet at each vertex, going
around each polygon once counts each edge 3 times, so there should be 18/3 = 6
polygons, not 5.
6.4.12 Parametrize the surface by σ (u, v) = (ρ(u) cos v, ρ(u) sin v, σ(u)), where γ (u) =
2
2
2
(ρ(u), 0, σ(u)).RRBy Example 6.1.3,
R the first fundamental form is du + ρ(u) dv ,
so the area is
ρ(u) dudv = 2π ρ(u) du.
R π/2
(i) Take ρ(u) = cos u, σ(u) = sin u, with −π/2 ≤ u ≤ π/2; so 2π −π/2 cos u du =
4π is the area.
(ii) For the torus, the profile curve is γ (θ) = (a + b cos θ, 0, b sin θ), but this is
not unit-speed; a unit-speed reparametrisation is γ̃γ (u) = a + b cos ub , 0, b sin ub
R 2πb
with 0 ≤ u ≤ 2πb. So 2π 0
a + b cos ub du = 4π 2 ab is the area.
6.4.13 Using the parametrization in Exercise 4.2.7, the first fundamental form is found
51
to be ((1 − κa cos θ)2 + τ 2 a2 ) ds2 + 2τ a2 dsdθ + a2 dθ 2 , so the area is
Z
s1
s0
Z
2π
0
a(1 − κa cos θ)dsdθ = 2πa(s1 − s0 ).
6.4.14 If E1 = λE2 , F1 = λF2 , G1 = λG2 , and if E1 G1 − F12 = E2 G2 − F22 , then λ2 = 1
so λ = 1 (since λ > 0).
6.4.15 (i) Any linear map R2 → R2 is obviously smooth so M is a diffeomorphism if
and only if M is bijective. This holds if and only if M takes a basis of R2 , such
u, v} is
as {ii, j}, to another basis. Hence, M is a diffeomorphism if and only if {u
2
u
v
a basis of R , i.e. if and only if and are linearly independent.
u + vvv. The first fundamental form of σ
(ii) M takes (u, v) ∈ R2 to σ (u, v) = uu
2
2
2
2
u.v)dudv+ k v k dv . This is equal to du2 + dv 2 if and only if
is k u k du + 2(u
k u k = k v k = 1 and u.v = 0, i.e. if and only if u and v are perpendicular unit
vectors.
u.v)dudv+ k v k2 dv 2 is a multiple of du2 + dv 2 if and only if
(iii) k u k2 du2 + 2(u
k u k = k v k and u.v = 0, i.e. if and only if u and v are perpendicular vectors of
equal length.
(iv) Since E = k u k2 , F = u.v, G =k v k2 , M is equiareal if and only if
u.v)2 = 1. From the proof of Proposition
EG − F 2 = 1, i.e. k u k2 k v k2 −(u
6.4.2, the left-hand side is equal to k u × v k2 . So M is equiareal if and only if
k u × v k = 1.
6.4.16 From the solution to Exercise 6.3.9, the first fundamental form of σ is (denoting
d/du by a dot)
(1+ f˙2 )du2 +u2 dv 2 . So σ is equiareal if and only if u2 (1+ f˙2 ) = 1,
√
2
df
= ± 1−u
. Thus,
i.e. du
u
f (u) = ±
Z √
p
1
1 − u2
−1
du = ±
1 − u2 − cosh
u
u
(up to adding an arbitrary constant). This is a parametrization of the tractrix,
so σ is equiareal ⇐⇒ σ is a reparametrization of the pseudosphere (see §8.3).
6.4.17 The first fundamental form of σ 3 is cos2 θdθ 2 + 2f˙ cos θdθdϕ + (1 + f˙2 )dϕ2 ,
where the dot denotes d/dθ. With E = cos2 θ, F = f˙ cos θ, G = 1 + f˙2 ,
EG−F 2 = cos2 θ. As this does not depend on f , the fact that the map σ 1 (θ, ϕ) 7→
σ 3 (θ, ϕ) is equiareal follows from the fact, proved in Theorem 6.4.6, that the map
σ 1 (θ, ϕ) 7→ σ 2 (θ, ϕ) is equiareal.
6.5.1 If the internal angles are equal to α, Theorem 6.4.7 gives 3α − π = 4π/4, so
α = 2π/3. Corollary 6.5.6 then gives the length of a side as A = cos−1 (−1/3).
6.5.2 Using the notation following Proposition 6.5.8, there is a rotation R1 of S 2 that
takes a′ to a; then a further rotation R2 around the diameter through a that
52
b′ ) coincide with the side through a and b. By
makes the side through a and R1 (b
Corollary 6.5.6 the two triangles have sides of the same length, so we must have
b′ ). If c and c′ are on the same side of the plane containing the side
b = R2 R1 (b
through a and b, we shall then have c = R2 R1 (cc′ ) and the isometry R2 R1 takes
the triangle with vertices a′ , b′ , c′ to the triangle with vertices a, b, c; if they are
on opposite sides the isometry R3 R2 R1 does this, where R3 is reflection in the
plane containing the side through a and b.
6.5.3 By applying an isometry of R3 , which leaves lengths and areas unchanged, we
can assume that p is the north pole (0, 0, 1). The spherical circle of radius R
and centre p is then the circle of latitude ϕ = π/2 − R (Example 4.1.4), which is
a circle of radius sin R. The area inside it is, by Example 6.1.3 and Proposition
R 2π R π/2
6.4.2, 0 π/2−R cos θdθdϕ = 2π(1 − cos R). The maximum value of R is π; if
π/2 ≤ R ≤ π, one replaces R by π − R in (i) and (ii).
′
′
w+b
′
6.5.4 (i) If M ′ (w) = ac′ w+d
′ is another unitary Möbius transformation, (M ◦ M )(w) =
Aw+B
, where A = a′ a + b′ c, B = a′ b + b′ d, C = c′ a + d′ c, D = c′ b + d′ d. Thus,
Cw+D
Ā = ā′ ā + b¯′ c̄ = d′ d + (−c′ )(−b) = D and similarly C = −B̄. Inverses are dealt
with similarly.
(ii) Denoting (x, y, z) ∈ R3 by (ξ, z) with ξ = x + iy ∈ C, the plane through
the origin perpendicular to (a, b) is w̄a + wā + 2bz = 0, and reflection in it
|w 2 −1
is F (ξ, z) = (ξ, z) − 2 w̄a+wā+2bz
(a, b). Taking ξ = |w|2w
2 +1 , z = |w|2 +1 , we
|a|2 +b2
2(|a|2 +b2 )w−2a(w̄a+wā+b(|w|2 −1))
, z′
(|w|2 +1)(|a|2 +b2 )
−ab|w|2 +b2 w−a2 w̄+ab
ξ′
giving w′ = 1−z
′ = b2 |w|2 +bāw+baw̄+|a|2
find that F (ξ, z) = (ξ ′ , z ′ ), where ξ ′ =
2
2
2
2
(|a| +b )(|w| −1)−2b(w̄a+wā+b(|w| −1))
,
(|w|2 +1)(|a|2 +b2 )
(−aw̄+b)(bw+a)
−aw̄+b
(bw̄+ā)(bw+a) = bw̄+ā .
=
=
(iii) By Proposition 6.5.7, if F is any isometry of S 2 ,
F∞ = (M1 ◦ J) ◦ (M2 ◦ J) ◦ · · · ◦ (Mk ◦ J) for some k.
Since J ◦ M ◦ J is easily seen to be a unitary Möbius transformation if M is one,
part (i) implies that F∞ is a unitary Möbius transformation if k is even, and of
the form M ◦ J with M unitary Möbius if k is odd.
aw+b
(iv) If a ∈ C, b ∈ R, call the unitary Möbius transformation M (w) = −bw+ā
special unitary. Then M = F∞ ◦ J where F is as in (ii). Since J = R∞ where R
is reflection in the yz-plane, M = (F ◦ R)∞ corresponds to the isometry F ◦ R of
S 2 . It therefore suffices to prove that every unitary Möbius transformation is a
composite of finitely-many special unitary Möbius transformations. If M ′ (w) =
Aw+B
is any unitary Möbius tranformation, where A, B ∈ C, let B = beiθ with
−B̄w+Ā
Aw+b
b, θ ∈ R. Then M ′ = ρ◦M ◦ρ−1 , where ρ(w) = eiθ and M (w) = −bw+
are both
Ā
aw+b
iθ/2
special unitary Möbius transformations (ρ(w) = −bw+ā with a = e
, b = 0).
6.5.5 The dilation v 7→ R−1v, followed by a suitable translation, takes the sphere S of
radius R to S 2 . It takes a triangle on S with sides of length A, B, C to a triangle
53
on S 2 with sides of length A/R, B/R, C/R and the same angles. Applying the
A
B
A
B
cosine rule to the triangle on S 2 gives cos C
R = cos R cos R + cos γ sin R sin R ,
where γ is the angle opposite the side of length C (and two similar formulas).
2
x
x
x
is approximately equal to 1 − 2R
If R is large, cos R
2 and sin R is approximately
x
equal to R
. Substituting into the cosine rule gives
C2
1−
=
2R2
B2
AB
A2
1−
+ 2 cos γ,
1−
2
2
2R
2R
R
2
2
which gives C 2 = A2 + B 2 − 2AB cos γ − A2RB2 . Letting R → ∞ gives the usual
cosine rule for a plane triangle with sides of length A, B, C and with γ the angle
opposite the side of length C. This happens because when R → ∞ and A, B, C
are fixed, the triangle becomes more and more nearly planar. (When standing
at a point on a sphere of very large radius, one seems to be standing on a plane
- this is why the Ancients thought the Earth is flat!)
6.5.6 Let n, a, b be the North Pole, Athens and Bombay, respectively, and let R =
6500km be the radius of the Earth. The angle at n of the spherical triangle with
vertices n, a, b is 73◦ − 24◦ = 49◦ , and the lengths of the sides opposite b and a
are obtained by multiplying by R the radian equivalents of 90◦ − 19◦ = 71◦ and
90◦ − 38◦ = 52◦ , respectively. If d is the spherical distance between Athens and
Bombay, applying the cosine rule in the preceding exercise thus gives
cos
d
= cos 52◦ cos 71◦ + sin 52◦ sin 71◦ cos 49◦ = 0.6892.
R
Hence, d/R = 0.8104 (radians) and d = 0.8104R = 5267km.
6.5.7 Drawing the diagonal of the square gives two spherical triangles with angles
α/2, α/2 and α, the sides opposite these angles being A, A and D (the length of
the diagonal). Applying the cosine rule to one of these triangles gives
1
cos A = cos A cos D + sin A sin D cos α,
2
2
2
cos D = cos A + sin A cos α.
Using the sine rule (Proposition 6.5.3(ii)) gives sin D = sin A sin α/ sin 21 α =
2 sin A cos 21 α. Substituting the formulas for cos D and sin D into the first equation above gives
1
cos A = cos3 A + cos A sin2 A cos α + 2 sin2 A cos2 α.
2
Hence, cos A sin2 A(1 − cos α) = 2 sin2 A cos2 12 α, and this gives the desired formula for cos A.
54
6.5.8 (i) This is obvious by the sine rule.
(ii) This follows immediately from Corollary 6.5.6. Similarly one proves that
cos β + cos α cos γ = λ sin γ sin A cos B.
(iii) By (ii),
(cos α + cos β)(1 + cos γ) = cos α + cos β cos γ + cos β + cos α cos γ
= λ sin γ sin B cos A + λ sin γ sin A cos B
by (ii), which = λ sin γ(sin A cos B + cos A sin B) = λ sin γ sin(A + B).
(iv) Writing the formula in (iii) in terms of half-angles gives
1
1
1
1
1
cos (α + β) cos (α − β) = λ tan γ sin (A + B) cos (A + B).
2
2
2
2
2
Similarly, the first of the two equations in (i) can be written
1
1
1
1
sin (α + β) cos (α − β) = λ sin (A + B) cos (A − B).
2
2
2
2
Dividing this equation by the preceding one gives the first formula in (iv). The
second follows similarly from (iii) and the second equation in (i).
(v) Applying the first formula in (iv) to the polar triangle, we get
cos 21 (π − α − π + β)
1
1
tan (π − A + π − B) =
cot (π − C),
1
2
2
cos 2 (π − α + π − β)
i.e.,
Similarly,
cos 12 (α − β)
1
1
tan C.
tan (A + B) =
1
2
2
cos 2 (α + β)
sin 12 (α − β)
1
1
tan (A − B) =
tan C.
1
2
2
sin 2 (α + β)
6.5.9 If a′ 6= ±aa a rotation through an angle 12 dS 2 (aa, a′ ) around the line through the
origin parallel to a × a′ takes a ′ to a . If a ′ = −aa the same thing is achieved by
v. Thus, in all cases there is an isometry of S 2
applying the isometry v 7→ −v
that takes a ′ to a. Since composites of isometries are isometries, it is sufficient
to prove the result when a′ = a .
If θ is the angle between the great circles passing through a and b and through a
and b′ , a rotation by θ about the line through the origin and a fixes a and takes
b′ to a point b′′ on the great circle passing through a and b. Since b and b′ are
equidistant from a, so are b and b′′ . It follows that either b′′ = b or b′′ is obtained
55
by reflecting b in the plane passing through the origin and a perpendicular to
the great circle passing through a and b. It is therefore sufficient to prove the
theorem when a = a′ and b = b′ .
The great circle passing through a and c makes the same angle with that passing
through a and b as does the great circle passing through a and c′ . Hence, either
these great circles coincide or one is obtained from the other by reflecting in the
plane containing a, b and the origin. It therefore suffices to prove the result
in the case where the two great circles coincide. Since c and c′ are the same
distance from a on the same great circle, either they coincide or c′ is obtained
from c by reflecting in the plane passing through the origin and a perpendicular
to the great circle passing through a and c.
Chapter 7
7.1.1 σ u = (1, 0, 2u), σ v = (0, 1, 2v), so N = λ(−2u, −2v, 1), where λ = √1+4u12 +4v 2 ;
σ uu = (0, 0, 2), σ uv = 0, σ vv = (0, 0, 2), so L = 2λ, M = 0, N = 2λ, and the
second fundamental form is 2λ(du2 + dv 2 ).
Nu = −σ
σ uu .N
N (since σ u .N
N = 0), so Nu .σ
σ u = 0; similarly, Nu .σ
σ v = Nv .σ
σu =
7.1.2 σ u .N
σ v = 0; hence, Nu and Nv are perpendicular to both σ u and σ v , and so
Nv .σ
are parallel to N . On the other hand, Nu and Nv are perpendicular to N since
N is a unit vector. Thus, Nu = Nv = 0, and hence N is constant. Then,
σ .N
N)u = σ u .N
N = 0, and similarly (σ
σ .N
N)v = 0, so σ .N
N is constant, say equal to
(σ
N = d.
d, and then σ is an open subset of the plane v.N
N = ±N
N, the sign being that of det(J). From σ̃
σ ũ = σ u ∂∂uũ + σ v ∂∂vũ ,
7.1.3 From §4.5, Ñ
∂v
σ ṽ = σ u ∂u
σ̃
∂ ṽ + σ v ∂ ṽ , we get
∂ 2u
∂ 2v
σ
σ
σ
σ̃ ũũ = u 2 + v 2 + σ uu
∂ ũ
∂ ũ
2
∂v
+ 2M ∂u
+N
So L̃ = ± L ∂u
∂ ũ
∂ ũ ∂ ũ
∂u
∂ ũ
2
∂v 2
∂ ũ
∂u ∂v
σ uv
+ 2σ
+ σ vv
∂ ũ ∂ ũ
∂v
∂ ũ
2
.
N = σ v .N
N = 0. This, to, since σ u .N
gether with similar formulas for M̃ and Ñ , are equivalent to the matrix equation
in the question.
7.1.4 Let σ be a surface patch, P a 3 × 3 orthogonal matrix, a ∈ R3 a constant vector,
σ = P σ +aa. Then, σ̃
σ u = P σ u , σ̃
σ v = P σ v , so σ̃
σ u ×σ̃
σ v = ±σ
σ u ×σ
σ v (Proposition
and σ̃
A.1.6), the sign being + if the isometry v 7→ P v +aa is direct and − if it is opposite.
It follows that (in the obvious notation), L̃ = ±L, M̃ = ±M, Ñ = ±N . The
v, where a is a non-zero constant, multiplies σ by a and hence
dilation v 7→ av
multiplies each of L, M, N by a.
7.1.5 σ u = (cos v, sin v, 0), σ v = (−u sin v, u cos v, 1), so σ u × σ v = (sin v, − cos v, u)
1
and N = √1+u
(sin v, − cos v, u). Then, σ uu = 0, σ uv = (− sin v, cos v, 0), σ vv =
2
56
1
(−u cos v, −u sin v, 0). This gives L = σ uu.N = 0, M = σ uv .N = − √1+u
,
2
√
.
N = σ vv .N = 0. So the second fundamental form is −2dudv
1+u2
7.1.6 The tangent developable is parametrized by σ (u, v) = γ (u) + vtt(u), where t =
dγγ /du. Using the standard notation and the Frenet-Serret equations, we find
v
n, σ v = t, σ u × σ v = −κvb
b, N = − |v|
σ u = t + κvn
b,
n+
σ uu = κn
dκ
n + κv(−κtt + τb
b), σ uv = κn
n, σ vv = 0.
vn
du
2
v
, M = N = 0. So the second fundamental form is −κτ |v|du2 .
Hence, L = −κτ |v|
This vanishes at all points of the surface if and only if τ = 0 at all points
of γ , which holds if and only if γ is planar (Proposition 2.3.3). In that case,
the tangent developable is part of a plane. Thus, for tangent developables, the
second fundamental form vanishes everywhere if and only if the surface is part
of a plane. This is a special case of Exercise 7.1.2.
7.2.1 The paraboloid is the level surface f = 0 where f (x, y, z) = z − x2 − y 2 and
(fx ,fy ,fz )
N = k(f
is the corresponding unit normal. So G(x, y, z) = (4x(−2x,−2y,z)
2 +4y 2 +1)1/2 .
x ,fy ,fz )k
7.2.2 This is obvious since changing the orientation changes the Gauss map G to −G.
7.2.3 The hyperboloid of one sheet x2 + y 2 − z 2 = 1 is obtained by rotating the
hyperbola x2 − z 2 = 1 in the xz-plane around the z-axis. For this hyperbola,
dz
= xz → ±1 as x → ±∞. The unit normal to the hyperbola therefore makes
dx
an angle θ with the x-axis, where the angle θ takes all values in the range
− π4 < θ < π4 . Rotating around the z-axis, it follows that the image of the Gauss
map of the hyperboloid is the region of S 2 for which the latitude θ satisfies
− π4 < θ < π4 . This is an annulus on S 2 .
The hyperboloid of two sheets x2 − y 2 − z 2 = 1 is obtained by rotating the same
hyperbola as above around the x-axis. The image of G is the union of√
two ‘caps’
2
2
2
2
on S , namely the parts of the sphere x + y + z = 1 with |x| > 1/ 2.
7.3.1 Let t be the parameter for γ , let s be arc-length along γ , and denote d/dt by a dot
2
ds 2 ′′
′ γ
γ + ddt2s γ ′ . By Proposition 7.3.5,
and d/ds by a dash. Then, γ̇γ = ds
dt γ , γ̈ = dt
1
1
γ̇γ , (ds/dt)
γ̇γ ii = hhγ̇γ , γ̇γ ii/(ds/dt)2 = hhγ̇γ , γ̇γ ii/hγ̇γ , γ̇γ i. For the
κn = hhγγ ′ , γ ′ ii = hh (ds/dt)
3 ′′
N × γ̇γ ) = ds
N × γ ′ ) = 0, we have γ̈γ.(N
N × γ ′) =
second part, since γ ′.(N
γ .(N
dt
hγ̇γ , γ̇γ i3/2 κg .
7.3.2 Let γ be a unit-speed curve on the sphere of centre a and radius r. Then,
(γγ − a).(γγ − a ) = r 2 ; differentiating gives γ̇γ .(γγ − a ) = 0, so γ̈γ .(γγ − a ) = −γ̇γ .γ̇γ =
−1. At the point γ (t), the unit normal of the sphere is N = ± r1 (γγ (t) − a), so
N = ± 1r γ̈γ .(γγ − a ) = ∓ 1r .
κn = γ̈γ .N
57
7.3.3 If the sphere has radius R, the parallel with latitude θ has radius r = R cos θ; if
p is a point of this circle, its principal normal at p is parallel to the line through
p perpendicular to the z-axis, while the unit normal to the sphere is parallel to
the line through p and the centre of the sphere. The angle ψ in Eq. 7.10 is
therefore equal to θ or π − θ so κg = ± 1r sin θ = ± R1 tan θ. Note that this is zero
if and only if the parallel is a great circle.
σ u )+v̇(F σ v −Gσ u ) ,
σ + v̇σ
σ , so by Exercise 6.4.7, N × γ̇γ = u̇(Eσ v −F√
7.3.4 We have γ̇γ = u̇σ
u
v
σ u + v̈σ
σ v + u̇2 σ uu + 2u̇v̇σ
σ uv + v̇ 2σ vv . Hence,
γ̈γ = üσ
N × γ̇γ ) = (u̇v̈ − v̇ü)
κg = γ̈γ .(N
where
p
EG−F 2
EG − F 2 + Au̇3 + B u̇2 v̇ + C u̇v̇ 2 + Dv̇ 3 ,
1
σ v − F σ u ) = E((σ
σu .σ
σ v )u − σ u .σ
σ uv ) − F (σ
σ u .σ
σ u )u
A = σ uu .(Eσ
2
1
1
= E(Fu − Ev ) − F Eu ,
2
2
with similar expressions for B, C, D.
p
p
If F = 0, we find by this method that A = − 12 Ev E/G, B = Gu E/G −
p
p
p
p
1
E
G/E, C = 12 Gv E/G − Ev G/E, D = 21 Gu G/E.
2 u
N1 .n
n, κ2 = κN
N2 .n
n, so
7.3.5 κ1 = κN
N1 .n
n)N
N2 − (N
N2 .n
n)N
N1 ) = κ(N
N1 × N 2 ) × n .
κ1 N2 − κ2 N1 = κ((N
Taking the squared length of each side, we get
N2 = κ2 k (N
N1 × N2 ) × n k2 .
κ21 + κ22 − 2κ1 κ2 N1 .N
N2 = cos α; γ̇γ is perpendicular to N1 and N2 , so N1 × N2 is parallel to
Now, N1 .N
N1 × N 2 ) × n k = k N 1 × N 2 kk n k = sin α.
γ̇γ , hence perpendicular to n; hence, k (N
7.3.6 A straight line has a unit-speed parametrization γ (t) = p + t q (with q a unit
N = 0. In general, κn = 0 ⇐⇒ γ̈γ is
vector), so γ̈γ = 0 and hence κn = γ̈γ .N
perpendicular to N ⇐⇒ N is perpendicular to n ⇐⇒ N is parallel to b (since N
is perpendicular to t).
√
7.3.7 The second fundamental form is (−du2 + u2 dv 2 )/u 1 + u2 , so a curve γ (t) =
σ (u(t), v(t)) is asymptotic if and only if −u̇2 + u2 v̇ 2 = 0, i.e. dv/du = v̇/u̇ =
±1/u, so ln u = ±(v + c), where c is a constant.
7.3.8 By Proposition 7.3.2, if κn = κg = 0 then κ = 0. But a curve with zero curvature
everywhere is part of a straight line.
58
7.3.9 We find that, at the point where u = v = 1, the first and second fundamental
√
, respectively. Taking u = v = t, so
forms are 32 du2 + 2dudv + 32 dv 2 and 2dudv
5
that u̇ = v̇ = 1, Exercise 7.3.1 gives the normal curvature as
2√
u̇v̇
5
3 2
2 u̇
+ 2u̇v̇ +
3 2
2 v̇
2
= √ .
5 5
7.3.10 From Examples 6.1.3 and 7.1.2, the first and second fundamental forms are
du2 + f (u)2 dv 2 and (f˙g̈ − f¨ġ)du2 + f ġdv 2 , respectively (a dot denoting d/du.
We use the formulas in Exercise 7.3.1.
(i) For a meridian we use u as the parameter, so the curve is γ (u) = σ (u, v)
(with v constant). Note that γ is unit-speed. The second formula in Exercise
7.3.1 gives κn = f˙g̈ − f¨ġ. Now, γ̇γ = (f˙ cos v, f˙ sin v, ġ), γ̈γ = (f¨ cos v, f¨ sin v, g̈),
N×γ̇γ ) = 0.
N = (−ġ cos v, −ġ sin v, f˙) (from Example 7.1.2 again), giving κg = γ̈γ. (N
(ii) For a parallel we use v as the parameter, so the curve is Γ(v) = σ (u, v)
Γ/dv =
(with u constant). Note that Γ is not unit-speed in general. Now, dΓ
2Γ
2
Γ.(N
N × Γ̇
Γ) = f 2 f˙.
(−f sin v, f cos v, 0), d /dv = (−f cos v, −f sin v, 0), giving Γ̈
2 ˙
˙
Hence, κg = ff 3f = ff .
7.3.11 The paraboloid z = x2 + y 2 is obtained by rotating the parabola z = x2 in the
xz-plane around the z-axis. We could therefore use the preceding exercise but we
would need a unit-speed parametrization of the parabola, which is complicated.
Proceeding directly, parametrize the paraboloid by σ (u, v) = (u cos v, u sin v, u2 ).
2
+2u2 dv 2
,
The first and second fundamental forms are (1+4u2 )du2 +u2 dv 2 and 2du√1+4u
2
respectively. A curve u = constant can be parametrized by γ (v) = σ (u, v).
2
Hence, κn =
√ 2u
1+4u2
u2
.
Now, denoting d/dv by a dot, we find that γ̇γ = (−u sin v, u cos v, 0), γ̈γ =
1
(−2u cos v, −2u sin v, 1), which gives
(−u cos v, −u sin v, 0), and N = √1+4u
2
N × γ̇γ ) =
γ̈γ .(N
2
√ u
.
1+4u2
Hence, κg =
1
u3
×
2
√ u
1+4u2
=
√ 1
.
u 1+4u2
7.3.12 σ u = t + vδ̇δ , σ v = δ , so along γ (where v = 0) we have σ u = t, σ v = δ and hence
×δ
N = tsin
θ . Hence,
δ − cos θtt
(tt × δ ) × t
N×t=
=
,
sin θ
sin θ
so
ṫt.δ
N × t) =
κg = ṫt.(N
,
sin θ
since ṫt.t = 0. Differentiating t.δ = cos θ gives ṫt.δ = − sin θ θ̇ − t.δ̇δ , hence the
stated formula for κg .
59
7.3.13 A curve v = constant is parametrized by γ (u) = σ (u, v), so (denoting d/du by
N × σ u ). Now, σ u.σ u = σ v .σ v = 1,
a dot) γ̇γ = σ u , γ̈γ = σ uu and κ′g = σ uu.(N
σ
u ×σ v
σ u.σ v = cos θ, so N = sin θ and
1
1
σu × σ v ) × σ u =
σ v − cos θσ
σ u ).
N × σu =
(σ
(σ
sin θ
sin θ
Hence,
σ uu.σ u
σ uu.σ v − cos θσ
.
κ′g =
sin θ
By Exercise 6.1.9, σ uu.σ u = 0, σ uu.σ v = −θu sin θ. Inserting these formulas in
the preceding equation for κ′g gives κ′g = −θu . The formula for κ′′g is proved
similarly.
7.3.14 From the solution to Exercise 7.3.4,
√
X
κg = EG(u̇v̈ − üv̇) + √
,
EG
where
1
1
1
1
3
2
EGv − GEv u̇v̇ 2 + GGu v̇ 3
X = − EEv u̇ + EGu − GEu u̇ v̇ +
2
2
2
2
1
1
= − Ev (1 − Gv̇ 2 )u̇ + EGu − GEu u̇2 v̇
2
2
1
1
+
EGv − GEv u̇v̇ 2 + Gu (1 − E u̇2 )v̇
2
2
1
1
1
= (Gu v̇ − Ev u̇) − G(Eu u̇ + Ev v̇)u̇v̇ + E(Gu u̇ + Gv v̇)u̇v̇.
2
2
2
Substituting into the formula for κg gives the desired result.
7.3.15 A parameter curve v = constant can be parametrized by γ (u) = σ (u, v), but this
curve will not be unit-speed in general. √Let s be the arc-length
of γ and denote
√
d/ds by a dot. Then, ds/du = k σ u k= E, so u̇ = 1/ E. Since v̇ = v̈ = 0, the
formula in the preceding exercise gives
Ev u̇
Ev
κ′g = − √
=− √ .
2 EG
2E G
′′
The formula for κg is proved similarly.
√
√
We have cos θ = u̇ E, sin θ = v̇ G, so
d
d
θ̇ = cos θ (sin θ) − sin θ (cos θ)
ds
ds
√
√
√
√
v̇
u̇
= u̇ E v̈ G + √ (Gu u̇ + Gv v̇) − v̇ G ü E + √ (Eu u̇ + Ev v̇)
2 G
2 E
!
r
r
√
1
E
G
= EG(u̇v̈ − üv̇) + u̇v̇
(Gu u̇ + Gv v̇) −
(Eu u̇ + Ev v̇) .
2
G
E
60
Using the preceding exercise and the first part of this exercise, we get
√
√
1
(Gu v̇ − Ev u̇) = κg + κ′g u̇ E + κ′′g v̇ G = κg − κ′g cos θ − κ′′g sin θ.
