This paper classifies connected s-homogeneous algebras with two relations using s-homogeneous triples, providing criteria for s-Koszulity and describing their Ext-algebras, thus advancing understanding of their structure and properties.
Contribution
It offers a classification of such algebras via pairs (A,M), characterizes their s-Koszulity, and describes their Ext-algebras, extending previous work with new criteria and realizations.
Findings
01
Classification of pairs (A,M) for these algebras
02
Criteria for checking s-Koszulity via Hilbert series or Koszul complex
03
Description of Ext-algebras for most such algebras
Abstract
In our preceding paper we have introduced the notion of an s-homogeneous triple. In this paper we use this technique to study connected s-homogeneous algebras with two relations. For such algebras, we describe all possible pairs (A,M), where A is the s-Veronese ring and M is the (s,1)-Veronese bimodule of the s-homogeneous dual algebra. For each such a pair we give an intrinsic characterization of algebras corresponding to it. Due to results of our previous work many pairs determine the algebra uniquely up to isomorphism. Using our partial classification, we show that, to check the s-Koszulity of a connected s-homogeneous algebras with two relations, it is enough to verify an equality for Hilbert series or to check the exactness of the generalized Koszul complex in the second term. For each pair (A,M) not belonging to one specific series of pairs, we check if there…
\begin{array}[]{c}(A,M,\varphi)\mbox{ is an $s$-homogeneous triple},\\
A={\bf k}\langle x,y\rangle/H,\mbox{ for some quadratic ideal $H$, with }A\not\cong A^{6},\mbox{ and }A\not\cong A^{7}(0)\end{array}
\begin{array}[]{c}(A,M,\varphi)\mbox{ is an $s$-homogeneous triple},\\
A={\bf k}\langle x,y\rangle/H,\mbox{ for some quadratic ideal $H$, with }A\not\cong A^{6},\mbox{ and }A\not\cong A^{7}(0)\end{array}
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TopicsCommutative Algebra and Its Applications · Algebraic structures and combinatorial models · Advanced Topics in Algebra
Full text
Homogeneous triples for homogeneous algebras with two relations.
Eduardo do Nascimento Marcos, Yury Volkov
111The first named author was supported by the tematic project of Fapesp number 2014/09310-5.
The second named author was supported by the RFBR research project number 18-31-20004
and had a trip to São Paulo paid by the Fapesp process number 2018/07046-0 during working on this paper.
Abstract
In our preceding paper we have introduced the notion of an s-homogeneous triple. In this paper we use this technique to study connected s-homogeneous algebras with two relations.
For such algebras, we describe all possible pairs (A,M), where A is the s-Veronese ring and M is the (s,1)-Veronese bimodule of the s-homogeneous dual algebra.
For each such a pair we give an intrinsic characterization of algebras corresponding to it. Due to results of our previous work many pairs determine the algebra uniquely up to isomorphism.
Using our partial classification, we show that, to check the s-Koszulity of a connected s-homogeneous algebras with two relations, it is enough to verify an equality for Hilbert series or to check the exactness of the generalized Koszul complex in the second term.
For each pair (A,M) not belonging to one specific series of pairs, we check if there exists an s-Koszulity algebra corresponding to it.
Thus, we describe a class of possible Ext-algebras of s-Koszul connected algebras with two relations and realize all of them except a finite number of specific algebras as Ext-algebras.
Another result that follows from our classification is that an s-homogeneous algebra with two dimensional s-th component cannot be s-Koszul for s>2.
1 Introduction
All the algebras under consideration are graded algebras of the form Λ=TkV/I, where k is an algebraically closed field, TkV is the tensor algebra of some finite dimensional space V
with the standard grading, and I is an ideal generated by some subspace W⊂TkV such that all the elements of W are homogeneous and the same degree.
The case dimkW=1 was studied in [10] (see also [6, 21]). This paper deals with the case dimkW=2 whose consideration was initiated in [21].
For this purpose we will use the technique of s-homogeneous triples introduced in the last mentioned paper.
The Ext-algebra of the algebra Λ as above is the graded algebra ⨁i≥0ExtΛi(k,k).
The notion of a Koszul algebra was introduced by S. Priddy in [27]. All Koszul algebras are quadratic and they appear in
pairs. The Ext-algebra of a Koszul algebra Λ, is another time a Koszul algebra and its Ext-algebra is the original
one. If we know that an algebra Γ is the Ext-algebra of a Koszul algebra Λ then one may
recuperate Λ from Γ by taking its quadratic dual. Note that even if a quadratic algebra is not Koszul, then it can be recovered from its quadratic dual algebra that in the non Koszul case is not anymore isomorphic to the Ext-algebra. Note also that if we have a quadratic algebra, then it is an Ext-algebra of some quadratic algebra if and only if it is Koszul.
The notion of a Koszul algebra was generalized to the s-homogeneous case by Berger in [5] for connected algebras.
Later the definition was rewritten for the case of an arbitrary quiver in [14].
This notion turned out to be important.
For example, it was shown in [7] that if Λ is an algebra defined by a homogeneous potential of degree s+1 and dimkW=dimkV, then Λ is 3-Calabi-Yau if and only if it is s-Koszul and \dim_{{\bf k}}\big{(}W\otimes_{{\bf k}}V\cap V\otimes_{{\bf k}}W\big{)}=1.
Calabi-Yau algebras were introduced in [12] while it is difficult to say where their twisted generalization that we will consider in the current paper was introduced. Twisted Calabi-Yau algebras were considered in many papers, see [7, 9, 13, 17, 19, 28]. The connections between Calabi-Yau algebras and algebras defined by a potential was noted in [12], where it was proved that 3-Calabi-Yau algebras that appear in some sense in a natural way are algebras defined by a potential. It was proved in [8] that any graded 3-Calabi-Yau algebra is defined by a potential. Recently, quadratic 3-Calabi-Yau algebras with three generators and cubic 3-Calabi-Yau algebras with two generators were classified in [22, 23]. All the algebras with the same number of generators and relations defined by a twisted potential were classified in [16]. Twisted potentials that will appear in this paper were considered in [9]. It is proved in [28] that twisted 3-Calabi-Yau algebras are exactly Artin-Schelter regular algebras of dimension 3. These algebras were introduced in [1] and were studied in many works since then. In particular, it was shown in [18] that such algebras are always domains.
In this paper an element of the free algebra TkV will be called a polynomial.
The main objective of our work [21] was to give a method of recovering an s-Koszul algebra from its Ext-algebra or, more generally, recover an s-homogeneous algebra from the associated s-homogeneous triple. The last notion was introduced in the same paper and gives the Ext-algebra with its A∞-structure in the s-Koszul case.
Using these ideas, we have shown in [21] that the algebra Λ=k⟨x1,…,xm⟩/(f), where f is a polynomial of degree s, is not s-Koszul if and only if f=gs for any linear polynomial g and there are some polynomial g1, g2 and h of nonzero degree such that f=g1h=hg2. We have also shown that, in the case s>2, the algebra Λ! is s-Koszul only in the case m=1 (i.e. in the case Λ=k[x]/(xs)). Moreover if Λ=k⟨x1,…,xm⟩/(f) is s-Koszul and f is not of the form gs for some linear polynomial g then the global dimension of Λ is two. If Λ=k⟨x1,…,xm⟩/(gs) with g a linear polynomial then Λ has infinite global dimension. All these results in fact can be deduced from [10], where the Poincaré series of algebras with one relation were computed and, in particular, it was shown that the global dimension of Λ is infinite in the not s-Koszul case.
In the current paper we obtain analogous results for an algebra Λ of the form Λ=k⟨x1,…,xm⟩/(f1,f2), where f1 and f2 are linearly independent polynomials of degree s. The study of such algebras was initiated in [21], where it was shown that the s-Veronese ring of Λ! is a quadratic algebra with two generators and at least two relations and the algebras Λ, for which this ring is isomorphic to k⟨x,y⟩/(xy,yx) or k⟨x,y⟩/(x2,y2) were classified. In the last mentioned cases Λ is s-Koszul of infinite global dimension and Λ! is not s-Koszul except the case s=2.
Here we will show that the last fact is not accidental and Λ! is not s-Koszul if s>2 and Λ is an s-homogeneous algebra defined by two relations (see Theorem 3.5.1). Our main result is the classification of pairs \big{(}(\Lambda^{!})^{(s)}),(\Lambda^{!})^{(s,1)}\big{)} that appear for s-homogeneous algebras Λ defined by two relations. Here Λ(s) and Λ(s,1) denote the s-Veronese ring and the (s,1)-Veronese bimodule of the graded algebra Λ.
For the remaining pairs we give an intrinsic description of s-homogeneous algebras corresponding to them. Also we discuss the Koszul property of the obtained s-homogeneous algebras.
Some of such pairs determine the corresponding algebra up to isomorphism and we give descriptions of corresponding algebras in such cases.
For the convenience of the reader we collect all our classification results in the last section, where we give also some interesting consequences and ask questions related to the subject. Thus, if one is interested in the final form of our results, he can go directly to Subsection 5.1.
One of the results we obtain here is the description of all possible Ext-algebras of s-Koszul algebras. Nevertheless for each s there is a finite number s-homogeneous algebras whose s-Koszulity we will not check in this paper. Each of these algebras has a pair \big{(}(\Lambda^{!})^{(s)}),(\Lambda^{!})^{(s,1)}\big{)} that determines the original algebra up to isomorphism, and thus we are not able to check if such a pair can be realized as an Ext-algebra.
During our manipulations, we will give many interesting examples of s-Koszul algebras. In particular, these examples include many 3-Calabi-Yau and twisted 3-Calabi-Yau algebras algebras defined by a twisted potential.
Note that the classification of Koszul algebras with two relations is presented in [31, Theorem 3.10.6] (see also [2, 30]) and essentially we make a big step towards the analogous classification in the case where s>2. Thus, though major part of our results can be applied to study Koszul algebras too, we will consider only the case s>2.
2 Preliminary results
In this section we recall some definitions and results concerning s-homogeneous algebras, s-homogeneous triples, s-Koszul algebras, 3-Calabi-Yau algebras and algebras defined by a potential, give the classification of quadratic algebras with two generators and recall the classification of indecomposable modules over them. In the last part of this section we describe the general ideas that we will use.
2.1 s-homogeneous algebras and s-homogeneous triples
We fix some notation during the paper. First of all, we fix some algebraically closed ground field k. All the algebras and vector spaces in this paper are over k. So we write simply ⊗ instead of ⊗k. If V is a graded space, then Vi denotes the subspace of V formed by the elements of degree i. Also V[t] denotes the shift of V by t, i.e. the graded space that coincides with V as a nongraded space and has the grading defined by the equality V[t]i=Vi+t. All the modules in this paper are right modules.
In fact, we will need often to deal with bimodules and with left and right module structures on them, but to talk about the left structure we will simply consider a module over the opposite algebra and will add op to the notation of the algebra. On the other hand, if M is an A-bimodule, then we will write MA to emphasize that we are talking about its right structure and AM for the left structure.
If A is an algebra, M is an A-module, and X is a subset of M, then ⟨X⟩A denotes the A-submodule of M generated by the set X. All graded algebras will be nonnegatively graded by Z.
All the modules that we will consider in this paper will be finitely generated and we will often assume this condition without mentioning.
We will use the standard grading on k⟨x1,…,xm⟩ defined by the equality deg(xi)=1 (1⩽i⩽m). Throughout the paper we consider the algebra Λ=k⟨x1,…,xm⟩/I, where I is some ideal of k⟨x1,…,xm⟩ generated by polynomials of the fixed degree s>2. Such an algebra is called an s-homogeneous connected algebra.
Let x1∗,…,xm∗ be the basis of Λ1∗=Homk(Λ1,k) dual to the basis
x1,…,xm of Λ1. Then the s-homogeneous dual of Λ is the algebra Λ!=k⟨x1∗,…,xm∗⟩/I!, where I! is generated by such f∈k⟨x1∗,…,xm∗⟩s≅k⟨x1,…,xm⟩s∗ that f(Is)=0.
Here we use the grading on k⟨x1∗,…,xm∗⟩ defined by the equalities deg(xi∗)=1 (1⩽i⩽m). Note that the algebras Λ and Λ! are graded in the natural way and that Λ! is an s-homogeneous connected algebra. Moreover, (Λ!)!≅Λ.
Definition 2.1.1**.**
Given a graded algebra A, we denote by A(r) the r-Veronese ring of A, i.e. the graded algebra A(r)=⊕i⩾0Ai(r), where Ai(r)=Ari and the multiplication of A(r) is induced by the multiplication of A.
Also we define the (r,t)-Veronese bimoduleA(r,t) as a graded A(r)-bimodule A(r,t)=⊕i⩾0Ai(r,t), where Ai(r,t)=Ari+t and the A(r)-bimodule structure on A(r,t) is induced by the multiplication of A.
Note that the multiplication of A induces an A(r)-bimodule homomorphism from (A(r,1))⊗A(r)r to A(r). We denote this homomorphism by ϕA(r).
**
Given a graded algebra A, a graded A-module M is called linear up to the n-th degree if there exists a projective resolution of M in the category of graded A-modules
[TABLE]
such that Pi is generated in degree i, i.e. Pi=(Pi)iA, for 0⩽i⩽n. We will say that M is linearly presented if it is linear up to the first degree.
We will denote by J(A) the ideal ⊕i>0Ai of A.
We define next the category of s-homogeneous triples, which will be useful for us.
Definition 2.1.2**.**
An s-homogeneous triple is a triple (A,M,φ), where A is a quadratic k-algebra, M is a graded A-bimodule linearly presented over A and over Aop, and φ:M⊗As→A[1] is a homomorphism of graded A-bimodules such that
Imφ=J(A)[1];
2. 2.
1M⊗Aφ=φ⊗A1M:M⊗As+1→M[1];
3. 3.
Ker(1M⊗Aφ)=Kerφ⊗AM+M⊗AKerφ;
4. 4.
Ker(φ⊗Aφ)=i=0∑sM⊗Ai⊗AKerφ⊗AM⊗As−i.
Let (A,M,φ) and (B,L,ψ) be s-homogeneous triples. A morphism from (A,M,φ) to (B,L,ψ) is a pair (f,g), where f:A→B is a morphism of graded algebras and g:M→L is a morphism of graded A-bimodules such that fφ=ψg⊗s. Here the A-bimodule structure on L is induced by the map f.
Note that if (A,M,φ) is an s-homogeneous triple and n⩾s is an integer, then the map 1M⊗Ai⊗Aφ⊗A1M⊗An−s−i:M⊗An→M⊗An−s[1] does not depend on 0⩽i⩽n−s due to the second item of Definition 2.1.2. For simplicity we denote this map by φ too. So, if n⩾ks, then we have a graded A-bimodule homomorphism φk:M⊗An→M⊗An−ks[s]. As usually, M⊗A0=A everywhere.
It was shown in [21] that the category of s-homogeneous triples is equivalent to the category of s-homogeneous algebras.
The corresponding equivalences F and G can be described on objects in the following way.
If Λ is an s-homogeneous algebra, then F(Λ)=(Λ(s),Λ(s,1),ϕΛ(s)).
Given an s-homogeneous triple (A,M,φ), one has {\mathcal{G}}(A,M,\varphi)=T_{\bf k}M_{0}/\big{(}(\operatorname{{Ker}}\varphi)_{0}\big{)}. Note also that G(A,M,φ)!=TkM0∗/(φ∗A1∗).
The main ingredient of our approach, which we state in the following lemma, is [21, Corollary 5.3].
Lemma 2.1.3**.**
If (A,M,φ) is an s-homogeneous triple, then there exists a graded A-bimodule S concentrated in degree [math] and an isomorphism of graded A-bimodules θ:M⊗As≅S⊕J(A)[1] such that φ equals to the composition
[TABLE]
where the second map is the canonical projection on the second summand and the third arrow is the canonical inclusion.
In particular, it can be deduced (see [21, Corollary 5.4]) that if the A-bimodule J(A)[1] does not contain nonzero direct summands concentrated in degree [math], then the triple
(A,M,φ) (and hence G(A,M,φ) too) can be recovered from the pair (A,M) modulo isomorphism.
From here on and until the end of the paper, (A,M,φ)=F(Λ!).
2.2 s-Koszulity and 3-Calabi-Yau property
In this subsection we recall and discuss the definitions of an s-Koszul algebra, of a (twisted) 3-Calabi-Yau algebra and of an algebra defined by a (twisted) potential.
