1 Definition of the Riemann curvature tensor

Let (Mn, g) be a Riemannian manifold and let χ (M) denote the space ofC∞ vector fields on M. The Riemann curvature tensor is the map

Rm : χ (M)× χ (M)× χ (M) −→ χ (M)

defined by

Rm (X,Y )Z + ∇X (∇Y Z)−∇Y (∇XZ)−∇[X,Y ]Z (1)

for X,Y, Z ∈ χ (M). Clearly we have the antisymmetry in X and Y :

Rm (Y,X)Z = −Rm (X,Y )Z. (2)

One may easily check that Rm is R-multilinear. That is,

Rm (X1 + cX2, Y )Z = Rm (X1, Y )Z + cRm (X2, Y )Z, (3)

Rm (X,Y1 + cY2)Z = Rm (X,Y1)Z + cRm (X,Y2)Z,

Rm (X,Y ) (Z1 + cZ2) = Rm (X,Y )Z1 + cRm (X,Y )Z2,

for c ∈ R and vector fields X’s, Y ’s and Z’s.For (Rn, gRn) , using that the Euclidean covariant derivative DXY =(

X(Y 1), . . . , X (Y n)

)is the directional derivative, one finds that the Rie-

mann curvature tensor of (Rn, gRn) is zero.

Lemma 1 For any X,Y, Z ∈ χ (M) and any C∞ function f :M→ R, wehave

Rm (fX, Y )Z = Rm (X, fY )Z = Rm (X,Y ) (fZ) = f Rm (X,Y )Z. (4)

By the characterization of tensors, (Rm (X,Y )Z)p ∈ TpM depends only onXp, Yp, Zp ∈ TpM. So we may consider Rm as a multilinear map

Rm p : TpM× TpM× TpM−→ TpM

for each p ∈M.

Proof. (1) We compute that

Rm (fX, Y )Z = ∇fX (∇Y Z)−∇Y (∇fXZ)−∇[fX,Y ]Z

= f∇X (∇Y Z)− (f∇Y (∇XZ) + Y (f)∇XZ)−∇f [X,Y ]−Y (f)XZ

= f Rm (X,Y )Z.

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(2) Since Rm (X,Y )Z is antisymmetric in X and Y , we have

Rm (X, fY )Z = f Rm (X,Y )Z.

(3) We also compute

Rm (X,Y ) fZ = ∇X (∇Y (fZ))−∇Y (∇X (fZ))−∇[X,Y ] (fZ)

= ∇X ((Y f)Z + f∇Y Z)−∇Y ((Xf)Z + f∇XZ)

− ([X,Y ] f)Z − f∇[X,Y ]Z

= X (Y f)Z + (Y f)∇XZ + (Xf)∇Y Z + f∇X (∇Y Z)

− Y (Xf)Z − (Xf)∇Y Z − (Y f)∇XZ − f∇Y (∇XZ)

− ([X,Y ] f)Z − f∇[X,Y ]Z

= f Rm (X,Y )Z,

where we used X (Y f)−Y (Xf)−[X,Y ] f = 0 and cancelled terms to obtainthe last equality.

2 Algebraic identities of the curvature and thefirst Bianchi identity

DefineRm (X,Y, Z,W ) + 〈Rm (X,Y )Z,W 〉 .

Some basic symmetries of the Riemann curvature tensor are

Rm (X,Y, Z,W ) = −Rm (Y,X,Z,W ) = −Rm (X,Y,W,Z) = Rm (Z,W,X, Y ) .(5)

The first equality in (5) is (2). We prove the second and third equalitiesbelow.

Proposition 2 The Riemann curvature tensor satisfies the first Bianchiidentity, i.e.,

Rm (X,Y )Z + Rm (Y, Z)X + Rm (Z,X)Y = 0.

Proof. To see the first Bianchi identity, we simply sum the following threeformulas (by definition)

Rm (X,Y )Z = ∇X (∇Y Z)−∇Y (∇XZ)−∇[X,Y ]Z,

Rm (Y,Z)X = ∇Y (∇ZX)−∇Z (∇YX)−∇[Y,Z]X,

Rm (Z,X)Y = ∇Z (∇XY )−∇X (∇ZY )−∇[Z,X]Y,

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while applying the torsion-free property ∇VW −∇WV = [V,W ], to obtain

Rm (X,Y )Z + Rm (Y,Z)X + Rm (Z,X)Y = ∇X [Y,Z] +∇Y (Z,X) +∇Z [X,Y ]

−∇[Y,Z]X −∇[Z,X]Y −∇[X,Y ]Z

= [X, [Y, Z]] + [Y, [Z,X]] + [Z, [X,Y ]]

= 0

by the Jacobi identity.We also have the following identities

Proposition 3 The Riemann curvature tensor is antisymmetric in the lasttwo components, i.e.,

〈Rm (X,Y )Z,W 〉 = −〈Rm (X,Y )W,Z〉 (6)

and is symmetric in the switch of the first pair with the second pair, i.e.,

〈Rm (X,Y )Z,W 〉 = 〈Rm (Z,W )X,Y 〉 . (7)

Proof. First we prove (6). A straightforward computation, where we movecovariant derivatives off of Z and onto W , yields

