Mat 4030

HW 3 Solution

20 Oct, 2007

1 do Carmo P.65 Q.2

S1 = {(x, y, z) ∈ R3|z = 0 and x2 + y2 ≤ 1} is not a regular surface. Forotherwise, there exists a coordinate patch x : D → S1 containing the point(1, 0, 0). One can easily observed that the connected component of x(D) mustbe homeomorphic to H = {(u, v) ∈ R2|u2 + v2 < 1 and v ≥ 0}. But H and Dare not homeomorphic: Suppose there is a homeomorphism f : H → D, thenwe have H − {(0, 0)} is homeomorphic to D − {f(0, 0)}, but the former set issimply connected while the latter is not. So there is a contradiction.

S2 = {(x, y, z) ∈ R3|z = 0 and x2 + y2 < 1} is a regular surface. LetU = {(u, v) ∈ R2|u2 + v2 < 1} ⊂ R2 is the unit open disc. Then the mapx : U → S2 given by x(u, v) = (u, v, 0) is a coordinate patch which covers S2,with xu = (1, 0, 0),xv = (0, 1, 0) and xu × xv = (0, 0, 1) 6= 0.

2 do Carmo P.66 Q.11

(a) x(u, v) = (u + v, u − v, 4uv), (u, v) ∈ R2. As (u + v)2 + (u − v)2 =4uv, x(R2) ⊂ S. Clearly, x is injective and differentiable with xu =(1, 1, 4v),xv = (1,−1, 4u) and xu × xv = (4u + 4v,−4u + 4v,−2) 6= 0. Sox is a parametrization for S.

For any point (x, y, z) ∈ S, ∃(x+y2 , x−y

2 ) ∈ R2 such that

x(x + y

2,x− y

2) = (

x + y

2+

x− y

2,x + y

2− x− y

2, 4

x + y

2x− y

2)

= (x, y, x2 − y2) = (x, y, z).

So x is surjective, i.e. x covers the whole S.

(b) x(u, v) = (u cosh v, u sinh v, u2), (u, v) ∈ R2, u 6= 0. As (u cosh v)2 +(u sinh v)2 = u2, x({(u, v) ∈ R2|u 6= 0}) ⊂ S. Clearly x is differentiablewith xu = (cosh v, sinh v, 2u),xv = (u sinh v, u cosh v, 0) and xu × xv =(−2u2 cosh v, 2u2 sinh v, u) 6= 0 as u 6= 0. To show that x is injective, letx(u1, v1) = x(u2, v2) for (u1, v1), (u2, v2) ∈ R2, ui 6= 0. Then u2

1 = u22 ⇒

1

u1 = ±u2. cosh2v1 = (u1 cosh v1)2

u21

= (u2 cosh v2)2

u22

= cosh2v2 ⇒ coshv1 =

coshv2 ⇒ u1 = u1 cosh v1cosh v1

= u2 cosh v2cosh v2

= u2 ⇒ sinh v1 = sinh v2 ⇒ ev1 =ev2 ⇒ v1 = v2. Therefore x is a parametrization for S.

We see that x covers the set {(x, y, z) ∈ S|z > 0} for any (x, y, z) ∈ Swith z > 0, ∃(

√z, log(x+y√

z)) ∈ R2 such that

x(√

z, log(x + y√

z)) = (

√z(

x+y√z

+√

zx+y

2),√

z(x+y√

z−

√z

x+y

2), (√

z)2)

= ((x + y)2 + z2

2(x + y),(x + y)2 − z2

2(x + y), z)

= ((x + y)2 + (x2 − y2)2

2(x + y),(x + y)2 − (x2 − y2)2

2(x + y), z)

= (x, y, z).

3 do Carmo P.67 Q.16

(a) The line joining N = (0, 0, 2) to the point (u, v, 0) on the xy-plane is givenby L(u,v)(t) = t(0, 0, 2) + (1 − t)(u, v, 0) = ((1 − t)u, (1 − t)v, 2t). WhenL(u,v) intersects the sphere S2, we have

[(1− t)u]2 + [(1− t)v]2 + (2t− 1)2 = 1

⇒(4 + u2 + v2)t2 − 2t(2 + u2 + v2) + u2 + v2 = 0

⇒(t− 1)[(4 + u2 + v2)t− (u2 + v2)] = 0

⇒ t = 0 or t =u2 + v2

4 + u2 + v2.