Definition 2.2.1**.**
The s-homogeneous algebra Λ is called s-Koszul if ExtΛi(k,k) is concentrated in degree −χs(i), where
[TABLE]
The 2-Koszul algebras are called simply Koszul algebras.
**
If s=2 and Λ is Koszul, then ExtΛ∗(k,k)≅(Λ!)op as a graded algebra. For s>2 the situation changes.
It is proved in [14] that if Λ is s-Koszul, then A is a Koszul algebra and M is linear over A and over Aop.
If Λ is s-Koszul and s>2, then ExtΛ∗(k,k)≅A⋉M as an algebra, where A⋉M is the trivial extension of A by M, i.e. its underlying space is A⊕M and the multiplication is given by the equality (a,x)(b,y)=(ab,ay+xb) for a,b∈A and x,y∈M. If we define the grading on A⋉M by the equalities (A⋉M)2n=An and (A⋉M)2n+1=Mn for n⩾0, then the isomorphism above will become degree preserving.
Let, as usually, Λ be an s-homogeneous algebra and F(Λ!)=(A,M,φ). Let us introduce the projective Λ-modules K2n=An∗⊗Λ and K2n+1=Mn∗⊗Λ for n⩾0.
Note that the map An⊗M0→Mn coming from the left A-module structure on M induces the map Mn∗→An∗⊗M0∗. Identifying of M0∗ with Λ1 and applying the multiplication Λ1⊗Λ→Λ we get the map d2n:K2n+1→K2n. Now we have also the map φ:Mn⊗M0⊗(s−1)→An+1 that induces the map An+1∗→Mn∗⊗(M0∗)⊗(s−1), and hence the map d2n+1:K2n+2→K2n+1. Then we get the complex of projective Λ-modules ⋯d2K2d1K1d0K0 with the augmentation map μ:K0=A0⊗Λ≅Λ→k induced by the factorization by J(Λ). The complex (K,d) is called the generalized Koszul complex of Λ.
It was noted in [21] that our definition of the generalized Koszul complex is equivalent to the one given in [5].
It is proved in [5] that Λ is s-Koszul if and only if K is exact in positive degrees. Let us recall that the Hilbert series of a graded space U=⊕i⩾0Ui is HU(t)=i⩾0∑(dimkUi)ti∈Z[[t]]. Calculating the Euler characteristic of K one gets immediately that
[TABLE]
in the case where Λ is s-Koszul. The opposite is not true in general even in the case s=2.
It is shown in [5] that the exactness of K implies the so-called extra condition
[TABLE]
where both sides of the equality are subspaces of (M0∗)⊗(2s−1).
In fact, this condition holds if and only if (Kerd1)i⊂Imd2 for i<2s.
Let us introduce the s-homogeneous triple (B,L,ψ)=F(Λ).
We denote by O(tn) the ideal of Z[[t]] generated by tn.
It follows from [21, Corollary 7.4] that Λ satisfies (2.2.3) if and only if
[TABLE]
where m=dimkL0 and lk=dimk(L⊗Ak)1 for 0⩽k⩽s−1.
It is proved in [21] that if I is generated by one polynomial f, then the following result holds.
Theorem 2.2.2**.**
Suppose that Λ=k⟨x1,…,xm⟩/(f), where f is some homogeneous polynomial of degree s. Then Λ is s-Koszul if and only if one of the following two conditions holds:
f=gs* for some linear polynomial g;*
2. 2.
if f=gh1=h2g for some polynomials g, h1 and h2, then degg∈{0,degf}.
Moreover, if s>2, then Λ! is s-Koszul if and only if m=1.
In the last theorem the case f=gs corresponds to the case A=k[x].
If f=gs for any linear polynomial g, then A=k[x]/(x2) and M=km. In particular, the s-Koszulity in the last mentioned case is equivalent to the equality gldimΛ=2.
Moreover, if Λ is not s-Koszul, then it does not satisfy the extra condition. The last assertion is a consequence of [11] that is mentioned in [6].
In fact, all of these facts can be obtained as consequences of the results of [10].
In this paper we will consider the case where Λ=k⟨x1,…,xm⟩/(f1,f2) for some linearly independent homogeneous polynomials f1 and f2 of degree s. It is shown in [21] that in this case A=k⟨x,y⟩/H for some quadratic ideal H such that dimkH2⩾2. The consideration of the cases H=(x2,y2) and H=(xy,yx) is presented in [21, Theorem 8.4]. Let us recall this result.
Theorem 2.2.3**.**
*Suppose that Λ=TkV/(f1,f2), where f1 and f2 are two linearly independent elements of V⊗s.
If (Λ!)(s)≅k⟨x,y⟩/(xy,yx), then either*
[TABLE]
for some m⩾0 or s=2t and
[TABLE]
*for some t⩾1 and m⩾0.
If (Λ!)(s)≅k⟨x,y⟩/(x2,y2), then s=2t+1 and*
[TABLE]
for some t⩾1 and m⩾0.
All the algebras Λ mentioned in this theorem are s-Koszul. Their s-homogeneous duals are not s-Koszul and it is a particular case of Theorem 3.5.1 presented below.
Let us recall that, for a Λ-bimodule M and an automorphism θ of Λ, θM denotes the Λ-bimodule that coincides with M as a right Λ-module and has the left Λ-module structure twisted by θ, i.e. defined by the equality a⋅x=θ(a)x for a∈Λ and x∈M.
Definition 2.2.4**.**
Let ν be an automorphism of an algebra Λ. Then the algebra Λ is called ν-twisted d-Calabi-Yau if the projective dimension of Λ as a Λ-bimodule equals d, ExtΛ⊗Λopi(Λ,Λ⊗Λop)=0 for i=d, and ExtΛ⊗Λopd(Λ,Λ⊗Λop)≅νΛ as a Λ-bimodule.
We will write CY instead of Calabi-Yau for short. We will call 1Λ-twisted d-CY algebras simply d-CY algebras. The automorphism ν is defined modulo the group of inner automorphisms by Λ. It is called the Nakayama automorphism of Λ
**
This definition makes sense for any algebra Λ and any automorphism ν, but in this paper we will use it only for graded (and even s-homogeneous) algebras and graded automorphisms.
Moreover, we will consider only the 3-CY property. In this case algebras defined by twisted potentials play a significant role. Let us recall the definition of these algebras and their connections with CY algebras.
In fact, we will define next a homogeneous twisted potential, but we omit the term homogeneous because we are not going to consider non homogeneous potentials in this paper.
Definition 2.2.5**.**
Let V be a finite dimensional vector space and σ be some automorphism of V. An element w∈V⊗N is called σ-twisted potential of degree N if ϕσ(w)=w, where ϕσ:V⊗N→V⊗N
acts on the tensor v1⊗⋯⊗vN∈V⊗N by the rule ϕσ(v1⊗⋯⊗vN)=σ(vN)⊗v1⊗⋯⊗vN−1. As before, we call 1V-twisted potentials simply potentials. The algebra defined by the σ-twisted potential w is by definition the algebra D(V,w)=TkV/I, where I is the ideal generated by the subspace R=\{\big{(}f\otimes 1^{\otimes(N-1)}\big{)}(w)\mid f\in V^{*}\}\subset V^{\otimes(N-1)}. We will write simply D(w) instead of D(V,w), because V can be always recovered from the context. It is clear from the definition that D(w) is an (N−1)-homogeneous algebra. Note also that σ:V→V induces an automorphism of the algebra D(w).
**
Remark 2.2.6**.**
Sometimes (for example, in [12]) a homogeneous potential is defined as a homogeneous element of the space TkV/[TkV,TkV]. This definition gives the same definition of an algebra defined by a potential in the case of zero characteristic. In the case of an arbitrary characteristic considered here, our definition is more appropriate.
It is shown in [9] that any graded ν-twisted 3-CY algebra is s-Koszul for some s and is isomorphic to an algebra defined by some ν-twisted potential of degree s+1.
Here ν denotes the automorphism ν:Λ→Λ and at the same time the induced automorphism of Λ1.
Let us recall one more definition.
Definition 2.2.7**.**
Let Λ=k⟨x1,…,xm⟩/I and Γ=k⟨y1,…,yk⟩/H be two s-homogeneous connected algebras. Their free product is by definition the algebra
[TABLE]
where I and H are included in k⟨x1,…,xm,y1,…,yk⟩ in the obvious way.
**
Example 2.2.8**.**
Let us consider the potentials w1=y1s+1+y2s+1 and w2=(y1y2)t+1+(y2y1)t+1 on k⟨y1,y2⟩s+1 (where s=2t+1 in the case of the second potential).
It is easy to see that
[TABLE]
are the algebras from Theorem 2.2.3. Both of them are s-Koszul, but obviously not 3-CY.
At the same time, w3=(y1y2)ty1+(y2y1)ty2∈k⟨y1,y2⟩s+1 (where s=2t) is a ν-twisted potential for ν interchanges y1 and y2. Then {\bf k}\langle x_{1},\dots,x_{m}\rangle*{\bf D}(w_{3})={\bf k}\langle x_{1},\dots,x_{m},y_{1},y_{2}\rangle/\big{(}(y_{1}y_{2})^{t},(y_{2}y_{1})^{t}\big{)} is the last algebra from Theorem 2.2.3 which is s-Koszul, but not ν-twisted 3-CY. Further we will meet (twisted) 3-CY algebras too (see Subsection 4.2).
2.3 Classification of quadratic algebras with two generators
We have not found an explicit classification of quadratic algebras with two generators modulo isomorphism. By this reason we present here this classification with a proof for completeness and convenience.
Lemma 2.3.1**.**
*Let A=k⟨x,y⟩/H, where H is generated by H2.
1. If dimkH2=0, then A=k⟨x,y⟩ is the free algebra on two generators.*
2.* If dimkH2=1, then A is isomorphic to one of the algebras*
•
Aˉ1=k⟨x,y⟩/(x2);
•
Aˉ2(q)=k⟨x,y⟩/(xy+qyx)(q∈k);
•
Aˉ3=k⟨x,y⟩/(x2+xy−yx).
We have the isomorphism Aˉ2(q)≅Aˉ2(q−1) for q=0,1,−1, all the other algebras from the list are pairwise non-isomorphic.
3.* If dimkH2=2, then A is isomorphic to one of the algebras*
•
A4=k⟨x,y⟩/(xy,yx);
•
A5=k⟨x,y⟩/(x2,y2);
•
A6=k⟨x,y⟩/(x2,yx);
•
A7(q)=k⟨x,y⟩/(x2,xy−qyx)(q∈k);
•
A8=k⟨x,y⟩/(xy,x2−y2);
•
A9(q)=k⟨x,y⟩/(xy,x2−yx−qy2)(q∈k).**
All of these algebras are pairwise non-isomorphic.
4.* If dimkH2=3, then A is isomorphic to one of the algebras*
•
A1=k⟨x,y⟩/(x2,xy,yx);
•
A2(q)=k⟨x,y⟩/(x2,y2,xy−qyx)(q∈k);
•
A3=k⟨x,y⟩/(x2,xy+yx,xy+y2).**
Estanislao Herscovich We have the isomorphism A2(q)≅A2(q−1) for q=0,1,−1, all the other algebras from the list are pairwise non isomorphic.
5.* If dimkH2=4, then A=k⟨x,y⟩/(x,y)2=A0.*
Proof.
Items 1 and 5 are clear. Note that item 4 follows from item 2 because (A1)!≅Aˉ1, A2(q)!≅Aˉ2(q), and (A3)!≅Aˉ3.
Let us prove item 2. In this case A=k⟨x,y⟩/(ax2+bxy+cyx+dy2) for some nonzero vector (a,b,c,d)∈k4.
Note that, for any t∈k, there is an isomorphism
[TABLE]
sending x to x+ty and y to y. Since k is algebraically closed, we may assume that A=k⟨x,y⟩/(ax2+bxy+cyx).
If b+c=0, then A≅k⟨x,y⟩/(bxy+cyx) via the isomorphism sending x to x and y to y−b+cax.
Now it is easy to turn A into one of the required forms by rescaling and interchanging (in the case b=0, c=0) x and y.
Direct calculations show that among the algebras Aˉ1, Aˉ2(q) and Aˉ3 only the algebra Aˉ1 has square zero element of degree 1, only the algebra Aˉ2(0) has two linearly independent elements of degree 1 whose product is zero, and, for q∈k∗, only the algebras Aˉ2(q) and Aˉ2(q−1) have such two linearly independent elements u and v of degree 1 that uv+qvu=0.
It remains to prove item 3. We have
A=k⟨x,y⟩/(r1,r2), where
[TABLE]
for some linearly independent vectors (a1,b1,c1,d1),(a2,b2,c2,d2)∈k4. One can show that t1r1+t2r2 can be decomposed into a product of two linear polynomials for t1,t2∈k if and only if the vectors
[TABLE]
are linearly dependent. Since this condition is equivalent to the vanishing of some homogeneous quadratic polynomial in t1 and t2, we may assume that either r1=x2 or r1=xy.
Suppose that r1=x2. Then we may assume that r2=bxy+cyx+dy2 for some b,c,d∈k. If d=0, then A can be put in the form of A6 or A7(q) by rescaling. In the opposite case we may assume that d=1.
If we also have b=c, then A≅A5 via the isomorphism sending x to x and y to y+bx. If b=c, then (y+bx)(y+cx) vanishes in A and we reduce to the case r1=xy.
Suppose now that r1=xy. We may assume that r2=ax2+cyx+dy2 for some a,c,d∈k. If a=d=0, then A=A4.
If a=c=0, then A≅A6 via interchanging x and y. If a=0 and c,d=0, then changing x by y and y by cx+dy we may assume that a=0.
Now it is easy to turn A into one of the required forms by rescaling x and y.
Let us prove that all the algebras from item 3 are pairwise nonisomorphic. First of all, A4 is the unique algebra from the list that is commutative and does not have square zero elements of degree 1. Only the algebra A5 has two linearly independent square zero elements of degree 1. Direct calculations show that the algebras A8 and A9(q) do not have square zero elements while the algebras A6 and A7(q) have unique modulo scalar square zero element of degree 1. Thus, any graded isomorphism from A7(q) to B∈{A7(q′)}q′∈k∪{A6} has to send x to x and y to kx+ly for some l=0. One can check that this map is an isomorphism if and only if B=A7(q).
Finally, suppose that there is an isomorphism from B∈{A9(q′)}q′∈k∪{A8} to A9(q). Using the fact that xy=0 in B, one can show that this isomorphism has to send x and y either to k1x and k2y or to k1(x−y) and k2(x+qy) correspondingly for some k1,k2∈k∗. One can show that k1=k2 and B=A9(q) in the first case and that q=−1, k1+qk2=0, and B=A9(q) in the second case.
∎
It is not difficult to show, using the results of [4], that the algebras Ai (1⩽i⩽7) are Koszul, we also may prove this by noting that the defining ideal for these algebras have quadratic Gröbner basis, which implies that they are Koszul. Algebras whose generating ideal have quadratic Gröbner basis are called strongly Koszul, see [15]. We also note that the algebras A8 and A9 are not Koszul.
2.4 Indecomposable modules
Let A be a quadratic algebra with two generators, i.e. A=k⟨x,y⟩/H, where H is a quadratic ideal.
In this case an A-module M with a linear presentation is determined by a pair (U,T), where U is some vector space and T is a subspace of U⊗A1.
Namely, the pair (U,T) corresponds to the cokernel of the map T⊗A→U⊗A induced by the inclusion of T into U⊗A1.
For instance, if M=k=A/(x,y), then
U=k and T=k⊗(kx⊕ky).
Two modules defined by (U1,T1) and (U2,T2) are isomorphic if and only if there is an isomorphism f:U1→U2 such that (f⊗1)(T1)=T2.
Note that we do not use the relations of A in this discussion. This means that the category of linear presented modules does not depend on the defining ideal of the algebra. Therefore the modules having linear presentation over A correspond bijectively in a natural way to the modules having linear presentation over A0.
Note that A0 is a string special biserial algebra and the classification of indecomposable modules over it can be found, for example, in [3] (see also [21]).
