Lemma 12.19.4669B72559DF9

Proof. For fixed $\delta $, choose a constant $M_\delta >0$ with $B_\delta \leq M_\delta \omega $; this shows that $C_\delta \leq M_\delta $. Suppose instead that $C_{\delta _ l}\leq C$ along a sequence $\delta _ l\to 0$. Then $\int _ X(C\omega -B_{\delta _ l})\wedge \Theta \geq 0$ for every admissible $\Theta $, and in particular for the squares of Gauduchon metrics. By theorem 11.6.1, there are quasi-plurisubharmonic functions $v_ l$ such that

By theorem 5.25.2, a subsequence converges in $L^1$ to a $C\omega $-plurisubharmonic function $v$, not identically $-\infty $. Since $\delta ^{-2}p(\zeta /\delta )\sigma _\zeta $ is an approximate identity in the variable $\zeta $,

On a coordinate ball where $\chi =1$, add to $v$ a smooth strictly plurisubharmonic function whose complex Hessian dominates $C\omega $. The resulting plurisubharmonic function has a current that dominates $[\zeta =0]$, so its Lelong number is at least one at every point of an open subset of this hypersurface. Smooth corrections do not change Lelong numbers, so the same holds for $v$. Siu's analyticity theorem, recalled in 5.34.2, shows that $Z=\{ \nu (v,\cdot )\geq 1\} $ is an analytic subset of $X$. It is a proper subset because $v$ is not identically $-\infty $.

Let $\widetilde Z$ be the inverse image of $Z$ in $\Omega $. It contains an open subset of $H_0\cap \Omega $, where $H_0=\{ w_0=z_0\} $. This surface is a projective plane with two points removed, so it is connected and irreducible, and the identity principle gives $H_0\cap \Omega \subset \widetilde Z$. Since $\widetilde Z$ is invariant under $g$, also

Now use the chart $z_1=1$ near the point $[0:1:1:0]\in \Omega $. There $H_ k$ is the graph $z_0=q^{-k}w_0$. For each fixed $(w_0,w_1)$ near $(1,0)$, every local holomorphic defining function of $\widetilde Z$ vanishes at the distinct values $z_0=q^{-k}w_0$, which tend to zero as $k\to +\infty $. The one-variable identity theorem makes each defining function vanish identically on a polydisk. Thus $\widetilde Z$ contains a nonempty open subset of the connected manifold $\Omega $, contradicting properness. Hence no subsequence of $C_\delta $ stays bounded as $\delta \downarrow 0$. $\square$