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 ! i j ∂ ∂ g = gijdx ⊗ dx , gij = g , , ∂xi ∂xj

where gij = gji and the matrix (gij) 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 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 = 5X Z + f 5Y Z,

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

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

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

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

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. Then there exists a unique U ∈ Γ(TM) such that α(Z) =< U, Z > for every Z ∈ Γ(TM).

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

2 < 5X Y, 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 define 5X Y use Koszul formula, we can verify the properties are satisfied. • Christoffel symbols. In terms of the coordinate chart (U, xi), we write m ∂ X k ∂ 5 ∂ = Γji , ∂xi j k ∂x k=1 ∂x k where the functions Γji are called the Christoffel symbols.

k k • The connection 5 is torsion free if and only if Γij = Γji, and if 5 is torsion free and compatible with the Riemannian metric g, then its Christoffel symbols are given by, from the Koszul formula above, ∂g ∂g ∂g 2Γk g = il + jl − ij , ij kl ∂xj ∂xi ∂xl or equivalently,

m ! k 1 X kl ∂gil ∂gjl ∂gij Γij = g j + i − l . 2 l=1 ∂x ∂x ∂x

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

m j p r 2 p j 0j X q ∂y ∂x ∂x ∂ x ∂y Γ ik = Γpr q i k + i k · p . p,q,r=1 ∂x ∂y ∂y ∂y ∂y ∂x

For every X,Y ∈ Γ(TM), write

m X i ∂ X = ξ i , i=1 ∂x

m X j ∂ Y = η j , j=1 ∂x then  k  X X i ∂η X k i j ∂ 5X Y =  ξ i + Γjiξ η  k . k i ∂x i,j ∂x

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

m ∂ X k ∂ 5 ∂ = Γji , ∂xi j k ∂x k=1 ∂x or m m m ∂ X X k i ∂ X k ∂ 5 j = Γjidx ⊗ k = ωj k , ∂x i=1 k=1 ∂x k=1 ∂x where m k X k i ωj = Γjidx . i=1 k The matrix ω = (ωi ) is called the connection matrix of 5 with respect to ∂/∂xi, 1 ≤ i ≤ m. So we see that we can regard 5 as the operator 5 : Γ(TM) → Γ(T ∗(M) ⊗ T (M)).

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

3 curve γ if and only if 5γ0(t)X = 0 for t ∈ [0, b]. In terms of the local coordinate (U, xi), write by xi(t) = xi ◦ γ(t), 1 ≤ i ≤ m and write

m X i ∂ X(t) = X (t) i |γ(t). α=1 ∂x Then i 0 X dx (t) ∂ γ (t) = i . i dt ∂x Hence

 k j  X dX (t) X k i dx (t) ∂ 0 5γ (t)X =  + ΓijX (t)  k |γ(t). k dt i,j dt ∂x Hence X(t) is parallel along the curve γ if and only if

k j dX (t) X k i dx (t) + ΓijX (t) = 0. dt i,j dt

A curve γ is geodesic in M if and only if the tangent vector of γ is parallel along the curve γ. Suppose that γ is given in a coordinate chart (U, xi) by xi(t) = xi ◦ γ(t) 1 ≤ i ≤ m. Then γ is geodesic if and only if 2 i m j k d x X i dx dx 2 + Γkj = 0 1 ≤ i ≤ m. dt j,k=1 dt dt • Covariant derivative(connection) for tensor fields. Note that for any (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 5X Y . 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 consider s = 0 case. Let f be a function (i.e. a (0, 0)-tensor field, then define 5X (f) = X(f). When s = 1, for any (0, 1)-tensor field ω (which is a differential 1-form), 5X (ω) is also a 1-form which is defined by, for any Y ∈ Γ(T (M)),

5X (ω)(Y ) = 5X (ω(Y )) − ω(5X Y ).

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

(5X Φ)(Y1,...,Ys) : = 5X (Φ(Y1,...,Ys)) s X − Φ(Y1,...,Yi−1, 5X Yi,Yi+1,...,Ys). i=1

5X Φ 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) := (5X Φ)(Y1,...,Ys).

