Notes on Differential Geometry

Page 1

Notes on Differential Geometry Peter Hislop Mingyang Sun November , 

CONTENTS 

Basic Idea

Examples of Manifold  Smooth manifold & regular surface  Tangent Vectors & Spaces  Orientable Manifolds  Tangent Bundle 

Riemannian Metric Introduction 



Lie Groups as Riemannian Manifolds 

Kozulor Affine Connections



Connections 

Geodesics



Definition of Geodesics  Geodesic Flow  Coordinates  Minimizing Properties 

Curvature

Geodesic Normal



Curvature of Geometric Objects  folds 

Curvature of Riemannian Mani-

1 BASIC IDEA Take a set M, and a Riemannian metric g. For all p ∈ V ⊂ M, M is an n-dimensional manifold if V looks like an open subset U ⊂ R. A manifold allows to measure distance between points in M.

. Examples of Manifold 1.1.1 Euclideon space Euclidean space M = Rn , Cartesian coordinates P ∈ Rn (x1 , …, xn ). Take a ball around P, then the ball is a manifold. To measure distance in such space, take P,Q ∈ Rn , and curve γ joining P and Q. Define infinitesimal distance (metric) n

ds2E = ∑ dx2i i=1

Then the length of γ is Q

Lγ =

∫P

Q

dsE =

n

∑ dxi

2

∫P

√ i=1

There are special “nice” curves that minimize Lγ . If we parametrize γ as γ(t) = tQ + (1 − t)P, in which 0 ≤ t ≤ 1, then γ is a straight line which is called a geodesic. Its length is Lγ = |P − Q|. In general, if d2 γi =0 dt2 then γ is a geodesic.

1.1.2 Hyperbolic space Now change to another metric. For M = {(x,y) | y > 0} which is the upper half plane, define hyperbolic space H2 , ds2H =

dx2 + dy2 y2

Consider a “straight line” γ from (0,1) to (0,t), 0 < t < 1, (0,1)

Lγ =

∫(0,t)

1

dsH =

dy = − log t > 0 ∫t y

If t → 0, Lγ goes to infinity. In this space, the geodesics are . .

Lines on which x is constant Semi circles whose center is on the x-axis Also, in this space, Euclid's th postulate (parallel postulate) fails. Now change notation a little, ds2E = ∑ gE ij dxi dxj i,j

Then [gE ij ] is the n × n identity matrix. 2

ds2H

= ∑ gH ij dxi dxj ,

[gH ij ] =

i,j=1

1 I2 y2

So for hyperbola space, [gij ] is not constant.

1.1.3 Sphere Consider n-sphere S ⊂ Rn+1 , n+1

Sn = {(x1 , …, xn+1 ) | F(x1 , …, xn+1 ) = 0},

F = ∑ x2i − 1 i=1

Using spherical coordinates {(θ,ϕ) | 0 < θ < π, 0 < ϕ < 2π}, we can map the -sphere on to a -plane (fig. (.)) with map ⎧Φ = {x1 = cos ϕ sin θ, x2 = sin ϕ sin θ, x3 = cos θ} ⎪ Φ : U ⊂ R2 → S 2 ⎨ ⎪ 2 ⎩Φ(U) = S \{(θ,ϕ) | ϕ = 0, 0 ≤ θ ≤ π} Note that we have to exclude the semicircle on which ϕ = 0, because on it ϕ can be either 0 or 2π. If we want to map the whole sphere on to a plane, a single map is not sufficient. ϕ 2π

0

π θ

Figure . A -sphere maps to a plane 3 2 2 2 2 ∑ dxi = sin θ(dϕ) + (dθ) i=1

Then the length of a curve can be calculated as Q

Lγ =

Q

1

ds2 = [sin2 θ(dϕ)2 + (dθ)2 ] 2 ∫P √ S ∫P

Suppose P and Q are both on the equator, then obviously the curve on which sin θ = 0 anywhere minimizes Lγ . Therefore the portion of equator between P and Q is a geodesic. Finally, note that in a hyperbolic space, x-axis is one of the infinities, while on a sphere, there is no infinity points. This is one of the differences between negative and positive curvature.

. Smooth manifold & regular surface An n-dimensional smooth (or C∞ or differentiable) manifold is a set M, together with a collection of n n {(xα ,Uα ) | Uα ⊂ R open and xα : Uα ⊂ R → M is an injective map (1 : 1)}

such that .

∪ xα (Uα ) = M α

.

Suppose Wαβ = xα (Uα ) ∩ xβ (Uβ ) ≠ ∅. The map −1 −1 x−1 β ∘ xα : xα (Wαβ ) → xβ (Wαβ )

.

is a differentiable (smooth, C∞ -) map. (fig. (.)) The set {(xα ,Uα )} is maximal (contains all compatible maps).

M

xα

x−1 α

x−1 β

xβ

x−1 β ◦ xα

Uα

Uβ

Figure . Explaining the definition of smooth manifolds Now consider map f : Rn → Rk and point u¯ = (u1 , … , un ) ∈ Rn ,

¯ = ( f1 (u), ¯ … , fk (u) ¯ ) f(u) and an open set U ⊂ Rn , u¯ 0 ∈ U. The derivative of f at u¯ 0 is a linear transformation Rn → Rk . Use matrix repersentation du¯ 0 f : Rn → Rk ,

j = 1, … , k,

∂f1 ⎛ ∂u1 ⎜ du¯ 0 f = ⎜ ⎜ ⎜ ∂fk ⎝ ∂u1

∂f1 ∂u2

⋯

∂fk ∂u2

⋯

l = 1, … , n ∂f1 ∂un ⎞ ⎟ ⎟ ⎟ ∂fk ⎟ ∂un ⎠

When n = k, det [du¯ 0 f] =

∂(f1 , … , fn ) ∂(u1 , … , un )

Now define a regular surface: S ⊂ Rn is a k-dimensional regular surface in Rn (k < n) if for every point P ∈ S ∃ neighborhood VP ⊂ Rn (open connected set) and a map x¯ : U ⊂ Rkopen → VP ∩ S such that . .

x¯ is homeomorphism (bijective & bicontinuous). dq x¯ : Rk → Rn is injective (1 : 1) for all q ∈ U. (x¯ is smooth.) A k-dimensional regular surface is a k-dimensional differentiable manifold. The definition of differentiable manifold require Uα and Uβ and their intersection on M because M is not in any space with structure, therefore differentiation of maps is meaningless. For example, S2 ⊂ R3 , S2+ = {(x1 ,x2 ,x3 ) | ∑ x2j = 1, x3 > 0}. (x1 ,x2 ) ∈ U1 ⊂ R2 with x21 + x22 < 1. ϕ1 : U1 ⊂ R2 → S2+ ,

ϕ1 (x1 ,x2 ) = (x1 , x2 ,

√

1 − x21 − x22 )

2 Then ϕ1 is continuous. ϕ−1 1 is the restriction to S of the projection

π (x ,x ,0). (x1 ,x2 ,x3 ) ⟶ 1 2

The matrix representation of dq ϕ1 is 3 × 2 ⎛ ⎜ dq ϕ1 = ⎜ ⎜

1 1 −x1

0 0 −x2

2

2

⎝ √1 − x 1 − x 2

√

⎞ ⎟ ⎟, ⎟

q ∈ U1

1 − x21 − x22 ⎠

This is a 1 : 1 map, therefore S2+ is regular. n+1 So, the  parametrizations of Sn : + ⊂R .

Spherical coordinates. For n = 2, (θ,ϕ),

0 < θ < π,

0 < ϕ < 2π

For n > 2, (θ,ϕ1 ,ϕ2 , … , ϕn−1 ),

0 < ϕj < 2π

Two parametrizations are needed. (Because of the ϕ = 0 line.) . Stereographic projection. Two parametrizations are needed (North pole). . Zero set of f(x1 ,x2 , … , xn ) = 1 − (x21 + ⋯ + x2n ). Six maps are needed.

Relation of regular surface and smooth manifold Theorem If S ⊂ Rn is a k-dimensional (k ≤ n − 1) regular surface, then S is a k-dimensional differentiable manifold. Remark  (differentiable map) Consider a map ϕ : U ⊂ Rk → Rn , ϕ(u1 , … , uk ) = (ϕ1 (u1 , … , uk ), … , ϕn (u1 , … , uk )) Map ϕ is differentiable if for all ℎ ∈ Rk , |ℎ| is small, ∃ a linear transformation T(ϕ,q) : Rk → Rn such that

|ϕ(q + ℎ) − ϕ(q) − T(ϕ, q)ℎ| =0 |ℎ| |h|→0 lim

Then T(ϕ,q) = dq ϕ = ϕ′ (q) is the differential of ϕ at q. In basis for Rk and Rn , ∂ϕ1 ⎛ ∂u1 ⎜ dq ϕ = ⎜ ⎜ ⎜ ∂ϕn ⎝ ∂u1

∂ϕ1 ∂u2

⋯

⋯

⋯

∂ϕ1 ∂uk ⎞ ⎟ ⎟ ⎟ ∂ϕn ⎟ ∂uk ⎠

Remark  Theorem (Inverse function theorem) Consider Rn → Rn , F is differentiable on U ⊂ Rn . If at q ∈ U, dq F is nonsingular (or invertible or injective), then ∃ open set Uq with q ∈ Uq , and open set Wq with F(q) ∈ Wq , such that F−1 : Wq → Uq is differentiable. Idea of proof Show that ∃ F which is an extension of ϕ1 , mapping open subset of Rn to Rn , such that dq F is nonsingular (this is an n × n matrix). Then apply the inverse function theorem to F to see that F−1 is differentiable. And finally at neighborhood in S, k k F−1 ∘ ϕ2 = ϕ−1 1 ∘ ϕ2 , R → R

and F−1 ∘ ϕ2 is also differentiable. Proof For point q ∈ U1 ⊂ Rk and map ϕ1 (u1 , … , uk ) = (v1 (u), … , vn (u)) The differential of ϕ1 at q is an n × k matrix

ϕ1 is not Rn → Rn , so we have to construct an n-dimensional version of it.

∂v1 ⎛ ∂u1 ⎜ dq ϕ1 = ⎜ ⎜ ⎜ ∂vn ⎝ ∂u1

∂v1 ∂u2

⋯

∂vn ∂u2

⋯

∂v1 ∂uk ⎞ ⎟ ⎟ ⎟ ∂vn ⎟ ∂uk ⎠

Since dq ϕ1 is injective, ∃ at least one k×k submatrix, such that its determinant is nonzero. Now consider map F(u1 , u2 , … , uk , tk+1 , … , tn ) = (v1 (u), v2 (u), … , vk (u), vk+1 (u) + tk+1 , vk+2 (u) + tk+2 , … , vn (u) + tn ) Then the differential of F can be evaluated on U1 as d(q,0,0,…) F =

[dq ϕ1 ]k×k ([dq ϕ1 ]k×(n−k)

0 I)

Coordinates may be rearranged so that the submatrix [dq ϕ1 ]k×k has nonzero determinant. Now apply the inverse function theorem to F. …

. Tangent Vectors & Spaces Consider a curve γ : I ⊂ R → R3 . γ(t) = (γ1 (t), γ2 (t), γ3 (t)) ,

γ′ (t) = (γ′1 (t), γ′2 (t), γ′3 (t))

Vector γ′ (t) is the tangent line to γ at γ(t) (translated to the origin). The length of the curve is 1

Lγ =



γ′ (t)| dt ∫0 |

1.3.1 Arc Length Parametrization of γ Define t

S(t) =

′

γ (u)| du = S(t) ∫0 |

Function S(t) : [0,1] → [0,Lγ ] is increasing and differentiable. dS = γ′ (t)| dt | If |γ′ (t)| > 0 ∀ t ∈ (0,1], S is invertible, we can treat t as a function of S, t(S). Then γ can be re-parametrized as 󾌃 γ(S) = (γ ∘ t)(S) Note that γ󾌃 and γ are the same curve. γ󾌃 ′ (S) = γ′ (t)

dt dS

For a map f that is locally invertible at b, (f−1 )′ (b) = [f′ (f−1 (b))]−1 Therefore ⇒ |γ󾌃 ′ (S)| = |γ′ (t)|

1 =1 |γ′ (t)|

(.)

