Pome27FE241587D2Strict positivity of the residual $\beta $ was used only to make the reference potential $\psi $ plurisubharmonic with positive non-pluripolar mass; the estimates for the divisorial orders do not use it. The section comparison therefore holds for semipositive residuals with positive top integral. In particular, it does not require the residual class to contain a positive form.
Corollary 18.39.19.AF408F696ADF In each of theorem 18.39.8, theorem 18.39.13, theorem 18.39.16, replace the condition $\beta >0$ by
where $M=B$ in the first two theorems and $M=Y$ in the third, and keep all the other hypotheses. Then all the conclusions remain valid. The same replacement is valid in corollary 18.39.9.
Proof. In each case the reference current is $\beta +\sum _ i c_ i[D_ i]$, which is positive. Locality off the union of the divisors, together with the fact that the non-pluripolar measure does not charge this union, gives the non-pluripolar measure $\beta ^ q$ on $M$. Its mass is finite because $\beta $ is smooth and $M$ is compact, and positive by assumption. In the case of a modified base, the potential descends and its measure is $\rho _*(\beta ^ q)$, by the same descent and measure identities as before.
The multiplier ideal on the base depends only on the logarithmic divisorial orders and the bounded smooth term. The lower inclusion uses the global lower bound for the selecting potential, and the upper inclusion uses the anisotropic threshold in a polydisk where the upper bound holds; neither uses strict positivity of $\beta $. The restriction estimates concern fixed line bundles, whatever their curvature, and the Jacobian and the fixed canonical twist in the case of a modified base are unchanged. Hence the inclusions and dimension bounds in all three proofs remain valid. The reverse inclusion for the exact anisotropic model uses only its local integrability test and the local boundedness off the pole set, so it also remains valid. $\square$
The condition on the top integral concerns the chosen reference: a semipositive residual may have zero top integral. Conversely, a semipositive residual with positive top integral need not become strictly positive after any smooth change of gauge. The following example shows this, with an exact computation of the selected spaces.
Example 18.39.20.C695A6659495 Let $q\geq 2$, let $\rho :B\to \mathbb P^ q$ be the blowup of $p=[1:0:\cdots :0]$, and let $E$ be its exceptional divisor. Let $\tau :B\to \mathbb P^{q-1}$ be the extension to $B$ of the projection from $p$, and put
where $\omega _{\mathbb P^ q}$ and $\omega _{\mathbb P^{q-1}}$ are the normalized Fubini–Study forms. Equip $\mathcal O_ B(E)=\rho ^*\mathcal O_{\mathbb P^ q}(1)\otimes \tau ^*\mathcal O_{\mathbb P^{q-1}}(-1)$ with the quotient metric, whose curvature is $\eta =\alpha -\gamma $, and let $s_ E$ be its canonical section. For an integer $k\geq 1$, equip
Let $X=B\times \mathbb P^1$, with projection $\pi $ to $B$. For a point $o\in \mathbb P^1$, use the normalized Fubini–Study metric on $\mathcal O_{\mathbb P^1}(o)$, and set
Then $\varphi \in \mathrm{PSH}(X,\omega )$, $\psi \in \mathrm{PSH}(B,\theta )$ and
The residual $k\alpha $ cannot be made strictly positive by a smooth change of gauge. Nevertheless, for every $m\geq 1$ the selected and reference spaces coincide as subspaces of $H^0(B,A^ m)$:
Proof. Since $(\rho ,\tau )$ is an embedding, $\kappa >0$. The quotient metric satisfies
as in example 18.39.17. The same calculation over a common denominator, now with coefficient $1/2$, gives
off the pole set $E\times \{ o\} $, and across it by the extension theorem. Hence $\omega +\mathrm{dd}^{\mathrm{c}}\varphi \geq \pi ^*\gamma +\tfrac 12\omega _{\mathbb P^1}\geq 0$. Along the smooth pole set $E\times \{ o\} $, the function $\varphi $ has the exact local model $\tfrac 12\log (|z|^4+|w|^6)+O(1)$, and it is locally bounded elsewhere. It also satisfies the global lower bound $\varphi \geq \log R$.
We may therefore take $c=1$, $u=0$, $p=2$, $b_1=3$ and $a=2/3$, with residual $\beta =\theta -\eta =k\alpha \geq 0$. Its top integral is $k^ q$, since the normalized Fubini–Study form has top integral one and $\rho $ is a modification. Corollary 18.39.19, Corollary 18.39.9 therefore give
Let $t\geq 0$ be an integer. A section of $\rho ^*\mathcal O(km)\otimes \mathcal O_ B(tE)$, divided by $s_ E^ t$, gives a holomorphic section of $\mathcal O(km)$ on $\mathbb P^ q\setminus \{ p\} $, which extends across $p$ by Hartogs' theorem since $q\geq 2$. Pulling the extension back and multiplying by $s_ E^ t$ recovers the original section off $E$, hence everywhere by the identity theorem. Thus every section of $A^ m$ has the form $s_ E^ m\rho ^*s$, and the natural embeddings of $W_ m$ and $V_ m$ into $H^0(B,A^ m)$ both have this image. Its dimension is the number $\binom {km+q}{q}$ of monomials of degree $km$.
Finally, let $C\subset E\simeq \mathbb P^{q-1}$ be a projective line; it is contracted by $\rho $. For every smooth real function $u$, Stokes' theorem gives $\int _ C(k\alpha +\mathrm{dd}^{\mathrm{c}}u)=0$, whereas a strictly positive form has positive integral over $C$. Changing the smooth metric on the divisor bundle changes the residual only by $\mathrm{dd}^{\mathrm{c}}$ of a smooth function, so this proves the obstruction. Moreover, $\varphi $ is finite on most of each fiber over $E$ and equals $-\infty $ at the point $o$ of that fiber, so it does not descend, even up to a bounded function. $\square$
Thus the selected and reference spaces agree, although the selecting potential does not descend and the reference has positive mass. The reason is that $E$ is a fixed component of every section of $A^ m$: all global sections vanish along $E$ to order $m$, more than the local integrability threshold requires. As before, the criterion presupposes the global bounds for the selecting potential and a residual with positive top integral; it does not produce them for arbitrary fiber singularities.
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