LECTURE 6: THE RIEMANN CURVATURE TENSOR. 1. The curvature tensor

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1 LECTURE 6: THE RIEMANN CURVATURE TENSOR 1. The curvature tensor Let M be any smooth manifod with inear connection, then we know that R(X, Y )Z := X Y Z + Y X Z + [X,Y ] Z. defines a (1, 3)-tensor fied on M, caed the curvature tensor of. Locay if we write R = Rijk dx i dx j dx k j, then the coefficients can be expressed via the Christoffe symbos of as R ijk = Γ s jkγ is + Γ s ikγ js i Γ jk + j Γ ik, Obviousy the curvature tensor for the standard connection on R n is identicay zero, since its Christoffe s symbos are a zero. Exampe. Consider M = S n. Last time we have seen that X Y = X Y X Y, n n. defines a (Levi-Civita) connection on S n, where is the standard connection on R n+1 : X i i (Y j j ) = X i i (Y j ) j. To cacuate its curvature tensor, we need rewrite it into a simper form. Since n = (x 1, x 2,, x n+1 ), one get X n = X i i (x j ) j = X. It foows X Y, n n = Y, X n n = X, Y n. So X Y = X Y + X, Y n. Thus Y X Z = Y X Z + Y, X Z n = Y ( X Z + X, Z n) + X Y, Z n X Y, Z n = Y X Z + Y ( X, Z ) n + X, Z Y + X( Y, Z ) n X Y, Z n. In view of the fact R = 0, we get R(X, Y )Z = X( Y, Z ) n Y, Z X Y ( X, Z ) n + Y X, Z n + Y ( X, Z ) n + X, Z Y + X( Y, Z ) n X Y, Z n + [X, Y ], Z n = X, Z Y Y, Z X. 1

2 2 LECTURE 6: THE RIEMANN CURVATURE TENSOR By definition one immediatey gets the foowing anti-symmetry: (1) R(X, Y )Z = R(Y, X)Z For the curvature tensor R, one has Proposition 1.1 (The First Bianchi identity). If is a torsion-free, then (2) R(X, Y )Z + R(Y, Z)X + R(Z, X)Y = 0. Proof. Reca that is torsion-free means X Y Y X [X, Y ] = 0. So we have R(X, Y )Z + R(Y, Z)X + R(Z, X)Y = X Y Z + Y X Z + [X,Y ] Z Y Z X + Z Y X + [Y,Z] X Z X Y + X Z Y + [Z,X] Y = X [Y, Z] Y [Z, X] Z [X, Y ] + [X,Y ] Z + [Y,Z] X + [Z,X] Y = [X, [Y, Z]] [Y, [Z, X]] [Z, [X, Y ]] = 0, where in the ast step we used the Jacobi identity for vector fieds. Obviousy one can then write (1) and (2) in oca coordinates as Rijk = R Rijk + Rjki + Rkij = 0. Reca that one can aways extend a inear connection on the tangent bunde to a inear connection on tensor bundes. In particuar, for the tensor fied R of type (1, 3), X R is aso a tensor fied of type (1, 3), given by ( X R)(Y, Z, W ) := X (R(Y, Z)W ) R( X Y, Z)W R(Y, X Z)W R(Y, Z) X W. Proposition 1.2 (The Second Bianchi Identity). Suppose is torsion free, then Proof. By definition, jik, ( X R)(Y, Z, W ) + ( Y R)(Z, X, W ) + ( Z R)(X, Y, W ) = 0. ( X R)(Y, Z, W ) + ( Y R)(Z, X, W ) + ( Z R)(X, Y, W ) = X (R(Y, Z)W ) R( X Y, Z)W R(Y, X Z)W R(Y, Z) X W + Y (R(Z, X)W ) R( Y Z, X)W R(Z, Y X)W R(Z, X) Y W + Z (R(X, Y )W ) R( Z X, Y )W R(X, Z Y )W R(X, Y ) Z W. Using the torsion-freeness and (1), one can simpify the midde two coumns to R([X, Z], Y )W + R([Y, X], Z)W + R([Y, X], Z)W.

