1 Riemannian Metric - UHminru/Riemann08/note1.pdf · Math 6396 Riemannian Geometry, Metric,...

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Math 6396 Riemannian Geometry, Metric, Connections, Curvature Tensors etc. By Min Ru, University of Houston 1 Riemannian Metric A Riemannian metric on a differentiable manifold M is a symmetric, positive-definite, (smooth) (0, 2)-tensor fields g on M , i.e., for any vec- tor field X, Y Γ(TM ), g(X, Y )= g(Y,X ),g(X, X ) 0 where the equality holds if and only if X = 0. In terms of the local coordinates, g can be expressed by g = g ij dx i dx j , g ij = g ∂x i , ∂x j , where g ij = g ji and the matrix (g ij ) is positive definite everywhere. There exists a Riemannian metric on a differentiable manifold with countable basis (for the topology). 2 Affine Connections A affine (or linear) connection is a map : Γ(TM ) × Γ(TM ) Γ(TM ) which satisfies, for all X,Y,Z Γ(TM ) and f C (M ), (a) X+fY Z = X Z + f Y Z , (b) X (Y + Z )= X Y + X Z , (c) X (fY )=(Xf )Y + f X Y . 1

Transcript of 1 Riemannian Metric - UHminru/Riemann08/note1.pdf · Math 6396 Riemannian Geometry, Metric,...

Page 1: 1 Riemannian Metric - UHminru/Riemann08/note1.pdf · Math 6396 Riemannian Geometry, Metric, Connections, Curvature Tensors etc. By Min Ru, University of Houston 1 Riemannian Metric

Math 6396 Riemannian Geometry, Metric, Connections, Curvature Tensors etc.By Min Ru, University of Houston

1 Riemannian Metric

• A Riemannian metric on a differentiable manifold M is a symmetric,positive-definite, (smooth) (0, 2)-tensor fields g on M , i.e., for any vec-tor field X, Y ∈ Γ(TM), g(X, Y ) = g(Y,X), g(X, X) ≥ 0 where theequality holds if and only if X = 0.

In terms of the local coordinates, g can be expressed by

g = gijdxi ⊗ dxj, gij = g

(∂

∂xi

,∂

∂xj

),

where gij = gji and the matrix (gij) is positive definite everywhere.

• There exists a Riemannian metric on a differentiable manifold withcountable basis (for the topology).

2 Affine Connections

• A affine (or linear) connection 5 is a map 5 : Γ(TM) × Γ(TM) →Γ(TM) which satisfies, for all X, Y, Z ∈ Γ(TM) and f ∈ C∞(M),

(a) 5X+fY Z = 5XZ + f 5Y Z,

(b) 5X(Y + Z) = 5XY +5XZ,

(c) 5X(fY ) = (Xf)Y + f 5X Y .

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• Levi-Civita connection. Let (M, g) be a Riemannian manifold.Then there exists a unique connection (called the Levi-Civita connec-tion) such that

(1) 5XY −5Y X = [X, Y ] (i.e. torsion free)

(2) X < Y, Z >=< 5XY, Z > + < Y,5XZ > .

Proof: We first make the following remark: Let α : Γ(TM) → C∞(M)be a C∞(M)-linear, regarding Γ(TM) as an C∞(M)-module. Thenthere exists a unique U ∈ Γ(TM) such that α(Z) =< U, Z > for everyZ ∈ Γ(TM).

Outline of the proof: First we can derive, from the conditions that theKoszul formula holds:

2 < 5XY, Z > = X < Y, Z > +Y (X, Z > −Z < X, Y >

− < Y, [X, Z] > − < X, [Y, Z] > − < Z, [Y,X] > .

This shows the uniqueness. Also, from the above discussion, if we define5XY use Koszul formula, we can verify the properties are satisfied.

• Christoffel symbols. In terms of the coordinate chart (U, xi), wewrite

5 ∂

∂xi

∂xj=

m∑k=1

Γkji

∂xk,

where the functions Γkji are called the Christoffel symbols.

• The connection 5 is torsion free if and only if Γkij = Γk

ji, and if 5is torsion free and compatible with the Riemannian metric g, then itsChristoffel symbols are given by, from the Koszul formula above,

2Γkijgkl =

∂gil

∂xj+

∂gjl

∂xi− ∂gij

∂xl,

or equivalently,

Γkij =

1

2

m∑l=1

gkl

(∂gil

∂xj+

∂gjl

∂xi− ∂gij

∂xl

).

