Pome

Proposition 19.3.9EF8E476D98F4

Proposition 19.3.9.EF8E476D98F4 Let $T$, $S$ be closed positive $(1,1)$-currents on $X$, $\lambda \in \mathbb {R}_{\geq 0}$. Then

  1. E67E9614A339

    if $T\preceq _{\mathcal{I}} S$, we have

    98CF17AA36AD 19.3.10
    \begin{equation} \nu _{Y_{\bullet }}(T)\geq _{\mathrm{lex}}\nu _{Y_{\bullet }}(S). \end{equation}
  2. BED1B9BF6DB3

    We have the following additivity property:

    61539109DEC3 19.3.11
    \begin{equation} \nu _{Y_{\bullet }}(T+S)=\nu _{Y_{\bullet }}(T)+\nu _{Y_{\bullet }}(S),\quad \nu _{Y_{\bullet }}(\lambda T)=\lambda \nu _{Y_{\bullet }}(T). \end{equation}

Proof. 1 We make an induction on $n\geq 0$. The case $n=0,1$ is trivial. Assume that $n\geq 2$ and the case $n-1$ is known. Observe that $\nu (T,Y_1)\geq \nu (S,Y_1)$, if the inequality is strict, we are done. So let us assume that $\nu (T,Y_1)=\nu (S,Y_1)$. By proposition 20.2.21, we find that

\[ \operatorname{Tr}_{Y_1}\Bigl (T-\nu (T,Y_1)[Y_1]\Bigr )\preceq _ P\operatorname{Tr}_{Y_1}\Bigl (S-\nu (T,Y_1)[Y_1]\Bigr ). \]

Hence the same comparison holds for $\preceq _{\mathcal{I}}$ by corollary 16.8.14. By the inductive hypothesis, we conclude (19.3.10).

2 We make an induction on $n\geq 0$. The cases $n=0,1$ are trivial. Assume that $n\geq 2$ and the case $n-1$ is known. By proposition 5.16.8, we have

\[ \nu (T+S,Y_1)=\nu (T,Y_1)+\nu (S,Y_1),\quad \nu (\lambda T,Y_1)=\lambda \nu (T,Y_1). \]

By proposition 20.2.21, we have

\[ \begin{split} \operatorname{Tr}_{Y_1}\Bigl (T+S-\nu (T+S,Y_1)[Y_1]\Bigr )\sim _ P& \operatorname{Tr}_{Y_1}\Bigl (T-\nu (T,Y_1)[Y_1]\Bigr )\\ & +\operatorname{Tr}_{Y_1}\Bigl (S-\nu (S,Y_1)[Y_1]\Bigr ),\\ \operatorname{Tr}_{Y_1}\Bigl (\lambda T-\nu (\lambda T,Y_1)[Y_1]\Bigr )\sim _ P& \lambda \operatorname{Tr}_{Y_1}\Bigl (T-\nu (T,Y_1)[Y_1]\Bigr ). \end{split} \]

Since $\sim _ P$ implies $\sim _{\mathcal{I}}$ by corollary 16.8.14, the inductive hypothesis and the independence of the valuation from the trace representative conclude (19.3.11). $\square$

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