Using the classification of indecomposable modules over the string special biserial algebra A0, we see that any indecomposable A-module is isomorphic either to a module of the form
[TABLE]
or to a module of the form
[TABLE]
For (α,β)∈k2∖{(0,0)} we choose some (α~,β~)∈k2 such that αβ~=βα~ and introduce
[TABLE]
It is not difficult to show that, for any nonzero (α,β), the module Bn(α,β) does not depend on the choice of (α~,β~) modulo isomorphism and is isomorphic to one of the modules listed above. Moreover, Bn(α,β)≅Bm(γ,δ) if and only if n=m and (α,β)=(cγ,cδ) for some c∈k∗. Thus, the classification of indecomposable A-modules can be rewritten in the following way. Any indecomposable A-module is isomorphic either to Bn(α,β) for some (α,β)∈P1 or to one of the modules
[TABLE]
There is analogous classification for left modules. Namely, we introduce
[TABLE]
and then any indecomposable Aop-module is isomorphic either to Bn′(α,β) for some (α,β)∈P1 or to one of the modules
Zn′, Wn′ (n⩾1). We remark that, given an A-module M isomorphic to B(α,β) and
(α′,β′)∈P1, there is an isomorphism of algebras A′→A, such that MA′ is isomorphic to B(α′,β′).
Now we give left and right module decompositions of J(A), where A is one of the algebras Ai (0⩽i⩽9) from Lemma 2.3.1. We collect this information in the next table.
Table. *Decompositions of J(A) for algebras with two generators.
AJ(A)[1]AAJ(A)[1]
A0Z1⊕Z1Z1′⊕Z1′
A1B1(1,0)⊕Z1B1′(1,0)⊕Z1′
A2(0)B1(0,1)⊕Z1B1′(1,0)⊕Z1′
A2(q) (q=0), A3Z2Z2′
A4, A5B1(1,0)⊕B1(0,1)B1′(1,0)⊕B1′(0,1)
A6B1(1,0)⊕B1(1,0)W1′⊕Z1′
A7(0)W1⊕Z1B1′(1,0)⊕B1′(1,0)
A7(q) (q=0)
B2(1,0)B2′(1,0)
A8B2(0,1)B2′(1,0)
A9(q)B1(1,−q)⊕B1(0,1)B1′(1,0)⊕B1′(1,1)
2.5 General ideas
Our general goal is to classify the s-homogeneous algebras whose defining ideal have two generators, and to distinguish which ones are s-Koszul.
Since we have proved that the category of s-homogenous algebras is equivalent to the category of s-homogenous triples by a specific equivalence, we will try to classify the corresponding s-homogeneous triples.
An s-homogenous triple is formed by a quadratic algebra, a graded bimodule and a morphism.
We observe that to to classify s-homogeneous algebras with two relations it is enough to classify s-homogeneous triples (A,M,φ) whose first component is a quadratic algebra with two generators.
Thus, the first step of our classification has been done above.
The second step of our approach is the classification of A-bimodules M such that the pair (A,M) can be completed to an s-homogeneous triple.
Lemma 2.1.3 giving restrictions on such a bimodule is the main ingredient of our second step.
The third step is the classification graded A-bimodule homomorphisms φ:M⊗As→A[1] such that (A,M,φ) is an s-homogeneous triple.
The nice side of this is that, for many algebras, if such a homomorphism exists, then, due to the remark after Lemma 2.1.3, it is uniquely determined by the pair (A,M), modulo isomorphism of s-homogeneous triples. In particular, φ is uniquely determined by A and M for all the algebras from Lemma 2.3.1 except A1 and A0.
The not so nice side is that if φ exists and is not uniquely determined, then usually there is no hope to classify all the maps φ. For example,
it is not done even for the algebra A=k[x]/(x2)! Thus, for A=A1 and A=A0 we will not get a full classification but only some facts about corresponding s-homogeneous algebras.
Note that due to [20, Theorem 2] if M and N are graded A-bimodules linearly presented as left and right A-modules, then M⊗AN has the same property.
Thus, the category lrLin(A) of graded A-bimodules linearly presented as left and right A-modules is a monoidal category. Note that k∈lrLin(A) and if A is a quadratic algebra, then J(A)[1]∈lrLin(A) too. Note that k⊗AM≅M⊗Ak≅kdimkM0 for any M∈lrLin(A). Thus, morphisms that can be factored though km for some m⩾0 constitute a tensor ideal T(A) in the category lrLin(A) and we can consider the tensor category lrLin+(A)=lrLin(A)/T(A).
Due to Lemma 2.1.3, if (A,M,φ) is an s-homogeneous triple, then M∈lrLin(A) satisfies the condition M⊗As≅J(A)[1] in lrLin+(A).
Note that M satisfies this condition if and only if M⊕km satisfies it for any m⩾0. Moreover, it is clear that if (A,M,φ) is an s-homogeneous triple, then (A,M⊕km,φπ⊗As) is an s-homogeneous triple, where π:M⊕km→M is the projection on the first summand.
3 First two components
Until the end of the paper we suppose that A is a quadratic algebra with two generators, i.e. A=k⟨x,y⟩/H, where H is a quadratic ideal.
This section is devoted to the description of bimodules M that can appear as the second component of an s-homogeneous triple with the first component A.
In particular, this will allow us to describe the s-homogeneous algebras corresponding to good algebras A, i.e. such that J(A)[1] does not contain direct A-bimodule summands concentrated in degree zero.
Also, using the obtained description, we will show that a connected s-homogeneous algebra with two dimensional s-th component cannot be s-Koszul if s>2.
For an A-bimodule N, a subspace X⊂N, and a an element a∈A, we introduce
[TABLE]
For u∈A1 and 1⩽t⩽s, let us introduce ru(t)=dimk(M0⊗tu) and lu(t)=dimk(uM0⊗t), where M0⊗t is considered as a subspace of M⊗At.
3.1 First restrictions on M
Here we prove the first lemma that restricts the number of left and right A-module structures that the bimodule M from an s-homogeneous triple (A,M,φ) can have.
Lemma 3.1.1**.**
Suppose that 1⩽t⩽s and (A,M,φ) is an s-homogeneous triple, where A=k⟨x,y⟩/H for some quadratic ideal H.
There are linearly independent elements u,v∈A1 such that ru(t),rv(t)⩽2.
2. 2.
If there is some Y∈M0⊗t such that dimkA1Y=1, then there exist linearly independent u,v∈A1 such that ru(t)⩽1 and rv(t)⩽2.
3. 3.
If the set of elements Y∈M0⊗t such that dimkA1Y⩽1 generate M0⊗t, then there are linearly independent elements u,v∈A1 such that ru(t),rv(t)⩽1.
Proof.
Suppose that u∈A1 can be represented in the form u=φ(X⊗Y) with X∈M0⊗(s−t) and Y∈M0⊗t. Let us set k=dimkA1Y.
Suppose that there is a set {Zi}1⩽i⩽k+1⊂M0⊗t such that the set {Ziu}1⩽i⩽k+1⊂(M⊗At)1 is linearly independent.
If there are αi∈k (1⩽i⩽k+1) not all zero such that (i=1∑k+1αiφ(Zi⊗X))Y=0, then we have
[TABLE]
which is a contradiction since {Ziu}1⩽i⩽k+1 is linearly independent. The obtained contradiction shows that ru(t)⩽dimkA1Y.
Since the map φ∣M0⊗s:M0⊗s→A1 is surjective, there are such Xu,Xv∈M0⊗(s−t) and Yu,Yv∈M0⊗t that u=φ(Xu⊗Yu) and v=φ(Xv⊗Yv) generate A1. Then we have ru(t)⩽dimkA1Yu⩽2 and rv(t)⩽dimkA1Yv⩽2.
2. 2.
Since φ(M0⊗s)Y=A1Y=0, there are such Z∈M0⊗t and X∈M0⊗(s−t) that Zφ(X⊗Y)=φ(Z⊗X)Y=0, and hence φ(X⊗Y)=0.
Then we may choose Yu=Y in the previous part of our proof and get ru(t)⩽dimkA1Y=1.
3. 3.
In this case M0⊗s is generated by elements of the form X⊗Y with X∈M0⊗(s−t), Y∈M0⊗t, and dimkA1Y⩽1. Thus, we can choose Yu and Yv above in such a way that dimkA1Yu⩽1 and dimkA1Yv⩽1.
∎
Note that we can apply Lemma 3.1.1 to the s-homogeneous triple (Aop,M,φ) and obtain the analogous assertion with ru(t) replaced by lu(t). The same argument works for all assertions that we will prove (with appropriate modifications). Thus,
whenever we prove some statement for right modules, we will propose the analogous fact for left modules.
Most of the time we will apply Lemma 3.1.1 with t=1. Let us give an immediate corollary of it.
Corollary 3.1.2**.**
In the settings of Lemma 3.1.1, dimk(M⊗At)1⩽4 and (M⊗At)A does not contain
any direct summands isomorphic to Bn(α,β)((α,β)∈P1), Zn+1 and Wn with n⩾3.
We are going to prove a series of claims that improve the assertion of Corollary 3.1.2, but first we need to introduce some notation.
The notation N⊂A⊕K means that N is an A-module direct summand of K. We will write N⊂A⊕K for the opposite assertion.
We claim that if W1⊂A⊕K⊗AN for an A-module K and an A-bimodule N, then W1⊂A⊕N.
To see the claim first observe that W1≅AA and choose some A-module epimorphism π:W1n→K. Then there is an epimorphism π⊗A1N:Nn→K⊗AN. Hence, if W1⊂A⊕K⊗AN, then there is an A-module epimorphism from Nn to W1. Since W1 is projective, W1⊂A⊕Nn, and hence W1⊂A⊕N.
This shows the claim.
Lemma 3.1.3**.**
In the settings of Lemma 3.1.1, we have W1⊂A⊕M⊗At if and only if A≅A7(0). In this case MA≅W1⊕Z1l for some l⩾1.
Proof.
By the argument above, we have W1⊂A⊕M if W1⊂A⊕M⊗Ak for some k⩾1. In particular, if A≅A7(0), then W1⊂A⊕J(A)[1]⊂A⊕M⊗As, and hence W1⊂A⊕M.
Suppose now that MA≅W1⊕K for some A-module K. Let us prove by induction that M⊂A⊕M⊗Ak for any k⩾1. This statement is obviously true for k=1. For k>1 we have the isomorphism
[TABLE]
and hence M⊂A⊕M⊗Ak follows from M⊂A⊕M⊗A(k−1).
In particular, W1⊂A⊕M⊗As≅J(A)[1]⊕S with S concentrated in degree zero by Lemma 2.1.3, and hence A≅A7(0) due to Table of decompositions presented in Subsection 2.4.
Suppose that A≅A7(0) and M≅W1⊕K. Since M⊂A⊕M⊗As in this case, we have
[TABLE]
Thus, in this case K is concentrated in degree zero.
∎
As usually, the statement dual to Lemma 3.1.3 is true too. Note that in such a statement A7(0) has to be replaced by A6.
We will continue the search of restrictions on the direct summands of M and its tensor powers, but due to Lemma 3.1.3 it is convenient to exclude the algebras A6 and A7(0) first.
3.2 s-homogeneous triples with A=A6 or A=A7(0)
In this subsection we will classify the s-homogeneous algebras Λ with two relations such that F(Λ!)=(A,M,φ) with A∈{A6,A7(0)}. Note that A7(0)=(A6)op, and hence, if F(Λ!)=(A7(0),M,φ), then {\mathcal{F}}\big{(}(\Lambda^{\rm op})^{!}\big{)}=(A^{6},M^{\rm op},\varphi^{\rm op}). Thus, one can consider the case A=A6 and then apply op to get the classification for A=A7(0). So we set A=A6 in our discussion.
Let us consider the s-homogeneous triple (A,M,φ). We have AM=W1′⊕K′ for some K′ concentrated in degree zero by Lemma 3.1.3. Then dimkM1=2, and so the right A-module structure of M can be recovered using the following lemma.
Lemma 3.2.1**.**
Suppose that (A6,M,φ) is an s-homogeneous triple. Then M0x=0.
Proof.
Note that we have a direct decomposition M1=xM0⊕yM0, where dimkxM0=dimkyM0=1. We have A1(xM0)=0 and A1(yM0)=M2, and hence, if M0x⊂xM0, then M2=A1M0x=M1x.
Let us denote by π:M⊗s↠M⊗As the canonical projection.
Then we have (M^{\otimes_{A}s})_{2}=\pi\big{(}M_{0}^{\otimes(s-1)}\otimes M_{2}\big{)}=\pi\big{(}M_{0}^{\otimes(s-1)}\otimes M_{1}x\big{)}\subset(M^{\otimes_{A}s})_{1}x=0 that is not true.
Thus, M0x⊂xM0, i.e. M0x=xM0 in the case M0x=0. On the other hand, if M0x=xM0, then xM0⊗s=M0⊗sx which is not true. Thus, M0x=0.
∎
Lemma 3.2.1 implies that MA≅B1(1,0)⊕B1(1,0)⊕K for some K concentrated in degree zero. Now we are ready to describe s-homogeneous algebras with two relations corresponding to A6 and A7(0).
Theorem 3.2.2**.**
*Suppose that Λ=TkV/(f1,f2), where f1 and f2 are two linearly independent elements of V⊗s.
If (Λ!)(s)≅k⟨x,y⟩/(x2,yx)=A6, then*
[TABLE]
*for some m⩾0.
If (Λ!)(s)≅k⟨x,y⟩/(x2,xy)=A7(0), then*
[TABLE]
for some m⩾0.
In particular, if (Λ!)(s)∈{A6,A7(0)}, then Λ is s-Koszul.
Proof.
As it was explained above, we will consider only the case where F(Λ!)≅(A6,M,φ) and then will get the result for A7(0) via the functor op. Let us set A=A6 and fix the A-module decomposition M=B1(1,0)⊕B1(1,0)⊕K with A-module K concentrated in degree zero. Our first step is to recover the A-bimodule structure on M.
First of all, note that A1K⊂lAnnM1(y)=0. Then rAnnM0(x)=rAnnM0(y) is a subspace of M0 of codimension one containing K. Let f1,f2 be such a basis of \big{(}B_{1}(1,0)\oplus B_{1}(1,0)\big{)}_{0}\subset M_{0} that A1f2=0. i.e. f2 is a complementary vector of K in rAnnM0(A1) and f1 is a complementary vector of ⟨f2⟩+K in M0. Note that f1yi,f2yi is a basis of Mi for any i⩾1. Since 0=xf1∈M1 and x(xf1)=0, it is easy to see that xf1=af2y for some a∈k∗. Now we may rescale f2 and turn a into one.
Thus, we have xf1=f2y and yf1=af1y+bf2y for some a,b∈k, a=0. Let us replace f1 by f1+abf2. Now we have xf1=f2y and yf1=af1y. Now one can show that tensors of the form v1⊗⋯⊗vs∈M0⊗s such that either vi∈K for some 1⩽i⩽s or vi=f2 for some 2⩽i⩽s generate an A-bimodule concentrated in degree zero. The subspace generated by these tensors has codimension two in M0⊗s and its complementary subspace is ⟨f1⊗s,f2⊗f1⊗(s−1)⟩. Since A1(f2⊗f1⊗(s−1))=0, we have φ(f2⊗f1⊗(s−1))=bx and φ(f1⊗s)=cx+dy for some b,d∈k∗, c∈k. Since df1y=f1φ(f1⊗s)=φ(f1⊗s)f1=adf1y+cf2y, we have a=1 and c=0. It is not difficult to show that we get an s-homogeneous triple (A6,J(A6)[1]⊕km,φ) with m=dimkK that does not depend on φ modulo isomorphism because J(A6)[1] does not contain direct A6-bimodule summands concentrated in degree zero.
Using the description of (Kerφ)0 we can conclude that \Lambda\cong{\mathcal{G}}(A^{6},J(A^{6})[1]\oplus{\bf k}^{m},\varphi)^{!}=T\big{(}(J(A^{6})_{1}\oplus{\bf k}^{m})^{*}\big{)}/\big{(}\varphi^{*}(A^{6})_{1}^{*}\big{)} has the required form.
∎
3.3 More restrictions on the one side module structure
Let us continue the study of s-homogeneous triples. Since we have fully considered s-homogeneous triples corresponding to the algebras A6 and A7(0), we will proceed our discussion in the following settings:
[TABLE]
Lemma 3.3.1**.**
In the settings (3.3.1), N⊂A⊕M for N∈{Z3,Z2⊕Z2}.
Proof.
Suppose that MA≅N⊕K for some A-module K and N∈{Z3,Z2⊕Z2}. For any nonzero w∈A1, we have
[TABLE]
Then it follows from Lemma 3.1.1 that dimkA1Y=1 for any
Y∈M0 and K0u=K0v=0 for two linearly independent u,v∈A1, i.e. K1=K0A1=0.
