Does the Riemannian connection commute with $(1, 1)$ tensors in the first slot?

Question

I know the Riemannian connection commutes with musical isomorphisms, i.e $\sharp\left( \nabla_X \xi \right) = \nabla_X \left( \sharp \xi \right)$, and similarly for $\flat$. My question is: does this generalize for $(1, 1)$ tensors, as in, is it true that given a $(1, 1)$ tensor $F: \Gamma\left( T M \right) \to \Gamma\left( T M \right)$ we have $$\nabla_{F(X)} Y = F\left(\nabla_X Y \right), \ \forall X, Y \in \Gamma\left( T M \right) \text{?}$$ I think it would be true if $F$ were in the second slot instead of the first one since $\nabla$ commutes with the musical isomorphisms, but from some computations I've been doing I suspect it will (in general) be false in this context. Can anyone think of either a proof or a counterexample? Thanks!

Answer 1

You can easily make a counterexample by taking (on a sphere, say) a geodesic whose orthogonal trajectories are not geodesics. Take $Y=X$ and $F(X)$ the rotation of $X$ through $\pi/2$. Indeed, on the sphere missing two points, you can easily make this global.