Proposition 22.1.84.A87B7BF0C3D5 Let $\Gamma \in \mathrm{PSH}^{\mathrm{NA}}(X,\theta )_{>0}$ with $\Gamma _{\max }\leq 0$. Let $\Psi $ be the complexification of $\Gamma ^*$. Then
B5F5AD50A5AD
22.1.85
\begin{equation} \Gamma ^{\mathrm{An}}(v)=-\sigma (v)(\Psi )\quad \forall v\in X^{\operatorname{div}}. \end{equation}
Proof.
Recall that
\[ \Psi (x,\delta )=\sup _{\tau \leq \Gamma _{\max }} (\Gamma _{\tau }(x)-\log |\delta |^2\tau )\quad \text{for }x\in X, \delta \in \Delta ^*. \]
By (22.1.69), we have $\sigma (v)(\log |\delta |^2)=1$ and $\sigma (v)(\Gamma _{\tau })=v(\Gamma _{\tau })$ for all $\tau \leq \Gamma _{\max }$. So we have
\[ \sigma (v)(\Gamma _\tau (x) - \log |\delta |^2 \tau ) = v(\Gamma _\tau ) - \tau . \]
Lastly, since $\sigma (v)$ is a divisorial valuation on $X \times \Delta $, by corollary 5.16.9, we conclude (22.1.85).
$\square$
Pome
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