23.2 Locally conformally Kähler geometry9B4394B392ED

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. $\square$

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. $\square$

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. $\square$

Proof. Since $\mathrm d\eta _ t=\lambda \wedge \omega $ and $\mathrm d\lambda ^{\mathrm c}=-\kappa _\lambda $, Stokes' theorem gives

Differentiating the wedge polynomial proves the formula. The integrand on the right is positive wherever $\lambda \ne 0$, including when $n=2$. Its integral is therefore strictly positive. Finally, $\kappa _\lambda $ is closed and de Rham exact, so

This removes the coefficient of degree $n$. $\square$

Proof. Let $r$ be the constant least relative eigenvalue. If $r<0$, put $t_0=-1/r$. Then $\eta _ t>0$ for $0\leq t<t_0$, whereas $\eta _{t_0}$ is semipositive with a kernel at every point. Thus $V(t_0)=0$. The preceding proposition and polynomial continuity instead give $V(t_0)\geq V(0)>0$. This proves $r\geq 0$ and semipositivity.

The nonnegative smooth form $\kappa _\lambda ^ n$ has integral zero by the same proposition, hence vanishes everywhere. Thus the least eigenvalue is zero at every point. Applying the Lee degree (23.2.8) to a conformal Gauduchon representative, as above, proves that the class is nonzero. The form $\kappa _\lambda $ itself represents it positively.

Constant spectrum gives constant rank. The positive pairing (23.2.8) excludes rank zero, and the vanishing top power excludes rank $n$. On the stated compact quotient, the relative matrix is constant in a left-invariant frame, which proves the last assertion. $\square$

Proof. Fix $0<\epsilon <1$ and put

The bounds $(1-\epsilon )\rho \leq \Phi \leq (1+\epsilon )\rho $ give positivity. Since $\mathrm{dd}^{\mathrm{c}}\Phi =\mathrm{dd}^{\mathrm{c}}\rho $ and both $\Phi $ and $\mathrm{dd}^{\mathrm{c}}\rho $ scale by $1/4$ under $z\mapsto z/2$, the metric and Lee form descend and satisfy the LCK equation. Polar coordinates identify the quotient with $S^{2n-1}\times (\mathbb R/(\log 2)\mathbb Z)$, so it is compact and connected. The Lee period from $z$ to $z/2$ is $2\log 2$, which proves strictness.

Here $\kappa _\lambda =\mathrm{dd}^{\mathrm{c}}\log \Phi $. With $F_ i=\bar z_ i+\epsilon z_1\delta _{i1}$, its complex Hessian is

Relative to $\omega $, the eigenvalues are $1$ with multiplicity $n-1$ and $h=1-|F|^2/\Phi $. Writing $z_1=x+\mathrm i y$ and $B=\sum _{j>1}|z_ j|^2$, one finds

At $(1,0,\ldots ,0)$ the spectrum is $-\epsilon ,1,\ldots ,1$, hence is indefinite. The bounded smooth function $u=\log \rho -\log \Phi $ is deck invariant, and

by the Cauchy–Schwarz inequality for the Hessian of $\log \rho $. The correction also gives the standard Hopf conformal representative: $\mathrm e^{-u}\omega =\mathrm{dd}^{\mathrm{c}}\rho /\rho $.

The first boundary occurs at $t_0=1/\epsilon $. At that value,

It vanishes only on the real $z_1$ rays with the remaining coordinates zero. The other relative eigenvalues of $\omega +t_0\kappa _\lambda $ equal $1+1/\epsilon >0$. The boundary form is therefore semipositive and its top power is positive on a nonempty open set, proving positive total volume. $\square$

Proof. Put $q=|z_1|^2$ and, for $0<\epsilon <1$, define

For $a=q/\rho \in [0,1]$, direct differentiation gives

The first two terms are positive semidefinite and $a^2\leq 1$, so $\mathrm{dd}^{\mathrm{c}}\Psi \geq (1-\epsilon )\mathrm{dd}^{\mathrm{c}}\rho >0$. The potential is smooth on the punctured space and scales by $1/4$ under the deck map. Hence $\omega _\epsilon $ descends, satisfies the LCK equation and has Lee period $2\log 2$.