3 LECTURE 6: THE RIEMANN CURVATURE TENSOR 3 Now expand each R using its definition, the whoe expression becomes a summation of 27 terms, the first 9 terms being X Y Z W + X Z Y W + X [Y,Z] W [X,Z] Y W + Y [X,Z] W + [[X,Z],Y ] W + Y Z X W Z Y X W [Y,Z] X W, the second and third 9 terms are simiar to the first 9 terms above: one just repace X, Y, Z by Y, Z, X and Z, X, Y respectivey. It is not hard to check that a those expressions containing three s (12 terms in tota) cance out triviay, a those expressions containing two s (aso 12 terms in tota) cance out by using the fact [X, Y ] = [Y, X], and the remaining three terms in view of the Jacobi identity. [[X,Z],Y ] W + [[Y,X],Z] W + [[Z,Y ],X] W = 0 In oca coordinates we can write n R = Rijk ;ndx i dx j dx k j, Then the second Bianchi identity can be written as Rijk ;n + Rjnk ;i + Rnik ;j = The Riemann curvature tensor Now suppose (M, g) be a Riemannian manifod and the Levi-Civita connection. As ast time, by using the Riemannian metric g one can convert the (1, 3)-tensor R to a (0, 4)-tensor Rm Γ( 4 T M): Rm(X, Y, Z, W ) = g(r(x, Y )Z, W ). We sha ca Rm the Riemann curvature tensor of (M, g). Locay if we write Rm = R ijk dx i dx j dx k dx, then R ijk = Rm( i, j, k, ) = g(rijk m m, ) = g m Rijk m. In other words, the Riemannian metric ower one of the the index. Obviousy one can rewrite the identities (1) and (2) using Rm as Rm(X, Y, Z, W ) + Rm(Y, X, Z, W ) = 0, Rm(X, Y, Z, W ) + Rm(Y, Z, X, W ) + Rm(Z, X, Y, W ) = 0, or in oca coordinates as R ijk = R jik, R ijk + R jki + R kij = 0. Simiary the second Bianchi identity can be written in terms of Rm as ( X Rm)(Y, Z, W, V ) + ( Y Rm)(Z, X, W, V ) + ( Z Rm)(X, Y, W, V ) = 0,

4 4 LECTURE 6: THE RIEMANN CURVATURE TENSOR and if we denote R ijk;n = ( n R)( i, j, k, ), then R ijk;n + R jnk;i + R nik;j = 0. Exampe. The Riemann curvature tensor for S n (equiped with the standard round metric) is Rm(X, Y, Z, W ) = X, Z Y, W Y, Z X, W. Note that this can be written as Rm = 1 2 g g where is the Kukarni-Nomizu product which takes 2 symmetric (0, 2)- tensor into one (0, 4)-tensor that has many nice symmetry properties: (T 1 T 2 )(X, Y, Z, W ) =T 1 (X, Z)T 2 (Y, W ) T 1 (Y, Z)T 2 (X, W ) T 1 (X, W )T 2 (Y, Z) + T 1 (Y, W )T 2 (X, Z).. By staring at the above exampe, one see that the Riemann curvature tensor Rm on the standard S n has even more (anti-)symmetries than the ones we have seen, e.g. one can exchange Z with W to get a negative sign, or even exchange X, Y with Z, W. In fact thest two (anti-)symmetries are consequneces of metric compatibiities, and thus hod in genera: Proposition 2.1. The Riemann curvature tensor Rm satisfies Rm(X, Y, Z, W ) = Rm(X, Y, W, Z), (3) Rm(X, Y, Z, W ) = Rm(Z, W, X, Y ). Proof. We first notice that if we denote f = Z, Z, then in other words, It foows So X Z, Z = Xf Z, X Z, X Z, Z = 1 2 Xf. X Y Z, Z = X Y Z, Z Y Z, X Z = 1 2 X(Y f) Y Z, X Z. Rm(X, Y, Z, Z) = R(X, Y )Z, Z = X Y Z + Y X Z + [X,Y ] Z, Z = 1 2 X(Y f) Y (Xf) + 1 [X, Y ]f = 0. 2 As a consequence, we get Rm(X, Y, Z, W ) + Rm(X, Y, W, Z) =Rm(X, Y, Z + W, Z + W ) Rm(X, Y, Z, Z) Rm(X, Y, W, W ) =0.

5 LECTURE 6: THE RIEMANN CURVATURE TENSOR 5 The second one is a consequence of the first one together with (1) and (2). In fact, by the first Bianchi identity (2) one has Rm(X, Y, Z, W ) + Rm(Y, Z, X, W ) + Rm(Z, X, Y, W ) = 0, Rm(Y, Z, W, X) + Rm(Z, W, Y, X) + Rm(W, Y, Z, X) = 0, Rm(Z, W, X, Y ) + Rm(W, X, Z, Y ) + Rm(X, Z, W, Y ) = 0, Rm(W, X, Y, Z) + Rm(X, Y, W, Z) + Rm(Y, W, X, Z) = 0, Summing the equations and using (1) as we as the first one in (3) that we just proved, the first two coumns cance out and we get Rm(Z, X, Y, W ) + Rm(W, Y, Z, X) = 0, which is equivaent to the second one in (3). Of course if one use oca coordinates, then the two identities in (3) can be rewritten as R ijk = R ijk, R ijk = R kij.