The projective-plane Beilinson monad for a framed torsion-free sheaf

Near a point of $L_{\mathrm{\infty }}$, let $R$ be the regular local surface ring, $t$ a local equation of $L_{\mathrm{\infty }}$, and $M$ the torsion-free stalk of $\mathcal{E}$. Multiplication by $t$ is injective on $M$. Lift a basis of $M\mathrm{/}tM\cong (R\mathrm{/}(t){)}^N$ to obtain a surjection $R^N\twoheadrightarrow M$ by Nakayama [F8]. If $K$ is its kernel, tensoring with $R\mathrm{/}(t)$ is exact on the left because $t$ is injective on $M$; the lifted basis makes $(R\mathrm{/}(t){)}^N\to M\mathrm{/}tM$ an isomorphism, so $K\mathrm{/}tK=0$. Nakayama [F8] gives $K=0$, proving local freeness of rank $N$ near the framed line.

Let $Q$ be the universal rank-two quotient bundle on the first factor. The section of $\mathcal{O}(1)$ on the second factor tensored with $Q$ on the first, induced by the second tautological line modulo the first, vanishes exactly on the diagonal. In affine coordinates transverse to the diagonal its two components are a regular parameter pair; hence its Koszul complex is exact and resolves ${\mathcal{O}}_{\mathrm{\Delta }}$. Since $\det Q=\mathcal{O}(1)$ and $Q^{\ast }={\mathrm{\Omega }}^1(1)$, the resolution is $0\to \mathcal{O}(-1)⊠\mathcal{O}(-2)\to {\mathrm{\Omega }}^1(1)⊠\mathcal{O}(-1)\to \mathcal{O}⊠\mathcal{O}\to {\mathcal{O}}_{\mathrm{\Delta }}\to 0$. The regularity check is local along the diagonal, and outside it one component is a unit; this verifies exactness at every point. This exactness also persists after tensoring by the pullback from the first factor of any coherent sheaf, including over a base scheme $S$. Indeed, near the relative diagonal choose the same affine chart on both factors over an affine base: if $M$ is its module on the first chart, its pullback is $M[y_1,y_2]$, and the diagonal components are $y_1-x_1,y_2-x_2$. Multiplication by $y_1-x_1$ is injective by its leading coefficient, and its quotient is $M[y_2]$; multiplication by $y_2-x_2$ is likewise injective, with quotient $M$. For these two commuting operators $a,b$, a relation $au+bv=0$ gives $v=aw$ after reduction modulo $a$, and then $u=-bw$ by injectivity of $a$; also $a$ injective makes the left Koszul map injective. This proves exactness of the tensored length-two complex. Away from the diagonal a section component is a unit, giving a contracting homotopy after every tensor operation. Thus no local freeness of the sheaf being tensored is required.

