一类32维半单Hopf代数的拟三角结构

杨士林; 宋嫒月

doi:10.11936/bjutxb2018090006

一类32维半单Hopf代数的拟三角结构

北京工业大学应用数理学院, 北京 100124

基金项目:

国家自然科学基金资助项目 11671024

北京市自然科学基金资助项目 1162002

详细信息

作者简介:
杨士林(1964-), 男, 教授, 博士生导师, 主要从事量子群及表示方面的研究, E-mail:slyang@bjut.edu.cn

中图分类号: O153.3
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出版历程
- 收稿日期: 2018-09-04
- 网络出版日期: 2022-08-03
- 发布日期: 2019-08-09
- 刊出日期: 2019-08-14

Quasitriangular Structure for a Class of 32-dimension Semisimple Hopf Algebra

College of Applied Sciences, Beijing University of Technology, Beijing 100124, China

摘要

摘要:
Kac和Paljutkin构造了一类非交换非余可换的半单Hopf代数K₈，后来Masuoka用提升方法重新构造了这类代数.Ore扩张方法是构造新的非交换非余可换Hopf代数的一类很重要的方法，通过它可以得到许多有意义的量子代数.人们用Ore扩张方法构造了更为广泛的非交换非余可换半单Hopf代数H_2n²，其余代数乘法由Drinfeld扭元及代数自同构所确定.推广了Hopf代数K₈，首先给出一类32维非交换非余可换的半单Hopf代数H₃₂的定义，此类Hopf代数可以通过给定域上的Abel群代数K[C₄×C₄]利用特殊的Ore扩张得到，它有一个子Hopf代数，恰好同构于8维非交换非余交换的唯一的半单Hopf代数K₈.然后，主要研究Hopf代数H₃₂的拟三角性.通过详细计算，精确地得到Hopf代数H₃₂的所有泛R-矩阵，结合Wakui得出的结论，得知H₈为极小拟三角，而H₃₂非极小拟三角.
- Hopf代数 /
- Ore扩张 /
- 半单Hopf代数 /
- 泛R-矩阵 /
- 拟三角Hopf代数 /
- 拟三角结构
Abstract:
Kac and Paljutkin constructed a class of non-commutative and non-cocommutative semisimple Hopf algebra K₈, then Masuoka reconstructed this algebra by lifting method. Ore extension is an important method to construct new examples of neither commutative nor cocommutative Hopf algebras. Many interesting quantum algebras can be obtained in this way. More extensive non-commutative and non-cocommutative semisimple Hopf algebra H_2n² was constructed by means of Ore extension. Its coalgebraic multiplication was determined by the Drinfeld twist element and some algebraic automorphism. In this paper, the definition of a class of non-commutative and non-cocommutative semisimple Hopf algebra H₃₂of dimension 32 was given, which can be obtained by the special Ore extension for the given K[C₄×C₄] of Abel group algebras of order 16. It has a sub-Hopf algebra of dimension 8, which is just isomorphic to the unique neither commutative nor cocommutative semisimple 8-dimension Hopf algebra K₈. Then, the main task was to study the quasitriangular structure of the Hopf algebra H₃₂. Through detailed calculation, all the universal R-matrices for this class of Hopf algebras were given. Combining with Wakui's conclution, we know that H₈ is a minimal quasitriangular Hopf algebra; however, H₃₂ is not.
- Hopf algebra /
- Ore extension /
- semisimple Hopf algebra /
- universal R-matrix /
- quasitriangular Hopf algebra /
- quasitriangular structure

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设$\mathbb{C} $为复数域，除特殊说明外文中的代数，Hopf代数和⊗均定义在复数域$\mathbb{C} $上.

拟三角Hopf代数是由Drinfeld^[1]引入.它提供了量子Yang-Baxter方程的解.文献[2-4]对其进行了大量的研究并得到了许多重要的结果.因此给出某类Hopf代数的所有拟三角结构具有十分重要的意义.

20世纪60年代，Kac等^[5]引入了一类非交换非余可换的半单Hopf代数，Masuoka^[6]用提升方法重新构造了这类代数. Ore扩张^[7-12]是一种重要的构造新的Hopf代数的方法. Pansera^[13]构造了一类非交换非余可换的2n²维的半单Hopf代数H_2n².本文研究一类32维非交换非余可换半单Hopf代数的拟三角结构，这类Hopf代数也可通过Ore扩张得到，此类代数包含8维的Hopf代数与K₈同构.作为Hopf代数的一种推广，文献[14]引入了弱Hopf代数的概念，进而得到了比较广泛的研究，有许多这类弱Hopf代数的新例子，例如苏冬等^[15]研究了2类特殊的有限维Δ-结合代数的Green环，这2类代数均是由唯一的非交换非余可换的8维半单Hopf代数K₈形变而得，主要是通过弱化Hopf代数的结构，使之成为弱Hopf代数或者Δ-结合代数.程诚等^[16]研究了A_n-型非标准量子群的弱Hopf代数的结构.本文给出这类32维半单Hopf代数H₃₂的所有泛R-矩阵，从而容易知道它不是极小拟三角的.

1. 预备知识

Hopf代数相关概念和基本知识可见文献[17].

首先，给出一类32维Hopf代数的定义.

定义1设ω是4次本原单位根，Hopf代数H₃₂作为代数由x、y、z生成，满足下面关系.

