Proposition 18.36.1.7E88736EF73B Pullback of sections induces an isomorphism
\[ H^0\bigl (B,A^ m\otimes \mathcal I(m\psi )\bigr ) \simeq H^0\bigl (X,L^ m\otimes \mathcal I(m\varphi )\bigr ). \]
Let $K\subseteq B$ be closed and non-pluripolar, let $w\in C(K)$, and let $\mu $ be a Bernstein–Markov probability measure for $(K,w)$ on $B$. Set $K_ X=\pi ^{-1}(K)$, $w_ X=w\circ \pi $ and $\mu _ X=\mathcal L\mu $. Then pullback is an isometry from the norm $\| \cdot \| _{m,w,\mu }$ on $B$ to the norm $\| \cdot \| _{m,w_ X,\mu _ X}$ on $X$, the measure $\mu _ X$ is Bernstein–Markov for $(K_ X,w_ X)$, and the partial kernel measures on $B$ and $X$, normalized with the same integer $r$, satisfy
\[ \beta _{X,m}^{(r)}=\mathcal L\bigl (\beta _{B,m}^{(r)}\bigr ). \]
Proof.
Let $s$ be a holomorphic section of $\pi ^*A^ m$. Over an open set of $B$ on which $A^ m$ is trivial, the coefficient of $s$ is a holomorphic function on the inverse image, and it is constant on each fiber, since the fibers are compact and connected. It is therefore the pullback of a function on the base, which is holomorphic because it is the composition of the coefficient with a local holomorphic section of the submersion. These local functions transform by the transition functions of $A^ m$, so they define a section of $A^ m$ whose pullback is $s$. This proves the identification without multiplier ideals.
By lemma 5.37.7, the multiplier ideal $\mathcal I(m\varphi )$ on $X$ is generated by the pullbacks of the germs in $\mathcal I(m\psi )$. A pulled-back germ belongs to it exactly when the base germ belongs to $\mathcal I(m\psi )$: pullback gives one implication, and restriction to a local holomorphic section of $\pi $ gives the other. This identifies the selected spaces.
Pointwise norms of corresponding sections agree and are constant along the fibers. Since each $\nu _ y$ has mass one,
\[ \int _{K_ X}|\pi ^*s|_{h^ m}^2\mathrm{e}^{-mw_ X}\, \mathrm{d}\mu _ X =\int _ K|s|_{h_ A^ m}^2\mathrm{e}^{-mw}\, \mathrm{d}\mu . \]
The supremum norms on $K_ X$ and $K$ also agree, because $\pi $ is surjective. Thus the Bernstein–Markov inequality transfers from $(K,w,\mu )$ to $(K_ X,w_ X,\mu _ X)$, and the integral norms on $X$ are nondegenerate. Pulling back an orthonormal basis gives $B_{X,m}=B_{B,m}\circ \pi $ on $K_ X$, where each kernel is set to zero outside its weighted set. Integrating a continuous test function first along the fibers proves the identity of measures.
$\square$
Pome
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