10.24 An incomplete metric in a big classE61E86F27D3F

Proof of theorem 10.24.1. The surface and the divisor. Use the incidence model

The projections $\pi :Y\to \mathbb P^2$ and $q:Y\to \mathbb P^1$ identify $Y$ with the blowup of $\mathbb P^2$ at $p$, with exceptional divisor $E=\pi ^{-1}(p)$. Let $\omega _1$ be the Fubini–Study form on $\mathbb P^1$ with the same normalization as $\omega _2$, and write

The form $\Omega $ is Kähler, being the restriction of a product Kähler metric.

The linear map $(Z_0,Z_1,Z_2)\mapsto (Z_1,Z_2)$ defines a section $\sigma $ of

because on $Y$ the vector $(Z_1,Z_2)$ lies in the tautological line $[W_1:W_2]$. The section vanishes simply along $E$, which identifies this line bundle with $\mathcal O_ Y(E)$. For the product of the Fubini–Study metric and the metric induced on the tautological bundle, its squared norm is

In a blowup chart $\pi (t,w)=(t,tw)$, $q(t,w)=[1:w]$, one has

The current identity follows either from this formula and the second blowup chart or from the Poincaré–Lelong formula.

The big class and its equation. Set

Thus $\rho \in \mathrm{PSH}(Y,\theta )$. The class $\alpha =\{ \theta \} $ is big, because the potential $\psi =(5/2)\log s$ satisfies

It is not nef, since $\int _ E\theta =-2$: the restriction of $\beta $ to $E$ vanishes and $q|_ E$ is an isomorphism onto $\mathbb P^1$. A nef class has nonnegative degree on every curve, by its defining smooth approximations.

On $Y\setminus E$ define $\mathrm{d}V_ Y=\beta ^2/s$. To check that it extends across $E$, put

and let $\mathrm{d}\lambda _{z_1,z_2}$ be the analogous Euclidean volume in affine coordinates of $\mathbb P^2$. The normalized Fubini–Study volume is

Since the complex Jacobian determinant of $(t,w)\mapsto (t,tw)$ is $t$, we obtain

The second expression is smooth and strictly positive at $t=0$. Interchanging the two homogeneous coordinates gives the extension on the other chart. The two extensions agree off $E$, hence everywhere on their overlap.

The non-pluripolar product of $T$ agrees with $\beta ^2$ off $E$ and gives no mass to $E$. The smooth measure $\beta ^2$ also gives no mass to $E$, so

The right side of (10.24.2) is defined on $Y\setminus E$ and extended by zero on $E$, so no product of zero and infinity occurs. This proves the equation and gives $\int _ Y\langle T^2\rangle =\int _{\mathbb P^2}\omega _2^2=1$.

Minimal singularities and energy. Let $u\in \mathrm{PSH}(Y,\theta )$. Since

the function $u-\rho $ on $Y\setminus E$ descends to an $\omega _2$-plurisubharmonic function on $\mathbb P^2\setminus \{ p\} $. By lemma 5.24.8, it is bounded above, so $u\leq \rho +C_ u$ there. This inequality holds almost everywhere on $Y$, and lemma 5.24.3 makes it hold everywhere. In particular $u=-\infty $ on $E$, and $\rho $ has minimal singularities.

Let $V_\theta $ be the normalized envelope of section 5.40. Since $s\leq 1$, we have $\rho \leq 0$, and minimality gives

Thus $\rho -V_\theta $ is bounded off $E$, and the energy is

The mixed measures have finite mass in the closed Kähler theory and do not charge $E$. Hence $E_\theta (\rho )>-\infty $, so $\rho \in \mathcal E^1(Y,\theta )$. A potential with minimal singularities has full mass; in this example that mass is $1$.

Uniqueness. Suppose $u\in \mathcal E^1(Y,\theta )$ is another solution, and let $\mathrm{d}\mu =s^{-1}\mathrm{d}V_ Y$ on $Y\setminus E$, extended by zero on $E$. This measure is smooth and positive off $E$, but has infinite total mass. The classical comparison principle for full-mass potentials in the closed theory gives

Both integrals are finite, since the two Monge–Ampère measures have mass $1$. The integrand on the left is strictly larger, which forces $\mu (\{ u<\rho \} )=0$. Interchanging $u$ and $\rho $ gives $\mu (\{ \rho <u\} )=0$ as well. Therefore $u=\rho $ almost everywhere on $Y\setminus E$ with respect to a smooth volume form, hence almost everywhere on $Y$, since $E$ has zero volume. By lemma 5.24.3, $u=\rho $ everywhere.

