Example 12.12.7D37A528BDFB8

Proof. Realize the blowup as $X=\{ (x,[u:v])\in \mathbb P^2\times \mathbb P^1:x_1v=x_2u\} $, with projections $p,q$ and exceptional curve $E=p^{-1}([1:0:0])$. Let $L=p^*\omega _{\mathrm{FS},\mathbb P^2}$, $Q=q^*\omega _{\mathrm{FS},\mathbb P^1}$ and $\Omega =L+Q$, with the Fubini–Study forms normalized to have integral one on projective lines. The incidence embedding makes $\Omega $ Kähler. The degree-one blowdown, a fiber of $q$, and the isomorphism $q|_ E:E\to \mathbb P^1$ give

Take $t=1$, $\alpha =L$, $\kappa _0=2L+Q$ and $\mu =\kappa _0^2/8$. With $r=1-s$, the core volumes are $V_ s=1+4r+3r^2$; in particular $V_0=8$ and $V=V_1=1$. Choose a nonconstant smooth function $0\leq \chi \leq 1$ supported away from $E$, and put

Since $\int _ XL\wedge \kappa _0=3$, we have $|f|\leq 1$ and $\int _ XfL\wedge \kappa _0=0$. Thus $B\geq L/2$ and $B|_{TE}=0$, so the curve $C=E$ satisfies the hypotheses of theorem 12.12.2. Where $\mathrm{d}\chi \ne 0$, the form $L$ is symplectic, and wedging with $L$ is injective on real one-forms in real dimension four. Hence $\mathrm{d}B=\tfrac 12\mathrm{d}f\wedge L\ne 0$.

At the endpoint $\theta _1=(2+f/2)L\geq 3L/2$, and $\lambda _\mu (L)=1$ by the closed nef formula; this gives the lower bound for $\lambda _\mu (\theta _1)$. Since $\kappa _0^2=8\mu $, the initial solution is $\gamma _0=\kappa _0$ with $u_0=0$. Therefore $D_\mu (\kappa _0)=\sqrt8$, $I_ B=3$, $\Delta _\mu (B)=3/\sqrt8$ and $U_\mu =11/\sqrt8$, so $\lambda _\mu (\theta _0)\leq 11/\sqrt8$. The cost identity and the lower bound at the endpoint give the asserted estimate for $\mathcal C$.

The pullback of $\{ x_1=0\} $ is $E$ plus a fiber of $q$, so $\mathcal O(E)=p^*\mathcal O_{\mathbb P^2}(1)\otimes q^*\mathcal O_{\mathbb P^1}(-1)$. Give this bundle its product metric $h_ E$, with curvature $L-Q$, and let $s_ E$ be its canonical section. The Poincaré–Lelong formula gives, for $\rho =\tfrac 12\log |s_ E|_{h_ E}^2$,

Finally, $\int _ X\Omega \wedge a_0=5$, so the right side of (12.12.6) is $8-10=-2$, although $V_1=1$. $\square$