For example, α(t) = (sin t, cos t), t ∈ (0,2π]. α′ (t) = (cos t, − sin t), |α′ (t)| = 1. Therefore α(t) is arc length parametrized. From now on, all curves are arc length parametrized. From eq. (.), γ′ (t) ⋅ γ′ (t) = 1

⇒ γ′′ (t) ⋅ γ′ (t) = 0

Therefore γ′′ (t) is perpendicular to γ′ (t). Write them as γ′ (t) = |γ′ (t)| ˆt,

ˆ γ′′ (t) = |γ′′ (t)| n



ˆ form a plane called osculating plane that locally contains the Basis ˆt and n curve. The curvature of γ is κ(t) = |γ′′ (t)|. The curvature of a straight line is ; the curvature of a circle is . ˆ ˆ (t). Basis Assume κ ≠ 0. In R3 , define binormal vector b(t) = ˆt(x) × n 3 ˆ ˆ ˆ , b} form R at γ(t). {t, n ˆ ′ (t) = ˆt′ (t) × n ˆ (t) + ˆt(t) × n ˆ ′ (t) = ˆt(t) × n ˆ ′ (t) b ˆ ′ is perpendicular to ˆt. Also b ˆ ′ is perpendicular to b. ˆ Therefore b ˆ′ Therefore b ˆ ′ as b ˆ ′ (t) = τ(t)ˆ ˆ . Write b is parallel to n n(t). The scalar τ(t) is called torsion.

Fundamental Theorem Theorem motions. Proof

Given κ>0 and τ ∈ R, a curved can be determined up to Euclidean

Use the Frenet formulas ′ ⎧ ⎪ ˆt ˆ′ b ⎨ ⎪ ′ ˆ ⎩n

= γ′′ = κˆ n = τˆ n ˆ = −κˆt − τb

1.3.2 Tangent Space Suppose two (or more) curves go through point p. The tangent vectors of these curves at p form a tangent spaces Tp S. Consider a neighborhood of p, Wp , in ambient space R3 , a function f ∈ C∞ 0 (Wp ), f ∘ γ : (−ϵ,ϵ) → R. The map δ:f↦

d f∘γ = γ′ (0)(∇f)(p) | dt t=0

is called derivation. Define δγ′ (0) (f) = γ′ (0)⋅(∇f)(γ(0)). Property of derivation:



δγ′ (0) ( f + g) = δγ′ (0) ( f) + δγ′ (0) (g) δγ′ (0) ( fg) = f(γ(0))ϕγ′ (0) (g) + g(γ(0))δγ′ (0) ( f) The set of all such derivations is an n-dimensional LVS. This definition does not require an ambient space to work. So it works for smooth manifolds. A formal definition is the following. Tangent space Consider an n-dimensional smooth manifold M, a point p ∈ M, a neighborhood Vp of p in M, a C∞ 0 (Vp ) smooth function f : Vp → R, (so that if we have a smooth map x¯ : Up ⊂ Rn → Vp , then f ∘ x¯ : Up → R is also smooth.) Then the set of all derivations on C∞ 0 (Vp ) at p is tangent space Tp M. So, suppose γ : (−ϵ,ϵ) ⊂ R → M is a smooth curve with γ(0) = p, then d ( f ∘ γ) = δp ( f) = γ′ (0)( f) |t=0 dt In this case, the tangent vector γ′ (0) is a functional of f. A local calculation of γ′ (0) 󾌃 Consider map x¯ : Up ⊂ Rn → Vp ⊂ M. Define γ(t) = x¯ −1 ∘ γ is a function of 󾌃 t, γ(0) = x¯ −1 (p). f ∘ x¯ : Up → R d d ¯ ∘ (x¯ −1 ∘ γ)](t) ( f ∘ γ)(t) = [( f ∘ x) dt dt −1 (x¯ ∘ γ) (t) = (x1 (t), … , xn (t))

Define q = (x1 , … , xn ) ∈ Up , ¯ ( f ∘ x)(q) = f(x1 , … , xn ) ⇒

d d ( f ∘ γ)(t) = f(x1 (t), … , xn (t)) dt dt



Therefore δp ( f) =

d ( f ∘ γ) = (∇f)(x¯ −1 (p)) ⋅ x′ (0) | dt t=0

= ∑ x′j (x¯ −1 (p)) j

∂f ∂xj

(t = 0)

In this case, γ′ (0) = ∑ x′j (x¯ −1 (0)) j

∂ ∂xj

Definition Differentiable map between manifolds with different dimensions n m φ : M1 → M 2 ¯ : U p ⊂ Rn → Mn For any point p ∈ Mn 1 and a local parametrization x 1, ∃ a parametrization y¯ of φ(p) such that map y¯ −1 ∘ φ ∘ x¯ : Up → Rm is smooth. The differential of φ at point p ∈ Mn 1 , δφp , is a linear transformation Tp M1 → Tφ(p) M2 Proof We need to show that (δφp )(ν) ⊂ Tφ(p) M2 in which φ : M1 → M2 is smooth, p ∈ M1 , ν ∈ Tp M1 . Let γ(t) : Iϵ = (−ϵ, ϵ) → M1 such that γ(0) = p, and γ′ (0) = ν. Define β(t) = (φ ∘ γ)(t) and β(0) = φ(γ(0)) = φ(p) β′ (0) = (δφp )(γ′ (0)),

δφp : Tp M1 → Tφ(0) M2

so that δφp maps ν to β′ (0). It is claimed that δφp is independent of curve γ.



Proof

Take two parametrizations x¯ : U ⊂ Rn → Mn 1,

¯ p ∈ x(U)

y¯ : V ⊂ Rm → Mm 2 ,

¯ φ(p) ∈ y(V)

Then y¯ −1 ∘ φ ∘ x¯ : U → Rm . Define φ󾌃 = y¯ −1 ∘ φ ∘ x¯ q = (x1 , … , xn ) ∈ U ν = (y1 , … , ym ) 󾌃 ¯ … , ym (x)), ¯ φ(q) = (y1 (x),

φ󾌃 : U → Rm

¯ ∘ (x¯ −1 ∘ γ) β = φ ∘ γ = (φ ∘ x) So (x¯ −1 ∘ γ)(t) = (x1 (t), … , xn (t)) ¯ is a curve in U. Define x(t) = (x1 (t), … , xn (t)), then ¯ ¯ (y¯ −1 ∘ β)(t) = (y1 (x(t)), … , ym (x(t))) is a curve in V. So m ′

β (0) = ∑ β′j (0) j=1

∂ ( ∂yj )0

as a curve in Tφ(p) Mm 2 . Also n

∂yi x′j (0) ∂x ( ) j j=1

β′ (0) = ∑

which is a m-component vector in Ty¯ −1 φ(p) V. Recall that x′j (0) = νi is the component of ν ∈ Tp Mn 1 . So (dφ)p (ν) =

∂yi (x′j (0))n×1 ∂x ( j )m×n



Therefore (dφ)p only depends on Tp Mn 1 . Note that if M2 = M1 , then ∂yi −1 ¯ p. ( ∂xj )m×m is d(y¯ ∘ x)

Example: Inverse of regular values of smooth surface S2 3

F(x1 , … , x3 ) = 1 − ∑ x2j = 0 j=1

The general setting is that map F : U ⊂ Rn → Rm is smooth, in which U is an open subset. A point p ∈ U is critical point of F if dFp : Tp U → TF(p) Rm is not surjective. Note that if n < m, every point is critical. If p ∈ U is a critical point, then F(p) ∈ F(U) is a critical value of F. ¯ = a} Theorem Let a ∈ F(U) be a regular value of F, F−1 ({a}) = {x¯ ∈ U|F(x) is a regular surface. x For example, for the S2 mentioned above, F : R3 → R, dFp =

∂F ∂F ∂F , , = (∇F)p = −2(x1 , x2 , x3 )| p ( ∂x1 ∂x2 ∂x3 )

Consider (dFp )(a,b,c)T = (−2)(p1 a + p2 b + p3 c). When (p1 ,p2 ,p3 ) = 0, dFp is not surjective, the critical value is F(0,0,0) = 1. Then 0 is a regular value, F−1 ({0}) is a regular surface. Proof Map F : U ⊂ Rn → Rm , n − m = k. Now extend F to a map URn → Rn . Pick p ∈ F−1 ({a}), then dFp : Rn → Rm is surjective. Define φ (y1 , … , ym , x1 , … , xk ) = ( f1 (q), … , fm (q), x1 , … , xk ) in which F(q) = ( f1 (q), … , fn (q)). Map φ maps the neighborhood of p to the neighborhood of φ(p), φ : Qp ⊂ Rn → Wφ(p) ⊂ Rn Then



dφp =

[dFp ]m×m 0 (

0 Ik×k )

,

det (dφp ) = 0

n×n

Apply inverse function theorem to φ. Now consider map x¯ : (x1 , … , xk ) ⊂ V → φ−1 (a1 , … , an , x1 , … , xk ) Map x¯ is a homeomorphism. Example

G : R3 → R2 , G(x1 ,x2 ,x3 ) = (x21 , x2 x3 ). dGp =

2x1 ( 0

0 x3

0 2p1 = x2 ) ( 0 p

0 p3

0 p2 )

is not surjective for p ∈ {(0,p2 ,p3 )} ∪ {(p1 ,0,0)}. Therefore (1,1,1) is a regular point, and (1,1) is a regular value.

. Orientable Manifolds Orientable manifolds Suppose {(x¯ α , Uα )} is the atlas of an n-dimensional −1 manifold M. If d(xα ∘ xβ )p∈M has positive determinent ∀ α and β, then M is orientable. And this gives the orientation of M. Example Consider a sphere S1 , x21 + x22 = 1. Use stereographic projection to map it onto a straight line on the equator. For the north semi-sphere, project from the north pole, 2u u2 − 1 ϕ1 (u) = , ( u2 + 1 u2 + 1 ) For the south semi-sphere, project from the south pole, ϕ2 (u) =

2u 1 − u2 , ( u2 + 1 u2 + 1 )



⇒ ϕ−1 1 (x1 ,x2 ) =

x1 , 1 − x2

U1 = S1 \{N},

ϕ−1 2 (x1 ,x2 ) =

x1 1 + x2

U2 = S1 \{S}

−1 ϕ−1 1 ∘ ϕ2 (u) = ϕ2 ∘ ϕ1 (u) =

1 u

d(ϕ−1 1 ∘ ϕ2 )u : Tu R → Tu R

Projective Space Projective space An n-dimensional real projective space Pn (R) is a set of (three equivalent definitions) . . . In the original notes, it's λ ∈ R\{v}??

all non-trivial lines through 0 in Rn+1 . all points on Sn with ω ∈ Sn identified with −ω. equivalent classes in Rn+1 \{0} from the equivalent relation x ∼ λx, x ∈ Rn+1 \{0}, λ ∈ R\{0}. Suppose [x] are equivalent classes. For each [x], if x ∈ [x], ∃ xi ≠ 0, so that we can define Xi =

Big Ui or small ui ??

x1 xi−1 xi+1 xn+1 ,⋯, , 1, ,⋯, ∈ [x] xi xi xi ) ( xi

for i = 1, … , n + 1. The set Vi = {[Xi ] = [x] | xi ≠ 0} is everything but the hyperplane determined by xi = 0. Then n+1

∪ Vi = yi (Ui ),

Ui ⊂ Rn

i=1

Given (u1 , … , ui ) ⊂ Rn , y¯ i maps (u1 , … , ui ) ⊂ Rn → pn (R) = [(u1 , … , ui−1 , 1, ui , … , un )] ∈ Rn+1 = y¯ i ((u1 , … , un )) WTF is all this???!!!