The first condition means that indecomposable direct summands of AM are isomorphic to Z1′ and Wn′ (n⩾1). Since dimkM1=dimkN1=2, we have AM≅W1′⊕K′ for some K′ concentrated in degree zero that contradicts Lemma 3.1.3. Thus, the required assertion is proved.
∎
Suppose that MA≅W2⊕K for some A-module K. Analogously to the proof of Lemma 3.3.1, one can show that K is concentrated in degree zero and indecomposable direct summands of AM are isomorphic to Z1′ and Wn′ (n⩾1).
Since dimkM1=dimk(W2)1, we have AM≅W2′⊕K′ for some K′ concentrated in degree zero. A direct calculation shows that dimk(W2⊗AW2′)1=4, and hence dimk(M⊗AM)1=4 too.
Thus, it follows from Lemma 3.1.1 that dimkfA1=1 for any f∈(M⊗AM)0. Then indecomposable direct summands of (M⊗AM)A are isomorphic to Z1 and Wn (n⩾1). Since dimk(M⊗AM)1=4, we have either W1⊕W1⊂A⊕M⊗AM that contradicts Lemma 3.1.3 or W3⊂A⊕M⊗AM that contradicts Corollary 3.1.2. Thus, W2⊂A⊕M.
∎
Lemma 3.3.3**.**
In the settings (3.3.1), if MA≅B2(α,β)⊕K for some (α,β)∈P1 and A-module K, then K is concentrated in degree zero and B2′(α′,β′)⊂Aop⊕M for some (α′,β′)∈P1.
Proof.
Suppose that MA≅B2(α,β)⊕K for some A-module K. Let us first consider the case where dimkA1f=1 for any
f∈M0. In this case indecomposable direct summands of AM are isomorphic to Z1′ and Wn′ (n⩾1). Which is impossible by Corollary 3.1.2 and Lemmas 3.1.3, and 3.3.2.
Suppose now that dimkA1f=1 for some f∈M0. Then by Lemma 3.1.1 there are u,v∈A1 such that ru(1)⩽1 and rv(1)⩽2.
It is easy to see that \dim_{\bf k}\big{(}B_{2}(\alpha,\beta)_{0}(\alpha x-\beta y)\big{)}=1 and \dim_{\bf k}\big{(}B_{2}(\alpha,\beta)_{0}w\big{)}=2 for any w∈A1 linearly independent with αx−βy. Since
r_{w}(1)=\dim_{\bf k}\big{(}B_{2}(\alpha,\beta)_{0}w\big{)}+\dim_{\bf k}(K_{0}w)
for any w∈A1, we have rαx−βy(1)⩾1 and rw(1)⩾2 for any w∈A1 linearly independent with αx−βy. Thus, we may assume that u=αx−βy and K0u=K0v=0, i.e. K1=K0A1=0 and K is concentrated in degree zero. Moreover, if B2′(α′,β′)⊂Aop⊕M for any (α′,β′)∈P1, then AM is a direct sum of modules of the form Z1′, Z2′, and B1′(α′,β′) ((α′,β′)∈P1). It is easy to see that in this case M0 is generated by such elements f∈M0 that dimkA1f⩽1 that contradicts to the fact that rw(1)⩾2 for any w∈A1 linearly independent with αx−βy.
∎
Suppose that N⊂A⊕M, where N≅B1(α,β)⊕Z2 or N≅B1(α,β)⊕B1(α,β) for some (α,β)∈P1. Then our standard argument (see, for example, the proof of Lemma 3.3.3) shows that rw(1)⩾2 for any w linearly independent with αx−βy. Then M0 cannot be generated by elements f∈M0 such that dimkA1f⩽1 due to item 3 of Lemma 3.1.1. Thus, AM has to contain a direct summand isomorphic to one of the modules Wn, Zn+2 or Bn+1(α′,β′) (n⩾1, (α′,β′)∈P1). Corollary 3.1.2 and Lemmas 3.1.3, 3.3.1, 3.3.2, and 3.3.3 imply that we get a contradiction, and hence N⊂A⊕M.
∎
Corollary 3.3.5**.**
In the settings (3.3.1), MA≅N⊕K, where K is concentrated in degree zero and N belongs to the set
[TABLE]
In particular, dimkM1⩽2.
Proof.
Let us represent MA in the form MA≅N⊕K, as in the statement, with N without direct summands concentrated in degree zero.
We have Wi⊂A⊕M for any i⩾1 by Corollary 3.1.2 and Lemmas 3.1.3 and 3.3.2.
If B2(α,β)⊂A⊕M for some (α,β)∈P1, then the required assertion follows from Lemma 3.3.3.
In the remaining case it follows from Corollary 3.1.2 and Lemmas 3.3.1 and 3.3.4 that if N≅Z2, then indecomposable direct summands of N are isomorphic to B1(α,β) ((α,β)∈P1) and that all of them are pairwise nonisomorphic.
It remains to show that dimkM1⩽2 in this case. This follows from item 3 of Lemma 3.1.1 and the fact that elements f∈M0 with dimkfA1⩽1 generate M0.
∎
3.4 Combining one side structures
Corollary 3.3.5 gives strong restrictions on possible one side A-module structures of M. Now we are going to combine possible right structures with possible left structures. Here, for each right A-module structure on M, we will determine the possible left A-module structures of M and then the left and right A-module structures on M⊗Ak for any k>1. Thus, in view of our Table of decompositions for J(A), we will be able to recover the algebra A from the left and right A-module structures of M. In particular, we will exclude some of the algebras from the list of candidates for the first position of an s-homogeneous triple.
Let us first prove a lemma that will be useful in our discussion.
Lemma 3.4.1**.**
Suppose that M0u=0 for any nonzero u∈A1 in the settings (3.3.1). Then \varphi\big{(}M_{0}^{\otimes(k-1)}\otimes r{\rm Ann}_{M_{0}}(A_{1})\otimes M_{0}^{\otimes(s-k)}\big{)}=0 for any 1⩽k⩽s.
Proof.
We will prove the required assertion using descending induction by k. Suppose that φ(X⊗f)=u=0 for some X∈M0⊗(s−1) and f∈rAnnM0(A1). Then we have gu=gφ(X⊗f)=φ(g⊗X)f=0 for any g∈M0, i.e. M0u=0. The obtained contradiction shows that the required assertion is valid for k=s.
Suppose that 1⩽k<s and we have already proved that
[TABLE]
Suppose that φ(X⊗f⊗Y⊗h)=u=0 for some X∈M0⊗(k−1), f∈rAnnM0(A1), Y∈M0⊗(s−k−1), and h∈M0. Then we have
[TABLE]
for any g∈M0 by the induction hypothesis, i.e. M0u=0. The obtained contradiction shows that the required assertion is valid for any 1⩽k⩽s.
∎
Lemma 3.4.2**.**
Suppose that in the settings (3.3.1) one has MA≅B2(α,β)⊕Z1m for some (α,β)∈P1 and m⩾0. Then (M⊗Ak)A≅B2(α,β)⊕Z1(m+2)k−2 and A(M⊗Ak)≅B2′(α,β)⊕(Z1′)(m+2)k−2 for any k⩾1.
Proof.
We have AM≅B2′(α′,β′)⊕(Z1′)m for some (α′,β′)∈P1 by Lemma 3.3.3. Note that φ(M0⊗(i−1)⊗(Z1′)m⊗M0⊗(s−i))=φ(M0⊗(i−1)⊗(Z1)m⊗M0⊗(s−i))=0 for any 1⩽i⩽s by Lemma 3.4.1. If Z1m=(Z1′)m, then the space Z1m+(Z1′)m⊂M0 has codimension less or equal to one, and hence 2=dimkImφ∣M0⊗s⩽1. The obtained contradiction shows that Z1m=(Z1′)m, and hence M≅N⊕km, where NA≅B2(α,β) and AN≅B2′(α′,β′). Moreover, φ factors through the canonical projection M⊗As↠N⊗As.
Suppose first that (α′,β′)=(α,β). Note that after a replacement of A by an isomorphic algebra we may assume that (α,β)=(1,0) and (α′,β′)=(0,1).
Let us denote by f1 and f2 the standard generators of B2(1,0) and by f1′ and f2′ the standard generators of B2(0,1).
Let us prove that φ(N0⊗(i−1)⊗f2′⊗N0⊗(s−i))⊂⟨x⟩ for any 1⩽i⩽s by descending induction on i. Suppose that φ(Y⊗f2′)=bxx+byy with by=0 for some Y∈N0⊗(s−1). Let us choose c1,c2∈k not all zero such that c1φ(f1⊗Y)+c2φ(f2⊗Y)∈⟨y⟩. Then we have
0=\varphi\big{(}(c_{1}f_{1}+c_{2}f_{2})\otimes Y\big{)}f_{2}^{\prime}=(c_{1}f_{1}+c_{2}f_{2})\varphi(Y\otimes f_{2}^{\prime})\not=0. The obtained contradiction shows that φ(N0⊗(s−1)⊗f2′)⊂⟨x⟩.
Suppose now that φ(N0⊗i⊗f2′⊗N0⊗(s−i−1))⊂⟨x⟩ for some 1⩽i<s. Let us take arbitrary Y∈N0⊗(i−1), Z∈N0⊗(s−i−1) and h∈N0. Let us choose c1,c2∈k not all zero such that c1φ(f1⊗Y⊗f2′⊗Z)+c2φ(f2⊗Y⊗f2′⊗Z)=0. It is possible because φ(f1⊗Y⊗f2′⊗Z),φ(f2⊗Y⊗f2′⊗Z)∈⟨x⟩ by induction hypothesis. We have
[TABLE]
and hence φ(Y⊗f2′⊗Z⊗h)∈⟨x⟩. In analogous way, one can show that φ(N0⊗(i−1)⊗f2⊗N0⊗(s−i))⊂⟨y⟩ for any 1⩽i⩽s.
If f2 and f2′ are linearly dependent, then φ(N0⊗(i−1)⊗f2⊗N0⊗(s−i))=0 for any 1⩽i⩽s, and hence we again get 2=dimkImφ∣M0⊗s⩽1. If f2 and f2′ are linearly independent, then
they constitute a basis of N0, and hence we have φ(f2⊗s)=cy for some c∈k∗. Then we have 0=c1f2′y=f2′φ(f2⊗s)=φ(f2′f2⊗(s−1))f2=0. The obtained contradiction shows that the case (α′,β′)=(α,β) is impossible.
Now we already have AM≅B2′(α,β)⊕(Z1′)m. We can assume again that (α,β)=(1,0). Direct calculations show that \dim_{\bf k}\big{(}B_{2}(1,0)\otimes_{A}B_{2}^{\prime}(1,0)\big{)}_{1}=2. Note that, for any u∈A1 linearly independent with x, M0u=M1, and hence (M⊗AM)0u=(M⊗AM)1 too. As usually, the analogous assertion holds for A(M⊗AM).
It follows that that ru(2)=2 for any u∈A1 linearly independent with x, and hence M0⊗M0 is not generated by elements f∈M0⊗M0 such that dimk(A1f)⩽1.
Since nonsemisimple direct summands of A(M⊗AM) can be B2′(1,0), B1′(1,0) and Z2′, the last assertion means that B2′(1,0)⊂Aop⊕M⊗AM. Then the dual argument shows that B2(1,0)⊂A⊕M⊗AM. Thus, we have (M⊗AM)A≅B2(1,0)⊕Z1(m+2)2−2 and A(M⊗AM)≅B2′(1,0)⊕(Z1′)(m+2)2−2.
In particular, this means that B2(1,0)⊗AM≅B2(1,0)⊕Z12m+2 and M⊗AB2(1,0)′≅B2′(1,0)⊕(Z1′)2m+2.
Now the required assertion follows by induction on k.
∎
Lemma 3.4.3**.**
Suppose that in the settings (3.3.1) one has AM≅B1(α,β)⊕B1(γ,δ)⊕Z1m for some (α,β),(γ,δ)∈P1 and m⩾0. Then (M⊗Ak)A≅B1(α,β)⊕B1(γ,δ)⊕Z1(m+2)k−2 and A(M⊗Ak)≅B1′(α,β)⊕B1′(γ,δ)⊕(Z1′)(m+2)k−2 for any k⩾1.
Proof.
We have AM≅B1′(α′,β′)⊕B1′(γ′,δ′)⊕(Z1′)m for some (α′,β′),(γ′,δ′)∈P1 by Lemma 3.3.3 and Corollary 3.3.5. Note that (α,β)=(γ,δ) and (α′,β′)=(γ′,δ′) by Lemma 3.3.4.
Note that M0 is generated by elements f′∈M0 such that dimk(A1f′)⩽1 and by elements f∈M0 such that dimk(fA1)⩽1.
Let us choose among the elements of the form φ(f⊗X⊗f′)∈A1 (f,f′∈M0, dimk(fA1),dimk(A1f′)⩽1, X∈M0⊗(s−2)) two linearly independent elements u and v.
Following the proof of Lemma 3.1.1, we obtain that ru(1)⩽1, and hence u∈⟨αx−βy⟩ or u∈⟨γx−δy⟩. Analogous inclusion for v and the dual argument show that
[TABLE]
and hence {(α′,β′),(γ′,δ′)}={(α,β),(γ,δ)}.
Now we already have AM≅B1′(α,β)⊕B1′(γ,δ)⊕(Z1′)m. After a replacement of A by an isomorphic algebra we may assume that (α,β)=(1,0) and (γ,δ)=(0,1). Now it is easy to see that
[TABLE]
Thus, dimk(B1(1,0)⊗AM)1=dimk(B1(0,1)⊗AM)1=1, i.e. (M⊗AM)A has two indecomposable direct summands with one dimensional first components. Since M(ax+by)=M for any a,b∈k∗, these direct summands are isomorphic to B1(1,0), B1(0,1) and Z2. Analogously A(M⊗AM) has two indecomposable direct summands isomorphic to B1′(1,0), B1′(0,1) or Z2′. Since M0⊗M0 is generated by such elements f∈M0⊗M0 that dimkA1f⩽1, it is easy to check using Lemma 3.1.1 that only the case B1(1,0)⊕B1(0,1)⊂A⊕M⊗AM is possible, i.e. (M⊗AM)A≅B1(1,0)⊕B1(0,1)⊕Z1(m+2)2−2.
In particular, this means that \big{(}B_{1}(1,0)\oplus B_{1}(0,1)\big{)}\otimes_{A}M\cong B_{1}(1,0)\oplus B_{1}(0,1)\oplus Z_{1}^{2m+2}.
Now the required formula for (M⊗Ak)A follows by induction on k. The proof of the formula for A(M⊗Ak) is dual.
∎
Lemma 3.4.4**.**
Suppose that in the settings (3.3.1) one has MA≅Z2⊕Z1m for some m⩾0. Then AM≅Z2′⊕(Z1′)m and M⊗Ak≅k(m+2)k for any k>1.
Proof.
First, suppose that AM≅B1′(α,β)⊕(Z1′)m+1 for some (α,β)∈P1.
Let us set K′:=rAnnM0(A1)=(Z1′)m+1. Then we have φ(M0⊗(k−1)⊗K′⊗M0⊗(s−k))=0 for any 1⩽k⩽s by Lemma 3.4.1.
Since K′ has codimension one in M0, we get 2=dimkImφ∣M0⊗s⩽1.
The obtained contradiction shows that AM does not have a direct summand of the form B1′(α,β), and hence AM≅Z2′⊕(Z1′)m. It is easy to see that (Z2⊗AZ2′)1=0, and thus M⊗Ak is concentrated in degree zero for any k>1.
∎
Lemma 3.4.5**.**
Suppose that in the settings (3.3.1) one has MA≅B1(α,β)⊕Z1m for some (α,β)∈P1 and m⩾0. Then AM≅B1(α′,β′)⊕(Z1′)m for some (α′,β′)∈P1. Moreover,
M⊗Ak≅k(m+1)k* for any k>1 if (α′,β′)=(α,β);*
2. 2.
(M⊗Ak)A≅B1(α,β)⊕Z1(m+1)k−1* and A(M⊗Ak)≅B1′(α,β)⊕(Z1′)(m+1)k−1 for any k>1 if (α′,β′)=(α,β).*
Proof.
The fact that AM has the form stated in the lemma follows from Lemma 3.4.4. Now the first part follows from the fact that \big{(}B_{1}(\alpha,\beta)\otimes_{A}B_{1}^{\prime}(\alpha^{\prime},\beta^{\prime})\big{)}_{1}=0 for (α′,β′)=(α,β).