θ̇ = κg − √
2 EG
7.3.16 Let γ (t) be a unit-speed parametrization of C, and let p = γ (t0 ). Let N be a unit
N. Denoting
normal of S at p. Then, a parametrization of C˜ is γ̃γ = γ − (γγ .N )N
¨
˙
N, so the curvature of γ̃γ
N, γ̃γ = γ̈γ − (γ̈γ .N)N
d/dt by a dot as usual, γ̃γ = γ̇γ − (γ̇γ.N)N
at p is
N) k = k γ̇γ (t0 ) × (N
N × (γ̈γ (t0 ) × N)) k,
κ̃ = k γ̇γ (t0 ) × (γ̈γ (t0 ) − (γ̈γ (t0 )..N)N
since γ̃γ˙ (t0 ) = γ̇γ (t0 ) is a unit vector. Now, γ̈γ (t0 ) = κg N × γ̇γ (t0 ) + κn N, where κg
and κn are the geodesic and normal curvatures of γ at p. Hence, γ̈γ (t0 ) × N =
N × γ̇γ (t0 )) × N = κg γ̇γ (t0 ), so
κg (N
κ̃ = k γ̇γ (t0 ) × (κg N × γ̇γ (t0 )) k = |κg | k γ̇γ (t0 ) kk N × γ̇γ (t0 ) k = |κg |
as N and γ̇γ (t0 ) are perpendicular unit vectors.
2
2
−dv
7.3.17 The second fundamental form is √du
, so a curve γ (t) = σ (u(t), v(t)) is
1+u2 +v 2
2
2
asymptotic if and only if u̇ − v̇ = 0, i.e. u + v or u − v is constant. If u + v = c,
say, where c is a constant, then σ (u, v) = (u, c − u, 12 c(2u − c)), which is a
parametrization of a straight line. Similarly if u − v is constant.
Geometrically, the surface is a hyperbolic paraboloid and is doubly ruled (Exercise 5.2.5). Both of the straight lines on the surface passing through a given
point are asymptotic curves (Exercise 7.3.6).
b) = t along γ ; σ v = n. Hence, along
7.3.18 In the usual notation, σ u = t + v(−κtt + τb
n + v τ̇b
b = κn
n
γ , σ u × σ v = b. Next, we find that σ uu = −v κ̇tt + (κ − (κ2 + τ 2 )v)n
along γ . The parameter curve v = 0 is asymptotic if and only if L = 0. But,
n.b = 0.
along γ , L = σ uu.N = κn
7.3.19 Parametrize the surface by σ (u, v) = (u2 cos v, u2 sin v, 2u). The second funda2
+2u2 dv 2
mental form is −2du√1+u
, so the asymptotic curves are given by
2
−2u̇2 + 2u2 v̇ 2 = 0.
This gives v = ± ln u + c, where c is an arbitrary constant. The projection of this
asymptotic curve on the xy-plane is given by x = u2 cos v, y = u2 sin v. Since
u2 = e±2(v−c) , this is a logarithmic spiral.
7.3.20 Let γ be a unit-speed parametrization of C. Then, γ is asymptotic ⇐⇒ γ̈γ .N =
0 ⇐⇒ n.N = 0, where N is a unit normal to S and n is the principal normal of
61
γ . Since γ̇γ = t is perpendicular to N, γ is asymptotic ⇐⇒ N is perpendicular to
both t and n. But the osculating plane is spanned by t and n.
7.3.21 If every curve on a surface is asymptotic, Lu̇2 + 2M u̇v̇ + N v̇ 2 = 0 for all curves
γ (t) = σ (u(t), v(t)) on the surface. This is possible only if L = M = N = 0
everywhere. By Exercise 7.1.2 this implies the surface is an open subset of a
plane.
7.3.22 N.n = cos ψ, N.t = 0, so N.b = sin ψ; hence, N = n cos ψ + b sin ψ and B =
n cos ψ + b sin ψ) = b cos ψ − n sin ψ. Hence,
t × (n
N = ṅ
n cos ψ + ḃ
b sin ψ + ψ̇(−n
n sin ψ + b cos ψ)
Ṅ
b) cos ψ − nτ sin ψ + ψ̇(−n
n sin ψ + b cos ψ)
= (−κtt + τb
b cos ψ − n sin ψ)
= −κ cos ψtt + (τ + ψ̇)(b
= −κn t + τg B.
B is proved similarly. Since {tt, N, B} is a right-handed orthonorThe formula for Ḃ
3
N, Ḃ
B in terms
mal basis of R , Exercise 2.3.6 shows that the matrix expressing ṫt, Ṅ
of t, N, B is skew-symmetric, hence the formula for ṫt.
N; then, B = t × N = ∓n
n. Hence,
7.3.23 By Exercise 7.3.6, b is parallel to N, so b = ±N
B = ∓ṅ
n = ∓(−κtt + τb
b) = ±κtt − τN
N; comparing with the formula for Ḃ
B in the
Ḃ
preceding exercise shows that τg = τ (and κn = ±κ).
7.4.1 ṽv is a smooth function of t and lies in Tγ (ϕ(t)) S = Tγ̃γ (t) S, so ṽv is a tangent vector
v dϕ ,
field along γ̃γ . The formula follows from Eq. (7.11) and the fact that ddtv = ddṽ
t̃ dt
v = 0 ⇐⇒ ∇γ v = 0.
where t̃ = ϕ(t). The last part follows since ϕ̇ 6= 0 so ∇γ̃γ ṽ
7.4.2 If p and q correspond to the parameter values t = a and t = b, respectively, let
qp
is
Γ(t) = γ (a + b − t) (thus, Γ is γ ‘traversed backwards’). We show that Π
Γ
pq
the inverse of Πγ . Let w ∈ Tp S and let v be the tangent vector field parallel
pq w
) = v(b). By Exercise 7.4.1, V (t) =
along γ such that v(a) = w. Then, Πγ (w
qp
qp
V(a)) = V(b) = v(a) = w .
v(b)) = Π (V
v(a + b − t) is parallel along Γ so Π (v
Γ
Γ
pq
qp
This proves that Π ◦ Πγ is the identity map on Tp S. One proves similarly
Γ
qp
pq
is the identity map
(or by interchanging the roles of γ and Γ) that Πγ ◦ Π
Γ
on Tq S.
7.4.3 Let α, β, γ be the internal angles of the triangle at p, q, r, respectively. Since the
arc through p and q is part of a great circle, the tangent vector of the arc is
parallel along the arc (Example 7.4.7). So the result of parallel transporting v0
to q along the arc pq through p and q is a vector v1 tangent to pq at q. Now v1
makes an angle π − β with the arc qr at q, so parallel transporting v1 along qr
62
to r gives a vector v2 which makes an angle (π − β) + (π − γ) with the arc rp at
r. Parallel transporting v2 along rp to p then gives a vector v3 which makes an
angle (π − β) + (π − γ) + (π − α) with v0 . Since v0 , v1 , v2 , v3 all have the same
length (Proposition 7.4.9(ii)), the result follows from Theorem 6.4.7.
7.4.4 Putting E = G = 1, F = cos θ in the formulas in Proposition 7.4.4 gives Γ111 =
θu cot θ, Γ211 = −θu cosec θ, Γ122 = −θv cosec θ, Γ222 = θv cot θ, Γ112 = Γ212 = 0.
7.4.5 In the notation of Proposition 7.4.5, the tangent vector σ v to the parameter
curves u = constant are given by α = 0, β = 1, and a curve v = constant
corresponds to v̇ = 0 (and u̇ 6= 0). From Eq. 7.13, σ v is parallel along the curves
v = constant if and only if Γ112 = Γ212 = 0; from the formulas in Proposition
7.4.4, this holds if and only if
(*)
GEv = F Gu and EGu = F Ev .
If the parameter curves form a Chebyshev net, Ev = Gu = 0 (Exercise 6.1.5), so
(*) holds. Conversely, if (*) holds than
(EG − F 2 )Gu = G(EGu ) − F (F Gu ) = G(F Ev ) − F (GEv ) = 0
and so Gu = 0; then GEv = 0 by (*) again, so Ev = 0. Hence, if (*) holds the
parameter curves form a Chebyshev net. This proves that (i) is equivalent to
(ii). That (i) is equivalent to (iii) is proved similarly.
.σ v = F/√EG, so A = √EG − EG cos2 θ = √EG sin θ.
7.4.6 We have cos θ = kσσuukk
σv k
Differentiating the first equation with respect to u and using the second equation
gives
√
2EGFu − F GEu − EF Gu
EG
F
√
−θu =
=
.
A
2AEG
EG u
On the other hand, the formulas in Proposition 7.4.4 give
Γ211
Γ112
1
2EFu − EEv − F Eu
GEv − F Gu
+
=
+
E
G
2A2
E
G
2EGFu − F GEu − EF Gu
=
.
2A2 EG
7.4.7 From A2 = EG − F 2 we get 2AAu = EGu + GEu − 2F Fu . On the other hand,
u −2F Fu
. By the
the formulas in Proposition 7.4.4 give Γ111 + Γ212 = GEu +EG
2A2
preceding formula, this is equal to 2AAu /2A2 = Au /A. The other formula is
proved similarly.
63
Chapter 8
8.1.1 Parametrize the surface by σ (x, y) = (x, y, f (x, y)). Then, σ x = (1, 0, fx), σ y =
(0, 1, fy ), N = (1 + fx2 + fy2 )−1/2 (−fx , −fy , 1), σ xx = (0, 0, fxx), σ xy = (0, 0, fxy ),
σ yy = (0, 0, fyy ). This gives E = 1 + fx2 , F = fx fy , G = 1 + fy2 and L =
(1 + fx2 + fy2 )−1/2 fxx , M = (1 + fx2 + fy2 )−1/2 fxy , N = (1 + fx2 + fy2 )−1/2 fyy . By
Corollary 8.1.3, K =
2
fxx fyy −fxy
,
(1+fx2 +fy2 )2
H=
(1+fy2 )fxx −2fx fy fxy +(1+fx2 )fyy
.
2(1+fx2 +fy2 )3/2
8.1.2 For the helicoid σ (u, v) = (v cos u, v sin u, λu),
σ u = (−v sin u, v cos u, λ),
σ v = (cos u, sin u, 0),
1
(−λ sin u, λ cos u, −v),
N= √
λ2 + v 2
σ uu = (−v cos u, −v sin u, 0),
σ uv = (− sin u, cos u, 0),
σ vv = 0.
√
This gives E = λ2 + v 2 , F = 0, G = 1 and L = N = 0, M = λ/ λ2 + v 2 . Hence,
K = (LN − M 2 )/(EG − F 2 ) = −λ2 /(λ2 + v 2 )2 .
For the catenoid σ (u, v) = (cosh u cos v, cosh u sin v, u),
σ u = (sinh u cos v, sinh u sin v, 1),
σ v = (− cosh u sin v, cosh u cos v, 0),
N = sech u(− cos v, − sin v, sinh u),
σ uu = (cosh u cos v, cosh u sin v, 0),
σ uv = (− sinh u sin v, sinh u cos v, 0),
σ vv = (− cosh u cos v, − cosh u sin v, 0).
This gives E = G = cosh2 u, F = 0 and L = −1, M = 0, N = 1. Hence,
K = (LN − M 2 )/(EG − F 2 ) = −sech4 u.
σ v is never zero, N = σ u ×σ
σ v / k σ u ×σ
σ v k is smooth.
8.1.3 Since σ is smooth and σ u ×σ
Hence, E, F, G, L, M and N are smooth. Since EG − F 2 > 0 (by the remark
following Proposition 6.4.2), the formulas in Corollary 8.1.3 show that H and K
are smooth.
N = 0 ⇐⇒ δ̇δ .((tt +vδ̇δ )×δδ ) = 0 ⇐⇒ δ̇δ .(tt ×δδ ) =
8.1.4 From Example 8.1.5, K = 0 ⇐⇒ δ̇δ .N
b, t × δ = b, so K = 0 ⇐⇒ τ = 0 ⇐⇒ γ is planar (by
0. If δ = n, δ̇δ = −κtt + τb
n, t × δ = −n
n, so again K = 0 ⇐⇒ τ = 0.
Proposition 2.3.3). If δ = b, δ̇δ = −τn
64
8.1.5 The dilation (x, y, z) 7→ (ax, ay, az), where a is a non-zero constant, multiplies
E, F, G by a2 and L, M, N by a, hence H by a−1 and K by a−2 (using Corollary
8.1.3).
8.1.6 This follows immediately from Definition 8.1.1 and the hint.
8.1.7 Suppose that the cone is the union of the straight lines joining points of a curve
C to a vertex v. It is clear that the Gauss map G is constant along the rulings of
the cone, so the image of the cone under G is the same as the image of C under
G, which is a curve.
8.1.8 By Eq. 8.2, the area of σ (R) is
ZZ
R
k Nu × Nv k dudv =
ZZ
R
|K| k σ u × σ v k dudv =
ZZ
R
|K|dAσ .
8.1.9 Using the parametrization σ in Exercise 4.2.5, we find that E = b2 , F = 0, G =
cos θ
(a+b cos θ)2 and L = b, M = 0, N = (a+b cos θ) cos θ. This gives K = b(a+b
cos θ) .
+
−
It follows that S and S are the annular regions on the torus given by
−π/2 ≤ u ≤ π/2 and π/2 ≤ u ≤ 3π/2,
respectively.
S+
S−
It is clear that as a point p moves
over S + RR
(resp. S − ), the unit normal at p
RR
covers the whole of S 2 . Hence, S + |K|dA = S − |K|dA = 4π by the preceding
exercise; since |K| = ±K on S ± , this gives the result.
wu.N)N
N so
8.1.10 ∇uw = wu − (w
wuv .N)N
N − (w
wu .Nv )N
N − (w
wu .N)N
Nv − (w
wuv .N)N
N
∇v (∇u w) = wuv − (w
wuv .N)N
N + (w
wu.N v )N
N + (w
wu.N)(N
Nv .N)N
N
+ (w
wuv .N)N
N − (w
wu.N)N
Nv .
= wuv − (w
65
Interchanging u and v and subtracting gives the first formula. Replacing w by
w in this formula gives
λw
wv .N)N
Nu −(w
wu.N)N
Nv }+λv (w
w.N)N
Nu −λu (w
w.N )N
Nv = λ{(w
wv .N )N
Nu −(w
wu.N)N
Nv }
λ{(w
since w.N = 0. It is also obvious that
w1 + w 2 )) − ∇u (∇v (w
w1 + w 2 )) = (∇v (∇uw 1 ) − ∇u (∇v w1 ))
∇v (∇u (w
+ (∇v (∇uw2 ) − ∇u (∇v w 2 ))
for any two tangent vector fields w1 , w2 .
σ uv .N)N
Nu − (σ
σ uu.N )N
Nv = MN
Nu − LN
Nv (in the
Now ∇v (∇uσ u ) − ∇u (∇v σ u ) = (σ
σ
σ
σ u + dσ
σv )
usual notation). Using Proposition 8.1.2, this = M (aσ u + bσ v ) − L(cσ
and using the explicit expressions for a, b, c, d in Proposition 8.1.2 this becomes
σv − F σ u ). Similarly, ∇v (∇uσ v ) − ∇u (∇v σ v ) = K(−Gσ
σ u + F σ v ).
K(Eσ
If
(*)
∇v (∇uw) − ∇u (∇v w) = 0
for all w, then taking w = σ u gives K = 0 since E 6= 0, σ v 6= 0. Conversely,
if K = 0 then (*) holds for w = σ u and σ v , and hence by the first part of the
σ u + βσ
σ v for all smooth functions α, β of (u, v). But every
exercise it holds for ασ
tangent vector field w is of this form.
8.1.11 σ u = (1, 1, v), σ v = (1, −1, u), σ uu = σ vv = 0, σ uv = (0, 0, 1). When√u = v = 1,
we find from this that E = 3, F = 1, G = 3 and L = N = 0, M = −1/ 2. Hence,
K =√(LN − M 2 )/(EG − F 2 ) = −1/16, H = (LG − 2M F + N G)/2(EG − F 2 ) =
1/8 2.
8.1.12 r
We write the equation of the quadric in the form z = f (x, y), where f (x, y) =
2
2
c 1 − xa − yb . The result of the calculation will give the value of the Gauss-
ian curvature K on the part of the quadric where z > 0. Similar calculations
(with the same result) give K on the parts of the surface where z < 0, x > 0,
x < 0, y > 0 and y < 0. Together these regions cover the whole surface.
r 2
2
Taking f (x, y) = c 1 − xa − yb , and applying the first formula in Exercise
8.1.1, we get, after straightforward calculation,
2
2
2
y
x
z
1
a + b + c
K=
2 .
abc x2
y2
z2
+ b2 + c2
a2
66
The vector ∇
x2
a
+
y2
b
+
z2
c
=2
x y z
a, b, c
(x, y, z).. xa , yb , zc
d= q
y2
x2
z2
a2 + b2 + c2
=q
This gives the stated formula for K.
For the paraboloid we take f (x, y) =
8.1.1 gives
K=
The vector
x y
2a , 2b , −1
(x, y, z)..
q
d=
x2
4a2
x2
a
4ab 1 +
+
1
x2
a2
y2
b
1
x2
4a2
+
+
y2
b2
+
z2
c2
.
. The first formula in Exercise
y2
4b2
2 .
is normal to the surface, so
x y
2a , 2b , −1
+
1
4
is normal to the surface, so
y2
4b2
+1
x2
2a
=q
+
x2
4a2
y2
2b
+
−z
y2
4b2
+1
=q
x2
4a2
|z|
+
y2
4b2
.
+1
Hence, K = d4 /4abz 4 .
8.1.13 Taking f (x, y) = 1/xy, we get fx = −z/x, fy = −z/y, fxx = 2z/x2 , fxy = z/xy,
fyy = 2z/y 2 . The first formula in Exercise 8.1.1 gives
K=
4z 2
x2 y 2
1+
−
z2
x2
z2
x2 y 2
+
z2
y2
2 =
3x−2 y −2 z −2
,
(x−2 + y −2 + z −2 )2
hence the stated formula for K. Similarly, the second formula in Exercise 8.1.1
gives
2z
z2
2z 3
2z
z2
1
+
−
+
1
+
x2
y2
x2 y 2
y2
x2
1
xyz(x2 + y 2 + z 2 )
p
H=
=
=
.
2
2 + y 2 + z 2 )3/2
2 + y 2 + z 2 )3/2
2 + y2 + z2
(x
(x
x
3
3
x y
We showed in Exercise 5.1.3 that the smallest value of X 2 + Y 2 + Z 2 subject to
the condition XY Z = 1 is 3, and that it is attained precisely when (X, Y, Z) =
(1, −1, −1), (−1, 1, −1), (−1, −1, 1) or (1, 1, 1). Taking X = 1/x, Y = 1/y,
Z = 1/z, it follows that the largest value of K is 3(1+1+1)−2 = 1/3, and that it
is attained precisely when (1/x, 1/y, 1/z) = (1, −1, −1), (−1, 1, −1), (−1, −1, 1)
or (1, 1, 1), i.e. when (x, y, z) = (1, −1, −1), (−1, 1, −1), (−1, −1, 1) or (1, 1, 1).
As noted in the solution to Exercise 5.1.3, these points lie at the vertices of a
regular tetrahedron.
67
8.1.14 If the circle in the xz-plane has radius a and touches the z-axis at the point
(0, 0, c), it can be parametrized by γ (u) = (a(1 + cos u), 0, a sin u + c). Rotating
γ (u) around the z-axis through an angle v gives the point
(a(1 + cos u) cos v, a(1 + cos u) sin v, a sin u + c).
Since the circle rotates at constant angular velocity and moves parallel to the
z-axis at constant speed, the distance it moves parallel to the z-axis in the time
taken to rotate through an angle v is a constant multiple of v, say bv. Hence,
after the circle has rotated through an angle v the point initially at γ (u) will
have moved to the point σ (u, v) in the statement of the exercise.
If a = b and c = 0 we have σ (u, v) = a((1 + cos u) cos v, (1 + cos u) sin v, sin u + v).
We find that the first fundamental form is
a2 (du2 + 2 cos ududv + ((1 + cos u)2 + 1)dv 2 ).
Let A = k σ u × σ v k =
fundamental form is
√
EG − F 2 , so that A2 = 2a4 (1 + cos u). The second
a3
((1 + cos u)du2 − 2 sin2 ududv + (1 + cos u)2 cos udv 2 ).
A
The first formula in Corollary 8.1.3 now gives
K=
a6 ((1 + cos u)3 cos u − sin4 u)
.
A4
Writing c = cos u we get
K=
c(1 + c) − (1 − c)2
3 cos u − 1
c(1 + c)3 − (1 − c2 )2
=
=
.
4a2 (1 + c)2
4a2
4a2
Since d = a(1 + cos u), this agrees with the formula in the statement of the
exercise.
8.1.15 Using the parametrization in Example
8.1.5, we have σ u × σ v = γ̇γ × δ + vδ̇δ × δ .
√
Recalling that k σ u × σ v k = EG − F 2 , the formula for K in that example
becomes
−(δ̇δ. (γ̇γ × δ ))2
K=
.
k γ̇γ × δ + vδ̇δ × δ k4
If δ̇δ ×δδ 6= 0, K → 0 as v → ∞ (with u fixed); thus, if K is constant we must have
K = 0. On the other hand, if δ̇δ × δ = 0, δ̇δ is parallel to δ , hence δ̇δ.(γ̇γ × δ ) = 0
and again K = 0.
68
8.1.16 We use the parametrization σ (s, θ) in Exercise 4.2.7, a dot to denote d/ds, and
the standard notation relating to the Frenet-Serret data for γ . Then, σ s =
n + τ a cos θb
b, σ θ = −a sin θn
n + a cos θb
b. Hence,
(1 − κa cos θ)tt − τ a sin θn
n + sin θb
b),
σ s × σ θ = −a(1 − κa cos θ)(cos θn
n + sin θb
b), E = (1 − κa cos θ)2 + τ 2 a2 , F = τ a2 , G = a2 and
so N = −(cos θn
EG − F 2 = a2 (1 − κa cos θ)2 . Next,
n
σ ss = (κτ a sin θ − κ̇a cos θ)tt + (κ(1 − κa cos θ) − τ̇ a sin θ − τ 2 a cos θ)n
b,
+ (τ̇ a cos θ − τ 2 a sin θ)b
n − τ a sin θb
b,
= κa sin θtt − τ a cos θn
σ sθ
n − a sin θb
b.
σ θθ = −a cos θn
Hence, L = aτ 2 − κ cos θ(1 − κa cos θ), M = τ a, N = a, Thus,
K=
−κa cos θ(1 − κa cos θ)
−κ cos θ
=
.
2
2
a (1 − κa cos θ)
a(1 − κa cos θ)
(i) We have dA = k σ s × σ θ k dsdθ = a(1 − κa cos θ)dsdθ, so
Z ℓZ
0
(ii) Since
R 2π
0
2π
KdA =
0
Z ℓZ
0
| cos θ|dθ = 2
R π/2
−π/2
Z ℓZ
0
2π
(−κ cos θ)dsdθ = 0.
0
cos θdθ = 4, we get
2π
0
|K|dA = 4
Z
ℓ
κ(s)ds.
0
8.1.17 It follows from Exercises 6.1.2 and 7.1.4 that applying a direct isometry of R3
leaves both H and K unchanged, while applying an opposite isometry leaves K
unchanged but changes the sign of H.
8.1.18 From the proof of Proposition 7.3.3, if γ is unit-speed its normal curvature is κn =
N.γ̇γ = W(γ̇γ )..γ̇γ . Hence, γ is asymptotic if and only if W(γ̇γ ) is perpendicular to
−Ṅ
γ̇γ at each point of γ . Since −W(γ̇γ ) is the tangent vector of the image of γ under
the Gauss map of S, the result follows.
N = −W(γ̇γ ), and that W is self-adjoint (Corollary 7.2.4), we have
8.1.19 Recalling that Ṅ
to prove that
hW 2 (γ̇γ ) − 2HW(γ̇γ ) + K γ̇γ , γ̇γ i = 0.
This follows from the preceding exercise.
69
We can assume that γ is unit-speed. As in the solution of Exercise 8.1.18, if γ
N.γ̇γ = 0 and the first part gives Ṅ
N.Ṅ
N = −K. But by Exercise
is asymptotic then Ṅ
b so Ṅ
N = ∓τn
n (where n is the principal normal
7.3.6, if γ is asymptotic, N = ±b
2
N.Ṅ
N=τ .
of γ and b is its binormal). Hence, Ṅ
8.2.1 For the helicoid σ (u, v) = (v cos u, v sin u, λu),
√ the first and second fundamental
2
2
2
2
forms are (λ + v )du + dv and 2λdudv/ λ2 + v 2 , respectively. Hence, the
−κ(λ2 + v 2 ) √λ2λ+v 2
= 0, i.e. κ =
principal curvatures are the roots of
√ λ
−κ
λ2 +v 2
±λ/(λ2 + v 2 ).
For the catenoid σ (u, v) = (cosh u cos v, cosh u sin v, u), the first and second fundamental forms are cosh2 u(du2 +dv 2 ) and −du2 +dv 2 , so the principal curvatures
−1 − κ cosh2 u
0
are the roots of
= 0, i.e. κ = ±sech2 u.
0
1 − κ cosh2 u
N.
8.2.2 This is obvious, since W(γ̇γ ) = −Ṅ
σ u + v̇σ
σ n is a principal vector for all t ⇐⇒
8.2.3 γ
is a line
of curvature
⇐⇒ γ̇γ = u̇σ
L M
E F
=κ
for some scalar κ. Writing this matrix equation as two
L N
F G
scalar equations and then eliminating κ gives the stated equation.
For the second part, if the second fundamental form is a multiple of the first, the
Weingarten map is a scalar multiple of the identity map, so every tangent vector
is principal and every curve on the surface is a line of curvature. If F = M = 0
the matrices FI and FII are diagonal, hence so is the matrix FI−1 FII of the
σ u , σ v }. This means that σ u and
Weingarten map with respect to the basis {σ
σ v are principal vectors, i.e. that the parameter curves v = constant and u =
constant are lines of curvature.
Conversely, if every parameter curve is a line of curvature, the stated equation
must hold if u̇ = 0 and if v̇ = 0. This gives EM = F L and F N = GM , which
imply that (EN −GL)F = EGM −EGM = 0 and so either EN = GL or F = 0.
If F = 0 then GM = 0 so M = 0 and (ii) holds. If EN = GL the equation
in the exercise implies that every curve is a line of curvature, so every tangent
vector is principal, so (i) holds.
Condition (i) implies that the two principal curvatures are equal everywhere,
i.e. every point is an umbilic, so σ is an open subset of a plane or a sphere by
Proposition 8.2.9.
From Examples 6.1.3 and 7.1.2, the first and second fundamental forms of a
surface of revolution are du2 + f (u)2 dv 2 and (f˙g̈ − f¨ġ)du2 + f ġdv 2 , respectively.
Since the terms dudv are absent, the vectors σ u and σ v are principal; but these
are tangent to the meridians and parallels, respectively.
70
N1.(tt × N 1 ) = 0. Since Ṅ
N1 is
8.2.4 Let N 1 be a unit normal of S. Then, K = 0 ⇐⇒ Ṅ
N1
perpendicular to N1 and N1 is perpendicular to t , this condition holds ⇐⇒ Ṅ
N1 = −λγ̇γ for some scalar λ. Now use Exercise 8.2.2.
is parallel to t, i.e. ⇐⇒ Ṅ
8.2.5 Let N 1 and N2 be unit normals of the two surfaces; if γ is a unit-speed parametrizaN1 = −λ1 γ̇γ for some scalar λ1 by Exercise 8.2.2. If C is a line
tion of C, then Ṅ
N2 = −λ2 γ̇γ for some scalar λ2 , and then (N
N1 .N
N2 )˙ =
of curvature of S2 , then Ṅ
N2 − λ2 γ̇γ .N
N1 = 0, so N1 .N
N2 is constant along γ , showing that the angle
−λ1γ̇γ .N
N2 is constant, then N1 .Ṅ
N2 = 0
between S1 and S2 is constant. Conversely, if N1 .N
N1 .N
N2 = −λ1 γ̇γ .N
N2 = 0; thus, Ṅ
N2 is perpendicular to N1 , and is also persince Ṅ
pendicular to N2 as N 2 is a unit vector; but γ̇γ is also perpendicular to N1 and
N2 must be parallel to γ̇γ , so there is a scalar λ2 (say) such that
N2 ; hence, Ṅ
N2 = −λ2 γ̇γ .
Ṅ
8.2.6 (i) Differentiate the three equations in (8.5) with respect to w, u and v, respectively; this gives
σ v + σ u .σ
σ vw = 0, σ uv .σ
σ w + σ v .σ
σ uw = 0, σ vw .σ
σ u + σ w .σ
σ uv = 0.
σ uw .σ
σ vw = 0,
Subtracting the second equation from the sum of the other two gives σ u .σ
σ uw = σ w .σ
σ uv = 0.
and similarly σ v .σ
σ w = 0, it follows that the matrix FI for the u = u0 surface is
(ii) Since σ v .σ
diagonal (and similarly for the others). Let N be the unit normal of the u = u0
surface; N is parallel to σ v × σ w by definition, and hence to σ u since σ u , σ v and
σ u = 0, hence σ vw .N
N = 0, proving that the
σ w are perpendicular; by (i), σ vw .σ
matrix FII for the u = u0 surface is diagonal.
(iii) By part (ii), the parameter curves of each surface u = u0 are lines of curvature. But the parameter curve v = v0 , say, on this surface is the curve of
intersection of the u = u0 surface with the v = v0 surface.
8.2.7 On the open subset of the ellipsoid
qwith z 6= 0, we can use the parametrization
2
2
σ (x, y) = (x, y, z), where z = ±r 1 − xp2 − yq 2 . By Proposition 8.1.2 and the
remarks following Proposition 8.2.1, the condition for an umbilic is that FII =
κFI for some scalar κ. The formulas in the solution of Exercise 8.1.1 lead
to the equations zxx = λ(1 + zx2 ), zxy = λzx zy , zyy = λ(1 + zy2 ), where
q
λ = κ 1 + zx2 + zy2 . If x and y are both non-zero, the middle equation gives λ =
−1/z, and substituting into the first equation gives the contradiction p2 = r 2 .
