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.
Best Answer
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.