10.24.13 An incomplete metric with compatible flow data76505E13E4B8

Proof. Barriers and the radial inverse. Set $L=1-\log h=1+\log (1+1/r)$ and $w=-2\log L$. Differentiation gives

The function $Q$ is positive, since $L\geq 1$ and $c>2$. Its limits are $2/c$ at $0$ and $1-2/c$ at infinity, so $0<q_-\leq Q\leq q_+<\infty $ for some constants $q_\pm $. Choose $a_0\leq \min (0,\log q_-)$ and $b_0\geq \max (0,\log q_+)$. Then $l=w+a_0$ and $H=w+b_0$ satisfy

Exhaust $\mathbb P^1\setminus \{ 0\} $ by the disks $|\zeta |<R$ in the coordinate $\zeta =1/z$, and solve the equation on them with boundary value $w$. For $q=|\zeta |^2$, the equation becomes

where $\Delta $ is the Euclidean real Laplacian.

For $k>0$, the regular radial solution $v=V(q)$ of $(-\Delta +k)v=F(q)$ satisfies

The $j$-fold iterate of the integral kernel applied to the constant $1$ is $q^ j/(4^ j(j!)^2)$, so successive approximation converges uniformly on every finite interval. The coefficient of $A_0$ in the solution is $\sum _{j\geq 0}(kq/4)^ j/(j!)^2>0$, so exactly one value of $A_0$ gives any prescribed value at $q=R^2$. Moreover,

and this formula gives smoothness at $q=0$ by induction. The maximum principle for $-\Delta +k$ shows that this radial inverse is unique and preserves order.

Choose $k\geq \sup f\mathrm{e}^{H}$ on the fixed disk, and let $T(v)$ be the radial solution of $(-\Delta +k)T(v)=kv-f\mathrm{e}^{v}+g$ with boundary value $w$; the right side is nondecreasing in $v$ on $[l,H]$. The barrier inequalities give $l\leq T(l)$ and $T(H)\leq H$, so the iterates of $T$ starting from $l$ increase and stay below $H$. The derivative formula gives a uniform bound for their first derivatives on the fixed interval. Compactness and monotonicity give uniform convergence; passing to the limit in (10.24.20) proves the equation, and the averaged derivative formula bootstraps the limit to smoothness.

The same barriers bound the solutions on every fixed disk in $\zeta $ as $R\to \infty $. For an exact solution $U(q)$ of the nonlinear equation,

and induction gives bounds for all derivatives on compact intervals. A diagonal limit is therefore smooth on the $\zeta $-plane, including the point $z=\infty $, and satisfies $u=-2\log (-\log r)+O(1)$ at $0$.

The asymptotics, mass and energy. In the variable $t=\log r$, the equation reads

Put $v(t)=u(t)+c\log (1+\mathrm{e}^ t)$ and $V(s)=v(-s)$. Then

Convexity and this sublinear decay imply $V'<0$ and $V'\to 0$ at infinity. The function $J=(V')^2-2k_0\mathrm{e}^{V}$ satisfies $J'=-2k_0'\mathrm{e}^{V}$ and tends to zero. Since $k_0'=O(\mathrm{e}^{-s})$ and $\mathrm{e}^{V}\asymp s^{-2}$,

Thus $(\mathrm{e}^{-V/2})'=\sqrt{c/2}(1+O(\mathrm{e}^{-s}))$, and integration gives $\mathrm{e}^{-V/2}=\sqrt{c/2}(s+A)+O(\mathrm{e}^{-s})$. This identity and the differential equation give all three asymptotic formulas.

The local potential $u+c\log (1+r)$ extends subharmonically across $0$ by lemma 5.24.7. Its radial slope tends to zero at the puncture, so the extension adds no point mass. The density of $g_1$ is asymptotic to

Stokes' theorem for currents gives mass $c$. The energy integral at the cusp is bounded by $C\int ^\infty (\log s)/s^2\, \mathrm{d}s$, and the integral against the smooth form $\alpha $ is finite as well. $\square$

Proof of theorem 10.24.16. The initial data and the resolution. Take $X=\mathbb P^1_ Z\times \mathbb P^1_ W$, with the pulled-back normalized Fubini–Study forms $\alpha ,\beta $ of the two factors, and put

Let $H=(Z_1=0)$ and $p=([0:1],[1:0])=(\infty ,0)$. Define

The zero sets of $h_ H$ and $\eta $ are $H$ and $\{ p\} $, which are disjoint. Let $\pi :Y\to X$ be the blowup at $p$, write $E$ for the exceptional curve, and set $H_ Y=\pi ^{-1}H$ and $D=H_ Y+E$. The surface $Y$ is smooth, compact, connected and projective, and $\pi $ is an isomorphism off $p$.

