• 沒有找到結果。

# the diffeomorphism ϕ : S → ¯S prserves the first fundamental form

N/A
N/A
Protected

Share "the diffeomorphism ϕ : S → ¯S prserves the first fundamental form"

Copied!
24
0
0

(1)

Isometries; Conformal Maps

Definition A diffeomorphism ϕ : S → ¯S is an isometryif

hw1, w2ip = hdϕp(w1), dϕp(w2)iϕ(p) ∀ p ∈ S, ∀ w1, w2 ∈ TpS.

The surfaces S and ¯S are then said to beisometric.

In other words, a diffeomorphism ϕ is an isometry if the differential dϕ preserves the inner product. It follows that, dϕp : TpS → Tϕ(p)S being an isometry,¯

Ip(w) = hw, wip = hdϕ(w), dϕ(w)iϕ(p)= Iϕ(p)(dϕp(w)) ∀ w ∈ TpS, i.e. the diffeomorphism ϕ : S → ¯S prserves the first fundamental form.

Conversely, if the diffeomorphism ϕ : S → ¯S prserves the first fundamental form, then 2hw1, w2i = Ip(w1+ w2) − Ip(w1) − Ip(w2) ∀ w1, w2 ∈ TpS

= Iϕ(p)(dϕp(w1+ w2)) − Iϕ(p)(dϕp(w1)) − Iϕ(p)(dϕp(w2))

= hdϕ(w1), dϕ(w2)i, and ϕ is, therefore, an isometry.

Definition A map ϕ : V → ¯S of a neighborhood V ⊂ S of p ∈ S is alocal isometry at p if there exists a neighborhood ¯V ⊂ ¯S of ϕ(p) ∈ ¯S such that ϕ : V → ¯V is an isometry. If there exists a local isometry into ¯S at every p ∈ S, the surface S is said to be locally isometric to ¯S.

It is clear that if ϕ : S → ¯S is a diffeomorphism and a local isometry for every p ∈ S, then ϕ is an isometry (globally).

However, a local isometry is not necessary an isometry globally, e.g. the xy-plane P = {(x, y, z) ∈ R3 | z = 0} and the cylinder S = {(x, y, z) ∈ R3 | x2+ y2 = 1} are locally isometric, but they are not homeomorphic, so P and S are not diffeomorphic or isometric globally.

Since any simple closed curve C ⊂ P in the plane P can be shrunk (deformed) continuously into a point without leaving the plane P, and this topological property in P is preserved by a homeomorphism ϕ : P → ϕ(P ).

Note that a parallel C0, e.g. C0 = {(cos u, sin u, 0) | u ∈ [0, 2π]} ⊂ S, of the cylinder S does not have that property while the corresponding unit circle C = {(x, y, 0) | x2+ y2 = 1} in P can be shrunk continuously into a point without leaving the plane P, so P and S are not homeomorphic.

(2)

Proposition Suppose that there exist parametrizations X : U → S and ¯X : U → S such that E = ¯E, F = ¯F , G = ¯G in U. Then the map ϕ = ¯X ◦ X−1 : X(U ) → ¯S is a local isometry.

Proof Let p ∈ X(U ) and w ∈ TpS. Then w is tangent to a curve X(α(t)) at t = 0, where α(t) = (u(t), v(t)) is a curve in U ; thus, w may be written (t = 0)

w = Xuu0+ Xvv0.

By definition, the vector dϕp(w) is the tangent vector to the curve ¯X ◦ X−1◦ X(α(t)) = ¯X(α(t)) at t = 0. Thus,

p(w) = ¯Xuu0 + ¯Xvv0. Since

Ip(w) = E(u0)2+ 2F u0v0+ G(v0)2, Iϕ(p)(dϕp(w)) = E(u¯ 0)2+ 2 ¯F u0v0+ ¯G(v0)2,

and the assumption E = ¯E, F = ¯F , G = ¯G in U, we conclude that Ip(w) = Iϕ(p)(dϕp(w)) for all p ∈ X(U ) and all w ∈ TpS; hence, ϕ is a local isometry.

Definition A diffeomorphism ϕ : S → ¯S is called a conformal mapif

hdϕp(w1), dϕp(w2)i = λ2(p) hw1, w2iw=⇒ |dϕ1=w2 p(w1)|2 = λ2(p) |w1|2 ∀ p ∈ S, ∀ w1, w2 ∈ TpS, where λ2 is a nowhere-zero differentiable function on S; the surface S and ¯S are then said to be conformal. A map ϕ : V → ¯S of a neighborhood V ⊂ S of p ∈ S is a local conformal map at p if there exists a neighborhood ¯V ⊂ ¯S of ϕ(p) ∈ ¯S such that ϕ : V → ¯V is a conformal map.

