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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2$\begingroup$ Yes, that's true for any Killing field (Kahler and holomorphic is not needed). $\endgroup$– Robert HaslhoferCommented Feb 28, 2013 at 2:31
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10$\begingroup$ 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 YangCommented Feb 28, 2013 at 3:04
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$\begingroup$ @Deane Yang: Thanks for your clear explanation. $\endgroup$– ModuliCommented Feb 28, 2013 at 15:13
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1 Answer
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$g\mapsto R(g)$ 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.
answered Feb 28, 2013 at 3:57