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With coordinate charts (U, xi) and (V, yi). On U ∩ V , the Christoffelsymbols satisfy the following transformation formula

Γ′jik =

m∑p,q,r=1

Γqpr

∂yj

∂xq

∂xp

∂yi

∂xr

∂yk+

∂2xp

∂yi∂yk· ∂yj

∂xp.

For every X, Y ∈ Γ(TM), write

X =m∑

i=1

ξi ∂

∂xi,

Y =m∑

j=1

ηj ∂

∂xj,

then

5XY =∑k

∑i

ξi ∂ηk

∂xi+∑i,j

Γkjiξ

iηj

∂xk.

• In terms of the coordinate chart (U, xi), we write

5 ∂

∂xi

∂xj=

m∑k=1

Γkji

∂xk,

or

5 ∂

∂xj=

m∑i=1

m∑k=1

Γkjidxi ⊗ ∂

∂xk=

m∑k=1

ωkj

∂xk,

where

ωkj =

m∑i=1

Γkjidxi.

The matrix ω = (ωki ) is called the connection matrix of 5 with

respect to ∂/∂xi, 1 ≤ i ≤ m. So we see that we can regard 5 as theoperator 5 : Γ(TM) → Γ(T ∗(M)⊗ T (M)).

• Parallel translation, geodesics. Suppose γ(t), 0 ≤ t ≤ b, is a curvein M . A (tangent) vector field X ∈ Γ(TM) (actually X can be thevector which is only defined along γ) is said to be parallel along the

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curve γ if and only if 5γ′(t)X = 0 for t ∈ [0, b]. In terms of the localcoordinate (U, xi), write by xi(t) = xi γ(t), 1 ≤ i ≤ m and write

X(t) =m∑

α=1

X i(t)∂

∂xi|γ(t).

Then

γ′(t) =∑

i

dxi(t)

dt

∂xi.

Hence

5γ′(t)X =∑k

dXk(t)

dt+∑i,j

ΓkijX

i(t)dxj(t)

dt

∂xk|γ(t).

Hence X(t) is parallel along the curve γ if and only if

dXk(t)

dt+∑i,j

ΓkijX

i(t)dxj(t)

dt= 0.

A curve γ is geodesic in M if and only if the tangent vector of γ isparallel along the curve γ. Suppose that γ is given in a coordinatechart (U, xi) by xi(t) = xi γ(t) 1 ≤ i ≤ m. Then γ is geodesic if andonly if

d2xi

dt2+

m∑j,k=1

Γikj

dxj

dt

dxk

dt= 0 1 ≤ i ≤ m.

• Covariant derivative(connection) for tensor fields. Note that forany (fixed) tangent vector field X, according to the definition above,the operator 5X sends (1, 0)-tensor field Y (i.e. tangent vector field)to (1, 0)-tensor field 5XY . Here we want to show that such 5 induces,for any (fixed) tangent vector field X, a map (still denoted by 5X

which sends (0, s)-tensor fields to (0, s)-tensor fields. We first considers = 0 case. Let f be a function (i.e. a (0, 0)-tensor field, then define5X(f) = X(f). When s = 1, for any (0, 1)-tensor field ω (which is adifferential 1-form), 5X(ω) is also a 1-form which is defined by, for anyY ∈ Γ(T (M)),

5X(ω)(Y ) = 5X(ω(Y ))− ω(5XY ).

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In general, let Φ be a (0, s)-tensor field (resp. a (1, s)-tensor field), andlet X ∈ Γ(TM). Then we define the Covariant derivative of Φ in thedirection of X by the formula, for every Y1, . . . , Ys ∈ Γ(T (M)),

(5XΦ)(Y1, . . . , Ys) : = 5X(Φ(Y1, . . . , Ys))

−s∑

i=1

Φ(Y1, . . . , Yi−1,5XYi, Yi+1, . . . , Ys).

5XΦ is then also a (0, s)-tensor (resp. a (1, s)-tensor), and 5Φ is a(0, s + 1)-tensor (resp. a (1, s + 1)-tensor) by means of the formula

5Φ(X, Y1, . . . , Ys) := (5XΦ)(Y1, . . . , Ys).

