Math 501: Differential Geometry

Homework 8

Lawrence Tyler Rush

<me@tylerlogic.com>

April 14, 2013

http://coursework.tylerlogic.com/courses/math501/homework08

1 do Carmo pg 237 problem 1

With F = 0, the Christoffel symbols simplify to

Γ 1 = Eu- 11 2E Γ 2 = - Ev- 11 2G Γ 1 = Ev- 12 2E Γ 2 = Gu- 12 2G 1 Gu- Γ22 = - 2E 2 Gv- Γ22 = 2G

From here, one would use the equation

2 2 1 2 2 2 2 2 1 2 - EK = (Γ12)u - (Γ11)v + Γ12Γ11 + Γ12Γ12 - Γ11Γ22 - Γ11Γ12

and expand/simplify appropriately to get the desired answer. However, I was not able to find the correct sequence of expansions/simplifications of the equation resulting from the combination of the above equations.

2 do Carmo pg 237 problem 2

For E = G = λ(u,v) and F = 0 we can use the previous problem to find K to be

(( ) ( ) ) (( ) ( ) ) -- 1- -λv-- -λu-- - 1 λv- λu- K = 2√λλ √ λλ v + √ λλ u = 2λ λ v + λ u

Thus since Δlog λ = ( )
λvλ _v + ( )
λλu _u then we have that

1 K = - 2λΔ (logλ)

Now let λ = (u² + v² + c)^-2. To make the following computation easier to follow, we’ll define γ = u² + v² + c, i.e. λ = γ^-2. We have the following value of K

K = - -1Δ (log λ) 2λ(( ) ( ) ) = - 1 λv- + λu- 2λ (λ( v λ) u ( ) ) - 1-2 - 2γ-3γv - 2γ-3γu- = 2 γ γ-2 v + γ-2 u - 1 (( ) ( ) ) = 2-γ2 - 2γ-1γv v + - 2γ-1γu u = γ2 ((γ- 1γ) + (γ- 1γ )) ( v v uu ) = γ2 (- 1)γ -2γ2v + γ-1γvv + (- 1)γ-2γ2u + γ- 1γuu = - γ2v + γγvv - γ2u + γγuu

With our definition of γ, γ_u = 4u, γ_v = 4v, and γ_uu = γ_vv = 4. Continuing the equation above we obtain

K = - γ2v +γ γvv - γ2u + γγuu = - 4v2 + (u2 + v2 +c)(2)- 4u2 + (u2 + v2 + c)(2) = 4c

Let S be a surface with geodesic coordinates X so that E = 1 and F = 0.

(a) Christoffel Symbols

Since E = 1 then E_u = E_v = 0 and thus with F = 0 most of the Christoffel symbols are zero. Making use of the equations in (2) on pg 232 of do Carmo, we have

Γ 1 = Γ 2 = Γ 1 = 0 11 11 12

as well as

Γ 2 = Gu, Γ 1 = --Gu , and Γ 2 = Gv 12 2G 22 2 22 G

(b) Gaussian Curvature

From the equations of the previous part of the problem, due to most of the Christoffel Symbols being zero, we have this small equation for the Gaussian curvature.

( 2 ) ( 2 )2 - K = (Γ12)u + Γ(12 ) Gu- Gu- 2 = 2G u + 2G ( ) 2 = 1 Gu- + (Gu)2- 2 ( G u 4)G = 1 Guu-- G2u- + (Gu-)2 2 G G2 4G2 Guu (Gu )2 = -2G-- -4G2- G (G )2 K = - -uu-+ --u2- 2G 4G

Let g =

. Then G = g², G_u = 2gg_u and G_uu = 2g_u² + 2gg_uu, so substituting these equations into the result of the previous part of the problem gets us

2g2+ 2gg 4g2g2 K = - --u---2-uu + ---u4- 2 2g 2 4g = - gu +-gguu-+ gu g2 g2 gu2--g2u +-gguu = g2 guu- = g

(a)

X = (- sinθ sinϕ,sinθcosϕ,0) ϕ

Xθ = (cosθcosϕ,cosθsinϕ,- sin θ)

