# 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.

• I don't have much to say about your question, I just wanted to point out that there's a rather coordinate free proof of the second Bianchi identity in these lecture notes by Salamon and Robbin in Thm 4.77 on p.210. In general, if you're looking for coordinate free proofs, I strongly recommend Gallot-Hulin-Lafontaine, but I'm not sure that answers your question.
– t.b.
Commented Jun 24, 2011 at 0:50

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*}

• Yikes, I'll show this my friend who thinks that going coordinate-free always means that the computations are short.
– Tom
Commented Dec 10, 2019 at 3:57
• @Tom After thinking about it for a while, I realized this computation is actually not coordinate free at all, we're just hiding the coordinates under the hood of a geodesic frame (whose proof of existence requires using normal coordinates). It's still very beautiful though! Commented Jun 9, 2022 at 4:57