12.12 Subtraction with a closed nef core75F31FAA4616

Proof. We first locate the boundary of $\mathscr P$ along the path. Let $s<t$. Since $\alpha $ is nef, for every $\delta >0$ there is a smooth function $v_\delta $ with $\alpha +\mathrm{dd}^{\mathrm{c}}v_\delta \geq -\delta \Omega $. Taking $\delta <t-s$ makes $a_ s+\mathrm{dd}^{\mathrm{c}}v_\delta $ Kähler, and adding $B\geq 0$ gives a positive representative of $\theta _ s$. The same approximations show that $\theta _ t$ is nef. On the other hand, the hypotheses on $C$ and Stokes' theorem on the compact curve give

for every smooth $v$. At $s=t$ the right side vanishes, which excludes a positive representative of $\theta _ t$. For $s>t$, a representative with $\theta _ s+\mathrm{dd}^{\mathrm{c}}v\geq -\delta \Omega $ and $\delta <s-t$ would contradict this identity, so $\theta _ s$ is not nef.

Since every $a_ s$ with $s\leq t$ is closed and nef, the closed nef formula following theorem 12.11.1 gives $\lambda _\mu (a_ s)=V_ s^{1/n}$, including at $s=t$. The difference $\theta _ t-a_ t=B$ is semipositive, so monotonicity gives $\lambda _\mu (\theta _ t)\geq V^{1/n}$, while any fixed Gauduchon test bounds $\lambda _\mu (\theta _ t)$ from above. Since $\theta _ s=\theta _ t+(t-s)\Omega $, theorem 12.11.1 with the Hermitian form $\Omega $ identifies $\lambda _\mu (\theta _ t)$ with the limit of $\lambda _\mu (\theta _ s)$ as $s\uparrow t$.

Write $f(s)=\lambda _\mu (\theta _ s)$ for $s<t$. The path has constant velocity $-\Omega $, so proposition 12.3.2 shows that $f$ is smooth and

In particular, the ratio $H$ is continuous and independent of the choice of minimizing metrics. For $r<t$, the fundamental theorem of calculus gives

Since $H\geq 0$, letting $r\uparrow t$ and using the limit just established yields the identity in (12.12.3).

It remains to bound $f(0)=\lambda _\mu (\theta _0)$ from above by two test metrics. The Kähler form $\kappa _0$ is Gauduchon, and since $\kappa _0$ is cohomologous to $a_0$, closed Stokes gives $\int _ X\theta _0\wedge \kappa _0^{n-1}=V_0+I_ B$; hence $\lambda _\mu (\theta _0)\leq U_\mu $. For the second test $\gamma _0$, the equation $\gamma _0^ n=V_0\mu $ and the cohomology class of $\gamma _0$ give

so its quotient for $\theta _0$ is $V_0^{1/n}+\Delta _\mu (B)$. Combined with the lower bound $\lambda _\mu (\theta _ t)\geq V^{1/n}$, these two tests prove the upper bound in (12.12.3). If $B\leq M\kappa _0$, closed Stokes yields $\int _ XB\wedge \gamma _0^{n-1}\leq M\int _ X\kappa _0\wedge \gamma _0^{n-1}=MV_0$, which gives the remaining cost bounds.

Finally, $V=\int _ X\alpha ^ n>0$, so the closed nef class of $\alpha $ is big by the nef-and-big criterion on compact Kähler manifolds recalled in [SGPT]. Hence it contains a Kähler current $\alpha +\mathrm{dd}^{\mathrm{c}}\rho \geq \varepsilon \Omega $. Adding $B\geq 0$ gives the last assertion. Unlike the first current, the second is not closed when $\mathrm dB\neq 0$, since its exterior derivative is $\mathrm{d}B$. $\square$

Proof. The factors of $S$ are closed, and $\gamma _0^{n-1}-\kappa _0^{n-1}=\mathrm{dd}^{\mathrm{c}}u_0\wedge S$. Integration by parts therefore gives (12.12.5). Since $\int _ X\mathrm{dd}^{\mathrm{c}}B\wedge S=0$, the right side does not change when a constant is added to $u_0$, so we may assume $\sup _ Xu_0=0$. Then $-\operatorname {osc}_ X(u_0)\leq u_0\leq 0$, and the lower bound for $\mathrm{dd}^{\mathrm{c}}B$ together with the positivity of the summands of $S$ bounds the correction term above by

The equality holds because the cohomologous Kähler forms $\gamma _0,\kappa _0$ have all mixed top intersections equal to $V_0$. Finally, restrict $\mathrm{dd}^{\mathrm{c}}B$ to the complex two-planes of $TX$ and normalize by $\kappa _0$; compactness of the bundle of these planes bounds the restrictions and gives $C_ B$. $\square$

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 $V_1=1$. $\square$