TITLE: Lie derivative of curvature
QUESTION [7 upvotes]: Let $M$ be a Kähler manifold, with Kähler metric $g$. Let $X$ be a holomorphic Killing vector field of $g$, i.e. $\mathcal{L}_{X} g = 0$, where $\mathcal{L}_{X}$ is the Lie derivative along $X$. Let $R$ be the Riemannian curvature tensor of $g$. Is $\mathcal{L}_{X} R = 0$?

REPLY [3 votes]: True. $X$ is Killing, $\mathcal{L}_{X} g = 0$, iff its flow, say $\phi_t$, is by isometries, ie $\phi_t^*g=g$ for all $t$. The Riemann curvature tensor is naturally associated with the metric, hence $\phi_t^*R=R$ for all $t$. Taking derivative wrt $t$ of both sides of the last equation gives $\mathcal{L}_{X} R = 0$. Note that the Kahler structure does not play any role.