Proposition 24.3.2.32F82EA89236 Let $h$, $\theta $ and $\mu $ be as above, and let $(\gamma _ j)$ be a sequence of Gauduchon metrics.
Suppose $n\geq 2$, $\lambda _\mu (\theta )>0$, and $\theta $ is not big. If
\[ A_ h(\gamma _ j)=1,\qquad A_\theta (\gamma _ j)\longrightarrow 0, \]
then $D_\mu (\gamma _ j)\to 0$. Such sequences exist.
More generally, fix a real number $\delta >0$. If $A_ h(\gamma _ j)=1$ and
\[ A_\theta (\gamma _ j)\longrightarrow \kappa _ h(\theta ), \qquad \liminf _ jD_\mu (\gamma _ j)\geq \delta >0, \]
then
\[ \kappa _ h(\theta )\geq \delta \lambda _\mu (\theta ). \]
In particular, positive root volume then implies bigness.
Proof.
Suppose first that $\theta $ is not big. Since $\theta $ is pseudoeffective, the threshold criterion of corollary 11.6.2 gives $\kappa _ h(\theta )=0$, so there are Gauduchon metrics whose quotients $A_\theta /A_ h$ tend to zero. Multiplying a Gauduchon metric by a positive constant multiplies $A_\theta $, $A_ h$ and $D_\mu $ by the same factor; for $n\geq 2$ this factor can be chosen to make $A_ h=1$, without changing either quotient. This gives the sequences in the first assertion. By the definition of the root volume, every Gauduchon metric satisfies
C672C5EDF4D5
24.3.3
\begin{equation} \lambda _\mu (\theta )D_\mu (\gamma )\leq A_\theta (\gamma ). \end{equation}
Dividing by the positive root volume proves the first assertion. Applying the same inequality to the sequence in the second assertion and taking lower limits gives $\delta \lambda _\mu (\theta )\leq \kappa _ h(\theta )$. If the root volume is positive, then $\kappa _ h(\theta )>0$, and the threshold criterion gives bigness.
$\square$
Pome
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