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

Computational Geometry

Description: 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, Inverse, Continuous, Two Metrics, Riemannian, Standard Di Erential Structure, Gaussian Curvature, Vector Eld, Satisfying
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Solutions for Midterm Exam Differential Geometry II April 22, 2010
Professor: Tommy R. Jensen
Question 1 Is the following mapping xa proper patch?
x:R2R3,(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 x1maps (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 x1is not differentiable, we would have to
find a patch yin Mso that x1(y) : R2R2is 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 xis 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 F1= (x, y, x2+y2).
If x:DPis any patch in P, then F(x) is a differentiable function
from Dto R2.So Fis a mapping of surfaces. Since F1is 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:RMbe a 2-segment, where R=
[0,1] ×[0,1].
Let φbe the 1-form on Msuch that φ(xu) = uvand φ(xv) = uv.
Calculate R Rxand Rxφdirectly from their definitions. Check that
your computations agree with Stokes’s Theorem.
Z Zx
=Z1
0Z1
0
(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
Zx
φ=Zα
φ+Zβ
φZγ
φZδ
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
(t0)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
(t1)dt =1
2,
Zδ
φ=Z1
0
φ(xv(0, t))dt =Z1
0
0dt = 0.
Altogether:
Zα
φ+Zβ
φZγ
φZδ
φ=1
2+1
2+1
2+ 0 = 3
2.
The calculations show
Z Zx
=Zx
φ=3
2.
This agrees with Stokes’s Theorem.
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