I'm working on a problem in do Carmo chapter 4. We have that a Riemannian manifold, M, is locally symmetric if $\nabla R = 0$, where R is the curvature tensor of M (i.e., $R(X,Y,Z,W) = \langle R(X,Y)Z, W \rangle$, and $R(X,Y)Z = {\nabla}_Y{\nabla}_XZ - {\nabla}_X{\nabla}_YZ - {\nabla}_{[X,Y]}Z$).
I'm trying to show that if M is locally symmetric, $\gamma: [0,l] \rightarrow M$ a geodesic, and $X,Y,Z$ are parallel vector fields along $\gamma$, then $R(X,Y)Z$ is a parallel field along $\gamma$.
So, Let $X,Y,Z,W$ be vector fields on M parallel along $\gamma$. Then since $\nabla R = 0$ we have in particular that $\nabla_\dot{\gamma}R = 0$. So then
$$0 = \dot{\gamma}(R(X,Y,Z,W)) - R(\nabla_\dot{\gamma}X, Y, Z, W) - R(X, \nabla_\dot{\gamma}Y, Z, W) - R(X, Y, \nabla_\dot{\gamma}Z, W) - R(X, Y, Z, \nabla_\dot{\gamma}W)$$
But since X,Y,Z,W are parallel along $\gamma$ this reduces to $$0 = \dot{\gamma}(R(X, Y, Z, W)) = \dot{\gamma}(\langle R(X,Y)Z, W \rangle) = \langle \nabla_{\dot{\gamma}}R(X,Y)Z, W\rangle + \langle R(X,Y)Z, \nabla_{\dot{\gamma}}W \rangle$$
And since W is parallel along $\gamma$, we have $$0 = \langle \nabla_{\dot{\gamma}}R(X,Y)Z, W\rangle$$
So I want to say this implies that $\nabla_{\dot{\gamma}}R(X,Y)Z = 0$ so that $R(X,Y)Z$ is parallel along $\gamma$, but I am thinking there is something else I need to show since there is the possibility, I think, that $\nabla_{\dot{\gamma}}R(X,Y)Z \neq 0$ but rather $\nabla_{\dot{\gamma}}R(X,Y)Z$ and $W$ are orthogonal. Is there any nice way to show this other than just by assumption?