If there exists a local conformal map into ¯S at every p ∈ S, the surface S is said to be locally conformalto ¯S.

The geometric meaning of the above definition is thatthe angles (but not necessarily the lengths) are preserved by conformal maps. In fact, let α : I → S and β : I → S be two curves in S which intersect at, say, t = 0. Their angle θ at t = 0 is given by

cos θ = hα0, β0i

0| |β0|, 0 ≤ θ ≤ π.

A conformal map ϕ : S → ¯S maps these curves into ϕ ◦ α : I → ¯S, ϕ ◦ β : I → ¯S, which intersect when t = 0, making an angle ¯θ given by

cos ¯θ = hdϕ(α0), dϕ(β0)i

|dϕ(α0)| |dϕ(β0)| = λ20, β0i

λ20| |β0| = cos θ.

Proposition Suppose that there exist parametrizations X : U → S and ¯X : U → ¯S such that E = λ2E, F = λ¯ 2F , G = λ¯ 2G in U, where λ¯ 2 is a nowhere-zero differentiable function in U. Then the map ϕ = ¯X ◦ X−1 : X(U ) → ¯S is a local conformal map.

Example For a > 0, let

X(u, v) = (a cosh v cos u, a cosh v sin u, av), (u, v) ∈ U = {0 < u < 2π, −∞ < v < ∞}

X(¯¯ u, ¯v) = (¯v cos ¯u, ¯v sin ¯u, a¯u), (¯u, ¯v) ∈ U = {0 < ¯u < 2π, −∞ < ¯v < ∞}

be parametrizations of the catenoid S and the helicoid ¯S, rspectively. Then the coefficients of the first fundamental forms are

(3)

E = a2cosh2v, F = 0, G = a2(1 + sinh2v) = a2cosh2v ∀ (u, v) ∈ U, E = a¯ 2+ ¯v2, F = 0,¯ G = 1¯ ∀ (¯u, ¯v) ∈ U.

Let us make the following change of parameters

¯

u = u, ¯v = a sinh v, ∀ (u, v) ∈ U, which is possible since the map is clearly one-to-one, and the Jacobian

∂(¯u, ¯v)

∂(u, v) = a cosh v 6= 0 ∀ (u, v) ∈ U.

Thus,

X(u, v) = (a sinh v cos u, a sinh v sin u, au),¯ (u, v) ∈ U, is a new parametrization of the helicoid with

E = a2cosh2v, F = 0, G = a2cosh2v ∀ (u, v) ∈ U.

We conclude that the catenoid and the helicoid are locally isometric.

Example Let S be the one-sheeted cone (minus the vertex) z = kp

x2+ y2, k > 0, (x, y) 6= (0, 0), and let U ⊂ R2 be the open set given in polar coordinates (ρ, θ) by

0 < ρ < ∞, 0 < θ < 2π sin α,

(4)

where 2α (0 < 2α < π) is the angle at the vertex of the cone (i.e., where cot α = k), and let F : U → S ⊂ R3 be the map

F (ρ, θ) =



ρ sin α cos

 θ sin α



, ρ sin α sin

 θ sin α



, ρ cos α

 . Then

• F (U ) ⊂ S, since

kp

x2+ y2 = cot α q

ρ2sin2α = ρ cos α = z,

• F : U → S \ {(ρ sin α, 0, ρ cos α) | 0 < ρ < ∞} is a diffeomorphism from U onto the cone minus a generator θ = 0, since F and dF are one-to-one in U,

and thus F (ρ, θ) is a parametrization of S with the coefficients of the first fundamental form being

E = hFρ, Fρi = 1, F = hFρ, Fθi = 0, G = hFθ, Fθi = ρ2, Also since U may be viewed as a regular surface parametrized by

X(ρ, θ) = (ρ cos θ, ρ sin θ, 0) ∈ R¯ 3, 0 < ρ < ∞, 0 < θ < 2π sin α, with the coefficients of the first fundamental form of U in this parametrization being

E = h ¯¯ Xρ, ¯Xρi = 1 = E, F = h ¯¯ Xρ, ¯Xθi = 0 = F, G = h ¯¯ Xθ, ¯Xθi = ρ2 = G, F : U → S is a local isometry.

The most important property of conformal maps is given by the following theorem, which we shall not prove.

Theorem Any two regular surfaces are locally conformal.

