11.43.1 Canonical models and their mass6439CBEB8575

Proof. The Legendre transform of $L$ is $L^*=\ell $ on $\Sigma $ and $+\infty $ outside it. Since $-\log (n+1)\leq \ell \leq 0$,

A finite convex function of the logarithmic moduli is plurisubharmonic: for a smooth convex function the complex Hessian is positive, and convolution gives the general case. On $T\cap \{ Z_ j\ne 0\} $, add the local Fubini–Study potential $\log \sum _ i|Z_ i/Z_ j|^2$; the resulting functions are again convex functions of the logarithmic moduli of the affine coordinates, and they are bounded above by that smooth potential because $h_ K,g_ K\leq L$. Lemma 5.24.7 therefore gives the extensions across the coordinate hyperplanes and the stated bounds. If $p\in \operatorname {int}K$, then $p\in \operatorname {int}\Sigma $, and at $x_ i=\log (p_ i/(1-\sum _ k p_ k))$ the supremum defining $L=\ell ^*$ is attained at $p\in K$, so $g_ K(x)=L(x)$. This proves $\sup \Phi _ K=0$.

For $C\geq \log (n+1)$, let $v_ C=P_{\omega _{\mathrm{FS}}}(\psi _ K+C,0)$, the envelope of the obstacle $\min (\psi _ K+C,0)$. Since $\Phi _ K$ is a competitor, $\Phi _ K\leq v_ C\leq 0$. The obstacle is upper semicontinuous, so $v_ C\leq \psi _ K+C$ everywhere. The compact torus acting by rotations of the coordinates preserves the obstacle, so $v_ C$ is invariant, and on $T$ the function $f=v_ C+L$ is a finite convex function of $x$ with $f\leq L$ and $f\leq h_ K+C$. The second inequality shows that the Legendre transform $f^*$ is $+\infty $ outside $K$, and the first gives $f^*\geq \ell $. Thus

Hence $v_ C\leq \Phi _ K$ on $T$, and since both functions are quasi-plurisubharmonic and $T$ is the complement of an analytic set, $v_ C=\Phi _ K$ on $\mathbb P^ n$. Letting $C\to \infty $ gives $P_{\omega _{\mathrm{FS}}}[\psi _ K]=\Phi _ K$. Since $\psi _ K\leq \Phi _ K\leq \psi _ K+\log (n+1)$, the envelopes of $\min (\Phi _ K+C,0)$ and $\min (\psi _ K+C,0)$ interlace as $C\to \infty $, so also $P_{\omega _{\mathrm{FS}}}[\Phi _ K]=\Phi _ K$.

For the final assertion put $\varepsilon =1-\delta $, $b=\sum _ i p_{*,i}$ and $\chi =p_*\cdot x-\varepsilon L$ on $T$. The inclusion $p_*+\delta \Sigma \subset \Sigma $ implies $p_{*,i}\geq 0$ and $b\leq \varepsilon $, and the extension of $\chi $ to $\mathbb P^ n$ satisfies

Since $h_ K\geq p_*\cdot x+\delta h_\Sigma $ and $L-h_\Sigma \leq \log (n+1)$, one has $\chi -\delta \log (n+1)\leq \Phi _ K$. Thus $\chi -\delta \log (n+1)$ is an $\varepsilon \omega _{\mathrm{FS}}$-plurisubharmonic competitor for $P_{\varepsilon \omega _{\mathrm{FS}}}(\Phi _ K)$, which proves the claim. $\square$

Proof. On $\mathbb P^ n$, the envelope identity is lemma 11.43.2. In the blowup case, every $\pi ^*\omega _{\mathrm{FS}}$-plurisubharmonic function descends to $\mathbb P^ n$. Indeed, off $E_ q$ add a local potential of $\omega _{\mathrm{FS}}$; the resulting plurisubharmonic function is bounded above near $q$, because the original function is bounded above near the compact fiber $E_ q$, and therefore extends across $q$. The pullback of the extension agrees with the original function off $E_ q$, hence everywhere. Pullback is therefore a bijection from $\mathrm{PSH}(\mathbb P^ n,\omega _{\mathrm{FS}})$ onto $\mathrm{PSH}(X,\Theta )$ preserving inequalities and envelopes, and positive scaling preserves them as well.

On projective space, $\Theta =\omega _{\mathrm{FS}}$ is Kähler. For bigness in the blowup case, choose unitary homogeneous coordinates $[W_0:\cdots :W_ n]$, used only in this paragraph, in which $q=[1:0:\cdots :0]$, set

and let $Q:X\to \mathbb P^{n-1}$ be the map induced by the linear projection from $q$. For $0<t<1$,

The smooth part is a Kähler form, since $(\pi ,Q)$ embeds $X$ as the incidence graph in $\mathbb P^ n\times \mathbb P^{n-1}$ and both coefficients are positive. This proves the assertions about the manifold and the form $\Theta $.

For a smooth convex $g$, the coordinates $z_ i=\mathrm{e}^{x_ i/2+\mathrm{i}\vartheta _ i}$ give

For a finite convex $g$, convolution and the continuity of bounded Monge–Ampère products under local uniform convergence identify $(\mathrm{dd}^{\mathrm{c}}g)^ n$ on $T$ in the same way: its image under $z\mapsto x$ is $n!M_ g$, where $M_ g$ is the Alexandrov measure, which assigns to a Borel set of $\mathbb R^ n$ the Euclidean volume of its subgradient image. For $g_ K$, the bound $g_ K\leq h_ K+\log (n+1)$ and a separation argument put every subgradient in $K$. Conversely, if $p\in \operatorname {int}K$ and $p+c\mathbb B\subset K$, where $\mathbb B$ is the closed unit ball of $\mathbb R^ n$ and $c>0$, then $g_ K(x)-p\cdot x\geq c|x|$; this function attains a minimum, so $p$ is a subgradient of $g_ K$ at some point. Consequently

since the boundary of a convex body has zero volume. The analytic complement of $T$ receives no non-pluripolar mass, so this is the full mass. Locality off the exceptional divisor proves the formula after pullback, since the divisor and its image carry no non-pluripolar mass, and homogeneity gives the factor $r^ n$. The reference is locally bounded off the coordinate hyperplanes and their inverse image, an analytic set, hence has small unbounded locus. Finally, scaling by $r$ and pulling back the competitor found at the end of the proof of lemma 11.43.2 gives the last assertion. $\square$