Proposition 11.43.15C9CD848ECB1A

Proof. On $Y=\operatorname {Bl}_ o\mathbb P^ n$, let $\pi _ o$ be the blowdown map and $Q_ o:Y\to \mathbb P^{n-1}$ the map induced by the linear projection from $o$. As in the proof of lemma 11.43.3, the incidence graph embedding shows that

is a Kähler form, where $\omega _{\mathrm{FS},n-1}$ is the Fubini–Study form of $\mathbb P^{n-1}$. Since the two centers are distinct, $Z$ is also the blowup $\sigma :Z\to Y$ of the point above $q$. Put $\alpha _ t=\sigma ^*\beta _ t$. The Poincaré–Lelong formula gives (11.43.16); there is no term along $E_ q$, since $\psi _ t$ is locally bounded near $q$.

In a blowup chart, $\sigma $ is $(s,w_2,\ldots ,w_ n)\mapsto (s,sw_2,\ldots ,sw_ n)$. At $s=0$ its differential has rank one and kills the $n-1$ tangent directions of $E_ q$. Positivity of $\beta _ t$ therefore gives the assertion about the rank; off $E_ q$ the map is biholomorphic.

Write $H=\pi _ o^*[\omega _{\mathrm{FS}}]$ and $E=[E_ o]$ for the cohomology classes on $Y$. The current identity gives $[\beta _ t]=H-tE$. Since $H|_{E_ o}=0$, all mixed top intersections vanish. The normal bundle of $E_ o\cong \mathbb P^{n-1}$ is $\mathcal O_{\mathbb P^{n-1}}(-1)$, so $\int _ YE^ n=(-1)^{n-1}$. Therefore

The first equality holds because $\sigma $ has degree one. Finally, if $u\sim \pi ^*\Phi _{K_ t}\sim \pi ^*\psi _ t$, then $u\circ \tau $ differs from the pulled-back prototype $\psi _ t\circ \pi \circ \tau $ by a bounded function off $E_ o$. By (11.43.16), after adding a local potential of $\alpha _ t$ this bounded difference becomes plurisubharmonic off $E_ o$, and it extends across $E_ o$ by lemma 5.24.7. No Monge–Ampère product of unbounded potentials is involved. $\square$