PomeD1551EE7322BTheorem 20.3.33.D1551EE7322B Suppose $Y$ is Kähler. Choose a Hermitian form $\Omega _0\geq \theta |_ Y$ and a Kähler form $\kappa \geq 2\Omega _0$ on $Y$. For every Hermitian form $\Omega \geq \kappa $,
for every prime divisor $E$ on every smooth compact modification $\pi :Z\to Y$.
Proof. The curvature errors $\epsilon _ m$ tend to zero, so $u_ m|_ Y\in \mathrm{PSH}(Y,\kappa )$ for all sufficiently large $m$. By theorem 20.3.27, applied with $\Omega =\kappa $ and $\delta _ m=0$, the envelope fixed points
decrease to $T=\mathcal T_\kappa (u;Y)$, after discarding finitely many indices. The normalized potential $v$ of proposition 20.3.24 belongs to $\mathrm{PSH}(Y,\Omega _0)\subset \mathrm{PSH}(Y,\kappa /2)$ and satisfies $v\leq T$, so $P_{\kappa /2}(T)\not\equiv -\infty $. Thus lemma 16.28.69 and Stokes' theorem give, with $r=\dim Y$,
For a decreasing sequence of envelope fixed points in a closed background, positive mass of the limit implies convergence of every divisorial Lelong number [Proposition 3.1.12, SGPT]. Applied on $Y$, this yields
Here the middle equality uses the ordinary equivalence of each $q_ j$ with $u_{2^ j}|_ Y$.
If $\Omega \geq \kappa $, then $T$ is a trace competitor for $\Omega $, so $T\leq T_\Omega =\mathcal T_\Omega (u;Y)$. Together with the defining inequalities of $T_\Omega $, this gives
This proves the assertion for every such $\Omega $. $\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.