Example 18.23.2.47BCB7DF51C8 Let $X=\mathbb P^1\times \mathbb P^1$, let $q$ be its first projection and let $r:\mathbb P^1\to \mathbb P^1$ be $r([z_0:z_1])=[z_0^2:z_1^2]$. Put $p=r\circ q$ and $\theta =p^*\omega _{\mathrm{FS}}$, with $\int _{\mathbb P^1}\omega _{\mathrm{FS}}=1$. There are two distinct analytic ideal $\theta $-models $U_+,U_-$ which do not descend through $p$. Their $q$-normalized polar distance is four and their $p$-normalized distance is two, yet
\[ p_*\lambda _{U_+}^\gamma =p_*\lambda _{U_-}^\gamma \]
for every Gauduchon form $\gamma $. Thus the pushforward of polar trace to the original base does not separate the models.
Proof.
The finite map $r$ has degree two and is branched at zero and infinity. In the affine coordinate it is $r(z)=z^2$, so the smooth semipositive form $\sigma =r^*\omega _{\mathrm{FS}}$ has area two. For $s\in \{ 1,-1\} $, set
\[ v_ s(z)=2\log |z-s|^2-\log (1+|z|^4)-\log 8. \]
This extends smoothly across infinity and has curvature $\sigma +\mathrm{dd}^{\mathrm{c}}v_ s=2\delta _ s$. The inequality $|z-s|^4\leq (|z|+1)^4\leq 8(1+|z|^4)$, with equality at $z=-s$, shows that its supremum is zero. The model of $2\delta _ s$ from proposition 13.29.1 has precisely this total curvature, since its polar measure already has the full mass two. Harmonic uniqueness and normalization identify it with $v_ s$. Write $v_+=v_1$ and $v_-=v_{-1}$. Both have analytic singularities, hence are ideal by proposition 18.4.8. The model pullback bijection makes $U_\pm =v_\pm \circ q$ analytic ideal $\theta $-models.
The function $U_+$ is $-\infty $ on $q^{-1}(1)$ and finite on $q^{-1}(-1)$, although both divisors are mapped by $p$ to the same base point $1$. It therefore cannot descend through $p$; the same argument with the divisors exchanged applies to $U_-$. The variation of $2\delta _1-2\delta _{-1}$ is four. The finite factor proposition gives the two distances and
\[ p_*\lambda _{U_\pm }^\gamma =2\kappa _ q^\gamma \delta _1. \]
The upstairs polar traces are distinct, while this common pushforward loses which component carries the pole.
$\square$
Pome
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