PomeLemma 12.18.6.5771D42397D7 For a fixed positive current $T$ and fixed $\mathrm dV$, the measure $\Gamma _{0,h}(T)$ is independent of the positive auxiliary metric $h$. This statement includes singular and degenerate currents.
12.18.5 Intrinsic determinant measures and the complete profile5613B7E2253B
Keep the smooth positive probability volume $\mathrm dV$ fixed. For a positive $(n-1,n-1)$-current $T$ of finite mass, let $\Gamma _{0,h}(T)$ denote the determinant measure of (12.7.1) computed with a positive auxiliary metric $h$. The current need not be smooth or strictly positive. The following coordinate calculation identifies the measure itself, including on singular coefficient measures.
Proof. In a holomorphic frame put $\lambda =\bigwedge _{j=1}^ n(i\, dz_ j\wedge d\bar z_ j)$, $h=i\sum H_{ij}\, dz_ i\wedge d\bar z_ j$ and $\mathrm dV=v\lambda $. Positive currents have measure coefficients. Choose a positive Radon measure $\nu $ dominating them. For a smooth real form $\xi =i\sum \Xi _{ij}\, dz_ i\wedge d\bar z_ j$, let $Q$ be the positive semidefinite Hermitian coefficient matrix characterized by $\xi \wedge T=\operatorname {tr}(\Xi Q)\nu $. Then
\[ \sigma _ h=\operatorname {tr}(HQ)\nu ,\qquad P_ h=\frac{nH^{1/2}QH^{1/2}}{\operatorname {tr}(HQ)}. \]
The formula holds on $Q\ne 0$; zero is assigned when $Q=0$. Since $h^ n=n!\det (H)\lambda $, determinant algebra gives
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12.18.7
\begin{equation} \Gamma _{0,h}(T) =n(n!)^{-1/n}v^{1/n}(\det Q)^{1/n}\nu . \end{equation}
This expression has no $H$. Replacing $\nu $ by $L\nu $, where $L>0$ almost everywhere, replaces $Q$ by $Q/L$, leaving the measure unchanged; a common dominating measure compares any two choices. The original geometric definition ensures compatibility under frame changes. Matrix AM–GM bounds it locally by a constant times a reference trace measure, proving local finiteness. Compactness gives a finite Radon measure. $\square$
Write $\Gamma _0(T)$ for this intrinsic measure.
Theorem 12.18.8.3EA495FF9D1A Let $\beta $ be smooth positive Hermitian with $\mathrm{dd}^{\mathrm{c}}(\beta ^ j)=0$ for $1\leq j\leq n-1$, $V=\int _ X\beta ^ n$, and let $h=\beta +\mathrm{dd}^{\mathrm{c}}u>0$ for a smooth global real function $u$. Then
\[ \mathscr G_ h=\mathscr G_\beta ,\qquad \{ \Gamma _{0,h}(T):T\in \mathscr G_ h\} =\{ \Gamma _{0,\beta }(T):T\in \mathscr G_\beta \} . \]
Every measure in this common family has a density $k_ T$ satisfying
\[ k_ T\geq 0,\qquad \| k_ T\| _{L^{n/(n-1)}(\mathrm dV)}\leq V^{-1/n}. \]
For every real or complex $p\in L^ n(\mathrm dV)$,
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12.18.9
\begin{equation} \sup _{T\in \mathscr G_ h}\int _ X|p|\, \Gamma _0(T) =V^{-1/n}\| p\| _{L^ n(\mathrm dV)}. \end{equation}
All suprema retain singular and degenerate tests.
Proof. Distributional integration by parts gives $\int h\wedge T=\int \beta \wedge T+\int u\mathrm{dd}^{\mathrm{c}}T =\int \beta \wedge T$. This proves equality of the actual current sets; Lemma 12.18.6 proves equality of their measure sets.
Take a smooth strictly positive real $r$ and put $M=\int r^ n \mathrm dV$. By theorem 14.2.1 there are a smooth potential $w$, a positive form $A=\beta +\mathrm{dd}^{\mathrm{c}}w$ and a positive coefficient in its prescribed-volume equation. The mass-invariance hypothesis and theorem 11.1.7 determine that coefficient, giving
In particular, the total mass of $A$ is $V$. No estimates uniform in a degenerating smooth density are required.
For every $T\in \mathscr G_\beta $, integration by parts gives $\int A\wedge T=1$. In an $A$-unitary frame, matrix AM–GM gives $\det (P_ A)^{1/n}\leq 1$, including singular coefficient densities. Thus as Radon measures
\[ \Gamma _0(T)\leq (\mathrm dV/A^ n)^{1/n}(A\wedge T). \]
Multiplying by $r$ and using (12.18.10) yields
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\begin{equation} \int r\, \Gamma _0(T)\leq V^{-1/n}\| r\| _{L^ n(\mathrm dV)}. \end{equation}
Replacing a smooth nonnegative $r$ by $r+\eta $ and letting $\eta \downarrow 0$ extends this bound to nonnegative smooth tests. For smooth real or complex $s$, smooth positive approximations $\sqrt{|s|^2+\eta ^2}$ imply $|\int s\, \Gamma _0(T)|\leq V^{-1/n}\| s\| _ n$. Consequently integration against $\Gamma _0(T)$ extends continuously from smooth functions to $L^ n$. Its positive dual representative is $k_ T\in L^{n/(n-1)}$ with the stated bound. Equality on smooth tests identifies the two Radon measures, proving absolute continuity. The integrals of arbitrary $L^ n$ roots are therefore well-defined.
For the reverse inequality with $r>0$ smooth, let $T_ r=A^{n-1}/V$. It is smooth and positive. Write $E=\mathrm{dd}^{\mathrm{c}}w$. For $j\leq n-1$, expansion and closedness of $E$ give
\[ \mathrm{dd}^{\mathrm{c}}(A^ j)=\sum _{l=0}^ j\binom jl\mathrm{dd}^{\mathrm{c}}(\beta ^{j-l})\wedge E^ l=0. \]
The term with $j=l$ differentiates the constant one. Thus $\mathrm{dd}^{\mathrm{c}}T_ r=0$, and smooth integration by parts gives
\[ \int _ X\beta \wedge A^{n-1} =\int _ X A^ n-\int _ X\mathrm{dd}^{\mathrm{c}}w\wedge A^{n-1}=V. \]
Consequently $T_ r\in \mathscr G_\beta $. Its positive root is $\gamma =V^{-1/(n-1)}A$, so
\[ k_{T_ r}=(\gamma ^ n/\mathrm dV)^{(n-1)/n} =V^{-1/n}\frac{r^{n-1}}{\| r\| _ n^{n-1}}. \]
It attains equality in (12.18.11). For a nonnegative $p\in L^ n$, take smooth positive $r_ j\to p$ in $L^ n$. The common dual bound controls $\int (p-r_ j)k_{T_{r_ j}}\mathrm dV$, proving the lower bound in (12.18.9); the upper bound already holds for every current. Apply the same argument to $|p|$ for complex roots. $\square$
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