E = ⟨X ϕ,X ϕ⟩ = sin2θsin2 ϕ+ sin2 θcos2 ϕ = sin2θ

F = ⟨X ϕ,Xθ⟩ = - sinθsin ϕcosθcosϕ + sinθ cos ϕcosθsin ϕ = 0

G = ⟨X θ,Xθ⟩ = cos2θcos2ϕ +cos2θsin2ϕ +sin2θ = cos2θ +sin2θ = 1

Eθ = 2 sinθ cosθ E ϕ = 0 Fθ = 0 F ϕ = 0 Gθ = 0 G ϕ = 0

1 2 1 2 Γ111sin θ = 22sinθ cos θ Γ 11 = 0 Γ11 = cotθ

Γ 12sin2θ = 0 Γ 2 = 0 Γ 12 = 0 12

Γ 122 = 0 Γ 222 = 0

(b)

Let W(s) = a(s)X_θ + b(s)X_ϕ. With

α(s) = (sinθcos(-s-) ,sinθ sin( -s-) ,cosθ) sinθ sinθ

we have u′ = -sin (-s-)
sinθ . Thus by equation (1) on page 239 of do Carmo, we have the following since all Christoffel Symbols are zero except for Γ₁₁¹

′ 1 ′ ′ ( ′ (-s--)) ′ (a + Γ11au )X θ + b Xϕ = a - acotθsin sin θ Xθ + b Xϕ

(c)

Solving the ODEs of

( ( s ) ) a′ = a 1 +cotθ sin sinθ- b′ = b

we find that

b = es + b(0)

and

a = es-sinθcotθcos(sisnθ) + a(0)

How much does the parallel transport of a vector rotate after one loop? ???

With the definition of α above we get

( ( s ) ( s ) ) α ′(s) = - sin sin-θ ,cos sinθ ,0

′ 1 ( ( s ) ( s ) ) α (s) = sin-θ - cos sinθ ,- sin sinθ ,0

Curvature

( ( ) ( )) k = |α ′(s)| = -1--- cos2 -s-- + sin2 -s-- = --1-- sin2θ sin θ sin θ sin2 θ

Normal Curvature Let N be the normal map on the sphere N(p) = p. Than at some p in the trace of α we have

k = k⟨n,N ⟩ n ⟨ ′′ ⟩ = k α-(s),α(s) k = ⟨α′′(s),α(s)⟩ 2( -s--) 2 (-s--) = - sinθ c(os s(inθ -) sin θsin( sin))θ = - sinθ cos2 -s-- - sin2 -s-- sin θ sin θ = - sinθ

Geodesic Curvature We have

( ) k2g = k2 - k2n = sin2 θ- sin4θ = sin2θ 1 - sin2θ = sin2θcos2θ

(a) Non-length-minimizing positive curvature geodesic

The sketch in Figure 1 has a geodesic between p and q that is not length minimizing. The curve that travels “the long way” along the great circle containing p and q will be a geodesic, but not length-minimizing. The geodesic traveling “the short way” will be length-minimizing though.

Figure 1: Sketch of positive curvature surface with a geodesic between p and q which is not length-minimizing.

(b) Non-length-minimizing zero curvature geodesic

The sketch in Figure 2 has a geodesic between p and q that is not length minimizing. Again going the long way, but this time along a horizontal circle of this cylinder.

Figure 2: Sketch of zero curvature surface with a geodesic between p and q which is not length-minimizing.

The sketch in Figure 3 has a geodesic between p and q that is not length minimizing. Again going the long way, but this time around the “waist” of this surface of revolution.

Figure 3: Sketch of negative curvature surface with a geodesic between p and q which is not length-minimizing.

(d)

(e) A surface where any two points can be joined by a geodesic, but the geodesic is only defined for a finite amount of time.

The sketch in Figure 4 has a geodesic between every two points, namely the straight line between them. However, since geodesics need to have constant velocity, then only finite time geodesics could exist.

Figure 4: Sketch of surface with only finite time geodesics.