Zero total covariant derivative of curvature tensor

Question

Let $M$ be a Riemannian manifold and $p \in M$. For some normal ball $B_r(p) \subset M$, let $s_p: B_r(p) \to B_r(p)$ be the local diffeomorphism given by $s_p(\exp_p(v)) = \exp_p(-v)$. Suppose that $s_p$ is an isometry in some geodesic ball $B_r(p)$. Prove that $(\nabla R)_p = 0$, where $R$ is the Riemannian curvature tensor.

So in normal coordinates I believe if $v = (v^1, \dots, v^n) \in T_p M$ we have that $\exp_p(v) = (v^1, \dots, v^n)$, and $s_p$ being an isometry means $\left< X, Y \right> = \left< (s_p)_* X, (s_p)_* Y \right>$, but wouldn't $(s_p)_*(v) = -v$, in which case it is obvious that $\left<-X, -Y \right> = (-1)(-1) \left< X, Y \right> = \left< X, Y \right>$? I recognize that $(\nabla_V R)_p = \nabla_V \left< R(X,Y)Z, W \right>$ which I can then expand in terms of covariant derivatives and hope for the best , but how does that use $s_p$?

Answer 1

Hint: The considerations that you start with are correct. But then you have to use that an isometry has to preserve $\nabla R$ as a tensor field. Viewing this as a $\binom05$-tensor field, you can directly compute how $(s_p)_*$ acts.

Answer 2

Thanks to Andreas Cap, the solution is quite obvious now: Since $s_p$ is an isometry, $$(\nabla R)(X,Y,Z,W,V) = (s_p)^*(\nabla R)(X,Y,Z,W,V) = (\nabla R)((s_p)_* X, \dots, (s_p)_* V) = (\nabla R)(-X,-Y,-Z,-W,-V) = (-1)^5 (\nabla R)(X,Y,Z,W,V) = - (\nabla R)(X,Y,Z,W,V)$$ thus $\nabla R = 0$.