Use $|\sigma _ H|^2=\pi ^*h_ H$ and $|\sigma _ E|^2=\pi ^*\eta $. Near $p$, put $x=Z_0/Z_1$ and $y=W_1/W_0$. In the blowup chart $x=a$, $y=ab$,

Thus $\pi ^*\eta $ defines a smooth Hermitian metric on $\mathcal O_ Y(E)$ across $E$. The second chart is obtained by interchanging $x$ and $y$, and the metric factors satisfy the transition rule.

The sections $Z_0W_0$, $\sqrt2Z_0W_1$, $Z_1W_1$ of $\mathcal O(1,1)$ have summed squared Fubini–Study norm $\eta $, and near $p$ they generate the ideal $(x,\sqrt2xy,y)=(x,y)$. Dividing their pullbacks by $\sigma _ E$ gives a morphism $F:Y\to \mathbb P^2$ with

where $\gamma $ is the normalized Fubini–Study form on $\mathbb P^2$. In the first chart $F=[1:\sqrt2ab:b]$, so $F|_ E$ is an isomorphism onto a line.

Near $p$,

Thus $\phi _0$ has analytic singularities. In the convention $\frac1{2m}\log \sum |f_ j|^2$, the coefficient $2$ is obtained by raising the numerator to the fourth power and expanding it as a sum of squared holomorphic monomials. The resolved identities are

The first two terms of $R_0$ are positive away from $E$, and along $E$ their kernel is $TE$, on which $F^*\gamma $ is positive. Hence $R_0$ is Kähler. The resolved current is $2[H_ Y]+2[E]+R_0$, which shows that $\phi _0$ is $\omega $-plurisubharmonic away from $p$; since it is bounded above, it extends across $p$. In the notation of (10.24.14), $m_ H=m_ E=2$ and $\ell _0=0$.

The Jacobian determinant of the blowup in the chart is $a$, so $b_ E=1$ and $b_ H=0$. Directly, $\mathrm{e}^{-q\phi _0}$ is integrable near $H$ exactly when $q<1/2$ and near $p$ exactly when $q<1$. Consequently $\lambda =\max \{ 2,1\} =2$, and

Since $\Theta _ E|_ E=-\omega _{\mathrm{FS}}$, the class $\{ \theta \} $ has degree $-2$ on $E$, so it is not nef. It is big, since the potential $3\log |\sigma _ E|^2$ gives the current

The prescribed volume. Define $\mathrm{d}V_ Y=\pi ^*\omega ^2/|\sigma _ E|^2$ off $E$. In the chart above,

since the factor $|a|^2$ from the Jacobian cancels exactly. The second chart gives the extension on the rest of $E$, and the extensions agree on overlaps. Thus $\mathrm{d}V_ Y$ is a smooth strictly positive volume form with $\pi ^*\mathrm{d}V_ X=|\sigma _ E|^2\mathrm{d}V_ Y$, as required in (10.24.15). Since $H_ Y$ and $E$ are disjoint, the measure $|\sigma _ H\sigma _ E|^{-2}\mathrm{d}V_ Y$ has a nonintegrable pole with smooth positive coefficient along each component.

The solution and the minimal reference. Take $u$ from lemma 10.24.18 with $c=6$, pulled back to $X$ from the first factor, and put $g_ X=(6\alpha +\mathrm{dd}^{\mathrm{c}}u)+6\beta $. This form is smooth and strictly positive near $p$, and

Set $P=2\log |\sigma _ E|^2$ and $\rho =P+\pi ^*u$. Then

Off $D$, the equation follows from the identity

Both measures extend by zero on $D$, which proves the global non-pluripolar equation.