其代数结构定义为

$$ {x^4} = 1,{y^4} = 1 $$

$$ xy = yx,zx = yz,zy = xz $$

$$ {z^2} = \frac{1}{2}\left[ {\left( {1 + {x^2}} \right) + \left( {1 - {x^2}} \right){y^2}} \right] $$

其余代数结构定义为

$$ \Delta \left( x \right) = x \otimes x $$

$$ \Delta \left( y \right) = y \otimes y $$

$$ \Delta \left( z \right) = \frac{1}{2}\left[ {\left( {1 + {x^2}} \right) \otimes 1 + \left( {1 - {x^2}} \right) \otimes {y^2}} \right]\left( {z \otimes z} \right) $$

$$ \varepsilon \left( x \right) = \varepsilon \left( y \right) = \varepsilon \left( z \right) = 1 $$

对极为

$$ S\left( x \right) = {x^3},S\left( y \right) = {y^3},S\left( z \right) = z $$

令

$$ J = \frac{1}{2}\left[ {\left( {1 + {x^2}} \right) \otimes 1 + \left( {1 - {x^2}} \right) \otimes {y^2}} \right] $$

则Δ(z)=J(z⊗z).直接验证可知，它是非交换非余可换的Hopf代数.

$$ t = \left( {\sum\limits_{i = 0}^3 {{x^i}} } \right)\left( {\sum\limits_{j = 0}^3 {{y^i}} } \right)\left( {1 + z} \right) $$

是H₃₂的左(右)积分.而且

$$ \varepsilon \left( {\int_H^l {} } \right) = \varepsilon \left( {\int_H^r {} } \right) \ne 0 $$

所以，H₃₂是半单Hopf代数.它有一组基

$$ \left\{ {{x^i}{y^j},{x^i}{y^j}z\left| {0 \le i,j \le 3} \right.} \right\} $$

对任意0≤i≤3，0≤j≤3，令

$$ {e_i} = \frac{1}{4}\sum\limits_{t = 0}^3 {{\omega ^{ - it}}{x^t}} $$

$$ {f_j} = \frac{1}{4}\sum\limits_{k = 0}^3 {{\omega ^{ - kj}}{y^k}} $$

易知

$$ J = \sum\limits_{i = 0}^3 {{e_i} \otimes {y^{2i}}} $$

令E_ij=e_if_j, 则有E_ijz=zE_ji且{E_ij, E_ijz|0≤i, j≤3}是H₃₂的一组基.易知

$$ x{E_{ij}} = {\omega ^i}{E_{ij}} $$

$$ y{E_{ij}} = {\omega ^j}{E_{ij}} $$

$$ {E_{ij}}{E_{kl}} = {\delta _{ik}}{\delta _{jl}}{E_{ij}} $$

$$ \sum\limits_{i,j = 0}^3 {{E_{ij}}} = 1 $$

拟三角Hopf代数的定义在许多文献中可以找到，这里为方便应用给出它的定义.

定义2 设H是Hopf代数, R是H⊗H中的可逆元，(H, R)称为拟三角Hopf代数，若

$$ {\Delta ^{{\rm{cop}}}}\left( h \right) = \mathit{\boldsymbol{R}}\Delta \left( h \right){\mathit{\boldsymbol{R}}^{ - 1}} $$

$$ \left( {\Delta \times id} \right)\left( \mathit{\boldsymbol{R}} \right) = {\mathit{\boldsymbol{R}}^{13}}{\mathit{\boldsymbol{R}}^{23}} $$

$$ \left( {id \times \Delta } \right)\left( \mathit{\boldsymbol{R}} \right) = {\mathit{\boldsymbol{R}}^{13}}{\mathit{\boldsymbol{R}}^{12}} $$

式中$\mathrm{R}=\sum\limits_{i}{{{a}_{i}}}\otimes {{b}_{i}}$，且

$$ {\mathit{\boldsymbol{R}}^{13}} = \sum\limits_i {{a_i} \otimes 1 \otimes {b_i}} $$

$$ {\mathit{\boldsymbol{R}}^{23}} = \sum\limits_i {1 \otimes {a_i} \otimes {b_i}} $$

$$ {\mathit{\boldsymbol{R}}^{12}} = \sum\limits_i {{a_i} \otimes {b_i} \otimes 1} $$

$$ \tau :H \otimes H \to H \otimes H $$

$$ \tau \left( {a \otimes b} \right) = b \otimes a $$

$$ {\Delta ^{{\rm{cop}}}} = \tau \circ \Delta $$

此时R称为泛R-矩阵.容易知道

$$ \left( {{\rm{id}} \otimes \varepsilon } \right)\left( \mathit{\boldsymbol{R}} \right) = \left( {\varepsilon \otimes {\rm{id}}} \right)\left( \mathit{\boldsymbol{R}} \right) = 1 $$

2. H₃₂的泛R-矩阵

引理1^[18]设G=〈g〉×〈h〉, 其中〈g〉和〈h〉是n阶循环群，ζ是n次本原单位根，则K[G]的任意泛R-矩阵均可表示为

$$ \mathit{\boldsymbol{R}}_{pqrs}^{K\left[ G \right]}: = \sum\limits_{i,j = 0}^{n - 1} {\sum\limits_{k,l = 0}^{n - 1} {{\zeta ^{\left( {pij + rkj} \right) + \left( {skl + qil} \right)}}{E_{ik}} \times {E_{jl}}} } $$

式中

$$ {E_{ik}} = \frac{1}{{{n^2}}}\sum\limits_{j,l = 0}^{n - 1} {{\zeta ^{ - ij - kl}}{g^j}{h^l}} $$

p、q、r、s均为0, 1，…，n－1.

对于Hopf代数H₃₂，下面是本文的主要定理.

定理1 H₃₂是拟三角Hopf代数，其所有泛R-矩阵可表示为

$$ {\mathit{\boldsymbol{R}}_{p,q}} = \sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {{\omega ^{pij + \left( {q + 2} \right)kj + pkl + qil}}{E_{ik}} \otimes {E_{jl}}} } $$

式中p、q均为0, 1, 2, 3.

证明：由前面讨论知

$$ \left\{ {{E_{ik}},{E_{ik}}z\left| {0 \le i,k \le 3} \right.} \right\} $$

是H₃₂的一组基.