Its Lee curvature is $\mathrm{dd}^{\mathrm{c}}\log \rho $, which is semipositive of rank $n-1$. The least relative eigenvalue is consequently zero everywhere. At $(1,0,\ldots ,0)$, the Hessian of $q^2/\rho $ is $\operatorname {diag}(1,-1,\ldots ,-1)$, giving the spectrum

At $(0,1,0,\ldots ,0)$ the term $q^2/\rho $ vanishes to fourth order, so its Hessian is zero and the spectrum is $0,1,\ldots ,1$. These spectra differ for every $\epsilon >0$. $\square$

Proof. The first identity follows from $\mathrm{d}\mathrm{dd}^{\mathrm{c}}u=0$ and $\mathrm{d}\omega =\lambda \wedge \omega $. The second follows from $\mathrm{d}_\lambda \omega =0$ and (23.2.19). $\square$

Proof. Using $\alpha =\mathrm{dd}^{\mathrm{c}}s$ and $\beta =\mathrm{d}s\wedge \mathrm d^{\mathrm c}s$, direct differentiation gives

Although $s$ is not a function on the quotient, $\mathrm{d}s$ descends to it. Its integral along the loop induced by $z\mapsto \mathrm{e}^{-1}z$ is $-2$, so the Lee form is not exact.

Let $g_{S^5}$ be the round metric on the unit sphere. Up to a constant factor, the Riemannian metric associated with $\omega $ on the cover is

The one-form $\mathrm{d}t$ is parallel for this product metric, which proves the Vaisman condition. Finally, $\kappa _\lambda =\mathrm{dd}^{\mathrm{c}}s=\alpha $ and $\lambda \wedge \lambda ^{\mathrm c}=\beta $, so $D_\lambda 1=\omega $ by (23.2.22). The unbounded masses are given by theorem 23.1.1. $\square$

Proof. The function $v$ is constant on each local plaque. On every transverse coordinate chart it satisfies

This is a linear elliptic equation without zeroth-order term. Let $c=\max _ Mv$. If $v=c$ at some point, the strong maximum principle on a connected transverse chart shows that $v=c$ throughout that chart, and since $v$ is basic, $v=c$ on the corresponding foliation chart. Hence $\{ v=c\} $ is nonempty, open and closed, and connectedness of $M$ proves the assertion. $\square$

of theorem 23.2.29. Replacing $\omega $ by $\tau \omega $ and $\mu $ by $\tau ^ n\mu $ does not change the Lee form or the assertion, and it makes the new constant $\tau $ equal to one. We may therefore assume $\tau =1$, so that $\omega =\eta +\lambda \wedge \lambda ^{\mathrm c}$. Write $\mu =f\omega ^ n$; since $\mu $ and $\omega ^ n$ are invariant under both flows, the positive smooth function $f$ is basic.

Existence. By theorem 14.2.1, there are a unique smooth function $u$ and a unique constant $b$ such that

Pullback by the flow of $A$ or of $B$ preserves the equation and the normalization, so uniqueness shows that $u$ is basic. Hence $\eta _ u=\eta +\mathrm{dd}^{\mathrm{c}}u$ is a transverse Kähler form, and

Integrating over $X$ and using (23.2.31) and $\int _ X\mu =\int _ X\omega ^ n$ gives $b=0$.

Set $\lambda _ u=\lambda -\mathrm{d}u$ and

Here $\eta _ u=\kappa _{\lambda _ u}$ has kernel $\mathcal F$, while $\lambda _ u(A)=1$ and $\lambda _ u(B)=0$; together these facts give $\omega _ u'>0$. Direct differentiation gives $\mathrm{d}\omega _ u'=\lambda _ u\wedge \omega _ u'$. Since $u$ is basic, $A$ is a holomorphic Killing field for the Riemannian metric $g_ u'$ of $\omega _ u'$, and (23.2.33) gives

Since $\lambda _ u$ is closed, $\nabla ^{g_ u'}A$ is symmetric, whereas the Killing equation makes it skew-symmetric. Thus $A$ and $\lambda _ u$ are parallel, so $\omega _ u'$ is Vaisman with Lee class $[\lambda ]$.

Finally, $\eta _ u^ m$ has top transverse degree, and every term of $\lambda _ u\wedge \lambda _ u^{\mathrm c}-\lambda \wedge \lambda ^{\mathrm c}$ contains a horizontal one-form. Therefore

Uniqueness. Let $\widehat\omega $ be another Vaisman solution, with Lee form $\widehat\lambda =\lambda +\mathrm{d}v$, and denote by $\widehat s,\widehat\tau ,\widehat A,\widehat B,\widehat\eta $ the corresponding objects for $\widehat\omega $. The form $\eta +\widehat\eta $ is de Rham exact and semipositive. Its top power is nonnegative and has zero integral, so it vanishes everywhere, and $\eta +\widehat\eta $ has a nonzero kernel at every point. The kernel of a sum of two semipositive Hermitian forms is the intersection of their kernels. Since both kernels are complex lines, they coincide, and both metrics define the same foliation $\mathcal F$.

