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vivo:departmentOrSchool "Science, Faculty of"@en, "Mathematics, Department of"@en ;
edm:dataProvider "DSpace"@en ;
ns0:degreeCampus "UBCV"@en ;
dcterms:creator "Hwang, Junho"@en ;
dcterms:issued "2016-04-28T19:36:41Z"@*, "2016"@en ;
vivo:relatedDegree "Doctor of Philosophy - PhD"@en ;
ns0:degreeGrantor "University of British Columbia"@en ;
dcterms:description "This dissertation studies stability of 3-dimensional quadratic AS-regular algebras and their moduli. A quadratic algebra defined by a regular triple (E, L, σ) is stable if there is no node or line component of E fixed by σ. We first prove stability of the twisted homogeneous coordinate ring B(E, L, σ), then lift stability to that of A(E, L, σ) by analyzing the central element c₃ where B = A/(c₃). We study a coarse moduli space for each type, A, B, E, H, S. S-equivalence of strictly semistable algebras is studied. We compute automorphisms of AS-regular algebras and of those that appear in the boundary of the moduli. We found complete DM-stacks for 2,3-truncated algebras. Type B algebra as Zhang twist of type A is studied. We found exceptional algebras which appear in the exceptional divisor of a blowing-up at a degenerate algebra in the moduli of 3-truncations. 2-unstable algebras are also studied."@en ;
edm:aggregatedCHO "https://circle.library.ubc.ca/rest/handle/2429/57948?expand=metadata"@en ;
skos:note "On the Stability and Moduli ofNoncommutative AlgebrasbyJunho HwangB.A. Math., Pohang University of Science and Technology, 2000B.A. Phys., Pohang University of Science and Technology, 2000M.S. Math., Pohang University of Science and Technology, 2002A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)April 2016© Junho Hwang 2016AbstractThis dissertation studies stability of 3-dimensional quadratic AS-regular al-gebras and their moduli.A quadratic algebra defined by a regular triple (E,L,σ) is stable if thereis no node or line component of E fixed by σ. We first prove stability of thetwisted homogeneous coordinate ring B(E,L,σ), then lift stability to thatof A(E,L,σ) by analyzing the central element c3 where B = A/(c3).We study a coarse moduli space for each type, A, B, E, H, S. S-equivalenceof strictly semistable algebras is studied. We compute automorphisms of AS-regular algebras and of those that appear in the boundary of the moduli.We found complete DM-stacks for 2,3-truncated algebras. Type B algebraas Zhang twist of type A is studied. We found exceptional algebras whichappear in the exceptional divisor of a blowing-up at a degenerate algebra inthe moduli of 3-truncations. 2-unstable algebras are also studied.iiPrefaceThis dissertation consists in mainly two parts.First, we study stability of AS-regular algebras. This is done in Chapter2. It is a paper in preparation and a joint work with Prof. Kai Behrend.I conjectured stability of 3-dimensional quadratic AS-regular algebras. Mycontributions are some computations of weights and description of moduli.Second, we study geometry of AS-regular algebras and their moduli.This is done in Chapter 3. It is mostly an independent work with help frommy advisor Prof. Dr. Kai Behrend.iiiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivList of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiList of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiiAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ixDedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Stability of algebras . . . . . . . . . . . . . . . . . . . . . . . . . . 32.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Stability of regular algebras . . . . . . . . . . . . . . . . . . . . 42.2.1 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2.2 Quadratic regular algebras of dimension 3 . . . . . . . 62.2.3 The main theorem . . . . . . . . . . . . . . . . . . . . . 72.2.4 Twisted homogeneous coordinate rings . . . . . . . . . 122.2.5 More analysis of c3 . . . . . . . . . . . . . . . . . . . . . 172.2.6 Non-special points . . . . . . . . . . . . . . . . . . . . . 222.3 Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.3.1 The main dimension estimate . . . . . . . . . . . . . . 242.3.2 The test configuration of B generated by a filtration . 252.3.3 Exploiting non-commutativity . . . . . . . . . . . . . . 282.3.4 Case: P off X . . . . . . . . . . . . . . . . . . . . . . . . 302.3.5 Case: P not a node, not a fixed point . . . . . . . . . 312.3.6 Case: P smooth fixed point . . . . . . . . . . . . . . . . 352.3.7 Case: P node . . . . . . . . . . . . . . . . . . . . . . . . 36ivTable of Contents2.3.8 Case: Y linear component of X . . . . . . . . . . . . . 382.4 Moduli stacks . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.4.1 Moduli stacks of stable regular algebras . . . . . . . . 432.4.2 Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463 Moduli of 3 dim’l quadratic AS regular algebras . . . . . . 493.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.2 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.2.1 GIT and S-equivalence . . . . . . . . . . . . . . . . . . . 503.2.2 Stability of free, symmetric, exterior algebras . . . . . 513.2.3 Zhang twist of polynomial algebra . . . . . . . . . . . . 543.2.4 Algebras and triple (E,L,σ) . . . . . . . . . . . . . . . 553.3 Type S algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.3.1 Type S1 algebras . . . . . . . . . . . . . . . . . . . . . . 573.3.2 Type S′1 algebras . . . . . . . . . . . . . . . . . . . . . . 663.3.3 Type S′′1 algebras . . . . . . . . . . . . . . . . . . . . . . 703.3.4 Type S2 algebras . . . . . . . . . . . . . . . . . . . . . . 723.3.5 Semi-stability, S-equivalence, and Polystability . . . . 753.3.6 Conclusion and description of moduli space . . . . . . 783.4 Type A algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . 793.4.1 Triple (E,L,σ) . . . . . . . . . . . . . . . . . . . . . . . 803.4.2 Isomorphisms and Automorphisms of type A algebras 803.4.3 Triangle E and the 12 lines . . . . . . . . . . . . . . . . 863.4.4 Singular E and their S-equivalence . . . . . . . . . . . 913.4.5 More singular than nodal . . . . . . . . . . . . . . . . . 923.4.6 Degenerate algebras and Exceptional algebras . . . . . 963.4.7 Conclusion and description of moduli . . . . . . . . . . 1063.5 Type B algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . 1073.5.1 Type B triple (E,L,σ) . . . . . . . . . . . . . . . . . . 1083.5.2 Isomorphisms and Automorphisms of type B algebras 1113.5.3 Nodal E and S-equivalence . . . . . . . . . . . . . . . . 1143.5.4 Degenerate algebras . . . . . . . . . . . . . . . . . . . . 1183.5.5 Type B as Zhang twist of type A. . . . . . . . . . . . . 1243.6 Type E and H algebras . . . . . . . . . . . . . . . . . . . . . . . 1313.6.1 Type E . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1323.6.2 Type H . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1333.7 Unstable algebras . . . . . . . . . . . . . . . . . . . . . . . . . . 1343.7.1 Central extension of non-regular triple . . . . . . . . . 1343.7.2 2-unstable algebras . . . . . . . . . . . . . . . . . . . . . 135vTable of ContentsBibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146viList of Tables3.1 Automorphisms of type S algebras . . . . . . . . . . . . . . . . 613.2 Elements of SL(2,F3) and their fixed points . . . . . . . . . . . 843.3 Automorphisms of type A algebras . . . . . . . . . . . . . . . . 853.4 Weights for U = ⟨x⟩ ⊂ V . . . . . . . . . . . . . . . . . . . . . . . 1353.5 Weights for W = ⟨x, y⟩ ⊂ V . . . . . . . . . . . . . . . . . . . . . 1383.6 Weights for U ⊂W ⊂ V . . . . . . . . . . . . . . . . . . . . . . . 140viiList of Figures3.1 Schematic picture of type A moduli . . . . . . . . . . . . . . . 106viiiAcknowledgementsFirst and foremost thanks are due to my supervisor Prof. Dr. Kai Behrendfor all his mathematical and financial support. Without his consistent sup-port, encouragement, and advice this works would not be started nor com-pleted. I would like to thank Prof. Dr. Jim Bryan and Prof. Dr. KalleKaru for their instructions in algebraic geometry, wonderful lectures, andhelp as committee members.Special thanks go to Prof. Dr. Bumsig Kim, who introduced me toalgebraic geometry and provided me with so much mathematical and finan-cial support and encouragement, including a great research environment inKIAS. I also would like to thank Prof. Dr. Kazushi Ueda for reading andsuggesting improvements for this dissertation.For noncommutative algebraic geometry, I received help from MSRI Non-commutative Algebraic Geometry Summer Graduate School, 2012.UBC algebraic geometry students and postdocs also deserve my thanks.I also thank UBC Mathematics department staff and especially graduatesecretaries for their help.Finally, but mostly, I give heartfelt thanks to my wife for her love andcare.ixDedicationFor my parents.xChapter 1Introduction3-dimensional AS-regular algebra was introduced in [3] and classified withstudy of algebraic aspects. Geometry of this algebra is first studied in [4].To an AS-regular algebra, they associated a triple (E,L,σ), where E is agenus 1 curve or P2, L is a very ample degree 3 line bundle, and σ is anautomorphism of E. Natural algebra related to the triple with elliptic E isthe twisted homogeneous coordinate ring B(E,L,σ) whose Hilbert series is(1 − t3)/(1 − t)3, implying linear growth. To have an algebra A(E,L,σ) ofHilbert series (1 − t)−3, i.e. of quadratic growth, we only consider relationsin degree 2 of B(E,L,σ). To have right Hilbert series, regularity, which isσ∗σ∗L⊗L ≅ σ∗L⊗ σ∗L ,is required.While algebraic properties of AS-regular algebras have been activelystudied, their moduli is less studied. Recent work [1] provides relations ofvarious moduli spaces related to the quadratic AS-regular algebras, follow-ing [10], [11]. However, our moduli and stability are different from theirs,as they do not distinguish non-isomorphic algebras related by the Zhangtwist. It is because their moduli space parametrizes categories of modules,not algebras themselves.Our research was originated from the work [6] on derived moduli ofnoncommutative algebras using DGLA in line with the work for coherentsheaves [5]. They provided GIT problem and numerical criterion for stabil-ity. Stability is given byF (n) > F (1)where F (n) is the Futaki-function which is the sum of dimensions of degreen part of filtration of two-sided ideals. This dissertation provides examplesof stable algebras and strictly semistable algebras.This dissertation consists of two main part. Chapter 2 is devoted to aproof of stability of 3-dimensional quadratic AS-regular algebras with ge-ometric criterion for stability. Some properties of moduli stack are alsoprovided. It is a paper in preparation which is a joint work with Prof. Dr.1Chapter 1. IntroductionKai Behrend. Chapter 3 contains study of triples and their automorphisms.Then we study S-equivalence and compactification of the moduli.Chapter 2 consists mainly of two parts. Section 2.2, 2.3 are devoted ina proof of stability of AS-regular algebra. We first compute Futaki functionfor B(E,L,σ) using the Riemann-Roch Theorem of curves and exploitingnon-commutativity. Using this we compute Futaki function for A(E,L,σ)by analysis of a weight of the central element c3.In Section 2.4 we prove that moduli stack of stable algebras is smoothDM-stack of finite type, consisting of 4 components, see Theorem 2.4.2. Itis also proven that the property ‘stable and regular‘ is an open property, seeProposition 2.4.3.Chapter 3 is study of algebras and their moduli.In Section 3.2, we first review GIT problem, including S-equivalence andpolystability. Then we see that free, symmetric, exterior algebras are strictlysemistable. It is due to a symmetry of them. We also see that algebras giveby Zhang twist of polynomial algebras are strictly semistable.In Section 3.3, we study type S algebras and their moduli.[(P1 × P1 × P1)/S3]parametrizes 2-cutoff of algebras. To parametrize 3-cutoff we need to blowup at the point of the cyclic monomial algebra. In Section 3.3, we studytype A algebras and their moduli[P2/G216]which parametrizes 2-cutoff. The Hesse group G216 acts via its quotientSL(2,F3). To parametrize 3-cutoff we need to blow up at two non-isomorphicpoints and their orbits corresponding to the two degenerate algebras (one iscyclic, the other is nilpotent).Section 3.5 contains detailed study of the triple for type B algebras.Their moduli is also studied. Type B algebras are parametrized by[P1/SL(2,F3)]where SL(2,F3) acts via its quotient PSL(2,F3). This description of type Bmoduli comes from our observation that type B algebra can be obtained bya Zhang twist of a type A algebra with σ2 = id.Section 3.6 contains study for type E, H algebras.Section 3.7 contains list of 2-unstable algebras and a family of algebraswhich we conjecture to be of Hilbert series (1 − t)−3 and unstable.2Chapter 2Stability of algebras2.1 IntroductionIn [6], a notion of stability was introduced for non-commutative gradedalgebras (connected and generated in degree 1). Moreover, quasi-projectivemoduli stacks of Deligne-Mumford type were constructed, for stable gradedalgebras of fixed Hilbert series. Here we are concerned with the Hilbertseries (1 − t)−3, which is the Hilbert series of the projective plane.Interesting examples of algebras with this Hilbert series are quadraticregular algebras of global dimension 3 (see [3], [4]). In this article, we de-termine exactly which of these algebras are stable, and then describe themoduli stack of stable regular algebras.In [4], it was proved that all quadratic regular algebras of dimension 3are associated to elliptic triples (X,σ,L), consisting of a scheme X, anautomorphism σ of X, and line bundle L, which embeds X into P2 as adivisor of degree 3. We prove that the regular algebra A determined by sucha triple is stable if and only if X does not contain any linear componentpreserved by σ, and it does not contain any singularity of X fixed by σ.This leaves only two stable cases: the case where X is non-singular, and thecase where X is a Neron triangle and the automorphism σ acts transitivelyon the set of its components/sides.See Section 2.2.3 for a more detailed outline of the proof of this result,which is the main contribution of this paper.The moduli stack of these stable algebras turns out to be smooth, andto decompose into 4 irreducible components, of dimensions 2, 1, 0, and 0,respectively, see Theorem 2.4.2. These components can be characterized bythe order of the automorphism which σ induces on Pic0(X). This ordercan be 1,2,3, or 4. In the classification of [3], these components correspondto Types A,B,E, and H, respectively. Unfortunately, neither of the twopositive-dimensional components is proper.We prove that the property ‘stable and regular’ is an open property, forflat families of graded algebras. Therefore, the moduli stack we construct isopen in the moduli stack of all stable algebras of Hilbert series (1− t)−3, and32.2. Stability of regular algebrashence also dense in each component which it intersects. We do not addressthe question of whether or not there are components of the moduli stack ofstable algebras which do not contain any regular algebras.Let us emphasize that our goal here is not to classify quadratic regularalgebras, or describe their moduli. This has been done elsewhere, see [4] forthe classification, and [1], and references therein, for moduli. Rather, thepurpose of this work is to determine how our notion of stability relates tothese known moduli spaces.The question of compactifying our moduli stack turns out to be rathersubtle: the geometric invariant theory considerations of [6] provide us withcompactifications using semi-stable algebras, but these compactifications de-pend on an integer q, because they are constructed in terms of algebras trun-cated beyond degree q. Preliminary considerations show that the boundariesof these compactifications grow, as q increases.There is a natural absolute notion of semi-stability, but it does not giverise to compact moduli. We conjecture that all quadratic regular algebras ofdimension 3 are semi-stable. Examples show that the converse is not true:there are semi-stable algebras which are degenerate, in the sense of [4]. Onthe other hand, we do not know any examples of stable algebras which arenot regular. For details on semi-stability, see Chapter 3.We will work over an algebraically closed field of characteristic avoiding 2and 3, and call it C. All our graded algebras A will be connected, i.e.,A =⊕n≥0Anwith A0 = C, and generated in degree 1.2.2 Stability of regular algebrasWe start by reviewing the notion of stability for graded algebras from [6], toset up notation. Then we review the the theory of quadratic regular algebrasof dimension 3, as far as it is relevant for us. We announce our main result,characterizing the stable regular algebras, and outline the proof.We prove some general facts that we will later need in the proof.2.2.1 StabilityLet A be a connected graded algebra, finitely generated in degree 1. Werecall some notions from [6].42.2. Stability of regular algebrasDefinition 2.2.1 The algebra A is q-stable, (for an integer q > 1), if forevery non-trivial test configuration for A, generated in degree 1, the Futakifunction satisfies F (q) > F (1). It is stable, if there exists an N > 0, suchthat it is q-stable for all q > N .The degree 1 part of a test configuration is a filtrationA1 =W (0) ⊃W (1) ⊃ . . . , (2.1)(the inclusions are not strict), such that W (k) = 0, for sufficiently large k.The test configuration B ⊂ A[t, t−1] generated by the filtration (2.1) is theC[t]-algebra defined byBn =⊕kt−k ∑k1+...+kn=kW(k1) . . .W (kn) . (2.2)Here, for fixed n > 0, and k ∈ Z, the sum is over all n-tuples of non-negativenumbers (k1, . . . , kn), such that k1 + . . . kn = k. If k < 0, the set of suchn-tuples is empty, and the coefficient of t−k is An = (A1)n, by definition.SettingI(k)n = ∑k1+...+kn=kW(k1) . . .W (kn)defines a descending sequence of two-sided graded ideals A ⊃ I(1) ⊃ I(2) . . .in A, satisfying I(k)I(`) ⊂ I(k+`), for all k, ` ≥ 0. The fibre over t = 0 of B isthe doubly graded algebraB/tB =⊕k≥0 I(k)/I(k+1) .The weight function of the test configuration (2.2) is given byw(n) = ∑k>0 dim I(k)n = ∑k>0 dim ( ∑k1+...+kn=kW (k1) . . .W (kn)) .The weight w(n) is equal to the total weight of the graded vector space(B/tB)n =⊕k≥0(B/tB)(k)n :w(n) = ∑k≥0k dim(B/tB)(k)n .The Futaki function of the test configuration (2.2) isF (n) = w(n)n dimAn.52.2. Stability of regular algebrasIn particular,F (1) = 1dimA1∑k>0 dimW (k) .To test for stability, we can restrict to test configurations generated byfiltrations such that W (1) /= A1. We will always restrict attention to suchtest configurations.2.2.2 Quadratic regular algebras of dimension 3Recall ([4], Definition 4.5) that an elliptic triple is a triple (X,σ,L), con-sisting of a scheme X, together with a very ample line bundle L, such thatV = Γ(X,L) is of dimension 3, and the closed immersion X → P(V ) makesX into a divisor of degree 3 inside P(V ). Moreover, σ ∶ X → X is anautomorphism of X.We will assume our triples to be regular, which means thatσ∗σ∗L⊗L ≅ σ∗L⊗ σ∗L ,but not linear, which is the stronger conditionσ∗L ≅ L .The quadratic algebra A associated to the regular elliptic triple (X,σ,L)is the quotient of the tensor algebra of V , by the two-sided ideal generatedby the kernel R ofV ⊗ V Ð→ Γ(X,L⊗ σ∗L) (2.3)x⊗ y z→ x⊗ σ∗y .As (2.3) is surjective, and dim Γ(X,L ⊗ σ∗L) = 6, the kernel R is of di-mension 3. Therefore, A has 3 generators in degree 1, and 3 relations indegree 2.By [4] Theorem 6.8.(ii), the algebra A is a regular algebra of global di-mension 3, in particular, its Hilbert series is given by (1− t)−3, equivalently,dimAn = 12(n + 1)(n + 2), for all n ≥ 0.The twisted coordinate ring B, associated to the triple (X,σ,L) is definedbyBn = Γ(X,Ln) ,whereLn = L⊗ σ∗L⊗ . . .⊗ (σn−1)∗L ,62.2. Stability of regular algebraswhich multiplication given by(x0⊗ . . .⊗xn−1) ⋅(y0⊗ . . .⊗ym−1) = x0⊗ . . .⊗xn−1⊗(σn)∗y0⊗ . . .⊗(σn)∗ym−1 .We have dimBn = 3n, for n > 0.There is a canonical morphism of graded C-algebrasAÐ→ B , (2.4)induced by the identification V = A1 = B1. By [4] Theorem 6.8, the mor-phism (2.4) is an epimorphism, whose kernel is equal to both c3A and Ac3,for an element c3 ∈ A3, which is both a left and a right non-zero divisor (ofcourse, c3 is unique up to multiplication by a non-zero scalar).2.2.3 The main theoremTheorem 2.2.2 (Stability of regular algebras) Let (X,σ,L) be a reg-ular elliptic triple, with associated regular algebra A. The following areequivalent:(i) A is stable,(ii) A is q-stable, for every q ≥ 3,(iii) A is 3-stable,(iv) X contains no linear component preserved by σ, and no singularitypreserved by σ.(In (iv), it is not required that the linear component be fixed pointwise byσ.)An elliptic triple without linear components preserved by σ, and withoutsingularities preserved by σ, is necessarily either smooth, or the union of aline and a smooth conic intersecting at two points, or a union of three linesintersecting transversally at three nodes. The second of these three casesis not regular, but exceptional, see [4], 4.9, and is therefore excluded in thetheorem.Thus, if X is smooth, or a triangle on which σ acts by cyclic permutationof the edges, A is stable, in all other cases A is at best semi-stable.Method of proofTo prove the theorem, we prove the following three propositions.72.2. Stability of regular algebrasProposition 2.2.3 Let U ⊂ V be a one-dimensional subspace, such that theline Y = Z(U) ⊂ P(V ) is contained in X ⊂ P(V ), and such that σ∣Y ∶ Y →Xfactors through Y ⊂ X. Then the test configuration for A generated by theflag V ⊃ U ⊃ 0 of A1 has constant Futaki function F̃ , i.e., F̃ (n) = F̃ (1), forall n ≥ 2.Proposition 2.2.4 Let W ⊂ V be a 2-dimensional subspace, such that thepoint P = Z(W ) ∈ P(V ) is contained in X, is a singular point of X, andsatisfies σ(P ) = P . Then the test configuration for A generated by the flagV ⊃W ⊃ 0 of A1 has constant Futaki function.Proposition 2.2.5 (Stability estimates) Consider the filtrationV ⊃W ⊃ . . . ⊃W´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶` times⊃ U ⊃ . . . ⊃ U´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶m times⊃ 0 (2.5)of A1, where at least one of the two integers ` ≥ 0, m ≥ 0 is positive, anddimU = 1, dimW = 2. Let P = Z(W ), and Y = Z(U).Suppose that if P ∈ X, and σ(P ) = P , then P is a non-singular pointof X, and assume that if Y ⊂ X, then X is a Neron triangle, whose sidesare permuted cyclically by σ. Then for the Futaki function F (n) of the testconfiguration generated by the flag (2.5), we have⎧⎪⎪⎨⎪⎪⎩F (n) ≥ F (1) , for n = 1,2F (n) > F (1) , for n ≥ 3 .Propositions 2.2.3 and 2.2.4 will be proved in the next section on twistedcoordinate rings.The stability estimatesLet us set up notation used in the proof of Proposition 2.2.5. Let (X,σ,L)be a regular elliptic triple, with associated regular algebra A and twistedcoordinate ring B = A/c3A. The flag (2.5) generates test configurations(I(k)n ) for B and (J(k)n ) for A. Let w(n) be the weight function of (I(k)n ) andw̃(n) the weight function of (J(k)n ). Let a(n) = dimAn, and b(n) = dimBn,soa(n) = 12(n + 1)(n + 2) , and b(n) = 3n .For future reference, let us also introduceã(n) = a(n − 3) = 12(n − 1)(n − 2) .82.2. Stability of regular algebrasLet F (n) be the Futaki function of (I(k)n ) and F̃ (n) the Futaki function of(J(k)n ).We haveF̃ (n) = w̃(n)12n(n + 1)(n + 2) ,andF̃ (1) = F (1) = 2` +m3,because w̃(1) = w(1) = 2` +m.For n = 2, we have w̃(2) = w(2), and our claim that F̃ (2) ≥ F̃ (1),amounts tow(2) ≥ 8` + 4m.For n ≥ 3 our claim that F̃ (n) > F̃ (1) amounts tow̃(n) > 2` +m6n(n + 1)(n + 2) . (2.6)To prove (2.6), we first prove that lower estimates for w(n), imply lower es-timates for w̃(n) via a bootstrapping method using the short exact sequence0 // An−3 c3 ⋅ // An // Bn // 0 .The larger the integer p such that c3 ∈ J(p)3 , the better the estimates arethat the bootstrapping method gives. For different cases, different valuesfor p apply. In many cases, it will be sufficient that c3 ∈ J(2`)3 , but in thecase where Y is a linear component of X, we will need c3 ∈ J(3`)3 . We willprove these facts in Theorem 2.2.29 and Proposition 2.2.30.Finally, the estimates for w(n) are postponed to the next section underthe heading ‘Estimates’.Our estimates for w(n) and w̃(n) will always be of the formw(n) ≥ `w`(n) +mwm(n) and w̃(n) ≥ ` w̃`(n) +mw̃m(n) , (2.7)with functions w`(n),wm(n), w̃`(n), w̃m(n), which do not depend on ` orm.The stability condition is3w̃(n) − (2` +m)na(n) > 0 ,which we can deduce from`(3w̃`(n) − 2na(n)) +m(3w̃m(n) − na(n)) > 0 .92.2. Stability of regular algebrasLet us therefore introduce notationG`(n) = 3w̃`(n) − 2na(n) = 3w̃`(n) − n(n + 1)(n + 2) ,andGm(n) = 3w̃m(n) − na(n) = 3w̃m(n) − 12n(n + 1)(n + 2) .With this notation, stability will be implied by`G`(n) +mGm(n) > 0 . (2.8)Definition 2.2.6 The one-dimensional subspace U ⊂ V corresponding tothe line Z = Z(U) ⊂ P(V ) is called special, if Z is contained in X, i.e., if Zis a component of X.In the non-special case, the zero locus of U in X is an effective Cartierdivisor D on X, such that L = O(D).Definition 2.2.7 The two-dimensional subspace W ⊂ V corresponding tothe point P = Z(W ) ∈ P(V ) is called special, if P lies on the smooth partof X and σ(P ) = P .One of the most useful facts about the non-special case is that it impliesVW +WV = B2, as we will see later.If W is special, then P is an effective Weil divisor on X, and we haveW = Γ(L(−P )).Bootstrapping: comparing A and B using c3Proposition 2.2.8 Suppose that c3 ∈ J(p)3 . Then we havew̃(n) ≥ ∑1≤n−3i≤n (w(n − 3i) + pa(n − 3i − 3)) ,where the sum is over all integers i satisfying 1 ≤ n − 3i ≤ n.Proof. By properties of test configurations, we have c3J(k)n−3 ⊂ J(k+p)n , forall k ≥ 0 and all n ∈ Z. Moreover, we have a surjection J(k+p)n ↠ I(k+p)n ,whose kernel contains c3J(k)n−3, which proves thatdimJ(k+p)n ≥ dim I(k+p)n + dim c3J(k)n−3 ,102.2. Stability of regular algebrasfor all k ≥ 0 and all n ∈ Z. As c3 is a non-zero divisor, we have dim c3J(k)n−3 =dimJ(k)n−3, and sodimJ(k+p)n ≥ dim I(k+p)n + dimJ(k)n−3 .Thus we havew̃(n) = ∞∑k=1 dimJ(k)n≥ ∞∑k=1 dim I(k)n + p dimAn−3 +∞∑k=1 dimJ(k)n−3= w(n) + pa(n − 3) + w̃(n − 3) ,because for k ≤ 0, we have J(k)n−3 = An−3. From this recursion, we deducew̃(n) ≥ ∞∑i=0w(n − 3i) + p∞∑i=1a(n − 3i) ,which is our claim. 2Corollary 2.2.9 Assume that c3 ∈ J(q`+sm)3 . Thenw̃(n) ≥ ⌊n−13 ⌋∑i=0 (w + (q` + sm)ã)(n − 3i) ,and hence we may usew̃`(n) = ⌊n−13 ⌋∑i=0 (w` + qã)(n − 3i)andw̃m(n) = ⌊n−13 ⌋∑i=0 (wm + sã)(n − 3i) .Both w`(n) and wm(n) will turn out to be polynomial functions of degree2 in n. So we will need the following sums later on.112.2. Stability of regular algebrasLemma 2.2.10ζ2(n) = 3 ⌊n−13 ⌋∑i=0 (n − 3i)2 = 16 n(n + 3)(2n + 3) +⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩0 if n ≡ 0 mod 3−13 if n ≡ 1 mod 3+13 if n ≡ 2 mod 3 ,ζ1(n) = 3 ⌊n−13 ⌋∑i=0 (n − 3i) = 12 n(n + 3) +⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩0 if n ≡ 0 mod 3+1 if n ≡ 1 mod 3+1 if n ≡ 2 mod 3 ,ζ0(n) = 3 ⌊n−13 ⌋∑i=0 1 = n +⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩0 if n ≡ 0 mod 3+2 if n ≡ 1 mod 3+1 if n ≡ 2 mod 3 .Remark 2.2.11 Note that, no matter the divisibility of n modulo 3,n(n + 1)(n + 2) = 3 ζ2(n) − 3 ζ1(n) + 2 ζ0(n) . (2.9)Because of this, it will turn out to be convenient to express all our estimatesfor G`(n) and Gm(n) in terms of ζ2(n), ζ1(n) and ζ0(n).2.2.4 Twisted homogeneous coordinate ringsSuppose (X,σ,L) is an elliptic triple, with associated twisted homogeneouscoordinate ringB =⊕nΓ(X,Ln) , Ln = L⊗Lσ ⊗ . . .⊗Lσn−1 .Let (Z, τ) be another scheme and automorphism, and let φ ∶ Z → X bea morphism, such that σ ○ φ = φ ○ τ . Let N = φ∗L. LetB̃ =⊕nΓ(Z,Nn) , Nn = N ⊗N τ ⊗ . . .⊗N τn−1be the twisted homogeneous coordinate ring of (Z, τ,N).We have a canonical ring morphismB → B̃ .In degree n it is given by the pullback map(φ∗)⊗n ∶ Γ(X,Ln)Ð→ Γ(Z,Nn) ,which exists becauseN τi = (φ∗L)τ i = φ∗(Lσi) .122.2. Stability of regular algebrasLine in XLet U ⊂ V be 1-dimensional, and Z = Z(U) ⊂ P(V ). Assume that Z ⊂ X,and that σ factors through Z. Then we get an associated algebra quotientB → B̃. The algebra B̃ is a twist of the polynomial ring in two variables, ora quantum projective line.In degree 2, we have a short exact sequence0 // Q // V /U ⊗ V /U // B̃2 // 0 .The dimension of B̃2 is 3, by the classification of quantum projective lines,and dimQ = 1. We consider the induced morphism of short exact sequences0 // R //V ⊗ V //B2 //00 // Q // V /U ⊗ V /U // B̃2 // 0 .Let T (V ) be the free or tensor algebra on V , and A = T (V )/R the regularalgebra associated to our elliptic triple. Let C3 ⊂ A3 be the one-dimensionalsubspace generated by c3.Lemma 2.2.12 The map R → Q is surjective.Proof. If not, it is zero. Then R is in the kernel of V ⊗ V → V /U ⊗ V /U ,which is U ⊗ V + V ⊗U . If we choose a basis (xi) for V , with x1 ∈ U , and abasis (fi) for R, then the matrix M , such that f =Mx is of the formM = ⎛⎜⎝m11 a12 x1 a13 x1m21 a22 x1 a23 x1m31 a32 x1 a33 x1⎞⎟⎠ .Because X is defined by detM = 0, we see that the double line x21 = 0 iscontained in X. Also, we see that along this line, the matrix M has rank atmost 1. This contradicts the assumption that A is regular, by Lemma 4.4of [4]. 