假定R∈H₃₂⊗H₃₂, 且(H₃₂, R)是拟三角Hopf代数，它的泛R-矩阵写成

$$ \mathit{\boldsymbol{R}} = \sum\limits_{s,t = 0}^1 {\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {\mathit{\boldsymbol{R}}_{jlt}^{iks}{E_{ik}}{z^s} \otimes {E_{jl}}{z^t}} } } ,\mathit{\boldsymbol{R}}_{jlt}^{iks} \in \mathbb{F} $$

由于$\left( {{\rm{id}} \otimes \varepsilon } \right){\mathit{\boldsymbol{R = }}}\left( {\varepsilon \otimes {\rm{id}}} \right){\mathit{\boldsymbol{R = }}}1$,从而

$$ \sum\limits_{s,t = 0}^1 {\mathit{\boldsymbol{R}}_{jlt}^{00s}{E_{jl}}{z^t}} = 1 $$

$$ \sum\limits_{s,t = 0}^3 {\mathit{\boldsymbol{R}}_{00t}^{iks}{E_{ik}}{z^s}} = 1 $$

即

$$ \begin{array}{*{20}{c}} {\mathit{\boldsymbol{R}}_{jl0}^{000} + \mathit{\boldsymbol{R}}_{jl0}^{001} = 1,\mathit{\boldsymbol{R}}_{jl1}^{000} + \mathit{\boldsymbol{R}}_{jl1}^{001} = 0}\\ {\mathit{\boldsymbol{R}}_{000}^{ik0} + \mathit{\boldsymbol{R}}_{001}^{ik1} = 1,\mathit{\boldsymbol{R}}_{jl1}^{000} + \mathit{\boldsymbol{R}}_{jl1}^{001} = 0} \end{array} $$

(1)

另一方面，对h=x、y、z，直接计算Δ^cop(h)R=RΔ(h)，当且仅当

$$ \begin{array}{*{20}{c}} {{\Delta ^{{\rm{cop}}}}\left( x \right)\mathit{\boldsymbol{R}} = }\\ {\left( {x \otimes x} \right)\sum\limits_{s,t = 0}^1 {\sum\limits_{i,j = 0}^3 {\sum\limits_{k,t = 0}^3 {\mathit{\boldsymbol{R}}_{jlt}^{iks}{E_{ik}}{z^s} \otimes {E_{jl}}{z^t}} } } = }\\ {\sum\limits_{s,t = 0}^1 {\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {\mathit{\boldsymbol{R}}_{jlt}^{iks}\left( {x \otimes x} \right)\left( {{E_{ik}}{z^s} \otimes {E_{jl}}{z^t}} \right)} } } = }\\ {\sum\limits_{s,t = 0}^1 {\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {{\omega ^{i + j}}\mathit{\boldsymbol{R}}_{jlt}^{iks}\left( {{E_{ik}}{z^s} \otimes {E_{jl}}{z^t}} \right)} } } } \end{array} $$

$$ \begin{array}{*{20}{c}} {\mathit{\boldsymbol{R}}\Delta \left( x \right) = }\\ {\sum\limits_{s,t = 0}^1 {\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {\mathit{\boldsymbol{R}}_{jlt}^{iks}{E_{ik}}{z^s} \otimes {E_{jl}}{z^t}\left( {x \otimes x} \right)} } } = }\\ {\sum\limits_{s,t = 0}^1 {\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {\mathit{\boldsymbol{R}}_{jlt}^{iks}\left( {{x^{{\delta _{0s}}}}{y^{{\delta _{1s}}}} \otimes {x^{{\delta _{0t}}}}{y^{{\delta _{1t}}}}} \right)\left( {{E_{ik}}{z^s} \otimes {E_{jl}}{z^t}} \right)} } } = }\\ {\sum\limits_{s,t = 0}^1 {\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {\mathit{\boldsymbol{R}}_{jlt}^{iks}{\omega ^{{\delta _{0s}}i + {\delta _{1s}}k + {\delta _{0t}}j + {\delta _{1t}}l}}\left( {{E_{ik}}{z^s} \otimes {E_{jl}}{z^t}} \right)} } } } \end{array} $$

同理可得

$$ {\Delta ^{{\rm{cop}}}}\left( y \right)\mathit{\boldsymbol{R}} = \sum\limits_{s,t = 0}^1 {\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {{\omega ^{k + l}}\mathit{\boldsymbol{R}}_{jlt}^{iks}\left( {{E_{jk}}{z^s} \otimes {E_{jl}}{z^t}} \right)} } } $$

$$ \mathit{\boldsymbol{R}}\Delta \left( y \right) = \sum\limits_{s,t = 0}^1 {\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {\mathit{\boldsymbol{R}}_{jlt}^{iks}{\omega ^{{\delta _{0s}}k + {\delta _{1s}}i + {\delta _{0t}}l + {\delta _{1t}}j}}\left( {{E_{ik}}{z^s} \otimes {E_{jl}}{z^t}} \right)} } } $$

$$ \begin{array}{*{20}{c}} {{\Delta ^{{\rm{cop}}}}\left( z \right)\mathit{\boldsymbol{R}} = }\\ {\frac{1}{2}\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {\mathit{\boldsymbol{R}}_{jl0}^{ik0}\left( {1 + {\omega ^{2l}} + {\omega ^{2i}} - {\omega ^{2i + 2l}}} \right)} } \cdot }\\ {\left( {z \otimes z} \right)\left( {{E_{ik}} \otimes {E_{jl}}} \right) + }\\ {\frac{1}{2}\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {\mathit{\boldsymbol{R}}_{jl1}^{ik0}\left( {1 + {\omega ^{2l}} + {\omega ^{2i}} - {\omega ^{2i + 2l}}} \right)} } \cdot }\\ {\left( {z \otimes z} \right)\left( {{E_{ik}} \otimes {E_{jl}}z} \right) + }\\ {\frac{1}{2}\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {\mathit{\boldsymbol{R}}_{jl0}^{ik1}\left( {1 + {\omega ^{2l}} + {\omega ^{2i}} - {\omega ^{2i + 2l}}} \right)} } \cdot }\\ {\left( {z \otimes z} \right)\left( {{E_{ik}}z \otimes {E_{jl}}} \right) + }\\ {\frac{1}{2}\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {\mathit{\boldsymbol{R}}_{jl1}^{ik1}\left( {1 + {\omega ^{2l}} + {\omega ^{2i}} - {\omega ^{2i + 2l}}} \right)} } \cdot }\\ {\left( {z \otimes z} \right)\left( {{E_{ik}}z \otimes {E_{jl}}z} \right)} \end{array} $$

