12.23 Strict changes under one blowupD0D47B7389D7

Proof. Step 1: the backgrounds. Put $W_0=\mathbb P^ p$, with Fubini–Study form $\omega $ normalized by $\int _{W_0}\omega ^ p=1$. Write $s=|z|^2$ on the affine chart of $B$. Choose a smooth function $Q(s)\geq 0$ that equals $s^ k$ near zero, is positive for $s>0$, and is constant for large $s$; it defines a smooth function $q$ on $B$. Put $n=m+p$, choose $b>0$, and define

The map $\Pi =\pi \times \mathrm{id}$ is the blowup of the smooth center $\{ 0\} \times W_0$, and $\widetilde\theta =\Pi ^*\theta $. Fix the Kähler perturbation directions $h=H+\omega $ and $\widetilde h=\kappa +\omega $, and the smooth positive probability densities $\mu =H^ m\wedge \omega ^ p$ and $\widetilde\mu =\kappa ^ m\wedge \omega ^ p/(2^ m-1)$. Both backgrounds are smooth and semipositive, and they are not closed, since $\mathrm d\theta =\mathrm dq\wedge \omega \neq 0$.

Step 2: explicit solutions. Write $\nu _{\rm down}=H^ m$ and $\nu _{\rm up}=\kappa ^ m/(2^ m-1)$ for the base forms; we show below that both are probability forms. For either product, write the perturbed background as $\xi =A+g\omega $, where downstairs $A=(b+\epsilon )H$, upstairs $A=(b+\epsilon )H+\epsilon L$, and in both cases $g=q+\epsilon $, with pullback understood upstairs. The form $A$ is Kähler on the relevant base. Let $\nu $ be the corresponding base probability form, and put $M=\int A^ m$ and $J=\int g^{-p}\nu $. The smooth equation of theorem 14.2.1, together with closed Stokes, gives a smooth $u$ on the base with

Let $\Xi (s)$ denote the mass of $\nu $ on the set $\{ |z|^2<s\} $. These cumulative masses are

To check these formulas and to compute $M$, write $t=\log |z|^2$ on the punctured affine chart. For a radial potential $F(t)$, polar integration of its complex Hessian gives

The potentials of $H$ and $L$ are $\log (1+\mathrm{e}^ t)$ and $t$. Thus the potential of $xH+yL$ has slopes $y$ and $x+y$ at the two ends, and the total mass of $xH+yL$ is $(x+y)^ m-y^ m$. The cases $(x,y)=(1,0)$ and $(1,1)$ give (12.23.2); the exceptional divisor and the hyperplane at infinity have zero smooth volume. In particular, $M=(b+\epsilon )^ m$ downstairs and $M=(b+2\epsilon )^ m-\epsilon ^ m$ upstairs. The form

solves the normalized equation, since $\nu \wedge \omega ^ p$ is the chosen probability density. The Gauduchon metric conformal to it is

Indeed, its $(n-1)$st power is

In the first term a function on the base multiplies a top-degree form on the base, and every factor in the second term is closed; hence both terms are closed, and $\gamma _*$ is balanced, in particular Gauduchon. Theorem 12.2.3 gives the exact identities

where $J_{\rm down}=\int _ B(q+\epsilon )^{-p}\nu _{\rm down}$ and $J_{\rm up}=\int _{\widetilde B}(q\circ \pi +\epsilon )^{-p}\nu _{\rm up}$, and shows that $\gamma _*$ attains the infimum.

Step 3: asymptotics of the denominators. Near zero, differentiation of (12.23.2) gives

For a density $(a s^{d-1}+O(s^ d))\, \mathrm{d} s$ near zero and $q=s^ k$, the integral of $(q+\epsilon )^{-p}$ over a fixed small interval $[0,s_0]$ has the following leading behavior as $\epsilon \downarrow 0$:

Here $\mathrm B(u,v)=\int _0^\infty t^{u-1}(1+t)^{-u-v}\, \mathrm{d} t$ is the Beta function. In the first case substitute $s=\epsilon ^{1/k}v$ and then $t=v^ k$. In the second case the integrand is asymptotic to $a/s$ on the range $\epsilon ^{1/k}\ll s\ll 1$. The error term $O(s^ d)$ is negligible: on a sufficiently small fixed interval its relative size is arbitrarily small, while the integral over the complementary interval stays bounded. For $kp<d$, monotone convergence applies, since the limiting density is integrable. The numerators in (12.23.4) tend to $\binom {n}{m}b^ m>0$, and the integrals over the complement of a neighborhood of the origin stay bounded. Hence (12.23.5) and (12.23.6), with $d=m$ downstairs and $d=1$ upstairs, give the stated orders.

Step 4: examples. For $m=k=2$ and $p=1$, the denominators satisfy $J_{\rm down}\sim \log (1/\epsilon )$ and $J_{\rm up}\sim (\pi /3)\epsilon ^{-1/2}$, so the volumes are asymptotic to $3b^2/\log (1/\epsilon )$ and $(9b^2/\pi )\epsilon ^{1/2}$, respectively. The order changes from $0$ to $1/2$. For $m=2$ and $k=p=1$, the downstairs denominator has a finite positive limit, while $J_{\rm up}\sim (2/3)\log (1/\epsilon )$. Thus a background of positive volume downstairs pulls back to one whose perturbed volumes are asymptotic to $9b^2/[2\log (1/\epsilon )]$, and both orders are zero. Finally, for an integer $j\geq 1$, choose $m=j+2$, $p=j+1$ and $k=1$. Then $kp<m$ gives a positive limit of the volumes downstairs, while the orders are $0$ and $j$. $\square$