Obtaining a Kähler metric from a Kähler form

Question

If I have a $(1,1)$ closed real form $\omega$ on a complex manifold $(M,J)$, it is known that if for all $v$ vector field on $M$ we have $\omega(v,Jv) > 0$ then there exists a Kähler metric whose Kähler form is $\omega.$

I do believe that the reason for this is that we can define such metric as: $h (u,v) := \omega(u,Jv).$ Here is where I stuck. It should be easy to show that $h(Ju,Jv) = h(u,v)$, i.e., $h$ satisfy what is required for being a hermitian metric. The problem is that I am not being able to prove this. All that I get is that $h(Ju,Jv) = h(v,u)$.

In fact:

$$h(Ju,Jv) = \omega(Ju,J^2v) = -\omega(Ju,v) = \omega(v,Ju) = h(v,u).$$

What am I missing?

Thanks

Answer 1

You haven't used the fact that $\omega(Ju, Jv) = \omega(u, v)$; i.e. that $\omega$ is a $(1, 1)$-form with respect to $J$.

Using this equality, we have

$$h(Ju, Jv) = \omega(Ju, J^2v) = \omega(u, Jv) = h(u, v).$$