$$ \begin{array}{*{20}{c}} {\mathit{\boldsymbol{R}}\Delta \left( z \right) = \frac{1}{2}\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {\mathit{\boldsymbol{R}}_{jl0}^{ik0}\left( {1 + {\omega ^{2k}} + {\omega ^{2j}} - {\omega ^{2k + 2j}}} \right)} } \cdot }\\ {\left( {z \otimes z} \right)\left( {{E_{ik}} \otimes {E_{jl}}} \right) + }\\ {\frac{1}{2}\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {\mathit{\boldsymbol{R}}_{jl1}^{ik0}\left( {1 + {\omega ^{2k}} + {\omega ^{2l}} - {\omega ^{2k + 2l}}} \right)} } \cdot }\\ {\left( {z \otimes z} \right)\left( {{E_{ik}} \otimes {E_{jl}}z} \right) + }\\ {\frac{1}{2}\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {\mathit{\boldsymbol{R}}_{jl0}^{ik1}\left( {1 + {\omega ^{2i}} + {\omega ^{2j}} - {\omega ^{2i + 2j}}} \right)} } \cdot }\\ {\left( {z \otimes z} \right)\left( {{E_{ik}}z \otimes {E_{jl}}} \right) + }\\ {\frac{1}{2}\sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {\mathit{\boldsymbol{R}}_{jl1}^{ik1}\left( {1 + {\omega ^{2i}} + {\omega ^{2j}} - {\omega ^{2i + 2l}}} \right)} } \cdot }\\ {\left( {z \otimes z} \right)\left( {{E_{ik}}z \otimes {E_{jl}}z} \right)} \end{array} $$

从而对h=x、y、z, Δ^cop(h)R=RΔ(h)当且仅当

$$ {\omega ^{i + j}}\mathit{\boldsymbol{R}}_{jlt}^{iks} = {\omega ^{{\delta _{0s}}i + {\delta _{1s}}k + {\delta _{0t}}j + {\delta _{1t}}l}}\mathit{\boldsymbol{R}}_{jlt}^{iks} $$

(2)

$$ {\omega ^{k + l}}\mathit{\boldsymbol{R}}_{jlt}^{iks} = {\omega ^{{\delta _{0s}}k + {\delta _{1s}}i + {\delta _{0t}}l + {\delta _{1t}}j}}\mathit{\boldsymbol{R}}_{jlt}^{iks} $$

(3)

$$ {\left( { - 1} \right)^{kj - il}}\mathit{\boldsymbol{R}}_{jl0}^{ik0} = \mathit{\boldsymbol{R}}_{lj0}^{ki0} $$

(4)

$$ {\left( { - 1} \right)^{kj - ij}}\mathit{\boldsymbol{R}}_{jl1}^{ik0} = \mathit{\boldsymbol{R}}_{lj1}^{ki0} $$

(5)

$$ {\left( { - 1} \right)^{ij - il}}\mathit{\boldsymbol{R}}_{jl0}^{ik1} = \mathit{\boldsymbol{R}}_{lj0}^{ki1} $$

(6)

$$ \mathit{\boldsymbol{R}}_{jl1}^{ik1} = \mathit{\boldsymbol{R}}_{lj1}^{ki1} $$

(7)

由式(2)~(7)可知，若j≠l，则R_jl1^ik0=0;若i≠k，则R_jl0^ik1=0;若i+j≠l+k(mod 4), 则R_jl1^ik1=0.因此，只要考虑R_jl0^ik0、R_jj1^ik0、R_jl0ⁱⁱ¹、R_jl1^{m－j, m－l, 1}那些不为零的值即可.

继续通过直接计算(Δ⊗id)R=R¹³R²³及(id⊗Δ)R=R¹³R¹².其中

$$ \begin{array}{*{20}{c}} {\left( {\Delta \otimes {\rm{id}}} \right)\mathit{\boldsymbol{R}} = }\\ {\sum\limits_{j,a,b = 0}^3 {\sum\limits_{l,p,q = 0}^3 {\mathit{\boldsymbol{R}}_{jl0}^{a + b,p + q,0}\left( {{E_{ap}} \otimes {E_{bq}} \otimes {E_{jl}}} \right)} } + }\\ {\sum\limits_{j,a,b = 0}^3 {\sum\limits_{l,p,q = 0}^3 {\mathit{\boldsymbol{R}}_{jl1}^{a + b,p + q,0}\left( {{E_{ap}} \otimes {E_{bq}} \otimes {E_{jl}}z} \right)} } + }\\ {\frac{1}{2}\sum\limits_{j,a,b = 0}^3 {\sum\limits_{l,p,q = 0}^3 {\mathit{\boldsymbol{R}}_{jl0}^{a + b,p + q,1}\left( {1 + {\omega ^{2a}} + {\omega ^{2q}} - {\omega ^{2a + 2q}}} \right)} } \cdot }\\ {\left( {{E_{ap}}z \otimes {E_{bq}}z \otimes {E_{jl}}} \right) + }\\ {\frac{1}{2}\sum\limits_{j,a,b = 0}^3 {\sum\limits_{l,p,q = 0}^3 {\mathit{\boldsymbol{R}}_{jl1}^{a + b,p + q,1}\left( {1 + {\omega ^{2a}} + {\omega ^{2q}} - {\omega ^{2a + 2q}}} \right)} } \cdot }\\ {\left( {{E_{ap}}z \otimes {E_{bq}}z \otimes {E_{jl}}z} \right)} \end{array} $$

