Let $R_{ij}$ denote the Ricci tensor . If $R_{ij,k}+R_{jk,i}+R_{ki,j}=0$ , prove that the scalar curvature $R$ is constant .
A straightforward calculation gives $$R_{ij,k}=\frac{\partial R_{ij}}{\partial x^k}-\Gamma^l_{ik}R_{lj}-\Gamma^l_{jk}R_{il} \\ R_{jk,i}=\frac{\partial R_{jk}}{\partial x^i}-\Gamma^l_{ji}R_{lk}-\Gamma^l_{ki}R_{jl} \\ R_{ki,j}=\frac{\partial R_{ki}}{\partial x^j}-\Gamma^l_{kj}R_{li}-\Gamma^l_{ij}R_{kl}$$ Since $R_{ij}$ is symmetric $$R_{ij,k}+R_{jk,i}+R_{ki,j}=0 \\ \implies \frac{\partial R_{ij}}{\partial x^k}+\frac{\partial R_{jk}}{\partial x^i}+\frac{\partial R_{ki}}{\partial x^j}=2\Gamma^l_{ik}R_{lj}+2\Gamma^l_{jk}R_{il}+2\Gamma^l_{ij}R_{kl}$$ I know that $R=g^{ij}R_{ij}$ , but multiplying above expression with any one of $g^{lj},g^{il},g^{kl}$ won't help .
How to approach this one ? Any help is appreciated .
The Contracted Bianchi Identity says:
By other hand, tracing the identity $$R_{ij;k}+R_{jk;i}+R_{ki;j}=0$$ we get $$R_{ij;}^j+R_{;i}+R_{;j}=0.\tag{2}$$ Therefore, applying (1) in (2) we obtain $$\frac{5}{2}R_{;i}=0,\quad\forall i.$$ (In previous equality we choose $j=i$, since (1) is valid for all $i,j,k$) Finally, we conclude $$R\equiv C.$$ (Is necessary suppose a connected manifold, otherwise, the scalar curvature can assume a different constant value in each connected component.)