Proposition 10.19.11.983D2EEDBE68 Let $a,b\geq 0$, $\theta =a\alpha +b\beta $ and $\Omega _\epsilon =\theta +\epsilon \omega $. For every $\epsilon >0$, the metric $\Omega _\epsilon $ is Gauduchon, and it minimizes the quotient defining $\lambda _{\mu _0}(\Omega _\epsilon )$. Put
\[ I_\epsilon =8\pi \bigl (ab+2(a+b)\epsilon +3\epsilon ^2\bigr ), \qquad T_\epsilon =\frac{\Omega _\epsilon ^2}{I_\epsilon }, \qquad \gamma _\epsilon =\frac{\Omega _\epsilon }{\sqrt{I_\epsilon }}. \]
Then $\int _ X\omega \wedge T_\epsilon =1$, $\mathrm{dd}^{\mathrm{c}}T_\epsilon =0$ and $T_\epsilon =\gamma _\epsilon ^2$. As $\epsilon \downarrow 0$, the forms $T_\epsilon $ converge smoothly to
\[ T_0=\begin{cases} (4\pi )^{-1}\alpha \wedge \beta ,& a,b>0,\\ (8\pi )^{-1}\alpha \wedge (\beta +D),& a>0,\ b=0,\\ (8\pi )^{-1}\beta \wedge (\alpha +D),& a=0,\ b>0,\\ (24\pi )^{-1}\omega ^2,& a=b=0. \end{cases} \]
In every case $\theta \wedge T_0=0$. If $q$ is the number of positive coefficients among $a,b$, the denominator of the normalized minimizing metric satisfies
\[ D_{\mu _0}(\gamma _\epsilon ) =\frac{\bigl (24\pi \epsilon (a+\epsilon )(b+\epsilon )\bigr )^{2/3}}{I_\epsilon } \asymp \epsilon ^{q/3}, \]
where the two comparison constants may depend on the fixed pair $(a,b)$.
Proof.
Write $x=a+\epsilon $, $y=b+\epsilon $ and $z=\epsilon $. Since $\alpha ^2=\beta ^2=D^2=0$,
\[ \Omega _\epsilon ^2 =2xy\alpha \wedge \beta +2xz\alpha \wedge D+2yz\beta \wedge D. \]
The forms $\alpha ,\beta $ are closed and $\mathrm{dd}^{\mathrm{c}}D=2\alpha \wedge \beta $. Hence the first term is closed, and $\mathrm{dd}^{\mathrm{c}}$ of each of the last two terms contains $\alpha ^2$ or $\beta ^2$ and therefore vanishes. Thus $\mathrm{dd}^{\mathrm{c}}(\Omega _\epsilon ^2)=0$. Since the zero potential solves the normalized equation for $\Omega _\epsilon $ and $\mu _0$, the characterization of minimizing metrics in theorem 12.2.3 shows that $\Omega _\epsilon $ is a minimizer.
Integration along the fibers gives $\int _ X\alpha \wedge \beta \wedge D=4\pi $. Hence
\[ \int _ X\omega \wedge \Omega _\epsilon ^2 =8\pi (xy+xz+yz)=I_\epsilon , \]
which is the trace normalization. Dividing the three coefficients in the wedge expansion of $\Omega _\epsilon ^2$ by $I_\epsilon $ gives each of the stated limits; the convergence holds in every smooth norm, because the forms $\alpha \wedge \beta $, $\alpha \wedge D$ and $\beta \wedge D$ are fixed. The identities $\alpha ^2=\beta ^2=0$ then give $\theta \wedge T_0=0$ in all four cases.
Set $c_\epsilon =24\pi \epsilon (a+\epsilon )(b+\epsilon )$. The equation for $\Omega _\epsilon $ gives
\[ \gamma _\epsilon ^3 =\frac{c_\epsilon }{I_\epsilon ^{3/2}}\mu _0. \]
Substituting this in the definition of $D_{\mu _0}$ and using $\int _ X\mu _0=1$ yields $D_{\mu _0}(\gamma _\epsilon )=c_\epsilon ^{2/3}/I_\epsilon $. For fixed $(a,b)$ one has $c_\epsilon \asymp \epsilon ^{3-q}$ and $I_\epsilon \asymp \epsilon ^{2-q}$, so the quotient is of order $\epsilon ^{q/3}$.
$\square$
Pome
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