10.27 Nonrigid classes on a quotient resolution96EA3D3C069F

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 $X$. $\square$

Proof. The form $\theta _ h$ is closed, because $\mathrm dh(y_1)$ is proportional to $\mathrm dy_1$, and it is semipositive. Since $h$ is even, $\theta _ h$ is invariant under the involution. At every fixed point, $y_1$ belongs to $\frac12\mathbb Z$ modulo $\mathbb Z$, so the form vanishes on a neighborhood of that point. Its descent to the complement of the singular points of $Y$ therefore extends by zero across their resolved neighborhoods. This proves smoothness and the asserted vanishing.

Let $h_1,h_2$ be admissible and put $k=h_1-h_2$. This function is even, has mean zero and vanishes near $\frac12\mathbb Z$. Choose $f$ with

The derivative $f'$ is periodic and odd, so its mean is zero, and $f$ can be chosen periodic and even. Evenness and the zero mean of $k$ give $\int _0^{1/2}k=0$, hence $f'=0$ near $\frac12\mathbb Z$. Therefore $f(y_1)$ descends, and it extends by constants near each exceptional curve to a smooth function $G$ on $X$. Since

Thus all the forms $T_ h$ have the same Bott–Chern class $\alpha $. Distinct admissible functions give distinct forms, so $\alpha $ is not rigid. The forms are semipositive and have square zero, so $\alpha ^2=0$. By lemma 10.27.1, adding any positive multiple of a Kähler form gives a Kähler representative of a nearby class; therefore $\alpha $ is nef.

By the mean-one condition, the periods of $\theta _ h$ on the real two-tori generated by $(\lambda _1,\lambda _3)$ and by $(\lambda _2,\lambda _3)$ are $1$ and $a$, respectively. Translate both cycles by $(0,\mathrm i/4)$. The translated cycles lie in $\{ y_2=1/4\} $ modulo $\mathbb Z$ and avoid the sixteen fixed points, and their periods are unchanged because the form is closed. Their images in $X$ lie in the complement of the exceptional curves, which is doubly covered without ramification by $A\setminus A[2]$. The first period shows that $\alpha \ne 0$. If a nonzero multiple $b\alpha $ had a rational de Rham class, then evaluating it on these cycles would make both $b$ and $ba$ rational, contradicting $a\notin \mathbb Q$. Finally, since $\alpha ^2=0$, Stokes' theorem shows that $\alpha $ itself has no Kähler representative, so it lies on the boundary of the nef cone. $\square$