This paper introduces a new family of Artin-Schelter Gorenstein Hopf algebras, explores their relationships with known Hopf algebras, and computes their Grothendieck rings, revealing non-isomorphic algebras with identical Grothendieck rings.
Contribution
It defines a new series of Hopf algebras $H_eta$, analyzes their properties, and determines their Grothendieck rings, connecting them to existing Hopf algebras.
Findings
01
Radford's and Gelaki's Hopf algebras are homomorphic images of $H_eta$.
02
The Grothendieck ring $G_0(H_eta)$ is explicitly determined.
03
Non-isomorphic Hopf algebras with isomorphic Grothendieck rings are identified.
Abstract
We define a series of Artin-Schelter Gorenstein Hopf algebras Hβ with injective dimensions 3. Radford's Hopf algebra and Gelaki's Hopf algebra are homomorphic images of Hβ. We determine its Grothendieck ring G0(Hβ). Meanwhile we can obtain Grothendieck rings of Gelaki's Hopf algebras and Radford's Hopf algebras U(N,ν,ω) in \cite{R}, and non-isomorphic Hopf algebras with isomorphic Grothendieck rings.
No public reviews on file for this paper yet. If you reviewed it on a platform where reviews are public (OpenReview, ICLR, NeurIPS, ICML), you can paste yours below so the community can read it here.
Videos
No videos yet. Explain this paper in a talk, walkthrough, or lecture? Add one.
Taxonomy
TopicsAlgebraic structures and combinatorial models · Advanced Topics in Algebra · Nonlinear Waves and Solitons
We give a series of infinite dimensional noncommutative and noncocommutative pointed Hopf algebras, which are Artin-Schelter Gorenstein Hopf algebras with injective dimensions 3. Radford’s Hopf algebra and Gelaki’s Hopf algebra are homomorphic images of these Hopf algebras.
We determine the irreducible representations of them. We describe the Grothendieck rings of them. We obtain that two non-isomorphic Hopf algebras could have isomorphic Grothendieck rings.
Hopf algebra, finite-dimensional representation, Grothendieck ring
2000 Mathematics Subject Classification:
Primary 17B37, 81R50, 16E30, 06B15, 16T05
The author Jing Wang is supported by NNSFC (No.11901034) and the Fundamental Research Funds for the Central Universities (No.BLX201721). The work of Zhixiang Wu and Yan Tan is sponsored
by ZJNSF (No. LY17A010015) and NNSFC (No.11871421).
1. Introduction
The tensor product of representations of a Hopf algebra is important in the representation theory of Hopf algebras and quantum groups. In particular, the decomposition of the tensor product of indecomposable modules into a direct sum of indecomposable modules has received enormous attention. However, little is known about how a tensor product of two indecomposable modules decomposes into a direct sum of indecomposable modules over a Hopf algebra or a quantum group. There are some results for the decompositions of tensor products of modules over a Hopf algebra or a quantum group [3, 5, 8, 9, 16]. Recently, Green rings and their homomorphic images, which are called Grothendieck rings, of various finite-dimensional Hopf algebras have attracted numerous attentions [4, 6, 13, 18]. But most of them are considered in the case of finite dimensional Hopf algebras or quantum groups.
In this paper, we describe an infinite-dimensional pointed Hopf algebra Hβ for any β∈(K∗)3 and its Grothendieck ring, which differs from the Green ring. The Hopf algebra is constructed by adding group-like elements to the restricted quantum enveloping algebra of sl2. The Hopf algebra Hβ has
many interesting finite dimensional images, such as Gelaki’s Hopf algebra U(n,N,ν,q,α,β,γ) and Radford’s Hopf algebra U(N,ν,ω). Here, all irreducible representations of Hβ and the decomposition of the tensor product of two irreducible Hβ-modules are determined as in [2], but we consider the decomposition in the sense of Grothendieck group. Thus we obtain the Grothendieck ring of the infinite-dimensional Hopf algebras Hβ. Meanwhile we obtain the Grothendieck rings
of Gelaki’s Hopf algebra U(n,N,ν,q,α,β,γ) and Radford’s Hopf algebra U(N,ν,ω).
This paper is organized as follows. In Section 2, we introduce the infinite dimensional noncommutative and noncocommutative Hopf algebra Hβ, which is a pointed AS-Gorenstein Hopf algebra of injective dimension 3. And we prove that the homological integral of Hβ is isomorphic to K as Hβ-bimodules. To construct irreducible representations of Hβ, we define a finite-dimensional quotient Hopf algebra Hα,β of Hopf algebra Hβ in Section 3, where α=(n,m,n1,n2,n3)∈N5,
β=(β1,β2,β3)∈K3. We prove that the algebra Hα,β could be decomposed into direct sum of matrix rings over some commutative local rings. Based on this result, all irreducible representations of Hβ are illustrated in Section 4. Besides, we point out that the category of all finite-dimensional representations of Hβ is not semisimple. Meanwhile we obtain all irreducible representations of Radford’s Hopf
algebra U(N,ν,ω) and Gelaki’s Hopf algebra U(n,N,ν,q,α,β,γ). In Section 5, the Grothendieck ring G0(Hβ) of Hβ is constructed. We illustrate the structure of the Grothendieck ring G0(Hβ) of Hβ in several cases. In details, Theorem 5.2(iv) describes the case in which β1=β2=β3=0. Theorems 5.5, 5.12 and 5.16 illustrate the cases that iff two the three equations β1=0, β2=0 and β3=0 hold. In theorems 5.10, 5.18 and 5.20, we give the cases in which iff one of the conditions β1=0, β2=0 and β3=0 holds. In additional, we also obtain the Grothendieck rings G0(U(N,ν,ω)) and G0(U(n,N,ν,q,α,β,γ)). We prove that there exists non-isomorphic Hopf algebras with isomorphic Grothendieck rings.
Throughout this paper, K denotes an algebraically closed field of characteristic zero, K∗ the group of all nonzero elements of K with the product of K and q∈K a primitive n-th root of unity, n≥2. For any z∈K∗, we denote ⟨z⟩ the subgroup of K∗ generated by z, the quotient group K∗/N of K∗ with respect to a normal subgroup N, and zˉ the element zN.
We simplify the tensor product V⊗KW of two vector spaces over K as V⊗W. The ring of integers is denoted by Z and (N1,⋯,Nk) denotes the greatest common divisor of N1,⋯,Nk, for any N1,⋯,Nk∈Z.
2. Definition of Hopf algebra Hβ
In this section, we give the definition of Hopf algebra Hβ. Then we prove that Hβ is an Artin-Schelter Gorenstein Hopf algebra with injective dimension 3.
Definition 2.1**.**
Let n1 be a positive integer such that 2∤(n,n1) and 1≤n1≤n. Assume that q∈K is a primitive n-th root of unity. Suppose that Hβ is an algebra generated by a,b,c,x,y with the relations
[TABLE]
[TABLE]
for β=(β1,β2,β3)∈K3. The coproduct Δ and counit ε of Hβ is determined by
[TABLE]
and
[TABLE]
respectively.
An anti-automorphism s of Hβ is determined by
[TABLE]
In the following theorem, we prove that Hβ is an infinite dimensional Hopf algebra with the above coproduct Δ, counit ε and antipode s.
Theorem 2.2**.**
The algebra (Hβ,Δ,ε,s) is a pointed Hopf algebra with a basis
[TABLE]
Proof.
First, we prove that Δ can determine a homomorphism of
algebras from Hβ to Hβ⊗Hβ.
By the definition of Hβ, we obtain that Δ(x)k=l=0∑k(lk)qn1blxk−l⊗a(k−l)n1xl, where (lk)qn1=(l)qn1!(k−l)qn1!(k)qn1! and (p)qn1!=(p)qn1(p−1)qn1⋯(1)qn1 for (p)qn1=1+qn1+⋯+q(p−1)n1.
Hence
[TABLE]
Similarly, we can prove
Δ(yn)=Δ(y)n. Moreover,
[TABLE]
It is easy to verify that
[TABLE]
[TABLE]
[TABLE]
Thus
Δ induces a homomorphism of algebras.
Second, it is easy to prove that ε is a
homomorphism from the algebra Hβ to K and
[TABLE]
for any K∈{a,b,c,x,y}.
Thus Hβ is a bialgebra.
At last, we need to prove that s is an antipode of the algebra Hβ.
By s(x)=−q−n1a−n1b−1x,
we get that
[TABLE]
It is not difficult to prove that s(x)n=s(xn). Indeed,
if n is odd then (−1)n+1q21n(n−1)n1=1.
If n is even then we obtain that n1 is odd by 2∤(n,n1) and
[TABLE]
Similarly, we have
s(yn)=s(y)n. Moreover,
[TABLE]
It is easy to verify that
[TABLE]
[TABLE]
[TABLE]
Thus s is an anti-automorphism of Hβ.
Hence, we only need to verify the antipode axiom on the generators x,y,a,b,c. It is easy to verify this.
Altogether, we have proved that Hβ is a Hopf algebra.
By [12, Lemma 1] or [7, Lemma 1.1], we obtain that the Hopf algebra Hβ is pointed and
the set {aibjck∣a,b,c∈Z} is of all group-like elements of Hβ.
Similarly to [7, 12], we can prove that
{aibjckxuyv∣i,j,j∈Z,0≤u,v≤n−1} is a basis
of Hβ by the Diamond Lemma [1].
∎
Remark 2.3**.**
Notice that if the generators of Hβ
satisfy the relations
[TABLE]
*then the Hopf algebra Hβ is a Gelaki’s Hopf algebra U(n,N,n1,q,β1,β2,β3), which is defined in [7].
*
Note that if ω is a primitive N-th root of unity and
N∤n12, then
[TABLE]
as Hopf algebras for any γ∈K∗.
Especially,
[TABLE]
Here, U(N,n1,ω) is called Radford’s Hopf algebra.
Let S={aknblct∣k,l,t are nonnegative integers}.
Then S is a multiplicatively closed set of K[an,b,c]. Since K[a±n,b±1,c±1] is the localization of K[an,b,c] with respect to S, we get that the
Gelfand-Kirillov dimensions of K[a±n,b±1,c±1] and K[an,b,c] are equal by [11, Proposition 8.2.13].
For simplification, we write the Gelfand-Kirillov dimension as GK dim in this section.
Recall that a Hopf algebra A over field K is Artin-Schelter Gorenstein (simply AS-Gorenstein) defined in [14, Definition 3.1] if
(AS1) injdim AA=d<∞,
(AS2) dimKExtAd(AK,AA)=1,ExtAi(AK,AA)=0 for all i=d,
(AS3) the right A-module versions of the conditions (AS1,AS2)
hold.
Theorem 2.5**.**
Hβ* is an AS-Gorenstein Hopf algebra of injective dimension 3.*
Proof.
Let T=K[a±n,b±1,c±1]. Then Hβ is finitely generated module over the Noetherian ring T. So Hβ is both right and left Noetherian. By [11, Corollary 13.1.13], Hβ is a PI Hopf algebra. Moreover, by [11, Propositions 8.2.9, 8.2.13 and 8.1.15] we obtain that
[TABLE]
By [14, Theorem 0.1], Hβ is an AS-Gorenstein Hopf algebra of injective dimension 3.