2Corollary 2.2.13 We have B̃ = T (V )/⟨U,R⟩, where ⟨U,R⟩ denotes thetwo-sided ideal in the tensor algebra T (V ), generated by the subspaces U ⊂ Vand R ⊂ V ⊗ V .132.2. Stability of regular algebrasProof. By construction, B̃ comes with a morphism T (V )/⟨U,R⟩→ B̃. Wehave to construct a morphism in the other direction.By the classification of quantum projective lines, we know that B̃ =T (V /U)/⟨Q⟩. The embedding V /U → T (V )/⟨U⟩ induces an algebra mor-phism T (V /U) → T (V )/⟨U⟩, hence a morphism T (V /U) → T (V )/⟨U,R⟩,which, by the lemma, annihilates Q. This gives us the morphism B̃ →T /⟨U,R⟩, which is the required inverse. 2Corollary 2.2.14 In A3, we have C3 ⊂ UV V + V UV + V V U .Proof. We have the succession of quotientsT (V ) // // A // // B // // B̃ ,which corresponds to the sequence of ideals in T (V )0 ⊂ ⟨R⟩ ⊂ ⟨R,C3⟩ ⊂ ⟨U,R⟩ .In particular, modulo ⟨R⟩, we have C3 ⊂ ⟨U⟩3 = UV V + V UV + V V U . 2Proof of Proposition 2.2.3. In B, we have that UV = V U . It followsthat this is also true in A, as it is a claim about A2 = B2. Then it followsthat UA = AU , because A is generated by V . From this, it follows that infact C3 ⊂ UV V and C3 ⊂ V V U . Let z be a generator of U . Then it followsthat z is a left and right regular element in A, because this is true for anygenerator c3 of C3.The test configuration in A, generated by the flag V ⊃ U ⊃ 0 in A1, isthe sequence of two-sided ideals J(k) = ⟨U⟩k. By what we just proved, wehave ⟨U⟩k = zkA ≅ A(−k) the shift of A by −k.Therefore, for the weight function w̃(n) of the test configuration (J(k)),we havew̃(n) = n∑k=1a(k − 1) ,where a(n) = dimAn = 12(n + 1)(n + 2). This implies thatw̃(n) = 13na(n) ,orF̃ (n) = w̃(n)na(n) = 13 = F̃ (1) .So we see that the Futaki function of (zk) is constant. This proves Propo-sition 2.2.3. 2142.2. Stability of regular algebrasLine with embedded pointConsider now P1 with a 2-dimensional embedded point. Call this schemeZ̃, and Z = Z̃red = P1. The scheme Z̃ is embedded into P(V ) by the choiceof a flag 0 ⊂ U ⊂W ⊂ V , with dimU = 1 and dimW = 2, as the intersectionof two quadrics Z̃ = Z(UW ). The homogeneous coordinate ring of Z̃ isC[V ]/(UW ), which is, in coordinates, C[x, y, z]/(x2, xy).Lemma 2.2.15 Any twisted homogeneous coordinate ring of Z̃ is quadratic.More precisely, if σ ∶ Z̃ → Z̃ is any scheme automorphism, then the associ-ated twisted homogeneous coordinate ring isB̃ = T (V )⟨U⟩⟨W ⟩ + ⟨W ⟩⟨U⟩ + ⟨Q⟩ ,where Q is defined by the exact sequence0 // Q //V ⊗ VU ⊗W +W ⊗U // B̃2 // 0 .Proof. We omit the proof, which is not difficult, as this result is not usedin the proof of the main theorem. 2Assume now that Z̃ ⊂X, and that σ factors through Z̃. This will be thecase, for example, if X is nodal, contains the line Z(U) ⊂X, invariant by σ,and a node Z(W ) ⊂X fixed by σ.Then we get an associated algebra quotient B → B̃. In degree 2, we havea short exact sequence0 // Q //V ⊗ VU ⊗W +W ⊗U // B̃2 // 0 .The dimension of B̃2 is 4, so dimQ = 2. We consider the induced morphismof short exact sequences0 // R //V ⊗ V //B2 //00 // Q //V ⊗ VU ⊗W +W ⊗U // B̃2 // 0 .Lemma 2.2.16 The map R → Q is surjective.152.2. Stability of regular algebrasProof. If not, the intersection of R and UW +WU has dimension 2. If wechoose a basis (xi) for V , with x1 ∈ U , and x2 ∈W , and a basis (fi) for R,such that f1, f2 ∈ UW +WU , then the matrix M , such that f = Mx is ofthe formM = ⎛⎜⎝m11(x1, x2) a12 x1 0m21(x1, x2) a22 x1 0m31 m32 m33⎞⎟⎠ .At the point ⟨0,0,1⟩, this matrix has rank 1, so this contradicts the non-degeneracy assumption, see Lemma 4.4 of [4]. 2Corollary 2.2.17 Let T (V ) be the free algebra on V . ThenB̃ = T (V )⟨U⟩⟨W ⟩ + ⟨W ⟩⟨U⟩ + ⟨R⟩ .Proof. By the lemma, modulo ⟨W ⟩⟨U⟩, we have Q = R. 2Corollary 2.2.18 In A3, we haveC3 ⊂ UWV +UVW + V UW +WUV + VWU +WV U .Proof. We have the succession of quotientsT (V ) // // A // // B // // B̃ ,which corresponds to the sequence of ideals in T (V )0 ⊂ ⟨R⟩ ⊂ ⟨R,C3⟩ ⊂ ⟨U⟩⟨W ⟩ + ⟨W ⟩⟨U⟩ + ⟨R⟩ .In particular, modulo ⟨R⟩, we have C3 ⊂ (⟨U⟩⟨W ⟩ + ⟨W ⟩⟨U⟩)3. 2Singular pointConsider now a point P = Z(W ) ∈ X, fixed by σ, and assume that P isa singularity of X. As σ ∶ X → X is a scheme automorphism, σ inducesan automorphism of the first order neighbourhood Z of P in X, which isisomorphic to SpecC[x, y]/(x, y)2, as P is a singular point of X. The schemeZ = SpecC[x, y]/(x, y)2 is embedded into P(V ) as Z = Z(WW ).Let B̃ the the twisted coordinate ring of (Z,σ∣Z). We have quotientsA→ B → B̃. Choose a vector z ∈ V , z /∈W . Then z̃, the image of z in B̃, is aleft and right regular element. Moreover, both left and right multiplicationby z̃ induce isomorphisms B̃n → B̃n+1, for all n ≥ 2.162.2. Stability of regular algebrasLet Q be the kernel defined by the exact sequence0 // Q //V ⊗ VW ⊗W // B̃2 // 0 .Then B̃ = T (V )/(⟨WW ⟩+ ⟨Q⟩), as can be seen by studying the structure ofB̃.As above, regularity implies that R maps onto Q, and hence thatB̃ = T (V )⟨W ⟩⟨W ⟩ + ⟨R⟩ ,and thatc3 ∈WWV +WVW + VWW . (2.10)In particular, we have that left or right multiplication by z ∈ A inducesan isomorphism (A/⟨W ⟩2)nzÐ→ (A/⟨W ⟩2)n+1 ,for all n ≥ 2.Consider now the (doubly) graded ringR = ∞⊕k=0⟨W ⟩n/⟨W ⟩n+1 .In fact, R is the central fibre of the test configuration generated by W , andA is S-equivalent to R.By the facts we proved, the subring of k-degree zero, R0 = A/⟨W ⟩ isa quantum projective line, and R is generated over R0 by one elementwhich quantum commutes with R0. From this, the graded dimension ofeach J(k)⟨W ⟩k can be computed, giving a proof of Proposition 2.2.4.2.2.5 More analysis of c3Let (X,σ,L) be a regular elliptic triple, such that L /≅ Lσ.Recall the notion of tame line bundle, defined in [4]. A line bundle on Xis tame if either H0 vanishes, or H1 vanishes, or if it is trivial. For tame linebundles, h0 and h1 can be calculated as deg or −deg, respectively. Bundlesgenerated by sections, duals of tame bundles, and bundles with non-negativedegree on each component are all tame. For example, every non-trivial tameline bundle of degree 0 has H0 =H1 = 0.As in [4], for a sheaf F , generated by global sections, we denote by F ′′ thekernel of the epimorphism Γ(X,F )⊗COX → F . The reason for introducingF ′′ is the following Proposition 7.17 of [4].172.2. Stability of regular algebrasProposition 2.2.19 If M is a coherent OX-module generated by global sec-tions, and N is locally free with H1(X,N) = 0, then the kernel and cokernelof the multiplication mapΓ(X,M)⊗k Γ(X,N)Ð→ Γ(X,M ⊗OX N)are given by H0(X,M ′′ ⊗OX N) and H1(X,M ′′ ⊗OX N), respectively.If we apply this to M = L and N = Lσ, we get the exact sequence0 // Γ(L′′ ⊗Lσ) // Γ(L)⊗ Γ(Lσ) // Γ(L⊗Lσ) // 0 .If we identify A2 with Γ(L ⊗ Lσ), this identifies Γ(L′′ ⊗ Lσ) with R, thekernel of multiplication V ⊗ V → A2. By the regularity of A, we can writeA3 as the cokernel of R⊗A1 → V ⊗A2. This gives the exactness of the lowerrow in the following diagram:Γ(L′′ ⊗Lσ)⊗ Γ(Lσ2) β //=Γ(L′′ ⊗Lσ ⊗Lσ2) // _C3 _// 0Γ(L′′ ⊗Lσ)⊗ Γ(Lσ2) // Γ(L)⊗ Γ(Lσ ⊗Lσ2) // A3 // 0 .(2.11)The exactness of the upper row comes from Lemma 7.29 of [4].The case where P lies on XConsider a 2-dimensional subspace W ⊂ Γ(L), which does not generate L.Let P ∈X be the point where W fails to generate L. (Mapping P into P2, viathe embedding X → P2, we obtain the point in P2, dual to the hyperplaneW in V .) We have a surjection of sheaves on XW ⊗C OX // // L⊗OX mP = mPL . (2.12)If P is a smooth point of X, then P is a Cartier divisor, and hence mPL =L(−P ), but we do not want to make this assumption.We note also that Γ(L ⊗OX mP ) = W , and that mPL is generated byglobal sections. We consider (mPL)′′:0 // (mPL)′′ //W ⊗OX // mP L // 0If P is a smooth point of X, then (mPL)′′ is locally free of rank 1, but nototherwise.182.2. Stability of regular algebrasLemma 2.2.20 Any non-zero element of Λ2W induces an isomorphism(mPL)′′ ∼ // (mPL)∨ .Proof. Let us choose a non-zero element in Λ2W . As Λ2W is one-dimensional, this element is a basis element, and also induced a basis el-ement in the dual space Λ2W ∨. Let us denote this element by κ, it isa non-degenerate alternating pairing on W , and induces a non-degeneratealternating pairing on the trivial bundle W ⊗OX .We claim that (mPL)′′ ⊂ W ⊗ OX is isotropic with respect to κ. As κtakes values in OX , and mP contains a non-zero divisor of OX , we can checkthis claim after removing P from X, where it becomes trivial, as all sheavesinvolved become locally free. Thus, κ induces a pairingκ ∶ (mPL)′′ ⊗OX mPL // OX .This pairing is non-degenerate, i.e., induces injections (mPL)′′ → (mPL)∨and mPL → (mPL)′′∨. To prove these facts, it is again enough to restrictto X − {P}, because both (mPL)′′ and mPL are submodules of locally freefinite rank OX -modules.By the snake lemma, the cokernel of (mPL)′′ → (mPL)∨ is equal to thekernel of mPL→ (mPL)′′∨, and hence vanishes. 2Lemma 2.2.21 We haveH1((mPL)′′ ⊗Lσ) = 0 , and H1((mPL)′′ ⊗Lσ ⊗Lσ2) = 0 .Proof. By the previous lemma, there exists a monomorphism L−1 →(mPL)∨ ≅ (mPL)′′, whose cokernel is supported over {P}. Hence we getinduced epimorphisms H1(L−1 ⊗ Lσ) → H1((mPL)′′ ⊗ Lσ) and H1(L−1 ⊗Lσ ⊗ Lσ2) → H1((mPL)′′ ⊗ Lσ ⊗ Lσ2). Now it suffices to remark that byLemma 7.18 of [4], both line bundles L−1 ⊗ Lσ and L−1 ⊗ Lσ ⊗ Lσ2 aretame, and so their respective H1 vanishes; in the first case because L−1⊗Lσis a non-trivial line bundle of degree 0, and in the second case becauseL−1 ⊗Lσ ⊗Lσ2 is of degree 3. 2Lemma 2.2.22 The sheaf mPLσ is also generated by its global sections.We have (mPLσ)′′ ≅ (mPLσ)∨, as well as H1((mPLσ)′′ ⊗Lσ2) = 0.Proof. The same proofs apply. 2192.2. Stability of regular algebrasChoosing a basis for V /W , we get the exact square(mPL)′′ //W ⊗OX //mPLL′′ //Γ(L)⊗OX //LmP // OX // OP(2.13)For N = Lσ and N = Lσ⊗Lσ2 , we consider the induced short exact sequence0 // (mPL)′′ ⊗N // L′′ ⊗N // mPN // 0 .In both cases, we get an induced short exact sequence of vector spaces0 // Γ((mPL)′′ ⊗N) // Γ(L′′ ⊗N) // Γ(mPN) // 0 ,because H1((mPL)′′ ⊗N) = 0, by Lemma 2.2.21.We obtain the morphism of short exact sequences0 // Γ((mPL)′′ ⊗Lσ)⊗ Γ(Lσ2) //αΓ(L′′ ⊗Lσ)⊗ Γ(Lσ2)β//0 // Γ((mPL)′′ ⊗Lσ ⊗Lσ2) // Γ(L′′ ⊗Lσ ⊗Lσ2) //// Γ(mPLσ)⊗ Γ(Lσ2) //γ0// Γ(mPLσ ⊗Lσ2) // 0 .By Lemma 2.2.22, γ is surjective. It follows that cokα → cokβ is an epi-morphism. Hence we get a morphism of exact sequencesΓ((mPL)′′ ⊗Lσ)⊗ Γ(Lσ2) α // _Γ((mPL)′′ ⊗Lσ ⊗Lσ2) // _cokα// 0Γ(L′′ ⊗Lσ)⊗ Γ(Lσ2) β // Γ(L′′ ⊗Lσ ⊗Lσ2) // C3 // 0(2.14)202.2. Stability of regular algebrasComposing (2.14) with (2.11), we obtain the morphism of exact se-quencesΓ((mPL)′′ ⊗Lσ)⊗ Γ(Lσ2) // _Γ((mPL)′′ ⊗Lσ ⊗Lσ2) // _cokα// 0Γ(L′′ ⊗Lσ)⊗ Γ(Lσ2) // Γ(L)⊗ Γ(Lσ ⊗Lσ2) // A3 // 0(2.15)By construction, Γ((mPL)′′ ⊗ Lσ ⊗ Lσ2) → Γ(L) ⊗ Γ(Lσ ⊗ Lσ2) factorsthrough W ⊗ Γ(Lσ ⊗Lσ2). We conclude:Proposition 2.2.23 If P = Z(W ) ∈X, then in A3, we have c3 ∈WV V .Smooth pointTo go further, let us now assume that P is a smooth point of X. Then wehave mPL = L(−P ), and (mPL)′′ = L−1(P ).Consider the lattice of subspacesΓ(L)⊗ Γ(Lσ ⊗Lσ2)W ⊗ Γ(Lσ ⊗Lσ2)OOΓ(L−1(P )⊗Lσ ⊗Lσ2)OOW ⊗ Γ(Lσ ⊗Lσ2(−T ))jjΓ(L−1(P )⊗Lσ)⊗ Γ(Lσ2)αOOΓ(L−1(P )⊗Lσ ⊗Lσ2(−T )) .OOjjHere T is an effective Cartier divisor of degree 1 or 2.Lemma 2.2.24 Let Q ∈ Xreg be the unique point such that Lσ(P ) = L(Q).Suppose that Q /∈ suppT , and that L−1(P ) ⊗ Lσ ⊗ Lσ2(−T ) is generated byglobal sections. Then we haveΓ(L−1(P )⊗Lσ ⊗Lσ2)= Γ(L−1(P )⊗Lσ)⊗ Γ(Lσ2) + Γ(L−1(P )⊗Lσ ⊗Lσ2(−T )) .212.2. Stability of regular algebrasProof. The line bundle L−1(P )⊗Lσ has degree 1, so it has an essentiallyunique non-zero section, which we shall call g ∈ Γ(L−1(P )⊗Lσ). It vanishesexactly at the point Q. We conclude that Γ(L−1(P ) ⊗ Lσ) = Γ(L−1(P ) ⊗Lσ(−Q)). Therefore the image of Γ(L−1(P )⊗Lσ)⊗ Γ(Lσ2) in Γ(L−1(P )⊗Lσ ⊗Lσ2) is contained in Γ(L−1(P )⊗Lσ ⊗Lσ2(−Q)).As the bundle L−1(P ) ⊗ Lσ ⊗ Lσ2(−T ) is generated by global sections,Γ(L−1(P )⊗Lσ ⊗Lσ2(−T )) is not contained in Γ(L−1(P )⊗Lσ ⊗Lσ2(−Q)).This implies the lemma for dimension reasons. 2Corollary 2.2.25 Under the same assumptions, as subspaces of W⊗Γ(Lσ⊗Lσ2), we haveΓ(L−1(P )⊗Lσ⊗Lσ2) ⊂ Γ(L−1(P )⊗Lσ)⊗Γ(Lσ2) +W ⊗Γ(Lσ⊗Lσ2(−T )) .Corollary 2.2.26 If P = Z(W ) is a smooth point of X such that σ(P ) = P ,then in A3, we have c3 ∈WVW .Proof. This follows from the above considerations, upon taking T = P , inparticular Sequence (2.14), and the fact that V ⊗W → Γ(Lσ ⊗Lσ2(−P )) issurjective. 2Corollary 2.2.27 If P = Z(W ) is a smooth point of X, and σ is a trans-lation, of order not equal to 2, then in A3, we have c3 ∈WWW .Proof. If σ is the translation by the point S ∈ Xreg, then Q = P − 3S, inthe group law on Xreg. Moreover, P σ = P − S, and P σ2 = P − 2S, so thelemma and its corollary apply to T = P + P σ, as 2S /= 0.We also use that fact that W ⊗W → Γ(Lσ(−P σ) ⊗ Lσ2(−P σ2)) is sur-jective, which follows from Lemma 2.3.1, below. 22.2.6 Non-special pointsThis is the case where if P ∈X, then σ(P ) /= P . We do not exclude the casethat P is a singular point of X.Lemma 2.2.28 Let W be a non-special 2-dimensional subspace of V = B1.In B2 we have WV + VW = V V = B2.Moreover, if P /∈X, we have the stronger result that B2 =WV .222.2. Stability of regular algebrasProof. If the base locus of W is empty, W generates L, and we have ashort exact sequence of vector bundles on X:0 //M //W ⊗OX // L // 0 .Here M is a line bundle which is isomorphic to L−1, but not canonically so.Tensoring with Lσ and taking global sections gives us the exact sequenceW ⊗ Γ(Lσ) // Γ(L⊗Lσ) // H1(M ⊗Lσ) .As L /≅ Lσ, we have M ⊗ Lσ /≅ OX , and hence H1(M ⊗ Lσ) = 0, and soW ⊗ Γ(Lσ) → Γ(L ⊗ Lσ) is surjective. This proves that WV = B2. (Indegenerate cases, this proof uses the fact that L−1 ⊗ Lσ is tame, in thelanguage of [4], which is proved in Lemma 7.18 of [ibid.].)For a line bundle M on X, and reduced point Q ∈X, the set of sectionsof M vanishing at Q is Γ(X,M ⊗ mQ) ⊂ Γ(X,M). (We do not use thenotation Γ(X,M(−Q)) to include the case where Q is a singularity, andhence not a Cartier divisor.) If M has sections which do not vanish at Q,then dim Γ(M ⊗mQ) = dim Γ(M) − 1.Now suppose that W has a base point P ∈ X. The base locus of W ′ =σ∗W ⊂ Γ(X,σ∗L) is P ′ = σ−1P . We have W = Γ(L ⊗ mP ) ⊂ Γ(L) andW ′ = Γ(Lσ ⊗ mP ′) ⊂ Γ(Lσ). Because P ′ /= P , the line bundle L ⊗ Lσ hassections which vanish at P but not at P ′, and sections which vanish at P ′but not at P . Thus Γ(L⊗ Lσ ⊗mP ) and Γ(L⊗ Lσ ⊗mP ′) are two distinctcodimension 1 subspaces of Γ(L ⊗ Lσ). Their sum is then necessarily thewhole space:Γ(L⊗mP ⊗Lσ) + Γ(L⊗Lσ ⊗mP ′) = Γ(L⊗Lσ) . (2.16)We now remark that the image of Γ(L⊗mP )⊗Γ(Lσ) in Γ(L⊗Lσ) is equalto Γ(L⊗mP ⊗Lσ). (This follows formally from the fact that Γ(L)⊗Γ(Lσ)→Γ(L ⊗ Lσ) is surjective, and that Γ(L ⊗mP ) ⊗ Γ(Lσ) ⊂ Γ(L) ⊗ Γ(Lσ) andΓ(L ⊗ Lσ ⊗ mP ) ⊂ Γ(L ⊗ Lσ) are both codimension 1 subspaces.) Thus areformulation of (2.16) is WV ′ + VW ′ = V V ′. This is our claim. 2An application to the study of c3Theorem 2.2.29 Let W ⊂ V be an arbitrary 2-dimensional subspace. InA3, we have c3 ∈WWV +WVW + VWW .Proof. Let P = Z(W ). If P /∈ X, we have V V = WV , and hence A3 =V V V = WV V = WWV , and the claim is trivial. If P ∈ X, but σ(P ) /= P ,232.3. Estimateswe use Proposition 2.2.23, to deduce c3 ∈ WV V = W (WV + VW ). Thisleaves the case where P ∈ X, and σ(P ) = P . If P is a non-singular point,our claim follows from Corollary 2.2.26. If P is a singular point, it followsfrom (2.10). 2We also need the following amplification:Proposition 2.2.30 Suppose X is a Neron triangle, and σ permutes thesides of X cyclically. Assume that P = Z(W ) is a point on X. Then in A3we have c3 ∈WWW .Proof. Fix a base point for X, making Xreg a group scheme, and σ atranslation. Then the order of σ is not equal to 2. So if P is a smooth point,the claim follows from Corollary 2.2.27. If P is a node, we can deduce theproof from the case where P is smooth by a degeneration argument.22.3 EstimatesHere we prove the stability estimates claimed in Proposition 2.2.5, by find-ing suitable values for the functions w`(n) and wm(n), and then provingFormula (2.8).We may assume that X is either smooth, or a Neron triangle, on whichσ acts by cyclically permuting the sides.We always use notation P = Z(W ) and Y = Z(U), and treat the followingcases in turn:(i) P lies off X,(ii) P lies on X, but is neither a node of X nor a fixed point of σ,(iii) P ∈Xreg is a fixed point of σ,(iv) P is a node of X,(v) Y is a linear component of X.In the cases (i) to (iv), we assume Y /⊂ X. Most of these cases are dividedinto two subcases, a case where ` is large compared to m, which includesthe case m = 0, and the converse case.2.3.1 The main dimension estimateLemma 2.3.1 Let L and M be line bundles on X, both generated by globalsections, and both of degree at least 2. The multiplication mapΓ(L)⊗ Γ(M)Ð→ Γ(L⊗M)242.3. Estimatesis surjective, unless degL = degM = 2 and L ≅M .Proof. Because of Proposition 2.2.19, an equivalent statement is thatH1(L′′ ⊗M) = 0 .The proof will proceed by induction on the integer degL+degM . The basecase is when degL = degM = 2. In this case, L′′ ≅ L−1 is a tame line bundle.Then L′′⊗M is also tame. Hence H1(L′′⊗M) vanishes if and only if L /≅M .Now suppose that degL > 2. Find a point P ∈ X, such that L(−P ) isstill generated by global sections, and such that L(−P ) /≅M . We can applythe induction hypothesis to the bundles L(−P ) and M , obtaining thatH1(L(−P )⊗M ′′) = 0 .But there is a surjectionH1(L(−P )⊗M ′′) // // H1(L⊗M ′′)proving that H1(L⊗M ′′) = 0. 2Corollary 2.3.2 Let L1, . . . , Ln be line bundles of degree at least 2 on X,all generated by global sections. The mapΓ(L1)⊗ . . .⊗ Γ(Ln)Ð→ Γ(L1 ⊗ . . .⊗Ln)is surjective, unless all Li are of degree 2 and pairwise isomorphic to eachother.Proof. Assume that the bundles are not all of degree 2 or not all isomorphicto one another. Then, after relabelling, we may assume that if degL1 =degL2 = 2, then L1 /≅ L2. Then we can apply the lemma successively toL1 ⊗ . . .⊗Li and Li+1, for i = 1, . . . , n − 1. 22.3.2 The test configuration of B generated by a filtrationWe will write {V γWαUβ} ⊂ Bγ+α+βdo denote the sum over all products W1 . . .Wγ+α+β, where γ of the Wi areequal to V , and α of the Wi are equal to W , and β of the Wi are equal toU . For example,{VWU} = VWU + V UW +WUV +WV U +UVW +UWV ⊂ B3 .252.3. EstimatesLemma 2.3.3 Suppose that ` > 0 and m > 0. Then we haveI(k)n =∑α,β{V n−α−βWαUβ} ,where (α,β) ranges over all pairs of non-negative integers satisfying(i) α + β ≤ n,(ii) α + β(` + 1) ≤ k ≤ α` + β(` +m).Proof. By construction, {V n−α−βV αUβ} partakes in I(k)n if and only ifthere exist integers 0 < i1, . . . , iα ≤ ` and ` < j1, . . . , jβ ≤ ` +m such thatk = i1 + . . .+ iα + j1 + . . .+ jβ. This immediately proves the stated bounds onk.Conversely, if (ii) is satisfied, we can write k = k1 + k2, with α ≤ k1 ≤ α`and β(` + 1) ≤ k2 ≤ β(` +m). Then we can write k1 = i1 + . . . + iα with0 < i1, . . . , iα ≤ ` and k2 = j1 + . . . + jβ with ` < j1, . . . , jβ ≤ ` +m. 2Lemma 2.3.4 Suppose that ` > 0 and m = 0. Then we haveI(k)n =∑α{V n−αWα} ,where α ranges over all integers such that α ≤ n and α ≤ k ≤ α`. 2Lemma 2.3.5 Suppose that ` = 0 and m > 0. Then we haveI(k)n =∑β{V n−βUβ} ,where β ranges over all integers satisfying β ≤ n and β ≤ k ≤ βm. 2Useful reformulations: the W -caseLemma 2.3.6 For all α = 1, . . . , n we have(i) if (α − 1)` < k ≤ α` then {V n−αWα} ⊂ I(k)n ,(ii) if n` + (α − 1)m < k ≤ n` + αm then {Wn−αUα} ⊂ I(k)n .Proof. The hardest claim is that n`+(α−1)m < k implies (n−i)+α(`+1) ≤k. This is proved by noting that under our assumptions0 ≤ (n − α)(` − 1) + (m − 1)(α − 1) .Reformulating gives(n − α) + α(` + 1) ≤ n` + (α − 1)m + 1 ,which is what we need. 2262.3. EstimatesThe following variation will also be needed:Lemma 2.3.7 For all i = 1, . . . , n − 1, and every k, such that(i − 1)` < k ≤ i` ,we have {V n−iW i} ⊂ I(k)n .For all i = 1, . . . , n − 1, and all k which satisfy(n − 1)` + (i − 1)m < k ≤ (n − 1)` + im (2.17)we have {VWn−1−iU i} ⊂ I(k)n .For (n− 1)(`+m) < k ≤ (n− 1)(`+m)+ `, we have {WUn−1} ⊂ I(k)n , andfor (n − 1)(` +m) + ` < k ≤ n(` +m), we have Un ⊂ I(k)n .Proof. Let us check the second claim. We have that0 ≤ (` − 1)(n − i − 1) + (m − 1)(i − 1) ,because each of the four quantities involved is individually non-negative, byour assumptions. Rewriting this inequality gives(n − 1 − i) + i(` + 1) ≤ (n − 1)` + (i − 1)m + 1 . (2.18)We also have (n − 1)` + im ≤ (n − 1 − i)` + i(` +m) . (2.19)Now suppose that k satisfies (2.17). Then, in fact,(n − 1)` + (i − 1)m + 1 ≤ k ≤ (n − 1)` + im .From (2.18) and (2.19) we conclude(n − 1 − i) + i(` + 1) ≤ k ≤ (n − 1 − i)` + i(` +m) .With α = n − 1 − i and β = i this isα + β(` + 1) ≤ k ≤ α` + β(` +m) ,which is the condition we need. 2272.3. EstimatesThe U-caseLemma 2.3.8 Suppose that m > 0. For all i = 1, . . . , n we have(i) if (i − 1)(` +m) < k ≤ i(` +m) −m then {V n−iWU i−1} ⊂ I(k)n ,(ii) if i(` +m) −m < k ≤ i(` +m) then {V n−iU i} ⊂ I(k)n .Proof. Similar. 22.3.3 Exploiting non-commutativityLet Z(U) /⊂X. The main dimension estimate Corollary 2.3.2, gives thatdimV n−α−βWαUβ ≥ 3(n − α − β) + 2α= 3n − α − 3β ,at least if n − α − β ≥ 1. We can improve this estimate for {V n−α−βWαUβ}.Proposition 2.3.9 We have(i) for 0 ≤ β ≤ n2 ,dim{V n−βUβ} ≥ 3n − 2β ,(ii) for n2 ≤ β ≤ n,dim{V n−βUβ} ≥ 4n − 4β .(iii) for 1 ≤ β ≤ n+12 ,dim{V n−βWUβ−1} ≥ 3n + 1 − 2β ,(iv) for n+12 ≤ β ≤ n,dim{V n−βWUβ−1} ≥ 4n + 2 − 4β .If P = Z(W ) is a smooth point of X, such that σ(P ) = P , we also have(i) for 0 ≤ β ≤ n2dim{Wn−βUβ} ≥ 2n − β ,(ii) for n2 ≤ β ≤ ndim{Wn−βUβ} ≥ 3n − 3β .282.3. EstimatesProof. Choosing a non-zero element s ∈ U defines an injection OX → L,which identifies L with O(D), where D is the Cartier divisor on X, definedby the vanishing of s. Let E be the greatest common divisor if D and Dσ.Because D /=Dσ, as L /≅ Lσ, the degree of E is at most 2. LetH = Γ(X,L⊗Lσ(−E)) ⊂ Γ(L⊗Lσ) .Then an argument counting dimensions, using that UV = Γ(L ⊗ Lσ(−D))and V U = Γ((L ⊗ Lσ(−Dσ)), proves dimH = 6 − degE ≥ 4, and that thatinside B2 we haveH = UV + V U .Now for 0 ≤ β ≤ n2 , we have{V n−βUβ} ⊃ {V βUβ}V n−2β ⊃HβV n−2β ,and hencedim{V n−βUβ} ≥ dimHβV n−2β≥ 3(n − 2β) + β dimH≥ 3(n − 2β) + 4β= 3n − 2β ,using Corollary 2.3.2.Similarly, for n2 ≤ β ≤ n, we have{V n−βUβ} ⊃ {V n−βUn−β}U2β−n ⊃Hn−βU2β−n ,and hencedim{V n−βUβ} ≥ dimHn−βU2β−n≥ 4(n − β)≥ 4n − 4β .For the next two claims, involving a factor of W , we proceed similarly.Let us explicate the case 1 ≤ β ≤ n+12 . Here we have{V n−βWUβ−1} ⊃ {V n−βUβ−1}W ⊃{V β−1Uβ−1}V n+1−2βW ⊃Hβ−1V n+1−2βW ,292.3. Estimatesand hencedim{V n−βWUβ−1} ≥ dimHβ−1V n+1−2βW≥ 3(n + 1 − 2β) + (β − 1)dimH + 2≥ 3n + 5 − 6β + 4(β − 1)= 3n + 1 − 2β ,using Corollary 2.3.2, or in the case that W is a node, Lemma 2.3.10, below.The last two claims have the same proof as the first two, except for wereplace L by L(−P ), throughout. This means we replace D by D − P , andthe greatest common divisor E of D −P and Dσ −P is of degree at most 1.The line bundle L(−1) ⊗ Lσ(−P )(−E) has degree at least 3, and thereforethe proof goes through, mutatis mutandis. 22.3.4 Case: P off XThis is the case where P = Z(W ) is not on X. Then Y = Z(U) cannot be acomponent of X, so it intersects X in a Cartier divisor.Let us first deal with the case where m = 0. By Lemma 2.3.6, for everyα = 1, . . . , n, we have ` instances of k where {V n−αWα} ⊂ I(k)n . For α < n, weuse Lemma 2.2.28 to conclude that V n−αWα = V n = Bn, which has dimen-sion 3n. For α = n, the estimate dimWn ≥ 2, will be sufficient. Therefore,we may usew`(n) = 3n(n − 1) + 2 (2.20)= 3n2 − 3n + 2 .Now by Theorem 2.2.29, we have that c3 ∈ J(2`)3 , and so we compute(w` + 2ã)(n) = 3n2 − 3n + 2 + (n − 1)(n − 2)= 4n2 − 6n + 4 ,and thenG`(n) = 4ζ2(n) − 6ζ1(n) + 4ζ0(n) − n(n + 1)(n + 2)= 4ζ2(n) − 6ζ1(n) + 4ζ0(n) − 3ζ2(n) + 3ζ1(n) − 2ζ0(n)= ζ2(n) − 3ζ1(n) + 2ζ0(n) .This is 0, for n = 2, and positive for n ≥ 3, which is what we needed to prove.302.3. EstimatesNow if m > 0, we use Lemma 2.3.7. Then for every α = 1, . . . , n − 1, wehave ` instances of k where {V n−αWα} ⊂ I(k)n . We also have ` instancesof k where WUn−1 ⊂ I(k)n , which has dimension 2. Thus, we get the sameestimate (2.20) we used above, leading to the same value for G`:G`(n) = ζ2(n) − 3ζ1(n) + 2ζ0(n) .We also get that for every β = 1, . . . , n−1, there arem instances of k where{VWn−1−βUβ} ⊂ I(k)n . Again by Lemma2.2.28, we have VWn−1−βUβ =V n−βUβ, which has dimension 3(n − β), by Corollary 2.3.2. We add an-other m instances of Un, which has dimension 1. This gives uswm(n) = n−1∑β=1(3n − 3β) + 1= 32n2 − 32n + 1 ,andGm(n) = 32ζ2(n) − 32ζ1(n) + ζ0(n) − 12n(n + 1)(n + 2)= 0 .Together with the above, we get1`(`G`(n) +mGm(n)) = G`(n) + m`Gm(n)= ζ2(n) − 3ζ1(n) + 2ζ0(n) ,which we have already remarked is 0, for n = 2, and positive for n ≥ 3. Thisfinishes the case where P is not on X.2.3.5 Case: P not a node, not a fixed pointNow we assume that P = Z(W ) is on X, but is neither a singularity of X,nor a fixed point of σ. We also assume that Y = Z(U) is not contained inX, so it defines a Cartier divisor on X.We distinguish to subcases: ` ≥m ≥ 0, and m > ` ≥ 0.In the subcase ` ≥ m ≥ 0, the main fact we use about W is that V V =VW +WV in B2 = A2, see Lemma 2.2.28. We also use the main estimateCorollary 2.3.2.For the subcasem > ` ≥ 0, we exploit the non-commutativity for {V n−βUβ}and {V n−βWUβ−1}, using Lemma 2.3.9.Finally, it will be important that c3 ∈ I(2`)3 .312.3. EstimatesSubcase ` ≥m ≥ 0Let us first deal with the case where m = 0. Then by Lemma 2.3.6, foreach α = 1, . . . , n, we have ` instances of k, such that {V n−αWα} ⊂ I(k)n . ByLemma 2.2.28, we have(i) for 1 ≤ α ≤ ⌊n2 ⌋, that {V n−αWα} = V n = Bn, which has dimension 3n,(ii) for ⌊n2 ⌋+1 ≤ α ≤ n−1, that {V n−αWα} ⊃ V 2n−2αW 2α−n, and the latterhas dimension at least 4n − 2α, by Corollary 2.3.2.We also use that dimWn ≥ 2, which can be deduced, for example, fromCorollary 2.3.2, upon choice of a suitable U . Altogether, we getw`(n) = ⌊n2 ⌋∑α=1 3n +n−1∑α=⌊n2⌋+1(4n − 2α) + 2 (2.21)= n−1∑α=1(4n − 2α) −⌊n2⌋∑α=1(n − 2α) + 2= 4n(n − 1) − n(n − 1) − n⌊n2 ⌋ + ⌊n2 ⌋(⌊n2 ⌋ + 1) + 2= 114n2 − 52n + 2 − 12{n2 } .We have c3 ∈ J(2`)3 , by Theorem 2.2.29, and so to calculate G`, we use(w` + 2ã)(n) = 114n2 − 52n + 2 − 12{n2 } + (n − 1)(n − 2)= 154n2 − 112n + 4 − 12{n2 } .This gives usG`(n) = 154ζ2(n) − 112ζ1(n) + (4 − 12{n2 })ζ0(n) − n(n + 1)(n + 2)= 154ζ2(n) − 112ζ1(n) + (4 − 12{n2 })ζ0(n) − 3ζ2(n) + 3ζ1(n) − 2ζ0(n)= 34ζ2(n) − 52ζ1(n) + (2 − 12{n2 })ζ0(n) .Thus G`(2) ≥ 0, and G`(n) > 0, for n ≥ 3, which is what we needed to prove.In the case that m > 0, we use Lemma 2.3.7, instead of Lemma 2.3.6.This gives us, for each α = 1, . . . , n − 1, altogether ` instances of k where{V n−αWα} ⊂ I(k)n , and another ` instances of k where WUn−1 ⊂ I(k)n . Using322.3. EstimatesLemma 2.2.28, as above, and the fact that dimWUn−1 = 2, we get the sameestimate (2.21), which we used above, and henceG`(n) = 34ζ2(n) − 52ζ1(n) + (2 − 12{n2 })ζ0(n) ,as above.Lemma 2.3.7 gives us, for each β = 1, . . . n − 1, also m instances of k forwhich {VWn−1−βUβ} ⊂ I(k)n . For such k, we get dim I(k)n ≥ 2n + 2 − 2β, for1 ≤ β ≤ n − 2 (because of V V = VW +WV ), and 3, for β = n − 1. We alsohave m instances of Un, for a contribution of 1. In total,wm(n) ≥ n−2∑β=1(2n + 2 − 2β) + 3 + 1= n2 + n − 2 .HenceGm(n) ≥ ζ2(n) + ζ1(n) − 2ζ0(n) − 12n(n + 1)(n + 2)= ζ2(n) + ζ1(n) − 2ζ0(n) − 32ζ2(n) + 32ζ1(n) − ζ0(n)= −12ζ2(n) + 52ζ1(n) − 3ζ0(n) .If we now assume that ` ≥m, then we get1m(`G`(n) +mGm(n)) ≥ 1m(mG`(n) +mGm(n))= G`(n) +Gm(n)= 14ζ2(n) − (1 + 12{n2 }) ζ0(n) .This is equal to 0 at n = 2, but positive for n ≥ 3, which is what we neededto prove.Subcase m > ` ≥ 0We use Lemma 2.3.8. This gives us, for every β = 1, . . . , n, the existenceof ` instances of k where V n−βWUβ−1 ⊂ I(k)n , and m instances of k whereV n−βUβ ⊂ I(k)n . We estimate these dimensions with Lemma 2.3.9 to obtain332.3. Estimatesw`(n) = ⌊n+12 ⌋∑β=1 (3n + 1 − 2β) +n∑β=⌊n+12⌋+1(4n + 2 − 4β) (2.22)= n∑β=1(4n + 2 − 4β) −⌊n+12⌋∑β=1 (n + 1 − 2β)= 2n2 − ⌊n+12 ⌋(⌈n+12 ⌉ − 1)= 74n2 + 12{n2 } .and (not forgetting dimUn = 1)wm(n) = ⌊n2 ⌋∑β=1(3n − 2β) +n∑β=⌊n2⌋+1(4n − 4β) + 1= n∑β=1(4n − 4β) −⌊n2⌋∑β=1(n − 2β) + 1= 2n2 − 2n + 1 − ⌊n2 ⌋(⌈n2 ⌉ − 1)= 74n2 − 32n + 1 − 12{n2 } .We have(w` + 2ã)(n) = 74n2 + 12{n2 } + (n − 1)(n − 2)= 114n2 − 3n + 2 + 12{n2 } ,which givesG`(n) = 114ζ2(n) − 3ζ1(n) + (2 + 12{n2 })ζ0(n) − n(n + 1)(n + 2)= −14ζ2(n) + 12{n2 }ζ0(n) .We also haveGm(n) = 74ζ2(n) − 32ζ1(n) + (1 − 12{n2 })ζ0(n) − 12n(n + 1)(n + 2)= 14ζ2(n) − 12{n2 }ζ0(n) .342.3. EstimatesWe observe that for any n ≥ 2, we have Gm(n) > 0. This proves the casewhere ` = 0. If we now assume that m > `, then we have, for n ≥ 2,`G`(n) +mGm(n) > `G`(n) + `Gm(n) = 0 ,which proves what we need.2.3.6 Case: P smooth fixed pointThis is the case where P = Z(W ) ∈ Xreg, and σ(P ) = P . In particular,Z(U) /⊂X.The previous proof of the casem > ` ≥ 0, which starts with Formula (2.22),applies here, too, and so we only need to deal with the case ` ≥m ≥ 0. Thiscase is different than before, because we cannot use V V = VW +WV in theestimate for w`, instead we exploit non-commutativity using Lemma 2.3.9,to improve the estimate on wm.So assume that ` ≥m ≥ 0.We use Lemma 2.3.6. Thus we have, for every α = 1, . . . , n, a total of` instances of k, where {V n−αWα} ⊂ I(k)n , and a total of m instances of kwhere {Wn−αUα} ⊂ I(k)n .We have dimV n−αWα ≥ 3n − α, by Corollary 2.3.2, even for α = n,because the line bundles L(−P ) and Lσ(−P ) are not isomorphic. Therefore,we can takew`(n) = n∑α=1(3n − α)= 52n2 − 12n .Then(w` + 2ã)(n) = 52n2 − 12n + (n − 1)(n − 2)= 72n2 − 72n + 2 ,andG`(n) = 72ζ2(n) − 72ζ1(n) + 2ζ0(n) − n(n + 1)(n + 2)= 12ζ2(n) − 12ζ1(n) .We observe that this is positive, for all n ≥ 2, proving the m = 0 case.352.3. EstimatesNow assume that m > 0. For β ≤ n2 we have dim{Wn−βUβ} ≥ 2n−β, andfor β ≥ n2 , we have dim{Wn−βUβ} ≥ 3n − 3β, by Lemma 2.3.9. Therefore,(not forgetting that dimUn = 1)wm(n) = ⌊n2 ⌋∑β=1(2n − β) +n∑β=⌊n2⌋+1(3n − 3β) + 1= n∑β=1(3n − 3β) −⌊n2⌋∑β=1(n − 2β) + 1= 32n2 − 32n + 1 − ⌊n2 ⌋(⌈n2 ⌉ − 1)= 54n2 − n + 1 − 12{n2 } .HenceGm(n) = 54ζ2(n) − ζ1(n) + (1 − 12{n2 })ζ0(n) − 12n(n + 1)(n + 2)= −14ζ2(n) + 12ζ1(n) − 12{n2 } ζ0(n) .As we are in the case ` ≥m, we get1m(`G`(n) +mGm(n)) ≥ G`(n) +Gm(n)= 14ζ2(n) − 12{n2 } ζ0(n) ,which we already determined to be positive, for all n ≥ 2.2.3.7 Case: P nodeNow let us deal with the case where P = Z(W ) is a node of X. We canstill use the arguments from the above subcase m > ` ≥ 0, which starts withEquation (2.22), and consider this case proved.So assume that ` ≥m ≥ 0.We remark that P is not fixed by σ, so we still have the convenient factV V = VW +WV . On the other hand, the estimate for the dimension ofV n−α−βWαUβ drops, but this is made up for by the fact that c3 ∈ I(3`)3 , byProposition 2.2.30.362.3. EstimatesWe use Lemma 2.3.6. Using Lemma 2.3.10, below, this gives the estimatew`(n) = ⌊n2 ⌋∑α=1 3n +n∑α=⌊n2⌋+1(5n + 1 − 4α)= n∑α=1(5n + 1 − 4α) −⌊n2⌋∑α=1(2n + 1 − 4α)= 3n2 − n − ⌊n2 ⌋(2⌈n2 ⌉ − 1)= 52n2 − 12n .But note that this estimate is not optimal. For example, we counted thecontribution of Wn with n+1, but in fact, it is larger than that, because wecannot have only one node involved. It will be sufficient to improve w`(n)by adding 1 to it:w`(n) = 52n2 − 12n + 1 .We have c3 ∈ J(3`)3 , and hence we compute(w` + 3ã)(n) = 52n2 − 12n + 32(n − 1)(n − 2) + 1= 4n2 − 5n + 3 + 1 .This gives usG`(n) = 4ζ2(n) − 5ζ1(n) + 3ζ0(n) − n(n + 1)(n + 2) + ζ0(n)= ζ2(n) − 2ζ1(n) + ζ0(n) + ζ0(n)= 16n(2n2 + 3n − 3) − {n3 } + ζ0(n) .This is positive, for all n ≥ 2, proving the stability bound in the case m = 0.For the case m > 0, we note that Lemma 2.3.6 also gives uswm(n) = n∑β=1(n + 1 − β)= 12n2 + 12n ,and henceGm(n) = 12ζ2(n) + 12ζ1(n) − 12n(n + 1)(n + 2)= −ζ2(n) + 2ζ1(n) − ζ0(n) .372.3. EstimatesThen we have1m(`G`(n) +mGm(n)) ≥ G`(n) +Gm(n)= ζ0(n)> 0 .2.3.8 Case: Y linear component of XWe assume now that Y = Z(U) ⊂ X. Then Y is a linear component of X,and X is a Neron triangle. Moreover, the automorphism σ acts transitivelyon the three edges of X. Any choice of non-singular base point of X turnsXreg, the non-singular part of X, into a commutative group scheme, suchthat σ becomes a translation. The point P = Z(W ) necessarily lies on X,but is not fixed by σ (of course, it may be a node).To formulate dimension estimates, let Y ′ = σ−1Y , and Y ′′ = σ−2Y be theother two edges of X. Let Q = Y ∩Y ′, Q′ = Y ′ ∩Y ′′ and Q′′ = Y ′′ ∩Y be thethree nodes of X. Let M be a line bundle on X, whose degrees on the threeedges, d, d′, d′′, are all non-negative. Denote by Γ(M(−βY )) = Γ(M⊗I βY ) ⊂Γ(M) the subspace of sections which vanish to order at least β on Y , andΓ(M(−αQ)) = Γ(M ⊗ mαQ) ⊂ Γ(M) the subspace of sections vanishing toorder at least α at Q. (The ideal sheaf of the closed subscheme Y ⊂ X islocally principal, generated by s, for any non-zero element s ∈ U .)Lemma 2.3.10 We have(i) if 2β ≤ d′ + d′′ + 1, then dim Γ(M(−βY )) = d′ + d′′ + 1 − 2β,(ii) if β + γ ≤ d′′ + 1, then dim Γ(M(−βY − γY ′)) = d′′ + 1 − β − γ,(iii) if 2β ≤ d + d′ + d′′ + 1, then dim Γ(M(−βQ)) = d + d′ + d′′ + 1 − 2β,(iv) if β + γ ≤ d′ + 1, and β + γ ≤ d′′ + 1, then dim Γ(M(−βY − γQ′)) =d′ + d′′ + 2 − 2β − 2γ,(v) if β + γ ≤ d′ + 1, and β + γ ≤ d + d′′ + 1, then dim Γ(M(−βQ − γQ′)) =d + d′ + d′′ + 2 − 2β − 2γ,(vi) if α+β ≤ d′+1, and β+γ ≤ d′′+1, and α+γ ≤ d+1, then dim Γ(M(−αQ−βQ′ − γQ′′)) = d + d′ + d′′ + 3 − 2α − 2β − 2γ.In call cases, it is important, that α,β, γ ≥ 1.Proof. These formulas follow easily by breaking up X into rational nodalcurves. 2382.3. EstimatesCorollary 2.3.11 For 1 ≤ β ≤ n+12 , we haveΓ(Ln(−βY )) + Γ(Ln(−βY ′)) = Γ(Ln(−βQ)) ,andΓ(Ln(−βQ)) + Γ(Ln(−βY ′′)) = Γ(Ln) ,and hence alsoΓ(Ln(−βY )) + Γ(Ln(−βY ′)) + Γ(Ln(−βY ′′)) = Γ(Ln) .Corollary 2.3.12 If 1 ≤ γ ≤ α ≤ n+12 , we haveΓ(Ln(−αY − γY ′)) + Γ(Ln(−αY − γY ′′)) = Γ(Ln(−αY − γQ′)) ,andΓ(Ln(−αY − γQ′)) + Γ(Ln(−αY ′ − γQ′′)) = Γ(Ln(−αQ − γQ′ − γQ′′)) ,and alsoΓ(Ln(−αQ− γQ′ − γQ′′))+Γ(Ln(−αY ′′ − γQ)) = Γ(Ln(−γQ− γQ′ − γQ′′)) .Corollary 2.3.13 For 1 ≤ β ≤ ⌊n3 ⌋, we have that{V n−βUβ} = Bn .For ⌊n3 ⌋ + 1 ≤ β ≤ 2⌊n3 ⌋, we have{V n−βUβ} ⊃ Γ(Ln(−iQ − iQ′ − iQ′′)) ,where i = β − ⌊n3 ⌋.If n ≡ 1 mod 3, and β = 2⌊n3 ⌋ + 1, we have{V n−βUβ} ⊃ Γ(Ln(−(⌊n3 ⌋ + 1)Y − ⌊n3 ⌋Q′)) ,if n ≡ 2 mod 3, and β = 2⌊n3 ⌋ + 1, we have{V n−βUβ} ⊃ Γ(Ln(−(⌊n3 ⌋ + 1)Q − ⌊n3 ⌋Q′ − ⌊n3 ⌋Q′′)) , (2.23)and if β = 2⌊n3 ⌋ + 2, we have{V n−βUβ} ⊃ Γ(Ln(−(⌊n3 ⌋ + 1)Y − (⌊n3 ⌋ + 1)Y ′)) .392.3. EstimatesProof. Note that(UV V ) . . . (UV V )´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶βV . . . V´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶n−3β = Γ(Ln(−βY )) ,andV UV . . . V UV´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶βV . . . V´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶n−3β = Γ(Ln(−βY ′)) ,andV V U . . . V V U´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶βV . . . V´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶n−3β = Γ(Ln(−βY ′′)) .The first claim now follows from Corollary 2.3.11.The second claim follows from Corollary 2.3.12 upon consideringUUV . . .UUV´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶β−⌊n3⌋UV V . . .UV V´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶2⌊n3⌋−βV . . . V´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶n−3⌊n3⌋and its 5 ‘cousins’.The last three claims follow by similar considerations. 2Subcase: ` = 0 .To prove the stability estimates, let us start with the case that ` = 0. FromLemma 2.3.8, we obtain, for every β = 1, . . . , n exactly m instances of kwhere {V n−βUβ} ⊂ I(k)n . By Corollary 2.3.13, we have (for n ≥ 1):wm(n) = ⌊n3 ⌋∑β=1 3n +2⌊n3⌋∑β=⌊n3⌋+1 (3n + 3 − 6(β − ⌊n3 ⌋))= ⌊n3 ⌋∑β=1(6n + 3 − 6β)= (6n + 3)⌊n3 ⌋ − 3⌊n3 ⌋(⌊n3 ⌋ + 1)= ⌊n3 ⌋(6n − 3⌊n3 ⌋)= 13(5n + 3{n3 })(n − 3{n3 })= 13⋅ ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩5n2 if n ≡ 0 mod 35n2 − 4n − 1 if n ≡ 1 mod 35n2 − 8n − 4 if n ≡ 2 mod 3 .402.3. EstimatesBut we have not been optimal, yet, if n ≡ 1 (3) or n ≡ 2 (3). By Corol-lary 2.3.13, if n ≡ 1 (3), and n ≥ 4, we can add 13(2n + 4), but for n = 1,we can only add 1 and not 2. In order to have uniform formulas, we willtherefore add only 13(2n + 1).If n ≡ 2 (3), and n ≥ 5, we can add n + 5 and 13(n + 1), but if n = 2, wecan only add 5 and 1, and not 7 and 1. Again, in the interest of uniformformulas, we add only n + 3 and 13(n + 1), for a total of 23(2n + 5).In sum, we havewm(n) = 13⋅ ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩5n2 if n ≡ 0 mod 35n2 − 2n if n ≡ 1 mod 35n2 − 4n + 6 if n ≡ 2 mod 3 .Hence,Gm(n) = 13⋅ ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩5ζ2(n) − 32n(n + 1)(n + 2) if n ≡ 0 mod 35ζ2(n) − 2ζ1(n) − 32n(n + 1)(n + 2) if n ≡ 1 mod 35ζ2(n) − 4ζ1(n) + 6ζ0(n) − 32n(n + 1)(n + 2) if n ≡ 2 mod 3 .= 16⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ζ2(n) + 9ζ1(n) − 6ζ0(n) if n ≡ 0 mod 3ζ2(n) + 5ζ1(n) − 6ζ0(n) if n ≡ 1 mod 3ζ2(n) + ζ1(n) + 6ζ0(n) if n ≡ 2 mod 3 .A direct calculation, using the formulas of Lemma 2.2.10, shows thatmGm(n) >0, for all n ≥ 2.Subcase: ` > 0 .Let us now deal with the case where ` > 0. Lemma 2.3.8 gives us, forevery β = 0, . . . , n−1 precisely m instances of k where {V n−β−1WUβ} ⊂ I(k)n .Using the two facts that V V = WV + VW and WUV + V UW = V UV , wecan replace the single factor of W by a factor of V , and still use the samearguments as before, in the following cases:(i) 0 ≤ β ≤ ⌊n3 ⌋,(ii) ⌊n3 ⌋ + 1 ≤ β ≤ 2⌊n3 ⌋ − 1, or n ≡ 2 mod 3 .412.3. EstimatesWe getw`(n) ≥ ⌊n3 ⌋∑β=0 3n +2⌊n3⌋−1∑β=⌊n3⌋+1 (3n + 3 − 6(β − ⌊n3 ⌋))= ⌊n3 ⌋∑β=1(6n + 3 − 6β) + 3n − (3n + 3 − 6⌊n3 ⌋)= ⌊n3 ⌋(6n − 3⌊n3 ⌋) + 6⌊n3 ⌋ − 3= 13⋅ ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩5n2 + 6n − 9 if n ≡ 0 mod 35n2 + 2n − 16 if n ≡ 1 mod 35n2 − 2n − 25 if n ≡ 2 mod 3 .If n ≡ 2 mod 3, we can add a summand of 3n+ 3− 6⌊n3 ⌋, and we get insteadw`(n) = 13⋅ ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩5n2 + 6n − 9 if n ≡ 0 mod 35n2 + 2n − 16 if n ≡ 1 mod 35n2 + n − 4 if n ≡ 2 mod 3 .These estimates are still not sufficient.For n ≡ 1 (3), we consider {V n−β−1WUβ} with β = 2⌊n3 ⌋. It contains thethree subspacesUUV . . .UUV´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶⌊n3 ⌋W , UV U . . .UV U´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶⌊n3 ⌋W , V UU . . . V UU´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶⌊n3 ⌋W .Considering the first two, we see that the factor W can be moved into aposition where it does not affect the dimension calculation. This meansthat we get a dimension estimate for the sum of these three spaces which isworse by 1, than the dimension estimate 3n + 3 − 6(β − ⌊n3 ⌋), which we usedfor {V n−βUβ}. We get an estimate of 3n + 2 − 6⌊n3 ⌋ = n + 4.For n ≡ 2 (3), consider the contribution (2.23), which gave us a dimensionestimate of n + 3, for n ≥ 2. In the presence of a factor of W , this estimatedrops by 2 to n + 1.This gives us the following improved formulas for w`:w`(n) = 13⋅ ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩5n2 + 6n − 9 if n ≡ 0 mod 35n2 + 5n − 4 if n ≡ 1 mod 35n2 + 4n − 1 if n ≡ 2 mod 3 .422.4. Moduli stacksWe have that c3 ∈ J(3`)3 , by Proposition 2.2.30, and so we add 32(n−1)(n−2)to get(w` + 3ã)(n) = 16⋅ ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩19n2 − 15n if n ≡ 0 mod 319n2 − 17n + 10 if n ≡ 1 mod 319n2 − 19n + 16 if n ≡ 2 mod 3 ,which gives, by subtracting 3ζ2 − 3ζ1 + 2ζ0,G`(n) = 16⋅ ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ζ2(n) + 3ζ1(n) − 12ζ0(n) if n ≡ 0 mod 3ζ2(n) + ζ1(n) − 2ζ0(n) if n ≡ 1 mod 3ζ2(n) − ζ1(n) + 4ζ0(n) if n ≡ 2 mod 3 .A direct calculation shows that `G`(n) > 0, for all n ≥ 2. This proves therequired stability estimate for m = 0.If both ` > 0, and m > 0, Lemma 2.3.8 shows that we may simply addour values for `G` and mGm to prove the stability estimate. Since they arepositive individually, this has already been observed.2.4 Moduli stacks2.4.1 Moduli stacks of stable regular algebrasRecall that we are working over an algebraically closed field C, of charac-teristic not equal to 2 or 3.Definition 2.4.1 A flat family of stable regular algebras, parametrizedby the C-scheme T , is a graded vector bundle A =⊕∞n=0An over T , endowedwith the structure of sheaf of graded OX -algebras, such that for every t ∈ T ,the fibre At is a stable regular algebra as in Theorem 2.2.2.Let us denote the C-stack of families of stable regular algebras by Ms,r.Theorem 2.4.2 The stack Ms,r of flat families of stable regular algebras isa smooth Deligne-Mumford stack of finite type. It has 4 components, Ms,rA ,Ms,rB , Ms,rE , and Ms,rH . Concretely,(A) Ms,rA = [U/G216], where U ⊂ P2 is the complement of 4 concurrent lines,and 4 points in P2. The group G216 is the group of automorphisms ofthe oriented affine plane over F3 (which has 216 elements)1. It acts Uvia its quotient SL2(F3),1See Section 3.4.2 for G216 and its action on P2432.4. Moduli stacks(B) Ms,rB = [V /Z4], where V ⊂ P1 is the complement of 3 points in P1, andZ4 acts via its quotient Z2.(E) Ms,rE = BZ3,(H) Ms,rH = BZ4.Proof. Let A be a flat family of stable regular algebras parametrized bythe scheme T . As all members of A are elliptic, the triple (X,σ,L) of Ais a flat family of elliptic triples. We have a short exact sequence of vectorbundles on T :0 // R // V ⊗ V // pi∗(L⊗Lσ) = A2 // 0 .Here pi ∶X → T is the structure morphism. The algebra A is recovered fromR → V ⊗OT V as the quotient of the tensor algebra on V divided by the twosided sheaf of ideals generated by R.Now σ ∶ X → X induces a group automorphism Pic0(σ) ∶ Pic0(X/T ) →Pic0(X/T ). The order of Pic0(σ) is necessarily finite, of order 1,2, 3, or 4,as we are avoiding characteristic 2 and 3. The order of Pic0(σ) is also locallyconstant over T , and so T breaks up into 4 open and closed subschemes Ti,where Ti ⊂ T is the locus where the order of Pic0(σ) is i. This proves thatMr,s also breaks up into 4 open and closed substacks Mr,si .Let us deal with each of the 4 components in turn, and assume that theorder of Pic0(σ) is constant.Case A. First, assume that the order of Pic0(σ) is 1, so that Pic0(σ)is the identity of Pic0(X/T ). By our classification of stable algebras, theautomorphism σ acts transitively on the set of components in every fibre.Therefore, by Theorem II 3.2 in [9], for any section P of Xreg, there existsa unique structure of generalized elliptic curve on X, having P as origin,such that σ acts as translation. In particular, Xreg is a commutative groupscheme, which acts on X, and there is another section S of Xreg, such thatσ(Q) = Q + S, for all sections Q of X.Using this, we prove that in the the fibered productZ //TLXQ↦O(Q+σQ+σ−1Q) // Pic3(X)the scheme Z is a form of the oriented affine plane over F3. In particular,Z is a finite e´tale cover Z → T of degree 9.442.4. Moduli stacksThen, at least e´tale locally, we can assume that P is a section of Z,which we use to turn X into a generalized elliptic curve. Then Z is equal tothe scheme of 3-division points in X, in particular, Z is an oriented vectorbundle over F3. We can choose, e´tale locally, an oriented basis Q1,Q2 forZ. We let S be the section of Xreg, such that σ is translation by S. Fromthe fact that the algebra A is elliptic, it follows that S avoids Z ⊂X.Thus, at least e´tale locally, we can associate to A a generalized ellipticcurve with full oriented level-3-structure, with an extra point S on it. Theambiguity is in the choice of the oriented coordinate system (P,Q1,Q2) forthe bundle of affine planes Z. The only restriction is that in singular fibres,S avoids the 3-division sections, that S stays away from Xsing, and that inthe singular fibres, S avoids the connected component of P .Thus, conversely, let M(3) be the moduli scheme of generalized ellipticcurves with full oriented level-3-structure. Let E(3)→M(3) be the universalcurve. It classifies quintuples (E,P,Q1,Q2, S), where (E,P ) is a generalizedelliptic curve, whose geometric fibres are smooth or triangles, Q1 and Q2are 3-division points on E, forming an oriented bases for E3, and S ∈ E isa further point on E. Let E(3)0 ⊂ E(3) be the open subscheme defined bythe conditions that S is not a node of E, not in the component of identityin any triangle, and not in E3. Using the Hesse family of elliptic curves, itis not hard to identify E(3)0 with the complement of 4 concurrent lines and4 points in P2.The scheme E(3)0 has the tautological triple (E, τS ,O(3P )) over it.Associated to this triple is a flat family of stable regular algebras AE(3)0 ,parametrized by E(3)0. The scheme of isomorphisms of AE(3)0 is canonicallyidentified with the transformation groupoid of G216 acting on E(3)0.We have seen above, that the algebra AE(3)0 is versal, i.e., every flatfamily of stable regular algebras is e´tale locally induced from AE(3)0 . Thisfinishes the proof that Ms,r1 ≅ [E(3)0/G216].Case B. Let M(2)0 be the moduli stack of (non-singular) elliptic curveswith a fixed 2-division point. It classifies triples (E,P,Q), where (E,P ) isa smooth elliptic curve, and Q is a non-zero section of E2. Note that theLegendre family of elliptic curves identifies M(2)0 with [V /Z4], where V ⊂P1 is the complement of 3 points in P1. To (E,P,Q) we associate the elliptictriple (X,σ,L), given by X = E, σ∶X → X is the involution R ↦ Q − R,whose fixed points are the 4 second roots of Q, and L = O(2S+Q) = O(3P ),where S is any second root of Q. The associated family of algebras is a flatfamily of stable regular algebras of Type B over M(2)0.Conversely, let A → T be a flat family of elliptic regular algebras, withassociated flat family of elliptic triples (X,σ,L). Assume that the order of452.4. Moduli stacksPic0(σ) is 2. Then Pic0(σ) is necessarily the multiplication by −1 auto-morphism. This implies that for any two local sections Q,Q′ of X, the linebundles OX(Q+σ(Q)) and OX(Q′+σ(Q′)) are isomorphic. Thus there existsa unique, and therefore global, section P , such that O(Q+σ(Q)+P ) ≅ L, forany Q. This makes (X,P ) into an elliptic curve, and the automorphism σinto R ↦ σ(P )−R. Regularity of the triple (X,σ,L) implies that Q = σ(P )is a 2-division point on (X,P ). We see that Ms,r2 ≅M(2)0.Cases E and H. Left to the reader. 22.4.2 DensityProposition 2.4.3 Suppose that A is a flat family of graded q-truncatedalgebras, parametrized by the finite type k-scheme T . Then the locus ofpoints t ∈ T , such that the fibre At is the q-truncation of a stable regularalgebra is open in T .Proof. By definition,A = q⊕n=0Aq ,is a direct sum of vector bundles, where A0 = OT . Let us assume that t0is a point of T , such that the fibre A ∣t0 of A over t0 is the truncation ofa non-singular elliptic regular algebra. We will prove that there exists anopen neighbourhood U of t0 in T , such that for every t ∈ U , the fibre A ∣t isa non-singular elliptic regular algebra as well.The rank of the vector bundle An is a locally constant function on T ,and so by restricting to an open neighbourhood of t0, we may assume thatit is constant. It is then equal to 12(n+1)(n+2), because it takes that valueat t0, by Formula (1.15) in [3].Let us write V = A1. This is now a vector bundle of rank 3 on T .Multiplication in A defines a homomorphism of vector bundlesV ⊗OT V → A2 . (2.24)The locus of points in T , over which (2.24) is not surjective is closed in T .Over the point t0, the homomorphism (2.24) is surjective, because A ∣t0 isgenerated in degree 1. So the locus of points in T where (2.24) is surjective isan open neighbourhood of t0, and be restricting to this open neighbourhood,we may assume that (2.24) is surjective. Then the kernel of (2.24) is a vectorbundle R of rank 3 on T .By the same reasoning, we may assume that A2⊗OT V → A3 and V ⊗OTA2 → A3 are epimorphisms of vector bundles, and that the respective kernels462.4. Moduli stacksK and K ′ are vector bundles as well. Both vector bundles K and K ′ haverank 8.We have the following commutative diagram of coherent sheaves of OT -modules:Wβ //αR⊗ V //K ′ //Q′V ⊗R //V ⊗ V ⊗ V //V ⊗A2K //A2 ⊗ V // A3QAll tensor products are over OT . All rows and columns are exact, whencompleted with zeros on all ends. By the snake lemma, Q′ = Q. Since A ∣t0is quadratic, the homomorphisms V ⊗R →K and R⊗V →K ′ are surjectivenear t0, and so we can assume they are surjective, and that Q′ = Q = 0.Then All sheaves in our diagram are locally free, and in particular W is avector bundle of rank 1.We may also assume that the images of α and β have full rank, i.e.,any non-zero section of ω of W induces isomorphisms β(ω)∗ ∶ R∗ → V andα(ω)∗ ∶ V ∗ → R, because this is the case at the point t0.We now consider the projective bundle P(V ) → T of one-dimensionalquotients of V . From the homomorphism of vector bundles R → V ⊗ V , weobtain homomorphismsΛ3R Ð→ Λ3V ⊗OT Sym3 V and Λ3R Ð→ Sym3 V ⊗OT Λ3V ,and henceΛ3V ∗ ⊗Λ3R Ð→ Sym3 V and Λ3R⊗Λ3V ∗ Ð→ Sym3 V . (2.25)The arrows (2.25) are both strict monomorphisms of vector bundles over T ,and hence X1, X2 are flat families of Cartier divisors of degree 3 in P(V ). Infact, by our assumption that both α and β have full rank, we have X1 =X2,i.e., we are in the semi-standard case.We will call this scheme X =X1 =X2.Moreover, R → V ⊗V defines a family of subschemes Γ in P(V )×T P(V ).By construction the projections factor pi1 ∶ Γ → X and pi2 ∶ Γ → X. At t0,both these morphisms are isomorphisms, and so by properness of Γ and X,we may assume that they are isomorphisms.472.4. Moduli stacksThis gives us an elliptic triple, from which we can construct a flat familyof elliptic regular algebras A ′, together with a morphism of graded algebrasA ′ → A . At the point t0 it is an isomorphism, so we may assume that it isan isomorphism.Let us finish by proving that there is an open neighbourhood of t0 wherethe triple (X,σ,L) is regular. First note that the condition of being regularor exceptional is open: it is the locus where R1pi∗(L ⊗ (Lσ)−⊗2 ⊗ Lσ2) hasrank 1.If At0 is of type B, E, or H, we can pass to the open set where X issmooth, to get rid of the exceptional locus.So assume that t0 is of type A. If X is smooth at t0, we simply passto the open neighbourhood where X is smooth. If X is not smooth at t0it is a triangle, which is being rotated by σ. We now restrict to the openneighbourhood of t0 where X is nodal and Pic0(σ) is the identity.We also exclude all points where the algebra is not stable, as this locusis closed. Then every fibre is either smooth, or a triangle on which σ actsby rotation, or exceptional, in which case σ acts by swapping components.Thus in every fibre, σ acts transitively on connected components, and by byLemme II 1.7 of [9], we may endow X with the structure of a generalizedelliptic curve. Then by Proposition II 1.15 of [9], the exceptional locus,which is the locus where X has two components, is closed in T , and so wecan remove it, and we are left with a family of stable regular algebras, asrequired. 2Corollary 2.4.4 For every q, the moduli stack of stable algebras Msq hasan open substack isomorphic toMs,rA ∐Ms,rB ∐Ms,rE ∐Ms,rH .Each stack Ms,ri , for i ∈ {A,B,E,H}, is an open dense substack in theirreducible component of Msq which contains it.We can unfortunately not prove that every stable algebra is regular.There may be further stable algebras in the boundary of the non-properstacks Ms,rA and Ms,rB , or there may be entire components consisting of stablealgebras which are not regular.In particular, it may happen that exceptional algebras are stable, astheir triples can satisfy our stability criterion, but we have not examinedthis possibility closer.For every q, we have a projective coarse moduli space of S-equivalenceclasses of semi-stable algebras, but we do not have a complete descriptionof these, except for low values of q.48Chapter 3Moduli of 3 dim’l quadraticAS regular algebras3.1 IntroductionIn this chapter we study 3-dimensional quadratic AS-regular algebras andtheir moduli. We compute automorphisms of algebras. Group action onthe parameter space is also studied. Section 3.3, 3.4, 3.5, which is the mainbody, has a short introduction in the beginning of each section.Notation 3.1.1 We abuse notation and (a, b, c), (x, y, z) also mean ho-mogeneous coordinates. Most of the coordinates we use are homogeneous.However it means affine coordinates when we study flat family around de-generate algebras, and when we use (xi), where xi ∈ Gm to compute auto-morphism of singular triple using geometry. In the notation of type S alge-bra S1(β1, β2, β3), it is not homogeneous coordinates. But for the algebraS̃1(β1∶β2∶β3), which is obtained by adding degree 3 relations to S1(0,0,0),we use homogeneous coordinates notation ‘∶’ for clarity.Defining equations for type A,B,E,H,S1,S′1,S2 algebras can be found in [7].But we modify coefficients into a form fitting for description of moduli.3.2 BasicsHere we present some basic materials. In Section 3.2.1 we recall the GITproblem from [6]. In Section 3.2.2 we study stability of free, symmetric, ex-terior algebras. In Section 3.2.3 we study stability of algebras given by triple(Pn,O(n), σ) where σ ∈ PGL(n + 1). In Section 3.2.4 we present observa-tion of algebra and triple. Readers who are interested in the 3-dimensionalquadratic AS regular algebras can skip this section. Only Corollary 3.2.5and Terminology 3.2.7 are required for understanding of our description of S-equivalence of AS-regular algebras and their automorphisms in Section 3.3,3.4, 3.5 .493.2. Basics3.2.1 GIT and S-equivalenceThe GIT problem we have is as following:Fix a graded vector space V = ⊕Vn with dimension vector v⃗ = (d1, . . . , dq).Let R = L1 = Homgr(V ⊗ V,V ) whose elements are µ = (µij) where µij ∶Vi ⊗ Vj → Vi+j . Let G = ⊕Gn where Gn = End(Vn). Then an element g ofG acts on µ as following:(g ∗ µ)ij = gi+j ○ µi+j ○ (g−1j ⊗ g−1j ).Then we introduce a stability parameter vector θ⃗ = (θ1, . . . , θq) ∈ Zq whichdefines a character χθ byχθ(g1, . . . , gq) = q∏i=1 det(gi)θi .We then lift the G action up to the trivial line bundle R×C by g∗(µ, s) = (g∗µ,χ(g)−1s). We can define a pairing between character χ and 1-parametersubgroup λ by ⟨χ,λ⟩ = p when χ(λ) = tp.Action of one-parameter subgroup λ on V gives weight decompositionof V on each degree. This induces a weight decomposition of µ such thatµ =⊕µk whereµkij ∶ V li ⊗ V mj → V l+m+ki+j ,and λ(t) ∗ µk = tkµk. Then limt→0 λ(t) ∗ µ exists when µk = 0 for k < 0,i.e. µ is a map of filtrated algebra filtered by k (i.e. there is no weightdecreasing part of the multiplication). When the limit exists we have thatlimt→0 λ(t) ∗ µ = µ0 which is the multiplications on the associated gradedalgebra of the filtered algebra. Let us consider the fiber over this limit pointin R. An λ orbit of (µ, s) has a limit point in R ×C ∖R × {0} if ⟨χ,λ⟩ = 0.In this case limt→0 λ(t) ∗ (µ, s) = (µ0, s). If ⟨χ,λ⟩ > 0 the orbit is closed. If⟨χ,λ⟩ < 0 we have limt→0 λ(t) ∗ (µ, s) = (µ0,0)An algebra is unstable if there is λ such that limt→0 λ(t)∗µ = µ0 existsand ⟨χ,λ⟩ < 0. We call A is semistable if it is not unstable. We denotethe set of semistable µ by Rss. A semistable algebra A is polystable if itsG-orbit is closed in Rss. There is unique polystable algebra for each closedG-orbit. Two semistable algebras A and B are S-equivalent if closuresof their G-orbits intersect. S-equivalent algebras have the same polystablealgebra which is S-equivalent to them.Lemma 3.2.1 Let A is an algebra defined by µ. We have that503.2. Basics(i) A is polystable if and only if µ = µ0 for any 1-parameter subgroup λwhich satisfies ⟨χ,λ⟩ = 0.