Let (U, xi) be a local coordinate. Let Φ be a (0, s)-tensor field. Write

Φ|U = Φj1···jsdxj1 ⊗ · · · ⊗ dxjs ,

then one can verify that

5 ∂

∂xiΦ = Φj1···js,idxj1 ⊗ · · · ⊗ dxjs ,

where

Φj1···js,i =∂Φj1···js

∂xi−

s∑b=1

Φj1···jb−1kjb+1···jsΓkjbi

.

• Connections over vector bundles. Let π : E → M be a vectorbundle. Denote by Γ(E) the set of smooth sections of E. Γ(E) is areal vector space; it is also a C∞(M)-module. A connection 5 over Eis a map 5 : Γ(T (M)× Γ(E) → Γ(E), (X, s) 7→ 5Xs, which satisfies,for all X, Y ∈ Γ(T (M)), s1, s2 ∈ Γ(E), c constant, and f ∈ C∞(M):

(a) 5X+fY s = 5Xs + f 5Y s,

(b) 5X(s1 + cs2) = 5Xs1 + c5X s2,

(c) 5X(fs) = (Xf)s + f 5X s.

5

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Note that we can also view 5 as the map Γ(E) → Γ(T ∗(M) ⊗ E) by(5s)(X) = 5Xs for every s ∈ Γ(M), X ∈ Γ(TM).

Note that if E = TM , then 5 is the connection we introduced earlier.Let 5 be a connection on TM , according to the discussion earlier, itinduces a connection on the cotangent bundle T ∗M defined by, forevery ω ∈ Γ(T ∗M),

(5Xω)(Y ) = X(ω(Y ))− ω(5XY ),

for every Y ∈ Γ(T (M)).

There exists a connection on every vector bundle π : E → M if M is adifferentiable manifold with countable basis (for the topology).

Let π : E → M be a vector bundle of rank k. Let 5 be a connectionon E, i.e. 5 : Γ(E) → Γ(T ∗(M)⊗E) Let (U, xi) be a local coordinateof M . Let sα, 1 ≤ α ≤ k be a basis for Γ(E|U) (they are called thelocal frame field). Then, for every s ∈ Γ(E), write

5sα =k∑

β=1

ωβα ⊗ sβ.

So, the connection 5 is associated to a k × k matrix ω = (ωβα) of

smooth differential one forms on U , with respect to the local framesα, 1 ≤ α ≤ k. It is easy to verify that

ωβα =

m∑k=1

Γβαkdxk,

where Γjik are the Christoffel symbols. The matrix ω is called the

connection matrix of 5 with respect to sα, 1 ≤ α ≤ k.Let s ∈ Γ(E) and write

s =k∑

α=1

λαsα,

Then

5s =k∑

α=1

dλα +k∑

β=1

λβωαβ

⊗ sα.

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For X ∈ Γ(TM),

5Xs =k∑

α=1

Xλα +k∑

β=1

λβωαβ (X)

⊗ sα.

• Curvature tensor. For every X, Y ∈ Γ(TM), define R(X, Y ) (calledthe curvature operator) as R(X, Y ) : Γ(TM) → Γ(TM) by, for everyZ ∈ Γ(TM),

R(X, Y )Z = 5X 5Y Z −5Y 5X Z −5[X,Y ]Z.

It satisfies the following properties:

(1) R(X, Y ) = −R(Y,X),

(2) For every f ∈ C∞(M), R(fX, Y ) = R(X, fY ) = fR(X, Y ),

(3) R(X, Y )(fZ) = fR(X, Y )Z,

(4) When5 is torsion free, we haveR(X, Y )Z+R(Y, Z)X+R(Z,X)Y =0 (it is called the first Bianchi identity).

From property (3) above, we see that the map R(X, Y ) : Γ(TM) →Γ(TM) is a C∞(M)-linear map, so it is a smooth (1, 1)-tensor field. Itis thus called the curvature tensor.

In addition, from (1) and (2) we see that R(X, Y ) is also C∞(M)-linearin X and Y . So for every v, w ∈ Tp(M), we can define the linear mapR(v, w) : Tp(M) → Tp(M). Further, we can define the map

R : Γ(TM)× Γ(TM)× Γ(TM) → Γ(TM)

by (Z,X, Y ) 7→ R(X, Y )Z for every X, Y, Z ∈ Γ(TM). From (1)-(4)above, we see that it is multi-C∞(M)-linear. So R is a (1, 3)-tensorfield.

In terms of the local coordinates (U, xi), write

R(∂/∂xi, ∂/∂xj)∂

∂xk= Rl

kij

∂xl,

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or

R =∑

i,j,k,l

Rlkijdxk ⊗ ∂

∂xl⊗ dxi ⊗ dxj.

