Elliptic theta, coherent classification and all-arity transversal comparison

Generic collections and finite cofinal refinement. Use only finite transversal words with no triple concurrence. Subwords, cyclic rotation and common tensoring preserve the required data; every new auxiliary point can avoid the finitely many concurrence equations as well as the forbidden positions in the arity/weight proof. Here is the cofinality construction without borrowing a geometric equivalence from the completion lemma. For a semistable bundle E, choose disjoint effective divisors D1,D2 of distinct sufficiently large positive degrees. The maps $a=(s_{D2},-s_{D1}),b=(s_{D1},s_{D2})$ give the actual exact sequence $0\to E(-D1-D2)\to E(-D1)$ plus $E(-D2)\to E\to 0$: locally one divisor section is a unit, and tensoring by the locally free E preserves the kernel calculation. Choose the divisor degrees so the three slopes, and all slopes in a finite list of such resolutions, avoid the finitely many previously used slopes. Moving one point of a positive divisor varies its Picard parameter surjectively; this makes the finitely many remaining position and triple-concurrence conditions avoidable while keeping supports disjoint. Their nilpotent local-system parts are unchanged. For rational bundle classes carry this choice through the finite isogeny classification proved below; on a common cover there are only finitely many additional lift-concurrence conditions. Direct sums and finite upper-triangular complexes use fresh choices for each occurrence. To index all finite choices, including an uncountable small object set, distinguish a chart's finite HISTORY record from its ACTIVE fresh resolution terms. The history records its finite target list and all terms of finitely many downward-closed ancestor charts; it need not be a generic collection. The active terms are chosen mutually generic and generic against every recorded ancestor term. A chart lies above its declared ancestors, and its history contains their histories and active terms. Two charts have a common upper chart: take the finite union of their histories, active terms and ancestry, and choose entirely new active resolutions avoiding that finite record. Incomparable old active terms may fail to be jointly generic; they remain history, and are never asserted jointly generic or used in a composable forward word. Along any composable ancestry chain, however, each later active collection avoids all earlier ones, so the active terms in that word form an actual generic collection. For any prescribed finite target list, the subposet of charts whose histories contain that list is cofinal, since every chart can be enlarged in this way. This is precisely the forward-chain finite refinement property needed for the completion's directed resolution category, rather than the false assertion that unions of arbitrary generic collections are generic. This proves the stated finite generic resolution/refinement property. It is a coherent exact-resolution construction; actual geometric quasi-isomorphisms and directed localization are proved separately in the completion, not assumed here.

Degree-r theta basis and absolute convergence. Fix τ in the upper half-plane, $q=\exp (2\pi i\tau )$, and $E=C\mathrm{/}(Z+\tau Z)$. For $r>0$ and a in Z/rZ define ${\mathrm{\Theta }}_a^r(z)=\sum$ over u in a+rZ $\exp (\pi i\tau u^2\mathrm{/}r+2\pi iuz)$. On every compact z-set, the term is bounded by $\exp (-c{u}^2+C\mathrm{\mid }u\mathrm{\mid })$, with $c=\pi \mathrm{Im}(\tau )\mathrm{/}r>0$. The series and every derivative converge uniformly there. Its transformations are ${\mathrm{\Theta }}_a^r(z+1)={\mathrm{\Theta }}_a^r(z),{\mathrm{\Theta }}_a^r(z+\tau )=\exp (-\pi ir\tau -2\pi irz){\mathrm{\Theta }}_a^r(z)$. Thus these are holomorphic sections of the degree-r factor-of-automorphy line bundle $L^r$. They are linearly independent by their disjoint Fourier congruence classes. They span: any 1-periodic entire section has a Laurent/Fourier expansion on C*, and the second transformation gives the coefficient recurrence $c_{u+r}=q^{u+r\mathrm{/}2}{c}_u$. Exactly r initial coefficients determine the expansion. Its solutions are precisely the above theta series. Translation $z\mapsto z+x$ gives the basis for the translated degree-r factor. The same bounds justify all regroupings, differentiation and finite nilpotent substitutions below. Here a representative a can be changed by r without changing Theta. This convention avoids the ambiguity of writing a characteristic without its degree.

Complete addition identity by integer reindexing. For positive r,s, a mod r, b mod s, put $t=r+s$ and $k_j=r$ b-s a+r $sj,Z=(r{z}_1+s{z}_2)\mathrm{/}t$. Then ${\mathrm{\Theta }}_a^r(z_1){\mathrm{\Theta }}_b^s(z_2)=$ sum over j in Z $\exp (\pi i\tau k_j^2\mathrm{/}(rst)+2\pi i{k}_j(z_2-z_1)\mathrm{/}t){\mathrm{\Theta }}_{a+b+sj}^t(Z)$. (T) Proof. Write $u=a+rl$ and $v=b+sm$ in the product double series. Set $j=m-l,w=u+v$ and $k=r$ v-s u. This is a bijection between (l,m) in $Z^2$ and (l,j) in $Z^2$, and gives $k=k_j,w=a+b+s$ j+t $l,u^2\mathrm{/}r+v^2\mathrm{/}s=w^2\mathrm{/}t+k^2\mathrm{/}(rst),u{z}_1+v{z}_2=wZ+k(z_2-z_1)\mathrm{/}t$. For fixed j the sum over l is exactly the theta function on the right. Absolute convergence of the original Gaussian double series permits this regrouping. The reorganized series is likewise normally convergent on compact sets: the underlying positive quadratic form is the same. This proves (T) without importing an addition theorem. If the $z_i$ contain commuting nilpotent endomorphisms, interpret each exponential and theta function by its finite Taylor expansion in those endomorphisms. The differentiated Gaussian bounds are polynomial times a decaying Gaussian; normal convergence still holds. Therefore (T) remains true after every such substitution. This is the version needed for nonsemisimple local systems.

Unipotent factors: actual spaces of sections. Let N be nilpotent on a finite-dimensional complex V. Let $F_N$ be the bundle whose sections on C* satisfy $w(qu)=\exp (N)w(u)$, with $u=\exp (2\pi iz)$. For a positive-degree scalar factor $\phi (u)=\text{constant}$ times $u^{-r}$, put $D=-u$ d/du. The operator $V_{\phi ,N}(f\otimes v)=\exp (DN\mathrm{/}r)fv$ is a bijection $H^0(L_{\phi })$ tensor $V\to H^0(L_{\phi }\otimes F_N)$. Nilpotence makes this a finite differential operator. Since $D(\phi f)=\phi (D+r)f$, its image satisfies exactly the factor φ exp(N). The inverse exp(-DN/r) removes that factor by the same calculation. Hence covariance AND invertibility are proved, rather than inferring a basis from covariance alone. For two bundles $B_i=L_{{\phi }_i}$ tensor $F_{N_i}$, the Hom bundle has the nilpotent operator $M_{12}(T)=N_2$ T-T $N_1$ on $\mathrm{Hom}(V_1,V_2)$. When $r=\mathrm{deg}{\phi }_2-\mathrm{deg}{\phi }_1>0$, the preceding construction and theta basis give a basis of $\mathrm{Hom}(B_1,B_2)$. If the degree difference is negative, Hom vanishes by negative-degree vanishing, and Serre duality identifies ${\mathrm{Ext}}^1$ with the dual of the reverse Hom. These are the actual smooth elliptic-curve duality/degree interfaces, not an HMS result. For equal degree the underlying line factors may differ. If their quotient is nontrivial in $Pi{c}^0(E)$, both Hom and ${\mathrm{Ext}}^1$ vanish. One can see this directly in the factor model: a 1-periodic matrix function has Fourier terms $T_k$ exp(2πikz), and covariance forces $\exp (2\pi ik\tau )T_k=\lambda \exp (M_{12})T_k$. For real unitary holonomy parameter λ, these scalars coincide only for $k=0$ and $\lambda =1$. Nilpotence cannot change that eigenvalue test. In the coincident-factor case only $k=0$ survives, giving $\mathrm{Hom}(B_1,B_2)=\ker M_{12},{\mathrm{Ext}}^1(B_1,B_2)=\mathrm{coker}{M}_{12}$. The second formula follows by the same Fourier resolution of the two-period cocycle, or by elliptic Serre duality and the trace identity $tr((N_{2}T-T{N}_1)S)=tr(T(S{N}_2-N_{1}S))$. These are intertwiners and coinvariants, NOT arbitrary vector-space maps when the nilpotent factors are nonzero.

