PomeExample 12.10.41.2A68BBAE5086 For every $n\geq 2$, there is a smooth nonnegative density $f$ on $\mathbb P^ n$, vanishing precisely on the coordinate hyperplanes, such that $b_\omega (f)=1$ and $D_{f^{1/n}}$ has a smooth semipositive maximizer strictly positive off those hyperplanes. The threshold solution is smooth, but $f^{1/n}$ is not differentiable at a general point of any coordinate hyperplane. There is no smooth strictly positive maximizing form. The normalized greatest exponential solutions converge uniformly to the normalized threshold solution.
12.10.40 A branched projective exampleBE9EC228849D
The squaring map of projective space gives a smooth threshold solution whose maximizing form degenerates along the coordinate hyperplanes. The density is smooth, but its $n$th root is not differentiable along these hyperplanes.
For this example take $X=\mathbb P^ n$, $n\geq 2$, with homogeneous coordinates $[Z_0:\cdots :Z_ n]$, and let $\omega $ be the Fubini–Study form normalized by $\int _ X\omega ^ n=1$. Set $\mathrm{d}V=\omega ^ n$.
Proof. Define the holomorphic map $F:\mathbb P^ n\to \mathbb P^ n$ and the semipositive form $A$ by
\[ F[Z_0:\cdots :Z_ n]=[Z_0^2:\cdots :Z_ n^2],\qquad A=\tfrac 12F^*\omega . \]
The homogeneous formula
\[ u=\tfrac 12\log \! \left(\sum _{i=0}^ n|Z_ i|^4\right) -\log \! \left(\sum _{i=0}^ n|Z_ i|^2\right) \]
defines a smooth function on projective space, and $A=\omega +\mathrm{dd}^{\mathrm{c}}u$. The form $A$ is semipositive, and it is strictly positive exactly where every homogeneous coordinate is nonzero. Define
\[ f=A^ n/\omega ^ n,\qquad T=A^{n-1}. \]
Since $\omega $ and $A$ are closed and cohomologous, Stokes' theorem gives $\int _ X f\omega ^ n=\int _ X\omega \wedge T=1$, and $\mathrm{dd}^{\mathrm{c}}T=0$; thus $T\in \mathscr G_\omega $. Every $v\in \mathrm{PSH}(X,\omega )$ has extended mass at most one: each bounded truncation $\max (v,-k)$ has mass one, and restricting its Monge–Ampère measure to $\{ v>-k\} $ can only decrease that mass. Consequently an inequality $\mathrm{MA}^{\mathrm{ext}}_\omega (v)\geq c f\omega ^ n$ forces $c\leq 1$. The smooth solution $u$ gives the reverse bound, so $b_\omega (f)=1$.
The determinant formula for smooth semipositive forms gives
\[ k_ T=f^{(n-1)/n},\qquad D_{f^{1/n}}(T)=\int _{\mathbb P^ n}f\omega ^ n=1. \]
By (12.10.2) and $\rho _\omega (f^{1/n})=b_\omega (f)^{-1/n}=1$, the form $T$ is a maximizer. It is strictly positive away from the coordinate hyperplanes because $A$ is; on these hyperplanes $A$ has deficient rank, and hence so does the coefficient matrix of $T=A^{n-1}$.
For the density root, use the chart $Z_0=1$ with coordinates $z_1,\ldots ,z_ n$. Pulling back the Fubini–Study volume by $(z_1,\ldots ,z_ n)\mapsto (z_1^2,\ldots ,z_ n^2)$ and including the factor $2^{-n}$ gives
\[ f(z)=2^ n\left(\prod _{i=1}^ n|z_ i|^2\right) \frac{(1+\sum _ i|z_ i|^2)^{n+1}}{(1+\sum _ i|z_ i|^4)^{n+1}}. \]
Near a point of $\{ z_1=0\} $ at which all the other coordinates are nonzero, $f^{1/n}$ is $|z_1|^{2/n}$ times a smooth strictly positive function. Restricting to a real line transverse to that hyperplane proves nondifferentiability, also in the case $n=2$, where $|z_1|^{2/n}=|z_1|$. By symmetry of the coordinates, the same holds on every coordinate hyperplane. The absence of a smooth strictly positive maximizer follows from corollary 12.10.14, since $f$ has zeros. Finally, $f$ is bounded, so corollary 12.10.31 gives the uniform convergence of the normalized exponential solutions. $\square$
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