According to (1), we have Rlkij = −Rl

kji.

Then Rlkij satisfies the transformation rule for components of type (1,3)

tensors, i.e. on U ∩ V 6= ∅,

R′lkij =

m∑p,q,r,s=1

Rlprs

∂yl

∂xq

∂xp

∂yk

∂xr

∂yi

∂xs

∂yj.

Hence Rlkij (thus R) is an (1,3)-tensor field.

Write

5 ∂

∂xi

∂xj= Γk

ji

∂xk,

where Γkji are Christoffel symbols, then

R(

∂xi,

∂xj

)∂

∂xk=

(∂Γl

kj

∂xi− ∂Γl

ki

∂xj+ Γh

kjΓlhi − Γh

kiΓlhj

)∂

∂xl.

Hence

Rlkij =

∂Γlkj

∂xi− ∂Γl

ki

∂xj+ Γh

kjΓlhi − Γh

kiΓlhj.

For X, Y ∈ Γ(TM), if

X = X i ∂

∂xi, Y = Y j ∂

∂xj,

then

R(X, Y ) = X iY jRlkijdxk ⊗ ∂

∂xl.

In particular, if (M, g) is a Riemannian manifold. Then there is aunique connection (Levi-Civita connection) associated to g. Let, forX, Y, Z, W ∈ Γ(TM),

R(X, Y, Z, W ) = g(R(Z,W )X, Y ).

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Then it is a map

R : Γ(TM)× Γ(TM)× Γ(TM)× Γ(TM) → C∞(M)

which is multi-C∞(M)-linear. Hence R is a (0, 4)-tensor field. It iscalled the Riemannian curvature tensor. It satisfies the followingproperty: for X, Y, Z, W ∈ Γ(TM),

(1) R(X, Y, Z, W ) = −R(Y,X, Z, W ),

(2) R(X, Y, Z, W ) = −R(X, Y, W, Z),

(3) First Bianchi identity:

R(X, Y, Z, W ) + R(Z, Y,W,X) + R(W, Y, X, Z) = 0,

(4) R(X, Y, Z, W ) = R(Z,W, X, Y ).

In terms of the local coordinates (U, xi), write R = Rklijdxk ⊗ dxl ⊗dxi ⊗ dxj, then

Rklij =1

2

(∂2gik

∂xj∂xl+

∂2gjl

∂xi∂xk− ∂2gil

∂xj∂xk− ∂2gjk

∂xi∂xl

)+Γh

ikΓpjlgph−Γp

ilΓpjkgph.

From (1)-(4), we get, (1) Rijkl = −Rjikl = −Rijlk,

(2) First Bianchi identity: Rijkl + Rljik + Rkjli = 0,

(3) Rijkl = Rklij.

• The connection matrix, and the curvature matrix.

The curvature tensor can also be introduced from the connec-tion matrix as follows: Let (U, xi) be a local coordinate of M . Then∂/∂xi, 1 ≤ i ≤ m is a local frame for Γ(U, TM). Let ω1, . . . , ωmbe its dual. Denote by ω = (ωj

i ) be the connection matrix of 5 withrespect to ∂/∂xi, 1 ≤ i ≤ m, i.e.

5 ∂

∂xi=

m∑j=1

ωji

∂xj.

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Theorem. Let Rjikl be components of R with respect to the local frame

∂/∂x1, . . . , ∂/∂xm, i.e.

R(∂/∂xk, ∂/∂xl)∂

∂xi= Rj

ikl

∂xj.

Then

dωji = ωh

i ∧ ωjh +

1

2Rj

iklωk ∧ ωl.

For this reason, we let Ω = dω − ω ∧ ω. We call Ω the curvaturematrix with respect to the frame field ∂/∂xi, 1 ≤ i ≤ m.

Write Ω = (Ωji ). Then, from the definition that Ω = dω − ω ∧ ω, we

have

Ωji = dωj

i +m∑

k=1

ωki ∧ ωj

k.

The above theorem implies that

Ωji =

1

2

m∑k,l=1

Rjikldxk ∧ dxl.

This means we have alternative way to define the curvature tensor,R,namely, from the connection matrix to get the connection matrix,then to get Rj

ikl, finally to get R.

We have the Bianchi identity:

dΩ = ω ∧ Ω− Ω ∧ ω.

