Solutions for Midterm Exam Diﬀerential Geometry II April 22, 2010

Professor: Tommy R. Jensen

Question 1 Is the following mapping xa proper patch?

x:R2→R3,(u, v)7→ (u3, u +v3, v).

Prove that your answer is correct. Is xa diﬀeomorphism?

It is easy to see that xis injective. The partial velocities are xu=

(3u2,1,0) and xv= (0,3v2,1).They are linearly independent, so xis

a patch. The inverse function x−1maps (p1, p2, p3) to ( 3

√p1, p3),which is

a continuous function, so xis proper. It is not a diﬀeomorphism, because

3

√xis not diﬀerentiable at x= 0.

Correction. To prove that x−1is not diﬀerentiable, we would have to

ﬁnd a patch yin Mso that x−1(y) : R2→R2is not diﬀerentiable in

the usual sense (Deﬁnition 5.1). This will not be possible, because in fact

the inverse of xis diﬀerentiable, which follows from Theorem 5.4. So the

correct answer is that xis actually a diﬀeomorphism!

Question 2 Calculate a parametrization of the surface obtained by revolving the curve

C:x= 2 −cos z(−π/2< z < 3π/2) around the z-axis.

x(u, v) = ((2 −cos u) cos v, (2 −cos u) sin v, u).

Question 3 Let Mbe a surface and let xbe a patch in M. Show that if x∗is the

tangent map of x,then x∗(U1) = xuand x∗(U2) = xv,where {U1, U2}

is the natural frame ﬁeld on R2,and xu,xvare the partial velocities of x.

x∗(U1)(p) = x∗(U1(p)) = x∗((1,0)p)

= (x(α))0(0) (where α(t) = p+ (1,0)t)

=d

dtx(p+ (1,0)t)|t=0

=d

dt(p1+t)∂

∂ux(p) + d

dt(p2)∂

∂v x(p)

=∂

∂ux(p)

=xu(p)

The calculation for x∗(U2)(p) is similar.

Question 4 The paraboloid is the surface P:z=x2+y2in R3.Prove that Pis

diﬀeomorphic to R2.

Show that Pis also diﬀeomorphic to Σ0(the sphere without the north

pole).

The function (p1, p2, p3)7→ (p1, p2) is clearly a bijection Ffrom Pto R2,

with inverse map F−1= (x, y, x2+y2).

If x:D→Pis any patch in P, then F(x) is a diﬀerentiable function

from Dto R2.So Fis a mapping of surfaces. Since F−1is diﬀerentiable,

it follows that Fis a diﬀeomorphism.

It again follows that Pis diﬀeomorphic to Σ0,since Σ0is also diﬀeomor-

phic to R2,as shown in the lectures.

Question 5 Let Mbe a surface and let x:R→Mbe a 2-segment, where R=

[0,1] ×[0,1].

Let φbe the 1-form on Msuch that φ(xu) = u−vand φ(xv) = uv.

Calculate R Rxdφ and R∂xφdirectly from their deﬁnitions. Check that

your computations agree with Stokes’s Theorem.

Z Zx

dφ =Z1

0Z1

0

dφ(xu, xv)dudv

=Z1

0Z1

0∂

∂u(φ(xv)) −∂

∂v (φ(xu))dudv

=Z1

0Z1

0

(v+ 1)dudv

= [1

2v2+v]1

0

=3

2.

To calculate the integral of φover the boundary of x,let α(u) = x(u, 0),

β(v) = x(1, v), γ(u) = x(u, 1) and δ(v) = x(0, v).Then

Z∂x

φ=Zα

φ+Zβ

φ−Zγ

φ−Zδ

φ.

The line integral of φover αis by deﬁnition

Zα

φ=Z1

0

φ(α0(t))dt

=Z1

0

φ(∂

∂tx(t, 0))dt

=Z1

0

φ(xu(t, 0))dt

=Z1

0

(t−0)dt

=1

2.

The similar calculations for β, γ and δgive

Zβ

φ=Z1

0

φ(xv(1, t))dt =Z1

0

tdt =1

2,

Zγ

φ=Z1

0

φ(xu(t, 1))dt =Z1

0

(t−1)dt =−1

2,

Zδ

φ=Z1

0

φ(xv(0, t))dt =Z1

0

0dt = 0.

Altogether:

Zα

φ+Zβ

φ−Zγ

φ−Zδ

φ=1

2+1

2+1

2+ 0 = 3

2.

The calculations show

Z Zx

dφ =Z∂x

φ=3

2.

This agrees with Stokes’s Theorem.

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