Degree-zero lines and the normalized kernel. Every degree-zero line bundle L is O(p−0) for a unique $p\in E$. Indeed L(0) has degree one, $H^1=0$ and $H^0$ dimension one by Riemann–Roch/duality. Its unique nonzero section has one simple zero p. If O(p)≅O(p'), their one-dimensional section spaces give the same zero divisor, so $p=p^{\mathrm{\prime }}$. This also shows the family O(Δ−E×0−0×E) parametrizes the degree-zero lines, normalized at the two axes (choose the harmless constant fibre trivialization at their intersection). Its group law agrees with addition on E. In the analytic factor description, degree-one theta has exactly one zero, at (1+τ)/2 before translating the origin: pairing its terms indexed by m and −m−1 proves that zero, and its divisor has degree one. Translation gives the factor description of O(p−0). Degree-zero factors multiply by adding p; the factors differing by lattice points are changed by the explicit exponential gauge. Thus O(p−0) tensor $O(p^{\mathrm{\prime }}-0)=O(p+p^{\mathrm{\prime }}-0)$. This calculation supplies the group assertion; it is not inferred merely from a bijection of points. Denote the normalized Poincaré kernel by P on E×E; either sign convention for p is allowed if retained throughout. Its restrictions $P_y$ are trivial exactly for $y=0$. Every nontrivial degree-zero line has $H^0=0$ (a section of degree zero would trivialize it) and $H^1=0$ by Serre duality. For $O_E$ both dimensions are one. For later use, the following normalized see-saw statement is sufficient, and has a direct proof. A line bundle on E×B, B smooth reduced, trivial on every E-fibre and rigidified along 0×B is trivial. Its universal cohomology complex has fibre ranks $H^0=H^1=1$ everywhere. Constant-rank matrix minors over reduced B give locally free $H^0$ with base change; evaluation is an isomorphism on every fibre and hence everywhere by Nakayama. The rigidification trivializes the $H^0$ line. Apply this with $B=E^2$ to compare the normalized line families P(x,y) tensor P(y,z) and P(y,x+z). Their fibre isomorphisms agree at $y=0$, and their quotient is therefore trivial. This proves the kernel biadditivity needed below.

The vector-bundle cohomology interfaces used in classification. The named Riemann–Roch and Serre suppliers state their interfaces for line bundles, so we derive the required vector-bundle versions here. A vector bundle $B$ on the smooth integral curve admits a saturated line subbundle: choose a nonzero vector in its generic fibre, clear its finitely many poles to obtain a nonzero map from a line bundle, and saturate its image. Over every local DVR a saturated rank-one submodule of a free module is free and its quotient is torsion-free and free. Repeating on the quotient yields a finite filtration by line bundles; it is not a claim that this filtration splits. Determinants and Euler characteristic are additive along this filtration, so line Riemann–Roch gives $\chi (B)=\mathrm{deg}(B)$ on $E$. The same filtration establishes finiteness of its cohomology. The curve has a two-affine cover: embed it as the plane cubic, choose two lines whose intersection lies off the cubic, and take their affine complements on the curve. Their intersection is affine by separatedness. The actual finite affine Čech supplier therefore gives $H^j(B)=0$ for $j>1$, for every coherent $B$, and computes the exact cohomology sequences used here. For a short exact sequence of vector bundles, dualizing and tensoring by ${\omega }_E$ is exact. Cup product followed by the fixed line-bundle trace pairs its cohomology sequence with the dual sequence in reverse order. The connecting maps are adjoint with the usual sign: on the two-open Čech complex, applying the differential to a product gives this identity directly by the Leibniz rule. Start with the perfect line-bundle pairings and induct up the filtration; exactness and the five lemma give perfect pairings $H^1(B)\times H^0(B^{\ast }\otimes {\omega }_E)\to \mathbb{C}$ and $H^0(B)\times H^1(B^{\ast }\otimes {\omega }_E)\to \mathbb{C}$. Here the canonical bundle is trivialized by $\eta =-2\pi idz$, while $z$ remains the coordinate used for theta functions. To reconcile this with the fixed trace of [F1], let $hdz$ be a principal part and take a cutoff $\chi$ equal to one near its pole. Its positive connecting class is represented by $dz\otimes \bar{\mathrm{\partial }}(\chi h)$, and Stokes on the punctured cutoff disk gives $\int dz\wedge \bar{\mathrm{\partial }}(\chi h)=2\pi i\mathrm{res}(hdz)$. Since [F1] fixes trace to negative residue, its Dolbeault realization is $t_E([dz\otimes \beta ])=-(2\pi i{)}^{-1}{\int }_{E}dz\wedge \beta$. Consequently $t_E([\eta \otimes \beta ])={\int }_{E}dz\wedge \beta$. This is a choice of canonical-bundle trivialization, not a rescaling of the fixed algebraic trace. For bundles $A,B$, local freeness makes the sheaf-Hom functor $\mathcal{H}om(A,-)=A^{\ast }\otimes (-)$ exact; applying the finite Čech resolutions gives ${\mathrm{Ext}}^j(A,B)=H^j(A^{\ast }\otimes B)$. Thus ${\mathrm{Ext}}^1(A,B)=\mathrm{Hom}(B,A{)}^{\ast }$ with its trace pairing and ${\mathrm{Ext}}^{j>1}(A,B)=0$. This proves exactly the vector duality, Euler and extension interfaces used below, without silently enlarging the line-bundle supplier Statements.

The actual derived pushforward $R{p}_{\ast }P$. Let $p:E\times E\to E$ project to the parameter. Outside 0 all fibre cohomology vanishes, so the universal perfect complex is acyclic there. Near 0 put that finite-free complex in minimal form by splitting each matrix entry which is a unit. The fibre at 0 has $H^0$ and $H^1$ both one and no other cohomology; hence its remaining model is $[O_U--h-\to O_U]$ in degrees 0 and 1. The function h vanishes at 0 and is nonzero away from 0. We prove its zero is SIMPLE. Restrict the parameter to the dual numbers at 0. The obstruction to lifting the unit section of $O_E$ is precisely the first-order deformation class of P in $H^1(E,O_E)$: this is the connector of $0\to O_E\to P_{\mathrm{d}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{s}}\to O_E\to 0$, computed by differences of any local lifts of 1. In a uniformizer z at 0 the moving-point divisor has transition $(z-\epsilon )\mathrm{/}z=1-\epsilon \mathrm{/}z$. Thus that class is represented by ±1/z on a punctured neighbourhood. Its Serre-residue pairing with the nonvanishing differential is ±ω(0), nonzero. Therefore the connector is nonzero, equivalently h has nonzero linear term. Thus h is a uniformizer times a unit and $R{p}_{\ast }P=O_0[-1]$. (F0) This argument computes the DERIVED pushforward: its derived fibre at 0 has two cohomology groups. It does not wrongly identify its ordinary $H^0$ sheaf with the fibre $H^0$.

Kernel convolution proves an autoequivalence. Define $S(F)=R{p}_{2\ast }(p_1^{\ast }F\otimes P)$ on Db(Coh E). Every bounded coherent complex is perfect: for a coherent sheaf a sufficiently negative finite free twist surjects onto it, and its kernel is torsion-free on the smooth curve, hence locally free; then take finite complexes. Derived tensors are therefore represented by finite vector-bundle complexes. The required tensor/projection and base-change identities here can be checked on those complexes with a finite affine cover of E. Over each affine parameter open, the relative Čech terms are sections on affine products; these are flat over the parameter for a line bundle. Their bounded Čech direction is K-flat. Tensoring a perfect complex, changing the parameter by a flat morphism, and taking two successive pushforwards give the same finite Čech double complex, with the same restrictions and signs. Consequently convolution of the two kernels represents $S^2$; no general integral-transform theorem is imported. By steps 5.1 and 6.2 the convolution kernel is the pullback, under the addition map $a:E^2\to E$, of $R{p}_{\ast }P$. Addition is smooth, hence flat, so (F0) gives $K_{S^2}=O_{a^{-1}(0)}[-1]=O_{graph(-1)}[-1]$. The normalization along the axes removes any extra line factor; an overall constant one-dimensional factor can equivalently be fixed by its unit at 0. Thus $S^2=(-1_E{)}^{\ast }[-1]$. The right side is an autoequivalence. It commutes with S, as is also clear on the kernel; composing S with its square's inverse supplies a two-sided inverse. Hence S is an ACTUAL autoequivalence of Db(Coh E). We will use two direct values: $S(O_E)=O_0[-1]$ from (F0), and $S(O_0)=O_E$ from restricting P to 0×E. These determine ranks/degrees below without a GRR theorem.

Harder–Narasimhan existence and splitting on E. For completeness fix a positive-degree line H. Eventual generation of $B^{\ast }$ tensor $H^N$ gives an injection $B\to (H^N{)}^a$. Taking exterior powers bounds the degree of every saturated rank-r subbundle above by rN deg H: a nonzero component of its determinant map is a section of $(detA{)}^{\ast }$ tensor $H^{rN}$. For the finitely many possible ranks, degrees are integers bounded above. Thus a maximal slope subbundle exists; among those choose largest rank. Saturating can only increase degree, so this choice is saturated and its quotient is a bundle. The degree/rank identities for intersections and sums show this maximal object is unique: the sum of two maximal-slope subbundles still has that slope, unless one obtains a subbundle of larger slope, which is impossible; largest rank therefore contains both. Every subbundle of its quotient has smaller slope, or its inverse image would contradict maximal slope or maximal rank. Induct on the quotient rank. This constructs the finite unique HN filtration, with semistable factors of strictly decreasing slope. Semistability also gives $\mathrm{Hom}(A,B)=0$ when $\mu (A)>\mu (B)$: the image's saturation is a quotient of A and a subbundle of B, which would have slope at least μ(A) and at most μ(B). On $E,{\omega }_E=O_E$. Therefore ${\mathrm{Ext}}^1(B_{\mathrm{l}\mathrm{o}\mathrm{w}},A_{\mathrm{h}\mathrm{i}\mathrm{g}\mathrm{h}})=\mathrm{Hom}(A_{\mathrm{h}\mathrm{i}\mathrm{g}\mathrm{h}},B_{\mathrm{l}\mathrm{o}\mathrm{w}}{)}^{\ast }=0$ for the HN factors. Induction splits the HN filtration. In particular EVERY indecomposable bundle is semistable. This splitting is special to genus one and is proved here, not presumed for arbitrary curves.

