12.10.16 A smooth maximizing form with an analytic degeneracy set8D604EFF48DF

Proof. Put $p=f^{1/n}$ and $A=\omega +\mathrm{dd}^{\mathrm{c}}u$, where $u$ is any threshold subsolution; one exists by theorem 15.28.4. Fix a smooth strictly positive $T_0\in \mathscr G_\omega $, and for real $\varepsilon >0$ set $T_\varepsilon =(T+\varepsilon T_0)/(1+\varepsilon )$. Each $T_\varepsilon $ is a smooth strictly positive member of $\mathscr G_\omega $, so proposition 12.10.11 gives

For a smooth semipositive form $T$ with coefficient matrix $P'$ relative to $\omega ^ n$, (12.7.1) gives $k_ T\, \mathrm{d}V=f_0(\det P')^{1/n}\omega ^ n$, so that $k_ T=f_0^{1-n}(\det P')^{1/n}$ is a continuous function of that matrix. Thus $k_{T_\varepsilon }\to k_ T$ uniformly on the compact manifold $X$. Since $p\in L^1$, passing to the limit gives

The left side has mass one by Stokes' theorem and $\mathrm{dd}^{\mathrm{c}}T=0$, and the right side has mass $b^{1/n}D_ p(T)=1$ because $T$ maximizes. Their difference is positive and has mass zero, hence

On $U=X\setminus Z$ write $T=\gamma ^{n-1}$, where $\gamma $ is smooth and strictly positive, and set $a=(bf\, \mathrm{d}V/\gamma ^ n)^{1/n}$. The trace equality becomes $A\wedge \gamma ^{n-1}=a\gamma ^ n$ on $U$. On relatively compact subsets of $U$, it controls every coefficient measure of $A$ by an $L^ n$ density. The trace inequalities for all Hermitian forms and the matrix arithmetic–geometric mean inequality, used exactly as in the proof of corollary 12.10.12, force $A=a\gamma $ on $U$. This step uses only the positivity of $\gamma $ on $U$, since the Hermitian test forms needed there can be taken to be global forms obtained by extension from smaller coordinate charts.

The local regularity lemma lemma 5.22.4 makes $u$ and $a$ smooth on $U$. Taking the top power proves the equation on $U$, and $f^{1/n}=b^{-1/n}a(\gamma ^ n/\mathrm{d}V)^{1/n}$ is smooth and nonnegative there. Both $\mathrm{MA}^{\mathrm{ext}}_\omega (u)$ and $f\, \mathrm{d}V$ put no mass on the pluripolar set $Z$. Hence the equality holds on $X$, and its total mass is $b\int _ Xf\, \mathrm{d}V<+\infty $.

It remains to prove uniqueness. Let $v$ be another threshold subsolution and put $w=\max (u,v)$. By proposition 13.14.6,

Here the pole sets carry no mass for any of the measures involved. Thus $w$ is also a threshold subsolution, and the preceding argument gives the same smooth current $\omega +\mathrm{dd}^{\mathrm{c}}u=\omega +\mathrm{dd}^{\mathrm{c}}v=\omega +\mathrm{dd}^{\mathrm{c}}w=a\gamma $ on $U$.

We apply lemma 5.38.2 to $u\leq w$ and to $v\leq w$: their complex Hessians agree on $X\setminus Z$, so $w-u$ and $w-v$ are constant. Hence $u-v$ is constant. $\square$