(ii) A and B are S-equivalent if there is a weight decomposition of V suchthat the weight 0 part of A and B are isomorphic.3.2.2 Stability of free, symmetric, exterior algebrasFormulation in terms of a weight functionFormulation in [6] is done in terms of vector spaces and filtrations withoutusing basis. But algebras are usually defined in terms of generators andrelations. So, here we discuss stability algebraically. However, as our proofof stability of AS-regular algebra used geometry, readers who are interestedin only geometry can skip this section.Let V1 be a vector space. We call a function w ∶ V1 → Z≥0 a weightfunction of a vector space if w satisfies(i) w(x + y) ≥ min{w(x),w(y)},(ii) w(ax) = w(x) for a non-zero scalar a,(iii) w(x) =∞ iff x = 0.Then w gives a weight decomposition V1 = ⊕m≥0V m1 where V m1 = w−1(m)and corresponding weight filtration V1 = V (0)1 ⊃ V (1)1 ⊃ . . ., where V (m)1 =⊕i≥m V i1 . For a basis B1 = {bj1}dimV1j=1 of V1 and a map w ∶ B1 → Z≥0 wecan extend w to V1 using above rules. Consider a graded algebra V = ⊕iVigenerated in degree 1. We call a function w ∶ V → Z≥1 is weight function ofa graded algebra if it also satisfies(iv) w(xy) ≥ w(x) +w(y).When the equality holds for any non-zero element then we say w is static.If proper inequality holds for some x,y then we say w is weight raising.When xy = 0 for non-zero x, y so that w(xy) = ∞, then we say this weightraising is improper. If (iv) holds for x, y which are not zero divisorsthen we say the weight raising is proper. We see that static weight is adiscrete valuation. Tensor algebras, symmetric algebras, exterior algebras,and quantum algebras (those with relations like xy−ayx) admit static weightfor any weight function in degree 1.For a weight function of a graded algebra V = ⊕iVi we can define for513.2. Basicseach i, a total weight, w(i) in degree i which is the sum of weights of basis,w(i) =∑kw(bk)=∑mm∣w−1(m)∣=∑mmdimV mi .This is the weight function of the filtration V (k) defined in [6](3.4). So wecan compute weight w(i) using a weighted basis.If there is a basis Bi = {bk} for Ai generated in degree 1 by B1 = {xj}dimA1j=1such that bk = ∏il=1 xjkl . Then for a static weight w, we have that w(bk) =w(∏il=1 xjkl) = ∑lw(xjkl), which givesw(i) =∑kw(bk) =∑k,lw(xjkl).So w(i) is determined by the number of times each xj appears in Bi. If oc-currence of each xj in Bi is the same then we call Ai is equally generatedby B1. We say an algebra A is equally generated by B1 if Ai is equally gener-ated for all i. We see that free algebra, symmetric algebra, exterior algebra,and quantum algebra has generators by which it is equally generated.But the sum of number of occurrence of all xjs is idimAi. So each xjappears idimAidimA1 many times. Thus we have thatw(i) =∑kw(bk) =∑k,lw(xjkl) = idimAidimA1 ∑j w(xj) = idimAidimA1 w(1) . (3.1)Therefore we have thatLemma 3.2.2 If Aq is equally generated by a basis B1, and A admits staticweight w generated by a weight function on B1 then the Futaki function isconstant w.r.t the test configuration generated by w. If for any B and aweight function on it A admits static weight then A is strictly semistable.Moreover, if this holds for any i ≤ q for a fixed B1 then A≤q is strictlysemistable for any standard parameter θ.Proof. We only need to prove the last statement. When Ai is equally523.2. Basicsgenerated by B1 for any i ≤ q then, (3.1) holds for any i ≤ q thenq∑i=1 θiwi =q∑i=1 θiidimAidimA1w1=w1d1q∑i=1 iθidi=0,and the test configuration generated by B1 gives strictly semistability. 2Weight raising relationsHere we present algebraic observations on stability.Consider the Sklyanin algebra whose defining equations aref1 = ax2+ b yz + c zyf2 = b zx + cxz +ay2f3 = bxy + c yx +az2 (3.2)with abc ≠ 0 and a weight function generated by w(x) = 0,w(y) = w(z) = 1.From f1 we have three replacements. First consider,x2 = −1/a(byz + czy).On the LHS we have w(x2) ≥ w(x) + w(x) = 0. But on the RHS we havew(yz) ≥ w(y) + w(z) = 2, and w(byz + czy) ≥ min{w(byz),w(czy)} =min{w(yz),w(zy)} = 2. So we have that w(x2) ≠ w(x) + w(x) which isweight raising. f1 raises the weight of x2 from 0 to 2. Now consider follow-ing replacement,yz = −1/b(ax2 + czy).On the LHS we have that w(yz) ≥ 2. On the RHS, w(ax2+czy) ≥ min{x2, zy} =0. Here we don’t see weight raising. So is the other replacement for zy.Now, consider f2. First, in a replacementy2 = −1/a(bzx + cxz),on the LHS, w(y2) ≥ 2 and on the RHS w(bzx+ cxz) ≥ 1. Even though LHShas higher lower bound, this is not a weight raising. Sum of elements can sitin a higher weight subspace in a weight increasing filtration. So this is justa case when strict inequality holds in w(x+y) > min{w(x),w(y)}. Considera replacementzx = −1/b(ay2 + cxz).533.2. BasicsOn the LHS, w(zx) ≥ 1 and on the RHS w(ay2 + cxz) ≥ 1. So LHS andRHS has the same lower bounds. Similar argument shows there is no weightraising in f3The reason why weight raising happens in only f1 with this weight on V1,is that the lowest weight part of f1 is a monomial ax2, while those of f2 andf3 are polynomials bzx+cxz and bxy+cyx respectively. So in a replacementthe term to be replaced has higher or equal weight and its weight is notraised.Each relation kills one monomial in it. So we can ask“Is there a naturalway to choose the monomial to be eliminated?” The answer is “Eliminatethe monomial whose weight is raised.” The monomial whose weight is raiseddoes not have a weight induced from degree 1, so it is natural that weeliminate it. When there is more than one monomial of lowest weight in therelation, then we can choose any of them.It is easy to see that if all relations are of equal weight then there is noweight raising and we have static weight. Therefore from Lemma 3.2.2 wehaveCorollary 3.2.3 Free, symmetric, exterior algebras are strictly semistablefor any standard parameter θ.For any weight on (3.2) one of x2, y2, z2 is eliminated giving higher weightfunction than the polynomial algebra which is strictly semistable. Thereforewe may guess that the algebra defined by (3.2) maybe stable. This is how wefirst guessed the Sklyanin algebra is stable, while our proof used geometrybecause we don’t know how the relations change for all of change of variables.Now we consider S-equivalence in terms of relations. The limit algebraµ0 = limt→0 λ(t)µ has no proper weight raising. So we haveLemma 3.2.4 The limit limt→0 λ(t)µ is the algebra whose relations are thelowest weight part of the relations of µ w.r.t. the weight determined by λ(t).Proof. When the limit exists µ has no negative weight part and limt→0 µ =µ0. And µ0 is weight preserving, so the relations are of equal weight. 2Corollary 3.2.5 Two algebras are S-equivalent if they have the same lowestweight part for some weight filtration.3.2.3 Zhang twist of polynomial algebraWe consider a triple (Pn,O(n), σ), where σ ∈ PGL(n + 1). We have543.2. BasicsProposition 3.2.6 A = A((Pn,O(n), σ)) is strictly semistable and S-equivalentto A((Pn,O(n), σd)) where σd is the diagonal matrix which has the samecharacteristic polynomial with σ.Proof. Choose a basis {xi}i of H0(O(n)) such that σ is represented byupper triangular matrix F = (fij)ij (up to scalar). We denote commutativemultiplication of xi and xj by xixj and noncommutative multiplication byxi ∗ xj . Then multiplication in A is xi ∗ xj = xiσ(xj) = xi(∑k≥j fjkxk) =∑k≥j fjkxixk. Let G = (gij)ij be the inverse of F which is also upper trian-gular. We have that xixj = ∑k≥j gjkxi ∗ xk. In A we have that0 = xixj − xjxi =∑k≥j gjkxi ∗ xk −∑k≥i gikxj ∗ xk= (gjjxi ∗ xj − giixj ∗ xi)´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶weight=w(i)+w(j)+∑k>j gjkxi ∗ xk −∑k>i gikxj ∗ xk´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶weight>w(i)+w(j)(3.3)where gii ≠ 0 for all i because σ is invertible. With a test configurationgenerated by a flag with w(xi) < w(xj) for any i < j, we have that hij ∶=gjjxi ∗xj −giixj ∗xi has the lowest weight in (3.3). So the associated weightgraded algebra G(A) is given by G(A) = C⟨xi⟩/(hij)j>i. G(A) is equallygenerated by {xi} and admits static weight, so it is strictly semistable. It iseasy to see that G(A) = A((Pn,O(n), σd)). 23.2.4 Algebras and triple (E,L,σ)While we study algebras and their moduli, our method is mainly geometric.Computation of Aut(A)Terminology 3.2.7 Any algebra generated in degree 1 has tautologicalautomorphism φt which is given by φt(a) = tdeg(a)a. Therefore it is good torestrict our study to an automorphism which is not tautological. LetAut(A) ∶= Aut(A)/(φt)be the group of non-tautological automorphisms.Automorphism of algebras defined by tripe (E,L,σ) can be computedusing geometric data.553.3. Type S algebrasMorphism between triples (X1, L1, σ1) and (X2, L2, σ2) is a morphismτ ∶X1 →X2 which satisfies L1 = τ∗L2 and makes the diagram commuteX1σ1τ // X2σ1X1τ // X2The following lemma shows automorphisms are preserved by the functorfrom the category of triples and category of algebras. Its proof can begeneralized for any X and very ample line bundle L.Lemma 3.2.8 Let τ is an automorphism of a genus 1 curve E. Then τ∗is an automorphism of A(E,L,σ) if and only if στ = τσ and τ∗L ≅ L.Proof. If τ∗ is an automorphism of A then τ is an automorphism of thepoint scheme2 as a cubic in P2. So τ is an automorphisms of the polarizedgenus 1 curve (E,L) and thus τ∗L ≅ L. Because it also means that τ∗ is anautomorphism of (commutative) homogeneous coordinate ring B(E,L, id),we have τ∗(a ⋅ b) = τ∗(a) ⋅ τ∗(b) for any a, b ∈ Γ(X,L), where ⋅ denotescommutative multiplication. We denote twisted product in B(E,L,σ) by⋅σ. We have τ∗(a ⋅σ b) = τ∗ (a ⋅ σ∗(b)) = τ∗(a) ⋅ τ∗ (σ∗(b)) = τ∗(a) ⋅ τ∗σ∗(b) =τ∗(a) ⋅ (στ)∗(b) and τ∗(a) ⋅σ τ∗(b) = τ∗(a) ⋅ σ∗(τ∗(b)) = τ∗(a) ⋅ (τσ)∗(b).Thus that τ∗ preserves twisted product ⋅σ implies (στ)∗ = (τσ)∗. Becausesections of L separates points this implies στ = τσ.Conversely, if τ is an automorphism of genus 1 curve with τ∗L ≅ Land τσ = στ then τ∗ preserves addition and twisted multiplication, so it isan homomorphism of algebras both B(E,L,σ) and A(E,L,σ). The sameholds for τ−1 and τ∗τ−1∗ = (ττ−1)∗ = id∗ so τ is an automorphism of algebraA(E,L,σ). 23.3 Type S algebrasHere we study type S1,S′1,S2 quadratic AS-regular algebras and their mod-uli. We first study the triples. We also compute their automorphisms.AS-regular algebras do not form a compact moduli. So for compactificationwe add algebras on the boundary of the parameter space. The algebras onthe boundary are degenerate ones and some of them are also linear. Thoughbeing not AS-regular, degenerate ones are of right Hilbert series, except two2 E is called the point scheme of A. Refer [16] for details.563.3. Type S algebras(isomorphic) algebras. They are cyclic monomial algebras, which are alsoin the boundary of type A algebras. To have a flat family of 3, and 4-cutoffalgebras we need to blow up at the cyclic monomial algebras. Type S1 isstudied in Section 3.3.1In Section 3.3.2 we study type S′1 algebra. It is S-equivalent to type S1.In the 3-dimensional moduli of type S1, the type S′1 algebras sit inside as ahyperplane.In Section 3.3.3 we also introduce 1-dimensional family of algebras whichare S-equivalent to the type S1 and S′1 algebras. The elliptic curve for themis an irreducible nodal. We call this algebra of type S′′1 . In the moduli oftype S1 algebras the line consisting of type S′′1 algebras is the intersection ofthe moduli of type S and type A algebras.In Section 3.3.4 we study type S2. Moduli of type S2 algebras intersectswith that of type B in a point.In Section 3.3.5 we study semistability of type S algebras. We also discussS-equivalence. Polystable algebras for the type S algebras are found.3.3.1 Type S1 algebrasThe defining equations of type S1 algebra aref1 = b1 yz + c1 zyf2 = b2 zx + c2 xz (3.4)f3 = b3 xy + c3 yxwhere bi, ci ∈ C. The algebra is AS-regular if bi, ci ≠ 0. However we allow 0for bi, ci for compact moduli. The matrix M defined by f =MX isM = ⎛⎜⎝0 c1 z b1 yb2 z 0 c2 xc3 y b3 x 0⎞⎟⎠ ,detM = (b1b2b3 + c1c2c3)xyz .When b1b2b3 + c1c2c3 ≠ 0 we have a triangle Z(xyz). On Z(x), we haveσ(0, y, z) = (0, b1 y,−c1 z). On Z(y), we have σ(x,0, z) = (−c2 x,0, b2 z).And on Z(z), we have σ(x, y,0) = (b3 ,−c3 y,0).Relations (3.4) define a family of algebras parametrized by (P1)3. It isflat except at (0,0,0) and (∞,∞,∞). We will discuss a flat family for thesepoints later (See (3.11) and Proposition 3.3.2).573.3. Type S algebrasTo simplify notation, by S1(β1, β2, β3) we mean the type S1 algebraS1(β1, β2, β3) = C⟨x, y, z⟩⎛⎜⎜⎝f1 = yz + β1 zy,f2 = zx + β2 xz,f3 = xy + β3 yx⎞⎟⎟⎠(3.5)where β1, β2, β3 ∈ P1. When β1 =∞ we have f1 = zy.Representing (P1)3 as a cube, S1(β1, β2, β3) is degenerate if (β1, β2, β3)lies on the boundary faces. It is linear, i.e. detM = detN = 0, if (β1, β2, β3)lies on the cubic surface β1β2β3 + 1 = 0, or if one of β1, β2, β3 is 0 andanother is ∞. The later is also degenerate. So S1(β1, β2, β3) is degenerateand linear if (β1, β2, β3) lies on the 6 edges which do not contain the verticesV0 = (0,0,0) or V∞ = (∞,∞,∞). These 6 edges are the boundary of thecubic surface β1β2β3 + 1 = 0 of non-degenerate linear algebras.Isomorphisms and AutomorphismsWe now study isomorphisms and automorphisms. Gm3 acts as a change ofvariables (u, v,w).(x, y, z) = (ux, v y,w z) and gives automorphisms. SoGm3 ⊂ Aut (S1(β1, β2, β3)) .For generic type S1 algebras there is no more automorphism.Consider a symmetric group S3 acting on {1,2,3}. We have an inducedaction on C{x1, x2, x3}⊗C{x1, x2, x3} defined by, for f = ∑i,j fij xi ⊗ xj , wehave σ(f) ∶= ∑i,j fij xσ(i) ⊗ xσ(j). For fi in (3.5) we have that σ(fi) is alsoin the form of (3.5). In fact we have that ifC⟨x1,x2,x3⟩(f1,f2,f3) = S1(β1, β2, β3) thenC⟨x1,x2,x3⟩(σ(f1),σ(f2),σ(f3)) = S1(σ(β1, β2, β3)), whereσ(β1, β2, β3) = ⎧⎪⎪⎪⎨⎪⎪⎪⎩(βσ−1(1), βσ−1(2), βσ−1(3)) if σ is even( 1βσ(1) , 1βσ(2) , 1βσ(3)) if σ is odd .Hence we have a S3 action onS1 ∶= {(β1, β2, β3)×S1(β1, β2, β3)}∣(β1,β2,β3)∈(P1)3 ,a family of type S1 algebras parametrized by (P1)3, given byσ ((β1, β2, β3) × S1(β1, β2, β3)) = σ(β1, β2, β3) × S1 (σ(β1, β2, β3))Now consider a projection pi =S1 → (P1)3. Then this family is S3-equivariant.Therefore we have a family of algebrasS1/S3 over the quotient stack [(P1)3/S3].583.3. Type S algebrasAs we have S1(β1, β2, β3) ≅ S1(β′1, β′2, β′3) if and only if (β′1, β′2, β′3) = σ(β1, β2, β3),[(P1)3/S3] parametrizes isomorphism class of S1(β1, β2, β3).In terms of the cube, an odd permutation interchanges two vertices V0and V∞ and stabilizes the 6 linear edges which do not contain V0 and V∞.In the stabilized edges the odd permutation gives an inversion t ↦ 1t fixing±1. An even permutation is a rotation fixing V0 and V∞. The 6 linear edgesform 2 orbits.Now we find fixed points of the S3 action. g0(x, y, z) = (x, z, y) givesisomorphism S1(β1, β2, β3) ≅ S1( 1β1 , 1β3 , 1β2 ). Its fixed point is given by β21 = 1and β2β3 = 1. In the cube the fixed points locus is 6 lines. Three of them,{(−1, β2, 1β2 )} and its permutations, lie on the surface of linear algebras.They are lines connecting (−1,0,∞) and (−1,∞,0), etc., and intersect in(−1,−1,−1). The other three lines, {(1, β2, 1β2 )}, intersect in (1,1,1).A change of variables g1(x, y, z) = (y, z, x) gives an isomorphism S1(β1, β2, β3) ≅S1(β2, β3, β1). Its fixed point is given by β1 = β2 = β3. In the cube the fixedpoints locus is the diagonal connecting V0 and V∞. We will see later thatthis is the line where the moduli of type S intersect with that of type A.Some linear algebra has a higher dimensional automorphism group thanthose from the S3 action. We will discuss degenerate cases later. Here weconsider triples of non-degenerate linear type S1 algebra S1(β1, β2, ,− 1β1β2 ),where β1, β2 ≠ 0,∞. The polarized scheme is (P2,O(1)) and the conditionτ∗L ≅ L is trivial. σ is given by σ(x, y, z) = (β1β2 x, y,−β1 z).The σ has one eigenvalue if (β1, β2) = (−1,−1). This is the commutativepolynomial algebra. Then σ = id, so any τ ∈ PGL(3) satisfies στ = τσ.ThereforeAut (S1(−1,−1,−1)) = PGL(3)If σ has two different eigenvalues, then (β1, β2, β3) is in the S3 orbit of(−1, β2, 1β2 ), where β2 ≠ −1. For A(−1, β2, 1β2 ), σ is a multiplication −β2on x. Then the condition στ = τσ means τ is of the form ( ∗ 0 000 M) whereM ∈ PGL(2). So we haveAut(S1(−1, β2, 1β2)) = {( 1 0 000 M)} ≅ GL(2)If none of β1, β2, β3 is equal to −1 then σ has three different eigenvaluesand στ = τσ requires τ is also diagonal, andAut(A(β1, β2,− 1β1β2)) = Gm3 if none of β1, β2, β3 is −1 .593.3. Type S algebrasNow we prove that elliptic type S1 algebras which have extra automor-phism are those who have non-trivial stabilizers of the S3 action. We com-pute Aut(S1(β1, β2, β3)) geometrically. The cubic is triangle X = X1 ∪X2 ∪ X3. Let xi ∈ Gm = Xi/{0,∞} be a smooth point. σ(x1, x2, x3) =(−β1 x1,−β2 x2,−β3 x3).First consider a multiplication on eachXi give by τ(x1, x2, x3) = (t1 x1, t2 x2, t3 x3).It commutes with σ because both are multiplications on xi. But τ∗L ≅ Lrequires t1t2t3 = 1. So the subgroup of Aut consisting of multiplication isisomorphic to Gm2.Second we consider S3 action which interchanges three sides. For a rota-tion given by τ(x1, x2, x3) = (x2, x3, x1) we have στ = τσ if and only if β1 =β2 = β3. When τ is odd permutation so that τ(x1, x2, x3) = ( 1x1 , 1x3 , 1x2 ), wehave that στ(x1, x2, x3) = σ ( 1x1 , 1x3 , 1x2 ) = (− ax1 ,− bx3 ,− cx2 ) and τσ(x1, x2, x3) =τ(−β2 x1,−β2 x2,−β3 x3) = (− 1β1 x1 ,− 1β3 x3 ,− 1β2 x2 ). So στ = τσ implies β21 =β2β3 = 1. This explains the table 3.1.We note that the table 3.1 is also true when β1, β2, β3 are 0 or ∞.S1(−ζi3,−ζi3,−ζi3) is linear but does not have an extra automorphism. Solinear does not necessarily mean a higher dimensional automorphism group.Degenerate type S1Now we consider type S1 algebra which has some of β1, β2, β3 is 0 or ∞.When (β1, β2, β3) ≠ (0,0,0), (∞,∞,∞) a degenerate algebra has the rightHilbert series.We first compute their dimension. We haveLemma 3.3.1 Degenerate type S1 algebra, except S1(0,0,0) and S1(∞,∞,∞)has Hilbert series (1 − t)−3.Proof. First consider S1(0,0,∞). The defining equations are yz = zx =yx = 0. We find a monomial basis for degree n. y and z cannot comebefore x because zx = yx = 0. And x and z cannot come after y becauseyz = yx = 0. Therefore the nonzero monomial should be in a form of xizjyk,where 0 ≤ i, j, k ≤ n and i+j+k = n. Hence dimAn = ∑ni=0∑n−ij=0 1 = (n+1)(n+2)2 .For S1(β1, β2,∞) we have the same monomial basis xizjyk and hence thesame dimension. Other cases are similar. This proof dose not work for(0,0,0) or (∞,∞,∞) because there is no initial or terminal variable. 2We first consider algebras which are both linear and degenerate. There603.3. Type S algebrasType S1(β1, β2, β3) ∈ (P1)3 stabilizer Aut(−1,−1,−1) S3 PGL(3)linear(−1, β2, 1β2 ) Z/2Z {( 1 0 000 M )} ≅ GL(2)(1,1,1) S3 ⟨( 1 0 00 0 10 1 0) , ( 0 1 00 0 11 0 0) , ( 1 0 00 ∗ 00 0 ∗ )⟩ ≅ S3 ×Gm2(β1, β1, β1), β1 ≠ ±1 A3 ⟨( 0 1 00 0 11 0 0) , ( 1 0 00 ∗ 00 0 ∗ )⟩ ≅ A3 ×Gm2 type A, linear if β31 = −1(1, β2, 1β2 ) Z/2Z ⟨( 1 0 00 0 10 1 0 ) , ( 1 0 00 ∗ 00 0 ∗ )⟩ ≅ Z/2Z ×Gm2otherwise id ⟨( 1 0 00 ∗ 00 0 ∗ )⟩ ≅ Gm2 linear if β1β2β3 = −1Type S′1(β1, β2) ∈ (P1)2 stabilizer Aut(−1,−1) Z/2Z {( 1 0 0∗∗ M )} ⊂ PGL(3) S-equi. to linear(−1,1) Z/2Z {( 1 0 000 M)} ≅ SL(2)(1,±1) Z/2Z ⟨( 0 1 00 0 11 0 0) ,( 1 0 00 u 00 0 u−1 )⟩ ≅ Z/2Z ×Gm(β1, β2), β21 , β22 ≠ 1 id ⟨( 1 0 00 u 00 0 u−1 )⟩ ≅ Gm linear if β1β22 = −1Type S2β1 ∈ P1 stabilizer Autβ1 = 0,∞ Z/2Z {( 1 0 00 u w0 w u) , ( 1 0 00 u w0 −w −u )} ⊂ PGL(3) degenerate linearβ1 ≠ 0,∞ id ⟨( 1 0 00 1 00 0 −1 ) , ( 1 0 00 u 00 0 u )⟩ ≅ Z/2Z ×GmTable 3.1: Automorphisms of type S algebras613.3. Type S algebrasare two non-isomorphic 1-dimensional families of such algebras S1(a,∞,0)and S1(a,0,∞), corresponding to the two orbits of the 6 edges in the cube.S1(β1,∞,0) has defining equation yz + β1 zy = xz = xy = 0. The degree2 part defines a union of two planes X1 and X2 whereX1 = {(0,−β1 y2, z2) × (x2, y2, z2)}, X2 = {(x1, y1, z1) × (1,0,0)}S1(β1,0,∞) has defining equation yz + β1 zy = zx = yx = 0. The degree 2part also defines a union of two planes X1 and X2 whereX1 = {(x1, y1, z1) × (0, y1,−β1 z1)}, X2 = {(1,0,0) × (x2, y2, z2)}In both S1(β1,∞,0) and S1(β1,0,∞), σ is well defined automorphism ofZ(x), given by σ(y1, y2) = (y1,−β1 z1). So we consider τ ∈ PGL(C{y, z}).If β1 ≠ ±1 then the condition στ = τσ requires τ is diagonal. If β1 = 1then τ(y, z) = (z, y) also gives automorphism. If β1 = −1 then any τ ∈PGL(C{y, z}) gives an automorphism. We note this is same with the non-degenerate linear case.Now we study degenerate type S1 algebras which are not linear. Thereare three non-isomorphic families of such algebras, S1(0, β2, β3), S1(0,0, β3),where β2β3 ≠ 0, and S1(0,0,0). They have dimension 2,1,0, respectively.In terms of the cube, S1(0, β2, β3) is in the face containing (0,0,0). It hasdefining equation yz = zx+β2 xz = xy +β3 yx. The degree 2 part determinesa 4-gon whose components areX1 = {(0, y1, z1) × (0,1,0)}, X2 = {(0,0,1) × (0, y2, z2)},X3,4 = {(x1,0, z1) × (−β2 x1,0, z1)},X5,6 = {(x1, y1,0) × (x1,−β3 y1,0)} .An automorphism should stabilize the 4-gon. It fixes (1,0,0) and (0,0,1).Any element of S3 does not give an automorphism because it does not sta-bilize X1 ∪X2 the set of non-slant lines. Rescaling coordinates is an auto-morphism. So Aut(S1(0, β2, β3)) ≅ Gm2.S1(0,0, β3) lies on the edge containing (0,0,0) in the cube picture. It hasdefining equations yz = zx = xy+β3 yx. It defines a 5-gon whose componentsareX1 = {(0, y1, z1) × (0,1,0)}, X2 = {(0,0,1) × (0, y2, z2)},X3 = {(x1,0, z1) × (0,0,1)}, X4 = {(1,0,0) × (x2,0, z2)},X5,6 = {(x1, y1,0) × (x1,−β3 y1,0)} .623.3. Type S algebrasAn automorphism should stabilize the 5 gon. It fixes (1,0,0), (0,1,0), and(0,0,1), so should be a subgroup of S3 ×Gm3. But S3 does not stabilize theunique slant line X5,6. So Aut(S1(0,0, β3)) ≅ Gm2.S1(0,0,0) has the defining equations xy = yz = zx = 0. The degree 2 partdetermines a 6-gon in P2 × P2 whose components areX1 = {(0, y1, z1) × (0,1,0)}, X2 = {(0,0,1) × (0, y2, z2)}, (3.6)X3 = {(x1,0, z1) × (0,0,1)}, X4 = {(1,0,0) × (x2,0, z2)}, (3.7)X5 = {(x1, y1,0) × (1,0,0)}, X6 = {(0,1,0) × (x2, y2,0)} . (3.8)An automorphism should stabilize the 6 gon. It fixes (1,0,0), (0,1,0), and(0,0,1), so should be a subgroup of S3×Gm3. But odd permutations do notstabilize X1 ∪X3 ∪X5, the horizontal lines. A3 action gives automorphisms.So Aut(S1(0,0,0)) ≅ A3 ×Gm2.Adding degree 3 relations to S1(0,0,0) and flat familyFor algebraic properties of this degenerate algebra S(0,0,0) one can refer[13], [16]. Its Hilbert series is 1+t1−2t . Therefore the family S1 is not flat at(0,0,0). Let us consider a method to find flat family at (0,0,0).In degree 3, S1(0,0,0) has dimension 12 instead of 10. So we have to addnon-quadratic relations. S1(a, b, c) satisfies β2 xf1+β3 yf2+β1 zf3 = β3 f1x+β1 f2y+β2 f3z ∈ V ⊗R+R⊗V , which makes dim(V ⊗R+R⊗V ) = 9+9−1 = 17and a(3) = 33 − 17 = 10. However, in S1(0,0,0) it splits into three equationsxf1 = f3z, yf2 = f1x, and zf3 = f2y which are three overlapsx(yz) = (xy)zy(zx) = (yz)x (3.9)z(xy) = (zx)ymaking dim(R ⊗ V + V ⊗ R) = 15 and a(3) = 12. So to find a flat familyaround S1(0,0,0) we have to compensate the effect of this split up. It canbe done by adding relations corresponding to the overlaps (3.9) from thealgebras in the neighborhood. In S1(β1, β2, β3) we havex(yz) − (xy)z = −β1 xzy + β3 yxzy(zx) − (yz)x = −β2 yxz + β1 zyx (3.10)z(xy) − (zx)y = −β3 zyx + β2 xzy .633.3. Type S algebrasSo we add the right hand sides to S1(0,0,0) and defineS̃1(γ1∶γ2∶γ3) ∶= C⟨x, y, z⟩⎛⎜⎜⎜⎝xy, yz, zx,g1 ∶= γ1 zyx − γ2 yxz,g2 ∶= γ3 yxz − γ1 xzy,g3 ∶= γ2 xzy − γ3 zyx .⎞⎟⎟⎟⎠(3.11)Note that the parameter (γ1∶γ2∶γ3) of S̃1 is a homogeneous coordinate be-cause S̃1(γ1∶γ2∶γ3) = S̃1(k γ1∶k γ2∶k γ3). So we have a two dimensional fam-ily of algebras S̃1(γ1∶γ2∶γ3).We consider a family of algebras near (0,0,0). Consider an affine neigh-borhood U = {(β1, β2, β3) ∣βi ∈ C} of (0,0,0), and a blow up at (0,0,0),Bl(0,0,0)U = {(β1, β2, β3) × (γ1∶γ2∶γ3) ∣βiγj = βjγi}. We construct a familypi ∶ T → Bl(0,0,0)U of algebras bypi−1 ((β1, β2, β3) × (γ1∶γ2∶γ3)) = ⎧⎪⎪⎨⎪⎪⎩S1((β1, β2, β3)) if (β1, β2, β3) ≠ (0,0,0)S̃1(γ1∶γ2∶γ3) if (β1, β2, β3) = (0,0,0) .We can also consider this as a family of subspaces of V ⊗ V ⊗ V overBl(0,0,0)U . First consider a family pi ∶W × (U × P2)→ U × P2, whereW = C{xifj , fjxi, gk} ⊂ V ⊗ V ⊗ Vwhere gk are the degree 3 added relations in (3.11) and fj are the defin-ing equations of S1(β1, β2, β3) (3.5). Then we have a family of subspacesspanned by 21 degree 3 noncommutative polynomials. Over a generic pointits fiber has rank is 20. But on Z((βiγj − βjγi)i,j) the rank is constant and17, and we have a flat family we want.Now consider growth of the algebra S̃1(γ1∶γ2∶γ3). When γ1γ2γ3 ≠ 0,S̃1(γ1∶γ2∶γ3) has right a(n) for n < 5, and a(5) = 24 instead of 21. Whenone of βi is 0 we have a(4) = 16, and if two of them are 0 then a(4) = 17 .This implies that to have a flat family we have to add more relations as qgrows.643.3. Type S algebrasWe define an algebra S1(γ1∶γ2∶γ3) with relationsyz = zx = xy = 0 ,x`+1z`y = 0y`+1x`z = 0 (3.12)z`+1y`x = 0 ,zy`x`+1 = 0xz`y`+1 = 0 (3.13)yx`z`+1 = 0 ,γ`2xz`y` = γ`3z`y`xγ`3yx`z` = γ`1x`z`y (3.14)γ`1zy`x` = γ`2y`x`z .Proposition 3.3.2 S1(γ1∶γ2∶γ3) has Hilbert series (1 − t)−3. It is strictlysemistable, polystable and S1(γ1∶γ2∶γ3)≤3 = S̃1(γ1∶γ2∶γ3)≤3. It is a naturallimit lim(β1,β2,β3)→(0,0,0) S1(β1, β2, β3) along the direction (γ1∶γ2∶γ3).Proof. We see that the (3.14) with ` = 1 is the degree 3 relations in (3.11).This proves S1(γ1∶γ2∶γ3)≤3 = S̃1(γ1∶γ2∶γ3)≤3 .. Now we prove the Hilbertseries. Consider a monomial za1yb1xc1za2yb2xc2 . . . zakybkxck , where a1, b1,bk ck can be 0. First consider the case where at least four of ai, bi, ci arenot zero. If not all of the nonzero ai, bi, ci are the same then by (3.12) and(3.13) the monomial is 0. If all of them are the same then by (3.14) we canmove the last variable to the first. So we can assume that at most threeof ai, bi, ci are not zero. Then by (3.12) and (3.13) the middle one shouldhave the highest power. When two of x, y, z have the same power then by(3.14) we can put the first has the lowest power. For any triple (a, b, c) ofnon-negative integers there is a unique triple obtained by even permutationsuch that the second number is the largest and if two of them are the samethen the first is the lowest. Therefore S1(γ1∶γ2∶γ3) has the same number ofmonomial basis with the polynomial algebra and hence has the same Hilbertseries.653.3. Type S algebrasIt is polystable and strictly semistable because the subspace of relationsis invariant under A3 action.Finally, we show that S1(γ1∶γ2∶γ3) = lim(β1,β2,β3)→(0,0,0) S1(β1, β2, β3)along the direction (γ1∶γ2∶γ3). It is easy to see that (3.14) holds in S1(β1, β2, β3).In S1(β1, β2, β3) we have thatx`+1z`y = β−`2 x`+1yz` = β`+13 β−`2 yx`+1z` .If we choose a direction β3 = aβ2 then take the limit β2 → 0 along thisdirection then we have limx`+1z`y = 0. Other equations of (3.13) and (3.13)can be shown similarly. 2An automorphism of S̃1(γ1∶γ2∶γ3) is that of S1(0,0,0). We rememberthat Aut(S1(0,0,0)) is generated by even permutations and scalar multipli-cations. Because the added degree 3 relations are homogeneous and cyclicevery automorphisms of S1(0,0,0) is also that of S̃1(γ1∶γ2∶γ3).Aut (S̃1(γ1∶γ2∶γ3)) ≅ A3 × (Gm)2The same reason shows thatAut (S1(γ1∶γ2∶γ3)) ≅ A3 × (Gm)2 .3.3.2 Type S′1 algebrasType S′1 algebra S′1(β1, β2) is defined byf1 = yz + β1 zy + x2f2 = zx + β2 xz (3.15)f3 = xy + β2 yxwhere β1, β2 ∈ P1. We put f1 = zy + x2 when β1 = ∞. The algebra is AS-regular if β1, β2 ≠ 0,∞. Type S′1 algebras are parametrized by P1 × P1, andwe represent it by a square. We first consider the cases where (β1, β2) is inthe interior of the square, i.e. β1, β2 ≠ 0,∞. The point scheme is cut by thedeterminant ofM = ⎛⎜⎝x β1 z yz 0 β2 xβ2 y x 0⎞⎟⎠and isZ (x ((1 + β1β22)yz − β2 x2)) = Z(x) ∪Z ((1 + β1β22)yz − β2 x2) .663.3. Type S algebrasOn Z(x) the automorphism is σ(0, y, z) = (0, y,−β1 z), it stabilizes the lineZ(x). If β1 = −1 then σ fixes the line pointwise. If β1 ≠ −1 then σ has onlytwo fixed points which are the intersection points (0,1,0) and (0,0,1).When β1β22 ≠ −1, the conic Z ((1 + β1β22)yz − β2 x2) is smooth and σ iswell defined and given by σ(x, y, z) = (−β2 x,β22 y, z). If β2 = −1 then σ fixesevery point on the conic. If β2 ≠ −1 then σ has two fixed points which arethe intersection points.The smooth conic collapses into a double line Z(x2) if β1β22 = −1.Isomorphisms and AutomorphismsConsider a (Gm)3 action given by (u, v,w).(x, y, z) = (ux, v y,w y). Then itpreserves f1 upto scalar if and only if u2 = vw. So we have two dimensionalsubgroup of automorphism group for generic type S′1 algebras. If we consideronly non-tautological ones, we have Gm ⊂ Aut(S′1(β1, β2)), where Gm actsby u.(x, y, z) = (x,uy, 1u z).S3 action should fix x, and only g0(x, y, z) = (x, z, y) is possible. Then wehave to rescale x→ β−1/21 x. So a change of variables (x, y, z)↦ (β−1/21 x, z, y)gives an isomorphism S′1(a, b) ≅ S′1( 1β1 , 1β2 ) and there is no more isomorphismfor generic type S′1 algebras. (β1, β2) is fixed by the Z/2Z action whenβ21 = β22 = 1.However when β1 = −1 we have β1β22 = −1 and the point scheme is atriple line Z(x3). Let us consider the general case when the point scheme isa triple line Z(x3) and σ(x, y, z) = (− 1β2 x, y,−β1 z), where β1β22 = −1.If β1 or β2 is not −1 then σ has three different eigenvalues and στ = τσrequires τ is diagonal.Now we assume one of β1, β2 is −1. Then we have two cases (β1, β2) =(−1,−1) or (−1,1), because β1β22 = −1. If (β1, β2) = (−1,−1) then anyτ ∈ PGL(3) commutes with σ. However the point scheme Z(x) is preservedby τ . So τ has a form ( p 0 0∗∗ M ). To preserve f1 we require that p2 = detM .If we divide τ by p then we have detM = 1. So M ∈ SL(2).If (β1, β2) = (−1,1) then σ(x, y, z) = (−x, y, z). Then στ = τσ and that τpreserves Z(x) requires τ to be of the form. ( p 0 000 M). As before we can putp = 1 and M ∈ SL(2).Now we give a geometric computation of the automorphism group forthe case of the union of a line and a smooth conic. X = X1 ∪ X2 whereXi, i = 1,2, is isomorphic to P1. Let xi ∈ Gm = Xi ∖ {0,∞}. Thenσ(x1, x2) = (−β1 x1,−β2 x2). First, consider a multiplication on both Xi.