Isometry invariance of Christoffel symbols

Question

An isometry of a Riemannian manifold is generated by a Killing vector field $X$ with Lie derivative of the metric $L_X g=0$. Does this immediately imply that the Lie derivatives of the Christoffel symbols and Riemann curvature tensors are zero as well?

Answer 1

The Lie derivative of the Riemann curvature tensor is zero, because it is invariant under the flow. (If $Rm$ denotes the Riemann tensor and $\phi_t$ denotes the time-$t$ flow of $X$, then $\phi_t^*Rm = Rm$. Plugging this into the definition of the Lie derivative shows that $L_X Rm = 0$.)

The "Lie derivatives of the Christoffel symbols" do not make sense, because the Christoffel symbols are not components of any invariantly-defined tensor. However, what is true is that the Levi-Civita connection is invariant under the flow, which implies for example that $L_X(\nabla T) = \nabla (L_X T)$ for every tensor field $T$.