The nowhere vanishing holomorphic vector field $A-\mathrm iB$ trivializes the holomorphic tangent bundle of $\mathcal F$. The analogous field $\widehat A-\mathrm i\widehat B$ is a holomorphic multiple of it, and the multiple is constant because $X$ is compact. Write $\widehat A=aA+bB$ with $a,b\in \mathbb R$. Since $\widehat\lambda (\widehat A)=1$ and $\widehat\lambda (\widehat B)=0$,

Both fields are complete and $v$ is bounded, so $v$ cannot grow linearly along their flows; hence these constants vanish, and $a=1$, $b=0$. Thus $\widehat A=A$, $\widehat B=B$, and $v$ is basic.

From (23.2.30) we obtain the contraction formulas

Since the two volume forms are equal, $\widehat\eta ^ m=\widehat\tau ^ n\eta _ u^ m$. Since $\widehat\eta =\eta -\mathrm{dd}^{\mathrm{c}}v$ and $v$ is basic, integrating against $\lambda \wedge \lambda ^{\mathrm c}$ and applying (23.2.31) twice gives $\widehat\tau ^ n=1$. Thus $\widehat\tau =1$ and $\widehat\eta ^ m=\eta _ u^ m$. Since $\widehat\eta -\eta _ u=-\mathrm{dd}^{\mathrm{c}}(v+u)$, proposition 23.2.32 shows that $v+u$ is constant. It follows that $\widehat\lambda =\lambda _ u$, $\widehat\eta =\eta _ u$ and $\widehat\omega =\omega _ u'$. $\square$

Proof. The differential identities are

They give $\mathrm d\omega =(C'-D)\mathrm dr\wedge \alpha =\lambda \wedge \omega $. The closed form $\lambda $ is the Lee form, since wedging one-forms with $\omega $ is injective. The integral of $C'/C$ over a period is zero, which proves the formula for $L$; since $L\neq 0$, the Lee form is not exact. Two radial Lee forms have the same de Rham class exactly when their periods agree: a difference with zero period has a periodic primitive, and the radial loop detects the converse.

For a smooth function $G$ of $r$, the chain rule gives $\mathrm{dd}^{\mathrm{c}}G=G'\alpha +G''\beta $. Also $\mathrm d^{\mathrm c}(G\beta )=-G\alpha \wedge \mathrm d^{\mathrm c}r$, so

Expanding the powers of $\omega $ therefore gives

and

Since $(C^ m)''-m(C^{m-1}D)'=m\bigl (C^{m-1}(C'-D)\bigr )'$, this vanishes exactly when $C^{m-1}(C'-D)$ is constant. Multiplying the identity $C^{m-1}(C'-D)=k$ by $C$ gives (23.2.38). Since $C^ mC'$ has zero integral over a period, integrating (23.2.38) gives the first identity in (23.2.39), which forces $k<0$. The formulas for $l$ and $L$ follow from the formula for the Lee form. $\square$

Proof. Suppose $k_2>k_1$, and put $R=C_2-C_1$. At a zero of $R$, the equation gives $R'=(k_2-k_1)C_1^{1-m}>0$. Such zeros are isolated, so there are finitely many on the circle. A periodic differentiable function cannot cross zero only upwards, since returning around the circle would require a downward crossing. Hence $R$ has no zero and a constant strict sign. The identity $\int _0^2C_ i\, \mathrm dr=-\int _0^2F\, \mathrm dr/k_ i$ shows that this sign is positive, and the formula for $L(C)$ in (23.2.39) then gives $L(C_2)>L(C_1)$.

For equal $k$, uniqueness for the smooth scalar equation $C'=kC^{1-m}+FC^{-m}$ on $C>0$ shows that two solutions which meet coincide. Two solutions which never meet are strictly ordered, which contradicts the equality of their integrals. This proves uniqueness for fixed $k$. $\square$

Proof. Fix $k<0$, put $c=-k$, and let $f_0=\min F>0$ and $f_1=\max F$. The constant functions $f_0/c$ and $f_1/c$ are a subsolution and a supersolution, respectively, of $C'=C^{-m}(F-cC)$. By comparison, the forward flow of this equation preserves the compact interval $I_ k=[f_0/c,f_1/c]$ of positive numbers, and existence and uniqueness for smooth ordinary differential equations give a continuous time-two map $I_ k\to I_ k$. Its difference from the identity is nonnegative at the left endpoint and nonpositive at the right one, so the intermediate value theorem gives a fixed point; if $f_0=f_1$, the constant $C=f_0/c$ is the solution. Since the equation is periodic, the solution through a fixed point extends to a smooth positive periodic solution. It is unique by lemma 23.2.40 and satisfies

By lemma 23.2.40, the period $L(k)=-\int _0^2F/C_ k^{m+1}\, \mathrm dr$ is strictly increasing in $k$. It is also continuous. Indeed, for $k$ in a compact neighborhood of any $k_0<0$, (23.2.42) gives uniform positive lower and finite upper bounds for $C_ k$, and the equation then bounds all derivatives. Any sequence $k\to k_0$ therefore has smoothly convergent subsequences, and the equation and uniqueness identify every limit with $C_{k_0}$. Thus $C_ k\to C_{k_0}$ uniformly, which gives continuity of the integral.