Finally, let us consider the case (α′,β′)=(α,β). Since
\dim_{\bf k}\big{(}B_{1}(\alpha,\beta)\otimes_{A}B_{1}^{\prime}(\alpha,\beta)\big{)}_{1}=1, we have dimk(M⊗AM)1=1. Since (M⊗AM)0(αx−βy)=(αx−βy)(M⊗AM)0=0, it is easy to see that the required assertion is true for k=2. In particular, this means that B1(α,β)⊗AM≅B1(α,β)⊕Z1m and M⊗AB1′(α,β)≅B1′(α,β)⊕(Z1′)m. Then the required assertion follows by induction on k.
∎
Now we can prove the next part of our classification.
Theorem 3.4.6**.**
Suppose that Λ=TkV/(f1,f2), where f1 and f2 are two linearly independent elements of V⊗s.
Then (Λ!)(s) is not isomorphic to A2(q), A3, A8 and A9(q) for any q∈k.
Proof.
Suppose that the assertion of the theorem is not true. Then there is an s-homogeneous triple (A,M,φ) with A∈{A2(q),A3,A8,A9(q)}q∈k.
One can use Corollary 3.3.5, Lemmas 3.4.2, 3.4.3, 3.4.4, and 3.4.5, and Table of decompositions for J(A) to show that M⊗As cannot have AJ(A)[1] as a left A-module direct summand and J(A)[1]A as a right A-module direct summand at the same time. This contradicts Lemma 2.1.3, and thus finishes the proof of the theorem.
∎
3.5 Non Koszulity of Λ!
Here, using the results about s-homogeneous triples (A,M,φ) with quadratic algebra A having two generators and the formula (2.2.4) for Hilbert series of s-Koszul algebras, we show that an algebra Γ=TkV/I with I generated by a subspace W⊂V⊗s of codimension two cannot be s-Koszul if s>2.
Theorem 3.5.1**.**
Suppose that s>2 and Γ=TkV/(W), where W is a subspace of V⊗s. If dimkΓs=2, then Γ is not s-Koszul.
Proof.
If Γ is s-Koszul, then it satisfies the extra condition, and hence (2.2.4) holds with m=dimkM0 and lk=dimk(M⊗Ak)1 for 0⩽k⩽s−1.
Let us calculate the coefficient of ts+2 in the formal power series on the left hand side of \eqrefHS. Observe that l0=2, so we obtain that Γ may be s-Koszul only if
[TABLE]
where l1=dimkM1 and l2=dimk(M⊗AM)1. We have l1⩽2 by Lemma 3.1.3 and Corollary 3.3.5. Moreover, it follows from the same results and Lemmas 3.4.2 and 3.4.3 that l2=2 in the case l1=2. Since m⩾2, we have 2m(m−l1)+l2>0, and thus Γ cannot be s-Koszul.
∎
3.6 s-homogeneous triples with A=A7(q), q=0
Theorem 3.4.6 says that the cases A=A2(q), A=A3, A=A8, and A=A9(q) are impossible.
The cases A=A4 and A=A5 are fully considered in Theorem 2.2.3 while the cases A=A6 and A=A7(0) are fully considered in Theorem 3.2.2. Thus, it remains to consider the cases A=A0, A=A1, and A=A7(q) (q=0). Among these algebras only A7(q) give s-homogeneous algebras that can be recovered from the pair (A,M). Because of this, the case A7(q) is easier to deal with, and we consider it now. This is the last case for which this paper gives a full classification. In the next section we will give some partial results on the classification in the cases A=A0 and A=A1.
Until the end of this section we fix some q∈k∗ and set A=A7(q). Let (A,M,φ) be an s-homogeneous triple.
Due to the previous results of the paper and the Table of decompositions for J(A), we have MA≅B2(1,0)⊕Z1m and AM≅B2′(1,0)⊕(Z1′)m for some m⩾0.
Thus, we need to recover the A-bimodule structure gluing the given one side structures on M. Then the triple (A,M,φ) can be recovered uniquely modulo isomorphism.
Let us introduce the graded A7(q)-bimodule M(λ) for λ∈k∗ by the equality
[TABLE]
As usually, f1 and f2 are the standard free generators. It is not difficult to show that M(λ) is an A7(q)-bimodule such that M(λ)A≅B2(1,0) and AM(λ)≅B2′(1,0). In particular, it has a linear presentation as a left and as a right module.
Lemma 3.6.1**.**
Suppose that there is an automorphism θ of A and an isomorphism of graded spaces γ:M(λ)≅M(μ) such that γ(ufv)=θ(u)γ(f)θ(v) for all u,v∈A, f∈M(λ).
Then λ=μ.
Proof.
For a,b∈k∗ let us consider the function κa,b:P1→Z defined by the equality
[TABLE]
It is easy to see that κa,λ=κa,μ∘g, where g:P1→P1 is the linear transformation induced by θ∣A1. One can show now that κa,b has nonzero values if and only if a∈{1,q−1,b}. Moreover, if q=1, then
κq−1,b has value 2 in some point if and only if b=q−1.
Thus, b can be recovered if we know the set of values of κa,b for all a∈k∗. Since these sets coincide for all a for the elements λ and μ, we have λ=μ.
∎
Lemma 3.6.2**.**
M(λ)⊗AM(μ)≅M(λμ)⊕k2.
Proof.
Let us denote by g1,g2 the standard generators of M(λ) and by h1,h2 the standard generators of M(μ). It is not difficult to check that
[TABLE]
i.e. there is a homomorphism from (Aop⊗A)2⊕k2 to M(λ)⊗AM(μ) sending the standard generators f1 and f2 of (Aop⊗A)2 to g1⊗h2 and g1⊗h1 and the basis e1,e2 of k2 to g2⊗h2 and g2⊗h1−μg1⊗h2 correspondingly. Direct calculations show that this homomorphism factors throw the natural projection (Aop⊗A)2⊕k2↠M2(λμ)⊕k2. The obtained map γ:M(λμ)⊕k2→M(λ)⊗AM(μ) is surjective. Since both bimodules are linearly presented and have the same dimensions of the first and the second components, γ is an isomorphism.
∎
Lemma 3.6.3**.**
If (A7(q),M,φ) is an s-homogeneous triple, then (A7(q),M,φ)≅(A7(q),M(p)⊕km,φp) for some m⩾0, p∈k∗ such that ps=q and some map φp.
Proof.
Due to the proof of Lemma 3.4.2, we may assume that M=N⊕km, where N is an A-bimodule isomorphic to B2(1,0) as an A-module. Let f1,f2 be the standard generators of N as an A-module B2(1,0), i.e. f1,f2∈N0, f1x=f2y and f2x=0. Note that the basis of Ni is f1yi,f2yi for any i⩾0. Since lAnnN1(x)=⟨f2y⟩, we have A1f2⊂⟨f2y⟩. We have also
[TABLE]
i.e. xf2=0 and yf2=af2y for some a∈k∗ (note that rAnnN0(y)=0). Moreover, xN0=rAnnN1(x), and hence xf1=pf2y for some p∈k∗. Finally, we have yf1=bf1y+cf2y for some b,c∈k, b=0, because A1N0=N1.
We have φ(M⊗Ai⊗Akm⊗AM⊗A(s−1−i))=0 for any 1⩽i⩽s by Lemma 3.4.1. Moreover, it is easy to show using the equality xf1=pf1x that x(fi1⊗⋯⊗fis)=0, and hence φ(fi1⊗⋯⊗fis)∈⟨x⟩ if one of the numbers i1,…,is∈{1,2} equals 2.
Since A1⊂Imφ, we have φ(f1⊗s)=αx+βy for some α,β∈k, β=0. Then we have
[TABLE]
and hence b=1. Now, using the equality 0=(xy−qyx)f1=(p−qap)f2y2, we get a=q−1.
Suppose that q=1. The map sending f1 to f1+q−1qcf1 and f2 to f2 is an automorphism of NA.
At the same time, we have y(f1+q−1qcf2)=(f1+q−1qcf2)y. Hence, we may assume c=0 in the case q=1.
Let us now consider the case q=1. If we have also p=1, then c=0 by (3.6.1). Suppose that p=1. Then we can change the A-bimodule structure of M via the automorphism of A7(q) sending x to x and y to y+1−pcx. Note that in this way we will not change the right structure. Since (y+1−pcx)f1=f1(y+1−pcx), we again may assume that c=0.
Thus, it remains to show that ps=q. But we have (M(p)⊕km)⊗As≅M(ps)⊕k(m+2)s−2 by Lemma 3.6.2. Since J\big{(}A^{7}(q)\big{)}\cong M(q), we have ps=q by Lemma 3.6.1.
∎
Theorem 3.6.4**.**
Suppose that Λ=TkV/(f1,f2), where f1 and f2 are two linearly independent elements of V⊗s.
Suppose that (Λ!)(s)≅k⟨x,y⟩/(x2,xy−qyx)=A7(q) for some q∈k∗. Then
[TABLE]
for some m⩾0 and some p∈k∗ such that ps=q. In particular, Λ is s-Koszul.
Proof.
We may assume F(Λ!)=(A7(q),M(p)⊕km,φp) by Lemma 3.6.3. It remains to describe the kernel of φp.
Let f1 and f2 be the standard generators of M(p).
As we have mentioned above, φp can have nonzero values only on the tensors of the form X=fi1⊗⋯⊗fis with i1,…,is∈{1,2}. Also if one of the indices ik (1⩽k⩽s) equals 2, then φp(X)∈⟨x⟩. Let us prove that φp(X)=0 if there are two different 1⩽k<l⩽s such that ik=il=2. It is enough to prove that A1X=0. We have already mentioned that xX=0. We may assume i1=i2=⋯=ik−1=1. Then we have
yX=f1⊗(k−1)⊗yf2⊗X′=q1f1⊗k⊗xX′=0, because the tensor X′ contains f2 by our assumption.
Let us now look at the equality f1φp(f1⊗(i−1)⊗f2⊗f1⊗(s−i))=φp(f1⊗i⊗f2⊗f1⊗(s−i−1))f2 for 1⩽i⩽s−1. Substituting
φp(f1⊗k⊗f2⊗f1⊗(s−k−1))=akx with ak∈k (0⩽k⩽s−1) to this equality, we get ai=p1ai−1.
Thus, we have proved that decomposable tensors with a component belonging to km, decomposable tensors with minimum two components equal to f2 and the tensors pf1⊗i⊗f2⊗f1⊗(s−i−1)−f1⊗(i−1)⊗f2⊗f1⊗(s−i) (1⩽i⩽s−1) belong to the kernel of φp. Since we have obtained a subspace of codimension two, the listed elements generate Kerφp.
Now it is easy to see that \Lambda\cong{\mathcal{G}}(A^{7}(q),M(p)\oplus{\bf k}^{m},\varphi_{p})^{!}=T\big{(}(M(p)_{0}\oplus{\bf k}^{m})^{*}\big{)}/\big{(}\varphi_{p}^{*}A^{7}(q)_{1}\big{)} has the required form.
The s-Koszulity of Λ follows from the fact that G={y1s,i=0∑s−1ps−i−1y1iy2y1s−i−1} is a Gröbner basis with respect to the lexicographical order such that y1>y2>x1>⋯>xm. Indeed, we have tip(G)={y1s,y1s−1y2} and the required assertion follows, for example, from [20, Corollary 12].
∎
4 Triples not determined by two components
The only cases that have not been considered yet are A=A0 and A=A1 (see the beginning of Subsection 3.6). The main difficulty of these cases is that J(A)[1] contains simple A-bimodule summands, and thus Λ cannot be recovered from the pair (A,M) while the classification of maps φ:M⊗As→A[1] turning (A,M,φ) into an s-homogeneous triple is at least as difficulty as the classification of all s-homogeneous algebras with one relation. Nevertheless, in these cases we can say something about the algebra Λ.
It follows from Corollary 3.3.5, Lemmas 3.4.2, 3.4.3, 3.4.4 and 3.4.5 and Table of decompositions for J(A) that all the pairs that we have to consider satisfy one of the conditions:
A=A0, MA≅B1(α,β)⊕Z1m, AM≅B1′(α′,β′)⊕(Z1′)m for some (α,β),(α′,β′)∈P1, (α,β)=(α′,β′). In this case after a change of basis we may assume that (α,β)=(1,0) and (α′,β′)=(0,1);
4. 4.
A=A0, M=km.
In this section we will give some addition information in each of these cases.
4.1 s-homogeneous triples with A=A1
Throughout this subsection we consider A:=A1. As it was mentioned above, we have MA≅B1(1,0)⊕Z1m and AM≅B1′(1,0)⊕(Z1′)m in this case. Let us first finish the description of the A-bimodule M.
Note that the bimodule A/(x) satisfies the conditions \big{(}A/(x)\big{)}_{A}\cong B_{1}(1,0) and {}_{A}\big{(}A/(x)\big{)}\cong B_{1}^{\prime}(1,0). Moreover, \big{(}A/(x)\big{)}^{\otimes_{A}k}\cong A/(x) for any k⩾1.
Lemma 4.1.1**.**
There is an isomorphism of A-bimodules M≅A/(x)⊕km.
Proof.
We know that MA≅B1(1,0)⊕Z1m. Let us recover the left A-module structure on B1(1,0)⊕Z1m corresponding to the left A-module structure of M. Note that lAnnM1(y)=0 and (yZ1m)y=0, and hence yZ1m=0. Hence, we have an isomorphism of graded A-bimodules M≅N⊕km, where N is a graded A-bimodule such that NA≅B1(1,0) and AN≅B1′(1,0). Let f∈N0 be a nonzero element. Then we have yf=αfy for some α∈k∗ and all that we need is to prove that α=1. Note that (f1⊗⋯⊗fs)y=0 for f1⊗⋯⊗fs∈M0⊗s such that fi∈km for some 1⩽i⩽s, and hence φ(f1⊗⋯⊗fs)∈⟨x⟩ in this case. Since y∈Imφ we have φ(f⊗s)=βx+γy for some β∈k and γ∈k∗. Then we have
[TABLE]
i.e. α=1.
∎
Theorem 4.1.2**.**
Suppose that Λ=TkV/(f1,f2), where f1 and f2 are two linearly independent elements of V⊗s.
If (Λ!)(s)≅k⟨x,y⟩/(x2,xy,yx)=A1, then Λ≅k⟨x1,…,xm,y1⟩/(f,y1s)
for some f∈k⟨x1,…,xm,y1⟩s.
Conversely, if Λ=k⟨x1,…,xm,y1⟩/(f,y1s), then (Λ!)(s)≅A1 if and only if there are no linear polynomial g∈k⟨x1,…,xm,y1⟩ and α,β∈k such that
either f=i=0∑s−1αiβs−i−1y1igy1s−i−1 or f=gs+αy1s.
Proof.
We have Λ≅TkM0∗/(φ∗A1∗), where (A,M,φ) is an s-homogeneous triple. If (Λ!)(s)≅A1, then we may take A=A1 and, by Lemma 4.1.1, M=A/(x)⊕km.
Let now x1,…,xm,y1∈M0∗ be the basis dual to a basis f1,…,fm+1 of M0 whose first m elements constitute a basis of the summand km and the last one is some nonzero element of the summand A/(x).
Let now fx,fy be the basis of A1∗ dual to the basis x,y of A1.
It follows from the proof of Lemma 4.1.1 that fyφ(fi1⊗⋯⊗fis)=0 if ik=m+1 for some 1⩽k⩽s, and hence φ∗fy=αy1s for some α∈k∗. Thus, φ∗A1∗=⟨f,y1s⟩ for some f∈k⟨x1,…,xm,y1⟩s.
Let us now assume that Λ=k⟨x1,…,xm,y1⟩/(f,y1s) for some f∈k⟨x1,…,xm,y1⟩s linearly independent with y1s. If f=gs+αy1s, then Λ=k⟨x1,…,xm,y1⟩/(gs,y1s), and hence (Λ!)(s)≅A4 (see Theorem 2.2.3). If f=i=0∑s−1αiβs−i−1y1igy1s−i−1, then (Λ!)(s)≅A7(αβ) if α=0 and (Λ!)(s)≅A6 if α=0 (see Theorems 3.2.2 and 3.6.4). Suppose now that f cannot be presented in the form f=i=0∑s−1αiβs−i−1y1igy1s−i−1 or in the form f=gs+αy1s.
Then it follows from Theorems 2.2.3, 3.2.2, 3.4.6 and 3.6.4 that (Λ!)(s)≅A2(q),A3,A4,A5,A6,A7(q),A8,A9 for any q∈k. On the other hand, it is true that dimkΛn!=0 for any n⩾0, and hence (Λ!)(s)≅A0. It follows from our classification of quadratic algebras with two generators that (Λ!)(s)≅A1.
