PomeTheorem 19.5.37.EF3A66F3D2C0 Assume that $T\in \mathcal{Z}_+(X,\alpha )_{>0}$ is a Kähler current. We have
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19.5.38
\begin{equation} \min _{\mathrm{lex}} \Delta _{Y_{\bullet }}(T)=\nu _{Y_{\bullet }}(T). \end{equation}
Theorem 19.5.37EF3A66F3D2C0
Proof. We argue by induction on $n\geq 0$. The case $n=0$ is of course trivial. Let us assume that $n>0$ and the case $n-1$ has been proved.
We first observe that by theorem 19.5.26,
\[ \Delta _{Y_{\bullet }}(T-\nu (T,Y_1)[Y_1])+(\nu (T,Y_1),0,\ldots ,0)\subseteq \Delta _{Y_{\bullet }}(T). \]
By proposition 18.1.27 5, removing the divisorial part does not change the volume. Comparing the volumes of both sides using theorem 19.5.26 and proposition 18.1.27, we find that equality holds:
\[ \Delta _{Y_{\bullet }}(T-\nu (T,Y_1)[Y_1])+(\nu (T,Y_1),0,\ldots ,0)=\Delta _{Y_{\bullet }}(T). \]
Replacing $T$ by $T-\nu (T,Y_1)[Y_1]$, we may therefore assume that $\nu (T,Y_1)=0$. It suffices to apply 19.5.34 and the inductive hypothesis. $\square$
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