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Lemma 5.38.111D91BF7CA736

Lemma 5.38.11.1D91BF7CA736 Let $v$ be quasi-plurisubharmonic on $X$, and let $a$ have analytic singularities on $X$. If $v\preceq _{\mathcal I}a$, then $v\preceq a$. More precisely, it suffices to assume the finitely many inequalities

\[ \nu (v\circ \pi ,E)\geq \nu (a\circ \pi ,E) \]

for the prime components $E$ of the divisor obtained by principalizing, on one log resolution $\pi $, the ideal of an analytic presentation of $a$.

Proof. Principalize the coherent ideal of an analytic presentation of $a$ by a log resolution $\pi :Y\to X$ with smooth $Y$. In suitable local coordinates $z_1,\ldots ,z_ n$ on $Y$,

\[ a\circ \pi =\sum _{i=1}^ r b_ i\log |z_ i|^2+B, \qquad b_ i\geq 0,\qquad B\in L^\infty _{\mathrm{loc}}, \]

where $r\leq n$. The finitely many inequalities in the statement give $\nu (v\circ \pi ,\{ z_ i=0\} )\geq b_ i$; if $v\preceq _{\mathcal I}a$, these inequalities follow from theorem 5.37.3. Add a smooth local function to $v\circ \pi $ to make it plurisubharmonic. Divisorial removal for this closed positive complex Hessian shows that

\[ v\circ \pi -\sum _ i b_ i\log |z_ i|^2 \]

has a quasi-plurisubharmonic representative; here the difference is first interpreted off the divisor, or as a locally integrable distribution. Such a representative is locally bounded above. Since $B$ is locally bounded, $v\circ \pi \leq a\circ \pi +C$ on each relatively compact chart of $Y$, away from the divisor. Since $\pi $ is proper, finitely many such charts cover the inverse image of a smaller coordinate neighborhood in $X$, which makes $C$ uniform there. Push the inequality down off the analytic exceptional set; as quasi-plurisubharmonic functions are determined by their values almost everywhere, it then holds everywhere. Finitely many neighborhoods cover the compact manifold, so $v\leq a+C$ globally.

The only input from resolution of singularities is the principalization of a coherent ideal. Divisorial removal is applied to the closed positive complex Hessian obtained after a local smooth correction, so no closed background on $X$ is needed. $\square$

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