∎
We recall the homological integral of H from [10]. Let H be an AS-Gorenstein Hopf algebra of injective dimension d over field K.
The vector space of left homological integrals of H is defined as ∫Hl:=ExtHd(HK,HH) and the right homological integrals of H as ∫Hr:=ExtHd(KH,HH).
Besides, homological integrals agree with the classical integrals for finite dimensional Hopf algebra in [10, section 1]. Then we get the following corollary.
Corollary 2.6**.**
ExtHβop3(KHβ,HβHβ)≃ExtHβ3(HβK,HβHβ)≃K* as two sided Hβ-modules.*
Proof.
Let H′=Hβ/(b−1),H′′=H′/(c−1) and H′′′=H′′/(an(n−1)−1). Then H′′′ is isomorphic to a Gelaki’s Hopf algebra U(n,n(n−1),n1,q,β1,β2,β3). Since
a(n−1)n−1,b−1,c−1 are in the center of Hβ, we get ∫Hβl=∫H′l=∫H′′l=∫H′′′l by [10, Lemma 2.6].
Since H′′′ is finite dimensional, by [7, Proposition 3.2] we obtain that ∫H′′′l=Kλ, where
λ=n(n−1)1(i=0∑n(n−1)−1ai)xn−1yn−1. It follows that Kλ≃K as Hβ-bimodules, since k⋅λ=ε(k)λ for any k∈Hβ the same action as on 1∈K.
∎
3. Properties of Hα,β
In this section, we construct and illustrate some properties of the Hopf algebra Hα,β, which is a quotient of Hβ.
It is necessary to determine all irreducible representations of Hβ in the next section.
Definition 3.1**.**
Assume that N=mn(n−1), where m≥1. Let α=(n,m,n1,n2,n3)∈N5, where 1≤n1<m(n−1), 0≤n2,n3<n−1. Let I be an ideal of Hβ generated by aN−1,b−amnn2 and c−amnn3.
Let
[TABLE]
Then Hα,β is a Hopf algebra
generated by a,x,y satisfying
[TABLE]
[TABLE]
The coalgebra structure of Hα,β is determined by
[TABLE]
[TABLE]
The antipode of Hα,β is determined by
[TABLE]
Besides, we can prove that λ=i=0∑N−1Naixn−1yn−1 is a two-sided integral of Hα,β. Indeed,
by yx−q−n1xy=β3(a2n1−amn(n2+n3)) we get that
[TABLE]
where uk=q−(k−1)n1+⋯+q−n1+1.
Let α=(n,m,n1,n2,n3),α′=(n′,m′,n1′,n2′,n3′),
β=(1,1,β3) and β′=(1,1,β3′). Suppose that β1β2=0. Then we can assume that β1=β2=1. Moreover, we assume that xn=0, yn=0 and yx=q−n1xy. Under these assumptions, we have the following proposition.
Proposition 3.2**.**
Hα,β≃Hα′,β′* if and only if α=α′ and β3=cβ3′,
where cn=1.*
Proof.
Let f be an isomorphism from Hα,β to Hα′,β′ and Hα′,β′ be a Hopf algebra generated
by g=a+I′, z=x+I′ and w=y+I′, where I′ is the ideal generated by an′(n′−1)m′−1,b−an′m′n2′ and c−am′n′n3′.
Notice that the subset G(Hα,β) of all group-like elements of Hα,β is equal to
{at∣0≤t≤n(n−1)m−1}, and the subset G(Hα′,β′) of all group-like elements of Hα′,β′ is equal to
{gt∣0≤t≤n′(n′−1)m′−1}. Since
f induces an isomorphism from G(Hα,β) to
G(Hα′,β′) and mn(n−1)=m′n′(n′−1). Moreover, we can assume that f(a)=g. Since f(a)f(x)=qf(x)f(a), f(x)=i=0∑N−1j=1∑n′−1xijgizjwj−1 for some xij∈K.
By Δ(f(x))=(f⊗f)Δ(x), we get f(x)=x01z. It is obvious that x01=0.
Similarly, we can prove that f(y)=uy for some nonzero u∈K. Since f(s(x))=s(f(x)), we have q−n1a−mnn2−n1x=q−n1′a−m′n′n2′−n1′x. Then mnn2=m′n′n2′ and n1=n1′. Since xn=ann1−amnn2, we have
[TABLE]
Since gnn1−mnn2=0, we have x01n=1 and n=n′. Thus m=m′ and n2=n2′. Similarly, we can prove that un(gn′n1′−gm′n′n3′)=gnn1−gmnn3.
Hence n3=n3′ and un=1. Notice that
by applying the
function f to the equation yx−q−n1xy=β3(a2n1−amn(n2+n3)), we obtain the equation
β3=ux01β3′.
∎
Remark 3.3**.**
The Hopf algebra Hα,β is finite dimensional and by Theorem 2.2 we get
[TABLE]
Especially, if n2=n3=0 then Hα,β is Gelaki’s Hopf algebra by Remark 2.3.
Next, we determine the algebraic structure of Hα,β. For this purpose, we construct some idempotent elements of Hα,β. Suppose that ω0 is a primitive m(n−1)-th root of unity and
[TABLE]
Note that j=0∑m(n−1)−1(ω0i)j=0 for any
i=0. Thus we have the following claim.
Lemma 3.4**.**
1=e0+e1+⋯+em(n−1)−1* is a sum of central idempotent elements.*
Proof.
Since anx=xan,any=yan, ei are in the center
of Hα,β. Hence
[TABLE]
and
[TABLE]
where δij is the
Kronecker symbol. The claim is true.∎
Let Ai=eiHα,β, where i∈{0,1,⋯,m(n−1)−1}. Since
eian=ω0−iei, we obtain that the set {eiajxlyt∣0≤j,l,t≤n−1} is
a basis of Ai and dimKAi=n3. It follows that Hα,β is a direct sum of algebras Ai by Lemma 3.4.
Lemma 3.5**.**
Let ω be a primitive N-th root of unity such that ωn=ω0. Then Ai is generated by g=ωieia,x′,y′,
satisfying the following relations
[TABLE]
and
[TABLE]
where \beta_{1}^{\prime}=\beta_{1}(\omega_{0}^{-n_{1}i}-\omega_{0}^{-mn_{2}i}),\\beta_{2}^{\prime}=\left\{\begin{array}[]{ll}\beta_{2}(\omega_{0}^{-n_{1}i}-\omega_{0}^{-mn_{3}i})&\beta_{3}=0\\
\beta_{2}\beta_{3}^{-n}(\omega_{0}^{n_{1}i}-\omega_{0}^{(2n_{1}-mn_{3})i})&\beta_{3}\neq 0\end{array}\right..
Proof.
Let x′=xei and y^{\prime}=\left\{\begin{array}[]{ll}ye_{i}&\beta_{3}=0\\
\beta_{3}^{-1}\omega^{2n_{1}i}ye_{i}&\beta_{3}\neq 0\end{array}\right..
It is obvious that Ai is generated by g,x′,y′. It is easy to
check the relations on these generators g,x′,y′ in this lemma.
∎
For the rest of this section, we illustrate the properties of Ai.
For simplification, we denote A as an algebra
generated by g,x,y, satisfying
[TABLE]
and
[TABLE]
Let fi=n1j=0∑n−1(qig)j,i=0,1,⋯,n−1. Then
[TABLE]
and fifj=δij. Thus
[TABLE]
as a direct sum of left ideals.
Lemma 3.6**.**
Suppose that either β1′=0 or β2′=0. Then A is isomorphic to
[TABLE]
where Ri are commutative local rings.
Proof.
It is easy to verify that fix=xfi−1 and
fiy=yfi+1. If β1′=0, then Afi is isomorphic to
Afi−1 as left A-modules by multiplying x from the right.
If β2′=0, then Afi is isomorphic to Afi+1 as
left A-modules by multiplying y from the right. Thus A is
isomorphic to HomA(A,A)≃Mn(EndA(Afi)). Since
[TABLE]
EndA(Afi)=fiAfi=span{fixtytfi∣t=0,1,⋯,n−1}.
Notice that
flyxfl−q−n1flxyfl=(q−2n1l−ω02n1i−m(n2+n3)i)fl when β3=0.
Let
γl=0 if β3=0 and γl=q−2n1l−ω02n1i−m(n2+n3)i if β3=0.
Then
[TABLE]
Similarly, we can prove that
(fixyfi)k=q−(k−1)n1fixkykfi+bkfixk−1yk−1fi+⋯+b1
for k≥2. Hence R=EndA(Afi) is generated by
fixyfi over the field K as an algebra. Since dimKR=n, there exists a minimal polynomial p(x) such that
p(fixyfi)=0. Therefore, R=R1⊕⋯⊕Rt is a
direct sum of local rings.∎
Lemma 3.7**.**
Suppose that β1′=β2′=β3=0. Then the Jacobson radical J(A) of A is
spanned by {gixjyk∣0≤i,j,k≤n−1,j+k>0}
and A/J(A)≅K[Zn].
Proof.
Since the ideal I=⟨x,y⟩ is nilpotent, we have I⊆J(A). Besides, there is a K algebra isomorphism A/I=K[g]/⟨gn−1⟩≅K[Zn]. It follows that J(A)=I.
∎
Lemma 3.8**.**
Suppose that β1′=β2′=0 and β3=0. Then A is generated by K,E,F with relations
[TABLE]
[TABLE]
where ρ=ω02n1i−mn(n2+n3)i.
Proof.
Let E=qn1−11ygn−n1,F=qn1+11x, K=g, and K−1=gn−1. It is obvious that
A is generated by E,F,K. Thus we have EF−FE=K2n1qn1−q−n1Kn1−ρK−n1 from yx−q−n1xy=g2n1−ω2n1i−mn(n2+n3)i.
By E=qn1−11ygn−n1, then
En=(qn1−11ygn−n1)n=(qn1−1)n1q2n(n−1)(n−n1)yng(n−n1)n. Hence En=0 by the relation (2) yn=β2′=0 of A.
The other relations hold trivially.
∎
Notice that a weak Hopf algebra related to the algebra Ai in Lemma 3.8 has been studied in [17]. Especially, if N∣[2n1i−mn(n2+n3)i] then the algebra A is isomorphic to Radford’s Hopf algebra U(n,n1,ω).
4. Irreducible representations of Hβ
In this section, we determine all irreducible representations of Hβ. The following key Lemma is necessary.
Lemma 4.1**.**
Every irreducible representation of Hβ is finite-dimensional.
Proof.
The proof is similar to that of [15, Corollary 1.2]. For the sake of completeness, we give a sketch. Let M be a simple module over Hβ and T:=K[a±n,b±1,c±1]. Since T is contained in the center
of Hβ, every element t of T induces an endomorphism of the
Hβ-module M. We denote φ(t) as the endomorphism
of M induced by t∈T. Suppose that K={φ(t)∣t∈T}. Then kM=M for any nonzero k∈K. Since Hβ is finitely
generated over T, M is a finitely generated K-module. Let m1,m2,⋯,mr be the generators of M.
By kM=M, we obtain
[TABLE]
for some aij∈K.
