Ample rank-two bundles on the quadric surface without Griffiths-positive metrics

1. An explicit bundle and ampleness for all $m$

The bundle $G$ of Proposition 2.3 has two surjections $G\twoheadrightarrow L_i$. After pullback by $f_m$, the product of their kernels is $\mathcal O_S(18m,18m)$. Proposition 3.1 incorporates this positive product into a symmetric-power filtration and proves that $\mathcal O_{\mathbf P(E_m)}(n)$ is very ample whenever $n$ is even and $n-1\ge3m$. Every $m\ge1$ admits such an $n$, giving ampleness for all $m$.

Proposition 2.3 · p. 6

Proposition 3.1 · pp. 7–8

2. Infinitely many positive metrics produce one limit

If positive metrics exist for infinitely many $m$, choose a sequence with $m\to\infty$. Proposition 4.1 gives a finite, nonnegative, nonzero psh function $q$ on $F=G^\vee$, with $q(x,v)=H_x(v,v)$ and $H_x\ge0$ at every $x\in S$. Rank may drop, but the function cannot vanish identically or cease to be quadratic. The pointwise statement allows the graph identities on the divisor to be evaluated directly in $q$.

Proposition 4.1 · pp. 9–12

3. From the measure contradiction to finitely many exceptions

Proposition 5.4 says that such a $q$ must be zero for the constructed bundle. Hence the set of integers $m$ admitting a positive metric is finite. Choose $m_0$ larger than its maximum, or $m_0=1$ if it is empty. The proof excludes an infinite sequence of metrics rather than computing the curvature obstruction at a specified $m$.

Proposition 5.4 · p. 16

§6, Proof of Theorem 1.1 · p. 16

1. Finite orders provide compatibility at intersections

The orders of $A,B,A^{-1}B$ in $\mathrm{SL}_2(\mathbf C)$ are $6,6,8$. Each pair of graphs meets transversely at the two distinct eigenlines of the relative matrix. There are no triple points because $A$ and $B$ have no common eigenline. Thus $D$ has bidegree $(3,3)$ and six nodes. At an intersection, $Nv=\lambda Mv$ and $\lambda^{24}=1$, so $M^{\otimes24}$ and $N^{\otimes24}$ agree on the actual fiber. The nodal local-ring sequence

turns this agreement into gluing on the whole divisor.

Lemmas 2.1–2.2 · p. 5

2. The local basis also proves surjectivity

Let $d$ be a local equation of $D$ and let the unit $h$ lift the local coefficient of $\beta$. Then $e=(1,h)$ and $f=(0,d)$ form a basis of $G$, including at nodes. This basis proves that both projections are surjective and their kernels are $L_i(-D)$. Dualizing gives $\iota_2\alpha=\iota_1$, the precise relation that will identify the seminorms coming from the two rulings.

Proposition 2.3 · p. 6

1. Alternating quotients group the two negative directions

Although $K_1=\mathcal O_S(21m,-3m)$ and $K_2=\mathcal O_S(-3m,21m)$, their product is $K=\mathcal O_S(18m,18m)$. Alternating the sequences $0\to K_i\to V\to\widetilde L_j\to0$ gives, for even $n$, a filtration of $\operatorname{Sym}^nV$ with line factors

Each step uses one locally split quotient sequence, so it remains valid where the two kernels have the same fiber image. For $n=2$, the manuscript checks local bases to retain locally free quotients at nodes and on nonreduced pullbacks of the divisor.

§3, equations (3.1)–(3.5) · pp. 7–8

2. The twist $P^{n-1}$ makes all factors globally generated

The line factors of $W=\operatorname{Sym}^nV\otimes P^{n-1}$ have bidegrees

The inequality $n-1\ge3m$ offsets the sole remaining negative term $-3m$. A line bundle of nonnegative bidegree is globally generated and has $H^1=0$. The long exact sequence therefore lifts sections through successive quotients, proving that $W$ is globally generated. The cohomology calculation uses Čech comparison for the standard affine cover of $\mathbf P^1$ and the Künneth formula on the product.

§3, Global generation after twisting · p. 8

3. One final factor $P$ embeds the base

Evaluation and the relative Veronese give $\mathbf P(V)\hookrightarrow\mathbf P(H^0(S,W))\times S$. The hyperplane bundle from the first factor pulls back to $\mathcal O_{\mathbf P(V)}(n)\otimes\pi^*P^{n-1}$. Embed $S$ by $P$ and compose with Segre. The resulting hyperplane pullback is

Separating the roles of $P^{n-1}$ and the final $P$ explains the passage from global generation to very ampleness.

