PomeExample 23.2.53.E37387C4675B There is a compact complex Kodaira surface with holomorphically trivial canonical bundle and a strictly LCK Chern–Ricci-flat metric $\omega $ that is not Gauduchon. No metric conformal to $\omega $ is both Gauduchon and Chern–Ricci flat, although the same conformal class contains a Vaisman metric.
Example 23.2.53E37387C4675B
Proof. We construct the quotient and the two conformal metrics explicitly. On $G=\mathbb R^4$ use the multiplication
\[ (x,y,z,t)(x',y',z',t') =(x+x',y+y',z+z'+xy',t+t'). \]
The integer subgroup $\Gamma =\mathbb Z^4$ acts on the left by
\[ (x,y,z,t)\longmapsto (x+a,y+b,z+c+ay,t+d). \]
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
\[ e^1=\mathrm dx,\quad e^2=\mathrm dy,\quad e^3=\mathrm dz-x\, \mathrm dy,\quad e^4=\mathrm dt \]
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
\[ \mathrm d\phi ^1=0,\qquad \mathrm d\phi ^2=-\tfrac {\mathrm i}{2}\phi ^1\wedge \overline{\phi ^1}, \]
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
\[ \lambda _1=-e^4+f'e^1,\qquad \eta _1=J\lambda _1=e^3+f'e^2,\qquad \omega _1=\lambda _1\wedge \eta _1+C e^1\wedge e^2. \]
This is positive and Hermitian, with Riemannian metric
\[ g_1=\lambda _1\otimes \lambda _1+\eta _1\otimes \eta _1 +C(e^1\otimes e^1+e^2\otimes e^2). \]
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
\[ \lambda =-e^4+\mathrm d(f+\log h). \]
Let $E_ i$ be dual to $e^ i$. Solving $g(A,\cdot )=\lambda $ gives
\[ A=h'E_1+(f'h'-h^{-1})E_4. \]
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
\[ f(x)=\frac{\cos (2\pi x)}{16\pi ^2},\qquad C(x)=1+\tfrac 14\cos (2\pi x)\geq \tfrac 34, \]
one has
\[ q(0)=-\frac{4\sqrt5\pi ^2}{25},\qquad q(1/2)=\frac{4\sqrt3\pi ^2}{9}. \]
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. $\square$
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