Then det(I−k(aij))=0, where I is the identity matrix with
order r. Thus, there exists h∈K such that kh=1. Hence K is a field. Notice that
K=K[φ(a±n),φ(b±1),φ(c±1)]. It is an algebraic extension of K.
Since K is an algebraically closed field, K=K. Consequently, M is finite-dimensional.
∎
From the proof of Lemma 4.1, we can construct an algebra homomorphism λM∈HomK(T,K) for any simple module M.
Let
λM(an)=γ1,λM(b)=γ2 and λM(c)=γ3.
Since a,b,c are invertible in T, γi=0 for i=1,2,3.
Then M can be
viewed as a module over HM:=Hβ/(an−γ1,b−γ2,c−γ3).
Let a′=nγ11a, b′=nγ1n11b, c′=nγ1n11c, x′=x and
y^{\prime}=\left\{\begin{array}[]{ll}y&\beta_{3}=0\\
\beta_{3}^{-1}\gamma_{1}^{-\frac{2n_{1}}{n}}y&\beta_{3}\neq 0\end{array}\right.. Then Hβ is generated by a′,b′,c′,x′,y′
with the relations
[TABLE]
[TABLE]
[TABLE]
where
β1′=γ1n1β1,\beta_{2}^{\prime}=\left\{\begin{array}[]{lll}\gamma_{1}^{n_{1}}\beta_{2}&&\beta_{3}=0\\
\beta_{3}^{-n}\gamma_{1}^{-n_{1}}\beta_{2}&&\beta_{3}\neq 0\end{array}\right.. Thus the generators a′,x′,y′ in
HM satisfy
[TABLE]
[TABLE]
In the following we illustrate all irreducible representations of Hβ.
Lemma 4.2**.**
Let β1′′=β1′(1−γ1n1γ2n),β2′′=β2′(1−γ1n1γ3n), and
[TABLE]
(I)
Suppose that β1′′=0. Then the irreducible
Hβ-module M has a basis {m0,m1,⋯,mn−1} such that the actions of a,b,c,x,y on M with respect to this basis are given by
[TABLE]
[TABLE]
[TABLE]
where k1,⋯,kn are determined by
[TABLE]
for 2≤l≤n and k1k2⋯kn=(γ1n1−γ3n)β2. We denote this irreducible module M by VI(γ1,γ2,γ3;i).
(II)
Suppose that β1′′=0 and β2′′=0. Then the irreducible
Hβ-module M has a basis
[TABLE]
such that the actions of a,b,c,x,y on M with respect to this basis satisfying
[TABLE]
[TABLE]
[TABLE]
where k1,⋯,kn are determined by
[TABLE]
for 1≤l≤n−1 and k1k2⋯kn=(γ1n1−γ2n)β1.
We denote this irreducible module M by VII(γ1,γ2,γ3;i).
Proof.
The proof of (I) is similar to that of (II). We only
give the proof of (II). Similarly to the proof of Lemma 3.6, we can
prove that HM is a direct sum of Mn(Rt), where Rt are
commutative local rings. Hence every irreducible module over HM
is of dimension n. Since Hβ module M can be viewed as a HM module, dim(M)=n. Let m0 be an eigenvector of a′ with
eignvalue qi for some 0≤i≤n−1. Since
y′nm0=β2′′m0=0,
{m0,y′m0,⋯,y′n−1m0} is a basis of M. Moreover
a′x′m0=q−1x′a′m0=qi−1x′m0. Therefore x′m0=kn′y′n−1m0
for some kn′∈K. Thus M is a Hβ module with a basis {m0,ym0,⋯,yn−1m0} and the action of a and
y on M with respect to this basis is given by (6) and (7) respectively. Meanwhile, xm0=knyn−1m0
for some kn∈K. Since xa=qax and
ylx−q−ln1xyl=β3∑j=0l−1q−jn1(q−(l−1)n1a2n1−bc)yl−1, the action of x can be
realized by
[TABLE]
where
[TABLE]
Since Xn=k1k2⋯knIn=β1(γ1n1−γ2n)In, kn is a root of the equation
[TABLE]
Hence, the claim is true.
∎
Lemma 4.3**.**
Suppose that β1′′=β2′′=β3′′(i)=0 for some 0≤i≤n−1. Then the
irreducible Hβ module M=K is determined by a⋅1=nγ1qi,b⋅1=γ2,c⋅1=γ3 and
x⋅1=y⋅1=0 for some 0≤i≤n−1. We denote this irreducible module M by V0(γ1,γ2,γ3;i) in the sequel.
Proof.
We first consider M as the irreducible HM module. Suppose that m0 is an eigenvector of a′. Since x′nm0=0, we can assume that
x′m0=0. If y′m0=0, then there exist j∈{1,⋯,n−1} such that y′jm0=0 and y′j+1m0=0. Hence {m0,y′m0,y′2m0,⋯,y′jm0} is a basis of the module M. <y′jm0> is the submodule of M by x′y′jm0=qjn1y′jx′m0=0. This gives a contradiction. Thus x⋅m0=y⋅m0=0 and M=Km0≅K.
∎
Suppose that β3(i)′′=β3(q2n1i−γ1−n2n1γ2γ3)=0. If there is an integer v such that
[TABLE]
then there is a minimal positive integer r such that q(2i−r+1)n1=γ1−n2n1γ2γ3. As β3(i)′′=0, we have q(2i−r+1)n1=q2n1i and q(1−r)n1=1. Thus 2≤r≤t, where t=∣qn1∣ the order of qn1.
If there exists no integer v such that q(2i−v+1)n1=γ1−n2n1γ2γ3,
then we define r:=t.
Lemma 4.4**.**
Suppose that β1′′=β2′′=0,
but β3′′(i)=0 for some 0≤i≤n−1. Then the irreducible Hβ module M has a basis {mi∣0≤i≤r−1} such that the actions of a,b,c,x,y are given by
[TABLE]
[TABLE]
[TABLE]
where k1,⋯,kr are determined by
[TABLE]
for 1≤l≤r−1.
Proof.
We first consider M as the irreducible HM module. Suppose that m0 is an eigenvector of a′. Since y′nm0=0, we can assume that
y′m0=0. Moreover, we can assume that x′jm0=0 for 0≤j≤r′−1 and x′r′m0=0. It is obvious that r′≤n and
a′m0=qim0 for some 1≤i≤n. It is easy to check that
[TABLE]
Since
[TABLE]
either qr′n1=1, or
q(2i−r′+1)n1=γ1−n2n1γ2γ3. If r′>t, then
span{x′tm0,⋯,x′r′−1m0} is a nonzero submodule of the simple module M. This is impossible.
Thus r′≤t. Similarly, we have r≤r′. Hence
r′=min{t,r}. Since
y′x′jm0=(v=0∑j−1q(2i−v)n1−v=0∑j−1qvn1γ1−n2n1γ2γ3)x′j−1m0
for all j≥1, y′(x′lm0)=kl′(x′l−1m0) for 1≤l≤r−1, where kl′=∑j=0l−1q−jn1(q(2i−l+1)n1−γ1−n2n1γ2γ3).
Finally, we prove that
M=span{x′vm0∣0≤v≤r−1} is a simple HM-module. Let V′ be a nonzero module of M. Then there exists
x′i−sm0∈V′ for some 0≤s≤r−1. If s=i, then m0∈V′. If s=i, then y′i−sx′i−sm0=k1′k2′⋯ki−s′m0∈V′. Thus m0∈V′
and V′=M.
Since a′=nγ11a,
b′=nγ1n11b,
c′=nγ1n11c, x′=x, and
y′=β3−1γ1−n2n1y, the action of a,b,c,x,y on M is given by (9)(10)(11).
∎
Recall the modules VI(γ1,γ2,γ3;i), VII(γ1,γ2,γ3;i) and V0(γ1,γ2,γ3;i) defined in Lemma 4.2 and Lemma 4.3.
We denote the irreducible module described in Lemma 4.4 by Vr(γ1,γ2,γ3;i).
By all the above lemmas, we obtain the following classification theorem.
Theorem 4.5**.**
Let M be an irreducible representation of Hβ. Then
M is isomorphic to one of the following types:
VI(γ1,γ2,γ3;i), VII(γ1,γ2,γ3;i), V0(γ1,γ2,γ3;i), and
Vr(γ1,γ2,γ3;i).
Proof.
It is easy to verify that VI(γ1,γ2,γ3;i),
VII(γ1,γ2,γ3;i),
V0(γ1,γ2,γ3;i), and
Vr(γ1,γ2,γ3;i) are irreducible
representations of Hβ. Let M be an irreducible representation of
Hβ. Then there exists an algebra homomorphism λ∈HomK(T,K) such that
λ(an)=γ1=0, λ(b)=γ2=0 and
λ(c)=γ3=0. Since γ1,γ2,γ3
lie in one of the above four cases, M can be viewed as a
representation of HM described by Lemma 4.2, Lemma 4.3 and Lemma
4.4. Being careful with the change of the generators, M is
isomorphic to one of the above four kinds of irreducible representations.
∎
Next proposition shows that the category of all Hβ modules is not semisimple.
Proposition 4.6**.**
Suppose that γ1−n2n1γ2γ3=qvn1
for any v and
(γ1n1−γ2n)β1=(γ1n1−γ3n)β2=0.
Then
ExtHβ(Vt(γ1,γ2,γ3;i−t+n),Vt(γ1,γ2,γ3;i))=0 for all 1≤i≤n−1, where t=∣qn1∣.
Proof.
Let
ku(i)=1−q−n11−q−un1(q(2i−u+1)n1−γ1−n2n1γ2γ3) and L be a vector
space with a basis {v1,⋯,v2t}. Set
[TABLE]
Then span{vt+1,⋯,v2t} is isomorphic to Vt(γ1,γ2,γ3;i−t+n) and
[TABLE]
Obviously, the sequence
0→Vt(γ1,γ2,γ3;i−t+n)→L→Vt(γ1,γ2,γ3;i)→0 is
not splitting. Hence
ExtHβ(Vt(γ1,γ2,γ3;i−t+n),Vt(γ1,γ2,γ3;i))=0.∎
Remark 4.7**.**
Since Hβ is a Hopf algebra, the dual of any
Hβ module is still a Hβ module. It is easy to prove that the dual
of any irreducible representation is irreducible. Therefore the dual
module VI(γ1,γ2,γ3;i)∗,
VII(γ1,γ2,γ3;i)∗,
V0(γ1,γ2,γ3;i)∗, and
Vr(γ1,γ2,γ3;i)∗ are also irreducible
representations of Hβ.
5. Grothendieck ring G0(Hβ)
We recall some notations. Let R be an algebra over field K.
Recall that the Grothendieck group G0(R) is the abelian group generated by the
set {[M]∣M is a left R-module} of isomorphic classes of finite-dimensional R-modules with relations [B]=[A]+[C] for every short exact sequence
0→A→B→C→0 of R-modules. Unlike ordinary algebras, the Grothendieck group G0(H) of a Hopf algebra H has a product given by [M][N]=[M⊗N]. With this product, Grothendieck group G0(H) becomes a ring, which is called the Grothendieck ring of the Hopf algebra H.