Hence, either x = 0 or y = 0. If x = 0, the equations have the four solutions
x = 0, y = ±q
s
q2
p2
−
, z = ±r
2
q − r2
s
r 2 − p2
.
r2 − q2
71
Similarly, one finds the following eight other candidates for umbilics:
s
s
2
2
p −q
r2 − q2
,
y
=
0,
z
=
±r
,
x = ±p
p2 − r 2
r 2 − p2
s
s
p2 − r 2
q2 − r2
x = ±p
,
y
=
±q
, z = 0.
p2 − q 2
q 2 − p2
Of these 12 points, exactly 4 are real, depending on the relative sizes of p2 , q 2
and r 2 .
If p = q 6= r, the only umbilics are the two points (0, 0, ±r). If p = q = r every
point of the ellipsoid (now a sphere) is an umbilic.
8.2.8 By Proposition
8.2.3, the principal curvatures are the roots of κ2 −2Hκ+K = 0,
√
i.e. H ± H 2 − K. If there are no umbilics, we must have H 2 > K, and then
the principal curvatures are smooth because H and K are (Exercise 8.1.3). The
second part follows from Exercises 6.1.4 and 7.1.3 and Proposition 8.2.6.
8.2.9 Setting x = r cos θ, y = r sin θ gives z = aθ so a parametrization of the surface is
σ (r, θ) = (r cos θ, r sin θ, aθ). Then σ r = (cos θ, sin θ, 0), σ θ = (−r sin θ, r cos θ, a),
so E = 1, F = 0, G = r 2 + a2 . Next, σ r × σ θ = (a sin θ, −a cos θ, r) so
N = G−1/2 (a sin θ, −a cos θ, r). From σ rr = 0, σ rθ = (− sin θ, cos θ, 0), σ θθ =
(−r cos θ, −r sin θ, 0) we get L = 0, M = −aG−1/2 , N = 0. The principal
curvatures are the roots of
−κ
−aG−1/2
i.e. κ = ±a/G =
±a
r2 +a2
=
−aG−1/2
= 0,
−κG
±a
.
x2 +y 2 +a2
8.2.10 In the notation of Proposition 8.2.3, H 2 −K = 41 (κ1 +κ2 )2 −κ1 κ2 = 14 (κ1 −κ2 )2 .
8.2.11 The condition H 2 = K gives x2 + y 2 + z 2 = 3. We showed in Exercise 5.1.3 that
the points (x, y, z) which satisfy this condition as well as xyz = 1 are precisely
the four points (1, −1, −1), (−1, 1, −1), (−1, −1, 1) and (1, 1, 1), and we found in
Exercise 8.1.13 that these are the points where the Gaussian curvature attains
its maximum value.
8.2.12 If K > 0, then since K = κ1 κ2 , either κ1 > 0 and κ2 > 0, or κ1 < 0 and κ2 < 0.
It follows from Euler’s theorem 8.2.4 that the normal curvature κn of any curve
γ on the surface is > 0 in the first case and < 0 in the second (since sin θ and
cos θ cannot be zero simultaneously). By Eq. 7.9, the curvature of γ is > 0.
8.2.13 Changing the sign of N changes the sign of the Weingarten map W. With
the notation in Proposition 8.2.1, (−W)(tt1 ) = (−κ1 )tt1 , (−W)(tt2 ) = (−κ2 )tt2 ,
which shows that −κ1 and −κ2 are the principal curvatures and t1 and t 2 are
corresponding principal vectors.
72
v) = P v + a be an isometry of R3 , where P
8.2.14 Let σ (u, v) be a surface patch, let T (v
σ = T (σ
σ ). Then, σ̃
σ u = P (σ
σ u ), etc.
is an orthogonal matrix and a ∈ R3 , and let σ̃
σ
.
σ
σ
.
σ
so in the obvious notation Ẽ = P (σ u ).P (σ u ) = u u (since P is orthogonal) so
N = εN
N, the sign ε
Ẽ = E. Similarly, F̃ = F, G̃ = G. By Proposition A.1.6, Ñ
being + if T is direct and − if T is opposite, so L̃ = εL, M̃ = εM , Ñ = εN .
The result now follows from Proposition 8.2.6.
8.2.15 Let κ1 and κ2 be the principal curvatures of a surface patch of S at p and θ the
oriented angle between a principal vector corresponding to κ1 and the tangent
vector of one of the m curves at p. By Euler’s Theorem 8.2.4, the sum of the
normal curvatures of the m curves at p is
m−1
X
r=0
rπ rπ κ1 cos2 θ +
+ κ2 sin2 θ +
.
m
m
Using cos θ = 21 (eiθ + e−iθ ), sin θ =
1
iθ
2i (e
− e−iθ ), etc., the sum becomes
m−1
m−1
X
1
1 X
(κ1 + κ2 ) + (κ1 − κ2 )
e2i(θ+rπ/m) + e−2i(θ+rπ/m) .
2 r=0
4
r=0
But,
Pm−1
r=0
±2iπ
1=e
e±2i(θ+rπ/m) = e±2iθ 1−e
±2iπ/m = 0, so the desired sum is simply
m−1
1 X
1
(κ1 + κ2 ) = m(κ1 + κ2 ) = mH,
2 r=0
2
where H is the mean curvature of S at p.
8.2.16 Using Proposition 6.2.5 and Exercise 7.1.6 (and the notation there), E = 1+v 2 κ2 ,
F = G = 1, L = −κτ v, M = N = 0 (we work in the part of the surface in which
v > 0; the region v < 0 is treated similarly). By Exercise 8.2.3 the lines of
curvature are given by κτ v u̇2 + κτ v u̇v̇ = 0, so u̇ = 0 or u̇ + v̇ = 0, i.e. u =
constant or u + v = constant; the first of these corresponds to the tangent lines
to the given curve.
N1 (u) in the notation of Exercise
8.2.17 The surface is parametrized by σ (u, v) = γ (u)+vN
N1 = −κγ̇γ , where κ is the principal
8.2.4. Since γ is a line of curvature, we have Ṅ
curvature corresponding to the principal vector γ̇γ .
(i) If the surface is a generalized cone, the rulings all pass through some fixed
point, so there is a function v(u) such that σ (u, v(u)) is independent of u. DifN1 − κvγ̇γ = 0. Since γ̇γ is perpendicular
ferentiation with respect to u gives γ̇γ + v̇N
to N1 , this gives 1 − κv = 0 and v̇ = 0, so κ = 1/v is a non-zero constant.
Conversely, if κ is constant, σ (u, 1/κ) is constant so the surface is a generalized
cone.
73
(ii) If the surface is a generalized cylinder, the rulings are all parallel, so N1 is
N1 = 0, and since Ṅ
N1 = −κγ̇γ we have κ = 0. Conversely,
constant along γ . So Ṅ
N1 = 0 so N1 is constant along γ and the rulings are parallel.
if κ = 0 then Ṅ
8.2.18 The image of γ under the Gauss map is the curve N, so the condition is that
N, i.e. Ṅ
N = λγ̇γ for some scalar λ. The result now follows from
γ̇γ is parallel to Ṅ
Exercise 8.2.2. The second part follows from the last sentence of Proposition
8.2.1.
8.2.19 If every curve on S is a line of curvature, every tangent vector of S at p is a
principal vector. By Proposition 8.2.6, it is enough to prove that every point
of S is an umbilic. Suppose for a contradiction that p ∈ S is not an umbilic,
and let κ1 6= κ2 be the principal curvatures at p with corresponding non-zero
principal vectors t 1 and t 2 . If W is the Weingarten map of S at p, we have
W(tt1 + t2 ) = κ1 t1 + κ2 t2 . Since every tangent vector is principal, there is a
scalar λ such that W(tt1 + t2 ) = λ(tt1 + t2 ); then (κ1 − λ)tt1 + (κ2 − λ)tt2 = 0.
But t 1 and t 2 are linearly independent, so this equation forces κ1 = λ = κ2 , a
contradiction.
N.(tt × N). By Exercise 8.2.2, if γ is
8.2.20 In the notation of Exercise 7.3.22, τg = Ṅ
N is parallel to t, so τg = 0. Conversely, if τg = 0 then
a line of curvature Ṅ
N × t ) = 0; but Ṅ
N and t are both perpendicular to N, so Ṅ
N × t is parallel to
N.(Ṅ
N × t = 0, so Ṅ
N is parallel to t and γ is a line of curvature by Exercise
N; hence, Ṅ
8.2.2 again.
8.2.21 We use the usual Frenet-Serret notation for γ and let N be the unit normal of S.
n.N = b.Ṅ
N. If κ is the principal curvature
We are given that b.N is constant, so τn
N = −κtt so b.N = 0. If τ 6= 0 at some point
of S corresponding to γ , we have Ṅ
γ (t0 ) of γ , say, then for some ǫ > 0 we have τ (t) 6= 0 if t0 − ǫ < t < t0 + ǫ. For
values of t in this interval, n is perpendicular to N, so b is parallel to N, hence
b = ±N
N. But then ḃ
b = ±Ṅ
N so τn
n = ±κtt; this forces τ = κ = 0, contradicting
assumption. It follows that τ = 0 at all points of γ , so γ is a plane curve.
The converse is a special case of Exercise 8.2.5, since every curve in a plane is a
line of curvature of the plane.
8.2.22 In the notation of the proof of Euler’s Theorem 8.2.4, t = cos θtt1 + sin θtt2 , so
N = −W(tt) = −κ1 cos θtt1 − κ2 sin θtt2 . Since t2 × N = t1 and t 1 × N = −tt2 , we
Ṅ
have t × N = − cos θtt2 + sin θtt1 so since t1 and t2 are perpendicular unit vectors,
N.(tt × N ) = −κ1 sin θ cos θ + κ2 sin θ cos θ = (κ2 − κ1 ) sin θ cos θ.
τg = Ṅ
8.2.23 By Euler’s Theorem 8.2.4, for an asymptotic curve we have κ1 cos2 θ +κ2 sin2 θ =
0, so κ1 κ2 = −κ22 tan2 θ ≤ 0 (if θ is a multiple of π/2 Euler’s Theorem gives
κ1 = 0 or κ2 = 0, and the condition again holds). If S is ruled, we must have
κ1 κ2 ≤ 0 because the rulings are asymptotic curves (Exercise 7.3.6).
If κ1 κ2 < 0, Euler’s theorem gives tan2 θ = −κ1 /κ2 , so there are two asymptotic
74
p
curves each of which makes an angle tan−1 −κ1 /κ2 with a principal vector
corresponding to κ1 , so the angle between them is as stated.
If κ1 = κ2 = 0 everywhere, every curve is asymptotic. If κ1 = 0 but κ2 6= 0,
say, Euler’s Theorem gives sin θ = 0 so there is one asymptotic curve through
each point, namely the line of curvature corresponding to κ1 (it is proved in
Proposition 8.4.2 that these curves are, in fact, straight lines).
8.2.24 Part (i) follows from the fact that hh , ii is symmetric; part (ii) follows from the
fact that it is linear in its second argument (since it is actually bilinear).
8.2.25 This is an immediate consequence of Proposition 7.3.3.
8.2.26 hhtt1 , t2 ii = hW(tt1 ), t2 i = κ1 htt1 , t2 i. But if κ1 6= κ2 then t1 is perpendicular to t2
(Proposition 8.2.1).
8.2.27 By Proposition 8.2.1, we can assume that t 1 is perpendicular to t2 (whether or
not κ1 and κ2 are distinct). Then t = cos θtt1 + sin θtt2 , t̃t = cos θ̃tt1 + sin θ̃tt2 , so
hhtt, t̃tii = cos θ cos θ̃hhtt1 , t1 ii + (cos θ sin θ̃ + cos θ̃ sin θ)hhtt1 , t2 ii + sin θ sin θ̃hhtt2 , t2 ii.
Now hhtti , tj ii = hW(tti ), tj i = κi htti , tj i = κi if i = j and = 0 otherwise. Hence,
hhtt, t̃tii = κ1 cos θ cos θ̃ + κ2 sin θ sin θ̃, from which the result follows.
8.2.28 Using the notation and result of the preceding exercise,
1
1
1
1
+
=
+
2
κn
κ̃n
κ1 cos2 θ + κ2 sin θ κ1 cos2 θ̃ + κ2 sin2 θ̃
1 + tan2 θ
1 + tan2 θ̃
=
+
κ1 + κ2 tan2 θ κ1 + κ2 tan2 θ̃
κ2
1 − κ1 tan2 θ
1 − κκ21
1
1
+
=
+
+
κ2
κ1
κ2
κ1 + κ2 tan2 θ
κ1 + 1 cot2 θ
κ2
1
1
+ .
=
κ1
κ2
v, γ̇γ ii = hv
v, W(γ̇γ )i = −hv
v, Ṅ
Ni = −v
v.Ṅ
N = v̇
v.N (since v.N = 0).
8.2.29 hhv
8.2.30 The formulas for E ∗ , F ∗ and G∗ were proved in Exercise 6.3.11, where it was
also shown that
σ u.σ )σ
σ
2(σ
σu
−
.
σ ∗u =
2
4
kσ k
kσ k
Hence,
σ ∗uu =
σ u.σ )σ
σ uu.σ + σ u.σ u )σ
σ u.σ )2σ
σu
σ 8(σ
4(σ
2(σ
σ uu
−
−
+
.
k σ k2
k σ k4
k σ k4
k σ k6
75
Using the formula for N∗ in Exercise 4.5.5, this gives
σ u.σ )σ
σ uu.σ + σ u.σ u )σ
σ u .σ )2 σ
σu
σ 8(σ
σ.N )σ
σ
4(σ
2(σ
2(σ
σ uu
∗
.
−
−
+
−N ,
L =
k σ k2
k σ k4
k σ k4
k σ k6
k σ k2
which simplifies to give
L∗ = −
σ .N
N)(σ
σ u .σ u )
σ .N
N)σ
σ
σ uu.N) 2(σ
L
2E(σ
(σ
−
=−
−
E.
2
4
2
4
kσ k
kσ k
kσk
kk
Similar calculations give the stated formulas for M ∗ and N ∗ .
In the usual notation, the preceding formulas can be written
FI∗ =
p.N)
1
1
2(p
∗
FI , FII
=−
FII −
FI .
4
2
kpk
kpk
k p k4
Hence, for any κ ∈ R,
∗
p.N))FI∗ = −
FII
+ (k p k2 κ + 2(p
1
(FII − κFI ).
k p k2
Parts (i) and (ii) follow immediately from this equation, and part (iii) is an
immediate consequence of (ii).
σ u + bσ
σv ,
8.2.31 In the notation of the proof of Proposition 8.1.2, we have Nu = aσ
σ u + dσ
σ v , so
Nv = cσ
Nu Nu .N
Nv
Nu .N
FIII =
Nv Nv .N
Nv
N u .N
Ea2 + 2F ab + Gb2
Eac + F (ad + bc) + Gbd
=
Eac + F (ad + bc) + Gbd
Ec2 + 2F cd + Gd2
a b
E F
a c
=
= (−FI−1 FII )t FI (−FI−1 FII )
c d
F G
b d
= FII FI−1 FI FI−1 FII = FII FI−1 FII .
8.2.32 We can assume that the profile curve of σ is unit-speed. From Examples 6.1.3
and 7.1.2, the principal curvatures are −f¨ġ + f˙g̈ = −f¨/ġ (if ġ 6= 0) and ġ/f .
(i) If ġ(u0 ) = 0 the first principal curvature must be non-zero when u = u0 ,
so g̈(u0 ) 6= 0. Since g̈ is a continuous function of u, there is an ǫ > 0 such
that g̈(u) 6= 0 if 0 < |u − u0 | < ǫ. Then, ġ(u) 6= 0 if 0 < |u − u0 | < ǫ by the
intermediate value theorem.
(ii) Assume that ġ is never zero. Then the first principal curvature must be
zero everywhere, so f¨ = 0, i.e. f (u) = au + b for some constants a, b. Then,
76
√
ġ 2 = 1 − f˙2 = 1 − a2 so (up to a sign) g(u) = 1 − a2 u + c for some constant c.
Since ġ 6= 0, a2 < 1. If a = 0 we have a circular cylinder; if a 6= 0 then making
the reparametrization ũ = au + b and a suitable√translation parallel to the z-axis
σ (ũ, v) = ũ(cos v, sin v, 1 − a2 ).
gives the circular cone σ̃
8.2.33 The first and second fundamental forms are
(1 + zx2 )dx2 + 2zx zy dxdy + (1 + zy2 )dy 2 ,
zxx dx2 + 2zxy dxdy + zyy dy 2
q
.
2
2
1 + zx + zy
At an umbilic we have FII = κFI , where κ is the principal curvature. This is
equivalent to
zxx = λ(1 + zx2 ), zxy = λzx zy , zyy = λ(1 + zy2 ),
q
where λ = κ 1 + zx2 + zy2 .
8.2.34 By Exercises 6.1.2 and 7.1.4, applying an isometry of R3 does not change the first
fundamental form, and either leaves the second fundamental form unchanged (if
the isometry is direct) or changes its sign (if the isometry is opposite). It follows
that the principal curvatures stay the same (if the isometry is direct) or both
change sign (if the isometry is opposite). All the assertions in the exercise follow
immediately from this.
8.2.35 (i) By applying an isometry of R3 , we can assume (Theorem 5.2.2) that the
hyperboloid is of the form
x2
y2
z2
+
=
− 1,
p2
q2
r2
where p, q, r are positive constants, and by the preceding exercise it suffices to
consider this case. Differentiating implicitly we find
zxx
r2 x
r2 y
zx = 2 , zy = 2 ,
p z
q z
2 2
2
r x
r2 y 2
r 4 xy
r2
r
1 − 2 2 , zxy = − 2 2 3 , zyy = 2
1− 2 2 .
= 2
p z
p z
p q z
q z
q z
By Exercise 8.2.33 the conditions for an umbilic are
r 2 x2
r 4 x2
r2
1− 2 2 =λ 1+ 4 2 ,
p2 z
p z
p z
r 4 xy
λr 4 xy
− 2 2 3 = 2 2 2,
p q z
p q z
2
2 2
r
r y
r4 y 2
1− 2 2 =λ 1+ 4 2 .
q2 z
q z
q z
77
The second equation gives x = 0, y = 0 or λ = −1/z; but substituting λ = −1/z
in the first equation leads to r 2 /p2 = −1, which is absurd; so x = 0 or y = 0.
If x = 0, the first equation gives λ = r 2 /p2 z and then the last equation gives
r4 y 2
r2 y 2
2
p 1− 2 2 =q 1+ 4 2 .
q z
q z
2
q
q
2 +r 2
z = ±r pq 2 +r
2 (with any
q
q
2
2
q 2 −p2
combination of signs). Similarly, y = 0 leads to x = ±p p2 +r2 , z = ±r pq 2 +r
+r2 .
Using y 2 /q 2 = z 2 /r 2 − 1, this leads to y = ±q
p2 −q 2
,
q 2 +r2
If p2 6= q 2 , this gives four distinct (real) points of the surface; if p2 = q 2 we get
the two points (0, 0, ±r).
(ii) We can take the surface to be
z=
x2
y2
+
.
p2
q2
Proceeding as in (i), we find that the conditions for an umbilic are
2
4λxy
4x2
2
4y 2
=λ 1+ 4 , 0= 2 2 , 2 =λ 1+ 4 .
p2
p
p q
q
q
The second equation gives x = 0, y = 0 or λ = 0, p
and λ = 0 is impossible by
the other two equations. If x = 0 we get y = ± 21 q p2 − q 2 , z = ± 14 (p2 − q 2 ).
p
If y = 0 we get x = ± 12 p q 2 − p2 , z = ± 14 (q 2 − p2 ). If p2 6= q 2 this gives two
(real) points of the surface; if p2 = q 2 we get the origin as the only umbilic.
(iii) The calculation is simplified by noting that these surfaces are ruled, so that,
if κ1 , κ2 are the principal curvatures, κ1 κ2 ≤ 0 everywhere (Exercise 8.2.23).
Hence, if κ1 = κ2 we must have κ1 = κ2 = 0, and hence FII = 0, so we are
looking for the points at which the second fundamental form is zero.
For the hyperboloid of one sheet
x2
y2
z2
+
=
+ 1,
p2
q2
r2
we can use x, y as parameters on the open subset of the surface where z 6= 0. The
formulas for zxx , zxy and zyy are the same as in part (i); the condition zxy = 0
forces x = 0 or y = 0, but neither of these is consistent with the conditions
zxx = zyy = 0. Hence, there are no umbilics for which z 6= 0.
To deal with the ‘waist’ z = 0 of the hyperboloid, we can use x, z as parameters
if x 6= 0 and y, z if y 6= 0. In the first case, for example, the condition for an
78
umbilic is yxx = yxz = yzz = 0; proceeding as above, we find that there are no
umbilics. Similarly in the last case.
Finally, for the hyperbolic paraboloid
z=
x2
y2
−
,
p2
q2
zxx = 2/p2 is never zero, so there are no umbilics.
8.2.36 We have to show that the umbilics on the surface xyz = 1 are the four points
(1, 1, 1), (1, −1, −1), (−1, 1, −1) and (−1, −1, 1). Using x, y as parameters, we
find as in the preceding exercise that the conditions for an umbilic are
1
2
=λ 1+ 4 2 ,
x3 y
x y
1
λ
= 3 3,
2
2
x y
x y
2
1
=λ 1+ 2 4 ,
xy 3
x y
for some λ. The second equation gives λ = xy and then the other equations give
x4 y 2 = x2 y 4 = 1; these equations force x2 = y 2 = 1, so x = ±1, y = ±1 (with
any combination of signs). Then z = 1/xy = ±1 and we obtain the four points
required.
8.2.37 On the open subset of S on which z 6= 0 we can use x, y as parameters. Differentiating implicitly we find that
zxx =
(y 2 + z 2 )(2x2 z 2 − y 2 z 2 − x2 y 2 )
,
z 3 (x2 + y 2 )2
so zxx = 0 when x2 = y 2 = z 2 = 1, and in particular at p. Similarly, zxy = zyy =
0 at p. It follows that the second fundamental form of S is zero at p (Exercise
8.2.33), i.e., p is a planar point. The argument actually proves that the eight
points (±1, ±1, ±1) are all planar, but this can also be deduced from Exercise
8.2.34 and the fact that S is preserved by the isometries (x, y, z) 7→ (±x, ±y, ±z)
of R3 (with any combination of signs).
(i) S is a level surface of f (x, y, z) = x2 y 2 + y 2 z 2 + z 2 x2 , so
∇ f = (2x(y 2 + z 2 ), 2y(x2 + z 2 ), 2z(x2 + y 2 ))
n is normal
is normal to S at (x, y, z). Taking x = y = 1 we find that (4, 4, 4) = 4n
to S at p, and hence so is n. Since t 1 and t2 are perpendicular to n, they are
tangent to S at p.
79
(ii) Since the vectors t1 , t2 , n are linearly independent, scalars X, Y, Z as in the
statement of the exercise exist, and are small if (x, y, z) is near p. When we
substitute x = 1 + X + Z, y = 1 − X + Y + Z, z = 1 − Y + Z in the equation
of S, we know that all quadratic and lower order terms will cancel, and the
quartic terms can be discarded as they will be small compared to the cubic
terms. Retaining only the cubic terms, we find after some algebra the equation
6Z − 3X 2 Y + 3XY 2 = 0, hence the stated equation.
√
√
(iii) Making the linear transformation X = −X̃ + 3Ỹ , Y = X̃ + 3Ỹ , Z̃ = Z,
we get Z̃ = X̃(X̃ 2 − 3Ỹ 2 ), which is a monkey saddle.
8.3.1 (i) Setting ũ = v, ṽ = w = e−u , we have u = − ln ṽ, v = ũ so, in the nota0 − ṽ1
tion of Exercise 6.1.4, J =
. Since J is invertible, (u, v) 7→ (v, w)
1 0
is a reparametrization
map.
w is
form in terms
of v,
Thefirst fundamental
1
1
0
0
−
E
F
0
1
Ẽ F̃
ṽ
=
given by
= Jt
J =
1
2
1
0
0
0
f
(u)
F
G
−
F̃
G̃
ṽ
1
0
w2
, so the first fundamental form is (dv 2 + dw2 )/w2 .
0 w12
(ii) We find that the matrix
∂v
∂v
1 2
(v − (w + 1)2 )
v(w + 1)
∂V
∂W
2
˜
J = ∂w ∂w =
,
v(w + 1)
− 21 (v 2 − (w + 1)2 )
∂V
∂W
1
0
2
t
w
so the first fundamental form in terms of V and W is J˜
J˜ =
0 w12
(v 2 +(w+1)2 )2
I
4w 2
= (1−U 24−V 2 )2 I, after some tedious algebra.
In (i), u < 0 and −π < v < π corresponds to −π < v < π and w > 1, a
semi-infinite rectangle in the upper half of the vw-plane.
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w
1
0
v
To find the corresponding region in (ii), it is convenient to introduce the complex
numbers z = v + iw, Z = U + iV . Then, the equations in (ii) are equivalent to
80
Z=
z−i
z+i ,
z=
Z+1
i(Z−1) .
The line v = π in the vw-plane corresponds to z + z̄ = 2π
(the bar denoting complex conjugate), i.e.
Z+1
i(Z−1)
= 2π, which simplifies
− i(Z̄+1
Z̄−1)
to |Z − (1 − πi )|2 = π12 ; so v = π corresponds to the circle in the U V -plane with
centre 1 − πi and radius π1 . Similarly, v = −π corresponds to the circle with
centre 1 + πi and radius π1 . Finally, w = 1 corresponds to z − z̄ = 2i, i.e.
Z+1
1 2
Z̄+1
1
i(Z−1) + i(Z̄−1) = 2i. This simplifies to |Z − 2 | = 4 ; so w = 1 corresponds to
the circle with centre 1/2 and radius 1/2 in the U V -plane. The required region
in the U V -plane is that bounded by these three circles:
V
U2 + V 2 = 1
11111
00000
00000
11111
00000
11111
O
00000
11111
00000
11111
U
For (iii) we follow the hint and make use of polar coordinates on the disc,
V = r cos θ, W = r sin θ, V̄ = r̄ cos θ̄, W̄ = r̄ sin θ̄. We find that r̄ = r22r+1 ,
θ̄ = θ. Suppose that the first fundamental form in terms of these param2(1−r2 )
eters is Edr̄ 2 + 2F dr̄dθ̄ + Gdθ̄ 2 . Since dr̄
dr = (1+r2 )2 , the first fundamental
2
4(1−r2 )2
)
Edr 2 + 4(1−r
F drdθ + Gdθ 2 .
(1+r2 )4
(1+r2 )2
(1+r2 )4
4(dr2 +r2 dθ2 )
,
we
get
E
=
= (1−r̄1 2 )2 , F
2
2
(1−r )
(1−r2 )4
form is
Equating this to
2
4(dV 2 +dW 2 )
=
(1−V 2 −W 2 )2
2
r̄
= 1−r̄
2 . Con2
4r
= 0, G = (1−r
2 )2
verting back to the parameters (V̄ , W̄ ), we have r̄dr̄ = V̄ dV̄ + W̄ dW̄ , r̄ dθ̄ =
81
V̄ dW̄ − W̄ dV̄ , so the first fundamental form becomes
(V̄ dV̄ + W̄ dW̄ )2 + (1 − r̄ 2 )(V̄ dW̄ − W̄ dV̄ )2
r̄ 2 (1 − r̄ 2 )2
(V̄ 2 + (1 − r̄ 2 )W̄ 2 )dV̄ 2 + 2r̄ 2 V̄ W̄ dV̄ dW̄ + (W̄ 2 + (1 − r̄ 2 )V̄ 2 )dW̄ 2
=
r̄ 2 (1 − r̄ 2 )2
(1 − W̄ 2 )dV̄ 2 + 2V̄ W̄ dV̄ dW̄ + (1 − V̄ 2 )dW̄ 2
.
=
(1 − V̄ 2 − W̄ 2 )2
8.3.2 √
The parametrisation is σ (u, v) = (f (u) cos v, f (u) sin v, g(u)), f (u) = eu , g(u) =
1 − e2u − cosh−1 (e−u ), −∞ < u < 0.
(i) A parallel u = constant is a circle of radius f (u) = eu , so has length 2πeu .
(ii) From Example 7.1.2, E = 1, F = 0, G = f (u)2 , so dAσ = f (u)dudv and the
R 2π R 0
area is 0 −∞ eu dudv = 2π.
(iii) From Examples 7.1.2 and 8.1.4, the principal curvatures are κ1 = f˙g̈ − f¨ġ =
−f¨/ġ = −(e−2u − 1)−1/2 , κ2 = f ġ/f 2 = ġ/f = (e−2u − 1)1/2 .
(iv) κ1 < 0, κ2 > 0.
8.3.3 Let γ (u) = (f (u), 0, g(u)) and denote d/du by a dot; by Example 8.1.4, f¨+Kf =
√
√
0. If K < 0, the general solution is f = ae− −Ku + be −Ku where a, b are
constants; the condition f (π/2) = f (−π/2) = 0 forces a = b = 0, so γ coincides
with the z-axis, contradicting the assumptions. If K = 0,√f = a + bu√and again
a = b = 0 is forced. So we must have K > 0 and f = a cos Ku+b sin Ku. This
time, f (π/2) = f (−π/2) = 0 and a, b not both zero implies that the determinant
√
√
cos √Kπ/2 sin √Kπ/2
= 0.
cos Kπ/2 − sin Kπ/2
√
This gives sin Kπ = 0, so K = n2 for some integer n 6= 0. If n = 2k is even,
f = b sin 2ku, but then f (0) = 0, contradicting the assumptions. If n = 2k + 1
is odd, f = a cos(2k + 1)u and f (π/2(2k + 1)) = 0, which contradicts the
assumptions
unless k = 0 or −1, i.e. unless K = (2k+1)2 = 1. Thus, f = a cos u,
q
p
ġ = 1 − f˙2 = 1 − a2 sin2 u. Now, γ̇γ = (f˙, 0, ġ) is perpendicular to the z√
axis ⇐⇒ ġ = 0. So the assumptions give 1 − a2 = 0, i.e. a = ±1. Then,
γ (u) = (± cos u, 0, ± sin u) (up to a translation along the z-axis) and S is the
unit sphere.
