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Example 12.6.3316E7687E1CC

Example 12.6.3.316E7687E1CC Let $\pi :X\to \mathbb P^2$ be the blowup at one point. Write $H$ for the pullback of the hyperplane class and $E$ for the exceptional divisor class. In the two-dimensional space they span, the cone of classes with a positive smooth representative is

\[ \mathscr C=\{ aH-bE:a>b>0\} . \]

These are closed classes, and we represent them by fixed smooth forms in order to apply proposition 12.6.1. Their intersection numbers are $H^2=1$, $H\cdot E=0$, $E^2=-1$. Consequently, for every smooth positive probability measure $\mu $,

\[ \lambda _\mu (aH-bE)=\sqrt{a^2-b^2}. \]

In the coordinates

\[ a=\mathrm{e}^ r\cosh s,\qquad b=\mathrm{e}^ r\sinh s, \qquad r\in \mathbb R,\quad s>0, \]

the Hessian metric is $\mathrm{d}r^2+\mathrm{d}s^2$. Its completion is therefore the Euclidean closed half-plane. The added boundary $s=0$ consists of the nef classes $\mathrm{e}^ rH$, each of positive volume $\mathrm{e}^{2r}$.

Proof. The projection from the blown-up point extends to a morphism $f:X\to \mathbb P^1$ with class $L=H-E$. The incidence map $(\pi ,f):X\to \mathbb P^2\times \mathbb P^1$ is an embedding. If $h$ and $\ell $ are pullbacks of Fubini–Study forms of integral one, then

\[ (a-b)h+b\ell \]

is positive for $a>b>0$, by the product embedding. Conversely, a positive representative has positive integrals over the exceptional curve and the strict transform of a line through the blown-up point. These integrals are respectively $b$ and $a-b$, proving the cone description. The same curves show that no nonzero linear combination in this two-dimensional space is smoothly $\mathrm{dd}^{\mathrm{c}}$-exact.

The intersection numbers follow from the hyperplane pullback and the normal bundle $\mathcal O_{\mathbb P^1}(-1)$ of the exceptional curve; equivalently $H^2=H\cdot L=1$ and $L^2=0$. Since the classes are closed, integrating the smooth equation gives $\mathcal V_\mu =a^2-b^2$, independently of $\mu $. Direct differentiation of $-\log \sqrt{a^2-b^2}$ gives

\[ g=\frac{(a^2+b^2)(\mathrm{d}a^2+\mathrm{d}b^2)-4ab\, \mathrm{d}a\, \mathrm{d}b}{(a^2-b^2)^2}. \]

Substituting the stated coordinates gives $\mathrm{d}r^2+\mathrm{d}s^2$. The coordinate domain is exactly $\mathbb R\times (0,\infty )$, so the intrinsic distance and completion are those of that Euclidean half-plane. Finally, $H$ has a semipositive pullback representative and is a limit of positive classes, hence is nef. Its square is one, but its integral over $E$ is zero, so it has no positive representative. This proves all assertions about the added boundary. $\square$

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