Assume that Hβ is the Hopf algebra defined in Section 2 and the modules
[TABLE]
are defined in Lemma 4.2, Lemma 4.3 and Lemma 4.4. In this section, we determine the Grothendieck ring G0(Hβ) of the Hopf algebra Hβ in several cases, which is generated by
[TABLE]
where 0≤i≤n−1, 2≤r≤t,
(γ1,γ2,γ3)∈(K∗)3 and t=∣qn1∣ is the order of qn1. The classification results of all cases are illustrated in several lemmas and theorems. In details, the Grothendieck rings in cases that iff two the three equations β1=0, β2=0 and β3=0 hold are stated in theorems 5.5, 5.12 and 5.16. To do that, lemmas 5.4, 5.7 and 5.14 are necessary. In theorems 5.10, 5.18 and 5.20, we give the cases in which iff one of the conditions β1=0, β2=0 and β3=0 holds. In additional, we also obtain properties of the Grothendieck rings G0(U(N,ν,ω)) and G0(U(n,N,ν,q,α,β,γ)) in several corollaries. We also observe that there exist non-isomorphic Hopf algebras with isomorphic Grothendieck rings.
In order to discuss G0(Hβ) in which β1=β2=β3=0, the analyze to the subring of G0(Hβ) generated by all [V0(γ1,γ2,γ3;i)] for all nonzero γi∈K and i∈Zn is crucial.
We present the results of this case in Theorem 5.2 and the following Lemma 5.1 is necessary.
Lemma 5.1**.**
The product of the Grothendieck ring G0(Hβ) satisfies
[TABLE]
for any i,j∈Zn and
γk,γk′∈K∗, k∈{1,2,3}.
Proof.
By the definition of V0(γ1,γ2,γ3;i) in Lemma 4.4, we obtain that the product
Let R1 be the subring of the Grothendieck ring G0(Hβ) generated by all [V0(γ1,γ2,γ3;i)] for all nonzero γi∈K and i∈Zn. Then we have the following.
(i)
If at most one of β1,β2,β3 equals to zero, then R1 is isomorphic to Z[Zn1×Zn2×K∗];
(ii)
If only β1=0 or β2=0, then R1 is isomorphic to Z[Zn1×Zn×(K∗)2];
(iii)
If only β3=0, then R1 is isomorphic to Z[Z2n1×Zn×(K∗)2];
(iv)
If β1=β2=β3=0, then R1 is isomorphic to Z[Zn×K∗×K∗×K∗].
Thus G0(Hβ)≅Z[Zn×K∗×K∗×K∗] if β=(β1,β2,β3)=0.
Proof.
We prove this claim by several cases.
(i)
First, we consider the case where β1β2β3=0. By β1′′=0 and β2′′=0, we obtain β1′′=β1′(1−γ1n1γ2n)=0 and β2′′=β2′(1−γ1n1γ3n)=0.
Hence γ1=γ2n1nωˉp=γ3n1nωˉl and γ3=quγ2 for some intergers p,l,u, where ωˉ is the primitive n1-th unitary root. Thus
[TABLE]
for some nonzero γ2 and i∈Zn. Then R1 is isomorphic to the group algebra Z[Zn1×Zn2×K∗], where K∗ is the multiplicative group of all nonzero elements of K.
If β2β3=0 and β1=0, then γ1=γ3n1nωˉp for some integer p. Since γ1−n2n1γ2γ3=q2n1i, γ2=q2n1i+uγ3 for some integer u. Thus
[TABLE]
Hence R1 is isomorphic to the group algebra Z[Zn1×Zn2×K∗]. Similarly, we can prove that R1 is also isomorphic to Z[Zn1×Zn2×K∗] if β1β3=0 and β2=0.
If β1β2=0 and β3=0, then γ1=γ3n1nωˉp and γ2=γ3qu for some integers p,u.
Thus
[TABLE]
Hence R1 is isomorphic to Z[Zn1×Zn2×K∗].
(ii)
If β1=0 and β2=β3=0, then γ1=γ2n1nωˉp for some integer p. Thus
[TABLE]
Hence R1 is isomorphic to Z[Zn1×Zn×(K∗)2]. Similarly, we can prove that R1 is isomorphic to Z[Zn1×Zn×(K∗)2] if β2=0 and β1=β3=0.
(iii)
If β1=β2=0 and β3=0, then γ1=(γ2γ3)2n1nωˉ2p for some integer p. Thus
[TABLE]
Hence R1 is isomorphic to Z[Z2n1×Zn×(K∗)2].
(iv)
Finally, if β=0, then \begin{array}[]{lll}V_{0}(\gamma_{1},\gamma_{2},\gamma_{3},i)&\cong&V_{0}(1,1,1;i)\otimes V_{0}(\gamma_{1},\gamma_{2},\gamma_{3};0).\end{array}
Hence R1 is isomorphic to Z[Zn×K∗×K∗×K∗].
∎
Since Hβ/(aN−1,b−1,c−1) is isomorphic to Gelaki’s Hopf algebra U(n,N,n1,q,β1,β2,β3), where n∣N, we obtain the following properties of the Gelaki’s Hopf alegbra.
Corollary 5.3**.**
Suppose that n∣N. Let q be a primitive N-th root of unity and S1 be the subring of R1 generated by [V0(γ,1,1;i)], where γnN=1. Then we have the following result.
(1)
If either β1β2=0 and β3=0, or β1=0 and β2=β3=0, or β2=0 and β1=β3=0, then S1=Z[g,h] for g=[V0(1,1,1;1)] and h=[V0(q(N/n,n1)N,1,1;0)]. Moreover,
[TABLE]
(2)
If β3=0 and β1=β2=0, then S1=Z[g,h1] for g=[V0(1,1,1;1)] and
[TABLE]
Moreover, gn=h1μ=1 and Z[g,h1]≃Z[Zμ×Zn], where μ=(n,2n1)(N,2n1).
(3)
If either β1β3=0 and β2=0, or β2β3=0 and β1=0, or β1β2β3=0, then S1=Z[g,h2] for g=[V0(1,1,1;1)] and h2=[V0(qnν,1,1;0)], where ν=(N,2n1,nn1)N. Moreover, gn=h2μ′=1 and Z[g,h2]≃Z[Zμ′×Zn], where μ′=(n,2n1)(N,2n1,nn1).
(4)
If either β1=β2=β3=0, then S1=Z[g,h3] for g=[V0(1,1,1;1)] and h3=[V0(qn,1,1;0)]. Moreover, gn=h3nN=1 and Z[g,h3]≃Z[ZN/n×Zn].
(5)
If β1=β2=β3=0, then the Grothendieck ring G0(U(n,N,n1,q,β1,β2,β3)) of the Gelaki’s Hopf algebra U(n,N,n1,q,0,0,0) is equal to Z[g,h3].
Proof.
If β1β2=0 and β3=0, then γ=ωˉp satisfying γnN=1, where ωˉ is a primitive n1-th root of unity. Thus n1∣nNp and (n1,nN)n1∣p.
Therefore γ∈⟨ωˉ(N/n,n1)n1⟩=⟨q(N/n,n1)N⟩, where q is a primitive N-th root of unity. Hence S1=Z[g,h]≃Z[Z(N/n,n1)×Zn] by (12), where g=[V0(1,1,1;1)], h=[V0(q(N,n1)N,1,1;0)]. It is obvious that gn=h(N/n,n1)=1.
If β3=0 and β1=β2=0, then γ−n2n1=1 for V0(γ,1,1;0). Since γnN=1, γ=qnr for some integer r, where q is a primitive N-th root of unity. Then q−2rn1=1 and N∣2rn1. Thus r=(N,2n1)Np for any integer p. Hence γ∈⟨q(N,2n1)Nn⟩. Thus
V0(γ,1,1;i)≅V0(q(N,2n1)Nn,1,1;0)⊗p⊗V0(1,1,1;1)⊗i.
Hence S1=Z[g,h1] for h1=[V0(q(N,2n1)Nn,1,1;0)]. It is obvious that S1 is isomorphic to the group algebra Z[Zμ×Zn], where μ=(n,2n1)(N,2n1).
If either β1β3=0, or β2β3=0, or β1β2β3=0, then γnN=γn1=γn2n1=1 for V0(γ,1,1;0). Thus γ=qnr for some integer r and qnrn1=q2n1r=1. Hence r=(N,nn1)Np=(N,2n1)Np′ for some integers p,p′. Let dˉ=((N,nn1)N,(N,2n1)N) and ν=dˉ(N,nn1)(N,2n1)N2=(N,2n1,nn1)N. Then γ∈⟨qnν⟩. Thus S1=Z[g,h2] for h2=[V0(qnν,1,1;0)]. It is obvious that S1 is isomorphic to the group algebra Z[Zμ′×Zn], where μ′=∣qnν∣=(n,2n1)(N,2n1,nn1).
If β1=0 and β2=β3=0, then γ=ωˉp for some integer p. Thus S1=Z[g,h], which is isomorphic to Z[Z(N/n,n1)×Zn]. Similarly, we can prove that S1 is isomorphic to Z[Z(N/n,n1)×Zn] if β2=0 and β1=β3=0.
Finally, if β=0, then \begin{array}[]{lll}V_{0}(\gamma,1,1,i)&\cong&V_{0}(1,1,1;i)\otimes V_{0}(\gamma,1,1;0).\end{array}
Hence S1=Z[g,h3], where h3=[V0(qn,1,1;0)]. Thus S1 is isomorphic to Z[ZN/n×Zn].
∎
Let q is a primitive N-th root of unity. Set
g:=[V0(1,1,1;1)],s:=p=0∑n−1gp,s′:=k=1∑ugkt,s′′:=k=0∑t−1gn−k,h:=[V0(q(N,n1)N,1,1;0)],h1:=[V0(q(N,2n1)Nn,1,1;0)],h2:=[V0(qnν,1,1;0)],h3:=[V0(qn,1,1;0)],
and
gγ1,γ2,γ3:=[V0(γ1,γ2,γ3;0)]∈G0(Hβ).
To determine the Grothendieck ring G0(Hβ) for β3=0, the following lemma is necessary.
Lemma 5.4**.**
Suppose that β3(q2n1i−γ1−n2n1γ2γ3)=0, β1(γ1′n1−γ3′n)=β2(γ2′n−γ3′n)=0 and
β3(γ1′−n2n1γ2′γ3′−q2n1j)=0. Then
[TABLE]
Proof.
As β3(γ1n2n1q2n1i−γ2γ3)=0, we have β3=0.
Moreover,
[TABLE]
Since β1(γ1n1−γ2n)=β2(γ1n1−γ3n)=0 and β1(γ1′n1−γ2′n)=β2(γ1′n1−γ3′n)=0, we have
[TABLE]
If γ1−n2n1γ2γ3=qn1(2i−r+1), then (γ1γ1′)−n2n1(γ2γ2′)(γ3γ3)=qn1(2(i+j)−r+1). If γ1n−2n1γ2γ3=qn1v for any integer v, then (γ1γ1′)−n2n1(γ2γ2′)(γ3γ3)=qn1v for any integer v.
