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23.2.28 Invariant volumes on Vaisman manifoldsA3211952AB85

The flows of the Lee and anti-Lee vector fields preserve a Vaisman metric. For a prescribed volume invariant under these flows, uniqueness for the smooth Hermitian equation forces the potential to be constant along them. Stokes' theorem in the transverse directions then fixes the normalizing constant, and changing the Lee form by the differential of the potential produces a Vaisman metric with the prescribed volume.

Theorem 23.2.29.19781C771E9D Let $\omega $ be a Vaisman metric on $X$ with Lee form $\lambda $, and let $\mu $ be a smooth positive volume form invariant under the flows of both the Lee field $\lambda ^\sharp $ and the anti-Lee field $J\lambda ^\sharp $. Assume

\[ \int _ X\mu =\int _ X\omega ^ n. \]

Then there is a unique Vaisman metric $\omega '$ with Lee class $[\lambda ]$ and $(\omega ')^ n=\mu $.

We first describe the transverse geometry and the integration of transverse forms. Let $g$ be the Riemannian metric of $\omega $. Since $\lambda $ is parallel and not identically zero, it has constant squared norm $s=|\lambda |_ g^2>0$. Write

\[ A=s^{-1}\lambda ^\sharp ,\qquad B=JA,\qquad \tau =\frac{s}{4\pi },\qquad \eta =\kappa _\lambda . \]

Thus $\lambda (A)=1$ and $\lambda (B)=0$. Since $\nabla \lambda =0$, the identity (23.2.4) gives

9A2166571118 23.2.30
\begin{equation} \begin{split} \tau \omega & =\eta +\lambda \wedge \lambda ^{\mathrm c},\\ \eta (U,JV)& =\frac1{4\pi }\bigl (sg(U,V)-\lambda (U)\lambda (V)-\lambda (JU)\lambda (JV)\bigr ). \end{split} \end{equation}

To obtain these identities, use $\lambda ^{\mathrm c}=-\lambda \circ J/(4\pi )$ and antisymmetrize $\nabla \lambda ^{\mathrm c}$; the terms containing $\nabla \lambda $ vanish. In particular, $\eta $ is a closed semipositive $(1,1)$-form whose kernel is the span $\mathcal F$ of $A$ and $B$.

The field $A$ is parallel, and the same connection formula yields

\[ \nabla _ U B=\tfrac 12\bigl (-JU+\lambda (U)B+\lambda (JU)A\bigr ), \qquad \nabla _ AJ=0. \]

The first identity shows that $B$ is a Killing field and that $[A,B]=0$. The second shows that $A$ is holomorphic, and integrability of $J$ then shows that $B=JA$ is holomorphic. Their flows preserve $g$, $J$, $\lambda $ and $\eta $. Consequently $\mathcal F$ is a holomorphic foliation by complex curves, and $\eta $ induces a Kähler form on each local transverse space. A form is called horizontal if its contractions with $A$ and $B$ vanish.

A smooth function $v$ is basic for $\mathcal F$ if $Av=Bv=0$. For such a function, $\mathrm{d}v$, $\mathrm d^{\mathrm c}v$ and $\mathrm{dd}^{\mathrm{c}}v$ are horizontal and invariant under both flows; for $\mathrm{dd}^{\mathrm{c}}v$ this follows from Cartan's formula. Put $m=n-1$ and $\eta _ v=\eta +\mathrm{dd}^{\mathrm{c}}v$. Then

E3B61067B040 23.2.31
\begin{equation} \int _ X\eta _ v^ m\wedge \lambda \wedge \lambda ^{\mathrm c} =\int _ X\eta ^ m\wedge \lambda \wedge \lambda ^{\mathrm c}. \end{equation}

Indeed, $\eta _ v^ m-\eta ^ m=\mathrm{dd}^{\mathrm{c}}v\wedge T$, where $T=\sum _{j=0}^{m-1}\eta _ v^ j\wedge \eta ^{m-1-j}$ is closed and horizontal. Since $\mathrm{d}(\lambda \wedge \lambda ^{\mathrm c})=\lambda \wedge \eta $, the differential of $\mathrm d^{\mathrm c}v\wedge T\wedge \lambda \wedge \lambda ^{\mathrm c}$ is $\mathrm{dd}^{\mathrm{c}}v\wedge T\wedge \lambda \wedge \lambda ^{\mathrm c}$ plus a term containing the horizontal form $\mathrm d^{\mathrm c}v\wedge T\wedge \eta $ of degree $2m+1$, which vanishes because the transverse real dimension is $2m$. Stokes' theorem therefore proves (23.2.31).

The following transverse maximum principle identifies the metric once its scale has been fixed.

Proposition 23.2.32.FCE8FB3BF729 Let $\mathcal F$ be a foliation of a connected compact manifold $M$ with a transverse holomorphic structure of complex dimension $m\geq 1$. Let $\eta _0,\eta _1$ be transverse Kähler forms, and let $v$ be a smooth real basic function, with $\mathrm{dd}^{\mathrm{c}}v$ computed in transverse holomorphic coordinates. Suppose

\[ \eta _1=\eta _0+\mathrm{dd}^{\mathrm{c}}v, \qquad \eta _1^ m=\eta _0^ m. \]

Then $v$ is constant and $\eta _1=\eta _0$.

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

\[ \mathrm{dd}^{\mathrm{c}}v\wedge \sum _{j=0}^{m-1}\eta _1^ j\wedge \eta _0^{m-1-j}=0. \]

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

\[ (\omega +\mathrm{dd}^{\mathrm{c}}u)^ n=\mathrm{e}^ b f\omega ^ n, \qquad \omega +\mathrm{dd}^{\mathrm{c}}u>0,\qquad \sup _ Xu=0. \]

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

\[ (\omega +\mathrm{dd}^{\mathrm{c}}u)^ n=n\eta _ u^ m\wedge \lambda \wedge \lambda ^{\mathrm c}. \]

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

1958C8BBF316 23.2.33
\begin{equation} \omega _ u'=\eta _ u+\lambda _ u\wedge \lambda _ u^{\mathrm c}. \end{equation}

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

\[ g_ u'(A,\cdot )=\frac1{4\pi }\lambda _ u. \]

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

\[ (\omega _ u')^ n =n\eta _ u^ m\wedge \lambda \wedge \lambda ^{\mathrm c} =(\omega +\mathrm{dd}^{\mathrm{c}}u)^ n=\mu . \]

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 $\widehat\lambda (\widehat B)=0$,

\[ \widehat A v=1-a,\qquad \widehat B v=b. \]

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

8455EE64C6A0 23.2.34
\begin{equation} \iota _ B\iota _ A\widehat\omega ^ n =\frac{n}{4\pi \widehat\tau ^ n}\widehat\eta ^ m, \qquad \iota _ B\iota _ A(\omega _ u')^ n=\frac{n}{4\pi }\eta _ u^ m. \end{equation}

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 $\widehat\omega =\omega _ u'$. $\square$

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