σ (ũ, ṽ) be a patch of S containing p = σ̃
σ (ũ0 , ṽ0 ). The Gaussian curvature
8.4.1 Let σ̃
K of S is < 0 at p; since K is a smooth function of (ũ, ṽ) (Exercise 8.1.3),
K(ũ, ṽ) < 0 for (ũ, ṽ) in some open set Ũ containing (ũ0 , ṽ0 ); then every point of
σ (Ũ ) is hyperbolic. Let κ1 , κ2 be the principal curvatures of σ̃
σ , let 0 < θ < π/2
σ̃
82
p
be such that tan θ = −κ1 /κ2 , and let e1 and e2 be the unit tangent vectors of
σ making angles θ and −θ, respectively, with the principal vector corresponding
σ̃
to κ1 (see Theorem 8.2.4). Applying Proposition 8.4.3 gives the result. For the
last part, put v̇ = 0 in the formula for κn in Proposition 7.3.5: this shows that
L = 0 if the parameter curves v = constant are asymptotic. Similarly N = 0 if
the parameter curves u = constant are asymptotic.
8.4.2 The first and √
second fundamental forms are (2 + v 2 )du2 + 2uvdudv + (2 + u2 )dv 2
and −4dudv/ 4 + 2u2 + 2v 2 . By Exercise 8.2.3, the lines of curvature are given
by
−(2 + v 2 )u̇2 + (2 + u2 )v̇ 2 = 0,
so
and hence
Z
dv
=±
2 + v2
Z
du
,
2 + u2
v
u
sinh−1 √ ± sinh−1 √ = constant.
2
2
We can take s = sinh−1 √v2 + sinh−1 √u2 , t = sinh−1 √v2 − sinh−1 √u2 , so u =
√
√
2 sinh 12 (s − t), v = 2 sinh 12 (s + t).
8.4.3 Assume that γ is unit-speed. Then, the ruled surface is σ (u, v) = γ (u) + vδδ (u),
N, t = γ̇γ (a dot denoting d/du), and N is the unit normal of (a patch
where δ = t ×N
N)..N = 0.
of) S. By Example 8.1.5, the flatness condition is δ̇δ.N = 0, i.e., (tt × Ṅ
N is tangent to S, so t × Ṅ
N is parallel to N . Hence, S̃ is flat if and only if
Now, Ṅ
N = 0, i.e. if and only if Ṅ
N is parallel to γ̇γ . By Exercise 8.2.2, this is precisely
t × Ṅ
the condition for γ to be a line of curvature of S.
8.5.1 By Corollary 8.1.3 and the fact that σ is conformal, the mean curvature of σ is
, so
H = L+N
2E
(*)
σ is minimal ⇐⇒ L + N = 0 ⇐⇒ (σ
σ uu + σ vv )..N = 0.
σ = σ uu +σ
σ vv = 0. For the converse, we have to
Obviously, then, σ is minimal if ∆σ
σ = 0 if (*) holds. It is enough to prove that ∆σ
σ.σ u = ∆σ
σ.σ v = 0,
show that ∆σ
σ u , σ v , N} is a basis of R3 . We compute ∆σ
σ.σ u = σ uu.σ u + σ vv .σ u =
since {σ
1
1
σ
σ
σ
σ
σ
2 (σ u .σ u )u + (σ v .σ u )v − (σ v .σ uv ) = 2 (σ u .σ u − σ v .σ v )u + (σ v .σ u )v . But, since
σ is conformal, σ u .σ u = σ v .σ v and σ u .σ v = 0. Hence, ∆σ
σ .σ u = 0. Similarly,
σ.σ v = 0.
∆σ
The first fundamental form of the given surface patch is
(1 + u2 + v 2 )2 (du2 + dv 2 ),
so it is conformal, and σ uu + σ vv = (−2u, 2v, 2) + (2u, −2v, −2) = 0.
83
8.5.2 Using the formula in Exercise 8.1.1 with f (x, y) = ln cos y − ln cos x gives
H=
sec2 x(1 + tan2 y) − sec2 y(1 + tan2 x)
= 0.
2(1 + tan2 x + tan2 y)3/2
Nu , Σv = σ v + wN
Nv , Σ w = N. Σ u .Σ
Σw = 0 since σ u .N
N = Nu .N
N = 0,
8.5.3 Σ u = σ u + wN
Σ
Σ
and similarly v .Σw = 0. Finally,
Σv = σ u .σ
σ v + w(σ
σ u .N
Nv + σ v .N
Nu ) + w2N u .N
Nv
Σ u .Σ
Nv = w2 Nu .N
Nv .
= F − 2wM + w2 Nu .N
L
N
σ u , Nv = − N
By the proof of Proposition 8.1.2, Nu = − E
G σ v , so N u .Nv =
LN
F = 0. Every surface u = u0 (a constant) is ruled as it is the union of
EG
the straight lines given by v = constant; by Exercise 8.2.4, this surface is flat
provided the curve γ (v) = σ (u0 , v) is a line of curvature of S, i.e. if σ v is a
principal vector; but this is true since the matrices FI and FII are diagonal.
Similarly for the surfaces v = constant.
8.5.4 We take the ruled surface to be σ (u, v) = γ (u) + vδδ (u), where γ is unit-speed
and δ is a unit vector. Denoting d/du by a dot and γ̇γ by t, the coefficients of
the first and second fundamental forms are E = k t + vδ̇δ k2 , F = t.δ , G = 1,
˙
L = (ṫt + vδ̈δ )..N, M = δ̇δ.N, N = 0, where N = (t+vδ˙ )×δ . Then, the mean
k(t +vδ )×δ k
curvature
LG − 2M F + N E
(ṫt − 2(tt.δ )δ̇δ + vδ̈δ )..N
H=
=
.
2
2(EG − F )
2(EG − F 2 )
If δ̇δ (u0 ) 6= 0 for some value of u0 , H(u0 , v) → 0 as v → ±∞, since the numerator
in the expression for H is independent of v while the denominator → ∞ as
v → ±∞. As H is constant, H = 0 everywhere, contradicting the assumption.
Hence, δ̇δ = 0 everywhere, δ is constant, and the surface is a generalized cylinder.
We can now assume that γ lies in a plane perpendicular to δ (Exercise 5.3.3).
Then, t × δ is a unit vector and the formula for the mean curvature simplifies to
H=
ṫt.(tt × δ )
.
2 k t × δ k2
n where κ is the curvature of γ and n is its principal normal. Also, t ×δδ
Now ṫt = κn
n.
is perpendicular to t and parallel to the plane containing γ ; hence, t × δ = ±n
It follows that H = ±κ/2, so if H is a non-zero constant then κ is constant and
the plane curve γ is a circle. Thus, the surface is a circular cylinder.
8.5.5 We use the notation and results of Proposition 8.5.2. If γ is a curve on σ , the
N. Now, γ λ
‘corresponding’ curve γ λ on the parallel surface σ λ is γ λ = γ + λN
84
λ
N = −κγ̇γ λ for some scalar κ (Exercise 8.2.2)
is a line of curvature on σ λ ⇐⇒ Ṅ
N = −κ(γ̇γ + λṄ
N). But this last equation holds if and only if Ṅ
N is a scalar
⇐⇒ ǫṄ
multiple of γ̇γ , i.e. if and only if γ is a line of curvature on σ . In this case,
N = (1 − κλ)γ̇γ is parallel to γ̇γ .
γ̇γ λ = γ̇γ + λṄ
σ (u, v) = σ (u, v)+λ(u, v)N
N(u, v) is
8.5.6 Let σ (u, v) be a patch of S. We are given that σ̃
a patch of S̃, for some function λ, where N(u, v) is the unit normal of σ at σ (u, v),
σ (u, v). Now σ̃
σ u = σ u + λN
Nu + λuN; since
and that N (u, v) is also normal to S̃ at σ̃
σ u and N u are all perpendicular to N, this implies that λu = 0; similarly
σ u , σ̃
λv = 0. Hence, λ is constant and S̃ = S λ is a parallel surface of S.
Chapter 9
9.1.1 By Exercise 4.1.3, there are two straight lines on the hyperboloid passing through
(1, 0, 0); by Proposition 9.1.4, they are geodesics. The circle z = 0, x2 + y 2 = 1
and the hyperbola y = 0, x2 − z 2 = 1 are both normal sections, hence geodesics
by Proposition 9.1.6.
N × γ̇γ ). Note that if Γ(t) = γ (ϕ(t)) is a reparametrization of γ ,
9.1.2 Let κ(γγ ) = γ̈γ. (N
3
Γ) = dϕ
Γ) = 0.
then κ(Γ
κ(γγ ). In particular, κ(γγ ) = 0 ⇐⇒ κ(Γ
dt
For (i), let γ be a pre-geodesic and let Γ be a geodesic reparametrization of γ .
Γ(t) = Γ(t/v) is
By Proposition 9.1.2 Γ has constant speed, say v, and then Γ̃
Γ) = 0, hence κ(Γ
Γ) = 0, hence
a unit-speed geodesic. By Proposition 9.1.3, κ(Γ̃
κ(γγ ) = 0. Conversely, if κ(γγ ) = 0 and if Γ is a unit-speed reparametrization of
Γ) = 0 so Γ is a geodesic by by Proposition 9.1.3.
γ , then κ(Γ
Part (ii) is obvious.
For (iii), let γ be a constant speed pre-geodesic, say with speed v. Then Γ(t) =
γ (t/v) is a unit-speed pre-geodesic, hence a geodesic by (i) and Proposition 9.1.3.
Γ, γ̈γ is perpendicular to the surface, so γ is a geodesic.
Since γ̈γ = v 2Γ̈
Finally, (iv) follows from (iii) and Proposition 9.1.2.
9.1.3 Let Πs be the plane through γ (s) perpendicular to t(s); the parameter curve
s = constant is the intersection of the surface with Πs . From the solution to
Exercise 4.2.7, the standard unit normal of σ is N = −(cos θ n + sin θ b). Since
this is perpendicular to t, the circles in question are normal sections.
2
2
2
9.1.4 Take the ellipsoid to be xp2 + yq 2 + zr2 = 1; the vector ( px2 , qy2 , rz2 ) is normal to the
ellipsoid by Exercise 5.1.2. If γ (t) = (f (t), g(t), h(t)) is a curve on the ellipsoid,
2
2
2
2
2
˙2
R = ( fp2 + ġq 2 + ḣr2 )−1/2 , S = ( fp4 + gq 4 + hr4 )−1/2 . Now, γ is a geodesic ⇐⇒ γ̈γ
is parallel to the normal ⇐⇒ (f¨, g̈, ḧ) = λ( f2 , g2 , h2 ) for some scalar λ(t). From
p
g2
f2
+
2
p
q2
2
+ hr2
= 1 we get
f f˙
g ġ
hḣ
p2 + q 2 + r2
= 0, hence
q
r
f˙2
ġ 2
f f¨ gg̈
ḣ2
hḧ
+
2
p
q 2 + r2 + p2 + q 2 + r2
= 0, i.e.
85
f˙2
p2
ġ 2
q2
ḣ2
r2
f2
p4
g2
q4
h2
r4
= 0, which gives λ = −S 2 /R2 . The curvature
2
1/2
2
2
S
k γ̈γ k = (f¨2 + g̈ 2 + ḧ2 )1/2 = |λ| fp4 + gq 4 + hr4
= |λ|
S = R2 . Finally,
!
!
1
f˙2
ġ 2 ḣ2
1 d
f f˙ g ġ hḣ
=
+ 4 + 4
+ 2+ 2
2 dt R2 S 2
p4
q
r
p2
q
r
!
˙¨
2
ff
g 2 h2
ġg̈ ḣḧ
f
+ 4 + 4
+ 2 + 2
+
p4
q
r
p2
q
r
!
!
1
λ f f˙ g ġ hḣ
f f˙ g ġ hḣ
= 2
+ 4 + 4 + 2
+ 4 + 4 = 0,
R
p4
q
r
S
p4
q
r
+
+
+λ
+
+
since λ = −S 2 /R2 . Hence, RS is constant.
9.1.5 Suppose that a geodesic γ lies in the plane v.a = b, where a and b are constants.
Then γ̇γ .a = γ̈γ.a = 0. Since γ̈γ is parallel to N (the unit normal of the surface),
N.a = 0, so Ṅ
N.a = 0. Since N, γ̇γ and Ṅ
N are all parallel to the plane and the last
two vectors are perpendicular to the first, they are parallel. Hence γ is a line
of curvature by Exercise 8.2.2. Conversely, if γ is both a geodesic and a line of
curvature, we may assume γ has unit-speed (for the unit-speed reparametrization
of γ would still be a geodesic and a line of curvature). Let a = N × γ̇γ . Then
N × γ̇γ +N
N × γ̈γ = 0 since the first term vanishes because γ is a line of curvature
ȧa = Ṅ
and the second because γ is a geodesic. So a is constant. And γ̇γ.a = 0 so γ .a is
a constant, say b. Hence γ lies in the plane v.a = b.
9.1.6 For (i) note that γ̈γ is a non-zero vector parallel to both N1 and N2 , so N 1 and N2
must be parallel. For an example, take S1 and S2 to be the sphere and cylinder
in Theorem 6.4.6.
(ii) Now suppose that S1 and S2 intersect perpendicularly. Then, N1 , N2 and γ̇γ
N2 = 0. If γ is
are perpendicular unit vectors. From γ̇γ.N2 = 0 we get γ̈γ.N2 + γ̇γ .Ṅ
N2 is perpendicular to
a geodesic on S1 , γ̈γ .N2 = 0 since γ̈γ is parallel to N1 , so Ṅ
γ
N
N
γ̇ . Since Ṅ2 is also perpendicular to 2 , it must be parallel to N1 . Conversely, if
N2 is parallel to N1 , then γ̇γ .Ṅ
N2 = 0 so γ̈γ is perpendicular to N 2 . Since γ̈γ is also
Ṅ
perpendicular to γ̇γ , it must be parallel to N 1 . Finally, if γ is a geodesic on both
N1 is parallel to N2 and Ṅ
N2 is parallel to N1 . It follows that
S1 and S2 , then Ṅ
N1 × N2 )˙ = Ṅ
N1 × N 2 + N1 × Ṅ
N2 = 0 so N1 × N2 is a constant vector. Since γ̇γ is
(N
a unit vector parallel to N1 × N2 , γ̇γ is constant, so C is part of a straight line.
9.1.7 The ellipses formed by the intersections of the ellipsoid with thethree coordinate
x y
z
planes are normal sections. Indeed, a normal to the ellipsoid is p2 , q 2 , r2 . The
plane z = 0 (say) has normal (0, 0,
1), so the intersection with the plane z = 0
is a normal section if and only if px2 , qy2 , rz2 .(0, 0, 1) = z/r 2 = 0 when z = 0,
which is true.
86
9.1.8 The intersections of the surface with each of the planes x = ±y, y = ±z and
z = ±x are normal sections. Indeed, by the solution to Exercise 8.2.37, a normal
to the surface is (x(y 2 + z 2 ), y(x2 + z 2 ), z(x2 + y 2 )). The planes x = ±y have
normals (1, ∓1, 0) so the intersection of the surface with x = ±y is a normal
section if and only if
x(y 2 + z 2 ) ∓ y(x2 + z 2 ) = 0 when x = ±y,
which is evidently true. Similarly for the other four planes.
9.1.9 Assume that γ is unit-speed with tangent vector t. Then (with the usual notan.a = 0. Since κ 6= 0, n is perpendicular
tion), t.a is constant, so ṫt.a = 0, i.e. κn
to a. As γ is a geodesic, ṫt is parallel to N, so n is parallel to N. Hence, a is
perpendicular to N , hence tangent to S.
9.1.10 From the solution to Exercise√7.3.3., a circle of radius r on a sphere of radius R
has geodesic curvature κg = R2 − r 2 /rR. Hence, κg = 0 ⇐⇒ r = R ⇐⇒ the
circle is a great circle.
9.1.11 The osculating plane is perpendicular to the binormal b of γ , and so is perpendicular to S at p if and only if b is perpendicular to N at p. Since t and N are
perpendicular, b is perpendicular to N ⇐⇒ n is parallel to N ⇐⇒ γ is a geodesic.
9.1.12 γ is asymptotic ⇐⇒ γ̈γ is perpendicular to N; γ is a geodesic ⇐⇒ γ̈γ is parallel
to N. Hence, if γ is both asymptotic and a geodesic, we must have γ̈γ = 0, so
γ (t) = taa + b for some constant vectors a, b, i.e. γ is a straight line.
9.1.13 γ is a geodesic on σ ⇐⇒ γ̈γ is parallel to σ u × σ v = (γ̇γ + vδ̇δ ) × δ when v = 0, i.e.
γ̈γ × (γ̇γ × δ ) = 0. Since γ̈γ is perpendicular to γ̇γ , γ is a geodesic ⇐⇒ γ̈γ .δ = 0 ⇐⇒ δ
n.
is perpendicular to γ̈γ = κn
9.1.14 Denoting by t, n and b the unit tangent vector, principal normal and binormal
N since Γ is a geodesic so
of Γ, and by κ its curvature, we have n = ±N
Γ.(N
N × Ṅ
N) = t.(n
n × ṅ
n) = t.(n
n × (−κtt + τb
b)) = t.(κb
b + τtt) = τ.
Γ̇
We can assume that γ is unit-speed; denote by a dash differentiation with respect
to the arc-length of γ . By Exercise 7.3.22, the geodesic torsion of γ is
N × N′ ).
τg = (γγ ′ × N)..N′ = γ ′ .(N
Γ at p, and we can assume that the sign is +
Since γ and Γ touch at p, γ ′ = ±Γ̇
by changing the sign of the parameter of γ if necessary. Since γ ′ = u′σ u + v ′ σ v ,
N′ = u′N u + v ′ Nv , etc., it follows that u′ = u̇ and v ′ = v̇ at p, and hence that
N at p. Hence, τg = Γ̇
Γ.(N
N × Ṅ
N) and the result now follows from the first
N′ = Ṅ
part of the exercise.
87
Since Γ touches itself at each of its points (!), its torsion is equal to its geodesic
torsion.
N so τ =
9.1.15 If γ is an asymptotic curve, then (with the usual notation) b = ±N
b.n = ∓Ṅ
N.n. Since n = b × t = ±N
N × t, we get
−ḃ
N. (N
N × t ) = t. (N
N × Ṅ
N).
τ = −Ṅ
The last part follows from the preceding exercise.
9.1.16 We use the notation of Exercise 9.1.14 and assume that γ is a line of curvature.
Γ = γ ′ and N′ = Ṅ
N at p. By Rodrigues’ formula, Ṅ
N = −κγ̇γ
We can assume that Γ̇
where κ is the principal curvature corresponding the line of curvature γ , so
N′ = −κγγ ′ . Hence, the torsion of Γ is
Γ. (N
N × Ṅ
N) = −κγγ ′ .(N
N × γ ′ ) = 0.
τ = Γ̇
9.1.17 (i) By Exercise 9.1.14, the torsion τ of a geodesic γ on a surface S is equal to its
geodesic torsion which, by Exercise 8.2.22, is equal to (κ2 − κ1 ) sin θ cos θ in the
notation there. At an umbilic κ1 = κ2 so τ = 0.
(ii) By (i), the torsion is given by τ = (κ2 − κ1 ) sin θ cos θ. If a second geodesic
γ̃γ intersects γ at right angles at p, then in the obvious notation θ̃ = θ ± π/2 and
hence τ̃ = −τ .
(iii) If γ is a geodesic, its curvature κ is equal, up to sign, to its normal curvature,
so κ = ±(κ1 cos2 θ + κ2 sin2 θ) by Euler’s Theorem 8.2.4. Taking the + sign,
κ − κ1 = (κ2 − κ1 ) sin2 θ, κ − κ2 = (κ1 − κ2 ) cos2 θ,
so
(κ − κ1 )(κ − κ2 ) = −(κ1 − κ2 )2 sin2 θ cos2 θ = −τ 2 .
Taking the − sign gives the other expression for τ 2 .
(iv) If the surface is flat, either κ1 = 0 or κ2 = 0 at p. If κ2 = 0, then
κ = ±κ1 cos2 θ, τ = −κ1 sin θ cos θ, so τ = ∓κ tan θ. The other formula results
if κ1 = 0 at p.
9.1.18 We can assume that γ is unit-speed as conditions (i) - (iii) are unchanged by
reparametrization. Let δ (u) be a non-zero vector parallel to the ruling through
γ (u). Then (i) says that γ is a geodesic, and by Exercise 9.1.13 this holds
⇐⇒ ṫt.δ = 0. By Exercise 5.3.4, (ii) holds ⇐⇒ t.δ̇δ = 0. And (iii) obviously holds
⇐⇒ t.δ is a constant. Everything now follows immediately.
9.1.19 Suppose that every geodesic on a surface S is a plane curve. By Proposition
8.2.9, it suffices to show that every point of S is an umbilic. Suppose for a
contradiction that p ∈ S is not an umbilic, let κ1 , κ2 be the distinct principal
88
curvatures of S at p, and let t be a non-zero principal vector corresponding to
the principal curvature κ1 at p. Choose an angle θ such that
(i) θ is not an integer multiple of π/2, and
(ii) κ1 cos2 θ + κ2 sin2 θ 6= 0.
Let γ be the unit-speed geodesic on S passing through p and making an angle
θ with t. By condition (ii) and Euler’s Theorem 8.4.2, γ has non-zero curvature
at p, and γ is a plane curve by the hypothesis. By Exercise 9.1.5, γ is a line of
curvature, but this contradicts property (i).
9.1.20 Assume that γ is unit-speed, and let r be the radius of the sphere. Then,
(γγ − c)..(γγ − c) = r 2 so γ̇γ .(γγ − c) = 0, γ̈γ .(γγ − c) = −γ̇γ .γ̇γ = −1. But, as γ is
N, where κ is the curvature of γ , so the last equation gives
a geodesic, γ̈γ = ±κN
N.(γγ − c) = ±1/κ, i.e.
1
,
κ=
N.(γγ − c)|
|N
as required.
9.2.1 If p and q lie on the same parallel of the cylinder, there are exactly two geodesics
joining them, namely the two circular arcs of the parallel of which p and q are
the endpoints. If p and q are not on the same parallel, there are infinitely-many
circular helices joining p and q (see Example 9.2.8).
9.2.2 Take the cone to be σ (u, v) = (u cos v, u sin v, u). By Exercise 6.2.1, σ is locally
isometric to an open subset of the xy-plane by
√
√
v
v
σ (u, v) 7→ u 2 cos √ , u 2 sin √ , 0 .
2
2
By Corollary 9.2.7, the geodesics on the cone correspond to the straight lines in
the xy-plane. Any such line, other than the axes x = 0 and y = 0, has equation
ax + by = 1, where a, b are constants; this line corresponds to the curve
v 7→
cos v
sin v
1
√
,√
,√
v
v
v
v
v
2(a cos √2 + b sin √2 )
2(a cos √2 + b sin √2 )
2(a cos √2 + b sin √v2 )
the x− and y-axes correspond to straight lines on the cone.
9.2.3 Parametrize the cylinder by σ (u, v) = (cos u, sin u, v). Then, E = G = 1, F = 0,
so the geodesic equations are ü = v̈ = 0. Hence, u = a + bt, v = c + dt, where
a, b, c, d are constants. If b = 0 this is a straight line on the cylinder; otherwise,
it is a circular helix.
!
;
89
9.2.4 For the first part,
(E u̇2 + 2F u̇v̇ + Gv̇ 2 )˙ = (Eu u̇ + Ev v̇) u̇2 + 2(Fu u̇ + Fv v̇)u̇v̇ + (Gu u̇ + Gv v̇)v̇ 2
+ 2E u̇ü + 2F (u̇v̈ + üv̇) + 2Gv̇v̈
= u̇(Eu u̇2 + 2Fu u̇v̇ + Gu v̇ 2 ) + v̇(Ev u̇2 + 2Fv u̇v̇ + Gv v̇ 2 )
+ 2E u̇ü + 2F (u̇v̈ + üv̇) + 2Gv̇v̈
= 2(Eu u̇ + F v̇)˙ u̇ + 2(F u̇ + Gv̇)˙ v̇ + 2(E u̇ + F v̇)ü
+ 2(F u̇ + Gv̇)v̈ by the geodesic equations
= 2[(E u̇ + F v̇)u̇]˙+ 2[(F u̇ + Gv̇)v̇]˙
= 2(E u̇2 + 2F u̇v̇ + Gv̇ 2 )˙.
Hence, (E u̇2 + 2F u̇v̇ + Gv̇ 2 )˙ = 0 and so k γ̇γ k2 = E u̇2 + 2F u̇v̇ + Gv̇ 2 is constant.
Suppose now that (i) and (ii) hold. Differentiating E u̇2 +2F u̇v̇ +Gv̇ 2 = constant
gives
(Eu u̇ + Ev v̇)u̇2 + 2(Fu u̇ + Fv v̇))u̇v̇ + (Gu u̇ + Gv v̇)v̇ 2
= −2(E u̇ + F v̇)ü − 2(F u̇ + Gv̇)v̈,
i.e.,
(Eu u̇2 + 2Fu u̇v̇ + Gu v̇ 2 )u̇ + (Ev u̇2 + 2Fv u̇v̇ + Gv v̇ 2 )v̇
= −2(E u̇ + F v̇)ü − 2(F u̇ + Gv̇)v̈.
Using (i) we get
2u̇
d
(E u̇ + F v̇) + 2(E u̇ + F v̇)v̈ = −2(F u̇ + Gv̇)v̈ − (Ev u̇2 + 2Fv u̇v̇ + Gv v̇ 2 )v̇.
dt
The left-hand side of this equation equals
2
d
d
d
(u̇(E u̇ + F v̇)) = −2 (F u̇v̇ + Gv̇ 2 ) = −2(F u̇ + Gv̇)v̈ − 2v̇ (F u̇ + Gv̇).
dt
dt
dt
Combining the last two equations gives (ii) provided v̇ 6= 0.
9.2.5 E = 1, F = 0, G = 1+u2 , so γ is unit-speed ⇐⇒ u̇2 +(1+u2 )v̇ 2 = 1. The second
a
d
((1 +u2 )v̇) = 0, i.e. v̇ = 1+u
equation in (9.2) gives dt
2 , where a is a constant. So
2
a
dv
v̇
a
2
u̇ = 1 − 1+u2 and, along the geodesic, du = u̇ = ± √
. If a = 0,
(1−a2 +u2 )(1+u2 )
√
then v = constant and we have a ruling. If a = 1, then dv/du = ±1/u 1 + u2 ,
which can be integrated to give v = v0 ∓ sinh−1 u1 , where v0 is a constant.
90
2 (u2 +1)(u2 +1−a2 )
so
=
For the last part, note that du
dv
a2
√
2
2
(i) if a > 1 then du/dv = 0 for u = ± a − 1 and this is the minimum distance
of the geodesic from the z-axis;
(ii) if a2 < 1 then |du/dv| > a−2 − 1 so u will decrease to zero and the geodesic
will cross the z-axis;
(iii) if a2 = 1 then du/dv = ±(u2 + 1) so u = ± tan(v + c) where c is a constant.
The information given implies that, when u = D, cos α = γ̇γ.σ u = u̇ (since
E = 1, F = 0) so a2 = (1 + D2 ) sin2 α. Then, a2 is > 1, < 1 or = 1 according as
D is >, < or = cot α.
9.2.6 This is straightforward algebra.
9.2.7 By Exercise 5.3.3 we can parametrize the generalized cylinder S by σ (u, v) =
γ (u) + vδδ , where γ is a unit-speed curve, δ is a constant unit vector parallel to
the rulings, and γ̇γ = dγγ /du is perpendicular to δ for all values of u. Then the
first fundamental form of σ is du2 + dv 2 , so the map σ (u, v) 7→ (u, v, 0) is a local
isometry from S to the xy-plane. The curves on S that make a constant angle
with δ correspond to the curves in the plane that make a constant angle with
the axes. These are, of course, exactly the straight lines in the plane, i.e. the
geodesics in the plane. It follows from Corollary 9.2.7 that the geodesics on S
are exactly the curves on S making a constant angle with δ .
9.2.8 Let γ be a parameter curve v = constant, and √assume that γ is unit-speed.
Then, in the usual notation, E u̇2 = 1 so u̇ = ±1/ E. By Theorem 9.2.1, γ is a
geodesic if and only if
(*)
1
d
1
d
(E u̇) = Eu u̇2 and
(F u̇) = Ev u̇2 .
dt
2
dt
2
The second equation in (*) holds ⇐⇒
i.e.
F
±√
E
1
√
E
u̇ =
u
F
√
E
u
=
Ev
,
2E
Ev
.
2E
This is equivalent to the stated condition. One checks in the same way that the
first equation in (*) holds identically.
