64c3814
==== TABLE OF CONTENTS ====
ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//01-introduction.tex
ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//02-construction.tex
ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//03-ampleness.tex
ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//04-limit.tex
ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//05-obstruction.tex
ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//06-conclusion.tex
ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//paper.tex
ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//references.bib
===========================

==== BEGIN ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//01-introduction.tex ====
\section{Introduction}
\label{sec:introduction}

A holomorphic vector bundle $E$ on a projective variety is ample when
the tautological quotient line bundle $\OO_{\PP(E)}(1)$ is ample;
here $\PP(E)$ parametrizes one-dimensional quotients
\cite[Proposition~3.2]{Hartshorne1966}.
A smooth Hermitian metric is strictly Griffiths-positive if its Chern
curvature is positive in every pair of nonzero tangent and fiber
directions. Such a metric implies ampleness. Griffiths' conjecture asks
for the converse on smooth complex projective varieties
\cite[Problem~(0.9)]{Griffiths1969}.

Set
\[
 S=\PP^1\times\PP^1,\qquad P=\OO_S(1,1),
\]
where $p_1,p_2:S\to\PP^1$ are the projections, and
$\OO_S(r,s)=p_1^*\OO_{\PP^1}(r)\otimes p_2^*\OO_{\PP^1}(s)$.
Let $f_m:S\to S$ raise the homogeneous coordinates in each factor
to their $m$th powers. Thus $\deg f_m=m^2$ and $f_m^*P\simeq P^m$.

\begin{theorem}\label{thm:main}
There is an algebraic vector bundle $G$ of rank two on $S$ such that,
for every positive integer $m$, the bundle
\[
 E_m=f_m^*G\otimes P
\]
is ample. There is an integer $m_0$ such that $E_m$ admits no smooth
Hermitian metric with strictly Griffiths-positive Chern curvature
for every $m\ge m_0$.
\end{theorem}

The bundle $G$ is given explicitly in Section~\ref{sec:construction}.
The ampleness assertion has a quantitative certificate:
Proposition~\ref{prop:ampleness} proves that $\OO_{\PP(E_m)}(n)$ is
very ample for every even $n$ with $n-1\geq3m$.
The threshold $m_0$ is obtained by compactness and is ineffective.
Thus Griffiths' conjecture fails in rank two on the quadric surface.

\subsection*{Historical context and related results}

Hartshorne developed the algebraic theory of ample vector bundles
through symmetric powers, global generation, and the tautological
line bundle \cite{Hartshorne1966}. Griffiths' original question used
the term ``cohomologically positive''; his definition of ``ample'' in
that article included a first-jet generation condition
\cite[Definitions~(0.2)--(0.3), Problem~(0.9)]{Griffiths1969}.
The conjecture above is the modern formulation of that question.
Its significance is the proposed passage from algebraic positivity
to a Hermitian curvature condition on the bundle itself, which can
then enter differential-geometric vanishing arguments.

For line bundles the algebraic and curvature notions agree
\cite[Proposition~1.15]{Umemura1973}. On complex curves, Hartshorne
characterized ampleness by positivity of the degrees of all nonzero
quotient bundles \cite[Theorem~2.4]{Hartshorne1971}.
Umemura proved the metric equivalence first for elliptic curves and
then for compact Riemann surfaces
\cite{Umemura1972,Umemura1973}; see in particular
\cite[Theorem~2.6]{Umemura1973} and the later work of Campana--Flenner
\cite{CampanaFlenner1990}. These results make a surface the first
possible base dimension for the phenomenon in Theorem~\ref{thm:main}.

Several positive results recover strong curvature after modifying the
bundle. Nakano positivity is a stronger curvature condition than
Griffiths positivity. Berndtsson proved Nakano positivity of $E\otimes\det E$ for
ample $E$ through direct-image curvature
\cite[Theorem~1.3]{Berndtsson2009}. Liu--Sun--Yang established
Nakano and dual-Nakano positivity for determinant-twisted symmetric
powers and for sufficiently large symmetric powers of ample bundles
\cite[Theorems~1.1--1.2 and Corollary~1.3]{LiuSunYang2013}.
The determinant twists and symmetric powers are part of these
conclusions; they do not supply a Griffiths-positive metric on $E$
itself.

Analytic approaches to that remaining question include Demailly's
Hermitian--Yang--Mills system and subsequent curvature functionals
\cite{Demailly2021,Demailly2022}. Pingali identified a limitation of
the initial system and solved a related system for direct sums of
ample line bundles on Riemann surfaces
\cite{Pingali2021,Pingali2023}. Murakami subsequently obtained an
analytic proof of the full curve case through the Demailly system
\cite[Theorem~1.2]{Murakami2026}. These developments concern smooth
Hermitian metrics and clarify the analytic difficulty of recovering
them from ampleness.

The requirement that the metric be Hermitian is essential to the
problem. Kobayashi characterized ampleness of $E$ by negatively curved
strongly pseudoconvex complex Finsler structures on $E^*$
\cite[Theorems~4.1 and~5.1]{Kobayashi1975}. His term ``convex''
means positivity of the complex fiber Hessian of the squared metric
\cite[Section~2]{Kobayashi1975}. Their squared norms are
homogeneous of degree two but need not be Hermitian quadratics in
the fiber variable. More recently, Lempert showed that an equivariant
fiberwise rule applied to a prescribed positive tautological metric
can fail to produce even a semipositive Hermitian metric
\cite[Theorem~2.1]{Lempert2026Construction}. His examples are direct sums of a
positive line bundle and also carry positive Hermitian metrics.
That result identifies an obstruction to the specified construction
rule; the obstruction below concerns every smooth strictly positive
metric on the indicated bundles.

In a preprint posted on September 22, 2026, Du--Xie state the
existence of a rank-two ample bundle on the abelian surface
$C\times C$, where $C$ is the elliptic curve $y^2=x^3-x$, without
a smooth Griffiths-semipositive metric
\cite[Theorem~1.1]{DuXie2026Counterexample}. Their accompanying
preprint gives a first-jet characterization of Griffiths positivity
in terms of an adjoint-Levi obstruction
\cite[Theorem~1.1]{DuXie2026FirstJet}.
The construction here is on the rational quadric and supplies a
specified power-map family, an all-$m$ ampleness certificate, and
the general limiting statement of Proposition~\ref{prop:limit}.
Its metric conclusion is the absence of smooth \emph{strictly}
Griffiths-positive metrics for all sufficiently large $m$.

A subsequent preprint of Liu--Wan, posted on September 29, 2026,
states the existence of ample rank-two bundles on every Hirzebruch
surface, including $\PP^1\times\PP^1$, that are slope-stable with
respect to suitable ample polarizations and admit no smooth
strongly pseudoconvex Finsler metric of semipositive Kobayashi curvature
\cite[Theorem~1.1]{LiuWan2026}. That curvature condition is imposed on
the original bundle; Kobayashi's ampleness characterization above uses
negative curvature on its dual. The result concerns a different
construction and a stronger metric exclusion than the strict
Hermitian conclusion proved here.

\subsection*{Proof strategy}

\paragraph{Compatible graph identifications.}
Choose two projective transformations $A,B$ of $\PP^1$ with order
three and no common fixed point, such that $AB$ is conjugate to
$z\mapsto\lambda z$ with $|\lambda|>1$. Their chosen linear lifts,
and the relative lift $A^{-1}B$, have finite order. The two properties
serve different purposes: finite relative orders allow line maps to
agree at graph intersections, whereas the expanding product will
prevent a common invariant measure.
For $M\in\{I,A,B\}$ put
$\Gamma_M=\{(x,Mx):x\in\PP^1\}$ and
$D=\Gamma_I\cup\Gamma_A\cup\Gamma_B$.
For a suitable common multiple $b$ of the finite lift orders, set
$L_1=\OO_S(b,0)$ and $L_2=\OO_S(0,b)$.
The lifted maps identify $L_1^{-1}|_D$ with $L_2^{-1}|_D$;
duality gives a corresponding identification of $L_1|_D$ and $L_2|_D$.
Define $G$ by the pairs of local sections of $L_1\oplus L_2$ that
agree under this identification on $D$.
This is an elementary transformation in the sense of
\cite[Introduction, (0.1)]{Maruyama1986}.
Section~\ref{sec:construction} checks compatibility at every node,
local freeness, and the two line-bundle quotients of $G$.
Their dual line subbundles in $F=G^\vee$ span away from $D$ and
coincide over $D$ according to the prescribed lifted transformations.

