# 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?

## 1 Answer

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.

I won't try to write the functions explicitly, partly because it's methodologically illogical. If one knows that a manifold is a Kähler manifold, then talking about the Kähler form and the complex structure should be "more fundamental" than talking about the metric. The metric is naturally calculated from the Kähler form, not the other way around. So in that case, it makes sense to assume that the Lie derivative of the Kähler form and the complex structure vanishes, and prove that the Lie derivative of the metric vanishes as well.

For a given field, if those claims are true, it should be easier to prove the vanishing of the Lie derivative of the Kähler form or the complex structure because these derivations use the simplifications of the complex geometry.