Theorem 23.2.2919781C771E9D

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