Lemma 18.18.1.7EF34A8A42E5 Every $\Phi _\sigma (q)$ is a normalized ordinary model, and
The background is not big if $f=0$ or some $B_ i=0$. If $f>0$ everywhere and all $B_ i>0$, it is big. It is not closed whenever $f$ is nonconstant.
7EF34A8A42E5Lemma 18.18.1.7EF34A8A42E5 Every $\Phi _\sigma (q)$ is a normalized ordinary model, and
The background is not big if $f=0$ or some $B_ i=0$. If $f>0$ everywhere and all $B_ i>0$, it is big. It is not closed whenever $f$ is nonconstant.
Proof. On $B=Y\times C_0$, let $\pi ,s_0$ be the projections and put
The case of the zero obstacle in proposition 18.16.2 gives $P_{\gamma _\sigma }[w]=w$. Replace an ordinary competitor for $w\circ \pi $ by its maximum with $w\circ \pi $. Its restriction to every section $Y\times \{ z\} $ is then a competitor for $P_{\gamma _\sigma }[w]$ that is not identically $-\infty $, so
Use the atomic Green representatives $g_ i$ constructed in the proof of proposition 18.16.2, which are smooth in the coordinates of zero area. Their sum $g=\sum _ i g_ i\circ \pi _ i\circ \pi $ is $\gamma _{\sigma ,f}$-plurisubharmonic and multiplier-ideal equivalent to $w\circ \pi $, and its complex Hessian is a closed countable divisor current of finite mass minus a smooth form. If $w_{\mathcal I}=\sum _ iq_{i,\mathcal I}\circ \pi _ i$, then $g\sim w_{\mathcal I}\circ \pi $. By lemma 5.38.5, every ideal competitor is ordinarily dominated by $g$, hence bounded by
Here the last equality holds because each ideal projection $q_{i,\mathcal I}$ on a curve is an ordinary model, so the argument with sections just given applies to $w_{\mathcal I}$. By pullback and addition, $w_{\mathcal I}\circ \pi $ is itself an ideal competitor, which proves the ideal identity on $B$. Now apply proposition 18.14.1 to $M\to B$ and the divisor $B\times D$ with weight one, whose induced zero obstacle is zero. The two identities upstairs and the independence of the factors give the assertion and normalization.
For the statements on bigness, put $n=m+1+k$ and $\Omega =p^*\beta +s^*\tau +r^*\kappa >0$, with the positive product form $\beta $ fixed earlier. For $f=0$, the closed positive test form
satisfies $\int _ M\Theta _{\sigma ,0}\wedge T_0=0$ and $\int _ M\Omega \wedge T_0>0$. If $B_ i=0$, use instead
Then $\int _ M\Theta _{\sigma ,f}\wedge T_ i=0$ for every $f$, since the only nonzero term contains the factor $\int _{C_ i}\sigma _ i=B_ i$, while $\int _ M\Omega \wedge T_ i>0$. In either case, Stokes' theorem against the closed test form shows that no current $\Theta _{\sigma ,f}+\mathrm{dd}^{\mathrm{c}}u$ with $u$ quasi-plurisubharmonic is bounded below by a positive multiple of $\Omega $, so the background is not big.
Finally, write $\sigma _ i=B_ i\eta _ i/A_ i+\mathrm{dd}^{\mathrm{c}}\chi _ i$ by lemma 5.29.22. The smooth potential $-\sum _ i\chi _ i\circ \pi _ i\circ p$ has strictly positive current when every $B_ i>0$ and $f>0$, which proves bigness in that case. The identity $\mathrm d\Theta _{\sigma ,f}=p^*(\mathrm df)\wedge s^*\tau $ shows that the background is not closed for nonconstant $f$. $\square$
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