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

j1 js Φ|U = Φj1···js dx ⊗ · · · ⊗ dx ,

then one can verify that

j1 js 5 ∂ Φ = Φj1···js,idx ⊗ · · · ⊗ dx , ∂xi where ∂Φ s Φ = j1···js − X Φ Γk . j1···js,i i j1···jb−1kjb+1···js jbi ∂x b=1 • Connections over vector bundles. Let π : E → M be a vector bundle. Denote by Γ(E) the set of smooth sections of E. Γ(E) is a real vector space; it is also a C∞(M)-module. A connection 5 over E is a map 5 : Γ(T (M) × Γ(E) → Γ(E), (X, s) 7→ 5X s, which satisfies, ∞ for all X,Y ∈ Γ(T (M)), s1, s2 ∈ Γ(E), c constant, and f ∈ C (M):

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

(b) 5X (s1 + cs2) = 5X s1 + c 5X s2,

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

5 Note that we can also view 5 as the map Γ(E) → Γ(T ∗(M) ⊗ E) by (5s)(X) = 5X s 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, it induces a connection on the cotangent bundle T ∗M defined by, for every ω ∈ Γ(T ∗M),

(5X ω)(Y ) = X(ω(Y )) − ω(5X Y ), for every Y ∈ Γ(T (M)).

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

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

k X β 5sα = ωα ⊗ sβ. β=1

β So, the connection 5 is associated to a k × k matrix ω = (ωα) of smooth differential one forms on U, with respect to the local frame {sα, 1 ≤ α ≤ k}. It is easy to verify that m β X β k ωα = Γαkdx , k=1

j where Γik 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 k X α s = λ sα, α=1 Then k  k  X α X β α 5s = dλ + λ ωβ  ⊗ sα. α=1 β=1

6 For X ∈ Γ(TM),

k  k  X α X β α 5X s = Xλ + λ ωβ (X) ⊗ sα. α=1 β=1

• Curvature tensor. For every X,Y ∈ Γ(TM), define R(X,Y ) (called the curvature operator) as R(X,Y ) : Γ(TM) → Γ(TM) by, for every Z ∈ 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) When 5 is torsion free, we have R(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. It is thus called the curvature tensor.

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

In terms of the local coordinates (U, xi), write ∂ ∂ R(∂/∂xi, ∂/∂xj) = Rl , ∂xk kij ∂xl

7 or X l k ∂ i j R = Rkijdx ⊗ l ⊗ dx ⊗ dx . i,j,k,l ∂x l l According to (1), we have Rkij = −Rkji. l Then Rkij satisfies the transformation rule for components of type (1,3) tensors, i.e. on U ∩ V 6= ∅,

m l p r s 0l X l ∂y ∂x ∂x ∂x R kij = Rprs q k i j . p,q,r,s=1 ∂x ∂y ∂y ∂y

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

Write ∂ k ∂ ∂ 5 j = Γji k , ∂xi ∂x ∂x k where Γji are Christoffel symbols, then

∂ ∂ ! ∂ ∂Γl ∂Γl ! ∂ R , = kj − ki + Γh Γl − Γh Γl . ∂xi ∂xj ∂xk ∂xi ∂xj kj hi ki hj ∂xl

Hence ∂Γl ∂Γl Rl = kj − ki + Γh Γl − Γh Γl . kij ∂xi ∂xj kj hi ki hj For X,Y ∈ Γ(TM), if

∂ ∂ X = Xi ,Y = Y j , ∂xi ∂xj then ∂ R(X,Y ) = XiY jRl dxk ⊗ . kij ∂xl

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

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

8 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 is called the Riemannian curvature tensor. It satisfies the following property: 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 ).

i k l In terms of the local coordinates (U, x ), write R = Rklijdx ⊗ dx ⊗ dxi ⊗ dxj, then

1 ∂2g ∂2g ∂2g ∂2g ! R = ik + jl − il − jk +Γh Γp g −Γp Γp g . klij 2 ∂xj∂xl ∂xi∂xk ∂xj∂xk ∂xi∂xl ik jl ph il jk ph

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, T M). Let {ω1, . . . , ωm} j be its dual. Denote by ω = (ωi ) be the connection matrix of 5 with respect to {∂/∂xi, 1 ≤ i ≤ m}, i.e.

m ∂ X j ∂ 5 i = ωi j . ∂x j=1 ∂x

9 j Theorem. Let Rikl be components of R with respect to the local frame {∂/∂x1, . . . , ∂/∂xm}, i.e.