The ample locus and incompleteness. The Kähler current in (10.24.5) is smooth off $E$. On the other hand, every $u\in \mathrm{PSH}(Y,\theta )$ satisfies $u\leq \rho +C_ u$, so its Lelong numbers at points of $E$ are at least those of $\rho =2\log s$, which are positive. Thus every Kähler current in $\alpha $ is singular at every point of $E$, and theorem 10.13.2 gives

The current $T$ equals $\beta $ there, and it is strictly positive because $\pi $ is a biholomorphism away from $E$. Its nonzero divisorial part $2[E]$ prevents smoothness at every point of $E$.

Let $U\subset Y\setminus E$ be a nonempty connected open set, and set $W=\pi (U)$. The map $\pi :(U,T|_ U)\to (W,\omega _2|_ W)$ is an isometry. Choose $x\in W$ and a smooth path $c:[0,1]\to \mathbb P^2$ from $x$ to $p$. Its first exit time

is positive, and $c([0,t_0))\subset W$. The path has finite Fubini–Study length. For $t_ k\nearrow t_0$, the lengths of its tails show that $c(t_ k)$ is a Cauchy sequence for the intrinsic metric of $W$. It has no limit in $W$: intrinsic convergence implies ambient convergence, and the ambient limit is $c(t_0)\notin W$. Thus $U$ is incomplete. Applying this argument to each connected component proves the assertion for a dense Zariski open set, which may be disconnected. $\square$

Proof. For $t\leq -2$ put $L(t)=-3\log (1-c\mathrm{e}^{t/2})$. With $q=c\mathrm{e}^{t/2}$,

This proves the left identity for every additive constant $B$.

Since $H'(p)=(p-1)p$, the increasing function

tends to $+\infty $ as $p\to 1$, because its integrand is asymptotic to $2/(1-p)^2$ there. Hence it has an inverse $p_ R(t)$, defined for $t\geq 2$ and on a neighborhood of $2$, with $p_ R(2)=3/4$ and $p_ R'=H(p_ R)/p_ R$. The function $F_ R=t+\log H(p_ R)$ has derivative $p_ R$ and satisfies the right identity. For $q=1-p_ R$,

Hence $q\sim 2/t$ and $H(p_ R)\sim 2/t^2$, and the asymptotics at the right end follow.

It remains to interpolate the derivatives. Set $d=p_ R(2)-p_ L(-2)>0$. Choose disjoint short collars in $[-2,2]$ at the two endpoints, and cutoffs $\alpha _ L,\alpha _ R$ equal to one on smaller collars. For small $\varepsilon >0$, the function

is positive and agrees with the derivative germs at the endpoints. The collars and $\varepsilon $ can be chosen so that $\int _{-2}^2Q_0<d$. Adding a nonnegative smooth bump in the interior, with integral $d-\int _{-2}^2Q_0$, gives a positive function $Q$, and we put

This function matches both endpoint germs, is increasing and takes values in $(0,1)$. Integrating $F'=p$ with $F(2)=F_ R(2)$ gives the required $F$ and fixes the constant $B$ at the left end. The left expansion follows directly. $\square$

Proof of theorem 10.24.7. The background and the pole. Use the incidence surface of theorem 10.24.1, with blowup map $\pi :Y\to \mathbb P^2$ and exceptional curve $E$ over $[1:0:0]$, and let $\beta ,\gamma $ be the pulled-back Fubini–Study forms as there. Put $D=\pi ^{-1}(Z_0=0)$ and $A=\pi ^*c_1(\mathcal O(1))$. Then $E\cap D=\varnothing $, $A^2=1$ and $A\cdot E=0$. The class $A$ is nef, since it has the smooth semipositive representative $\beta $. Moreover $\{ \gamma \} =A-[E]$, and $\beta +\varepsilon \gamma $ is Kähler for $\varepsilon >0$. Thus

is a Kähler current in $A$, which proves bigness.

