Show that $(divR) (X, Y,Z) = (\nabla_X Ric) (Y,Z) − (\nabla_Y Ric) (X,Z).$

Question

In course of solving Riemannian Geometry By Peter Petersen Chap. 2, I stuck on the following problem:

Show that in a Riemmanian manifold if $R$ is the $(1, 3)$ curvature tensor and $Ric$ the $(0, 2)$ Ricci tensor, then $(div~R) (X, Y,Z) = (\nabla_X Ric) (Y,Z) − (\nabla_Y Ric) (X,Z).$

I am absolutely clueless. Can anyone help me to solve it?

Answer 1

We have to prove $$ \nabla^i R_{ijkl} =\nabla_j R_{kl} -\nabla_k R_{jl} $$

Proof : From the second Bianchi identity, we have $$ \nabla_a R_{ijkl} + \nabla_i R_{jakl} + \nabla_j R_{aikl} =0 $$

Hence $$g^{al}\nabla_a R_{ijkl} + g^{al}\nabla_i R_{jakl} + g^{al}\nabla_j R_{aikl} =0 $$

$$g^{al}\nabla_a R_{kl ij} - \nabla_i R_{jk} + \nabla_j R_{ik} =0 $$

Hence we have $$-\nabla^l R_{lk ij} - \nabla_i R_{jk} + \nabla_j R_{ik} =0$$

Answer 2

Consider an orthonormal frame $\{E_i\}.$ Consider $X,Y,Z$ such that at a given point $p$ it is $\nabla_{E_i}E_j=\nabla_{E_i}X=\nabla_{E_i}Y=\nabla_{E_i}Z=0.$ Then, at $p,$ by definition,

$$ (\mathrm{div} R)(X,Y,Z) = \sum_{i=1}^n g((\nabla_{E_i}R)(X,Y,Z),E_i).$$ Now, having in mind the second Bianchi identity, $$(\nabla_UR)(V,W)+(\nabla_VR)(W,U)+(\nabla_WR)(U,V)=0,$$ one has

$$g((\nabla_{E_i}R)(X,Y,Z),E_i)=-g((\nabla_{X}R)(Y,E_i,Z),E_i)-g((\nabla_{Y}R)(E_i,X,Z),E_i).$$ But, since $\nabla$ commutes with traces, it is

$$\sum_{i=1}^ng((\nabla_{Y}R)(E_i,X,Z),E_i)=(\nabla_YRc)(X,Z)$$ and $$\sum_{i=1}^ng((\nabla_{X}R)(E_i,Y,Z),E_i)=-(\nabla_XRc)(Y,Z).$$ So, we have obtained the desired equality.

Answer 3

For completeness, a one-line coordinate computation $$(\delta R)_{ijk} = R_{ijk\phantom{\ell};\ell}^{\phantom{ijk;}\ell} = g^{\ell r}R_{ijkr;\ell}\stackrel{(\ast)}{=} - g^{\ell r}R_{j\ell kr;i} - g^{\ell r}R_{\ell i kr;j} = R_{jk;i} - R_{ik;j},$$where in $(\ast)$ we have used the second Bianchi identity.