§3, The projective embedding and its line bundle · p. 8

1. Tracing over the power map removes the twisting transition

The squared dual norm is psh on the total space of $f_m^*F\otimes P^{-1}$. For a local frame $t$ of $P^{-1}$, use $f_m^*P^{-1}\simeq P^{-m}$ to define

The sum counts multiplicities, and different root choices differ by a unit-modulus constant, leaving the squared norm unchanged. Off the branch locus it is a sum of psh pullbacks by holomorphic inverse branches; across that locus it is continuous and locally bounded. The inverse image of the branch locus in the total space is closed and pluripolar, so the removable-singularities theorem applies. A frame change $t'=gt$ gives $q_{m,t'}=|g|^{2/m}q_{m,t}$, whose extra factor tends uniformly to one on compact subsets.

Proposition 4.1, equations (4.1)–(4.5) · pp. 9–10

[Dem12, Chapter I, Theorem 5.24, p. 45]

2. Convert a normalized maximum into nonvanishing of the limit

Using background metrics, normalize the global quadratic $N_m=q_{m,t}|t|_0^{-2/m}$ to have maximum one on the unit sphere bundle. Its local coefficient matrices $H_m$ are uniformly bounded; a subsequence converges distributionally to $H\in L^\infty_{\mathrm{loc}}$. Let maximizing points $(x_m,v_m)$ converge and replace the nearby fiber vectors by a fixed vector $c$. The submean inequality for the psh function $c^*H_mc$ gives a fixed nonnegative test function $\chi$ with

Passing this estimate to the distributional limit proves $H\not\equiv0$. This integral bound supplies the noncollapse that normalization of the maximum alone would not establish.

Proposition 4.1, equation (4.6) · p. 11

3. Recover a psh representative usable on the divisor

Convolve the coefficients in the base variables with a radial kernel, setting $q_\delta(x,v)=v^*(H*\rho_\delta)(x)v$. These are smooth psh functions and decrease as $\delta\downarrow0$. Their local bounds give a finite psh limit $q_U$. Polarization recovers all matrix entries from the limiting values at the finite list $e_j,e_j+e_k,e_j+ie_k$, proving quadraticity at every point. On overlaps the local psh representatives agree almost everywhere, and the submean inequality together with upper semicontinuity gives equality everywhere.

Proposition 4.1, equations (4.7)–(4.8); Remark 4.2 · pp. 11–12

[Dem12, Chapter I, Theorem 5.4, p. 39]

The representatives thus glue to a global $q$. The graph identities are evaluated in this single everywhere-defined function, rather than in a raw pointwise limit or a limit of restrictions to the divisor. Remark 4.2 explicitly makes no assertion of convergence of those restrictions.

Proposition 4.1, equations (4.7)–(4.8); Remark 4.2 · pp. 11–12

1. Combine two rulings and three graphs into one seminorm

Suppose $q\not\equiv0$. In $q\circ\iota_i$, hold the $i$th coordinate and the fiber vector fixed. The resulting finite subharmonic function on the complementary $\mathbf P^1$ is constant. Each ruling therefore yields a psh squared seminorm $a_i$ on $\mathcal O_{\mathbf P^1}(-b)$. The identity along $\Gamma_I$ gives $a_1=a_2=:a_\bullet$; those along $\Gamma_A,\Gamma_B$ give

If $a_\bullet=0$, the two spanning lines over $S\setminus D$ and Cauchy–Schwarz for positive semidefinite forms force $q=0$ there. The submean inequality extends the vanishing to the total space. Thus $a_\bullet$ is nonzero.

Lemma 5.1 · p. 13

2. Include the zeros when computing curvature mass

In a local frame write $a_\bullet(z,we(z))=u_e(z)|w|^2$. Total-space plurisubharmonicity gives $u_\delta(u_\delta)_{z\bar z}-|(u_\delta)_z|^2\ge0$ for a positive smoothing, so $\log u_\delta$ is subharmonic. Nontriviality and the submean inequality bound its negative part in $L^1$; the manuscript thereby establishes $\log u_e\in L^1_{\mathrm{loc}}$, including at zeros. Consequently

is a global Radon measure. Frame changes add harmonic terms, and the lifts of $A,B$ are isometries, so $A^*\mu=B^*\mu=\mu$. Subtracting the reference logarithmic weight of $(1+|z|^2)^b$ on $\mathcal O(-b)$ gives a globally integrable potential whose $i\partial\bar\partial$ has integral zero. Hence

Retaining zeros and atoms is essential to preserving the nonzero mass.

Lemma 5.2 · pp. 14–15

3. An expanding product is incompatible with order three

The eigenvalues of $T=AB$ have different moduli; in projective coordinates it acts as $z\mapsto\lambda z$ with $|\lambda|=R>1$. The disjoint half-open annuli $C_n=\{R^n\le|z|<R^{n+1}\}$ have equal measure, so finiteness forces each to have measure zero. The nonempty support of $\mu$ is therefore contained in the two fixed points of $T$. Both $A$ and $B$ preserve this support. Since their projective orders are three, their permutations of a set of at most two points are trivial. Every support point would be fixed by both transformations, contradicting the absence of a common eigenline in Lemma 2.1. This excludes $q\not\equiv0$.

Lemma 5.3 · p. 16