PomeB32BAEE663D4We return to a connected compact complex manifold $X$ of dimension $n$ and fix a Hermitian form $\omega $ on $X$. A weight with trivial multiplier ideals leaves every integrable section space unchanged. When $X$ fails the bounded mass property of definition 13.14.14, theorem 13.14.18 provides such a weight $U\in \mathrm{PSH}(X,\omega )$ with infinite extended mass. For the trivial line bundle with constant metric, $\mathcal H_ m(U)$ consists of constants and every Bergman current vanishes. Thus section growth in a closed curvature class does not determine the mass in an unrelated Hermitian background.
Proposition 19.8.15.6400AAE4BA1E Let $\omega $ be a Hermitian metric on $X$ with $v_+(\omega )=+\infty $. There is a potential $U\in \mathrm{PSH}(X,\omega )$ such that, with $L=\mathcal O_ X$ and its constant metric,
All Bergman currents of $U$ are identically zero.
Proof. Take $U$ from theorem 13.14.18; its extended mass is infinite, and $\mathcal I(tU)=\mathcal O_ X$ for every $t>0$. Global holomorphic functions on $X$ are constant, and the constant function $1$ lies in $\mathcal H_ m(U)$ because $\mathcal I(mU)=\mathcal O_ X$ and $X$ is compact. Hence every weighted section space is one-dimensional, and its orthonormal basis consists of a nonzero constant. Each Bergman potential $\Phi _ m$ is therefore constant. Since the trivial metric has curvature zero, every Bergman current is zero. $\square$
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