The root volume and the bigness threshold divide the same pairing by different quantities. Positive volume can fail to imply bigness only if these denominators separate along tests approaching the threshold, and the following calculation isolates this degeneration. For the radial background of theorem 12.14.6, the normalized $(n-1)$-st powers of a family of balanced test metrics (definition 11.4.1) converge weakly to the integration current of the degenerate fiber (proposition 12.14.13). The volume-minimizing family approaches the bigness threshold only in dimension three, by proposition 12.14.10.
Fix a Hermitian form $h$ on $X$, and let $\theta $ be a smooth pseudoeffective real $(1,1)$-form. For a Gauduchon metric $\gamma $, abbreviate the pairings by
\[ A_\theta (\gamma )=\int _ X\theta \wedge \gamma ^{n-1}, \qquad A_ h(\gamma )=\int _ Xh\wedge \gamma ^{n-1}. \]
The threshold $\kappa _ h(\theta )$ of corollary 11.6.2 is the infimum of the quotient $A_\theta /A_ h$ over Gauduchon metrics, and it is positive exactly when $\theta $ is big. The root volume $\lambda _\mu (\theta )$ is the infimum of $A_\theta /D_\mu $, with the determinant denominator $D_\mu (\gamma )$ of definition 12.2.1.
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
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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.
$\square$
The quotient $D_\mu /A_ h$ is always bounded above. Indeed, the smooth equation for $h$, or (24.3.3) with $\theta =h$, gives
\[ D_\mu (\gamma )\leq \lambda _\mu (h)^{-1}A_ h(\gamma ). \]
What is missing for the converse direction is a lower bound for $D_\mu $ along suitable sequences minimizing the threshold. Compactness of normalized potentials does not give compactness of these Gauduchon metrics.
In dimension one this normalization issue does not arise: the root volume is $\int _ X\theta $, and $\kappa _ h(\theta )=\int _ X\theta /\int _ Xh$. Positive volume and bigness are therefore equivalent by the same threshold criterion.
One way to rule out this collapse is to deform the test metrics. Solving a Gauduchon equation replaces the volume density of a test metric while keeping the Aeppli class of $\gamma ^{n-1}$. When $\theta $ is closed, this leaves the numerator $A_\theta $ unchanged. For nonclosed $\theta $, integration by parts leaves derivatives of the background, so such a deformation must also control the change of the pairing; this is the additional estimate needed to use the smooth equation against collapse.
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