PomeExample 18.14.9.75FBCB1F7EC3 There is a surjective holomorphic submersion $p:M\to Y$ from a connected compact Kähler surface to an elliptic curve, with four components in every fiber, for which both identities in (18.14.2) fail. The pole divisor is smooth and maps onto $Y$, and the obstacle upstairs is smooth.
Example 18.14.975FBCB1F7EC3
Proof. Let $E=\mathbb C/(\mathbb Z+\mathrm i\mathbb Z)$, let $\beta $ be its flat probability form, and let $q:E\to E$ be multiplication by two. On $M=E\times \mathbb P^1$, denote the projections by $a,r$ and put $p=q\circ a$. Then $q$ is an unramified covering of degree four and $q^*\beta =4\beta $. The manifold $M$ is connected, whereas every fiber of $p$ is the disjoint union of four projective lines.
On $\mathbb P^1$ take the probability Fubini–Study form $\sigma $, a point $D$ and a logarithmic potential $\psi $ with $\sigma +\mathrm{dd}^{\mathrm{c}}\psi =[D]$ and $\sup \psi =0$. Put
\[ H=E\times D,\qquad \Psi =\psi \circ r,\qquad \Theta =p^*\beta +r^*\sigma =4a^*\beta +r^*\sigma . \]
Thus $\Theta +\mathrm{dd}^{\mathrm{c}}\Psi =p^*\beta +[H]$, which corresponds to the base potential $w=0$. Choose $\epsilon >0$ so small that
\[ f(\zeta )=\epsilon \cos (2\pi \operatorname {Re}\zeta ) \in \mathrm{PSH}(E,4\beta ), \]
and set $F=f\circ a$. The function $f$ is well defined and smooth on $E$. Since it changes sign under translation by $1/2$, it is not constant on the fibers of $q$.
Apply proposition 18.14.1 to the projection $a$, which has connected fibers, with base background $4\beta $ and base potential zero. Its induced obstacle is $f$, since $\sup \psi =0$. The bounded potential $f$ is a competitor for both base envelopes and equals their obstacle, so
\[ P_{\Theta ,F}[\Psi ]=P_{\Theta ,F}[\Psi ]_{\mathcal I} =\Psi +f\circ a. \]
For the map $p$, the fiber obstacle of proposition 18.14.1 would instead be
\[ \widehat F(y)=\min _{q(\zeta )=y}f(\zeta ). \]
Every expression $\Psi +b\circ p$ is invariant under translation of the first factor by points of the kernel of $q$. The computed envelope is not invariant under translation by $1/2$, because $\Psi $ is unchanged and $f$ changes sign. It therefore cannot equal either right side of (18.14.2) for $p$. $\square$
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