Example 18.14.8.68E397A3AF5C There is a Hermitian background $\Theta $ on a compact fourfold that fails the bounded mass property and admits a potential $u$ without an analytic ordinary type such that
\[ P_\Theta [u]=P_\Theta [u]_{\mathcal I}=\Psi , \qquad u\not\sim \Psi . \]
Both envelopes agree below every smooth obstacle, and their common model for the zero obstacle has both singularity volumes zero.
Proof.
Let $(Y,\omega )$ be the Hopf threefold of theorem 23.1.1. On $\mathbb P^1$, let $\sigma $ be the probability Fubini–Study form, choose a point $D$ and a logarithmic potential $\psi $ with $\sigma +\mathrm{dd}^{\mathrm{c}}\psi =[D]$ and $\sup \psi =0$. On $M=Y\times \mathbb P^1$, with projections $p,r$, put
\[ \Theta =p^*\omega +r^*\sigma ,\qquad H=Y\times D,\qquad \Psi =\psi \circ r. \]
Take the potential $U\in \mathcal E_{\mathrm{env}}(Y,\omega ,0)$ of infinite mass from proposition 16.28.72, which satisfies $P_\omega [U]=0$. Its multiplier ideals are trivial, so also $P_\omega [U]_{\mathcal I}=0$. Set $u=\Psi +U\circ p$. Here $\widehat0=0$, so proposition 18.14.1, corollary 18.14.5 give the envelope identities, the equality below all obstacles and the vanishing of the singularity volumes. Moreover $u\in \mathcal E_{\mathrm{env}}(M,\Theta ,\Psi )$ by proposition 18.14.4. Since $U$ has infinite extended mass, it is unbounded; restriction to a section outside $H$ therefore proves $u\not\sim \Psi $. If a potential $a$ with analytic singularities were ordinarily equivalent to $u$, proposition 18.4.8 would give $u\sim a\sim P_\Theta [a]=P_\Theta [u]=\Psi $, a contradiction.
Finally, let $V_ j$ be the smooth potentials of theorem 23.1.1. Their pullbacks are bounded and $\Theta $-plurisubharmonic, and proposition 20.3.59 gives
\[ \int _ M\mathrm{MA}_\Theta (V_ j\circ p) =4\int _ Y\mathrm{MA}_\omega (V_ j)\longrightarrow +\infty . \]
Thus $\Theta $ itself fails the bounded mass property.
$\square$
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