\paragraph{Ampleness from the two quotients.}
The two quotient sequences give two filtrations of symmetric powers.
The product of the two kernel line bundles is positive in both rulings.
Alternating the sequences groups the negative twists into powers of
this product; each remaining quotient loses at most one additional
copy of the graph divisor. After pullback by $f_m$, twisting
$\Sym^n(f_m^*G)$ by $P^{n-1}$ offsets that loss for sufficiently
large even $n$.
The line factors are then globally generated and have
vanishing first cohomology. Section~\ref{sec:ampleness} lifts their
sections through the filtration. A relative Veronese map, the base
embedding supplied by one further factor $P$, and a Segre embedding
then give the stated very-ampleness certificate. This
follows the symmetric-power framework of \cite{Hartshorne1966},
while using the two quotients separately also where their kernels
coincide.

\paragraph{A nonzero quadratic limit.}
Suppose that infinitely many $E_m$ have positive metrics. Their dual
squared norms are plurisubharmonic. Average these norms over $f_m$,
using a local $m$th root of a pulled-back \emph{twisting frame} to
absorb the factor $P$. Each average remains a Hermitian quadratic;
the twisting transition factor tends to one as $m\to\infty$.
After a scalar normalization, the coefficient matrices are locally
uniformly bounded. Proposition~\ref{prop:limit} produces a finite,
nonnegative, nonzero plurisubharmonic function $q$ on $F$, quadratic
on every fiber. Convolution of negative singular Hermitian metrics is
developed
in \cite{BerndtssonPaun2008,Raufi2015,PaunTakayama2018}.
Here noncollapse and the assertion on \emph{every} fiber require
separate arguments: the graph relations lie over a divisor and
cannot be read from an almost-everywhere representative.

\paragraph{Incompatible invariances.}
Restricting $q$ to either line subbundle makes its squared seminorm
independent of the other $\PP^1$ coordinate, by the maximum principle.
The graph relations give a single nonzero squared seminorm on
$\OO_{\PP^1}(-b)$, invariant under the lifts of $A$ and $B$.
Total-space plurisubharmonicity and quadratic fiber dependence imply
subharmonicity of the logarithmic weights. Applying
$i\partial\bar\partial$ to them gives a positive measure whose
nonzero total mass is fixed by the degree
$-b$, even when the seminorm has zeros.
Invariance under $AB$ forces this measure to be supported on its
two fixed points. Both $A$ and $B$ preserve this nonempty support.
Since a permutation of at most two points whose order divides three
is trivial, each would fix every support point, giving a common fixed
point. Section~\ref{sec:obstruction} proves this
contradiction, and Section~\ref{sec:conclusion} assembles the theorem.

\subsection*{Conventions and notation}

For a vector bundle $V$, $\PP(V)$ parametrizes one-dimensional
quotients, and $\OO_{\PP(V)}(1)$ is the tautological quotient line.
The projective line $\PP^1$ used to describe the graph transformations
is also identified with the lines in $\CC^2$; its tautological
subbundle is $\OO_{\PP^1}(-1)$.

We use the curvature convention of \cite{Demailly2021}. At a point,
in local coordinates and a frame orthonormal at that point, write
\[
 \Theta_{E,h}
 =i\sum_{j,k,\alpha,\beta}
 c_{jk\alpha\beta}\,dz_j\wedge d\bar z_k
 \otimes e_\alpha^*\otimes e_\beta.
\]
Strict Griffiths positivity means
\[
 \sum_{j,k,\alpha,\beta}
 c_{jk\alpha\beta}\xi_j\bar\xi_k v_\alpha\bar v_\beta>0
 \quad\text{for all }\xi\ne0,\ v\ne0.
\]
The dual has the opposite Griffiths sign. In particular its squared
norm is plurisubharmonic on the total bundle; this standard criterion
is recalled where it is used. Throughout, ``psh'' means
plurisubharmonic, and
$\ddc u=i u_{z\bar z}\,dz\wedge d\bar z$ on a complex curve.
A Hermitian quadratic form is allowed to be positive semidefinite
when this is stated explicitly.
==== END ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//01-introduction.tex ====

==== BEGIN ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//02-construction.tex ====
\section{A bundle defined by three graphs}\label{sec:construction}

We construct $G$ together with two line-bundle quotients. Their dual
inclusions will span away from three graphs and agree on those graphs
through specified linear maps. The quotient sequences will prove
ampleness; the graph identities will constrain a quadratic limit on
$G^\vee$.

We choose matrices $A,B$ with two different properties. The relative
lifts $A$, $B$, and $A^{-1}B$ must have finite order, so that their
tensor powers agree at graph intersections. The projective actions of
$A$ and $B$ will have order three and no common fixed point, while
$AB$ will have eigenvalues of distinct moduli. Together these properties
will exclude an invariant measure in Section~\ref{sec:obstruction}. As in the
introduction, $S=\PP^1\times\PP^1$, and we view $\PP^1$ as the lines
in $\CC^2$, with tautological bundle $\OO_{\PP^1}(-1)$.

Put
\begin{equation}\label{eq:matrices}
 a=e^{i\pi/3},\qquad
 s=\frac12-\frac{i(1+2\sqrt2)}{2\sqrt3},\qquad
 A=\begin{pmatrix}a&0\\0&a^{-1}\end{pmatrix},\qquad
 B=\begin{pmatrix}s&1\\ \sqrt2/3&1-s\end{pmatrix}.
\end{equation}
We use the same letters for their induced automorphisms of $\PP^1$.

\begin{lemma}\label{lem:matrix-data}
The matrices $A,B$ belong to $\SL_2(\CC)$ and satisfy
\[
 \tr A=\tr B=1,\qquad
 \tr(A^{-1}B)=-\sqrt2,\qquad
 \tr(AB)=1+\sqrt2.
\]
The orders of $A$, $B$, and $A^{-1}B$ in $\SL_2(\CC)$ are
$6$, $6$, and $8$, respectively. The induced projective orders of $A$ and $B$
are both $3$. The matrix $AB$ has two positive real reciprocal
eigenvalues of distinct moduli, and $A,B$ have no common eigenline.
\end{lemma}

\begin{proof}
The value of $s$ gives
\[
 s(1-s)=1+\frac{\sqrt2}{3},\qquad
 a^{-1}s+a(1-s)=-\sqrt2.
\]
These identities prove the determinant and trace assertions except
for the trace of $AB$. The characteristic polynomial of each of
$A,B$ is $t^2-t+1$, with roots $e^{\pm i\pi/3}$, while that of
$A^{-1}B$ is $t^2+\sqrt2t+1$, with roots $e^{\pm3i\pi/4}$.
The roots in each pair are distinct, proving the claimed orders;
in particular $A^3=B^3=-\Id$.

Cayley--Hamilton gives $A+A^{-1}=\Id$. Multiplication by $B$ and
taking traces therefore yield
\[
 \tr(AB)=\tr B-\tr(A^{-1}B)=1+\sqrt2>2.
\]
Since $\det(AB)=1$, its eigenvalues are positive real reciprocals,
one greater than one. Finally, the eigenlines of $A$ are the two
coordinate lines, and neither is preserved by $B$, whose two
off-diagonal entries are nonzero.
\end{proof}

For $M\in\{I,A,B\}$ let
\[
 \Gamma_M=\{(x,Mx):x\in\PP^1\},\qquad
 D=\Gamma_I\cup\Gamma_A\cup\Gamma_B,
\]
where $D$ has the reduced structure.

\begin{lemma}\label{lem:graph-divisor}
The divisor $D$ has bidegree $(3,3)$ and exactly six singular points,
all ordinary transverse intersections of two components. There are
no triple intersections.
\end{lemma}

\begin{proof}
Each graph is a smooth divisor of bidegree $(1,1)$. An intersection
of $\Gamma_M$ and $\Gamma_N$ corresponds to a fixed point of
$M^{-1}N$, hence to an eigenline of that matrix. The relative
matrices are $A,B,A^{-1}B$ or their inverses. Each has two distinct
eigenvalues, and the multiplier at either projective fixed point is
the ratio of those eigenvalues in one order or the other. This
ratio is not one. The fixed points are therefore simple, so each
pair of graphs meets transversely at two points. A point on all
three graphs would give a common eigenline of $A$ and $B$, contrary
to Lemma~\ref{lem:matrix-data}. Thus the six pairwise intersections
are distinct. Their local equations are products of two regular
parameters, and the asserted description follows.
\end{proof}

Set
\begin{equation}\label{eq:construction-lines}
 b=24,\qquad L_1=p_1^*\OO_{\PP^1}(b),\qquad
 L_2=p_2^*\OO_{\PP^1}(b).
\end{equation}
The exponent $24$ is a common multiple of the orders $6,6,8$ of
the specified relative linear lifts $A,B,A^{-1}B$. It makes the
induced tensor-power maps on the two branches of each graph
intersection agree on their common fiber, as used in the descent
argument below.

