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?
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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 $T$.
Jack Lee
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Thanks, Jack. Could you state the definition you are using for the Lie derivative? – Meer Ashwinkumar Nov 20 '15 at 02:11
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The Lie derivative with respect to $X$ of a covariant tensor field $T$ is $\lim_{h\to 0} (1/h) (\phi_t^* T - T)$, where $\phi$ is the flow of $X$. – Jack Lee Nov 24 '15 at 05:12