τ(x1, x2) = (t1x1, t2x2) permutes with σ, and τ∗L ≅ L implies x1x22 = 1 so673.3. Type S algebrasGm ⊂ Aut(X,σ).Because L∣X1 and L∣X2 have different degrees, τ cannot interchange X1andX2, and only inversion xi ↦ 1xi is allowed. Then στ(x1, x2) = σ ( 1x1 , 1x2 ) =(−β1x1 ,−β2x2 ) and τσ(x1, x2) = τ(−β1 x1,−β2 x2) = (− 1β1 x1 ,− 1β2 x2 ). So στ = τσimplies β21 = β22 = 1. But if β1 = −1 then we have β1β22 = −1 which case weexclude. In sum we haveAutS′1(β1, β2) =⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩Gm if βi ≠ ±1Gm ×Z/2Z if β1 = 1, and β2 = ±1{( 1 0 000 M)} ≅ SL(2) if β1 = −1, and β2 = 1{( 1 0 0∗∗ M )} if β1 = β2 = −1 .Degenerate type S′1Now we consider degenerate cases, i.e. (β1, β2) lies on the boundary of thesquare. The Z/2Z action gives isomorphism between algebras for (β1, β2)and ( 1β1 , 1β2 ). So the Z/2Z action rotates the square by 180 degree. There aretwo non-isomorphic edges (β1,0), (0, β2) and two non-isomorphic vertices(0,0) and (∞,0).Among these only (∞,0) gives linear algebra. We recall that type S1 al-gebra S1(∞,0,0) is linear, we will study this when we discuss S-equivalence.S′1(∞,0) has defining equationszy + x2 = zx = xy = 0Degree 2 part determines a union of two planes in X1 ∪X2 ⊂ P2 ×P2, whereX1 = {(x, y, z) × (0,0,1)}, X2 = {(0,1,0) × (x, y, z)}Automorphism should fix (0,0,1) and (0,1,0). As the points (0,0,1) and(0,1,0) is fixed by τ , this algebra has no extra automorphism, Aut(S′1(∞,0)) =Gm. This algebra has the right Hilbert series and in our moduli.S′1(0, β2) has defining equationsyz + x2 = zx + β2 xz = xy + β2 yx .It has detM = detN = x(−β2 x2 + yz). But it is degenerate at one node.Degree 2 part determines a union of two lines X1, X2 and a bidegree (2,2)curve where,X1 = {(0, y1, z1) × (0,1,0)}, X2 = {(0,0,1) × (0, y2, z2)},X3,4,5,6 = {(x1, y1, z1) × (x1y1,−β2 y21,−x21) ∣ y1z1 = β2 x2} .683.3. Type S algebrasS′1(β1,0) has defining equationsyz + β1 zy + x2 = zx = xy .Degree 2 part determines a 4-gon having componentsX1,2 = {(0, y1, z1) × (0, y1, β1 z1)},X3 = {(x1,0, z1) × (0,0,1)},X4,5 = {(x1, y1,0) × (−y1,0, x1)}, X6 = {(0,1,0) × (x2, y2,0)} .This algebra has no extra automorphism. It has right Hilbert series and inour moduli.S′1(0,0) has defining equationsyz + x2 = zx = xy .Degree 2 part determines a 5-gon having componentsX1 = {(0, y1, z1) × (0,1,0)}, X2 = {(0,0,1) × (0, y2, z2)},X3 = {(x1,0, z1) × (0,0,1)}, (3.16)X4,5 = {(x1, y1,0) × (−y1,0, x1)}, X6 = {(0,1,0) × (x2, y2,0)} .This algebra has no extra automorphism. But it has a(3) = 11 so we haveto add one or two degree 3 relations.S̃′1 = C⟨x, y, z⟩(yz + x2, zx, xy, zyx − xzy)This algebra has the right dimension upto degree 4. a(5) = 23. We can alsoadd 0 = xyz − xyz = x(yz) − (xy)z = x(−x2 − β1 zy) − (−β2 yx)z. This givesa(3) = 10. But when a = 0 or b = 0 we have a(4) = 17.Growth of degenerate type S′1 algebras is given byLemma 3.3.3 Degenerate type S′1 algebra, except S′1(0,0) and S′1(∞,∞),has the Hilbert series (1 − t)−3.Proof. First we consider S′1(0, β2). As we exclude the case S′1(0,0), weassume that β2 ≠ 0. Then for each degree n, S′1(0, β2) has the same monomialbase with S1(0,∞,∞). Hence they have the same Hilbert series.For S′1(β1,0) we can also assume that β1 ≠ 0. Then from f1 = yz +β1 zy +x2we put zy = − 1β1 (x2 + yz). While −x2 − yz is not a monomial we still havethe same monomial basis with S1(∞,0,0), which is {yixjzk}. S′1(∞,0) alsohas the same monomial basis. Hence they have the same Hilbert series. Theclaim follows from Lemma 3.3.1. 2693.3. Type S algebras3.3.3 Type S′′1 algebrasThe term type S′′1 does not appear [3] or [4] and it is our own definition. Letus consider an algebra defined byf1 =x2 + b yz + c zyf2 =y2 + b zx + cxz (3.17)f3 =bxy + c yx .where b, c ∈ C. The algebra, appearing [3](7.21), is AS-regular if b, c,≠ 0.When bc = 0 it has a wrong Hilbert series. We haveM = ⎛⎜⎝x c z b yb z y cxc y bx 0⎞⎟⎠detM = (b3 + c3)xyz − bc(x3 + y3) .detM = 0 when b3 + c3 = bc = 0, but then we have f3 = 0. So we excludethis case. When b3 + c3 = 0 we have detM = bc(x3 + y3), and when bc = 0 wehave detM = (b3+c3)xyz. Otherwise, detM is an irreducible nodal with thenode (0,0,1). By cross producting the third and first rows of M we haveσ(x, y, z) = (b2 xy,−bc y2,−bx2 + c2yz)σ has unique fixed point, the node. The tangent cone at the node is xy = 0and σ preserves two branches.For simplicity of notation we define S′′1 (β) asS′′1 (β) ∶= C⟨x, y, z⟩⎛⎜⎜⎝f1 = yz + β zy + x2 ,f2 = zx + β xz + y2 ,f3 = xy + β yx⎞⎟⎟⎠(3.18)where β ∈ P1 and when β =∞, we have f1 = zy + x2 and f2 = xz + y2.M = ⎛⎜⎝x β z yz y β xβ y x 0⎞⎟⎠detM = (1 + β3)xyz − β(x3 + y3) .703.3. Type S algebrasIsomorphisms and AutomorphismsA change of variable (x, y, z)↦ (y, x, z) gives isomorphismS′′1 (β) ≅ S′′1 ( 1β )Automorphisms are given byAutS′′1 (β) =⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩⟨( 1 0 00 ζ3 00 0 ζ23)⟩ ≅ Z/3Z if β ≠ ±1⟨( 0 1 01 0 00 0 1) ,( 1 0 00 ζ3 00 0 ζ23)⟩ ≅ S3 if β = 1⟨( 0 1 01 0 00 0 1) ,( 1 0 00 ζ3 00 0 ζ23) , ( 1 0 00 1 0∗ ∗ 1 )⟩ if β = −1Let E be an irreducible nodal with a node P . σ must preserve P andhave no smooth fixed point as it is type A. Therefore, for x ∈ X we canconsider x as an element of Gm and σ(x) = −λx, where λ ∈ Gm. We callthis triple by Ts′′1(λ). When λ3 + 1 = 0 this elliptic triple satisfies σ∗L ≅ L.We compute automorphisms for a non-linear type A irreducible nodal.If τ(x) = t x then τ commutes with σ. To preserve L we should have t3 = 1.So τ(x) = ζi3 x. Let us call this τ1. If τ(x) = 1x then στ = τσ holds onlywhen λ = ±1. But λ = −1 is linear which we exclude. For λ = 1 we call thisinversion τ2. Then τ1τ2 ≠ τ2τ1 so the automorphism group is not Abelian.There is no more automorphism. In sum we haveAut (Ts′′1(λ)) = ⎧⎪⎪⎨⎪⎪⎩Z/3Z if λ2 ≠ 1S3 if λ = 1.This explains the automorphism group of S′′1 (β). The case β = −1 will beconsidered in 3.4.5.Degenerate type S′′1Now we consider the cases β = 0,∞. Algebras for them are isomorphic toeach other, so we only consider the case β = 0. S′′1 (0) is degenerate and thedegree 2 part determines 4-gon whose components areX6,1 = {(0, y1, z1) × (−y1, z,0)}, X2 = {(0,0,1) × (0, y2, z2)},X3 = {(x1,0, z1) × (0,0,1)}, (3.19)X4,5 = {(x1, y1,0) × (−y1,0, x1)} .713.3. Type S algebrasThis 4-gon is made from the 5-gon of S′1(0,0), (3.19), by identifying X1 andX6 to have a graph of σ(0, y1, z1) = (−y1, z1,0). We further identify X2 andX3 to get a graph of σ(x1,0, z1) = (0, z1,−x1), so σ is a rotation of a trianglecomposed with multiplication by −1. It is the type A algebra A(1,1,0) ofthe rotating triangle, which we will study later.So far we have two kinds of non-isomorphic 4-gons and 5-gons, related tothe 6-gon S1(0,0,0). S1(0,0,0) is a 6-gon whose components are Xi, wherei = 1,2,3,4,5,6. One kind is type S1 algebras: S1(0,0, c) gives a 5-gon madefrom S1(0,0,0) by identifying X5 and X6. Then S1(0, b, c) gives a 4-gon byfurther identifying X3 and X4. If we further identify X1 and X2 then it is theregular algebra S1(a, b, c). This collapsing process is continuous as movingin the moduli of type S1 algebras. The other kind is between different types.S′1(0,0) is a 5-gon made from S1(0,0,0) by identifying X4 and X5. S′′1 (0)gives a 4-gon which is obtained by further identifying X6 and X1. They areS-equivalent so give one point in the moduli. If we further identify X2 andX3 then it is the type A algebra A(1,1,0). So this identification of the lasttwo lines is a jump in the moduli.3.3.4 Type S2 algebrasType S2 algebra S2(a) has defining equationsf1 =y2 − z2f2 =zx − axz (3.20)f3 =axy + yxS2(a) is AS-regular if a ≠ 0,∞. We haveM = ⎛⎜⎝0 y −zz 0 −axy ax 0⎞⎟⎠ .The point scheme is given byZ(ax(y2 + z2)) = Z(x) ∪Z(y − i z) ∪Z(y + i z)which is a union of three lines intersecting at (1,0,0), (0,1, i), (0,1,−i). OnZ(x) the automorphism is σ(0, y, z) = (0, z, y), the inversion having two fixedpoints (1,1,0) and (1,−1,0). On Z(y−i z) the automorphism is σ(x, i z, z) =(ax,−i z, z), and on Z(y+i z) the automorphism is σ(x,−i z, z) = (ax, i x, z).The node (1,0,0) is fixed. So σ interchanges Z(y − i z) and Z(y − i z), and723.3. Type S algebrasstabilizes Z(x). σ has order 2 on Z(x) but on the other two lines σ2 isstretching x by a2. So when a = ±1 we have σ2 = id.With a change of variablesy → y + zz → i(y − z)we have new relationsf ′1 = y2 + z2f ′2 = yx + axz (3.21)f ′3 = zx + axygiving M = ⎛⎜⎝0 y zy 0 axz ax 0⎞⎟⎠ . The point scheme is given byZ(2axyz).On Z(x) the automorphism is σ(0, y, z) = (0, z,−y) which has two fixedpoint (0,1, i) and (0,1,−i). On Z(y) we have σ(x,0, z) = (−ax, z,0), andon Z(z), σ(x, y,0) = (−ax,0, y). We will use this form of S2(a) when westudy type B algebras.Isomorphisms and AutomorphismsNow we compute automorphisms. From now on we will use the origi-nal defining equations (3.20). A Gm3 action given by (u, v,w).(x, y, z) =(ux, v y,w z) preserves (3.20) if v2 = w2. SoGm2×Z/2Z = ⟨(u, v, v), (1,1,−1)⟩acts as automorphisms making moduli of type S2 algebras Gm2×Z/2Z-gerbe.A change of variables (x, y, z)↦ (x, z, y) gives an isomorphismS2(a) ≅ S2(−a) .S2(0) has an extra automorphism.Now we obtain Aut(S2(a)) geometrically. The cubic is a triangle X =X1 ∪X2 ∪X3. Let xi ∈ Gm = Xi/{0,∞}. σ(x1, x2, x3) = (− 1x1 ,− ax3 ,− 1ax2 ).Consider τ(x1, x2, x3) = (t1 x1, t2 x2, t3 x3). Then στ(x1, x2, x3) = σ(t1 x1, t2 x2, t3 x3) =(− 1t1 x1 ,− at3 x3 ,− 1a t2 x2 ). And τσ(x1, x2, x3) = τ( 1−x1 ,− ax3 ,− 1ax2 ) = (−t1 1x1 ,−t2 ax3 ,−t3 1ax2 ).So we have t1 = ±1 and t3 = 1t2 . Then t1t2t3 = 1 implies τ(x1, x2, x3) =(x1, t x2, 1t x3). So Gm ⊂ Aut(S2(a)).733.3. Type S algebrasWe observe that σ has two non-singular fixed points on X1 and τ mustpreserve them, thus should fix X1, not necessarily pointwise. A scalar multi-ple on x2, x3 has already been discussed, so τ is inversion and τ(x1, x2, x3) =( 1x1 , 1x3 , 1x2 ). Then στ(x1, x2, x3) = σ ( 1x1 , 1x3 , 1x2 ) = (−x1,−ax2,− 1a x3) andτσ(x1, x2, x3) = τ(− 1x1 ,− ax3 ,− 1ax2 ) = (−x1,−ax2, 1ax3). So στ = τσ. In con-clusion we haveAut (S2(a)) = Z/2Z ×GmDegenerate type S2There are two degenerate type S2 algebras, S2(0) and S2(∞). Both arelinear.The S2(0) has defining equationsy2 − z2 = zx = yx = 0 .The degree 2 part determines a union of two planes X1 and X2 whereX1 = {(x1, y1, z1) × (0, z1, y1)}, X2 = {(1,0,0) × (x2, y2, z2)} .The S2(∞) has defining equationsy2 − z2 = xz = xy = 0The degree 2 part determines a union of two planes X1 and X2 whereX1 = {(0, z2, y2) × (x2, y2, z2)}, X2 = {(x,y1, z1) × (1,0,0)} .In both S2(0) and S2(∞), restricted to Z(x), we have an automorphismσ(0, y, z) = (0, z, y) as the non-degenerate S2(a) has. An automorphismof S2(0) and S2(∞) should fix (1,0,0) and Z(x), so it is a (1,2) blockdiagonal. And on Z(x) it should commute with σ(y, z) = (z, y) up-to scalarmultiplication. So we haveAut(S2(0)) = Aut(S2(∞)) = {( 1 0 00 u w0 w u) , ( 1 0 00 u w0 −w −u )} ⊂ PGL(3) .This type S2 degenerate linear algebra S2(0) looks similar to the typeS1 degenerate linear algebra S1(−1,0,∞), in their defining equations, andthe two planes X1 and X2 determined by them. But they differ. In bothcases σ is well-defined on Z(x). But σ∣Z(x) = id for the S1(−1,0,∞), andσ∣Z(x)(t) = 1t for S2(0).Finally, we compute the growth of S2(0) and S2(∞).743.3. Type S algebrasLemma 3.3.4 S2(0) and S2(∞) have the Hilbert series (1 − t)−3.Proof. They are isomorphic, so we only prove for S2(0) which has definingequations y2 − z2 = zx = yx = 0. Because zx = yx = 0, we have that yand z cannot come before x. If a monomial of degree n starts with x thenby eliminating the first x we have a monomial of degree n − 1. So a(n) =a(n − 1) + b(n) where b(n) is the number of monomials not having x. Fromz2 = y2, we have that z appears only as z not as zi, i ≥ 2. In degree 3 wehave zy2 = y2z from z3 = z(z2) = (z2)z. So if yi, i ≥ 2 appears in the middlewe can move it to the left, even to the first place. Therefore a monomialnot having x should be a form of yizyzy . . ., and there are n + 1 many ofsuch degree n monomials because it is determined by i where i = 0,1, . . . , n.So b(n) = n + 1. Hence we have a(n) = a(n − 1) + n + 1 and a(1) = 3. Thisimplies a(n) = (n+1)(n+2)2 . 23.3.5 Semi-stability, S-equivalence, and PolystabilitySemistabilityProposition 3.3.5 Type S1 algebra S1(β1, β2, β3) is strictly semistable. S′1(β1, β2)is strictly semistable. S′′1 (β) is strictly semistable.Proof. When S1(β1, β2, β3) is not degenerate, a test configuration cor-responding to a node or line component gives the constant Futaki func-tion for all q > 0. When S1(β1, β2, β3) is degenerate but not S1(0,0,0) orS1(∞,∞,∞), i.e. some of β1, β2, β3 is 0 or ∞ but not all of them, we canchoose the same degree 2 eliminated monomials with non-degenerate typeS1 algebra. Because it has right Hilbert polynomial it is strictly semistable.Defining equations of S1(0,0,0) and S1(∞,∞,∞) are homogeneous in x, y, zso the algebras are equally generated in x, y, z and strictly semistable. 2Proposition 3.3.6 For any (γ1∶γ2∶γ3) ∈ P2∖Z(γ1γ2γ3), we have that S̃1(γ1∶γ2∶γ3)is strictly semistable for all q. It has the correct Hilbert polynomial upto de-gree 4.Proof. When any of βi is not 0 then S̃1(γ1∶γ2∶γ3) is equally generated byx, y, z so it is strictly semistable. We checked by computer that it has a rightdimension upto degree 4. 2Here we present algebraic observations on strict semistability of type Salgebras.753.3. Type S algebrasLemma 3.3.7 If noncommutative graded algebra A with dimAn = (n+1)(n+2)2has a degree 1 element x which is not a zero divisor and UA1 = A1U whereU = ⟨x⟩, then A has a constant Futaki function with respect to U .Proof. From the assumption UV = V U we have thatJ(i)n = ∑permutationsV n−iU i = V n−iU i,which givesdimJ(i)n = dimV n−iU i = dimV n−i = a(n − i)because x is not a zero divisor. Soq(n) = n∑i=1 dimJ(i)n =n∑i=1a(n − i) = n(n + 1)(n + 2)6 .For a constant Futaki function we needq(n) − n ⋅ a(n)q(1)3 ⋅ 1 = n(n + 1)(n + 2)6 − n(n + 1)(n + 2)2 13 = 0which is true. 2Lemma 3.3.8 Type S1,S′1,S2 quadratic AS-regular algebra has a constantFutaki function with respect to the test configuration generated by a linecomponent.Proof. With U = ⟨x⟩ it is easy to see from the defining equations thatUV = UV . Then the claim follows from Lemma 3.3.7. 2Proposition 3.3.9 Type S1,S′1,S2 quadratic AS-regular algebra is strictly3-semistable.Proof. Test configurations not covered in Section 2.3 are those with P asa fixed node or a interchanged node and those with Y as a linear componentwhich is either fixed or interchanged. The Futaki function for the case whenP is a fixed node (whether there is no U or U is any line passing throughP ) can be computed by a similar method in Section 2.3. So we need tocompute the Futaki function for the case P is a node of order 2, which istype S2. We use the defining equations (3.21) which have Z(xyz) as E. IfY is the fixed line then we have U = ⟨x⟩ and W = ⟨x, y⟩. We give weightsw(x) = ` +m, w(y) = `, w(z) = 0. Then the terms of the lowest weight off ′1, f ′2, f ′3 are z2, xz, zx respectively. Then w(2) = 10`+4m. For semistabilitywe need w(2) ≥ 8` + 4m which is true. Similarly we can compute thatF (3) > F (1). 2763.3. Type S algebrasS-equivalenceProposition 3.3.10 S′1(β1, β2) is S-equivalent to S1(β1, β2, β2)Proof. With a test configuration generated by U = ⟨x⟩, the lowest weightpart of f1 of S′l(β1, β2) is yz+β1 zy which is the same as f1 of S1(β1, β2, β2).f2 and f3 of S′1(β1, β2) are the same as f2, f3 of S1(β1, β2, β2), respectively.So S1(β1, β2, β2) is the limit of S′1(β1, β2) along the 1-parameter subgroupgenerated by U , and they are S-equivalent. 2Proposition 3.3.11 The S′′1 (β) is S-equivalent to S1(β,β, β).Proof. With a flag with no U and W = ⟨x, y⟩, the lowest weight part of f1is yz + β zy which is same with the f1 of S1(β,β, β). The same is true forf2. And f3 for both algebras are the same. So the S1(β,β, β) is the limitalgebra for S′′1 (b, c) with respect to the 1 parameter subgroup generated byW . 2Type S1 is polystableWe have thatProposition 3.3.12 Type S1 algebras are polystable.Proof. S1(a, b, c) is strictly semistable with test configuration generatedby a line component, or a vertex, or both the line and a vertex on it. Andsuch 1 parameter subgroups fixes S1(a, b, c). And 1 parameter subgroupscorresponding to other test configuration do not have a limit. So the orbitof S1(a, b, c) is closed 2Polystable algebra for type S2Recall that type S2 algebras are given byf1 =y2 − z2f2 =zx − axz (3.22)f3 =axy + yx.The monomials in f2 have the same degrees in x, y, z. The same is true forf3. But f1 is not. A test configuration U = ⟨z⟩ will make f ′1 = y2. Then we773.3. Type S algebrashave M = ⎛⎜⎝0 y 0z 0 −axy ax 0⎞⎟⎠ with detM = xy2. The degree 2 part of algebradetermines a triple point whose components areX1 = {(x1,0, z1) × (ax1,0, z1)},X2 = {(0, y, z) × (0,0,1)}, X3 = {(0,0,1) × (0, y, z)}meeting in (0,0,1) × (0,0,1). This algebra has a(3) = 11, and we need toadd more relations. Because only f1 has been changed, we have to considera overlap from the f1, that is y3. We have y3−y3 = y(y2)−(y2)y = yz2−z2y.So we defineS̃2(a) ∶= C⟨x, y, z⟩(zx − axz, axy + yx,y2, yz2 − z2y ).Then we haveProposition 3.3.13 An algebra S̃2(a) is the polystable algebra S-equivalentto S2(a).Proof. Two terms of yz2 − z2y have the equal weight so the algebra ispolystable. It has the right Hilbert series because it has the same eliminatedmonomials with S2(a). 23.3.6 Conclusion and description of moduli spaceIn concluding this chapter we describe the moduli of type S algebras.Proposition 3.3.14 All type S quadratic AS-regular algebras are strictlysemistable for any q. To compactify moduli we need to add degenerate al-gebras which are limits of type S algebras. They have the Hilbert series(1 − t)−3, except the S1(0,0,0), which is isomorphic to S1(∞,∞,∞), andS′1(0,0), which is isomorphic to S′1(∞,∞).We have 1-1 correspondence between isomorphism classes of 2-cutoffS1(β1, β2, β3)≤2 of type S1 algebras and points of the quotient stack[(P1 × P1 × P1) /S3]and between isomorphism classes of 3-cutoff and points of[Bl(0,0,0),(∞,∞,∞) (P1 × P1 × P1) /S3] .783.4. Type A algebraswhere the exceptional divisor consists of S̃1(γ1∶γ2∶γ3). Moduli stacks of 2-cutoff and 3-cutoff of type S1 algebras are Gm3-gerbes over the quotientstacks.The moduli of type S′1 algebras is a hyperplane cut by the equation β2 = β3in the moduli of type S1 algebras, as S′1(β1, β2) is S-equivalent to S1(β1, β2, β2).There is 1-1 correspondence between isomorphism classes of type S2 al-gebras S2(a) and points of the quotient stackP1/(Z/2Z) .The moduli stack of type S2 algebras is a Gm2-gerbe over the quotient stack.3.4 Type A algebrasIn this chapter we study type A algebras. In Section 3.4.1 we first seebasics of the triple, and in Section 3.4.2 compute the isomorphisms andautomorphisms. Type A-algebras form a 2-dimensional moduli. In themoduli there are two lines L1 and L6 of singular elliptic curves which westudy in Section 3.4.3. L1 is where type A moduli meets with the moduliof type S and algebras on it are strictly semistable. Nodal cubics are S-equivalent to the triangle on L1. The other line L6 consists of trianglesrotated by σ and their algebras are stable. There is another line of interest,L13, on which σ2 = id. Algebras having additional automorphism lay on L1and L13. Type B algebra is a Zhang twist of an algebra on L13. So in themoduli of Z-graded AS-regular algebras the L13 is where the type B algebrassit in.In Section 3.4.4 we study nodal curves and their S-equivalence. A lineartype A algebra is S-equivalent to the algebra with a triple point and tripleline. We found other singular cubics, cusp, tacnode, etc. which are S-equivalent to the linear type A algebras. This is done in Section 3.4.5.2-cutoff algebras are parametrized by P2/SL(2,F3). There are two iso-morphic classes of degenerate algebras. To find a flat family around themwe need to blow-up. We can form the exceptional divisor of this blow upby degenerate algebras with additional degree 3 relations or by exceptionalalgebras. This is done in Section 3.4.6.793.4. Type A algebras3.4.1 Triple (E,L,σ)By A(a, b, c) we mean the algebra with defining equationsf1 = ax2 + b yz + c zyf2 = ay2 + b zx + cxz (3.23)f3 = az2 + bxy + c yxwhere a, b, c ∈ C. The matrix M,N which is given by f =MX =XtN whereX = (x y z)t isM = N = ⎛⎜⎝ax c z b yb z a y cxc y bx a z⎞⎟⎠ .The determinant of M is the Hesse cubicabc(x3 + y3 + z3) − (a3 + b3 + c3)xyzand determines a cubic curve E = Z(detM) in P2. E = P2 if abc = a3+b3+c3 =0. This is called the linear case. σ(x, y, z) is the null-vector of M evaluatedat (x, y, z). By cross producting the second and third rows of M we haveσ(x, y, z) = (a2 yz − bcx2, c2 xy − ab z2, b2 xz − ac y2) .The σ is not well defined if two of a, b, c is equal to 0 or a3 = b3 = c3 = 1.Such A(a, b, c) is called degenerate.The point P0 = (0,1,−1) is a flex and L ≅ O(3P0). And σ(P0) =(a, b, c) =∶ Q. So if we define a group law on E by choosing P0 as theorigin then σ is a translation by Q.The j invariant is given byj(X) = 27µ3(µ3 + 8)3(µ3 − 1)3where µ = a3+b3+c33abc . This µ ∈ P1 is a parameter of the Hesse pencil.3.4.2 Isomorphisms and Automorphisms of type A algebrasBefore we study moduli of type A algebras, let us first study automorphismsand isomorphisms.803.4. Type A algebrasHesse groupHere we summarize well known facts about isomorphisms and automor-phism of the Hesse cubic. The Hesse group G216 ⊂ PGL(3,C) which isisomorphic to (Z/3Z)2⋊SL(2,F3) is the group of projective transformationswhich preserve the Hesse pencil [8]. There is an exact sequence of grouphomomorphisms [2]0Ð→ ker ΦH Ð→ G216 ΦHÐ→ Aut(P1)where P1 is the parameter space of the Hesse pencil. The kernel of ΦH ,which has order 18, is the group of automorphisms of a Hesse cubic. It isgiven byker ΦH = ⟨g0, g1, g2⟩ ,whereg0(x, y, z) =(x, z, y)g1(x, y, z) =(y, z, x)g2(x, y, z) =(x, ζ3 y, ζ23 z).The image of ΦH which represents isomorphisms of different Hesse cubics isgenerated byg3 = ⎛⎜⎝1 0 00 ζ3 00 0 ζ3⎞⎟⎠ , g4 =⎛⎜⎝1 1 11 ζ3 ζ231 ζ23 ζ3⎞⎟⎠ .Im(ΦH) ≅ A4 where g3, g4 correspond to the generators of order 2, 3 of A4respectively. It is also known that⟨g0, g3, g4⟩ ≅ SL(2,F3)and it is a central extension of degree two of A4. We haveG216 = ⟨g0, g1, g2, g3, g4⟩ .Isomorphisms and automorphisms for generic Sklyanin algebrasFrom the G216, g0, g1, g2 fixes the Hesse cubic. With a choice of (0,1,−1) asthe identity and σ is a translation by (a, b, c), we have that g0 is a multipli-cation by −1, and g1, g2 are translations by 3 division points. So g0 gives anisomorphism between A(a, b, c) and A(a, c, b). g1 and g2 are automorphisms.813.4. Type A algebrasTo find all automorphisms and isomorphisms of A(a, b, c), we note thatif τ gives a non-tautological isomorphism between two type A algebras thenτ also gives an isomorphism of their Hesse cubics and thus τ ∈ G216. Sowe study how the action of G216 on P3(x, y, z), the ambient space of pointschemes, induces an action of G216 on the P3(a, b, c), the parameter space ofSklyanin algebras. In other words we find the kernel and image of Φ whereΦ ∶ G216 Ð→ Aut (P2(a, b, c)) .First, g0(x, y, z) = (x, z, y) givesf1(a, b, c)↦ f1(a, c, b)f2(a, b, c)↦ f3(a, c, b)f3(a, b, c)↦ f2(a, c, b).So Φ(g0) = h0 where h0(a, b, c) = (a, c, b).Second, g1(x, y, z) = (y, z, x) givesf1(a, b, c)↦ f2(a, b, c)f2(a, b, c)↦ f3(a, b, c)f3(a, b, c)↦ f1(a, b, c),and we have that g1 ∈ ker Φ.Third, g2(x, y, z) = (x, ζ3y, ζ23z) givesf1(a, b, c)↦ f1(a, b, c)f2(a, b, c)↦ ζ23f2(a, b, c)f3(a, b, c)↦ ζ3f3(a, b, c),so we have that g2 ∈ ker Φ.Fourth, g3(x, y, z) = (x, ζ3y, ζ3z) givesf1(a, b, c)↦ f1(a, ζ23b, ζ23c)f2(a, b, c)↦ ζ23f2(a, ζ23b, ζ23c)f3(a, b, c)↦ ζ23f3(a, ζ23b, ζ23c).So Φ(g3) = h′3 where h′3(a, b, c) = (a, ζ23b, ζ23c).Finally, g4(x, y, z) = (x + y + z, x + ζ3y + ζ23z, x + ζ23y + ζ3z) givesf1(a, b, c)↦ (f1 + f2 + f3)(a + b + c, a + ζ2b + ζc, a + ζb + ζ2c)f2(a, b, c)↦ (f1 + ζ2f2 + ζf3)(a + b + c, a + ζ2b + ζc, a + ζb + ζ2c)f3(a, b, c)↦ (f1 + ζf2 + ζ2f3)(a + b + c, a + ζ2b + ζc, a + ζb + ζ2c),823.4. Type A algebrasSo, Φ(g4) = h′4 where h′4(a, b, c) = (a + b + c, a + ζ2b + ζc, a + ζb + ζ2c).In sum we haveProposition 3.4.1 For generic A(a, b, c) we haveAut(A(a, b, c)) = ⟨g1, g2⟩ ≅ (Z/3Z)2 .And g0, g3, g4 gives isomorphisms between A and h0(A), h′3(A), h′4(A) re-spectively. ⟨h0, h′3, h′4⟩ ≅ SL(2,F3).Algebras with a nontrivial stabilizer of SL(2,F3) actionHere we give Table 3.2 the list of element of SL(2,F3) and points on P2(a, b, c)fixed by it. If more than one element of SL(2,F3) has the same set of fixedpoints then they are written in the same cell of the table.Table 3.3 shows point (a, b, c) ∈ P2 and their stabilizer in SL(2,F3) withnon-tautological automorphisms. The column titled Aut shows that whenj(X) = ∞ we have that Aut/(Z/3Z)2 is not isomorphic to the stabilizerand has a higher dimension. But not all algebras with j(X) =∞ have thisproperty. The algebra A(a, b,0) = C⟨x, y, z⟩/(ax2+b yz, a y2+b zx, a z2+bxy)has a triangle as E but its automorphism group is (Z/3Z)2. These algebrasare stable. The algebras on the row of j(X) = ∞ lie on the 4 lines and 4points in Theorem 2.4.2We also note that type A algebras with an extra automorphism lie ontwo lines a = 0 and b = c. We will see that the line a = 0 is the intersection ofthe moduli of type A and type S1. It consists of strictly semistable algebras.The line b = c consists of algebras with σ2 = id. An algebra on this line is aZhang twist of a type B algebra.We see that when A(a, b, c) is stable it has finite automorphisms.Not all triples with j = 1728 appear. The j invariant for A(a,1,1) isgiven byj(A(a,1,1)) = (a12 + 8a9 + 240a6 + 464a3 + 16)3a3 (a3 − 1)6 (a3 + 8)3a rational function of order 36. j = 1728 has 18 different solutions, and 6 ofthem appear on the above table. It means among 3 two-division points onlyone appears as a translation point of σ.Using lemma 3.2.8 we now compute automorphisms of type A triple. WehaveProposition 3.4.2 Let (E,L,σ) be a non-linearsmooth type A triple, i.e.(E,P0) is a smooth elliptic curve, L ≅ O(3P0) , σ(P ) = P ⊕Q where Q is833.4. Type A algebrasElement of SL(2,F3) Fixed points (a, b, c) ∈ P2s2 = ( 1 0 00 0 10 1 0) (a, b, b), (0,1,1), (0,1,−1)s3 = ( 1 0 00 ζ 00 0 ζ) , s5 = ( 1 0 00 ζ2 00 0 ζ2) (1,0,0), (0, b, c)s4 = ( 1 0 00 0 ζ0 ζ 0) , s6 = ( 1 0 00 0 ζ20 ζ2 0) (1,0,0), (0,1,1), (0,1,−1)s7 = ( 1 1 11 ζ ζ21 ζ2 ζ) , s8 = ( 1 1 11 ζ2 ζ1 ζ ζ2) (0,1,−1), (1 −√3,1,1), (1 +√3,1,1)s9 = ( 1 1 1ζ 1 ζ2ζ ζ2 1) (0,1,−1), (−ζ(b + c), b, c), (−2ζ,1,1), (ζ,1,1)s10 = ( 1 1 1ζ ζ2 1ζ 1 ζ2) (0,1,−1), (−2ζ,1,1), (ζ,1,1)s11 = ( 1 1 1ζ2 1 ζζ2 ζ 1) (a, b,−ζa − b), (1, ζ, ζ)s12 = ( 1 1 1ζ2 ζ 1ζ2 1 ζ) (0,1,−1), (1, ζ, ζ), (2,−ζ,−ζ)s13 = ( 1 ζ ζ1 1 ζ21 ζ2 1) (−ζ2(b + c), b, c), (ζ2,1,1), (−2ζ2,1,1)s14 = ( 1 ζ ζ1 ζ2 11 1 ζ2) (0,1,−1), (ζ2,1,1), (−2ζ2,1,1)s15 = ( 1 ζ ζζ 1 ζζ ζ 1) (0,1,−1), (1,1,1), (1,1,−2), (a, b,−a − b)s16 = ( 1 ζ ζζ ζ 1ζ 1 ζ) (0,1,−1), (1,1,1), (1,−1/2,−1/2)s17 = ( 1 ζ ζζ2 ζ2 ζζ2 ζ ζ2) , s18 = ( 1 ζ ζζ2 ζ ζ2ζ2 ζ2 ζ) (0,1,−1), (1 −√3, ζ2, ζ2), (1 +√3, ζ2, ζ2)s19 = ( 1 ζ2 ζ21 1 ζ1 ζ 1) (−ζ(b + c), b, c), (ζ,1,1), (−2ζ,1,1)s20 = ( 1 ζ2 ζ21 ζ 11 1 ζ) (0,1,−1), (ζ,1,1), (−2ζ,1,1)s21 = ( 1 ζ2 ζ2ζ ζ ζ2ζ ζ2 ζ) , s22 = ( 1 ζ2 ζ2ζ ζ2 ζζ ζ ζ2) (0,1,−1), ((1 +√3)ζ2,1,1), ((1 −√3)ζ2,1,1)s23 = ( 1 ζ2 ζ2ζ2 1 ζ2ζ2 ζ2 1) (0,1,−1), (1,1,1), (a, b,−a − b), (1,1,−2)s24 = ( 1 ζ2 ζ2ζ2 ζ2 1ζ2 1 ζ2) (0,1,−1), (1,1,1), (2,−1,−1)Table 3.2: Elements of SL(2,F3) and their fixed points843.4. Type A algebrasj-inv# oforbits∣σ∣ Point (a, b, c) ∈ P2 stabilizer, (i) in si Aut∞ 1 1 (0,1,−1) SL(2,F3) PGL(3) linear∞ 2 2(1,0,0), (0,1,1), (1,2,3,4,5,6)S3 ×Gm2 degenerate(1,1,1), (−2,1,1), (1,2,15,16,23,24)(1, ζ, ζ), (−2, ζ, ζ), (1,2,11,12,13,14)(1, ζ2, ζ2) (−2, ζ2, ζ2) (1,2,9,10,19,20)∞ P1 any(0, b, c), (1,3,5)A3 ×Gm2 Type S1(−(b + c), b, c), (1,15,23)(−ζ(b + c), b, c), (1,9,19)(−ζ2(b + c), b, c) (1,11,13)1728 1 2(1 −√3,1,1), (1 +√3,1,1), (1,2,7,8)Z/4Z × (Z/3Z)2(1 −√3, ζ, ζ), (1 +√3, ζ, ζ), (1,2,21,22)(1 −√3, ζ2, ζ2), (1 +√3, ζ2, ζ2) (1,2,17,18)any P1 2 (a, b, b) (1,2) Z/2Z × (Z/3Z)2 generic σ2 = idTable 3.3: Automorphisms of type A algebras853.4. Type A algebrasnot a 3-division point. When j(X) = 1728, let R be the 2-division point fixedby the multiplication by ζ4. Then we have thatAut (A(E,L,σ)) = ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩(Z/3Z)2 if Q⊕Q ≠ P0(Z/3Z)2 ×Z/4Z if j(X) = 1728 and Q = R(Z/3Z)2 ×Z/2Z otherwisewhere (Z/3Z)2 is a group of translation by 3-division points, Z/2Z is gener-ated by the multiplication by −1, and Z/4Z is generated by multiplication byζ4.Proof. First, consider τ which is a translation. Then τ satisfies στ = τσ.