# Zero total covariant derivative of curvature tensor

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 $$s_p$$?

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.
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 $$\nabla R = 0$$.