Conversely given



x1 xi−1 xi+1 xn+1 ,⋯, , 1, ,…, ∈ Vi xi xi xi )] [( xi the map Πi : Vi → Rn is Πi [Xi ] =

x1 xi−1 xi+1 xn+1 ,⋯, , 1, ,⋯, xi xi xi ) ( xi

¯ i (Rn ) = Pn (R). Note that y¯ i are bijective, and ⋃n+1 i=1 y Assume j > i, ¯ i = Rn → Rn , y¯ −1 j ∘y

u ∈ Rn

y¯ i (u) = [(u1 , …ui−1 , 1, ui , ui+1 , … , un )] ∈ Vi ∩ Vj =

uj−2 uj u1 ui−1 1 ui un ,⋯, , , ,⋯, , 1, ,⋯, uj−1 uj−1 uj−1 uj−1 uj−1 uj−1 )] [( uj−1

Note that 1 is the jth component in the (n + 1)-tuple. ¯ i )(u) = (y¯ −1 j ∘y

u1 1 un ,⋯, ,…, uj−1 uj−1 ) ( uj−1

¯ i ) is an n × n matrix, and Pn (R) is orientable iff n is odd. Transform d(y¯ −1 j ∘y Example

If n = 2, 1

u1 1 u2 , →⎛ 0 ( u2 u2 ) ⎝

1 −u u2

⎞ − u12 2⎠

det [d(y−1 j ∘ yi )] = −

2

1 u32

So P2 (R) is not orientable. If n = 3,



1 u3

u1 u2 1 ⎛ , , →⎜ ⎜0 ( u3 u3 u3 ) 0 ⎝

1 −u u2

0 1 u3

0

det [d(y−1 j ∘ yi )] = −

3 ⎞ 0 ⎟ ⎟ 1 − u2 3⎠

1 u42

P3 (R) is orientable.

. Tangent Bundle 1.5.1 Tangent bundle Tangent bundle of a smooth manifold M is defined as

Definition

TM = ∪ Tp M = {(p,v) | p ∈ M, v ∈ Tp M} p

Theorem

TM is a 2n-dimensional differentiable orientable manifold.

Proof Construct parametrization {(x¯ α , Uα )} of M, and parametrization (yα ,Uα × Rn ) of TM, yα : Uα × Rn → TM ⊃ {(p,v) | p ∈ x¯ α (Uα ), v ∈ Tp M} Define notation n

(

x¯ α (q), ∑ vj j=1

∂ = yα (x⏟⏟⏟⏟⏟⏟⏟ 1 , … , xn , v1 , … , vn ) ( ∂xi )p ) q∈Uα

and n

x¯ α (q) ∈ M,

∑ vj j=1



∂ ∈ Tp M ( ∂xi )p

Then n n âˆŞ yÎą (UÎą Ă— R ) = âˆŞ (xÂŻ Îą (UÎą ), d(xÂŻ Îą )q (R )) Îą

(ďœą.ďœ˛)

Îą

Since xÂŻ Îą : UÎą → M, the dierentiate d(xÂŻ Îą )q : Tq UÎą → TxÂŻ Îą (q) M. Therefore the union in eq. (ďœą.ďœ˛) contains everything in TM, â‹ƒÎą yÎą (UÎą Ă— Rn ) = TM. Suppose ω ∈ Tp M, and (p,ω) ∈ yÎą (UÎą Ă— Rn ) ∊ yβ (Uβ Ă— Rn ) then (p, ω) = (xÂŻ Îą (qÎą ), d(xÂŻ Îą )qÎą (vÎą )),

vÎą ∈ Rn , qÎą ∈ UÎą

= (x¯ β (qβ ), d(x¯ β )qβ (vβ )),

vβ ∈ Rn , qβ ∈ Uβ

−1 −1 y−1 β ∘ yÎą (qÎą ,vÎą ) = (qβ ,vβ ) = (xβ ∘ xÎą (qÎą ), d(xβ ∘ xÎą )|

qÎą

(vÎą ))

We can deďŹ ne a vector ďŹ eld X : M → TM, for p ∈ M, v ∈ Tp M, X(p) ∈ Tp M. TM locally looks like UÎą Ă— Rn . We require X to be smooth. (xÎą ,UÎą ) → (yÎą , UÎą Ă— Rn ). For y−1 β ∘ X ∘ xÎą to be smooth, n

X = ∑ vi (xÎą (q)) i=1

∂ ( ∂xi )q

n

yÎą (q,v) =

(

xÎą (q), ∑ vi (xÎą (q)) i=1

∂ ( ∂xi )q )

vi (xÎą (q)) has to be smooth. The set of all smooth vector ďŹ elds on M is đ?”›(M). To composite vector ďŹ elds X,Y ∈ đ?”›(M), locally at p ∈ Vp ⊂ M, X(p) = ∑ aj (p)

∂ , ∂xj

Y(p) = ∑ bj (p)

∂ ∂xj

′ Suppose f ∈ C∞ 0 (Vp ), p ∈ Vp ,

ďœ˛ďœą

X(f)(p′ ) = ∑ aj (p′ )

∂f ′ (p ) ∈ C∞ 0 (Vp ) ∂xj

We can apply Y on it. But Y(X(f)) is not a vector ďŹ eld because there are second order terms in it. Y(X(f))(q) = ∑ bi (q) i,j

∂ ∂f aj (q) (q) ∂xi ( ∂xj )

∂aj ∂f ∂2 f = ∑ bi (q) (q) (q) + bi (q)aj (q) (q) ∂xi ∂xj ∂xi ∂xj ] ij [ YX = ∑ bi (q) i,j

∂aj ∂ ∂2 + ∑ bi (q)aj (q) ∉ đ?”›(M) ∂xi ∂xj ∂xi ∂xj i,j

So, deďŹ ne composition operation Lie bracket to form Lie algebra, XY = ∑ ai (q) i,j

∂bj ∂ ∂2 + ∑ bi (q)aj (q) ∂xi ∂xj ∂xi ∂xj i,j

∂bj ∂aj ∂ [X,Y] = ∑ ∑ ai (q) − bi (q) ( ∂xi )q ( ∂xi )q ]} ( ∂xj )q j { i [ DeďŹ ne ∂bj ∂aj cj (q) = ∑ ai (q) − bi (q) ( ∂xi )q ( ∂xi )q ] i [ Then [X,Y] = ∑ cj (q) j

∂ ( ∂xj )q

Therefore, given X,Y ∈ đ?”›(M), ∃ Z ∈ đ?”›(M) such that Z = [X,Y] locally. Also, đ?”›(M) is a linear vector ďŹ eld, X,Y ∈ đ?”›(M) ⇒ ÎťX + Y ∈ đ?”›(M) Lie bracket identites:

ďœ˛ďœ˛

. . . .

[X,Y] = −[Y,X] Jacobian identity: [[X,Y], Z] + [[Y,Z], X] + [[Z,X], Y] = 0 [X + λY,Z] = [X,Z] + λ[Y,Z] Suppose f ∈ C∞ (M), (fX)(p) = ∑ f(p)aj (p) j

∂ ∂xj

[X,fY] = f[X,Y] + X(f)Y [A,BC] = B[A,C] + [A,B]C ∂ ∂ ∂ ∂f , f(x) = f−f = ∂xi ∂xi [ ∂xi ] ∂xi .

f,g ∈ C∞ (M), [fX, gY] = f[X, gY] + [f, gY]X = f(g[X, Y] + [X, g]Y) = g[f, Y]X + [f, g]YX, and [f, g] = 0 = fg[X, Y] + fX(g)Y − gY(f)X Lie algebra is the tangle bundle of Lie group.

WTF???

1.5.2 Vector bundle Vector bundle Consider two topological spaces: total space E and base space B. A vector bundle is a projection π : E → B which is a continuous surjection, such that . .

∀ b ∈ B, π−1 (b) is a linear vector space. And they satisfy the local triviality condition ∀ b0 ∈ B, ∃ open set Ub0 ⊂ B, b0 ∈ Ub For homeomorphism ℎ : Ub0 × Rn → π−1 (Ub0 ) ⊂ E, the map (for fixed b ∈ Ub0 ) x ∈ Rn toℎ(b,x) is an isomorphism of the linear vector space.



Example B Ă— Rn is a trivial vector bundle, so S1 Ă— Rn is a trivial line bundle over S1 . However the MĂśbius band is a nontrivial line bundle. A section X of a vector bundle is a map B → E. X(b) ∈ Ď€âˆ’1 ({b}). Also, TS1 is trivial, while TP1 (R) is nontrivial.

1.5.3 Local ow of vector ďŹ eld Consider X ∈ đ?”›(M). X(p) is in the form of n

∑ aj (p) j=1

∂ ( ∂xi )p

X generates a local ow through any point of M. Construct a curve (locally) at p, and a dierential equation on it. φ(t,p) : (âˆ’Ďľ,Ďľ) → M,

φ(0,p) = p ∈ M

d φ(t,p) = X(φ(t,p)) dt Note that ddt φ(t,p) gives an element in Tφ(t,p) M. Locally X can be pulled back to Rn . There exists at least one local solution to it. This reduces the problem to Rn : ďŹ nd X(t) ∈ Rn such that dX(t) = f(X(t)), dt

X(t = 0) = X0

Example If the equation is linear, the dierential operator can be written as a n × n matrix A, and X(t) = eAt X0 .

ďœ˛ďœ´

2 RIEMANNIAN METRIC ďœ˛.ďœą Introduction Inner product in a linear vector space Consider linear vector space V over R, and v,w ∈ V. Then the inner product of v and w is a map â&#x;¨v,wâ&#x;Š : V Ă— V → R such that ďœą. ďœ˛. ďœł.

It has positive deďŹ nition â&#x;¨v,vâ&#x;Š ≼ 0, and â&#x;¨v,vâ&#x;Š = 0 ⇔ v = 0. It is Bilinear. Suppose z ∈ V, then â&#x;¨v,Îąw + βzâ&#x;Š = Îąâ&#x;¨v,wâ&#x;Š + βâ&#x;¨v,zâ&#x;Š. It is symmetric. â&#x;¨v,wâ&#x;Š = â&#x;¨w,vâ&#x;Š Riemannian metric is a map that assigns to each p ∈ M an inner product on Tp M, such that p ∈ M → â&#x;¨, â&#x;Šp : Tp M Ă— Tp M → R is dierentiable in the sense that ∀ X,Y ∈ đ?”›(M), p ∈ M → â&#x;¨X(p), Y(p)â&#x;Šp ∈ R is dierentiable. Remark For a function M → R to be dierentiable, consider the atlas of M, {(xÎą ,UÎą )}, UÎą ⊂ Rn . Function f : M → R is dierentiable i f ∘ xÎą : UÎą ⊂ Rn → R is dierentiable. Locally, consider atlas {(xÎą ,UÎą )}, xÎą (q) = p, X(p) = ∑ aj (xÎą (q)) j

∂ ( ∂(xÎą )j )q

Write q = (x1 ,x2 , ‌ , xn ), X(p) = ∑ aj (xÎą (q)) j

∂ , ( ∂xj )q

Y(p) = ∑ bj (xÎą (q))

â&#x;¨X(p), Y(p)â&#x;Šp = ∑ ai (xÎą (q))bj (xÎą (q)) i,j

j

∂ ( ∂xj )q

∂ ∂ , ⌊( ∂xi )q ( ∂xj )q âŒŞ

q

DeďŹ ne

ďœ˛ďœľ

gij (q) = gji (q) = gij (x1 , … , xn ) =

∂ ∂ , 〈( ∂xi )q ( ∂xj )q 〉

q

This is the local representation of the “metric tensor”. Example

For H2 , ds2H = (dx21 + dx22 )/x22 . 1

dx1 x2 gH (x1 ,x2 )ij dxi dxj = , ⎛ 2 〈(dx2 ) ⎝ 0 Uα = R × R+ ,

0

dx1 1 ⎞ (dx2 )〉 x2 2⎠

xα = I

The length of a tangent vector X(p) ∈ Tp M is |X(p)|p = [⟨X(p), X(p)⟩p ]1/2 . The angle between two tangent vectors is cos θ(x,y)(p) =