∎
4.2 s-homogeneous triples with A=A0 and Z2⊂A⊕M
Let us now consider algebras Λ with F(Λ!) of the form (A0,M,φ) with MA≅Z2⊕Z1m. As it was mentioned above, we have AM≅Z2′⊕(Z1′)m in this case. Here and until the end of this section A denotes the algebra A0. Let us introduce the bimodule
[TABLE]
where we again denote by f1 and f2 the standard generators of (Aop⊗A)2. Let us now prove the following lemma recovering the A-bimodule structure on M.
Lemma 4.2.1**.**
If (A0,M,φ) is an s-homogeneous triple with MA≅Z2⊕Z1m, then M≅νD⊕km for some graded automorphism ν of A0. Moreover, the map φ:M⊗As→A[1] can be factored through the canonical projection M⊗As↠(νD)⊗As
Proof.
It is not difficult to see that any two dimensional graded A-bimodule K such that KA≅Z2 and AK≅Z2′ is isomorphic to νD for some graded automorphism ν of A.
Thus, for the first part it is enough to prove that M has an A-bimodule summand isomorphic to km. Let us consider the right A-module decomposition MA=Z2⊕Z1m.
We will prove that in fact Z1m is an A-bimodule summand isomorphic to km. Let us take some f∈Z1m. Suppose that uf=0 for some u∈A1. Note that then we have 0=φ(X)f=(1M⊗Aφ)(X⊗f) for some X∈M0⊗s; in particular, φ(Y⊗f)=0 for some Y∈M0⊗(s−1). On the other hand, φ(Y1⊗f⊗Y2)=0 for any 1⩽i⩽s, Y1∈M0⊗(i−1), Y2∈M0⊗(s−i), and f∈Z1m by Lemma 3.4.1. Note that the last mentioned assertion implies the second statement of the lemma too.
∎
Theorem 4.2.2**.**
Suppose that Λ=TkV/(f1,f2), where f1 and f2 are two linearly independent elements of V⊗s.
If (Λ!)(s)≅k⟨x,y⟩/(x,y)2=A0 and (Λ!)(s,1)≅Z2⊕Z1m as a right A0-module, then Λ≅k⟨x1,…,xm⟩∗D(w) for some twisted potential w of degree s+1 on a two dimensional space.
2. 2.
Suppose that Λ=k⟨x1,…,xm⟩∗D(w) for a ν-twisted potential w of degree s+1 on the two dimensional space U. Then (Λ!)(s)≅A0 and (Λ!)(s,1)≅Z2⊕Z1m as a right A0-module if and only if the space {(f⊗1U⊗s)w∣f∈U∗} is two dimensional and there are no nonzero f1,f2∈U∗ and u1,u2∈U such that one of the following conditions holds:
•
s=2t+1, (f1⊗1U⊗s)w=(u1⊗u2)⊗t⊗u1, and (f2⊗1U⊗s)w=(u2⊗u1)⊗t⊗u2;
•
s=2t, (f1⊗1U⊗s)w=(u1⊗u2)⊗t, and (f2⊗1U⊗s)w=(u2⊗u1)⊗t.
Proof.
Due to Lemma 4.2.1, we may set M=ν−1D⊕km for some automorphsm ν of A. As usually, we have Λ=TkM0∗/(φ∗A1∗). Let f1,f2 be the standard basis of ν−1D and g1,…,gm be a basis of km considered as direct summands of M. Let y1,y2,x1,…,xm be the basis of M0∗ dual to the obtained basis of M0 and U be the subspace of M0∗ spanned by y1 and y2. It follows from Lemma 4.2.1 that φ∗A1∗⊂U⊗s. Thus, Λ=k⟨x1,…,xm⟩∗TkU/(φ∗A1∗) and it remains to show that TkU/(φ∗A1∗)=D(w) for some twisted potential w of degree s+1 on U. Let fx,fy be the basis of A1∗ dual to x,y. Let us define the twisted potential w by the equality w=y1φ∗(fx)+y2φ∗(fy). We can see that
[TABLE]
and hence, to finish the proof of the first part, it remains to show that w is a ν-twisted potential, where ν:U→U is the composition UθA1νA1θ−1U with the isomorphism θ:U→A1 defined by the equalities θ(y1)=x and θ(y2)=y. To show that w is a ν-twisted potential it suffices to check the equality
y1φ∗(fx)+y2φ∗(fy)=φ∗(fx)ν−1(y1)+φ∗(fy)ν−1(y2). To see this, let us pick some X∈(ν−1D)0⊗(s+1) and apply both parts of the required equality to it. Direct calculations show that
[TABLE]
i.e. the applications of y1φ∗(fx)+y2φ∗(fy) and φ∗(fx)ν−1(y1)+φ∗(fy)ν−1(y2) give the same result on any element X∈(ν−1D)0⊗(s+1). Thus, the required equality is proved and w is a ν-twisted potential.
Let us now suppose that Λ=k⟨x1,…,xm⟩∗D(w) for a ν-twisted potential w of degree s+1 on the two dimensional space U. Then Λ is an s-homogeneous algebra with defining ideal generated by dimk{(f⊗1U⊗s)w∣f∈U∗} elements of degree s. Thus, the quadratic algebra (Λ!)(s) has two generators if and only if the last mentioned dimension equals 2. Note that one of the conditions listed in the theorem is satisfied for some f1,f2∈U∗ and linearly dependent u1,u2∈U if and only if Λ≅k⟨x1,…,xm,y1,y2⟩/(y2s,f) for some homogeneous polynomial f∈k⟨y1,y2⟩s. Then it follows from Theorems 2.2.3, 3.2.2, 3.6.4, and 4.1.2 that f1,f2∈U∗ and u1,u2∈U satisfying the conditions listed in the theorem exist if and only if (Λ!)(s) is isomorphic to one of the algebras A1, A4, A5, A6, and A7(q) (q∈k). Since (Λ!)(s)≅A2(q),A3,A8,A9(q) for any q∈k by Theorem 3.4.6, we have (Λ!)(s)≅A0 if and only if the space {(f⊗1U⊗s)w∣f∈U∗} is two dimensional and there are no f1,f2∈U∗ and u1,u2∈U satisfying one of the conditions listed in the theorem.
It remains to show that if Λ=k⟨x1,…,xm⟩∗D(w) for a ν-twisted potential w of degree s+1 on the two dimensional space U and (Λ!)(s)=A0, then M=(Λ!)(s,1) cannot be isomorphic to B1(α,β)⊕Z1m+1 ((α,β)∈P1) or Z1m+2 as a right A0-module. Let y1,y2 be a basis of U and f1,f2 be the basis of U∗ dual to it. Note that by definition D(w)=k⟨y1,y2⟩/(R), where R is the subspace of U⊗s generated by r1=(f1⊗1U⊗s)w and r2=(f2⊗1U⊗s)w. Note that w=y1r1+y2r2=r1ν−1(y1)+r2ν−1(y2)∈(U⊗R)∩(R⊗U). On the other hand, MA≅Z1m+2 means Λs+1!=0, and hence U∗⊗R⊥+R⊥⊗U∗=(U∗)⊗(s+1) and (U⊗R)∩(R⊗U)=0. Here, as usually, R⊥={f∈(U∗)⊗s∣f(R)=0}. At the same time, MA≅B1(α,β)⊕Z1m+1 means that there is f∈(U∗)⊗s such that f(r1) or f(r2) is nonzero, but g⊗f∈U∗⊗R⊥+R⊥⊗U∗ for any g∈U∗. Since h\big{(}(U\otimes R)\cap(R\otimes U)\big{)}=0 for any h∈U∗⊗R⊥+R⊥⊗U∗, we have 0=(g⊗f)(w)=g(y1)f(r1)+g(y2)f(r2) for any g∈U∗. This obviously contradicts to the fact that f(r1) and f(r2) cannot be zero simultaneously, and thus MA≅B1(α,β)⊕Z1m+1. The theorem is proved.
∎
Let us now describe the automorphisms ν of the algebra A0 for which there is an s-homogeneous triple with (A,M)=(A0,ν−1D). As it was proved above, this is equivalent to the existence of ν-twisted potential w of degree s+1 on a two dimensional space with the basis y1,y2 that does not satisfy conditions listed in item 2 of Theorem 4.2.2.
Modulo an automorphism of A0 that induces an isomorphism of the corresponding s-homogeneous triples, we may assume that either ν(y1)=λ1y1 and ν(y2)=λ2y2 for some λ1,λ2∈k∗ or ν(y1)=λy1 and ν(y2)=λ(y1+y2) for some λ∈k∗.
As usually, we consider only the case s>2 while there are no such potentials in the case s=2, for example, due to the classification [31, Theorem 3.10.6].
Lemma 4.2.3**.**
Suppose that ν(y1)=λ1y1 and ν(y2)=λ2y2 for some λ1,λ2∈k∗. An s-homogeneous algebra Λ such that F(Λ!)=(A0,ν−1D,φ) for some φ exists if and only if one of the following conditions holds:
•
there is some 2⩽k⩽s−1 such that λ1kλ2s+1−k=1;
•
λ1λ2s=λ1sλ2=1;
•
λ1=1, λ2s=1 (or λ1s=1, λ2=1).
Proof.
Suppose that w is a ν-twisted potential of degree s+1. Note first that w=ϕνs+1(w)=ν⊗(s+1)(w). Thus, ν⊗(s+1) has to have an eigenvalue equal to 1. Thus, λ1kλ2s+1−k=1 for some 0⩽k⩽s+1. Note that λ1s+1 and λ2s+1 correspond to the eigenvectors y1s+1 and y2s+1 respectively. Note also that the vector y1s+1 can appear only in the case where λ1=1 while y2s+1 can appear only in the case where λ2=1. Thus, if none of the conditions of the lemma holds, then either w=ay1s+1+bi=0∑s+1λ1iy1iy2y1s−i or w=ay2s+1+bi=0∑s+1λ2iy2iy1y2s−i. In both cases one can show that w satisfies one of the conditions listed in item 2 of Theorem 4.2.2, and hence does not determine an s-homogeneous triple of the required form.
If we have λ1kλ2s+1−k=1 for some 2⩽k⩽s−1, then the required ν-twisted potential is w=i=0∑k−1λ1iy1iy2s+1−ky1k−i+λ1ki=0∑s+1−kλ2iy2iy1ky2s+1−k−i.
If λ1λ2s=λ1sλ2=1 and λ1kλ2s+1−k=1 for any 2⩽k⩽s−1, then the required ν-twisted potential is w=i=0∑s+1λ1iy1iy2y1s−i+i=0∑s+1λ2iy2iy1y2s−i.
If λ1=1 and λ2s=1, then the required ν-twisted potential is w=y1s+1+i=0∑sλ2iy2iy1y2s−i. It is easy to see that the listed potentials do not satisfy the conditions listed in item 2 of Theorem 4.2.2.
∎
Lemma 4.2.4**.**
Suppose that ν(y1)=λy1 and ν(y2)=λ(y1+y2) for some λ∈k∗. If s>3, then an s-homogeneous algebra Λ such that F(Λ!)=(A0,ν−1D,φ) for some φ exists if and only if λs+1=1. A 3-homogeneous algebra Λ satisfying the same condition exists if and only if λ2=1.
Proof.
Suppose that w is a ν-twisted potential of degree s+1. Note that all the eigenvalues of ν⊗(s+1) are equal to λs+1, and hence λs+1=1 (see the proof of Lemma 4.2.3).
Let us consider the case s=3. If λ=1,−1, then the required ν-twisted potential is
[TABLE]
Suppose now that λ2=−1=1. Direct calculations show that any eigenvector of ν⊗4 is a linear combination of y14, i=0∑3aiy1iy2y13−i with i=0∑3ai=0, y12y22+y22y12−y1y22y1−y2y12y2 and y1y22y1+y2y12y2−(y1y2)2−(y2y1)2.
In particular, (y1y2)2 and (y2y1)2 have equal coefficients in w and the coefficient of (y2y1)2 in ϕν(w) equals to the coefficient of (y1y2)2 in w multiplied by λ. Thus, the coefficient of y1y22y1+y2y12y2−(y1y2)2−(y2y1)2 in the linear combination mentioned above is zero. Then the coefficient of y2y12y2 in w equals to the coefficient of y12y22 in w multiplied by −1 and, at the same time, the coefficient of y2y12y2 in ϕν(w) equals to the coefficient of y12y22 in w multiplied by λ. Thus, w is a linear combination of y14, i=0∑3aiy1iy2y13−i with i=0∑3ai=0. Then it is easy to see that w=a\big{(}y_{1}^{4}+(1-\lambda)\sum\limits_{i=0}^{3}\lambda^{i}y_{1}^{i}y_{2}y_{1}^{3-i}\big{)} for some a∈k, and hence satisfies one of the conditions listed in item 2 of Theorem 4.2.2.
Suppose now that s>3 and λs+1=1. To define a ν-twisted potential, we will consider three cases
If λ2+1=0, then
[TABLE]
2. 2.
If chark=2 and λ=1, then
[TABLE]
3. 3.
If λ2+1=0 and chark=2, then we have λ3+1=0. Note also that in this case 4∣s+1, i.e. s⩾7. In this case we define
[TABLE]
∎
4.3 s-homogeneous triples with A=A0 and B1(α,β)⊂A⊕M
Let us now consider algebras Λ with F(Λ!) of the form (A0,M,φ) with MA≅B1(α,β)⊕Z1m ((α,β)∈P1).
As it was mentioned above, we may assume in this case that (α,β)=(1,0) and AM≅B1′(0,1)⊕(Z1′)m. Note that m>0 follows from the definition of an s-homogeneous triple.
Let us introduce the A-bimodules B1=(Aop⊗A)/⟨f1x,yf1,f1y−xf1⟩Aop⊗A and B2=(Aop⊗A)2/⟨f1x,xf1,yf1,yf2,f2x,f2y,f1y−xf2⟩Aop⊗A, where f1,f2 as usually denote the standard generators. It is not difficult to see that there are only two possible A-bimodule structures on M; namely, we may have M=B1⊕km or M=B2⊕km−1. This reduces the first step that we usually fulfill in our proofs, and thus we can go directly to the description of algebras Λ corresponding to the s-homogeneous triple (A0,M,φ).
Theorem 4.3.1**.**
Suppose that Λ=TkV/(f1,f2), where f1 and f2 are two linearly independent elements of V⊗s.
If (Λ!)(s)≅k⟨x,y⟩/(x,y)2=A0 and (Λ!)(s,1)≅B1⊕km as an A0-bimodule, then Λ≅k⟨x1,…,xm,y1⟩/(y1f,fy1) for some f∈k⟨x1,…,xm,y1⟩s−1. Conversely, if Λ=k⟨x1,…,xm,y1⟩/(y1f,fy1), then (Λ!)(s)≅A0 and (Λ!)(s,1)≅B1⊕km as an A0-bimodule except the case where s=2t and f=(gy1)t−1g for some g∈k⟨x1,…,xm,y1⟩1.
2. 2.
If (Λ!)(s)≅k⟨x,y⟩/(x,y)2=A0 and (Λ!)(s,1)≅B2⊕km as an A0-bimodule, then Λ≅k⟨x1,…,xm,y1,y2⟩/(y1f,fy2) for some f∈k⟨x1,…,xm,y1,y2⟩s−1. Conversely, if Λ=k⟨x1,…,xm,y1,y2⟩/(y1f,fy2), then (Λ!)(s)≅A0 and (Λ!)(s,1)≅B2⊕km as an A0-bimodule except the case where s=2t+1 and f=α(y2y1)t for some α∈k∗ and the case where f=αy1s−1 or f=αy2s−1 for some α∈k∗.
Proof.
Suppose first that F(Λ!)=(A0,Bk⊕km,φ) for some homomorphism φ and some k∈{1,2}. Let now x1,…,xm,y1,…,yk∈M0∗ be the basis dual to a basis g1,…,gm,f1,…,fk of M0 whose first m elements constitute a basis of the summand km and the last k are standard generators of the summand Bk. We will denote k⟨x1,…,xm,y1,…,yk⟩ by C for short during this proof.
As usually, we have Λ≅C/(φ∗A1∗).
Let now fx,fy be the basis of A1∗ dual to the basis x,y of A1. Note that if fxφ(r⊗X)=0 for some r∈M0 such that rx=ry=0 and some X∈M0⊗(s−1), then 0=φ(r⊗X)fk=rφ(X⊗fk)=0. Hence, φ∗fx=y1f for some f∈Cs−1. Analogously, φ∗fy=f′yk for some f′∈Cs−1. For any X∈M0⊗(s−1), we have
f(X)xfk=φ(f1⊗X)fk=f1φ(X⊗fk)=f′(X)f1y, and hence f′=f.
