PomeProposition 11.30.3.A8FAF022D4CA The first integer $j\geq 0$ for which some $\widetilde G_{j,B}$, with $B$ an arbitrary Hermitian matrix, is strictly plurisubharmonic off the zero section is $j=q-1$. At every lower level, zero belongs to $\operatorname {conv}(\widetilde\Psi _ j(K))$; one atom witnesses this at $j=0$, and two atoms suffice at each $1\leq j<q-1$.
11.30.2 The weighted family and its first successful level5B78AD45B21A
Keep the root metric $\mathrm e^{-\phi }$ on $L$, and choose the sphere bundle $S$ and the compact set $K$ for $F$ as in section 11.29. For each integer $j\geq 0$, use the complete space
\[ \widetilde V_ j=H^0(X,L^{j+1}) \]
and pair its basis with $v\in F_ x$ to get a local column $p_ j(v)$, written in a local frame of $L^ j$. For every Hermitian coefficient matrix $B$, define
\[ \widetilde G_{j,B}(v)=\mathrm e^{-j\phi (x)}p_ j(v)^*B p_ j(v). \]
These are globally defined functions, quadratic on each fiber. Their Levi matrices are
\[ \begin{gathered} \widetilde\Psi _ j(v,\xi )=\mathrm e^{-j\phi } (r_ jr_ j^*-j\kappa p_ jp_ j^*),\\ r_ j=\mathrm dp_ j(\xi )-j\phi _ u p_ j,\qquad \kappa =\phi _{u\bar u}, \end{gathered} \]
where $u$ is the projection of $\xi $ to the base. As in (11.29.3), direct differentiation gives $\mathcal L\widetilde G_{j,B}(\xi )=\operatorname {tr}(B\widetilde\Psi _ j)$, and $\widetilde\Psi _ j$ transforms in the same way under changes of frames.
Proof. At $j=q-1$, choose the basis $\sigma _ i$ that defines $\phi _ A$ and take $B=\operatorname {Id}$. Since $p_ j(v)=v(\sigma _ i(x))_ i$ in the chosen frames,
\[ \widetilde G_{q-1,\operatorname {Id}}(v) =\mathrm e^{-(q-1)\phi }|v|^2\mathrm e^{q\phi } =\mathrm e^\phi |v|^2. \]
This function is strictly plurisubharmonic by lemma 11.30.1.
Now suppose that $0\leq j<q-1$. Then $j+1<q$, so by lemma 11.30.1 every section in $\widetilde V_ j$ comes from the first summand of the decomposition, which does not involve $w$. At a ramification point of $C$, choose a local coordinate $t$ on $C$, and use frames of the bundles pulled back from $\mathbb P^1\times \mathbb P^{n-1}$. Writing $\zeta $ for the coordinates of the factor $\mathbb P^{n-1}$, the column has the form $p_ j(v)=vU_ j(\pi (t),\zeta )$. It is nonzero for $v\ne 0$, since $L^{j+1}$ is globally generated, and since $\pi $ ramifies at the point, its derivative in the $t$ direction vanishes there when $v$ is held fixed.
Choose such a fiber vector $v$ on the sphere bundle $S$. Write tangent vectors to the total space of $F$ by their components in the $t$ direction, in the $\zeta $ directions and in the fiber direction. Let $\xi _ v=(0,0,v)$ be the vertical radial vector, let $u=\partial /\partial t$, and let $\xi _ h=(u,0,j\phi _ u v)$. At this point $\mathrm dp_ j(\xi _ v)=p_ j$ and $r_ j(\xi _ h)=0$. Thus
\[ \begin{aligned} \widetilde\Psi _ j(v,\xi _ v)& =\mathrm e^{-j\phi }p_ jp_ j^*,\\ \widetilde\Psi _ j(v,\xi _ h)& =-j\kappa \mathrm e^{-j\phi }p_ jp_ j^*, \qquad \kappa >0. \end{aligned} \]
Let $a_ v$ and $a_ h$ be the squared norms of $\xi _ v$ and $\xi _ h$ for the form $b$, and divide each tangent vector by its norm to obtain points of $K$. At $j=0$ the second matrix is zero, which gives the witness with one atom. For $j>0$, the matrix at the second normalized point is $-c$ times the matrix at the first, with $c=j\kappa a_ v/a_ h>0$. The probability measure giving these two points the weights $c/(1+c)$ and $1/(1+c)$ has zero matrix integral, which excludes strict positivity for every Hermitian $B$. This proves failure at all lower levels, so the first successful level is $j=q-1$. $\square$
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