The proof is based on the possibility of parametrizing a neighborhood of any point of a regular surface in such a way that the coefficients of the first fundamental form are

E = λ2(u, v), F = 0, G = λ2(u, v).

Such a coordinate system is called isothermal. Once the existence of an isothermal coordinate system of a regular surface S is assumed, S is clearly locally conformal to a plane, and by composition locally conformal to any other surface.

(5)

The Gauss Theorem and the Equations of Compatibility

Let X : U ⊂ R2 → S be a parametrization in the orientation of S. At each p ∈ X(U ), since Xu, Xv, N ∈ R3 are linearly independent, we may express vectors Xuu, Xuv, Xvu, Xvv, Nu, Nv ∈ R3 in the basis {Xu, Xv, N } and obtain

Xuu = Γ111Xu+ Γ211Xv+ eN Xuv = Γ112Xu+ Γ212Xv+ f N Xvu = Γ121Xu+ Γ221Xv+ f N Xvv = Γ122Xu+ Γ222Xv+ gN

Nu = a11Xu + a21Xv Nv = a12Xu + a22Xv

where the aij, i, j = 1, 2, were obtained in Chapter 3 and the coefficients Γkij, i, j = 1, 2, are called the Christoffel symbols of S in the parametrization X. Since Xuv= Xvu, we conclude that Γ112 = Γ121 and Γ212 = Γ221; that is, the Christoffel symbols are symmetric relative to the lower indices.

To determine the Christoffel symbols, we take the inner product of the first four relations with Xu and Xv, obtaining the system

 E F

F G

 Γ111 Γ211

=

hXuu, Xui hXuu, Xvi

=

 1 2Eu

Fu− 1 2Ev

=⇒

 Γ111 Γ211

=

 E F

F G

−1

 1 2Eu

Fu− 1 2Ev

 Γ112 Γ212

=

hXuv, Xui hXuv, Xvi

=

 1 2Ev 1 2Gu

=⇒

 Γ121 Γ221

=

 Γ112 Γ212

=

 E F

F G

−1

 1 2Ev 1 2Gu

 E F

F G

 Γ122 Γ222

=

hXvv, Xui hXvv, Xvi

=

Fv− 1 2Gu 1 2Gv

=⇒

 Γ122 Γ222

=

 E F

F G

−1

Fv− 1 2Gu 1 2Gv

where we have used hXuu, Xui = 1

2

∂uhXu, Xui = 1

2Eu, hXuu, Xvi = ∂

∂uhXu, Xvi−hXu, Xvui = Fu−hXu, Xuvi = Fu−1 2Ev

In particular, if X is an orthogonal parametrization, i.e. F = hXu, Xvi = 0 at each p ∈ X(U ), then

111 Γ211



= 1

2(EG − F2)

 GEu

−EEv



112 Γ212



= 1

2(EG − F2)

 GEv EGu



122 Γ222



= 1

2(EG − F2)

GGu EGv



(6)

Thus, it is possible to solve the above system and to compute the Christoffel symbols in terms of the coefficients of the first fundamental form, E, F, G and their derivatives. Hence, all geo- metric concepts and properties expressed in terms ofthe Christoffel symbols are invariant under isometries.

Example Let S be a surface of revolution parametrized by

X(u, v) = (f (v) cos u, f (v) sin v, g(v)), f (v) 6= 0.

Since

E = (f (v))2, F = 0, G = (f0(v))2+ (g0(v))2, we obtain

Eu = 0, Ev = 2f f0, Fu = Fv = 0, Gu = 0, Gv = 2(f0f00+ g0g00), and

Γ111= 0, Γ211 = − f f0

(f0)2 + (g0)2, Γ112= f f0

f2 , Γ212= 0, Γ122 = 0, Γ222= f0f00+ g0g00 (f0)2+ (g0)2. Since X : U ⊂ R2 → R3 is differentiable,

(Xuu)v = (Xuv)u,

⇐⇒ Γ111Xu+ Γ211Xv+ eN

v = Γ112Xu+ Γ212Xv+ f N

u

⇐⇒ Γ111Xuv+ Γ211Xvv+ eNv+ (Γ111)vXu+ (Γ211)vXv+ evN

112Xuu+ Γ212Xvu+ f Nu+ (Γ112)uXu+ (Γ212)uXv + fuN (∗) By equating the coefficients of Xv, and using

