Let $(X_0,h_0)$ be a connected compact Hermitian manifold of dimension $m\geq 1$. Set
\[ \begin{gathered} X=X_0\times Y,\qquad N=m+1,\qquad H=h_0+\kappa ,\\ V_0=\int _{X_0}h_0^ m,\qquad \nu =M\frac{h_0^ m}{V_0}\otimes \mu ,\qquad c=\frac{NV_0}{M}, \end{gathered} \]
where $M>0$ and factor forms and potentials are pulled back to $X$ when used there. Thus $\nu (X)=M$.
Proposition 15.31.13.BE532EB6E98C If $\mu $ is non-polar, then $\nu $ is non-pluripolar and
\[ u=(Q_\mu -\sup _ YQ_\mu )\circ \pi _ Y,\qquad \mathrm{MA}^{\mathrm{ext}}_ H(u)=c\nu ,\qquad \int _ X\mathrm{MA}^{\mathrm{ext}}_ H(u)=NV_0. \]
This is the unique supremum-zero solution among pullbacks of $\kappa $-plurisubharmonic functions. It is bounded exactly when $Q_\mu $ is bounded, and continuous exactly when $Q_\mu $ is continuous. If $v_-(h_0)=0$, then $v_-(H)=0$ and $H$ is not closed.
Proof.
A nontrivial local plurisubharmonic function on a product chart has a nontrivial subharmonic restriction to the curve for smooth-volume-almost every point of $X_0$, by local integrability and Fubini's theorem. Non-polarity of $\mu $ and a second application of Fubini give its pole set product measure zero. A countable cover proves that $\nu $ is non-pluripolar.
For bounded $p\in \mathrm{PSH}(Y,\kappa )$, bounded multilinearity and lemma 5.28.4 give
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15.31.14
\begin{equation} \mathrm{MA}_ H(p\circ \pi _ Y) =N h_0^ m\otimes (\kappa +\mathrm{dd}^{\mathrm{c}}p). \end{equation}
Only one curve factor survives in the expansion. For arbitrary $p$, apply this to $p_ j=\max (p,-j)$, restrict to the pullback of $\{ p>-j\} $ and increase $j$. The defining extended limit gives the same formula with $\mathrm{MA}^{\mathrm{ext}}_\kappa (p)$ on the right. Lemma 15.31.3 now proves the equation, finite mass and separated uniqueness. Boundedness and continuity follow from the same constant-difference identity.
If $v_-(h_0)=0$, pull back bounded potentials whose $h_0$-masses tend to zero. Their $H$-masses are $N$ times those masses, because $\int _ Y\kappa =1$, so $v_-(H)=0$. A closed $h_0$ would give every bounded potential mass $V_0>0$ by Stokes. Hence $h_0$ is not closed; restricting $H$ to a factor slice shows that $H$ is not closed either.
$\square$
For this fixed product measure define its subsolution threshold by
\[ b_ H(\nu )=\sup \{ a\geq 0:\exists p\in \mathrm{PSH}(X,H), \mathrm{MA}^{\mathrm{ext}}_ H(p)\geq a\nu \} . \]
For $\nu =f\, \mathrm dV$ this agrees with the individual-density threshold of theorem 15.28.4.
Theorem 15.31.15.5F0343E9A338 Suppose $Q=Q_\mu $ is finite and continuous. Then $b_ H(\nu )=c$, attained by the bounded continuous $u$ of proposition 15.31.13. For every $\lambda >0$, the continuous potential
\[ U_\lambda =\lambda ^{-1}\log c+w_\lambda \circ \pi _ Y \]
satisfies
\[ \mathrm{MA}_ H(U_\lambda )=\mathrm e^{\lambda U_\lambda }\nu , \qquad \int _ X\mathrm{MA}_ H(U_\lambda )=NV_0. \]
It dominates every $p\in \mathrm{PSH}(X,H)$ with $\mathrm{MA}^{\mathrm{ext}}_ H(p)\geq \mathrm e^{\lambda p}\nu $, and it is the unique bounded solution. For $v_\lambda =U_\lambda -\sup _ XU_\lambda $ and $c_\lambda =\mathrm e^{\lambda \sup _ XU_\lambda }$,
\[ \begin{gathered} \mathrm{MA}_ H(v_\lambda )=c_\lambda \mathrm e^{\lambda v_\lambda }\nu , \qquad c_\lambda =c\, \mathrm e^{\lambda \sup _ Yw_\lambda },\\ \| v_\lambda -u\| _\infty =O(\lambda ),\qquad \lambda ^{-1}\log (c_\lambda /c)\longrightarrow \sup _ YQ-\int _ YQ\, \mathrm d\mu . \end{gathered} \]
All the constructed potentials belong to $\mathcal E_{\mathrm{env}}(X,H,0)$. The same coefficient $c$ is attained if the defining supremum for $b_ H(\nu )$ is restricted to bounded potentials or to this envelope class.
Proof.
