HOMEWORK 4 - SOLUTIONS-2012

ANDRÉ NEVES

Exercise 8 of Chapter 4.2 of Do Carmo:

The first step is to show that G is a linear map. Fix a point p and consider
α1 (t) = p + tX, where X is any vector in R3 . Set F (t) = G ◦ α1 (t) − G(q).
By differentiation in the t variable when t = 0 we have

|F (t)|2 = |tX + p − q|2 =⇒ F 0 (0).F (0) = X.(p − q)

=⇒ DGp (X).(G(p)−G(q)) = X.(p−q) =⇒ X.(DGp )> (G(p)−G(q)) = X.(p−q)
for all X and all q. In the last line we use the fact that if A is a 3 × 3
matrix with transpose AT , then A(X).Y = X.AT (Y ) for every vectors X, Y .
Because the identity we derived is valid for all X it is simple to conclude
that
(DGp )> (G(p) − G(q)) = p − q for every q ∈ R3 .
In particular the linear map (DGp )T is surjective and thus injective. Denot-
ing by B its inverse we have from the above formula that
G(p) = G(q) + B(p − q)for every p, q ∈ R3 .
Using again the hypothesis we obtain that
|B(p − q)| = |G(p) − G(q)| = |p − q|.
Using Exercise 7 of Chapter 4.2 (more precisely, that b) is equivalent to a)
– which is simple to show) it follows at once that G is a linear isometry of R3 .

Exercise 11 of Chapter 4.2 of Do Carmo:

First a). From the previous exercise we know that F is a linear isometry
of R3 , which means that DFp = DF0 for all p in R3 (i.e. F is linear) and
DF0 (X).DF0 (Y ) = X.Y for all vectors X, Y in R3 . In particular for every
p ∈ S we have that DFp = DF0 and DFp (X).DFp (Y ) = X.Y for all X, Y
in Tp S, which means that F is an isometry of S. Technically speaking one
should show that if F is an ambient map of R3 , then the restriction of F to
S has the property that for every X ∈ Tp S, then DFp (X) as it was defined
in class is nothing but DFp (X) as it was defined in multivariable calculus.
This is just the chain rule and so I will not say anything else to avoid the
risk of just ending up making it more confusing.
1
2 ANDRÉ NEVES

Now b). The orthogonal linear transformations are just those 3 × 3 ma-
trices which have AT A = Id, i.e., A−1 = AT . In this case we have
|A(X)|2 = A(X).A(x) = X.AT A(X) = X.X = |X|2
and so A send the unit sphere into the unit sphere. Moreover |A(X) −
A(Y )| = |A(X − Y )| = |X − Y |, and so its distance preserving in the sense
of Exercise 8. Thus a) can be applied to conclude A is an isometry of the
sphere.
Finally c). Take F (x, y, 0) = (cos x, sin x, y) which is an isometry from
part of the plane {z = 0} into part of the cylinder {x2 + y 2 = 1}. F is
not distance preserving because |F (0, 0, 0) − F (θ, 0, 0)| = 2(1 − cos θ) and
|(0, 0, 0) − (θ, 0, 0)| = |θ|.

Exercise12 of Chapter 4.2 of Do Carmo:

Consider F (x, y, z) = (x, −y, −z). This is an isometry of the cylinder

S = {x2 + y 2 = 1} because F send the S into S and F is an orthogonal
linear transformation. Finally F (x, y, z) = (x, y, z) then z = 0 and y = 0.
But x2 + y 2 = 1, which means x = ±1, i.e., the only fixed points are (1, 0, 0)
and (−1, 0, 0).