a11 a21 a12 a22



= − e f f g

 E F

F G

−1

= −1

EG − F2

 eG − f F −eF + f E f G − gF −f F + gE

 . we obtain the following formula for the Gaussian curvature K

Γ111Γ212+ Γ211Γ222+ ea22+ (Γ211)v = Γ112Γ211+ Γ212Γ212+ f a21+ (Γ212)u

⇐⇒ (Γ212)u− (Γ211)v+ Γ112Γ211+ Γ212Γ212− Γ111Γ212− Γ211Γ222= ea22− f a21

⇐⇒ (Γ212)u− (Γ211)v+ Γ112Γ211+ Γ212Γ212− Γ111Γ212− Γ211Γ222= −E eg − f2 EG − F2

⇐⇒ (Γ212)u− (Γ211)v+ Γ112Γ211+ Γ212Γ212− Γ111Γ212− Γ211Γ222= −EK,

THEOREMA EGREGIUM (Gauss) The Gaussian curvature K of a surface is invariant by local isometries.

Remarks

• By equating the coefficients of Xuin equation(∗), we obtain another formula of the Gaussian curvature K.

112)u− (Γ111)v+ Γ112Γ111+ Γ212Γ112− Γ111Γ112− Γ211Γ122 = (Γ112)u− (Γ111)v + Γ212Γ112− Γ211Γ122= F K.

• By equating the coefficients of N in equation (∗), we obtain

ev − fu = eΓ112+ f (Γ212− Γ111) − gΓ211 (†).

(7)

• By equating the coefficients of N in equation (Xvv)u− (Xvu)v = 0, we obtain fv− gu = eΓ122+ f (Γ222− Γ112) − gΓ212 (††).

Equations(†) and (††)are called Mainardi-Codazzi equations.

Theorem (Bonnet) Let E, F, G, e, f, g be differentiable functions, defined in an open set V ⊂ R2, with E > 0 and G > 0. Assume that

• the given functions satisfy formally the Gauss and Mainardi-Codazzi equations,

• and that EG − F2 > 0.

Then,

• for every q ∈ V there exists a neighborhood U ⊂ V of q,

• and a diffeomorphism X : U → X(U ) ⊂ R3

such that the regular surface X(U ) ⊂ R3 has E, F, G and e, f, g as coefficients of the first and second fundamental forms, respectively.

Furthermore, if U is connected and if

X : U → ¯¯ X(U ) ⊂ R3

is another diffeomorphism satisfying the same conditions, then there exist a translation T and a proper linear orthogonal transformation ρ in R3 such that

X = T ◦ ρ ◦ X.¯

Remark In the following, we shall calculate the Christoffel symbols and Gaussian curvatures in terms of the metric tensor (gij) and its partial derivatives.

Let U be an open subset in the u1u2-plane, and X : U ⊂ R2 → S be a parametrization in the orientation of S. At each p ∈ X(U ), let X1 = Xu1, X2 = Xu2, and let

g11 = hX1, X1i = E, g12 = g21= hX1, X2i = F, g22 = hX2, X2i = G ⇐⇒ g11 g12 g21 g22



=E F

F G

 , and

g11 g12 g21 g22



= (gij) = (gij)−1 = 1 det(gij)

 g22 −g12

−g21 g11



= 1

EG − F2

 G −F

−F E

 . Note that

2

X

k=1

gmkgk` (†)= δm` =

(1 if m = ` 0 if m 6= `.

Since X1, X2, N ∈ R3 are linearly independent, we may express vectors Xij = Xuiuj ∈ R3 and Ni = Nui ∈ TpS as

Xij (∗)=

2

X

k=1

ΓkijXk+ hijN, i, j = 1, 2, where h11 h12 h21 h22



= e f f g



Ni (∗∗)=

2

X

j=1

ajiXj, i = 1, 2,

(8)

where

a11 a21 a12 a22



= −h11 h12 h21 h22

 g11 g12 g21 g22



= 1

EG − F2

h11 h12 h21 h22

 −g22 g21 g12 −g11

 ,

as obtained in Chapter 3 and the coefficients Γkij, i, j = 1, 2, are called theChristoffel symbols of S in the parametrization X. Since Xij = Xji, we conclude that Γkij = Γkji; that is, the Christoffel symbols are symmetric relative to the lower indices.