Lemma 15.31.4 gives non-polarity, and theorem 15.31.8 constructs $w_\lambda $. Substituting it in (15.31.14) proves the exponential equation and its finite mass. Supremum normalization gives the displayed $c_\lambda $, and the uniform curve estimate gives both limits. For any bounded $H$-plurisubharmonic $p$ and $t\geq 1$, the constant $a_ t=t\inf _ Xp$ is an $H$-plurisubharmonic competitor below $tp$. Hence $P_ H(tp)$ is genuine for every $t$, which proves the envelope-class assertion.
Greatestness. Write $s=U_\lambda $. If $b$ is a bounded exponential subsolution, then on $\{ s<b-\epsilon \} $,
\[ \mathrm{MA}_ H(s)\leq \mathrm e^{-\lambda \epsilon }\mathrm{MA}_ H(b-\epsilon ). \]
The bounded strict domination theorem, theorem 14.19.10, gives $s\geq b-\epsilon $. Letting $\epsilon \downarrow 0$ proves $s\geq b$, and applying the argument in both directions proves bounded uniqueness.
For an arbitrary exponential subsolution $r$, fix $A>0$ and put $p=s-A$, $W=\max (r,p)$. These maxima are $H$-plurisubharmonic by lemma 5.24.1, and $W$ is bounded. Choose $j$ with $-j<\inf _ Xp$ and set $r_ j=\max (r,-j)$; then $W=\max (r_ j,p)$. Plurifine locality on $\{ r_ j>p\} $ and lemma 5.28.3, applied to $p\leq W$ on their contact set, give
\[ \mathrm{MA}_ H(W)\geq \mathbf1_{\{ r_ j>p\} }\mathrm{MA}_ H(r_ j) +\mathbf1_{\{ r_ j\leq p\} }\mathrm{MA}_ H(p). \]
On the first set $r_ j=r>-j$, so strict-superlevel locality identifies the measure with $\mathrm{MA}^{\mathrm{ext}}_ H(r)$. On the second, $W=p$ and $\mathrm{MA}_ H(p)=\mathrm e^{\lambda s}\nu \geq \mathrm e^{\lambda p}\nu $. Thus $W$ is a bounded exponential subsolution, and $r\leq W\leq s$. This uses bounded products before every restriction and applies also when the measure of $r$ has infinite mass.
The maximal coefficient. Suppose $\mathrm{MA}^{\mathrm{ext}}_ H(r)\geq a\nu $ with $a>0$, and put $B=\sup _ Xr$. For every $\lambda >0$ the potential $r+\lambda ^{-1}\log a-B$ is an exponential subsolution, since its exponential is at most $a$. Greatestness gives
\[ r+\lambda ^{-1}\log a-B \leq \lambda ^{-1}\log c+w_\lambda \circ \pi _ Y. \]
Evaluate at a fixed finite point of $r$, multiply by $\lambda $, and let $\lambda \downarrow 0$. The uniform barriers for $w_\lambda $ give $a\leq c$. The equation for $u$ attains $c$ in each stated class.
$\square$
The order characterization of the zero-exponent limit is confined to the separated sector. Bounded uniqueness at a positive exponent and maximality of the coefficient do not determine all normalized critical potentials on the nonclosed product.
Proposition 15.31.16.DD2C5D7B64EC Suppose, for some $C>0$ and $\delta >0$,
\[ \mu (B_{\mathrm{ch}}(x,r))\leq Cr^\delta \qquad (x\in Y,\ 0<r\leq 1). \]
Then $Q_\mu $ is finite and continuous, so theorem 15.31.15 applies. In particular it applies to
\[ \mathrm d\mu (\vartheta )=(1+\epsilon \cos \vartheta ) \frac{\mathrm d\vartheta }{2\pi },\qquad 0<\epsilon <1, \]
on the circle $[1:\mathrm e^{\mathrm i\vartheta }]$. This measure and its product $\nu $ are singular with respect to smooth volume, and $Q_\mu $ is nonconstant on its support.
Proof.
Since $K=-\log d_{\mathrm{ch}}^2$, layer-cake integration gives
\[ \sup _ x\int _{\{ K(x,y)>L\} }K(x,y)\, \mathrm d\mu (y) \leq C\left(L+\frac2\delta \right)\mathrm e^{-\delta L/2}. \]
Consequently the continuous truncated-kernel potentials converge uniformly to $Q_\mu $. Weighted arclength on the circle has the ball bound with $\delta =1$. On that circle the kernel is $1+\log \sin ^2((\vartheta -\varphi )/2)$, whose first cosine coefficient is $-2$. Integrating its $L^1$ Fourier series against $1+\epsilon \cos \varphi $ gives
\[ Q_\mu ([1:\mathrm e^{\mathrm i\vartheta }]) =1-2\log 2-\epsilon \cos \vartheta . \]
This proves nonconstancy, and the support has smooth curve volume zero. Its product with $X_0$ also has smooth volume zero.
$\square$
Pome
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