PomeB7138A0B3C69Integration against $w^2$ also measures the obstruction to using the unweighted cofactor, that is, the case $h=1$. For smooth potentials, the error is exactly the pairing of $w^2$ with $\mathrm{dd}^{\mathrm{c}}T$. On a surface, the Gauduchon condition makes this error vanish for every pair of potentials; in dimension three, $\mathrm{dd}^{\mathrm{c}}T$ contains an additional term depending on the potentials.
Proposition 14.22.13.F563CACF5D7D Let $n\geq 2$, and let $u,v$ be smooth $\omega $-plurisubharmonic functions with $A^ n=B^ n=f\, \mathrm{d}V$ for a smooth function $f\geq 0$. Then
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14.22.14
If $\mathrm{dd}^{\mathrm{c}}T=0$ and $\{ f>0\} $ is dense in $X$, then $u-v$ is constant.
Proof. By (14.20.1), $\mathrm{dd}^{\mathrm{c}}w\wedge T=0$. Integrating (14.22.5) against $T$, and applying Stokes' theorem twice to the left-hand side, yields
Substitution proves (14.22.14).
For the uniqueness assertion, apply 14.22.1(1) with $h=1$. At every point where $f>0$, the form $B$ is positive definite, so the term $B^{n-1}$ makes the cofactor $T$ strictly positive. The set where $T$ is strictly positive is therefore dense, as required. $\square$
Corollary 14.22.15.EAC3CF41BD99 Assume $n=2$ and $\mathrm{dd}^{\mathrm{c}}\omega =0$. Let $f\geq 0$ be smooth, with a zero set of empty interior, and let $c>0$. Then any two smooth $\omega $-plurisubharmonic solutions of
normalized by $\sup _ Xu=0$ coincide.
Proof. Here $T=A+B$ and $\mathrm{dd}^{\mathrm{c}}T=2\mathrm{dd}^{\mathrm{c}}\omega =0$. The set $\{ f>0\} $ is dense, so the proposition, applied to the density $cf$, makes the difference of the two solutions constant, and the normalization makes this constant zero. $\square$
In dimension three, direct expansion gives
The Gauduchon condition cancels the first term but need not cancel the second. For instance, take both potentials equal to the perturbation $sv$ in example 11.23.5. The common form $A=B$ is positive and satisfies $\mathrm{dd}^{\mathrm{c}}(A^2)\ne 0$, so $\mathrm{dd}^{\mathrm{c}}T=3\mathrm{dd}^{\mathrm{c}}(A^2)\ne 0$. Thus even for two equal positive solution forms, the cofactor need not be $\mathrm{dd}^{\mathrm{c}}$-closed.
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