To determine the Christoffel symbols, we take the inner product of the Xij with Xk and use the definition of (gij) and (gij) to obtain

hXij, Xki = ∂

∂ujhXi, Xki − hXi, Xkji = ∂gik

∂uj − hXi, Xkji = gik,j− hXi, Xkji, where gik,j = ∂gik

∂uj

⇐⇒(∗) h

2

X

`=1

Γ`ijX`, Xki = gik,j− hXi, Xkji

⇐⇒

2

X

`=1

Γ`ijg`k = gik,j− hXi, Xkji

⇐⇒(†) 2

X

k=1 2

X

`=1

Γ`ijgmkg`k =

2

X

k=1

gmkgik,j

2

X

k=1

gmkhXi, Xkji, m = 1, 2

⇐⇒(†) 2

X

`=1

δm`Γ`ij =

2

X

k=1

gmkgik,j

2

X

k=1

gmkhXi, Xkji, m = 1, 2

⇐⇒(†) Γmij =

2

X

k=1

gmkgik,j

2

X

k=1

gmkhXi, Xkji, m = 1, 2

Γmijmji

⇐⇒ Γmji =

2

X

k=1

gmkgjk,i

2

X

k=1

gmkhXj, Xkii, m = 1, 2

=⇒ 2Γmij =

2

X

k=1

gmk(gik,j + gjk,i) −

2

X

k=1

gmk

∂ukhXi, Xji =

2

X

k=1

gmk(gik,j + gjk,i− gij,k), m = 1, 2

=⇒ Γmij = 1 2

2

X

k=1

gmk(gik,j+ gjk,i− gij,k), m, i, j = 1, 2.

Since X : U ⊂ R2 → R3 is differentiable, (Xii)j = (Xij)i, 1 ≤ i 6= j ≤ 2

⇐⇒

2

X

k=1

ΓkiiXk+ hiiN

!

j

=

2

X

k=1

ΓkijXk+ hijN

!

i

⇐⇒

2

X

k=1

kii)jXk+

2

X

k=1

ΓkiiXkj+ hii,jN + hiiNj =

2

X

k=1

kij)iXk+

2

X

k=1

ΓkijXki+ hij,iN + hijNi,

where (Γkii)j = ∂Γkii

∂uj, hij,i = ∂hij

∂ui

(9)

⇐⇒

2

X

k=1

kii)jXk+

2

X

k, `=1

ΓkiiΓ`kjX`+

2

X

k=1

ΓkiihkjN + hii,jN +

2

X

k=1

hiiakjXk

=

2

X

k=1

kij)iXk+

2

X

k, `=1

ΓkijΓ`kiX`+

2

X

k=1

ΓkijhkiN + hij,iN +

2

X

k=1

hijakiXk

⇐⇒

2

X

k=1

kii)jXk+

2

X

`, k=1

Γ`iiΓk`jXk+

2

X

k=1

ΓkiihkjN + hii,jN +

2

X

k=1

hiiakjXk k ↔ ` in double sum

=

2

X

k=1

kij)iXk+

2

X

`, k=1

Γ`ijΓk`iXk+

2

X

k=1

ΓkijhkiN + hij,iN +

2

X

k=1

hijakiXk k ↔ ` in double sum

=⇒ 0 =

2

X

k=1

"

kij)i− (Γkii)j +

2

X

`=1

Γ`ijΓk`i

2

X

`=1

Γ`iiΓk`j + hijaki− hiiakj

# Xk

+ hij,i− hii,j+

2

X

k=1

Γkijhki

2

X

k=1

Γkiihkj

! N

⇐⇒ (Γkij)i− (Γkii)j +

2

X

`=1

Γ`ijΓk`i

2

X

`=1

Γ`iiΓk`j = hiiakj− hijaki 1 ≤ i 6= j ≤ 2,

and hij,i− hii,j +

2

X

k=1

Γkijhki

2

X

k=1

Γkiihkj = 0 1 ≤ i 6= j ≤ 2 called Mainardi-Codazzi equations

Since hij = hji, 1 ≤ i 6= j ≤ 2,

h11 h12 h21 h22



=h11 h12 h21 h22

−1

= 1

eg − f2

 h22 −h21

−h12 h11



= 1

eg − f2

 h22 −h12

−h21 h11

 , we have

hiiakj− hijaki = (eg − f2)[hjjakj+ hjiaki] = (eg − f2)

2

X

`=1

hj`ak`

= (eg − f2) ×h11 h12 h21 h22



·a11 a21 a12 a22



jk

(the jk-entry of (hmn) · (apq))

= eg − f2

EG − F2 ×h11 h12 h21 h22



·h11 h12 h21 h22



·−g22 g21 g12 −g11



jk

= K ×−g22 g21 g12 −g11



jk

= K ×−G F F −E



jk

and the Gauss curvature formulas (Γkij)i− (Γkii)j +

2

X

`=1

Γ`ijΓk`i

2

X

`=1

Γ`iiΓk`j = hiiakj− hijaki = K ×−g22 g21 g12 −g11



jk

,

where Γkij = 1 2

2

X

`=1

gk`(gi`,j+ gj`,i− gij,`), i, j, k = 1, 2.

