15.30 Exact equations on products with zero lower massFC44870E2A2C

Proof. In homogeneous coordinates, $\kappa +\mathrm{dd}^{\mathrm{c}}\psi =\mathrm{dd}^{\mathrm{c}}\log \sum _{i=1}^ k|z_ i|^2\geq 0$ on $Y\setminus \{ o\} $. The function $\chi (t)=1-(1-t)^\alpha $, $t\leq 0$, is increasing and convex, with $0<\chi '\leq \alpha $. Hence, off $o$,

Since $u_\alpha $ is bounded above, it extends across $o$ as a $\kappa $-plurisubharmonic function, with value $-\infty $ at $o$. On the hyperplane $\{ z_0=0\} $ we have $\psi =0$, hence $u_\alpha =0$, so $\sup _ Yu_\alpha =0$.

To compute the measure, work in the affine chart $z_0=1$ and put $r=\sum _{i=1}^ k|z_ i|^2$. On this chart $\kappa =\mathrm{dd}^{\mathrm{c}}\log (1+r)$ and $\kappa +\mathrm{dd}^{\mathrm{c}}u_\alpha =\mathrm{dd}^{\mathrm{c}}F$, where $s(r)=1-\log (r/(1+r))$ and

For a radial function, the Levi form has the tangential eigenvalue $F'(r)=Q(r)/r$, with multiplicity $k-1$, and the radial eigenvalue

Integrating in the radial variable gives $\int _{\{ a<r<b\} }(\mathrm{dd}^{\mathrm{c}}F)^ k=Q(b)^ k-Q(a)^ k$; the constant of proportionality is fixed by the case $F=\log (1+r)$, for which $(\mathrm{dd}^{\mathrm{c}}F)^ k=\kappa ^ k$ has mass one on the chart and $Q(r)=r/(1+r)$. Since again $Q(0)=0$ and $Q(\infty )=1$, the punctured chart has mass one. The complement of the chart is a hyperplane, which has zero volume, and the non-pluripolar measure does not charge $o$. Hence $\int _ Yq_\alpha \kappa ^ k=1$.

Relative to $\kappa ^ k$, whose tangential and radial eigenvalues are $1/(1+r)$ and $1/(1+r)^2$, the density on this chart is

Near $r=0$ the smooth volume is comparable to $r^{k-1}\mathrm{d}r$ in the radial variable, so $q_\alpha $ is integrable. For $p>1$, the substitution $L=1-\log r$ turns the integral of $q_\alpha ^ p$ into the integral of $\mathrm{e}^{k(p-1)L}$ times a power of $L$, which diverges. The same asymptotics show that $q_\alpha (r)\to +\infty $ uniformly as $r\downarrow 0$, so every truncation $\min (q_\alpha ,s)$ is continuous at $o$. Finally, $q_\alpha \geq (1-\alpha )^ k>0$ by the lower bound $\kappa +\mathrm{dd}^{\mathrm{c}}u_\alpha \geq (1-\alpha )\kappa $. $\square$

Proof. Suppose first that $g$ is smooth. Since $\kappa $ is Kähler, theorem 14.2.1 on $Y$ gives a smooth $\phi _ g$ with $A_ g=\kappa +\mathrm{dd}^{\mathrm{c}}\phi _ g>0$ and $A_ g^ k=(g/M_ g)\kappa ^ k$; the scalar is determined by integration, since both sides have mass one. On $X$, since $h_0^{m+1}=0$ and $A_ g^{k+1}=0$ for dimensional reasons,

Uniqueness of the scalar gives $c_ H(Rg)=CV_0/M_ g$. For continuous $g>0$, choose smooth $g_ l>0$ converging uniformly to $g$. Since $g$ has a positive minimum, $(1-\varepsilon _ l)g\leq g_ l\leq (1+\varepsilon _ l)g$ for some $\varepsilon _ l\in (0,1)$ tending to zero. The order and homogeneity of the scalar in lemma 15.27.1 give

Since $M_{g_ l}\to M_ g$, the smooth case proves the formula. $\square$

Proof. Step 1: the threshold. Write $a=\min _ XR>0$, $A=\max _ XR$, and $M_ s=\int _ Y\min (q_\alpha ,s)\kappa ^ k$ for $s>0$. By lemma 15.30.1, the truncations $\min (q_\alpha ,s)$ are continuous and strictly positive, and $M_ s\uparrow 1$ as $s\to \infty $. The truncations of $f$ satisfy

Hence lemma 15.30.2 and the monotonicity of the scalar give

Both bounds tend to $CV_0$ as $j\to \infty $, so $b_ H(f)=CV_0$, whatever the reference volume $\mathrm{d}V$.

