Although no class $a\{ \alpha \} +b\{ \beta \} $ is big, its small Hermitian perturbations have positive Gauduchon variational volume $\mathcal V_\mu $ (definition 12.2.1). For the volume form of the balanced metric, the zero potential solves the normalized Monge–Ampère equation for every positive combination of $\alpha $, $\beta $ and $D$, so the volume is an explicit cubic polynomial in the perturbation parameter. Its order of vanishing at the nef boundary records how many of the two horizontal coefficients vanish.
Proposition 10.19.8.141C0ACF5576 On the elliptic threefold, let $a,b\geq 0$, put $\theta =a\alpha +b\beta $, and let $\mu _0=\omega ^3/(24\pi )$. Then for every $\epsilon >0$,
E7BC529546A7
10.19.9
\begin{equation} \mathcal V_{\mu _0}(\theta +\epsilon \omega ) =24\pi \epsilon (a+\epsilon )(b+\epsilon ). \end{equation}
In particular, $\mathcal V_\mu (\theta )=0$ for every smooth positive probability measure $\mu $. Let $q$ be the number of positive coefficients among $a,b$. Then the degeneration orders of proposition 12.20.1 are
\[ \underline a(\theta )=\overline a(\theta )=3-q, \]
for every smooth positive probability measure and every Hermitian perturbation direction.
Proof.
The forms $\alpha ,\beta ,D$ are semipositive of rank one, so their squares vanish, and their sum $\omega $ is strictly positive. Thus
\[ \Omega _\epsilon =\theta +\epsilon \omega =(a+\epsilon )\alpha +(b+\epsilon )\beta +\epsilon D>0. \]
Since $\omega ^3=6\alpha \wedge \beta \wedge D$ and $\int _ X\omega ^3=24\pi $, the wedge expansion gives
\[ \Omega _\epsilon ^3 =6\epsilon (a+\epsilon )(b+\epsilon )\alpha \wedge \beta \wedge D =24\pi \epsilon (a+\epsilon )(b+\epsilon )\mu _0. \]
Hence the zero potential solves the smooth normalized equation for $\Omega _\epsilon $ and $\mu _0$. By theorem 12.2.3, the scalar of this equation is $\mathcal V_{\mu _0}(\Omega _\epsilon )$, which proves (10.19.9). The continuity of the volume at nef classes in theorem 12.11.1 and comparison of smooth densities then give $\mathcal V_\mu (\theta )=0$.
As $\epsilon \downarrow 0$, the leading term of (10.19.9) is $24\pi ab\epsilon $ if $a$ and $b$ are both positive, $24\pi (a+b)\epsilon ^2$ if exactly one of them is positive, and $24\pi \epsilon ^3$ if both vanish. The logarithmic quotients defining the orders therefore have limits $1,2,3$, respectively. Their independence of the density and of the perturbing metric is part of proposition 12.20.1.
$\square$
The same number $q$ governs the growth of sections. For nonnegative integers $a,b$, the line bundle $L=\pi ^*\mathcal O_ B(a,b)$ satisfies
\[ h^0(X,L^ m)=(am+1)(bm+1), \]
since the descent along compact fibers and the monomial count in the proof of proposition 10.19.6 also apply when one or both integers vanish. The growth degree of these spaces is exactly $q$, whereas the boundary order of the smooth curvature form $a\alpha +b\beta $ of $L$ is $3-q$. In this explicit family, the boundary order thus records the loss of algebraic growth dimension, even though the three-dimensional Monge–Ampère measures in the unperturbed background vanish.
Formula (10.19.9) concerns one fixed smooth density; it does not assert that a total mass is conserved over all potentials. Indeed, by theorem 12.19.1 every Hermitian form on $X$, in particular each $\Omega _\epsilon $, has zero lower mass, while its upper mass is finite because $X$ has the bounded mass property (definition 13.14.14, proposition 12.19.18). Whether comparable identifications hold beyond such explicit families is part of 12.33.4.
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