Degree-zero indecomposables. Let B be semistable of degree zero. If $H^0(B\otimes P_y)=0$ for all y, duality and $\chi =0$ make $H^1$ vanish for all y too. The universal perfect complex for S(B) would then have zero derived fibre everywhere and hence be acyclic, by minimal free complexes and Nakayama. This contradicts the autoequivalence unless $B=0$. Thus some B tensor $P_y$ has a nonzero section. Its saturation is a line O(D) of nonnegative degree. Semistability forces deg $D=0$, so this is a degree-zero line subbundle. The quotient remains semistable of degree zero. Inducting on rank gives a filtration by degree-zero lines. Different degree-zero lines have $\mathrm{Hom}={\mathrm{Ext}}^1=0$. Induction on such filtrations then gives the same vanishing between any two objects whose line factors are distinct. Hence an indecomposable has every line factor equal to one L; it is L tensor U with U filtered by $O_E$. S sends every such U to T[-1], where T is a finite-length sheaf supported at 0: the assertion holds for $O_E$ by (F0), and an extension gives the same assertion by the long exact cohomology sequence of its transformed triangle. Conversely $S^{-1}(T[-1])$ is a bundle filtered by $O_E$, by induction on the length of T. This is an equivalence between these extension categories. Finite-length modules over the smooth local DVR at 0 decompose into $O\mathrm{/}(z^k)$: complete at 0, use the nilpotent multiplication operator z on the finite-dimensional C-vector space, put it in Jordan form, and note that completion changes no finite-length module. Thus there is exactly ONE indecomposable $U_k$ of each rank k in the unipotent category. Its explicit analytic factor is $\exp (J_k)$, with $J_k$ a single nilpotent Jordan block. That factor has a filtration by $O_E$; its endomorphisms are exactly the constant matrices commuting with $J_k$ (the equal-slope Fourier calculation in the theta construction above), namely $C[J_k]$. This is a local ring and has no nontrivial idempotent, so that bundle is indecomposable and is $U_k$. In particular $\dim \mathrm{End}(U_k)=k$.

Euclidean reduction and coprime stability. If B is indecomposable with positive degree d, then $H^1(B\otimes P_y)=0$ for every y by semistability and duality. Riemann–Roch and the actual cohomology/base-change criterion make S(B) a vector bundle of rank d. It is indecomposable because S is an equivalence, hence semistable by step 8.1. Its degree is -rank B. To see this without GRR, use Euler-pairing preservation and $S(O_0)=O_E$: deg $S(B)=\chi (O_E,S(B))=\chi (O_0,B)=-\mathrm{rank}B$. The last equality follows from the local length-one resolution of $O_0:\mathrm{Hom}(O_0,B)=0$ and ${\mathrm{Ext}}^1(O_0,B)$ has dimension rank B. Twisting by a line of degree one sends (r,d) to (r,d+r), and S sends a positive-degree indecomposable bundle to one with (d,-r). Normalize $0\le d<r$ by twists; if $d>0$, apply S and normalize again. The rank strictly decreases. This Euclidean process ends with an indecomposable degree-zero bundle of rank $h=\gcd (r,d)$, up to a uniform shift. Its form is L tensor $U_h$ by step 9.1. All transformations are exact derived autoequivalences. A stable bundle is simple: a nonzero endomorphism is invertible by the slope argument, and subtracting an eigenvalue (the characteristic coefficients are global functions and hence constants) makes any nonscalar endomorphism noninvertible. Thus $\mathrm{End}=C$. The Euclidean process and step 9.1 then force $h=1$ for a stable bundle. Conversely a semistable bundle with coprime rank and degree is stable, because a proper same-slope subbundle would have rank a nontrivial multiple of the denominator of d/r. Thus every stable bundle has coprime rank and degree, and the transforms of an indecomposable preserve the same primitive slope data.

Explicit finite-isogeny form. Let ${\pi }_r:E_{r\tau }\to E_{\tau }$ be the degree-r cyclic étale isogeny; its deck translations are kτ, k mod r. Finite étale trace identifies the dual of ${\pi }_{\ast }O$ with itself. Affine finite base change and its trace pairing therefore give BOTH adjunctions and ${\pi }^{\ast }{\pi }_{\ast }T=⨁$ over deck g of $g^{\ast }T$. A finite étale pullback preserves semistability here: its unique maximal destabilizing HN subbundle, if any, is invariant under the deck action, so its submodule descends by the explicit finite faithfully flat equalizer to a destabilizing subbundle downstairs. Conversely ${\pi }_{\ast }L$ is semistable for a line L upstairs: its pullback is a direct sum of same-degree lines, hence semistable, and a downstairs destabilizer would pull back. Finite pushforward preserves χ, so deg ${\pi }_{\ast }L=\mathrm{deg}L$ on genus-one curves. Let S be stable of rank r and degree d, necessarily coprime. The bundle ${\pi }_r^{\ast }S$ has integral slope d and is semistable. Tensor it by any line of degree -d; step 9.1 supplies a degree-zero line QUOTIENT (apply the filtration, or dualize it). Thus ${\pi }_r^{\ast }S$ has a line quotient L of degree d. Adjunction gives a nonzero map $S\to {\pi }_{r\ast }L$. Its target is semistable of rank r and degree d, hence stable by coprimality. A nonzero map between stable objects of the same slope and rank is an isomorphism. Therefore $S={\pi }_{r\ast }L$. For a general indecomposable B with slope d/r in lowest terms, choose its stable Jordan–Hölder factors among semistable bundles of that slope. Their existence follows by repeatedly taking a minimal-rank same-slope subbundle until stable; quotients stay semistable and rank decreases. Their stable ranks/degrees are (r,d) by step 10.1. Distinct stable factors have $\mathrm{Hom}={\mathrm{Ext}}^1=0$, by stability and elliptic duality. Hence indecomposable B is an iterated extension of one fixed stable S, of some length k. Apply the Euclidean autoequivalence taking S to a degree-zero line, up to a uniform shift. By the cohomology sequence of each transformed extension, it identifies this entire extension category with the degree-zero unipotent line block of step 9.1. There is thus just one indecomposable extension of length k. Write $S={\pi }_{r\ast }L$. The bundle ${\pi }_{r\ast }(L\otimes U_k)$ is such an iterated extension and is indecomposable. Indeed adjunction and the deck decomposition give its endomorphisms as the sum of $\mathrm{Hom}(g^{\ast }(L\otimes U_k),L\otimes U_k)$. For g nontrivial, $g^{\ast }L\mathrm{\ne }L$, since otherwise $\mathrm{End}({\pi }_{\ast }L)$ would have $\text{dimension}>1$, contradicting stability. Filtration by these distinct degree-d line factors makes those Hom groups zero. The identity deck term is $\mathrm{End}(U_k)=C[J_k]$, a local ring. Therefore the pushforward is indecomposable and is the unique length-k extension. We have proved the precise form B ≅ ${\pi }_{r\ast }(L\otimes F(C^k,\exp J_k))$. (C) No external Atiyah classification output was consumed.

Torsion sheaves and bounded complexes. A coherent sheaf on the smooth curve has its finite-length torsion subsheaf and a vector-bundle quotient. The extension splits because ${\mathrm{Ext}}^1(\text{bundle},\text{torsion})=H^1({\text{bundle}}^{\ast }\otimes \text{torsion})=0$; the sheaf has finite support and its higher cohomology vanishes. Each torsion sheaf decomposes by support, and the DVR/Jordan argument of step 9.1 gives the indecomposables $O_{kp}$, equivalently the source jet modules S(p,V,N) with a single Jordan block. The coherent category is hereditary. A bundle-to-anything ${\mathrm{Ext}}^j$ vanishes for $j>1$ by curve cohomological dimension. For torsion as the first argument the local length-one resolution has sheaf ${\mathrm{Ext}}^1$ of finite support, whose $H^1$ is zero; the local-to-global Hom double complex therefore also has ${\mathrm{Ext}}^j=0$ for $j>1$. This argument can be expressed using the actual two-term vector-bundle resolutions and their finite affine Čech complex, so it does not presume a hereditary classification theorem. Every bounded derived object consequently splits as the finite direct sum of its cohomology sheaves shifted. Use the standard truncation triangle and induction on its nonzero cohomology interval; its attaching maps are ${\mathrm{Ext}}^{\ge 2}$ and vanish. Thus (C), the torsion jets and integer shifts cover every object of Db(Coh E).

Plane algebraic/analytic bridge. The factor $\exp (J_k)$ used above is holomorphic. The coherent-side distinction can be bridged using the ACTUAL complete plane-projective GAGA construction thm-qftx-projective-gaga-for-monad-data, whose Statement concerns $P^2$ and $P^3$, not all projective schemes. Its exact application to $E\subset P^2$ is as follows. Push a coherent analytic E-sheaf to $P^2$ and algebraize there. The defining ideal of E annihilates it analytically; full faithfulness on $P^2$ makes that multiplication zero algebraically. It is therefore the pushforward of a coherent algebraic E-sheaf. Morphisms and inverse isomorphisms are identified in the same way. This gives an exact equivalence of the two coherent categories on E, hence termwise of their bounded derived categories. It does NOT identify ${\mathrm{Ext}}_E$ with ${\mathrm{Ext}}_{P^2}$. The existing GAGA carrier is draft/current-local proof, not asserted published or re-imported here; its exact recorded proof/status must be retained by the adopting owner.

