In "Riemannian Manifolds: An Introduction to Curvature," John Lee states the following lemma:
Lemma 7.7 (Contracted Bianchi Identity): The covariant derivatives of the Ricci and scalar curvatures satisfy $\text{div} Rc = \frac{1}{2}\nabla S,$ where $\text{div} Rc$ is the 1-tensor obtained from $\nabla Rc$ by raising one index and contracting. In components, this is $R_{ij};^j = \frac{1}{2}S_{;i}.$
Lee then proves the coordinate form of the statement. He does this by (metric) contracting the differential Bianchi identity in coordinates $R_{ijkl;m} + R_{ijlm;k} + R_{ijmk;l} = 0.$
I have two questions:
Is there a more coordinate-free proof of this fact? I suppose one can argue (in words) that contractions are coordinate-invariant and such, but I would prefer seeing a proof in symbols nevertheless.
Can we prove the identity directly from the symmetry of the Ricci tensor?
My second question was inspired by the following computation:
Evaluating the left-hand side at a vector field $X$: $(\text{div}Rc)(X) = (\text{tr}_g\nabla Rc)(X),$ while similarly on the right-hand side: $\frac{1}{2}(\nabla S)(X) = \frac{1}{2}\nabla_XS = \frac{1}{2}\nabla_X(\text{tr}_gRc) = \frac{1}{2}\text{tr}_g(\nabla_XRc).$ So, we can prove the Contracted Bianchi Identity if we can show that $(\text{tr}_g\nabla Rc)(X) = \frac{1}{2}\text{tr}_g(\nabla_XRc),$ which might somehow follow from the symmetry of $Rc$.
This question is in some sense related to a previous question of mine, in which I ask for a means of computing traces/contractions explicitly.