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Lemma 11.43.359D09F5BD0FB

Lemma 11.43.3.59D09F5BD0FB The manifold $X$ is Kähler, and $\Theta $ is smooth, closed, semipositive and big. For $r>0$, the reference $r\pi ^*\Phi _ K$ satisfies $P_{r\Theta }[r\pi ^*\Phi _ K]=r\pi ^*\Phi _ K$, has small unbounded locus, and satisfies

C1218CE2B784 11.43.4
\begin{equation} \int _ X\mathrm{MA}_{r\Theta }(r\pi ^*\Phi _ K) =r^ n n!\operatorname {Vol}_ n(K). \end{equation}

If $p_*+\delta \Sigma \subset K$ with $0<\delta <1$, then $P_{(1-\delta )r\Theta }(r\pi ^*\Phi _ K)\not\equiv -\infty $.

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

\[ \psi _ q=\log \frac{\sum _{i=1}^ n|W_ i|^2}{\sum _{i=0}^ n|W_ i|^2}, \]

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

\[ \Theta +\mathrm{dd}^{\mathrm{c}}(t\pi ^*\psi _ q) =(1-t)\pi ^*\omega _{\mathrm{FS}}+tQ^*\omega _{\mathrm{FS},n-1}+t[E_ q]. \]

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

\[ \mathrm{d}^{\mathrm c}g=\frac1{2\pi }\sum _ i g_ i\, \mathrm{d}\vartheta _ i, \qquad \int _ T(\mathrm{dd}^{\mathrm{c}}g)^ n=n!\int _{\mathbb R^ n}\det (D^2g)\, \mathrm{d}x. \]

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

\[ \operatorname {int}K\subset \partial g_ K(\mathbb R^ n)\subset K, \qquad M_{g_ K}(\mathbb R^ n)=\operatorname {Vol}_ n(K), \]

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. $\square$

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