Prove the diff Bianchi identity $d^\nabla R^\nabla=0$ using that $(d^\nabla\circ d^\nabla)\circ d^\nabla = d^\nabla\circ (d^\nabla\circ d^\nabla)$

Question

I found on page 25 of Arthur Besse’s “Einstein Manifolds” that the differential Bianchi identity

$$d^\nabla R^\nabla=0$$

follows from the alternate definition of the Riemann curvature tensor as the second covariant exterior derivative

$$R_{X,Y}^\nabla s = -d^\nabla(d^\nabla s) (X,Y)$$

and the fact that $(d^\nabla\circ d^\nabla)\circ d^\nabla = d^\nabla\circ (d^\nabla\circ d^\nabla)$. However, the proof of the differential Bianchi identity that I know is the one given by Robert Wald on pp. 39 and 40 of his book “General Relativity” and is more involved (in particular, it invokes the algebraic Bianchi identity). How do I see that $d^\nabla R^\nabla=0$ only from $(d^\nabla\circ d^\nabla)\circ d^\nabla = d^\nabla\circ (d^\nabla\circ d^\nabla)$?

Answer 1

I don't know a lot about Riemannian geometry, I only work on complex manifolds but I believe the idea is the same. A connection is a linear map $d^\nabla : \Omega^k(X,E) \rightarrow \Omega^{k + 1}(X,E)$ that verifies a Leibnitz relation. Here, $E$ is a vector bundle (maybe the tangant bundle in your case ?).

Then, we can prove that $d^\nabla \circ d^\nabla : \Omega^0(X,E) \rightarrow \Omega^2(X,E)$ is $\Omega^0(X,\mathbb{C})$-linear thus there is a unique $R^\nabla \in \Omega^2(X,\mathrm{End}(E))$ such that $d^\nabla(d^\nabla s) = R^\nabla s$ for all section $s$. This is the curvature.

Therefore, by the Leibnitz relation, for all smooth section $s$, $$ d^\nabla(d^\nabla(d^\nabla s)) = d^\nabla(R^\nabla s) = (d^\nabla R^\nabla)s + R^\nabla d^\nabla s = (d^\nabla R^\nabla)s + d^\nabla(d^\nabla(d^\nabla s)). $$ It is true for all $s$ hence $d^\nabla R^\nabla = 0$.