Theorem 16.29.9.E94DE568F212 Fix a Hermitian form $h$ on $X$, and let $j$ range over the positive integers. For every $\boldsymbol \Gamma \in \mathscr S_{a,c}(\theta )$ and every $\tau <c$, the sequence $\Gamma ^{h/j}_\tau $ decreases, and its limit
\[ (\mathsf R\boldsymbol \Gamma )_\tau =\lim _{j\to \infty }\Gamma ^{h/j}_\tau \]
is $\theta $-plurisubharmonic, in particular not identically $-\infty $. These limits define an element $\mathsf R\boldsymbol \Gamma $ of $\mathscr T_{a,c}(\theta )$, which does not depend on $h$. The maps $\mathsf E$ and $\mathsf R$ satisfy
\[ \mathsf R\mathsf E=\mathrm{id}, \qquad \mathsf E\mathsf R(\boldsymbol \Gamma ) \leq \boldsymbol \Gamma . \]
Both maps preserve order. More precisely,
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16.29.10
\begin{equation} \mathsf E(\Gamma )\leq \boldsymbol \Gamma \quad \Longleftrightarrow \quad \Gamma \leq \mathsf R\boldsymbol \Gamma . \end{equation}
Thus $\mathsf E\mathsf R(\boldsymbol \Gamma )$ is the greatest system below $\boldsymbol \Gamma $ that is realized in the original background.
Proof.
Step 1: the limit. Fix $\tau <c$ and put $q_ j=\Gamma ^{h/j}_\tau $. By compatibility, $q_ j=P_{\theta +h/j}[q_{j+1}]_{\mathcal I}$, and $q_{j+1}\leq 0$ is a competitor for this envelope; hence $q_ j\geq q_{j+1}$. The functions $q_ j$ have supremum zero and a common smooth lower curvature bound. By theorem 5.25.2, corollary 5.25.3, their decreasing limit $q$ is not identically $-\infty $ and has supremum zero. Passing to the limit in the distributional inequalities
\[ \theta +h/j+\mathrm{dd}^{\mathrm{c}}q_ j\geq 0 \]
gives $q\in \mathrm{PSH}(X,\theta )$.
The defining family of an ideal envelope grows when the background increases. Together with $q\leq q_ j$ and order preservation, this gives
\[ q\leq P_\theta [q]_{\mathcal I} \leq P_{\theta +h/j}[q]_{\mathcal I} \leq P_{\theta +h/j}[q_ j]_{\mathcal I}=q_ j. \]
Taking the infimum over $j$ proves $P_\theta [q]_{\mathcal I}=q$. Concavity, decreasingness and constancy on the left half-line pass to the pointwise limits, so these limits form a curve in $\mathscr T_{a,c}(\theta )$.
Step 2: independence of $h$. For every $\eta >0$ there is $j$ with $h/j\leq \eta $, and compatibility then gives $\Gamma ^\eta _\tau \geq \Gamma ^{h/j}_\tau $. Consequently
\[ (\mathsf R\boldsymbol \Gamma )_\tau =\inf _{\eta \in \mathscr H(X)}\Gamma ^\eta _\tau , \]
which proves that $\mathsf R\boldsymbol \Gamma $ does not depend on $h$. This formula also shows that $\mathsf R$ preserves order.
Step 3: the two compositions. Write $q=(\mathsf R\boldsymbol \Gamma )_\tau $. Since $q\leq \Gamma ^\eta _\tau $ and $\Gamma ^\eta _\tau $ is fixed by its envelope, order preservation gives $P_{\theta +\eta }[q]_{\mathcal I}\leq \Gamma ^\eta _\tau $. This proves $\mathsf E\mathsf R\leq \mathrm{id}$.
Now start with a curve $\Gamma \in \mathscr T_{a,c}(\theta )$. Its slice $\Gamma _\tau $ is a competitor in every larger background, so
\[ \Gamma _\tau \leq (\mathsf R\mathsf E\Gamma )_\tau \leq P_{\theta +\eta }[\Gamma _\tau ]_{\mathcal I} \quad (\eta >0). \]
The two outer terms have the same multiplier ideals, so the two ideal inclusions given by these inequalities show that the middle term has those ideals as well. Since both the middle term and $\Gamma _\tau $ are $P_\theta [\cdot ]_{\mathcal I}$-fixed, they are equal. This proves $\mathsf R\mathsf E=\mathrm{id}$.
Step 4: the adjunction. By the transition proposition, $\mathsf E$ preserves order. If $\mathsf E(\Gamma )\leq \boldsymbol \Gamma $, applying $\mathsf R$ gives $\Gamma =\mathsf R\mathsf E(\Gamma )\leq \mathsf R\boldsymbol \Gamma $. Conversely, if $\Gamma \leq \mathsf R\boldsymbol \Gamma $, applying $\mathsf E$ and using $\mathsf E\mathsf R\leq \mathrm{id}$ gives $\mathsf E(\Gamma )\leq \boldsymbol \Gamma $. This proves (16.29.10), and the final maximality assertion follows from it.
$\square$
Pome
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