# If $S$ is the $(1,1)$ version of the Hessian of $f$, then $L_{\nabla f} S = \nabla_{\nabla f} S$

Let $$(M,g)$$ be a Riemannian manifold and let $$\nabla$$ be the Levi-Civita connection on $$M$$. Let $$f: M \to \mathbb{R}$$ be a smooth function and let $$S(X) = \nabla_{X} \nabla f$$ be the $$(1,1)-$$tensor version of $$\mathrm{Hess} f.$$ How do we show that $$L_{\nabla f}S = \nabla_{\nabla f}S,$$ where $$L_{\nabla f}$$ is the Lie derivative along $$\nabla f$$ and $$\nabla_{\nabla f}S$$ is the covariant derivative of $$S$$ along $$\nabla f$$, which is a $$(1,1)-$$tensor for which $$(\nabla_{\nabla f}S )(X) = \nabla_{\nabla f}(S(X)) - S(\nabla_{\nabla f} X).$$

This can be done in coordinates, but it is a long and tedious approach. I have tried to do this in a coordinate-independent way, but I am not able to. As $$S$$ is a $$(1,1)-$$tensor, I don't know any "nice" formula for $$L_{\nabla f}S$$, besides its definition.

By a "nice" formula I mean a formula similar to the following for a covariant $$k-$$tensor $$T$$: $$(\nabla_X T)(Y_1, \cdots, Y_k) = X(T(Y_1, \cdots, Y_k)) - \sum_{j} T(Y_1, \cdots \nabla_X Y_j, \cdots, Y_k),$$ but this does not apply for $$(1,k)-$$tensors.

Remark: this is exercise $$3.4.3$$ from Petersen's Riemannain Geometry, third edition.

The Lie derivative is a derivation, so $$(L_{\nabla f}S)(X) = L_{\nabla f}(S(X)) - S(L_{\nabla f}X) .$$ You can now complete the problem using the identity $$\nabla_{S(X)}\nabla f - S(\nabla_X\nabla f) = 0,$$ which follows immediately from the definition of $$S$$, and the fact that $$\nabla$$ is torsion-free.