Example 18.15.4.6D425566CDE7 In the product setting, take $C=\mathbb P^1$ and $Z=\mathbb P^ k$, $k\geq 1$, with probability Fubini–Study forms and a hyperplane $D$. The positive closed background $\Theta $ admits normalized ordinary models $a\leq b$ with
\[ \begin{gathered} P_\Theta [a]=a,\qquad P_\Theta [a]_{\mathcal I}=P_\Theta [b]=P_\Theta [b]_{\mathcal I}=b,\\ a\sim _{\mathcal I}b,\qquad a\not\sim b,\\ \underline{\operatorname {vol}}_\Theta (a) =\overline{\operatorname {vol}}_\Theta (a) =\underline{\operatorname {vol}}_\Theta (b) =\overline{\operatorname {vol}}_\Theta (b)=0. \end{gathered} \]
Thus equal singularity volumes of a type and of its ideal model do not imply equality of the envelopes at zero volume.
Proof.
Use the normalized diffuse polar curve model $w$ of example 13.28.14 on $\mathbb P^1$ with its probability Fubini–Study form. It satisfies $P_\eta [w]=w$ and $P_\eta [w]_{\mathcal I}=0$, and it is unbounded. Take the normalized hyperplane potential on $\mathbb P^ k$ and put
\[ a=\Psi +w\circ p,\qquad b=\Psi . \]
The identities for the zero obstacle in proposition 18.15.1 give all four envelope identities. Both functions have supremum zero, and $w\leq 0$ gives $a\leq b$. Ideal projection preserves the ideal type by theorem 18.4.2, so $a\sim _{\mathcal I}b$. Restriction to $\mathbb P^1\times \{ z\} $, for any $z\notin D$, gives the unbounded difference $w$, so $a\not\sim b$. Finally, proposition 18.14.1 gives zero extended mass for every ordinary representative of either model, which proves all four volume identities.
$\square$
Pome
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