τ∗L ≅ L if and only if τ is a translation by a 3 division point. So we havethat (Z/3Z)2 ⊂ Aut(A).Now consider the case where τ fixes the origin P0. Then τ∗L ≅ L, so weonly need to check στ = τσ.When τ is a multiplication by −1, we have στ(P ) = σ(−P ) = (−P )⊕Q,and τσ(P ) = τ(P ⊕Q) = −(P ⊕Q) = (−P )⊕ (−Q). So στ = τσ is equivalentto Q ⊕ Q = P0. Therefore when Q is a 2-division point, we have Z/2Z ⊂Aut ((E,L,σ)).When j(X) = 0 and τ is a multiplication by ζi3, i = 1,2, it is easy tocheck that στ = τσ holds if and only if σ is a translation by fixed points ofτ . Then σ is a translation by a 3-division point and σ∗L ≅ L.When j(X) = 1728 and τ is a multiplication by ζi4, i = 1,2,3, we havethat στ = τσ holds if and only if σ is a translation by a 2-division point fixedby τ .Now we consider the cases where τ is a rotation by θ around R ≠ P0.Then we have τ∗L ≅ L only when τ preserves the lattice of 3-division points.This happens only in following cases: (a) j(X) = 1728, θ = pi2k and R⊕6 = P0,(b) j(X) = 0, θ = pi3k, and R⊕3 = P0, (c) j(X) = 0, θ = 2pi3 k, and R⊕6 = P0.It is tedious but one can show that in these cases στ = τσ holds only in thecases discussed above. 23.4.3 Triangle E and the 12 linesIsomorphisms and automorphisms on the 12 linesHere we study SL(2,F3) action on the 12 lines of the Hesse configuration.The Hesse pencil {Eµ}µ∈P with Eµ(a, b, c) ∶= Z(a3 + b3 + c3 − 3µabc) has 4863.4. Type A algebrassingular members and they are all triangles,E∞ ∶ Z (abc)E1 ∶ Z ((a + b + c)(a + ζ3b + ζ23c)(a + ζ23b + ζ3c))Eζ3 ∶ Z ((a + ζ23b + ζ23c)(a + ζ3b + c)(a + b + ζ3c))Eζ23∶ Z ((a + ζ3b + ζ3c)(a + ζ23b + c)(a + b + ζ23c))from which we have 12 lines. The base locus of the Hesse pencil consists of9 points which are the inflection points of the Hesse cubics,p0 = (0,1,−1), p1 = (0,1,−ζ3), p2 = (0,1,−ζ23),p3 = (1,0,−1), p4 = (1,0,−ζ3), p5 = (1,0,−ζ23),p6 = (1,−1,0), p7 = (1,−ζ3,0), p8 = (1,−ζ23 ,0) .Each line contains 3 of the 9 points. Recall that the point scheme is givenbyx3 + y3 + z3xyz= a3 + b3 + c3abcwhich is symmetric to (x, y, z) and (a, b, c). So when (a, b, c) ∈ Eµ(a, b, c)the point scheme of A(a, b, c) is Eµ(x, y, z) for the same µ. This impliesA(a, b, c) is singular(triangle) if and only if (a, b, c) lines on the 12 lines.Let us name the 12 linesl1 = Z(a), l2 = Z(a + b + c),l3 = Z(a + ζ2b + ζ2c), l4 = Z(a + ζb + ζc),l5 = Z(b), l6 = Z(c), (3.24)l7 = Z(a + ζb + ζ2c), l8 = Z(a + ζ2b + ζc),l9 = Z(a + b + ζc), l10 = Z(a + ζb + c),l11 = Z(a + b + ζ2c), l12 = Z(a + ζ2b + c) .Also recall thath0(a, b, c) = (a, c, b)h′3(a, b, c) = (a, ζ23b, ζ23c)h′4(a, b, c) = (a + b + c, a + ζ2b + ζc, a + ζb + ζ2c) .873.4. Type A algebrasLemma 3.4.3 We have the following:(i) An algebra on lk, k = 1, . . . ,4 is isomorphic to an algebra on l1. Andan algebra on lk, k = 5, . . . ,12 is isomorphic to an algebra on l6.(ii) ⟨h0, h′3⟩ ≅ Z/6Z preserves l1(ii-a) ⟨g3⟩ ≅ Z/3Z ⊂ Aut (A(0, b, c))(ii-b) ⟨g0⟩ ≅ Z/2Z gives an isomorphism between A(0, b, c) and A(0, c, b)(ii-c) ⟨g0, g3⟩ ≅ Z/6Z ⊂ Aut (A(0,1,±1))(iii) ⟨h′3⟩ ≅ Z/3Z preserves l6(iii-a) g3 gives an isomorphism between A(a, b,0) and A(a, ζ2c,0)(iii-b) ⟨g3⟩ ⊂ Aut (A(0,1,0)) and ⟨g0, g3⟩ ≅ Z/6Z ⊂ Aut (A(1,0,0)).Proof. (i) Points on l7 are of the form (−ζb − ζ2c, b, c), and h′3(−ζb −ζ2c, b, c) = (−ζb−ζ2c, ζ2b, ζ2c). But (−ζb−ζ2c)+ζ2(ζ2b)+(ζ2c) = −ζb−ζ2c+ζb + ζ2c = 0. So h′3 maps l7 to l12.From multiplication by s16 = ( 1 ζ ζζ ζ 1ζ 1 ζ) we obtain s16(0, b, c) = (ζb+ ζc, ζb+c, b + ζc). And ζb + ζc + ζ2(ζb + c) + ζ2(b + ζc) = ζb + ζc + b + ζ2c + ζ2b + c =(ζ + 1 + ζ2)b + (ζ + ζ2 + 1)c = 0. So s16 maps l1 to l3.We have h′4(0, b, c) = (b + c, ζ2b + ζc, ζb + ζ2c), and (b + c) + (ζ2b + ζc) +(ζb + ζ2c) = (1 + ζ2 + ζ)b + (1 + ζ + ζ2)c = 0. So h′4 maps l1 to l2.Similarly, h′4(a,0, c) = (a+c, a+ζc, a+ζ2c), and (a+c)+ζ(a+ζc)+ζ2(a+ζ2c) = (1 + ζ + ζ2)a + (1 + ζ2 + ζ4)c = 0. So h′4 maps l5 to l7.Using similar methods, we can show that h′4 gives cycles (l1, l2), (l3, l4),(l5, l7, l6, l8), and (l9, l11, l10, l12). But we observed that s16 ∶ l1 ↦ l3 andh′3 ∶ l7 ↦ l12. So we have orbits (l1, l2, l3, l4), (l5, l7, l6, l8, l9, l11, l10, l12).These two orbits are disjoint because there is no element of SL(2,F3) whichmaps l1 to l5. This can be easily seen from the fact that the sum of thesecond and third column elements of the second row of si matrices is never0.(ii) si preserves l1 if its second and third column of the first row is 0which is satisfied when i = 1,2,3,4,5,6. It is generated by h0 and h′3.On l1 we have h′3(0, b, c) = (0, ζ2b, ζ2c) = (0, b, c). So ⟨g3⟩ gives extraautomorphisms. h0(0, b, c) = (0, c, b) so it is an inversion, whose fixed pointsare (0,1,1) and (0,1,−1). Therefore g0 ∈ Aut (A(0,1,±1)).(iii) l6 is preserved by si if the first and second column of the third rowis 0, which is satisfied by s3, s5. It is generated by h′3.On l6 we have h′3(a, b,0) = (a, ζ2c,0) which has order 3. Its fixed pointsare (1,0,0) and (0,1,0). So ⟨g3⟩ ⊂ Aut (A(0,1,0)). As h0(1,0,0) = (1,0,0)883.4. Type A algebraswe further have ⟨g0, g3⟩ ⊂ Aut (A(1,0,0)) 2Now we give a geometric explanation on the automorphism group ofA(0, b, c) which has a higher dimension than that of generic Sklyanin algebra.From this lemma and Proposition 3.4.1 we have ⟨g1, g2, g3⟩ ≅ (Z/3Z)3 ⊂Aut (A(0, b, c)) and ⟨g0, g1, g2, g3⟩ ≅ S3 × (Z/3Z)2 ⊂ Aut (A(0,1,1)). Thecubic for A(0, b, c) is a triangle X = X1 ∪X2 ∪X3. Let xi ∈ Xi ∖ {0,∞} andregard xi ∈ Gm. Then σ is given by σ(x1, x2, x3) = (− cb x1,− cb x2,− cb x3).Then τ ∈ ⟨g2, g3⟩ ≅ (Z/3Z)2 acts as τ(x1, x2, x3) = (ζi3x1, ζj3x2, ζk3x3),where i, j, k = 0,1,2 and i + j + k ≡ 0 mod 3. The condition is added sothat τ∗L ≅ L. This action can be extended to Gm2 action which is given byτ(x1, x2, x3) = (t1 x1, t2 x2, t3 x3) where t1t2t3 = 1, because it commutes withσ. This explains why we have Gm2 instead of (Z/3Z)2 for the automorphismof A(0, b, c).Now we explain the finite part of the automorphism. The g1 acts asg1(x1, x2, x3) = (x2, x3, x1) and commutes with σ. And in the case ofA(0,1,1) the σ is multiplication by −1. So g0 which is given by g0(x1, x2, x3) =( 1x1 , 1x3 , 1x2 ), commutes with σ and is also an automorphism of A(0,1,1).This provides an explanation forAut(A(0,1,1)) ≅ S3 ×Gm2andAut(A(0, b, c)) ≅ A3 ×Gm2in Table 3.3.Stable triangles on l6Now we consider algebras on the line l6. They have defining relationsf1 = ax2 + b yzf2 = ay2 + b zxf3 = az2 + bxyanddetM = (a3 + b3)xyz .We have that σ(0, y, z) = (ay,−b z,0), σ(x,0, z) = (0, a z,−bx), σ(x, y,0) =(b y,0,−ax). To observe this triple geometrically let us write X =X1 ∪X2 ∪X3 where X1 = Z(x), X2 = Z(y), X3 = Z(z). We regard xi ∈ Xi as anelement of Gm. Then σ(x1, x2, x3) = (− bax3,− bax1,− bax2).893.4. Type A algebrasIf τ(x1, x2, x3) = (t1x1, t2x2, t3x3) then στ = τσ requires t1 = t2 = t3.And τ∗L ≅ L requires t31 = 1 which implies τ = g2. Therefore ⟨g2⟩ ≅ Z/3Z ⊂Aut(A(a, b,0)).When τ = g1 is a rotation then it permutes with σ so ⟨g1⟩ ≅ Z/3Z ⊂Aut(A(a, b,0)).If τ = g0 which interchanges two sides then it does not permute with σ.So there is no other automorphism. In sum we haveAut (A(a, b,0)) = ⟨g1, g2⟩ ≅ (Z/3Z)2 .This triangle has the same automorphism group as generic smooth cubics.From Theorem 2.2.2 we have that A(a, b,0) is stable if ab ≠ 0 (in this caseit is degenerate) and a3 + b3 = 1 (in this case it is linear).Proposition 3.4.4 In the compactification of the moduli of type A algebrasthose with triangle E appear on L1 and L6.(i) L1 = P1/(Z/2Z) where Z/2Z action is generated by h0 which is an in-version t↦ 1t . L1 parametrizes strictly semistable type A algebras andthe stack of algebras is a A3×Gm2-gerbe over L1. Algebras correspond-ing to the two orbifold points are A(0,1,−1), the polynomial algebrawhich is also linear, and A(0,1,1). They have the automorphism groupPGL(3) and S3 ×Gm2 respectively.(ii) L6 = P1/(Z/3Z) where Z/3Z action is generated by h′3, the multiplica-tion by ζ3. L6 parametrizes stable type A algebras whose point schemeis a triangle. The stack of algebras is a (Z/3Z)2-gerbe over L6. L6has two orbifold points whose corresponding algebras are A(1,0,0) andA(0,1,0). They are both degenerate algebras.(iii) L1 ∩ L2 = {(0,1,0), (0,1,−ζ3)}. A(0,1,−ζ3) is unique, upto isomor-phism, linear type A algebra which is not the polynomial algebra.Proof. (ii) and (ii) is from 3.4.3. We only need to show (iii). It is clear thatl1∩l6 = {(0,1,0)}. As l1 is mapped to l2, l3, l4 by the SL(2,F3) action we needto find the intersections of them with l6. They are (1,−1,0), (1,−ζ3,0), (1,−ζ23 ,0).These three points are identified by h′3 and s8(1,−1,0) = (0,1,−ζ3).2A(0,1,−ζ3) is not an orbifold point and is in the intersection of L1 andL6.The algebra A(0,1,1) which is on an orbifold point in L1 looks similarto the exterior algebra but is not the same one. x2 ≠ 0 in A(0,1,1).903.4. Type A algebrasThe algebraA(1,0,0) is the degenerate algebraDg1 = ⟨x, y, z⟩/(x2, y2, z2).We haveOrb((1,0,0)) = {(1,0,0), (1,1,1), (1, ζ3, ζ3), (1, ζ23 , ζ23)} . (3.25)Four lines l1, l2, l3, l4 and four points (1,0,0), (1,1,1), (1, ζ3, ζ3), (1, ζ23 , ζ23)are the four lines and four points in Theorem 2.4.2Now we consider another line of interest. The linel13 = {(a, b, b)}consists of algebras with σ2 = id. This line l13 is preserved by SL(2,F3). Suchalgebras have extra automorphism g0(x, y, z) = (x, z, y). So the SL(2,F3)action descent to the PSL(2,F3) = SL(2,F3)/Z(SL(2,F3)) action, whereZ(SL(2,F3)) is the center of SL(2,F3) generated by ( 2 00 2 ) which correspondsto the g0.PSL(2,F3) ≅ ⟨g3, g4⟩ .g3 gives isomorphism A(a, b, b) ≅ A(ζ3 a, b, b) giving extra automorphisms forA(1,0,0) and A(0,1,1). And g4 gives isomorphism A(a, b, b) = A(a+2b, a−b)giving an extra automorphism for A(1 −√3,1,1). This explains the ∣σ∣ = 2part of Table 3.3.Proposition 3.4.5 We have that(i) The regular type A algebras with σ2 = id are parametrized by L13 =P1/PSL(2,F3). The stack of algebras is a Gm × (Z/3Z)2 × (Z/2Z)-gerbe over this quotient stack.(ii) Algebras with extra dimensions are those of non-trivial stabilizers.There are three of them, (1,0,0), (0,1,1), (1 −√3,1,1).(i) Aut(A(1,0,0)) ≅ Aut(A(0,1,1)) ≅ S3 ×Gm2(ii) Aut(A) = (Z/4Z) × (Z/3Z)2.(iii) L1 ∩L13 = {(0,1,1)} and L6 ∩L13 = {(1,0,0)}.(iv) Type A algebra with an extra automorphism only appears on L1 andL13(v) Type A algebra on L13 is a Zhang twist of type B algebra.Proof. All except (v) are trivial. And (v) will be proven in Section 3.5.5. 23.4.4 Singular E and their S-equivalenceWe now study triples (E,L,σ) where E is singular but A(E,L,σ) is notstable.913.4. Type A algebrasIrreducible Nodal, Non-linear type S′′1In Section 3.3.3, a triple where E is irreducible nodal and σ has no fixedpoint is studied. The corresponding algebra was S′′1 (β), where β3 ≠ −1,which is S-equivalent to S1(β,β, β) which is also A(0,1, β), if β ≠ ∞. Ifβ =∞ it is S-equivalent to A(0,0,1).Nodal union of line and smooth conic, Non-linear type S′1We studied in Section 3.3.2, a triple where E is a nodal curve with twocomponents of degree 1 and 2 and σ has no smooth fixed point. The corre-sponding algebra was S′1(β1, β2), where β1β22 ≠ −1, which is S-equivalent toS1(β1, β2, β2). When β2 = β1 = β, we have that S′1(β,β) is S-equivalent toA(0,1, β). We have that β3 ≠ −1, because such β gives a curve more singularthan nodal.We haveProposition 3.4.6 Type A algebra with nodal E is S-equivalent to an al-gebra on the line L1 ∖ {(0,1,−1), (0,1,−ζ3)}.3.4.5 More singular than nodalNow we study the case when E is worse than nodal. We conjecture thatConjectrue 3.4.7 For a triple with E which is more singular than nodal,regular triples are linear. Moreover unless E is a triple line or a triple point,only σ = id is allowed.If this conjecture is true then we can say that we found all regular tripleswith more singular than nodal E and they are S-equivalent to the linear orpolynomial algebra.We can verify this conjecture for the case of triple point, i.e. threelines intersecting in one point. Let E = X1 ∪X2 ∪X3 where Xis intersectin P . Pic3(X) ≅ Gm. We choose a degree 3 line bundle L0 in Pic3(X)as the identity such that the embedded cubic in P(L∗0) is x3 + y3 = (x +y)(x + ζ3y)(x + ζ23y). For an automorphism σ of E we regard σ ∈ PGL(3)which fixes (0,0,1). Then σ∗(x + y) = x + s1(σ)y, σ∗(x + ζ3y) = x + s2(σ)y,σ∗(x + ζ23y) = x + s1(σ)y. Let us defineλ(σ) = s2(σ)2s1(σ)s3(σ)923.4. Type A algebrasWe have that λ(id) = 1, and σL0 ≅ L0 if and only if λ(σ) = 1. The regularitycondition (σ2)∗L0 ⊗L0 ≅ σ∗L0 ⊗ σ∗L0 is identical toλ(σ2) = λ(σ)2 (3.26)which is a degree 12 equation on the coefficients of the matrix representingσ. Then all solutions of (3.26) satisfy λ(σ) = 1.From now on we present algebras whose E is a singular curve which isnot nodal.Triple lineWhen c1c22+b1b22 = 0 but b2c2 ≠ 0, the type S′1 algebra with defining equationsf1 =x2 + b1 yz + c1 zy,f2 =b2 zx + c2 xz, (3.27)f3 =b2 xy + c2 yxhas a triple line Z(x3) as a point scheme. By cross producting the 1st andthe 3rd rows of ⎛⎜⎝x c1 z b1 yb2 z 0 c2 xc2 y b2 x 0⎞⎟⎠we haveσ(p + q 2, y, z) = (b1b2 (p + q 2)y,−b1c2 y2,−b2p2 2 + c1c2 yz)where 3 = 0. By putting b1 = b2 = b and c1 = c2 = c we have a type A algebrawith b3 + c3 = 0 which is S-equivalent to A(0,1,−ζi3) which is linear.Triple pointS′′1(−ζ i3) has a triple point Z(x3 + y3) as E, consisting of three lines Z(x +y), Z(x+ζ3y), Z(x+ζ23y) intersecting in (0,0,1). The automorphism is givenby σ(x, y, z) = (xy,−c y2,−x2 + c2 yz). On the line Z(x + y), σ(−y, y, z) =(y, c y, y−c2 z), on the line Z(x+ζ23y), σ(−ζ23y, y, z) = (ζ23y, cy, ζ3y−c2z), andon the line Z(x+ ζ3y), σ(−ζ3y, y, z) = (ζ3y, cy, ζ23y − c2z). So when c = −1, σstabilizes three lines, and when c = −ζ3, ζ23 it rotates three lines. (0,0,1) isthe only fixed point of σ.933.4. Type A algebrasNow we consider the automorphisms. We already mentionedAut(b,−ζi3b) =⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩⟨( 1 0 00 ζ3 00 0 ζ23)⟩ ≅ Z/3Z if i = −1,−2⟨( 0 1 01 0 00 0 1) ,( 1 0 00 ζ3 00 0 ζ23) , ( 1 0 00 1 0∗ ∗ 1 )⟩ if i = 0Now we give a geometric explanation on the automorphisms. Let E = X1 ∪X2 ∪X3 and xi ∈Xi. We consider xi as an element of Gm.When c = −1 we have σ = id, so any τ commutes with σ so we onlyneed to check τ∗L ≅ L. When τ(x1, x2, x3) = (t1 x1, t2 x2, t3 x3), the condi-tion imposes t1t2t3 = 1. This corresponds to the two dimensional matrices( 1 0 00 1 0∗ ∗ 1 ). When τ rotates Xis then it satisfies the condition. This corre-sponds to ( 1 0 00 ζ3 00 0 ζ23). When τ is a composition of inversion on each side thecondition τ∗L ≅ L requires that τ should also interchange X2 and X3. Thiscorresponds to ( 0 1 01 0 00 0 1).Union of a double line and a lineIn the commutative case, a union of a double line and a line is given by x2y,corresponding to w210. So let us consider the potential aw210+bw111+cw′111.The defining equations, after rescaling x → 1/√ax, y → 1/√ay, z → √a/bz,and c/b→ c aref1 = yz + czy + xy + yxf2 = zx + cxz + x2 (3.28)f3 = xy + cyxThe corresponding matrix is M = ⎛⎜⎝y x + c z yx + z 0 cxc y x 0⎞⎟⎠ giving the pointscheme cut bydetM = (1 − c + c2)xy (x + (c + 1)z) ,and σ(x, y, z) = (−cx, c2 y, x + z) . When c = −ζ3,−ζ23 the E = P2, and whenc = −1 the point scheme is cut by xy2 and σ(x, y, z) = (x, y, x + z). (0,0,1)is the only fixed point of σ.With a test configuration U = ⟨x⟩, this algebra is S-equivalent to thepolynomial algebra so it is strictly semistable.943.4. Type A algebrasThe non-tautological automorphism group of this algebra is 3 dimen-sional.Aut = ⟨( 1 0 00 s 0t u 1)⟩ ≅ C ×Gm ⋊Cwhere C is an additive group. Neither g1 nor g2 is an element of this auto-morphism group.CuspIn the commutative case, a cusp is given by x2y + z3, corresponding tow210 + w003. So let us consider the potential aw210 + bw111 + cw′111 + dw003.After rescaling x, y, z, a, b, c, d the defining equations aref1 = yz + czy + xy + yx ,f2 = zx + cxz + x2 , (3.29)f3 = xy + cyx + z2 .The corresponding linear matrix M = ⎛⎜⎝y x + c z yx + z 0 cxc y x z⎞⎟⎠ giving the pointscheme cut bydetM = (1 − c + c2)x2y − x2z + (1 + c3)xyz − (1 + c)xz2 − dz3,which has the singular point (0,1,0). σ(x, y, z) = (−cx2, c2 xy −xz − z2, x2 +xz). The singular point (0,1,0) is also the fixed point of σ which is node ifc ≠ −1 and cusp if c = −1. When c = −1 the point scheme is Z(3x2y−x2z+z3),and σ(x, y, z) = (x2, xy − xz − z2, x2 + xz).Consider a test configuration generated by W = I(2)1 = ⟨x, z⟩, U = I(3)1 =⟨x⟩. With this test configuration the algebra is S-equivalent to the polyno-mial algebra.This algebra has only the tautological automorphism,Aut = {id} .Reducible TacnodeIn the commutative case a line and conic intersecting in one point is byx2y + xz2, corresponding to w201 + w120. So consider a potential aw210 +953.4. Type A algebrasbw111 + cw′111 + dw210 + dw102. After rescaling the relations aref1 = yz + c zy + xy + yx + z2 ,f2 = zx + cxz + x2 , (3.30)f3 = xy + c yx + xz + zx .The corresponding linear matrix M = ⎛⎜⎝y x + c z y + zx + z 0 cxc y + z x z⎞⎟⎠ the givingpoint scheme cut bydetM = x (−x2 + (1 − c + c2)(xy + z2) + (1 + c3)yz) ,a union of line and conic intersecting in (0,1,0), and (0,−1,1 + c). Theautomorphism is given by σ(x, y, z) = (−cx, c2y + d(c − 1)z, x + z), so itstabilizes two components and fixes two nodes. When c = −1 two nodes areidentified, and the tangent cone at (0,1,0) is Z(x2). So the point scheme isa union of a conic and line meeting tangentially.Consider a test configuration generated by W = I(1)1 = ⟨x, z⟩, U = I(2)1 =⟨x⟩. With this test configuration the algebra is S-equivalent to the polyno-mial algebra.The non-tautological automorphism group of this algebra is 2 dimen-sional.Aut = ⟨( 1 0 0−t2 1 −2tt 0 1)⟩ ≅ Cwhere C is an additive group with the identity 0. Neither g1 nor g2 is anelement of this automorphism group.3.4.6 Degenerate algebras and Exceptional algebrasA(a, b, c) is degenerate if (a, b, c) belongs to the orbit of (1,0,0) or (0,1,0)under SL(2,F3) action.Orb((1,0,0)) = {(1,0,0), (1,1,1), (1, ζ3, ζ3), (1, ζ23 , ζ23)}Orb((0,1,0)) = {(0,1,0), (0,0,1), (1,1, ζ3), (1, ζ3,1),(1,1, ζ23), (1, ζ23 ,1), (1, ζ3, ζ23), (1, ζ23 , ζ3)} .Terminology 3.4.8 Let us give names to the two type A degenerate alge-brasDg1 ∶= A(1,0,0)Dg2 ∶= A(0,1,0).963.4. Type A algebrasDg1 and Dg2 both have Hilbert series (1 + t)/(1 − 2t)−1. So a(n) = 3 ⋅ 2n−1which does not give right dimension for n ≥ 3. Therefore we need to findassociated algebras with the right Hilbert series. We have several methods.First we add degree 3 relations to the Dg1, second, we consider several limitscombined with change of variables. Using the limit method, we find a familyof exceptional algebras in the case of Dg1. We also find degenerate algebraswhich are less degenerate than Dg1, but as they also have a wrong Hilbertseries we need to add degree 3 relations.Nilpotent monomial algebra Dg1 ∶= A(1,0,0)Dg1 has nilpotent defining equations x2 = y2 = z2 = 0. We have that detM =xyz but σ is not well defined on the vertices. It is neither one-to-one on thesides. Degree 2 part of Dg1 determines a 6-gon in P2×P2 whose componentsareX1 = {(1,0,0) × (0, y2, z2)}, X2 = {(x1,0, z1) × (0,1,0)},X3 = {(0,0,1) × (x2, y2,0)}, X4 = {(0, y1, z1) × (1,0,0)},X5 = {(0,1,0) × (x2,0, z2)}, X6 = {(x1, y1,0) × (0,0,1)} .Let us first find algebras of the right Hilbert series by adding degree 3relations. Recall that for type A algebras we have a potential xf1+yf2+zf3 =f1x+f2y+f3z, which makes a(3) = 10 instead of 9. But in Dg1 this splits intothree relations xf1 = f1x, yf2 = f2y, and zf3 = f3z, making a(3) = 12. So tofind a flat family of algebras we have to add relations lost by this splitting.This can be done by adding expressions which are equal to xf1 − f1x, etc.,in the algebras in its neighborhood.We consider an affine neighborhood of (1,0,0) ∈ P2 consisting ofA(1, b, c),such that A(1, b, c) is regular and thus of the right Hilbert series in theneighborhood. In A(1, b, c) we have f1 = x2 + b yz + c zy. So xf1 − f1x =x(b yz + c zy) − (b yz + c zy)x. Therefore we have to add [x, b yz + c zy], etc.,to the relations of Dg1 to find proper the limit (a, b) → (0,0) and thus aflat family. We define Ã1(b, c), for (a, b) ∈ P1,Ã1(b, c) ∶= C⟨x, y, z⟩⎛⎜⎜⎜⎝x2, y2, z2,[x, b yz + c zy],[y, b zx + cxz],[z, b xy + c yx]⎞⎟⎟⎟⎠(3.31)973.4. Type A algebrasLemma 3.4.9 Ã1(b, c) is strictly semistable for any (b, c). When bc ≠ 0 ithas the Hilbert series (1− t)−3. When bc = 0 it has the right dimension uptodegree 4.Proof. The 6 defining equations are weight invariant under A3 actionwhich acts on the generators x, y, z. Therefore it is strictly semistable. 2Now we consider isomorphisms and automorphisms.Recall that Aut(A(1,0,0)) ≅ S3 × Gm2 is generated by g0(x, y, z) =(x, z, y), g1(x, y, z) = (y, z, x), and g5(x, y, z) = (x, t1y, t2z). So to find auto-morphisms of Ã1(b, c) we only need to see how these generators act on theadded degree 3 relations.It is easy to see that g1, g5 preserves them. And g0 gives an isomorphismbetween Ã1(b, c) and Ã1(c, b), and an automorphism when c = ±b.Ã1(b, c) ≅ Ã1(c, b) .Aut (Ã1(b, c)) = ⎧⎪⎪⎪⎨⎪⎪⎪⎩⟨( 1 0 00 ∗ 00 0 ∗ ) , ( 0 1 00 0 11 0 0 ) , ( 1 0 00 0 10 1 0 )⟩ ≅ Gm2 × S3 if (b, c) = (1,±1)⟨( 1 0 00 ∗ 00 0 ∗ ) , ( 0 1 00 0 11 0 0 )⟩ ≅ Gm2 ×A3 otherwise1-dim’l family of Degenerate algebras at Dg1Now we consider several limits limk→0A(1, k b, k c) to A(1,0,0) along a direc-tion (0, b, c), combined with a change of variables. Consider a neighborhoodof (1,0,0) consisting of algebras A(1, b, c). We fix direction (a, b) ∈ P1 andintroduce new variable k ∈ C for a limit limk→0.f1 =x2 + k(b yz + c zx) (3.32)f2 =y2 + k(b zx + cxz) (3.33)f3 =z2 + k(bxy + c yx) (3.34)First by a change of variables z ↦√k z the defining equations becomef1 = x2 + k3/2 (b yz + c zy)f2 = y2 + k3/2 (b zx + cxz)f3 = k(z2 + bxy + c yx) .Then we take limit k → 0 to obtainf1 = x2f2 = y2f3 = z2 + bxy + c yx .983.4. Type A algebrasWe call this algebra Dg1′(b, c).Dg1′(b, c) ∶= C⟨x, y, z⟩(x2, y2, z2 + bxy + c yx)This algebra has detM = xyz. However σ is not well-defined, so we needto consider correspondence in P2 × P2 determined by the degree 2 part ofDg1′(b, c).When bc ≠ 0, we have a 4-gon whose components areX1,2 = {(x1,0, z1) × (0, z1,−bx1)},X3 = {(0,0,1) × (x2, y2,0)},X4,5 = {(0, y1, z1) × (z2,0,−c y1)}, X6 = {(x1, y1,0) × (0,0,1)} .So from the 6-gon of Dg1, X1 and X2 are identified to X1,2 giving thegraph of σ(x1,0, z1) = (0, z1,−bx1), a multiplication by −b. And X4 andX5 are identified to X4,5 giving the graph of σ(0, y1, z1) = (z2,0,−c y1), amultiplication by −1c .When c = 0, we have a 5-gon consisting of,X1,2 = {(x1,0, z1) × (0, z1,−bx1)},X3 = {(0,0,1) × (x2, y2,0)}, X4 = {(0, y1, z1) × (1,0,0)},X5 = {(0,1,0) × (x2,0, z2)}, X6 = {(x1, y1,0) × (0,0,1)} .So from the 6-gon of Dg1, X2 and X1 are identified to form X′1 the graph ofσ ∶ Z(y1)→ Z(x2) where σ(x1,0, z1) = (0, z1,−bx1), a multiplication by −b.Similarly b = 0 also gives a 5-gon. Here we see that the direction c = 0,b = 0 gives a more degenerate algebra.Dg1′(b, c) has growth (3,6,11, . . .) for any a, b. So we add lost relationsas Dg1 and define Ã′1(b, c),Ã′1(b, c) ∶= C⟨x, y, z⟩(x2, y2, z2 + bxy + c yx,[x, b yz + c zy] ).Then Ã′1(b, c) has the right Hilbert series for all degree if bc ≠ 0, and uptodegree 4 if bc = 0.We have isomorphismsÃ′1(b, c) ≅ Ã′1(c, b) .and automorphisms,Aut(Dg1′(b, c)) = Aut(Ã′1(b, c)) = {( s 0 00 t 00 0√st) ,( s 0 00 t 00 0 −√st) ∣ s, t ∈ Gm}993.4. Type A algebras1-dim’l family of Exceptional algebras at Dg1Now we find exceptional algebras which appear as a limit lim(b,c)→(0,0)A(1, b, c).Rescale variable x↦ 1kx in (3.32) to obtainf1 =x2 + k3(b yz + c zx)f ′2 =y2 + b zx + cxzf ′3 =z2 + bxy + c yx.When k ≠ 0 the algebra defined by f1, f ′2, f ′3 is the A(1, kb, kc). Now takelimit k → 0 and obtainf ′1 =x2f ′2 =y2 + b zx + cxzf ′3 =z2 + bxy + c yx.Let us call this algebra E1(b, c).E1(b, c) ∶= C⟨x, y, z⟩(x2, y2 + b zx + cxz, z2 + bxy + c yx) . (3.35)This algebra can be considered as a limit of type A algebras to (1,0,0) alonga direction (0, b, c). Here (b, c) is a homogeneous coordinate, and E1(b, c)are parametrized by (b, c) ∈ P1. Let us study the triple (E,L,σ) of E1(b, c)M = ⎛⎜⎝x 0 0b z y cxc y bx z⎞⎟⎠ , N =⎛⎜⎝x c z b y0 y cx0 bx z⎞⎟⎠ .detM = detN = −x (bcx2 − yz).When bc ≠ 0, the point scheme is a union of line C1 = Z(x) and conicC2 = Z (bcx2 − yz), intersecting in P1 = (0,1,0) and P2 = (0,0,1).σ(x, y, z) = (bcx2 − yz, b z2 − c2 xy, c y2 − b2 xz).So on Z(x) we have σ(0, y, z) = (−yz, b z2, c y2) and on Z (bcx2 − yz) we haveσ(x, y, z) = (0, b z2 − c2 xy, c y2 − b2 xz). Therefore σ interchanges two com-ponents. It also interchanges two nodes P1 with P2, so it has no fixed point.On (0, y, z) we have σ2(0, y, z) = (0, c y, b z). So we have a 1 dimensionalfamily of exceptional algebras E1(b, c), where (b, c) ∈ P1 ∖ {(0,1), (1,0)}.1003.4. Type A algebrasWhen (b, c) = (1,0) or isomorphically (0,1) then detM = xyz, but thealgebra is degenerate. However, it has the right Hilbert series. The degree2 part determines a 4-gon with the following componentsX1,2 = {(x1,0, z1) × (0,−z1, x1)}, X3,4 = {(0, y1, z1) × (y1,−z1,0)},X5 = {(0,1,0) × (x2,0, z2)}, X6 = {(x1, y1,0) × (0,0,1)} .So from the 5-gon of Dg1′(1,0), which is from the 6-gon of Dg1, two linesX3 and X4 are identified to form X3,4, the graph of σ ∶ Z(x1)→ Z(z2) whereσ(0, y1, z1) = (y1,− z1,0).If we further identify X5 and X6 then we have a graph of σ which rotatesthree sides of a triangle, the corresponding algebra is A(1,1,0) which is onthe line connecting (0,1,0) and (1,0,0). So we can regard it as a processthat A(1,1,0) is degenerated to first E1(1,0) then to Dg1′(1,0) and thento Dg1 by puffing up one side at one time.E1(1,0) has the right dimension upto degree 5. It is 4-strict semistableand 5-unstable w.r.t. U = ⟨x⟩.Proposition 3.4.10 Let U = ⟨x⟩ and consider the test configuration gener-ated by U . Then the associated graded algebra Gr(Ã1(b, c)≤3) is E1(b, c)≤3.Proof. We show that the degree 2,3 relations of Ã1(b, c) are the lowestweight part of E1(b, c). In degree 2, Ã1(b, c) has f1 = x2, f2 = y2, f3 = z3,and E1(b, c) has g1 = x2, g2 = y2 + b zx + cxz, g3 = z2 + bxy + c yx. With thefiltration U = ⟨x⟩, the lowest weight part of g2 is y2 = f2 and that of g3 isz2 = f3.The degree 3 part of E1(a, b) is cut byxg1 = g1x, xg2, g2x, xg3, g3x,yg1, g1y, yg2, g2y, zg2, g2z,zg1, g1z, zg2, g2z, zg3, g3zand the degree 3 part of Ã1(b, c) is cut byxf1 = f1x, xf2, f2x, xf3, f3x,yf1, f1y, yf2 = f2y, zf2, f2z,zf1, f1z, zf2, f2z, zf3 = f3z,[y, b zx + cxz], [z, b xy + c yx] .The lowest weight part of xg1 is xf1, and the similar is true for y, z andg2, g3. However, both of yg2 and g2y have the lowest term y3 = yf2 = f2y.1013.4. Type A algebrasSo two relations in G(Ã1(b, c)) given by yg2, g2y are y3 and [y, b zx + cxz].Similarly the two relations given by zg3, g3z are z3 and [z, b xy + c yx]. Thisshows the claim. 