The framing and local freeness from step 1.1 give $\det (\mathcal{E}){\mathrm{\mid }}_{L_{\mathrm{\infty }}}\cong {\mathcal{O}}_{L_{\mathrm{\infty }}}$, so $\mathrm{deg}(\mathcal{E})=\mu (\mathcal{E})=0$. Let $\mathcal{F}\subset \mathcal{E}$ have positive rank and replace it by its saturation ${\mathcal{F}}^{\mathrm{s}\mathrm{a}\mathrm{t}}$. The induced nonzero morphism of determinant line bundles $\det (\mathcal{F})\to \det ({\mathcal{F}}^{\mathrm{s}\mathrm{a}\mathrm{t}})$ has an effective zero divisor: at every codimension-one point, the local ring is a discrete valuation ring, both modules are finite free of the same rank, and the determinant of their inclusion has nonnegative valuation. For an effective divisor $D$, write $D=m{L}_{\mathrm{\infty }}+D^{\mathrm{\prime }}$ with $m\ge 0$ and $D^{\mathrm{\prime }}$ effective without $L_{\mathrm{\infty }}$ as a component. The canonical section of $\mathcal{O}(D^{\mathrm{\prime }})$ restricts nontrivially to $L_{\mathrm{\infty }}$, so its restriction has nonnegative degree: by [F5], $H^0({\mathbb{P}}^1,\mathcal{O}(d))=0$ for $d<0$. The hyperplane transition functions give $\mathcal{O}(L_{\mathrm{\infty }}){\mathrm{\mid }}_{L_{\mathrm{\infty }}}={\mathcal{O}}_{{\mathbb{P}}^1}(1)$, of degree one. Tensor products add line-bundle degrees, hence $\mathrm{deg}(\mathcal{O}(D){\mathrm{\mid }}_{L_{\mathrm{\infty }}})=m+\mathrm{deg}(\mathcal{O}(D^{\mathrm{\prime }}){\mathrm{\mid }}_{L_{\mathrm{\infty }}})\ge 0$. Applying this to the determinant zero divisor gives $\mathrm{deg}(\mathcal{F})\le \mathrm{deg}({\mathcal{F}}^{\mathrm{s}\mathrm{a}\mathrm{t}})$. At the generic point of $L_{\mathrm{\infty }}$, use the chart $Z_1\mathrm{\ne }0$ with coordinates $x=Z_0\mathrm{/}{Z}_1$, $y=Z_2\mathrm{/}{Z}_1$; the local ring $R=\mathbb{C}[x,y{]}_{(x)}$ is a discrete valuation ring with uniformizer $x$. The modules ${\mathcal{E}}_R$ and ${\mathcal{F}}_R^{\mathrm{s}\mathrm{a}\mathrm{t}}$ are free, and ${\mathcal{E}}_R\mathrm{/}{\mathcal{F}}_R^{\mathrm{s}\mathrm{a}\mathrm{t}}$ is torsion-free by saturation, hence free by [F7]. Therefore the inclusion splits over $R$, and its reduction modulo $x$ is injective. Writing $r=\mathrm{rk}(\mathcal{F})$, the $r$th exterior power gives a generically nonzero map $\det ({\mathcal{F}}^{\mathrm{s}\mathrm{a}\mathrm{t}}){\mathrm{\mid }}_{L_{\mathrm{\infty }}}\to {\bigwedge }^r(\mathcal{E}{\mathrm{\mid }}_{L_{\mathrm{\infty }}})\cong {\mathcal{O}}_{L_{\mathrm{\infty }}}^{\left(\genfrac{}{}{0px}{}{N}{r}\right)}$. At least one component is a nonzero map to ${\mathcal{O}}_{L_{\mathrm{\infty }}}$. The determinant map is defined across the finite non-locally-free set by taking double duals; [F9] gives the codimension-one extension. Since a nonzero map ${\mathcal{O}}_{L_{\mathrm{\infty }}}(d)\to {\mathcal{O}}_{L_{\mathrm{\infty }}}$ requires $d\le 0$, we have $\mathrm{deg}({\mathcal{F}}^{\mathrm{s}\mathrm{a}\mathrm{t}})\le 0$ and thus $\mathrm{deg}(\mathcal{F})\le 0$. This proves the slope inequality for every positive-rank subsheaf, hence slope-semistability of $\mathcal{E}$ with slope zero. For any torsion-free inclusion $\mathcal{Q}\subseteq \mathcal{V}$ of equal positive rank, the determinant inclusion has an effective zero divisor by the same discrete-valuation calculation, so $\mathrm{deg}(\mathcal{Q})\le \mathrm{deg}(\mathcal{V})$. No such bound for unequal ranks and an arbitrary target is asserted.

A nonzero section of $\mathcal{E}(-d)$ for $d=1,2$ is injective as a map $\mathcal{O}(d)\to \mathcal{E}$, because $\mathcal{E}$ is torsion-free on the integral surface. Its rank-one image has determinant degree at least $d>0$ by the saturation inequality in step 2.1, a contradiction. Hence $H^0(\mathcal{E}(-1))=H^0(\mathcal{E}(-2))=0$. A nonzero map $\mathcal{E}\to \mathcal{O}(-d)$ for $d=1,2$ has torsion-free rank-one image $Q$. If $N>1$, its kernel has rank $N-1$ and degree at most zero by step 2.1, so determinant additivity in the exact sequence with torsion-free quotient gives $\mathrm{deg}Q\ge \mathrm{deg}\mathcal{E}=0$; if $N=1$, torsion-freeness forces the kernel to vanish and $\mathrm{deg}Q=0$. But $Q\subseteq \mathcal{O}(-d)$ has degree at most $-d$ by the same saturation inequality, a contradiction. By coherent Serre duality [F4], these Hom vanishings give $H^2(\mathcal{E}(-1))=H^2(\mathcal{E}(-2))=0$.

The Hilbert polynomial gives $\chi (\mathcal{E}(-1))=\chi (\mathcal{E}(-2))=-k$. With the vanishings of step 3.1, $h^1(\mathcal{E}(-1))=h^1(\mathcal{E}(-2))=k$. The Euler sequence is locally split, so tensoring it by $\mathcal{E}$ remains exact; its first map therefore injects $H^0(\mathcal{E}\otimes {\mathrm{\Omega }}^1)$ into $H^0(\mathcal{E}(-1){)}^3$, which is zero by step 3.1.