Exercise 2 of Chapter 4.3 of Do Carmo:

Use the formula which is in Exercise 1 of same chapter and which was proven
in class. Set λ = e2u for convenience. Then we have
√ ∂x2 E ∂x1 G
EG = e2u , 1/2
= 2∂x2 u, = 2∂x1 u.
(EG) (EG)1/2
Thus, recalling that ∆h = ∂x1 x1 h + ∂x2 x2 h, we obtain
1
K = −e−2u ∆u = − ∆ ln λ.
2λ
√ 1
(Recall that ln λ = 2 ln λ).
To be consistent with the rest of the notation I should use λ = (x21 + x22 +
−2
c) for the last part. Then
4xi
∂xi ln λ = −2∂xi ln(x21 + x22 + c) = − 2
x1 + x22 + c
and
4 8x2i
∂x2i xi ln λ = − + .
x21 + x22 + c (x21 + x22 + c)2
Thus
8 8(x21 + x22 ) −8c
∆ ln λ = − 2 2 + 2 2 = 2 = −8cλ
x1 + x2 + c (x1 + x2 + c) 2 (x1 + x22 + c)2
and so
1
K=− ∆ ln λ = 4c.
2λ
 HOMEWORK 4 - SOLUTIONS-2012 3

Exercise 3 of Chapter 4.3 of Do Carmo:

We have
∂x ∂x
= (cos v, sin v, u−1 ), = (−u sin v, u cos v, 0)
∂u ∂v
and so
∂x ∂x ∂x ∂x ∂x ∂x
E= . = 1 + u−2 , F = . = 0, G = . = u2 .
∂u ∂u ∂u ∂v ∂v ∂v
Because ∂v E = 0 we obtain from Exercise 1
 
1 ∂ 2u 1
K̄ = − √ √ =−
2 1+u 2 ∂u 1+u 2 (1 + u2 )2
We have
∂ x̄ ∂ x̄
= (cos v, sin v, 0), = (−u sin v, u cos v, 0)
∂u ∂v
and so
∂ x̄ ∂ x̄ ∂ x̄ ∂ x̄ ∂ x̄ ∂ x̄
Ē = . = 1, F̄ = . = 0, G = . = 1 + u2 .
∂u ∂u ∂u ∂v ∂v ∂v
Using the formula in Exercise 1 we have
 
1 ∂ 2u 1
K̄ = − √ √ =− .
2 1+u 2 ∂u 1+u 2 (1 + u2 )2
Therefore the surfaces have the same curvature.
If the map x̄ ◦ x−1 were an isometry then by Proposition 1, page 220, we
would have E = Ē, F = F̄ , and G = Ḡ. Because this is false the surfaces
are not isometric.

Exercise 6 of Chapter 4.3 of Do Carmo:

If there was a surface with E = G = 1, F = 0 and e = −g = 1, f = 0, then

from the definition of Gaussian curvature we would have K = (eg −f )(EG−
F )−1 = −1. But from Gauss’s Theorem we know that K is intrinsic, i.e.,
depends only on E, F and G. Thus K must be zero because we can surely
put coordinates on a plane which have E = G = 1, F = 0 and a plane has
Gaussian curvature zero.

Exercise: Let S be a compact surface in R3 with no boundary which has

positive Gaussian curvature and denote the Gauss map by
N : S −→ {x2 + y 2 + z 2 = 1}.
Show that if γ is a geodesic in S which divides S into a set A and another
set B (i.e., S = A ∪ B and ∂A = ∂B = γ), then area(N (A)) = area(N (B)).
4 ANDRÉ NEVES

Hint: If you have a diffeomorphism

F : S −→ {x2 + y 2 + z 2 = 1} ⊂ R3 ,
what is the formula for area(F (A)) in terms of DF and the set A?

Because K > 0, we have from Gauss-Bonnet that S is diffeomorphic to a

sphere. Thus A andB are both diffeomorphic to a disc and ∂A = ∂B = γ is
a geodesic. Hence, if ν denotes the interior unit normal to ∂A we have
Z Z Z
KdA + k.νdσ = 2π =⇒ KdA = 2π.
A ∂A A
Likewise Z
KdA = 2π.
B
I will show now that
Z
area(N (A)) = KdA = 2π
A
and this completes the exercise.
The first remark is that the Gaussian map N must be a diffeomorphism.
Why? Well, because K = detA = −detdN , we have from the inverse func-
tion theorem that N is locally a diffeomorphism. And it happens that locally
diffeomorphisms from a sphere into a sphere must be global diffeomorphism.
The second remark is to show the hint, namely that
Z
area(F (A)) = |detdF |dA.
A
Suppose that φ : D ⊂ R2 −→ A is a chart for A. Then ψ = F ◦ φ is a
chart for N = {x2 + y 2 + z 2 = 1} because F is a diffeomorphism. From the
lectures we know that
Z  
 ∂ψ ∂ψ 
area(ψ(D)) =  ×  dxdy.
D ∂x ∂y 

