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  浙江大学学报(理学版)  2016, Vol. 43 Issue (6): 672-675  DOI:10.3785/j.issn.1008-9497.2016.06.008
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陈笑缘, 贾玲. 弱广义对角交叉积的表示范畴[J]. 浙江大学学报(理学版), 2016, 43(6): 672-675. DOI: 10.3785/j.issn.1008-9497.2016.06.008.
[复制中文]
CHEN Xiaoyuan, JIA Ling. The representation category of weak generalized diagonal crossed product[J]. Journal of Zhejiang University(Science Edition), 2016, 43(6): 672-675. DOI: 10.3785/j.issn.1008-9497.2016.06.008.
[复制英文]

基金项目

山东省自然科学基金资助项目(ZR2012AL02)

作者简介

陈笑缘(1963-), ORCID:http://orcid.org/0000-0003-2898-9976, 女, 教授, 主要从事代数研究

文章历史

收稿日期:2015-10-22
弱广义对角交叉积的表示范畴
陈笑缘1 , 贾玲2     
1. 浙江商业职业技术学院, 浙江 杭州 310053;
2. 鲁东大学 数学与统计科学学院, 山东 烟台 264025
摘要: 引入弱(HA)-Yetter Drinfeld模和弱广义对角交叉积代数,证明了弱广义对角交叉积的表示范畴同构于弱(HA)-Yetter Drinfeld模范畴.
关键词: 弱Hopf代数    弱(H, A)-Yetter Drinfeld模    弱广义对角交叉积    
The representation category of weak generalized diagonal crossed product
CHEN Xiaoyuan1 , JIA Ling2     
1. Zhejiang Business College, Hangzhou 310053, China;
2. School of Mathematics and Statistics Science, Ludong University, Yantai 264025, Shandong Province, China
Abstract: The notions of weak (H, A)-Yetter Drinfeld datum and weak generalized diagonal cross product algebra are introduced. We prove that the category of modules over a weak (H, A)-Yetter Drinfeld datum is isomorphic to the representation category of ${H^*}\bowtie A - $, which is an induced associative algebra called weak generalized diagonal cross product.
Key words: weak Hopf algebra    weak (H, A)-Yetter Drinfeld module    weak generalized diagonal cross product    

BOHM等[1]引入的弱Hopf代数是Hopf代数重要的推广结构之一,随着Hopf代数理论体系的日臻完善,其在数学物理、量子群等领域的应用日渐广泛;Yetter-Drinfeld是Hopf代数理论中的重要结构,文献[2-3]引入的Yetter-Drinfeld数组进一步推广了结论,得到其上的模范畴同构于对角交叉积表示范畴;文献[4]讨论了Yetter-Drinfeld群模的表示范畴;文献[5]将Yetter-Drinfeld结构在弱Hopf群余代数环境下进行了重新构建.

本文讨论弱广义对角交叉积代数的表示范畴,推广了Hopf代数理论中的相应内容.关于弱Hopf代数的基本概念请参考文献[1, 5-6].

注1 文中k为域, idA为从k-空间A到自身的恒等映射.记双线性映射μ:A⊗2Aμ(x, y)=xy,Δ:AA⊗2为Δ(x)=∑x1x2, xA,用Sweedle符号表示左(右)余模,${\rho ^l}:M \to H \otimes M, \mapsto \sum {{m_{(- 1)}} \otimes {m_{(0)}}({\rho ^r}:M \to M \otimes H, m \mapsto \sum {{m_{[0]}} \otimes {m_{[1]}}} )} $.规定$(h\rightharpoonup\alpha )(g) = \alpha (gh), (\alpha \leftharpoonup h)(g) = \alpha (hg), h, g \in H, \alpha \in {H^*}$.除特殊声明,文中均指有限维Hopf代数.