2Corollary 3.4.11 E1(b, c)≤3 is strictly semistable.Conjectrue 3.4.12 E1(b, c), where (b, c) ≠ (1,0), (0,1), has the Hilbertseries (1 − t)−1 and is strictly semistable. The test configuration generatedby U = ⟨x⟩ has a constant Futaki function.I checked this conjecture is true upto degree 7.Now we consider isomorphisms and automorphisms. It is easy to seethat g2(x, y, z) = (x, ζ3y, ζ23z) is an automorphism. And g0(x, y, z) = (x, z, y)gives an isomorphism between E1(b, c) and E1(c, b), thus an automorphismfor (b, c) = (1,1). When (b, c) = (1,−1), g′0(x, y, z) = (x,−y,−z) is an auto-morphisms. The degenerate ones E1(0,1) and E1(1,0) do not have addi-tional automorphism. In sum we haveE1(b, c) ≅ E1(c, b) .Aut (E1(b, c)) =⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩⟨( 1 0 00 ζ 00 0 ζ2) , ( 1 0 00 0 10 1 0)⟩ ≅ S3 if (b, c) = (1,1)⟨( 1 0 00 ζ 00 0 ζ2) , ( 1 0 00 0 −10 −1 0 )⟩ ≅ S3 if (b, c) = (1,−1)⟨( 1 0 00 ζ 00 0 ζ2)⟩ ≅ A3 if (b, c) ≠ (1,±1), bc ≠ 0Note that we can think E1(a, b) as obtained from Dg1 by adding termsof higher weight w.r.t. the U = ⟨x⟩. Adding terms of higher weight does notchange the lowest weight terms and therefore preserves S-equivalence class.Cyclic monomial algebra Dg2 ∶= A(0,1,0)A(0,1,0) has cyclic defining equations yz = zx = xy = 0. The point schemeof Dg2 is a 6-gon in P2 × P2 whose components areX1 = {(x1,0, z1) × (0,0,1)}, X2 = {(0,0,1) × (0, y2, z2)},X3 = {(0, y1, z1) × (0,1,0)}, X4 = {(0,1,0) × (x2, y2,0)},X5 = {(x1, y1,0) × (1,0,0)}, X6 = {(1,0,0) × (x2,0, z2)} .Now we consider various limits of lima→0A(a,1,0) along a direction(1,0,0). First by a change of variables x ↦ √a−1x the defining equations1023.4. Type A algebrasbecomef1 = x2 + yzf2 = a3/2 y2 + zxf3 = a3/2 z2 + xy .Then we take limit b↦ 0 to obtainf1 = x2 + yzf2 = zxf3 = xyWe call the algebra defined by these equations Dg2′. It is the S′1(0,0). Thepoint scheme as a correspondence in P2 × P2 is a 5-gon whose componentsareX1 = {(x1,0, z1) × (0,0,1)}, X2 = {(0,0,1) × (0, y2, z2)},X3 = {(0, y1, z1) × (0,1,0)}, X4 = {(0,1,0) × (x2, y2,0)},X5,6 = {(x1, y1,0) × (y1,0,−x1)} .So from the correspondence of Dg2 a vertical line X6 and a horizontal lineX5 are joined to form X′5 the graph of σ ∶ Z(z1)→ Z(y2) where σ(x1, y1,0) =(y1,0,−x1).Second, we do a change of variables z ↦ az to obtainf1 = a(x2 + yz)f2 = a(y2 + zx)f3 = a3 z2 + xy .Then we take limit b↦ 0 to obtainf1 = x2 + yzf2 = y2 + zxf3 = xyWe call this algebra Dg2′′. It is also S′′1 (0). It is also degenerate and thepoint scheme is a 4 gon-in P2 × P2 which has following components.X1 = {(x1,0, z1) × (0,0,1)}, X2 = {(0,0,1) × (0, y2, z2)},X ′3 = {(0, y1, z1) × (y1,−z1,0)},X ′5 = {(x1, y1,0) × (y1,0,−x1)} .1033.4. Type A algebrasSo from the correspondence of Dg2′ a vertical line X4 (or equivalently ahorizontal line X3) is contracted to form X′3 the graph of σ ∶ Z(x1)→ Z(z2)where σ(0, y1, z1) = (y1,−z1,0).If we further collapse X2 (or equivalently X1) then we have a graphof σ which rotates three sides of a triangle, the corresponding algebra isA(1,1,0). So we can regard it as a process that A(1,1,0) is degenerated tofirst Dg2′′ then to Dg2′ and then to Dg2 by puffing up one side at one time.Adding degree 3 relations to Dg2We add degree 3 relations to Dg2 to find a flat family of 3-truncated algebrasaround (0,1,0). For Dg2 = A(0,1,0), we have xf1 = f3z, yf2 = f1x, andzf3 = f2y. So we have to recover lost degree relations from a neighborhoodcorresponding to these overlaps. Consider an affine neighborhood of (0,1,0)consisting of A(a,1, c) whose defining equations arexy + az2 + c yx, etc.Degree 3 overlaps arexyz = x(yz) = (xy)z, etc.which givesax3 + cxzy = az3 + c yxz, etc.Therefore we add these and define Ã2(a, c), for (a, c) ∈ P1,Ã2(a, c) ∶= C⟨x, y, z⟩⎛⎜⎜⎜⎝xy, yz, zx,ax3 + cxzy = az3 + c yxz,a y3 + c yxz = ax3 + c zyx,a z3 + c zyx = ay3 + cxzy⎞⎟⎟⎟⎠. (3.36)When (a, c) ≠ (1,0), (0,1) the Hilbert series of Ã2(a, c) is (1 − t)−3.Now we compute automorphisms. Recall thatAut(Dg2) = ⟨( 1 0 00 t 00 0 u) , ( 0 1 00 0 11 0 0)⟩ (3.37)The A3 action cycling x, y, z preserves degree 3 relations. When a = 0 theadded degree 3 relations are homogeneous in x, y, z, so any scalar multipli-cation is an automorphism. When c = 0, we need t3 = u3 = 1 to preserve the1043.4. Type A algebrasadded degree 3 relations. When (a, c) ≠ (1,0), (0,1) we need t3 = u3 = tu,which is gk2 .Aut(Ã2(a, c)) =⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩⟨( 1 0 00 ζ 00 0 ζ2) , ( 0 1 00 0 11 0 0)⟩ ≅ (Z/3Z) ×A3, if (a, c) ≠ (1,0), (0,1)⟨( 1 0 00 ζ 00 0 ζ) ,( 1 0 00 ζ 00 0 ζ2) , ( 0 1 00 0 11 0 0)⟩ ≅ (Z/3Z)2 ×A3, if (a, c) = (1,0)⟨( 1 0 00 t 00 0 u) , ( 0 1 00 0 11 0 0)⟩ ≅ Gm2 ⋊A3, if (a, c) = (0,1)We note that when (a, c) ≠ (1,0), (0,1) the automorphisms group of Ã2(a, c)is the same as that of generic A(a, b, c).One might ask whether we can add terms to the Dg2 to obtain strictlysemistable quadratic algebras of the right Hilbert polynomial. But the an-swer is negative.Proposition 3.4.13 There is no quadratic algebra A such that Gr(A) =Dg2.Proof. By symmetry of relations of Dg2 we only need to consider a flaggenerated by W = ⟨y, z⟩ and U = ⟨z⟩. We see that a test configurationsatisfying 0 < w(y) < w(z) < 2w(y) minimizes the number of monomials ofequal weight. Thus it allows the most terms in the relations of the A. Thisis because the monomials of equal weight of xy, yz, zx must not appear inthe relation which contains them. So the relations are of the following form:f1 =A1xy +A2xz +A3zx +A4y2 +A5zy +A6yz +A7z2f2 =B3zx +B4y2 +B5zy +B6yz +B7z2f3 =C6yz +C7z2which givesM = ⎛⎜⎝A3z A1x +A4y +A5z A2x +A6yt +A7zB3z B4y +B5z B6y +B7z0 0 C6y +C7z⎞⎟⎠ .Then at (1,0,0) we haveM = ⎛⎜⎝0 A1 A20 0 00 0 0⎞⎟⎠ ,so M has rank 1, and σ is not well defined and the algebra A has a wrongHilbert polynomial. 21053.4. Type A algebras3.4.7 Conclusion and description of moduliFigure 3.1: Schematic picture of type A moduliStable type A AS-regular algebras are classified by the quotient stack[(P2 ∖ T )/SL(2,F3)]where T is a union of 4 lines l1, l2, l3, l4 (3.24) and 4 points P1, P2, P3, P4(3.25). The stack of algebras is a Gm × (Z/3Z)2-gerbe over this quotientstack (Theorem 2.4.2).This moduli space can be compactified by adding strictly semistablealgebras. The SL(2,F3) action identifies the 4 lines, l1, . . . , l4, to L1 =l1/(Z/2Z). The type S1 moduli and type A moduli intersect in the lineL1. Algebras on L1 are AS-regular except the degenerate algebra Dg2which is the cyclic monomial algebra. This algebra lies on the intersec-tion of the line L1 of the strictly semistable triangles and the line L6 ofstable rotated triangles. The other intersection point of L1 and L6 hasC⟨x, y, z⟩/(xy−ζ3 yx, yz−ζ3 zy, zx−ζ3 xz) as its corresponding algebra. Thisalgebra is the linear AS-regular algebra which is not commutative. The fam-ily of elliptic triples (E,L,σ) where σ is a translation by a 3-division point,corresponds to this point. The other linear algebra is the commutative poly-1063.5. Type B algebrasnomial algebra which also lies on L1. The family of elliptic triples with σ = idcorresponds to this algebra.The SL(2,F3) action identifies the 4 points P1, . . . , P4. The correspond-ing algebra is the degenerate algebra Dg1 which is the nilpotent monomialalgebra. This algebra lies on the intersection of the line L6 and the lineL13 = {(a, b, c) ∈ P2 ∣ b = c}. Algebras on L13 has σ2 = id. In Section 3.5.5 wewill see that the Zhang twist of an algebra on L13 by (x, y, z) ↦ (x, z, y) istype B. There is unique intersection point of L1 and L13 whose correspondingalgebra is the anticommutative algebra C⟨x, y, z⟩/(xy + yx, yz + zy, zx+xz).Algebras with extra automorphisms appear on the two lines of orbifoldpoints, i.e. L1 and L13.The compactification is degree sensitive. For higher degrees we need toadd more strictly semistable algebras. The 2-cutoff algebras A(a, b, c)≤2 canbe compactified asP2/SL(2,F3) .The 3-cutoff algebras A(a, b, c)≤3 can be compactified asBl12 points P2/SL(2,F3) ,where the 12 points are those in the orbits of Dg1 and Dg2. The exceptionaldivisors of the blow-up consist of algebras obtained by adding degree 3 rela-tions to the degenerate algebras (3.31), (3.36). The exceptional divisor overthe point corresponding to Dg1 also can be formed by exceptional algebras(3.35).Remark 3.4.14 In the moduli Mtri of triples (E,L0, L1) in [1], L1 and L6are identified. It is because Mtri requires further (Z/3Z) action which is givenby Zhang twists. Explicitly the Zhang twist by (x, y, z)↦ (y, z, x) identifiesL1 and L2. Dg1 and Dg2 are identified by this. So are the commutativepolynomial and the other linear algebra. As type B algebras are obtainedby Zhang twist we find the place of type B algebras in Mtri. We don’t knowwhether type E and H can be obtained by Zhang twist of type A.3.5 Type B algebrasIn this chapter we study the type B algebras. In Section 3.5.1 we givea detailed description of the triple as we are not aware of any literaturethat studies it. In Section 3.5.2 we study isomorphisms and automorphismsof type B algebras. In Section 3.5.3 we study strictly semistable algebrasand S-equivalence. As in the type A case, algebras for nodal cubics are1073.5. Type B algebrasS-equivalent to that of a triangle which is also type S2. Moduli of type Bintersect with the moduli of type S2. In Section 3.5.4 we present a detailedstudy of the degenerate algebra. In Section 3.5.5 we study type B algebraas a Zhang of type A and their moduli.3.5.1 Type B triple (E,L,σ)Type B algebras have Q = ( 1 0 00 1 00 0 −1 ), where M = NQt, f =MX = XtN . Thepotential w of type B quadratic AS-regular algebras is given byw = Aw012 +Dw102 +Bw120 +C w210 =Xtf =XtMXwherew012 = zyz − yz2 − z2yw102 = zxz − xz2 − z2xw120 = yxy + xy2 + y2xw210 = xyx + yx2 + x2y .From this we havef1 = C(xy + yx) +B y2 −Dz2,f2 = B(yx + xy) +C x2 −Az2, (3.38)f3 = A(yz − zy) +D(xz − zx).M = ⎛⎜⎝C y C x +B y −DzC x +B y B x −Az−Dz −Az Dx +Ay⎞⎟⎠anddetM = −B2Dx3 −B(AB +CD)x2y −C(AB +CD)xy2 −AC2 y3+D(2AB −CD)xz2 −A(AB − 2CD) yz2.When ABCD ≠ 0, scaling coordinatesx→√C/Ax,y →√B y,z →√Dz,1083.5. Type B algebrasturns the relations to the well known formf1 = xy + yx + y2 − z2,f2 = −az2 + x2 + yx + xy, (3.39)f3 = a(yz − zy) + xz − zx,where a = ACDB . Let us call the algebra defined by (3.39) B(a). AndM = ⎛⎜⎝y x + y −zx + y x −az−z −az ay + x⎞⎟⎠anddetM = −a2yz2 − a (x2y + x (y2 − 2z2) + y3 − 2yz2) − x (x2 + xy + y2 + z2) .So we have a pencil of cubics parametrized by a. This pencil of type Bcubics has three base points (0,0,1), (ζ3,1,0), and (ζ23 ,1,0). (0,0,1) is alsoa flex for any a. Let us name the base points together with (−a,1,0).P1 =(0,0,1)P2 =(−a,1,0)P3 =(ζ3,1,0)P4 =(ζ23 ,1,0) .With a choice of P1 as the identity, P2, P3, P4 are 2-division points.When a ≠ −ζ23 ,−ζ3,∞, σ is well defined andσ(x, y, z) = {−axz − ayz + xz, ayz − xz − yz,−x2 − xy − y2} .There are four fixed points of σ.Q1 =(√a2 − a + 1 + a − 1,1,√2√a2 − a + 1 + 2a − 1) ,Q2 =(√a2 − a + 1 + a − 1,1,−√2√a2 − a + 1 + 2a − 1) ,Q3 =(−√a2 − a + 1 + a − 1,1,√−2√a2 − a + 1 + 2a − 1) ,Q4 =(−√a2 − a + 1 + a − 1,1,−√−2√a2 − a + 1 + 2a − 1) .1093.5. Type B algebrasThe σ preserves {Pi} andσ(P1) = P2, σ(P2) = P1,σ(P3) = P4, σ(P4) = P3 . (3.40)Again consider the group law with the identity P1. Then we have2Qi = P2 = P −12 , for i = 1,2,3,43Q1 = Q−11 = Q2, 3Q2 = Q−12 = Q1, (3.41)3Q3 = Q−13 = Q4, 3Q4 = Q−14 = Q3 .So Qi are four square roots of P2 with respect to the group law with identityP1. Therefore we can consider type B algebra as a pair ((E,P1), P2) of anelliptic curve with origin P1 and a 2-torsion point P2.j-invariantTo compute the j invariant, we change the cubic into the Weierstrass normalform. We follow [12] p23. Let C be the cubic cut by−x3 − (1 + a)x2y − (1 + a)xy2 − ay3 + (2a − 1)xz2 + (2a − a2)yz2 .We choose a non-flex P2 = (−a,1,0). Then the tangent line at P2 is givenby Z = x + ay. It intersects with C at P0 = (0,0,1). Then the tangent lineat P0 is given by X = (−1 + 2a)x + (2a − a2)y. We can choose a line Y = zwhich passes through P2 but not equal to Z. With this change of variablesthe cubic becomes(9a5 − 27a4 + 27a3 − 9a2)XY 2 + (−a3 + 2a2 − 2a + 1)X2Z+ (−2a4 + 5a3 − 5a + 2)XZ2 + (−a5 + 3a4 − 5a3 + 5a2 − 3a + 1)Z3 .And let x = XZ and y = YZ and multiply x and replace xy by y, then the cubicbecomesy2 = (−a3 + 2a2 − 2a + 1)x3 + (−2a4 + 5a3 − 5a + 2)x2 + (−a5 + 3a4 − 5a3 + 5a2 − 3a + 1)x9a5 − 27a4 + 27a3 − 9a2 .After scaling y by the coefficient of x3 and scaling x by one root of thequadratic in x we have that the other root isλ = 2a7 − 7a6 − 3√3√−(a − 1)6a6 + 5a5 + 5a4 − 7a3 + 2a22a7 − 7a6 + 3√3√−(a − 1)6a6 + 5a5 + 5a4 − 7a3 + 2a2 .1103.5. Type B algebrasand j-invariant is therefore given byj(a) = −28 (a3 − 6a2 + 3a + 1)3 (a3 + 3a2 − 6a + 1)333(a − 1)2a2 (a2 − a + 1)6 .From this we see thatj(a) =∞ when a = 0,1,−ζ23 ,−ζ3,∞ .We see that j(a) is a rational function of degree 18 which agrees that typeB algebra corresponds to a pair of an elliptic curve and a 2-torsion point.18 = 6 × 3 where 6 comes from the order of SL(2,F2) which acts on P1(a)isomorphisms (See next section), and 3 comes from the choice of three 2-torsion points.3.5.2 Isomorphisms and Automorphisms of type B algebrasProposition 3.5.1 SL(2,F2) acts on the parameter space P1(a) of type Balgebras giving isomorphisms.Proof. We prove that B(a) ≅ B(1 − a), and B(a) ≅ B( 1a). Let us do achange of variablesx→ x + y,y → −y,z → izwhich changes the relations as following:f1(a) = xy + yx + y2 − z2 ↦(x + y)(−y) + (−y)(x + y) + (−y)2 − (iz)2= −xy − y2 − yx − y2 + y2 + z2= −xy − yx − y2 + z2= −f1(1 − a),f2(a) = −az2 + x2 + xy + yx↦ − a(iz)2 + (x + y)2 + (x + y)(−y) + (−y)(x + y)= az2 + x2 + y2 + xy + yx − xy − y2 − yx − y2= az2 + x2 − y2= −(1 − a)z2 + x2 + xy + yz − (xy + yx + y2 − z2)= f2(1 − a) − f1(1 − a),1113.5. Type B algebrasf3(a) = a(yz − zy) + xz − zx↦a(−y)(iz) − a(iz)(−y) + (x + y)(iz) − (iz)(x + y)= −aiyz + aizy + ixz + iyz − izx − izy= i ((1 − a)yz − (1 − a)zy + xz − zx)= if3(1 − a).So B(a) and B(1 − a) are isomorphic.And a change of variablesx→ y,y → x,z →√azturns the relations as following:f1(a) = xy + yx + y2 − z2 ↦yx + xy + x2 − ( 1√az)2= yx + xy + x2 − 1az2= f2(1/a),f2(a) = −az2 + x2 + xy + yx↦ − a( 1√az)2 + y2 + yx + xy= −z2 + y2 + yx + xy= f1(1/a),f3(a) = a(yz − zy) + xz − zx↦ax( 1√az) − a( 1√az)x + y( 1√az) − ( 1√az)y= √a(xz − zx + 1a(yz − zy))= √af3(1/a).So B(a) and B(1/a) are isomorphic. 2The first and second change of variables give automorphisms of the al-gebra with a = 1/2 and a = ±1 respectively. j (B(1)) =∞ and B(1) ≅ B(0) ≅B(∞). We will study this algebra with singular E in Section 3.5.3. Wehave B(1/2) ≅ B(2) ≅ B(−1). For B(2) the change of variables (x, y, z) ↦(x,−x − y, iz) is an automorphism.1123.5. Type B algebrasNow we find automorphisms for all B(a). In (3.39) we observe that zappears in f1 and f2 as z2 and all terms of f3 have z. So the change ofvariablesx↦ xy ↦ yz ↦ −z.is an automorphism of B(a). This is the inverse map with respect to thegroup structure with the identity P1. So τB(x, y, z) ∶= (x, y,−z) has fourfixed points Pi, i = 1,2,3,4.In sum we haveAut (B(a)) = ⎧⎪⎪⎨⎪⎪⎩Z/4Z if a = −1,2,12Z/2Z if a ≠ −1,2, 12 ,0,1,∞We have that j(2) = j(−1) = j(1/2) = 1728.While the SL(2,F2) action has the same form with the case of moduliof elliptic curves from the Legendre family, a is not the λ of the Legendreform.Geometric computation of AutFor type B triple (E,L,σ) we have that σ is a multiplication by −1 andL ≅ O(2O + P ) where P is a point of order 4.Let Q = P ⊕ P , and Pi, i = 1,2,3,4 be the four square roots of Q. Wealso rename Q1 = O, Q2 = Q, and call the other 2-division points Q3 andQ4. So Qi, i = 1,2,3,4 is fixed by σ. Then σ(τ(Qi)) = τσ(Qi) = τ(Qi) andwe have that τ(Qi) is fixed by σ. In other words, τ preserves {Qi}.In type B, a fixed point of σ is a 2-division point. Then τ(O)⊕τ(O) = Oand τ∗L ≅ L requires τ(P ) = P . Therefore τ is a rotation by θ around P .For any j(X) we have that θ = pi gives an automorphism of E as a genus1 curve. It is easy to check that στ = τσ. So τ ∈ Aut(A), and Z/2Z ⊂ Aut(A)for any j(X). In this case, τ fixes four square-roots of Q.When j(X) = 0, we have that θ = 2pi3 or 2pi3 also gives an automorphismof E as a genus 1 curve. But then τ does not preserve {Qi} and is not anautomorphism of A.When j(X) = 1728, we have that θ = pi2 or 3pi2 also gives an automorphismof E as a genus 1 curve. When Q = P ⊕ P is not fixed by multiplicationby ζ4, τ does not preserve {Qi}. If Q is fixed by multiplication by ζ4,,1133.5. Type B algebrasthen τ rotates Qis. It is easy to check that στ = τσ. So σ ∈ Aut(A), andZ/4Z ⊂ Aut(A). We have no more automorphisms.In sum we proved,Proposition 3.5.2 Let (E,L,σ) be a smooth type B triple, i.e. E a smoothelliptic curve, L ≅ O(2O+P ) where P is a 4-division point, σ multiplicationby −1. Let Q = P ⊕ P and ζ4Q is the image of Q by multiplication by ζ4.Then we have thatAut (A(E,L,σ)) = ⎧⎪⎪⎨⎪⎪⎩Z/4Z if j(X) = 1728 and ζ4Q = QZ/2Z otherwisewhere Z/2Z is generated by the automorphism of order 2 fixing four squareroots Pi of Q, and Z/4Z is generated by the rotation by pi2 around P .3.5.3 Nodal E and S-equivalenceNow we go back to the general form (with coefficients A,B,C,D) to studylimits. From w it is easy to see that interchanging x with y interchanges Bwith C, and A with D. So we only need to consider the case where A → 0and C → 0. When more than one of A,B,C,D vanishes the resulting algebrahas more degenerate E.TriangleWhen A = C = 0, by scaling of coordinates z → √B/Dz the defining equa-tions becomef1 =y2 − z2f2 =yx + xy (3.42)f3 =xz − zxwhich is the type S2 algebra S2(1).Irreducible nodalWhen C = 0 the defining equations becomef1 = B y2 −Dz2,f2 = B (yx + xy) −Az2, (3.43)f3 = A (yz − zy) +D (xz − zx).1143.5. Type B algebrasafter scalingx→ Ax,y →Dy,z →√BDz,the algebra has relationsf1 =y2 − z2,f2 =yx + xy − z2, (3.44)f3 =yz − zy + xz − zx.giving M = ⎛⎜⎝0 y −zy x −z−z −z x + y⎞⎟⎠ . This algebra, appearing in [3] (7.22), is a typeB algebra with an irreducible nodal curve point scheme E cut bydetM = −y3 − xy2 + 2yz2 − xz2with a node (1,0,0). For all points of E, M has rank 2, so σ is well definedon E. By cross producting the 2nd and the 3rd rows of M we see that theautomorphism σ is given byσ(x, y, z) = (x2 + xy − z2,−xy − y2 + z2, xz − yz).It has three fixed points(1,0,0), (1,2,2), (1,2,−2).As C → 0 two 2-torsion points collapse to (1,0,0) the node. At the node(0,1,0), the tangent cone is cut by y2 + z2 giving two branches z = iy andz = −iy. It is easy to see that σ interchanges two branches.Let us call this algebra S′′2 and say of type S′′2 .S′′2 ∶= C⟨x, y, z⟩(y2 − z2, yx + xy − z2, yz − zy + xz − zx) .Proposition 3.5.3 S′′2 , the type B algebra with an irreducible nodal pointscheme, is strictly semistable and S-equivalent to the B(0) = S2(1).1153.5. Type B algebrasProof. The flag W = ⟨y, z⟩ makes the two terms of f1 = y2 − z2 of equalweight. In f2 = yx + xy − z2, the yx + xy is the terms of the lowest weight.In f3 = xz − zx + yz − zy, the xz − zx has the lowest weight. And the lowestterms are the defining equations of S2(1). So S′′2 is S-equivalent to S2(1)and strictly semistable. 2Now we discuss geometrically. Let Xsm = Gm, and x ∈ Gm. σ fixesthe identity P1 = 1 ∈ Gm and interchanges 0 and ∞ so σ(x) = 1x . Theregularity means L = O(2P1 + Q) where Q is a 4-division point not fixedby σ. So Q = ±i. The triples with Q = i and Q = −i are isomorphic byan isomorphism −1. So there is a unique type B triple with an irreduciblenodal E.Let us consider an automorphism τ of the triple. τ should fix the set{1,−1}. The condition τ∗L ≅ L requires that τ fixes Q = i. The only non-identity automorphism satisfying both condition is τ(x) = − 1x . In terms ofalgebra this map is (x, y, z)↦ (x, y,−z).Aut = ⟨( 1 0 00 1 00 0 −1 )⟩ ≅ Z/2ZSo this algebra has the same automorphism as the smooth type B algebrawith j ≠ 1728.Line and conicWhen A = 0, by the same scaling of coordinates the algebra is the type Balgebra with a = 0.f1 =xy + yx + y2 − z2f2 =x2 + yx + xy (3.45)f3 =xz − zxwithM = ⎛⎜⎝y x + y −zx + y x 0−z 0 x⎞⎟⎠ .Its point scheme isZ (x(x2 + xy + y2 + z2))a union line and smooth conic intersecting at (0,1, i), (0,1,−i). The au-tomorphism σ is well defined on all points of X as M has rank 2 for allx ∈ X. σ stabilizes the line and the conic. On the line Z(x) we have1163.5. Type B algebrasσ(0, y, z) = (0, z, y), having two fixed points (0,1,1), (0,1,−1). And on theconic Z(x2 +xy + y2 + z2) we have σ(x, y, z) = (x,−x− y, z) having two fixedpoints (2,−1, i√3), (2,−1,−i√3). We have four fixed points which are alsosmooth.Proposition 3.5.4 B(0) is strictly semi-stable and S-equivalent to the S2(1).Proof. With a test configuration generated by U = ⟨x⟩, the lowest weightpart of f1 is y2 − z2, and that of f2 is xy + yx, and the terms of f3 are ofequal weight. These are the defining equations of S2(1). Therefore it isS-equivalent to S2(1). The first claim follows from this. 2With U = ⟨x⟩ we have UV = C{x2, xy, xz} and V U = C{x2, yx, zx}. But,C{x2, yx, zx} = C{x2,−x2 − xy, xz} = C{x2, xy, xz}. Therefore UV = V U .So if x is not a zero divisor then Lemma 3.3.7 implies that this algebra isstrictly semistable.We call this algebra S′2 and be of type S′2.Now we study geometrically. Let E = X1 ∪X2 where Xi is isomorphicto P1. Let xi ∈ Xi/{0,∞} and we regard as xi ∈ Gm. L has bidegree (2,1)on X1 and X2. Let P1 = 1 ∈ Gm =Xsm1 be the identity of the group law. Wealso choose a base point P2 ∈ X2. Then Xns has a group scheme structure.P2 is a 2-division point. Then −P2 also lies on X2 and is a 2-division point.Let L = O(2P1 +Q2) where Q2 ∈ X2. Then regularity requires that P2 bea 4-division point not fixed by σ. There are two such Q2 but −1 gives anisomorphism between triples with each Q. We call this triple Ts′2.Let us consider a non-trivial automorphism of the triple. As the irre-ducible nodal case, τ fixes the set {P2,−P2} ⊂ X2 and fixes a point Q2.Because τ ∣X2 has a fixed point we have τ ∣X2(x2) = λ2x2 . The λ2 is uniquelydetermined by {P2,−P2}. If we put P2 = 1 ∈ Gm then τ ∣X2(x2) = 1x2 . Thenτ ∣X1(x1) = λ2x1 as τ ∣X2 interchanges nodes. Because τ ∣X1 also fixes {1,−1} wehave that τ ∣X1(x1) = ± 1x1 . Therefore we haveAut(Ts′2) ≅ (Z/2Z)2In terms of algebra we haveAut(S′2) = ⟨( 1 0 00 1 00 0 −1 ) , ( 1 0 0−1 −1 00 0 1)⟩ ≅ (Z/2Z)2 .When the smooth elliptic curve with level 2-structure degenerates to thetriple with A = C = 0, there are two ways: 1) It first degenerates to an1173.5. Type B algebrasirreducible nodal with two 2-division points collapsed into the node, thenit further degenerates to a triangle where two sides are the tangent cone ofthe irreducible cubic at the node and two 2-division points sit in the otherside. 2) Second, it first degenerates to a union of line and conic where eachcomponent has two 2-division points, then the conic further degenerates toa union of two lines and the two 2-division points on the conic collapse tothe new node. However they are all S-equivalent and our moduli schemedoes not distinguish them.3.5.4 Degenerate algebrasB(a) is degenerate if a = ∞,−ζ3,−ζ23 . Among them B(−ζ3) ≅ B(−ζ23). Sowe have two non-isomorphic degenerate type B algebras.Type B algebra with a =∞When a =∞, (3.39) becomesf1 = xy + yx + y2 − z2,f2 = −z2, (3.46)f3 = yz − zy .M = ⎛⎜⎝y x + y −z0 0 −z0 −z y⎞⎟⎠ . So the point scheme is Z(yz2). On Z(y), we have thatσ(x,0, z) = (1,0,0), the line collapses to a point. So σ is not a graph. Let usconsider a correspondence Γ in P2 ×P2. Γ consists of three lines X1,X2,X3,where X1 is a double line, intersecting at (1,0,0) × (1,0,0).X1 = {(x1, y1,0) × (x1 + y1,−y1,0)}X2 = {(x1,0, z1) × (1,0,0)}X3 = {(1,0,0) × (x2,0, z2)} .As a →∞, the two 2-torsion points P3, P4 approach to the intersectionpoint (1,0,0), and P1 and P2 approach to (1,−2,0).Let us consider I3 = V1⊗R2∩R2⊗V1. We have two linearly independentrelationszf2 = f2zyf2 + zf3 = f2y − f3z .1183.5. Type B algebrasSo dim I3 = V1⊗R2 ∩R2⊗V1 = 2 and the algebra’s Hilbert polynomial is not1(1−t)3 . Therefore the algebra is not in our moduli.Even though B(∞) has a wrong Hilbert polynomial, due to the isomor-phism between B(a) and B( 1a), the place for B(∞) in the moduli is B(0)which is regular and strictly semistable.However let us find a algebra with the right Hilbert polynomial by addinga degree 3 element to B(∞).From the f2 of (3.39) and an overlap z3 = (z2)z = z(z2) we have, afterdividing by a,zx2 + zyx + zxy = x2z + yxz + xyz . (3.47)Let us define a new algebra by adding (3.47) to (3.46),B̃(∞) ∶= C⟨x, y, z⟩⎛⎜⎝xy + yx + z2 + y2,z2, yz − zy,zx2 + zyx + zxy − x2z − yxz − xyz⎞⎟⎠.Because y2 = −xy−yx in B(∞), (3.47) can be rewritten as zx2−x2z+y2z−zy2,soB̃(∞) = C⟨x, y, z⟩⎛⎜⎝xy + yx + y2,z2, yz − zy,[z, x2] − [z, y2]⎞⎟⎠,or equivalentlyB̃(∞) = C⟨x, y, z⟩⎛⎜⎜⎜⎝xy + yx + y2,z2, yz − zy,∑σ∈S3 sign(σ) ⋅ σ([x, y2])⎞⎟⎟⎟⎠.The last form is due to∑σ∈S3 sign(σ) ⋅ σ([x, y2]) =[x, y2] + [y, z2] + [z, x2] − [y, x2] − [z, y2] − [x, z2]=[x, y2] + [z, x2] − [y, x2] − [z, y2]=[x,−xy − yx] + [z, x2] − [y, x2] − [z, y2]=[z, x2] − [z, y2] − [y, x2] + [x,−xy − yx]=[z, x2] − [z, y2] − (yx2 − x2y) − (x2y + xyx − xyx − yx2)=[z, x2] − [z, y2].1193.5. Type B algebrasBy computer I checked that B̃(∞) has the Hilbert series (1−t)−3 upto degree8.Algebra with a = −ζ23B(−ζ23) is isomorphic to B(−ζ3) by both the relation B(a) ≅ B( 1a) andB(a) ≅ B(1 − a). The cubic for this algebra is−2x3− i√3x2y−3x2y− i√3xy2−3xy2+2√3ixz2− i√3y3−y3+√3iyz2+3yz2which has 3 singular points(ζ23 ,1,0) , (ζ3,1,− 4√−3) , (ζ3,1, 4√−3) (3.48)on which M has rank 1. Thus by Lemma 4.4 of ATV this algebra is degen-erate, and the Γ ⊂ P2 × P2, the zero locus of multilinearization of the degree2 part of the algebra, has p × l.As a → −ζ23 , the two 2-torsion points P1 and P3 approach to one of theintersection points (ζ3,1,− 4√−3) and P2 and P4 approach to (ζ3,1, 4√−3)another intersection point.Now consider a change of variables which sends the three singular points(3.48) to (1,0,0), (0,1,0), (0,0,1). A change of variablesx↦ζ23x + ζ3y + ζ3zy ↦x + y + zz ↦ 4√−3y − 4√−3zchanges the defining equations (3.39) intof1 ↦ − 2x2 + yz + zyf2 ↦4ζ3x2 − (a + ζ23)(y2 + z2) + (a − ζ23)(yz + zy) (3.49)f3 ↦ − (a + ζ23)(xy − yx + zx − xz) + (a + ζ3)(yz − zy)up to scalar. The linear matrix is then⎛⎜⎝−2x z y−4 − ζ3x −(a + ζ23)y + (a − ζ23)z (a − ζ23)y − (a + ζ23)z(a + ζ23)(y − z) −(a + ζ23)x − (a + ζ3)z (a + ζ23)x + (a + ζ3)y⎞⎟⎠ ,whose determinant is−(a + ζ23)2(y3 − y2z − yz2 + z3) − 4i√3(a + ζ23)(x2y + x2z)+2(a2 − a + 1)(xy2 + xz2) − 4(a + ζ3) (a − 12(1 − 3i√3))xyz.1203.5. Type B algebrasThis pencil of cubics has three base points (1,0,0), (0,1,1), (0,1,−1).When a = −ζ23 the relations are linearly equivalent tof1 =x2f2 =yzf3 =zy .Let us call this algebra Dg3Dg3 ∶= C⟨x, y, z⟩(x2, yz, zy) .This monomial algebra is a Zhang twist of the degenerate type A algebras,C⟨x, y, z⟩(x2, y2, z2), and C⟨x, y, z⟩(xy, yz, zy). Let θ(x) = x, θ(y) = z, θ(z) =y, then x ∗ x = xθ(x) = xx = 0, y ∗ y = yθ(y) = yz = 0, z ∗ z = zθ(z) = zy = 0.By the result of [15], this algebra has equivalent graded module categoriesto the degenerate type A algebras Dg1 and Dg2.The Ufnarovskii graph for Dg3 isxxxyy 99y??z zeez__ , (3.50)whose incidence matrix is ⎛⎜⎝0 1 11 1 01 0 1⎞⎟⎠ .Its fourth power is ⎛⎜⎝6 5 55 6 55 5 6⎞⎟⎠ ,which is the same with the fourth power of the incident matrix for the quiverof C⟨x, y, z⟩(x2, y2, z2), the Dg1. So by [14] they have equivalent categoriesof graded right modules.For Dg3, the cubic is xyz, a triangle. When x = 0 the null space of Mis spanned by (1,0,0), and when y = 0 by (0,0,1), when z = 0 by (0,1,0).So the σ collapses one side to the opposite vertex and the other two sides tothe vertex on itself, and puffs up a vertex to the side which collapses to it.So we have to consider a correspondence in P2 × P2. We will give the 6-gonfor Dg3 in the next section, (3.59).1213.5. Type B algebrasWe have x3 = x(x2) = (x2)x, yzy = y(zy) = (yz)y, and zyz = z(yz) =(zy)z. So dim(V ⊗R ∩R ⊗ V ) = 3. The Hilbert polynomial is not equal tothat of regular algebras, and this algebra is not a member of our moduli.Let us add degree 3 elements to find an algebra which will be placed in themoduli instead of Dg3.We recover lost relations of Dg3 in the limit process. From f2 + 2ζ3f1 of(3.49) we have(a − 2ζ3 − ζ23)(yz + zy) = (a + ζ23)(y2 + z2) .From this and f3 we have2a + ζ23 yz =xy − yx + zx − xzζ3 + a − y2 + z2ζ23 + 2ζ3 − a ,2a + ζ23 zy = − xy − yx + zx − xzζ3 + a − y2 + z2ζ23 + 2ζ3 − a .From this and an overlap (yz)y = y(zy) we havexy2 − yxy + zxy − xzyζ3 + a − y3 + z2yζ23 + 2ζ3 − a = −yxy − y2x + yzx − yxzζ3 + a − y3 + yz2ζ23 + 2ζ3 − a .And with an overlap (zy)z = z(yz) we have−xyz − yxz + zxz − xz2ζ3 + a − y2z + z3ζ23 + 2ζ3 − a = zxy − zyx + z2x − zxzζ3 + a − zy2 + z3ζ23 + 2ζ3 − a .We can rewrite the last two as(xy2 − y2x + zxy − xzy + yzx − yxz) + µ(yz2 − z2y) =0 (3.51)(z2x − xz2 + zxy − zyx + xyz − yxz) + µ(zy2 − y2z) =0where µ = (ζ3+a)/(ζ23 +2ζ3−a). When a = −ζ23 we have yz = zy = 0 so (3.51)becomesxy2 − y2x + zxy − yxz =0 (3.52)z2x − xz2 + zxy − yxz =0 .This can be rewritten asxy2 + xz2 = y2x + z2x (3.53)z2x − xz2 + zxy − yxz =0 .1223.5. Type B algebrasFrom f1 we have 2x2 = yz + zy and the overlap 2x3 = x(2x2) = (2x2)x, wehavexyz + xzy − yzx − zyx = 0but when yz = zy = 0 the left hand side is 0. So this overlap gives noadditional relation. Hence (3.52) is all the additional relations. When x2 =yz = zy = 0 we can rewrite (3.52) as[x, y2] + ∑σ∈S3 signσ ⋅ σ(xyz) = 0[z2, x] + ∑σ∈S3 signσ ⋅ σ(xyz) = 0 .or equivalently∑σ∈A3 σ([x, y2]) + ∑σ∈S3 signσ ⋅ σ(xyz) = 0∑σ∈A3 σ([x2, y]) + ∑σ∈S3 signσ ⋅ σ(xyz) = 0 .or equivalently∑σ∈S3 signσ ⋅ σ([x, y2]) = 0∑σ∈A3 σ([x2, y]) + ∑σ∈S3 signσ ⋅ σ(xyz) = 0 .So letD̃g3 = C⟨x, y, z⟩⎛⎜⎜⎝x2, yz, zy, ∑σ∈S3 signσ ⋅ σ([x, y2]),∑σ∈A3 σ([x2, y]) + ∑σ∈S3 signσ ⋅ σ(xyz)⎞⎟⎟⎠. (3.54)Then we have that D̃g3 has the Hilbert series (1 − t)−3 and completes themoduli of type B algebras at Dg3 for all degree.We have thatAut(Dg3) = Gm2 ⋊Z/2Z = ⟨( 1 0 00 ∗ 00 0 ∗ ) , ( 1 0 00 0 10 1 0 )⟩ (3.55)Because Aut(D̃g3) ⊂ Aut(Dg3), to find Aut(D̃g3) we only need to findelements of (3.55) which preserve (3.53). It is easy to see that only ( 1 0 00 0 10 1 0)preserves (3.53). SoAut(D̃g3) = Z/2Z = ⟨( 1 0 00 0 10 1 0)⟩1233.5. Type B algebrasAnd corresponding automorphism of B(−ζ23) to ( 1 0 00 0 10 1 0 ) is ( 1 0 00 1 00 0 −1 ). There-fore we have that B(−ζ23) has the same automorphism group with genericB(a).Proposition 3.5.5 B(a) is stable if a ∉ T = {0,1,∞,−ζ23 ,−ζ3}. The Iso-morphism classes are parametrized by (P1∖T )/SL(2,F2). To compactify weadd two strictly semistable algebras B(0) and D̃g3. The later is in place ofB(−ζ23) which is degenerate and of the wrong Hilbert series. The stack ofalgebras is a Gm × Z/2Z-gerbe over this quotient stack. Only B(−1), (andisomorphically B(1/2),B(2)) has an extra Z/2Z automorphism.3.5.5 Type B as Zhang twist of type A.Consider a Zhang twist of Sklyanin algebras by (x, y, z) ↦ (x, z, y), andinterchange f2 and f3,f1 =ax2 + b y2 + c z2,f2 =azy + bxz + c yx,f3 =ayz + b zx + cxy.Call this algebra ZA(a, b, c). For generic (a, b, c) this algebra is not reg-ular and has the Hilbert polynomial 3n. When b = c we have M = NQtwhere Q = ( 1 0 00 0 10 1 0) whose Jordan form is a diagonal matrix with a diago-nal (1,1,−1) as the type B algebras. ZA(a, b, b) is regular unless (a, b, c) =(1,0,0), (1,1,1), (ζ3,1,1), (ζ23 ,1,1) in which case it is degenerate. So let usput b = 1 and consider 1-dim’l family parametrized by a ∈ C.So BZ(a) ∶= ZA(a,1,1) is defined byC⟨x, y, z⟩⎛⎜⎜⎝f1 =ax2 + y2 + z2,f2 =azy + xz + yx,f3 =ayz + zx + xy⎞⎟⎟⎠. (3.56)Then we havedetM = detN = (a3 + 2)xyz − a(x3 + y3 + z3),σ(x, y, z) = (ay2 − xz,−a2xy + z2, ax2 − yz).1243.5. Type B algebrasσ has four fixed points,Q1 = (2a,−1 +√1 − a3 −√−2 − a3 + 2√1 − a3,−1 +√1 − a3 +√−2 − a3 + 2√1 − a3) ,Q2 = (2a,−1 +√1 − a3 +√−2 − a3 + 2√1 − a3,−1 +√1 − a3 −√−2 − a3 + 2√1 − a3) ,Q3 = (2a,−1 −√1 − a3 −√−2 − a3 − 2√1 − a3,−1 −√1 − a3 +√−2 − a3 − 2√1 − a3) ,Q4 = (2a,−1 −√1 − a3 +√−2 − a3 − 2√1 − a3,−1 −√1 − a3 −√−2 − a3 − 2√1 − a3) .Recall for a Hesse cubic x3 + y3 + z3 + λxyz the j-invariant is given byj(λ) = −λ3(λ3 − 216)3(λ3 + 27)3 .So the j-inv of the point scheme of BZ(a) isj(a) = (a3 + 2)3 (a9 + 6a6 + 228a3 + 8)3a3 (a3 − 1)6 (a3 + 8)3 .a degree 36 rational function. a appears as a3 in j(a).Now we choose P1 = (0,1,−1) as the identity to define a group law, then(x1, y1, z1)⊕ (x2, y2, z2) = (y1z1x22 −x21y2z2, x1y1z22 − z21x2y2, x1z1y22 − y21x2z2)(x, y, z)⊕ (x, y, z) = (xz3 − xy3, y3z − x3z, x3y − yz3).