Object and Hom correspondence for integral slopes. Choose real ${\alpha }_i,{\beta }_i$ and integer $n_i$, and set ${\phi }_i=t_{{\alpha }_i\tau +{\beta }_i}^{\ast }{\phi }_0$ times ${\phi }_0^{n_i-1},{\phi }_0(u)=q^{-1\mathrm{/}2}{u}^{-1}$. Associate $B_i=L_{{\phi }_i}$ tensor $F_{N_i}$ to the graded straight line ${\mathrm{\Lambda }}_i:y=n_{i}x-{\alpha }_i$ with connection $(-2\pi i{\beta }_{i}Id+N_i)dx$. Its monodromy eigenvalues have modulus one, while its Jordan blocks are permitted. Choose the source slope lift; integer grading shifts are retained. For $n_1<n_2$ set $r=n_2-n_1,{\alpha }_{12}=({\alpha }_2-{\alpha }_1)\mathrm{/}r,{\beta }_{12}=({\beta }_2-{\beta }_1)\mathrm{/}r$. There are r distinct intersections $e_a$, a mod r, with $x_a={\alpha }_{12}+a\mathrm{/}r,y_a=(n_1{\alpha }_2-n_2{\alpha }_1+n_{1}a)$/r. The Hom theta basis is $f_a(z)={\mathrm{\Theta }}_a^r(z+{\alpha }_{12}\tau +{\beta }_{12})$. Define its map to the intersection generator by ${\mathrm{\Phi }}_{12}(V_{{\phi }_2\mathrm{/}{\phi }_1,M_{12}}(f_a\otimes T))=$ $\exp (-\pi i\tau r{\alpha }_{12}^2)\exp ({\alpha }_{12}(M_{12}-2\pi ir{\beta }_{12}))T$ · $e_a$. (P) Both exponential factors are invertible, so this is an actual vector-space isomorphism. For parallel objects use the actual geometric category constructed independently in [F6], not an unproved clean/Morse model from [F3]. Distinct parallel fluxes have acyclic Hom. On one physical parent its actual copy-localized Hom is the two-term complex with differential $D-1$, where $D(T)=M_{2}T{M}_1^{-1}$. Distinct scalar holonomies give no eigenvalue one and hence acyclicity. In a matching scalar-holonomy block write $M_{12}(T)=N_{2}T-T{N}_1$; the left and right nilpotent actions commute, so $D=\exp (M_{12})$. Put $U(x)={\sum }_{j\ge 0}{x}^j\mathrm{/}(j+1)!$, truncated at the nilpotence order. Then $D-1=M_{12}U(M_{12})$, and U is invertible because its constant term is one. The chain map from the complex with differential $M_{12}$ to that with differential $D-1$ is identity in degree zero and $U(M_{12})$ in degree one. It is an isomorphism. On the kernel and coinvariant groups U induces identity, since $U(x)-1$ is divisible by x. Thus the actual geometric Hom groups are exactly the coherent $\ker M_{12}$ and $\mathrm{coker}{M}_{12}$ computed in step 4.1. The geometric degree-zero maps are flat coefficient intertwiners, and [F6] proves their actual triangle products and their actions on coinvariants are coefficient composition; two degree-one classes multiply to zero. Its perfect kernel/coinvariant trace pairing agrees with cyclic rotation of actual triangles. Consequently this is also the binary parallel dictionary, including nontrivial Jordan parts, before any nontransversal higher mirror functor is claimed. The theta formula (P) commutes with flat intertwiners because they commute with the left/right nilpotent operators and their finite exponentials. Hence its mixed transverse actions agree with [F6]'s actual coefficient actions; products of two transverse classes with a parallel output follow by the same perfect trace pairing. This supplies the analytic and binary bridge locally through an earlier independent supplier, without a dependency on the completion consumer. The reverse transverse degree follows by the trace/Serre pairings and grading shift. This gives the explicit Hom correspondence on this object class. It does not identify the whole derived categories.

Source orientations and constant triangles. For the source parallel-spin convention orient a disk determinant by attaching real strip caps and gluing it to the constant real-boundary disk; orient its kernel by the real constant section and its marked-domain parameters by their ordered real coordinates. Determinant gluing is the wedge orientation of the exact kernel/cokernel sequence and is associative, so these are coherent orientations. The source numbered corner path is clockwise, opposite to the positive traversal of the holomorphic boundary; reverse that word and carry its suspended orientation lines before comparing to a positive-order convention. In this source word, the cap-order Koszul sign is plus for arity k even and sign(x(p0)-x(pk)) for k odd; fixing one parallel-spin triangle reference fixes its generator normalization. This is also the convention in the explicit source coefficient formulas below. At a concurrent constant triangle the three-line compact doubled operator is the O(-1) real-boundary model: its kernel is zero by negative-degree vanishing and its real index is zero, hence it is onto. The implicit equation under translations of those three lines has a unique small triangle, including its constant limit; orientation transport gives its reference count one. Thus binary formulas extend through such isolated constant triangles. No regularity of higher constant polygons is assumed: all higher comparison collections avoid triple concurrence and their full nontransversal meaning comes from the separate fixed-copy completion.

Rigid-triangle coefficients and binary compatibility. Take $n_1<n_2<n_3$, write $r=n_2-n_1,s=n_3-n_2$ and $t=r+s$. Lift $e_a$ in ${\mathrm{\Lambda }}_1$ intersection ${\mathrm{\Lambda }}_2$ to the plane and lift $e_b$ in ${\mathrm{\Lambda }}_2$ intersection ${\mathrm{\Lambda }}_3$ by its additional horizontal integer j. The resulting triangle has output index $c=a+b+sjmodt$. Put $k_j=r$ b-s a+r s j and $\delta \alpha ={\alpha }_{23}-{\alpha }_{12}$. Its horizontal displacements along the three edges satisfy $l_2=\delta \alpha +k_j\mathrm{/}(rs),l_1=(s\mathrm{/}t)l_2,l_3=-(r\mathrm{/}t)l_2$. Its symplectic area in unit-area coordinates is $Are{a}_j=(r\mathrm{/}2)l_1{l}_2=(k_j+rs\delta \alpha {)}^2\mathrm{/}(2rst)$. (A) This derivation fixes the area directly; an intermediate inverted slope fraction cannot be used in place of (A). The boundary parallel transports are exactly the exponentials of $(-2\pi i{\beta }_i+N_i)$ times the signed horizontal edge lengths. Their composition is contraction over $V_2$ of $\exp (-l_1(N_1^{\ast }-2\pi i{\beta }_1)+l_2(N_2-2\pi i{\beta }_2)+l_3(N_3-2\pi i{\beta }_3))$. For every chosen homotopy class these data describe one affine Euclidean triangle. The elementary rigid-triangle enumeration has the following precise analytic interface to the Floer root: for transverse straight lines in a complex torus, each rigid holomorphic three-pointed disk lifts to this simple triangle, its conformal parametrization is unique, and its orientation is the source positive one. The actual corner/divisor and convex-polygon construction [F3], together with the source cap and word orientation fixed in step 15.1, supplies this precise triangle enumeration. Consequently its binary structure coefficient is the sum of $\exp (2\pi i\tau Are{a}_j)$ times the displayed boundary transport. The nilpotent part grows polynomially in j and scalar holonomy has unit modulus, so this sum converges absolutely. Substituting (A), using $l_1,l_2,l_3$, and dividing the input/output normalizations (P) reduces this coefficient to $\exp (\pi i\tau k_j^2\mathrm{/}(rst)+2\pi i{k}_j(z_2-z_1)\mathrm{/}t)$ with exactly the commuting nilpotent translations of step 3.1. Formula (T) therefore proves $\mathrm{\Phi }(m_2)=m_2(\mathrm{\Phi },\mathrm{\Phi })$. This is an equality of convergent coefficients including the nilpotent transports, not merely a match of dimensions. Under contraction over $V_2$, operators $N_2$ and $N_2^{\ast }$ are transposed by the trace pairing; operators on the other tensor factors commute. Thus the nilpotent formal substitution in (T) is legitimate even though the matrices originally belong to different Hom spaces. Here is the explicit nilpotent translation, with no unspecified shift. On the composable tensor set $M_{12}=N_2-N_1^{\ast }$ and $M_{23}=N_3-N_2^{\ast }$, and after contraction put $M_{13}=M_{12}+M_{23}$. Define $R=(s{M}_{12}-r{M}_{23})\mathrm{/}(2\pi it)$. Applying the inverse output section operator to the product of the two input section operators gives the two theta arguments $z_1=z+{\alpha }_{12}\tau +{\beta }_{12}-R\mathrm{/}r,z_2=z+{\alpha }_{23}\tau +{\beta }_{23}+R\mathrm{/}s$. Indeed the input shifts are $-M_{12}\mathrm{/}(2\pi ir),-M_{23}\mathrm{/}(2\pi is)$, while the inverse output operator shifts both by $+M_{13}\mathrm{/}(2\pi it)$. Their weighted average is $z+{\alpha }_{13}\tau +{\beta }_{13}$. Formula (T) therefore contributes the nilpotent factor $\exp (2\pi i{k}_{j}R\mathrm{/}(rs))$. The geometric transport's nilpotent factor is $\exp (2\pi i{l}_{2}R)$. Its constant exp(2πiδαR) is precisely the ratio $\exp ({\alpha }_{13}{M}_{13}-{\alpha }_{12}{M}_{12}-{\alpha }_{23}{M}_{23})$ from (P); the remaining factor is the required $\exp (2\pi i{k}_{j}R\mathrm{/}(rs))$. All of this is computed on separate tensor factors before the trace contraction. For the remaining normalization algebra, $t{\alpha }_{13}=r{\alpha }_{12}+s{\alpha }_{23}$ and $t{\beta }_{13}=r{\beta }_{12}+s{\beta }_{23}$. Expanding $(k_j+rs\delta \alpha {)}^2$ leaves a $k_j$-independent factor. The identity $r{\alpha }_{12}^2+s{\alpha }_{23}^2-t{\alpha }_{13}^2=(rs\mathrm{/}t)({\alpha }_{23}-{\alpha }_{12}{)}^2$ cancels precisely that factor against the three Gaussian normalizations in (P). The identical bilinear polarization identity with α and β cancels the scalar holonomy constant. The linear identities for $l_i$ cancel the remaining nilpotent normalizations. The surviving $k_j$-linear term is exactly the one in (T).