(10)

In particular, we obtain the following when j = k = 2, i = 1 =⇒ (Γ212)1− (Γ211)2+

2

X

`=1

Γ`12Γ2`1

2

X

`=1

Γ`11Γ2`2 = −h12a21+ h11a22= −EK,

j = k = 1, i = 2 =⇒ (Γ121)2− (Γ122)1+

2

X

`=1

Γ`21Γ1`2

2

X

`=1

Γ`22Γ1`1 = h22ak1− h21ak2 = −GK,

j 6= k = 1, i = 1 =⇒ (Γ112)1− (Γ111)2+

2

X

`=1

Γ`12Γ1`1

2

X

`=1

Γ`11Γ1`2 = −h12a11+ h11a12= F K.

Example Let X(u1, u2) be an orthogonal parametrization (that is, F = g12 = g21 = 0) of a neighborhood of an oriented surface S. Let g,mik = ∂gik

∂um and gk`,m = ∂gk`

∂um. Since

2

X

k=1

gikgkj = δij =

(1 if i = j,

0 if i 6= j., g12 = g21= 0 and g12= g21= 0,

and using Γkij = 1 2

2

X

k=1

gk`(gj`,i+ g`i,j− gij,`) = 1

2gkk(gjk,i+ gki,j− gij,k), we have

2

X

k=1

gik,mgk`+

2

X

k=1

gikgk`,m= 0 =⇒

2

X

`=1

g`j

2

X

k=1

g,mikgk`+

2

X

`=1

g`j

2

X

k=1

gikgk`,m= 0

=⇒ gij,m=

2

X

k=1

gik,mδjk = −

2

X

k, `=1

gikgk`,mg`j =⇒ g,mij = −giigij,mgjj =⇒ gii,m= −giigii,mgii,

=⇒ Γ212 = 1

2g22g22,1, Γ211 = −1

2g22g11,2, Γ112 = 1

2g11g11,2, Γ111= 1

2g11g11,1, Γ222= 1

2g22g22,2 and

Γ212

1− Γ211

2 +

2

X

`=1

Γ`12Γ2`1

2

X

`=1

Γ`11Γ2`2= −Kg11

⇐⇒ 1

2 g22g22,1

1+1

2 g22g11,2

2 −1

4 g11g11,2g22g11,2 +1

4 g22g22,1g22g22,1

−1

4 g11g11,1g22g22,1 +1

4 g22g11,2g22g22,2 = −Kg11

⇐⇒

 g22,1 2√

g11g22

√g11

√g22



1

+

 g11,2 2√

g11g22

√g11

√g22



2

+ (g22,1)2+ g11,2g22,2

4 (g22)2 − (g11,2)2+ g11,1g22,1 4 g11g22

= −Kg11

⇐⇒

 g22,1 2√

g11g22



1

√g11

√g22 +

 g11,2 2√

g11g22



2

√g11

√g22 = −Kg11

⇐⇒ K = − 1

2√ g11g22

 g22,1

√g11g22



1

+

 g11,2

√g11g22



2



Parallel Transport. Geodesics.

Definition Let w : U → R3 be a differentiable tangent vector field in an open set U ⊂ S and p ∈ U. Let y ∈ TpS. Consider a parametrized curve

α : (−ε, ε) → U, with α(0) = p and α0(0) = y,

(11)

and let w(t) = w(α(t)) ∈ Tα(t)S, t ∈ (−ε, ε), be the restriction of the vector field w to the curve α.

Then the covariant derivative at p of the vector field w relative to the vector y, denoted Dw dt (0) or Dyw(p), is defined to be the normal projection of dw

dt(0) onto the plane TpS, i.e.

Dw

dt (0) = dw

dt (0) − hdw

dt (0), N (p)iN.

In terms of a parametrization X(u1, u2) of U ⊂ S at p, let X(u1(t), u2(t)) = α(t) ⊂ S and

w(t) =a1(u1(t), u2(t))Xu1 + a2(u1(t), u2(t))Xu2 = a1(t)X1+ a2(t)X2 =

2

X

i=1

aiXi ∈ Tα(t)S

be the expression of α(t) and w(t) in the parametrization X(u, v), respectively. Then dw

dt =

2

X

i, j=1

aiXiju0j+

2

X

i=1

a0iXi =

2

X

i, j,k=1

aiu0jΓkijXk+

2

X

i, j=1

aiu0jhijN +

2

X

k=1

a0kXk

and the covariant derivative of w at t is given by Dw

dt =

2

X

i, j,k=1

aiu0jΓkijXk+

2

X

i=1

a0iXi

=

2

X

k=1

a0k+

2

X

i, j=1

Γkijaiu0j

!

