Theorem 10.24.18A2E8673E914

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. $\square$