PomeCorollary 19.5.44.E7520E0649C0 Assume that $T\in \mathcal{Z}_+(X,\alpha )_{>0}$. We have
4DF707F26C95
19.5.45
\begin{equation} \min _{\mathrm{lex}} \Delta _{Y_{\bullet }}(T)=\nu _{Y_{\bullet }}(T). \end{equation}
Corollary 19.5.44E7520E0649C0
Proof. By theorem 19.5.40, it is clear that
\[ \min _{\mathrm{lex}} \Delta _{Y_{\bullet }}(T)\leq _{\mathrm{lex}}\nu _{Y_{\bullet }}(T). \]
On the other hand, we clearly have
\[ \Delta _{Y_{\bullet }}(T)\subseteq \Delta _{Y_{\bullet }}(T+\omega ) \]
for any Kähler form $\omega $ on $X$. It follows that
\[ \min _{\mathrm{lex}} \Delta _{Y_{\bullet }}(T)\geq _{\mathrm{lex}} \min _{\mathrm{lex}} \Delta _{Y_{\bullet }}(T+\omega ). \]
By theorem 19.5.37, the right-hand side is just $\nu _{Y_{\bullet }}(T+\omega )=\nu _{Y_{\bullet }}(T)$. We conclude the proof. $\square$
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