# When is a Form a Kähler Form?

Let $M$ be a complex manifold, and $\omega$ a closed $2$-form. When is $\omega$ a Kähler form? I mean, when does there exist a Kähler metric for which $\omega$ is the corresponding form.

I would (wildly) guess that necessary and sufficient conditions might be got from the Kähler identities.

-
The obvious condition is that it should be a positive $(1,1)$. That is in local coordinates $$\omega = \frac{\sqrt{-1}}{2}\sum h_{ij} dz\wedge d\bar z_j$$ where is $h_{ij}$ is positive definite. –  Donu Arapura Aug 20 '11 at 22:03
I was hoping for something global –  Jean Delinez Aug 20 '11 at 22:16
Sorry, I don't know of anything like that. –  Donu Arapura Aug 20 '11 at 22:21
Closedness is the global condition. The only other requirement is that it be a positive (1,1) form. Donu is correct. –  Spiro Karigiannis Aug 21 '11 at 0:20

I decided to make my comment into a more detailed answer. When $M$ has an almost complex structure $J$, then one can talk about smooth complex-valued differential forms of type $(p,q)$ in the usual way. A complex valued $2$-form $\omega$ is type $(1,1)$ if and only if it satisfies $\omega(JX,JY) = \omega(X,Y)$ for all smooth vector fields $X$ and $Y$ on $M$. If $\omega$ is a real $2$-form of type $(1,1)$, which means that $\overline \omega = \omega$, and if we define $g(X,Y) = \omega(X, JY)$, then it is easy to show that $g$ is a smooth, symmetric bilinear form on $M$. So it is a Riemannian metric if and only if it is positive definite. This is the definition of a positive $(1,1)$-form (that the associated $g$ is positive definite.)

The triple of data $(J, \omega, g)$, where $J$ is an almost complex structure, $\omega$ is a real positive $(1,1)$-form, and $g$ is the associated Riemannian metric as defined above together define an almost Hermitian manifold. Now the condition for $M$ to be Kaehler is that $M$ be complex ($J$ is integrable) and that $d\omega = 0$. (These two conditions can be packaged together as $\nabla \omega = 0$ or $\nabla J = 0$, where $\nabla$ is the Levi-Civita connection of $g$.) Hence, if one is starting out with a complex manifold $M$, together with a closed real $2$-form, the only additional condition required to ensure that it defines a Kaehler metric is that it be a positive $(1,1)$-form.

-

It's possible to approach the question in a slightly different way if $X$ is compact. Donu and Spiro are of course right in that the condition for a smooth closed $(1,1)$-form $\omega$ to be a Kahler metric is that $\omega$ be positive. This is a pointwise condition on the manifold $X$, which can be difficult to check unless we have quite explicit expressions for $\omega$.

Suppose for a minute that $\omega$ is a Kahler metric. If $Y \subset X$ is a $p$-dimensional closed subspace, then we see that $$\int_Y \omega^p = \int_Y [\omega]^p > 0,$$ where $[\omega]$ is the cohomology class that $\omega$ defines. This holds because $\omega$ induces a Kahler metric on (the smooth part of) $Y$ and $\omega^p$ is $p!$ times the volume form of the induced metric. (I'm skipping over some subtle things that involve integration over singular subspaces, but one can make sense of all of this.)

Demailly and Paun proved that the converse holds. That is, if we have a cohomology class $[\omega]$ such that its integral over any subspace of $X$ (including $X$ itself) is positive, then $[\omega]$ contains a Kahler metric. (This is a big generalization of the Nakai-Moishezon criterion in algebraic geometry.)

This doesn't help if what you're interested in is knowing that the precise form you have is Kahler, because given a Kahler form we can always take a nonconstant real smooth function $f$ and add $i\partial\bar\partial f$ to the form; this doesn't change the cohomology class and adding (or subtracting) giant such multiples will break the positivity of the resulting form. But for some questions, all that matters is the cohomology class of the Kahler form, and for those you can check positivity of the class by integration.

-