$$ \begin{array}{*{20}{c}} {{\mathit{\boldsymbol{R}}^{13}}{\mathit{\boldsymbol{R}}^{23}} = }\\ {\sum\limits_{j,a,b = 0}^3 {\sum\limits_{l,p,q = 0}^3 {\left[ {\mathit{\boldsymbol{R}}_{jl0}^{ap0}\mathit{\boldsymbol{R}}_{jl0}^{bq0} + } \right.} } }\\ {\left. {\frac{1}{2}\left( {1 + {\omega ^{2j}} + {\omega ^{2l}} - {\omega ^{2j + 2l}}} \right)\mathit{\boldsymbol{R}}_{jl1}^{ap0}\mathit{\boldsymbol{R}}_{jl1}^{bq0}} \right] \cdot }\\ {\left( {{E_{ap}} \otimes {E_{bq}} \otimes {E_{jl}}} \right) + }\\ {\sum\limits_{j,a,b = 0}^3 {\sum\limits_{l,p,q = 0}^3 {\left( {\mathit{\boldsymbol{R}}_{jl0}^{ap0}\mathit{\boldsymbol{R}}_{jl0}^{bq0} + \mathit{\boldsymbol{R}}_{jl0}^{ap0}\mathit{\boldsymbol{R}}_{jl1}^{bq0}} \right)} } \cdot }\\ {\left( {{E_{ap}} \otimes {E_{bq}} \otimes {E_{jl}}z} \right) + }\\ {\sum\limits_{j,a,b = 0}^3 {\sum\limits_{l,p,q = 0}^3 {\left[ {\mathit{\boldsymbol{R}}_{jl0}^{ap1}\mathit{\boldsymbol{R}}_{jl0}^{bq1} + \frac{1}{2}\left( {1 + {\omega ^{2j}} + {\omega ^{2l}} - } \right.} \right.} } }\\ {\left. {\left. {{\omega ^{2j + 2l}}} \right)\mathit{\boldsymbol{R}}_{jl1}^{ap1}\mathit{\boldsymbol{R}}_{jl1}^{bq1}} \right] \cdot }\\ {\left( {{E_{ap}}z \otimes {E_{bq}}z \otimes {E_{jl}}} \right) + }\\ {\sum\limits_{j,a,b = 0}^3 {\sum\limits_{l,p,q = 0}^3 {\left( {\mathit{\boldsymbol{R}}_{jl1}^{ap1}\mathit{\boldsymbol{R}}_{jl0}^{bq1} + \mathit{\boldsymbol{R}}_{jl0}^{ap1}\mathit{\boldsymbol{R}}_{jl1}^{bq1}} \right)} } \cdot }\\ {\left( {{E_{ap}}z \otimes {E_{bq}}z \otimes {E_{jl}}z} \right)} \end{array} $$

$$ \begin{array}{*{20}{c}} {\left( {{\rm{id}} \otimes \Delta } \right)\mathit{\boldsymbol{R}} = }\\ {\sum\limits_{s,t = 0}^1 {\sum\limits_{i,a,b = 0}^3 {\sum\limits_{k,p,q = 0}^3 {\mathit{\boldsymbol{R}}_{a + b,p + q,t}^{i,k,s}\left( {1 \otimes {J^{{\delta _{1t}}}}} \right)} } } \cdot }\\ {\left( {{E_{ik}} \otimes {E_{ap}} \otimes {E_{bp}}} \right)\left( {{z^s} \otimes {z^t} \otimes {z^t}} \right)}\\ {{\mathit{\boldsymbol{R}}^{13}}{\mathit{\boldsymbol{R}}^{12}} = \sum\limits_{s,s',t,t' = 0}^1 {\sum\limits_{i,a,b = 0}^3 {\sum\limits_{k,p,q = 0}^3 {\mathit{\boldsymbol{R}}_{bqt}^{iks}\mathit{\boldsymbol{R}}_{apt'}^{iks'}} } } \cdot }\\ {\left( {{E_{ik}}{z^{s + s'}} \otimes {E_{ap}}{z^{s'}} \otimes {E_{bq}}{z^{t + t'}}} \right)} \end{array} $$

并比较等式两边的系数得

$$ \mathit{\boldsymbol{R}}_{jl0}^{ap0}\mathit{\boldsymbol{R}}_{jl0}^{bp1} + {\left( { - 1} \right)^{jl}}\mathit{\boldsymbol{R}}_{jl1}^{ap0}\mathit{\boldsymbol{R}}_{jl1}^{bq1} = 0 $$

(8)

$$ \mathit{\boldsymbol{R}}_{jl1}^{ap0}\mathit{\boldsymbol{R}}_{lj0}^{bq1} + \mathit{\boldsymbol{R}}_{jl0}^{ap0}\mathit{\boldsymbol{R}}_{jl1}^{bq1} = 0 $$

(9)

$$ \mathit{\boldsymbol{R}}_{jl0}^{ap1}\mathit{\boldsymbol{R}}_{jl0}^{bq0} + {\left( { - 1} \right)^{jl}}\mathit{\boldsymbol{R}}_{jl1}^{ap1}\mathit{\boldsymbol{R}}_{lj1}^{bq0} = 0 $$

(10)

$$ \mathit{\boldsymbol{R}}_{jl1}^{ap1}\mathit{\boldsymbol{R}}_{lj0}^{bq0} + \mathit{\boldsymbol{R}}_{jl0}^{ap1}\mathit{\boldsymbol{R}}_{jl1}^{bq0} = 0 $$

(11)

$$ \mathit{\boldsymbol{R}}_{jl0}^{a + b,p + q,0} = \mathit{\boldsymbol{R}}_{jl0}^{ap0}\mathit{\boldsymbol{R}}_{jl0}^{bq0} + {\left( { - 1} \right)^{jl}}\mathit{\boldsymbol{R}}_{jl1}^{ap0}\mathit{\boldsymbol{R}}_{lj1}^{bq0} $$

(12)

$$ \mathit{\boldsymbol{R}}_{jl1}^{a + b,p + q,0} = \mathit{\boldsymbol{R}}_{jl1}^{ap0}\mathit{\boldsymbol{R}}_{lj0}^{bq0} + \mathit{\boldsymbol{R}}_{jl0}^{ap0}\mathit{\boldsymbol{R}}_{jl1}^{bq0} $$

(13)

$$ \begin{array}{*{20}{c}} {{{\left( { - 1} \right)}^{aq}}\mathit{\boldsymbol{R}}_{jl0}^{a + b,p + q,1} = \mathit{\boldsymbol{R}}_{jl0}^{ap1}\mathit{\boldsymbol{R}}_{jl0}^{bq1} + }\\ {{{\left( { - 1} \right)}^{jl}}\mathit{\boldsymbol{R}}_{jl1}^{ap1}\mathit{\boldsymbol{R}}_{lj1}^{bq1}} \end{array} $$