Xk ∈ Tα(t)S

Note that the covariant differentiation Dw

dt depends only on the vector (u01, u02), the coordinates of α0(t) in the basis {X1, X2}, and not on the curve α. Also since it depends only on the Christoffel symbols, that is, the first fundamental form of the surface, the covariant differentiation Dw

dt is a concept of intrinsic geometry.

(12)

Definition A vector field w ∈ TαS along a parametrized curve α : I → S is said to be parallel if Dw

dt = 0 for every t ∈ I.

Example In a plane P, since Γkij = 0, 1 ≤ i, j, k ≤ 2, the notion of parallel field w = a1X1+ a2X2 along a parametrized curve α ⊂ P reduces to that of a constant field, i.e. a01 = a02 = 0, along α;

that is, the length of the vector and its angle with a fixed direction are constant.

Those properties are partially reobtained on any surface as the following proposition shows.

Proposition Let w, v ∈ TαS be parallel vector fields along α : I → S. Then hw(t), v(t)i is constant for all t ∈ I. In particular, the lengths |w(t)| and |v(t)| are constant, and the angle

∠(v(t), w(t)) between w(t), v(t) ∈ Tα(t)S is constant for all t ∈ I.

Proof Since w(t), v(t) ∈ Tα(t)S and Dw

dt = Dv

dt = 0, we have d

dthw(t), v(t)i = hdw

dt , v(t)i + hw(t),dv

dti = hDw

dt , v(t)i + hw(t), Dv dt i = 0,

and hw(t), v(t)i = constant for all t ∈ I and for any parallel vector fields w and v along α.

Proposition Let α : I → S be a parametrized curve in S and let w0 ∈ Tα(t0)S, t0 ∈ I. Then there exists a unique parallel vector field w(t) = a1(t)X1(u1(t), u2(t))+a2(t)X2(u1(t), u2(t)) along α(t), with w(t0) = w0, i.e. there is a unique solution to the initial-value problem

a0k+

2

X

i, j=1

Γkijaiu0j = 0, k = 1, 2, with a1X1+ a2X2|t=t0 = w(t0) = w0.

Definition Let α : I → S be a parametrized curve and w0 ∈ Tα(t0)S, t0 ∈ I. Let w be a paralle vector field along α, with w(t0) = w0. The vector w(t1), t1 ∈ I, is called the parallel transport of w0 along α at the point t1.

Definition A nonconstant, parametrized curve γ : I → S is said to be geodesic at t ∈ I if the field of its tangent vectors γ0(t) is parallel along γ at t; that is

0(t) dt = 0;

γ is a parametrized geodesic if it is geodesic for all t ∈ I, i.e. γ(t) = X(u1(t), u2(t)), t ∈ I is a geodesic if γ0(t) = u01X1+ u02X2 satisfies the geodesic equations

0(t)

dt = 0 ⇐⇒ u00k+

2

X

i,j=1

Γkiju0iu0j, k = 1, 2. (∗)

Examples

(13)

(1) If S is a plane, then S can be parametrized by X(u1, u2) with Xij = Xuiuj = 0 ∈ R3 everywhere in S, 1 ≤ i, j ≤ 2. This implies that X1 = Xu1 and X2 = Xu2 are constant vector in S, Γkij = 0 for all 1 ≤ i, j, k ≤ 2, and γ(t) = X(u1(t), u2(t)) is a geodesic in a plane S if

u00k(t) = 0, ∀ t ∈ I =⇒ u0k(t) = ck (a constant) ∀ t ∈ I =⇒ uk(t) = ckt+dk ∀ t ∈ I, k = 1, 2 Hence γ is a geodesic in a plane S if and only if γ is a straight line in S.

(2) Let γ(u2) = (f (u2), 0, g(u2)), f (u2) 6= 0, a < u2 < b, be a regular curve and S be a surface of revolution with the parametrization

X(u1, u2) = (f (u2) cos u1, f (u2) sin u1, g(u2)), 0 < u1 < 2π, a < u2 < b.

