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Let $f$ be a smooth function such on a compact kahler manifold $(M, w)$, and the component of $w$ is denoted by $g_{ij}$, assume there is a constant $s$ such that $sf = -g^{ij}\sqrt{-1}\partial_{j}\bar\partial_{i}f$, want to prove the following:

$s \nabla_j f = -\nabla_j \nabla_p \nabla_{\bar p} f$ suppresing the mertic,

where $\nabla \nabla \nabla$ is intepreted as in the first answer in here.

Thoughts: the left hand side is simply $\partial_j sf = \partial_j-g^{ik}\sqrt{-1}\partial_{j}\bar\partial_{k}f = - \partial_j g^{ik} \sqrt{-1}\partial_{j}\bar\partial_{k}f -\partial_j\sqrt{-1}\partial_{j}\bar\partial_{k}f g^{ik}$

How do you realize that $- \partial_j g^{ik} \sqrt{-1}\partial_{j}\bar\partial_{k}f -\partial_j\sqrt{-1}\partial_{j}\bar\partial_{k}f g^{ik} = -\nabla_j \nabla_p \nabla_{\bar p} f$ immediately?

When doing calculations, can you just naively replace partial derivative by covariant derivative?

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  • $\begingroup$ Your question is basically why $g^{ij} \sqrt{-1} \partial_j \bar{\partial}_i f = \nabla_p \nabla_{\bar{p}} f$, no? $\endgroup$
    – Willie Wong
    Commented Oct 31, 2019 at 15:29
  • $\begingroup$ No. Are you trying to suggest that $\nabla_j \nabla_p \nabla_p f = \nabla_j( \nabla_p \nabla_p f )$? According to the thread I posted, it should be interpreted as $\nabla \nabla \nabla f (\frac{\partial}{\partial z_j}, \frac{\bar \partial}{\partial z_p}, \frac{\partial}{\partial z_j})$. $\endgroup$
    – Keith
    Commented Oct 31, 2019 at 17:52
  • $\begingroup$ Yes, I am suggesting exactly that. $\nabla_p \nabla_{\bar{p}} f$ has the implicit summation over $p$ and is a scalar. $\endgroup$
    – Willie Wong
    Commented Oct 31, 2019 at 17:55
  • $\begingroup$ If you think of $\nabla_j \nabla_p \nabla_{\bar{p}} f$ as you wrote, as a component of a rank three object, it makes no sense to even think that can be equal to the component of a rank one object ($s \nabla_j f$)... $\endgroup$
    – Willie Wong
    Commented Oct 31, 2019 at 17:58
  • $\begingroup$ Well originally it is something like $\int_{M}s \nabla_j f \nabla_{\bar j} f w_{f}^n = \int _{M} \nabla_j \nabla_p \nabla_{\bar p} f \nabla_{\bar j} f w_{f}^n$, maybe I intepreted it wrong? $\endgroup$
    – Keith
    Commented Oct 31, 2019 at 20:33

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