Proposition 12.13.6.172CF285DEA0 The endpoint $\theta _ t$ is big exactly when $a$ is positive everywhere. If $a$ has a zero, there is no decomposition
\[ \theta _ t+\mathrm{dd}^{\mathrm{c}}f=\chi +\beta \]
with $f$ smooth and real, $\chi $ smooth and closed with nef class and $\int _ X\chi ^ n>0$, and $\beta \geq 0$ smooth.
Proof.
Suppose that $\theta _ t+\mathrm{dd}^{\mathrm{c}}u\geq \delta (\alpha +w)$ for a quasi-plurisubharmonic $u$ and $\delta >0$. Pull back an arbitrary smooth nonnegative top-degree form $\zeta $ on $B$. It is $\mathrm{dd}^{\mathrm{c}}$-closed by dimension, so Stokes' theorem for distributions gives
\[ 0\leq \int _ X(\theta _ t+\mathrm{dd}^{\mathrm{c}}u-\delta (\alpha +w))\wedge \zeta =\int _ B(a-\delta )\zeta . \]
A nonzero $\zeta $ supported where $a<\delta $ contradicts this inequality. Thus a zero excludes every strict current representative; no singular potential has been restricted to its zero fiber. Conversely, if $a>0$ everywhere, compactness makes $\theta _ t$ itself Hermitian, so it is big.
For a proposed decomposition, the closed nef-and-big criterion used in theorem 12.12.2 supplies a global quasi-plurisubharmonic $v$ with $\chi +\mathrm{dd}^{\mathrm{c}}v\geq \delta (\alpha +w)$ for some $\delta >0$. Then
\[ \theta _ t+\mathrm{dd}^{\mathrm{c}}(f+v)=\chi +\mathrm{dd}^{\mathrm{c}}v+\beta \geq \delta (\alpha +w), \]
contradicting the first part. This excludes any such closed core, including after a smooth change of gauge.
$\square$
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