Blowing up two points instead of one, we obtain a residual that is semipositive, cannot be made strictly positive, and still has positive top integral. Unlike the example with one blown-up point, the gap between the two divisorial thresholds now produces a nonzero defect of the largest possible order $m^{q-1}$. Counting the spaces requires the following elementary interpolation of jets at several points.
Lemma 18.39.24.822490631827 Let $x_1,\ldots ,x_ N$ be distinct points of $\mathbb P^ q$, $q\geq 1$, and let $d,j_1,\ldots ,j_ N$ be nonnegative integers with $\sum _ i j_ i\leq d+1$. Then the map from $H^0(\mathbb P^ q,\mathcal O(d))$ to the jets of order less than $j_ i$ at the points $x_ i$ is surjective. In particular,
\[ h^0\left(\mathbb P^ q,\mathcal O(d)\otimes \bigcap _ i\mathfrak m_{x_ i}^{j_ i}\right) =\binom {d+q}{q}-\sum _ i\binom {j_ i+q-1}{q}, \]
where $\binom {q-1}{q}=0$.
Proof.
We may discard the points with $j_ i=0$; if no point remains, the target is zero. Choose an affine chart containing all the remaining points, and use the corresponding trivialization of $\mathcal O(d)$. For each pair $i\ne k$ choose an affine linear function $L_{ik}$ vanishing at $x_ k$ and nonzero at $x_ i$, and set
\[ Q_ i=\prod _{k\ne i}L_{ik}^{j_ k}. \]
Given a jet $b_ i$ at $x_ i$, divide it by the invertible jet of $Q_ i$ at $x_ i$, and let $T_ i$ be the Taylor polynomial of degree at most $j_ i-1$ of the quotient. Then $Q_ iT_ i$ has the jet $b_ i$ at $x_ i$ and vanishes to order at least $j_ k$ at every other point $x_ k$, and its degree is at most $\sum _ kj_ k-1\leq d$. Hence $\sum _ iQ_ iT_ i$, homogenized to degree $d$, realizes all the prescribed jets. In $q$ variables there are $\binom {j_ i+q-1}{q}$ monomials of total degree less than $j_ i$, which gives the dimension formula.
$\square$
Example 18.39.25.47ABA9EAB173 Let $q\geq 2$ and let $\rho :B\to \mathbb P^ q$ be the blowup of the two points $x_1=[1:0:\cdots :0]$ and $x_2=[0:1:0:\cdots :0]$, with exceptional divisors $E_1,E_2$. Let $\tau _ i:B\to \mathbb P^{q-1}$ be the extension to $B$ of the projection from $x_ i$. With the normalized Fubini–Study forms, put
\[ \alpha =\rho ^*\omega _{\mathbb P^ q},\qquad \gamma _ i=\tau _ i^*\omega _{\mathbb P^{q-1}},\qquad \kappa =\alpha +\gamma _1+\gamma _2>0. \]
Equip $\mathcal O_ B(E_ i)=\rho ^*\mathcal O(1)\otimes \tau _ i^*\mathcal O(-1)$ with the quotient metric, whose curvature is $\eta _ i=\alpha -\gamma _ i$, and let $s_ i$ be its canonical section. Equip
\[ A=\rho ^*\mathcal O(2)\otimes \mathcal O_ B(E_1+E_2), \qquad \theta =2\alpha +\eta _1+\eta _2 \]
with the product metric. On $X=B\times \mathbb P^1$, with projection $\pi $ and a point $o\in \mathbb P^1$, set
\[ \begin{gathered} R_ i=\| \pi ^*s_ i\| ^2,\qquad T=\| s_ o\| ^2,\\ \varphi =\log (R_1^2+T^3)+\log (R_2^2+T^3),\\ \omega =4\pi ^*\kappa +7\omega _{\mathbb P^1},\\ \psi =2\log \| s_1\| ^2+2\log \| s_2\| ^2, \qquad \beta =\gamma _1+\gamma _2. \end{gathered} \]
Then $\varphi \in \mathrm{PSH}(X,\omega )$, $\psi \in \mathrm{PSH}(B,\theta )$, and
\[ \mathrm{MA}_\theta (\psi )=\beta ^ q,\qquad \int _ B\beta ^ q=2^ q-2>0. \]
No smooth change of gauge makes $\beta $ strictly positive. Nevertheless, writing $V_ m$ and $W_ m$ for the selected and reference spaces as before, their images in $H^0(B,A^ m)$ are, for every $m\geq 1$,
\[ \begin{aligned} V_ m& =s_1^ ms_2^ m\rho ^*H^0\left(\mathbb P^ q,\mathcal O(2m) \otimes \mathfrak m_{x_1}^{m-1}\mathfrak m_{x_2}^{m-1}\right),\\ W_ m& =s_1^ ms_2^ m\rho ^*H^0\left(\mathbb P^ q,\mathcal O(2m) \otimes \mathfrak m_{x_1}^{m}\mathfrak m_{x_2}^{m}\right). \end{aligned} \]
The defect and the common leading term of the dimensions are
\[ \begin{gathered} \dim V_ m-\dim W_ m=2\binom {m+q-2}{q-1} \sim \frac{2}{(q-1)!}m^{q-1},\\ \dim V_ m=\frac{2^ q-2}{q!}m^ q+O(m^{q-1}),\qquad \dim W_ m=\frac{2^ q-2}{q!}m^ q+O(m^{q-1}). \end{gathered} \]
Proof.
