12.14 Positive volume without bignessE4AB7B03C20D

Proof. In the affine chart $Z_0=1$, write $s=|z|^2$. Then $x=s/(1+s)$ and $\omega =\mathrm{dd}^{\mathrm{c}}\log (1+s)$. Polar integration gives

for every integrable function $b$ on $[0,1]$, since the normalized radial density of $\omega ^ m$ in the variable $s$ is $m s^{m-1}(1+s)^{-m-1}\, \mathrm{d}s$. In particular $\int _ B a^{-1}\omega ^ m=mt/(m-1)=1$, and monotone convergence in (12.14.1) gives $H_\epsilon \to 1$.

Define $q_\epsilon \geq 0$ and $\varphi _\epsilon $ on $[0,1]$ by

Near zero,

so $q_\epsilon $ is $x$ times a smooth positive function; moreover $q_\epsilon (1)=1$. Hence both apparent singularities of $\varphi _\epsilon '$, at $0$ and at $1$, are removable, and $\varphi _\epsilon $ is smooth on $[0,1]$. Composing with the smooth function $x$ gives a smooth function on $B$, again denoted $\varphi _\epsilon $.

To compute $A_\epsilon $, recall that for a local potential of the form $F(s)$, with $q=sF'(s)$, the complex Hessian has the tangential eigenvalue $q/s$, of multiplicity $m-1$, and the radial eigenvalue $\mathrm{d}q/\mathrm{d}s$. For $F_\epsilon =\log (1+s)+\varphi _\epsilon (x)$ one finds $sF_\epsilon '=q_\epsilon (x)$. Differentiating the integral defining $q_\epsilon $ shows that both eigenvalues are positive, also at $s=0$ by the smooth factorization above. Their product relative to the Fubini–Study determinant is

At a point of the hyperplane $\{ Z_0=0\} $, the smooth form $A_\epsilon $ is semipositive, as a limit of the positive matrices in the affine chart. Since the determinant just computed extends there with a strictly positive value, every eigenvalue is positive. This proves strict positivity on all of $B$ and (12.14.3).

We now pass to the limit. For $x>0$, $q_\epsilon (x)$ tends to $x^ t$. To bound $q_\epsilon $ uniformly, note that $q_\epsilon ^ m$ is the distribution function of the probability density $\frac{m}{H_\epsilon }\frac{y^{m-1}}{y/t+\epsilon }$ on $[0,1]$. Relative to the probability density $my^{m-1}$, this density has a decreasing weight; relative to $(m-1)y^{m-2}$, its weight is proportional to $y/(y+t\epsilon )$ and hence increasing. If one probability density on $[0,1]$ has a decreasing weight relative to another, its average weight on $[0,x]$ is at least one and its distribution function is the larger of the two; an increasing weight reverses the inequality. Consequently

The function

is bounded and continuous on $[0,1]$, since $t>0$, and smooth on $(0,1]$; its derivative extends smoothly at one. The integrable function $(y^ t-y)/[y(1-y)]$ dominates the derivatives of the functions $\varphi _\epsilon $, so dominated convergence gives uniform convergence $\varphi _\epsilon \to \varphi $ on $B$. Distributional convergence shows that $A$ is a closed positive current. On each coordinate chart, a smooth local potential of $\omega $ plus $\varphi $ is a uniform limit of the corresponding plurisubharmonic functions built with $\varphi _\epsilon $, including at $p$. Hence $\varphi $ is $\omega $-plurisubharmonic. Off $p$, the limit $q=x^ t$ gives

Smoothness of $\varphi $ off $p$ and the same determinant argument at the hyperplane $\{ Z_0=0\} $ prove (12.14.4). $\square$

Proof. The coefficient $a$ is nonnegative, and $\theta +\eta h>0$ for every $\eta >0$; this proves semipositivity and nefness. Since $a$ is nonconstant, $\mathrm{d}\theta =\mathrm{d}a\wedge w\ne 0$.

Suppose that $\theta +\mathrm{dd}^{\mathrm{c}}v\geq \delta h$ for a quasi-plurisubharmonic function $v$ and some $\delta >0$. Choose a nonzero smooth function $f\geq 0$ on $B$ supported in $\{ a<\delta \} $. Its pullback $S=f\omega ^ m$ is a smooth semipositive $(m,m)$-form, closed because it has top degree on the base. Positivity and distributional Stokes would give

This contradiction shows that $\theta $ is not big, since every Hermitian form on $X$ dominates a positive multiple of $h$.

For the lower bound on the volume, pull back the bounded potential $\varphi $ of lemma 12.14.2. The positive current $\Omega =A+a w=\theta +\mathrm{dd}^{\mathrm{c}}\varphi $ is a smooth strictly positive form off $Z=\{ p\} \times \mathbb P^1$, where

Let $\gamma $ be a Gauduchon metric. Since $\varphi $ is bounded, distributional Stokes lets us replace $\theta $ by $\Omega $ in the numerator of its quotient. The set $Z$ has zero smooth volume and $\Omega \wedge \gamma ^{n-1}$ is a positive measure, so the pointwise arithmetic–geometric mean inequality off $Z$ gives

This lower bound holds for every test in definition 12.2.1; no power of the current $\Omega $ along $Z$ is needed.