Step 2: the equation. For $u_{\alpha ,j}=\max (u_\alpha ,-j)$, the binomial expansion of bounded local products, with factor forms of degrees $m$ and $k$, gives

Multiplying by $\mathbf1_{\{ u_\alpha >-j\} }$ and letting $j\to \infty $, as in the definition (13.14.4) of the extended measure, we obtain

a measure of mass $CV_0$. The bounds $a\leq R\leq A$ transfer the integrability properties of $q_\alpha $ to $f$.

Step 3: lower mass and nonclosedness. Pull back bounded $h_0$-plurisubharmonic functions whose masses on $X_0$ tend to zero. The same expansion shows that their $H$-masses are $C$ times their masses on $X_0$, so $v_-(H)=0$. If $h_0$ were closed, Stokes' theorem would give every bounded $h_0$-plurisubharmonic function the mass $V_0>0$, contradicting $v_-(h_0)=0$. Hence $h_0$, and therefore $H$, is not closed.

Step 4: the envelopes. Choose $\alpha <\beta <1$. For $b\geq 1$ and $\varepsilon >0$ with $\varepsilon \beta \leq 1$, the number

is finite, since $\beta >\alpha $. The function $g_ b=-\varepsilon (1-\psi )^\beta -K_ b$ on $Y$ satisfies $\kappa +\mathrm{dd}^{\mathrm{c}}g_ b\geq (1-\varepsilon \beta )\kappa \geq 0$, by the chain-rule computation in the proof of lemma 15.30.1, applied to the increasing convex function $t\mapsto -\varepsilon (1-t)^\beta $, whose derivative is at most $\varepsilon \beta $ for $t\leq 0$. With $L=1-\psi \geq 1$, the inequality $g_ b\leq bu_\alpha $ reads $bL^\alpha -\varepsilon L^\beta -b\leq K_ b$, so it holds on $Y\setminus \{ o\} $, and also at $o$, where both sides are $-\infty $. Thus the pullback of $g_ b$ belongs to $\mathrm{PSH}(X,H)$ and lies below $bu_\alpha $. Since $b\geq 1$ is arbitrary, $u_\alpha \in \mathcal E_{\mathrm{env}}(X,H,0)$ by definition 16.28.61. For $0<\delta <1$, take $b=1$ and $\varepsilon \beta \leq \delta $. Then $g_1$ is $\delta H$-plurisubharmonic and lies below $u_\alpha $, so $P_{\delta H}(u_\alpha )\geq g_1$ is not identically $-\infty $. $\square$

Proof. Theorem 12.19.1 gives $v_-(\omega )=0$, and proposition 12.19.18 gives $V_0=24\pi $. The product theorem applies with $m=3$, $k=2$ and $C=\binom 53=10$. $\square$

Proof. We first compute the scalars of bounded densities, then use the closed factor's mass bound to identify the equation for an unbounded density.

Step 1: bounded densities. If $g$ is bounded and nonzero, theorem 14.19.2 on $Y$ gives a bounded continuous $v$ with $\mathrm{MA}_\kappa (v)=c_\kappa (g)g\kappa ^ k$. Closed Stokes gives $c_\kappa (g)=1/M_ g$. The bounded product expansion gives

Uniqueness of the bounded normalization scalar therefore gives $c_ H(Rg)=c_0/M_ g$, including densities with zeros.

Step 2: the threshold. For general $g\in L^1$, the original truncations on $X$ satisfy

By Step 1 and lemma 15.27.1,

Both bounds tend to $c_0/M_ g$.

Step 3: the exact equation. By proposition 15.28.7, $b_\kappa (g)=1/M_ g$. Thus theorem 15.28.4 on $Y$ gives $v\in \mathrm{PSH}(Y,\kappa )$ with supremum zero and

For $v_ j=\max (v,-j)$, closed Stokes gives $\int _ Y\mathrm{MA}_\kappa (v_ j)=1$. The defining non-pluripolar truncations consequently give $\int _ Y\mathrm{MA}^{\mathrm{ext}}_\kappa (v)\leq 1$. The prescribed lower measure has mass one, so equality holds both in mass and in the measure inequality.

The pullback $u$ is $H$-plurisubharmonic, and $\max (u,-j)$ is the pullback of $v_ j$. Multiplying the bounded product expansion by the pullback of $\mathbf1_{\{ v>-j\} }$ and passing to the defining non-pluripolar limit gives

Its mass is $c_0$. Finally, when $v_-(h_0)=0$, pulling back bounded potentials from $X_0$ whose masses tend to zero proves $v_-(H)=0$ by the same expansion. Closedness of $h_0$ would contradict that vanishing by Stokes' theorem, so $H$ is not closed. $\square$