# Relation between Kähler potential and Hermitian metric

Let $(M,\omega)$ be a Kähler 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 Kähler 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 $h$

• It is not so clear what $h$ is. If $h$ is a Hermitian form, it should be a matrix locally, so what is $\log h$? – user99914 Apr 25 '14 at 0:31
• Do you mean $h$ to be the Hermitian metric on the canonical line bundle of $M$? BTW, if $M$ is compact, your statement follows from the maximum principal. – Vicky Cheung Apr 25 '14 at 3:28

I agree with the comments that you must be working with a line bundle in order to make any sense out of that expression. So let's suppose that it the case (and hence $\omega$ is the curvature of the associated Chern connection). Also, I think that you have to assume, for the following argument to hold, that $M$ is compact and connected. I know how to prove this if $f$ is a real function with $\omega=i\partial \bar{\partial}f$, which you can always find locally (maybe there is a similar argument for the case in hand).
So suppose that $\phi$ is a real local function on $M$ such that $\partial\bar{\partial}\phi=0$. Hence $\partial\phi$ is a closed and holomorphic (1,0)-form (use $d\partial=\bar{\partial}\partial=-\partial\bar{\partial}$), and then the $\partial\bar{\partial}$-lemma implies that $\partial \phi=\bar{\partial}\psi$ for some $0$-form. But this forms have different type, and hence must be both equal zero. Hence $\partial \phi=0$ and also $\bar{\partial}\phi=0$ (since $\phi$ is real), and therefore you get $d\phi = 0$, i.e $\phi$ is constant ($M$ connected).
In your case, apply to $\phi=logf-logh$ (assuming $f,h$ real).