(14)

$$ {\left( { - 1} \right)^{aq}}\mathit{\boldsymbol{R}}_{jl1}^{a + b,p + q,1} = \mathit{\boldsymbol{R}}_{jl1}^{ap1}\mathit{\boldsymbol{R}}_{lj0}^{bq1} + \mathit{\boldsymbol{R}}_{jl0}^{ap1}\mathit{\boldsymbol{R}}_{jl1}^{bq1} $$

(15)

$$ \mathit{\boldsymbol{R}}_{bq0}^{ik0}\mathit{\boldsymbol{R}}_{ap1}^{ik0} + {\left( { - 1} \right)^{ik}}\mathit{\boldsymbol{R}}_{bq0}^{ik1}\mathit{\boldsymbol{R}}_{ap1}^{ki1} = 0 $$

(16)

$$ \mathit{\boldsymbol{R}}_{bq0}^{ik10}\mathit{\boldsymbol{R}}_{ap1}^{ki0} + \mathit{\boldsymbol{R}}_{bq0}^{ik0}\mathit{\boldsymbol{R}}_{ap1}^{ik1} = 0 $$

(17)

$$ \mathit{\boldsymbol{R}}_{bq1}^{ik0}\mathit{\boldsymbol{R}}_{ap0}^{ik0} + {\left( { - 1} \right)^{ik}}\mathit{\boldsymbol{R}}_{bq1}^{ik1}\mathit{\boldsymbol{R}}_{ap0}^{ki1} = 0 $$

(18)

$$ \mathit{\boldsymbol{R}}_{bq1}^{ik1}\mathit{\boldsymbol{R}}_{aq0}^{ki0} + \mathit{\boldsymbol{R}}_{bq1}^{ik0}\mathit{\boldsymbol{R}}_{aq0}^{ik1} = 0 $$

(19)

$$ \mathit{\boldsymbol{R}}_{a + b,p + q,0}^{i,k,0} = \mathit{\boldsymbol{R}}_{bq0}^{ik0}\mathit{\boldsymbol{R}}_{ap0}^{ik0} + {\left( { - 1} \right)^{ik}}\mathit{\boldsymbol{R}}_{bq0}^{ik1}\mathit{\boldsymbol{R}}_{ap0}^{ik1} $$

(20)

$$ \mathit{\boldsymbol{R}}_{a + b,p + q,0}^{ik1} = \mathit{\boldsymbol{R}}_{bq0}^{ik1}\mathit{\boldsymbol{R}}_{ap0}^{ki0} + \mathit{\boldsymbol{R}}_{bq0}^{ik0}\mathit{\boldsymbol{R}}_{ap0}^{ik1} $$

(21)

$$ \begin{array}{*{20}{c}} {{{\left( { - 1} \right)}^{aq}}\mathit{\boldsymbol{R}}_{a + b,p + q,1}^{ik0} = \mathit{\boldsymbol{R}}_{bq1}^{ik0}\mathit{\boldsymbol{R}}_{ap1}^{ik0} + }\\ {{{\left( { - 1} \right)}^{ik}}\mathit{\boldsymbol{R}}_{bq1}^{ik1}\mathit{\boldsymbol{R}}_{ap1}^{ki1}} \end{array} $$

(22)

$$ {\left( { - 1} \right)^{aq}}\mathit{\boldsymbol{R}}_{a + b,p + q,1}^{ik1} = \mathit{\boldsymbol{R}}_{bq1}^{ik1}\mathit{\boldsymbol{R}}_{ap1}^{ki0} + \mathit{\boldsymbol{R}}_{bq1}^{ik0}\mathit{\boldsymbol{R}}_{ap1}^{ik1} $$

(23)

情形一:

若对任意i、k、j、l、(s, t)=(0, 1)及(s, t)=(1, 0)，有R_jlt^iks=0.

此时由方程式(23)得到R_jl1^ik1=0.这样R-矩阵可表示为

$$ \mathit{\boldsymbol{R}} = \sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {\mathit{\boldsymbol{R}}_{jl0}^{ik0}{E_{ik}} \otimes {E_{jl}}} } $$

根据引理1可假定

$$ \mathit{\boldsymbol{R}}: = {\mathit{\boldsymbol{R}}_{pqrs}} = \sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {{\omega ^{\left( {pij + rkj} \right) + \left( {skl + qil} \right)}}{E_{ik}} \otimes {E_{jl}}} } $$

此时

$$ {\Delta ^{{\rm{cop}}}}\left( x \right)\mathit{\boldsymbol{R}} = \mathit{\boldsymbol{R}}\Delta \left( x \right) $$

$$ {\Delta ^{{\rm{cop}}}}\left( y \right)\mathit{\boldsymbol{R}} = \mathit{\boldsymbol{R}}\Delta \left( y \right) $$

还要满足

$$ {\Delta ^{{\rm{cop}}}}\left( z \right)\mathit{\boldsymbol{R}} = \mathit{\boldsymbol{R}}\Delta \left( z \right) $$

注意到

$$ {\Delta ^{{\rm{cop}}}}\left( z \right) = {J^\tau }\left( {z \otimes z} \right) = \left( {z \otimes z} \right)J $$

$$ z{E_{ik}} = {E_{ki}}z $$

因此有

$$ \begin{array}{*{20}{c}} {{\Delta ^{{\rm{cop}}}}\left( z \right)\mathit{\boldsymbol{R = }}\left( {z \otimes z} \right)\mathit{\boldsymbol{J}}{\mathit{\boldsymbol{R}}_{pqrs}}\mathit{\boldsymbol{ = }}}\\ {{{\left( { - 1} \right)}^{jk}}{\omega ^{\left( {pkl + ril} \right) + \left( {sij + qkj} \right)}}\left( {{E_{ik}} \otimes {E_{jl}}} \right)\left( {z \otimes z} \right)} \end{array} $$

$$ \begin{array}{*{20}{c}} {{\mathit{\boldsymbol{R}}_{pqrs}}\Delta \left( z \right)\mathit{\boldsymbol{ = }}{{\left( { - 1} \right)}^{il}}{\omega ^{\left( {pij + rkj} \right) + \left( {skl + qil} \right)}} \cdot }\\ {\left( {{E_{ik}} \otimes {E_{jl}}} \right)\left( {z \otimes z} \right)} \end{array} $$

因此

$$ {\left( { - 1} \right)^{il - kj}} = {\omega ^{pij + rkj + skl + qil - pkl - ril - sij - qkj}} $$

即

$$ 1 = {\omega ^{\left( {p - s} \right)ij + \left( {r - q - 2} \right)kj + \left( {s - p} \right)kl + \left( {q - r + 2} \right)il}} $$

对所有的i、j、k、l成立，从而有p=s, r=q+2.