The map $(\rho ,\tau _1,\tau _2)$ is an embedding: away from the exceptional divisors $\rho $ is already an embedding, and along $E_ i$ the map $\tau _ i$ separates the tangent directions of $E_ i$ while $\rho $ detects the normal direction. Hence $\kappa >0$. The identity for the quotient metric is the one-point identity of example 18.39.20, applied at each center. For each $i$, the calculation over a common denominator writes $R_ i^2+T^3$ as a sum of squared moduli of holomorphic coefficients over the metric denominator for $\mathcal O(2,3)$. Together with the extension theorem across the pole set, this gives
\[ \mathrm{dd}^{\mathrm{c}}\varphi \geq -4\pi ^*\alpha -6\omega _{\mathbb P^1}. \]
It follows that $\omega +\mathrm{dd}^{\mathrm{c}}\varphi \geq 4\pi ^*\beta +\omega _{\mathbb P^1}\geq 0$.
By the Poincaré–Lelong formula,
\[ \theta +\mathrm{dd}^{\mathrm{c}}\psi =\beta +2[E_1]+2[E_2]\geq 0. \]
The non-pluripolar product does not charge the divisors and equals $\beta ^ q$. The class of $\beta $ is $2H-[E_1]-[E_2]$, where $H=\rho ^*c_1(\mathcal O(1))$. As in example 18.39.17, $H^ q=1$, the mixed intersection numbers involving $H$ and some $[E_ i]$ vanish, and $[E_ i]^ q=(-1)^{q-1}$; moreover, the two exceptional divisors are disjoint. Expanding gives $\int _ B\beta ^ q=2^ q-2$.
Let $C$ be the strict transform of the line through $x_1$ and $x_2$. Both maps $\tau _ i$ are constant on $C$, so $\int _ C\beta =0$. By Stokes' theorem, $\int _ C(\beta +\mathrm{dd}^{\mathrm{c}}u)=0$ for every smooth real function $u$, whereas a strictly positive form has positive integral over $C$. This proves the obstruction, although the top integral is positive.
The selecting potential satisfies $\varphi \geq \pi ^*\psi $. Its pole set is the disjoint union of the sets $E_ i\times \{ o\} $. Near either of them the other summand is smooth, and
\[ \varphi =\log (|z|^4+|w|^6)+O(1),\qquad E_ i=\{ z=0\} ,\quad o=\{ w=0\} . \]
Thus $c_ i=p_ i=2$, $b_{i1}=3$ and $a_ i=2/3$. By theorem 18.39.13, corollary 18.39.19, $W_ m=H^0(B,A^ m(-2m(E_1+E_2)))$, and $V_ m$ is contained in $H^0(B,A^ m(-(2m-1)(E_1+E_2)))$; for an integer $\ell $, the strict local integrability condition $\ell +1>2m-2/3$ means $\ell \geq 2m-1$. Conversely, by lemma 18.39.7, divisibility by these orders gives integrability in every chart of the pole set, and off the pole set the selecting potential is locally bounded. This proves equality for $V_ m$.
Dividing a section of $A^ m$ by $s_1^ ms_2^ m$ gives a section of $\mathcal O(2m)$ away from the two centers, which extends across them by Hartogs' theorem since $q\geq 2$. The pullback of the extension, multiplied by $s_1^ ms_2^ m$, agrees with the original section everywhere, by the identity theorem. If the extended section vanishes to order $j_ i$ at $x_ i$, the corresponding section of $A^ m$ vanishes to order $m+j_ i$ generically along $E_ i$. The two divisorial conditions therefore give exactly the vanishing orders at $x_1$ and $x_2$ displayed in the statement. Since the two point ideals have disjoint supports, their product equals their intersection.
Apply lemma 18.39.24 with $d=2m$ and orders $(m,m)$ or $(m-1,m-1)$; in both cases their sum is at most $d+1$. This gives
\[ \begin{aligned} \dim W_ m& =\binom {2m+q}{q}-2\binom {m+q-1}{q},\\ \dim V_ m& =\binom {2m+q}{q}-2\binom {m+q-2}{q}. \end{aligned} \]
Pascal's identity gives the exact defect, and the leading terms give both growth laws. For $m=1$ the vanishing orders for $V_1$ are zero, and the defect is already two. For $q=2$ the dimensions are $\dim W_ m=(m+1)^2$ and $\dim V_ m=m^2+4m+1$.
$\square$
In this example the residual $\beta $ has zero integral over the curve $C$, so no smooth change of gauge makes it positive, and yet its top integral is positive. Thus the bound $O(m^{q-1})$ of corollary 18.39.19 cannot be improved to $o(m^{q-1})$ for semipositive residuals either. As in the previous examples, the selecting potential is finite on most of each fiber over $E_ i$ and equals $-\infty $ at the point $o$ of that fiber, so it does not descend, even up to a bounded function.
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