3 Expression in local coordinates

Differential Geometry 2017–2018 ENS de Lyon

Memo curvatures

Let(M, g) be a Riemannian manifold and ∇denote its Levi–Civita connection.

1 Definitions

Riemann curvature. TheRiemann curvature of(M, g) is the ³₁

-tensor R defined by:

∀X, Y, Z ∈Γ(T M), R(X, Y)Z =∇_X∇_YZ− ∇_Y∇_XZ− ∇_[X,Y]Z. (1) We denote by Re the fully-covariant version of the Riemann tensor, that is the ⁴₀

-tensor defined by:

∀X, Y, Z, T ∈Γ(T M), R(X, Y, Z, Te ) =g(R(X, Y)Z, T). (2) On other terms, R(X, Y, Z,e ·) = (R(X, Y)Z)^[, where ^[ : TxM → T_x^∗M is the isomorphism induced by the metric.

Ricci curvature. TheRicci curvatureof(M, g)is obtained by contracting the first covariant variable inR with the only contravariant variable. That is,Ricis a ²₀

-tensor defined by:

∀Y, Z ∈Γ(T M), Ric(Y, Z) = Tr (X 7→R(X, Y)Z) = Contr¹₁(R)(Y, Z). (3) Scalar curvature. The scalar curvature S of (M, g) is the trace of Ric seen as a bilinear map onT_xM. That is, the trace of its matrix in any orthonormal basis of T_xM. In order to define it intrinsincally, we first need to use one of the musical isomorphisms defined by the metric in order to get a ¹₁

-tensor (i.e. a section ofEnd(T M)) and then take the only possible contraction. Thus,

S= Contr¹₁(Ric^]) = Tr(Y 7→Ric(Y,·)^]), (4) where ^] : T_x^∗M → T_xM is the isomorphism induced by g and we applied it to one of the variables inRic(which one not important since Ricand gare symmetric).

Sectional curvature. If two vectorsu, v∈TxMare linearly independent, then thesectional curvature of the planeP spanned byu andv is:

K(P) = R(u, v, v, u)

g(u, u)g(v, v)−g(u, v)². (5)

It depends only onP and not on a choice of basis.

2 Symmetries

The Riemann curvature is skew-symmetric in the first two variables and satisfies the first Bianchi identity. That is, for any vector fieldsX, Y and Z, we have:

R(X, Y)Z =−R(Y, X)Z, (6)

R(X, Y)Z+R(Y, Z)X+R(Z, X)Y = 0. (7)

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Its fully-covariant version presents additional symmetries. For any vector fieldsX, Y, ZandT, we have:

R(X, Y, Z, Te ) =−R(Y, X, Z, Te ) =−R(X, Y, T, Ze ) =R(Z, T, X, Ye ), (8) R(X, Y, Z, Te ) +R(Y, Z, X, Te ) +R(Z, X, Y, Te ) = 0. (9) Finally, the Ricci tensor is symmetric. For anyY, Z∈Γ(T M), we have:

Ric(Y, Z) = Ric(Z, Y). (10)

3 Expression in local coordinates

Let(x₁, . . . , x_n) denote local coordinates on some open subset ofM. As usual, we denote by ∂

∂x1, . . . ,_∂x^∂

n

and (dx¹, . . . , dxⁿ) the associated local frames ofT M and T^∗M respectively.

The matrix ofg in these coordinates is(gij)_16i,j6n, where gij =g ∂

∂xi,_∂x^∂

j

. We denote by (g^ij)_16i,j6n the coefficients of the inverse matrix.

Christoffel symbols. The Christoffel symbols (Γ^k_ij)_16i,j,k6n of the connection ∇ in these coordinates are defined by the following relations:

∀i, j∈J1, nK, ∇ ∂

∂xi

∂

∂x_j =

n

X

k=1

Γ^k_ij ∂

∂x_k. (11)

Note that, since the Levi–Civita connection is torsion-free, we have Γ^k_ij = Γ^k_ji for any i, j, k.

The Christoffel symbols are given in local coordinates by:

∀i, j, k∈J1, nK, Γ^k_ij = 1 2

n

X

l=1

g^kl ∂g_il

∂x_j +∂gjl

∂x_i −∂gij

∂x_l

. (12)

Riemann tensor. Let us define the coefficients(R^l_ijk)16i,j,k,l6n by:

∀i, j, k ∈J1, nK, R ∂

∂x_i, ∂

∂x_j ∂

∂x_k =

n

X

l=1

R^l_ijk ∂

∂x_l. (13)

By Eq. (6) and (7), for anyi, j, k and l, we have:

R^l_ijk=−R^l_jik (14)

R^l_ijk+R^l_jki+R^l_kij = 0. (15) Then we can writeR locally as:

R = X

16i,j,k,l6n

R^l_ijkdxⁱ⊗dx^j⊗dx^k⊗ ∂

∂x_l = X

16i<j6n 16k,l6n

R^l_ijk(dxⁱ∧dx^j)⊗dx^k⊗ ∂

∂x_l. (16)

2

We proved that for anyi, j, k andl∈J1, nK, we have:

