Locally symmetric spaces and the curvature tensor Let $M$ be a Riemannian manifold. Suppose $\nabla R=0$ where $R$ is the curvature tensor (we then say $M$ is locally symmetric). Then if $\gamma$ is a geodesic of $M$ and $X,Y,Z$ are parallel vector fields along $\gamma$ then $R(X,Y)Z$ is a parallel field along $\gamma$.


However, I do not fully understand what does it mean $\nabla R=0$. Is it to say that $\nabla_ZR(X,Y,W,T)=0$ for all vector fields $Z,X,Y,W,T$?

How can we use this to prove $\dfrac{D}{dt}R(X,Y)Z=0$?

The only thing I can think is maybe we can use that if $V(t)=Y(c(t))$ (i.e. $V$ is induced by a vector field) then $\dfrac{DV}{dt}=\nabla_{c'}Y$.

Any hints would be appreciated.
$R$ is a $2$-tensor, $\nabla R=0$ is equivalent to saying that for every vector field $\nabla_ZR=0$.

$\nabla_ZR(X,Y)=\nabla_Z(R(X,Y))-R(\nabla_ZX,Y)-R(X,\nabla_ZY)$.
$R$ is a $2$-tensor, $\nabla R=0$ is equivalent to saying that for every vector field $\nabla_ZR=0$.

$\nabla_ZR(X,Y)=\nabla_Z(R(X,Y))-R(\nabla_ZX,Y)-R(X,\nabla_ZY)$.