定义1[1] H是域k上的弱Hopf代数.一个代数A被称为左H-余模代数指A是左H-余模且满足:

$ \sum {{{(ab)}_{(-1)}} \otimes } {(ab)_{(0)}} = \sum {{a_{(-1)}}{b_{(-1)}} \otimes } {a_{(0)}}{b_{(0)}}; $ (1)
$ \sum {a_{(-1)}^r \otimes } {a_{(0)}} = \sum {{1_{(-1)}} \otimes } a{1_{(0)}}. $ (2)

类似地,右H-余模代数是一个k-代数A且是右H-余模, 满足:

$ \sum {{{(ab)}_{[0]}} \otimes } {(ab)_{[1]}} = \sum {{a_{[0]}}{b_{[0]}} \otimes } {a_{[1]}}{b_{[1]}}; $ (3)
$ \sum {{a_{[0]}} \otimes } a_{[1]}^l = \sum {{1_{[0]}}a \otimes } {1_{[1]}}. $ (4)

另外,如果A既是右H-余模代数又是左H-余模代数,且对任意aA,满足:

$ \sum {{a_{[0]( - 1)}} \otimes {a_{[0](0)}} \otimes {a_{[1]}} = \sum {{a_{( - 1)}} \otimes {a_{(0)[0]}} \otimes {a_{(0)[1]}}, } } $ (5)

则称其为H-双余模代数.

注2 条件(2)和(4)可分别被下列式子替换:

$ (2')\;\sum {{1_1} \otimes {1_2}1{'_{(-1)}} \otimes 1{'_{(0)}} = } \sum {{1_{(-1)1}} \otimes {1_{(-1)2}} \otimes {1_{(0)}};} $
$ (4')\;\sum {1{'_{[1]}}{1_1} \otimes {1_2} \otimes 1{'_{[0]}} = } \sum {{1_{[-1]1}} \otimes {1_{[-1]2}} \otimes {1_{[0]}}.} $

例1[1] Hk上的弱Hopf代数,则HH-双余模代数可通过余乘法运算.

定义2 Hk上的弱Hopf代数.AH-双余模代数.一个弱左-右-(H, A)-Yetter-Drinfeld模MM是左A-模且是右H-模,且对任意aA, mM,满足以下等价条件之一:

$ \begin{array}{l} \sum {{{(a \cdot m)}_{[0]}} \otimes {{(a \cdot m)}_{[1]}} = \sum {{a_{[0](0)}} \cdot {m_{[0]}}} } \\ {a_{( - 1)}}{m_{[1]}}{S^{ - 1}}({a_{[0](1)}}); \end{array} $ (6)
$ \begin{array}{l} \sum {{a_{[0]}} \cdot {m_{[0]}} \otimes {a_{[1]}} \cdot {m_{[1]}} = \sum {{{({a_{(0)}} \cdot m)}_{[0]}} \otimes } } \\ {({a_{(0)}} \cdot m)_{[1]}}{a_{( -1)}}. \end{array} $ (7)

AYDH表示弱(H, A)-Yetter-Drinfeld模范畴.

引理1 Hk上的弱Hopf代数.AH-双余模代数.则k-空间H*A的如下乘法是结合的:

$ \begin{array}{l} (a \otimes a)(\beta \otimes b) = \sum \alpha ({a_{[0]( - 1)}}\rightharpoonup\beta \leftharpoonup{S^{ - 1}}({a_{[1]}})) \times \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{a_{[0](0)}}, \;\;b, \alpha, \beta \in {H^*}, a, b \in A. \end{array} $