The same bounds imply $L(k)\to 0$ as $k\uparrow 0$ and $L(k)\to -\infty $ as $k\to -\infty $. Hence exactly one $k$ has period $L_0$. For this $k$, set $D=F/C_ k^ m$. The identities of lemma 23.2.37 show that $C_ k\alpha +D\beta $ has the required properties. Conversely, every metric of the form (23.2.36) with these properties gives a positive periodic solution of the same equation, so the ordering of the periods also proves uniqueness. $\square$

Proof. Write $l=kC^{-m}<0$ and $f(r)=\int _0^ rl(s)\, \mathrm ds$ for real $r$. Then $f(r+2)=f(r)+L_0$. The solution of $U'-lU=C$ given by

is positive on $[0,2]$ and satisfies $U(2)=U(0)$. Since the equation is periodic, all derivatives also match at the endpoints, so $U$ defines a positive smooth function on $H_ n$. By (23.2.19), its twisted Hessian is

This proves that the twisted class is zero. The identity $\mathrm e^{-a}D_{\lambda _2}U_2=D_{\lambda _1}(\mathrm e^{-a}U_2)$ of subsection 23.2.25 gives the formula for the difference. For two metrics of the form (23.2.36), equal volume forms give the same function $F$, and equal Lee classes give the same period, so theorem 23.2.41 applies. This uniqueness concerns the metric, not the potentials representing it. $\square$

Proof. The radial complex line is orthogonal to $\ker \mathrm dr\cap \ker \mathrm d^{\mathrm c}r$. On a radial complex line, use a logarithmic coordinate $w=x+\mathrm i y$, so that $r=2x+\text{constant}$ and $\beta =\pi ^{-1}\mathrm dx\wedge \mathrm dy$. The Riemannian metric of $\omega $ restricts there to $D\pi ^{-1}(\mathrm dx^2+\mathrm dy^2)$, whence

A parallel Lee form has constant norm. Since $k\ne 0$, the Vaisman condition would therefore imply $D=Q C^{-2m}$ for a constant $Q>0$. At a maximum and at a minimum of $C$, where $C'=0$, the Gauduchon identity would then give $kC^{m+1}+Q=0$. This equation has only one positive root, so the maximum and the minimum of $C$ coincide. Thus $C$, $D$ and $F=C^ mD$ would all be constant, a contradiction. $\square$

Proof. Since $1/2<C<3/2$ and $m\geq 1$, we have $D\geq (3/2)^{1-m}-\pi \varepsilon >\frac34(3/2)^{1-m}>0$. The Gauduchon identity holds with $k=-1$, giving the Lee form and the positive density $F=C^ mD$. Its values $F(0)=1+\varepsilon $ and $F(1)=1-\varepsilon $ differ, so proposition 23.2.44 applies. The integral of $C^ mC'$ over a period is zero and the average of $C$ is one, so integration along the fibers of $q$ gives $\int _{H_ n}F\, \mathrm dV=1$. $\square$

Proof. We keep the divergence terms in the Ricci tensor and compare the Bianchi and Bochner identities before imposing the Gauduchon condition. All norms, traces and contractions use $g$. Put

Closedness of $\lambda $ makes $S$ symmetric, and $\operatorname {tr}_ gS=-q$. The tensors $S^+$ and $S^-=S-S^+$ are orthogonal. Write $J\lambda =-\lambda \circ J$, so that $\lambda ^{\mathrm c}=J\lambda /(4\pi )$.

Step 1: the Ricci tensor. We first show that

Locally write $\lambda =\mathrm df$ and put $\widetilde g=\mathrm e^{-f}g$. The real conformal Ricci formula in dimension $2n$ is

The local metric $\widetilde g$ is Kähler. Since the Chern–Ricci form of $g$ vanishes, the Ricci form of $\widetilde g$ is $n\mathrm i\partial \bar\partial f=(n/2)\mathrm d(J\lambda )$. Applying (23.2.4) to $(\nabla _ UJ\lambda )(V)=-S(U,JV)-\lambda ((\nabla _ UJ)V)$ and antisymmetrizing gives

Evaluating the Kähler Ricci form on $(U,JV)$ and comparing with the conformal formula proves (23.2.48). Taking its trace gives the scalar curvature

Step 2: the integral identity. Write $\| \cdot \| $ for the $L^2(g)$ norm and set

Since $S(A,A)=A(u)/2$ and $\operatorname {div}A=-q$, integration by parts gives

Pair the contracted Bianchi identity $\delta \operatorname {Ric}_ g+\mathrm ds/2=0$ with $\lambda $ and integrate. Here $\delta $ is minus covariant contraction, the formal adjoint of the symmetric covariant derivative. Thus

Substitution of equation 23.2.48, equation 23.2.49 gives

For the closed one-form $\lambda $, the integrated Bochner formula reads

The quartic terms in (23.2.48) cancel on $(A,A)$, and $J\lambda (A)=0$, so

Using (23.2.49), we obtain

Eliminating $E_ A,E_ B$ between the two identities gives $n(\| S^+\| ^2-\| S\| ^2)+n\| q\| ^2/2=0$. Orthogonality proves the asserted identity for $S^-$.