Scope and algebraic conventions. On the graded category of line bundles on E, use Hom plus ${\mathrm{Ext}}^1$, ordinary composition m2, and elliptic Serre pairing. A pair is transversal when the corresponding straight circles are distinct. Collections are pairwise transversal; distinct parallel circles have zero morphisms. At each fixed degree, the position parameter can avoid finitely many prohibited real values, including their finitely many inverse images under doubling. In particular all auxiliary line bundles and their tensor products used below can be chosen transversal to a prescribed finite list. An admissible structure has $m1=0$, ordinary m2, strict compatibility with tensoring all objects by one line bundle, and cyclicity for the Serre pairing. Express its equations on the suspended bar coalgebra: the coderivation b has degree one and $b^2=0$. A functor with first component identity is a coalgebra automorphism $F=Id$ plus higher-length components; its required equation is $Fb=b^{\mathrm{\prime }}F$. This convention fixes every sign by suspension, rather than an unspecified signed version of an induction. All polygon formulas below use the source's NEGATIVELY ordered corner word: its numbered vertex path $p_0,p_1,\mathrm{.}\mathrm{.}\mathrm{.},p_k,p_0$ is clockwise. The real symplectic orientation is positive dx wedge dy, so this is not the positive boundary traversal of a holomorphic disk. For nonvertical lines the source coefficient is '+' when k is even and $\mathrm{sign}(x(p_0)-x(p_k))$ when k is odd, and it is zero unless the cyclic sum of the adjacent Hom degrees is k-1. Matrix transports use the signed horizontal edge lengths in that same word. The generic Fukaya supplier's positive-boundary order must therefore be converted by reversing the corner word and transporting its suspended orientation/generator lines. This packet compares the coherent structure with that precisely specified source-order algebra; it does not silently identify the clockwise word with the generic positive-order definition. The shared torus polygon orientation root is responsible for their actual matching.

A multiplication presentation with all relations proved. Let L have degree $d\ge 3$. Let S be the allowed intermediate positive line bundles; assume the generic-choice property above. The map ${⨁}_{AB=L,\mathrm{deg}A,\mathrm{deg}B>0,B\in S}$ H0(A) tensor H0(B) $\to$ H0(L) is surjective and its kernel is generated by associativity relations $(s_1{s}_2)$ tensor $s_3$ - $s_1$ tensor $(s_2{s}_3)$, where the three factors have positive degree and both right intermediate factors belong to S. Here is a uniform proof, including $d=3$ and $d=4$. It suffices to prove the universal dual assertion: any compatible family $b_{A,B}(s,t)$, satisfying the displayed associativity relation, factors uniquely through st in H0(L). Choose distinct generic p,q with O(p),O(q),O(p+q) in S. The subspaces $H0(L(-p))s_p$ and $H0(L(-q))s_q$ span H0(L); their intersection is $H0(L(-p-q))s_p{s}_q$. These facts follow from the two evaluation sequences and Riemann–Roch: $d\ge 3$, and the common kernel has dimension d-2. Associativity makes b on these subspaces agree on that intersection. It defines a unique functional phi on H0(L). For any B of degree at least two, H0(B) is spanned by $H0(B(-p))s_p$ and $H0(B(-q))s_q$ unless deg $B=2$ and $B=O(p+q)$. Indeed the obstruction is H1(B(-p-q)), zero for positive degree, or for a nontrivial degree-zero line. For a spanning term, associativity gives $b_{A,B}(s,t^{\mathrm{\prime }}{s}_p)=b_{AB(-p),O(p)}(s{t}^{\mathrm{\prime }},s_p)=\varphi (s{t}^{\mathrm{\prime }}{s}_p)$. For B of degree one choose generic a,b so B(a),B(b) belong to S, neither is O(p+q), and if deg $A=2$ also A differs from O(a+b). Then H0(A) is spanned by sections $s^{\mathrm{\prime }}{s}_a$ and $s^{\mathrm{\prime }\mathrm{\prime }}{s}_b$, and $b_{A,B}(s^{\mathrm{\prime }}{s}_a,t)=b_{A(-a),B(a)}(s^{\mathrm{\prime }},s_{a}t)=\varphi (s^{\mathrm{\prime }}{s}_{a}t)$, using the degree-two case already proved. Finally for the exceptional $B=O(p+q)$, if deg $A\ge 2$ use the same factorization of sections of A, now with B(a) of degree three. If deg $A=1$, choose a different initial pair p',q' with O(p'+q') different from B. The functional defined using this pair equals phi because its defining right factors have degree one, a case already proved. The nonexceptional argument for that pair now handles B. This proves the assertion in every degree. The same proof holds for vector-valued b by testing linear functionals on its finite-dimensional target. All point choices exclude finitely many divisors or finitely many forbidden position values; the transversality hypothesis supplies the necessary infinitely many choices. No claim of surjectivity of a SINGLE degree-one product is used.

Conventions and explicit equations. Work in the suspended convention $V=s\mathrm{Hom}$, with |s $a\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }-1$. Each $b_j$ has cohomological degree one. Tensor evaluation always uses the Koszul rule. For a degree-d multilinear map f of arity r define its insertion operator by $(f\circ g)(x_1,\mathrm{.}\mathrm{.}\mathrm{.},x_{r+t-1})={\sum }_{i=0}^{r-1}(-1{)}^{\mathrm{\mid }g\mathrm{\mid }{\sum }_{j\le i}\mathrm{\mid }{x}_j\mathrm{\mid }}f(x_1,\mathrm{.}\mathrm{.}\mathrm{.},x_i,g(x_{i+1},\mathrm{.}\mathrm{.}\mathrm{.},x_{i+t}),\mathrm{.}\mathrm{.}\mathrm{.},x_{r+t-1})$. This definition, with empty prefix sign +1, specifies every sign below. Write $[f,g]=f$ circle $g-(-1{)}^{\mathrm{\mid }f\mathrm{\mid }\mathrm{\mid }g\mathrm{\mid }}g$ circle f. The Hochschild differential at fixed arity is $\delta f=[b_2,f]$. Associativity gives delta squared zero. Cyclicity means preserving the suspended Serre pairing, with its Koszul rotation sign; equivalently the scalar multilinear form obtained by pairing the output with one additional cyclic slot is invariant under that signed rotation. All cyclic slots refer to an actual composable cyclic sequence of line bundles; the tensor product of their Hom line factors is trivial. Suppose operations of arity less than n have already been matched. Put $D_n=b_n^{\mathrm{\prime }}-b_n$, where b' denotes the structure after the previously applied changes of coordinates. Subtract the length-(n+1) parts of b' squared zero and b squared zero. All terms not involving $b_2$ and $D_n$ cancel, leaving $[b_2,D_n]=0$. (1) There are no suppressed lower-arity correction terms: b' has been conjugated at every preceding stage. $D_n$ is cyclic. We seek a cyclic degree-zero map ${\xi }_{n-1}$ with $[b_2,{\xi }_{n-1}]=D_n$. (2) Then replace b' by $\exp ({\xi }_{n-1})$ b' $\exp (-{\xi }_{n-1})$. Its arity-n term changes by $[{\xi }_{n-1},b_2]=-D_n$. An arity-r map decreases bar length by r-1; hence the exponential is finite on every word. Its second commutator with $b_2$ has arity $2n-2>n$ for $n\ge 3$. It changes no smaller arity. The exponential of a cyclic coderivation preserves the pairing: differentiating the paired expression along exp(t xi) gives the cyclic infinitesimal identity zero, and this is a polynomial identity on each word. Thus no unproved nonlinear cyclicity condition on a bare coalgebra map is required.