Suppose now that Λ=C/(y1f,fyk), where C is as above and y1f and fyk are linearly independent s-homogeneous polynomials.
Let us prove first that (Λ!)(s)≅A0 except the cases listed in the theorem. Suppose that it is not true. Then due to Lemma 2.3.1 and Theorems 2.2.3, 3.2.2, 3.4.6, 3.6.4, and 4.1.2 there are some nonzero (but maybe linearly dependent) g1,g2∈C1 such that (g1g2)t,(g2g1)t∈(y1f,fyk) in the case s=2t and (g1g2)tg1,(g2g1)tg2∈(y1f,fyk) in the case s=2t+1. Let us recall that the tensor w∈V⊗s is called decomposable if w=w1⊗⋯⊗ws for some wi∈V (1⩽i⩽s).
Suppose that the tensor βy1f+γfyk is decomposable for some β,γ∈k∗. Let us present f in the form f=y1pf′ykq, where f′∈Cs−1−p−q can be presented neither in the form f′=y1f′′ nor in the form f′=f′′yk with f′′∈C. The tensor βy1f′+γf′yk has to be decomposable.
If p+q=s−1, then it is easy to see that there is some h1∈Cs−1−p−q∗ such that h1(f′)=1 and h1(f′′yk)=0 for any f′′∈Cs−2−p−q. Then we have (1C1⊗h1)(βy1f′+γf′yk)=βy1. On the other hand, there is some h2∈Cs−1−p−q∗ such that (1C1⊗h2)(βy1f′+γf′yk) is linearly independent with y1, and hence βy1f′+γf′yk cannot be decomposable. Thus, βy1f+γfyk can be decomposable only for f of the form f=αy1pyks−1−p with α∈k∗. It is easy to see that if f has such a form and is not equal to αy1s−1 or αyks−1, then there are no nonzero linear g1,g2 that satisfy the condition mentioned above. On the other hand, the cases f=αy1s−1 and f=αyks−1 are excluded by our assumptions. Our argument shows that if (Λ!)(s)≅A0, then f is decomposable. Since f is not linearly dependent with a power of y1 or yk, one of the elements g1,g2 mentioned above is linearly dependent with y1 and, in the case k=2, the second one is linearly dependent with y2. Then it is easy to see that one of the conditions prohibited by the theorem holds. Thus, we have proved that (Λ!)(s)≅A0.
Let us denote by M the A-bimodule (Λ!)(s,1).
Let us denote by R⊂Cs the subspace generated by y1f and fyk. Note that y1(fyk)=(y1f)yk∈RC1∩C1R, and hence M≅km+k (see the proof of Theorem 4.2.2). Let us now prove that M cannot contain right A-module summand isomorphic to Z2. Due to Theorem 4.2.2 it is enough to show that there is no twisted potential in two variables w∈Cs+1 such that {(h⊗1Cs)w∣h∈C1∗}=⟨y1f,fyk⟩.
Here we consider the cases k=1 and k=2 separately. If k=2, then w,f∈k⟨y1,y2⟩. Using the conditions above, we get w=y1(α1y1f+β1fy2)+y2(α2y1f+β2fy2) for some α1,α2,β1,β2∈k. Since w is a ν-twisted potential for some automorphism ν of k⟨y1,y2⟩1, we have
[TABLE]
for some α1′,α2′,β1′,β2′∈k. Let us represent f in the form
[TABLE]
Choosing maximal j such that there is ci1,…,it,j1,…,jt=0 with jt=j and looking at the equality above, we get β2′=0. In a similar way, one can show that α1=β2=α1′=0, and hence
α2y2y1f=β1′fy2y1+(α2′−β1)y1fy2. Since all monomials in the decomposition of y1fy2 start with y1 and end with y2, it follows that α2′=β1. If α2=0, then w=β1y1fy2, and hence {(h⊗1Cs)w∣h∈C1∗} is one dimensional. Thus, we have y2y1f=γfy2y1 for some γ∈k. Hence, s=2t+1 and f equals (y2y1)t in this case.
Let us now consider the case k=1. In this case w,f∈k⟨y1,z⟩ for some z∈C1 and, analogously to the case k=2, we get
[TABLE]
for some α1,α2,β1,β2mα1′,α2′,β1′,β2′∈k. Analogously to the case k=2, one can show that α1=β2=β1′=α2′=0, and hence
α2zy1f=β2′fy1z+(α1′−β1)y1fy1. Now, as before, we get α1′=β1 and zy1f=γfy1z for some γ∈k. Hence, s=2t and f equals (zy1)t−1z in this case. Thus, we have proved that Λ is not an algebra defined by a twisted potential on a two dimensional space.
It remains to prove that M cannot be isomorphic to B3−k⊕km+2k−3. Due to the facts that we have already proved, it suffices to show that there are no nonzero
f,f′∈k⟨x1,…,xm,y1,y2⟩s−1 and linear polynomial g such that ⟨gf,fg⟩=⟨y1f′,f′y2⟩. Suppose that we have found f, f′ and g that satisfy the just mentioned conditions.
We may assume without loss of generality that g and y1 are linearly independent. Indeed, in the opposite case g is linearly independent with y2 and one can simply apply the functor op and interchange y1 and y2.
Then obviously f′=y1f′′ for any f′′∈k⟨x1,…,xm,y1,y2⟩s−2. It is easy to check that f′=αy2s−1 for any α∈k, and hence βy1f′+γf′y2 cannot be presented in the form g′f′′ for some linear g′ if β,γ∈k∗ (see the proof that this tensor cannot be decomposable above).
Thus, we may assume that f′=gf′′ and f=f′′y2 for some f′′∈k⟨x1,…,xm,y1,y2⟩s−2. Then we have ⟨gf′′y2,f′′y2g⟩=⟨y1gf′′,gf′′y2⟩. It is clear that f′′=gf′′′ for any f′′′∈k⟨x1,…,xm,y1,y2⟩s−3. Then βgf′′y2+γf′′y2g cannot be presented in the form g′f′′′ for some linear g′ if β,γ∈k∗. Thus, we have f′′y2g=αy1gf′′ for some α∈k∗. The obtained equality is not valid for any f′′, and hence we get a contradiction that finishes the proof.
∎
4.4 s-homogeneous triples with A=A0 and M=km
Now we are ready to give a criterion for s-homogeneous algebra Λ having the s-homogeneous triple F(Λ!) of the form (A0,km,φ). Note that if Λ is an s-homogeneous algebra with two relations, then the last condition is equivalent to the equality Λs+1!=0. To achieve our purpose we simply exclude all other situations. Thus, we get the following theorem.
Theorem 4.4.1**.**
Suppose that Λ=TkV/(f1,f2), where f1 and f2 are two linearly independent elements of V⊗s. Then (Λ!)(s)≅k⟨x,y⟩/(x,y)2=A0 and (Λ!)(s,1)≅km as an A0-bimodule if and only if non of the following conditions holds:
•
There are some nonzero u,v∈V such that either s=2t and (u⊗v)⊗t,(v⊗u)⊗t∈⟨f1,f2⟩ or s=2t+1 and (u⊗v)⊗t⊗u,(v⊗u)⊗t⊗v∈⟨f1,f2⟩.
•
There is a two dimensional subspace U of V and some twisted potential w∈U⊗(s+1) such that ⟨f1,f2⟩={(f⊗1U⊗s)w∣f∈U∗}.
•
There are some f∈V⊗(s−1) and u,v∈V such that ⟨f1,f2⟩=⟨u⊗f,f⊗v⟩.
4.5 About the s-Koszulity
In this subsection we discuss the s-Koszulity of algebras considered in this paper, i.e. connected s-homogeneous algebras with two relations. It follows from what is already proved in this paper that Λ can be not s-Koszul only in the cases where A=A1 or A=A0.
We will show that the algebras Λ with A=A0 and M=B1⊕km are not s-Koszul and moreover do not satisfy the extra condition.
For all the remaining algebras we will prove criteria that simplify the verification of s-Koszulity.
Namely, we will show that the s-Koszulity follows from the equality (2.2.2) for the Hilbert series and is equivalent to the exactness of the generalized Koszul complex in the second term. Note that it was shown in [26, 29] that the equality for Hilbert series does not imply the Koszul property even for quadratic algebras. Moreover, it was shown in [24] that there exist two algebras one of which is Koszul and another is not Koszul that have the same Hilbert series.
Let us start with a not difficult consideration that the algebras Λ with (Λ!)(s)≅k⟨x,y⟩/(x,y)2=A0 and (Λ!)(s,1)≅B1⊕km are not s-Koszul.
Proposition 4.5.1**.**
Suppose that Λ≅k⟨x1,…,xm,y1⟩/(y1f,fy1) for some f∈k⟨x1,…,xm,y1⟩s−1 such that y1f and fy1 are linearly independent.
Then Λ does not satisfy the extra condition and, in particular, is not s-Koszul except the case where s=2t and f=(gy1)t−1g for some g∈k⟨x1,…,xm,y1⟩1.
Proof.
We have M0∗=i=1⨁mkxi⊕ky1 and φ∗A1∗=kfy1⊕ky1f. Our assumptions and Theorem 4.3.1 guarantee also that (φ∗A1∗)⊗M0∗∩M0∗⊗(φ∗A1∗)=ky1fy1. Note that fy1f∈(φ∗A1∗)⊗(M0∗)⊗(s−1)∩(M0∗)⊗(s−1)⊗(φ∗A1∗), and hence it is enough to show that fy1f∈y1fy1k⟨x1,…,xm,y1⟩s−2. But the last assertion is clear because f and y1s−1 are linearly independent.
∎
Let us now prove the main result of this subsection.
Theorem 4.5.2**.**
Suppose that Λ=TkV/(f1,f2), where f1 and f2 are two linearly independent elements of V⊗s. Then Λ is s-Koszul if and only if its generalized Koszul complex is exact in the second term and if and only if \Big{(}{\mathcal{H}}_{A}(t^{s})-t{\mathcal{H}}_{M}(t^{s})\Big{)}{\mathcal{H}}_{\Lambda}(t)=1, where A=(Λ!)(s) and M=(Λ!)(s,1).
Proof.
Note that the grading on Λ induces a grading on the terms of its generalized Koszul complex ⋯d2K2d1K1d0K0.
We will prove the following assertion. If there is some homogeneous u∈Ki, i>2 such that di−1(u)=0 and u∈Imdi, then there is sum u0∈K2 such that d1(u0)=0, u0∈Imd2 and deg(u0)<deg(u).
Let us first suppose that this fact is valid. Then the exactness in the second term obviously implies the exactness of K in all terms i⩾2 while the generalized Koszul complex is always exact in the first term and has homology isomorphic to k in the zero term.
Suppose now that K is not exact. Let us denote by Ki,j the j-degree component of Ki with respect to the grading induced by the grading of Λ. Let us consider minimal k>0 such that the complex ⋯d2∣K3,kK2,kd1∣K2,kK1,kd0∣K1,kK0,k is not exact. Then this complex is exact in all terms except the second one, and hence the coefficient of tk in \Big{(}{\mathcal{H}}_{A}(t^{s})-t{\mathcal{H}}_{M}(t^{s})\Big{)}{\mathcal{H}}_{\Lambda}(t) equals to dimkKer(d1∣K2,k)−dimkIm(d2∣K3,k)=0.
Note that the assertion of the theorem is obvious if either A≅A0,A1 or A=A0, M=km since in these cases K is automatically exact in degrees i>2. If A=A0 and M=B1⊕km, than Λ does not satisfy the extra condition by Proposition 4.5.1, and hence it is not s-Koszul, K is not exact in the second term, and \Big{(}{\mathcal{H}}_{A}(t^{s})-t{\mathcal{H}}_{M}(t^{s})\Big{)}{\mathcal{H}}_{\Lambda}(t)\not=1.
Let us now consider the case where A=A1 and M=A/(x)⊕km, i.e. Λ=k⟨x1,…,xm,y1⟩/(f,y1s) and f does not satisfy conditions listed in Theorem 4.1.2. In this case the generalized Koszul complex has the form
[TABLE]
where f=y1fy1+i=1∑mxifxi.
If there is some element g∈K2i,j=Λ(−is)j=Λj−is with i>1 such that d2i−1(g)=0 and g∈Im(d2i), then we have y1s−1g=0 and g∈y1Λ,
and hence (g,0)∈K2,j−(i−1)s=Λ2(−s)j−(i−1)s satisfies the conditions d1(g,0)=0 and (g,0)∈Im(d2). Suppose now that g∈K2i+1,j=Λ(−is−1)j=Λj−is−1 with i⩾1 is such that d2i(g)=0 and g∈Im(d2i+1), i.e. y1g=0 and g∈y1s−1Λ. Let us choose maximal 0⩽l<s−1 such that g∈y1lΛ, i.e. g=y1lg′ and g′∈y1Λ. Then we have y1s−1g′=y1s−1−lg=0, and hence (g′,0)∈K2,j−(i−1)s−k−1=Λ2(−s)j−(i−1)s−k−1 satisfies the conditions d1(g′,0)=0 and (g′,0)∈Im(d2).
Suppose now that A=A0 and M=ν−1D⊕km for some graded automorphism ν of A0, i.e. Λ=k⟨x1,…,xm⟩∗D(w) for some ν-twisted potential w on a two dimensional space U with the basis y1,y2 that does not satisfy conditions listed in Theorem 4.2.2.
Let us introduce fij=(fyi⊗fyj⊗1⊗(s−1))w for 1⩽i,j⩽2. Then the definition of a ν-twisted potential implies f1=y1f11+y2f12=f11ν−1(y1)+f21ν−1(y2) and f2=y1f21+y2f22=f12ν−1(y1)+f22ν−1(y2).
In particular, we have D(w)=k⟨y1,y2⟩/(f1,f2).
In this case the generalized Koszul complex has the form
[TABLE]
Suppose that there is some nonzero element g∈K3,j=Λ(−s−1)j=Λj−s−1 such that d2(g)=0, i.e. y1g=y2g=0. Let us pick such an element g with minimal j.
We need to show that there exist homogeneous h1,h2∈D(w) such that d1(h1,h2)=0 and (h1,h2)∈Imd2, i.e. f11h1+f21h2=f12h1+f22h2=0 and there is no h∈D(w) such that h1=ν−1(y1)h and h2=ν−1(y2)h, and deg(h1)=deg(h2)⩽deg(g).
Suppose that the last mentioned assertion is not satisfied. We will show by induction on i that for any ϕ∈D(w)i∗ there is a unique gϕ∈D(w)deg(g)−i such that ugϕ=ϕ(u)g for any u∈D(w)i. This will immediately yield a contradiction with the fact that D(w) has infinite dimension.
The case i=0 is trivial, because D(w)0=k, and hence one has gϕ=ϕ(1)g. Suppose now that we have proved the required assertion for functionals of degree k less than i. Let us pick some ϕ∈D(w)i∗⊂D(w)i−1∗⊗U∗ and write it in the form ϕ=ϕ1r1+ϕ2r2 with ϕ1,ϕ2∈D(w)i−1∗, where r1,r2 is the basis of U∗ dual to ν−1(y1),ν−1(y2). Let us prove that f11gϕ1+f21gϕ2=f12gϕ1+f22gϕ2=0.
If i+s−2>deg(g), then f11gϕ1+f21gϕ2 and f12gϕ1+f22gϕ2 belong to (y1,y2)g=0 by the definition of gϕ1 and gϕ2. Suppose that i+s−2⩽deg(g).
By our induction hypothesis, it is enough to show that u(f11gϕ1+f21gϕ2)=u(f12gϕ1+f22gϕ2)=0 for any u of degree deg(g)−i−s+2. We have
[TABLE]
The second equality can be proved in the same way. Our assumptions imply that there is some gϕ such that gϕ1=ν−1(y1)gϕ and gϕ2=ν−1(y2)gϕ. Then an arbitrary element u∈D(w)i can be presented in the form u=u1ν−1(y1)+u2ν−1(y2), and hence we get
[TABLE]
It remains to show that gϕ is unique that is equivalent to the fact that if h∈D(w)deg(g)−i and uh=0 for any u∈D(w)i, then h=0. Let us pick some u of maximal degree such that uh=0. Then we have y1uh=y2uh=0 and deg(uh)<deg(g) by our assumptions that contradicts the minimality assumption on the degree of g. Thus, we have proved the required assertion in the case where A=A0 and M=ν−1D⊕km.