Step 3: rigidity. By proposition 23.2.7, the Gauduchon condition is equivalent to $q=0$. In that case the integral identity forces $S=S^+$, hence $S(B,B)=S(A,A)$. Equation (23.2.49) gives $E_ A=E_ B=0$, and (23.2.51) implies $S=0$. Thus $\omega $ is Vaisman.

Equation (23.2.4) gives $\nabla _ AJ=0$. If $T$ is the symmetric endomorphism defined by $g(TU,V)=S(U,V)$, then

Consequently $A$ is real holomorphic exactly when $S=S^+$. The integral identity then gives $q=0$, hence $S=0$. Conversely, a parallel Lee form gives $q=0$ and $\mathcal L_ AJ=0$. Strictness ensures that the parallel Lee form is nonzero. $\square$

Proof. We construct the quotient and the two conformal metrics explicitly. On $G=\mathbb R^4$ use the multiplication

The integer subgroup $\Gamma =\mathbb Z^4$ acts on the left by

Reducing $x,y,t$ and then $z$ to unit intervals gives a compact fundamental set, so $X=\Gamma \backslash G$ is connected and compact. The forms

descend to $X$, with $\mathrm de^3=-e^1\wedge e^2$ and the other derivatives zero. Declare $\phi ^1=e^1+\mathrm i e^2$ and $\phi ^2=e^3+\mathrm i e^4$ to be of type $(1,0)$. Since

there is no $(0,2)$ component and the complex structure is integrable. The form $\Sigma =\phi ^1\wedge \phi ^2$ is closed and nowhere vanishing, hence is a holomorphic trivialization of the canonical bundle. In particular $c_1^{\mathrm{BC}}(X)=0$. Projection to $x+\mathrm i y$ exhibits this quotient as the Kodaira elliptic bundle over $\mathbb C/(\mathbb Z+\mathrm i\mathbb Z)$.

Choose a smooth unit-periodic function $f(x)$ with $C=1-f''>0$, and put

This is positive and Hermitian, with Riemannian metric

The identities $\mathrm d\lambda _1=0$ and $\mathrm d\eta _1=-C e^1\wedge e^2$ give $\mathrm d\omega _1=\lambda _1\wedge \omega _1$. The Lee period along the $t$-circle is $-1$, so this metric is strictly LCK. Its Lee dual is $-\partial _ t$, which is Killing because all coefficients are independent of $t$. Closedness of the Lee form and the Killing equation make $\nabla ^{g_1}\lambda _1$ both symmetric and skew-symmetric; thus $\omega _1$ is Vaisman.

Write $e^{1234}=e^1\wedge e^2\wedge e^3\wedge e^4$. The volume is $\omega _1^2/2=C e^{1234}$, whereas $\Sigma \wedge \overline\Sigma $ has constant density in this coframe. For $h=C^{-1/2}$, the conformal metric $\omega =h\omega _1$ has volume $e^{1234}$ and therefore is Chern–Ricci flat. Its Lee form is

Let $E_ i$ be dual to $e^ i$. Solving $g(A,\cdot )=\lambda $ gives

The metric volume is $\mathrm dx\, \mathrm dy\, \mathrm dz\, \mathrm dt$, and the fields $E_ i$ have zero divergence for this volume. Hence $\operatorname {div}A=h''$ and $q=-h''$.

For the choice

one has

Thus $\omega $ is not Gauduchon, and theorem 23.2.47 also excludes the Vaisman condition for this flat metric.

Finally, if $\mathrm e^ v\omega $ is also Chern–Ricci flat, then the conformal Ricci formula gives $\mathrm{dd}^{\mathrm{c}}v=0$. The real pluriharmonic function $v$ is constant on the connected compact complex manifold by the maximum principle. A constant multiple of $\omega $ still has nonzero divergence of its Lee field, so no flat metric in this conformal class is Gauduchon. The metric $\omega _1$ already constructed is a Vaisman representative of the class. $\square$

Proof. If $j_ X([\lambda ])=0$, write $\mathrm d^{\mathrm c}\lambda =\mathrm{dd}^{\mathrm{c}}f$ for a smooth real function $f$. Since $\mathrm d^{\mathrm c}\mathrm{d}=-\mathrm{d}\mathrm d^{\mathrm c}$, the form $\lambda +\mathrm{d}f$ is both $\mathrm{d}$-closed and $\mathrm d^{\mathrm c}$-closed. Since the two equations separate the $\partial $- and $\bar\partial $-derivatives, its $(1,0)$-part is holomorphic and closed. Thus $\lambda +\mathrm{d}f$ is the real part of the closed holomorphic form $2(\lambda +\mathrm{d}f)^{1,0}$. Conversely, the real part of a closed holomorphic one-form is closed for both operators, so its image under $j_ X$ is zero.

