PomeProposition 12.34.2.635FE1D11798 Let $\theta $ be nef with $\lambda _\mu (\theta )>0$. If $T\in \mathscr G_ h$ satisfies $\int _ X\theta \wedge T=0$, then the determinant-weighted measure of $T$, computed as in (12.7.1) with reference metric $h$ and with $\mu $ in place of $\mathrm e^ F\mathrm dV$, vanishes.
Proposition 12.34.2635FE1D11798
Proof. In dimension one the hypotheses cannot hold, since $\int _ X\theta \wedge T=\lambda _\mu (\theta )/\int _ Xh>0$. Suppose $n\geq 2$, and write $D_\mu (T)$ for the total mass of this determinant-weighted measure; for $T=\gamma ^{n-1}$ it is the denominator $D_\mu (\gamma )$ of definition 12.2.1. If $\eta \in \mathscr P$, its solution form $\Omega $ satisfies $\Omega ^ n=\lambda _\mu (\eta )^ n\mu $, so $\Omega /\lambda _\mu (\eta )$ is an admissible test in the infimum of lemma 12.7.2. Together with Stokes' theorem, this gives
\[ \int _ X\eta \wedge T=\int _ X\Omega \wedge T \geq \lambda _\mu (\eta )D_\mu (T). \]
Apply this to $\eta =\theta +\varepsilon h$. The left side equals $\varepsilon $, while theorem 12.11.1 gives $\lambda _\mu (\eta )\to \lambda _\mu (\theta )>0$. Hence $D_\mu (T)=0$. The determinant-weighted measure is positive and has total mass zero, so it vanishes. $\square$
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