〈Rm (X,Y )Z,W 〉 =⟨∇X (∇Y Z)−∇Y (∇XZ)−∇[X,Y ]Z,W

⟩= X 〈∇Y Z,W 〉 − 〈∇Y Z,∇XW 〉 − Y 〈∇XZ,W 〉

+ 〈∇XZ,∇YW 〉 − [X,Y ] 〈Z,W 〉+⟨Z,∇[X,Y ]W

⟩= XY 〈Z,W 〉 − Y X 〈Z,W 〉 − [X,Y ] 〈Z,W 〉

+ 〈∇XZ,∇YW 〉 −X 〈Z,∇YW 〉− 〈∇Y Z,∇XW 〉+ Y 〈Z,∇XW 〉+

⟨Z,∇[X,Y ]W

⟩= −〈Z,∇X∇YW 〉+ 〈Z,∇Y∇XW 〉+

⟨Z,∇[X,Y ]W

⟩= −〈Rm (X,Y )W,Z〉 ,

which is (6).Next we prove (7). By the first Bianchi identity we have

〈Rm (X,Y )Z,W 〉[1] + 〈Rm (Y, Z)X,W 〉[2] + 〈Rm (Z,X)Y,W 〉 = 0,

〈Rm (Y, Z)W,X〉[2] + 〈Rm (Z,W )Y,X〉[3] + 〈Rm (W,Y )Z,X〉 = 0,

〈Rm (Z,W )X,Y 〉[3] + 〈Rm (W,X)Z, Y 〉[4] + 〈Rm (X,Z)W,Y 〉 = 0,

〈Rm (W,X)Y,Z〉[4] + 〈Rm (X,Y )W,Z〉[1] + 〈Rm (Y,W )X,Z〉 = 0,

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where the superscripts [♥], ♥ = 1, 2, 3, 4, denote pairs of terms which cancelby applying (6) when the four equations above are added together. Hence

〈Rm (Z,X)Y,W 〉+ 〈Rm (W,Y )Z,X〉 = 0,

which, by one final application of (6), yields (7).

2.1 Riemann curvature tensor in local coordinates

Given a parametrization (U , x), the components of the (3, 1)-tensorRm are defined by

Rm

(∂

∂xi,∂

∂xj

)∂

∂xk+

n∑`=1

R`ijk

∂

∂x`

for i, j, k = 1, . . . , n. The components of Rm as a (4, 0)-tensor are

Rijk` +

⟨Rm

(∂

∂xi,∂

∂xj

)∂

∂xk,∂

∂x`

⟩=

n∑m=1

g`mRmijk.

In a parametrization, (5) may be written as

Rijk` = −Rjik` = −Rij`k = Rk`ij .

Lemma 4 (Components of the Riemann curvature (3, 1)-tensor) Wehave

R`ijk = ∂iΓ

`jk − ∂jΓ`

ik +n∑

p=1

(ΓpjkΓ`

ip − ΓpikΓ`

jp

), (8)

where ∂i + ∂∂xi .

Proof. Using (1, the definition of the Christoffel symbols, and [∂i, ∂j ] = 0,

4

we compute

R`ijk∂` = Rm (∂i, ∂j) ∂k

= ∇∂i

(∇∂j∂k

)−∇∂j (∇∂i∂k)

= ∇∂i

(n∑

`=1

Γ`jk∂`

)−∇∂j

(n∑

`=1

Γ`ik∂`

)

= ∂iΓ`jk − ∂jΓ`

ik +n∑

p=1

(Γpjk∇∂i∂p − Γp

ik∇∂j∂p

)

=

∂iΓ`jk − ∂jΓ`

ik +

n∑p=1

(ΓpjkΓ`

ip − ΓpikΓ`

jp

) ∂`.

In a parametrization, the first Bianchi identity is

Rijk` +Rjki` +Rkij` = 0, (9)

Extra material (not discussed in class):

Exercise 5 Let ϕ : (N n, h)→ (Mn, g) be an isometry. Let Rm g and Rmh

denote the Riemann curvature tensors of g and h, respectively. Prove that

Rm h (X,Y )Z = (dϕ)−1 (Rm g (dϕ (X) , dϕ (Y )) dϕ (Z)) . (10)

It is convenient to define the second covariant derivative of a vectorfield by

∇2 : χ (M)× χ (M)× χ (M) −→ χ (M) ,

where∇2 (X,Y, Z) + ∇2

X,Y Z + ∇X (∇Y Z)−∇∇XY Z. (11)

Using ∇XY −∇YX = [X,Y ], we see that Rm is the commutator of thesecond covariant derivative:

Rm (X,Y )Z = ∇2X,Y Z −∇2

Y,XZ. (12)

We shall later see why this definition is natural from the point of view of‘tensors’. One computes that for f ∈ C∞ (M),

∇2fX,Y Z = ∇2

X,fY Z = f∇2X,Y Z. (13)

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By the characterization of tensors, ∇2 (X,Y, Z)p depends only on Xp, Yp ∈TpM and Z ∈ χ (M). We also compute that

∇2X,Y (fZ) = f∇2

X,Y Z + Y (f)∇XZ +X (f)∇Y Z (14)

− ((∇XY ) f)Z +X (Y (f))Z.

Note that (4) also follows from (13), (14), and (12).

Exercise 6 Prove (13) and derive Rm (X,Y ) (fZ) = f Rm (X,Y )Z from(14).

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