PomeCorollary 18.26.11.179E3F0D00D8 Let $u,v\in \mathrm{PSH}(X,\omega )$. For every $\pi \in \mathscr R$, the polar curvature currents of $u\circ \pi $ and $v\circ \pi $, defined by coefficientwise restriction, are closed, and they agree if and only if $\mu _\pi (u)=\mu _\pi (v)$. Consequently, $\delta _\omega ^{\mathrm{pol}}(u,v)=0$ exactly when all these polar currents agree.
Corollary 18.26.11179E3F0D00D8
Proof. By proposition 5.43.18, the pullbacks belong to $\mathrm{PSH}(Y_\pi ,\pi ^*\omega )$, and proposition 5.46.7, applied with background $\pi ^*\omega $, shows that their polar currents are closed and positive. Both currents are carried by the union of the two Borel pole sets, which has Lebesgue measure zero. Equality of their $\eta _\pi $-traces therefore gives equality of the currents by lemma 5.46.4, and the converse is immediate. Since every $A_\pi $ is positive, the last assertion follows. $\square$
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