18.40.5 The domination envelope6419DC6AC605

We first compute the envelope on the base, using the affine chart $Z_0\ne 0$. Set $z_ i=Z_ i/Z_0$, $s=|z|^2$, $l=k-c$, $s_0=l/c$ and $C=k\log k-c\log c-l\log l$. In this chart the reference weight of $A$ is $\phi _ A=k\log (1+s)$. Put

As a function of $\log s$, $F_0$ is increasing and convex: its derivative increases to $l$ and then stays constant. Hence $F_0$ is plurisubharmonic. The two branches match to first order at $s=s_0$, and the tangent-line inequality gives $F_0\leq \phi _ A$. On a chart $Z_ j\ne 0$ with $j\geq 1$, write $x=Z_0/Z_ j$ and denote the remaining ratios by $\zeta $. Near $Z$ the corresponding weight is

Therefore $F_0$ and the $F_ j$ define a global plurisubharmonic metric weight $F$ on $A$, with a divisor pole of coefficient $c$ along $Z$ and a smooth remainder.

Let $v\leq \phi _ A$ be a competing plurisubharmonic weight, that is, $v=\phi _ A+u$ with $u\in \mathrm{PSH}(B,\theta )$ and $u\leq 0$, and suppose that $v\leq c\log |x|^2+O(1)$ near $Z$. Write $v_0$ for its expression in the chart $Z_0\neq 0$. The obstacle gives $v_0\leq F_0$ on $\{ s\leq s_0\} $. On each complex line $z=t\xi $ with $|\xi |=1$, either $v_0(t\xi )\equiv -\infty $, or $v_0(t\xi )-l\log |t|^2$ is subharmonic on $|t|>\sqrt{s_0}$ and bounded above near infinity. In the coordinate $1/t$ the puncture at infinity is then removable, and the maximum principle bounds this difference by its boundary bound $C$. Varying $\xi $ proves $v\leq F$ on $B$. Consequently $F=\phi _ A+u_ c$ is the largest such weight, and its contact set $\{ F=\phi _ A\} $ is $D_ c$.

We now return to $X$. By proposition 5.43.5, every $\pi ^*\theta $-plurisubharmonic function has the form $u\circ \pi $ for a $\theta $-plurisubharmonic function $u$ on $B$. Restriction to the local section $w_1=\cdots =w_ r=0$ shows that $u\circ \pi \preceq \varphi $ implies $u\leq c\log |x|^2+O(1)$ near $Z$. Conversely, this divisorial bound and the inequality $|x|^2\leq |x|^2+\sum _ j|w_ j|^2$ give the domination $u\circ \pi \leq \varphi +O(1)$ near $Y$. A finite cover of $Y$ gives one constant there. On the compact complement of a smaller neighborhood of $Y$, the function $\varphi $ is bounded below and $u\leq 0$, so one constant works on all of $X$. The two competitor families therefore correspond to each other, which proves $\mathcal P(\varphi )=u_ c\circ \pi $. The pole of this envelope fills $\pi ^{-1}(Z)$, although $\varphi $ is finite on $\pi ^{-1}(Z)\setminus Y$.