Let $M$ be a Kahler manifold, with Kahler metric $g$. Let $X$ be a holomorphic Killing vector field of $g$, i.e. $L_{X} g = 0$, where $L_{X}$ is the Lie derivative along $X$. Let $R$ be the Riemannian curvature tensor of $g$. Is $L_{X} R = 0$?

2$\begingroup$ Yes, that's true for any Killing field (Kahler and holomorphic is not needed). $\endgroup$ – Robert Haslhofer Feb 28 '13 at 2:31

10$\begingroup$ In particular, if you integrate a Killing vector field, you get a 1parameter family of isometries. Pulling back the curvature tensor by an isometry gives the original curvature tensor. So if you differentiate the 1parameter family of curvature tensors obtained by pulling back with the 1parameter family of isometries, you get the zero tensor. $\endgroup$ – Deane Yang Feb 28 '13 at 3:04

$\begingroup$ @Deane Yang: Thanks for your clear explanation. $\endgroup$ – Moduli Feb 28 '13 at 15:13
$g\mapsto R(g)$ is a natural (nonlinear) operation: it commutes with pullbacks by diffeomorphisms. This is just a transcription of: curvature transforms correctly under chart changes. Thus we get $L_X R(g) = dR(g)(L_X g)$ which implies "Yes"; two earlier comments also said this.