Kähler form ($\omega (u,v)=g(Ju,v)$)

For a Kähler metric $g$ with complex structure $J$, the associated real two-form is $\omega (u,v)=g(Ju,v)$. It has type $(1,1)$, is positive in the sense $\omega (v,Jv)=g(v,v)>0$ for $v\ne 0$, and is closed. Conversely a real closed positive $(1,1)$ form determines a Kähler metric. A local geometric Kähler potential writes it as $i\partial \bar{\partial }\phi$.

On a nonempty compact Kähler manifold of complex dimension $n$, the Kähler form is closed and ${\int }_X{\omega }^n>0$. If ${\omega }^k=d\eta$ for $1\le k\le n$, then ${\omega }^n=d(\eta \wedge {\omega }^{n-k})$; Stokes theorem contradicts the positive integral. Thus the de Rham cohomology is nonzero in every even degree through $2n$.

For a closed real $(1,1)$ form, the Poincaré lemma gives a real one-form $\eta$ with $d\eta =\omega$. Its $(0,1)$ part is $\bar{\partial }$-closed, so the Dolbeault-Poincaré lemma gives ${\eta }^{0,1}=\bar{\partial }\psi$. Reality gives ${\eta }^{1,0}=\partial \bar{\psi }$, hence $\omega =i\partial \bar{\partial }(2\mathrm{Im}\psi )$. Positivity is a separate requirement for this form to define a metric.

A smooth real rotation-invariant potential is constant on every positive-radius circle, so its logarithmic-radius function $u$ is smooth on $\mathbb{R}$. Off the origin, $f_{z\bar{z}}=e^{-t}{u}^{\prime \prime }(t)$. Since $f$ is smooth on the entire plane, this coefficient extends smoothly through the origin. A finite Taylor expansion, rather than a convergent series assumption, gives $f(z)=f(0)+b|z{|}^2+O(|z{|}^4)$.

For a potential already smooth on all of $\mathbb{C}$, the associated form is a Kähler form exactly when the displayed two positivity conditions hold. The first gives a positive metric coefficient off the origin; the second is positivity of $f_{z\bar{z}}(0)$. The associated real metric is $2f_{z\bar{z}}(d{x}^2+d{y}^2)$. Positivity away from the origin alone permits a degenerate metric at the origin.