On compact Kahler manifold, Lie derivative of Killing vector field commutes with $dd^c$

Question

Let $(M,\omega)$ be a compact Kahler manifold, $X$ a Killing vector field, $f \in C^\infty(M,\mathbb{R}). $ I would like to show that: $L_X(dd^cf) = dd^cL_Xf$.

If I write it out locally, $X = X^i \partial_{z_i}$, $X^i$ are pluriharmonic functions. Then locally, we have:

$L_X(dd^cf) = (f_{ki\bar{j}} X^i + f_{i\bar{j}} X^i_k) \space dz_k \wedge \bar{dz_j}$

On the other hand, we have:

$dd^cL_Xf = (f_iX^i_{k\bar{j}} + f_{ik}X^i_{\bar{j}} + f_{ki\bar{j}}X^i+f_{i\bar{j}}X^i_k) \space dz_k \wedge \bar{dz_j} = (f_{ik}X^i_{\bar{j}} + f_{ki\bar{j}}X^i+f_{i\bar{j}}X^i_k) \space dz_k \wedge \bar{dz_j}$

because $X^i$ pluriharmonic so $dd^c X^i = 0$ and $X^i_{k\bar{j}} = 0$.

The only difference between the 2 expressions is the term $f_{ik}X^i_{\bar{j}}$.

Answer 1

Let $X$ be a real holomorphic vector field on $M$, i.e. a smooth section of the (real) tangent bundle of $M$ such that $\mathcal{L}_XJ=0$. Every Killing vector field on a compact Kähler manifold has this property, see for example Theorem 4.81 in Ballmann's notes (note that he calls these vector fields automorphic, rather than real holomorphic).

We can show something more than your claim, namely that for any $p$-form $\alpha$ on $M$, one has $\mathcal{L}_X\mathrm{d}\mathrm{d}^c\alpha=\mathrm{d}\mathrm{d}^c\mathcal{L}_X\alpha$. It is not necessary to assume that $M$ is Kähler. To do this, we only need three ingredients:

1. if $\Phi_t$ is the flow of $X$, then for any tensor field $V$ on $M$ the Lie derivative is defined as $\mathcal{L}_XV=\partial_{t=0}\Phi_t^*V$;
2. the exterior derivative is natural, i.e. for any diffeomorphism $\Phi$ and any $p$-form $\alpha$ one has $\Phi^*(\mathrm{d}\alpha)=\mathrm{d}(\Phi^*\alpha)$;
3. a vector field is real holomorphic if and only if its flow is made of biholomorphisms of $M$.

With these in place, we can then proceed as follows, where $\Phi_t$ is the flow of $X$. First, by property $1$ we have $$\mathcal{L}_X\mathrm{d}\mathrm{d}^c\alpha=\partial_{t=0}\Phi_t^*\left(\mathrm{d}\mathrm{d}^c\alpha\right)=\partial_{t=0}\Phi_t^*\left(\mathrm{d}(J^{-1}\mathrm{d}(J\alpha)\right).$$ Now, using $2$ a couple of times we get $$\Phi_t^*\left(\mathrm{d}(J^{-1}\mathrm{d}(J\alpha)\right)=d\left((\Phi_t^*J^{-1})d\left((\Phi_t^*J)(\Phi_t^*\alpha)\right)\right)$$ but by $3$ we know $\Phi_t^*J=J$ (and $\Phi_t^*J^{-1}=J$), so putting everything together $$\mathcal{L}_X\mathrm{d}\mathrm{d}^c\alpha=\partial_{t=0}\left(\mathrm{d}(J^{-1}\mathrm{d}(J(\Phi_t^*\alpha))\right)=\partial_{t=0}\left(\mathrm{d}\mathrm{d}^c(\Phi_t^*\alpha)\right)=\mathrm{d}\mathrm{d}^c\mathcal{L}_X\alpha.$$

So we just need to show the three things above. $1$ and $2$ are found in any differential geometry book, and $3$ is a consequence of $1$ and the definition of real holomorphic vector field. Let me know if you need help to show this, but I think it would be a good exercise to go through on your own.