Lemma 12.17.473E13B1F9B7A

Proof. We first check that the determinant measure in (12.7.1) does not depend on the Hermitian form used to express it. Let $T$ be a positive $(2,2)$-current, and let $\sigma _ h=h\wedge T$ and $P=P_ h$ be its trace measure and coefficient matrix relative to $h$, as in section 12.7. In an $h$-unitary frame let $G$ be the matrix of $g$, and set

Comparing the defining trace pairings shows that $\sigma _ g=g\wedge T$ and $P_ g$ are the trace measure and coefficient matrix of $T$ relative to $g$, with $\operatorname {tr}P_ g=3$. Since $G>0$ and $\operatorname {tr}P=3$, one has $a>0$ even where $P$ is singular. Moreover $\det P_ g=a^{-3}\det G\det P$ and $g^3=\det G\, h^3$, so the factors $a$ and $\det G$ cancel in $\Gamma _0(T)$. The calculation applies to the matrix density on the whole trace measure, including its singular part.

For $T\in \mathscr G_ h$ put $m=\int _ Xg\wedge T$. Then $0<m\leq c^{-1}$ and $U=T/m\in \mathscr G_ g$. By homogeneity, $k_ T=m k_ U$. Applying theorem 12.9.2 to $g$ gives the norm bound, and applying it again to $h$ gives the lower-mass bound. The supremum here is taken over all of $\mathscr G_ h$, including singular and degenerate currents. $\square$