12.21 Elliptic surface volumes and order spectraE4B29689FB79

Proof. Define

Since $(a+\epsilon )g-|r|^2>0$, we have $c>0$, and the definition of $c$ makes $A-a-\epsilon $ have zero mean with respect to $\omega _1$. Dividing its nonconstant Fourier coefficients by the corresponding nonzero eigenvalues of the flat Laplacian produces a smooth real function $u$ on $E$ with $\mathrm{dd}^{\mathrm{c}}u=(A-a-\epsilon )\omega _1$. Pull it back to $X$. The form $\Omega =\theta +\epsilon h+\mathrm{dd}^{\mathrm{c}}u$ has coefficient matrix $\left(\begin{smallmatrix}A& r\\ \bar r& g\end{smallmatrix}\right)$, which has determinant $c$ and positive diagonal entries. Hence $\Omega >0$ and $\Omega ^2=2c\mu $, and theorem 12.2.3 gives (12.21.2).

The metric $\gamma _*=\Omega /g$ has constant coefficient $1$ in the fiber direction, and its remaining coefficients depend only on $z$. Applying $\mathrm{dd}^{\mathrm{c}}$ to its base term or its mixed terms produces a repeated $\mathrm{d} z$ or $\mathrm{d}\bar z$, so $\mathrm{dd}^{\mathrm{c}}\gamma _*=0$. Thus $\gamma _*$ is a smooth Gauduchon metric conformal to $\Omega $, and by theorem 12.2.3 it attains the infimum. The two special cases follow by direct substitution. $\square$

Proof. In proposition 12.21.1 take $a=b>0$ and $r=0$. Fix a point $z_0\in E$ and a flat coordinate $\zeta =z-z_0$ on a small disk about $z_0$. Choose $q$ smooth, positive outside $z_0$, and equal to $|\zeta |^{2m}$ on that disk; a positive smooth extension exists away from the disk. For $m=1$, polar coordinates give

and (12.21.2) gives the asymptotic formula of the statement. If $m>1$, the substitution $s=|\zeta |^2=\epsilon ^{1/m}t$ gives

so the common order is $1-1/m$. In the logarithmic case, $\lambda _\mu (\theta )\geq 0$ by semipositivity, while $\lambda _\mu (\theta )\leq \lambda _\mu (\theta +\epsilon h)\to 0$ by monotonicity; hence $\mathcal V_\mu (\theta )=0$. Moreover, the quotient of the volume by $\epsilon ^\delta $ diverges for every $\delta >0$, so both orders vanish. Finally, $\mathrm{d}\theta =\mathrm{d} q\wedge \omega _2\ne 0$, so these backgrounds are not closed. $\square$

Proof. We use the rank-one formula (12.21.3), with coefficients that vanish to infinite order at one point at a prescribed oscillating rate.

Step 1: the coefficients. For all sufficiently large $t$, define

Choose $T$ so large that $0<c_0\leq P'(t)\leq d_0<1$ for $t\geq T$; this is possible because

and $\alpha \leq p\leq \beta $. Fix $z_0\in E$ and a flat coordinate $\zeta $ centered at $z_0$. On $E\setminus \{ z_0\} $ choose a smooth function $t(z)\geq T$ that equals $|\zeta |^{-2}$ near $z_0$ and is bounded outside every neighborhood of $z_0$. Put

and set all three equal to zero at $z_0$. These functions are smooth and vanish to infinite order at $z_0$: we have $P(t)\geq \alpha t$, and every fixed derivative of $P$ grows at most polynomially in $t$, so by the chain rule each coordinate derivative is bounded by a polynomial in $|\zeta |^{-1}$ times $\mathrm{e}^{-c/|\zeta |^2}$ for some $c>0$. Since $|r|^2=aq$, the resulting $\theta $ is semipositive. It is not closed: the component $\mathrm{d} q\wedge \omega _2$ of $\mathrm{d}\theta $ contains both fiber differentials, and derivatives of the remaining terms cannot cancel it.

Step 2: asymptotics of $I$ and $K$. Write $\epsilon =\mathrm{e}^{-L}$, and let $T_0\geq T$ be such that $t=|\zeta |^{-2}$ on the disk $\{ |\zeta |^2<1/T_0\} $. Outside this disk the integrands of $I$ and $K$ in (12.21.3) are bounded, so their contributions are $O(1)$, while polar coordinates on the disk give

The part of $I_{\mathrm{loc}}$ over $t\geq L$ is at least a fixed multiple of $\mathrm{e}^ L/L$. For an upper bound, use $(\mathrm{e}^{-t}+\mathrm{e}^{-L})^{-1}\leq \mathrm{e}^ t$ for $t\leq L$ and $(\mathrm{e}^{-t}+\mathrm{e}^{-L})^{-1}\leq \mathrm{e}^ L$ for $t\geq L$, and split $[T_0,L]$ at $L/2$. Thus

The derivatives of $P$ and of $t-P(t)$ are positive and bounded away from zero. Splitting the integral for $K$ at $t=L$ therefore gives $K\leq C\mathrm{e}^{L-P(L)}$. On $[L,L+1]$ the denominator is at most $2\mathrm{e}^{-L}$ and $P(t)\leq P(L)+1$, so

the complementary $O(1)$ term is absorbed because $P(L)\leq \beta L$. Consequently $\log K=L-P(L)+O(\log L)$, and $K\to \infty $.

Step 3: the orders. Combining these estimates with (12.21.3) and (12.21.6) yields

Since $\sin (\log \log L)$ oscillates between $-1$ and $1$, the lower and upper limits of $p(L)$ are $\alpha $ and $\beta $, which proves the first assertion. If instead $P(t)=\sigma t$, the same smoothness and integral estimates give the common order $1+\sigma $. In particular, $\sigma =\sqrt2-1$ gives an irrational common order. $\square$