Q:

No global Kähler potential on a compact Kähler manifold

I am required to show that there cannot exist a global Kähler potential $u$ on a compact Kähler manifold, say $(M, J, \Omega)$. I know that $M$ is Kähler if and only if $\mathrm{d} \Omega = 0$ and $J$ is a complex structure, and that locally it is always possible to write $$\Omega = i \partial \overline{\partial}u.$$ How does compactness obstruct $u$ being global?

A:

If $\Omega = i\partial\bar{\partial}u$, then by the $\partial\bar{\partial}$-lemma, $\Omega = dv$ for some one-form $v$. In fact, as Arctic Char points out in the comments below, we don't even need the $\partial\bar{\partial}$-lemma: $$\Omega = i\partial\bar{\partial}u = \partial(i\bar{\partial}u) = d(i\bar{\partial}u) = dv$$ where $v = i\bar{\partial}u$. Then
$$\operatorname{Vol}(M) = \frac{1}{n!}\int_M\Omega^n = \frac{1}{n!}\int_M\Omega\wedge\Omega^{n-1} = \frac{1}{n!}\int_Mdv\wedge\Omega^{n-1} = \frac{1}{n!}\int_Md(v\wedge\Omega^{n-1}) = 0$$
by Stokes' Theorem, but this is a contradiction as $\operatorname{Vol}(M) > 0$. 
In general, if $\Omega$ is a Kähler form on a compact manifold, or even just a symplectic form, $\Omega$ is not exact; in particular $[\Omega] \in H^2_{\text{dR}}(M)$ is non-zero.

Note, if $M$ is non-compact, there are obstructions to the existence of a global Kähler potential. A sufficient condition is that $H^1(M, \mathcal{O}) = 0$ and $[\Omega] \in H^2(M; \mathbb{R})$ is zero. See this MathOverflow question for more details.

