HOMEWORK 4 - SOLUTIONS-2012

ANDRE 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 considerα1(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).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 × 3matrix 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 concludethat

(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 isometryof R3, which means that DFp = DF0 for all p in R3 (i.e. F is linear) andDF0(X).DF0(Y ) = X.Y for all vectors X,Y in R3. In particular for everyp ∈ S we have that DFp = DF0 and DFp(X).DFp(Y ) = X.Y for all X,Yin TpS, which means that F is an isometry of S. Technically speaking oneshould show that if F is an ambient map of R3, then the restriction of F toS has the property that for every X ∈ TpS, then DFp(X) as it was definedin 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 therisk of just ending up making it more confusing.

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2 ANDRE NEVES

Now b). The orthogonal linear transformations are just those 3 × 3 ma-trices which have ATA = Id, i.e., A−1 = AT . In this case we have

|A(X)|2 = A(X).A(x) = X.ATA(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 senseof Exercise 8. Thus a) can be applied to conclude A is an isometry of thesphere.

Finally c). Take F (x, y, 0) = (cosx, sinx, y) which is an isometry frompart of the plane {z = 0} into part of the cylinder {x2 + y2 = 1}. F isnot 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 cylinderS = {x2 + y2 = 1} because F send the S into S and F is an orthogonallinear transformation. Finally F (x, y, z) = (x, y, z) then z = 0 and y = 0.But x2 + y2 = 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 provenin class. Set λ = e2u for convenience. Then we have

√EG = e2u,

∂x2E

(EG)1/2= 2∂x2u,

∂x1G

(EG)1/2= 2∂x1u.

Thus, recalling that ∆h = ∂x1x1h+ ∂x2x2h, we obtain

K = −e−2u∆u = − 1

2λ∆ lnλ.

(Recall that ln√λ = 1

2 lnλ).

To be consistent with the rest of the notation I should use λ = (x21 +x2

2 +c)−2 for the last part. Then

∂xi lnλ = −2∂xi ln(x21 + x2

2 + c) = − 4xix2

1 + x22 + c

and

∂2xixi lnλ = − 4

x21 + x2

2 + c+

8x2i

(x21 + x2

2 + c)2.

Thus

∆ lnλ = − 8

x21 + x2

2 + c+

8(x21 + x2

2)

(x21 + x2

2 + c)2=

−8c

(x21 + x2

2 + c)2= −8cλ

and so

K = − 1

2λ∆ lnλ = 4c.

HOMEWORK 4 - SOLUTIONS-2012 3

Exercise 3 of Chapter 4.3 of Do Carmo:

We have

∂x

∂u= (cos v, sin v, u−1),

∂x

∂v= (−u sin v, u cos v, 0)

and so

E =∂x

∂u.∂x

∂u= 1 + u−2, F =

∂x

∂u.∂x

∂v= 0, G =

∂x

∂v.∂x

∂v= u2.

Because ∂vE = 0 we obtain from Exercise 1

K = − 1

2√

1 + u2

∂

∂u

(2u√

1 + u2

)= − 1

(1 + u2)2

We have∂x

∂u= (cos v, sin v, 0),

∂x

∂v= (−u sin v, u cos v, 0)

and so

E =∂x

∂u.∂x

∂u= 1, F =

∂x

∂u.∂x

∂v= 0, G =

∂x

∂v.∂x

∂v= 1 + u2.

Using the formula in Exercise 1 we have

K = − 1

2√

1 + u2

∂

∂u

(2u√

1 + u2

)= − 1

(1 + u2)2.

Therefore the surfaces have the same curvature.If the map x ◦ x−1 were an isometry then by Proposition 1, page 220, we

would have E = E, F = F , and G = G. Because this is false the surfacesare 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, thenfrom 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 surelyput coordinates on a plane which have E = G = 1, F = 0 and a plane hasGaussian curvature zero.

Exercise: Let S be a compact surface in R3 with no boundary which haspositive Gaussian curvature and denote the Gauss map by

N : S −→ {x2 + y2 + z2 = 1}.Show that if γ is a geodesic in S which divides S into a set A and anotherset B (i.e., S = A∪B and ∂A = ∂B = γ), then area(N(A)) = area(N(B)).

