Example 18.14.6.8119F7E76588 On $M=\mathbb P^1\times \mathbb P^1$, let $p,r$ be the projections and let $\sigma $ be the probability Fubini–Study form. With homogeneous coordinates $[y_0:y_1]$ and $[z_0:z_1]$, the divisor
\[ H=\{ y_0z_0^2+y_1z_1^2=0\} \]
is smooth, meets every fiber properly and ramifies over the first factor. Give $\mathcal O_ M(H)=\mathcal O_ M(1,2)$ its product metric, with logarithmic divisor potential $\Psi $ and curvature $\kappa _ H=p^*\sigma +2r^*\sigma $. Put $\Theta =2p^*\sigma +2r^*\sigma $ and choose $w\in \mathrm{PSH}(\mathbb P^1,\sigma )$ with $\sigma +\mathrm{dd}^{\mathrm{c}}w=\delta _{[0:1]}$. Both obstacle envelopes of $u=\Psi +w\circ p$ reduce to the base, but the product formula for its multiplier ideals fails at a ramification point.
Proof.
For each $[z_0:z_1]$ there is a unique point $[y_0:y_1]=[-z_1^2:z_0^2]$ on $H$, so $H$ is a smooth graph over the second factor, and its projection to the base has degree two. In particular no fiber of $p$ lies in $H$. Since $\Theta =p^*\sigma +\kappa _ H$, proposition 18.14.1 gives both envelope identities.
Near $([0:1],[1:0])$, use $y=y_0/y_1$ and $z=z_1/z_0$. Up to smooth additions,
\[ \Psi =\log |y+z^2|^2,\qquad w=\log |y|^2, \qquad u=\log |y(y+z^2)|^2. \]
At $t=3/4$, each of the two weights separately has trivial multiplier ideal, so the right side of (18.14.3) would be $\mathcal O_ M$. However, $1\notin \mathcal I(3u/4)$ at this point. Indeed, consider the region where $y=cz^2$ with $|c-1|<1/4$ and $0<|z|<\epsilon $; for fixed $z$, it has area comparable to $|z|^4$ in the $y$ variable. The weighted integral of $1$ is therefore bounded below by a positive constant times
\[ \int _{0<|z|<\epsilon }|z|^{-6}|z|^4\, \mathrm d\lambda _ z =2\pi \int _0^\epsilon \frac{\mathrm dr}{r}=+\infty , \]
where $\mathrm d\lambda _ z$ is planar area. Thus the multiplier ideal is proper. The envelope identities follow instead from the behavior of multiplier-ideal equivalence under addition, and do not require the product formula.
$\square$
Pome
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