Example 18.34.24.00C65C3DCEEB Let $Y$ be a connected compact complex manifold of positive dimension with a smooth positive probability volume $\nu _ Y$. On $X=\mathbb P^1\times Y$, use the pullback of $\mathcal O(1)$ with its Fubini–Study metric, the selecting potential zero, and the volume $\nu =\nu _{\mathrm{FS}}\otimes \nu _ Y$, where $\nu _{\mathrm{FS}}$ is the Fubini–Study area of mass one. Choose a nonconstant $q\in C(Y,\mathbb R)$ and put
\[ v_ m(b,y)=q(y)/m,\qquad Z_ q=\int _ Y\mathrm{e}^{-q}\, \mathrm{d}\nu _ Y,\qquad \nu _ q=Z_ q^{-1}\mathrm{e}^{-q}\nu _ Y. \]
Then $N_ m=m+1$ and, for every $m$, the kernel and the normalized kernel probability at the weight $v_ m$ are
53B83B2F665D
18.34.25
\begin{equation} B_{m,v_ m}(b,y)=(m+1)\frac{\mathrm{e}^{-q(y)}}{Z_ q}, \qquad \mu _{m,v_ m}=\nu _{\mathrm{FS}}\otimes \nu _ q. \end{equation}
Thus $v_ m\to 0$ uniformly, but the kernel measures with $r=1$ converge in total variation to $\nu _{\mathrm{FS}}\otimes \nu _ q$, whereas at the weight zero the limit is $\nu _{\mathrm{FS}}\otimes \nu _ Y$. In both cases the potentials $m^{-1}\log B_{m,v}$ tend uniformly to zero. With the normalization by the dimension of $X$, both sequences of kernel measures tend to zero in total variation.
Proof.
Every section descends to $\mathbb P^1$, because the fibers $\{ b\} \times Y$ are compact and connected, as in proposition 18.36.1. Hence the selected space is $H^0(\mathbb P^1,\mathcal O(m))$, of dimension $m+1$. Its unweighted Bergman kernel for $\nu _{\mathrm{FS}}$ is constant, since the unitary group acts transitively on $\mathbb P^1$ and preserves the metric and the volume; its integral $m+1$ determines the constant. For a section pulled back from $\mathbb P^1$, the norm on $X$ at the weight $v_ m$ is $Z_ q$ times the norm on $\mathbb P^1$, which proves (18.34.25). At the weight zero the normalized kernel probability is $\nu _{\mathrm{FS}}\otimes \nu _ Y$. These probabilities differ: equality would force $\mathrm{e}^{-q}/Z_ q=1$ almost everywhere, and then everywhere, because $q$ is continuous and every nonempty open set has positive volume; this contradicts the choice of $q$. The kernel measures with $r=1$ are $(m+1)/m$ times these probabilities, and the normalization in dimension $1+\dim Y\geq 2$ gives measures tending to zero in total variation. Finally, $m^{-1}\log B_{m,v_ m}=m^{-1}\bigl (\log (m+1)-q-\log Z_ q\bigr )$ and $m^{-1}\log B_{m,0}=m^{-1}\log (m+1)$ both tend uniformly to zero.
$\square$
Pome
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