Example 18.14.7A467F501617F

Proof. Use homogeneous coordinates $[y_0:y_1:y_2]$ and $[z_0:z_1]$, and take

Outside $E$, the divisor is the graph $[z_0:z_1]=[-y_1:y_0]$. Its closure contains the whole fiber over $E$, so $H$ is prime. It is smooth outside that fiber by the graph description, and smooth along it because the derivatives in the $y_0$ and $y_1$ directions are $z_0$ and $z_1$, which cannot both vanish. Its product metric has curvature $p^*\sigma _2+r^*\sigma _1$, hence

Writing $S=|y_0|^2+|y_1|^2$, the Cauchy–Schwarz inequality gives the fiber obstacle

It has no finite continuous extension at $E$.

We claim that the largest $\sigma _2$-plurisubharmonic potential below $G$ is

To see that $q$ is $\sigma _2$-plurisubharmonic, work on $y_2\ne 0$ with $x=(y_0/y_2,y_1/y_2)$ and $t=\log |x|^2$. Adding the Fubini–Study potential $L(x)=\log (1+|x|^2)$ gives

This function and its first derivative are continuous at $t=0$. For $t>0$, its first two derivatives are $(\mathrm e^ t-1)/(\mathrm e^ t+1)\geq 0$ and $2\mathrm e^ t/(1+\mathrm e^ t)^2>0$. It is therefore a convex nondecreasing function of $t$, hence plurisubharmonic in $x$, across the unit sphere as well. The expression is constant near $x=0$, and $q=G$ is smooth near $y_2=0$, so $q$ is globally $\sigma _2$-plurisubharmonic and bounded.

If $b\in \mathrm{PSH}(\mathbb P^2,\sigma _2)$ lies below $G$, then $b+L\leq \log 4$ on $|x|=1$, and the maximum principle on the unit ball gives $b\leq q$ inside it. Outside the ball, the obstacle already gives $b\leq G=q$. For $0<R=|x|^2\leq 1$, the inequality $q\leq G$ reads $4R\leq (1+R)^2$, and at $E$ it is automatic. The formula for $q$ also gives the asserted bounds, which proves the claim.

For every ordinary or ideal competitor $v$ for $\Psi $, the ordinary order or theorem 5.37.3 gives $\nu (v,H)\geq 1$. Divisorial removal and descent along the fibers, lemma 5.38.1, proposition 5.43.5, give $v=\Psi +b\circ p$ with $b\in \mathrm{PSH}(\mathbb P^2,\sigma _2)$. The bound $v\leq 0$ gives $b\leq G$ off $E$, hence everywhere with the extended value at $E$. Thus $v\leq \Psi +q\circ p$. Conversely, this upper bound is a competitor for both envelopes, since it is nonpositive and differs boundedly from $\Psi $. This proves the envelope identities directly.

On $M\setminus H$, bounded local truncations of $b$ have products pulled back from a two-dimensional base, so their third products vanish by lemma 5.28.4, and plurifine locality gives zero extended mass. Divisorial removal puts every ordinary representative of the model in this form, which proves both volume assertions. Finally, since the model is ordinarily equivalent to $\Psi $, proposition 16.28.62 gives $\Psi \in \mathcal E_{\mathrm{env}}(M,\Theta ,P_\Theta [\Psi ]_{\mathcal I})$, and proposition 18.12.2 gives equality below every smooth obstacle. $\square$