Serre duality identifies $H^2(\mathcal{E}\otimes {\mathrm{\Omega }}^1{)}^{\ast }$ with $\mathrm{Hom}(\mathcal{E},T_{{\mathbb{P}}^2}(-3))$. The twisted Euler sequence is $0\to \mathcal{O}(-3-d)\to \mathcal{O}(-2-d{)}^3\to T_{{\mathbb{P}}^2}(-3-d)\to 0$. For $d\ge -1$, both $H^0$ terms on the left vanish and $H^1(\mathcal{O}(-3-d))=0$ by [F5], hence $H^0(T_{{\mathbb{P}}^2}(-3-d))=0$. Any rank-one torsion-free subsheaf of $T_{{\mathbb{P}}^2}(-3)$ of degree at least $-1$ would extend by double dual to a nonzero map $\mathcal{O}(d)\to T_{{\mathbb{P}}^2}(-3)$ for some $d\ge -1$, a contradiction. Taking determinants in the Euler sequence gives $\mathrm{deg}{T}_{{\mathbb{P}}^2}(-3)=3(-2)-(-3)=-3$; its rank is two and slope is $-3\mathrm{/}2$. Integrality of degree makes each rank-one subsheaf have degree at most $-2$, hence smaller slope; this proves stability, meaning strict slope inequality for every proper positive-rank subsheaf. If a nonzero map $\mathcal{E}\to T_{{\mathbb{P}}^2}(-3)$ has image $Q$ of rank $r=1$ or $2$, its kernel is torsion-free, has degree at most zero by step 2.1 when nonzero, and has degree zero when zero. Determinant additivity gives $\mathrm{deg}Q\ge 0$. For $r=1$ the preceding rank-one bound gives $\mathrm{deg}Q\le -2$; for $r=2$ the equal-rank determinant bound in step 2.1 gives $\mathrm{deg}Q\le -3$. Both contradict $\mathrm{deg}Q\ge 0$. Consequently $H^2(\mathcal{E}\otimes {\mathrm{\Omega }}^1)=0$. From the Euler sequence and the Hilbert polynomial, $\chi (\mathcal{E}\otimes {\mathrm{\Omega }}^1)=3\chi (\mathcal{E}(-1))-\chi (\mathcal{E})=-(2k+N)$, so $h^1(\mathcal{E}\otimes {\mathrm{\Omega }}^1)=2k+N$.

Using the tensor exactness proved in step 1.2, tensor this resolution on its first factor by $\mathcal{E}(-1)$ and take derived pushforward to the second factor. By [F10], the ordered Čech complex of the three standard affine charts computes the coherent-sheaf cohomology here: ${\mathbb{P}}^2$ is quasi-compact and separated, the charts and all their intersections are affine, and the sheaves are quasi-coherent. Choose a Cartan–Eilenberg resolution of the resulting bounded complex. DC, obtained from AC by [F3], supplies compatible resolution comparisons. Filtering this one total complex in the two orders gives the first and second spectral sequences [F3]; the identity comparison of the total complex identifies their abutment and canonical edge maps. Thus no arbitrary splitting is used. By steps 3.1 and 5.1, all rows except $q=1$ vanish. The surviving row is $H^1(\mathcal{E}(-2))\mathcal{O}(-2)\to H^1(\mathcal{E}\otimes {\mathrm{\Omega }}^1)\mathcal{O}(-1)\to H^1(\mathcal{E}(-1))\mathcal{O}$. Its only cohomology is $\mathcal{E}(-1)$ in the middle because the diagonal resolution transforms $\mathcal{E}(-1)$ to itself. Twisting by $\mathcal{O}(1)$ gives the displayed monad. The dimensions are exactly those from steps 4.1 and 5.1.

Let $S,\mathcal{E},\mathrm{\Phi },{\mathcal{G}}_i$ satisfy the family hypotheses in the Statement. Work locally on affine open subsets of $S$ and, for the diagonal pushforward, on affine opens of its target. On each such open, [F10] supplies the standard three-chart Čech computation. Restrictions commute with its differentials and the diagonal maps, so these local constructions agree on overlaps and glue. Tensor the relative diagonal Koszul resolution with this $S$-flat framed family; step 1.2 proves exactness even for non-locally-free fibers and nonreduced bases. For each closed point $s$, the specified maps ${\beta }_{i,s}^q$ identify the fibers of the direct-image terms with the cohomology groups of the framed fiber ${\mathcal{E}}_s$; no base-change outside those maps is used [A1]. The direct-image terms are vector bundles of ranks $k,2k+N,k$, and the finite filtered complex commutes with residue-field restriction. On every closed fiber, step 6.1 gives the monad with framing ${\mathrm{\Phi }}_s$. The right end map is surjective on every fiber, hence surjective by Nakayama [F8], and its kernel $K$ is locally free. Locally on $S$, write the remaining exact sequence as $0\to P^{-1}\to K\to \mathcal{E}\to 0$, with $P^{-1}$ and $K$ flat over the base. Tensoring by any residue field $k(s)$ stays injective at the left because the fiber complex is a monad on the whole projective plane. Thus for every $y\in {\mathbb{P}}^2\times \{s\}$, ${\mathrm{Tor}}_1^{{\mathcal{O}}_{S,s}}(\kappa (s),{\mathcal{E}}_y)=0$, equivalently the maximal-ideal multiplication map on the finite local module ${\mathcal{E}}_y$ is injective. Since $\mathcal{E}$ is finite over the local ring of ${\mathbb{P}}^2\times S$, [F6] proves flatness over $S$. The same base-change maps identify the relative middle cohomology with the given family. This is precisely the family range stated and does not assert base change outside it.