Proposition 14.9.1CA9F2EF1D149

Proof. Integration of the equation and (23.1.5) give

Substituting the first equality into (23.1.39) proves (14.9.3).

If $D(F)=0$, then $(u_ F)_ w=0$ everywhere, because $u_ F$ is smooth. Since $u_ F$ is real, it is constant on each connected fiber, so it descends to a smooth real function $q$ on $\mathbb P^2$, and $\omega _{\mathrm{FS}}+\mathrm{dd}^{\mathrm{c}}q>0$ as in the proof of proposition 23.1.38. Since $\beta ^2=0$ and $\omega =\pi ^*\omega _{\mathrm{FS}}+\beta $, we have

Dividing by $\omega ^3=3\pi ^*\omega _{\mathrm{FS}}^2\wedge \beta $ and taking logarithms proves that $F$ is a smooth base pullback.

Conversely, suppose $F=\pi ^*H$. Apply theorem 14.2.1 on $(\mathbb P^2,\omega _{\mathrm{FS}})$ to obtain a smooth real $q$ and a scalar $c$ with

Since the base form is closed, Stokes' theorem gives $\mathrm{e}^ c\int _{\mathbb P^2}\mathrm{e}^ H\omega _{\mathrm{FS}}^2=1$. The preceding cube identity shows that $\pi ^*q$, after subtraction of its $\nu $-mean, solves the equation on $X$ with scalar $c$, and uniqueness gives $b(F)=c$. Since the base measure of $\nu $ is $\omega _{\mathrm{FS}}^2$, it follows that $D(F)=0$. $\square$