Let {m0,m1,⋯,mr−1} be the basis of Vr(γ1,γ2,γ3;i) with the action given by (9)-(11). Then {m0⊗1,m1⊗1,⋯,mr−1⊗1} is a basis of Vr(γ1,γ2,γ3;i)⊗V0(γ1′,γ2′,γ3′;j) and the action on this basis is as follows:
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
Let ul:=γ1′nn1lqjn1lml⊗1, then {u0,⋯,ur−1} is a new basis of Vr(γ1,γ2,γ3;i)⊗V0(γ1′,γ2′,γ3′;j). Under this basis,
[TABLE]
[TABLE]
[TABLE]
Hence Vr(γ1,γ2,γ3;i)⊗V0(γ1′,γ2′,γ3′;j)≅Vr(γ1γ1′,γ2γ2′,γ3γ3′;i+j).
Similarly, we get
[TABLE]
∎
Next, we discuss the case in which there is an integer v such that q(2i−v+1)n1=γ1−n2n1γ2γ3. Let r be the minimal positive integer such that q(2i−r+1)n1=γ1−n2n1γ2γ3. From Lemma 5.4, we obtain
[TABLE]
By Equations (10) and
(11), the action of Hβ on Vr(1,1,1;0) is given by a⋅mj=q21(r−1)−jmj, b⋅mj=c⋅mj=mj, where k1,⋯,kr are determined by kl=j=0∑l−1q−jn1β3(q(r−l)n1−1)
for 1≤l≤r−1.
In the case that γ1n2n1qvn1=γ2γ3 for any integer v, we have
[TABLE]
where γ1′=(γ3γ2)−2n1nγ1=ωˉ2p(n,n1) for any integer p.
Let R2 be the subring of the Grothendieck ring G0(Hβ) for β3=0 generated by
R1 and Vr(γ1,γ2,γ3;i).
Then R2 is isomorphic to a quotient ring of R1[zr,zξ′∣2≤r≤t,ξ∈K0], where
K0:=(K∗/{ωˉ2p(n,n1)∣p∈Z})∖{1}, zr=[Vr(1,1,1;0)] for 2≤r≤t and zξ′=[Vt(ξ,1,1;0)] for ξ∈K0. Moreover, if β3=0 and β1=β2=0, then R2=G0(Hβ).
Let rp=(r−2p)mod(t) and 1≤rp≤t, where r is the minimal positive integer such that ξξ′=q(−r+1)n1. Then we obtain the following theorem about the Grothendieck ring G0(Hβ).
Theorem 5.5**.**
Suppose that β1=β2=0, β3=0. Then the Grothendieck ring G0(Hβ) is isomorphic to R2:=R1[z2,zξ′∣ξ∈K0] with relations
[TABLE]
[TABLE]
and
[TABLE]
where zr=v=0∑[2r−1](−1)v(vr−1−v)g(n−1)vz2r−1−2v for 3≤r≤t.
Proof.
(1) Suppose that {m0,⋯,mr−1} and {m0′,m1′} are the basis of Vr(1,1,1;0) and V2(1,1,1;0), respectively. Then {ml⊗mk′,0≤l≤r−1,0≤k≤1} is a basis of Vr(1,1,1;0)⊗V2(1,1,1;0). With this basis, we have
[TABLE]
[TABLE]
[TABLE]
Since
Δ(xk)=l=0∑k(lk)qn1blxk−l⊗a(k−l)n1xl in Hβ, where (lk)qn1=(l)qn1!(k−l)qn1!(k)qn1! and (p)qn1!=(p)qn1(p−1)qn1⋯(1)qn1 for (p)qn1=1+qn1+⋯+q(p−1)n1, we have
[TABLE]
Thus xk′(m0⊗m0′)=0 only if k′≥l0:=min{t,r+1}.
If 2≤r≤t−1, then xr+1(m0⊗m0′)=0. In addition,
[TABLE]
Let ul=xl(m0⊗m0′). Then {u0,⋯,ur} is a basis of Hβ(m0⊗m0′), which is isomorphic to Vr+1(1,1,1;0) for r≤t−1. Let
A1:=Hβ(m0⊗m1′)/Hβ(m0⊗m0′). Then
[TABLE]
It is easy to prove that xp(m0⊗m1′)=q2pn1mp⊗m1′∈/Hβ(m0⊗m0′) for 0≤p≤r−2 and xr−1(m0⊗m1′)∈Hβ(m0⊗m0′). Let ul=xl(m0⊗m1′). Then {u0,⋯,ur−2} is a basis of Hβ(m0⊗m1′). Since a(m0⊗m1′)=q21(r−2)m0⊗m1′,Hβ(m0⊗m1′)≅Vr−1(1,1,1;0)⊗V0(1,1,1,n−1).
Thus zrz2=zr+1+gn−1zr−1 for 2≤r≤t−1. In particular, z22=z3+gn−1.
So zr+1=zrz2−gn−1zr−1 for 2≤r≤t−1.
If r=t, then yx(m0⊗m0′)=0 by (17). Since ax(m0⊗m0′)=q21(t−2)x(m0⊗m0′),
Hβx(m0⊗m0′)≅Vt−1(1,1,1;0)⊗V0(1,1,1;n−1). Thus Hβ(m0⊗m0′)/Hβx(m0⊗m0′)≅V0(1,1,1;0) and
[Hβ(m0⊗m0′)]=gn−1zt−1+1, where g=[V0(1,1,1;1)]. Let
A1:=Hβ(m0⊗m1′)/Hβ(m0⊗m0′). Then
y(m0⊗m1′)=0 and
[TABLE]
only if k′≥t. Since yxt−1(m0⊗m1′)=0. We have Hβxt−1(m0⊗m1′)≅V0(1,1,1;n−t) and Hβ(m0⊗m1′)/Hβxt−1(m0⊗m1′)≅Vt−1(1,1,1;n−1)≅Vt−1(1,1,1;0)⊗V0(1,1,1;n−1). Hence ztz2=2gn−1zt−1+gn−t+1.
Consequently, we have zr=v=0∑[2r−1](−1)v(vr−1−v)g(n−1)vz2r−1−2v for 3≤r≤t and v=0∑[2t](−1)vt−vt(vr−v)g(n−1)vz2t−2v−gn−t−1=0.
(2) Now we consider the product z2zξ′. Similar to (1), we get Hβ(m0⊗m0′)≅Vt(ξq2n,1,1;0) and Hβ(m0⊗m1′)≅Vt(ξq2n,1,1;n−1)≅Vt(ξq2n,1,1;0)⊗V0(1,1,1;n−1). Thus z2zξ′=zξq2n′+gn−1zξq2n′.
(3) Here, we determine the product zξ′zξ′′. Similar to the proof of (1), we get lp=t and Vt(ξ,1,1;0)⊗Vt(ξ′,1,1;0)=p=0∑t−1Hβ(m0⊗mp′). Suppose that ξξ′∈K0. Hence we get that
[TABLE]
Thus zξ′zξ′′=p=0∑t−1gn−pzξξ′′, where g=[V0(1,1,1;1)]∈R1.
(4) Finally, we assume that ξξ′∈{ωˉ2p(n,n1)∣p∈Z}.
Let r be the minimal positive integer such that (ξξ′)n2n1q(−r+1)n1=1. Then (ξξ′)n2n1q(−2p−rp+1)n1=1 for 1≤rp≤t satisfying rp≡(r−2p)mod(t). Hence
Hβ(m0⊗mp′)≅Vt(ξξ′,1,1;n−p)≅Vt(ξξ′,1,1;0)⊗V0(1,1,1;n−p) if rp=t. Otherwise, we have Hβxrp(m0⊗mp′)≅Vt−rp(ξξ′,1,1;n−p−rp)≅Vt−rp(ξξ′,1,1;0)⊗V0(1,1,1;n−p−rp) and Hβ(m0⊗mp′)/Hβxrp(m0⊗mp′)≅Vrp(ξξ′,1,1;n−p)≅Vrp(ξξ′,1,1;0)⊗V0(1,1,1;n−p).
Thus zξ′zξ′′=p=0∑t−1W′(p), where
W′(p)=gn−pzt if rp=t and
W′(p)=gn−p−rpzt−rp+gn−pzrp if rp<t.
The proof is completed.
∎
Especially, we obtain the structure of the Grothendieck ring of the Gelaki’s Hopf alegbra where β3=0.
Corollary 5.6**.**
Suppose that β3=0.
(1)
If N∤2n1t, then one of the following satisfies.
(1.1)
If either 2n∣N, or 2n∤N and 2(n,n1)∤n, then the Grothendieck ring G0(U(n,N,n1,q,0,0,β3)) is isomorphic to S2:=Z[g,h1,z2,z′] with relations (14), z2z′=(1+gn−1)z′ and
[TABLE]
where z0=0, z1=1, zp=v=0∑[2p−1](−1)v(vp−1−v)g(n−1)vz2p−1−2v for 3≤p≤t and v0=(2n1t,N)N.
(1.2)
If 2n∤N and 2(n,n1)∣n, then the Grothendieck ring G0(U(n,N,n1,q,0,0,β3)) is isomorphic to S2:=Z[g,h1,z3,z′] with relations (18), z3z′=(1+gn−1+gn−2)z′ and
[TABLE]
where z0=0, z1=1, v0=(2n1t,N)N and z2l+1=v=0∑[2l−1](−1)v(vl−1−v)g(n−2)v(z3−gn−1)l−1−2v for 3≤2l+1≤t−1.
(2)
If N∣2n1t, then the Grothendieck ring G0(U(n,N,n1,q,0,0,β3)) is isomorphic to
[TABLE]
which is a subring of S2.
Proof.
Since γnN=1, γ=qnp for some integer p. If γn2n1=qvn1 for some integer v, then q2n1p=qvn1. Hence q2n1pt=1, i.e. N∣2n1pt. Thus (N,2n1t)N∣p.
Then zγ′∈S2 only if γ∈⟨qn⟩ and γ∈/⟨qnv0⟩ for v0=(2n1t,N)N. If either 2n∣N, or 2n∤N and 2(n,n1)∤n, then we have generator z2=[V0(1,1,1;0)]. If v0=1, that is N(n,n1)∣2nn1, then S2=S1[z2]. Otherwise, S2=S1[z2,z′] for z′=[Vt(qn,1,1;0)]. If 2n∤N and 2(n,n1)∣n, then [Vr(γ,1,1;0)]∈U(n,N,n1,q,0,0,β3) for γn2n1∈⟨qn1⟩ implies that r is an odd integer. Let rp≡(r−2p)mod(t) for 1≤p≤t and z′=[Vt(qn,1,1;0)]. We further assume that 1≤rp≤t. Then (z′)v0=z′(z′)v0−1=s′′v0−2z′zqv0−1′ by Theorem 5.5. Since
[TABLE]
for some 2≤r≤t, z′zqv0−1′=p=0,rp<t∑t−1(gn−p−rpzt−rp+gn−pzrp)+i=0,rp=t∑t−1gn−pzt, where
[TABLE]
for 3≤p≤t.
Since qnq2n=qn, z2z′=(gn−1+1)z′ by Theorem 5.5. Similarly to the proof of Theorem 5.5, we have z3z′=(gn−2+gn−1+1)z′,
[TABLE]
and z2l+1=v=0∑[2l−1](−1)v(vl−1−v)g(n−2)v(z3−gn−1)l−1−2v for 3≤2l+1≤t−1.
∎
We always assume that z0=0, z1=1, z2=[V2(1,1,1;0)] and
[TABLE]
To determine the Grothendieck ring G0(Hβ) for β1=0, we need the following lemma.