Proof.

dΩ = −dω ∧ ω + ω ∧ dω

= −(Ω + ω ∧ ω) ∧ ω + ω ∧ (Ω + ω ∧ ω)

= −Ω ∧ ω + ω ∧ Ω.

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Recall operator T (X, Y ) = 5XY −5Y X − [X, Y ] is called the torsionoperator. We have the following theorem:

dωi − ωj ∧ ωij =

1

2ωi(T (∂/∂xj, ∂/∂xk))ωj ∧ ωk.

The equations

dωi = ωj ∧ ωij +

1

2ωi(T (∂/∂xj, ∂/∂xk))ωj ∧ ωk,

dωji = ωk

i ∧ ωjk +

1

2Rj

iklωk ∧ ωl

are called the structure equations for (M,5). In particular, if 5 istorsion free, then the structure equations becomes:

dωi = ωj ∧ ωij,

dωji = ωk

i ∧ ωjk +

1

2Rj

iklωk ∧ ωl.

Note that if 5 is the Levi-Civita connection of (M, g), then wealso have (from the torsion free condition and the metric compatiablecondition) that

dgij =m∑

k=1

(ωki gkj + ωk

j gik),

where gij = g(∂/∂xi, ∂/∂xj).

• Cartan’s moving frame method. The above results can be gener-alized to arbitrary local frames. On a smooth m-dimensional manifoldM , in addtion to the (standard) local frame ∂/∂xi, 1 ≤ i ≤ m, wealso often use an arbitrary local frame e1, . . . , em for Γ(U, T (M)) (forexample, we take the orthonormal basis after the Gram-Schmidt pro-cess). Let θ1, . . . , θm be its dual. We still write 5ei

ej = Γkjiek, where Γk

ji

are the Christoffel symbol with respect to the frame e1, . . . , em.Let

θji = Γj

ikθk.

Then 5ei = θji ej. The matrix θ = (θj

i ) is called the connection matrixof 5 with respect to the local frame e1, . . . , em.

11

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Theorem. Let Rjikl be components of R with respect to the local frame

e1, . . . , em, i.e.R(ek, el)ei = Rj

iklej,

or equivalently, Rjikl = θj(R(ek, el)ei). Then

dθji = θh

i ∧ θjh +

1

2Rj

iklθk ∧ θl.

Proof.

(dθji −

m∑h=1

θhi ∧ θj

h)(ek, el) = ek(θji (el))− el(θ

ji (ek))

−θji ([ek, el])−

m∑h=1

[θhi (ek)θ

jh(el)− θh

i (el)θjh(ek)]

= ek(Γjil)− el(Γ

jik)−

m∑h=1

θh([ek, el])Γjih

−m∑

h=1

[ΓhikΓ

jhl − Γh

ilΓjhk].

On the other hand,

R(ek, el)ei = 5ek5el

ei −5el5ek

ei −5[ek,el]ei

=m∑

j=1

5ek(Γj

ilej)−5el(Γj

ikej)−m∑

h=1

θh([ek, el])Γjihej

=m∑

j=1

5ek(Γj

il)−5el(Γj

ik) +m∑

h=1

(ΓhilΓ

jhk − Γh

ikΓjhl − θh([ek, el])Γ

jih)ej

Hence we have

R(ek, el)ei =m∑

j=1

[(dθji −

m∑h=1

θhi ∧ θj

h)(ek, el)]ej.

Thus

Rjikl = θj(R(ek, el)ei) = (dθj

i −m∑

h=1

θhi ∧ θj

h)(ek, el).

12

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The identity is thus proved.

By the same proof, we also have the following theorem:

dθi = θj ∧ θij +

1

2θi(T (ej, ek))θ

j ∧ θk.

Proof.

(dθi − θj ∧ θij)(ek, el) = ek(θ

i(el))− el(θi(ek))

−θi([ek, el])− [θj(ek)θij(el)− θj(el)θ

ij(ek)]

= θil(ek)− θi

k(el)− θi([ek, el])

= Γilk − Γi

kl − θi([ek, el])

= θi(5ekel −5el

ek − [[ek, el])

= θi(T (ej, ek).

This proves the theorem.

Let Ω = dθ− θ ∧ θ. Ω is called the curvature matrix with respect tothe frame field e1, . . . , em. Write Ω = (Ωj

i ). Then, from the definition

that Ω = dθ − θ ∧ θ, we have

Ωji = dθj

i +m∑

k=1

θki ∧ θj

k.