Along $\Gamma_M$, the map $M$ on tautological lines induces an
isomorphism from $L_1^{-1}$ to $L_2^{-1}$ by taking its $b$-th
tensor power.

\begin{proposition}\label{prop:construction}
These three isomorphisms glue to an algebraic isomorphism
\begin{equation}\label{eq:alpha}
 \alpha:L_1^{-1}|_D\xrightarrow{\sim}L_2^{-1}|_D,
 \qquad \alpha|_{\Gamma_M}=M^{\otimes b}.
\end{equation}
Let $\beta=(\alpha^\vee)^{-1}:L_1|_D\to L_2|_D$ and define
\begin{equation}\label{eq:equalizer}
 G=\bigl\{(u_1,u_2)\in L_1\oplus L_2:
          u_2|_D=\beta(u_1|_D)\bigr\}.
\end{equation}
Then $G$ is an algebraic vector bundle of rank two. Both projections
$\rho_i:G\to L_i$ are surjective, with exact sequences
\begin{equation}\label{eq:quotient-sequences}
 0\longrightarrow L_i(-D)\longrightarrow G
   \xrightarrow{\rho_j}L_j\longrightarrow0,
 \qquad \{i,j\}=\{1,2\}.
\end{equation}
For $F=G^\vee$, the dual inclusions
$\iota_i=\rho_i^\vee:L_i^{-1}\hookrightarrow F$ span each fiber
over $S\setminus D$ and satisfy
\begin{equation}\label{eq:dual-gluing}
 \iota_2\circ\alpha=\iota_1\qquad\text{on }D.
\end{equation}
\end{proposition}

\begin{proof}
The bundle will be an elementary transformation of $L_1\oplus L_2$
along the effective Cartier divisor $D$, as in
\cite[Introduction, (0.1)]{Maruyama1986}. We verify the gluing and the
local basis directly, since the resulting quotient maps are needed
in both subsequent arguments.

At a point of $\Gamma_M\cap\Gamma_N$, let $v$ span the
first-coordinate line. For some eigenvalue $\lambda$ of $M^{-1}N$
we have $Nv=\lambda Mv$. By Lemma~\ref{lem:matrix-data},
$\lambda^{24}=1$. Consequently
\[
 (Nv)^{\otimes b}=(Mv)^{\otimes b}.
\]
Thus the two maps agree as maps of the actual line fibers. This
also covers the reversed pair, for which the eigenvalue is
$\lambda^{-1}$.

We check descent at a node. If $R$ is the local ring of $S$ there
and $(x),(y)$ are the branch ideals, transversality gives
$(x)\cap(y)=(xy)$ and the exact sequence
\[
 0\longrightarrow R/(xy)\longrightarrow R/(x)\oplus R/(y)
   \longrightarrow R/(x,y)\longrightarrow0,
\]
whose last map is the difference of the residue values. Tensoring
this sequence by the locally free rank-one sheaf
$\operatorname{Hom}(L_1^{-1},L_2^{-1})|_D$ shows that the equal
fiber values are precisely the compatibility needed to descend
the two branch sections. Their common value is nonzero, so the
descended scalar is a unit in the nodal local ring. The component
maps therefore glue to the isomorphism~\eqref{eq:alpha}.

Near a point of $D$, choose frames of $L_1,L_2$ and a local equation
$d$ of $D$. The scalar defining $\beta$ modulo $d$ lifts, after
shrinking, to a regular unit $h$. In these frames the module
\eqref{eq:equalizer} has basis
\begin{equation}\label{eq:local-basis}
 e=(1,h),\qquad f=(0,d).
\end{equation}
Indeed, every element has a unique expression
$u_1e+kf$, since $u_2-hu_1$ is divisible by the non-zero-divisor
$d$. Away from $D$ the sheaf is simply $L_1\oplus L_2$, proving
local freeness everywhere.

The first projection maps $e$ to $1$, and the second maps $e$ to the
unit $h$, so both are surjective. Setting the $j$-th coordinate
equal to zero in~\eqref{eq:equalizer} identifies its kernel with
$L_i(-D)$, proving~\eqref{eq:quotient-sequences}. Their duals are
line subbundles of $F$, and away from $D$ they span. On $D$, if
$v\in L_1^{-1}$ and $(u_1,u_2)\in G$, then
\[
 \bigl(\iota_2(\alpha v)\bigr)(u_1,u_2)
 = (\alpha v)(\beta u_1)=v(u_1)
 =\bigl(\iota_1(v)\bigr)(u_1,u_2),
\]
because $\alpha^\vee\beta=\Id$. This is~\eqref{eq:dual-gluing}.
\end{proof}

For a positive integer $m$, define
\begin{equation}\label{eq:power-maps}
 f_m\bigl([z_0:z_1],[w_0:w_1]\bigr)
   =\bigl([z_0^m:z_1^m],[w_0^m:w_1^m]\bigr).
\end{equation}
This is finite of degree $m^2$, and
$f_m^*\OO_S(r,t)\simeq\OO_S(mr,mt)$. In particular,
$f_m^*P\simeq P^{\otimes m}$. The bundles in the theorem are
\begin{equation}\label{eq:family}
 E_m=f_m^*G\otimes P.
\end{equation}
==== END ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//02-construction.tex ====

==== BEGIN ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//03-ampleness.tex ====
\section{Ampleness of every member of the family}\label{sec:ampleness}

For each $m$, we construct enough symmetric-power sections to embed
$\PP(E_m)$. The two quotient sequences are used alternately: their
kernels individually have one negative degree, but their tensor product
is positive in both rulings. The resulting filtration gives an explicit
very-ampleness range within Hartshorne's symmetric-power framework
\cite[Corollary~2.5 and Proposition~3.2]{Hartshorne1966}.

\begin{proposition}\label{prop:ampleness}
For every positive integer $m$, the bundle
$E_m=f_m^*G\otimes P$ is ample. More precisely,
$\OO_{\PP(E_m)}(n)$ is very ample for every even integer $n$ with
$n-1\geq3m$.
\end{proposition}

\begin{proof}
Fix $m$ and write
\[
 V=f_m^*G,\qquad \widetilde L_i=f_m^*L_i,\qquad
 \widetilde D=f_m^*D,\qquad
 K_i=\widetilde L_i(-\widetilde D),\qquad K=K_1\otimes K_2.
\]
Pullback of~\eqref{eq:quotient-sequences} gives
\begin{equation}\label{eq:pullback-quotients}
 0\longrightarrow K_i\longrightarrow V\longrightarrow
 \widetilde L_j\longrightarrow0,
 \qquad\{i,j\}=\{1,2\}.
\end{equation}
These sequences remain exact because their original quotients
are locally free. Here
\[
 K_1=\OO_S(21m,-3m),\qquad K_2=\OO_S(-3m,21m),
\]
and hence
\begin{equation}\label{eq:kernel-product}
 K=\OO_S(18m,18m),\qquad
 \OO_S(\widetilde D)=\OO_S(3m,3m).
\end{equation}
We write powers of line bundles for tensor powers throughout this
proof.