9.2.9 Let γ be the unit-speed curve on the surface corresponding to u + v = c, where
c is a constant. Then, u̇ + v̇ = 0 so the unit-speed condition gives
(1 + u2 + 2uv + 1 + v 2 )u̇2 = 1,
91
i.e., (2 + c2 )u̇2 = 1, so u̇ (and hence also v̇) is constant. The geodesic equations
are
d
((1 + u2 )u̇ − uv v̇) = u̇(uu̇ − v v̇),
dt
d
(−uv u̇ + (1 + v 2 )v̇) = v̇(v v̇ − uu̇).
dt
Using v̇ = −u̇ the first equation becomes
d
((1 + cu)u̇) = cu̇2 ,
dt
i.e. (1 + cu)ü = 0. Similarly, the second geodesic equation is equivalent to
(1 + cv)v̈ = 0. Both equations are obviously satisfied when u̇ and v̇ are constant.
9.2.10 The geodesic equations
ü =
1
d
1
Gu v̇ 2 ,
(Gv̇) = Gv v̇ 2
2
dt
2
must be satisfied if v = cu and c is any constant. It is easy to see that these
equations are satisfied if c = 0 (since in that case u̇ = ±1 from the unit-speed
condition), so we assume that c 6= 0 from now on. Then the second equation is
1
d
2
dt (Gu̇) = 2 cGv u̇ , i.e.
1
Gü + (Gu + cGv )u̇2 = 0.
2
Using the first geodesic equation and writing c = v/u we get the stated equation
for G.
Substituting G = f (u)/v 2 leads to the ordinary differential equation
(f + 2u2 )
df
= 2uf,
du
which can be solved for the function f (u). (The solution is 2u3 = 3f 3 ln af ,
where a is an arbitrary constant.)
9.2.11 Assume that the geodesic is unit-speed. The second geodesic equation in Theorem 9.2.1 gives F u̇+Gv̇ = Ω, a constant. Using this and the unit-speed condition
E u̇2 + 2F u̇v̇ + Gv̇ 2 = 1 gives
r
G − Ω2
u̇ = ±
.
EG − F 2
Hence, either u = constant or, along the geodesic,
r
v̇
Ω − F u̇
F
Ω EG − F 2
dv
= =
=− ±
.
du
u̇
Gu̇
G G
G − Ω2
92
b(u), where b is the binormal
9.2.12 The surface is parametrized by σ (u, v) = γ (u) + vb
n, σ v = b and the first
of γ (we assume that γ is unit-speed). Then σ u = t − τ vn
fundamental form is
(1 + τ v 2 )du2 + dv 2 .
Let Γ(t) be a unit-speed geodesic on S, and denote d/dt by a dot. The first
geodesic equation in Theorem 9.2.1 gives
(1 + τ 2 v 2 )u̇ = c,
where c is a constant. To determine c, note that, at the intersection point p
Γ.t = cos α, which gives
(say), Γ̇
n) + v̇b
b)..t = cos α,
(u̇(tt − τ vn
i.e. u̇ = cos α at p. Since v = 0 at p, we have c = cos α and so
u̇ =
cos α
.
1 + τ 2v2
The unit-speed condition (1 + τ 2 v 2 )u̇2 + v̇ 2 = 1 now gives
v̇ 2 =
sin2 α + τ 2 v 2
,
1 + τ 2v2
and hence
du
= ±q
dv
cos α
(sin2 α + τ 2 v 2 )(1 + τ 2 v 2 )
.
If τ 6= 0, |du/dv| decreases as |v| increases and, for large |v|, |du/dv| is approximately cos α/τ 2 v 2 . It follows that u approaches limits u± (say) as v → ±∞ and
that u is always between these limits.
If τ is always zero, then du/dv = ± cot α. If cot α 6= 0 then u → ±∞ as v → ±∞
and then γ is not contained between two rulings. If cot α = 0 then u is constant
and γ is a ruling.
√
√
9.2.13 Writing E = (U + V )P , G = (U + V )Q, we have cos θ = u̇ E, sin θ = v̇ G so
we have to prove that U Gv̇ 2 − V E u̇2 is constant along γ . Now
2
d
d
d
d
d
(U Gv̇ 2 − V E u̇2 ) = 2U v̇ (Gv̇) − 2V u̇ (E u̇) + 2Gv̇ (U v̇) − 2E u̇ (V u̇).
dt
dt
dt
dt
dt
93
Using the geodesic equations in Theorem 9.2.1, this becomes
d
d
(U v̇) − 2E u̇ (V u̇)
dt
dt
2
2
= u̇ v̇(U Ev − 2EVv ) + u̇v̇ (2GUu − V Gu ) + 2GU v̇v̈ − 2EV u̇ü
v̇U (Ev u̇2 + Gv v̇ 2 ) − u̇V (Eu u̇2 + Gu v̇ 2 ) + 2Gv̇
+ Gv U v̇ 3 − Eu V u̇3
= u̇2 v̇(U Ev − 2EVv ) + u̇v̇ 2 (2GUu − V Gu )
d
d
+ U (Gv̇ 2 ) − U Gu u̇v̇ 2 − V (E u̇2 ) + V Ev u̇2 v̇
dt
dt
= u̇2 v̇((U + V )Ev − EVv ) + u̇v̇ 2 (GUu − (U + V )Gu )
d
d
+ U (Gv̇ 2 ) + Uu Gu̇v̇ 2 − V (E u̇2 )Vv E u̇2 v̇
dt
dt
d
= (U Gv̇ 2 − V E u̇2 ) + u̇2 v̇((U + V )Ev − EVv ) + u̇v̇ 2 (GUu − (U + V )Gu ).
dt
Thus,
d
(U Gv̇ 2 − V E u̇2 ) = u̇2 v̇((U + V )Ev − EVv ) + u̇v̇ 2 (GUu − (U + V )Gu ).
dt
But, since U and P depend only on u, Ev = Vv P and so (U + V )Ev − EVv =
(U + V )(Vv P − Vv P ) = 0. Similarly, GUu − (U + V )Gu = 0.
9.2.14 The first part can be verified directly. Alternatively, this parametrization can be
deduced from that in (5.12) by putting a = p2 − u, b = q 2 − u, c = r 2 − u and
then reparametrizing by ũ = u − v, ṽ = u − w.
Straightforward algebra gives the first fundamental form as
vdv 2
udu2
.
(u − v)
−
(a + u)(b + u)(c + u) (a + v)(b + v)(c + v)
Thus, the quadric is a Liouville surface with U = u, V = −v. The result now
follows from the preceding exercise.
9.3.1 They are normal sections.
2
2
9.3.2 (i) Let the spheroid be obtained by rotating the ellipse xp2 + zq 2 = 1 around
the z-axis, where p, q > 0. Then, p is the maximum distance of a point of the
spheroid from the z-axis, so the angular momentum Ω of a geodesic must be ≤ p
(we can assume that Ω ≥ 0). If Ω = 0, the geodesic is a meridian. If 0 < Ω < a,
the geodesic is confined
to the annular region on the spheroid contained between
q
2
the circles z = ±q 1 − Ωp2 , and the discussion in Example 9.3.3 shows that the
geodesic ‘bounces’ between these two circles (see diagram below).
94
If Ω = p, Eq. 9.10 shows that the geodesic must be the parallel z = 0.
(ii) Let the torus be as in Exercise 4.2.5. If Ω = 0, the geodesic is a meridian (a
circle). If 0 < Ω < a − b, the geodesic spirals around the torus. If Ω = a − b,
the geodesic is either the parallel of radius a − b or spirals around the torus
approaching this parallel asymptotically (but never crossing it):
0 <Ω< a−b
Ω=a−b
If a − b < Ω < a + b, the geodesic is confined to the annular region consisting
of the part of the torus a distance ≥ Ω from the axis, and bounces between the
two parallels which bound this region:
95
If Ω = a + b, the geodesic must be the parallel of radius a + b.
q
9.3.3 The two solutions of Eq. 9.14 are v = v0 ± Ω12 − w2 , so the condition for a
q
self-intersection is that, for some w > 1, 2 Ω12 − w2 = 2kπ for some integer
q
k > 0. This holds ⇐⇒ 2 Ω12 − 1 > 2π, i.e. Ω < (1 + π 2 )−1/2 . In this case, there
q
are k self-intersections, where k is the largest integer such that 2kπ < 2 Ω12 − 1.
9.3.4 From the solution to Exercise 8.3.1, Z = U + iV = z−i
z+i , where z = v + iw.
This is a Möbius transformation, so it takes lines and circles to lines and circles
and preserves angles (Appendix 2). Since the geodesics on the pseudosphere
correspond to straight lines and circles in the vw-plane perpendicular to the vaxis, they correspond in the V W -plane to straight lines and circles perpendicular
to the image of the V -axis under the transformation z 7→ z−i
z+i , i.e. the unit circle
2
2
V + W = 1.
A straight line aV̄ + bW̄ = c in the V̄ W̄ -plane (where a, b, c are constants)
corresponds to the curve 2aV + 2bW = c(V 2 + W 2 + 1) in the V W -plane. If
c = 0 this is a straight line through the origin, which corresponds to a geodesic
on the pseudosphere by the first part. If c 6= 0 it is the equation of a circle with
centre (a/c, b/c) and squared radius (a2 + b2 − c2 )/c2 . This circle intersects the
boundary circle V 2 + W 2 = 1 orthogonally because the square of the distance
between the centres of the two circles is equal to the sum of the squares of their
radii. Hence this circle also corresponds to a geodesic on the pseudosphere. This
proves that every straight line in the V̄ W̄ -plane corresponds to a geodesic on the
pseudosphere. That every geodesic on the pseudosphere arises from a straight
line in the V̄ W̄ -plane in this way can be proved by similar arguments, or by
noting that there is a straight line in the V̄ W̄ -plane passing through any point
of the disc V̄ 2 + W̄ 2 < 1 in any direction and using Proposition 9.2.4.
9.3.5 By Proposition 9.3.1(ii), a surface of revolution for which every parallel is a
geodesic would have df /du = 0 for all values of u, and hence f = constant.
96
Then the surface would be a circular cylinder.
9.3.6 By Exercise 6.1.6(iii), the first fundamental form is cosh2 u(du2 + dv 2 ), so the
result is a special case of Exercise 9.2.11.
9.3.7 By Clairaut’s Theorem 9.3.2, ρ is constant along every geodesic. It follows that
ρ is actually a constant (and hence the surface is a circular cylinder). For if not,
then dρ/dz 6= 0 for some value of z, say z = z0 . But then a geodesic passing
through a point on the parallel z = z0 and not tangent to that parallel would
have non-constant ρ, contrary to hypothesis.
9.3.8 By Clairaut’s Theorem, a geodesic on the unit cylinder (for which ρ is a constant)
must have ψ = constant, i.e. the geodesic must intersect the meridians (i.e. the
rulings) of the cylinder at a constant angle. By Exercise 4.1.6, the curves on the
cylinder that have this property are the straight lines (rulings), circles (parallels)
and helices.
9.3.9 The first fundamental form is du2 + k 2 cos2 udv 2 . By Clairaut’s Theorem,
(k 2 cos2 u)v̇ = Ω,
where Ω is a constant. As in Exercise 9.2.11, we find that, along any geodesic
that is not a parallel,
dv
du
2
=
Ω2
.
k 2 cos2 u(k 2 cos2 u − Ω2 )
The geodesic γ makes an angle α with the parallel u = 0 at the point σ (0, 0) =
(k, 0, 0). Hence,
γ̇γ .σ v
= cos α,
k
which gives v̇ = cos α/k at (k, 0, 0). On the other hand, v̇ = Ω/k 2 cos2 u = Ω/k 2
at (k, 0, 0). Hence, Ω = k cos α and we get
dv
du
2
=
cos2 α
.
k 2 cos2 u(cos2 u − cos2 α)
This gives
k
Integrating,
dv
sec2 u
= ±√ 2
.
du
tan α − tan2 u
kv = ± sin
−1
tan u
tan α
,
97
and hence tan u = ± tan α sin kv. Since | sin kv| ≤ 1, the maximum value of u
along the geodesic is α, so the maximum height above the xy-plane is
Z
α
0
p
1 − k 2 sin2 θ dθ.
9.3.10 (i) We note that all geodesics γ (t) on the pseudosphere, except meridians, are
defined only for t in some finite interval α ≤ t ≤ β, say, whereas meridians are
defined on a semi-infinite interval. Since f (γγ (t)) is a geodesic defined on the
same interval as γ (t), it follows that f takes meridians to meridians, i.e. if v is
constant, so is ṽ. Hence, ṽ does not depend on w.
(ii) f preserves angles and takes meridians to meridians, so must take parallels
to parallels. Hence, w̃ does not depend on v.
(iii) The parallel w = constant has length 2π/w by Exercise 8.3.2(i) (w = e−u ).
As f preserves lengths, part (ii) implies that 2π/w = 2π/w̃, so w = w̃.
σ(u, v)) = σ (F (v), w)for some smooth (iv) We now know that f (σ
function F (v).
2
The first fundamental form of σ (F (v), w) is w−2 dF
dv 2 + dw2 ; since f is an
dv
isometry, this is equal to w−2 (dv 2 + dw2 ), hence dF/dv = ±1, so F (v) = ±v + α,
where α is a constant. If the sign is +, f is rotation by α around the z-axis; if
the sign is −, f is reflection in the plane containing the z-axis making an angle
α/2 with the xz-plane.
9.4.1 From Exercise 6.2.1, the cone is√ isometric to the ‘sector’ S of the plane with
vertex at the origin and angle π 2:
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Geodesics on the cone correspond to possibly broken line segments in S: if a line
segment meets the boundary of S at a point A, say, it may continue from the
point B on the other boundary line at the same distance as A from the origin
and with the indicated angles being equal:
98
B
O
A
(i) TRUE: if two points P and Q can be joined by a line segment in S there is no
problem; otherwise, P and Q can be joined by a broken line segment satisfying
the conditions above:
q1
Q
q2
O
p1
S
R
P
p2
To see that this is always possible, let p1 , p2 , q1 and q2 be the indicated distances,
and let R and S be the points on the boundary of the sector at a distance
(p2 q1 + p1 q2 )/(p2 + q2 ) from the origin. Then, the broken line segment joining
P and R followed by that joining S and Q is the desired geodesic.
(ii) FALSE:
Q
P
(iii) FALSE: many meet in two points, such as the two geodesics joining P and
Q in the diagram in (ii).
(iv) TRUE: the meridians do not intersect (remember that the vertex of the cone
has been removed), and parallel straight lines that are entirely contained in S
99
do not intersect.
(v) TRUE: since (broken) line segments in S can clearly be continued indefinitely
in both directions.
(vi) TRUE: a situation of the form
O
in which the indicated angles are equal is clearly impossible. But the answer
to this part of the question depends on the angle of the cone: if the angle is α,
instead of π/4, lines can self-intersect if α < π/6, for then the corresponding
sector in the plane has angle < π:
O
9.4.2 We consider the intersection of S 2 with the plane passing through p and q and
making an angle θ with the xy-plane, where −π/2 < θ < π/2. This intersection
is a circle Cθ but it is not a great circle unless θ = 0. Hence, if θ 6= 0, the short
segment of Cθ joining p and q is not a geodesic and so has length > π/2 (the
length of the shortest geodesic joining p and q). Since the length of Cθ is ≤ 2π,
the length of the long segment of Cθ joining p and q has length < 3π/2 if θ 6= 0,
i.e. strictly less than the length of the long segment of the geodesic C0 joining
p and q. So the long geodesic segment is not a local minimum of the length of
curves joining p and q.
2
2
9.4.3 (i) This is obvious if n ≥ 0 since e−1/t → 0 as t → 0. We prove that t−n e−1/t →
0 as t → 0 by induction on n ≥ 0. We know the result if n = 0, and if n > 0
100
t−n
2
e1/t
we can apply L’Hopital’s rule: limt→0
= limt→0
nt−n−1
2 1/t2
3e
= limt→0
t
n t−(n−2)
2 e1/t2 ,
which vanishes by the induction hypothesis.
(ii) We prove by induction on n that θ is n-times differentiable with
dn θ
=
dtn
Pn (t) −1/t2
t3n e
0
if t 6= 0,
if t = 0,
where Pn is a polynomial in t. For n = 0,the assertion holds with
P0 = 1.
Pn′
2Pn
−3nPn
dn+1 θ
−1/t2
if t 6=
Assuming the result for some n ≥ 0, dtn+1 = t3n+1 + t3n + t3n+3 e
0, so we take Pn+1 = (2 −3nt2 )Pn +t3 Pn′ . If t = 0,
Pn (0) limt→0
−1/t2
e
t3n+1
dn+1 θ
dtn+1
= limt→0
Pn (t) −1/t2
e
t3n+1
=
= 0 by part (i). Parts (iii) and (iv) are obvious.
9.4.4 For the circular cylinder, the answers are: (i) true; (ii) false; (iii) false; (iv) true;
(v) true; (vi) false; (vii) true.
For the sphere: (i) true; (ii) false; (iii) false; (iv) false; (v) true; (vi) false; (vii)
true.
RR
σ r dr = R. Differentiating with
σ r = 1, so 0 σ r .σ
9.5.1 Since γ θ is unit-speed, σ r .σ
RR
σ rθ dr = 0, and then integrating by parts gives
respect to θ gives 0 σ r .σ
σ r |r=R
σ θ .σ
r=0
−
Z
R
σ rr dr = 0.
σ θ .σ
0
Now σ (0, θ) = p for all θ, so σ θ = 0 when r = 0. So we must show that the
integral in the last equation vanishes. But, σ rr = γ̈γ θ , the dot denoting the
derivative with respect to the parameter r of the geodesic γ θ , so σ rr is parallel
N = 0, it follows that σ θ .σ
σ rr = 0. The first
to the unit normal N of σ ; since σ θ .N
σ r = 1 and σ r .σ
σ θ = 0.
fundamental form is as indicated since σ r .σ
R1q
R1q
2
2
˙
γ
p
q
9.5.2 (i) The length of the part of between and is 0 f + Gġ dt ≥ 0 f˙2 dt =
f (1) − f (0) = R.
(ii) Use the hint, noting that the length of the part of γ between p and q′ is ≥ R.
(iii) If the part of γ between p and q has length R, then it must stay inside
theqgeodesic circle with
centre p and radius R by (ii), and then we must have
R1q
R1
f˙2 + Gġ 2 dt =
f˙2 dt. Then Gġ = 0 for all t ∈ (0, 1), so ġ = 0 (as G > 0)
0
0
and so g is a constant which must be α as γ passes through q. This means that
γ is a parametrization of the radial line θ = α.
9.5.3 The first fundamental form is dr 2 + Gdθ 2 and a geodesic circle is a parameter
curve r = constant. By Exercise 7.3.15, its geodesic curvature is Gr /2G. We are
101
given that this is a function of r only, say A(r). Then,
∂
(ln G) = A(r)
∂r
Z
ln G =
∴
∴
R
where f (r) = e
A(r)dr
A(r)dr + B(θ) (say)
G = f (r)g(θ),
and g(θ) = eB(θ) .
Chapter 10
σ u , σ v } is FI−1 FII =
10.1.1 The matrix of the Weingarten map with respect to the basis {σ
−1 cos2 v 0
− cos2 v 0
=
= −I, so Nu = σ u , Nv = σ v . Thus, N =
0
1
0
−1
σ −aa, where a is a constant vector. Hence, k σ −aa k = 1, showing that the surface
is an open subset of the sphere S of radius 1 and centre a. The standard latitudelongitude parametrization σ (u, v) of S 2 has first and second fundamental forms
both given by du2 + cos2 udv 2 , so the parametrization σ (v, u) + a of S has the
given first and second fundamental forms (the second fundamental form changes
σ u × σ v ).
sign because σ v × σ u = −σ
10.1.2 Γ122 = sin u cos u and the other Christoffel symbols are zero; the second Codazzi–
Mainardi equation is not satisfied.
10.1.3 The Christoffel symbols are Γ111 = 0, Γ211 = 1/w, Γ112 = −1/w, Γ212 = 0, Γ122 = 0,
Γ222 = −1/w. Using the first equation in Proposition 10.1.2 we get K = −1.
The Codazzi-Mainardi equations are Lw = −(L + N )/w, Nv = 0. Hence, N
depends only on w, and since −1 = K = LN/EG, we have LN = −1/w4 so
L also depends only on w; the first Codazzi-Mainardi equation gives dL/dw =
−L/w + 1/Lw5 , which is the stated differential equation. Putting P = Lw2 we
2
2
get dP/dw = (1 + P 2 )/wP
√ which integrates to give 1 + P = Cw , where C > 0
is a constant, i.e. L = ± Cw2 − 1/w. Hence, the second fundamental form is
only defined for w ≥ C −1/2 or w ≤ −C −1/2 .
The first fundamental form is this exercise is the same as that of a suitable
parametrization of the pseudosphere (see Exercise 8.3.1(i)). We saw that the
pseudosphere corresponds to (part of) the region w > 1.
10.1.4 The Christoffel symbols are Γ111 = Eu /2E, Γ211 = −Ev /2G, Γ112 = Ev /2E,
Γ212 = Gu /2G, Γ122 = −Gu /2E, Γ222 = Gv /2G. The first Codazzi–Mainardi
equation is
LEv
1
−Ev
L
N
Lv =
= Ev
,
−N
+
2E
2G
2
E
G
102
and similarly for the other equation. Finally,
LEv
Ev
N
Ev N
L
Ev L
− 2 =
=
+
−
(κ2 − κ1 ),
(κ1 )v =
2E E
G
E
2E G
E
2E
and similarly for (κ2 )u .
10.1.5 If E, F, G, L, M, N are constant, the Christoffel symbols are all zero so Eqs. 10.1
are obviously satisfied, and the Gauss equations (Proposition 10.1.2) are satisfied
⇐⇒ K = 0, i.e. LN = M 2 . Of course, we must also have E > 0, G > 0 and
EG − F 2 > 0.
a b
E F
L M
Let C be the constant matrix
, and let FI =
, FII =
.
c d
F G
M N
By Exercises 6.1.4 and 7.1.3, the reparametrization (u, v) 7→ (ũ, ṽ) leads to the
stated first and second fundamental forms if and only if
κ 0
t
t
C FI C = I, C FII C =
.
0 0
To prove the existence of an invertible matrix C satisfying these conditions,
note first that since
FI is symmetric, there is an orthogonal matrix P such
λ
0
that P t FI P =
, where λ, µ are the eigenvalues of FI (see Appendix
0 µ
0). Since the trace E + G and the determinant EG − F 2 of FI are both > 0,
it follows that the sum and product
−1/2 of λ, µ are
both > 0, and hence that λ, µ
λ
0
are both > 0. Letting A =
, we have At P t FI P A = I. The
0
µ−1/2
matrix At P t FII P A issymmetric,
so there is an orthogonal matrix Q such that
κ
0
1
Qt (At P t FII P A)Q =
, for some (constants) κ1 , κ2 . The matrix C =
0 κ2
κ
0
1
P AQ has the property that C t FI C = I, C t FII C =
.
0 κ2
By Proposition 8.2.6, the principal curvatures are the roots of
det(FII − κFI ) = 0.
Since C is invertible, these are the same as the roots of
det(C t FII C − κC t FI C) = 0,
i.e. κ1 , κ2 . Since K = κ1 κ2 , one of κ1 , κ2 is zero,
and we might as well assume
0 1
that κ2 = 0 (if κ1 = 0, we can replace C by C
). This gives the stated
1 0
result, with κ1 = κ.
103
If κ = 0, the surface has the same first and second fundamental forms as a plane,
so by Theorem 10.1.3 the surface can be obtained from a plane by applying a
direct isometry of R3 . It follows that the surface is a plane.
If κ 6= 0, then by the reparametrization v 7→ |κ|−1/2 v we can assume that κ = ±1.
By Examples 6.1.4, 7.1.2 and 10.1.4, in both cases there is a parametrization of
the unit cylinder with these first and second fundamental forms. By Theorem
10.1.3 again, the surface is obtained by applying to this cylinder a direct isometry
of R3 , and so is a circular cylinder.
Ev
u
v
u
, Γ211 = − 2G
, Γ112 = E
, Γ212 = G
,
10.1.6 (i) The Christoffel symbols are Γ111 = E
2E
2E
2G
Gv
Gu
2
1
Γ22 = − 2E , Γ22 = 2G . Eqs. 10.1 (with L = N = 0, M = 1) now state that
(E/G)u = 0 and (E/G)v = 0. Hence, E/G is constant.
(ii) If E/G = λ4 , the reparametrization ũ = λu, ṽ = λ−1 v leads to the first
and second fundamental forms Ẽ(dũ2 + dṽ 2 ) and 2dũdṽ, respectively, where
Ẽ = λ−2 E.
(iii) If E = G, the first of the Gauss equations in Proposition 10.1.2 gives
1
− =−
E
i.e.
Ev
2E
v
−
Eu
2E
+
u
Eu
2E
Eu
2E
2
−
Ev
2E
+
u
v
Ev
2E
=
2
+
Ev
2E
2
−
Eu
2E
2
,
2
,
E
which is the stated equation.
10.1.7 The assumption is that L = N = 0. The Codazzi-Mainardi equations are
−Mu = M (Γ212 − Γ111 ),
Mv = M (Γ222 − Γ112 ).
10.1.8 (i) By Proposition 8.4.1, there is a patch σ (U, V ) of S containing p such that
F = M = 0. If κ1 = L/E is a constant κ, Exercise 10.1.4 implies that EV = 0
(since the principal curvatures
are distinct by assumption). So E is a function
R√
of U only. Define u =
E dU , v = V . Then the first and second fundamental
2
forms are of the form du + Gdv 2 and κdu2 + N dv 2 , respectively.
(ii) The Christoffel symbols are Γ111 = Γ211 = Γ112 = 0, Γ212 = Gu /2G, Γ122 =
−Gu /2G, Γ222 = Gv /2G.
(iii) The parameter curve v = constant is parametrized by Γ(u) = σ (u, v). DeΓ = σ u , Γ̈
Γ = σ uu ...
N, by Proposition 7.4.4
noting d/du by a dot, we have Γ̇
= κN
Nu = −κ2 σ u . Since
and part√(ii). By the proof of Proposition 8.1.2, Γ = κN
σ u , σ v / G, N} is a right-handed orthonormal basis of R3 ,
{σ
Γ × Γ̈
Γk
κ
k Γ̇
= k √ σ v k = |κ|.
3
Γk
G
k Γ̇
104
The torsion of Γ is
...
√
Γ .(Γ̇
Γ × Γ̈
Γ)
σ v / G)
(−κ2 σ u )..(−κσ
= 0,
=
κ2
Γ × Γ̈
Γ k2
k Γ̇
since σ u.σ v = 0. It follows that Γ is a circle of radius r = 1/|κ|.
Nu = −κ2 σ u , so σ uu + r12 u is a function of v only. Hence, σ (u, v) =
(iv) σ uuu = κN
γ (v) + C (v) cos ur + D(v) sin ur for some smooth functions C , D, γ of v. We may
assume that the curve γ is unit-speed, since replacing v by a smooth function
of v does not change the coefficients E, F, L, M (as F = M = 0). By (iii),
k σ (u, v) k2 = r 2 for all u, v. This implies that C and D are perpendicular
vectors of length r. Let c = r −1C , d = r −1D.
(v) Using the fact that c(v) and d(v) are perpendicular unit vectors for all values
of v, we get
dcc
u
u
dγγ + r .d
. −cc sin + d cos
σ u .σ v =
dv
r
r
dv
dγ
dγ
for all u, v. Hence,
.c =
.d = 0 (= dc .d). It follows that, for some angle α
dv
dv
(possibly depending on v), we have
dv
d = −n
n sin α + b cos α,
c = n cos α + b sin α, ±d
where n and b denote the principal normal and binormal of γ , respectively.
dγ
(The condition dv .d = 0 implies that dα/dv = −τ , where τ is the torsion of γ .)
Writing s = v, θ = α ± κu shows that σ is a reparametrization of the tube of
radius r around γ (Exercise 4.2.7).
10.1.9 Since p−1Σ u , q −1Σ v and r −1Σ w are perpendicular unit vectors,
Σu + bΣ
Σv + cΣ
Σw ,
Σuu = aΣ
for some scalars a, b, c. We have
p2 a = Σ uu.Σ u =
1
1
Σu.Σ u )u = (p2 )u = ppu ,
(Σ
2
2
so a = pu /p. Next,
1
Σu.Σ v )u − Σ u.Σuv = − (Σ
Σu.Σu )v = −ppv ,
q 2 b = Σuu.Σv = (Σ
2
so b = −ppv /q 2 . Similarly, c = −ppw /r 2 , hence the stated formula for Σuu . The
other formulas for the second derivatives of Σ are proved similarly.
Σuu )v =
We now use these formulas to compute both sides of the equation (Σ
Σuv )u ; both sides are then linear combinations of Σu , Σ v and Σw . We find that
(Σ
105
the coefficients of Σu are automatically equal, while equating coefficients of Σv
and Σw on the two sides yields the third and fourth of Lamé’s equations. The
others are proved similarly.
10.2.1 By Corollary 10.2.3(i),
1
K=− λ
2e
∂
∂u
Ẽ
F̃
(eλ )u
eλ
F̃
G̃
∂
+
∂v
(eλ )v
eλ
E
F
F
G
t
=−
1
(λuu + λvv ).
2eλ
∂r
∂u
∂θ
∂u
∂r
∂v
∂θ
∂v
=
10.2.2 By Exercise 6.1.4,
= J
J, where J =
u
v
r
r
. By Exercise 9.5.1, E = 1, F = 0, and we get the stated foru
−v
r2
r2
2
2
mulas for Ẽ, F̃ , G̃. From Ẽ − 1 = vr2 rG2 − 1 , G̃ − 1 = ur2 rG2 − 1 , we get
u2 (Ẽ − 1) = v 2 (G̃ − 1). Since Ẽ and G̃ are smooth functions of (u, v), they have
P
P
i j
2
i j
2
Taylor expansions Ẽ =
i+j≤2 eij u v + o(r ), G̃ =
i+j≤2 gij u v + o(r ),
where o(r k ) denotes terms such that o(r k )/r k → 0 as r → 0. Equating coefficients on both sides of u2 (Ẽ − 1) = v 2 (G̃ − 1) shows that all the e’s and g’s
are zero except e02 = g20 = k, say. Then, Ẽ = 1 + kv 2 + o(r 2 ), √which implies
2
that G = r 2 + kr 4 + o(r 4 ). By Corollary 10.2.3(ii), K = − √1G ∂∂r2G . From the
√
first part, G = r + 21 kr 3 + o(r 3 ), hence K = −3k + o(1). Taking r = 0 gives
p) = −3k.