Finally, suppose that A=A0 and M=B2⊕km, i.e. Λ=k⟨x1,…,xm,y1,y2⟩/(y1f,fy2) and f does not satisfy conditions listed in item 2 of Theorem 4.3.1. In this case the generalized Koszul complex has the form
[TABLE]
where f=y1fy1+y2fy2+i=1∑mxifxi.
Suppose that there is some nonzero element g∈K3,j=Λ(−s−1)j=Λj−s−1 such that d2(g)=0, i.e. y2g=0. Then (0,g)∈K2,j−1=Λ2(−s)j−1 satisfies the conditions d1(0,g)=0 and (0,g)∈Im(d2).
∎
Our next goal is to verify what pairs (A,M) from the s-homogeneous triples corresponding to s-homogeneous algebras with two relations admit an s-Koszul algebra Λ such that (Λ!)(s))=A and (Λ!)(s,1)=M that is equivalent to the isomorphism ExtΛ∗(k,k)≅A⋉M (see Subsection 2.2). We will solve this question for all pairs (A,M) except pairs of the form (A0,ν−1D⊕km), where ν(y1)=λy1 and ν(y2)=λ−sy2 for λ∈k such that λs+1 is a primitive root of unity of degree s−1.
•
Let A=A1 and M=A/(x)⊕km, i.e. Λ=k⟨x1,…,xm,y1⟩/(f,y1s) and f does not satisfy conditions listed in Theorem 4.1.2. It is not difficult to see that Λ is s-Koszul for f=x1y1s−2x2. Note that we have given an example of s-Koszul Λ in the case m>2. If m=2 and s=3, then Λ can not be s-Koszul, because s-Koszulity implies
[TABLE]
that is impossible. If m=2 and s>3, then Λ is s-Koszul for f=y1x1s−1−x1y1s−2x1. To show this, let us choose the lexicographical order on monomials induced by the order y1>x1. Then one can show that the elements y1s, y1x1s−1−x1y1s−2x1 and y1s−1(x1y1s−3)ky1x1 (k⩾1) constitute a Gröbner of the ideal (y1s,f). Now one can directly show that the generalized Koszul complex is exact in the second degree or calculate the Hilbert series of the algebra Λˉ=k⟨x1,y1⟩/(y1s,y1x1s−1−x1y1s−2x1,{y1s−1(x1y1s−3)ky1x1}k⩾1) using the Anick resolution and get that
[TABLE]
•
Let A=A0 and M=ν−1D⊕km, where ν is an automorphism satisfying one of the conditions of Lemmas 4.2.3 and 4.2.4, i.e. Λ=k⟨x1,…,xm⟩∗D(w) for some ν-twisted potential w on a two dimensional space with the basis y1,y2 that does not satisfy conditions listed in Theorem 4.2.2.
In this case Λ is s-Koszul if and only if the algebra D(w) is ν-twisted 3-Calabi-Yau.
If ν(y1)=λy1 and ν(y2)=λ(y1+y2) for some λ∈k such that either s>3 and λs+1=1 or s=3 and λ2=1, then the twisted potential w from the proof of Lemma 4.2.4 defines a twisted 3-Calabi-Yau algebra. Indeed, (f1⊗1⊗s)w and (f2⊗1⊗s)w, where f1,f2 is the basis of V∗ dual to the basis y1,y2, have largest monoms y22y1s−2 and y2y1s−1 respectively. Here we use the lexicographical order induced by the order y2>y1. Then one can check that these elements constitute a Gröbner basis of the ideal \big{(}(f_{1}\otimes 1^{\otimes s})w,(f_{2}\otimes 1^{\otimes s})w\big{)}, and hence Λ has the required Hilbert series.
The case ν(y1)=λ1y1 and ν(y2)=λ2y2 with λ1kλ2s+1−k=1 for some 2⩽k⩽s−1 can be considered in the same way using the potential defined in the proof of Lemma 4.2.3. In this case (f1⊗1⊗s)w and (f2⊗1⊗s)w have largest monoms y2s+1−ky1k−1 and y2s−ky1k respectively.
Suppose now that λ1=1, λ2s=1. Then the potential w=y1s+1+i=0∑sλ2iy2iy1y2s−i defines a ν-twisted 3-Calabi-Yau algebra. Really, we have D(w)=k⟨y1,y2⟩/(y1s+y2s,i=0∑s−1λ2iy2iy1y2s−i−1). Setting y1>y2 and choosing the corresponding lexicographical order on monoms, we get the Gröbner basis
[TABLE]
Then the Hilbert series of D(w) equals the Hilbert series of the algebra k⟨y1,y2⟩/(y1s,y1y2s−1,y1s−1y2y1y2s−2). It can be easily verified that the Hilbert series of the last mentioned algebra is 1−2t+2ts−ts+11.
Finally, in the case λ1λ2s=λ1sλ2=1 and λ1kλ2s+1−k=1 for any 2⩽k⩽s−1 one has λ1=λ and λ2=λ−s for some λ∈k such that λs+1 is a primitive root of unity of degree s−1.
Due to the proof of Lemma 4.2.3, the algebra Λ! has an s-homogeneous triple with the first two components (A0,ν−1D⊕km) if and only if Λ≅k⟨x1,…,xm,⟩∗D(w) with w=i=0∑s+1λiy1iy2y1s−i+i=0∑s+1λ−isy2iy1y2s−i. Thus, the algebra A^{0}\ltimes\big{(}{}_{\nu^{-1}}D\oplus{\bf k}^{m}\big{)} can be Ext-algebra of some s-Koszul algebra if and only if the algebra
[TABLE]
is s-Koszul. The verification of this is a tedious task, because the Gröbner basis of this algebra grows fast while the degree increase. We believe that D(w) is an s-Koszul (and, in particular, twisted CY) algebra with infinite Gröbner basis, but the proof of this is out of this work.
•
Let A=A0 and M=B2⊕km, i.e. Λ=k⟨x1,…,xm,y1,y2⟩/(y1f,fy2) and f does not satisfy conditions listed in item 2 of Theorem 4.3.1. It is not difficult to show that Λ is s-Koszul for f=y1y2s−2.
•
Let A=A0 and M=km, i.e. Λ=k⟨x1,…,xm⟩/(f1,f2), where and f1 and f2 do not satisfy conditions listed in Theorem 4.4.1. It is not difficult to show using Gröbner bases that Λ is s-Koszul for f1=x1s−1x2, f2=x1s−1x3. If s⩾5, then Λ is s-Koszul also for f1=x1x2x1x2s−3, f2=x12x2s−2. Thus, we have shown that Λ can be s-Koszul if m⩾3 or s⩾5. If m=2 and s=3,4, then Λ is s-Koszul if and only if its Hilbert series equals
[TABLE]
or
[TABLE]
that is impossible.
5 Main result, consequences and open problems
In this section, for the convenience of the reader, we give a summary of our classification.
Let Λ=TkV/(f1,f2), where f1 and f2 are two linearly independent elements of V⊗s. Let us denote by A the algebra (Λ!)(s) and by M the A-bimodule (Λ!)(s,1). Then modulo isomorphism one of the following conditions holds:
(A,M)=\big{(}{\bf k}\langle x,y\rangle/(x^{2},xy-qyx),M(p)\oplus{\bf k}^{m}\big{)}, where p,q∈k∗, ps=q and M(p) is defined in Subsection 3.6. In this case
[TABLE]
7. 7.
(A,M)=\big{(}{\bf k}\langle x,y\rangle/(x^{2},xy,yx),A/(x)\oplus{\bf k}^{m}\big{)}. Such pairs correspond to algebras of the form Λ≅k⟨x1,…,xm,y1⟩/(y1s,f) with f∈k⟨x1,…,xm,y1⟩s, where f=i=0∑s−1αiβs−i−1y1igy1s−i−1 and f=gs+αy1s for any linear polynomial g and α,β∈k.
8. 8.
(A,M)=\big{(}{\bf k}\langle x,y\rangle/(x,y)^{2},{}_{\nu^{-1}}D\oplus{\bf k}^{m}\big{)}, where D is the bimodule defined in Subsection 4.2 and ν is a graded automorphism of k⟨x,y⟩/(x,y)2 defined in one of the following ways:
(a)
ν(x)=λ1x, ν(y)=λ2y, where λ1kλ2s+1−k=1 for some 2⩽k⩽s−1;
2. (b)
ν(x)=λ1x, ν(y)=λ2y, where λ1λ2s=λ1sλ2=1 and λ1kλ2s+1−k=1 for any 2⩽k⩽s−1;
3. (c)
ν(x)=λ1x, ν(y)=λ2y, where λ1=1, λ2s=1 and λ2k=1 for any 1⩽k⩽s−1;
4. (d)
ν(x)=λx, ν(y)=λ(x+y), where λs+1=1 if s>3 and λ=±1 if s=3.
Such pairs correspond to algebras of the form Λ≅k⟨x1,…,xm⟩∗D(w) with a ν-twisted potential w∈k⟨y1,y2⟩s+1 on the two dimensional space U with the basis y1,y2, where w=g1gs+g2f, w=(g1g2)t+1+α(g2g1)t+1, and w=(g1g2)tg1+α(g2g1)tg2 for any g1,g2,g∈U, f∈k⟨y1,y2⟩s and α∈k∗.
Here ν:U→U is the composition UθA1νA1θ−1U with the isomorphism θ:U→A1 defined by the equalities θ(y1)=x and θ(y2)=y.
In the cases 8a and 8d one has as minimum one algebra Λ such that (Λ!)(s)=A0 and (Λ!)(s,1)=ν−1D⊕km. In the cases 8b and 8c one has exactly one such an algebra. Namely, in the case 8b the unique ν-twisted potential algebra satisfying the required properties is
[TABLE]
while in the case 8c the unique algebra satisfying the required properties is
[TABLE]
9. 9.
(A,M)=\big{(}{\bf k}\langle x,y\rangle/(x,y)^{2},B_{1}\oplus{\bf k}^{m}\big{)}, where B1 is the bimodule defined in Subsection 4.3. Such pairs correspond to algebras of the form Λ≅k⟨x1,…,xm,y1⟩/(y1f,fy1) with f∈k⟨x1,…,xm,y1⟩s−1, where f=(gy1)t−1g for any linear polynomial g.
10. 10.
(A,M)=\big{(}{\bf k}\langle x,y\rangle/(x,y)^{2},B_{2}\oplus{\bf k}^{m}\big{)}, where B2 is the bimodule defined in Subsection 4.3. Such pairs correspond to algebras of the form Λ≅k⟨x1,…,xm,y1,y2⟩/(y1f,fy2) with f∈k⟨x1,…,xm,y1,y2⟩s−1, where f=α(y2y1)t, f=αy1s−1, and f=αy2s−1 for any α∈k∗.
11. 11.
(A,M)=\big{(}{\bf k}\langle x,y\rangle/(x,y)^{2},{\bf k}^{m}\big{)}. Such pairs correspond to all algebras that do not appear in one of the previous items. In more details this condition is written down in Theorem 4.4.1.
Let us now collect the results on the s-Koszulity of s-homogeneous algebras with two relations.
•
If one of Conditions 1–6 above is satisfied, then Λ is an s-Koszul algebra.
•
If Conditions 7 is satisfied, then Λ can be s-Koszul in the case where m>2 or s>3. In the case (m,s)=(2,3) the algebra Λ is not s-Koszul.
•
If Conditions 8 is satisfied, then Λ can be s-Koszul in the cases 8a, 8c and 8d. It is not known if Λ is s-Koszul or not in the case 8b
•
if Condition 9 above is satisfied, then Λ does not satisfy the extra condition, and hence it is not s-Koszul.
•
If Conditions 10 is satisfied, then Λ can be s-Koszul for any m⩾2 and s⩾3.
•
If Conditions 11 is satisfied, then Λ can be s-Koszul in the case where m>2 or s>4. In the cases (m,s)=(2,3) and (m,s)=(2,4) the algebra Λ is not s-Koszul.
•
If one of Conditions 7, 8, 10, 11 is satisfied, then the s-Koszulity of Λ is equivalent to the exactness of the generalized Koszul complex in the second term and, at the same time, to the equality {\mathcal{H}}_{\Lambda}(t)=\Big{(}{\mathcal{H}}_{A}(t^{s})-t{\mathcal{H}}_{M}(t^{s})\Big{)}^{-1}, where the right hand side of the last mentioned equality can be easily calculated in each case.
Note that if Λ is s-Koszul, then, as it was mentioned above, ExtΛ∗(k,k) is the trivial extension of A by M with an appropriate grading. Thus, we have described a set of possible Ext-algebras of connected s-Koszul algebras with two relations and show that all of these algebras are possible except the algebras arising in the case 8b.
5.2 Some consequences and problems
Here we mention some results that follow from our classification and state some problems that seem to be interesting for us.
Note first of all that the number of different quadratic algebras that can appear as first components of an s-homogeneous triple F(Λ!) is very restricted: only the algebras A0, A1, A4, A5, A6 and A7(q) (q∈k) are possible. Moreover, all of this algebras are Koszul and have minimum two relations.
Problem 1**.**
Given a quadratic algebra A, what are the sufficient and necessary conditions for A to be isomorphic to Λ(s) for some s-homogeneous algebra Λ? Does A have to be Koszul? In the case where the number of generators of A is given, how many relations can A have?
As it was stated above, an Ext-algebra of an s-Koszul algebra with two relations is isomorphic to A⋉M, where the pair (A,M) is described in one of Conditions 1–11 of the previous subsection. For each such a pair it is determined if it really can be Ext-algebra with only one exception. This exception motivates the next question.
Problem 2**.**
Given λ∈k such that λs+1 is a primitive root of unity of degree s−1, is the algebra
[TABLE]
s-Koszul?
If the answer to this problem is positive, then we will get the following result. If there exists an s-homogeneous algebra Λ such that \big{(}(\Lambda^{!})^{(s)}),(\Lambda^{!})^{(s,1)}\big{)}=(A^{0},{}_{\nu^{-1}}D), then there exists an s-Koszul algebra Γ such that \big{(}(\Gamma^{!})^{(s)}),(\Gamma^{!})^{(s,1)}\big{)}. This assertion has a generalization to an arbitrary number of relations.
Problem 3**.**
Suppose that there exists a ν-twisted potential w on an n-dimensional space V such that the generalized Koszul complex of D(w) has the form
[TABLE]
Does it follows from this that there exists a ν-twisted potential w′ such that D(w′) is twisted 3-CY?
In our classification, the algebras are divided into several concrete algebras and several series of algebras. It would be interesting to give for each of this series a criterion for s-Koszulity like in Theorem 2.2.2.
Problem 4**.**
For the pairs (A,M) satisfying one of the Conditions 7, 8, 10 and 11, give some easily verifiable criterion of s-Koszulity.
Our result allow to compute all possible Hilbert series of s-Koszul algebras with two relations that can be computed using (2.2.2). Namely, the possible Hilbert series are:
[TABLE]
All of these series are rational functions and this lead to the next problem (the same problem for Koszul algebras is stated in [25, Conjecture 1]).
Problem 5**.**
Is the Hilbert series rational for any s-Koszul algebra?
All the Hilbert and Poincaré series of s-homogeneous algebras with one relation were computed in [10].
Problem 6**.**
Find all the possible Hilbert and Poincaré series of s-homogeneous algebras with two relations.
The computations of [10], in particular, imply that the global dimension of an s-homogeneous algebra with one relation can be two of infinity. Our results show that the global dimension of an s-Koszul algebra with two relations can be two, three or infinity.
Problem 7**.**
Is the global dimension of an s-homogeneous restricted by some function depending on the number of relations in the case where it is finite?
Is the number of possible Hilbert Series of s-homogeneous algebras finite for given numbers of generators and relations? Are these assertion true for s-Koszul algebras in the case of a negative answer for arbitrary s-homogeneous algebras?
Finally, we have shown that if the s-homogeneous algebra Λ has m generators and ms−2 relations, then it cannot be s-Koszul. On the other hand, if s>4 or m>2, then a generic s-homogeneous algebra with m generators and two relations is s-Koszul. One can show that in the case where s=3,4 and m=2, a generic algebra is not s-Koszul.
Problem 8**.**
For which triples (s,m,k) of integer numbers there exists an s-Koszul algebra with m generators and k relations? For which triples (s,m,k) of integer numbers a generic algebra with m generators and k relations is s-Koszul?
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