Choose an even nonnegative smooth function $\zeta $, supported in $[-1,1]$, positive on its interior and of integral one. The smooth convex function

satisfies $U=0$ for $t\leq -1$, $U=t$ for $t\geq 1$, and $U',U''\geq 0$. On $Z_0\ne 0$, write $z=(Z_1/Z_0,Z_2/Z_0)$, $s=|z|^2$ and $t=\log s$, and give the section $Z_0$ of $\mathcal O(1)$ the squared norm $\mathrm{e}^{-U(t)}$; this norm is constant near the affine origin. Near the line at infinity, use

The section $Z_1$ then has squared norm $\mathrm{e}^{-U}/|w|^2=1/h$ near $w=0$. These formulas define a smooth Hermitian metric on $\mathcal O(1)$. Pull it back to $\mathcal O_ Y(D)$ and let $\sigma =\pi ^*Z_0$. Its curvature form $\theta $ and the norm of $\sigma $ are

The radial Hessian shows that $\theta $ is semipositive. It vanishes near $E$ and has rank one near $D$. In particular $V_\theta =0$, and the nef class has volume $A^2=1$.

Extension of the potential. On $Y\setminus (E\cup D)\simeq \mathbb C^2\setminus \{ 0\} $, set

so that $\theta +\mathrm{dd}^{\mathrm{c}}\rho =\Omega $. The eigenvalues of the Hermitian matrix of $\Omega $ are $p/s$ and $p'/s$, which are positive. Extend $\rho $ to $E$ by $B$ and to $D$ by $-\infty $.

Near the affine origin, the smooth functions $F(\log (|z|^2+\varepsilon ))$ are plurisubharmonic and decrease to the continuous extension of $F(\log |z|^2)$ as $\varepsilon \downarrow 0$. Indeed, their two Hessian eigenvalues are $p/(s+\varepsilon )$ and $(s p'+\varepsilon p)/(s+\varepsilon )^2$, with $p,p'$ evaluated at $\log (s+\varepsilon )$. Since $\theta =0$ there, pulling back to $Y$ proves positivity near $E$.

Near $D$ put $t_\varepsilon =\log h-\log (|w|^2+\varepsilon )$ and $g(t)=F(t)-t$. The local potentials $g(t_\varepsilon )+\log h$ decrease to $\rho +\log h$, and their Hessians are

Thus $\rho $ is $\theta $-plurisubharmonic on $Y$. Since $\rho $ is bounded near $E$ and grows like a double logarithm near $D$, all its Lelong numbers vanish.

The volume form and full mass. With Euclidean area $\mathrm{d}A$ in each complex variable, the radial determinant is

Define $\mathrm{d}V_ Y=\mathrm{e}^{-F}\Omega ^2$ off the two divisors. In an exceptional chart $z=(u,uv)$, the real Jacobian is $|u|^2$ and $s=|u|^2h$, so the left profile identity gives

At infinity, the map $z=(1/w,v/w)$ has real Jacobian $|w|^{-6}$, and the right profile identity gives

The analogous second charts cover each divisor. The extensions are smooth and strictly positive, and they agree on overlaps because they agree on the punctured overlaps. The last expression exhibits the nonintegrable pole with smooth positive coefficient.

On the complement of the two divisors, $\mathrm{e}^{\rho }|\sigma |^{-2}\mathrm{d}V_ Y=\Omega ^2$; we extend this density by zero on $D$. Both measures give zero mass to the two divisors. In polar coordinates on $\mathbb C^2$, the Euclidean volume is $\pi ^2s\, \mathrm{d}s$ times the normalized angular measure, so

This proves the global equation and shows that $\rho $ has full mass.

Bounded truncations and energy. Near $D$ write $\alpha =\mathrm{dd}^{\mathrm{c}}t=\mathrm{dd}^{\mathrm{c}}\log h$ and $\beta _ t=(\mathrm{i}/2\pi )\partial t\wedge \bar\partial t$. These forms have rank one, and

Integration of $\alpha \wedge \beta _ t$ over the angular and projective variables gives $\mathrm{d}t$. Since $|\rho |=O(\log t)$ and $p'=O(t^{-2})$, the integrals of $|\rho |$ against $\Omega ^2$ and $\Omega \wedge \theta $ near $D$ are finite. Near $E$ the potential is bounded and $\theta =0$, and the remaining region is compact with smooth data.