\smallskip\noindent\textbf{Alternating the quotient sequences.}
For an extension $0\to K'\to V'\to L'\to0$ with $K',L'$ line
bundles, multiplication gives
\[
 0\longrightarrow K'\otimes\Sym^{r-1}V'
   \longrightarrow\Sym^r V'\longrightarrow(L')^r
   \longrightarrow0\qquad(r\geq1).
\]
Indeed, a local splitting gives a kernel generator $k'$ and a lift
$e'$ of a quotient generator. The monomials containing $k'$ form the
kernel of the symmetric quotient, and the remaining monomial $(e')^r$
maps to a quotient generator. This is the usual symmetric-power
filtration of an extension; compare the proof of
\cite[Corollary~3.4]{Hartshorne1966}.

Apply this construction first to $V\to\widetilde L_2$ and then to
$V\to\widetilde L_1$. For $r\geq2$ it gives
\begin{align}
 0&\longrightarrow K_1\otimes\Sym^{r-1}V
   \longrightarrow\Sym^r V\longrightarrow\widetilde L_2^{\,r}
   \longrightarrow0,\label{eq:symmetric-first}\\
 0&\longrightarrow K\otimes\Sym^{r-2}V
   \longrightarrow K_1\otimes\Sym^{r-1}V
   \longrightarrow K_1\otimes\widetilde L_1^{\,r-1}
   \longrightarrow0.\label{eq:symmetric-second}
\end{align}
Thus each pair of steps leaves the residual bundle
$K\otimes\Sym^{r-2}V$, with one line-bundle quotient at each step.
For even $n$, iteration gives a filtration of $\Sym^nV$ with line
factors
\begin{equation}\label{eq:graded-lines}
 \begin{gathered}
 K^j\otimes\widetilde L_2^{\,n-2j},\qquad
 K^j\otimes\widetilde L_1^{\,n-2j}(-\widetilde D),
 \qquad 0\leq j<n/2,\\
 K^{n/2}.
 \end{gathered}
\end{equation}
Indeed, after $j$ pairs the residual bundle is
$K^j\otimes\Sym^{n-2j}V$, whose second quotient is
$K^jK_1\widetilde L_1^{\,n-2j-1}
 =K^j\widetilde L_1^{\,n-2j}(-\widetilde D)$.
Only the second quotient contains an additional twist by
$-\widetilde D$ beyond the factors absorbed into $K^j$.

Each step uses just one of the locally split sequences
\eqref{eq:pullback-quotients}, so it remains valid when the two kernel
lines have the same fiber image. Concretely, in the pulled-back basis
$e,f$ of~\eqref{eq:local-basis}, the kernels are generated by
$k=de-hf$ and $f$, with $h$ a unit. For $n=2$ the filtration is
\[
 \OO(kf)\subset kV\subset\Sym^2V.
\]
Since $e,k$ are also a basis of $V$, the module $kV$ has basis
$ke,k^2$ and is a direct summand of $\Sym^2V$, with complement
generated by $e^2$. Moreover,
\[
 kf=\frac{d}{h}ke-\frac1h k^2.
\]
The coefficient $-1/h$ is a unit, so $kf$ extends to a basis of $kV$.
Both inclusions therefore have locally free quotients.
This verification uses no smoothness of $d=0$ and applies at nodes
and to nonreduced pullbacks of the divisor.

\smallskip\noindent\textbf{Global generation after twisting.}
Choose $n$ even with $n-1\geq3m$, and set
\begin{equation}\label{eq:generated-symmetric-power}
 W=\Sym^n V\otimes P^{n-1}.
\end{equation}
By~\eqref{eq:kernel-product}, the bidegrees of its filtration
factors, in the order of~\eqref{eq:graded-lines}, are
\begin{align*}
 &(18mj+n-1,\;18mj+24m(n-2j)+n-1),\\
 &(18mj+24m(n-2j)-3m+n-1,\;18mj-3m+n-1),\\
 &(9mn+n-1,\;9mn+n-1).
\end{align*}
Every entry is nonnegative: the only possibly negative contribution
is $-3m$, and it is offset by $n-1$.

For $c,t\geq0$, products of homogeneous monomials generate
$\OO_S(c,t)$. Moreover $H^1(\PP^1,\OO(c))=H^1(\PP^1,\OO(t))=0$, as
computed on the two standard affine charts, so the K\"unneth formula
gives $H^1(S,\OO_S(c,t))=0$
\cite[Tags~01XD and~0BED]{StacksProject}.
If $0\to U\to Z\to Q\to0$ is an extension of globally generated
bundles with $H^1(S,U)=H^1(S,Q)=0$, the cohomology sequence gives
$H^1(S,Z)=0$ and lifts every global section of $Q$ to $Z$.
The lifted generators of $Q$, together with generators of $U$,
span each fiber of $Z$. Starting with the innermost line factor and
applying this observation successively proves that $W$ is globally
generated.

\smallskip\noindent\textbf{The projective embedding and its line bundle.}
We record the tautological twists in the quotient-projectivization
and Segre constructions
\cite[Tags~01OA, 01N1, and~01WD]{StacksProject}.
Let $\pi:\PP(V)\to S$ and $H=H^0(S,W)$. Evaluation
$H\otimes\OO_S\to W$ induces a closed immersion
\[
 \PP(W)\hookrightarrow\PP(H)\times S
\]
whose projective hyperplane line pulls back to
$\OO_{\PP(W)}(1)$. The relative Veronese closed immersion
$\PP(V)\to\PP(\Sym^n V)$ sends a quotient line $\ell$ to
$\ell^{\otimes n}$. Under the natural identification
$\PP(\Sym^n V)\simeq\PP(W)$, its tautological pullback is
\[
 \OO_{\PP(V)}(n)\otimes\pi^*P^{n-1}.
\]
Consequently their composition embeds $\PP(V)$ into
$\PP(H)\times S$ with this hyperplane pullback from the first
factor. Embedding $S$ by $P=\OO_S(1,1)$ and then applying the Segre
embedding gives a closed immersion into projective space whose
hyperplane line pulls back to
\[
 \bigl(\OO_{\PP(V)}(n)\otimes\pi^*P^{n-1}\bigr)
       \otimes\pi^*P
 =\bigl(\OO_{\PP(V)}(1)\otimes\pi^*P\bigr)^n.
\]
This line is therefore very ample. Finally,
$\PP(E_m)=\PP(V\otimes P)\simeq\PP(V)$ identifies
$\OO_{\PP(E_m)}(1)$ with
$\OO_{\PP(V)}(1)\otimes\pi^*P$, proving the proposition.
\end{proof}
==== END ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//03-ampleness.tex ====

==== BEGIN ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//04-limit.tex ====
\section{A necessary quadratic limit}\label{sec:limit}

The power maps allow the fixed twist by $P$ to disappear in a limit.
We prove that the quadratic structure survives at every base point,
even where a raw pointwise limit need not be plurisubharmonic.
The argument has four steps: trace the dual norms, normalize and
extract their coefficient limit, prove that the limit is nonzero,
and recover its intrinsic plurisubharmonic representative.
We establish this statement for an arbitrary vector bundle on $S$.

\begin{proposition}\label{prop:limit}
Let $G$ be a holomorphic vector bundle of positive rank on
$S=\PP^1\times\PP^1$, and put $F=G^\vee$.  Suppose that
$f_m^*G\otimes P$ admits a smooth strictly Griffiths-positive Hermitian
metric for infinitely many positive integers $m$.  Then there is a
function
\[
q:F\longrightarrow [0,\infty)
\]
which is plurisubharmonic on the total space, is not identically zero,
and has the following property: for every $x\in S$ there is a finite
positive semidefinite Hermitian form $H_x$ on $F_x$ such that
\[
q(x,v)=H_x(v,v)\qquad(v\in F_x).
\]
In particular, this fiberwise assertion holds at every base point.
\end{proposition}

\begin{proof}
Put $r=\operatorname{rank}G$.  Choose an unbounded sequence of the
indicated integers and let
$\|\cdot\|_m$ denote the smooth dual metric on
\[
(f_m^*G\otimes P)^\vee=f_m^*F\otimes P^{-1}.
\]
Its squared norm is plurisubharmonic on the total space.  We recall the
sign in this standard implication.  For a smooth Hermitian metric with
matrix $h$, choose a holomorphic frame normal at the point in question,
so that $h=\Id$ and its first derivatives vanish there.  Along a
holomorphic disk with coordinate $\zeta$, write the Hermitian curvature
as $iC\,d\zeta\wedge d\bar\zeta$.  In the convention used here,
\[
C=-\frac{\partial^2h}{\partial\zeta\,\partial\bar\zeta}
\]
at the normal point.  Thus, for a holomorphic section $s$ along the disk,
\begin{equation}\label{eq:limit-dual-hessian}
\frac{\partial^2}{\partial\zeta\,\partial\bar\zeta}\|s\|^2
=\|D'_{\partial/\partial\zeta}s\|^2-\langle Cs,s\rangle.
\end{equation}
Duality reverses Griffiths positivity, so the second term on the right
is nonnegative for the dual metric.  Applying this calculation to the
section of the pulled-back bundle defined by any disk in its total
space proves the assertion. For this classical norm criterion and
duality, see \cite[Remark~1.3 and Lemma~1.7]{Umemura1973} and
\cite[Section~2]{Raufi2015}; the same implication is recorded in
\cite[Proposition~2.6]{ForstnericKusakabe}.