⟨X(p), X(p)⟩p |x(p)|p |Y(p)|p

Example In R3 , for Cartesian coordinates, gij (x1 ,x2 ,x3 ) = δij . For spherical polar coordinates, 1 ⎛ g = ⎜0 ⎝0

0 r2 0

0 ⎞ 0 ⎟ 2 2 r sin θ⎠

Note that dV = √det g dr dθ dϕ Consider a curve γ : I ⊂ R → M. γ′ (t) ∈ Tγ(t) M,

γ′ (t) = ∑(γ ∘ xα )′i (t) Then define the length of the curve as



1

2 |γ′ (t)|γ(t) = ⟨γ′ (t), γ′ (t)⟩γ(t)

∂ ( ∂xi )γ(t)

l(γ) =

∫I

|γ′ (t)|γ(t) dt =

1 2

γ′ (t)γ′j (t)gij (γ(t)) dt ] ∫I [ i

Isometries of M Consider R3 , euclidean space. SO(3) acts on R3 preserves the length of vectors. ⃗ = |⃖x|. ⃗ If x⃖⃗ ∈ R3 , R ∈ SO(3), then |R⃖x| Isometric A map f : M → N between two Riemannian manifolds is an isometric if ∀ p ∈ M, the differential dfp : Tp M → Tf(p) N satisfies ∀ up ,vp ∈ Tp M, ⟨up ,vp ⟩p = ⟨dfp (up ), dfp (vp )⟩f(p) . A Riemannian metric can be defined on every paracompact smooth manifold. Proof Use partition of unity. Suppose {(xα ,Uα )} → xα (Uα ) = Vα is a family of functions {fα } such that fα ≥ 0, the support of fα , supp fα ⊂ Vα , and fα : M → R is smooth and ∑α fα = 1. “Support” means the number of α such that fα (p) ≠ 0 is finite. Suppose p ∈ xα (Uα ),

vp ,wp ∈ Tp M

d(xα )q : Tq Uα → Txα (q) M = Tp M −1 ⟨vp ,wp ⟩(α) = ⟨d(x−1 p α (vp ), d(xα )q (wp )⟩q

is an inner product on Rn . Then (α)

∑ fα (p)⟨vp ,wp ⟩p α

is an inner product on Tp M. Given a Riemannian metric on a manifold, a distance function can be defined as dM (p,q) = inf(Lγ | γ(0) = p & γ(1) = q)

p,q ∈ M



. Lie Groups as Riemannian Manifolds 2.2.1 Lie group Suppose G is a group. Then . . . .

G × G → G is the group operation. Identity e. ex = xe = x ∀ x ∈ G Inverse. x ∈ G, ∃ x−1 ∈ G such that xx−1 = x−1 x = e. Associativity. x,y,z ∈ G, ⇒ (xy)z = x(yz). Lie group is a group G with smooth structure, so that G → G given by −1 x → x is smooth, and G × G → G by (x,y) → xy is smooth. General linear n × n real nonsingular (det A ≠ 0) matrices group GL(n,R) and special (det A = 1) linear n × n real matrices group SL(n,R) < GL(n,R) are 2 Lie groups. GL(n,R) is a homeomorphic to an open subset of Rn ; SL(2,R) is an -dimensional Riemannian manifold. Suppose Ψ : SL(2,R) → R4 by mapping a b to (a,b,c,d) ∈ R4 ( c d) Φ : R4 → R,

Φ(a,b,c,d) = ab − cd

Ψ(SL(2,R)) = {(v1 ,v2 ,v3 ,v4 ) ∈ R4 | Φ(v1 v2 ,v3 ,v4 ) = 1} Ψ(SL(2,R)) is a regular surface. Proof

Differential dΦa : Tx R4 → TΦ(x) R, a ∈ Ψ(SL(2,R)). dΦa = (a4 , −a3 , −a2 ,a1 ) : R4 → R dΦa (x1 ,x2 ,x3 ,x4 )T = (a4 , −a3 , −a2 ,a1 ) ⋅ (x1 ,x2 ,x3 ,x4 )



(a4 , −a3 , −a2 ,a1 ) = 0 is the only critical point of ÎŚa . Therefore 1 is a regular value, ÎŚâˆ’1 ({1}) is a regular surface. Suppose G is a Lie group, then G has actions on itself. For x ∈ G, Lx : G → G by Lx y = xy Rx : G → G by Rx y = yx Lx and Rx are dieomorphisms of G: (Lx )−1 = L(x−1 ) . Now suppose Ď• : M → N is a dieomorphism, m ∈ M, then dĎ•m : Tm M → TĎ•(m) N. If f ∈ C∞ (VĎ•(m) ) in which VĎ•(m) is a neighborhood of Ď•(m) on N, then f ∘ Ď• ∈ C∞ (Vm ). Suppose X ∈ đ?”›(M), then ((dĎ•)(X))Ď•(m) (f) = X(f ∘ Ď•)m Also note that dĎ•(X) ∈ đ?”›(N). Go back to Lie group G. In the discussion above, Ď• corresponds to Lx , m corresponds to y. For ďŹ xed x ∈ G, d(Lx )y : Ty G → TLx y G = Txy G Left invariance ďœą.

A Riemannian metric on G is left invariant if ∀ y ∈ G, ∀ vy ,wy ∈ Ty G, for each x ∈ G, â&#x;¨vy ,wy â&#x;Šy = â&#x;¨d(Lx )y vy , d(Lx )y wy â&#x;ŠLx y

ďœ˛.

Note that d(Lx )y vy ∈ Txy G. X ∈ đ?”›(G) is left invariant if ∀ x ∈ G, d(Lx )X = X. Right invariance can be deďŹ ned similarly. Every Lie group admits a left invariant Riemannian metric and has left invariant vector ďŹ elds. Proof Take a tangent vector Xe ∈ Te G in which e is the identity element. Construct map Xx = d(Lx )e Xe for x ∈ G, d(Lx )e : Te G → Tx G. Claim: {Xx } ∀ x ∈ G is a left invariant vector ďŹ eld on G.

ďœ˛ďœš

(d(Lx )e Xe )(f) = Xe (f ∘ Lx ),

f ∘ Lx ∈ C∞ (Ve )

(d(Ly )x Xx )(f) = (d(Ly )x Xe )(f ∘ Lx ) = Xe (f ∘ Ly x) = Xyx (f) ⇒ d(Ly )x Xx = Xyx Now construct a left invariant Riemannian metric on G. v,w ∈ Tx G. Apply d(Lx−1 )x v ∈ Te G, â&#x;¨v,wâ&#x;Šx = â&#x;¨d(Lx−1 )x v, d(Lx−1 )x wâ&#x;Še This gives a left invariant Riemannian metric on G, where we choose â&#x;¨â‹…, â‹…â&#x;Še to be an inner product on Te G.

2.2.2 Observations on Lie bracket Suppose X,Y ∈ đ?”›(M), then [X,Y] ∈ đ?”›(M). If X and Y are also left invariant, then [X,Y] is left invariant. Proof

Suppose p,z ∈ G, f ∈ C∞ (Vz,p ), d(Lz )p [X, Y](f) = d(Lz )p (XY − YX)(f) = X(d(Lz )p Y(f)) − Y(d(Lz )p X(F)) = XY(f ∘ Lz ) − YX(f ∘ Lz ) = [X, Y](f ∘ Lz )

If X and Y are left invariant and X = d(Lx )Xe , Y = d(Lx )Ye , for Xe ,Ye ∈ Te G, then [X,Y] can be restrited to Te G, [X,Y]e ∈ Te G. Tangent space Te G with this product is a Lie algebra. Lie algebra A Lie algebra (real) â„’ is a real linear vector space with a product â„’ Ă— â„’ → â„’ called (X,Y) ↌ [X,Y] such that

ďœłďœ°

. . .

[α1 X1 + α2 X2 , Y] = α1 [X1 ,Y] + α2 [X2 ,Y]. [X,Y] = −[Y,X] [[X,Y], Z] + [[Z,X], Y] + [[Y,Z], X] = 0 Theorem G is a compact Lie group, then G admits a bi-invariant Riemannian structure. Proof G,

Construct this by averaging a left invariant Riemannian metric over

⟨u,v⟩y =

∫G

⟨(dRx )y u, (dRx )y v⟩xy dw(x)

in which the inner products are left invariant. Example SU(2) is 2 × 2 unitary complex matrices group. Suppose A ∈ SU(2), then A* A = I, det A = 1. SU(2) is a 3-dimensional real Lie group, λe−iθ −iδ (−ρe

ρeiδ , λeiθ )

λ2 + ρ2 = 1,

θ,δ ∈ R

The generators of 3-dimensional real Lie groups are Pauli matrices σx =

1 0 2 (1

1 , 0)

σy =

1 0 2 (i

−i , 0)

σz =

1 1 2 (0

0 −1)

with [σi ,σj ] = iϵijk σk . Elements in Lie algebra gives a curve on the manifold of Lie group whose tangent vector at the identity is the elements themselves. For example, curve ϕ(t), ddϕt = iσj ϕ, with ϕ(0) being the identity. For j = 3, ϕ(t) =

eit (0

0 e

−it

∈ SU(2) )

SU(2) is a simply connected compact Lie group. If Z2 = {1, −1}, then SU(2)\Z2 = SO(3). Topologically, SO(3) ≈ PR3 . ∂ ∂ ∂ If the space is infinitely dimensional, T0 R3 is spanned by { ∂x , ∂y , ∂z }, and



∂ ∂ Lx = −i y − z , ∂y ) ( ∂z

∂ ∂ Ly = −i z −x , ∂z ) ( ∂x

∂ ∂ Lz = −i x −y ∂x ) ( ∂y

with [Li ,Lj ] = iϵijk Lk . Lj ∈ 𝔛(R3 ), (Uθ f)(X) = f(R(θ)X), Uψ,z = e−iψLz .



3 KOZULOR AFFINE CONNECTIONS ďœł.ďœą Connections 3.1.1 DeďŹ nition and properties Ë™ Consider a curve c : I → M, c(t) = ddct ∈ Tc(t) M. If c is arc length parametrized, Ë™ ¨ ¨ the velocity |c(t)| = 1. The question arises whether c(t) exists, whether c(t) ∈ Tc(t) M, and how to dierentiate a vector ďŹ eld along a curve. Vector ďŹ eld along a curve A vector ďŹ eld V(t) along a curve c : I → M is a smooth map on I so that ∀t ∈ I, V(t) ∈ Tc(t) M, and f ∈ C∞ (M,R), t → V(t)(f) ∈ R is dierentiable. Ë™ For example, c(t) is a vector ďŹ eld on c. If X ∈ đ?”›(M), deďŹ ne V(t) = X(c(t)) =

∂ ∑ bj (c(t)) ∂x j |c(t) ) ( j

is a vector ďŹ eld along c(t). We will ďŹ nd a notion of a derivative of a vector ďŹ eld V(t) along c(t) called the dV ′ covariant derivative DdV t of V along c. It replaces dt and gives when V(t) = c (t) a notion of “intrinsic accelerationâ€? as viewed from M. Example Consider regular surface S = {(x1 ,x2 ,0) | x1 ,x2 ∈ R} ⊂ R3 . Suppose curve c(t) = (x1 (t), x2 (t), 0), Ë™ c(t) = (xË™ 1 (t), xË™ 2 (t), 0),

¨ c(t) = (x¨ 1 (t), x¨ 2 (t), 0)

Ë™ ¨ â‹… c(t) Ë™ All of them live in the plane S. Also c(t) â‹… c(t) = 0, c(t) = 0. The normal ˆ (t) = eˆ3 is a constant normal to Tc(t) S. vector n Example

Consider ďœ˛-sphere S2 . Take a meridian as parametrized curve,

ďœłďœł

c(t) = (sin t, 0, cos t), ˙ c(t) = (cos t, 0, − sin t),

c(0) = (0,0,1),

I = (0,Ď€),

|c(t)| = 1

¨ c(t) = (− sin t, 0, − cos t) = −c(t)

Ë™ c(t) â‹… c(t) = 0,

ˆ (t) = c(t) n

¨ c(t) is not a vector ďŹ eld along the curve because it is not in the tangent space. We want to distinguish these two extreme examples by ďŹ nding what is the acceleration like if we are on the manifold. Therefore we project the acceleration to the tangent space, Ë™ D[c(t)] ¨ = Î Tc(t) S c(t) dt In general if V(t) is a vector ďŹ eld along c(t), then DV(t) dV = Î Tc(t) S dt ( dt ) is a vector ďŹ eld along c(t). It is called covariant derivative of V along c. If DV(t) DV D dt = 0, c is a geodesic. dt ∈ Tc(t) M. dt is a derivative in the sense that for vector ďŹ elds V,W along c, D DV DW (V + W) = + dt dt dt D DV (fV) = f + fË™ V, dt dt

f:I→R

Connection A connection ∇ on a dierentiable manifold M is a map ∇ : đ?”›(M) Ă— đ?”›(M) → đ?”›(M), ∇ : (X,Y) ↌ ∇X Y ∈ đ?”›(M) such that ∀ X,Y,Z ∈ đ?”›(M) and for f,g ∈ đ?’&#x;(M) in which đ?’&#x;(M) = C∞ (M,R), ďœą. ďœ˛. ďœł.