(x, y, z)−1 = (x, z, y).The three 2-division points areP2 =(a,1,1)P3 =(−a2+√2a+ a24,1,1)P4 =(−a2−√2a+ a24,1,1) .Therefore the same equations hold as (3.41), and (3.40).1253.5. Type B algebrasIsomorphisms and AutomorphismsFor any BZ(a) the change of variables (x, y, z) ↦ (x, z, y) is an automor-phism of order 2, so the moduli stack is a Gm×(Z/2Z) gerbe. Pi, i = 1,2,3,4are the fixed points of this automorphism.Proposition 3.5.6 SL(2,F3) acts on P2(x, y, z) as a change of variablesfor BZ algebra. Its center gives an automorphism and PSL(2,F3) gives anisomorphism. Two generators of PSL(2,F3) give isomorphisms of BZ(a) toBZ(ζ23a) and to BZ(a+2a−1).Proof. Because an isomorphism of algebras gives an isomorphism of pointschemes, isomorphisms of BZ(a) are elements of G216. But the image ofBZ(a) by g1(x, y, z) = (y, z, x) and g2(x, y, z) = (x, ζ3y, ζ23z) is no longer inthe form of BZ(a′). So we only need to check SL(2,F3). It is trivial tosee that (x, y, z)↦ (x, z, y) gives an automorphism. It is straightforward tocompute that g3(x, y, z) = (x, ζ3y, ζ3z) changes BZ(a) to BZ(ζ23a), and thatg4(x, y, z) = (x + y + z, x + ζ3y + ζ23z, x + ζ23y + ζ3z) changes it to BZ(a+2a−1). 2Corollary 3.5.7 Isomorphism classes of BZ(a) are parametrized by P1/PSL(2,F3).This PSL(2,F3) action is identical to the same with the action of thesame group on l13 on the parameter space of type A algebras. The fixedpoints of this action and their orbits can be found in Table 3.3. However wepresent here again.Orb(0) ={0,−2,−2ζ3,−2ζ23}Orb(∞) ={∞,1, ζ3, ζ23}Orb(1 +√3) ={1 +√3, ζ3(1 +√3), ζ23(1 +√3),1 −√3, ζ3(1 −√3), ζ23(1 −√3)} .BZ(0) is the type S2 algebra S2(1). When a is not in the orbits of 0,∞,1+√3we have Aut(BZ(a)) = Z/2Z. The stabilizer of SL(2,F3) action at 0 is Z/3Z,but Aut(BZ(0)) = Gm × Z/2Z as we will see below. So the automorphismgroup for BZ(0) is bigger than that from the SL(2,F3) action. The similarhappens for BZ(∞). We have AutBZ(a) = Z/4Z for a ∈ Orb(1 +√3). Thiscorresponds to AutB(a) = Z/4Z for a = −1,2,1/2. And the automorphismsfor B(1 +√3) are exactly those which come from the SL(2,F3) action.Proposition 3.5.8 The moduli of stable type B algebras is a quotient stack[(P1 ∖ 8 points) /SL(2,F3)]1263.5. Type B algebraswhere the 8 points are those in the orbits of 0 and ∞, and SL(2,F3) acts viaits quotient PSL(2,F3).We give the list of elements of PSL(2,F3) as a subgroup of Aut(P1).z0(a) =a, z1(a) =ζ3a, z2(a) =ζ23a,z3(a) =a + 2a − 1 , z4(a) =ζ3a + 2a − 1 , z5(a) =ζ23 a + 2a − 1 ,z6(a) =ζ3a + 2ζ3a − 1 , z7(a) =ζ3 ζ3a + 2ζ3a − 1 , z8(a) =ζ23 ζ3a + 2ζ3a − 1 ,z9(a) =ζ23a + 2ζ23a − 1 , z10(a) =ζ3 ζ23a + 2ζ23a − 1 , z0(a11) =ζ23 ζ23a + 2ζ23a − 1 .Singular EThis pencil of Hesse cubics is singular when a = 0,1, ζ3, ζ23 ,−2,−2ζ3,−2ζ23 .Algebras for these a values are isomorphic to that of either a = 0 or ∞.However let us consider a slightly general form to study degeneration. Wecan consider a family of algebras defined byf1 =A1 x2 + y2 + z2f2 =A2 zy + xz + yxf3 =A2 yz + zx + xy .This algebra also satisfies M = NQt. When A1,A2 ≠ 0 this algebra isisomorphic to BZ(a)TriangleWhen A1 = A2 = a = 0,f1 =y2 + z2,f2 =xz + yx,f3 =zx + xy.M = ⎛⎜⎝0 y zy 0 xz x 0⎞⎟⎠, and detM = xyz. σ(0, y, z) = (0, z,−y), σ(x,0, z) =(x,−z,0), σ(x, y,0) = (x,0,−y). So σ has three fixed points (1,0,0), (0,1, i),(0,1,−i). This algebra is (3.21) the type S2 algebra after a change of vari-ables.This algebra has an additional automorphism (x, y, z) ↦ (x, ty, tz). Sothe non-tautological automorphism group is Gm ×Z/2Z.1273.5. Type B algebrasIrreducible nodalWhen A1 = 0 we can put A2 = a. But rescaling y ↦ 1ay and z ↦ 1az we canput a = 1. Then the defining equations aref1 =y2 + z2f2 =zy + xz + yxf3 =yz + zx + xy,M = ⎛⎜⎝0 y zy z xz x y⎞⎟⎠ , N =⎛⎜⎝0 z yy x zz y x⎞⎟⎠ ,detM = 2xyz − (y3 + z3).So the point scheme is an irreducible nodal curve with a node (1,0,0).σ(x, y, z) = (xy − z2, yz,−y2)which has three fixed points (1,0,0), (−1 + i,−2i,2), (−1 − i,2i,2). The lasttwo approach to (0,1, i) and (0,1,−i) respectively as a→ 0.A base change (x, y, z)↦ (4ix+ iy,−3iy + 3z,−3iy − 3z) gives an isomor-phism of this algebra to the irreducible nodal type B algebra (3.44).With a test configuration generated by U = ⟨x⟩ this algebra is S-equivalentto the BZ(0) ≅ S2(1)This algebra has no additional automorphisms. So it is natural to con-sider this algebra as a limit lima→0BZ(a) because it has no additional au-tomorphisms and is S-equivalent to the BZ(0) ≅ S2(1).Line and conicWhen A2 = 0 we can put A1 = a ≠ 0. But by a change of variables y ↦ y/a,z ↦ z/a we can also put a = 2. Then the defining equations aref1 =2x2 + y2 + z2f2 =xz + yxf3 =zx + xy ,M = ⎛⎜⎝2x y zy 0 xz x 0⎞⎟⎠ , N =⎛⎜⎝2x z yy x 0z 0 x⎞⎟⎠ ,−M = N = 2x(x2 − yz) .1283.5. Type B algebrasSo the point scheme is a reducible nodal curve with the node (0,1,0), (0,0,1).σ(x, y, z) = (xy,−2x2 + yz,−y2) .So on Z(x) we have σ(0, y, z) = (0, z,−y), and on Z(x2 − yz) we haveσ(x, y, z) = (x,−z,−y), which has four fixed points (0,1, i), (0,1,−i), (i,1,−1), (i,−1,1).The last two collapse to (1,0,0) as a→ 0.A base change τ1(x, y, z) = (√6x, ix + 2iy + 2z, ix + 2iy − 2z) gives anisomorphism of this algebra to the reducible nodal type B algebra (3.45).This algebra has an additional automorphism τ2(x, y, z) = (x,−y,−z).So the non-tautological automorphism group is (Z/2Z)2.With a test configuration generated by W = ⟨y, z⟩ this algebra is S-equivalent to the BZ(0) ≅ S2(1).Degenerate algebra BZ(∞) =Dg3If BZ(a) is degenerate then it is isomorphic to BZ(∞). The defining equa-tions f1, f2, f3 of BZ satisfy xf1 + yf2 + zf3 = f1x+ f2z + f3y. But in BZ(∞)this splits into three overlaps, xf1 = f1x, yf2 = f3y, and zf3 = f3y, whichmakes a(3) = 12. So as in the case of Dg1, to find flat a family we have toadd relations from its neighborhood corresponding to these overlaps. Thethree overlaps of BZ(∞) arex3 =x(x2) = (x2)xyzy =y(zy) = (yz)yzyz =z(yz) = (zy)z ,which are in the neighborhood a ≠ 0, equivalent tox(y2 + z2) =(y2 + z2)xy(xz + yx) =(zx + xy)y (3.57)z(zx + xy) =(xz + yx)z .It is easy to see that the sum of the first and the second equations gives thethird.LetB̃Z(∞) ∶= ⟨x, y, z⟩(x2, yz, zy, [y2, x] + [z2, x], yxz − zxy + [y2, x]) . (3.58)It is easy to see that (3.52) is equivalent to (3.58). So we have thatD̃g3 = B̃Z(∞) .1293.5. Type B algebrasDegree 2 part of BZ(∞) = Dg3 determines a 6-gon in P2 × P2 whosecomponents areX1 = {(0, y1, z1) × (1,0,0)}, X2 = {(0,0,1) × (x2,0, z2)} (3.59)X3 = {(x1,0, z1) × (0,0,1)}, X4 = {(1,0,0) × (0, y2, z2)} (3.60)X5 = {(x1, y1,0) × (0,1,0)}, X6 = {(0,1,0) × (x2, y2,0)}. (3.61)Now consider some limits of a →∞. By a change of variables x ↦ 1√axthe defining equations becomef1 = x2 + y2 + z2f2 = a3/2 zy + xz + yxf3 = a3/2 yz + zx + xythen take limit a↦ 0 to obtainf1 = x2 + y2 + z2f2 = zyf3 = yz .We call the algebra defined by these equations Dg3′. The point scheme is a4-gon in P2 × P2 whose components areX1 = {(0, y1, z1) × (1,0,0)},X ′3 = {(x1,0, z1) × (z1,0,−x1)}, X4 = {(1,0,0) × (0, y2, z2)}X ′5 = {(x1, y1,0) × (y1,−x1,0)} .So from the 6-gon of Dg3, X2(or equivalently X3) is collapsed to give a graphof the automorphism σ∣Z(y1) given by σ(x1,0, z1)× (z1,0,−x1), an inversioncomposed by multiplication by −1, andX6(or equivalentlyX5) is collapsed togive a graph of the automorphism σ∣Z(z1) given by σ(x1, y1,0) = (y1,−x1,0),also an inversion composed by multiplication by −1.Exceptional algebraWe find an exceptional algebra which can be put in place of Dg3 in themoduli using the limit method as before. First by a change of variablesx↦ ax we havef1 =a3x2 + y2 + z2,f2 =azy + axz + ayx,f3 =ayz + azx + axy .1303.6. Type E and H algebrasThen take limit a→∞ and obtainf1 =x2,f2 =zy + xz + yx, (3.62)f3 =yz + zx + xy.Let us call this algebra E3.E3 = C⟨x, y, z⟩(x2, zy + xz + yx, yz + zx + xy) .We haveM = ⎛⎜⎝x 0 0y z xz x y⎞⎟⎠ , N =⎛⎜⎝x z y0 x z0 y x⎞⎟⎠detM = −detN = xyz − x3 = x(yz − x2)σ(x, y, z) = (yz − x2, xz − y2, xy − z2) .The point scheme is the union of a line and a conic intersecting in (0,1,0),(0,0,1). σ interchanges two components and fixes two intersection points,and satisfies σ2 = id. This exceptional algebra satisfies (3.57).We have thatAut(E3) = ⟨( 1 0 00 0 10 1 0)⟩ ≅ Z/2Z .So if we choose E3 in place of BZ(∞) and the irreducible nodal one forBZ(0) then all members have Z/2Z as the non-tautological automorphismgroup.We have thatProposition 3.5.9 B̃Z(∞) has the Hilbert series (1−t)−3. The exceptionalalgebra E3 is S-equivalent to B̃Z(∞).Proof. They have the same lowest weight part for each relations by U =⟨x⟩. 23.6 Type E and H algebrasIn this chapter we study type E and H algebras. We study the triples forthem and their automorphisms.1313.6. Type E and H algebras3.6.1 Type EType E algebra has defining equationsf1 = yz + ζ29zy + ζ9x2f2 = zx + ζ89xz + ζ49y2f3 = xy + ζ59yx + ζ79z2 ,withM = ⎛⎜⎝ζ9x ζ29z yz ζ49y ζ89xζ59y x ζ79z⎞⎟⎠ ,which givesdetM = −(x3 + y3 + z3) .By cross producting the 2nd and 3rd rows of M we haveσ(x, y, z) = (−ζ89x2 + ζ29yz, ζ49xy − ζ79z2,−y2 + xz) .x3+y3+z3 is a Hesse cubic so its automorphism is an element of G216 whichis generated by gi, i = 0,1,2,3,4 where g0(x, y, z) = (x, z, y), g1(x, y, z) =(y, z, x), g2(x, y, z) = (x, ζ3y, ζ23z), g3(x, y, z) = (x, ζ3y, ζ3z), g4(x, y, z) =(x + y + z, x + ζ3y + ζ23z, x + ζ23y + ζ3z). One can check by computation thatg0, g1, g3, g4 do not preserve J2 = C{f1, f2, f3}, and g2 does. So we haveAut(A) = ⟨g2⟩ ≅ Z/3Z .Now we compute the automorphisms using geometry. For the type Etriple (E,L,σ) we have j(E) = 0 and σ is an automorphism of order 3, themultiplication by ζ3, and L ≅ O(2O + P ) where P is a point of order 3 notfixed by σ.Let Q1, Q2 be the 3-division points fixed by σ. Then from στ = τσwe have that τ preserves {O,Q1,Q2}. When τ(O) = O we also have thatτ(P ) = P from τ∗L ≅ L. But there is no automorphism of an elliptic curvewhich fixes both the origin and 3-division point which is not fixed by themultiplication by ζ3. So we have τ(O) = Qi, and without loss of generalitywe can put τ(O) = Q1. Then τ(O) ⊕ τ(O) = Q1 ⊕Q1 = Q2. And τ∗L ≅ Lrequires that τ(P ) = P ⊕ (−Q2) = P ⊕Q1. This with τ(O) = Q1 = O ⊕Q1implies that τ is a translation by Q1. It is easy to check that στ = τσ. Soτ ∈ Aut(X) and Z/3Z ⊂ Aut(X). There is no more automorphism.We have proved1323.6. Type E and H algebrasProposition 3.6.1 Let (E,L,σ) be the type E triple, i.e. j(X) = 0 and L ≅O(2O+P ) where P is a 3-division point not fixed by σ. Let A = A(E,L,σ),then we haveAut(A) = Z/3Zwhich is generated by a translation by the 3-division point fixed by σ.3.6.2 Type HType H algebra has defining equationsf1 = x2 − y2f2 = xy − yx + i z2f3 = yz − i zy ,withM = ⎛⎜⎝x −y 0−y x i z0 −i z y⎞⎟⎠ , N =⎛⎜⎝x y 0−y −x z0 i z −i y⎞⎟⎠ .Therefore we havedetM = −idetN = x2y − y3 − xz2 .This cubic has P1 ∶= (0,0,1) as a flex, and L ≅ O(3P1). By cross productingthe 1st and the 2nd rows of M we haveσ(x, y, z) = (i yz, i xz, y2 − x2) ,and by cross producting the 2nd and the 3rd rows of M we haveσ(x, y, z) = (xy − z2, y2, i yz) .σ has two fixed points P3 ∶= (1,1,0), P4 ∶= (1,−1,0). Tangent lines at thesetwo points pass through P1. This means L ≅ O(2P3 + P1) ≅ O(2P4 + P1). Ifwe choose P3 or P4 as the identity of group law then P1 is a 2-division pointnot fixed by σ. And if we choose a flex (0,0,1) as the identity of group law,then the two fixed points of σ are 2-division points. So σ is the rotation bypi/2 around the 2-division point (1,1,0) or (1,−1,0). The other 2-divisionpoint is P2 ∶= (1,0,0) which is fixed by non-tautological automorphisms ofA (see below).Let τH1 (x, y, z) = (x,−y, iz). Then τH2 ∶= (τH1 )2(x, y, z) = (x, y,−z), τH3 ∶=(τH1 )3(x, y, z) = (x,−y,−iz), and τH4 ∶= (τH1 )4 = id. τH1 and τH3 have {P1, P2}as a fixed points set and τH2 has {P1, P2, P3, P4} as its fixed points set.1333.7. Unstable algebrasNow we find the automorphisms using geometry. For type E triple(E,L,σ) we have j(E) = 1728 and σ is an automorphism of order 4, multi-plication by ζ4. And L ≅ O(2O +P ) where P is a point of order 2 not fixedby σ. Let Q be the 2-division point fixed by σAs the type B and E case, στ = τσ requires that τ preserves the set offixed points of σ. So τ(O) is a 2-division point and τ(O)⊕ τ(O) = O. Thenτ∗L ≅ L requires τ(P ) = P . Thus τ is a rotation by pi2 i, i = 1,2,3,4 aroundP . It is easy to check στ = τσ. Therefore τ ∈ Aut(A), and Z/4Z = Aut(X).We have provedProposition 3.6.2 Let (E,L,σ) be the type H triple, i.e. j(X) = 1728and L ≅ O(2O + P ) where P is a 2-division point not fixed by σ. Let A =A(E,L,σ) thenAut(A) = Z/4Zwhich is generated by a rotation by pi2 around P .3.7 Unstable algebrasHere we study 2-unstable algebras. Most of them grow faster than (1− t)−3.However we found a family of algebras which we conjecture to have theHilbert series (1 − t)−3 and to be unstable as the Futaki function has thelimit n→∞F (n) = 0. See Conjectrue 3.7.3 for details.3.7.1 Central extension of non-regular tripleWe first present an example of unstable algebra.Proposition 3.7.1 Let A = C⟨x, y, z⟩/(f1, f2, f3) be an algebra of Hilbertseries (1 − t3)/(1 − t)3 generated in degree 1 andCt(A) ∶= A⟨t⟩(txi − xit)i=1,2,3 = C⟨x, y, z, t⟩(f1, f2, f3, txi − xit)i=1,2,3 , (3.63)where t has degree 3. Then Ct(A) is an algebra of the Hilbert series (1−t)−3and is unstable.Proof. It is unstable because it is not generated in degree 1, by Proposition3.14 of [6]. To prove dimCt(A)i = (i+1)(i+2)2 note that if a ∈ Ct(A)i then ais uniquely expressed asa = ⌊i/3⌋∑j=0 tjbj , bi ∈ Ai−3j .1343.7. Unstable algebrasSodimCt(A)i = ⌊i/3⌋∑j=0 dimAi−3j .When i = 3n we havedimCt(A)i = dimA0 + n−1∑j=0 3(3n − 3j)= 1 + 9(n + 1)n/2= (9n2 + 9n + 2)/2= (i + 1)(i + 2)/2 .When i = 3n + k where k = 1,2 we havedimCt(A)i = n∑j=0 3(3n + k − 3j)= 3 ((3n + k)(n + 1) − 3(n + 1)n/2)= 3 (k(n + 1) + 3(n + 1)n/2)= (3n + 3)(3n + 2k)/2= (i + 3 − k)(i + k)/2= (i + 1)(i + 2)/2 .2Corollary 3.7.2 Let T = (E,L,σ) be a non-regular triple such that A(T ) =B(T ) has the Hilbert series (1− t3)/(1− t)3. Then Ct(A(T )) is an unstablealgebra with the Hilbert series (1 − t)−3.3.7.2 2-unstable algebrasIn this section we find degree 2 defining equations for each test configurationwhich make the algebra 2-unstable.Unstabilized by U = ⟨x⟩With a test configuration generated by U = ⟨x⟩ with w(x) = α, an algebraA is 2-unstable if w(2) < 4α.0 α 2αz2, y2, zy, yz xz, zx, yx, xy x2Table 3.4: Weights for U = ⟨x⟩ ⊂ V1353.7. Unstable algebrasCase I. w(2) = 2α: If z2 survives then all of weight α monomials,xz, zx, yz, zy should vanish. It requires 4 degree 2 relations making dimA2 =5. So we have z2 = 0 and two of xz, zx, yz, zy is eliminated. Relations are offollowing form:f1 =x2f2 =a1zx + b1xz + c1xy + d1yxf3 =a2zx + b2xz + c2xy + d2yx,withM = ⎛⎜⎝x 0 0d1y + a1z c1x b1xd2y + a2z c2x b2x⎞⎟⎠ , N =⎛⎜⎝x c1y + b1z c2y + b2z0 d1x d2x0 a1x a2x⎞⎟⎠ ,givingdetM = (b2c1 − b1c2)x3, detN = (a2d1 − a1d2)x3and the correspondence Γ ⊂ P2 × P2 isΓ = ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩{(0, y1, z1) × (0, y2, z2)}, if b2c1 − b1c2 ≠ 0, and a2d1 − a1d2 ≠ 0P2 × {(0, y2, z2)}, if b2c1 − b1c2 = 0, and a2d1 − a1d2 ≠ 0{(0, y1, z1)} × P2, if b2c1 − b1c2 ≠ 0, and a2d1 − a1d2 = 0 .These algebras have the Hilbert series (1 + t)(1 − 2t)−1.Case II. w(2) = 3α:Subcase II-a: If x2 survives, then three of xz, zx, yx, xy should be elim-inated to have w(2) = 3α. This gives defining equations of the followingform:f1 =a1zx + b1xz + c1xy + d1yx + e1x2f2 =a2zx + b2xz + c2xy + d2yx + e2x2f3 =a3zx + b3xz + c3xy + d3yx + e3x2 .detM = x2h1, detN = x2h2 where h1, h2 are generically linear in x, y, z, and0 for some coefficients. The generic one hasP1 × P1 = {(0, y1, z1) × (0, y2, z2)}as a connected component of Γ. The generic one has growth (3,6,10,17,31, . . .).1363.7. Unstable algebrasSubcase II-a: If x2 is eliminated then one of xz, zx, yx, xy and one ofy2, z2, yz, zy should be eliminated to have w(2) = 3α. This gives relations offollowing form:f1 =x2f2 =a1zx + b1xz + c1xy + d1yxf3 =a2zx + b2xz + c2xy + d2yx + i2y2 + j2yz + k2zy + l2z2 .Then we haveM = ⎛⎜⎝x 0 0d1y + a1z c1x b1xd2y + a2z c2x + i2y + k2z b2x + j2y + l2z⎞⎟⎠anddetM = x2 ((c1b2 − b1c2)x + (c1j2 − b1i2)y + (c1l2 − b1k2)z) .N = ⎛⎜⎝x 0 0c1y + b1z d1x a1xc2y + b2z d2x + i2y + j2z a2x + k2y + l2z⎞⎟⎠tanddetN = x2 ((d1a2 − a1d2)x + (d1k2 − a1i2)y + (d1l2 − a1j2)z)This algebra is degenerate. The correspondence Γ is a union of two P1 andone double P1. Γ = Γ1 ∪ Γ2 ∪ Γ3, whereΓ1 = {(0, y, z) × (0,−(j2y + l2z), i2y + k2z)} = {(0,−(k2y + l2z), i2y + j2z) × (0, y, z)}Γ2 = {(x, y, (c1b2 − b1c2)x + (c1j2 − b1i2)y−(c1l2 − b1k2) ) × (0,−b1, c1)}Γ3 = {(0,−a1, d1) × (x, y, (d1a2 − a1d2)x + (d1k2 − a1i2)y−(d1l2 − a1j2) )} .With a test configuration generated by U = ⟨x⟩ the limit algebra hasrelationsf1 =x2f2 =a1zx + b1xz + c1xy + d1yx (3.64)f3 =i2y2 + j2yz + k2zy + l2z2 .1373.7. Unstable algebrasThen we haveM = ⎛⎜⎝x 0 0d1y + a1z c1x b1x0 i2y + k2z j2y + l2z⎞⎟⎠anddetM = x2 ((c1j2 − b1i2)y + (c1l2 − b1k2)z) .N = ⎛⎜⎝x 0 0c1y + b1z d1x a1x0 i2y + j2z k2y + l2z⎞⎟⎠tanddetN = x2 ((d1k2 − a1i2)y + (d1l2 − a1j2)z) .This algebra is degenerate. And Γ of the generic one is a union of two P1and one double P1. Γ = Γ1 ∪ Γ2 ∪ Γ3, whereΓ1 = {(0, y, z) × (0,−(j2y + l2z), i2y + k2z)} = {(0,−(k2y + l2z), i2y + j2z) × (0, y, z)}Γ2 = {(x,−(c1l2 − b1k2)y, (c1j2 − b1i2)y) × (0,−b1, c1)}Γ3 = {(0,−a1, d1) × (x,−(d1l2 − a1j2)y, (d1k2 − a1i2)y)} .If we restrict Z(x) we have a graph of σ ∈ End(Z(x)) given by σ(y, z) =(−j2y − l2z, i2y + k2z). But lines containing points {(1,0,0) × (0,−b1, c1)},and {(0,−a1, d1)×(1,0,0)} respectively, are attached to make it degenerate,which corresponds to Γ2 and Γ3 respectively.Conjectrue 3.7.3 The generic one of the algebras defined by (3.64) hasthe Hilbert series (1 − t)−3 and is unstable as w(n) = a(n − 1) = n(n + 1)/2with the test configuration generated by U = ⟨x⟩.When a∶d = b∶ c its growth is (6,11,19,32, . . .).Unstabilized by W = ⟨x, y⟩With a test configuration generated by W = ⟨x, y⟩ with w(x) = w(y) = α, analgebra A is 2-unstable if w(2) < 8α.0 α 2αz2 zy, yz, xz, zx x2, xy, yx, y2Table 3.5: Weights for W = ⟨x, y⟩ ⊂ V1383.7. Unstable algebrasCase w(2) = 6α: The minimal possible weight is 6α = 0+α+α+α+α+2αand for this weight 3 of x2, xy, yx, y2 should be eliminated.f1 =a1x2 + b1xy + c1yx + d1y2f2 =a2x2 + b2xy + c2yx + d2y2f3 =a3x2 + b3xy + c3yx + d3y2 .Because relations have no z, the third column of M and N t is 0 and detM =detN = 0. The generic one has growth (6,13,28,60, . . .).Case w(2) = 7α: w(7) = 7α = 0 + α + α + α + 2α + 2α is achieved whentwo of x2, xy, yx, y2 and one of yz, zy, zx, xz are eliminated .f1 =a1x2 + b1xy + c1yx + d1y2f2 =a2x2 + b2xy + c2yx + d2y2f3 =a3x2 + b3xy + c3yx + d3y2 + i3yz + j3zy + k3zx + l3xz .Then we haveM = ⎛⎜⎝a1x + c1y b1x + d1y 0a2x + c2y b2x + d2y 0a3x + c3y + k3z b3x + d3y + j3z l3x + i3y⎞⎟⎠anddetM = (l3x+i3y) ((a1b2 − b1a2)x2 + (a1d2 − d1a2 + c1b2 − b1c2)xy + (c1d2 − d1c2)x2) .N = ⎛⎜⎝a1x + b1y c1x + d1y 0a2x + b2y c2x + d2y 0a3x + b3y + l3z c3x + d3y + i3z k3x + j3y⎞⎟⎠tanddetN = (k3x+j3y) ((a1c2 − c1a2)x2 + (a1d2 − d1a2 + b1c2 − c1b2)xy + (b1d2 − d1b2)x2) .detM and detN determines three lines intersecting in one point. Genericone has growth (3,6,9, . . .).Unstabilized by W = ⟨x, y⟩ ⊃ U = ⟨x⟩With a test configuration generated by W = C⟨x, y⟩ ⊃ U = ⟨x⟩ with w(y) = α,w(x) = β an algebra A is 2-unstable if w(2) < 4α + β. The weights ofmonomials in degree 2 are1393.7. Unstable algebras0 α β 2α α + β 2βz2 zy, yz xz, zx y2 xy, yx x2Table 3.6: Weights for U ⊂W ⊂ VCase I. 4α+2β: The minimal weight is 4α+2β = 0+α+α+β+β+2α. Thenthree defining equations are xy, yx, x2. This algebra has detM = detN = 0and the growth is (6,14,31,70,157, . . .).Case II. 5α + 2β: 4α + 4β − (5α + 2β) = 2β − α > 0. So w(2) = 5α + 2βis 2-unstable. 5α + 2β = 0 + α + α + β + (α + β) + 2α. Then three definingequations aref1 =x2f2 =a1xy + b1yxf3 =a2xy + b2yx + c2y2 + d2xz + e2zx .And we haveM = ⎛⎜⎝x 0 0b1y a1x 0b2y + e2z a2x + c2y d2x⎞⎟⎠anddetM = a1d2x3.N = ⎛⎜⎝x 0 0a1y b1x 0a2y + d2z b2x + c2y e2x⎞⎟⎠tanddetN = b1e2x3.Γ has two components{(0, y, z) × (0,0,1)} ∪ {(0,0,1) × (0, y, z)} .The generic ones has growth (6,11,21,40,76 . . .).Case III. 6α + 2β: 4α+ 4β − (6α+ 2β) = 2β − 2α > 0. So w(2) = 6α + 2βis 2-unstable. We have that 6α+2β = 0+α+α+ (α+β)+ (α+β)+2α. Thenthree defining equations aref1 =x2f2 =a1xy + b1yx + c1y2 + d1xz + e1zxf3 =a2xy + b2yx + c2y2 + d2xz + e2zx .1403.7. Unstable algebrasThen we haveM = ⎛⎜⎝x 0 0b1y + e1z a1x + c1y d1xb2y + e2z a2x + c2y d2x⎞⎟⎠anddetM = x2 ((a1d2 − d1a2)x + (c1d2 − d1c2)y) .N = ⎛⎜⎝x 0 0a1y + d1z b1x + c1y e1xa2y + d2z b2x + c2y e2x⎞⎟⎠tanddetN = x2 ((b1e2 − e1b2)x + (c1e2 − e1c2)y) .The generic one has its Γ as{(0, y, z) × (0,0,1)} ∪ {(0,0,1) × (0, y, z)} .The generic one has growth (6,10,17, . . .).So far it was of the form nα + 2β. And nα + 3β. 2α + 3β is not possible.Case IV. 3α + 3β: We have that 3α + 3β = 0 + α + α + β + β + (α + β).Then three defining equations aref1 =x2f2 =a1xy + b1yxf3 =a2xy + b2yx + c2y2 + d2xz + e2zxwhere d2 = e2 = 0 when 2α > β. This algebra is identical to the case II when2α ≤ β. If 2α > β then detM = detN = 0 and the generic one has growth(6,13,28,60,129, . . .).Case V. 4α + 3β: We have that 4α + 3β = 0 +α +α + β + 2α + 2β. Thenthree defining equations aref1 =a1xy + b1yx + c1x2f2 =a2xy + b2yx + c2x2f3 =d3xz + e3zx + i3y2 + a3xy + b3yx + c3x2 .And we haveM = ⎛⎜⎝c1x + b1y a1x 0c2x + b2y a2x 0c3x + b3y + e3z a3x + i3y d3x⎞⎟⎠1413.7. Unstable algebrasdetM = −d3x2 ((a1c2 − c1a2)x + (a1b2 − b1a2)y) .N = ⎛⎜⎝c1x + a1y b1x 0c2x + a2y b2x 0c3x + a3y + d3z b3x + i3y e3x⎞⎟⎠tanddetN = −e3x2 ((b1c2 − c1b2)x + (b1a2 − a1b2)y) .The Γ of the generic one is{(0, y, z) × (0,0,1)} ∪ {(0,0,1) × (0, y, z)} .The generic one has growth (6,10,17, . . .).Case VI. 5α + 3β: 4α + 4β − (5α + 3β) = β −α > 0. So w(2) = 5α + 3β is2-unstable. We have that 5α + 3β = α +α + β + β + 2α + (α + β). Then threedefining equations aref1 =x2f2 =a2xy + b2yxf3 =d3xz + e3zx + i3y2 + j3yz + k3zy + z2 + a3xy + b3yx .Then we haveM = ⎛⎜⎝x 0 0b2y a2x 0b3y + e3z a3x + i3y + k3z d3x + j3y + z⎞⎟⎠detM = a2x2 (d3x + j3y + z) .N = ⎛⎜⎝x 0 0a2y b2x 0a3y + d3z b3x + i3y + j3z e3x + k3y + z⎞⎟⎠tanddetN = b2x2 (e3x + k3y + z) .The Γ of the generic one has three componentsΓ1 = {(0, y, z) × (0,−(j3y + z), i3y + k3z)} = {(0,−(k3y + z), i3y + j3z) × (0, y, z)}Γ2 = {(x, y,−d3x − j3y) × (0,0,1)}Γ3 = {(0,0,1) × (x, y,−e3x − k3y)}The generic one has growth (6,11,19, . . .).1423.7. Unstable algebrasCase VII. 6α + 3β: When 2α < β we have that 6α + 3β < 4α + 4β. Sow(2) = 6α + 3β is 2-unstable. 6α + 3β = α + α + β + 2α + (α + β) + (α + β).Then three defining equations aref1 =x2f2 =a2xy + b2yx + d2xz + e2zxf3 =d3xz + e3zx + i3y2 + j3yz + k3zy + z2 + a3xy + b3yx .This is the Subcase II-a of only U .So far it was of the form nα + 3β. It is easy to see that nα + 4β. α + 4βis not possible.Case VIII. 2α+4β: We have that 2α+4β = 0+α+α+β +β +2β. Thenwe have following defining equationsf1 =a1xy + b1yx + c1x2f2 =a2xy + b2yx + c2x2f3 =d3xz + e3zx + i3y2 + a3xy + b3yx + c3x2 .This is the case V above. If 2β > α then d3 = e3 = 0 and detM = detN = 0as the relations have no z. Its growth is (6,13,28,60,129 . . .).Case IX. 3α+ 4β: We have that 3α+ 4β = 0+α+β +β + 2α+ 2β. Thendefining equations aref1 =a1xy + b1yx + c1x2f2 =a2xy + b2yx + c2x2f3 =j3yz + k3zy + d3xz + e3zx + i3y2 + a3xy + b3yx + c3x2 .Then we haveM = ⎛⎜⎝c1x + b1y a1x 0c2x + b2y a2x 0c3x + b3y + e3z a3x + i3y + k3z d3x + j3y⎞⎟⎠detM = −x(d3x + j3y) ((a1c2 − c1a2)x + (a1b2 − b1a2)y) .N = ⎛⎜⎝c1x + a1y b1x 0c2x + a2y b2x 0c3x + a3y + d3z b3x + i3y + j3z e3x + k3y⎞⎟⎠tanddetN = −x(e3x + k3y) ((b1c2 − c1b2)x + (b1a2 − a1b2)y) .1433.7. Unstable algebrasThe Γ of the generic one has four componentsΓ1 = {(0, y, z) × (0,−j3y, i3y + k3z)} = {(0,−k3y, i3y + j3z) × (0, y, z)}Γ2 = ⎧⎪⎪⎨⎪⎪⎩(x, a1c2 − c1a2a1b1 − a1b2x, z)× (−a1x, c1x + b1y, a1x(c3x + b3y + e3z) − (c1x + b1y)(a3x + i3y + k3z)d3x + j3y )⎫⎪⎪⎬⎪⎪⎭Γ3 = {(j3x,−d3x, z) × (0,0,1)}Γ4 = {(0,0,1) × (k3x,−e3x, z)} .The generic one has growth (6,10,16, . . .).So far it was of the form nα + 4β. Now nα + 5β. α + 5β is not possible.Case X. 2α + 5β: 4α + 4β − (2α + 5β) = −β + 2α > 0 if 2α > β. Sow(2) = 2α + 5β is 2-unstable when 2α > β. We have that 2α + 5β = 0 + α +β + β + (α + β) + 2β. Then three defining equations aref1 =a1xy + b1yx + c1x2f2 =y2 + a2xy + b2yx + c2x2f3 =d3xz + e3zx + i3y2 + j3yz + k3zy + a3xy + b3yx + c3x2 .Then we haveM = ⎛⎜⎝c1x + b1y a1x 0c2x + b2y a2x + y 0c3x + b3y + e3z a3x + i3y + k3z d3x + j3y⎞⎟⎠detM = (d3x + j3y) ((c1a2 − a1c2)x2 + (b1a2 − a1b2 + c1)xy + b1y2)N = ⎛⎜⎝c1x + a1y b1x 0c2x + a2y b2x + y 0c3x + a3y + d3z b3x + i3y + j3z e3x + k3y⎞⎟⎠tanddetN = (e3x + k3y) ((c1b2 − b1c2)x2 + (a1b2 − b1a2 + c1)xy + a1y2) .The Γ of the generic one has three componentsΓ1 = degree(2,2)Γ2 = {(x,−d3j3x, z) × (0,0,1)}Γ3 = {(0,0,1) × (x,−e3k3x, z)} .1443.7. Unstable algebrasThe Γ1 is a graph of conic parts of detM and detN given byσΓ1(x, y, z) = (−a1x, c1x + b1y, a1x(c3x + b3y + e3z) − (c1x + b1y)(a3x + i3y + k3z)d3x + j3y ) .The generic one has growth (6,9,14, . . .).So far it was of the form nα+ 5β because 4α+ 4β − (3α+ 5β) = α−β < 0.Now we consider nα+6β. α+6β can not be achieved by adding weights of 6monomials. And 4α+4β−(2α+6β) = 2(α−β) < 0 so 2α+6β is not unstable.So there is no 2-unstable algebras of weight nα + 6β. This completes theexploration of generic 2-unstable algebras by test configuration and w(2).145Bibliography[1] T. Abdelgadir, S. Okawa, K. Ueda, Compact moduli of noncom-mutative projective planes, arXiv preprint arXiv:1411.7770 (2014).[2] M. Artebani, I. Dolgachev, The Hesse pencil of plane cubic curves,LEnseign. Math. 55 (2009), 235 - 273.[3] M. Artin and W. F. Schelter, Graded algebras of global dimension3, Adv. in Math., 66 (1987), no. 2, 171 - 216.[4] M. Artin, J. Tate, and M. Van den Bergh, Some algebras asso-ciated to automorphisms of elliptic curves, In The GrothendieckFestschrift, Vol. I, volume 86 of Progr. Math., pages 33 - 85.Birkhauser Boston, Boston, MA, 1990.[5] K. Behrend, I. Ciocan-Fontanine, J. Hwang, and M. Rose, Thederived moduli space of stable sheaves, Algebra Number Theory, 8(2014), no. 4, 781 - 812.[6] K. Behrend and B. 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Smith, Degenerate 3-dimensional Sklyanin algebras are mono-mial algebras, J. Algebra, 358 (2012), 74 - 86.[14] S.P. Smith, equivalences involving graded modules over path alge-bras of quivers, Adv. Math. 230 (2012), 1780 - 1810.[15] J.J. Zhang, Twisted graded algebras and equivalences of graded cat-egories, Proc. Lond. Math. Soc., 72 (1996), 281 - 311.[16] C. Walton, Degenerate Sklyanin algebras and generalized twistedhomogeneous coordinate rings, Doctoral dissertation, University ofMichigan (2011).147"@en ;
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