By proposition 10.11.2, the pseudoeffective Bott–Chern cone contains no line. Thus $j_ X(\xi )$ and $-j_ X(\xi )$ are both pseudoeffective exactly when $j_ X(\xi )=0$, which is the identity $\mathcal C_ X\cap (-\mathcal C_ X)=K_ X$. $\square$

Proof. Let $\mathcal Z_ X$ be the real vector space of smooth $(n-1,n-1)$-forms $\Gamma $ satisfying $\mathrm{dd}^{\mathrm{c}}\Gamma =0$, and set

We first show that $q_ X(\mathcal Z_ X)=W_ X$. The inclusion into $W_ X$ follows from (23.2.58). If $[\lambda ]$ annihilates the image of $q_ X$, then by (23.2.58) the class $\{ \kappa _\lambda \} $ has zero pairing with every Gauduchon form. The generalized Lamari criterion, theorem 11.6.1, implies that both $\{ \kappa _\lambda \} $ and $-\{ \kappa _\lambda \} $ are pseudoeffective, so $[\lambda ]\in K_ X$ by lemma 23.2.56. Thus the image of $q_ X$ and $W_ X$ have the same annihilator in $V_ X$, and since the Poincaré pairing between $H^1_{\mathrm{dR}}$ and $H^{2n-1}_{\mathrm{dR}}$ is nondegenerate, they are equal.

Let $\mathcal Z_ X^+$ be the set of strictly positive forms in $\mathcal Z_ X$. By the linear-algebra construction used in the proof of theorem 11.6.1, every such form is the $(n-1)$st power of a unique Hermitian form, which is then Gauduchon. Hence

By Gauduchon's theorem, $\mathcal Z_ X^+$ is nonempty, and it is a convex cone; hence so is $\operatorname {LG}(X)$.

For openness, choose forms $\Gamma _1,\ldots ,\Gamma _ r\in \mathcal Z_ X$ whose images form a basis of $W_ X$. If $\Gamma \in \mathcal Z_ X^+$, then $\Gamma +\sum _ i t_ i\Gamma _ i$ remains strictly positive for all sufficiently small real $t_ i$, and the images of these forms cover a neighborhood of $q_ X(\Gamma )$ in $W_ X$. This includes the case $W_ X=\{ 0\} $, with the empty basis. Since a nonempty open subset spans the ambient vector space, the assertion about the span follows. $\square$

Proof. The first equality is theorem 11.6.1 applied to the smooth closed form $\kappa _\lambda $, together with (23.2.58). It expresses $\mathcal C_ X$ as an intersection of closed half-spaces, so this cone is closed.

If $a=[\mathrm d^{\mathrm c}(\gamma ^{n-1})]\in \operatorname {LG}(X)$ and $\xi =[\lambda ]\in \mathcal C_ X\setminus K_ X$, choose a closed positive current $T=\kappa _\lambda +\mathrm{dd}^{\mathrm{c}}u$. It is nonzero because it represents the nonzero class $j_ X(\xi )$. Therefore

Conversely, suppose that $a\in W_ X\setminus \operatorname {LG}(X)$. Separating the point $a$ from the nonempty open convex cone $\operatorname {LG}(X)$ in the finite-dimensional space $W_ X$ gives a nonzero linear functional $\ell $ on $W_ X$ such that

Indeed, in the usual separating inequalities the constant can be taken to be zero, by scaling the elements of the cone. By Poincaré duality, $\ell =\langle \xi ,\cdot \rangle $ for a class $\xi \in V_ X$ that is unique modulo $K_ X$, and $\xi \notin K_ X$ because $\ell \neq 0$. The first equality of the theorem gives $\xi \in \mathcal C_ X$, and then $\langle \xi ,a\rangle \leq 0$ shows that $a$ does not belong to the right side of the second equality. This proves the second equality; if $\operatorname {LG}(X)=W_ X$, there is nothing to prove. $\square$

Proof. By proposition 10.11.2, both pullback and pushforward preserve pseudoeffectivity; the pullback of potentials is the one used in lemma 5.43.19. Since $\pi $ has degree one,

Indeed, the identity of currents $\pi _*\pi ^*\theta =\theta $ follows by integration over the locus where $\pi $ is an isomorphism, whose complement has volume zero. This left inverse proves injectivity and the reverse implication for pseudoeffectivity. $\square$

Proof. For a proper analytic subset $A$ of a connected complex manifold $M$, the restriction map $H^1(M,\mathbb R)\to H^1(M\setminus A,\mathbb R)$ is injective, and it is an isomorphism if $A$ has complex codimension at least two. Indeed, choose a Whitney stratification of $A$. General position moves every one-cycle off the strata, which have real codimension at least two. If those codimensions are at least four, it also moves every two-chain with boundary in $M\setminus A$ off the strata, relative to its boundary. Thus the map $H_1(M\setminus A,\mathbb R)\to H_1(M,\mathbb R)$ is surjective in the first case and bijective in the second, and dualizing over $\mathbb R$ gives the assertions about restriction.

