# Lie Derivative of Kahler 2-form

Suppose there is a Killing vector $k$ on a Kahler manifold $M$. By definition, $k$ generates isometries of the metric. That is, $L_kg=0$, where $L$ is the Lie derivative. At the same time, there is a holomorphic complex structure which satisfies $L_kJ=0$. I have two questions:

1. Is $L_kJ=0$ generic for Kahler manifolds or is that specific to my particular example where I wanted $J$ to commute with the SUSY generators (apparently this has something to do with making $J$ holomorphic)?

2. More importantly, the fundamental 2-form on the Kahler manifold is defined as $\omega_{ab} = g_{ac} J^c_b$. How can I show that $L_k \omega=0$?

In some sense it is obvious: $L_k \omega_{ab} = L_k (g_{ac} J^c_b) = (L_k g_{ac}) J^c_b + g_{ac} (L_k J^c_b) = 0$ since Lie derivative obeys product rule.

However, I should be able to recover it in components also. For a general rank 2 tensor field, $$(L_k T)_{ab}= k^c \nabla_c T_{ab} + T_{ac} \nabla_b k^c + T_{cb} \nabla_a k^c.$$ If I use that $\omega_{ab} = g_{ac}J^c_b$ in this general formula, I get: $$(L_k \omega)_{ab} = k^c \nabla_c (g_{ad} J^d_b) + g_{ad} J^d_c \nabla_b k^c + g_{cd} J^d_b \nabla_a k^c$$ and I don't know how to show this is equal to zero?

The claims aren't true in complete generality. For example, $R^{2n}$ is a Kähler manifold and any vector $k$ is a Killing vector. But the Lie derivatives of $J$ and $\omega$ don't vanish for a general $k$.
However, this example was special because it was insufficiently curved. For generic and curved enough Kähler manifold, the objects $J,\omega$ may be calculated basically uniquely out of an existing metric tensor field $g$. Because they are functions of $g$, their Lie derivative must vanish because the Lie derivative of the metric $g$ vanishes.