∇fX+gY Z = f ∇X Z + g∇Y Z ∇X (Y + Z) = ∇X Y + ∇X Z ∇X (fY) = X(f)Y + f ∇X Y

ďœłďœ´

Suppose X ∈ đ?”›(M), X(t) = X ∘ c(t) is a vector ďŹ eld along c, then DX(t) = ∇c(t) X(t) Ë™ dt Example Suppose M = Rn , X = (X1 (u), ‌ , Xn (u)), Y = (Y1 (u), ‌ , Yn (u)). 󞌃 DeďŹ ne map ∇, n 󞌃∇V X = ∑ Vj ∂ X = (V â‹… ∇)X = (V â‹… ∇X1 (u), ‌ , V â‹… ∇Xn (u)) ( ∂uj ) j=1

in which ∇ is the gradient operator. Now verify that ∇󞌃 is a connection. ďœą.

Suppose f,g : Rn → R. ∇󞌃 fV+gW X = (fV + gW) ⋅ ∇X = fV ⋅ ∇X + gW ⋅ ∇X = f∇󞌃 V X + g∇󞌃 W X

ďœ˛. ďœł.

∇󞌃 V (X + Y) = (V ⋅ ∇)(X + Y) = ∇󞌃 V X + ∇󞌃 V Y ∇󞌃 V (fX) = (V ⋅ ∇)(fX) = (V ⋅ (∇f))X + f∇󞌃 V X = V(f)X + f∇󞌃 V X. Note that V = ∑ Vj

∂ , ∂uj

V(f) = ∑ Vj ∂j f = V ⋅ ∇f

Therefore ∇󞌃 is a connection. Connections are local. Consider two local expressions for X,Y ∈ đ?”›(M) X = ∑ xj Xj ,

Y = ∑ yj Xj

where at p ∈ M, Xj (p) ∈ Tp M form a basis. Suppose ∇ is a connection on M, ∇X Y = ∇∑ xj Xj Y = ∑ xj ∇Xj Y j

∇Xj Y = ∑ ∇Xj (yk Xk ) = ∑ [Xj (yk )Xk + yk ∇Xj Xk ] k

k

ďœłďœľ

Recall that gkl (p) = ⟨Xk (p), Xl (p)⟩p if M is Riemannian, and that if {Xk (p)} is a basis from Tp M, then ∇Xj Xk = δjk Xk in Rn , ∇Xj Xk a section of TM, and n l ∇Xj Xk = ∑ Γjk Xl l=1 l in which Γjk is Christoffel symbol of the connection ∇. Then l ∇Xj Y = ∑ Xj (yk )Xk + yk Γjk Xl [ ] k

Now exchange l and k, k ⇒ ∇X Y = ∑ ∑ xj Xj (yk ) + ∑ xj yl Γjl Xk k [ j l,j ]

The expression in the bracket is a function that only depends on where it is evaluated on M. This formula shows the existence and uniqueness of ∇. Now proof the uniqueness and existence of the covariant derivative. ¯ U), x(U) ¯ Proof Take a parametrization of M, (x, ∩ c(t) ≠ ∅. Define x¯ −1 ∘ c(t) = (x1 (t), … , xn (t)) V(t) = ∑ vj (t)

∂ , ∂xj (t)

Xj =

∂ ∂xj (t)

dvj DXj DV D = ∑ vj (t)Xj = ∑ Xj + vj (t) dt dt dt ) dt j j ( Get it??

(.)

Now use the connection between connections and covariant derivatives, the ˙ Xj are Xj on U resitricted to x(t) = (x¯ −1 ∘ c)(t), noting that x(t) is the tangent vector along x(t), DXj = ∇x(t) Xj | ˙ x(t) dt Expand x˙ ∈ Tx(t) U,



(.)

l ˙ x(t) = ∑ x˙ k Xk ⇒ ∇x(t) Xj = ∑ x˙ k (t)∇xk Xj = ∑ x˙ k (t)Γkj Xl ˙ k

k

Substitute this into eq. (.), exchange l and j, and substitute into eq. (.). Note that the first term in eq. (.) is free of this exchange of index. ⇒

dvj DV j =∑ + vl (t)x˙ k (t)Γkl Xj ∈ Tx(t) U dt dt ] |x(t) j,k,l [

˙ If V is the velocity vector field x(t), (using Einstein summation notation,) Dx˙ j = x¨ j (t) + x˙ j (t)x˙ k (t)Γkl Xj [ ] dt in which the second term is only determined by the Riemannian metric. Parallelism

A vector field along a curve c is paralell to c if

DV dt

For ∀ c, ∃ a unique vector field V along c such that V(t0 ) = V0 , Proof

= 0, t ∈ I. DV dt

= 0.

¯ U) and x(t) = x¯ −1 ∘ c(t). Suppose parametrization (x, ∂ j V0 = ∑ v0 ∂xj (t0 )

We look for V(t) = ∑ vj (t)

∂ ∂xj (t)

such that DV j = ∑ (v˙ j + x˙ j vl Γlk ) Xj = 0 dt j which is equivalent to say ∀ j, j v˙ j + vl (x˙ k Γlk ) = 0,

⇒

j

vj (t0 ) = v0

d (v1 , … , vn )T = M(v1 , … , vn )T dt



j

in which M is an n × n matrix, Mlj = x˙ n Γlk is known. This linear system always has a solution. ˙ If V = c(t), there is no acceleration on the manifold. For example, the velocity of a geodesic is parallel to the geodesic. Now we will let metric play a role in this. Consider a Riemannian structure (M,g), glm = ⟨Xl ,Xm ⟩, and a connection ∇ on M. The idea is to take two vector fields V and W along c, and look at function t ∈ I → ⟨V(t), W(t)⟩c(t) . We expect something like d dV dW ⟨V(t), W(t)⟩c(t) = , W + V, dt 〈 dt 〉 〈 dt 〉 But the right-hand side is not always a vector field, therefore use covariant derivatives Compatibility Given a Riemannian manifold (M,g), a connection ∇ is comaptible with the matric if ∀ c : I → M, ∀ two parallel vector fields P and P′ along c, ⟨P,P′ ⟩ is constant. Theorem Given (M,g), a connection ∇ on M is compatible with the metric iff ∀ curve c : I → M and ∀ two vector fields X and Y along c, d DX DY ⟨X,Y⟩ = , Y + X, dt 〈 dt 〉 〈 dt 〉 Proof (Forward) Take a curve c(t), t0 ∈ I. In Tc(t0 ) M, choose an orthonormal basis {p01 , … , p0n }. For each j, ∃ a parallel vector field Pj (t) along c. Because ∇ is compatible with the metric, ⟨Pj (t), Pk (t)⟩c(t) = ⟨P0j ,P0k ⟩c(t0 ) = δjk Expand the two vector fields X and Y along c in {Pj (t)}, X = ∑ xj Pj , Since



DPj dt

= 0,

Y = ∑ yl Pl

DX = ∑ x˙ j (t)Pj , dt ⇒

DY = ∑ y˙ i (t)Pi dt

DX ,Y = xË™ j (t)yk (t)â&#x;¨Pj ,Pk â&#x;Š = ∑ xË™ j yj , ⌊ dt âŒŞ ∑ ⇒

DY X, = xj yË™ j ⌊ dt âŒŞ ∑

DX DY d d , Y + X, = xj yj = â&#x;¨X,Yâ&#x;Š ∑ dt ⌊ dt âŒŞ ⌊ dt âŒŞ dt

(Backward) is trivial. Take X = P and Y = P′ parallel along c, then d ′ dt â&#x;¨P,P â&#x;Š = 0. Theorem: Fundamental theorem of Riemannian Geometry Consider Riemannian manifold (M,g), a unique connection ∇ : đ?”›(M) Ă— đ?”›(M) → đ?”›(M) compatible with the metric g and symmetric. This connection is called Riemannian connection or Levi-Cevita connection. A connection ∇ is symmetric if ∀ pairs of vector ďŹ elds X,Y ∈ đ?”›(M), ∇X Y − k k ∇Y X = [X,Y], ⇔ Γij = Γji . Compatibility (alternative deďŹ nition) ∀ X,Y,Z ∈ đ?”›(M),

∇ is compatible with the metric if

Xâ&#x;¨Y,Zâ&#x;Š := X(â&#x;¨Y,Zâ&#x;Š) = â&#x;¨âˆ‡X Y, Zâ&#x;Š + â&#x;¨Y, ∇X Zâ&#x;Š Proof of fundamental theorem

Uniqueness: Take X,Y,Z ∈ đ?”›(M),

Xâ&#x;¨Y, Zâ&#x;Š = â&#x;¨âˆ‡X Y, Zâ&#x;Š + â&#x;¨Y, ∇X Zâ&#x;Š

(ďœł.ďœł)

Yâ&#x;¨Z, Xâ&#x;Š = â&#x;¨âˆ‡Y Z, Xâ&#x;Š + â&#x;¨Z, ∇Y Xâ&#x;Š

(ďœł.ďœ´)

Zâ&#x;¨X, Yâ&#x;Š = â&#x;¨âˆ‡Z X, Yâ&#x;Š + â&#x;¨X, ∇Z Yâ&#x;Š

(ďœł.ďœľ)

Take eq. (ďœł.ďœł) + eq. (ďœł.ďœ´) − eq. (ďœł.ďœľ), the right-hand side is â&#x;¨âˆ‡X Y,Zâ&#x;Š + â&#x;¨Z, ∇Y Xâ&#x;Š + â&#x;¨âˆ‡Y Z,Xâ&#x;Š − â&#x;¨âˆ‡Z Y,Xâ&#x;Š + â&#x;¨Y, ∇X Zâ&#x;Š − â&#x;¨Y, ∇Z Xâ&#x;Š = 2â&#x;¨âˆ‡X Y,Zâ&#x;Š + â&#x;¨[Y,Z], Xâ&#x;Š + â&#x;¨[Z,X], Yâ&#x;Š + â&#x;¨[X,Y], Zâ&#x;Š noting that â&#x;¨a,bâ&#x;Š = â&#x;¨b,aâ&#x;Š.

ďœłďœš

1 ⇒ ⟨∇X Y, Z⟩ = (X⟨Y, Z⟩ + Y⟨Z, X⟩ − Z⟨X, Y⟩ 2 − ⟨[X, Y], Z⟩ − ⟨[Y, Z], X⟩ − ⟨[Z, X], Y⟩)

(.)

The right-hand side is only determined by the metric, therefore it is unique. Existence: Define X,Y → ∇X Y by eq. (.) with arbitrary Z. We can proof this is a connection.