For the upper bound, put $c_\epsilon =n/H_\epsilon $ and $\Omega _\epsilon =A_\epsilon +(a+\epsilon )w=\theta _\epsilon +\mathrm{dd}^{\mathrm{c}}\varphi _\epsilon $. Then $\Omega _\epsilon ^ n=c_\epsilon \mu $. The metrics in (12.14.8) satisfy

The first term has top degree on the base and the second is a product of closed forms, so $T_\epsilon $ is closed; that is, $\gamma _\epsilon $ is balanced, and in particular Gauduchon. Smooth Stokes and (12.14.5) give

so the quotient of $\gamma _\epsilon $ at $\theta _\epsilon $ equals $c_\epsilon ^{1/n}$. The same arithmetic–geometric mean argument, applied to the smooth representative $\Omega _\epsilon $, bounds every Gauduchon quotient at $\theta _\epsilon $ below by $c_\epsilon ^{1/n}$. This proves that $\gamma _\epsilon $ minimizes, and gives the formula for $\mathcal V_\mu (\theta _\epsilon )$. Since $\theta \leq \theta _\epsilon $, the metrics $\gamma _\epsilon $, used as tests for $\theta $, also give $\lambda _\mu (\theta )\leq c_\epsilon ^{1/n}$. Letting $\epsilon \downarrow 0$ proves (12.14.7). By theorem 12.11.1, this value is also the limit of the volumes of the Hermitian perturbations $\theta +\epsilon h$. $\square$

Proof. By the power formula in the proof of theorem 12.14.6 and (12.14.3),

Since $A_\epsilon $ is cohomologous to $\omega $, closed Stokes gives $\int _ B\omega \wedge A_\epsilon ^{m-1}=1$. Consequently the two terms have $h$-traces $I_\epsilon /H_\epsilon $ and $m$, which proves the trace normalization of $g_\epsilon ^ m=T_\epsilon /M_\epsilon $. Subtracting $\epsilon w$ from $\theta _\epsilon $ subtracts $\epsilon I_\epsilon /H_\epsilon $ from the pairing $\int _ X\theta _\epsilon \wedge T_\epsilon =n$. Rescaling $\gamma _\epsilon $ by $M_\epsilon ^{-1/m}$ divides both the linear pairing and $D_\mu $ by $M_\epsilon $, proving (12.14.11).

Polar integration gives

When $m=2$, put $q=\epsilon /2$. Direct integration yields

Hence $I_\epsilon \sim \tfrac 12\log (1/\epsilon )$ and $\epsilon I_\epsilon \to 0$. Since $H_\epsilon \to 1$, the asymptotics in (12.14.12) follow from (12.14.11).

For the weak limit when $n=3$, note that the mixed positive term $2A_\epsilon \wedge w$ has $h$-trace two; divided by $M_\epsilon \to \infty $, it tends weakly to zero. The horizontal probability measure

on $B$ concentrates at $p$: outside any fixed neighborhood of $p$ its unnormalized density stays bounded, while $I_\epsilon \to \infty $. For a smooth $(1,1)$-test form on $X$, integration along the fibers $\{ b\} \times \mathbb P^1$ gives a smooth function on $B$. The convergence of these probability measures to the Dirac mass at $p$ therefore shows that the horizontal term of $g_\epsilon ^2$ converges to $[Z]$, because its coefficient $I_\epsilon /(H_\epsilon M_\epsilon )$ tends to one.

For $m\geq 3$, monotone convergence instead gives

In both cases $\epsilon I_\epsilon \to 0$. Using $H_\epsilon \to 1$ in (12.14.11) proves the remaining limits, including that of the volume quotients. $\square$

Proof. Choose a nonzero smooth function $f_\rho \geq 0$ on $B$, supported in $\{ x<\rho \} $ and normalized by $\int _ Bf_\rho \omega ^ m=1$. The form

is smooth, strictly positive and closed: the first summand has top degree on the base, and $h$ is Kähler. The positive-root construction in the proof of theorem 11.6.1 gives a unique smooth metric $b_\rho $ with $b_\rho ^ m=\Gamma _\rho $, and $b_\rho $ is balanced because $\Gamma _\rho $ is closed. Since $\int _ Xh^{m+1}=n$, its $h$-trace is one.

Put $C_\theta =\int _ X\theta \wedge h^ m$. Since $a<\rho /t$ on the support of $f_\rho $,

The lower bound in the proof of theorem 12.14.6 gives $D_\mu (b_\rho )\leq n^{-1/n}\int _ X\theta \wedge \Gamma _\rho $, so the determinant integrals also tend to zero. The probability measures $f_\rho \omega ^ m$ converge to the Dirac mass at $p$ because their supports shrink to that point. Fiber integration, as in the preceding proof, gives the current limit $[Z]$, while the term $\rho h^ m$ tends to zero. $\square$