Choose $M$ so that $\phi =\rho -M\leq 0$, and set $\phi _ k=\max (\phi ,-k)$. For large $k$ the truncation changes $\phi $ only in the cusp near $D$, where $\phi =\log H(p(t))-M$ is strictly decreasing. If $\phi (T_ k)=-k$, then $k=2\log T_ k+O(1)$. On $\mathbb C^2\setminus \{ 0\} $, the radial local potential of $\theta +\mathrm{dd}^{\mathrm{c}}\phi _ k$ is

Near $D$, in the frame used above, the local potential of $\theta +\mathrm{dd}^{\mathrm{c}}\phi _ k$ is $G_ k(t)+\log |w|^2$, which equals the smooth function $\log h-k$; the term $\log |w|^2$ cancels the divisorial part of $\mathrm{dd}^{\mathrm{c}}G_ k(t)$. At $t=T_ k$, the derivative of $G_ k$ jumps upward from $p(T_ k)$ to $1$.

Convolving $G_ k$ in $t$ with an even nonnegative smooth kernel gives convex functions decreasing to $G_ k$ as the scale decreases. The corresponding bounded local potentials converge uniformly on compact sets, so their Bedford–Taylor products converge. Integrating the smooth radial measures through the transition shows that the transition carries mass $1-p(T_ k)^2$ for $(\theta +\mathrm{dd}^{\mathrm{c}}\phi _ k)^2$ and mass $1-p(T_ k)$ for $(\theta +\mathrm{dd}^{\mathrm{c}}\phi _ k)\wedge \theta $. Beyond the transition, $G_ k$ is linear and both top-degree measures vanish. Since $\phi _ k=-k$ on the transition, its energy contributions have absolute values $k(1-p(T_ k)^2)$ and $k(1-p(T_ k))$, and both tend to zero because $1-p(T_ k)\sim 2/T_ k$. The contributions of the remaining region converge absolutely by the bounds above. At $E$, bounded potentials give no mass to the curve, in accordance with the vanishing radial boundary mass $p(-\infty )^2=0$. Consequently

has a finite limit, and in particular is bounded below. This proves $\rho \in \mathcal E^1(Y,\theta )$.

Uniqueness follows from the full-mass comparison argument in the proof of theorem 10.24.1, now with $\mathrm{d}\mu =|\sigma |^{-2}\mathrm{d}V_ Y$ off $D$, extended by zero on $D$. Its total mass is infinite, but each exponential density $\mathrm{e}^{\rho }\, \mathrm{d}\mu $ and $\mathrm{e}^{u}\, \mathrm{d}\mu $ is a finite measure of mass $1$. Comparison on the two strict sublevel sets forces equality almost everywhere, and lemma 5.24.3 gives equality of the potentials everywhere.

The two metric ends. Fix $v=v_0$ in an exceptional chart and write $h_0=1+|v_0|^2$. The restriction of $\Omega $ to the normal line $\{ v=v_0\} $ is

Its coefficient is asymptotic to $(3c\sqrt{h_0}/4)|u|^{-1}$, so the current has no smooth extension at any point of $E$. The radial path $u=r>0$ has finite remaining length, bounded by $C\int _0^{r_0}r^{-1/2}\, \mathrm{d}r$. At $D$ the normal coefficient is asymptotic to $2/(|w|^2t^2)$ and also diverges. Thus every open set on which $\Omega $ is smooth and strictly positive misses both curves.

Let $W$ be a dense Zariski open set on which $\Omega $ is smooth and strictly positive. Its complement $Y\setminus W$ is a proper analytic set containing $E$, and by local finiteness of irreducible components and compactness it has finitely many components, which are curves or points. Each curve other than $E$ meets $E$ in finitely many points. Choose a point of $E$ that avoids all these intersections and all point components. In a sufficiently small chart around it, the complement of $W$ is exactly $E$, so a tail of the radial path above lies in $W$. Its points approaching $E$ form a Cauchy sequence for the intrinsic metric with no limit in $W$: the smooth positive metric induces the manifold topology of $W$, and the only ambient limit is the removed point. Hence $W$ is incomplete.

Finally, on the cusp $\Omega =p\mathrm{dd}^{\mathrm{c}}t+p'\, \mathrm{d}t\wedge \mathrm d^{\mathrm c}t$, both terms are nonnegative, and $p'\geq c_0/t^2$ for large $t$ and some $c_0>0$. The length of any path with $t\to \infty $ is therefore bounded below by a positive multiple of $\int |\mathrm{d}t|/t$, which diverges. Thus the end at finite distance is the null curve $E$, disjoint from the pole. $\square$

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. $\square$