Proposition 11.34.183A41CC697C2A

Proof. On $X\setminus D$, the function $w=v-\psi $ is bounded, because $v\sim \varphi \sim \psi $, and satisfies $\beta +\mathrm{dd}^{\mathrm{c}}w\geq 0$. After adding local potentials of the closed form $\beta $, lemma 5.24.7 extends $w$ across $D$; the extensions agree on overlaps and give a bounded $\beta $-plurisubharmonic function on $X$. Conversely, for every bounded $w\in \mathrm{PSH}(X,\beta )$, the function $\psi +w$ is $\theta $-plurisubharmonic with current $\beta +\mathrm{dd}^{\mathrm{c}}w+[D]$, and it has the ordinary type of $\varphi $. The decomposition is unique, since two such functions $w$ agree off $D$ and hence everywhere.

Put $T=\beta +\mathrm{dd}^{\mathrm{c}}w$. The bounded-product construction gives

and Stokes' theorem against the constant function yields $\int _ XT^2=4$. The product gives no mass to $D$ by proposition 5.28.2; we also give a direct cutoff estimate, which will be used again below. Shift $w$ so that $0\leq w\leq M$, and set $f_ t=\max \{ \psi /t,-1\} $ for $t>0$. Choose smooth convex functions $\sigma _\epsilon $ with derivative in $[0,1]$, equal to $-1$ below $-1-\epsilon $ and to the identity above $-1+\epsilon $, converging uniformly to $\max \{ \cdot ,-1\} $. Then $f_{t,\epsilon }=\sigma _\epsilon (\psi /t)$ is globally smooth, constant near $D$, and

Testing the distributional identity against $f_{t,\epsilon }$, using positivity of $T+\beta $, and letting $\epsilon \to 0$, we obtain

Here the last integral equals four by Stokes' theorem. The right-hand side tends to zero as $t\to \infty $, since $f_ t\to 0$ off $D$ and $\beta ^2(D)=0$; as $-f_ t=1$ on $D$, this gives $T^2(D)=0$. Off $D$ the currents $\theta +\mathrm{dd}^{\mathrm{c}}v$ and $T$ agree, so locality and the fact that the extended measure puts no mass on the pole set prove (11.34.19).

For the Hermitian background $\beta $, theorem 14.19.2 with density $\mathrm e^ F$ gives a bounded solution $w$ and a constant $c>0$, and the correspondence above turns it into a solution $\psi +w$ in the ordinary type of $\varphi $. Every potential of that type has mass four, so integration forces $c=4/\int _ X\mathrm e^ F\mathrm dV$, which is the formula for $B_\varphi $. Jensen's inequality for the probability $\nu _ F=\mathrm e^ F\mathrm dV/\int _ X\mathrm e^ F\mathrm dV$ gives

Thus $\nu _ F$ is a supporting probability. For any supporting probability, testing its inequality at $G=F\pm th$ and differentiating at $t=0$ shows that its integral against every $h\in C(X)$ equals that of $\nu _ F$.

For a smooth real $h$ and sufficiently small $|t|$, we have $\beta +t\mathrm{dd}^{\mathrm{c}}h>0$, so $\psi +th$ stays in the same ordinary type and its residual measure is $(\beta +t\mathrm{dd}^{\mathrm{c}}h)^2$, with derivative $2\mathrm{dd}^{\mathrm{c}}h\wedge \beta $ at $t=0$. Since $\beta $ is closed, the constant weight $\rho =1/4$ annihilates this derivative after integration, and $Q=\rho \beta $ gives the asserted probability. This normalization uses the residual mass: since $\int _ X\theta \wedge Q=3/2$, the cofactor $Q$ is not a normalized Gauduchon current for $\theta $ in the sense of section 11.7. $\square$