The following two examples of closed classes show that the unit-volume hypersurface may be complete or may have a boundary at finite distance. The first also shows why the derivative is needed in proposition 12.5.1: knowing the volume for every smooth probability measure need not identify the class.
For the first example take $X=\mathbb P^1\times \mathbb P^1$, and let $\alpha ,\beta $ be the pulled-back Fubini–Study forms of integral one on their respective factors. They satisfy
\[ \alpha ^2=\beta ^2=0,\qquad \int _ X\alpha \wedge \beta =1. \]
For $a,b>0$ the form $\theta _{a,b}=a\alpha +b\beta $ is Kähler. Since it is closed, integrating the smooth equation gives, for every smooth positive probability measure $\mu $,
\[ \mathcal V_\mu (\theta _{a,b})=2ab, \qquad \lambda _\mu (\theta _{a,b})=\sqrt{2ab}. \]
Consequently, for real $t$, all the forms $\theta _{\mathrm{e}^ t,\mathrm{e}^{-t}}$ have volume $2$ for every $\mu $. Their background classes are nevertheless distinct: for any fixed point $p$ of the second factor, integration over $\mathbb P^1\times \{ p\} $ gives $\mathrm{e}^ t$, and this integral is unchanged by the addition of $\mathrm{dd}^{\mathrm{c}}f$.
By direct differentiation, the restriction of (12.4.4) to this two-dimensional cone is
\[ \mathfrak g=\frac12\left(\frac{\mathrm{d}a^2}{a^2}+\frac{\mathrm{d}b^2}{b^2}\right). \]
The coordinates $((\log a)/\sqrt2,(\log b)/\sqrt2)$ identify this metric with the Euclidean metric on $\mathbb R^2$, so the metric on this family is complete. The curve $a=\mathrm{e}^ t$, $b=\mathrm{e}^{-t}$ has unit speed. All distances here are intrinsic to this two-dimensional family.
In the second example, the blowup of $\mathbb P^2$ at a point, the cone is incomplete, and the boundary added by the completion consists of classes of positive volume.
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.
$\square$
Pome
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