Lemma 5.7**.**
Suppose that β1(γ1n1−γ2n)=0. Then
[TABLE]
Proof.
Since (γ1n1−γ2n)β1=0, β1=0. Thus γ1′n1=γ2′n and γ2′=γ1′nn1qs for some 0≤s≤n−1. Hence ((γ1γ1′)n1−(γ2γ2′)n)β1=(γ1′)n1(γ1n1−γ2n)β1=0.
Let {m0,m1,⋯,mn−1} be the basis of VI(γ1,γ2,γ3;i) with the action given by Equations (3)-(5). Then {m0⊗1,m1⊗1,⋯,mn−1⊗1} is a basis of VI(γ1,γ2,γ3;i)⊗V0(γ1′,γ2′,γ3′;j). With this basis, we have
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
Let ul=γ1′nn1lqjn1lml⊗1. Then {u0,⋯,un−1} is a new basis of VI(γ1,γ2,γ3;i)⊗V0(γ1′,γ2′,γ3′;j). Under this basis, we have
[TABLE]
[TABLE]
[TABLE]
Thus VI(γ1,γ2,γ3;i)⊗V0(γ1′,γ2′,γ3′;j)≅VI(γ1γ1′,γ2γ2′,γ3γ3′;i+j).
Similarly, we get
where γ1′=γ1γ2−n1n satisfying γ1′n1=1, γ3′=γ3γ2−1q−2n1i.
If β2=β3=0, then VI(γ1′,1,γ3′;0)≅VI(γ1′,1,1;0)⊗V0(1,1,γ3′;0). If β2=0, β3=0 and γ3′=qu for some integer u, then VI(γ1′,1,γ3′;0)≅VI(γ1′,1,1;0)⊗V0(1,1,γ3′;0). If β2=0, β3=0 and (γ3′)2n=1, then VI(γ1′,1,γ3′;0)≅VI(γ1′(γ3′)−2n1n,1,1;0)⊗V0((γ3′)2n1n,1,γ3′;0). If β2β3=0 and (γ3′)2n=γ3′n=1, then VI(γ1′,1,γ3′;0)≅VI(γ1′(γ3′)−2n1n,1,1;0)⊗V0((γ3′)2n1n,1,γ3′;0).
Remark 5.9**.**
Let R3 be the subring of the Grothendieck ring G0(Hβ) for β1=0 generated by
R2 and {[VI(γ1,γ2,γ3;i)]∣γ1n1=γ2n,i∈Zn}. Then R3 is isomorphic to a quotient ring of R2[xζ1,ζ2∣ζ2∈K∗,ζ1∈K0], where
K0=(K∗/{ωˉr∣r∈Z})∖{1} and xζ1,ζ2=[VI(ζ1,1,ζ2;0)]. Moreover, if β2=0 and β1β3=0, then R3 is the Grothendieck ring of G0(Hβ).
Suppose that β2=β3=0 and β1=0. Then the Grothendieck ring G0(H) is isomorphic to a quotient ring of the algebra R2′=R1[xζ,1∣ζ∈K0].
Let rp=(r−2p)mod(t) and 1≤rp≤t, where r is the minimal positive integer such that ζ2ζ2′(ζ1ζ1′)−n2n1=q(−r+1)n1. Then rp is the minimal positive integer such that
[TABLE]
Theorem 5.10**.**
Suppose that β1β3=0, β2=0 and K∗∗=(K∗/{ζ∈K∣ζ2n=1})∖{1ˉ}. Let zξ′′=[Vt(1,1,ξ;0)] for ξ∈K∗:=(K∗/⟨qn1⟩)∖{1ˉ}. Then the Grothendieck ring R3:=G0(Hβ) is isomorphic to the commutative ring
R1[z2,xζ1,ζ2,zξ′′∣ζ1∈K0,ζ2∈K∗∗,ξ∈K∗] with relations (14),
[TABLE]
[TABLE]
[TABLE]
for r′ is the minimal positive integer such that ξξ′=qr′n1, and
[TABLE]
where s′=k=1∑ugkt∈R1,
u=tn, zr=v=0∑[2r−1](−1)v(vr−1−v)g(n−1)vz2r−1−2v for 3≤r≤t and η=[V0(q2n,1,qn1;0)].
Proof.
Since β2=0, G0(Hβ) is generated by [VI(γ1,γ2,γ3;i)],[V0(γ1,γ2,γ3;i)] and [Vr(γ1,γ2,γ3;i)] for r≤t. Suppose that γ1n1=γ2n and γ1−n2n1γ2γ3=qn1v for any integer v.
Then
[TABLE]
where γ2−1γ3=qn1v for any integer v. In addition, we have
[TABLE]
Therefore G0(Hβ) is generated by R1, z2, xζ1,ζ2 and zζ′′=[Vt(1,1,ξ;0)] for ξ∈(K∗/⟨qn1⟩)∖{1ˉ}.
In the following, we determine the relations of the generators in several subcases.
(1) Similar to the proof of Theorem 5.5, we have Equation (14),
[TABLE]
and
[TABLE]
for ξξ′=qr′n1, where zr=v=0∑[2r−1](−1)v(vr−1−v)g(n−1)vz2r−1−2v for 3≤r≤t, η=[V0(q2n,1,qn1;0)], and rp′=(r′−2p)(modt) for 1≤p≤t, r′ is the minimal positive integer such that ξξ′=qr′n1.
(2) Let {m0,⋯,mn−1} (resp. {m0′,m1′}) be the basis of VI(ζ1,1,ζ2;0) (resp. V2(1,1,1;0)) with the actions given by Equations (3)-(5)(resp. Equations (9)-(11), where ki is replaced by ki′). Then {ml⊗mv′,0≤l≤n−1,0≤v≤1} is a basis of VI(ζ1,1,ζ2;0)⊗V2(1,1,1;0). With this basis, we have
[TABLE]
[TABLE]
[TABLE]
and
[TABLE]
If (n1,n)=u, then ∣qn1∣=∣qu∣=un. Hence n=ut and
[TABLE]
Let v0=m0⊗m0′+α1mn−1⊗m1′ , v1=m0⊗m1′ and yv0=θxn−1v0. Then
[TABLE]
and
[TABLE]
Thus
[TABLE]
Hence we can determine the α1. In addition, Hβv0≅VI(ζ1q2n,1,ζ2;0) is an irreducible submodule of VI(ζ1,1,ζ2;0)⊗V2(1,1,1;0) by the proof of Lemma 4.2.
By the same method, we prove that
[TABLE]
is an irreducible submodule of VI(ζ1,1,ζ2;0)⊗V2(1,1,1;0)/Hβv0.
Hence
[TABLE]
where g=[V0(1,1,1;1)]. Likewise, we have
z2xζ1,ζ2=xζ1q2n,ζ2+gn−1xζ1q2n,ζ2.
(3) Similarly, we obtain
[TABLE]
Hence xζ1,ζ2zξ′′=p=0∑t−1gn−pxζ1,ζ2ξ.
(4) Let {m0,⋯,mn−1} (resp. {m0′,⋯,mn−1′}) be a basis of VI(ζ1,1,ζ2;0) (resp. VI(ζ1′,1,ζ2′;0)) with the action given by (3)-(5) (resp. (3)-(5) with ki replaced by ki′). Then {ml⊗mv′,0≤l,v≤n−1} is a basis of VI(ζ1,1,ζ2;0)⊗VI(ζ1′,1,ζ2′;0) and the action on this basis is given by
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
(4.1) If ((ζ1ζ1′)n1−1)β1=0, let vp=m0⊗mp′+l=p+1∑n−1αlmn+p−l⊗ml′ for 0≤p≤n−1. Assume that yv0=θxn−1v0, similar to (1), we can determine αl,1≤l≤n−1.
Hence, A0:=Hβv0≅VI(ζ1ζ1′,1,ζ2ζ2′;0) is an irreducible submodule of VI(ζ1,1,ζ2;0)⊗VI(ζ1′,1,ζ2′;0).
By using the same method, we can determine that
[TABLE]
is an irreducible submodule of VI(ζ1,1,ζ2;0)⊗VI(ζ1′,1,ζ2′;0)/∑l=0p−1Hβvl for 1≤p≤n−1.
Hence [VI(ζ1,1,ζ2;0)][VI(ζ1′,1,ζ2′;0)]=p=0∑n−1[VI(ζ1ζ1′,1,ζ2ζ2′;n−p)]=p=0∑n−1[VI(ζ1ζ1′,1,ζ2ζ2′;0)][V0(1,1,1;n−p)].
Consequently, xζ1,ζ2xζ1′,ζ2′=p=0∑n−1gn−pxζ1ζ1′,ζ2ζ2′=p=0∑n−1gpxζ1ζ1′,ζ2ζ2′.
(4.2) We discuss the case in which ((ζ1ζ1′)n1−1)β1=0. Since β2=0, we have yn=0. There exists an element v′∈VI(ζ1,1,ζ2;0)⊗VI(ζ1′,1,ζ2′;0) such that yv′=0. We assume that v′=l=0∑dαlmd−l⊗ml′+l=d+1∑n−1αlmn+d−l⊗ml′ for ay=qya. Let
[TABLE]
then det(C)=0.
Let vp=l=0∑pαlmp−l⊗ml′+l=p+1∑n−1αlmn+p−l⊗ml′ for 0≤p≤n−1,
[TABLE]
[TABLE]
where
[TABLE]
[TABLE]
Since det(Dp)=det(C)=0 and Dp−λBp=0 for any λ, there exists αl for 0≤l≤n−1 such that yvp=0 and xn−1vp=0. In fact, Hvi∩Hvj=0 for any 0≤i<j≤n−1. If there is a λ∈K such that vj=λxj−ivi. Then xn−1v1=λxn−1+j−iv0=λβ1((ζ1ζ1′)n1−1)xj−i−1vi=0, which is a contradiction. Thus, VI(ζ1,1,ζ2;0)⊗VI(ζ1′,1,ζ2′;0)=p=0⨁n−1Hβvp.
Notice that
[TABLE]
(4.2.1) In the case that there exists an integer l such that ζ2ζ2′−(ζ1ζ1′)n2n1q(−2p−l+1)n1=0.
Let rp be the minimal positive integer such that ζ2ζ2′(ζ1ζ1′)−n2n1=q(−2p−rp+1)n1. Then 1≤rp≤t and rp=(r−2p)mod(t), where r is the minimal positive integer such that ζ2ζ2′(ζ1ζ1′)−n2n1=q(−r+1)n1.
