Pome

Theorem 14.27.1E16A6F31A061

Theorem 14.27.1.E16A6F31A061 There are a smooth nonclosed Hermitian form $\omega $ on $\mathbb P^3$ and a smooth nonnegative density $g$, positive off two disjoint smooth real hypersurfaces, such that the equation

\[ \mathrm{MA}_\omega (u)=g\omega _ X^3 \]

has uncountably many distinct smooth solutions normalized by $\sup _ Xu=0$. The background $\omega $ has bounded mass.

of theorem 14.27.1. Let $\eta _0$ be as in lemma 14.27.3, and for $0<\eta <\eta _0$ denote by $h_\eta $ the solution with initial value $sp(0)-\eta $. Extend $h_\eta $ by zero, which gives a smooth function by the preceding lemma, and take a primitive $G_\eta $. It is constant near each end of the real line, so $G_\eta (r)$ extends smoothly to all of $\mathbb P^3$, including the two omitted projective lines. Set

\[ \widetilde u_\eta =G_\eta (r)-\rho . \]

On the open band,

\[ \omega +\mathrm{dd}^{\mathrm{c}}\widetilde u_\eta =(x+h_\eta )P+(y-h_\eta )Q+(z+h_\eta ')R. \]

The coefficients of $P$ and $Q$ are strictly positive by lemma 14.27.3, while the equation gives

\[ z+h_\eta '=\frac{k_ s}{(x+h_\eta )(y-h_\eta )}>0. \]

Outside the band the form is $\omega _0$, and on the two boundary hypersurfaces it is also the rank-one form $\omega _0$, since $h_\eta $ vanishes to infinite order there. It is therefore semipositive on $X$, and by (14.26.4) its cube equals $6k_ sP\wedge Q\wedge R=(\omega _0+sH)^3$ on the band and $\omega _0^3=(\omega _0+sH)^3$ elsewhere. Thus its cube is everywhere the measure $g\omega _ X^3$ of (14.27.2). Finally, normalize by subtracting $\sup _ X\widetilde u_\eta $.

Distinct values of $\eta $ give different slopes at $r=0$, so the corresponding potentials do not differ by a constant, and their normalized versions are distinct. This gives an uncountable family. The background is that of theorem 14.26.1, so it has bounded mass. $\square$

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