K(p
R 2π
R 2π √
R 2π
p)R3 + o(R3 ) dθ =
G dθ = 0 R − 16 K(p
10.2.3 (i) CR = 0 k σ θ k dθ = 0
p)R3 + o(R3 ) .
2π R − 16 K(p
√
R R R 2π √
G drdθ is equal to
(ii) Since dAσ = G drdθ, the area AR = 0 0
2π
Z
0
R
p) 2
1
K(p
3
3
2
2
p)r + o(r ) dr = πR 1 −
r − K(p
R + o(R ) .
6
12
If S = S 2 , Exercise 6.5.3 gives CR = 2π sin R = 2π(R = 61 R3 + o(R3 )), AR =
1
R4 + o(R4 )). Since K = 1 these formulas agree
2π(1 − cos R) = 2π( 21 R2 − 24
with those in (i) and (ii).
10.2.4 (i) Let s be the arc-length of γ , so that ds/dθ = λ, and denote d/ds by a dot. The
first of the geodesic equations (9.2) applied to γ gives r̈ = 12 Gr θ̇ 2 . Since r = f (θ),
′
this gives λ1 λ1 f ′ = 2λ1 2 Gr . This simplifies to give the stated equation.
(ii) Since σ r and γ̇γ are unit vectors, cos ψ = σ r .γ̇γ = λ1 σ r .(f ′σ r + σ θ ) = f ′ /λ.
′ ′
√
√
σ r ×σ
σ θ ) = λG N, so sin ψ = G/λ. Hence, fλ = −ψ ′ sin ψ =
Also, σ r ×γ̇γ = λ1 (σ
√
√
′ ′
∂ G
=
−
.
− λG ψ ′ , and so ψ ′ = − √1G f ′′ − f λλ = − 2√1G ∂G
∂r
∂r
106
(iii) Using the formula for K in Corollary 10.2.3(ii) and the expression for the
first fundamental form of σ in Exercise 9.5.1, we get
ZZ
T
√
1 ∂2 G √
−√
G drdθ
KdAσ =
G ∂r 2
0
0
√
Z α
Z α √ r=f (θ)
∂ G
∂ G
′
dθ =
ψ +
=−
∂r
∂r
0
0
Z
α
Z
f (θ)
r=0
r=0
!
dθ.
√
√
By Exercise 10.2.2,
G = r +o(r) so ∂ G/∂r = 1 at r = 0, where o(r)/r → 0 as
RR
r → 0. Hence, ABC KdAσ = ψ(α) −ψ(0) +α = γ −(π −β) +α = α +β +γ −π.
(
6
A)
C
(0)
B
A
10.2.5 With the notation of Example 4.5.3 we have, on the median circle t = 0,
θ
θ
θ
σ t = − sin cos θ, − sin sin θ, cos
, σ θ = (− sin θ, cos θ, 0),
2
2
2
hence E = 1, F = 0, G = 1 and N = − cos 2θ cos θ, − sin 2θ sin θ, − sin 2θ ;
σ tt = 0, σ tθ =
θ
θ
1
θ
θ
1
θ
1
− cos cos θ + sin sin θ, − cos sin θ − sin cos θ, − sin
2
2
2
2
2
2
2
2
giving L = 0, M = 21 . Hence, K = (LN −M 2 )/(EG−F 2 ) = −1/4. Since K 6= 0,
the Theorema Egregium implies that the Möbius band is not locally isometric
to a plane.
10.2.6 The catenoid has first fundamental form cosh2 u(du2 + dv 2 ) and its Gaussian
curvature is K = −sech4 u (Exercise 8.1.2). If f is an isometry of the catenoid, let
σ(u, v)) = σ (ũ, ṽ). By the Theorema Egregium, sech4 u = sech4 ũ, so ũ = ±u;
f (σ
reflecting in the plane z = 0 changes u to −u, so assume that the sign is +. Let
ṽ = f (u, v); the first fundamental form of σ (u, f (u, v)) is
(cosh2 u + fu2 )du2 + 2fu fv dudv + fv2 cosh2 udv 2 .
,
107
Hence, cosh2 u = cosh2 u + fu2 , fu fv = 0 and fv2 cosh2 u = cosh2 u. So fu = 0,
fv = ±1 and f = ±v + α, where α is a constant. If the sign is + we have a
rotation by α about the z-axis; if the sign is − we have a reflection in the plane
containing the z-axis and making an angle α/2 with the xz-plane.
10.2.7 Using Corollary 10.2.3(i), we find that
1
K=
4
m(2 − m) n(2 − n)
+
un v 2
u2 v m
.
Hence, the surface is flat if and only if (m, n) = (0, 0), (2, 0), (0, 2) or (2, 2).
If m = n = 0 the first fundamental form is du2 + dv 2 so the surface is locally
isometric to a plane (Corollary 6.2.3). If (m, n) = (2, 0), we get v 2 du2 + dv 2 ,
which is the first fundamental form pf a cone (Example 6.1.5), and this is locally
isometric to a plane by Exercise 6.2.5. Similarly if (m, n) = (0, 2). The case
in which m = n = 2 is less obvious, but the reparametrization U = 2−1/2 uv,
V = ln uv transforms the first fundamental form v 2 du2 +u2 dv 2 into dU 2 +U 2 dV 2 ,
which is the first fundamental form of a cone.
10.2.8 Using the Christoffel symbols calculated in Exercise 7.4.4, the second of the
Gauss equations (Proposition 10.1.2) gives
K cos θ = 0 − (θu cot θ)v + 0 − (−θu cosec θ)(−θv cosec θ) = −θuv cot θ,
hence the result.
10.2.9 By Exercise 6.2.5 and the Theorema Egregium, K = 0 for a generalized cylinder
and a generalized cone. But K 6= 0 for a sphere so by the Theorema Egregium
again, there can be no local isometry between a sphere and a generalized cylinder
or cone.
σ are
10.2.10 The first fundamental forms of σ and σ̃
1
1 + 2 du2 + u2 dv 2 and du2 + (u2 + 1)dv 2 ,
u
σ (u, v) is not an
respectively. Since these are different, the map σ (u, v) 7→ σ̃
isometry. Nevertheless, by Corollary 10.2.3(i), the Gaussian curvatures are equal:
∂
1
K = K̃ = − √
2 u2 + 1 ∂u
2u
√
1 + u2
=−
1
.
(1 + u2 )2
1
σ (ũ, ṽ) is an isometry, the Theorema Egregium tells us that − (1+u
If σ (u, v) 7→ σ̃
2 )2
1
= − (1+ũ2 )2 , so ũ = ±u; let ṽ = f (u, v). The first fundamental form of
σ (±u, f (u, v)) is (1 + (1 + u2 )fu2 )du2 + 2(1 + u2 )fu fv dudv + (1 + u2 )fv2 dv 2 ; this
σ̃
108
is equal to the first fundamental form of σ (u, v) ⇐⇒ 1 + (1 + u2 )fu2 = 1 + 1/u2 ,
fu fv = 0 and (1 + u2 )fv2 = u2 . The middle equation gives fu = 0 or fv = 0, but
these are both impossible by the other two equations. Hence, the isometry does
not exist.
10.2.11 From the solution of Exercise 8.1.9, the Gaussian curvature of the torus (in the
parametrization of Exercise 4.2.5) is
K=
cos θ
1
a
= 2− 2
.
b(a + b cos θ)
b
b (a + b cos θ)
If σ (θ, ϕ) 7→ σ (θ̃, ϕ̃) is an isometry of the torus, the Theorema Egregium implies
that we must have cos θ = cos θ̃, and hence θ̃ = ±θ (up to adding integer
multiples of 2π). Since the first fundamental form is b2 dθ 2 + (a + b cos θ)2 dϕ2 ,
we must therefore have dϕ̃2 = dϕ2 , and hence ∂ ϕ̃/∂θ = 0, ∂ ϕ̃/∂ϕ = ±1. Hence,
ϕ̃ = ±ϕ + constant.
10.2.12 Since all parameter curves are pre-geodesics we have, by Exercise 9.2.8,
EEv + F Eu = 2EFu , GGu + F Gv = 2GFv .
If the parameter curves intersect orthogonally, F = 0 so these equations give
2
2
Ev = Gu = 0,
p the first fundamental
R p form is E(u)du +G(v)dv . ReparametrizR so
E(u)du, V =
G(v)dv, the first fundamental form becomes
ing by U =
dU 2 + dV 2 . Hence, the surface is locally isometric to the plane, and so is flat by
the Theorema Egregium.
The result is not true without the assumption of orthogonality. Example 10.4.3
gives a parametrization σ (u, v) of S 2 (actually of a hemisphere) with the property
that the pre-geodesics on S 2 correspond exactly to the straight lines in the uvplane. In particular, all the parameter curves are pre-geodesics.
10.3.1 Arguing as in the proof of Theorem 10.3.4, we suppose that J attains its maximum value > 0 at some point p ∈ S contained in a patch σ of S. We can assume
that the principal curvatures κ1 and κ2 of σ satisfy κ1 > κ2 > 0 everywhere.
Since H = 12 (κ1 + κ2 ), κ1 > H and J = 4(κ1 − H)2 . Thus, J increases with κ1
when κ1 > H, so κ1 must have a maximum at p, and then κ2 = 2H − κ1 has
a minimum there. By Lemma 10.3.5, K ≤ 0 at p, contradicting the assumption
that K > 0 everywhere.
10.3.2 We start with the parametrization
σ (U, V ) = (f (U ) cos V, f (U ) sin V, g(U )), where
R√
U
2U
1 − e dU . The first and second fundamental forms are
f (U ) = e , g(U ) =
√
U
−e
2
2U
2
dU + e dV and √ 2U dU 2 + eU 1 − e2U dV 2 , respectively. In the notation
1−e
√
√
of the proof of Proposition 10.3.2, κ1 = −1/eU√ 1 − e2U , κ2 = e−U 1 − e2U . So
we are in case (ii) of the proof and tan ω = e−2U − 1. We find that e(U ) =
109
E/ sin2 ω =
1
1−e2U
, g(V ) = G sec2 ω = 1. So Ṽ = V and Ũ =
R
√ dU
1−e2U
=
− cosh−1 (e−U ) − c for some constantqc. So U = − ln(cosh(Ũ + c)). Hence, θ =
√
2ω = 2 tan−1 e−2U − 1 = 2 tan−1 cosh2 (Ũ + c) − 1 = 2 tan−1 sinh(Ũ + c).
Finally, u = 12 (Ũ + Ṽ ), v = 12 (Ṽ − Ũ ), so Ũ = u − v.
√
10.3.3 We have E = G = 1, F = cos θ and θuv = sin θ. Hence, EG − F 2 = θuv .
Consider the quadrilateral bounded by the parameter curves u = u0 , u = u1 ,
v = v0 and v = v1 , and let
α1 be the angle between the parameter curves u = u0 and v = v0 ,
α2 be the angle between the parameter curves u = u0 and v = v1 ,
α3 be the angle between the parameter curves u = u1 and v = v0 ,
α4 be the angle between the parameter curves u = u1 and v = v1 .
The area of the quadrilateral is
Z
v1
v0
Z
u1
θuv dudv =
u0
Z
v1
v0
(θv (u1 , v) − θv (u0 , v))dv
= θ(u1 , v1 ) − θ(u1 , v0 ) − θ(u0 , v1 ) + θ(u0 , v0 )
= α4 − (π − α3 ) − (π − α2 ) + α1
= α1 + α2 + α3 + α4 − 2π.
10.4.1 Let f : S → S̃ be a local diffeomorphism that takes unit-speed geodesics to unitspeed geodesics. Let p ∈ S and let 0 6= v ∈ Tp S. There is a unique geodesic γ (t)
on S such that γ (0) = p and γ̇γ (0) = v/ k v k. Then γ is unit-speed so γ̃γ = f ◦ γ
v/ k v k) is a unit vector, i.e.
is a unit-speed geodesic. In particular, γ̃γ˙ = Dp f (v
v) k = k v k. This means that Dp f : Tp S → Tf (p) S̃ is an isometry, so f
k Dp f (v
is a local isometry.
10.4.2 Local isometries take geodesics to geodesics by Corollary 9.2.7. If we apply a
v, where a 6= 0 is a constant, to a surface, the first fundamental
dilation v 7→ av
form is multiplied by a2 so the Christoffel symbols are unchanged (see Proposition 7.4.4). By Proposition 9.2.3, the geodesic equations are unchanged. It
follows that dilations take geodesics to geodesics. Hence, any composite of local
isometries and dilations also takes geodesics to geodesics.
The converse is false: the map from the xy-plane to itself given by (x, y) 7→ (x, 2y)
takes geodesics to geodesics (as it takes straight lines to straight lines) but is not
the composite of a dilation and a local isometry.
10.4.3 (i) This is true because F is conformal.
110
(ii) The parameter curve u 7→ σ (u, v0 ) is a geodesic on σ for any fixed v0 by
construction of the geodesic patch σ . Since F is a geodesic local diffeomorphism,
σ (u, v0 )) is a pre-geodesic on σ̃
σ . Hence, for some smooth function u(t),
u 7→ F (σ
σ (u(t), v0 )) is a geodesic on σ̃
σ . The second geodesic equation in Theorem
t 7→ F (σ
2
9.2.1 gives λv u̇ = 0, so λv = 0 and λ is independent of v.
(iii) Let γ (t) = σ (u(t), v(t)); we can assume that γ is unit-speed. Since the first
fundamental form of σ is du2 + Gdv 2 , the parameter curves v = constant and
u =√
constant intersect orthogonally and unit vectors parallel to them are σ u and
σ v / G, respectively. If the oriented angle between γ̇γ and the curve v = constant
σ u + G−1/2 sin θσ
σ v = u̇σ
σ u + v̇σ
σ v . Hence, u̇ = cos θ and
is θ, we have γ̇γ = cos θσ
sin
θ
v̇ = √G . The first geodesic equation in Theorem 9.2.1 gives ü = 12 Gu v̇ 2 , i.e.
√
dθ
u v̇
θ̇ sin θ = 21 Gu v̇ 2 . Substituting for v̇ gives dv
= v̇θ̇ = 2Gsin
= Gu /2 G.
θ
(iv) Apply (iii) to F ◦ γ and use the fact that F is conformal.
(v) Parts (iii) and (iv) imply (λG)u = λGu , hence λu G = 0, hence λu = 0, i.e.
λ is independent of u. By (ii), λ is constant.
(vi) If Dλ−1/2 is the dilation by a factor λ−1/2 , the composite Dλ−1/2 ◦F preserves
the first fundamental form and so is a local isometry, say G. Then, F = Dλ1/2 ◦G.
Chapter 11
11.1.1 Let l meet the real axis at b and suppose that Re(a) > b (the case Re(a) < b is
similar). The semicircle with centre c on the real axis and radius |a − d| passes
through a and does not meet l provided that |a − d| ≤ |d − b|, i.e. provided
d ≥ (|a|2 − b2 )/(Re(a) − b).
11.1.2 Suppose that a and b lie on a half-line geodesic,
say a = r + is, b = r + it, where
Rt
d
=
ln(t/s) = d, say, so t/s
r, s, t ∈ R and t > s. Then, dH (a, b) = s dw
w
=e .
On the other hand, the formula in Proposition 11.1.4 gives 2tanh−1
d 2tanh−1 eed −1
= 2tanh−1 (tanh d2 ) = d.
+1
t−s
t+s
=
11.1.3 The required hyperbolic line cannot be a half-line, so must be a semicircle with
centre the origin, and it must have radius |a|.
11.1.4 By Proposition 11.1.4, z ∈ Ca,R ⇐⇒ 2tanh−1
2
2
2
z−a
z−ā
= R. If λ = tanh(R/2),
this is equivalent to (1 − λ )|z| − (ā − λ a)z − (a − λ2 ā)z̄ + (1 − λ2 )|a|2 = 0.
According to Proposition A.2.3, this is the equation of a circle provided that
λ2 < 1, which is obvious, and |ā − λ2 a|2 > (1 − λ2 )2 |a|2 . This condition reduces
to 2|a|2 > a2 + ā2 . Writing a = |a|eiθ this becomes cos 2θ < 1, which is true
because a ∈ H implies 0 < θ < π.
Cic,R will be a circle with centre on the imaginary axis, say at ib. Then Ca,R
intersects the imaginary axis at the points i(b ± r), so these two points must
111
be a hyperbolic distance 2R apart, i.e. 2R = 2tanh−1 2ir
2ib . This gives r =
1
b+r
b tanhR, which is equivalent to R = 2 ln b−r . Next, the points i(b ± r) are
the same hyperbolic distance from ic, so
i(b+r)−ic
i(b+r)+ic
=
i(b−r)−ic
i(b−r)+ic
. This gives
(b + r − c)(b − r + c) = (c + r − b)(c + r + b), which simplifies to c2 = b2 − r 2 .
Parametrizing Cic,R by v = r cos
θ, w = b+r sin θ, and denoting d/dθ by a dot, the
R 2π √v̇ 2 +ẇ2
R 2π
circumference of Cic,R is 0
dθ = 0 b+rrsin θ dθ = √b2πr
= 2π sinh R.
2 −r 2
w2
R
RR
= Cic,R dv
The area inside Cic,R is int(Cic,R ) dvdw
w
w by Green’s theorem, which
R 2π −r sin θ
= 0 b+r sin θ dθ = √b2πb
− 2π = 2π(cosh R − 1).
2 −r 2
11.1.5
11.1.6
11.2.1
11.2.2
11.2.3
The circumference is 2π sinh R = 2π(R+ 61 R3 +o(R3 )) and the area is 2π(cosh R−
1
1) = 2π( 21 R2 + 24
R4 + o(R4 ))). Since K = −1 these formulas are consitent with
those in Exercise 10.2.7. (See Exercise 10.2.4 for the o( ) notation.)
Suppose that l is the imaginary axis for simplicity. If a = reiθ , with 0 < θ < π/2,
the distance of a from l is the distance of a from the point ir on l. From the
proof of Proposition 11.1.4, this distance is − ln tan θ2 . Hence, the points at a
fixed distance from l are those with a constant value of θ, i.e. the straight halflines passing through the origin. Of course, there are two such half-lines for any
given distance, each being the (Euclidean) reflection of the other in l. These
lines are not geodesics (unless the distance is zero, in which case they coincide
with l), as they are not perpendicular to the real axis.
The region {(u, v) | u < 0, −π < v < π} in the parametrization of the pseudosphere corresponds to the region {(v, w) | − π < v < π, w > 1} in H, a
semi-infinite rectangle in the upper half of the vw-plane.
By Proposition 11.2.3, there is an isometry that takes a to i and b to ir, say,
where r > 0. Since isometries leave distances unchanged, we need only prove
the result for a = i, b = ir. Assume that r > 1 (the case r < 1 is similar). Then,
dH (a, b) = ln r. On the other hand, if γ (t) = v(t) + iw(t) is any curve in H
with√γ (t0 ) = i, γ (t1 ) = ir, say, the length of the part of γ between a and b is
R t1 ẇ
R r dw
R t1 v̇ 2 +ẇ2
dt
≥
dt
=
= ln r.
w
1 w
t0 w
t0
By applying an isometry we can assume that l is a half-line, and then the result
was proved in Exercise 11.1.1.
By applying an isometry, we can assume that l is the imaginary axis. Then,
m must be the semicircle with centre the origin and radius |a|. Let a = ρeiθ ,
where ρ > 0, −π < θ < π, and let c = it, t > 0. Since tanh−1 x is a strictly
increasing function of x, we have to show that a−it
> a−iρ
if t 6= ρ. The
a+it
a+iρ
2
θ
sin θ
second expression equals 1−sin
, and the difference is 2(ρ−t)
, which is
1+sin θ
|a+it|2 1+sin θ
> 0 if t 6= ρ.
11.2.4 (i) If a ∈ H, let l′ and m′ be the unique hyperbolic lines passing through a
and perpendicular to l and m, respectively (Exercise 11.2.3). Let b and c be
112
the intersections of l′ and m′ with l and m, respectively. We are given that
F (b) = b and F (c) = c, so F (l′ ) = l′ and F (m′ ) = m′ as l′ and m′ are the unique
hyperbolic lines passing through b and c and perpendicular to l and m. Since a
is the unique point of intersection of l′ and m′ , we must have F (a) = a.
(ii) Either F or I0,1 ◦ F fixes l, m and the interior of the semicircle m. Next,
either F , I0,1 ◦ F , R0 ◦ F or R0 ◦ I0,1 ◦ F fixes l, m, the interior of the semicircle
m and the region H>0 = {z ∈ H | Re(z) > 0} to the right of l. Let G be this
isometry. Then, G fixes each point of m because there is a unique point of m at
any given distance d > 0 from i in the region H>0 . Similarly, G fixes each point
of l. Hence, G is the identity by (i).
(iii) Let F be any isometry of H. By the proof of Proposition 11.2.3, there is an
isometry G that is a composite of elementary isometries and which takes F (i)
to i and F (l) to l. Then, G ◦ F is an isometry that fixes l and i. As m is the
unique hyperbolic line intersecting l perpendicularly at i, G ◦ F fixes m. By (iii),
G ◦ F is one of four composites of elementary isometries. It follows that F is a
composite of elementary isometries.
(iv) By (iii) it suffices to prove that every elementary isometry is a composite
of reflections and inversions in lines and circles perpendicular to the real axis.
For reflections and inversions there is nothing to prove, so we need only consider
translations and dilations. But, if a ∈ R, Ta = R0 ◦R1 , where R1 is the reflection
in the line Re(z) = a/2; and if a > 0, D√a = I0 ◦ I, where I is inversion in the
circle with centre the origin and radius a.
11.2.5 (i) This is obvious from the proof of Proposition A.1.2(ii).
ad−bc
If
(ii) If a, b, c, d ∈ R, a calculation shows that Im(M (z)) = |cz+d|
2 Im(z).
ad − bc > 0 this is > 0 whenever Im(z) > 0. Conversely, suppose that M
takes H to itself. Assume that c 6= 0 and d 6= 0 (the cases in which c = 0
or d = 0 are similar but easier). Then, M must take the real axis to itself
(as it must take the lower half-plane −H to itself), i.e. az+b
cz+d ∈ R if z ∈ R.
Taking z = 0 gives b/d = λ ∈ R, say. Letting z → ∞ gives a/c = µ ∈ R,
λ−µ
say. Then, az+b
cz+d = µ + dc z+1 . This is real whenever z is real, so c/d = ν ∈ R,
say. Hence, a, b, c, d are, up to an overall multiple, equal to the real numbers
µν, λ, ν, 1, respectively. The condition ad − bc > 0 now follows from the previous
calculation.
(iii) Following the proof (and the notation) of Proposition A.2.2, we have M =
Tb/d ◦ Da/d if c = 0 (and then d 6= 0), while if c 6= 0,
M = Ta/c ◦ D(ad−bc)/c2 ◦ (−K) ◦ Td/c .
Hence, it suffices to show that −K is a composite of elementary isometries. But
−K = R0 ◦ I0,1 in the notation of Exercise 11.2.4.
(iv) J is reflection in the imaginary axis and hence an isometry of H, so the
result follows from (i).
113
(v) This follows from the fact that, if M is a real Möbius transformation, so is
J ◦ M ◦ J. For example, if M1 , M2 are real Möbius transformations, then
(M1 ◦ J) ◦ (M2 ◦ J) = M1 ◦ (J ◦ M2 ◦ J)
is a composite of real Möbius transformations, hence a real Möbius tranformation
by (i).
(vi) By (v) and Exercise 11.2.4(iii), it suffices to prove that every elementary
isometry of H is a Möbius isometry. If a ∈ R, Ta is real Möbius and Ra = T2a ◦J.
If a > 0 then Da is real Möbius. Finally, if a ∈ R, r > 0 then
Ia,r = Ta ◦ Dr2 ◦ I0,1 ◦ T−a
(see the proof of Proposition 11.2.1), so it suffices to prove that I0,1 is a Möbius
isometry; but I0,1 = (−K) ◦ J, where −K(z) = −1/z is real Möbius.
11.2.6 By Proposition 11.1.3(i), there is a unique hyperbolic line l, say, passing through
a and b. By Proposition 11.2.3, by applying a suitable isometry of H we can
assume that l is the imaginary axis. Let C and C ′ be the hyperbolic circles with
centres a and b, respectively, which pass through c (Exercise 11.1.4). These are
Euclidean circles with centres on the imaginary axis.
Suppose for a contradiction that the stated inequality is false, and let a = iw,
b = iw′ , where w′ > w (interchanging the roles of a, b if necessary). Then,
C intersects the imaginary axis at a point ir, say, with r > w; we must have
r < w′ since r ≥ w′ implies that d(a, b) ≤ d(a, ir) = d(a, c), which contradicts
our assumption. Similarly, C ′ intersects the imaginary axis at a point is, say,
between a and b. Since C and C ′ intersect, we must have s ≤ r; in fact, s < r
for if s = r then C and C ′ touch, contradicting the fact that a, b, c do not lie on
a geodesic. Then,
d(a, c) + d(c, b) = d(a, ir) + d(is, b) > d(a, ir) + d(ir, b) = d(a, b),
a contradiction.
11.2.7 Let F be an isometry of H that takes l to the imaginary axis. By Exercise
11.1.5, the equidistant curves are then a pair of Euclidean half-lines l1 , l2 passing
through the origin. The isometry F −1 of H, being a composite of reflections and
inversions (Exercise 11.2.4), takes Euclidean lines and circles to Euclidean lines
and circles (Appendix 2). Since the real axis is the only Euclidean straight line
passing through a and b, F −1 must take l1 , l2 to a pair of circular arcs passing
through a and b. We observed in Exercise 11.1.5 that these are not geodesics.
11.2.8 By applying a suitable isometry, we can assume that a and d are on the imaginary
axis. The centre of the semicircle geodesic passing through a and b is (R, 0),
114
where a = iR cot 12 α, and its radius is Rcosec 21 α. Let d = it. We must have
Rcosec 12 α < R + t. Then,
d(a, d) = ln
1
R
cot α
t
2
< ln
cot 12 α
cosec 12 α − 1
= f (α).
11.2.9 As in the proof of Theorem 11.2.4, we can assume that a = a′ , that a, c and
c′ are all on the imaginary axis, that b and b′ are on the same side of the
imaginary axis, and that c and c′ are either both ‘above’ or both ‘below’ a on
the imaginary axis. Then, clearly, c = c′ . Since the angles of the two triangles at
a are equal, the geodesic through a and b is the same as that through a and b′ .
Since d(a, b) = d(a, b′ ) and since b and b′ are on the same side of the imaginary
axis, we must have b = b′ .
11.2.10 The proof is the same as in Euclidean geometry. Let a, b ∈ H, let d be the
mid-point of the geodesic l passing through a and b, and let m be the geodesic
passing through d perpendicular to l. If c is any point on m then, by the
preceding exercise, the triangles with vertices a, d, c and b, d, c are congruent. In
particular, d(a, c) = d(b, c). Hence, m is the set of points equidistant from a and
b.
11.3.1 The distance we want is 2tanh−1
The algebra is straightforward.
P −1 (b)−P −1 (a)
P −1 (b)−P −1 (a)
. Now use P −1 (z) =
z+1
.
i(z−1)
11.3.2 By Proposition 11.2.3, there is an isometry F that takes l to the real axis and
the point of intersection of l and m to the origin. Then F must take m to the
imaginary axis as this is the unique hyperbolic line through the origin perpendicular to the real axis. The number of such isometries is the number of isometries
that take the real axis to itself and the imaginary axis to itself. If G is such an
isometry, then either G, R0 ◦ G, R1 ◦ G or R1 ◦ R0 ◦ G fix the real and imaginary
axes and also each quadrant into which the disc DP is divided by the axes. If H
is this isometry, the argument used in the solution of Exercise 11.2.4(ii) shows
that H must be the identity map. Hence, G = R0 , R1 , R0 ◦ R1 or the identity
map.
11.3.3 Let us call a Möbius transformation of the type in the statement of the exercise
a hyperbolic Möbius transformation. Since P is a Möbius transformation, the
Möbius transformations that take DP to itself are those of the form PM P −1 ,
where M is a Möbius transformation that takes H to itself, i.e. a real Möbius
, where a, b, c, d ∈ R and
transformation (Exercise 11.2.5). If M (z) = az+b
cz+d
ad − bc > 0, we find that PM P −1 (z) =
(a+d+i(b−c))z+a−d−i(b+c)
(a−d+i(b+c))z+a+d−i(b−c) .
Since
|a + d + i(b − c)|2 − |a − d − i(b + c)|2 = 4(ad − bc) > 0,
115
PM P −1 is hyperbolic. Conversely, we have to show that if M is a hyperbolic
Möbius transformation, then P −1 M P is a real Möbius transformation. The
calculation is similar to that already given.
11.3.4 By Exercise 11.2.5(iii), the isometries of H are of the form M or M ◦ J where
M is real Möbius and J(z) = −z̄. Hence, the isometries of DP are PM P −1 and
−1
−1
−1
−1
−1
P(M
◦ J)P = PM
P ◦ PJP . But PM P is hyperbolic and PJP (z) =
z̄+1
z+1
= P i(z̄−1)
= P(P −1 (z̄)) = z̄.
