Invariant proof of the Contracted Bianchi Identity 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.
 A: Well, two and a half years later, I stumbled across an answer.  The following is a paraphrasing of Petersen's "Riemannian Geometry," page 40.
We work on a Riemannian manifold $(M,\langle \cdot, \cdot \rangle)$ with Levi-Civita connection $\nabla$.  Let $\text{Ric}$ denote the symmetric $(1,1)$-tensor $\text{Ric}(v) = \sum R(v,e_i)e_i$, where $\{e_i\}$ is an orthonormal basis.  Let $S$ denote the scalar curvature, i.e. $S = \text{tr}(\text{Ric}) = \sum \langle \text{Ric}(e_i), e_i \rangle$.  We note that $\nabla S = dS$.

Prop: $dS = 2\,\text{div}(\text{Ric})$.

Proof: Let $\{E_i\}$ be an orthonormal frame at $p \in M$ with $\nabla E_i|_p = 0$. Let $X$ be a vector field with $\nabla X |_p = 0$.  From the second Bianchi identity:
$$\begin{align*}
dS(X)(p) = XS(p) & = X \sum \langle \text{Ric}(E_i), E_i \rangle \\
& = X \sum \langle R(E_i, E_j)E_j, E_i\rangle \\
& = \sum \langle \nabla_X [R(E_i, E_j)E_j], E_i \rangle \\
& = \sum \langle (\nabla_X R)(E_i, E_j)E_j, E_i \rangle \\
& = -\sum \langle (\nabla_{E_j}R)(X, E_i)E_j, E_i \rangle - \sum \langle (\nabla_{E_i}R)(E_j,X)E_j, E_i \rangle \\
& = -\sum (\nabla_{E_j}R)(X, E_i, E_j, E_i) - \sum (\nabla_{E_i}R)(E_j, X, E_j, E_i) \\
& = \sum (\nabla_{E_j}R)(E_j, E_i, E_i, X) + \sum (\nabla_{E_j} R)(E_i, E_j, E_j, X) \\
& = 2 \sum (\nabla_{E_j}R)(E_j, E_i, E_i, X) \\
& = 2 \sum \nabla_{E_j}(R(E_j, E_i, E_i, X) \\
& = 2 \sum \nabla_{E_j} \langle \text{Ric}(E_j), X \rangle \\
& = 2 \sum \nabla_{E_j} \langle \text{Ric}(X), E_j \rangle \\
& = 2 \sum \langle \nabla_{E_j}(\text{Ric}(X)), E_j) \rangle \\
& = 2 \sum \langle (\nabla_{E_j}\text{Ric})(X), E_j \rangle \\
& = 2\,\text{div}(\text{Ric})(X)_p.
\end{align*}$$