\smallskip\noindent\textbf{Trace quadratics and normalization.}
Fix, for each $m$, the holomorphic isomorphism
$f_m^*P^{-1}\simeq P^{-m}$.  If $t$ is a nonvanishing holomorphic frame
of $P^{-1}$ on an open set $U\subset S$, define on $F|_U$
\begin{equation}\label{eq:limit-trace}
q_{m,t}(x,v)
=\frac{1}{m^2}\sum_{y\in f_m^{-1}(x)}
\bigl\|v\otimes(f_m^*t)^{1/m}(y)\bigr\|_{m,y}^2,
\end{equation}
where the sum counts multiplicities.  The root in this expression is
local on the source.  Indeed, in a local frame $s$ of $P^{-1}$ the
section $f_m^*t$ has the form $\eta(y)s^m$, with $\eta$ holomorphic and
nowhere zero.  A local holomorphic logarithm of $\eta$ gives the root
$\exp((\log\eta)/m)s$.  Two choices differ by an $m$-th root of unity,
which does not change the squared norm.  The expression
\eqref{eq:limit-trace} is consequently well defined and is a positive
definite Hermitian quadratic in $v$.

We verify its regularity through the branch locus.  Off that locus,
every local inverse branch of $f_m$ gives a holomorphic map
\[
(x,v)\longmapsto
\bigl(y(x),v\otimes(f_m^*t)^{1/m}(y(x))\bigr)
\]
into $f_m^*F\otimes P^{-1}$.  Hence each summand, and their sum, is
plurisubharmonic.  The unordered multiset of roots of $y^m=x$ varies
continuously with $x$, counting coalescing roots with their
multiplicities.  Applied in the two factors, and using local source
frames for the norm in \eqref{eq:limit-trace}, this proves continuity
of the sum across the branch locus, including a crossing of its two
components.

The continuous trace is plurisubharmonic across the branch locus as
well. In a local total-space chart, the inverse image of the branch locus
under the bundle projection $F|_U\to U$ is an analytic set, contained in
$\{z_1z_2=0\}$ after choosing affine
coordinates on the two base factors. It is therefore pluripolar.
The trace is locally bounded above there by continuity, so the
removable-singularities theorem for plurisubharmonic functions
\cite[Chapter~I, Theorem~5.24]{DemaillyCADG} gives a unique
plurisubharmonic extension. Its upper-limit values agree with the
continuous trace. This proves joint plurisubharmonicity of $q_{m,t}$
on all of $F|_U$.

If $t'=gt$ is another nonvanishing frame, then
\begin{equation}\label{eq:limit-frame-law}
q_{m,t'}(x,v)=|g(x)|^{2/m}q_{m,t}(x,v).
\end{equation}
Choose smooth background Hermitian norms $|\cdot|_0$ on $F$ and
$P^{-1}$.  By \eqref{eq:limit-frame-law}, the continuous quadratic
\[
N_m(x,v)=q_{m,t}(x,v)|t(x)|_0^{-2/m}
\]
is globally defined. We use $N_m$ to normalize the traces globally;
the local functions $q_{m,t}$ retain the plurisubharmonicity used
below. The background unit sphere bundle is compact,
and $N_m$ is positive there.  Multiplying the dual metric's squared
norm by a positive constant, we may therefore arrange
\begin{equation}\label{eq:limit-normalization}
\max_{x\in S,\ |v|_0=1}N_m(x,v)=1.
\end{equation}
This rescaling does not change the curvature.  Homogeneity gives
\begin{equation}\label{eq:limit-local-bound}
0\le q_{m,t}(x,v)\le |t(x)|_0^{2/m}|v|_0^2.
\end{equation}

\smallskip\noindent\textbf{A measurable coefficient limit.}
In a holomorphic trivialization of $F|_U$, write
\[
q_{m,t}(x,v)=v^*H_m(x)v.
\]
On each relatively compact subset of $U$, the positive matrices $H_m$
are uniformly bounded, by \eqref{eq:limit-local-bound}.  Weak compactness
in $L^2$ and a diagonal subsequence on a countable collection of charts
and compact exhaustions give distributional limits $H$ for all the
coefficient matrices.  These limits belong to $L^\infty_{\mathrm{loc}}$,
are Hermitian, and are positive semidefinite almost everywhere.  For
example, positivity and the same local upper bound pass to
$c^*Hc$ against every nonnegative test function for each
$c\in(\mathbb Q+i\mathbb Q)^r$; density then gives the matrix
inequalities on a single set of full measure.

The coefficient limits have the untwisted transition law of a form
on $F$.  More explicitly, let $v=T(x)w$ be a holomorphic change of
bundle coordinates and let $t'=gt$.  The original matrices satisfy
\[
H'_m=|g|^{2/m}T^*H_mT.
\]
On a compact subset $K$ of the overlap, the error obtained by deleting
the scalar factor has, against a smooth compactly supported matrix
test function $\Phi$, the bound
\[
\left|\left\langle
\bigl(|g|^{2/m}-1\bigr)T^*H_mT,\Phi
\right\rangle\right|
\le C_K\|\Phi\|_{L^1}
\sup_K\bigl||g|^{2/m}-1\bigr|\longrightarrow0.
\]
Here the local matrix bound and the boundedness of $T$ are absorbed
into $C_K$.  Multiplication by the fixed smooth matrices $T,T^*$
preserves distributional convergence, as does a change of base
coordinates.  Consequently $H'=T^*HT$ almost everywhere.

\smallskip\noindent\textbf{Nonvanishing of the measurable limit.}
The limiting measurable form is not zero.  Choose maximizing unit
vectors $(x_m,v_m)$ in \eqref{eq:limit-normalization}.  After a further
subsequence they converge in the compact unit sphere bundle to
$(x_*,v_*)$.  This further subsequence retains all the distributional
limits already obtained.  In one fixed coordinate and frame chart
around $x_*$, let $c_m\to c$ be their fiber coordinates.  Then $c\ne0$,
and a local operator bound $\|H_m\|\le C$ gives
\[
\bigl|c^*H_m(x_m)c-c_m^*H_m(x_m)c_m\bigr|
\le C|c-c_m|(|c|+|c_m|)\longrightarrow0.
\]
On the other hand,
\[
c_m^*H_m(x_m)c_m=|t(x_m)|_0^{2/m}\longrightarrow1,
\]
since the norm of the nonvanishing frame is bounded above and away
from zero on a fixed compact neighborhood.  Thus the nonnegative
plurisubharmonic function
$u_m(x)=c^*H_m(x)c$ satisfies $u_m(x_m)\ge\tfrac12$ for large $m$.
Choose $\varrho>0$ so that $B(x_*,4\varrho)$ lies in the coordinate
chart, and choose a fixed smooth nonnegative function $\chi$ supported
in $B(x_*,3\varrho)$ and equal to one on $B(x_*,2\varrho)$.
For all sufficiently large $m$, the ball $B(x_m,\varrho)$ lies in
that latter ball.  The submean inequality yields
\begin{equation}\label{eq:limit-noncollapse}
\int\chi(x)c^*H_m(x)c\,dx
\ge\int_{B(x_m,\varrho)}u_m(x)\,dx
\ge\frac12\operatorname{Vol}(B(0,\varrho)).
\end{equation}
The left side is a pairing against the fixed smooth matrix test
function $\chi cc^*$, so the inequality passes to $H$.  It proves
that $H$ is nonzero as a measurable form.