Now let {e1 , e2 } be a basis for Tφ(p) A which has dFp (e1 ) = λ1 e1 and dFp (e2 ) =
λ2 e2 , i.e., {e1 , e2 } is an eigenbasis for dF . Then
∂φ ∂φ
(p) = a1 e1 + b1 e2 (p) = a2 e1 + b2 e2
∂x ∂y
and thus  
 ∂φ ∂φ 
 ∂x × ∂y  = |a1 b2 − a2 b1 ||e1 × e2 |.
 

Furthermore
 
∂ψ ∂φ
(p) = dFφ(p) (p) = λ1 a1 e1 + λ2 b1 e2
∂x ∂x
and  
∂ψ ∂φ
(p) = dFφ(p) (p) = λ1 a2 e1 + λ2 b2 e2 .
∂y ∂y
 HOMEWORK 4 - SOLUTIONS-2012 5

Thus
   
 ∂ψ ∂ψ   ∂φ ∂φ 
 ∂x × ∂y  = |λ1 λ2 ||a1 b2 − a2 b1 ||e1 × e2 | = |detdFφ(p) |  ∂x × ∂y  .
   

(This formula was long to derive but it should be highly expected).

Therefore we have
Z  
 ∂ψ ∂ψ 
area(F (φ(D))) = area(ψ(D)) =  ×  dxdy.
D ∂x ∂y 

Z   Z
 ∂φ ∂φ 
= |detdFφ(p) | 
 ×  dxdy = |detdF |dA.
D ∂x ∂y  φ(D)

Covering A with a finite number of charts we arrive at

Z
area(F (A)) = |detdF |dA.
A
Using this formula with F = N the Gauss map we obtain
Z Z Z
area(N (A)) = |detdN |dA = |K|dA = KdA = 2π.
A A A

Exercise: Let S be the cylinder {x2 + y 2 = 1} in R3 , with the chart

φ : [0, 2π] × R −→ S, φ(θ, z) = (cos θ, sin θ, z).
and let γ be a simple closed curve in S.

(1) Let γ be a simple closed curve in S for which there is a path

α : [0, 1] −→ [0, 2π] × R
so that α(0) = (0, z0 ), α(1) = (2π, z0 ), and γ = φ ◦ α. In other
words, γ loops once around the z-axis. If ~ν denotes a unit normal
vector to γ show that
Z
~k.~ν dσ = 0.
γ

(2) Let γ be a simple closed curve in S for which there is a path

α : [0, 1] −→ [0, 2π] × R
so that α(0) = (θ0 , z0 ) = α(1). In other words, γ is the boundary of
a disc D in S. Show that γ is not a geodesic.
If the path α(t) = (θ(t), z(t)), choose z1 so that z(t) > z1 for all t ∈ [0, 1]
and consider the path β(t) = (2πt, z1 ). Note that φ ◦ β is a geodesic and
there is a region R in S so that ∂S = γ ∪ φ ◦ β. To recognize this last
fact just let R = φ(A), where A is the region in [0, 2π] × R below α and
6 ANDRÉ NEVES

above β. Note that R is diffemorphic to a cylinder and thus has zero Euler
characteristic. Using Gauss-Bonnet we obtain that
Z Z Z
~
KdA + k.~ν dσ + ~k.~ν dσ = 0,
R γ φ◦β
where ~ν is the interior unit normal. Because K = 0 and φ ◦ β is a geodesic
we obtain that Z
~k.~ν dσ,
γ
which is what we wanted to show.

(1) Let γ be a simple closed curve in S for which there is a path

α : [0, 1] −→ [0, 2π] × R
so that α(0) = (θ0 , z0 ) = α(1). In other words, γ is the boundary of
a disc D in S. Show that γ is not a geodesic.

The Euler characteristic of D is one and so by the Gauss-Bonnet Theorem

we have Z Z Z
~
KdA + k.~ν dσ = 2π =⇒ ~k.~ν dσ = 2π.
D γ γ
Thus, γ cannot be a geodesic.