Degree types, weights, and the cyclic overlap. Rotate and tensor-normalize a potentially nonzero n-operation so its output is in H0. Its unshifted degree 2-n then forces exactly two H0 inputs and n-2 H1 inputs. Positive-degree line factors have only H0, negative-degree ones only H1, and a generic degree-zero factor has neither. Let w be the sum of the two positive line degrees. Positive output line degree gives $n\le w+1$. Pairing the output with its Serre-dual slot describes $D_n$ as a cyclic form with TWO positive slots and n-1 negative slots. In contrast, a potentially nonzero correcting map ${\xi }_{n-1}$, of unshifted degree 2-n, has a cyclic form with exactly ONE positive slot and n-1 negative slots. Use that unique positive slot to anchor its cyclic word. Define the form in this anchored order; its values on all rotations are then the Koszul rotation of that value. This has no overlap ambiguity: a rotation returning the positive slot to the anchor is the full rotation. Its total Koszul sign is one, since each pair of slot degrees crosses twice. Thus the usual phrase “extend cyclically” is an actual single-valued instruction. Perfect Serre duality recovers the unique output map from this cyclic form. Common tensoring leaves the Hom factors unchanged, so this instruction is strictly tensor-compatible. Induct on w at fixed n. At the current weight assume $D_n$ is zero at every smaller positive weight. The following bookkeeping also proves that later corrections leave this property intact. A correcting cyclic form of weight w has one positive line factor of degree w. In its Hochschild differential, merging two positive slots produces weight exactly w and requires those slots to be adjacent. Merging a positive and negative slot can contribute only when it leaves one other positive slot of degree w; the unmerged positive contributes a strictly positive additional degree. Such a term has input weight strictly greater than w. Other merges vanish by their cohomological types. Consequently its differential at weight w is ONLY the adjacent-positive merge, and it vanishes on separated-positive words of that weight. These are exhaustive cases: two negative slots multiply to $\mathrm{Ext}2=0$; a positive-negative product lies in H0 or H1 according to its line degree, with degree-zero generic output absent. For separated positive slots with $w>2$, factor a section of a positive line of degree $d>1$ as products $s_{p}s$ of degrees 1,d-1. Such products span: the two evaluation kernels for distinct generic points p,q span H0 of the line when $d\ge 2$, choosing q so its degree-one kernel differs from the first when $d=2$. Insert these two positive factors into (1). Among all terms, their recombination is the desired $D_n$. Every other nonzero term has smaller positive weight: an outer factor has been removed, or a positive factor has absorbed a strictly negative line. There is no other adjacent-positive pair because the two original positive slots were separated. Hence the desired value is zero. At $w=2$ the degree bound forces $n=3$ and the separated value is exactly the fundamental degree-(1,-1,1) product, already matched. For adjacent positive slots A,B of total degree w, fix all negative classes and a Serre-dual output class. Define ${\beta }_{A,B}(s,t)=$ paired $D_n(s,t,e_1,\mathrm{.}\mathrm{.}\mathrm{.},e_{n-2})$. Take three consecutive positive factors and the same negative word in (1). The two positive-positive merges give exactly the associativity relation for beta. Every other term has smaller positive weight by the preceding case enumeration. The two displayed merges have the relative signs given by the insertion formula; after pairing and placing the unique positive slot at the anchor this is the ordinary relation ${\beta }_{AB,C}(st,u)={\beta }_{A,BC}(s,tu)$. (3) This sign assertion can be read without a convention guess: positive sections have suspended degree -1, so moving the inserted $b_2$ across the first positive section changes its insertion sign by -1; moving the two terms to opposite sides gives (3). For $w\ge 3$ the multiplication presentation gives a unique linear functional gamma on H0(A B), with the fixed negative cyclic slots, whose value on st is ${\beta }_{A,B}(s,t)$. Choose the anchored cyclic form of xi to be minus gamma in the displayed adjacent-positive order: the corresponding term of $[b_2,\xi ]$ is -xi circle $b_2$, with empty-prefix insertion sign +1. Thus (2) holds in that order. Signed cyclic rotation gives it in every adjacent order. The separated values were zero and the correction leaves them zero. This removes the entire weight-w discrepancy.

The fundamental scalar Massey coefficient. Let $\theta (z)={\sum }_n\exp (\pi i\tau n^2+2\pi inz)$, let $a=\mathrm{Im}\tau$, and translate its unique zero to $z_0=(1+\tau )\mathrm{/}2$. Let $L(u)=t_u^{\ast }L(0)$, deg $L(0)=1$. Consider the transversal quadruple O, L(t), $L(t)L(0{)}^{-1}$, L(t+u), where the position components $t_2,u_2$ are not integers. For the analytic calculation one needs only t,u outside Z+τZ. The middle adjacent Hom complexes have degree-zero NONTRIVIAL line factors, hence are acyclic. The triple product is therefore a well-defined univalued Massey product, independent of contraction/metric choices. Normalize the degree-one Serre-dual class by $\alpha (z)=i\mathrm{/}\sqrt{2a}$ times conjugate(theta(z)) $\exp (-2\pi a{z}_2^2)$ d(conjugate z). Its pairing with theta(z), using the trivialization $\eta =-2\pi idz$ and the fixed negative-residue trace reconciled in step 6.1, is one: the resulting scalar pairing is exactly the raw integral ${\int }_{E}dz\wedge (\theta \alpha )$. With $dz$ itself the same fixed-trace pairing would instead be $-1\mathrm{/}(2\pi i)$. Integrating first in $z_1$ kills unequal Fourier modes; the remaining integral is ${\int }_R$ $\exp (-2\pi a{s}^2)ds=1\mathrm{/}\sqrt{2a}$, and dz wedge $d(\bar{z})=-2iad{z}_1$ wedge $d{z}_2$. This gives the stated normalization. For t nonlattice there is a unique $h_t$ with dbar $h_t=\theta (z+t)\alpha$. Thus $h_t(z)\theta (z+u)-h_u(z)\theta (z+t)=H(t,u)\theta (z+t+u)$. The expression on the left is dbar-closed and the output line has degree one, proving the scalar factorization. The acyclic degree-zero family has a local holomorphic inverse: in a periodic trivialization its Fourier eigenvalues are (π/a)(mτ-n+t), with just the zero eigenvalue at $t=0$. All nonzero modes have uniformly invertible eigenvalues on a small parameter neighbourhood, with the usual rapid Fourier coefficient estimates. Consequently H is meromorphic, with only simple lattice poles. Its residue at $t=0$ is -1/(2πi): the constant Fourier coefficient of theta alpha is i/(2a), divided by the small eigenvalue πt/a. Its automorphies are $H(t+1,u)=H(t,u),H(t+\tau ,u)=\exp (2\pi iu)H(t,u)$. They follow from the theta transformations and uniqueness of $h_t$. Define $F(t,u)={\theta }^{\mathrm{\prime }}(z_0)\theta (t-u+z_0)$ /[2πi $\theta (t+z_0)\theta (-u+z_0)]$. It has the same automorphies and residue +1/(2πi). Thus H+F is entire in t with those automorphies. A 1-periodic entire function has a Laurent expansion in exp(2πit); its coefficients satisfy $(q^n-\exp (2\pi iu))c_n=0$. Since u is nonlattice, all coefficients vanish. Therefore $H=-F$. This proves the actual Dolbeault Massey coefficient.

The exceptional weight-two start. Only $n=3$ remains at $w=2$. Write M for the negative line of degree -1, and choose a generic degree-one line J with a nonzero section v. Multiplication $H1(M{J}^{-1})\to$ H1(M), e' $\to$ e' v, is onto: its Serre-dual map is multiplication by v from $H0(M^{-1})$ to $H0(M^{-1}J)$, which is injective. Choose a linear right inverse $e\to$ e'. The length-four instance (1), evaluated on $(s_1,s_2,e^{\mathrm{\prime }},v)$, has just two possibly unmatched terms, giving $D_3(s_1,s_2,e^{\mathrm{\prime }}v)=-D_3(s_1{s}_2,e^{\mathrm{\prime }},v)$. (4) Here the other separated term is the already equal (1,-1,1) product. The outer terms have generic degree-zero output and vanish. These statements also specify the sign: the middle insertion $b_2$ in $(s_1,s_2,e^{\mathrm{\prime }},v)$ crosses $s_1$ of suspended degree -1, whereas the first insertion crosses no slot. Define the correcting value on H0(L), $L=AB$ of degree two, by the right side of (4), with the sign prescribed by (2). This definition is independent of J,v,e' and linear in e. Indeed (4) fixes it on EVERY product $s_1{s}_2$ from degree-one factorizations of L. Those products span H0(L): choose two generic points so the two degree-one vanishing subspaces are distinct lines in the two-dimensional H0(L). Any two choices therefore agree on a spanning set, and a fixed linear right inverse proves linearity. It depends only on L, not on its chosen factorization. The cyclic anchoring above again has only one positive slot, so this correction has no rotational overlap. Its differential is the weight-two adjacent discrepancy and has no separated contribution at weight two. The preceding reduction now applies to separated weight three; there is no assumption that those values vanished before this correction.