Since $X$ is smooth, there is an analytic subset $A\subset X$ of complex codimension at least two such that $\pi $ is an isomorphism over $U=X\setminus A$. Restriction from $X$ to $U$ is therefore an isomorphism, whereas restriction from $Y$ to $\pi ^{-1}U$ is injective. Every class on $\pi ^{-1}U\simeq U$ extends to $X$ and hence pulls back to a class on $Y$. The commutative diagram of restriction maps now shows that $\pi ^*$ is an isomorphism, and Poincaré duality and the projection formula give the assertion for $\pi _*$. $\square$

Proof. First let $\pi :Y\to X$ be a modification. Holomorphic pullback commutes with $\mathrm d^{\mathrm c}$, so

By lemma 23.2.62, a class $\xi $ lies in $\mathcal C_ X$ if and only if $\pi ^*\xi $ lies in $\mathcal C_ Y$, and likewise for $K_ X$ and $K_ Y$, since the pullback is injective on Bott–Chern classes. By lemma 23.2.63, every degree-one class on $Y$ is of the form $\pi ^*\xi $. Thus the isomorphism $\pi ^*$ in degree one identifies the two cones and the two kernels. Its dual $\pi _*$ identifies $W_ Y$ with $W_ X$, and theorem 23.2.60 gives

For a bimeromorphic map between $X$ and $Y$, resolve its graph to obtain modifications $p:Z\to X$ and $q:Z\to Y$ from a compact complex manifold $Z$. The preceding argument applies to both maps and identifies the cones and kernels via $(q^*)^{-1}p^*$ and its Poincaré dual. These identifications do not depend on the resolution: they are compatible with restriction to a common open set on which the original map is an isomorphism, and by the injectivity of degree-one restriction proved above, this determines each class. The dual identifications are therefore canonical as well. $\square$

Proof. An open cone in a finite-dimensional vector space that contains zero is the entire space, since scaling a neighborhood of zero reaches every point. The two dual descriptions in theorem 23.2.60 now prove the equivalences, including the case $W_ X=\{ 0\} $. If $b$ is a balanced metric, the form $b^{n-1}$ is closed and of pure type, hence also $\mathrm d^{\mathrm c}$-closed. Thus $b$ is Gauduchon and gives the element $[\mathrm d^{\mathrm c}(b^{n-1})]=0$ of $\operatorname {LG}(X)$.

On a compact Kähler manifold the $\partial \bar\partial $-lemma gives $j_ X=0$, since every $\kappa _\lambda $ is closed and de Rham exact. Thus $K_ X=V_ X$ and $W_ X=\{ 0\} $, and since $\operatorname {LG}(X)$ is nonempty, $\operatorname {LG}(X)=\{ 0\} $. By theorem 23.2.64, this conclusion extends to Fujiki's class $\mathcal C$. $\square$

Proof. Put $\Gamma =\gamma ^{n-1}$ and use $\beta $ as above. Then

Thus $[\mathrm d^{\mathrm c}\Gamma ]=0$.

Next, let $S$ be a real current of degree one such that $T=\mathrm{d}S$ is a nonzero positive $(1,1)$-current, and let $\Theta $ be a closed real form with strictly positive $(n-1,n-1)$-part $\Gamma $. By type considerations and Stokes' theorem for currents,

But the first integral is strictly positive because $T\ne 0$ and $\Gamma >0$, a contradiction. Thus a strongly Gauduchon metric excludes nonzero positive exact currents.

Conversely, suppose that no strongly Gauduchon metric exists. In the real Fréchet space of smooth $(2n-2)$-forms, let $\mathcal Z$ be the closed subspace of closed forms, and let $\mathcal P$ be the open convex cone of forms with strictly positive $(n-1,n-1)$-part. This cone is nonempty, since it contains $h^{n-1}$ for any Hermitian metric $h$. The two sets are disjoint, because the positive $(n-1)$st root of the $(n-1,n-1)$-part of a form in their intersection would be a strongly Gauduchon metric. The Hahn–Banach separation theorem therefore gives a nonzero real current $T$ of degree two such that

Fix $\Theta _0\in \mathcal P$. If a real form $\Psi $ has zero $(n-1,n-1)$-part, then $\Theta _0+t\Psi \in \mathcal P$ for every $t\in \mathbb R$, so $T(\Psi )=0$. Thus $T$ is of type $(1,1)$. For any semipositive $(n-1,n-1)$-form $\Gamma $, we have $\Gamma +\varepsilon \Theta _0\in \mathcal P$ for $\varepsilon >0$, and letting $\varepsilon \downarrow 0$ gives $T(\Gamma )\geq 0$; thus $T$ is positive.

