Does Ricci form has a globally smooth potential? (On a Kahler manifold)

Question

Let $\left(M,\omega\right)$ be a Kahler manifold. We know that the Ricci form $Ric\left(\omega\right)$ is locally by $$ Ric\left(\omega\right)=dd^{c}\log\left(\det g_{j\overline{k}}\right). $$ Is it true that there is a smooth function $h\in C^{\infty}\left(M\right)$ such that $Ric\left(\omega\right)=dd^{c}h?$ That is, $Ric\left(\omega\right)$ has a globally smooth potential $h.$

Answer 1

For the compact case, the existence of global function $h$ implies that $$[Ric(\omega)] = [d(d^c h)] =0$$ in $H^2(M, \mathbb R)$. But this class is the first Chern class $c_1(M)$ of $M$, so the global function $h$ does not exists whenever $c_1(M)\neq 0$.

On the other hand, if $c_1(M) = 0$, then $Ric(\omega) = d\alpha$ for some one form $\alpha$. Since $Ric(\omega)$ is a $1, 1$-form and is $d$-exact, the $\partial \bar\partial$-lemma implies that $Ric(\omega)$ is $\partial\bar\partial$-exact. Thus there is a function $h$ so that

$$Ric(\omega) = \frac{i}{2\pi} \partial \bar\partial h = d d^c h.$$