Pome

Proposition 16.6.9708B2E9B2EF0

Proposition 16.6.9.708B2E9B2EF0 Let $\theta '=\theta +\mathrm{dd}^{\mathrm{c}}g$ for some $g\in C^{\infty }(X)$. Then for any $\varphi \in \mathrm{PSH}(X,\theta )$, we have $\varphi '\coloneqq \varphi -g\in \mathrm{PSH}(X,\theta ')$ and

\[ P_{\theta }[\varphi ]_{\mathcal{I}}\sim P_{\theta '}[\varphi ']_{\mathcal{I}}. \]

Proof. By symmetry, it suffices to show that

\[ P_{\theta }[\varphi ]_{\mathcal{I}}\preceq P_{\theta '}[\varphi ']_{\mathcal{I}}. \]

After replacing $g$ by $g+C$ for a suitable constant $C$, which only changes $\varphi '$ by an additive constant, we may assume that $g\geq 0$. Let $\psi \in \mathrm{PSH}(X,\theta )$ be such that $\psi \leq 0$ and $\psi \sim _{\mathcal{I}}\varphi $. Then $\psi '\coloneqq \psi -g$ lies in $\mathrm{PSH}(X,\theta ')$, satisfies $\psi '\leq 0$, and is $\mathcal{I}$-equivalent to $\varphi '$ since adding a smooth function does not change multiplier ideals. Hence

\[ \psi -g=\psi '\leq P_{\theta '}[\varphi ']_{\mathcal{I}}. \]

It follows that

\[ \psi -\sup _ X g\leq P_{\theta '}[\varphi ']_{\mathcal{I}}. \]

Taking the upper envelope over all such $\psi $ gives

\[ P_{\theta }[\varphi ]_{\mathcal{I}}-\sup _ X g\leq P_{\theta '}[\varphi ']_{\mathcal{I}}, \]

which proves the desired comparison. $\square$

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