Since $T$ annihilates every exact test form, it is closed. By the de Rham theorem for currents and Poincaré duality, its cohomology class is determined by its pairings with closed forms of complementary degree. All these pairings vanish, so $T=\mathrm{d}S$ for a real current $S$. Thus $T$ is a nonzero positive exact current, which proves the converse. $\square$

Proof. Suppose that $\gamma $ is strongly Gauduchon, and choose a smooth closed real form $\Theta $ whose $(n-1,n-1)$-part is $\gamma ^{n-1}$. By compactness, $h^{n-1}\leq A\gamma ^{n-1}$ for some $A>0$. Positivity and type considerations then give

The last pairing is a fixed linear functional on the finite-dimensional de Rham group, which proves (23.2.71). Conversely, this bound forces every closed positive de Rham-exact current to vanish, and proposition 23.2.69 gives a strongly Gauduchon metric.

For the compactness assertion, the bound gives uniformly bounded trace masses. A weak limit of such currents is positive and closed, and its de Rham class is the limit of their classes, as one sees by pairing with a finite basis of smooth closed forms of complementary degree and using Poincaré duality. Thus the family is weakly closed, and the usual weak compactness of positive currents with bounded mass proves the assertion.

Let $c_ j$ be a convergent sequence in the image cone, and choose closed positive currents $T_ j$ with $[T_ j]_{\mathrm{dR}}=c_ j$. By the compactness assertion, a subsequence converges to a closed positive current representing the limit of the classes. Its Bott–Chern class is pseudoeffective, so the image cone is closed. Convexity follows from addition and nonnegative scaling. Finally, if both $c$ and $-c$ lie in the image, choose closed positive currents $T$ and $S$ representing them. Their sum is de Rham exact, so (23.2.71) gives $T+S=0$, and positivity gives $T=S=0$, hence $c=0$. $\square$

Proof. Consider the holomorphic action of the additive group $\mathbb C$ on

The action is free, since $\mathrm{e}^ t=\mathrm{e}^{\mathrm i t}=1$ forces $t=0$. It is also proper. Indeed, if both $(z,w)$ and $t\cdot (z,w)$ range in compact subsets of $\widetilde X$, then the norms of the four vectors $z$, $w$, $\mathrm e^ tz$ and $\mathrm e^{\mathrm it}w$ in $\mathbb C^2$ are bounded above and away from zero. Writing $t=a+\mathrm i b$, the ratios of these norms bound $\mathrm{e}^ a$ and $\mathrm{e}^{-b}$ above and away from zero, so $a$ and $b$ stay bounded. A free proper holomorphic action has local holomorphic slices, and its quotient $X$ is a Hausdorff complex manifold of dimension three.

Every orbit meets $S^3\times S^3$ exactly once, at the image of $(z,w)$ under the group element with

These expressions depend smoothly on $(z,w)$, so the slice $S^3\times S^3$ identifies the quotient with $S^3\times S^3$ as a smooth manifold. In particular, $X$ is compact and connected, and

It follows that $V_ X=K_ X=W_ X=0$ and $\operatorname {LG}(X)=\{ 0\} $; the latter cone is nonempty by Gauduchon's theorem and lies in the zero group $H^5_{\mathrm{dR}}(X,\mathbb R)$.

The map $(z,w)\mapsto ([z],[w])$ is invariant under the action and descends to a holomorphic map

It is a submersion, because the map on $\widetilde X$ is a submersion and the orbits of the action lie in its fibers. Let $\eta $ be the sum of the Fubini–Study forms on the two factors. The smooth form

is closed, semipositive and of rank two everywhere. It is nonzero and de Rham exact, since $H^2_{\mathrm{dR}}(X,\mathbb R)=0$. By proposition 23.2.69, $X$ admits no strongly Gauduchon metric.

The Bott–Chern class of $\alpha $ is nonzero: otherwise $\alpha =\mathrm{dd}^{\mathrm{c}}f$ for a smooth real function $f$, and the integral of $\alpha $ against any Gauduchon form would vanish, contradicting $\alpha \geq 0$ and $\alpha \ne 0$. The class is pseudoeffective, since $\alpha $ is a positive representative. Finally, $j_ X$ has zero domain, so this class is not in its image. $\square$