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 $L_{X} R = 0$?

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    $\begingroup$ Yes, that's true for any Killing field (Kahler and holomorphic is not needed). $\endgroup$ – Robert Haslhofer Feb 28 '13 at 2:31
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    $\begingroup$ In particular, if you integrate a Killing vector field, you get a 1-parameter family of isometries. Pulling back the curvature tensor by an isometry gives the original curvature tensor. So if you differentiate the 1-parameter family of curvature tensors obtained by pulling back with the 1-parameter 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 (non-linear) 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.

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