18.47 Three monomial bands from a multiplier-ideal selection435C250083A1

Proof. Near $p_+$ use the coordinates $\zeta _1=z_1$, $\zeta _2=1/z_2$ and the frame $T_1Z_2$. In this frame the two pencil sections $T_1T_2$ and $Z_1Z_2$ have coefficients $\zeta _2$ and $\zeta _1$, so

The same calculation applies at $p_-$. The formula for the ideals is lemma 5.37.1 with normal dimension two and coefficient one, including the case of equality in the integrability condition.

Now suppose that $v\preceq _{\mathcal I}\psi $. By theorem 5.37.3, the Lelong numbers satisfy $\nu (v,p_\pm )\geq 1$; here $\nu (v,p)$ is the Lelong number of section 5.34, not the measure $\nu $. Let $\phi $ be the plurisubharmonic local potential of $v$ in a frame near $p_\pm $, and consider its radial maximum

inside a fixed coordinate ball. This function is finite and convex: by the three-circles theorem, the corresponding maximum on each complex line through the pole is convex, and $M$ is the supremum of these functions; lines on which $\phi $ is identically $-\infty $ do not affect the supremum. The limiting slope of $M$ as $a\to -\infty $ is the Lelong number. For $a<a_0$, convexity therefore gives

Consequently $v\leq \psi +C$ near both poles. After $C$ is increased, the same bound holds on the compact complement, because $v$ is bounded above and $\psi $ is finite and continuous there. Conversely, domination $v\leq \psi +C$ implies the inclusions of multiplier ideals at all weights, by the definition through integrability. Thus the two competitor classes coincide before any obstacle is imposed. $\square$

Proof. For $\xi \in \mathbb C^*$, the curve of the pencil

is a smooth rational curve of bidegree $(1,1)$ through both poles, and its $\theta $-degree is two. The restriction of $v$ to this connected curve is either identically $-\infty $ or $\theta |_{C_\xi }$-plurisubharmonic. In the latter case, its curvature measure has total mass two by Stokes' theorem. The local upper bounds of lemma 18.47.1 give Lelong numbers at least one at the two poles, hence, by the identification of Lelong numbers with atoms in lemma 5.37.18, atoms of mass at least one there. These atoms exhaust the measure. The curvature of $\psi |_{C_\xi }$ consists of the same two unit atoms. Thus $v-\psi $ is harmonic, in the sense of distributions, on the compact curve $C_\xi $, and therefore constant.

On the torus use the holomorphic coordinates $(z_1,\xi =z_1z_2)$. Local integrability and Fubini's theorem show that, for almost every $\xi $, the restriction of $v$ to $C_\xi $ is not identically $-\infty $; by the preceding paragraph it is then finite off the poles. Hence the restriction of $v$ to the transverse curve $S=\{ z_1=1\} \simeq \mathbb P^1$ is not identically $-\infty $. Put $u=v|_ S+\log 2$. Since $\theta |_ S=\omega _{\rm FS}$ and $\psi _{\rm FS}|_ S=-\log 2$, the constants on the curves $C_\xi $ give the asserted identity almost everywhere on the torus. Both sides have plurisubharmonic local potentials, so they agree everywhere off the poles. The logarithmic upper bounds give the value $-\infty $ at the poles.

Conversely, $\theta +\mathrm{dd}^{\mathrm{c}}(\psi _{\rm FS}+u\circ f)=f^*(\omega _{\rm FS}+\mathrm{dd}^{\mathrm{c}}u)\geq 0$ off the poles. In a chart at a pole, the local potential is bounded above by $\log (|\zeta _1|^2+|\zeta _2|^2)+C$, since $u$ is bounded above. It therefore extends across the pole as a plurisubharmonic function with value $-\infty $ there, which gives a global $\theta $-plurisubharmonic function dominated by $\psi +C$. $\square$

Proof. For an arbitrary competitor $v$, lemma 18.47.2 gives, in terms of the affine coordinate $\xi $ of $\mathbb P^1$,

On the interior of $K$, the obstacle condition, which holds quasi-everywhere, becomes an inequality everywhere, by the submean inequality and the continuity of $w+\varphi _ X$. Evaluating at $x_1=x_2=s/2$, with $s=\log |\xi |^2$, gives

for every argument of $\xi $. At $s=0$ this bounds $\chi $ by zero on $|\xi |=1$, and the maximum principle for the subharmonic function $\chi $ bounds it by zero on the whole disk $|\xi |\leq 1$. At infinity use the local potential

The obstacle at $s=2$ bounds it by $-1$ on $|\eta |=\mathrm e^{-1}$, and the same maximum principle gives $\chi (\xi )\leq s-1$ for $s\geq 2$. Together with the obstacle for $0\leq s\leq 2$, this proves $\chi \leq g$, without any invariance assumption on the competitor.