∂ ∂ R(∂/∂xk, ∂/∂xl) = Rj . ∂xi ikl ∂xj Then 1 dωj = ωh ∧ ωj + Rj ωk ∧ ωl. i i h 2 ikl

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

j Write Ω = (Ωi ). Then, from the definition that Ω = dω − ω ∧ ω, we have m j j X k j Ωi = dωi + ωi ∧ ωk. k=1 The above theorem implies that

m j 1 X j k l Ωi = Rikldx ∧ dx . 2 k,l=1

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

10 Recall operator T (X,Y ) = 5X Y − 5Y X − [X,Y ] is called the torsion operator. We have the following theorem: 1 dωi − ωj ∧ ωi = ωi(T (∂/∂xj, ∂/∂xk))ωj ∧ ωk. j 2 The equations 1 dωi = ωj ∧ ωi + ωi(T (∂/∂xj, ∂/∂xk))ωj ∧ ωk, j 2 1 dωj = ωk ∧ ωj + Rj ωk ∧ ωl i i k 2 ikl are called the structure equations for (M, 5). In particular, if 5 is torsion free, then the structure equations becomes:

i j i dω = ω ∧ ωj, 1 dωj = ωk ∧ ωj + Rj ωk ∧ ωl. i i k 2 ikl Note that if 5 is the Levi-Civita connection of (M, g), then we also have (from the torsion free condition and the metric compatiable condition) that m X k k dgij = (ωi gkj + ωj gik), k=1 i j where gij = g(∂/∂x , ∂/∂x ). • Cartan’s moving frame method. The above results can be gener- alized to arbitrary local frames. On a smooth m-dimensional manifold M, in addtion to the (standard) local frame {∂/∂xi, 1 ≤ i ≤ m}, we also often use an arbitrary local frame e1, . . . , em for Γ(U, T (M)) (for example, we take the orthonormal basis after the Gram-Schmidt pro- 1 m k k cess). Let θ , . . . , θ be its dual. We still write 5ei ej = Γjiek, where Γji are the Christoffel symbol with respect to the frame {e1, . . . , em}. Let j j k θi = Γikθ . j j Then 5ei = θi ej. The matrix θ = (θi ) is called the connection matrix of 5 with respect to the local frame {e1, . . . , em}.

11 ˜j Theorem. Let Rikl be components of R with respect to the local frame {e1, . . . , em}, i.e. ˜j R(ek, el)ei = Riklej, ˜j j or equivalently, Rikl = θ (R(ek, el)ei). Then 1 dθj = θh ∧ θj + R˜j θk ∧ θl. i i h 2 ikl

Proof.

m j X h j j j (dθi − θi ∧ θh)(ek, el) = ek(θi (el)) − el(θi (ek)) h=1 m j X h j h j −θi ([ek, el]) − [θi (ek)θh(el) − θi (el)θh(ek)] h=1 m j j X h j = ek(Γil) − el(Γik) − θ ([ek, el])Γih h=1 m X h j h j − [ΓikΓhl − ΓilΓhk]. h=1 On the other hand,

R(ek, el)ei = 5ek 5el ei − 5el 5ek ei − 5[ek,el]ei m m X j j X h j = {5ek (Γilej) − 5el (Γikej) − θ ([ek, el])Γih}ej j=1 h=1 m m X j j X h j h j h j = {5ek (Γil) − 5el (Γik) + (ΓilΓhk − ΓikΓhl − θ ([ek, el])Γih)}ej j=1 h=1

Hence we have

m m X j X h j R(ek, el)ei = [(dθi − θi ∧ θh)(ek, el)]ej. j=1 h=1 Thus m ˜j j j X h j Rikl = θ (R(ek, el)ei) = (dθi − θi ∧ θh)(ek, el). h=1

12 The identity is thus proved.

By the same proof, we also have the following theorem: 1 dθi = θj ∧ θi + θi(T (e , e ))θj ∧ θk. j 2 j k

Proof.

i j i i i (dθ − θ ∧ θj)(ek, el) = ek(θ (el)) − el(θ (ek)) i j i j i −θ ([ek, el]) − [θ (ek)θj(el) − θ (el)θj(ek)] i i i = θl (ek) − θk(el) − θ ([ek, el]) i i i = Γlk − Γkl − θ ([ek, el]) i = θ (5ek el − 5el ek − [[ek, el]) i = θ (T (ej, ek).

This proves the theorem.

Let Ω = dθ − θ ∧ θ. Ω is called the curvature matrix with respect to j the frame field {e1, . . . , em}. Write Ω = (Ωi ). Then, from the definition that Ω = dθ − θ ∧ θ, we have

m j j X k j Ωi = dθi + θi ∧ θk. k=1 The above theorem implies that

m j 1 X ˜j k l Ωi = Riklθ ∧ θ , 2 k,l=1

˜j where R(ek, el)ei = Riklej. Similarily, we have the structure equations: 1 dθi = θj ∧ θi + θi(T (e , e ))θj ∧ θk, j 2 j k

13 m m j X k j j 1 X ˜j k l dθi − θi ∧ θk = Ωi = Riklθ ∧ θ . k=1 2 k,l=1 In particular, if 5 is torsion free, then have the structure equations:

i j i dθ = θ ∧ θj,

m m j X k j 1 X ˜j k l dθi − θi ∧ θk = Riklω ∧ ω . k=1 2 k,l=1

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 have the torsion free condition implies that

m i X j i dθ = θ ∧ θj; j=1 and the metric compatiable condition implies that

m X k k dgij = (θi gkj + θj gik), k=1

j k where gij = g(ei, ej). Now let θi = θ gji, θij = θi gkj, then the above two equations become

j dθi = θi ∧ θj, dgij = θij + θji.