PJ i(z−1)
11.3.5 We know that every isometry of H is the composite of reflections Ra and inversions Ia,r with a ∈ R, r > 0 (Exercise 11.2.4(iv)). Now, Ia,r = Ta ◦Dr2 ◦I0,1 ◦T−a
and any translation Ta (a ∈ R) is the composite of reflections R0 ◦Ra/2 . It therefore suffices to show that, if F is any isometry of H of the form Ra (a ∈ R), Da
(a > 0) or I0,1 , then P ◦ F ◦ P −1 is a composite of isometries of DP of the types
in Proposition 11.3.3. We find that (a) if a 6= 0, P ◦ Ra ◦ P −1 = Ib,r , where
−1
is reflection in the real axis; (c) P ◦I0,1 ◦P −1
b = 1+ia
ia , r = 1/|a|; (b) P ◦R0 ◦P
is reflection in the imaginary axis; (d) if a > 0, P ◦ Da ◦ P −1 is the composite of
two inversions of the type in Proposition 11.3.3(i), namely Ic,√c2 −1 ◦ Ib,√b2 −1 ,
where b, c are real numbers such that b2 > 1, c2 > 1 and a = f (c)/f (b) where
x+1
. (We can take b to be any real number > 1 and distinct from 1/f (a),
f (x) = x−1
then choose c = f (af (b)); then, f (c) = f (f (af (b))) = af (b) using the property
f (f (x)) = x for all x 6= 1.)
11.3.6 Suppose first that γ = π/2. The cosine rule gives cosh C = cosh A cosh B and
cosh A = cosh B cosh C − sinh B sinh C cos α. Eliminating cosh C gives cos α =
cosh A sinh B/ sinh C. Hence,
sin2 α sinh2 C = sinh2 C − cosh2 A sinh2 B
= cosh2 A cosh2 B − 1 − cosh2 A sinh2 B
= cosh2 A − 1 = sinh2 A,
so
sin α
sinh A
=
1
sinh C .
Interchanging the roles of A and B gives
sin β
sinh B
=
1
sinh C .
In the general case suppose that the hyperbolic line through the vertex of the
triangle with angle α intersects the opposite side at a point which divides that
side into segments of lengths A′ and A′′ , so that β is the angle between the
sides of lengths C and A′ . Suppose also that this hyperbolic line segment has
length D. Then the original triangle is divided into two triangles, one with angles
π/2, β, α′ and sides of lengths C, D, A′ and the other with angles π/2, γ, α′′ and
sides of lengths B, D, A′′ . Applying the first part to each of these triangles gives
sin β/ sinh D = 1/ sinh C and sin γ/ sinh D = 1/ sinh B. Hence, sin β/ sinh B =
sin γ/ sinh C. The other equation is proved by interchanging the roles of A and
B (for example). The case in which the hyperbolic line through a vertex meets
116
the hyperbolic line through the other two vertices in a point outside the triangle
is similar.
11.3.7 (i) This was established in the solution of the preceding exercise.
B cosh C−cosh A
B
and cos α = coshsinh
.
(ii) Using the sine and cosine rules, sin β = sinh
sinh C
B sinh C
2
cos α
cosh B cosh C−cosh A
cosh B cosh A−cosh A
Hence, sin β =
=
= cosh A.
sinh2 B
sinh2 B
(iii) Using sin β = sinh B/ sinh C and the cosine rule for cos β, we get cot β =
sinh2 A cosh B
sinh A
cosh A cosh C−cosh B
=
sinh A sinh B
sinh A sinh B = tanhB .
11.3.8 If γ = π/2 the formula we want is that in Exercise 11.3.7(ii). In the general case,
we use the method (and notation) of the solution of Exercise 11.3.6. In the case
where the hyperbolic line through a vertex perpendicular to the opposite side
meets that side at a point inside the triangle, applying Exercise 11.3.7(ii) to the
two right-angled triangles gives cosh A′ = cos α′ / sin β, cosh A′′ = cos α′′ / sin γ,
so
cosh A = cosh(A′ + A′′ ) = cosh A′ cosh A′′ + sinh A′ sinh A′′
=
cos α′ cos α′′
tanh2 D
+
sin β sin γ
tan β tan γ
using Exercise 11.3.7(iii). By Exercise 11.3.7(ii), cosh D =
cos β
sin α′
=
cos γ
sin α′′
so
cosh A sin β sin γ = cos α + sin α′ sin α′′ + tanh2 D cos β cos γ
= cos α + sech2 D cos β cos γ + tanh2 D cos β cos γ
= cos α + cos β cos γ.
The case in which the perpendicular meets the opposite side at a point outside
the triangle is similar.
11.3.9 Let γ (t) = (x(t), y(t), z(t)) be a curve on S 2 . Then, Π(γγ (t)) = (u(t), v(t)), where
ẋ+xż
y
x
u = 1−z
, v = 1−z
. Denoting d/dt by a dot, u̇ = (1−z)
with a similar formula
(1−z)2
for v̇, which give
4
u̇2 + v̇ 2
1
2 2
2
2
2 2
=
(1
−
z)
(
ẋ
+
ẏ
)
+
(x
+
y
)
ż
+
2(x
ẋ
+
y
ẏ)
ż(1
−
z)
.
1 + u2 + v 2
(1 − z)2
Using x2 +y 2 = 1−z 2 , which implies xẋ+y ẏ = −z ż, this expression simplifies to
ẋ2 +ẏ 2 +ż 2 . Hence,
the length ofpΠ◦γγ calculated using the given first fundamental
R 2√u̇2 +v̇ 2
R
2
form on R is 1+u2 +v 2 dt =
ẋ2 + ẏ 2 + ż 2 dt, which is the length of γ . Hence,
Π is an isometry.
11.3.10 Since P : H → DP is an isometry, P −1 (l1 ) and P −1 (l2 ) are geodesics in H. By
Proposition 11.2.3, there is an isometry F of H that takes P −1 (l1 ) to P −1 (l2 )
117
and P −1 (z1 ) to P −1 (z1 ). Then, PF P −1 is an isometry of DP that takes l1 to l2
and z1 to z2 .
11.3.11 Draw the radii of DP given by arg(z) = 2πk/n with k = 0, 1, . . . , n − 1, and let
a0 , a1 , . . . , an−1 be the points at a Euclidean distance r (with 0 < r < 1) from
the origin on these radii. For each value of r, we then have a regular hyperbolic
n-gon with vertices a0 , a1 , . . . , an−1 . It is clear that the interior angle α of this
n-gon is an increasing function of r and depends continuously on r. When r
tends to zero, the area of the n-gon also tends to zero, so α must approach its
Euclidean value (Theorem 11.1.5), namely (n − 2)π/n. On the other hand, as r
tends to 1, α tend to zero. By the intermediate value theorem, when 0 < r < 1,
α take every value in the interval 0 < α < (n − 2)π/n.
By Exercise 11.3.8,
cosh A =
∴
2
cos 2π
n + cos
sin2 α2
α
2
2 cos2 πn
sin2 α2
cos nπ
A
cosh =
.
2
sin α2
1 + cosh A =
∴
11.3.12 Let D be the length of the diagonal of the quadrilateral, and let β be the angle
between the diagonal and the side of length B. By Corollary 11.3.7,
cosh D = cosh A cosh B,
and by the cosine rule (Theorem 11.3.6),
cosh C = cosh A cosh D − sinh A sinh D sin β.
By the sine rule (Exercise 11.3.6), sinh A = sinh D sin β so
cosh C = cosh A cosh D − sinh2 A = cosh2 A cosh B − sinh2 A.
By the sine rule, the other two (equal) angles α of the quadrilateral are given by
sin α =
sinh D cos β
.
sinh C
By the cosine rule,
cosh A = cosh B cosh D − sinh B sinh D cos β.
118
Hence,
sin α =
cosh A sinh B
cosh B cosh D − cosh A
=
.
sinh B sinh C
sinh C
11.3.13 Let n be one of the geodesics bisecting the angle between l and m, and let a be
a point on n. We will show that a is equidistant from l and m. By applying an
isometry we can assume that n is the real axis in DP .
Let j and k be the geodesics through a meeting l and m at right angles, and let
b and c be the points in which j meets l and k meets m, respectively. Reflection
R in the real axis is an isometry of DP that interchanges l and m and fixes each
point of n. If R(c) 6= b, the hyperbolic triangle with vertices a, b and R(c) would
have two angles equal to π/2, contradicting Theorem 11.1.5. It follows that R
interchanges b and c, and hence that d(a, b) = d(a, c).
Consider a hyperbolic triangle with vertices a, b, c and let l, m, n be the geodesics
bisecting the angles at a, b, c, respectively. Let d be the point of intersection of l
and m. Then, d is equidistant from all three sides of the triangle, and hence lies
on n.
11.3.14 By applying an isometry of DP , we can assume that l is the imaginary axis, m
is the real axis, a = 0 and c = 1. By Proposition 11.1.4, the hyperbolic circle
with centre b > 0 and radius R is
|z − b|
R
= tanh .
|1 − bz|
2
If this circle passes through the origin, b = tanh R
2 so the hyperbolic circle is
|z − b| = b|1 − bz|,
which can be written as
|z|2 −
b
(z + z̄) = 0.
1 + b2
As b → 1 this tends to the circle |z|2 = 12 (z + z̄), i.e.
1
z−
2
2
=
1
,
4
which is a circle Γ touching C at 1.
A circle Γ̃ that intersects C perpendicularly at 1 also intersects Γ perpendicularly
at 1, and hence the other intersection of Γ and Γ̃ is also at right angles.
11.4.1 Let l and m be two distinct hyperbolic lines in H that do not intersect at any
point of H. If l and m are both half-lines they are parallel as they do not have
119
a common perpendicular. If at least one of l and m is a semicircle, then l and
m are parallel if they intersect at a point of the real axis, and ultra-parallel
otherwise.
11.4.2 We work in H and assume that l is the imaginary axis (by applying a suitable
isometry). If a = v + iw, the semicircle geodesic through a intersects l at
√
i v 2 + w2 = ir, say. The distance of a from l is 2tanh−1 ir−v−iw
ir−v+iw . Setting this
equal to a constant, say D, gives (after some algebra) v 2 /w2 = 2 sinh2 (D/2).
This is the equation of a pair of lines passing through the origin. As they are
not perpendicular to the real axis (unless D = 0), they are not hyperbolic lines.
11.4.3 We work in DP . By applying an isometry we can assume that a is the origin
and b > 0. Suppose that the hyperbolic triangle with vertices a, b, c has internal
angles α, β, γ, and assume that Re(c) > 0.
The hyperbolic line through b and c is part of a circle Γ with centre d and radius
r, say. The line through b and d makes an angle π/2 − β with the real axis,
so d = b + r sin β + ir cos β. Similarly, d = c + r sin(α + γ) − ir cos(α + γ) =
c + r sin(A + β) + ir cos(A + β) using A = π − α − β − γ. Writing c = v + iw we
get v = b − r sin A cos β + r(1 − cos A) sin β, w = r(1 − cos A) cos β + r sin A sin β.
Then, v(1 − cos A) + w sin A = (b + 2r sin β)(1 − cos A). But, since γ intersects
the boundary C of DP perpendicularly, r 2 + 1 = |d|2 = (b + r sin β)2 + r 2 cos2 β =
A
r 2 + b2 + 2br sin β, so 2br sin β = 1 − b2 . Hnece, v(1 − cos A) + w sin A = 1−cos
.
b
This is the equation of a straight line that interescts the real axis at 1/b and
makes an angle A/2 with the (negative) real axis. The set of points c for which
the triangle with vertices a, b, c has area A is the union of this line together with
its reflection in the real axis. These lines are not hyperbolic lines as they do not
pass through the origin.
11.4.4 In the first two parts of this exercise it is slightly easier to work in H (by applying
P −1 ).
(i) By applying a suitable isometry of H, we can assume that one of the sides
of the triangle is the imaginary axis. Then the other sides must be a semicircle
passing through the origin and some point a ∈ R, together with the half-line
Re(z) = a. By further applying a suitable translation parallel to the real axis
and a positive dilation (which are isometries of H), we can assume that the three
sides are the half-lines Re(z) = ±1 and the semicircle |z|2 = 1. Then, the area
of the triangle is
Z
1
−1
dv
Z
∞
√
1−v 2
dw
=
w2
Z
1
−1
√
dv
= π.
1 − v2
(ii) Let a be the vertex of the biasymptotic triangle in H, and let l be the geodesic
through a perpendicular to the opposite side, meeting it at a′ , say. By applying
an isometry of H, we can assume that l is the imaginary axis and that a′ = i.
120
Then, the side opposite a is the semicircle |z|2 = 1. The side joining 1 and a is
an arc of a circle with centre b < 0, say, where
cos
α
−b
=
.
2
1−b
The equation of the circle is (v − b)2 + w2 = (1 − b)2 , so the area of the triangle
is
!
Z 1
1
1
√
−p
dv.
2
1 − v2
(1 − v)(1 + v − 2b)
0
The second integral can be evaluated using the substitution u = v − b. This
gives the area as
π−2
Z
1−b
−b
p
du
−b
du = 2 sin−1
1−b
(1 − b)2 − u2
−b
−1
= π − α.
= π − 2 cos
1−b
That a bisymptotic triangle with angle α exists for any 0 < α < π is proved
using the method of Exercise 11.3.11.
(iii) We now switch back to DP . Assume that the vertex at which the angle is α
is 0, let b be the other vertex of the triangle in DP . Let the side of the triangle
passing through 0 and b meet C at c, and let the other side passing through 0
meet C at d (these sides are radii of C). The area of the asymptotic triangle with
vertices 0, b, d is the difference between the areas of the biasymptotic triangles
with vertices 0, c, d and b, c, d, which by part (ii) is
(π − α) − (π − (π − β)) = π − α − β.
Let a be a point on the radius joining 0 and d and consider the triangle with
vertices 0, a, b. Let the angle of this triangle at b be β ′ and the angle at a be α′ .
Thus, α′ tends to zero and β ′ tends to β as a tends to d along the radius (‘tends
to’ in the Euclidean sense). By Exercise 11.3.8, the length A of the side joining
0 and b is given by
cos α′ + cos α cos β ′
cosh A =
.
sin α sin β ′
Letting a tend to c, we obtain
cosh A =
1 + cos α cos β
.
sin α sin β
121
11.4.5 (i) Let a, b ∈ DP be the vertices of one of the triangles, let l be the geodesic
passing through a, b and let m be the other side of the triangle passing through
a; define a′ , b′ , l′ , m′ similarly for the other triangle. By applying an isometry to
one of the triangles, we can assume that a = a′ , m = m′ and that l and l′ are
on the same side of m. As the triangles have the same angle at a, l = l′ and by
Exercise 11.4.4(iii) the lengths of the finite sides of the two triangles are equal.
It follows that b = b′ , and since the two triangles have the same angle at b their
third sides must also coincide.
(ii) As in (i), by applying an isometry to one of the triangles we can assume that
their vertices in DP coincide, as do the two sides passing through this vertex.
But then the two triangles coincide.
(iii) This was established in the solution of Exercise 11.4.4(i).
11.4.6 We consider the biasymptotic triangle with vertices 0, 1 and e3iπ/4 , where the
two sides meeting at the origin are radii of DP and the side joining 1 and e3iπ/4
is an arc of a circle meeting C at right angles at 1 and e3iπ/4 . The ‘altitude’ of
this triangle passing through 1 is a circular arc that meets C at −i, and that
through e3iπ/4 is a circular arc meeting C at e5iπ/4 . These two geodesics are
clearly ultraparallel (one is confined to the region v > 0, the other to the region
v < 0). If we now replace the vertices 1 and e3iπ/4 with nearby points a, b ∈ DP ,
the altitudes of the hyperbolic triangle with vertices 1, a, b passing through a and
b will still be ultraparallel.
2w
v 2 +w 2 −1
2v
−1
11.5.1 From Example 6.3.5, Π (v, w) = v 2 +w2 +1 , v 2 +w2 +1 , v 2 +w2 +1 , so K(v, w) =
2(v,w)
.
v 2 +w 2 +1
11.5.2 From Appendix 2, every Möbius transformation is a composite of transformations
of the form z 7→ 1/z, z 7→ z + λ, z 7→ λz (where λ 6= 0 in the last case). Hence, it
suffices to establish Eq. 11.11 when M is of this form. For the last two types this
(a−1 −c−1 )(b−1 −d−1 )
is obvious; for the first, (a−1 , b−1 ; c−1 , d−1 ) = (a
−1 −d−1 )(b−1 −d−1 ) = (a, b; c, d)
on multiplying numerator and denominator by abcd. This argument is only
valid provided none of a, b, c, d is 0 or ∞, but a similar argument works in the
other cases. For example, if a = ∞ but b, c, d 6= 0, we have to show that
(0, b−1 ; c−1 , d−1 ) = (∞, b; c, d). This is proved by multiplying numerator and
−c−1 (b−1 −d−1 )
denominator of (0, b−1 ; c−1 , d−1 ) = −d
−1 (b−1 −c−1 ) by bcd.
If M : C∞ → C∞ is a bijection satisfying Eq. 11.11, let b, c, d ∈ C∞ be
such that M (b) = 1, M (c) = 0, M (d) = ∞. Then, M (z) = (M (z), 1; 0, ∞) =
(M (z), M (b); M (c), M (d)) = (z, b; c, d), so M is a Möbius transformation.
11.5.3 It is enough to prove the existence when a′ = ∞, b′ = 0, c′ = 1. For if M
and M ′ are Möbius transformations taking (a, b, c) and (a′ , b′ , c′ ) to (∞, 0, 1),
−1
then M ′ ◦ M is a Möbius transformation taking (a, b, c) to (a′ , b′ , c′ ). But
M (z) = (a, b; c, z) is a Möbius transformation that takes (a, b, c) to (∞, 0, 1). For
122
the uniqueness, note that if M is a Möbius transformation that takes (a, b, c) to
(∞, 0, 1), then M (z) = (∞, 0; 1, M (z)) = (a, b; c, z).
2
(a−b)(− 1 + 1 )
(a−b)(b̄−ā)
a−b
b̄
= − tan2 21 d by
= (1+āb)(1+a
= − 1+āb
11.5.4 (a, −1/ā; b, −1/b̄) = (a+ 1 )(−ā1 −b)
b̄)
ā
b̄
Proposition 6.5.2.
11.5.5 The reflection in the line through the origin making an angle θ with the real axis
2e2iθ z̄
2iθ
is R(z) = e2iθ z̄. Then, K(R(z)) = |z|
K(z) = R(K(z)).
2 +1 = e
11.5.6 This follows from Exercises 11.3.5 and 11.5.5 and Proposition 11.5.4.
11.5.7 Since K = pr ◦ Π−1 , and since Π is conformal (Example 6.3.5), it suffices
2
to show that pr : S−
→ D preserves the angles between curves intersecting at pr(0, 0, −1) = p
(0, 0, 0). But the first fundamental form of the patch
2
is
pr−1 (x, y, 0) = (x, y, − 1 − x2 − y 2 ) of S−
1
((1 − y 2 )dx2 + 2xydxdy + (1 − x2 )dy 2 ),
1 − x2 − y 2
which reduces to dx2 + dy 2 when x = y = 0.
11.5.8 Suppose first that a, b, c, d lie on a Circle C. Let M be the Möbius transformation
such that M (a) = ∞, M (b) = 0, M (c) = 1 (Exercise 11.5.3), and let M (d) = z.
˜ say (Proposition A.2.4); since C˜ passes through ∞ it
Now M (C) is a Circle C,
must be a straight line, and as this line passes through 0 and 1 it must be the
real axis; hence z ∈ R. But (a, b; c, d) = (∞, 0; 1, z) = z (Exercise 11.5.2).
Conversely, assume that (a, b; c, d) ∈ R, and let M be as above. Then, M (d) ∈ R
and 0, 1, ∞, M (d) lie on a Circle C, namely the real axis. But then a, b, c, d lie
on the Circle M −1 (C).
11.5.9 Let M be the Möbius transformation taking ∞, 0, 1 to a, b, c, respectively. Then,
M (d) = λ. By Exercise 11.5.2, if we subject a, b, c, d to any permutation, the
resulting cross-ratio will be the cross-ratio of the points obtained by subjecting
∞, 0, 1, λ to the same permutation. It therefore suffices to check that the six
stated values are the values of the cross-ratios of the points ∞, 0, 1, λ taken in
any order. This is straightforward.
Chapter 12
12.1.1 κ1 + κ2 = 0 =⇒ κ2 = −κ1 =⇒ K = κ1 κ2 = −κ21 ≤ 0. K = 0 ⇐⇒ κ21 = 0 ⇐⇒
κ1 = κ2 = 0 ⇐⇒ the surface is an open subset of a plane (by Proposition 8.2.9).
σ u × σ v , where κ1 , κ2 are the
12.1.2 From Eq. 8.15, σ λu × σ λv = (1 − λκ1 )(1 − λκ2 )σ
principal curvatures of σ . Since σ is minimal, κ2 = −κ1 so
ZZ
Aσ λ (U ) =
(1 − λ2 κ21 ) k σ u × σ v k dudv
U
ZZ
2
κ21 k σ u × σ v k dudv.
= Aσ (U ) − λ
U
123
Since the integrand in the last integral is ≥ 0 everywhere, the stated inequality
follows. Equality holds ⇐⇒ the last integral vanishes, which happens ⇐⇒ the
integrand vanishes everywhere, i.e. ⇐⇒ κ1 = 0 everywhere. In that case κ2 =
−κ1 = 0 also, and σ is an open subset of the plane by Proposition 8.2.9.
12.1.3 By Proposition 8.6.1, a compact minimal surface would have K > 0 at some
point, contradicting Exercise 12.1.1.
12.1.4 The first part follows from Exercises 6.1.2 and 7.1.4. The map which wraps the
plane onto the unit cylinder (Example 6.2.4) is a local isometry, but the plane
is a minimal surface and the cylinder is not.
12.1.5 If κ1 , κ2 are the principal curvatures at the point, we have κ1 + κ2 = 0 as the
surface is minimal, and κ1 = κ2 as the point is an umbilic. Hence, κ1 = κ2 = 0.
12.1.6 Let Σ0 be the minimal surface and let S = Σλ in the notation of Definition 8.5.1.
Let k1 , k2 be the principal curvatures of Σ. By Proposition 8.5.2(ii), we have
1
1 − λk1
1 − λk2
2λ
1
+
=
+
=− ,
κ1
κ2
ǫk1
ǫk2
ǫ
since
1
k1
+
1
k2
= 0 (because k1 + k2 = 0).
12.1.7 The condition for a minimal surface is LG − 2M F + N E = 0 (Corollary 8.1.3).
From the solution of Exercise 8.1.16, E = (1−κa cos θ)2 +τ 2 a2 , F = τ a2 , G = a2 ,
L = aτ 2 − κ cos θ(1 − κa cos θ), M = τ a, N = a, and we find that the condition
for the tube to be a minimal surface is
a(1 − κa cos θ)(1 − 2a cos θ) = 0,
which obviously cannot hold at all points of the surface.
12.1.8 Let t1 , t2 be orthonormal principal vectors of S at p corresponding to principal
curvatures κ1 , κ2 (Corollary 8.2.2). Then, t = a1t 1 + a2t2 for some a1 , a2 ∈ R,
so W(tt) = κ1 a1t1 + κ2 a2t2 (where W is the Weungarten map of S at p). Hence,
hW(tt), W(tt)i
κ2 a2 + κ22 a22
hhhtt, tiii
.
=
= 1 12
htt, ti
htt, ti
a1 + a22
But, as S is a minimal surface, κ2 = −κ1 , so
hhhtt, tiii
= κ21 = −κ1 κ2 = −K.
htt, ti
Alternatively, this exercise can be deduced from Exercise 8.1.6.
12.2.1 By the solution of Exercise 8.1.2, the helicoid σ (u, v) = (v cos u, v sin u, λu) has
124
E = λ2 + v 2 , F = 0, G = 1, L = 0, M = λ/(λ2 + v 2 )1/2 , N = 0, so H =
LG−2M F +NG
= 0.
2(EG−F 2 )
12.2.2 A straightforward calculation shows that the first and second fundamental forms
of σ t are cosh2 u(du2 + dv 2 ) and − cos t du2 − 2 sin t dudv + cos t dv 2 , respectively,
2
u+cos t cosh2 u
= 0.
so H = − cos t cosh
2 cosh4 u
12.2.3 From Example 5.3.1, the cylinder can be parametrized by σ (u, v) = γ (u) + vaa,
where γ is unit-speed, k a k = 1 and γ is contained in a plane Π perpendicular to
a. We have σ u = γ̇γ = t (a dot denoting d/du), σ v = a, so E = 1, F = 0, G = 1;
n, σ uv = σ vv = 0, so L = κn
n.(tt × a), M = N = 0. Now
N = t × a, σ uu = ṫt = κn
t × a is a unit vector parallel to Π and perpendicular to t, hence parallel to n;
so L = ±κ and H = ±κ/2. So H = 0 ⇐⇒ κ = 0 ⇐⇒ γ is part of a straight line
⇐⇒ the cylinder is an open subset of a plane.
12.2.4 The first fundamental form is (cosh v + 1)(cosh v − cos u)(du2 + dv 2 ), so σ is
conformal. By Exercise 8.5.1, to show that σ is minimal we must show that
σ vv = 0; but this is so, since σ uu = (sin u cosh v, cos u cosh v, sin u2 sinh v2 ) =
σ uu +σ
σ vv .
−σ
(i) σ (0, v) = (0, 1 − cosh v, 0), which is the y-axis. Any straight line is a geodesic.
(ii) σ (π, v) = (π, 1 + cosh v, −4 sinh v2 ), which is a curve in the plane x = π
such that z 2 = 16 sinh2 v2 = 8(cosh v − 1) = 8(y − 2), i.e. a parabola. The
d
d
geodesic equations are dt
(E u̇) = 12 Eu (u̇2 + v̇ 2 ), dt
(E v̇) = 12 Ev (u̇2 + v̇ 2 ), where
a dot denotes the derivative with respect to the parameter t of the geodesic
and E = (cosh v + 1)(cosh v − cos u). When u = π, the unit-speed condition is
E v̇ 2 = 1, so v̇ = 1/(cosh v + 1). Hence, the first geodesic equation is 0 = 12 Eu v̇ 2 ,
which holds because Eu = sin u(cosh v + 1) = 0 when u = π; the second geodesic
d
(cosh v + 1) = (cosh v + 1) sinh v v̇ 2 = sinh v v̇, which obviously
equation is dt
holds.
(iii) σ (u, 0) = (u −sin u, 1 −cos u, 0), which is the cycloid of Exercise 1.1.7 (in the
xy-plane, with a = 1 and with t replaced by u). The second geodesic equation
is satisfied because Ev = sinh v(2 cosh v + 1 − cos u) = 0 when v = 0. The
unit-speed condition is 2(1 − cos u)u̇2 = 1, so u̇ = 1/2 sin u2 . The first geodesic
d
d
equation is dt
(4 sin2 u2 u̇) = sin uu̇2 , i.e. dt
(2 sin u2 ) = cos u2 u̇, which obviously
holds.
12.2.5 Using Exercise 8.1.1, the surface is minimal ⇐⇒
2
(1 + g ′ )f¨ + (1 + f˙2 )g ′′ = 0,
where a dot denotes d/dx and a dash denotes d/dy; hence the stated equation.
Since the left-hand side of the stated equation depends only on x and the righthand side only on y, we must have
f¨
1 + f˙2
= a,
g ′′
= −a,
1 + g′2
125
for some constant a. Suppose that a 6= 0. Let r = f˙; then f¨ = rdr/df and
the first equation is rdr/df = a(1 + r 2 ), which can be integrated to give af =
1
2
corresponds to translating
2 ln(1 + r ), up to adding an arbitary constant (which
√
the surface parallel to the z-axis). So df /dx = ± e2af − 1, which integrates to
give f = − a1 ln cos a(x + b), where b is a constant; we can assume that b = 0
by translating the surface parallel to the x-axis. Similarly, g = a1 ln cos ay, after
translating the surface parallel to the y-axis. So, up to a translation, we have
cos ay ,
az = ln
cos ax
which is obtained from Scherk’s surface by the dilation (x, y, z) 7→ a(x, y, z). If
a = 0, then f¨ = g ′′ = 0 so f = b + cx, g = d + ey, for some constants b, c, d, e,
and we have the plane z = b + d + cx + ey.
−1
12.2.6 Using
p Exercise 8.1.1, taking f (x, y) = sin (sinh x sinh y), and writing λ =
1/ 1 − sinh2 x sinh2 y, we find that fx = λ cosh x sinh y, fy = λ cosh y sinh x,
fxx = λ3 sinh x sinh y cosh2 y, fxy = λ3 cosh x cosh y, fyy = λ3 sinh x sinh y cosh2 x.
It is easy to check that (1 + fy2 )fxx − 2fx fy fxy + (1 + fx2 )fyy = 0.
12.3.1 (i) From the proof of Theorem 12.3.2, the Gauss map is conformal ⇐⇒ W 2 =
λ..id, where λ is a smooth function on S. If H 6= 0 at p, then H 6= 0 on an open
2
+K
subset O of S containing p. By Exercise 8.1.6, W = λ 2H
.id, so every point of
O is an umbilic and O is an open subset of a plane or a sphere by Proposition
8.2.9. Since H 6= 0 the planar case is impossible.
Part (ii) is now obvious.
For (iii), assume that S is not minimal. Then, there is a point p ∈ S at which
H 6= 0, say H = µ. The argument in (i) and (ii) shows that the set Sµ of points
of S at which H = µ is a (non-empty) open subset of S; it is also a closed subset
because H is a continuous function on S. Since S is connected, Sµ = S. Hence,
every point of S is an umbilic, and so S is an open subset of a sphere (the planar
case is impossible as µ 6= 0).