When t = 0, L(u,v)(0) = N . So we have

π−1(u, v) = L(u,v)(u2 + v2

u2 + v2 + 4) = (

4u

u2 + v2 + 4,

4v

u2 + v2 + 4,

2(u2 + v2)u2 + v2 + 4

).

(b) Note that the stereographic projection π : S2 − {N} → R2 defines acoordinate patch x = π−1 : R2 → S2 − {N} omitting the north pole N .By the same method, we can use it to define another coordinate patchy R2 → S2 − {S = (0, 0, 0)} omitting the south pole S by Y = ρ ◦ π−1

where ρ is any rotation in R3 around the point (0, 0, 1) which interchangesN and S. Then {x,y} together cover the sphere.

2

4 Oprea 2.1.16

x(u, v) = (u− u3

3+ uv2, v − u3

3+ vu2, u2 − v2)

xu = (1− u2 + v2, 2uv, 2u)

xv = (2uv, 1− v2 + u2,−2v)

xu × xv = (−2u(u2 + v2 + 1), 2v(u2 + v2 + 1), 1− (u2 + v2)2)

By putting, u = r cos θ, v = sin θ, we have

x(r, θ) = x(r cos θ, r sin θ) = (r cos θ−r3 cos3 θ

3+r3 cos θ sin2 θ, r sin θ−r3 sin3 θ

3+r3 sin θ cos2 θ, r2(cos2 θ−sin2 θ))

and

x2 + y2 +43z2 =(u− u3

3+ uv2)2 + (v − u3

3+ vu2)2 +

43(u2 − v2)2

=(r cos θ − r3 cos3 θ

3+ r3 cos θ sin2 θ)2

+ (r sin θ − r3 sin3 θ

3+ r3 sin θ cos2 θ)2

+43r4(cos2 θ − sin2 θ)2

=r2 + 2r4 cos2 θ(cos2 θ

3+ sin2 θ) + r6 cos2 θ(−cos2 θ

3+ sin2 θ)2

+ 2r4 sin2 θ(sin2 θ

3+ cos2 θ) + r6 sin2 θ(− sin2 θ

3+ cos2 θ)2

+43r4(cos2 θ − sin2 θ)2

=r2 +r4

3+

r6

9

=19r2(3 + r2)2.

From this, we know that when r takes different values, the value of x must bedifferent. So for x(r1, θ1) = x(r2, θ2), we must have r1 = r2 = r. Now assumer <

√3. Then by simplifying (x)(r, θi), we get from the third coordinate that

cos 2θ1 = cos 2θ2, cos2 θ1 = cos2 θ2, sin2 θ1 = sin2 θ2. From the first and secondcoordinates we obtain

cos theta1(r +4r3

3sin2 θ1 −

r3

3cos 2θ1) = cos theta2(r +

4r3

3sin2 θ1 −

r3

3cos 2θ1)

sin theta1(r +4r3

3cos2 θ1 −

r3

3cos 2θ1) = sin theta2(r +

4r3

3cos2 θ1 −

r3

3cos 2θ1).

It remains to show that |r+ 4r3

3 sin2 θ1− r3

3 cos 2θ1| and |r+ 4r3

3 cos2 θ1− r3

3 cos 2θ1|always greater than 0 so that we can deduce cos θ1 = cos θ2, sin θ1 = sin θ2, and

3

thus theta1 = θ2. But they follows from the estimates

|r +4r3

3sin2 θ1 −

r3

3cos 2θ1| ≥ r − r3

3= r(1− r2

3) > r(1−

√32

3) = 0,

|r +4r3

3cos2 θ1 −

r3

3cos 2θ1| ≥ r − r3

3= r(1− r2

3) > r(1−

√32

3) = 0.

5 Oprea 2.1.20

Let x(u, v) = (u, v, uv) , which is a parametrization of the saddle surface.Then x(u, v) = (u, 0, 0) + v(0, 1, u), where we have β(u) = (u, 0, 0), δ(u) =(0, 1, u),which are curves.Moreover, we can let y(u, v) = (u+v, u, u(u+v)) to be another parametrization.Then y(u, v) = (u, u, u2) + v(1, 0, u) with β(u) = (u, u, u2), δ(u) = (1, 0, u) asthe directrix and ruling.Having two different ruled patches y(u, v),x(u, v), the saddle surface z = xy isdoubly ruled.

