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 $k=0$.
$\square$
Pome
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