So rp, which satisfies 1≤rp≤t, is the minimal positive integer such that yxrpvp=0. Thus yxkt+rpvp=0 for 0≤k≤u−1, where u=tn. Notice that yxktvp=0 for 0≤k≤u−1. Hence
Ap,2k+1:=Hβx(u−k−1)t+rpvp/Hβx(u−k)tvp≅Vt−rp(ζ1ζ1′,1,ζ2ζ2′;n−p+(k+1)t−rp)≅Vt−rp(1,1,1;0)⊗V0(ζ1ζ1′,1,ζ2ζ2′;n−p+(k+1)t−rp)≅Vt−rp(1,1,1;0)⊗V0(ζ1ζ1′,1,ζ2ζ2′;0)⊗V0(1,1,1;1)⊗(n−p+(k+1)t−rp),
Ap,2(k+1):=Hβx(u−k−1)tvp/Hβx(u−k−1)t+rpvp≅Vrp(ζ1ζ1′,1,ζ2ζ2′;n−p+(k+1)t)≅Vrp(1,1,1;0)⊗V0(ζ1ζ1′,1,ζ2ζ2′;n−p+(k+1)t)≅Vrp(1,1,1;0)⊗V0(ζ1ζ1′,1,ζ2ζ2′;0)⊗V0(1,1,1;1)⊗(n−p+(k+1)t)
for 0≤k≤u−1. Therefore [Ap,2k+1]=zt−rpg(k+1)t−p−rpgζ1ζ1′,1,ζ2ζ2′ and [Ap,2(k+1)]=zrpg(k+1)t−pgζ1ζ1′,1,ζ2ζ2′.
Consequently,
[TABLE]
where s′=k=1∑ugkt.
(4.2.2) If ζ2ζ2′=(ζ1ζ1′)n2n1qvn1 for any integer v, then yxktvp=0 for 0≤k≤u−1, where u=tn. Thus Ap,v:=Hβx(u−v)tvp/Hβx(u−v+1)tvp≅Vt(ζ1ζ1′,1,ζ2ζ2′;n−p+vt)≅V0(ζ1ζ1′,1,1;n−p+vt)⊗Vt(1,1,ζ2ζ2′;0) for 1≤v≤u. Hence [Ap,v]=gζ1ζ1′,1,1gn−p+vtzζ2ζ2′′′.
Consequently, xζ1,ζ2xζ1′,ζ2′=p=0∑n−1v=1∑u[Ap,v]=usgζ1ζ1′,1,1zζ2ζ2′′′, where u=tn.
∎
Especially, we obtain the structure of the Grothendieck ring of the Gelaki’s Hopf alegbra where β1β3=0.
Corollary 5.11**.**
(1)*
Suppose that β1β3=0 and (N,nn1,2n1t)<(nn1,N).*
(1.1)* If either 2n∣N, or 2n∤N and 2(n,n1)∤n, then the Grothendieck ring*
[TABLE]
is isomorphic to the commutative ring
S2[x∗]=Z[g,h2,z2,z′′,x∗] with relations Equation (14),
[TABLE]
z2z′′=(1+gn−1)z′′,* z2x∗=(1+gn−1)x∗ and x∗(N/n,n1)N=us(N/n,n1)N−1z′′.*
(1.2)* If 2n∤N and 2(n,n1)∣n, then the Grothendieck ring S3:=G0(U(n,N,n1,q,β1,0,β3)) is isomorphic to the commutative ring
S2[x∗]=Z[g,h2,z3,z′′,x∗] with relations (19), (21), z3z′′=(1+gn−1+gn−2)z′′,z3x∗=(1+gn−1+gn−2)x∗ and x∗(N/n,n1)N=us(N/n,n1)N−1z′′.*
(2)* Suppose that β1β3=0 and (N,nn1,2n1t)=(nn1,N). Then the Grothendieck ring*
[TABLE]
is a subring of S3.
Proof.
Suppose that VI(γ,1,1;0) is an irreducible representation of
U(n,N,n1,q,β1,0,β3). Then γnN=1 and γn1=1. Assume that γnN=γn1=1.
Then γ=qnr for some integer r and N∣nn1r. Thus γ∈⟨q(N,nn1)Nn⟩. We further assume that γn2n1∈⟨qn1⟩, then γ∈⟨q(N,nn1)Nn⟩∩⟨q(N,2n1t)Nn⟩=⟨q(N,nn1,2n1t)Nn⟩.
Let v1=(nn1,2n1t,N)(nn1,N) and q(N,nn1,2n1t)2n1N=q(r−1)n1 for some 2≤r≤t. Then
[TABLE]
z2z′′=(1+gn−1)z′′ and z3z′′=(1+gn−1+gn−2)z′′, where z′′=[Vt(q(N,nn1)Nn,1,1;0)]. Let x∗=[VI(qn,1,1;0)]. Then z2x∗=(1+gn−1)x∗, z3x∗=(1+gn−1+gn−2)x∗, and x∗(N/n,n1)N=us(N/n,n1)N−1z′′
by Theorem 5.10.
∎
Theorem 5.12**.**
Suppose that β1=0 and β2=β3=0. Then the Grothendieck ring G0(Hβ)=R2′:=R1[xζ,1∣ζ∈K0] with relations
[TABLE]
Proof.
In the case that (ζ1ζ1′)n1−1=0, according to the proof of Theorem 5.10, we have
[TABLE]
Hence, xζ,1xζ′,1=p=0∑n−1gn−pxζζ′,1=sxζζ′,1.
Next we discuss the case that (ζ1ζ1′)n1=1. Let {m0,⋯,mn−1} and {m0′,⋯,mn−1′} be the basis of VI(ζ1,1,1;0) and VI(ζ1′,1,1;0) with the actions given by Equations (3)-(5) respectively. Then {ml⊗mk′,0≤l,k≤n−1} is a basis of VI(ζ1,1,1;0)⊗VI(ζ1′,1,1;0). From the proof of Theorem 5.10, we have VI(ζ1,1,1;0)⊗VI(ζ1′,1,1;0)=p=0⨁n−1Hβvp, where vp=l=0∑pαlmp−l⊗ml′+l=p+1∑n−1αlmn+p−l⊗ml′ in the proof of Theorem 5.10 for 0≤p≤n−1. Since β3=0, we obtain that yxvp=0 for 0≤p≤n−1. Hence
[TABLE]
Since gn=1, p=0∑n−1v=1∑ngn−v−p=p=0∑n−1gn−pv=1∑ngn−v=np=0∑n−1gp. Thus
[TABLE]
∎
Especially, we obtain the structure of the Grothendieck ring of the Gelaki’s Hopf alegbra where β1=0.
Corollary 5.13**.**
Suppose that β1=0. Then the Grothendieck ring G0(U(n,N,n1,q,β1,0,0)=S1[x∗]=Z[g,h,x∗] with relations
[TABLE]
Proof.
Suppose that VI(γ,1,1;0) is an irreducible representation of U(n,N,n1,q,β1,0,0). Then γnN=1 and γn1=1.
Thus γ∈⟨qn⟩∖⟨q(N/n,n1)N⟩ by the proof of Corollary 5.11. Let x∗=[VI(qn,1,1;0)].
Then G0(U(n,N,n1,q,β12,0,0))=Z[g,h,x∗].
By q(N/n,n1)Nn1=1, by Theorem 5.12 we get
[TABLE]
∎
To determine the Grothendieck ring G0(Hβ) for β2=0, we need the following lemma.
Lemma 5.14**.**
Suppose that β1(γ1n1−γ2n)=0, β2(γ1n1−γ3n)=0 and β1(γ1′n1−γ2′n)=β2(γ1′n1−γ3′n)=β3(γ1′−n2n1γ2′γ3′−q2n1j)=0. Then
[TABLE]
Proof.
Since β2(γ1n1−γ3n)=0, β2=0. Thus γ1′n1=γ3′n and γ3′=γ1′nn1qs for some integer 1≤s≤n−1. Hence β2((γ1γ1′)n1−(γ3γ3′)n)=β2(γ1n1−γ3n)(γ3′)n=0.
Since β1(γ1n1−γ2n)=0 and β1(γ1′n1−γ2′n)=0, β1((γ1γ1′)n1−(γ2γ2′)n)=0.
Let {m0,m1,⋯,mn−1} be the basis of VII(γ1,γ2,γ3;i) with the actions given by Equations (6)-(8). Then {m0⊗1,m1⊗1,⋯,mn−1⊗1} is a basis of VII(γ1,γ2,γ3;i)⊗V0(γ1′,γ2′,γ3′;j). With this basis, we have
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
Let ul=γ1′nn1lqjn1lml⊗1. Then {u0,⋯,un−1} is also a basis of VII(γ1,γ2,γ3;i)⊗V0(γ1′,γ2′,γ3′;j).
Then we have
where γ1′=γ1γ3−n1n satisfying γ1′n1=1, γ2′=γ2γ3−1q−2n1i. If β1=β3=0, then VII(γ1,γ2,1;0)≅VII(γ1,1,1;0)⊗V0(1,γ2,1;0). If β1=0, then VII(γ1,γ2,1;0)=VII(γ1,γ1nn1qu,1;0)=VII(γ1,γ1nn1,1;0)⊗V0(1,qu,1;0) for some integer u. If β1=0, β3=0 and γ2=qu for some integer u, then VII(γ1,γ2,1;0)≅VII(γ1,1,1;0)⊗V0(1,γ2,1;0).
Let yϵ1,ϵ2=[VII(ϵ1,ϵ2,1;0)] for ϵ1∈K0:=(K∗/{ωˉv∣v∈Z})∖{1}, ϵ2∈K∗. If β2=0 and β1=β3=0, then the Grothendieck ring of G0(Hβ) is isomorphic to a quotient ring of R2′′=R1[yϵ,1∣ϵ∈K0]. Similar to the proof of Theorem 5.12, we have the following theorem.
Theorem 5.16**.**
Suppose that β2=0 and β1=β3=0. Then the Grothendieck ring G0(Hβ)=R2′′:=R1[yϵ,1∣ϵ∈K0] with the relations
[TABLE]
Especially, we obtain the structure of the Grothendieck ring of the Gelaki’s Hopf alegbra where β2=0.
Corollary 5.17**.**
Suppose that β2=0. Then the Grothendieck ring G0(U(n,N,n1,q,0,β2,0))=S2′′:=S1[y∗] with the relation
Suppose that β1β2=0 and β3=0. Then the Grothendieck ring R3′ of Hβ is equal to R1[xζ1,ζ2,yϵ1,ϵ2∣(ζ1,ζ2)∈K0×K∗,ϵ1∈K0,ϵ2=ϵ1nn1] with relations
[TABLE]
[TABLE]
[TABLE]
where ζ=ζ2ζ2′.
Proof.
By the proof of Theorem 5.12 and Theorem 5.16, we have the Equation (27) except for the case of (ζ1ζ1′)n1−1=0,(ζ1ζ1′)n1−(ζ2ζ2′)n=0 and Equation (28).
(1) Here, we discuss the case in which ((ζ1ζ1′)n1−1)β1=0 and ((ζ1ζ1′)n1−(ζ2ζ2′)n)β2=0. Then (ζ1ζ1′)n1=1 and (ζ2ζ2′)n=(ζ1ζ1′)n1.
Let {m0,⋯,mn−1} and {m0′,⋯,mn−1′} be the basis of VI(ζ1,1,ζ2;0) and VI(ζ1′,1,ζ2′;0) with the actions given by Equations (3)-(5) respectively. Then {ml⊗mk′,0≤l,k≤n−1} is a basis of VI(ζ1,1,ζ2;0)⊗VI(ζ1′,1,ζ2′;0). Let vp=m0⊗mp′+l=p+1∑n−1αlmn+p−l⊗ml′ for 0≤p≤n−1. Assume that xv0=θyn−1v0. Similar to the proof of Theorem 5.12, we can determine αl,1≤l≤n−1.
