Pome

Theorem 22.1.1119B811DBD89E3

Proof. The map (22.1.102) is an order preserving injection by 22.1.105. It remains to prove that it is surjective. Let $\phi \in \mathrm{PSH}(L^{\mathrm{An}})$. We want to construct $\Gamma \in \mathrm{PSH}^{\mathrm{NA}}(X,\theta )$ with $\Gamma ^{\mathrm{An}}=\phi $.

Let $H$ be an ample line bundle and $(\epsilon _ i)_ i$ be a decreasing sequence of rational numbers with limit $0$, $\phi _ i\in \mathcal{H}^{\mathrm{gf}}_{\mathbb {Q}}((L+\epsilon _ i H)^{\mathrm{An}})$ such that

\[ \phi =\inf _{i>0}\phi _ i. \]

The existence of these data is guaranteed by theorem 22.1.75. Fix a Kähler form $\omega \in c_1(H)$.

Thanks to 22.1.107, we can find $\Gamma ^ i\in \mathrm{PSH}^{\mathrm{NA}}(X,\theta +\epsilon _ i \omega )$ with $(\Gamma ^ i)^{\mathrm{An}}=\phi _ i$. It follows from 22.1.105 that

\[ \Gamma ^ i\geq P_{\theta +\epsilon _ i\omega }\left[ \Gamma ^{i+1} \right]_{\mathcal{I}} \geq \Gamma ^{i+1}. \]

Therefore, for any $\omega '\in \textup{Käh}(X)$, after discarding finitely many $i$ so that $\omega '\geq \epsilon _ i\omega $, the sequence $(P_{\theta +\omega '}\left[ \Gamma ^{i} \right]_{\mathcal{I}})_ i$ is decreasing. This does not change its infimum. We let

\[ \Gamma ^{\theta +\omega '}=\inf _{i\gg 1}P_{\theta +\omega '}\left[ \Gamma ^{i} \right]_{\mathcal{I}}\in \mathrm{PSH}^{\mathrm{NA}}(X,\theta +\omega '). \]

Note that the infimum is defined thanks to (22.1.104). It follows from proposition 22.1.109 that

\[ \left(\Gamma ^{\theta +\omega '} \right)^{\mathrm{An}}=\phi . \]

From this, we conclude that for $\omega ',\omega ''\in \textup{Käh}(X)$ with $\omega '\leq \omega ''$, we have

\[ P_{\theta +\omega ''}\left[ \Gamma ^{\theta +\omega '}\right]_{\mathcal{I}}=\Gamma ^{\theta +\omega ''}. \]

Indeed, both sides have analytification $\phi $, by (22.1.103) and (22.1.102). The injectivity part of 22.1.105 gives our assertion.

It follows that $(\Gamma ^{\theta +\omega '})_{\omega '\in \textup{Käh}(X)}$ defines an element $\Gamma $ in $\mathrm{PSH}^{\mathrm{NA}}(X,\theta )$ and $\Gamma ^{\mathrm{An}}=\phi $. $\square$

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