Quadrilateral lattice, normalization, sign and Kronecker coefficient. We need the following extension of [F3], whose Statement itself assumes an entirely pairwise-transverse collection. Let a finite collection of distinct primitive straight circles have every pair transverse or disjoint and no triple concurrence. Use cyclic words with distinct labels, and give a word with a disjoint consecutive pair zero generator space. For every remaining word all prescribed corners are transverse. Here is a proof that the required regularity, rigid enumeration and area-cutoff compactification still hold. The corner expansion in the supplier's Proof 1.1 uses only the two circles meeting at that corner: after rotation it is $u(z)=c{z}^{\delta +k}+\cdots$, with $0<\delta <1$ and $k\ge 0$. Its Proofs 2.1–6.1 then apply to this actual map without using intersections between nonadjacent boundary circles. In particular, for $N$ corners the doubled tangent-quotient line has degree $d=2-N+e$, where $e=2B_{int}+B_{edge}+\sum k_i$. The weighted cokernel equals the compact-model cokernel. If $d\ge -1$ it vanishes by the doubled-line adjoint calculation; otherwise $K\le e$ and $N-K\ge 3$, since $N-K\le 2$ would force $d=2+e-N\ge 0$. Choose three simple corners as anchors; releasing all other simple corners raises the degree to $e-K-1\ge -1$, where $K$ is the number of corners with positive $k_i$. These released real conditions are exactly the marked-domain variations. Thus the universal operator is onto and its actual dimension is $e$. Every rigid nonconstant polygon has $e=0$. The supplier's Proof 11.1 applies to its affine boundary lines: the derivative quotient reflects to a polynomial, the zero count gives total turning $2\pi$, and the image is a simple strictly convex Euclidean polygon. Opposite supporting lines may be parallel; the argument uses only the nonzero adjacent angles. Conversely its Proof 12.1 constructs the unique conformal parametrization of each such strictly convex polygon, including the boundary and corner extension, modulo positive disk automorphisms. This proves the needed rigid enumeration, rather than assuming that adjacent transversality is the supplier's interface. For compactification, the torus has no sphere bubbles, and an essential circle has injective fundamental group, so a one-label disk bubble has zero area and is constant. Stable constants with three distinct labels are excluded by no triple concurrence. At a new boundary node, the two incident labels are inherited distinct labels of the original word. They must meet at its matching image point. Hence they cannot be a disjoint pair and their tangent lines are transverse. More explicitly, a neck whose edges lay on a disjoint pair cannot occur in a bounded-energy sequence: their distance is some $\epsilon >0$, every transverse strip slice has length at least $\epsilon$, and Cauchy–Schwarz gives ${\int }_0^1\mathrm{\mid }{u}_t{\mathrm{\mid }}^{2}dt\ge {\epsilon }^2$. Holomorphic energy on a strip of length $R$ is therefore at least ${\epsilon }^{2}R$. Every actual long neck instead has a transverse matching intersection and the supplier's proved weighted end inverse and exponential extraction apply. Its compact-core estimates and positive-energy extraction are unchanged, and its Proof 10.1 constructs the regular collars using precisely these onto component operators and transverse matching ends. Thus area-cutoff rigid finiteness and the one-dimensional broken-polygon boundary formula, with zero terms for disjoint intermediate pairs, follow by the same explicit local estimates. Determinant caps and the given coherent orientations at every existing corner and node are unchanged, so this extension also retains the source-word sign convention already fixed above. It proves all the polygon inputs used below for the Statement's transverse-or-disjoint collections without introducing a Hamiltonian perturbation or a continuation assertion. Write $t=t_1+\tau t_2$ and $u=u_1+\tau u_2$, with real components, and first assume $0<t_2,u_2<1$. For the quadruple in step 20.2 choose successive supporting lifts $y=0,y=x-t_2,y=-t_2+m,y=x-t_2-u_2-n$, where m,n are integers. Their consecutive vertices, in the source clockwise word, are $p_0=(t_2,0),p_1=(m,m-t_2),p_2=(m+n+u_2,m-t_2),p_3=(t_2+n+u_2,0)$. The four horizontal edge increments are $a=m-t_2,b=n+u_2$, -a, -b. Thus the word is a clockwise strictly convex parallelogram exactly when $ab>0$, and its positive symplectic area is ab. These are all lifts: fixing $p_0$ fixes the first two supporting lines, while the remaining horizontal and diagonal translates are independently indexed by m,n. The two horizontal circles are disjoint because $t_2\mathrm{\notin }\mathbb{Z}$, and the two slope-one circles are disjoint because $u_2\mathrm{\notin }\mathbb{Z}$. Each adjacent pair is transverse and no three circles concur. Thus the local disjoint-pair extension proved in this paragraph applies to this quadruple: every displayed strictly convex parallelogram has one regular index-zero conformal map modulo disk automorphisms, and every rigid map is one of these parallelograms. In particular the unextended pairwise-transverse interface of [F3] is not being invoked for this collection. The scalar connection coefficients on the four lines are respectively $0,t_1,t_1,t_1+u_1$ in the convention -2pi i beta dx. Hence the boundary holonomy in this source word is $\exp (2pii(u_{1}a-t_{1}b))$. Multiplying it by the area factor gives exp(2pi i(tau mn+mu-nt)) times $\exp (-2pii(tau{t}_2{u}_2+t_2{u}_1+t_1{u}_2))$. In the normalization (P), the three input factors are $P_{0}1=\exp (-\pi i\tau t_2^2-2\pi i{t}_2{t}_1),P_{1}2=1$ (the Serre-dual reverse of the zero-position positive intersection), and $P_{2}3=\exp (-\pi i\tau u_2^2-2\pi i{u}_2{u}_1)$; the output factor is $P_{0}3=\exp (-\pi i\tau (t_2+u_2{)}^2-2\pi i(t_2+u_2)(t_1+u_1))$. Their ratio $P_{0}1P_{1}2P_{2}3\mathrm{/}{P}_{0}3$ is $\exp (2pii(tau{t}_2{u}_2+t_2{u}_1+u_2{t}_1))$, precisely the inverse of the preceding constant. The source odd-arity sign is $\mathrm{sign}(x(p_0)-x(p_3))=-\mathrm{sign}(b)=-\mathrm{sign}(a)$, because $ab>0$. Consequently the normalized quadrilateral coefficient is -K(t,u), where $K(t,u)={\sum }_{(m-t_2)(n+u_2)>0}\mathrm{sign}(m-t_2)$ exp(2pi i(tau mn+mu-nt)). This proves the sign and every scalar normalization, rather than borrowing the triangle calculation. In this chamber, $m\ge 1$ permits exactly $n\ge 0$, and $m\le 0$ permits exactly $n\le -1$. Each geometric series is absolutely convergent; summing it gives $K(t,u)={\sum }_{m\in Z}{U}^m\mathrm{/}(1-q^m\mathrm{/}T)$, with $T=\exp (2\pi it)$, $U=\exp (2\pi iu)$. Normal convergence follows because the positive tail decays like $U^m$ and the negative tail like $(q\mathrm{/}U{)}^{-m}$, uniformly on compact pole-free t-sets; thus K extends meromorphically in t. Reindexing gives $K(t+\tau ,u)=U$ K(t,u), and $K(t+1,u)=K(t,u)$. Its residue at $t=0$ is 1/(2pi i). The function F of step 20.2 has exactly the same automorphies and residues. Their difference is entire; its 1-periodic Laurent coefficients satisfy $(q^k-U)c_k=0$. For nonlattice u these all vanish. Hence $K=F$ and the normalized polygon coefficient is $-F=H$, the actual Dolbeault Massey coefficient. Lattice translations of t,u and the corresponding relabeling of m,n extend this computation to every chamber with nonintegral reduced position components. The same equality holds on complex neighbourhoods away from the poles, so every finite holonomy derivative agrees as well. Together with the arity/weight construction this supplies its required scalar degree-(1,-1,1) input; a separate verification of that construction and its Artin/descent bindings is still required before native acceptance.

Termination, uniqueness, and naturality. Increasing w removes $D_n$ completely, and then increasing n gives the comparison. At fixed n a given finite cyclic word can be affected only by weights no larger than the sum of its positive line degrees. For a fixed bar word there are finitely many consecutive subwords, so the compositions of exponentials involve only finitely many nonzero corrections on that word. This is an algebraic conilpotent construction, not analytic completion. For uniqueness, let ${\xi }_r$ be the first nonzero logarithmic component of a cyclic tensor-compatible automorphism inducing identity in arity one. It satisfies $[b_2,{\xi }_r]=0$. Its cyclic form has one positive slot. Induct on that slot's positive degree w. Evaluate this equation on two adjacent positive slots factoring that line. All other terms have smaller positive weight and vanish. Their products span H0 of the degree-w line for $w\ge 2$, so the cyclic form vanishes. For $w=1$ there are at least two negative slots when $r\ge 2$, whose degree sum cannot be -1; it is zero by line degree. Hence ${\xi }_r=0$, a contradiction. The logarithm is finite on each word just as the exponential is. This proves uniqueness within the stated cyclic, strict-tensor, identity-first-component class; it does not assert uniqueness of arbitrary A-infinity functors. The same construction works over a local Artin C-algebra R for transversal reductions. Every cohomology space here is finite free with base change; the multiplication presentation is exact over R by the nilpotent Nakayama argument already given in Proof 22.2. Degree-two spanning is witnessed by unit determinants modulo the maximal ideal and therefore over R. The surjection in the exceptional start has finite free target, so it splits; its choice disappears by the spanning argument. The pairing is perfect, all signs are units, and the exponential denominators are nonzero complex constants. Thus no “zero mod maximal ideal implies zero” shortcut for an obstruction is used: each obstruction is killed by the actual construction over R. Uniqueness makes every component commute with Artin base change. For nilpotent local systems use independent object variables and evaluate monomials at their coefficient vertices, moving an intermediate nilpotent between the two half-edges by the trace identity. Exponents add under gluing, so this evaluates each finite bar equation term by term; it does not replace noncommuting Hom matrices by a common scalar. The split Dolbeault contraction identities survive the same evaluation, identifying the resulting model with the actual unipotent bundle complex. For descent, the uniqueness just proved gives a conditional naturality criterion: whenever a finite group acts on two already constructed cyclic tensor-compatible structures, preserving their pairing and first theta map, conjugation of their comparison lies in the same uniqueness class. It must therefore equal that comparison in every arity. Invariant inputs then have invariant outputs, and inverse pullback on those outputs descends the bar equations. Likewise, if two covers have already been shown to preserve the underlying structures, passing to a common cover and applying uniqueness on the entire intermediate line/unipotent category makes the two descended comparisons equal. This criterion assumes invariant structures; it does not construct them or use the later descent conclusion as a premise. The pullback-compatible metrics and orthogonal contractions are constructed independently in steps 26.1–27.1, where this criterion is then applied.