3.1.2 Local formula for the Christoffel symbol ¯ U), U ⊂ Rn . Define Consider parametrization (x, Xi =

∂ , ∂xi

gij = ⟨Xi ,Xj ⟩,

l ∇Xi Xj = Γij Xl

In this case X = Xi , Y = Xj , Z = Xk . l l ⟨∇X Y,Z⟩ = ⟨∇Xi Xj ,Xk ⟩ = ∑ Γij ⟨Xl ,Xk ⟩ = ∑ Γij glk l

l

Since we are working in U ⊂ Rn , [Xi ,Xj ] = 0. X⟨Y,Z⟩ = Xi ⟨Xj ,Xk ⟩ =

∂ gik ∂xi

1 ∂gik ∂gki ∂gij l ∑ Γij glk = 2 ∂x + ∂x + ∂x ( i j k) l

(.)

The matrix representation of g, [g], is a n × n nonsingular matrix. Define gkm = [g−1 ]km , glk gkm = δlm . Multiply eq. (.) by gkm and sum over k, l

∑ Γij glk g k,l

km

=∑ k

1 ∂gjk ∂gki ∂gij + − gkm 2 ( ∂xi ∂xj ∂xk )

1 m ⇒ Γij = gmk (∂i gjk + ∂j gik − ∂k gij ) 2



3.1.3 Vector ďŹ elds and ows Consider dieomorphism Ď• : M → N. Then dĎ•p : Tp M → TĎ•(p) M, dĎ• : đ?”›(M) → đ?”›(N), (dĎ•)Xp = XĎ•(p) . Suppose f ∈ C∞ (VĎ•(p) , then ((dĎ•)X)(f) = XĎ•(p) (f) = Xp (f ∘ Ď•), f ∘ Ď• ∈ C∞ (Vp ). DeďŹ ne notation Xp (f ∘ Ď•) = (Ď•âˆ— X)p (f) = ((dĎ•)X)Ď•(p) (f), Ď•âˆ— X ∈ đ?”›(N). Given a smooth ďœą-parameter family of dieomorphisms t ∈ R ↌ θt , i.e. θ0 = 1, θt ∘ θs = θt+s , θ−1 t = θ−t , (note that one of these dieomorphisms is −tX θt = e ,) ∃! vector ďŹ eld X ∈ đ?”›(M) such that Symbol ∃! means “exists uniqueâ€?.

Xp (f ∘ θt ) − Xp (f) 1 Xp (f) = lim = [(θt∗ X)p (f) − Xp (f)] t→0 t t Then X is the inďŹ nitesimal generator of θt . Conversely, X ∈ đ?”›(M), for each p ∈ M, ∃ δ > 0 such that dθt = X ∘ θt , dt

θt=0 (p) = p

has a unique solution on (−δ,δ). Suppose θt is a ow with inďŹ nitesimal generator X. X is invariant under θt , θt∗ X = X, (θt∗ X)p = Xθt (p) . (This can be checked using (θt∗ X)(f) = Xp (f ∘ θt ), f ∈ C∞ (Vθt (p) .) Lie derivative

LX Y is deďŹ ned as 1 [Yp (f) − (θt )∗ Yθ−t (p) (f)] , t→0 t

(LX Y)p (f) = lim

f ∈ C∞ (Vp )

in which θt is the local ow generated by X. Lemma

X(f ∘ θt ) = X(f) + tX(gt ).

With this lemma, we can write Yp (f) − θt∗ Yθ−t (p) (f) = Yp (f) − Yθt (p) (f ∘ θt ) = Yp (f) − Yθ−t (p) (f + tgt ) = Yp (f) − Yθ−t (p) (f) − tYθ−t (p) (gt )

ďœ´ďœą

1 [Yp (f) − Yθ−t (p) (f)] − Yθ−t (p) (gt ) t→0 { t }

(LX Y)p (f) = lim

The ďŹ rst term gives XY, the second gives YX. This suggests that đ?”›(M) is a inďŹ nite dimensional linear vector space over R, and it has Lie brackets in it. Therefore đ?”›(M) is a Lie algebra. Note that in contrast, Tp M is a Lie algebra of a ďŹ nite dimensional Lie group.

ďœ´ďœ˛

4 GEODESICS . Definition of Geodesics Geodesic Consider (M,g, ∇), ∇ is compatible to the metric. γ : I ⊂ R → M is a geodesic at t0 ∈ I if D dγ =0 dt dt |t=t0 And (γ,I) is geodesic if this hold ∀ t ∈ I. In another word, dγ (t) ∈ Tγ(t) M ( dt ) The acceleration vector field D dγ (t) ∈ Tγ(t) M dt ( dt ) If γ is a geodesic, then this vector field along γ is zero. Till now, only a connection is needed. Metric has not play any role. If we have a Riemannian structure (M,g), dγ 2 dγ dγ =2 , = gij (γ(t))γ˙ i (t)γ˙ j (t) | dt |γ(t) 〈 dt dt 〉γ(t) ∑ ij If ∇ is compatible, d dγ 2 D dγ dγ =2 , =0 dt | dt |γ(t) 〈 dt dt dt 〉 Remark Recall that we can always scale and shift t so that t starts from zero, and ends at 1. Consider curve c : [a,b] ⊂ I → M,



t

s(t) =

∫a

˙ |c(u)| c(u) du

˜ Define c(t) = c(t + a), then c˜ : [0,b − a] ⊂ I → M. t

s˜ (t) =

∫0

˜˙ |c(u)| du = s(t + a) ˜ c(u)

˙ If γ : I = [a,b] → M is a geodesic, s(t) = (t − a)cγ , in which cγ = |γ(t)|. ˜ ˆ (t) = γ((t + a)/cγ ). s˜ (t) = cγ t. Define γ(t) = γ(t + a), γ

Local Equation for Geodesics ¯ p ∈ x(U), ¯ Consider (U, x), q = x¯ −1 (p). Locally q ∈ U, γ ∘ x¯ −1 (t) = (x1 (t), … , xn (t)). The covariant derivative of a vector field V along γ is DV dvk dvi dxi k =∑ + Γ Xk dt dt dt ij ) ( dt

with V = ∑ vj (x) j

∂ = vj Xj ∂xj |x(t) ∑

Apply to a geodesic in local coordinates V=

dγ dx = , dt dt

0=

D dγ d2 xk dxi dxj k ∂ =∑ + Γij 2 dt dt dt dt dt ) ∂xk t (

˙ 0 ) = v0 . Then ∃ local solution x(t). with initial conditions x(t0 ) = x0 , x(t

. Geodesic Flow Note that any ODE can be written as a set of first order ODEs. Therefore the n geodesic equations can be written as 2n first order ODEs living in the tangent bundle TM called geodesic flow. Geodesic flow Locally, take (x1 , … , xn , x˙ 1 , … , x˙ n ) ∈ TU as variables. (x1 , … , xn ) ∈ U, (x˙ 1 , … , x˙ n ) ∈ Tx U = Rn . Then the geodesic flow is



G(⃖x, ⃖⃗ x) ⃖⃗˙ =

x⃖⃗˙ x⃖⃗˙ = k ¨ (x⃖⃗ ) (−Γij x˙ i x˙ j )

Note that this is a vector with 2n components. Index k runs from 1 to n. The geodesic flow G is a vector field on TM, G(⃖x, ⃖⃗ x) ⃖⃗˙ ∈ T(⃖x, ⃖⃗ x) ⃖⃗˙ (TM). Now we will introduce the exponential map on TM. It maps a small vector v in the tangent space in the neighborhood of 0 ∈ Tp M to a point on M by taking a geodesic on M starting from p with initial velocity v and let the geodesic run for a specific time. The geodesic flow on TM is d γ = G(γ,γ) ˙ dt (γ˙ ) ∃ solution ϕ(t,q,v) to d ϕ(t,q,v) = Gϕ(t,q,v) dt where ϕ(t,q,v) ∈ TM, ϕ(t = 0,q,v) = (q,v) ∈ TM, v ∈ Tq M. It is a result of uniformity that if fix point (p,0) ∈ TM, p ∈ M, 0 ∈ Tp M, ∃ neighborhood Vp ⊂ ˙ TM, (p,0) ∈ Vp , δ > 0, such that ∀ (q,v) ∈ Vp , the ODE ϕ(t,q,v) = Gϕ(t,q,v) , ϕ(t = 0,q,v) = (q,v) has an unique solution for t ∈ (−δ,δ). Informally, we can relate this to geodesics on M by taking p ∈ M. For any ϵ > 0, define the cylinder Up,ϵ = {(q,v) | q ∈ any open subset of ΠVp ⊂ M where Vp is the neighborhood of (p,0) in the result and v ∈ Tq M with |v|q < ϵ}. Formally, we have this preposition: Given p ∈ M, ∃ open neighborhood of p V󾌃p ⊂ M, ϵ > 0 and δ > 0, such that ∀ (q,v) ∈ Up,ϵ = {(q,v) | q ∈ V󾌃p , |v|q < ϵ}, ∃! geodesic γ(t,q,v) for t ∈ (−δ,δ) and γ(0,q,v) = q and γ′ (0,q,v) = v. Theorem: Scaling relation Suppose γ(t,q,v) is a geodesic on (−δ,δ), then γ(at,q,v) is a geodesic on (−δ/a, δ/a), and γ(at,q,v) = γ(t,q,av) for a > 0.



Proof Define ℎ(t) = γ(at,q,v). ℎ(0) = γ(0,q,v) = q. The the relation is equivalent to ℎ′ (0) = aγ′ (0,q,v) = av. ∇h′ (t) ℎ′ (t) = ∇aγ′ (at,q,v) (aγ′ (at,q,v)) = a2 ∇γ′ (at,q,v) γ′ (at,q,v). Since γ is a geodesic, ∇γ′ (at,q,v) γ′ (at,q,v) = 0. Therefore ℎ(t) = γ(t,q,av) by uniqueness. Exponential map Given p ∈ M, v ∈ Tp M, choose ϵ as in the corollary, so for any (q,v) ∈ Up,ϵ , define exponential map v exp(q,v) = γ(1,q,v) = γ |v|, q, |v| ) ( Note that the domain is a priory only Bϵ (0) ⊂ Tq M in which Bϵ (0) is the ball with radius ϵ at 0 ∈ Tq M. This is call the normal neighborhood in M, or the geodesic ball in M. Theorem: Local diffeomorphism theorem Suppose ϕ : M → N is differentiable. Manifolds M and N have the same dimension. Also suppose ∃ p ∈ M such that dϕp : Tp M → Tϕ(p) N is a isomorphism. Then ϕ is a local diffeomorphism. Theorem Suppose q ∈ M, ∃ ϵ > 0 such that expq : Bϵ (0) ⊂ Tq M → M is a diffeomorphism with an open neighborhood of q in M. Proof The differential of exponential map is d(expq )0 : Tq˜ (Tq M) → Texpq (q) ˜ M, in which q˜ is a point in Tq M. Note that Tq˜ (Tq M) = Tq M. d(expq )0 v =

d expq (tv) , |t=0 dt

v ∈ Tq M

expq (tv) = γ(1,q,tv) ⇒



d expq (tv) = γ′ (1,q,0) = v |t=0 dt

⇒ d(expq )0 = 1 which is a isomorphism. Therefore expp is a diffeomorphism. Normal neighborhood If V0 ⊂ Tq M is a neighborhood of 0 ∈ Tq M, such that expq : V0 → M is a diffeomorphism, then expq (V0 ) is a normal neighborhood of q in M. Define notation expq (Bϵ (0)) = Bϵ (q) ⊂ M. If ϵ > 0 is small so that Bϵ (0) ⊂ V0 ⊂ Tq M then Bϵ (q) ⊂ M is a geodesic/normal ball about q ∈ M. As an example, consider Bϵ (N) on M = S2 where N is the north pole. The boundary of Bϵ (N), ∂Bϵ (N) ⊂ M is a circle. Bπ (N) = expN (BN (0)) = S2 \{S} where S is the south pole. As shown later in Gauss’ lamma, the radial geodesics are perpendicular to ∂Bϵ (N). Theorem: Hopf-Rinow Theorem alent.

