Schur's theorem in DoCarmo's "Riemannian Geometry"

Question

The exercise 8 of chapter 4 of Do Carmo's "Riemannian Geometry" ask to prove the Schur's Theorem. I don't understand a step in the hint (the "hint" is essentially the proof of the theorem).

Schur's Theorem: Let $M^n$ be a connected Riemannian manifold with $n\ge3$. Suppose $M$ is isotropic, i.e. for each point $p \in M$ the sectional curvature $K(p, \sigma)$ does not depend on $\sigma \subset T_pM$. Prove that $M$ has constant sectional curvature, i.e. $K(p, \sigma)$ also doesn't depend on $p$.

Here is the beginning of the hint provided by the book.

Define a tensor $R'$ of order $4$ by $$ R'(W,Z,X,Y) = \langle W, X \rangle \langle Z, Y \rangle - \langle Z, X \rangle \langle W, Y \rangle. $$

If $K(p, \sigma) = K$ does not depend on $\sigma$, we know (thanks to a previous lemma) that $R = KR'$, where $R$ is the curvature tensor of $M$. Then the book says: Therefore, for every smooth vector field $U$, $\nabla_UR = (UK)R' $. I can't understand this passage. Shouldn't be $\nabla_UR = (UK)R' + K(U(R'))$?

Answer 1

That's because the metric itself is nabla-flat, so by Leibniz rule you get what you want.

Answer 2

I don't quite understand why the fact that the metric is nabla flat gives the answer.

What I saw is that, starting off from the definition of the covariant differential, if we expand the 2nd term we obtain

$$R(\nabla_U W,Z,X,Y) = K(p)R'(\nabla_U W,Z,X,Y) = K(p)\big(\langle \nabla_U W, X \rangle \langle Z,Y \rangle - \langle Z,X \rangle \langle \nabla_U W, Y \rangle\big),$$

and if we expand the first term of $K(p)U\big(R'(W,Z,X,Y)\big)$ we obtain

$$K(p)U\big(\langle W,X \rangle\big) \langle Z,Y \rangle = K(p)\big(\langle \nabla_U W, X \rangle + \langle W, \nabla_U X \rangle\big) \langle Z,Y \rangle,$$

and thus we will get that all of those terms cancel each other, i.e.

$$\nabla_U R(W,Z,X,Y) = \nabla R(W,Z,X,Y,U) = U(K)R'(W,Z,X,Y).$$