# Smooth - Differential Geometry - Solved Exam, Exams for Computational Geometry. Aligarh Muslim University

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This is the Solved Exam of Differential Geometry which includes Normal Vector, Normal Vector, Binormal Vector, Curvature, Torsion, Binormal Vector, Speed Space Curve etc. Key important points are: Smooth, Bijective, Inv...
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Solutions for Midterm Exam Differential Geometry II April 22, 2010

Professor: Tommy R. Jensen

Question 1 Is the following mapping x a proper patch?

x : R2 → R3, (u, v) 7→ (u3, u+ v3, v).

It is easy to see that x is injective. The partial velocities are xu = (3u2, 1, 0) and xv = (0, 3v

2, 1). They are linearly independent, so x is a patch. The inverse function x−1 maps (p1, p2, p3) to ( 3

√ p1, p3), which is

a continuous function, so x is proper. It is not a diffeomorphism, because 3 √ x is not differentiable at x = 0.

Correction. To prove that x−1 is not differentiable, we would have to find a patch y in M so that x−1(y) : R2 → R2 is not differentiable in the usual sense (Definition 5.1). This will not be possible, because in fact the inverse of x is differentiable, which follows from Theorem 5.4. So the correct answer is that x is 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− cosu) cos v, (2− cosu) sin v, u).

Question 3 Let M be a surface and let x be a patch in M. Show that if x∗ is the tangent map of x, then x ∗ (U1) = xu and x ∗ (U2) = xv, where {U1, U2} is the natural frame field on R2, and xu,xv are the partial velocities of x.

x ∗ (U1)(p) = x ∗ (U1(p)) = x ∗ ((1, 0)p) = (x(α))′(0) (where α(t) = p+ (1, 0)t)

= d

dt x(p+ (1, 0)t)|t=0

= d

dt (p1 + t)

∂u x(p) +

d

dt (p2)

∂v x(p)

= ∂

∂u x(p)

= xu(p)

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

Question 4 The paraboloid is the surface P : z = x2 + y2 in R3. Prove that P is diffeomorphic to R2. Show that P is also diffeomorphic to Σ0 (the sphere without the north pole).

The function (p1, p2, p3) 7→ (p1, p2) is clearly a bijection F from P to R2, with inverse map F−1 = (x, y, x2 + y2).

If x : D → P is any patch in P, then F (x) is a differentiable function from D to R2. So F is a mapping of surfaces. Since F−1 is differentiable, it follows that F is a diffeomorphism.

It again follows that P is diffeomorphic to Σ0, since Σ0 is also diffeomor- phic to R2, as shown in the lectures.

Question 5 Let M be a surface and let x : R → M be a 2-segment, where R = [0, 1]× [0, 1]. Let φ be the 1-form on M such that φ(xu) = u− v and φ(xv) = uv. Calculate

∫ ∫ x dφ and

∫ ∂x φ directly from their definitions. Check that

your computations agree with Stokes’s Theorem.

∫ ∫ x dφ =

∫ 1 0

∫ 1 0

dφ(xu, xv)dudv

=

∫ 1 0

∫ 1 0

( ∂

∂u (φ(xv))−

∂v (φ(xu))

) dudv

=

∫ 1 0

∫ 1 0

(v + 1)dudv

= [ 1

2 v2 + v]10

= 3

2 .

To calculate the integral of φ over 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∫

∂x φ =

∫ α

φ+

∫ β

φ− ∫ γ

φ− ∫ δ

φ.

The line integral of φ over α is by definition∫ α

φ =

∫ 1 0

φ(α′(t))dt

=

∫ 1 0

φ( ∂

∂t x(t, 0))dt

=

∫ 1 0

φ(xu(t, 0))dt

=

∫ 1 0

(t− 0)dt

= 1

2 .

The similar calculations for β, γ and δ give∫ β

φ =

∫ 1 0

φ(xv(1, t))dt =

∫ 1 0

tdt = 1

2 ,

∫ γ

φ =

∫ 1 0

φ(xu(t, 1))dt =

∫ 1 0

(t− 1)dt = −1 2 ,∫

δ

φ =

∫ 1 0

φ(xv(0, t))dt =

∫ 1 0

0dt = 0.

Altogether: ∫ α

φ+

∫ β

φ− ∫ γ

φ− ∫ δ

φ = 1

2 +

1

2 +

1

2 + 0 =

3

2 .

The calculations show ∫ ∫ x dφ =

∫ ∂x φ =

3

2 .

This agrees with Stokes’s Theorem.