4 ANDRE NEVES

Hint: If you have a diffeomorphism

F : S −→ {x2 + y2 + z2 = 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 asphere. Thus A andB are both diffeomorphic to a disc and ∂A = ∂B = γ isa geodesic. Hence, if ν denotes the interior unit normal to ∂A we have∫

AKdA+

∫∂Ak.νdσ = 2π =⇒

∫AKdA = 2π.

Likewise ∫BKdA = 2π.

I will show now that

area(N(A)) =

∫AKdA = 2π

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 locallydiffeomorphisms from a sphere into a sphere must be global diffeomorphism.

The second remark is to show the hint, namely that

area(F (A)) =

∫A|detdF |dA.

Suppose that φ : D ⊂ R2 −→ A is a chart for A. Then ψ = F ◦ φ is achart for N = {x2 + y2 + z2 = 1} because F is a diffeomorphism. From thelectures we know that

area(ψ(D)) =

∫D

∣∣∣∣∂ψ∂x × ∂ψ

∂y

∣∣∣∣ dxdy.Now let {e1, e2} be a basis for Tφ(p)A which has dFp(e1) = λ1e1 and dFp(e2) =λ2e2, i.e., {e1, e2} is an eigenbasis for dF . Then

∂φ

∂x(p) = a1e1 + b1e2

∂φ

∂y(p) = a2e1 + b2e2

and thus ∣∣∣∣∂φ∂x × ∂φ

∂y

∣∣∣∣ = |a1b2 − a2b1||e1 × e2|.

Furthermore

∂ψ

∂x(p) = dFφ(p)

(∂φ

∂x(p)

)= λ1a1e1 + λ2b1e2

and∂ψ

∂y(p) = dFφ(p)

(∂φ

∂y(p)

)= λ1a2e1 + λ2b2e2.

HOMEWORK 4 - SOLUTIONS-2012 5

Thus∣∣∣∣∂ψ∂x × ∂ψ

∂y

∣∣∣∣ = |λ1λ2||a1b2 − a2b1||e1 × e2| = |detdFφ(p)|∣∣∣∣∂φ∂x × ∂φ

∂y

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

Therefore we have

area(F (φ(D))) = area(ψ(D)) =

∫D

∣∣∣∣∂ψ∂x × ∂ψ

∂y

∣∣∣∣ dxdy.=

∫D|detdFφ(p)|

∣∣∣∣∂φ∂x × ∂φ

∂y

∣∣∣∣ dxdy =

∫φ(D)|detdF |dA.

Covering A with a finite number of charts we arrive at

area(F (A)) =

∫A|detdF |dA.

Using this formula with F = N the Gauss map we obtain

area(N(A)) =

∫A|detdN |dA =

∫A|K|dA =

∫AKdA = 2π.

Exercise: Let S be the cylinder {x2 + y2 = 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 otherwords, γ loops once around the z-axis. If ~ν denotes a unit normalvector to γ show that ∫

γ

~k.~νdσ = 0.

(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 ofa 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 andthere is a region R in S so that ∂S = γ ∪ φ ◦ β. To recognize this lastfact just let R = φ(A), where A is the region in [0, 2π] × R below α and

6 ANDRE NEVES

above β. Note that R is diffemorphic to a cylinder and thus has zero Eulercharacteristic. Using Gauss-Bonnet we obtain that∫

RKdA+

∫γ

~k.~νdσ +

∫φ◦β

~k.~νdσ = 0,

where ~ν is the interior unit normal. Because K = 0 and φ ◦ β is a geodesicwe obtain that ∫

γ

~k.~νdσ,

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]× Rso that α(0) = (θ0, z0) = α(1). In other words, γ is the boundary ofa disc D in S. Show that γ is not a geodesic.

The Euler characteristic of D is one and so by the Gauss-Bonnet Theoremwe have ∫

DKdA+

∫γ

~k.~νdσ = 2π =⇒∫γ

~k.~νdσ = 2π.

Thus, γ cannot be a geodesic.