PomeProposition 19.3.7.FE522E15C1BA Let $T$ be a closed positive $(1,1)$-current on $X$. Then $\nu _{Y_{\bullet }}(T)\in \mathbb {R}_{\geq 0}^ n$ defined in definition 19.3.6 is independent of the choices of the trace operators in the definition. Moreover, $\nu _{Y_{\bullet }}(T)$ depends only on the $\mathcal{I}$-equivalence class of $T$.
Proposition 19.3.7FE522E15C1BA
Proof. We will prove both statements at the same time by induction on $n\geq 0$. The cases $n=0,1$ are trivial.
Let us consider the case $n>1$ and assume that the result is known in dimension $n-1$. We first observe that $\nu _{Y_{\bullet }}(T)$ is independent of the choice of the trace operator: Different choices of $\operatorname{Tr}_{Y_1}(T-\nu (T,Y_1)[Y_1])$ are $\mathcal{I}$-equivalent by proposition 20.2.8. Therefore, by induction, its valuation is well-defined.
Next, let $T'$ be another closed positive $(1,1)$-current such that $T\sim _{\mathcal{I}}T'$. Using proposition 16.6.2, we know that $\nu (T,Y_1)=\nu (T',Y_1)$. Therefore,
\[ T- \nu (T,Y_1)[Y_1]\sim _{\mathcal{I}} T'-\nu (T',Y_1)[Y_1]. \]
It follows by induction that
\[ \nu _{Y_1\supseteq \cdots \supseteq Y_ n}\left(\operatorname{Tr}_{Y_1}(T-\nu (T,Y_1)[Y_1])\right)=\nu _{Y_1\supseteq \cdots \supseteq Y_ n}\left(\operatorname{Tr}_{Y_1}(T'-\nu (T',Y_1)[Y_1])\right). \]
This completes the proof. $\square$
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