Lemma 10.27.122EA1C593ACD

Proof. Blow up the sixteen fixed points on $A$ and take the quotient by the lifted involution. Near one exceptional line, a blowup chart has coordinates $(u,v)$ with $z_1=u$ and $z_2=uv$, and the involution is $(u,v)\mapsto (-u,v)$. The quotient chart has smooth coordinates $(s,v)=(u^2,v)$. On the other chart, $v'=1/v$ and $s'=v^2s$. These charts describe the total space of $\mathcal O_{\mathbb P^1}(-2)$ near its zero section. The quotient is therefore smooth and resolves the double point, with the stated exceptional curve. Doing this at all fixed points gives the compact surface $X$.

Choose a flat Kähler form on $A$ that equals $\mathrm{dd}^{\mathrm{c}}\rho $, where $\rho =|z_1|^2+|z_2|^2$, on disjoint balls centered at the fixed points. Away from those points it descends to $Y$. On the resolution chart,

Choose a smooth function $F$ on $[0,+\infty )$ with $F(\rho )=c\rho ^2$ near zero, where $c>0$, with $F'>0$ on $(0,+\infty )$ and $F''\geq 0$, and with $F(\rho )=\rho -C$ for $\rho \geq \delta $, for a sufficiently small $\delta >0$. Such a function is obtained by choosing $F'$ nondecreasing, linear near zero and equal to one beyond $\delta $. Replace the flat form in each ball by $\mathrm{dd}^{\mathrm{c}}F(\rho )$. This agrees with the flat form near the boundary of the ball and extends smoothly across the exceptional line. The resulting closed form $\eta $ is positive off the exceptional set, while along it $\eta $ is positive in the normal direction and zero in the tangent direction.

To restore positivity in the tangent direction, choose a smooth cutoff $\chi $ equal to one for $\rho \leq \delta $ and zero for $\rho \geq 2\delta $. For $s\ne 0$, set

Near $s=0$ this equals $\log (1+|v|^2)$, so it extends smoothly. On the overlap of the two charts, $q_{v'}-q_ v=-2\log |v|$ is pluriharmonic, so their $\mathrm{dd}^{\mathrm{c}}$ forms agree. They define a smooth closed form $R$ supported in the resolved ball, positive in the tangent direction along the exceptional line and semipositive where $\chi =1$. Hence $\eta +\varepsilon R$ is positive where $\chi =1$, including along the line. On the compact transition annulus $\eta $ is already positive, so $\eta +\varepsilon R$ remains positive there for sufficiently small $\varepsilon >0$, and outside the ball $R=0$. Adding these corrections in all the disjoint balls produces a Kähler form on $X$. $\square$