PomeA8F518936B0AIn the conformal class of an LCK metric, the Gauduchon representative is the one whose Lee form is harmonic. A closed $(1,1)$-form built from the Lee form pairs positively with this Gauduchon metric when the structure is strictly LCK, even though the form is de Rham exact. It therefore defines a nonzero Bott–Chern class, which obstructs Kähler modifications.
Set
so that $\lambda ^{\mathrm c}=\mathrm d^{\mathrm c}f$ wherever $\lambda =\mathrm{d}f$. The closed real $(1,1)$-form
29D45DC05342
23.2.6
connects the Lee class with Bott–Chern cohomology. It is de Rham exact, but it need not be $\mathrm{dd}^{\mathrm{c}}$-exact.
Proposition 23.2.7.11EBB7AD1165 An LCK metric is Gauduchon if and only if its Lee form is co-closed for that metric. Consequently, the Gauduchon representative in its conformal class has harmonic Lee form. Moreover, for every LCK metric,
35851308DD2B
23.2.8
If $\lambda \not\equiv 0$, the right side is strictly positive.
Proof. On a chart where $\omega $ is conformally Kähler,
Let $*$ be the real Hodge star of the Riemannian metric associated with $\omega $, and let $\mathrm d^*$ be the formal adjoint of the exterior derivative. By the pointwise Hodge-star identities for Hermitian metrics, $\lambda ^{\mathrm c}\wedge \omega ^{n-1}$ is a fixed nonzero constant multiple of $*\lambda $. Consequently, the Gauduchon condition $\mathrm{dd}^{\mathrm{c}}(\omega ^{n-1})=0$ is equivalent to $\mathrm{d}*\lambda =0$, that is, to $\mathrm{d}^*\lambda =0$. Since $\mathrm{d}\lambda =0$, co-closedness is equivalent to harmonicity. The Gauduchon representative exists by theorem 11.4.2, and conformal changes preserve the LCK condition.
Stokes' theorem and $\mathrm{d}(\omega ^{n-1})=(n-1)\lambda \wedge \omega ^{n-1}$ give
For a real one-form, $\lambda \wedge \lambda ^{\mathrm c} =\mathrm i\lambda ^{1,0}\wedge \lambda ^{0,1}/(2\pi )$ is semipositive, and its trace with respect to $\omega $ is positive wherever $\lambda \ne 0$. This proves the last assertion. $\square$
The same construction applies to every closed real one-form and defines a linear map
624B1D55AC6E
23.2.9
which we call the Lee map. It is well defined, since replacing $\lambda $ by $\lambda +\mathrm{d}a$ replaces $\kappa _\lambda $ by $\kappa _\lambda -\mathrm{dd}^{\mathrm{c}}a$. Consequently, the pairing of $\{ \kappa _\lambda \} $ with any Gauduchon form $\gamma ^{n-1}$ is well defined. For a strictly LCK structure, the Gauduchon representative in its conformal class gives a strictly positive pairing by (23.2.8). In particular $\{ \kappa _\lambda \} \ne 0$, although its image in de Rham cohomology is zero.
Corollary 23.2.10.F6950D5704C9 If $X$ carries a strictly LCK metric, then no class in $H^{1,1}_{\mathrm{BC}}(X,\mathbb R)$ is big. In particular, $X$ does not belong to Fujiki's class $\mathcal C$.
Proof. If there were a big class, the modification criterion recalled in section 10.10 would give a modification $\pi :Y\to X$ with $Y$ compact Kähler. The form $\pi ^*\kappa _\lambda $ is closed and de Rham exact, so the $\partial \bar\partial $-lemma on $Y$ gives a smooth real function $h$ on $Y$ with $\pi ^*\kappa _\lambda =\mathrm{dd}^{\mathrm{c}}h$. Since $\pi $ has degree one, pushing forward gives the current identity $\kappa _\lambda =\mathrm{dd}^{\mathrm{c}}(\pi _*h)$. By Stokes' theorem, the pairing of this current with a Gauduchon form is zero, which contradicts (23.2.8) for the Gauduchon representative of the LCK structure. $\square$
For the Hopf manifold considered below, $\kappa _\lambda =\alpha $ is a smooth semipositive form, so its class is nef. Its non-nef locus is empty, whereas its non-Kähler locus is all of $X$, by the convention for classes that are not big in section 10.10. These statements concern the closed class $\{ \alpha \} $; the Monge–Ampère equations on the Hopf manifold use instead the strictly positive, nonclosed form $\omega $.
Proposition 23.2.11.26993B1C8C74 Suppose that $X$ carries a strictly LCK metric with Lee form $\lambda $ and that $\{ \kappa _\lambda \} $ is pseudoeffective. Then $X$ admits no balanced metric.
Proof. Let $T=\kappa _\lambda +\mathrm{dd}^{\mathrm{c}}u\geq 0$ be a closed positive current representing the class. It is nonzero, since the class is nonzero, and it is de Rham exact. If a Hermitian metric $b$ were balanced, then $\mathrm{d}(b^{n-1})=0$, and Stokes' theorem for currents would give
This is impossible, since the mass of a nonzero positive $(1,1)$-current against a smooth strictly positive $(n-1,n-1)$-form is strictly positive. $\square$
The balanced obstruction motivates the following conjecture.
Conjecture 23.2.12.4EA8A1D43A8C If $X$ carries a strictly LCK metric with Lee form $\lambda $, then the Bott–Chern class $\{ \kappa _\lambda \} $ is pseudoeffective.
The positivity in (23.2.8) tests the class only against the Gauduchon representative of the LCK metric. By theorem 11.6.1, pseudoeffectivity requires a nonnegative pairing with every Gauduchon metric.
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.