此时H₃₂具有如下泛R-矩阵

$$ {\mathit{\boldsymbol{R}}_{p,q}} = \sum\limits_{i,j = 0}^3 {\sum\limits_{k,l = 0}^3 {{\omega ^{pij + \left( {q + 2} \right)kj + pkl + qil}}{E_{ik}} \otimes {E_{jl}}} } $$

(24)

式中p、q均为0, 1, 2, 3.

情形二:

若对某些j₀≠l₀, 有R_j₀l₀0⁰⁰¹≠0.

根据式(10)，当a=p=0时，对任意b、q, 有R_j₀l₀0⁰⁰¹R_j₀l₀0^bq0=0.因而必有R_j₀l₀0^bq0=0.特别地，有R_j₀l₀0⁰¹⁰=0.若存在某个数对j′≠l′使得R_j′l′0⁰⁰¹=0, 则R_j′l′0⁰⁰⁰=1.由式(12)可得

$$ \mathit{\boldsymbol{R}}_{j'l'0}^{ap0} = {\left( {\mathit{\boldsymbol{R}}_{j'l'0}^{010}} \right)^a}{\left( {\mathit{\boldsymbol{R}}_{j'l'0}^{100}} \right)^p} $$

有

$$ \mathit{\boldsymbol{R}}_{j'l'0}^{000} = {\left( {\mathit{\boldsymbol{R}}_{j'l'0}^{010}} \right)^4}{\left( {\mathit{\boldsymbol{R}}_{j'l'0}^{100}} \right)^4} = 1 $$

可知R_j′l′0¹⁰⁰≠0, R_j′l′0⁰¹⁰≠0.又根据式(20)有

$$ 0 \ne \mathit{\boldsymbol{R}}_{j'l'0}^{010} = {\left( {\mathit{\boldsymbol{R}}_{100}^{010}} \right)^{j'}}{\left( {\mathit{\boldsymbol{R}}_{010}^{010}} \right)^{l'}} $$

所以R₁₀₀⁰¹⁰≠0, R₀₁₀⁰¹⁰≠0.此时0 =R_j₀l₀0⁰¹⁰=(R₁₀₀⁰¹⁰)^j₀(R₀₁₀⁰¹⁰)^l₀≠0矛盾.因此对所有j≠l及任意b、q, 有R_jl0⁰⁰¹≠0, R_jl0^bq0=0.特别地，R_jl0ⁱⁱ⁰=0, R_jl0⁰⁰⁰=0.因而根据式(1)得R_jl0⁰⁰¹=1.

进一步地，由式(14)可得

$$ {\omega ^{2a}}\mathit{\boldsymbol{R}}_{jl0}^{a + 1,a + 1,1} = \mathit{\boldsymbol{R}}_{jl0}^{aa1}\mathit{\boldsymbol{R}}_{jl0}^{111} $$

进而有

$$ \begin{array}{*{20}{c}} {\mathit{\boldsymbol{R}}_{jl0}^{a + 1,a + 1,1} = {\omega ^{ - 2a}}\mathit{\boldsymbol{R}}_{jl0}^{aa1}\mathit{\boldsymbol{R}}_{jl0}^{111} = }\\ {{\omega ^{ - a\left( {a + 1} \right)}}{{\left( {\mathit{\boldsymbol{R}}_{jl0}^{111}} \right)}^{a + 1}}} \end{array} $$

因此

$$ \mathit{\boldsymbol{R}}_{jl0}^{aa1} = {\omega ^{a\left( {1 - a} \right)}}{\left( {\mathit{\boldsymbol{R}}_{jl0}^{111}} \right)^a} $$

从而对所有i及j≠l, 有R_jl0ⁱⁱ¹≠0，且

$$ 1 = \mathit{\boldsymbol{R}}_{jl0}^{001} = \mathit{\boldsymbol{R}}_{jl0}^{441} = {\left( {\mathit{\boldsymbol{R}}_{jl0}^{111}} \right)^4} $$

则存在某s=0, 1, 2, 3使得

$$ \mathit{\boldsymbol{R}}_{jl0}^{111} = {\omega ^s} $$

进而有

$$ \mathit{\boldsymbol{R}}_{jl0}^{ii1} = {\omega ^{i\left( {1 - i} \right)}}{\omega ^s},s = 0,1,2,3 $$

根据

$$ \mathit{\boldsymbol{R}}_{jl0}^{ii1} = {\left( { - 1} \right)^{ij - il}}R_{lj0}^{ii1} $$

最后适当调整s可得

$$ \mathit{\boldsymbol{R}}_{jl0}^{ii1} = {\omega ^{i\left( {1 - i} \right)}}{\omega ^{\left( {sj - \left( {2 - s} \right)l} \right)i}} $$

$$ \mathit{\boldsymbol{R}}_{lj0}^{ii1} = {\omega ^{i\left( {1 - i} \right)}}{\omega ^{\left( {sl - \left( {2 - s} \right)j} \right)i}} $$

式中s可取0, 1, 2, 3.

根据式(20)，当a≠p时，可得R_ap0ⁱⁱ¹R_bb0ⁱⁱ¹=0.因为这时R_ap0ⁱⁱ¹≠0，必有R_bb0ⁱⁱ¹=0.

若R₀₀₁^ik0=0, 对所有i≠k.根据式(22)，当i≠k, a=p=0, b、q任意时，可得R_bq1^ik0=0.再由式(17), 当b≠q, a, p任意时，可得R_bq0ⁱⁱ¹R_ap1ⁱⁱ⁰=0. 因为这时R_bq0ⁱⁱ¹≠0，则必有R_ap1ⁱⁱ⁰=0.所以可得对任意的i、k、j、l, 有R_jl1^ik0=0.