R_ijk^l = ∂Γ^l_jk

∂x_i −∂Γ^l_ik

∂x_j +

r

X

m=1

Γ^m_jkΓ^l_im−

r

X

m=1

Γ^m_ikΓ^l_jm. (17) Similarly, we define(Rijkl)16i,j,k,l6nby:

∀i, j, k, l∈J1, nK, R_ijkl=Re ∂

∂x_i, ∂

∂x_j, ∂

∂x_k, ∂

∂x_l

. (18)

By Eq. (8) and (9), for anyi, j, k and k, we have:

R_ijkl=−R_jikl =−R_ijlk=R_klij (19)

Rijkl+Rjkil+Rkijl= 0. (20)

ThenRe can be written locally as:

Re= X

16i,j,k,l6n

R_ijkldxⁱ⊗dx^j⊗dx^k⊗dx^l= X

16i<j6n 16k<l6n

R_ijkl(dxⁱ∧dx^j)⊗(dx^k∧dx^l). (21)

By Eq. (13) and (18), we see that, for anyi, j, k andl, we have:

Rijkl=

n

X

m=1

R^m_ijkgml and R^l_ijk =

n

X

m=1

Rijkmg^ml. (22)

Ricci tensor. As above, we can define the coefficient ofRicby:

∀j, k∈J1, nK, Ric_jk = Ric ∂

∂xj

, ∂

∂xk

, (23)

so that in local coordinates we have:

Ric = X

16j,k6n

Ric_jkdx^j ⊗dx^k. (24)

We deduce from the definition ofRic(cf. Eq. (3)) and Eq. (22) that, for anyj andk, we have:

Ric_jk =

n

X

i=1

Rⁱ_ijk= X

16i,m6n

R_ijkmg^mi. (25)

Scalar curvature. Since g

∂

∂xi,·

=P_n

j=1g_ijdx^j for any i∈J1, nK, we have:

∀i∈J1, nK, ∂

∂xi

[

=

n

X

j=1

g_ijdx^j and ∀j∈J1, nK, dx^j]

=

n

X

i=1

g^ji ∂

∂xi

. (26) Then we get:

Ric^]= X

16j,k6n

Ric_jkdx^j ⊗(dx^k)^]= X

16i,j,k6n

Ric_jkg^kidx^j⊗ ∂

∂x_i, (27)

so that

S= X

16j,k6n

Ric_jkg^kj. (28)

3

In nice coordinates. Let us now assume that our coordinates are such that ∂

∂x1, . . . ,_∂x^∂

n

is orthonormal atx= 0. Then we have(gij(0))_16i,j6n =In= (g^ij(0))_16i,j6n, whereIn is the identity matrix of sizen. In these coordinates we can simplify Eq. (12), (22), (25) and (28) in the following way:

∀i, j, k ∈J1, nK, Γ^k_ij(0) = 1 2

∂g_ik

∂xj

(0) +∂g_jk

∂xi

(0)−∂g_ij

∂x_k(0)

, (29)

∀i, j, k, l∈J1, nK, R^l_ijk(0) =Rijkl(0), (30)

∀j, k∈J1, nK, Ric_jk(0) =

n

X

i=1

Rⁱ_ijk(0) =

n

X

i=1

R_ijki(0), (31)

S(0) =

n

X

j=1

Ricjj(0). (32)

In normal coordinates. If we assume that(x1, . . . , xn)are normal coordinates around the point of coordinatesx= 0, then we have:

(gij(x))_16i,j6n=In+O

kxk²

= (g^ij(x))_16i,j6n. (33) It is then possible to simplify further Eq. (12) and (17) to get:

∀i, j, k∈J1, nK, Γ^k_ij(0) = 0, (34)

∀i, j, k, l∈J1, nK, R_ijk^l (0) = 1 2

∂²g_jl

∂xi∂x_k(0)− ∂²g_jk

∂xi∂x_l(0)− ∂²gil

∂xj∂x_k(0) + ∂²gik

∂xj∂x_l(0)

. (35)

4 The case of surfaces

Whenn= 2, we have Re=R₁₂₁₂(dx¹∧dx²)⊗(dx¹∧dx²) for any choice of local coordinates (cf. Eq. (21)). For anyp∈M, letκ(p) =K(TpM)denote the sectional curvature of the tangent plane. In any local coordinates centered at p and such that

∂

∂x1(0),_∂x^∂

2(0)

is orthonormal we haveκ(p) =Re

∂

∂x1(0),_∂x^∂

2(0),_∂x^∂

2(0),_∂x^∂

1(0)

=−R₁₂₁₂(0), so that at the point p:

Rep =−κ(p)(dx¹∧dx²)⊗(dx¹∧dx²). (36) A computation gives that, for any vector fieldsX, Y, Z, T onM, we have:

R(X, Y, Z, Te ) =κ(g(Y, Z)g(X, T)−g(X, Z)g(Y, T)). (37) By definition ofR, this means that:e

R(X, Y)Z =κ(g(Y, Z)X−g(X, Z)Y), (38) for anyX, Y and Z∈Γ(T M). The definition ofRic as a trace yields:

∀Y, Z∈Γ(T M), Ric(Y, Z) =κg(Y, Z). (39) Finally, we get:

S= 2κ. (40)

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