Return to an LCK metric $\omega $ on $X$ with Lee form $\lambda $. Chern–Ricci flatness expresses the Riemannian Ricci tensor through $\lambda $ and its covariant derivative. The Bianchi and Bochner identities then relate the part of this derivative that is not $J$-invariant to the divergence of the Lee field. When the metric is Gauduchon, this divergence vanishes by proposition 23.2.7, and the identities force the Lee form to be parallel.
We write $\operatorname {Ric}^{\mathrm{Ch}}=\operatorname {Ric}/(2\pi )$ for the normalized Chern–Ricci form, which represents the first Bott–Chern class $c_1^{\mathrm{BC}}(X)$, with $\operatorname {Ric}$ as in 12.34.3.
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
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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
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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
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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
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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.
$\square$
In the theorem, both conditions are imposed on the same metric. If $c_1^{\mathrm{BC}}(X)=0$, each conformal class of Hermitian metrics contains a Chern–Ricci-flat metric: if a smooth real function $h$ satisfies $\operatorname {Ric}^{\mathrm{Ch}}(\omega )=\mathrm{dd}^{\mathrm{c}}h$, then
\[ \operatorname {Ric}^{\mathrm{Ch}}(\mathrm{e}^{h/n}\omega ) =\operatorname {Ric}^{\mathrm{Ch}}(\omega )-\mathrm{dd}^{\mathrm{c}}h=0. \]
By Gauduchon's theorem, the same conformal class also contains a Gauduchon metric. The rigidity theorem applies when these two metrics coincide. The example below shows that changing the metric within its conformal class cannot always arrange this coincidence; the manifold existence question follows it.
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