The above theorem implies that

Ωji =

1

2

m∑k,l=1

Rjiklθ

k ∧ θl,

where R(ek, el)ei = Rjiklej.

Similarily, we have the structure equations:

dθi = θj ∧ θij +

1

2θi(T (ej, ek))θ

j ∧ θk,

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dθji −

m∑k=1

θki ∧ θj

k = Ωji =

1

2

m∑k,l=1

Rjiklθ

k ∧ θl.

In particular, if 5 is torsion free, then have the structure equations:

dθi = θj ∧ θij,

dθji −

m∑k=1

θki ∧ θj

k =1

2

m∑k,l=1

Rjiklω

k ∧ ωl.

We also have the Bianchi identity:

dΩ = θ ∧ Ω− Ω ∧ θ.

Note that if 5 is the Levi-Civita connection of (M, g), then we havethe torsion free condition implies that

dθi =m∑

j=1

θj ∧ θij;

and the metric compatiable condition implies that

dgij =m∑

k=1

(θki gkj + θk

j gik),

where gij = g(ei, ej). Now let θi = θjgji, θij = θki gkj, then the above

two equations become

dθi = θji ∧ θj, dgij = θij + θji.

We also letΩij = Ωk

i gkj.

Then we have

(1) Ωij = 12Rijklθ

k ∧ θl, where Rijkl =< R(ek, el)ei, ej >,

(2) Ωij + Ωji = 0

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Page 15: 1 Riemannian Metric - UHminru/Riemann08/note1.pdf · Math 6396 Riemannian Geometry, Metric, Connections, Curvature Tensors etc. By Min Ru, University of Houston 1 Riemannian Metric

(3) θi ∧ Ωij = 0,

(4) dΩij = θki ∧ Ωkj + Ωik ∧ θk

j = θki ∧ Ωkj − θk

j ∧ Ωik.

In particular, if e1, . . . , em (or θi, 1 ≤ i ≤ m) is an orthonormal(w.r.t. the Riemannian metric), then we have

dθi =m∑

j=1

θj ∧ θij;

θij + θj

i = 0.

• The sectional curvature and its calculation. Let (M, g) be a Rie-mannian metric and 5 is the Levi-Civita connection. Let E ⊂ Tp(M)be a two-dimensional subspace, and X, Y be two linearly independentvector fields in E, then Kp(E), the sectional curvature of M at pwith respect to E, is defined by

Kp(E) = −R(X, Y, X, Y )

g(X, Y, X, Y )

where g(X, Y, Z, W ) = g(X, Z)g(Y,W )−g(X, W )g(Y, Z). Let e1, . . . , em

be an orthonormal(with respect to the Riemannian metric) localframe field for T (M) on U with e1, e2 ∈ E. Let θ1, . . . , θm be its dual(they are one-forms). Then g =

∑mi=1 θi ⊗ θi. So g(e1, e2, e1, e2) = 1.

ThusKp(E) = −R(e1, e2, e1, e2).

where R(e1, e2, e1, e2) = g(R(e1, e2)e1, e2). Write

R(e1, e2)e1 =m∑

j=1

θj(R(e1, e2)e1)ej.

Then g(R(e1, e2)e1, e2) = θ2(R(e1, e2)e1). Hence

Kp(E) = −θ2(R(e1, e2)e1).

Recall the (second) structure equation (taking i = 1, j = 2):

dθ21 =

m∑h=1

θh1 ∧ θ2

h +1

2

m∑k,l=1

θ2(R(ek, el)e1)θk ∧ θl.

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Page 16: 1 Riemannian Metric - UHminru/Riemann08/note1.pdf · Math 6396 Riemannian Geometry, Metric, Connections, Curvature Tensors etc. By Min Ru, University of Houston 1 Riemannian Metric

This means the sectional curvature Kp(E) can calculated bycalculating the connection forms θk

j , calculating dθ21 and using

the (second) structure equation.

Example: On Rm, define

g =4

(1 +∑m

i=1(xi)2)2

m∑i=1

dxi ⊗ dxi.

Calculate its sectional curvature.

Solution: Let A = 1 +∑m

i=1(xi)2, then gij = g(∂/∂xi, ∂/∂xj) = 0 if

i 6= j and gii = 2A. So let ei = (A/2)∂/∂xi. Then e1, . . . , em are

the orthonormal basis (with respect to this Riemannina metric). Letθ1, . . . , θm be its dual basis. So have θi = 2

Adxi. We can write

g =m∑

i=1

θi ⊗ θi.