6 Oprea 2.1.21

From example 2.1.14, we learn that the parametrization of helicoid is x(u, v) =(av cos u, bv sinu, bu), and so

x(u, v) = (0, 0, bu) + v(a cos u, a sinu, 0).

We have directrix β(u) = (0, 0, bu) and ruling δ(u) = (a cos u, a sinu, 0) thathelicoid is a ruled surface.

7 Oprea 2.1.27

• Example 2.1.5(the Monge Patch)

x(u, v) = (u, v, f(u, v))

xu = (1, 0,∂f

∂u)

xv = (0, 1,∂f

∂v)

Thus , E = 1 + (∂f∂u )2, F = (∂f

∂u )(∂f∂v ), G = 1 + (∂f

∂v )2.

• Still within the example 2.1.5, we have parametrization of paraboloid.

x(u, v) = (u, v, u2 + v2)xu = (1, 0, 2u)xv = (0, 1, 2v)

Thus,E = 1 + 4u2, F = 4uv,G = 1 + 4v2.

4

• Example 2.1.9(Geographical Coordinates)

x(u, v) = (R cos u cos v,R sinu cos v,R sin v)xu = (−R sinu cos v,R cos u cos v, 0)xv = (−R cos u sin v,−R sinu sin v,R cos v)

Thus, E = R2 cos2 v, F = 0, G = R2.

• Example 2.1.10 (Surface of Revolution)

x(u, v) = (g(u), h(u) cos v, h(u) sin v)

xu = (∂g

∂u,∂h

∂ucos v,

∂h

∂usin v)

xv = (0,−h(u) sin v, h(u) cos v)

Thus, E = ( ∂g∂u )2 + (∂h

∂u )2, F = 0, G = (h(u))2.

• Example 2.1.13 (Torus)

x(u, v) = ((R + r cos u) cos v, (R + r cos u) sin v, r sinu)xu = (−r sinu cos v,−r sinu sin v, r cos u)xv = (−(R + r cos u) sin v, (R + r cos u) cos v, 0)

Thus, E = r2, F = 0, G = (R + r cos u)2.

• Example 2.1.14 (Helicoid)

x(u, v) = (av cos u, av sinu, bu)xu = (−av sinu, av cos u, b)xv = (a cos u, a sinu, 0)

Thus, E = a2v2 + v2, F = 0, G = a2.

• Exercise 2.1.16 (Enneper’s Surface)

x(u, v) = (u− u3

3+ uv2, v − u3

3+ vu2, u2 − v2)

xu = (1− u2 + v2, 2uv, 2u)

xv = (2uv, 1− v2 + u2,−2v)

Thus, E = (1 + u2 + v2)2, F = −2uv,G = (1 + u2 + v2)2.

• Exercise 2.1.22 (Hyperboloid)

x(u, v) = (a coshu cos v, b coshu sin v, c sinhu)xu = (a sinhu cos v, b sinhu sin v, c coshu)xv = (−a coshu sin v, b coshu cos v, 0)

Thus, E = a2 cosh2 u cos2 v + b2 cosh2 u sin2 v + c2 sinh2 u,F = 0, G = a2 cosh2 u sin2 v + b2 cosh2 u cos2 v.

5

• Exercise 2.1.23 (Another patch for hyperboloid)

x(u, v) = (au− v

u + v, b

1 + uv

u + v, c

uv − 1u + v

)

xu = (a2v

(u + v)2, b

v2 − 1(u + v)2

, cv2 + 1

(u + v)2)

xv = (a2u

(u + v)2, b

u2 − 1(u + v)2

, cu2 + 1

(u + v)2)

Thus, E =1

(u + v)4(4a2v2 + b2(v2 − 1)2 + c2(v2 + 1)2)

F =1

(u + v)4(4a2uv + b2(v2 − 1)(u2 − 1) + c2(v2 + 1)(u2 + 1)

G =1

(u + v)4(4a2u2 + b2(u2 − 1)2 + c2(u2 + 1)2)

From the examples, we see that

• E ≡ 1 if and only if all u-parameter curves are parametrized by arc length.

• G ≡ 1 if and only if all v-parameter curves are parametrized by arc length.

• F ≡ 0 if and only it every pair of u-,v-parameter curves are orthogonal toeach other at every point inside the coordinate patch.

When r =√

3, we have (r, θ) = (√

3, 0), (√

3, π) takes the same point of thesurface via x.

6