Assignment 8 for Differential Geometry | MATH 423, Assignments of Geometry

Material Type: Assignment; Class: Differential Geometry; Subject: Mathematics; University: University of Illinois - Urbana-Champaign; Term: Fall 2006;

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Math 423 Differential Geometry Fall 2006
Homework 8: More on Curvature
Due Thursday Nov. 9
โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”
1. Prove that โˆ‡ยฏxuโˆ‡ยฏxvU=โˆ‡ยฏxvโˆ‡ยฏxuU
2. Prove that a surface of revolution with K= 0 is part of a cone or a plane. Hint:
model your proof after our proof that a surface of revolution with H= 0 is part
of a catenoid or a plane.
3. Compute the Gaussian curvature of the sphere of radius Rusing the formula from
Gaussโ€™s Theorem Egregium:
K=โˆ’1
2โˆšEG ๎˜’โˆ‚
โˆ‚v ๎˜’Ev
โˆšEG ๎˜“+โˆ‚
โˆ‚u ๎˜’Gu
โˆšEG ๎˜“๎˜“.
โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”โ€”
Bonus Questions for those taking the course for four credits or for those who
want more of a challenge.
4. Show that the Uterms in the equations xuuv โˆ’xuvu = 0 and xvvu โˆ’xuvu = 0 give
Lvโˆ’Mu=M๎˜’Gu
2Gโˆ’Eu
2E๎˜“+EvH=LEv
2E+M๎˜’Gu
2Gโˆ’Eu
2E๎˜“+NEv
2G
and
Nuโˆ’Mv=M๎˜’Ev
2Eโˆ’Gv
2G๎˜“+GuH=LGu
2E+M๎˜’Ev
2Eโˆ’Gv
2G๎˜“+NGu
2G.
What do these equations say in the case when H= 0 and the parameterization
satisfies E=G?
1

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Math 423 Differential Geometry Fall 2006

Homework 8: More on Curvature

Due Thursday Nov. 9

  1. Prove that โˆ‡ยฏxu โˆ‡xยฏv U = โˆ‡xยฏv โˆ‡ยฏxu U
  2. Prove that a surface of revolution with K = 0 is part of a cone or a plane. Hint:

model your proof after our proof that a surface of revolution with H = 0 is part of a catenoid or a plane.

  1. Compute the Gaussian curvature of the sphere of radius R using the formula from

Gaussโ€™s Theorem Egregium:

K = โˆ’

EG

โˆ‚v

Ev โˆš EG

โˆ‚u

Gu โˆš EG

Bonus Questions for those taking the course for four credits or for those who

want more of a challenge.

  1. Show that the U terms in the equations xuuv โˆ’ xuvu = 0 and xvvu โˆ’ xuvu = 0 give

Lv โˆ’ Mu = M

Gu

2 G

Eu

2 E

  • Ev H = L

Ev

2 E

+ M

Gu

2 G

Eu

2 E

+ N

Ev

2 G

and

Nu โˆ’ Mv = M

Ev

2 E

Gv

2 G

  • GuH = L

Gu

2 E

+ M

Ev

2 E

Gv

2 G

+ N

Gu

2 G

What do these equations say in the case when H = 0 and the parameterization satisfies E = G?