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.
$\square$
Pome
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