23.2.1 The Lee form and the Kähler coveringE0F12265EE84

A Hermitian form is locally conformally Kähler when its exterior derivative is controlled by a closed one-form.

Fix an LCK metric $\omega $ with Lee form $\lambda $. On a compact manifold a nonzero parallel one-form cannot be exact: the differential of a function vanishes at a maximum, whereas a parallel form has constant length. Thus a nonzero parallel Lee form is automatically nowhere zero and not exact.

The Lee form is determined by $\omega $, because the map $a\mapsto a\wedge \omega $ is injective on one-forms: after wedging with $\omega ^{n-2}$, this follows from the bijectivity of the Lefschetz map $a\mapsto a\wedge \omega ^{n-1}$ on one-forms.

Locally write $\lambda =\mathrm{d}f$. Then

so $\Omega $ is a Kähler form. Choose a connected regular covering $\pi :\widetilde X\to X$ on which $\pi ^*\lambda =\mathrm{d}F$, for example the universal covering. The quotient of $\widetilde X$ by its deck group is $X$. The same construction gives a Kähler form

For a deck transformation $\sigma $, the difference $\sigma ^*F-F=c_\sigma $ is constant, and

Thus $\chi $ is a homomorphism from the deck group to the multiplicative group of positive real numbers, which records how the Kähler metric is rescaled. Forms on $\widetilde X$ transforming by $\chi $ may equivalently be viewed as forms on $X$ with values in the flat real line bundle defined by $\chi $, the weight bundle. In this sense an LCK form is closed as a form with values in the weight bundle, OV24LCK].

Let $g$ be the Riemannian metric of $\omega $, let $J$ be the complex structure, and let $\lambda ^\sharp $ be the vector field dual to $\lambda $ with respect to $g$. On a chart where $\lambda =\mathrm df$, the metric $\widetilde g=\mathrm{e}^{-f}g$ is Kähler. For real vector fields $U,V$, the Levi–Civita connections $\nabla $ of $g$ and $\widetilde\nabla $ of $\widetilde g$ are related by

Substituting $JV$ for $V$ and using $\widetilde\nabla J=0$ gives

Both sides are global tensors, so this identity holds on all of $X$.

For $a\in C^\infty (X,\mathbb R)$, the conformal change $\omega '=\mathrm{e}^ a\omega $ replaces the Lee form by $\lambda +\mathrm{d}a$. In particular, the de Rham class $[\lambda ]$ depends only on the conformal class of $\omega $. If $\lambda $ is exact, the construction above gives a Kähler form on $X$ itself.