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Let $(M,\omega)$ be a Kaehler manifold and $h$ be its Hermitian form, then in local sense we can write $$\omega=\partial\bar\partial\log h$$ and also if $f$ be the kaehler potential then we can write $$\omega=\partial\bar\partial\log f$$. So, my question is can say $f$ is equal to $h$ up to additional constant? if we have $f$ then how can we find $h$

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What's the Hermition (sic?) form of a Kahler metric? – Gunnar Þór Magnússon Apr 24 '14 at 19:21
Gunnar , I mean was hermitian form – Hassan Jolany Apr 24 '14 at 20:19
But what is the "Hermitian form" of a Kahler manifold? When I google it, it is defined to be the $(1,1)$-form $\omega$. – Deane Yang Apr 24 '14 at 23:03
Here you can see in equation 1.11. $h$ is hermitian structure – Hassan Jolany Apr 25 '14 at 13:47
I guess $h$ here is $g=h+i\omega$, – Hassan Jolany Apr 25 '14 at 13:55
up vote 2 down vote accepted

The function $\psi:=\log(fh^{-1})$ satisfies $\partial\bar\partial \psi=0$, because $ \partial\bar\partial\log f=\partial\bar\partial\log h =\omega$. Such functions are called pluriharmonic. Locally a pluriharmonic function is a real part of a holomorphic function, by Poincare-Dolbeault-Grothendieck lemma. This fact is true globally, because any real-valued holomorphic function vanishes, and therefore the local holomorphic functions can be glued together. Then, $h$ is unique (up to a constant multiplier) if and only if the manifold has no non-constant global holomorphic functions.

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The question here is if $h(s_1,s_2)$ (with sections $s_i$)be hermitian structure then how can we find $h$ by using kahler potential $f$? – Hassan Jolany Apr 25 '14 at 14:31
If $h$ is the Hermitian form, what is the meaning of $\log h$? – Paul Reynolds Apr 26 '14 at 17:41
It makes no sense, of course; I assumed that the poster confused a metric with its potential – Misha Verbitsky Apr 27 '14 at 16:14
See in equation 1.11, IN FACT , $h$ is hermitian metric corresponding to hermitian form, so $\log h$ here must means $\log \hat h$ which $\hat h$ is hermitian metric corresponding to hermitian form in local trivialization – Hassan Jolany Apr 27 '14 at 18:24
I don't have access to this article; but anyway, $dd^c$ cannot be applied to metric, or to its logarithm. – Misha Verbitsky Apr 27 '14 at 23:38

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