Then the matrix (gij) and its inverse (gij) of the first fundamental form

2

X

i,j=1

giju0iu0j are given by

g11 g12 g21 g22



=f2 0

0 (f0)2+ (g0)2



⇐⇒ g11 g12 g21 g22



=f−2 0

0 (f0)2+ (g0)2−1



where f and g are functions of u2and the Christoffel symbols Γkij = 1 2

2

X

`=1

gk`(gj`,i+g`i,j−gij,`) are given by

Γ1ij = 1

2f−2(gj1,i+ g1i,j − gij,1) and Γ2ij = 1

2(f0)2+ (g0)2−1

(gj2,i+ g2i,j − gij,2) and

111 Γ112 Γ121 Γ122



=

0 f f0 f2 f f0

f2 0

 and Γ211 Γ212 Γ221 Γ222



=

− f f0

(f0)2+ (g0)2 0 0 f0f00+ g0g00

(f0)2+ (g0)2

 this implies that X(u1(t), u2(t)) is a geodesic of the surface of revolution S if u1, u2 satisfy the system ofequations

u001 +2f f0

f2 u01u02 = 0 and u002 − f f0

(f0)2+ (g0)2(u01)2+ f0f00+ g0g00

(f0)2+ (g0)2(u02)2 = 0, (††)

where u0k = duk

dt , f0 = df

du2 and g0 = dg du2.

If the meridian γ(s) = {X(u1, u2) | u1 = constant, u2 = u2(s)} is parametrized by arc length s, then the 1st equation of (††)holds, and, since γ0(s) = X1u01+ X2u02 = X2u02, 1 = hγ0(s), γ0(s)i = Ip0(s)) = hX1u01+X2u02, X1u01+X2u02i = g22(u02)2 =(f0)2+(g0)2(u02)2, we have

(u02)2 (∗)= 1

(f0)2+ (g0)2 =⇒ u02u002 = − f0f00+ g0g00

(f0)2+ (g0)22 u02

(14)

by differentiating both sides with respect to s and using the Chain Rule to get d

dsf0 = f00u02 and d

dsg0 = g00u02. Multiplying both sides by u02, we get (u02)2u002 = − f0f00+ g0g00

(f0)2+ (g0)22 (u02)2 =⇒(∗) u002 = −f0f00+ g0g00

(f0)2+ (g0)2(u02)2 the 2nd equation of (††) and this implies that arc length parametrized meridians are geodesics.

If the parallel γ(s) = {X(u1, u2) | u2 = constant, u1 = u1(s)} is parametrized by arc length s, since 1 = Ip0(s)) = (f (u2))2(u01)2, we have (u01)2 = 1/f (u2) = constant 6= 0 which implies that 2u01u001 = 0 =⇒ u001 = 0, i.e. the 1st equation of (††) holds, so the arc length parametrized parallels are geodesics if it satisfies the 2nd equation of (††)

f f0

(f0)2+ (g0)2(u01)2 = 0 =⇒ f0 = 0 since f 6= 0, u01 6= 0.

Definition Let w be a differentiable field of unit vectors along a parametrized curve α : I → S on an oriented surface S. Sincew(t), t ∈ I, is a unit vector field,

dw

dt (t) ⊥ w(t) =⇒ Dw

dt = Dw dt



(N ∧ w(t)),

where the real number  Dw dt



is called the algebraic value of the covariant derivative of w at t.

Definition Let C be an oriented regular curve contained in an oriented surface S, and let α(s) be a parametrization of C, in a neighborhood of p ∈ S, by the arc length s. The algebraic value of the covariant derivative of α0(s) at p,  Dα0(s)

ds



= kg is called the geodesic curvature of C at p.

Remark The geodesics which are regular curves are thus characterized as curves whose geodesic curvature is zero and note that the geodesic curvature of C ⊂ S changes sign when we change the orientation of either C or S.

Thus the given improper integral is convergent and, since the integrand is positive, we can interpret the value of the integral as the area of the shaded region in Figure

In order to establish the uniqueness of a prime factorization, we shall use the alternative form of the Principle of Mathematical Induction.. For the integer 2, we have a unique

interpretation of this result, see the opening paragraph of this section and Figure 4.3 above.) 2... (For

6 《中論·觀因緣品》，《佛藏要籍選刊》第 9 冊，上海古籍出版社 1994 年版，第 1

Population: the form of the distribution is assumed known, but the parameter(s) which determines the distribution is unknown.. Sample: Draw a set of random sample from the

volume suppressed mass: (TeV) 2 /M P ∼ 10 −4 eV → mm range can be experimentally tested for any number of extra dimensions - Light U(1) gauge bosons: no derivative couplings. =&gt;

• Formation of massive primordial stars as origin of objects in the early universe. • Supernova explosions might be visible to the most

In this study, we compute the band structures for three types of photonic structures. The first one is a modified simple cubic lattice consisting of dielectric spheres on the