\smallskip\noindent\textbf{A quadratic psh representative on every fiber.}
The measurable form is defined only up to a set of measure zero in
the base. Even a
pointwise limit of the traces may fail to be plurisubharmonic: in one
base variable, tracing $(1+|y|^2)|v|^2$ through $x=y^m$ gives
\[
 (1+|x|^{2/m})|v|^2.
\]
Its pointwise limit has coefficient $1$ at $x=0$ and $2$ elsewhere,
whereas its plurisubharmonic representative has coefficient $2$
everywhere. We now recover the intrinsic plurisubharmonic
representative of our measurable form and show that it remains a
Hermitian quadratic on every fiber.

Convolution regularization of
negative singular Hermitian metrics appears in
\cite[Section~3, Proposition~3.1]{BerndtssonPaun2008} and
\cite[Proposition~1.3(i)]{Raufi2015}; the use of polarization to recover
an everywhere-defined representative is also discussed in
\cite[Remark~2.2.3]{PaunTakayama2018}.
We supply the argument for the present possibly rank-deficient limit,
including the compatibility of the representatives on overlaps.
Work in a coordinate and bundle
chart $U$, and let $\rho$ be a nonnegative smooth radial function on
$\CC^2$, supported in the unit ball and of integral one.  Put
\[
\rho_\delta(y)=\delta^{-4}\rho(y/\delta),\qquad
H_\delta=H*\rho_\delta,\qquad
q_\delta(x,v)=v^*H_\delta(x)v,
\]
on the interior domain where this convolution is defined.
Convolving $H_m$ in the same way gives
\[
q_{m,\delta}(x,v)
=\int q_{m,t}(x-y,v)\rho_\delta(y)\,dy.
\]
It is plurisubharmonic jointly in $(x,v)$, since each base translation
$(x,v)\mapsto(x-y,v)$ is holomorphic and the average has nonnegative
weights.  For fixed $\delta$, distributional convergence of $H_m$
implies convergence of the convolved coefficients and all their
derivatives at every interior point: these are pairings against
translates of derivatives of the fixed smooth kernel.  The Levi
matrices of $q_{m,\delta}$ therefore converge pointwise to that of
$q_\delta$.  Hence $q_\delta$ is smooth and plurisubharmonic, and on
each smaller chart
\begin{equation}\label{eq:limit-convolution-bound}
0\le q_\delta(x,v)\le C|v|^2
\end{equation}
for all sufficiently small $\delta$.

These quadratics decrease as $\delta$ decreases.  To check this for
any fixed nonnegative radial kernel, write $\rho(y)=\phi(|y|)$ and
let $M_u(x,s)$ denote the spherical mean of a subharmonic function
$u$ on $\RR^4$.  Polar coordinates give
\begin{equation}\label{eq:limit-radial-means}
(u*\rho_\delta)(x)
=|\mathbb S^3|\int_0^1
M_u(x,\delta s)\phi(s)s^3\,ds.
\end{equation}
The spherical mean is nondecreasing in its radius.  The integration
weight in \eqref{eq:limit-radial-means} is nonnegative and does not
depend on $\delta$, so the convolution is nondecreasing in $\delta$.
Apply this to $u(x)=q_{m,t}(x,v)$ for each constant $v$, then pass to
the limit in $m$ at two fixed radii.  It follows that
\[
H_{\delta_1}(x)\le H_{\delta_2}(x)
\qquad(0<\delta_1<\delta_2)
\]
in the order of Hermitian forms.

Thus $q_\delta$ decreases, as $\delta\downarrow0$, to a finite
nonnegative plurisubharmonic function $q_U$ on the total-space chart.
The decreasing-limit assertion follows also by passing to the
submean inequalities; see \cite[Chapter~I, Theorem~5.4]{DemaillyCADG}.
Finiteness is supplied by \eqref{eq:limit-convolution-bound}.
At every fixed base point $x$, this limit is a Hermitian quadratic.
Indeed, all entries of $H_\delta(x)$ are determined by the finite
set of values at $e_j$, $e_j+e_k$, and $e_j+i e_k$, through
\begin{align*}
2\operatorname{Re}(H_\delta)_{jk}
&=q_\delta(x,e_j+e_k)-q_\delta(x,e_j)-q_\delta(x,e_k),\\
2\operatorname{Im}(H_\delta)_{jk}
&=q_\delta(x,e_j)+q_\delta(x,e_k)-q_\delta(x,e_j+i e_k).
\end{align*}
Each value on the right has a finite limit.  Hence the entries
converge to one Hermitian matrix $\widehat H_U(x)$, and passage to the
finite quadratic polynomial gives
\[
q_U(x,v)=v^*\widehat H_U(x)v\qquad\text{for every }v\in\CC^r.
\]
This matrix is positive semidefinite at every $x$, by nonnegativity.
At almost every $x$, approximate-identity convergence also gives
$\widehat H_U(x)=H(x)$.

Finally these local representatives agree on overlaps.  After the
holomorphic change of bundle coordinates, their measurable matrices
agree outside a base-null set, by the transition law already proved.
The quadratic functions therefore agree for every fiber vector
outside that set, hence almost everywhere in the total-space chart
by Fubini's theorem.  Both functions are locally bounded and
plurisubharmonic.  Such a function is determined everywhere by its
almost-everywhere values: the submean inequality and upper
semicontinuity give, in local Euclidean coordinates,
\[
u(p)=\lim_{\varepsilon\downarrow0}
\frac{1}{\operatorname{Vol}(B(p,\varepsilon))}
\int_{B(p,\varepsilon)}u.
\]
Thus the local $q_U$ glue to a single function $q$ on $F$.  Its
coefficients represent the original measurable form almost
everywhere, so \eqref{eq:limit-noncollapse} implies $q\not\equiv0$.
All the required pointwise and plurisubharmonic properties now hold.
\end{proof}

\begin{remark}\label{rmk:canonical-limit}
The function in Proposition~\ref{prop:limit} represents a
distributional limit; no convergence of restrictions to a divisor is
asserted. The graph identities will instead be evaluated directly in
this single everywhere-defined function $q$.
\end{remark}
==== END ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//04-limit.tex ====

==== BEGIN ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//05-obstruction.tex ====
\section{The invariant measure obstruction}
\label{sec:obstruction}

We now exclude the quadratic function produced by
Proposition~\ref{prop:limit} for the graph-defined bundle $G$.
The two rulings and the three graph identities force a nonzero squared seminorm on $\OO_{\PP^1}(-b)$
that is invariant under the lifts of $A$ and $B$.
We then turn that seminorm into a positive measure of mass $2\pi b$
and show that the two transformations cannot preserve such a measure.

\subsection{From the two rulings to one invariant seminorm}

\begin{lemma}\label{lem:ruling-reduction}
Let $G$ be the bundle of Proposition~\ref{prop:construction}, and put
$F=G^\vee$. Suppose that $q:F\to[0,\infty)$ is finite,
plurisubharmonic, not identically zero, and a Hermitian quadratic form
on every fiber. Then there is a finite, nonnegative, nonzero
plurisubharmonic function $a_\bullet$ on $\OO_{\PP^1}(-b)$, a Hermitian
positive semidefinite quadratic on each fiber, such that
\[
 a_\bullet(Mx,M^{\otimes b}v)=a_\bullet(x,v)
 \qquad(M=A,B)
\]
for every $x\in\PP^1$ and $v\in\OO(-b)_x$.
\end{lemma}

\begin{proof}
Write
$q(x,v)=H_x(v,v)$; nonnegativity says that every $H_x$ is positive
semidefinite.  Pull $q$ back by the line inclusions
\[
 \iota_i:L_i^{-1}=p_i^*\OO_{\PP^1}(-b)\longrightarrow F,
 \qquad i=1,2.
\]
For fixed $x_i\in\PP^1$ and $v\in\OO(-b)_{x_i}$, the function obtained
by holding $v$ fixed while varying the other base coordinate is finite
and subharmonic on the complementary $\PP^1$.  It is constant by the
maximum principle.  Consequently $q\circ\iota_i$ descends to a finite
nonnegative squared seminorm $a_i$ on $\OO_{\PP^1}(-b)$ over the
$i$-th factor.  Fixing any point in the other factor realizes $a_i$ as a
holomorphic pullback of $q$, so $a_i$ is plurisubharmonic on the total
line bundle.