We also let k Ωij = Ωi gkj. Then we have

1 k l (1) Ωij = 2 Rijklθ ∧ θ , where Rijkl =< R(ek, el)ei, ej >,

(2) Ωij + Ωji = 0

14 i (3) θ ∧ Ωij = 0,

k k k k (4) dΩij = θi ∧ Ωkj + Ωik ∧ θj = θi ∧ Ωkj − θj ∧ Ωik.

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

m i X j i dθ = θ ∧ θj; j=1

i j θj + θi = 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 independent vector fields in E, then Kp(E), the sectional curvature of M at p with respect to E, is defined by R(X,Y,X,Y ) K (E) = − p 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) local 1 m frame field for T (M) on U with e1, e2 ∈ E. Let θ , . . . , θ be its dual Pm i i (they are one-forms). Then g = i=1 θ ⊗ θ . So g(e1, e2, e1, e2) = 1. Thus Kp(E) = −R(e1, e2, e1, e2).

where R(e1, e2, e1, e2) = g(R(e1, e2)e1, e2). Write m X j R(e1, e2)e1 = θ (R(e1, e2)e1)ej. j=1

2 Then g(R(e1, e2)e1, e2) = θ (R(e1, e2)e1). Hence

2 Kp(E) = −θ (R(e1, e2)e1). Recall the (second) structure equation (taking i = 1, j = 2):

m m 2 X h 2 1 X 2 k l dθ1 = θ1 ∧ θh + θ (R(ek, el)e1)θ ∧ θ . h=1 2 k,l=1

15 This means the sectional curvature Kp(E) can calculated by k 2 calculating the connection forms θj , calculating dθ1 and using the (second) structure equation.

Example: On Rm, define

m 4 X i i g = Pm i 2 2 dx ⊗ dx . (1 + i=1(x ) ) i=1 Calculate its sectional curvature.

Pm i 2 i j Solution: Let A = 1 + i=1(x ) , then gij = g(∂/∂x , ∂/∂x ) = 0 if 2 i i 6= j and gii = A . So let ei = (A/2)∂/∂x . Then {e1, . . . , em} are the orthonormal basis (with respect to this Riemannina metric). Let 1 m i 2 i {θ , . . . , θ } be its dual basis. So have θ = A dx . We can write

m g = X θi ⊗ θi. i=1

m Let E ⊂ TpR be a subspace of dimension two, without loss of gener- ality, we assume that e1, e2 ∈ E. From above, to calculate the sectional 2 curvature, we only need to calculate θ (R(e1, e2)e1). To do this, we use the structure eqautions,

m m 2 X h 2 1 X 2 k l dθ1 = θ1 ∧ θh + θ (R(ek, el)e1)θ ∧ θ . h=1 2 k,l=1

j We first find the connection forms θi (using the fundamental theorem of Riemannian geometry). In fact,

i i dx 4 X j i j X j i j dθ = 2 2 ∧ dA = 2 x dx ∧ dx = x θ ∧ θ . A A j j

On the other hand, by structure equation

i X j i i j dθ = θ ∧ θj, θj + θi = 0. j

16 i i To find θj, write (since θ , 1 ≤ i ≤ m, is a basis)

m i X i,j k θj = Ak θ . k=1

i,j i j We need to determine Ak . First of all, from θj +θi = 0, We know that i i,j j,i θi = 0 and we know that Ak = −Ak . From the structure equation, we have

m X j j i i X j i X j X i,j k θ ∧ (−x θ ) = dθ = θ ∧ θj = θ ∧ ( Ak θ ). j j j k=1

i,j j i,j By combining both sides, we get Ai = −x for all j 6= i, and Ak = 0 i,j j,i i if k 6= i, k 6= j. If k = j, then Aj = −Aj = x . Hence we get the connection forms

i ij i ij j i j j i θj = Ai θ + Aj θ = x θ − x θ .

