# Lie derivative and taking trace

Let $(M,\omega)$ be a complex Kahler manifold, and $g$ is a smooth function such that $\int_Mg\omega^n=0$. It is obvious that there exists a smooth function $f$ such that $\triangle_\omega f=g$. Furthermore, we assume $X$ is a holomorphic vector field over $M$.

My question is whether we can represent $X(f)$ in terms of $g$, $X$ and $\omega$.

While $\alpha=\sqrt{-1}\partial\bar{\partial}f$, we know $\sqrt{-1}\partial\bar{\partial}X(f)=L_X\alpha$, where $L_X$ is the Lie derivative with respect to $X$. We know less information about $f$ from $\triangle_\omega f=g$, but the solution to such equation is still unique modulo a constant. So I wonder whether we can know some information about $X(f)$ from this equation, too. Or what about the special case of $g=\text{trace}_\omega\alpha$, and $\int_M\alpha\wedge\omega^{n-1}=0$?

• Yes, if you allow non-local operators in your formula, such as the Green operator of $\omega$ (and then it is trivial to derive a formula for $X(f)$). If you insist on a purely local formula, then I am afraid the answer is no, because $X(f)$ involves only one derivative of $f$, while $g$ involves two derivatives. On the other hand, if all you care about is an integral relation between these quantities, then often you can do this, by using integration by parts (in particular you can do it if your vector field $X$ has a zero, in which case $X$ has a holomorphy potential). – YangMills Jun 17 '15 at 7:58
• Thank you for your reply! But I can not really understand the last sentence "(in particular you can do it if your vector field X has a zero, in which case X has a holomorphy potential)". Would you like to give some details? In my opinion, the difficulty of this problem is that we get less information from $g=\triangle f$, in particular, the Hessian of $f$. – Daniel Jun 18 '15 at 9:19
• If $X$ has a zero then there is a complex-valued function $u$ such that $X=g^{i\bar{j}}\partial_{\bar{j}}u \partial_i$, and so $\int_M X(f) \omega^n=-\int_X u\Delta f\omega^n=-\int_X ug\omega^n$. – YangMills Jun 22 '15 at 10:41
• Yes! But what I care more is the pointwise property of $X(f)$. And it seems impossible if we only know the Laplace of $f$. – Daniel Jun 24 '15 at 1:17