# 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).
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.