Lemma 22.1.89.6784060E951F Let $A$ be a piecewise linear curve in $\mathrm{PSH}(X,\theta )$. Let $(A'_{\tau })_{\tau \in \mathbb {R}}$ be the upper concave envelope of $\tau \mapsto A_{\tau }$. Define $\tilde A$ by
\[ \tilde{A}^{\theta +\omega }_{\tau }\coloneqq P_{\theta +\omega }[A'_{\tau }]_{\mathcal I}, \quad \omega \in \textup{Käh}(X),\ \tau <\tau _0. \]
Then $\tilde A\in \mathrm{PSH}^{\mathrm{NA}}(X,\theta )$, and
F46F1129051C
22.1.90
\begin{equation} A^{\mathrm{An}}=\tilde{A}^{\mathrm{An}}\quad \text{on }X^{\operatorname{div}}. \end{equation}
Proof.
We continue to use the notations in definition 22.1.86. The fact that $\tilde{A}\in \mathrm{PSH}^{\mathrm{NA}}(X,\theta )$ follows from 22.1.88 and the cocycle property of the $\mathcal{I}$-projection. In order to prove (22.1.90), we fix $v\in X^{\operatorname{div}}$ and $\omega \in \textup{Käh}(X)$. The inductive construction in 22.1.88 uses only finite maxima and positive affine combinations of the potentials $A_{\tau _ i}$. On divisorial valuations, analytification commutes with both operations. Moreover, $P_{\theta +\omega }[\bullet ]_{\mathcal I}$ does not change divisorial valuations. Hence
\[ \tau \mapsto \left(P_{\theta +\omega }[A'_\tau ]_{\mathcal I}\right)^{\mathrm{An}}(v)= (A'_\tau )^{\mathrm{An}}(v) \]
is the ordinary upper concave envelope of $\tau \mapsto A_\tau ^{\mathrm{An}}(v)$. The Legendre-type supremum over $\tau $ is therefore unchanged by passing from $A$ to $A'$, which proves (22.1.90).
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.