The following three statements are equiv-

.

The exponential map expp is defined on all of Tp M ∀ p ∈ M.

. .

Any geodesics can be extended to all t ∈ R (M is geodesically complete). (M,q) as a topological space is complete.

. Geodesic Normal Coordinates Consider expq : Tq M → M, q ∈ M. expq (v) = γ(1,q,v). Tv (Tq M) = Tq M → Texpq (v) M, v ∈ Tq M. Theorem: Gauss lemma defined, then

d(expq )v :

If v,w ∈ Tq M, so that expq (v) and expq (w) are

〈d(expq )v (v), d(expq )v (w)〉expq (v) = ⟨v,w⟩q Proof Write w = wN + wT , where wT = λv and ⟨wN ,v⟩q = 0. Consider two extreme conditions.



wT is parallel to v, and wN is perpendicular to v.

.

If w = λv, λ ≠ 0. 1 d(expq )v (v) = lim [(expq )(v + λv) − (expq )(v)] λ→0 λ d = γ((1 + λ)v, q, v) |λ=0 dλ = λγ′ (1, q, v) = λd(expq )v (v) ⇒ ⟨d(expq )v (v), d(expq )v (λv)⟩ = λ⟨γ′ (1,q,v), γ′ (1,q,v)⟩ = λ⟨v,v⟩ = ⟨v,λv⟩ = ⟨v,w⟩

. So v(s) is a curve on a sphere with radius |v|.

If w = wN , ⟨v,w⟩q = 0. Consider curve v(s) in Tq M, v(0) = v, v′ (0) = wN , |v(s)| = |v|. Let u(t,s) = tv(s), 0 ≤ t ≤ 1, |s| < ϵ, so that f(t,s) := expq u(t,s) is defined. |u(t,s)| ≤ |v|. Note that if we fix s to be s = s0 , then f(t,s0 ) is a geodesic, expq (tv(s0 )) = γ(1,q,tv(s0 )) = γ(t,q,v(s0 )) Then ∂f ∂f , (t, s) = d(expq )v (v), d(expq )v (wN )〉 expq (v) 〈 ∂s ∂t 〉 |t=1,s=0 〈 d ∂f ∂f D ∂f ∂f ∂f D ∂f , = , + , dt 〈 ∂s ∂t 〉 〈 dt ∂s ∂t 〉 〈 ∂s dt ∂t 〉 The last term D ∂f =0 dt ∂t

From some lemma…

since f(t,s) with s fixed is a geodesic. Also, D ∂f D ∂f = dt ∂s ds ∂t



(.)

which by compatibility condition of connections leads to D ∂f ∂f D ∂f ∂f 1 d ∂f ∂f 1 d|v|2 , = , = , = =0 〈 dt ∂s ∂t 〉 〈 ds ∂t ∂t 〉 2 ds 〈 ∂t ∂t 〉 2 ds ∂f Therefore ⟨ ∂s ,

⇒

∂f ∂t ⟩

(.)

is independent of t.

∂f (t,s = 0) = d(expq )tv (tv), ∂s

∂f (t = 0,s = 0) = 0 ∂s

∂f ∂f , (t = 0,s = 0) = 0 〈 ∂s ∂t 〉 Therefore from eq. (.) and eq. (.), 〈d(expq )v (v), d(expq )v (wN )〉expq (v) = 0 = ⟨v,w⟩ Since d(expq )v is linear, Gauss lemma holds for arbitrary w.

. Minimizing Properties The exponential map expp is a diffeomorphism from Bϵ (0) ⊂ Tp M to Bϵ (p) ⊂ M. If we start from p ∈ Bϵ (p), construct a curve c to q ∈ Bϵ (p), then l(c) ≥ l(γ) in which γ is the geodesic connecting p and q. Proof

Gauss lemma states that for v,w ∈ Bϵ (0) ⊂ Tp M, ⟨v,w⟩p = 〈d(expp )v (v), d(expp )v (w)〉 expp (v)

This implies that radial geodesics give curvilinear coordinates in Bϵ (p). For v ∈ Bϵ (0), construct geodesics γ(1,p,tv) = γ(t,p,v) : t ↦ expp (tv), d expp (tv) =v |t=0 dt The tangent vector γ′ (t,p,v) = d(expp )v (v),



1

⇒ l(γ) =

∫0

| γ′ (t)|dt = |v|

Theorem: Minimizing Theorem Let c : [0,1] → M is a piece-wise differentiable curve, c(0) = γ(0) = p, c(1) = γ(1). Note that c([0,1]) needs not stay in Bϵ (p). Then l(c) > l(γ) and l(c) = l(γ) iff c([0,1]) = γ([0,1]). Note that . .

There are geodesics that are not minimizers. Let q ∈ Bϵ (p), ∃ wp ∈ Bϵ (0) ⊂ Tp M such that γ(1,p,wp ) = expp (wq ) = q. So γ(t) = expq (twq ) is a geodesic connecting p to q. Then given any q ∈ Bϵ (0), ∃! geodesic γ connecting p to q that minimizes the length. Theorem (global version of minimizing theorem) If γ : [a,b] → M is a piece-wise differentiable curve in arc length parametrization such that ∀ other ˜ ˜ ˜ then γ curves γ˜ : [a,b] → M with γ(a) = γ(a) and γ(b) = γ(b), l(γ) ≤ l(γ), is a geodesic. Proof (local) Suppose c([0,1]) ⊂ Bϵ (p). Parametrize c(t) with polar coordinates in Tp M: γ(t) > 0, v(t) ∈ Sn−1 i.e. |v(t)| = 1. So r(t)v(t) ∈ Bϵ (0). c(t) = expp (r(t)v(t)) = γ(1,p,r(t)v(t)) = γ(r(t), p,v(t)) = f(r(t), t) c′ (t) = Note that

∂f ∂v

is in

∂f ′ ∂f r (t) + ∂r ∂t

∂f ∂t .

|c′ (t)|2 = ⟨∂r fr′ + ∂t f, ∂r fr′ + ∂t f⟩ Use Gauss lemma, ⟨∂r f, ∂t f⟩ = 0 = |r′ (t)|2 ∂f 2 with | ∂r | = 1. Therefore



∂f 2 ∂f 2 + | ∂r | | ∂t |

1

l(c) =

∫0

1

|c′ (t)| dt ≥

∫0

1

|r′ (t)| dt ≥

∫0

r′ (t)dt = r(1) = l(γ)

Now consider c([0,1]) not all in Bϵ (p), bla bla… Example Take M = H2 , ds2 = (dx2 +dy2 )/y2 . Consider curve γ(t) = (0,t), t ∈ [a,b], b > a > 0. Suppose c(t) = (x(t), y(t)), b

l(c) =

∫a √

x′2 (t)

+

y′2 (t)

b b dt dy 1 dy b ≥ dt = = ln = l(γ) ∫a y y(t) ∫a | dt | y(t) (a)





5 CURVATURE . Curvature of Geometric Objects 5.1.1 Curvature of curves in R3 Suppose α : I ∈ R → R3 , and α(t) is arc length parametrized, |α′ (t)| = 1. Then (

α′ ,

α′′ α′′ ′ , α × |α′′ | |α′′ | )

forms a triad. |α′′ (t)| = κ(t) is the curvature.

5.1.2 Curvature of regular surfaces Consider regular surface S ⊂ R3 , x¯ : U ⊂ R2 → R3 , (u,v) ∈ U ⊂ R2 . Then ¯ x(u,v) = (x1 (u,v), x2 (u,v), x3 (u,v)) ¯ is differentiable and a homeomorphism. ∃ V ⊂ R3 such that x(U) = V ∩ S. For q ∈ U, dx¯ q : Tq U → Tx(q) S is injective. Suppose S is orientable, so that for ¯ ¯ β : Wαβ ⊂ R2 → Wαβ ⊂ R2 x¯ −1 α ∘x ¯ β )] > 0. ∃ a field of unit normal vectors p ∈ S ↦ the determinant det [d(x¯ −1 α ∘x N(p) such that N(p) ⊥ Tp S, |N(p)| = 1. Note that N(p) ∈ S2 , N : p ∈ S ↦ N(p) ∈ S2 is the Gauss map for S. In principle, dNp : Tp S → TN(p) S2 The hyperplane Tp S and TN(p) S are parallel, therefore we can identify dNp as dNp : Tp S → Tp S.



Note that the ďŹ rst fundamental form is just the metric.

Fundamental forms of S ∀ p ∈ S, the ďŹ rst fundamental form is Ip (v) = â&#x;¨v,vâ&#x;Šp for v ∈ Tp S. The second fundamental form is IIp (v) = −â&#x;¨dNp v, vâ&#x;Šp . Suppose curve on S Îą : I ⊂ R → S ⊂ R3 , Îą(0) = p, ι′ (0) = vp ∈ Tp S, ι′ (t) ∈ TÎą(t) S, N(Îą(t)) ∈ S2 , |N(Îą(t))| = 1, N′ (Îą(t)) = (dNÎą(t) )(ι′ (t)) So dNÎą(t) contains how the normal N(Îą(t)) is changing along Îą(t). We can choose a basis for Tp S. Assume we choose {e1 ,e2 }, then dNp is a 2 Ă— 2 real symmetric diagonalizable matrix. Suppose −k1 and −k2 are the eigenvalues of dNp , i.e. −dNp =

k1 (p) 0 , k2 (p)) ( 0

k1 ≼ k2

then k1 (p) and k2 (p) are the principle curvature of S at p. Gauss curvature Tr(−dNp ) = k1 (p)+k2 (p) is the mean curvature. det (−dNp ) = k1 (p)k2 (p) is the Gauss curvature.

ďœľ.ďœ˛ Curvature of Riemannian Manifolds Consider Riemannian structure (M,g, ∇), in which ∇ is symmetric. For X,Y ∈ đ?”›(M), ∇X Y − ∇Y X = [X,Y] Curvature operator R(X,Y) is a linear transformation deďŹ ned for any pair of X,Y ∈ đ?”›(M), R(X,Y) : đ?”›(M) → đ?”›(M),

Z ∈ đ?”›(M) ↌ R(X,Y)Z ∈ đ?”›(M)

so that R(X,Y)Z = ∇Y ∇X Z − ∇X ∇Y Z + ∇[X,Y] Z

ďœľďœ´

Suppose X = (xËœ 1 , xËœ 2 , â‹Ż), Y = (y1 ,y2 , ‌), and Z = (z1 ,z2 , â‹Ż). ∇Y ∇X Z| = ∇Y (X â‹… ∇)Z| = Y â‹… ∇(xËœ l ∂l zk ) = ym ∂m (xËœ l ∂l )zk k

k

= ym x˜ l ∂m ∂l zk + ym (∂m x˜ l )(∂l zk ) [X, Y] = [x˜ l ∂l , ym ∂m ] = x˜ l [∂l , ym ∂m ] + [x˜ l , ym ∂m ]∂l = x˜ l ym 0 + x˜ l (∂l ym )∂m If {Xi } is a basis of TM, [Xi ,Xj ] = 0, R(Xi ,Xj )Xk = (∇Xj ∇Xi − ∇Xi ∇Xj )Xk This being nonzero means lack of commutation of directional derivatives.

5.2.1 properties of R ďœľ.ďœ˛.ďœą.ďœą General properties ďœą. ďœ˛.

R(X,Y) = −R(Y,X) For any X,Y ∈ đ?”›(M), R(X,Y) is a linear map, R(X,Y)(gZ + W) = gR(X,Y)Z + R(X,Y)W Proof

Recall that ∇X (fZ) = X(f)Z + f∇X Z. ∇Y ∇X (fZ) = YX(f)Z + X(f)∇Y Z + Y(f)∇X Z + f∇Y ∇X Z ∇[X,Y] (fZ) = [X,Y](f)Z + f∇[X,Y] Z

ďœł.