Let E ⊂ TpRm be a subspace of dimension two, without loss of gener-

ality, we assume that e1, e2 ∈ E. From above, to calculate the sectionalcurvature, we only need to calculate θ2(R(e1, e2)e1). To do this, we usethe structure eqautions,

dθ21 =

m∑h=1

θh1 ∧ θ2

h +1

2

m∑k,l=1

θ2(R(ek, el)e1)θk ∧ θl.

We first find the connection forms θji (using the fundamental theorem

of Riemannian geometry). In fact,

dθi = 2dxi

A2∧ dA =

4

A2

∑j

xjdxi ∧ dxj =∑j

xjθi ∧ θj.

On the other hand, by structure equation

dθi =∑j

θj ∧ θij, θi

j + θji = 0.

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Page 17: 1 Riemannian Metric - UHminru/Riemann08/note1.pdf · Math 6396 Riemannian Geometry, Metric, Connections, Curvature Tensors etc. By Min Ru, University of Houston 1 Riemannian Metric

To find θij, write (since θi, 1 ≤ i ≤ m, is a basis)

θij =

m∑k=1

Ai,jk θk.

We need to determine Ai,jk . First of all, from θi

j +θji = 0, We know that

θii = 0 and we know that Ai,j

k = −Aj,ik . From the structure equation,

we have

∑j

θj ∧ (−xjθi) = dθi =∑j

θj ∧ θij =

∑j

θj ∧ (m∑

k=1

Ai,jk θk).

By combining both sides, we get Ai,ji = −xj for all j 6= i, and Ai,j

k = 0if k 6= i, k 6= j. If k = j, then Ai,j

j = −Aj,ij = xi. Hence we get the

connection forms

θij = Aij

i θi + Aijj θj = xiθj − xjθi.

In particular,θ21 = x2θ1 − x1θ2.

To calculate the curvature, we use the second structure equation. To

do so, we need to calculate dθ21 −

m∑h=1

θh1 ∧ θ2

h. In fact,

dθ21 −

m∑h=1

θh1 ∧ θ2

h = (dx2 ∧ θ1 + x2dθ1 − dx1 ∧ θ2 − x1dθ2)

−m∑

h=1

(xhθ1 − x1θh) ∧ (x2θh − xhθ2)

=

(Aθ2 ∧ θ1 +

m∑k=1

x2xkθ1 ∧ θk −m∑

k=1

x1xkθ2 ∧ θk

)

−m∑

h=1

(x2xhθ1 ∧ θh + x1xhθh ∧ θ2 + (xh)2θ2 ∧ θ1)

=

(A−

m∑k=1

(xk)2

)θ2 ∧ θ1 = −θ1 ∧ θ2

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Page 18: 1 Riemannian Metric - UHminru/Riemann08/note1.pdf · Math 6396 Riemannian Geometry, Metric, Connections, Curvature Tensors etc. By Min Ru, University of Houston 1 Riemannian Metric

From the structure equation, we have

dθ21 −

m∑h=1

θh1 ∧ θ2

h =1

2

m∑k,l=1

θ2(R(ek, el)e1)θk ∧ θl.

Hence

−θ1 ∧ θ2 =1

2

m∑k,l=1

θ2(R(ek, el)e1)θk ∧ θl.

Which means that θ2(R(e1, e2)e1) = −1. we have

Kp(E) = −θ2(R(e1, e2)e1) = 1.

Hence its sectional curavture is constant. Q.E.D.

Remark: We can also use the above method to calculate the Christofellsymbol. In fact, from above, ∂

∂xi = 2Aei. Hence

5 ∂

∂xi= 5

(2

Aei

)= 5

(2

1 +∑m

i=1(xi)2

ei

)

= d

(2

1 +∑m

i=1(xi)2

)⊗ ei +

2

1 +∑m

i=1(xi)2

θji ⊗ ej

=4

(1 +∑m

i=1(xi)2)2

−∑j

xjdxj ⊗ ei +∑j

(xjdxi − xidxj)⊗ ej

=

2

1 +∑m

i=1(xi)2

∑j,k

(−xjδik − xiδij)dxj ⊗ ∂

∂xk.

Hence, we get

Γkij =

2

1 +∑m

i=1(xi)2

∑j,k

(−xjδik − xiδij).