On $\Gamma_M$, the identity $\iota_2\alpha=\iota_1$ from
Proposition~\ref{prop:construction} and the definition of $\alpha$ give
\begin{equation}
 \label{eq:seminorm-graph-identities}
 a_1(x,v)=a_2(Mx,M^{\otimes b}v),
 \qquad M\in\{\Id,A,B\}.
\end{equation}
The case $M=\Id$ identifies $a_1$ and $a_2$ as a single function
$a_\bullet$ on $\OO(-b)$.  This function is not identically zero.
Indeed, if it were zero, then for every $x\notin D$, the form $H_x$
would vanish on each of two spanning lines.  The inequality
\[
 |H_x(v,w)|^2\leq H_x(v,v)H_x(w,w)
\]
would then force $H_x=0$.  Hence $q$ would vanish on the full-measure
complement of the inverse image of $D$ in $F$.  Nonnegativity and the
submean inequality on small total-space coordinate balls would give $q=0$
everywhere, a contradiction.

The cases $M=A,B$ of \eqref{eq:seminorm-graph-identities} give the
asserted invariances under the holomorphic lifts $A^{\otimes b}$ and
$B^{\otimes b}$.
\end{proof}

\subsection{Logarithmic weights and their measure}

The seminorm in Lemma~\ref{lem:ruling-reduction} may vanish in some
fibers. The next lemma shows that its logarithmic weights still define
a positive measure with the mass dictated by the negative line bundle.
This is the rank-one squared-norm/logarithmic-norm principle from
singular Hermitian metric theory
\cite[Section~3]{BerndtssonPaun2008} and
\cite[Definition~1.2 and Section~2]{Raufi2015}.
We prove local integrability at zeros before taking distributional
curvature, so all possible atoms remain in the mass calculation.

\begin{lemma}[Logarithmic curvature of a squared seminorm]
\label{lem:seminorm-curvature}
Let $L$ be a holomorphic line bundle on a connected Riemann surface $C$.
Suppose that $a_\bullet:L\to[0,\infty)$ is finite, plurisubharmonic, and
not identically zero, and that its restriction to every fiber is a
Hermitian positive semidefinite quadratic form.  In a nonvanishing
holomorphic frame $e$ on $U\subset C$, write
\[
 a_\bullet(z,w e(z))=u_e(z)|w|^2.
\]
Then $\log u_e$, with the value $-\infty$ allowed, is subharmonic and
locally integrable.  The local measures $\ddc\log u_e$ define a positive
Radon measure $\mu$ on $C$.

If a holomorphic bundle automorphism $\widetilde T:L\to L$ covers an
automorphism $T:C\to C$ and satisfies
$a_\bullet\circ\widetilde T=a_\bullet$, then $T^*\mu=\mu$.
If $C=\PP^1$ and $L=\OO_{\PP^1}(-b)$ with $b\geq1$, then
\begin{equation}
 \label{eq:seminorm-curvature-mass}
 \mu(\PP^1)=2\pi b.
\end{equation}
\end{lemma}

Where $u_e$ is smooth and strictly positive, $\ddc\log u_e$ is the
negative of the Chern curvature of the corresponding metric in our
convention.
This accounts for the positive mass when $L=\OO(-b)$.

\begin{proof}
Fix a coordinate disk $U$ and a frame $e$, and abbreviate $u_e$ to $u$.
The function $u$ is finite, nonnegative, subharmonic, and locally bounded,
by restriction to the holomorphic section $e$.  Choose a nonnegative
smooth radial function $\rho$ supported in the unit disk, with integral
one, and put
\[
 \rho_\delta(z)=\delta^{-2}\rho(z/\delta),
 \qquad u_\delta=u*\rho_\delta+\delta.
\]
These functions are defined on the interior of $U$ for sufficiently
small $\delta>0$.  Averaging the holomorphic base translations
$(z,w)\mapsto(z-\zeta,w)$ shows that
$(u*\rho_\delta)(z)|w|^2$ is plurisubharmonic.  Adding
$\delta|w|^2$ preserves this property, so $u_\delta>0$ and
$u_\delta(z)|w|^2$ is smooth and plurisubharmonic.  Its Levi matrix at
$w=1$ is
\[
 \begin{pmatrix}
  (u_\delta)_{z\bar z} & (u_\delta)_z\\
  (u_\delta)_{\bar z} & u_\delta
 \end{pmatrix}.
\]
Its nonnegative determinant gives
\[
 (\log u_\delta)_{z\bar z}
 =\frac{u_\delta(u_\delta)_{z\bar z}
             -|(u_\delta)_z|^2}{u_\delta^2}\geq0.
\]
Thus $f_\delta:=\log u_\delta$ is subharmonic.

The radial convolutions are nonnegative weighted averages of circular
means of $u$.  These means increase with their radii, and converge to
$u$ as the radii tend to zero.  Consequently
\[
 u_\delta\downarrow u,
 \qquad f_\delta\downarrow\log u
 \quad\text{as }\delta\downarrow0.
\]
We establish local integrability before taking distributional
derivatives of the limit.  Suppose that $u(z_0)>0$, and choose $R>0$
with $\overline{B(z_0,R)}\subset U$.  For all sufficiently small
$\delta$, local boundedness supplies a constant $C_0\geq0$ such that
$f_\delta\leq C_0$ on this disk.  Moreover
$f_\delta(z_0)\geq\log u(z_0)$.  Integrating the circular submean
inequality over the radius gives
\[
 \int_{B(z_0,R)} f_\delta\,dA
 \geq \pi R^2\log u(z_0).
\]
Writing $f^-:=\max\{-f,0\}$, we obtain the explicit bound
\begin{equation}
 \label{eq:log-negative-part}
 \int_{B(z_0,R)} f_\delta^-\,dA
 \leq \pi R^2\bigl(C_0-\log u(z_0)\bigr).
\end{equation}
The negative parts increase as $\delta$ decreases, whereas the positive
parts are bounded by $C_0$.  Monotone convergence therefore proves
$\log u\in L^1(B(z_0,R))$.  It also gives
$f_\delta\to\log u$ in $L^1$ on this disk.

This conclusion propagates throughout $C$.  Indeed, let $\Omega$
consist of the points near which a local logarithmic weight is
integrable.  A frame change $e'=g e$ gives
\begin{equation}
 \label{eq:log-frame-change}
 \log u_{e'}=\log u_e+\log|g|^2,
\end{equation}
so $\Omega$ is independent of the chosen frame.  It is open and
nonempty: nontriviality of $a_\bullet$ supplies a point with $u>0$,
to which \eqref{eq:log-negative-part} applies.  If $x$ is in the
boundary of $\Omega$, choose a coordinate disk containing
$\overline{B(x,4r)}$.  There is a small open disk contained in
$\Omega\cap B(x,r)$.  On that disk integrability of $\log u$ implies
$u>0$ almost everywhere, so it contains a point $z_0$ with $u(z_0)>0$.
The disk $B(z_0,2r)$ has closure in the original chart and contains a
neighborhood of $x$.  Applying \eqref{eq:log-negative-part} on this
larger disk proves $x\in\Omega$.  Thus $\Omega$ is also closed, and
connectedness gives $\Omega=C$.

On every relatively compact disk, $C_0-f_\delta$ increases to the
integrable function $C_0-\log u$.  Hence the local convergence is in
$L^1$, and the decreasing limit $\log u$ is subharmonic.  In particular,
$\ddc\log u$ is a positive distribution and thus a positive Radon
measure.  Since $g$ in \eqref{eq:log-frame-change} is holomorphic and
nowhere zero, $\log|g|^2$ is harmonic.  The local measures therefore
glue to $\mu$.

