Proposition 12.19.20.3290D6E85075 Let $\pi :X\to B=\mathbb P^1\times \mathbb P^1$ be the elliptic fibration above, and put $L=4\pi $. Consider the smooth potentials
\[ u=f(t)+\pi ^*v,\qquad \omega +\mathrm{dd}^{\mathrm{c}}u>0, \]
where $f:\mathbb R\to \mathbb R$ is smooth and $L$-periodic and $v\in C^\infty (B,\mathbb R)$. The infimum of $\int _ X(\omega +\mathrm{dd}^{\mathrm{c}}u)^3$ over these potentials is $8$. Every such mass is strictly greater than $8$, and the infimum is approached already with $v=0$.
Proof.
Write $p=f'$, an $L$-periodic function with mean zero. The chain rule gives
\[ \omega +\mathrm{dd}^{\mathrm{c}}u =(1-p)\alpha +(1+p)\beta +\pi ^*\mathrm{dd}^{\mathrm{c}}v+(1+p')D. \]
Positivity on the vertical line implies $p'>-1$. On the complex plane $\ker \partial t$, the form $D$ vanishes and the projection to the base is an isomorphism. For each value of $t$ modulo $L$, every base point has a point above it with that value. Hence, writing $\alpha _ B,\beta _ B$ for the two Fubini–Study forms on $B$, the form $(1-p(s))\alpha _ B+(1+p(s))\beta _ B+\mathrm{dd}^{\mathrm{c}}v$ is positive on $B$ for every $s$. Integration over the two rulings gives $1-p>0$ and $1+p>0$. Thus $|p|<1$ and $p'>-1$; when $v=0$, these inequalities also suffice for strict positivity.
In a logarithmic fiber coordinate $x+\mathrm{i}y$, both real periods are $2\pi $, $t=2x$ plus a constant, and $D|_ E=\pi ^{-1}\mathrm{d}x\wedge \mathrm{d}y$. Hence $\int _ E F(t)D=\int _0^ L F(s)\, \mathrm{d}s$ for every $L$-periodic function $F$. Only the derivative of $u$ along the fibers contributes to (12.19.19), and it equals $p(t)$ times the derivative of $t$. Fiber integration therefore gives
87EE58E6037C
12.19.21
\begin{equation} \int _ X(\omega +\mathrm{dd}^{\mathrm{c}}u)^3 =6\left(L-\int _0^ L p(s)^2\, \mathrm{d}s\right). \end{equation}
We now bound $\int _0^ Lp^2$. Suppose first that $p\not\equiv 0$, and put $m_+=\max p>0$, $m_-=-\min p>0$ and $r=m_++m_-<2$. The function $G(s)=(m_+-s)(s+m_-)$ is nonnegative on $[-m_-,m_+]$. Choose an interval $I$ of length at most $L$ that runs from a maximum point of $p$ forward to the next minimum point. Since $p'\geq -1$, we have $G(p)\geq -G(p)p'$. Using that $p$ has mean zero, we obtain
\[ \begin{split} m_+m_-L-\int _0^ L p^2\, \mathrm{d}s & =\int _0^ L G(p)\, \mathrm{d}s \geq -\int _ I G(p)p'\, \mathrm{d}s\\ & =\int _{-m_-}^{m_+}G(s)\, \mathrm{d}s=\frac{r^3}{6}. \end{split} \]
As $m_+m_-\leq r^2/4$, it follows that
\[ \int _0^ L p^2\, \mathrm{d}s \leq \frac{Lr^2}{4}-\frac{r^3}{6}<L-\frac43, \]
where the last inequality holds because the middle expression is strictly increasing in $r\in (0,2]$, as $L>2$. The inequality $\int _0^ Lp^2<L-4/3$ holds trivially when $p=0$. Equation (12.19.21) now gives the strict lower bound $8$.
To prove sharpness, put $a=(L-2)/2$ and define an $L$-periodic function $p_0$ by
\[ p_0(s)= \begin{cases} 1,& 0\leq s<a,\\ 1-(s-a),& a\leq s<a+2,\\ -1,& a+2\leq s<L. \end{cases} \]
It has mean zero, $|p_0|\leq 1$ and $\int _0^ L p_0^2\, \mathrm{d}s=L-4/3$. Its distributional derivative is $-1$ on the descending interval and zero on the two constant intervals, and it has a positive atom where the period closes up. Thus $p_0'\geq -1$ as distributions on the circle. Let $\rho _\varepsilon $ be a nonnegative smooth periodic approximate identity of integral one, and set $p_\varepsilon =(1-\varepsilon )(p_0*\rho _\varepsilon )$ for $0<\varepsilon <1$. Then $p_\varepsilon $ is smooth, has mean zero, and
\[ |p_\varepsilon |\leq 1-\varepsilon , \qquad p_\varepsilon '\geq -1+\varepsilon , \qquad p_\varepsilon \longrightarrow p_0\quad \hbox{in }L^2. \]
Its primitive $f_\varepsilon $ is smooth and periodic, so $u_\varepsilon =f_\varepsilon (t)$ is admissible. By (12.19.21), its mass tends to $8$.
$\square$
Pome
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