F. Schur’s Theorem. Assuem that dim M ≥ 3. If for every p ∈ M ,KP (E) is constant, i.e. it is independent of E. Then K is constant,i.e. p is independent of p.

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Page 19: 1 Riemannian Metric - UHminru/Riemann08/note1.pdf · Math 6396 Riemannian Geometry, Metric, Connections, Curvature Tensors etc. By Min Ru, University of Houston 1 Riemannian Metric

Proof. Let ei1≤i≤m be a local frame, and let θ1, . . . , θm be its dual.By definition, for E = spanei, ej,

Kp(E) = −< R(ei, ej)ei, ej >

giigjj − g2ij

.

So, by the assumption, we have < R(ei, ej)ei, ej > |p = −Kp(giigjj−g2ij)

where K is a smooth function on M . This will implies that (the detialis omitted) that Rijkl =< R(ek, el)ei, ej >= −K(gikgjl−gilgjk). By thetheorem above, we get

Ωij =1

2Rijklθ

k ∧ θl = −K

2(gikgjl − gilgjk)θ

k ∧ θl = −Kθi ∧ θj.

Our goal is to show that K is constant. To do so, we use Bianchi-identity

dΩ = θ ∧ Ω− Ω ∧ θ.

So we first calculate dΩ:

dΩij = −dK ∧ θi ∧ θj −Kdθi ∧ θj + Kθi ∧ dθj.

By the structure equation, the above identity becomes

dΩij = −dK ∧ θi ∧ θj −Kθki ∧ θk ∧ θj + Kθi ∧ θk

j ∧ θk.

On the other hand,

ω ∧ Ω− Ω ∧ ω = θki ∧ Ωkj + Ωik ∧ θk

j

= Kθki ∧ θk ∧ θj −Kθi ∧ θk ∧ θk

j .

Hence the Bianchi-identity implies that

dK ∧ θi ∧ θj = 0.

If we let dK =∑

k Kkθk, then we get∑k

Kkθk ∧ θi ∧ θj = 0.

Hence Kk = 0. This implies that K is a constant.

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Page 20: 1 Riemannian Metric - UHminru/Riemann08/note1.pdf · Math 6396 Riemannian Geometry, Metric, Connections, Curvature Tensors etc. By Min Ru, University of Houston 1 Riemannian Metric

• Ricci curvature and scalar curvature. Denote by Kp(E) = Kp(X∧Y ). Let X ∈ Tp(M) be a unit vector. We take an orthonormal basise1 = X, e2, . . . , em in Tp(M) and consider the following “averages”:

Ricp(X) :=m∑

j=2

Kp(X ∧ ej),

ρ(P ) :=∑

i

Ricp(ei),

which are called the Ricci curvature in the direction X and the scalarcurvature at p, respectively. It is easy to prove that they do not dependon the choice of the orthnormal basis.

In order to explain these facts,we define a bilinear form r on Tp(M)as follows: let X, Y ∈ Γ(TM) and put r(X, Y ) = the trace of themap: Z 7→ R(Z,X)Y, which is called the Ricci tensor. In the aboveorthonormal basis e1 = X, e2, . . . , em, we have

r(X, Y ) =∑j

< R(ej, X)Y, ej >=∑j

< R(ej, Y )X, ej >= r(Y,X),

i.e. r is symmetric and

r(X, X) =∑j

R(X, ej, X, ej >=∑j

K(X ∧ ej) = Ric(X).

Also, the bilinear form r induces a linear self-adjoint map Lr such that< Lr(X), Y >= r(X, Y ). In an orthonormal basis e1, e2, . . . , em,, wehave

ρ =∑

i

Ricp(ei) =∑

i

r(ei, ei) =∑

i

< Lr(ei), ei >= trace of Lr.

In the local coordinates (U, xi), these can be expressed as r(X, Y ) =RijX

iY j with Rij = gklRikjl,

ρ = gikRik = gikgjlRijkl.

If M is of constant sectional curvature c, then Ric(X) = (m− 1)c andits scalar curvature is ρ = m(m− 1)c.

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Page 21: 1 Riemannian Metric - UHminru/Riemann08/note1.pdf · Math 6396 Riemannian Geometry, Metric, Connections, Curvature Tensors etc. By Min Ru, University of Houston 1 Riemannian Metric

A Riemannian manifold (M, g) is called an Einstein manifold if thereis a constant λ on M such that r = λg. In such casse, g is called theEinstein metric, and the scalar curvature ρ = nλ is constant.

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