In particular, 2 2 1 1 2 θ1 = x θ − x θ . To calculate the curvature, we use the second structure equation. To m 2 X h 2 do so, we need to calculate dθ1 − θ1 ∧ θh. In fact, h=1

m 2 X h 2 2 1 2 1 1 2 1 2 dθ1 − θ1 ∧ θh = (dx ∧ θ + x dθ − dx ∧ θ − x dθ ) h=1 m − X(xhθ1 − x1θh) ∧ (x2θh − xhθ2) h=1 m m ! = Aθ2 ∧ θ1 + X x2xkθ1 ∧ θk − X x1xkθ2 ∧ θk k=1 k=1 m − X(x2xhθ1 ∧ θh + x1xhθh ∧ θ2 + (xh)2θ2 ∧ θ1) h=1 m ! = A − X(xk)2 θ2 ∧ θ1 = −θ1 ∧ θ2 k=1

17 From the structure equation, we have

m m 2 X h 2 1 X 2 k l dθ1 − θ1 ∧ θh = θ (R(ek, el)e1)θ ∧ θ . h=1 2 k,l=1 Hence m 1 2 1 X 2 k l −θ ∧ θ = θ (R(ek, el)e1)θ ∧ θ . 2 k,l=1 2 Which means that θ (R(e1, e2)e1) = −1. we have

2 Kp(E) = −θ (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 Christofell ∂ 2 symbol. In fact, from above, ∂xi = A ei. Hence ∂  2  5 = 5 e ∂xi A i 2 ! = 5 Pm i 2 ei 1 + i=1(x ) ! 2 2 j = d Pm i 2 ⊗ ei + Pm i 2 θi ⊗ ej 1 + i=1(x ) 1 + i=1(x )   4 X j j X j i i j = Pm i 2 2 − x dx ⊗ ei + (x dx − x dx ) ⊗ ej (1 + i=1(x ) ) j j

2 X j i j ∂ = Pm i 2 (−x δik − x δij)dx ⊗ k . 1 + i=1(x ) j,k ∂x Hence, we get

k 2 X j i Γij = Pm i 2 (−x δik − x δij). 1 + i=1(x ) j,k

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.

18 1 m Proof. Let {ei}1≤i≤m be a local frame, and let {θ , . . . , θ } be its dual. By definition, for E = span{ei, ej},

< R(ei, ej)ei, ej > Kp(E) = − 2 . giigjj − gij

2 So, by the assumption, we have < R(ei, ej)ei, ej > |p = −Kp(giigjj −gij) where K is a smooth function on M. This will implies that (the detial is omitted) that Rijkl =< R(ek, el)ei, ej >= −K(gikgjl − gilgjk). By the theorem above, we get 1 K Ω = R θk ∧ θl = − (g g − g g )θk ∧ θl = −Kθ ∧ θ . ij 2 ijkl 2 ik jl il jk i j 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

k k dΩij = −dK ∧ θi ∧ θj − Kθi ∧ θk ∧ θj + Kθi ∧ θj ∧ θk. On the other hand,

k k ω ∧ Ω − Ω ∧ ω = θi ∧ Ωkj + Ωik ∧ θj k k = Kθi ∧ θk ∧ θj − Kθi ∧ θk ∧ θj . Hence the Bianchi-identity implies that

dK ∧ θi ∧ θj = 0.

P If we let dK = k Kkθk, then we get X Kkθk ∧ θi ∧ θj = 0. k

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

19 • Ricci curvature and scalar curvature. Denote by Kp(E) = Kp(X∧ Y ). Let X ∈ Tp(M) be a unit vector. We take an orthonormal basis {e1 = X, e2, . . . , em} in Tp(M) and consider the following “averages”:

m X Ricp(X) := Kp(X ∧ ej), j=2 X ρ(P ) := Ricp(ei), i which are called the Ricci curvature in the direction X and the scalar curvature at p, respectively. It is easy to prove that they do not depend on 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 the map: Z 7→ R(Z,X)Y, which is called the Ricci tensor. In the above orthonormal basis {e1 = X, e2, . . . , em}, we have X X r(X,Y ) = < R(ej,X)Y, ej >= < R(ej,Y )X, ej >= r(Y,X), j j i.e. r is symmetric and X X r(X,X) = R(X, ej, X, ej >= K(X ∧ ej) = Ric(X). j j

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},, we have X X X ρ = Ricp(ei) = r(ei, ei) = < Lr(ei), ei >= trace of Lr. i i i In the local coordinates (U, xi), these can be expressed as r(X,Y ) = i j kl RijX Y with Rij = g Rikjl,

ik ik jl ρ = g Rik = g g Rijkl.

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

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

21