R(X,Y) is bilinear, R(gX + Z,Y) = gR(X,Y) + R(Z,Y). Theorem: Bianchi Identity R(X,Y)Z + R(Y,Z)X + R(Z,X)Y = 0 Proof (∇Y ∇X Z−∇X ∇Y Z+∇[X,Y] Z)+(∇Y ∇X Z−∇X ∇Y Z+∇[X,Y] Z)+(∇Y ∇X Z−∇X ∇Y Z+∇[X,Y] Z)

ďœľďœľ

= ∇Y [X, Z] + ∇X [Z, Y] + ∇Z [Y, X] + ∇[X,Y] Z + ∇[Y,Z] X + ∇[Z,X] Y = (∇[Z,X] Y − ∇Y [Z, X]) + (∇[X,Y] Z − ∇Z [X, Y]) + (∇[Y,Z] X − ∇X [Y, Z]) = [[Z, X], Y] + [[X, Y], Z] + [[Y, Z], X] =0 ďœľ.ďœ˛.ďœą.ďœ˛ Symmetry properties of R Sometimes I just write this as (XYZT ).

ďœą. ďœ˛. ďœł. ďœ´.

For X,Y,Z,T ∈ đ?”›(M), deďŹ ne scalar (X,Y,Z,T ) = â&#x;¨R(X,Y)Z, T â&#x;Š, then (X,Y,Z,T ) + (Y,Z,X,T ) + (Z,X,Y,T ) = 0 (Bianchi identity) (X,Y,Z,T ) + (Y,X,Z,T ) = 0 (X,Y,Z,T ) + (X,Y,T,Z) = 0 (X,Y,Z,T ) = (Z,T,X,Y) Proof of property ďœł

We will show that (XY W W) = 0, then

(X,Y, Z + T, Z + T ) = 0 = (X,Y,Z, Z + T ) + (X,Y,Z, Z + T ) = (XYZZ) + (XYZT ) + (XYTZ) + (XY T T ) Consider (XYZZ) = â&#x;¨âˆ‡Y ∇X Z − ∇X ∇Y Z + ∇[X,Y] Z, Zâ&#x;Š Since X(â&#x;¨Y,Zâ&#x;Š) = â&#x;¨âˆ‡X Y,Zâ&#x;Š + â&#x;¨Y, ∇X Zâ&#x;Š, â&#x;¨âˆ‡Y ∇X Z, Zâ&#x;Š = Y(â&#x;¨âˆ‡X Z, Zâ&#x;Š) − â&#x;¨âˆ‡X Z, ∇Y Zâ&#x;Š â&#x;¨âˆ‡X ∇Y Z, Zâ&#x;Š = X(â&#x;¨âˆ‡Y Z, Zâ&#x;Š) − â&#x;¨âˆ‡Y Z, ∇X Zâ&#x;Š 1 â&#x;¨âˆ‡[X,Y] Z, Zâ&#x;Š = [X,Y](â&#x;¨Z,Zâ&#x;Š) − â&#x;¨Z, ∇[X,Y] Zâ&#x;Š = [X,Y](â&#x;¨Z,Zâ&#x;Š) 2 Since the metric is symmetric, â&#x;¨âˆ‡X Z, ∇Y Zâ&#x;Š + â&#x;¨âˆ‡Y Z, ∇X Zâ&#x;Š = 0

ďœľďœś

1 ⇒ (XYZZ) = Y(⟨∇X Z, Z⟩) − X(⟨∇Y Z, Z⟩) + [X, Y](⟨Z, Z⟩) 2 1 1 1 =Y X(⟨Z, Z⟩) − X Y(⟨Z, Z⟩) + [X, Y](⟨Z, Z⟩) (2 ) (2 ) 2 =0 Use Bianchi identity four times,

Proof of property 

(XYZT) + (YZXT) + (ZXYT) = 0 (YZTX) + (ZTYX) + (TYZX) = 0 (ZTXY) + (TXZY) + (XZTY) = 0 (TXYZ) + (XYTZ) + (YTXZ) = 0 Add together, ⇒ 2(ZXYT) + 2(TYZX) = 0 ⇒ (ZXYT) = −(TYZX) ⇒ (ZXYT) = (YTZX)

5.2.2 Local formulas ∂ ¯ Suppose p ∈ X(U) ⊂ M, Xi = ∂x is a basis of Tx¯ −1 (p) U. R(Xi ,Xj )Xk is a point i in the tangent space. Write the components of this point,

R(Xi ,Xj )Xk = Rlijk Xl k Note a similar relation ∇Xi Xj = Γij Xk . Define notation

gij,k =

∂ gij ∂xk

Then 1 m Γij = (gik,j + gki,j − gij,k )gkm , 2

gkm = [g−1 ]km



Define X = ∑ ui Xi ,

Y = ∑ yj Xj ,

Z = ∑ wk Xk

R(X,Y)Z = ui yj wk R(Xi ,Xj )Xk R(Xi ,Xj )Xk = ∇Xj ∇Xi Xk − ∇Xi ∇Xj Xk + ∇[Xi ,Xj ] Xk Since [Xi ,Xj ] = 0, and l l l l m ∇Xj (∇Xi Xk ) = ∇Xj (Γik Xl ) + Γik ∇Xl Xl = Xj (Γik )Xl + Γik Γjl Xm l l m ∇Xi (∇Xj Xk ) = Xi (Γjk )Xl + Γjk Γil Xm l l l m l m ⇒ R(Xi ,Xj )Xk = Xj (Γik ) − Xi (Γjk ) Xl + (Γik Γjl − Γjk Γil ) Xm = Rlijk Xl [ ] l s l s s s ⇒ Rsijk = Γik Γjl − Γjk Γil + Γik,j − Γjk,i

Riemannian curvature tensor

Rijks = Rlijk gls

The Riemannian curvature tensor satisfies several symmetry properties, Rijkl = −Rjikl ,

Rijkl = −Rijlk ,

Rijkl = Rklij ,

Rijks + Rjkis + Rkijs = 0. We can then define Ricci curvature tensor as Rij = Rkikj = glk Rlikj , and scalar curvature R = gij glk Rlikj . The Einstein field equation for metric [gij ] is 1 Rij − gij R = −8πGTij , 2 in which Tij is the momentum-energy tensor.



5.2.3 Sectional curvature Suppose p ∈ M, and a 2-dimensional subspace of Tp M, σ ⊂ Tp M ≈ Rn . Vectors u,v ∈ Tp M. Define the magnitude of outter product as |u ∧ v|p = Sectional curvature is defined by

√

⟨u,u⟩p ⟨v,v⟩p − ⟨u,v⟩2p .

κ(u,v,p) for linearly independent vectors u,v ∈ Tp M

κ(u,v,p) =

⟨R(u, v)u, v⟩p . |u ∧ v|2p

Claim Sectional curvature κ(u,v,p) does not depend on the choice of u and v. So the sectional curvature of M at p for σ ⊂ Tp M can be written as κ(σ,p). Proof

Write κ as κ(u,v,p) =

(uvuv)p . |u ∧ v|2p

Suppose {u,v} and {u′ ,v′ } are basis of σ, and u′ = α1 u + α2 v,

v′ = β1 u + β2 v.

Then the Jacobian det

α1 (β1

α2 = det S ≠ 0. β2 )

Then |u′ ∧ v′ | = det S(u ∧ v). (u′ v′ u′ v′ ) = ((α1 u + α2 v), (β1 u + β2 v), (α1 u + α2 v), (β1 u + β2 v)) = (det S)2 (uvuv), ⇒ κ(u′ ,v′ ,p) = κ(u,v,p).



5.2.4 Spaces of constant curvature The sectional curvature can be scaled by a constant, therefore we can simply write ⎧ +1 κ= 0 ⎨ ⎩ −1 Proposition by

Sn Rn Hn

.

Define R′ : Tp M × Tp M × Tp M → Tp M to be a trilinear map ⟨R′ (X,Y,W), Z⟩ = ⟨X,Y⟩⟨Y,Z⟩ − ⟨Y,W⟩⟨X,Z⟩.

(.)

Then the Riemannian manifold M has a constant sectional curvature κ0 iff ∀ X,Y,W,Z, ⟨R(X,Y)W,Z⟩ = κ0 ⟨R′ (X,Y,W), Z⟩. Corollary Let p ∈ M, and {e1 , … , en } to be orthonormal basis of Tp M. Then for all -dimensional subspace σ ⊂ Tp M, Rijkl (p) = ⟨R(ei ,ej )ek ,el ⟩p ,

κ(p,σ) = κ0

iff Rijkl (p) = κ0 (δik δjl − δil δjk ). Note that this is a local restatement of eq. (.). Proof

Define (XYWZ)′ = ⟨R′ (X,Y,W), Z⟩. Then (XYWZ)′ = −(YXWZ)′ = −(XYZW)′ = (WZXY)′ . κ(p,σ) =

(XYXY) (XYXY) = , 2 − ⟨X, Y⟩ (XYXY)′

|X|2 |Y|2

⇒ (XYXY)′ = κ(p,σ)(XYXY). This implies that sectional curvatures determine the curvature tensor. We can reconstruct (XYZT) from (XYZT)′ . Proposition Suppose V is a linear vector space with dimension greater or equal to 2. Define R : V × V → V, R′ : V × V → V,



(xyzt) = ⟨R(x,y)z,t⟩,

and (xyzt)′ = ⟨R′ (x,y)z,t⟩.

Suppose they satisfy all the symmetries of curvature. Define κ(σ) and κ′ (σ) on -dimensional subspace σ ⊂ V spanned by x and y. If κ(σ) equals κ′ (σ) for all σ, then R = R′ .

5.2.5 Ricci curvature The Ricci curvature is an average of all the sectional curvatures. Ricci curvature Take p ∈ M, choose x ∈ Tp M, and let {z1 , … , zn−1 } to be a orthonormal basis of Tp M minus the space spanned by {x}. The Ricci curvature at p ∈ M corresponding to x ∈ Tp M is then Ricp (x) =

1 n−1 ⟨R(x, zi )x, zi ⟩ 1 n−1 = κ(p,σ(x,zi )). n−1 ∑ |x ∧ zi |2 n−1 ∑ i=1 i=1

5.2.6 Scalar curvature Scalar curvature Take p ∈ M, and an orthonormal basis {z1 , … , zn } of Tp M. The scalar curvature at p is κ(p) =

1 n Ricp (zi ). n∑ i=1

Ricci curvature and scalar curvature are independent of the choice of basis of Tp M. Proof Take x,y ∈ Tp M, and orthonormal basis {zi }n i=1 of Tp M. Define bilinear map n

Q : Tp M × Tp M → R,

Q : (x,y) ↦ ∑⟨R(x,zi )y,zi ⟩. i=1



n

Q(y,x) = ∑⟨R(y,zi )x,zi ⟩ = ∑(xzi yzi ) = Q(x,y) i

i=1

Therefore there exits a linear transformation κ : Tp M → Tp M such that Q(x,y) equals ⟨κx,y⟩ in which κ is symmetric. n

Q(x,x) = ∑⟨R(x,zi )x,zi ⟩ = (n − 1)Ricp (x) i=1

Remark The Ricci curvature Tensor Rij is defined by Q(x,y) = ∑ xi yj Rij , and it is symmetric. For x,y ∈ Tp M, define 󾌃 R(x,y) : z ∈ Tp M ↦ R(x,z)y ∈ Tp M Then 󾌃 󾌃 TrR(x,y) = ∑⟨R(x,y)z j , zj ⟩ = ∑⟨R(x,zj )y, zj ⟩ = Q(x,y) j

j

󾌃 TrR(x,x) = Q(x,x) = (n − 1)Ricp (x) Trace is independent of choice of basis.



Turn static files into dynamic content formats.

Create a flipbook
Issuu converts static files into: digital portfolios, online yearbooks, online catalogs, digital photo albums and more. Sign up and create your flipbook.