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
\[ B=JA,\qquad S=\nabla \lambda ,\qquad S^+(U,V)=\tfrac 12\bigl (S(U,V)+S(JU,JV)\bigr ),\qquad u=|\lambda |_ g^2. \]
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
C15BADA6DBC4
23.2.48
\begin{equation} \operatorname {Ric}_ g =nS^+-(n-1)S+\tfrac {q-u}{2}g +\tfrac 12\lambda \otimes \lambda +\tfrac n2 J\lambda \otimes J\lambda . \end{equation}
Locally write $\lambda =\mathrm df$ and put $\widetilde g=\mathrm e^{-f}g$. The real conformal Ricci formula in dimension $2n$ is
\[ \operatorname {Ric}_{\widetilde g} =\operatorname {Ric}_ g+(n-1)S +\tfrac {n-1}{2}\lambda \otimes \lambda +\tfrac 12\{ -q-(n-1)u\} g. \]
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
\[ \mathrm d(J\lambda )(U,JV) =2S^+(U,V)-u g(U,V) +\lambda (U)\lambda (V)+J\lambda (U)J\lambda (V). \]
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
\[ s=(n-1)q-\tfrac {n-1}{2}u. \]
Step 2: the integral identity. Write $\| \cdot \| $ for the $L^2(g)$ norm and set
\[ E_ A=\int _ X S(A,A)\, \mathrm d\operatorname {vol}_ g, \qquad E_ B=\int _ X S(B,B)\, \mathrm d\operatorname {vol}_ g. \]
Since $S(A,A)=A(u)/2$ and $\operatorname {div}A=-q$, integration by parts gives
B65427234DA6
23.2.49
\begin{equation} \int _ Xu q\, \mathrm d\operatorname {vol}_ g=2E_ A. \end{equation}
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
\[ 0=\int _ X\langle \operatorname {Ric}_ g,S\rangle \, \mathrm d\operatorname {vol}_ g +\tfrac 12\int _ Xs q\, \mathrm d\operatorname {vol}_ g. \]
Substitution of equation 23.2.48, equation 23.2.49 gives
900FA1D2D218
23.2.50
\begin{equation} 0=n\| S^+\| ^2-(n-1)\| S\| ^2+\tfrac {n-2}{2}\| q\| ^2 +\tfrac {4-n}{2}E_ A+\tfrac n2E_ B. \end{equation}
For the closed one-form $\lambda $, the integrated Bochner formula reads
\[ \| q\| ^2=\| S\| ^2 +\int _ X\operatorname {Ric}_ g(A,A)\, \mathrm d\operatorname {vol}_ g. \]
The quartic terms in (23.2.48) cancel on $(A,A)$, and $J\lambda (A)=0$, so
\[ \operatorname {Ric}_ g(A,A) =(1-\tfrac n2)S(A,A)+\tfrac n2 S(B,B)+\tfrac 12q u. \]
Using (23.2.49), we obtain
1F4612E67B59
23.2.51
\begin{equation} \| q\| ^2=\| S\| ^2+\tfrac {4-n}{2}E_ A+\tfrac n2E_ B. \end{equation}
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
\[ (\mathcal L_ AJ)V=JTV-TJV. \]
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.
$\square$
Pome
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