The arity/weight construction over Artin rings. The argument of the fundamental Massey calculation above works over every local Artin C-algebra R, not just over C. Here line bundles mean invertible sheaves on $E_R$ with specified reductions; transversality is tested on their reduced positions. For a positive-degree line, its H0 is free of the degree rank and H1 is zero; for a negative-degree line the reverse holds. For a nontrivial reduced degree-zero line both vanish. These statements follow by the actual nonnegative universal cohomology complex: split invertible matrix entries and reduce modulo the nilpotent maximal ideal. The remaining H0 or H1 modules are finite free and commute with every R-base change. The trace on $H1(O_E)$ tensor R gives the relative Serre pairing. It is perfect: both modules being paired are finite free, its reduction is the actual elliptic Serre pairing, and the determinant is a unit by Nakayama. No general duality theorem over a singular base is presumed. The multiplication-presentation lemma lifts as well. Its map onto H0(L) is surjective modulo the maximal ideal, hence surjective over R by nilpotent Nakayama. The target is free, so that map splits; its kernel is projective and commutes with reduction. The associativity relations surject onto that kernel modulo the maximal ideal. Their cokernel C satisfies $C=\mathfrak{m}C$, hence $C={\mathfrak{m}}^{N}C=0$. This last argument needs NO finite-generation assumption on the direct sum of all generic intermediate factors. Thus the same exact presentation and universal factorization hold over R. One may also repeat the two-point evaluation proof directly; the requisite evaluation matrices have unit determinants after choosing the same generic reduced points. All bar-length/weight induction steps, the weight-two Massey base case, and uniqueness are therefore valid over R. There is no division other than by nonzero complex constants. All operations and the resulting unique admissible functor commute with Artin base change, because their defining factorizations are unique.

Actual cyclic Dolbeault contraction and nilpotent perturbation. For a line Hom factor L on E take $P=\overline{-}d$ acting from smooth sections to smooth (0,1)-forms. By [F5] its L2 range is closed, its kernel H0 and adjoint kernel H1 are finite-dimensional and smooth, and augmenting the range by H1 has an actual bounded right inverse at every Sobolev order. Let p0,p1 be the orthogonal projections and define h on (0,1)-forms to be the unique solution of $Phx=(1-p1)x$ orthogonal to H0, and zero on degree-zero sections. Concretely obtain a solution from the augmented right inverse, then subtract p0; uniqueness makes these constructions compatible at all orders. Smooth input has smooth output by the higher-range/regularity interface. On the two-term complex this gives P h+h $P=1-ip,h2=hi=ph=0$. No inverse of a nonexistent surjective P is taken. Choose dual metrics for L and its inverse, with canonical-bundle trivialization $\eta =-2\pi idz$ as in step 6.1 and the flat volume in coordinate z. The Serre scalar pairing is therefore $B(x,y)={\int }_{E}dz\wedge \mathrm{tr}(xy)$; the fixed algebraic trace has not been replaced. In a local holomorphic frame of metric g, the formal adjoint is a nonzero fixed metric factor times $-g^{-1}{\mathrm{\partial }}_z(gf)$ on f dbar-z; hence its harmonic forms correspond to the holomorphic dual sections by f dbar-z $\to g$ conjugate(f). Integration by parts and orthogonality therefore give, for degree-one x,y in dual factors, $B(h_{L}x,y)=-B(x,h_{L^{-1}}y)$, where B is the Serre integral; harmonic summands vanish on both sides. This proves the needed cyclic adjoint identity for the actual smooth h, not an arbitrary algebraic splitting. Common tensor line factors cancel in the Hom bundle; assign the same operator and metric to that identical Hom factor. For a local Artin algebra R with nilpotent maximal ideal m, trivialize the smooth line deformation and write $d_R=d+\delta$, with delta in m and the dual deformation -transpose(delta). The formulas $h_R=h(1+\delta h{)}^{-1},i_R=(1+h\delta {)}^{-1}i,p_R=p(1+\delta h{)}^{-1},d_H=p$ $\delta (1+h\delta {)}^{-1}i$ are FINITE geometric sums because m is nilpotent. Expand them to verify the contraction identities. On every transversal Hom its residual cohomology is free in just one degree or zero by step 22.2, so $d_H=0$. The dual adjoint identity persists termwise using $(1+h\delta {)}^{-1}h=h(1+\delta h{)}^{-1}$. Finite planar-tree transfer with multiplication at vertices, $h_R$ on internal edges, $i_R$ at leaves and $p_R$ at the root now gives actual cyclic all-arity products and quasi-isomorphism components. Expanding $d_R{h}_R+h_R{d}_R=1-i_R{p}_R$ on each internal edge proves the bar identities by cancellation of the contracted-edge and composed-tree terms. The same root-edge adjoint calculation proves cyclicity. Thus these are genuine admissible structures over R; no merely specified enhancement, formal Hodge inverse or unproved enhancement uniqueness is used. For any second admissible transversal structure over R with the same binary product and fundamental Massey coefficient, the arity/weight construction of steps 18.1, 19.1, 20.1, 21.1, 22.1, 20.2, 21.2 and 22.2 gives its unique admissible comparison with this coherent model, naturally in R. This application requires the second structure to have actually been constructed; no later polygon or matrix conclusion is a premise for the coherent contraction. Here is an explicit all-arity convergence bound for these rational straight polygons. Fix the corner word and an output intersection lift, modulo $Z^2$. Write every supporting line as $n_i$ dot $x=c_i+k_i$ with primitive integer normal $n_i$ and integer $k_i$. A vertex is the intersection of two such lines, so for every vertex family and every edge family its signed distance from that edge lies in ONE of finitely many affine lattices (1/D)Z+c, divided by the fixed norm of $n_i$; choose a common denominator D from the finitely many nonzero integer determinants. The positive distances in this finite list of discrete sets have a uniform lower bound $\delta >0$. A nondegenerate convex polygon has a longest edge of length at least its diameter/N, and a vertex off that edge at height at least delta. Therefore its area is at least delta times diameter/(2N). Once $\mathrm{diameter}\le R$, every vertex lies in a fixed ball of radius R about the chosen output lift, and each affine vertex lattice has at most $O((R+1{)}^2)$ points. The number of candidate N-gons is consequently bounded by a fixed polynomial in R. The exponential exp(-2π Im(tau) Area) dominates that polynomial and every finite number of length derivatives. Thus the positive-area polygon sums and their nilpotent derivatives converge ABSOLUTELY over C, not merely formally in a Novikov parameter. More explicitly, write a finite nilpotent holonomy transport as an ordered product over the edges of its word. If $N_i^{k_i}=0$, each edge exponential is a finite sum of terms $l_i^a{N}_i^a\mathrm{/}a!,0\le a<k_i$, times its scalar transport. For this fixed word the norm of that ordered product is bounded by $C(1+\mathrm{diameter}{)}^K$ for fixed C,K; no commutation between coefficient maps is required. The same bound holds for each finite Taylor coefficient in the independent object parameters. Scalar theta derivatives obey the earlier Gaussian estimate $\exp (-c{u}^2+C\mathrm{\mid }u\mathrm{\mid })$ times a polynomial in |u|. Thus both theta sums and polygon sums admit every finite derivative and nilpotent substitution used here. Formal area coefficients remain finite under the actual area-cutoff polygon supplier; the displayed absolute bound justifies their complex specialization and all finite regroupings. This convergence statement uses the affine polygon classification of [F3], not any later comparison. Finally, the matrix preservation of the contraction can be proved here without a forward premise. At each coefficient vertex introduce an independent truncated scalar ${ϵ}_i$ and replace ${ϵ}_i^a$ by $N_i^a$ on that fibre. The equality $\xi (N_i^{a}v)=((N_i^{\ast }{)}^{a}xi)$(v) moves the insertion between its two half-edges. Gluing adds exponents at the glued vertex, exactly as scalar monomial multiplication. Consequently this rule preserves the polynomial identities $d_R{h}_R+h_R{d}_R=1-i_R{p}_R,p_R{i}_R=1$ and their side conditions, before taking cohomology. Evaluated transition functions and differentials are precisely those of the unipotent bundles. The residual differential remains zero on transversal Hom factors. Hence these evaluated identities give an actual split Dolbeault contraction even for coefficient modules that are not flat over the parameter ring. This argument establishes the matrix-cochain preservation now; the later identification of the resulting full comparison is its application.