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Lemma 22.1.105CB69CB164255

Lemma 22.1.105.CB69CB164255 The map (22.1.102) is order preserving. Moreover, suppose that $\Gamma ,\Gamma '\in \mathrm{PSH}^{\mathrm{NA}}(X,\theta )$ satisfies that $\Gamma ^{\mathrm{An}}\leq \Gamma '^{\mathrm{An}}$, then $\Gamma \leq \Gamma '$.

In particular, the map (22.1.102) is injective.

Proof. The map (22.1.102) is order preserving by definition. Let us take $\Gamma ,\Gamma '\in \mathrm{PSH}^{\mathrm{NA}}(X,\theta )$ with $\Gamma ^{\mathrm{An}}\leq \Gamma '^{\mathrm{An}}$. Fix a Kähler form $\omega $ on $X$.

Let $v\in X^{\operatorname{div}}$ and $t \in \mathbb {Q}_{>0}$. Then, using (22.1.83) and the homogeneity of divisorial valuations, we notice that

99B9B27CF451 22.1.106
\begin{equation} t \Gamma ^\mathrm{An}\left(t^{-1} v \right) = \sup _{\tau \in \mathbb {R}}\left( \left(\Gamma _{\tau }^{\theta +\omega }\right)^{\mathrm{An}}(v)+ t \tau \right). \end{equation}

A similar equality holds for $\Gamma '$. Therefore, by corollary 21.11.26, we have

\[ \left(\Gamma _{\tau }^{\theta +\omega }\right)^{\mathrm{An}}\leq \left(\Gamma '{}_{\tau }^{\theta +\omega }\right)^{\mathrm{An}} \]

for all $\tau \in \mathbb {R}$. It follows that

\[ \Gamma _{\tau }^{\theta +\omega }\preceq _{\mathcal{I}} \Gamma '{}_{\tau }^{\theta +\omega } \]

by proposition 16.8.9. Since both sides are $\mathcal{I}$-model, proposition 16.8.17 gives

\[ \Gamma _{\tau }^{\theta +\omega }\leq \Gamma '{}_{\tau }^{\theta +\omega } \]

for all $\tau \in \mathbb {R}$. Hence $\Gamma \leq \Gamma '$ by the definition of the order. $\square$

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