Hence, Hβv0≅VII(ζ1ζ1′,1,ζ2ζ2′;0)≅V0((ζ2ζ2′)n1n,ζ2ζ2′,ζ2ζ2′;0)⊗VII(ζ1ζ1′(ζ2ζ2′)−n1n,(ζ2ζ2′)−1,1;0) and Hβv0 is an irreducible submodule of VI(ζ1,1,ζ2;0)⊗VI(ζ1′,1,ζ2′;0).
Similarly, we can show that
[TABLE]
and (Hβvp+∑l=0p−1Hβvl)/l=0∑p−1Hβvl is an irreducible submodule of VI(ζ1,1,ζ2;0)⊗VI(ζ1′,1,ζ2′;0)/l=0∑p−1Hβvl for 1≤p≤n−1.
Hence [VI(ζ1,1,ζ2;0)][VI(ζ1′,1,ζ2′;0)]=p=0∑n−1[VII(ζ1ζ1′,1,ζ2ζ2′;n−p)]. Consequently,
[TABLE]
where c2=p=0∑n−1[V0((ζ2ζ2′)n1n,ζ2ζ2′,ζ2ζ2′;n−p)]=p=0∑n−1gp[V0((ζ2ζ2′)n1n,ζ2ζ2′,ζ2ζ2′;0)].
(2) Let {m0,⋯,mn−1} and {m0′′,⋯,mn−1′′} be the basis of VI(ζ1,1,ζ2;0) and VII(ϵ1,ϵ2,1;0) with the actions given by Equations(3)-(5) and (6)-(8) respectively. Then {ml⊗mk′′,0≤l,k≤n−1} is a basis of VI(ζ1,1,ζ2;0)⊗VII(ϵ1,ϵ2,1;0). Let vp′=v=0∑n−p−1αvmp+v⊗mv′′ for α0=1.
Similar to the proof of Theorem 5.10, we can prove
[TABLE]
and ∑l=0pHβvl′′/∑l=0p−1Hβvl′′ is an irreducible submodule of VI(ζ1,1,ζ2;0)⊗VII(ϵ1,ϵ2,1;0)/l=0∑p−1Hβvl′′. Hence
[TABLE]
Similarly, we have VII(ϵ1,ϵ2,1;0)⊗VI(ζ1,1,ζ2;0)=p=0∑n−1VI(ζ1,1,ζ2ϵ2−1;0)⊗V0(ϵ1,ϵ2,ϵ2;n−p). Consequently, we obtain xζ1,ζ2yϵ1,ϵ2=c3xζ1,ζ2ϵ2−1=yϵ1,ϵ2xζ1,ζ2, where c3=p=0∑n−1[V0(ϵ1,ϵ2,ϵ2;n−p)].
∎
Especially, we obtain the structure of the Grothendieck ring of the Gelaki’s Hopf alegbra where β1β2=0.
Corollary 5.19**.**
Suppose that β1β2=0. Then the Grothendieck ring G0(U(n,N,n1,q,β1,β2,0))=S3′:=S1[x∗]=Z[g,h,x∗] with relation
[TABLE]
Proof.
Since β1′′=0 implies γn1=1, U(n,N,n1,q,β1,β2,0) has no irreducible representation VII(γ,1,1;0). Thus
S3′=S1[x∗]. The rest of the proof is
similar to that of Corollary 5.13.∎
Let {m0′,⋯,mr−1′} be the basis of Vr(γ1′,γ2′,γ3′;j) with the actions given by Equations (9)-(11). Since kl′=0 for 1≤l≤r−1, {ml′′:=∏v=r−lr−1kv′mr−l−1′∣0≤l≤r−1} is also a basis of Vr(γ1′,γ2′,γ3′;j). The action of Hβ on this basis is given by
aml′′=v=r−l∏r−1kv′amr−l−1′=(γ1′)n1qj−r+1+lml′′, for 0≤l≤r−1;
xml′′=kr−l′ml−1′′, for 1≤l≤r−1; xm0′′=0;
yml′′=ml+1′′, for 0≤l≤r−2; ymr−1′′=0.
Let K0:=(K∗/⟨q⟩)∖{1ˉ}. By the same method as Theorem 5.10, we can get the following theorem.
Theorem 5.20**.**
Suppose that β2β3=0 and β1=0. Let zξ=[Vt(1,ξ,1;0)], where ξ∈K∗.
Then the Grothendieck ring R3′′ of Hβ is isomorphic to the commutative ring R1[z2,z~ξ,yϵ1,ϵ2∣(ϵ1,ϵ2)∈K0×K0,ξ∈K∗] with relations
[TABLE]
[TABLE]
[TABLE]
for ξξ′=qrn1,
[TABLE]
where η′=[V0(q2n,qn1,1;0)] and rp=(r−2p)mod(t), r is the minimal positive integer such that ϵ1ϵ2=qrn1.
Especially, we obtain the structure of the Grothendieck ring of the Gelaki’s Hopf alegbra where β2β3=0.
Corollary 5.21**.**
Suppose that β2β3=0.
(1)
Suppose (N,nn1,2n1t)<(nn1,N).
If either 2n∣N, or 2n∤N and 2(n,n1)∤n, then the Grothendieck ring S3′=G0(U(n,N,n1,q,0,β2,β3)) is isomorphic to the commutative ring
S2[y∗]=Z[g,h2,z2,z′′,y∗] with relations (14), (21), z2z′′=(1+gn−1)z′′,
z2y∗=(1+gn−1)y∗ and y∗(N/n,n1)N=us(N/n,n1)N−1z′′.
If 2n∤N and 2(n,n1)∣n, then the Grothendieck ring S3′:=G0(U(n,N,n1,q,0,β2,β3)) is isomorphic to the commutative ring
S2[y∗]=Z[g,h2,z3,z′′,y∗] with relations (19), (21), z3z′′=(1+gn−1+gn−2)z′′, z3y∗=(1+gn−1+gn−2)y∗ and y∗(N/n,n1)N=us(N/n,n1)N−1z′′.
(2)
Suppose that β2β3=0 and (N,nn1,2n1t)=(nn1,N). Then the Grothendieck ring
[TABLE]
is a subring of S3′.
Similar to the proof of Theorem 5.5-Theorem 5.20, we have the following theorem.
Theorem 5.22**.**
Suppose that β1β2β3=0. Then the Grothendieck ring R4 of Hβ is equal to R1[z2,zξ′,xζ1,ζ2,yϵ1,ϵ1∣ξ∈K0,(ζ1,ζ2)∈K0⊗K∗,(ϵ1,ϵ2)∈K0×K0,ϵ1n1=ϵ2n] with relations (14), (30),
[TABLE]
[TABLE]
[TABLE]
and
[TABLE]
where ζ=(ζ2ζ2′)n1n, ζ∗=ζ2ζ2′, ζ′′=ζ1ζ1′,
rp=(r−2p)mod(t) and 1≤rp≤t, r is the minimal positive integer such that ζ2ζ2′=(ζ1ζ1′)n2n1q(−r+1)n1.
Especially, we obtain the structure of the Grothendieck ring of the Gelaki’s Hopf alegbra where β1β2β3=0.
Corollary 5.23**.**
Suppose that β1β2β3=0.
(1)
Suppose (N,nn1,2n1t)<(nn1,N). If either 2n∣N, or 2n∤N and 2(n,n1)∤n, then the Grothendieck ring S3′′=G0(U(n,N,n1,q,β1,β2,β3)) is isomorphic to the commutative ring
S2[x∗]=Z[g,h2,z2,z′′,x∗] with relations in Corollary 5.11(1)(I).
If 2n∤N and 2(n,n1)∣n, then the Grothendieck ring S3′′:=G0(U(n,N,n1,q,β1,β2,β3)) is isomorphic to the commutative ring
S2[x∗]=Z[g,h2,z3,z′′,x∗] with relations in Corollary 5.11(1)(II).
(2)
Suppose that β1β2β3=0 and (N,nn1,2n1t)=(nn1,N). Then the Grothendieck ring
[TABLE]
is a subring of S3′′.
Let ω be a primitive N-th root of unity. If
N∤n12, then U(N,n1,ω)=U(N/(N,n1),N,n1,ωn1,0,0,1) and U(N/(N,n1),N,n1,ωn1,0,0,γ)≃U(N,n1,ω) as Hopf algebras for any γ∈K∗. Hence V0(γ1,1,1,i) and Vr(γ1,1,1,i) are irreducible representations of U(N,n1,ω), where γ1(N,n1)=1. Thus, the Grotendieck ring of U(N,n1,ω) is the same as the Grothendieck ring G0(U(n,N,n1,q,0,0,β3)) with β3=0 in Corollary 5.6.
Corollary 5.24**.**
Let h′=[V0(ωn′,1,1;0)] and z′=[Vt(ωn,1,1;0)]. Then gn=h′(N,n′)N=1, where n=(N,ν)N and n′=(N2,Nν,2ν2)N2.
(1)
Suppose that (ν2,N)∤2ν.
If either 2∣(N,ν), or 2∤(N,ν) and 2(N,ν2)∤N, then the Grothendieck ring S=G0(U(N,ν,ω))≅Z[g,h′,z2,z′] with relations (14), (18) and z2z′=(1+gn−1)z′.
If 2∤(N,ν) and 2(N,ν2)∣N, then the Grothendieck ring S=G0(U(N,ν,ω))≅Z[g,h′,z3,z′] with relations (18), (19) and z3z′=(1+gn−1+gn−2)z′.
(2)
Suppose that 2ν=(ν2,N). Then the Grothendieck ring
But U(n,N,n1,q,β1,β2,β3)≆U(n,N,n1,q,β1,0,β3) by [7, Proposition 3.2].
Bibliography18
The reference list from the paper itself. Each links out to its DOI / PubMed record.
1[1] G. Bergman, The diamond lemma for ring theory , Adv. Math., 29(1978), 172-218.
2[2] H. Chen, Irreducible representations of a class of quantum doubles , J. Algebra, 225(1)(2000), 391-409.
3[3] H. Chen, The Green ring of Drinfeld doubles D ( H 4 ) 𝐷 subscript 𝐻 4 D(H_{4}) , Algebr. Represent. Theory 17(2014), 1457-1483.
4[4] H. Chen, V. Oystaeyen Fred and Y. Zhang, The Green rings of Taft algebras , Proceed. AMS, (3)142(2014), 765 - 775.
5[5] H. Chen, H. S. E. Mohammed, H. Sun, Indecomposable decomposition of tensor products of modules over Drinfeld Doubles of Taft algebras , Journal of Pure and Applied Algebra 221(2017), 2752-2790.
6[6] K. Erdmann, et al, Stable Green Ring of the Drinfeld Doubles of the Generalised Taft Algebras (Corrections and New Results) , Algebr Represent Theor., 1-27(2018).
7[7] S. Gelaki, Pointed Hopf Algebras and Kaplansky’s 10th Conjecture , J. Algebra, 209(1998), 635-657.
8[8] E. Gunnlaugsdóttir, Monoidal structure of the category of u + q superscript subscript absent 𝑞 {}_{q}^{+} -modules , Linear Algebra and its Applications, 365(2003), 183-199.