For the invariance assertion, choose frames $e$ on $U$ and $e'$ on $V$
with $T(U)\subset V$, and write
\[
 \widetilde T(e(z))=c(z)e'(Tz),
\]
where $c$ is holomorphic and nowhere zero.  The isometry condition gives
\[
 u_e(z)=|c(z)|^2u_{e'}(Tz),
 \qquad
 \log u_e=\log|c|^2+(\log u_{e'})\circ T.
\]
Taking distributional $\ddc$, which commutes with pullback by the
holomorphic diffeomorphism $T$, gives $\mu=T^*\mu$.  Here pullback is
that of a top-degree current: for a test function $\varphi$,
\begin{equation}
 \label{eq:measure-pullback}
 \int_C\varphi\,d(T^*\mu)
   =\int_C(\varphi\circ T^{-1})\,d\mu.
\end{equation}
Equivalently, $(T^*\mu)(E)=\mu(T(E))$ for Borel sets $E$.  This is
ordinary measure invariance under the bijection $T$; in particular,
$T^*\delta_p=\delta_{T^{-1}p}$ with unchanged atom mass.

Finally suppose $C=\PP^1$ and $L=\OO(-b)$.  Let $h_0$ be the smooth
Hermitian metric on $\OO(-b)$ induced by the standard norm on $\CC^2$.
In the tautological affine frame $e(z)=(1,z)^{\otimes b}$ its weight is
\[
 u_0(z)=h_0(e(z),e(z))=(1+|z|^2)^b.
\]
The local functions $\log u_e-\log h_0(e,e)$ define a single globally
integrable function $\Phi$ on $\PP^1$, by the common frame law.
If $\mu_0=\ddc\log h_0(e,e)$, then
\[
 \mu-\mu_0=\ddc\Phi,
 \qquad
 \int_{\PP^1}(\mu-\mu_0)
   =\langle\ddc\Phi,1\rangle=0.
\]
The measure $\mu$ is finite because it is locally finite on the compact
sphere.  With the convention
$\ddc f=i f_{z\bar z}\,dz\wedge d\bar z$, the smooth reference measure
has mass
\[
 \int_{\PP^1}\mu_0
 =\int_{\CC}\frac{b\,i\,dz\wedge d\bar z}{(1+|z|^2)^2}
 =4\pi b\int_0^\infty\frac{r\,dr}{(1+r^2)^2}
 =2\pi b.
\]
This proves \eqref{eq:seminorm-curvature-mass}.  Zeros of $u_e$ and any
atoms of $\mu$ are included in the distributional calculation; no
points have been excised.
\end{proof}

\subsection{The incompatible invariances}

We now use the special dynamics of the two matrices. Concentration
of invariant measures on proper projective subsets is a classical
mechanism; see \cite[Lemma~2 and the proof of Lemma~3]{Furstenberg1976}.
The following elementary argument gives precisely the rank-two
statement needed here.

\begin{lemma}
\label{lem:no-invariant-measure}
The projective transformations induced by the matrices $A$ and $B$ in
\eqref{eq:matrices} preserve no nonzero finite positive
Borel measure on $\PP^1$.
\end{lemma}

\begin{proof}
Suppose that $\mu$ is such a measure.  It is invariant under $T=AB$.
The two eigenvalues of $AB$ have distinct moduli, so after a projective
change of coordinate $T$ acts as $z\mapsto\lambda z$ with
$R:=|\lambda|>1$.  The Borel sets
\[
 C_n=\{z\in\CC^*:R^n\leq|z|<R^{n+1}\},\qquad n\in\ZZ,
\]
are pairwise disjoint, cover $\CC^*$, and satisfy $T(C_n)=C_{n+1}$.
Each has the same measure.  Finiteness and positivity imply that this
common measure is zero, and countable additivity gives
$\mu(\CC^*)=0$.  The half-open annuli assign every boundary circle to
exactly one set, so this argument applies equally to atomic and diffuse
measures and to nonreal $\lambda$.

Thus the nonempty support of $\mu$ consists of one or two of the fixed
points of $T$.  Invariance under $A$ and $B$ preserves this support as a
set: the inverse image of an open neighborhood of the image of a
support point has positive measure.  Both projective transformations
have order three.  Their induced permutations on a set of at most two
points are therefore trivial.  Each point of the support would be
fixed by both $A$ and $B$, and hence would be a common eigenline of the
two matrices. Lemma~\ref{lem:matrix-data} excludes such a line.
\end{proof}

The preceding results rule out exactly the object supplied by
Proposition~\ref{prop:limit}.

\begin{proposition}
\label{prop:obstruction}
For the bundle $G$ of Proposition~\ref{prop:construction}, every finite
nonnegative plurisubharmonic function on the total space of $F=G^\vee$
whose restriction to each fiber is a Hermitian quadratic form is
identically zero.
\end{proposition}

\begin{proof}
If such a function were nonzero, Lemma~\ref{lem:ruling-reduction}
would give a nonzero squared seminorm on $\OO_{\PP^1}(-b)$ invariant
under the lifts of $A$ and $B$. Lemma~\ref{lem:seminorm-curvature}
would then produce a finite positive measure on $\PP^1$ of mass
$2\pi b>0$, invariant under both projective transformations.
This contradicts Lemma~\ref{lem:no-invariant-measure}.
\end{proof}
==== END ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//05-obstruction.tex ====

==== BEGIN ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//06-conclusion.tex ====
\section{Proof of the main theorem}
\label{sec:conclusion}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
Take the algebraic rank-two bundle $G$ of
Proposition~\ref{prop:construction}. By
Proposition~\ref{prop:ampleness}, every $E_m=f_m^*G\otimes P$ is ample
in the quotient convention.

If the set of positive integers $m$ for which $E_m$ has a smooth
strictly Griffiths-positive metric were infinite, it would contain
a sequence tending to infinity. Proposition~\ref{prop:limit} would
then give a finite, nonnegative, nonzero psh function on the total
space of $G^\vee$ that is a Hermitian positive semidefinite quadratic
form on every fiber. Proposition~\ref{prop:obstruction} excludes
precisely such a function. Consequently that set of integers is
finite. Choose $m_0$ larger than all its elements, taking $m_0=1$ if
the set is empty. Every $m\ge m_0$ gives the required counterexample.
\end{proof}

The limit in the proof is permitted to lose rank, but not to vanish
identically or to lose its quadratic structure at any base point.
These are exactly the properties needed to transfer the
graph relations to a nonzero invariant measure. The argument requires no uniform estimates for the input metrics or
their curvature, and it does not determine an effective value of $m_0$.
==== END ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//06-conclusion.tex ====

==== BEGIN ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//paper.tex ====
\newcommand{\DoNotLoadEpstopdf}{}
\documentclass[11pt]{article}
\pdftrailerid{}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amssymb,amsthm,mathtools}
\usepackage{enumitem}
\usepackage{microtype}
\usepackage{xcolor}
\definecolor{linkblue}{RGB}{25,62,115}
\usepackage[colorlinks=true,linkcolor=linkblue,citecolor=linkblue,urlcolor=linkblue]{hyperref}
\hypersetup{pdftitle={Ample rank-two bundles on the quadric surface without Griffiths-positive metrics},pdfauthor={OpenAI}}
\newcommand{\PP}{\mathbb P}
\newcommand{\OO}{\mathcal O}
\newcommand{\CC}{\mathbb C}
\newcommand{\RR}{\mathbb R}
\newcommand{\ZZ}{\mathbb Z}
\newcommand{\NN}{\mathbb N}
\newcommand{\ddc}{i\partial\bar\partial}
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\numberwithin{equation}{section}
\setlist{itemsep=3pt,topsep=5pt}
\title{Ample rank-two bundles on the quadric surface\\
without Griffiths-positive metrics}
\author{OpenAI}
\date{September 24, 2026}
\begin{document}
\maketitle
\begin{abstract}
We give counterexamples to Griffiths' conjecture in rank two on the quadric
surface $\PP^1\times\PP^1$. We construct an explicit bundle $G$ whose
coordinatewise power pullbacks, tensored with $\OO(1,1)$, are ample for every
positive power, but admit no smooth strictly Griffiths-positive Hermitian
metric for all sufficiently large powers.
\end{abstract}
\tableofcontents
\input{sections/01-introduction}
\input{sections/02-construction}
\input{sections/03-ampleness}
\input{sections/04-limit}
\input{sections/05-obstruction}
\input{sections/06-conclusion}
\begingroup
\small
\setlength{\emergencystretch}{1em}
\bibliographystyle{abbrv}
\bibliography{references}
\endgroup
\end{document}
==== END ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//paper.tex ====

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==== END ample-rank-two-bundles-on-the-quadric-surface-without-griffiths-positive-metrics//references.bib ====