Conversely, $g$ is convex with slopes in $[0,1]$, so $g(\log |\xi |^2)-\log (1+|\xi |^2)$ is an $\omega _{\rm FS}$-plurisubharmonic function on $\mathbb P^1$. By lemma 18.47.2, it defines the global function $P$. Directly,

Thus $P$ belongs to the competitor class of lemma 18.47.1 and satisfies the obstacle condition everywhere on $K$. Since it dominates every competitor, it is the envelope, including its values $-\infty $ at the poles. $\square$

Proof. At $p_+$ the monomial with exponents $(a,b)$ has local exponents $(a,m-b)$, and at $p_-$ it has local exponents $(m-a,b)$. By lemma 18.47.1, both total orders must be at least $m-1$. Distinct exponent pairs cannot cancel, so these conditions select exactly the monomials with $|a-b|\leq 1$. The three bands $a=b$, $a=b+1$ and $a=b-1$ have $m+1$, $m$ and $m$ elements. The bound $|\sigma |_{h^ m}^2\leq C\mathrm e^{m\psi }$ would require total order $m$ at both poles and leave only the diagonal. The product of two sections with offset $a-b=1$ has offset two and lies outside the selected space of the sum degree; such products fail for every pair of positive degrees. Thus the linear growth of $N_ m$ is not the growth of a graded section ring.

Integration over the angles makes all global monomials orthogonal, including those outside the selected space. For $t\in [0,1]$ set

The squared norm of the monomial with exponents $(a,b)$ is

With $c=\int _{-1}^1\mathrm e^{-y^2/2}\, \mathrm dy$, uniformly in $t\in [0,1]$,

For the lower bound, integrate over the interval of radius $m^{-1/2}$ centered at $t$; for the last assertion, note that each of the two omitted Gaussian tails is at most $\mathrm e^{-m/2}/m$. Hence

For $x\in [-1,2]^2$, completing squares bounds the weighted contribution of each normalized monomial by $9m/c^2$. Summing over all $(m+1)^2$ global monomials and using the evaluation inequality gives

This polynomial bound proves the Bernstein–Markov property. On the selected space the same argument gives the factor $Cm$. $\square$

Proof. The diagonal monomials belong to $V_ m$. The upper bound for their norms in (18.47.6) gives

For $0\leq s\leq 2$, the grid point $j/m$ nearest to $s/2$ loses at most $1/(4m^2)$ compared with $g(s)$; outside this interval the maximizing endpoint lies on the grid. This proves (18.47.8) on the torus.

For a monomial with offset one, $(a,b)=(j+1,j)$, the lower bound in (18.47.6) shows that the exponent of its normalized contribution, apart from $-m\varphi _ X$ and constants, is at most both

The band with $(a,b)=(j,j+1)$ is obtained by interchanging the indices. Put

The lower bound for the norms and the count of monomials give

The function $R$ extends continuously across the coordinate divisors away from the poles. In a chart at $p_+$, put $A=|\zeta _1|^2$ and $B=|\zeta _2|^2$. The possibly large term of $R$ satisfies

and the other term is bounded; the same holds at $p_-$. Thus $R\leq C(1+r^{-2})$ near either pole, where $r=|\zeta |$ is the local radius, and $\log R$ belongs to $L^ p$ for every finite $p$ with respect to smooth volume forms.

For $m\geq 2$, all selected sections vanish at the poles. The two diagonal sections $T_1^ mT_2^ m$ and $Z_1^ mZ_2^ m$ have no other common zeros, so $\mathcal B_ m>0$ off the poles. Away from the poles, $P$ and $U_ m$ are continuous, so both inequalities extend from the torus to $X\setminus \{ p_+,p_-\} $. They give the asserted local uniform convergence and convergence in $L^ p$, and differentiation of the $L^1$ limit in the sense of distributions gives the weak convergence of the currents. At the poles both $U_ m$ and $P$ equal $-\infty $ for $m\geq 2$, which also proves the pointwise statement on $X$.

For a fixed $m\geq 2$, the lowest-order terms of the selected sections at a pole include every binary monomial of degree $m-1$, and the sum of their squared moduli is comparable to $r^{2(m-1)}$ in all directions. Hence

so the difference $U_ m-P$ tends to $+\infty $ near each pole. Nevertheless, (18.47.9) gives, uniformly in $m\geq 2$ and $r<1$,

Combined with uniform convergence on compact subsets away from the poles, these bounds give uniform convergence of $\mathrm e^{U_ m}$.

The contribution of each normalized monomial to $B_ m\nu $ is exactly the product of two truncated Gaussian probability measures in $x_1,x_2$, centered at $(a/m,b/m)$, with the Haar probability measures in the angles. Completing squares in (18.47.5) and using its uniform bounds proves concentration at these centers, uniformly over all three bands; for example, the probability of $|x_1-a/m|\geq \eta $ is at most $C_\eta \sqrt m\, \mathrm e^{-m\eta ^2/2}$. Uniform continuity of a test function then reduces the sum to three lattice Riemann sums. Each band contains asymptotically one third of the $N_ m$ monomials, and in each band the centers $(a/m,b/m)$ approach the diagonal point $(t,t)$, $t\in [0,1]$. This proves both positive limits. The total mass of $2m^{-2}B_ m\nu $ is $2(3m+1)/m^2\to 0$, which proves the convergence to zero in total variation.

Off the poles, $\theta +\mathrm{dd}^{\mathrm{c}}P$ is the pullback by $f$ of the current on $\mathbb P^1$ defined by $g$. The function $g$ is $C^{1,1}$, and the local complex Hessian of $g(s)$ has rank at most one. Smooth convex approximations and the product of locally bounded potentials show that its square vanishes, as in the Hessian argument of section 18.45. The non-pluripolar product assigns no mass to the two poles, so the global square is zero. In particular, the positive-mass hypothesis of theorem 18.35.5 fails in this model. $\square$