Let $\psi $ be $\theta $-plurisubharmonic. Off $E$, the function $\psi -P$ descends to an $\omega $-plurisubharmonic function on $X\setminus \{ p\} $, which by lemma 5.24.8 extends across $p$ and is bounded above on $X$. Thus $\psi \leq P+C_\psi $ off $E$, and lemma 5.24.3 gives the inequality everywhere. Since $P$ itself is $\theta $-plurisubharmonic, it has minimal singularities.

The current of $P$ has non-pluripolar mass $\int _ X\omega ^2=72$, which is therefore the volume of the class, whereas the cup-product square of the class is $72+4E^2=68$. The cusp has mass $6$ and no atom, so the current of $\rho $ also has mass $\int _ Xg_ X^2=72$, and $\rho $ has full mass. Near $E$, $\pi ^*u$ is smooth, so $\rho $ has the Lelong numbers of $P$; along $H_ Y$ the cusp has zero Lelong number. The solution $\rho $ does not itself have minimal singularities, because $u$ is unbounded below near $H$.

Relative energy and uniqueness. Choose $M$ with $u-M\leq 0$ and use the potential $P+M$, which has minimal singularities, as reference. The canonical approximants of $\rho $ are

and their difference from the reference off $E$ is $\pi ^*u_ k$. On the curve let $b(t)=6\log (1+\mathrm{e}^ t)$ and $V(t)=b(t)+u(t)$. For large $k$ there is a unique level $t=T_ k$ in the cusp with $u(T_ k)-M=-k$. If $S_ k=-T_ k$, the asymptotics of lemma 10.24.18 give $S_ k\asymp \mathrm{e}^{k/2}$.

The local potential of the truncated curve current is $b+u_ k=\max (V-M,b-k)$. On $\{ t<T_ k\} $ its mass is $b'(T_ k)$, and at $t=T_ k$ its derivative jumps upward by $u_ t(T_ k)$. Smooth convex approximation and continuity of Bedford–Taylor products of bounded potentials therefore give total mass $V'(T_ k)=O(S_ k^{-1})$ on the cutoff region $\{ t\leq T_ k\} $. There $u_ k=-k$, so the absolute energy contribution of this region is

Outside it the density and the potential are unchanged, and $\int |u|g_1$ is finite. Consequently $\int |u_ k|g_{1,k}$ has a finite limit, where $g_{1,k}=6\alpha +\mathrm{dd}^{\mathrm{c}}u_ k$, and $\int |u_ k|\alpha $ converges by dominated convergence.

The mixed measures of the product current and of its reference are

Their pullbacks are exactly the mixed non-pluripolar measures on $Y$, which do not charge $E$. All three relative energy integrals of the canonical approximants therefore have finite limits. This proves $\rho \in \mathcal E^1(Y,\theta )$; replacing $P+M$ by the normalized envelope $V_\theta $ changes the reference by a bounded amount and does not affect this conclusion.

Uniqueness follows from the full-mass comparison argument in the proof of theorem 10.24.1. The measure $|\sigma |^{-2}\mathrm{d}V_ Y$ has infinite mass, but the two exponential densities being compared have mass $72$. Comparison on the strict sublevel sets forces equality almost everywhere, and hence equality of the quasi-plurisubharmonic potentials everywhere.

The missing metric boundary. The set on which the current is smooth and strictly positive is $Y\setminus (H_ Y\cup E)$: the cusp coefficient diverges at $H_ Y$, and the nonzero current $2[E]$ excludes smoothness along $E$. The smooth part $\pi ^*g_ X$ also degenerates along $TE$.

Let $U$ be a dense Zariski open set on which the current is smooth and strictly positive. As in the proof of theorem 10.24.7, choose a point of $E$ avoiding every other component of the analytic set $Y\setminus U$; in a small chart around it the complement of $U$ is exactly $E$. A transverse path $a=\xi >0$, $b=b_0$ lies in $U$ for small $\xi $ and approaches that point. Since $g_ X$ is smooth near $p$ and $\pi $ is smooth, the coefficients of $\pi ^*g_ X$ are bounded in the chart, so the path has bounded speed and finite remaining length. Its tails give an intrinsic Cauchy sequence with no limit in $U$, since its only ambient limit is the removed point and a smooth positive Riemannian metric induces the manifold topology. Hence every such $U$ is incomplete. $\square$