The pulled-back curvature of a descended potential has no component in the fiber direction; the form $\zeta $ supplies it. Since $\zeta ^2=0$, the Hermitian top power of a descended potential consists of a single mixed term, and fiber integration of this term determines the mass and reduces the Monge–Ampère equation for descended potentials to the base.
For a positive Borel measure $\nu $ on $\mathbb P^{n-1}$, we write $\pi ^*\nu \wedge \zeta $ for the measure on $X_ n$ defined, on nonnegative Borel functions $F$, by
\[ \int _{X_ n}F\, \pi ^*\nu \wedge \zeta =\int _{\mathbb P^{n-1}} \left(\int _{\pi ^{-1}(y)}F\zeta \right)\mathrm{d}\nu (y). \]
When $\nu $ is given by a smooth top-degree form, this agrees with the pullback of the form wedged with $\zeta $.
Theorem 21.14.4.94243AEE3360 For a quasi-plurisubharmonic function $v$ on $\mathbb P^{n-1}$, the pullback $u=\pi ^*v$ belongs to $\mathrm{PSH}(X_ n,\omega )$ if and only if $v\in \mathrm{PSH}(\mathbb P^{n-1},\sigma )$. In this case
\[ \mathrm{MA}^{\mathrm{ext}}_\omega (u) =n\, \pi ^*\mathrm{MA}^{\mathrm{ext}}_\sigma (v)\wedge \zeta . \]
Consequently
\[ \int _{X_ n}\mathrm{MA}^{\mathrm{ext}}_\omega (u) =2n\int _{\mathbb P^{n-1}}\mathrm{MA}^{\mathrm{ext}}_\sigma (v) \leq 2n, \]
with equality for bounded $v$. For an integer $k\geq 0$ and $v\in \mathrm{PSH}(\mathbb P^{n-1},k\sigma )$,
\[ \mathrm{MA}^{\mathrm{ext}}_{k\alpha }(\pi ^*v)=0 \]
as a measure on the $n$-dimensional manifold $X_ n$.
Proof.
If $v$ is $\sigma $-plurisubharmonic, then
\[ \omega +\mathrm{dd}^{\mathrm{c}}u=\pi ^*(\sigma +\mathrm{dd}^{\mathrm{c}}v)+\zeta \geq 0. \]
Conversely, suppose that the left-hand side is positive. Wedge it with the smooth semipositive form $\zeta $ and push the result forward to the base. Since $\zeta ^2=0$ and the fiber integral of $\zeta $ is two,
\[ \pi _*\bigl ((\omega +\mathrm{dd}^{\mathrm{c}}u)\wedge \zeta \bigr ) =2(\sigma +\mathrm{dd}^{\mathrm{c}}v)\geq 0; \]
this identity of currents follows by testing against smooth forms on the base and integrating over the fibers. It proves the converse.
For the measure formula, first let $v\in \mathrm{PSH}(\mathbb P^{n-1},\sigma )$ be bounded. By lemma 5.28.4, its bounded products commute with $\pi ^*$. Since $\zeta ^2=0$ and the base has dimension $n-1$, expanding the power gives
\[ (\omega +\mathrm{dd}^{\mathrm{c}}u)^ n =n\, \pi ^*(\sigma +\mathrm{dd}^{\mathrm{c}}v)^{n-1}\wedge \zeta , \]
and fiber integration identifies the right-hand side with the measure defined above.
For general $v$, normalize $v\leq 0$ and set $v_ j=\max (v,-j)$. Since $\sigma $ is positive, these are bounded $\sigma $-plurisubharmonic functions, and their pullbacks are exactly $u_ j=\max (u,-j)$. Multiply the bounded identity by the common indicator function $\mathbf1_{\{ u>-j\} }=\pi ^*\mathbf1_{\{ v>-j\} }$ and pass to the increasing limits defining the extended measures. Fiber integration of nonnegative functions justifies the limit on the right-hand side, including over the poles.
The mass formula follows because each fiber has $\zeta $-mass two. The extended mass on the base is at most one: on $\{ v>-j\} $, its truncated measure is bounded above by the full bounded product, whose mass is $\int \sigma ^{n-1}=1$ by Stokes' theorem for the closed form $\sigma $. For bounded $v$ nothing is removed, and equality holds.
For the last assertion, take bounded truncations of $v\in \mathrm{PSH}(\mathbb P^{n-1},k\sigma )$. By lemma 5.28.4, the $n$th powers of their pulled-back currents vanish. Restricting to the truncation levels and taking the increasing limit shows that the extended measure is zero; this includes the case $k=0$.
$\square$
If an $\omega $-plurisubharmonic potential is constant on the fibers, the local-section argument in proposition 5.43.5 shows that it descends to a quasi-plurisubharmonic function on the base. The theorem therefore bounds its Hermitian mass by $2n$. On the Hopf threefold this bound is six, so the unbounded masses of section 23.1 require potentials that vary along the fibers.
For the equation, let $p>1$ and let $f\geq 0$ belong to $L^ p(\mathbb P^{n-1},\sigma ^{n-1})$, with $\int f\sigma ^{n-1}>0$. For potentials $u=\pi ^*v$ with bounded $v\in \mathrm{PSH}(\mathbb P^{n-1},\sigma )$, the equation
D50C510DA3A9
21.14.5
\begin{equation} \index{Monge--Ampere equation@Monge--Amp\` ere equation!Hopf reduction} (\omega +\mathrm{dd}^{\mathrm{c}}u)^ n=c(f\circ \pi )\, \mathrm{d}V,\qquad c>0, \end{equation}
is equivalent to the base equation
\[ (\sigma +\mathrm{dd}^{\mathrm{c}}v)^{n-1}=\frac{c}{2n}f\sigma ^{n-1}. \]
Indeed, the base equation implies (21.14.5) by theorem 21.14.4 and the formula for $\mathrm{d}V$. Conversely, pushing (21.14.5) forward to the base gives $2n(\sigma +\mathrm{dd}^{\mathrm{c}}v)^{n-1}=cf\sigma ^{n-1}$.
Integrating the base equation determines the constant:
\[ c=\frac{2n}{\int _{\mathbb P^{n-1}}f\sigma ^{n-1}}. \]
Since $\sigma $ is closed, it satisfies the Guan–Li condition (14.25.1). Applied to $(\mathbb P^{n-1},\sigma )$, 14.25.2 gives a continuous solution of the base equation with this constant, unique up to additive constants among bounded $\sigma $-plurisubharmonic functions. Consequently (21.14.5) has a continuous solution, unique up to additive constants among bounded descended potentials. For smooth positive $f$, theorem 14.2.1 on the base gives a smooth solution $v$, and the constant is again determined by Stokes' theorem.
For a real parameter $\lambda $, the same calculation shows that the equation with right-hand side $c\mathrm{e}^{\lambda u}(f\circ \pi )\mathrm{d}V$ is equivalent to the base equation with right-hand side $(c/(2n))\mathrm{e}^{\lambda v}f\sigma ^{n-1}$.
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