根据式(23), 当i≠k, b、q任意时，可得R_bq1^ik1=0.再由式(16)，当b≠q，可得R_bq0ⁱⁱ¹R_aa1ⁱⁱ¹=0.

因为这时R_bq0ⁱⁱ¹≠0，则对所有a、b、i有R_aa1ⁱⁱ¹=0.所以对任意的i、k、j、l有R_jl1^ik1=0.

另一方面，由式(14)当j≠l时，取a≠p, b≠q，但a+b=p+q.这种情况总是可以取到.这时等式左边为R_jl0^{a+b, a+b, 1}≠0，而右边值为0, 矛盾, 这表明一定存在某些i₀≠k₀, 有R₀₀₁^i₀k₀0≠0.此时用上面同样方法可以得到对任意i≠k，有R₀₀₁^ik0≠0，且对任意的j, R_jj0^ik0=0.

$$ \mathit{\boldsymbol{R}}_{jj1}^{ik0} = {\omega ^{j\left( {1 - j} \right)}}{\omega ^{\left( {ti - \left( {2 - t} \right)k} \right)j}} $$

$$ \mathit{\boldsymbol{R}}_{jj1}^{ki0} = {\omega ^{j\left( {1 - j} \right)}}{\omega ^{\left( {tk - \left( {2 - t} \right)i} \right)j}} $$

其中t=0, 1, 2, 3.

对任意j，一定存在j=a+b=p+q且a≠p, b≠q.根据式(22)，可得R_jj1ⁱⁱ⁰=0.再由式(21)得

$$ {\omega ^{2a}} = \mathit{\boldsymbol{R}}_{a + 1,a + 1,1}^{ik0} = \mathit{\boldsymbol{R}}_{111}^{ik0}\mathit{\boldsymbol{R}}_{ap1}^{ik0} + {\left( { - 1} \right)^{ik}}\mathit{\boldsymbol{R}}_{111}^{ik1}\mathit{\boldsymbol{R}}_{ap1}^{ki1} $$

$$ {\omega ^{2\left( {j - 1} \right)}}\mathit{\boldsymbol{R}}_{jj1}^{ik0} = \mathit{\boldsymbol{R}}_{111}^{ik0}\mathit{\boldsymbol{R}}_{j - 1,j - 1,1}^{ik0} + {\left( { - 1} \right)^{ik}}\mathit{\boldsymbol{R}}_{111}^{ik1}\mathit{\boldsymbol{R}}_{j - 1,j - 1,1}^{ki1} $$

当i≠k时得

$$ {\omega ^{2\left( {j - 1} \right)}}\mathit{\boldsymbol{R}}_{jj1}^{ik0} = \mathit{\boldsymbol{R}}_{111}^{ik0}\mathit{\boldsymbol{R}}_{j - 1,j - 1,1}^{ik0} = {\left( {\mathit{\boldsymbol{R}}_{111}^{ik0}} \right)^j}\mathit{\boldsymbol{R}}_{001}^{ik0} $$

ω^2(j－1)R_jj1^ik0=R₁₁₁^ik0R_{j－1, j－1, 1}^ik0=(R₁₁₁^ik0)^jR₀₀₁^ik0

进一步地，由式(20)可得对所有j, R_jj0ⁱⁱ⁰≠0.下面计算R_jj0ⁱⁱ⁰的值.由式(12)及式(20)

$$ \begin{array}{*{20}{c}} {\mathit{\boldsymbol{R}}_{jj0}^{a + 1,a + 1,0} = \mathit{\boldsymbol{R}}_{jj0}^{aa0}\mathit{\boldsymbol{R}}_{jj0}^{110} = }\\ {\mathit{\boldsymbol{R}}_{jj0}^{a - 1,a - 1,0}{{\left( {\mathit{\boldsymbol{R}}_{jj0}^{110}} \right)}^2} = {{\left( {\mathit{\boldsymbol{R}}_{jj0}^{110}} \right)}^{a + 1}}} \end{array} $$

所以R_jj0ⁱⁱ⁰=(R₁₁₀¹¹⁰)ⁱ=(R₁₁₀¹¹⁰)^ij.

当i=1, j=4时, 有

$$ \mathit{\boldsymbol{R}}_{440}^{110} = \mathit{\boldsymbol{R}}_{000}^{110} = {\left( {\mathit{\boldsymbol{R}}_{110}^{110}} \right)^4} = 1 $$

得R₁₁₀¹¹⁰=ω^t′，其中t′=0, 1, 2, 3.所以R_jj0ⁱⁱ⁰=ω^t′ij.

最后考虑R_jl1^ik1的值.根据式(15)可知, ${\mathit{\boldsymbol{R}}}_{jj1}^{a{\rm{ }} + b,a + b,1} = 0$ 从而对所有R_jj1ⁱⁱ¹=0.当j≠l, a≠p，b≠q，同时a+b=p+q时，由式(14)可知, 0≠R_jl0^{a+b, a+b, 1}=R_jl1^ap1R_lj1^bq1, 所以R_jl1^ap1≠0.

然而总可以取到a+b≠p+q, 同时a≠p, b≠q (因为a、b、p、q均∈$\mathbb{Z}$₄总可以取到这种情形.例如取a=1, b=0, p=2, q=1).此时，a≠p, b≠q, a+b≠p+q.

此时由式(13)知

$$ 0 \ne \mathit{\boldsymbol{R}}_{jj1}^{a + b,p + q,0} = \mathit{\boldsymbol{R}}_{jj1}^{ap0}\mathit{\boldsymbol{R}}_{jj0}^{bq0} + \mathit{\boldsymbol{R}}_{jj0}^{ap0}\mathit{\boldsymbol{R}}_{jj1}^{bq0} = 0 $$

这是一个矛盾.

实际上由式(21)也能得出矛盾，从而此种情形不存在H₃₂的泛R-矩阵.

情形三:

若对某些i≠k，R₀₀₁^ik0≠0，同理可得相应结论.

综上所述，Hopf代数H₃₂只有式(24)形式的泛R-矩阵，定理得证.

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