12.3.2 (i) From Example 12.1.4, N = (−sech u cos v, −sech u sin v, tanh u). Hence, if
N(u, v) = N (u′ , v ′ ), then u = u′ since u 7→ tanh u is injective, so cos v = cos v ′
and sin v = sin v ′ , hence v = v ′ ; thus, N is injective. If N = (x, y, z), then
x2 +y 2 = sech2 u 6= 0, so the image of N does not contain
p the poles. Given a point
−1
2
(x, y, z) ∈ S other than the poles, let u =p±sech
x2 + y 2 , thepsign being that
of z, and let v be such that cos v = −x/ x2 + y 2 , sin v = −y/ x2 + y 2 ; then,
N(u, v) = (x, y, z).
(ii) By the solution of Exercise 8.1.2, N = (λ2 + v 2 )−1/2 (−λ sin u, λ cos u, −v).
Since N(u, v) = N (u + 2kπ, v) for all integers k, the infinitely-many points
σ (u + 2kπ, v) = σ (u, v) + (0, 0, 2kπ)
126
of the helicoid all have the same image under the Gauss map. (Of course, this is
geometrically obvious because the helicoid itself is left unchanged by the translation by 2π parallel to the z-axis.) If N = (x, y, z), then x2 +y 2 = λ2 /(λ2 +v 2 ) 6= 0,
2
2
2
so the image of
pN does not contain the poles. If (x, y, z) ∈ Spand x + y 6= 0,
let vp= −λz/ x2 + y 2 and let u be such that sin u = −x/ x2 + y 2 , cos u =
−y/ x2 + y 2 ; then N(u, v) = (x, y, z).
12.4.1 By Proposition 12.3.2, if K 6= 0 then the Gauss map G : S → S 2 is a conformal
p)
local diffeomorphism. Let R be a rotation of R3 about the origin that takes G(p
to the south pole of S 2 (or any point other than the north pole). There is an
open subset O of S containing p such that G(O) does not contain the north pole.
By Example 6.3.5, Π ◦ R ◦ G is a conformal diffeomorphism from O to an open
subset U of R2 . The inverse of this diffeomorphism is the desired surface patch
σ.
12.4.2 Writing x = r cos θ, y = r sin θ gives z = aθ, so we have the parametrization
σ (r, θ) = (r cos θ, r sin θ, aθ). The first fundamental form of σ is dr 2 +(r 2 +a2 )dθ 2 ,
so σ is not a conformal parametrization. However, noting that
Z
r
dr
√
= sinh−1 ,
2
2
a
r +a
we let ϕ = sinh−1
r
a
so that r = a sinh ϕ. The reparametrization
σ (θ, ϕ) = σ (a sinh ϕ, θ) = (a sinh ϕ cos θ, a sinh ϕ sin θ, aθ)
σ̃
has first fundamental form
(r 2 + a2 )(dϕ2 + dθ 2 ) = a2 cosh2 ϕ(dθ 2 + dϕ2 ),
σ is a conformal parametrization of the surface.
so σ̃
12.5.1 We have
σv
ϕ = σ u − iσ
= (1 − u2 + v 2 − 2iuv, 2uv − i(1 − v 2 + u2 ), 2u + 2iv)
= (1 − ζ 2 , −i(1 + ζ 2 ), 2ζ).
So the conjugate surface is, up to a translation,
Z
σ (u, v) = Re (i(1 − ζ 2 ), 1 + ζ 2 , 2iζ) dζ
σ̃
ζ3 2
ζ3
, ζ + , iζ
= Re i ζ −
3
3
3
v
u3
2
2
= −v + u v − , u +
− uv , −2uv .
3
3
127
√
√
σ (U, V ) = σ (u, v); then,
Let U = (u − v)/ 2, V = (u + v)/ 2, σ̃
1 3 1 3
1
2
2
σ (U, V ) = √
U − V + UV − U V + V − U ,
σ̃
3
3
2
1
1 3 1 3
2
2
2
2
√
.
U + V + UV + U V − V − U , U − V
3
3
2
σ (U, V ) then
Applying the π/4 rotation (x, y, z) 7→ √12 (x + y), √12 (y − x), z to σ̃
gives U − 31 U 3 + U V 2 , V − 31 V 3 + U 2 V, U 2 − V 2 , which is Enneper’s surface
again.
12.5.2 ϕ = 12 (1 − ζ −4 )(1 − ζ 2 ), 2i (1 − ζ −4 )(1 + ζ 2 ), ζ(1 − ζ −4 ) , so
2
1
ζ3
i
ζ3
ζ
ζ −3
ζ −3
ζ −2
−1
−1
σ = Re
ζ−
,
ζ+
,
−ζ +
+ζ +
+
2
3
3
2
3
3
2
2
i
1
1
= Re − (ζ − ζ −1 )3 , (ζ + ζ −1 )3 , (ζ + ζ −1 )2 ,
6
6
2
σ (ũ, ṽ), where
up to a translation. Put ζ = eζ̃ , ζ̃ = ũ + iṽ. Then, σ (u, v) = σ̃
4
4i
σ (ũ, ṽ) = Re − sinh3 ζ̃, cosh3 ζ̃, 2 cosh2 ζ̃
σ̃
3
3
1
= 4 sinh ũ cos ṽ(cosh2 ũ sin2 ṽ − sinh2 ũ cos2 ṽ) ,
3
1
4 sinh ũ sin ṽ( sinh2 ũ sin2 ṽ − cosh2 ũ cos2 ṽ),
3
2(cosh2 ũ cos2 ṽ − sinh2 ũ sin2 ṽ) .
12.5.3 The first part is obvious.
σ qv = a(σ
σ u − iσ
σ v ) implies that σ a = aσ
σ + a, where
(i) If a ∈ R, the identity σ au − iσ
a
a is a constant vector. Hence, σ is obtained from σ by applying the dilation Da
followed by the translation Ta (Appendix 1).
(ii) If f and g are the functions in the Weierstrass representation of σ (Proposition
12.5.4), those in the Weierstrass representation of σ a are af and g (see Eq.
12.23). By Eq. 12.25, replacing f by af leaves the first fundamental form
unchanged, so the map σ (u, v) 7→ σ a (u, v) is an isometry, and by Eq. 12.26 N
does not depend on f , so the tangent planes of σ and σ a at corresponding points
are parallel.
it
it
σ ev = eit (σ
σ u − iσ
σ v ). Since σ (u, v) = cosh u cos v, cosh u sin v, u),
12.5.4 We have σ eu − iσ
we get
it
σ eu = (cos t sinh u cos v − sin t cosh u sin v, sin t cosh u cos v + cos t sinh u sin v, cos t),
it
σ ev = (− cos t cosh u sin v − sin t sinh u cos v, cos t cosh u cos v − sin t sinh u sin v, − sin t).
128
Integrating gives
it
σ e (u, v) = cos t(cosh u cos v, cosh u sin v, u) + sin t(− sinh u sin v, sinh u cos v, −v)
σ (u, v) + sin tσ̂
σ (u, v),
= cos tσ
σ (sinh u, π2 + v) =
say (up to a translation). In the notation of Exercise 6.2.3, σ̃
π
(− sinh u sin v, sinh u cos v, 2 +v). Reflecting in the xy-plane and then translating
σ (u, v).
σ (sinh u, π2 + v) to σ̂
by π/2 along the z-axis takes σ̃
12.5.5 (i) ϕ is never zero since we have arranged that F ′ and G′ are never both zero.
Condition (ii) in Theorem 12.5.2 is obvious.
∂
d
(ii) When v = 0, F ′ (z) = ∂u
F (u, 0) = Fu , etc, so du
σ (u, 0) = (f˙, ġ, 0) = γ̇γ (a
dot denoting d/du). This proves
q (ii).
(−ġ,f˙,0)
(iii) When v = 0, ϕ = (f˙, ġ, i f˙2 + ġ 2 ). Using Eq. 12.26, N = √
. Then,
f˙2 +ġ 2
q
N × γ̇γ ) = 0. It follows that γ is a
N × γ̇γ = (0, 0, − f˙2 + ġ 2 ) and finally γ̈γ .(N
pre-geodesic on σ (Exercise 9.1.2).
σv = ϕ =
(iv) If γ is the cycloid, F (z) = z − sin z, G(z) = 1 − cos z, so σ u − iσ
(1 − cos z, sin z, 2i sin z). This gives
u
v
sinh ),
2
2
v
u
σ v = (− sin u sinh v, − cos u sinh v, −2 sin cosh ).
2
2
σ u = (1 − cos u cosh v, sin u cosh v, −2 cos
Integrating gives σ (u, v) = (u − sin u cosh v, − cos u cosh v, −4 sin u2 sinh v2 ), up to
a translation. Translating by (0, 1, 0) gives Catalan’s surface.
ϕ = ( 21 if (1 − g 2 ), 2i if (1 + g 2 ), if g), which
12.5.6 ϕ = ( 21 f (1 − g 2 ), 2i f (1 + g 2 ), f g) =⇒ iϕ
corresponds to the pair if and g.
12.5.7 From Example 12.5.3, for the catenoid we have ϕ1 = sinh ζ, ϕ2 = −i cosh ζ,
ϕ3 = 1, so f = ϕ1 − iϕ2 = sinh ζ − cosh ζ = −e−ζ , g = ϕ3 /(ϕ1 − iϕ2 ) = −eζ .
Since the helicoid is the conjugate surface of the catenoid (Example 12.5.3 again),
the preceding exercise gives f = −ie−ζ , g = −eζ for the helicoid.
12.5.8 By Example 6.3.5, Π(x, y, z) = (x + iy)/(1 − z). From Eq. 12.26, G(ζ) =
1
2
|g|2 +1 (g + ḡ, −i(g − ḡ), |g| − 1), so
π(G(ζ)) =
g + ḡ + g − ḡ
= g(ζ).
|g|2 + 1 − (|g|2 − 1)
12.5.9 By Eq. 12.22,
i 2
1 2
−2
−2
ϕ = − (ζ − ζ ), (ζ + ζ ), 1 ,
2
2
129
so the parametrization is, up to a translation,
Z
σ (u, v) = Re ϕ(ζ)dζ
(u + iv)3
u − iv
i(u + iv)3
i(u − iv)
= Re −
−
,
−
, u + iv
6
2(u2 + v 2 )
6
2(u2 + v 2 )
u
1
v
1 3
2
2
3
, (−3u v + v ) −
,u .
= − (u − 3uv ) −
6
2(u2 + v 2 ) 6
2(u2 + v 2 )
By Proposition 12.5.5, the Gaussian curvature is
K=
−64|ζ|6
,
(1 + |ζ|4 )4
√
and this tends to zero as |ζ| = u2 + v 2 → ∞.
12.5.10 Using the conformal parametrization in Exercise 12.4.2, we get
σϕ
ϕ = σ θ − iσ
= a(− sinh ϕ sin θ − i cosh ϕ cos θ, sinh ϕ cos θ − i cosh ϕ sin θ, 1)
= a(−i cos ζ, −i sin ζ, 1),
where ζ = θ + iϕ. Then,
f = ϕ1 − iϕ2 = −iae−iζ ,
ϕ3
g=
= ieiζ .
ϕ1 − iϕ2
By Proposition 12.5.5,
16|eiζ |2
a2 |e−iζ |2 (1 + |eiζ |2 )4
16e−2ϕ
= − 2 2ϕ
a e (1 + e−2ϕ )4
16
=− 2 ϕ
a (e + e−ϕ )4
K=−
= −a−2 sech4 ϕ.
130
Chapter 13
RR
13.1.1 If γ is a simple closed geodesic, Theorem 13.1.2 gives int(γ ) KdA = 2π; since
K ≤ 0, this is impossible. The parallels of a cylinder are not the images under a
π)
surface patch σ : U → R3 of a simple closed curve π in the plane such that int(π
is contained in U . Note that the whole cylinder can actually be covered by a
single patch (see Exercise 4.1.4) in which U is an annulus, but that the parallels
correspond to circles going ‘around the hole’ in the annulus.
n, the geodesic curvature κg of n
13.1.2 Since the unit normal N of S 2 is equal to ±n
′′ . n
′
n
n
is, up to a sign,
(n × ). Let t be the arc-length
of γ and denote d/dt by
√
2
2
n k = k −κtt + τb
b k = κ + τ = R, say, where t =
a dot. Then, ds/dt = k ṅ
γ̇γ .
−κt +τ b
d
b)/R, n × n′ = (κb
b + τtt)/R, and n′′ = R1 dt
Then, n′ = (−κtt + τb
=
R
b − R−2 (κ2 + τ 2 )n
n. These formulas give
−R−1 (κ/R)˙tt + R−1 (τ /R)˙b
n × n′ ) = −R−2 τ (κ/R)˙+ R−2 κ(τ /R)˙ = (κτ̇ − τ κ̇)/R3 .
n′′. (n
′
′
d
κ
= ± ds
tan−1 κτ . Applying Theorem 13.1.2
Since κ̇ = Rκ′ , etc, κg = ± κτκ2 −τ
+τ 2
R ℓ(n)
κg dt = 0
to the curve n on S 2 , and noting that K = 1 for S 2 and that 0
n)-periodic function (where ℓ(n
n) is the length
because κg is the derivative of an ℓ(n
of the closed curve n), we get that the area inside n is 2π.
13.1.3 Before smoothing K = 0 so the total curvature is zero. To compute the total
curvature after smoothing, we apply Theorem 13.1.2 to the smoothed cone with
γ being the circle on the cone given by z = a for some constant a > 0. The
radius of this circle is a tan α so its curvature is κ = 1/a tan α. Since the angle
of the cone is α, the geodesic curvature of γ is κg = κ sin α = a−1 cos α. The
part of the cone with z > a still has K = 0 so does not contribute to the total
curvature. Integrating K over the part of the cone with z < a, Gauss-Bonnet
gives the total curvature as
ZZ
Z
KdA = 2π −
κg ds = 2π − (a−1 cos α)(2πa tan α) = 2π(1 − sin α).
γ
13.2.1 The parallel u = u1 is the circle γ 1 (v) = (f (u1 ) cos v, f (u1 ) sin v, g(u1 )); if s
is the arc-length of γ 1 , ds/dv = f (u1 ). Denote d/ds by a dot and d/du by
a dash. Then, γ̇γ = (− sin v, cos v, 0), γ̈γ = − f (u1 1 ) (cos v, sin v, 0), and the unit
normal of the surface is N = (−g ′ cos v, −g ′ sin v, f ′ ). This gives the geodesic
R ℓ(γ 1 )
′
(u1 )
γ
N × γ̇γ ) = ff (u
κg ds =
.
Since
ℓ(γ
)
=
2πf
(u
),
curvature of γ as κg = γ̈γ .(N
1
1
0
1)
RR
′
′′
2πf (u1 ). Similarly for γ 2 . By Example 8.1.4, K = −f /f , so R KdAσ =
R 2π R u2 f ′′
− f f dudv = 2π(f ′ (u1 ) − f ′ (u2 )). Hence,
0
u1
ZZ
Z ℓ(γ 2 )
Z ℓ(γ 1 )
κg ds =
KdAσ .
κg ds −
0
0
R
131
This equation is the result of applying Theorem 13.2.2 to the curvilinear polygon
shown below.
B
u = u2
A u = u1
13.2.2 Applying Theorem 12.2.2 to the curvilinear n-gon, and noting the −K ≥ 1, we
get
X
A(int(γγ )) ≤ (n − 2)π −
αi .
i
If n ≤ 2 the right-hand side is < 0, which is absurd. If n = 3 the right hand-side
is < π.
13.2.3 Consider a small curvilinear quadrilateral on S bounded by parameter curves
u = a, u = a + h, v = b, v = b + k, say, where h and k are small. The internal
angles are α, α, π − α, π − α, where α is the angle between the parameter curves
u = a and v = b. Let A be the (small) area of the quadrilateral. Since the sides
of the quadrilateral are geodesics, Theorem 13.2.2 gives
K(a, b)A = (4 − 2)π − (α + α + π − α + π − α) = 0.
13.3.1 This can be proved by expressing a, b, c in component form and computing both
sides. Alternatively, one may observe that both sides of the equation are linear
in each of a, b, c (separately), and change sign when any two of a, b, c are
interchanged. This means that it is enough to prove the formula when a = i, b =
j, c = k, when both sides are obviously equal to 1.
13.3.2 Define ϕk as in the hint. The stated properties are easily checked.
RR
13.4.1 By Corollary 13.4.8, S KdA = 4π(1 − g), and g = 1 since S is diffeomorphic
to T1 . By Proposition 8.6.1, K > 0 at some point of S.
RR
13.4.2 K > 0 =⇒ S KdA > 0 =⇒ g < 1 by Corollary 13.4.8; since g is a non-negative
integer, g = 0 so S is diffeomorphic to a sphere. The converse is false: for
132
example, a ‘cigar tube’ is diffeomorphic to a sphere but K = 0 on the cylindrical
part.
13.4.3 The ellipsoid is diffeomorphic to S 2 by the map (x, y, x) 7→ (x/a, y/a,
RR z/b), so
the genus of the ellipsoid is zero. Hence, Corollary 13.4.8 gives S KdA =
4π(1 − 0) = 4π.
Parametrising the ellipsoid by σ (θ, ϕ) = (a cos θ cos ϕ, a cos θ sin ϕ, b sin θ) (cf.
the latitude-longitude parametrization of S 2 ), the first and second fundamental
forms of σ are
(a2 sin2 θ+b2 cos2 θ)dθ 2 +a2 cos2 θdϕ2 and p
ab
a2
2
sin θ +
b2
cos2
θ
(dθ 2 +cos2 θdϕ2 ),
respectively. This gives
b2
,
(a2 sin2 θ + b2 cos2 θ)2
p
dAσ = a cos θ a2 sin2 θ + b2 cos2 θdθdϕ.
K=
Hence,
ZZ
S
KdAσ =
Z
0
2π
Z
π/2
−π/2
ab2 cos θ dθdϕ
.
(a2 sin2 θ + b2 cos2 θ)3/2
13.4.4 The surface in Exercise 5.4.1(ii) is diffeomorphic to S 2 , hence its Euler number
is 2.
13.5.1 Assuming that every country has ≥ 6 neighbours, the argument in the proof of
Theorem 13.5.1 gives E ≥ 3F and 2E ≥ 3V , so V − E + F ≤ 2E
− E + E3 = 0,
3
contradicting V − E + F = 2.
13.5.2 3F = 2E because each face has three edges and each edge is an edge of two
faces. From χ = V − E + F , we get χ = V − E + 32 E, so E = 3(V − χ).
133
Since each edge has two vertices and two edges cannot intersect in more than
one vertex, E ≤ 12 V (V − 1); hence, 3(V − χ) ≤ 21 V (V − 1), which is equivalent
√
1
to V 2 − 7V + 6χ ≥ 0.
The
roots
of
the
quadratic
are
7
±
49
−
7χ
, so
2
√
√
1
1
V ≤ 2 7 − 49 − 7χ or V ≥ 2 7 + 49 − 7χ . Since χ = 2, 0, −2, . . . , the
first condition gives V ≤ 3, which would allow only one triangle; hence, the
second condition must hold.
13.5.3 This is obvious since a diffeomorphism S → S̃ takes an n-coloured map on S to
an n-coloured map on S̃.
13.5.4 Since three edges meet at each vertex and each edge has two vertices, we have
3V = 2E. As each edge is an edge of two countries,
2E =
∞
X
ncn .
n=2
Hence, E =
1
2
P
P
P
ncn , V = 31
ncn , and since F =
cn we get
∞
∞ X
1
1X
1
(6 − n)cn .
n − n + 1 cn =
χ=
3
2
6 n=2
n=2
13.5.5 (i) In the notation of the preceding exercise, for a soccer ball we have cn = 0
unless n = 5 or 6 so the formula in the preceding exercise gives c5 = 12 (since
χ = 2 for a sphere).
(ii) This time we are given that cn = 0 unless n = 4 or 6, so the formula gives
2c4 = 12.
13.6.1 The circle θ = θ0 is a circle in the plane z = b sin θ0 with centre on the zaxis, so its principal normal is a unit vector perpendicular to the circle and in
this plane, hence equal (up to a sign) to (cos ϕ, sin ϕ, 0). The unit normal of σ
is N = (− cos θ cos ϕ, − cos θ sin ϕ, − sin θ), so the angle between N and n at a
point of the circle θ = θ0 is θ0 . The radius
of the circle is a + b cos θ0 , so its
R
sin θ0
geodesic curvature is a+b cos θ0 . Hence, κg ds = 2π sin θ0 and the holonomy is
2π − 2π sin θ0 . The circles ϕ = constant are geodesics (as they are meridians on
a surface of revolution) so κg = 0 and the holonomy is 2π − 0 = 2π.
13.6.2 For the circle v = 1 on the cone, the radius is 1 and the angle between the
principal normal n of the√circle and√the unit normal N of the cone is π/4, so
the holonomy is 2π − 2π/ 2 = (2 − 2)π. The cone is flat so if the converse of
Proposition 13.6.5 were true the holonomy around any closed curve on the cone
would be zero.
R ℓ(γ ) dϕ
13.6.3 For a general closed curve, 0
ds ds is an integer multiple of 2π, say 2nπ.
Then, (13.28) is replaced by
Z ℓ(γ )
κg ds.
2nπ −
0
134
13.6.4 As in the proof of Proposition 13.6.1, v = cos ϕtt + sin θtt1 , where t1 = N × t, so
v = ϕ̇(− sin ϕtt + cos ϕtt1 ) + cos ϕṫt + sin ϕṫt1 .
v̇
By Exercise 7.3.22, ṫt = κn N + κg t1 , ṫt1 = −κg t + τg N, so
N = (κn cos ϕ + τg sin ϕ)N
N,
v̇v = (ϕ̇ + κg )(− sin ϕtt + cos ϕtt1 ) + (κn cos ϕ + τg sin ϕ)N
by Proposition 13.6.1, and hence
k v̇v k2 = (κn cos ϕ + τg sin ϕ)2 .
13.7.1 Take the reference tangent vector field to be ξ = (1, 0), and take the simple
closed curve γ (s) = (cos s, sin s). At γ (s), we have V = (α, β), where
(cos s + i sin s)k
if k > 0,
α + iβ =
−k
(cos s − i sin s)
if k < 0.
By de Moivre’s theorem, α = cos ks, β = sin ks in both cases. Hence, the angle ψ
between V and ξ is equal to ks, and Definition 13.7.2 shows that the multiplicity
is k.
σ (ũ, ṽ), where (ũ, ṽ) 7→ (u, v) is a reparametrization map, then
13.7.2 If σ (u, v) = σ̃
∂ ṽ
∂ ṽ
σ u + βσ
σ v = α̃σ̃
σ ũ + β̃σ̃
σ ṽ =⇒ α̃ = α ∂∂uũ + β ∂u
, β̃ = α ∂∂vũ + β ∂v
. Hence, α̃
V = ασ
and β̃ are smooth if α and β are smooth. Since the components of the vectors
σ u and σ v are smooth, if V is smooth so are its components. If the components
σ v + βσ
σ v are smooth, then V.σ
σ u and V.σ
σ v are smooth functions, hence
of V = ασ
V
σ
V
σ
V
σ
V
σ
. u)
u )−F ( . v )
v )−F (
α = G( . EG−F
, β = E( . EG−F
are smooth functions, so V is
2
2
smooth.
13.7.3 If ψ̃ is the angle between V and ξ̃ξ , we have ψ̃ − ψ = θ (up to multiples of 2π); so
R ℓ(γ )
θ̇ ds = 0 (a dot denotes d/ds). This is not obvious since
we must show that 0
θ is not a well defined smooth function of s (although dθ/ds is well defined).
However, ρ = cos θ is well defined and smooth, since ρ = ξ .ξ̃ξ / k ξ kk ξ̃ξ k. Now,
R ℓ(γ ) ρ̇
√
ρ̇ = −θ̇ sin θ, so we must prove that 0
ds = 0. Using Green’s theorem,
2
1−ρ
this integral is equal to
Z
Z
ρu du + ρv dv
p
=
1 − ρ2
int(π )
π
∂
∂u
ρv
p
1 − ρ2
!
∂
−
∂v
ρu
p
1 − ρ2
!!
dudv,
π (s)); and this line integral vanishes
where π is the curve in U such that γ (s) = σ (π
because
!
! ∂
ρuv (1 − ρ2 ) + ρρu ρv
∂
ρv
ρu
p
p
=
=
.
∂u
∂v
(1 − ρ2 )3/2
1 − ρ2
1 − ρ2
135
σ be
13.8.1 Let F : S → R be a smooth function on a surface S, let p ∈ S, let σ and σ̃
σ (ũ0 , ṽ0 ) = p, and let f = F ◦ σ and
patches of S containing p, say σ (u0 , v0 ) = σ̃
∂u
∂v
˜
˜
˜
σ
f = F ◦ σ̃ . Then, fũ = fu ∂ ũ + fv ∂ ũ , fṽ = fu ∂u
+ fv ∂∂vṽ , so if fu = fv = 0 at
∂ ṽ
(u0 , v0 ), then f˜ũ = f˜ṽ = 0 at (ũ0 , ṽ0 ).
2
2
+ 2fuv ∂∂uũ ∂∂vũ + fvv ∂∂vũ , with
Since fu = fv = 0 at p, we have f˜ũũ = fuu ∂u
∂ ũ
˜
similar expressionsfor f˜ũṽ and
fṽṽ . This gives, in an obvious notation, H̃ =
J t HJ, where J =
∂u
∂ ũ
∂v
∂ ũ
∂u
∂ ṽ
∂v
∂ ṽ
is the Jacobian matrix of the reparametrization
map (ũ, ṽ) 7→ (u, v). Since J is invertible, H̃ is invertible if H is invertible.
Since the matrix H is real and symmetric, it has eigenvectors v1 , v2 , with eigenvalues λ1 , λ2 , say, such that vtivj = 1 if i = j and = 0 if i 6= j. Then, if
v = α1v1 + α2v2 is any vector, where α1 , α2 are scalars, vt Hvv = λ1 α21 + λ2 α22 ;
hence, vt Hvv > 0 (resp. < 0) for all v 6= 0 ⇐⇒ λ1 and λ2 are both > 0 (resp. both
< 0) ⇐⇒ p is a local minimum (resp. local maximum); and hence p is a saddle
point ⇐⇒ vt Hvv can be both > 0 and < 0, depending on the choice of v. Since
vt J t HJ ṽv = vt Hvv.
J is invertible, a vector ṽv 6= 0 ⇐⇒ v = J ṽv 6= 0; and ṽvt H̃ṽv = ṽ
The assertions in the last sentence of the exercise follow from this.
13.8.2 (i) fx = 2x − 2y, fy = −2x + 8y,
so fx = fy = 0 at the origin. fxx = 2, fxy =
2 −2
−2, fyy = 8, so H =
. H is invertible so the origin is non-degenerate;
−2 8
√
of H are
and the eigenvalues 5 ± 13 both > 0, so it is a local minimum.
2 4
(ii) fx = fy = 0 and H =
at the origin; detH = −16 < 0, so the
4 0
eigenvalues of H are of opposite sign and the origin is a saddle point.
(iii) fx = fy = 0 and H = 0 at the origin, which is therefore a degenerate critical
point.
13.8.3 Using the parametrization σ in Exercise 4.2.5 (with a = 2, b = 1) gives f (θ, ϕ) =
σ (θ, ϕ)) = (2+cos θ) cos ϕ+3. Then, fθ = − sin θ cos ϕ, fϕ = −(2+cos θ) sin ϕ;
F (σ
since 2 + cos θ > 0, fϕ = 0 =⇒ ϕ = 0 or π, and then fθ = 0 =⇒ θ = 0 or π;
p = (3, 0, 0), q = (1, 0, 0), r= (−1,
so there are four critical points,
0, 0) and
− cos θ cos ϕ
sin θ sin ϕ
−1 0
s = (−3, 0, 0). Next, H =
=
−(2 + cos
0 −3
sin θ sin
ϕ
θ) cosϕ
1 0
−1 0
1 0
at p, =
at q, =
at r, and =
at s; hence, p is a
0 −1
0 1
0 3
local maximum, q and r are saddle points, and s is a local minimum (all of which
is geometrically obvious).
13.8.4 Since the torus is compact, any smooth function f on the torus must achieve
its maximum and minimum values, and these are of course critical points of
multiplicity 1. Since the Euler number of the torus is zero, Theorem 13.8.6
implies that f must have at least two saddle points (more if f has other local
136
maxima or minima).
13.8.5 The position of the two rods can be specified by giving the angles θ and ϕ made
by the rods with the vertical. The bijection takes this configuration of the rods
to the point σ (θ, ϕ) of the torus, in the notation of the solution of Exercise 4.2.5.
The potential energy is E = 21 aα cos θ + (a + 21 b)β cos ϕ + constant, where a, b
are the lengths of the two rods and α, β are positive constants; write this as
E = λ cos θ + µ cos ϕ + ν for simplicity (where λ, µ > 0 and ν are constants). For
a critical point we must have ∂E/∂θ = ∂E/∂ϕ = 0, which gives sin θ = sin ϕ = 0.
Thus, there are four critical points, namely (θ, ϕ) = (0, 0), (π, 0), (0, π), (π, π).
In the notation of Definition 13.8.3,
H=
−λ cos θ
0
0
−λ cos ϕ
,
which is obviously invertible if θ, ϕ = 0 or π, so all the critical points are nondegenerate.
At (0, 0) the matrix H has two negative eigenvalues, at (π, 0) and (0, π) it has
one, and at (π, π) it has none. So (0, 0) is a local maximum, (π, 0) and (0, π) are
saddle points and (π, π) is a local minimum. The left-hand side of the equation
in the statement of Theorem 13.8.6 is 1 − 2 + 1 = 0, which is indeed the Euler
number of the torus (Example 13.8.7).
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