证明 事实上,对任意α, β, γH*, a, b, cA,

$ \begin{array}{l} [(\alpha \otimes a)(\beta \otimes b)](\gamma \otimes c) = \sum \alpha ({a_{[0]( - 1)}}\rightharpoonup\beta \leftharpoonup{S^{ - 1}}({a_{[1]}})) \times \\ ({a_{[0](0)[0]( - 1)}}{b_{[0]( - 1)}}\rightharpoonup\gamma \leftharpoonup{S^{ - 1}}({a_{[0](0)[1]}}{b_{[1]}})) \otimes \\ {a_{[0](0)[0](0)}}{b_{[0](0)}}c\mathop = \limits^{(5)} \alpha ({a_{[0]( - 1)1}}\rightharpoonup\beta \leftharpoonup\\ {S^{ - 1}}({a_{[1]2}}))({a_{[0]( - 1)2}}{b_{[0]( - 1)}}\rightharpoonup\gamma \leftharpoonup{S^{ - 1}}({a_{[1]1}}{b_{[1]}})) \otimes \\ {a_{[0](0)}}{b_{[0](0)}}c = \sum \alpha ({a_{[0]( - 1)}}\rightharpoonup\\ ({b_{[0]( - 1)}}\rightharpoonup\beta \leftharpoonup{S^{ - 1}}({b_{[1]}}))\rightharpoonup{S^{ - 1}}({a_{[-1]}})) \otimes \\ {a_{[0](0)}}{b_{[0](0)}}c = (\alpha \otimes a)[(\beta \otimes b)(\gamma \otimes c)]. \end{array} $

引理2 Hk上的弱Hopf代数.AH-双余模代数,则形如αa-(αa)(ε⊗1)和αa-(ε⊗1)(αa), αH*, aA的元素所生成的k-空间IH*A的双边理想.

证明 实际上,对任意α, βH*, a, bA,

$ \begin{array}{l} (\alpha \otimes a)(\beta \otimes b)- (\varepsilon \otimes 1)(\alpha \otimes a)(\beta \otimes b) = \\ \sum {\alpha ({\alpha _{[0]( - 1)}}\rightharpoonup\beta \leftharpoonup{S^{ - 1}}({a_{[-1]}}))} \otimes {a_{[0](0)}}b - \\ ({1_{[0]( - 1)}}\rightharpoonup\alpha \leftharpoonup{S^{ - 1}}({1_{[-1]}})) \times \\ ({1_{[0](0)[0]( - 1)}}\rightharpoonup\beta \leftharpoonup{S^{ - 1}}({1_{[0](0)[1]}}{a_{[1]}})) \otimes \\ {1_{[0](0)[0](0)}}{a_{[0](0)}}b\mathop = \limits^{(5)} \sum \alpha ({a_{[0]( - 1)}}\rightharpoonup\beta \leftharpoonup\\ {S^{ - 1}}({a_{[1]}})) \otimes {a_{[0](0)}}b - ({1_{[0]( - 1)1}}\rightharpoonup\alpha \leftharpoonup\\ {S^{ - 1}}({1_{[1]2}})) \times ({1_{[0]( - 1)2}}{a_{[0](0)}}b)\mathop = \limits^{(2')(4')} \\ \sum \alpha ({a_{[0]( - 1)}}\rightharpoonup\beta \leftharpoonup{S^{ - 1}}({a_{[1]}})) \otimes \\ {a_{[0](0)}}b - (1{'_1}\rightharpoonup\alpha \leftharpoonup{S^{ - 1}}(1{_2})) \times \\ (1{'_2}{1_{[0]( - 1)}}{a_{[0]( - 1)}}\rightharpoonup\beta \leftharpoonup{S^{ - 1}}({1_{[1]}}{a_{[1]}}){S^{ - 1}}(1{_{[1]}})) \otimes \\ {1_{[0](0)}}{a_{[0](0)}}b\mathop = \limits^{(1)(3)} \sum \alpha ({a_{[0]( - 1)}}\rightharpoonup\beta \leftharpoonup{S^{ - 1}}({a_{[1]}})) \otimes \\ {a_{[0](0)}}b - \alpha ({a_{[0]( - 1)}}\rightharpoonup\beta \leftharpoonup{S^{ - 1}}({a_{[1]}})) \otimes {a_{[0](0)}}b = 0.\\ (\beta \otimes b)(\alpha \otimes a) - (\beta \otimes b)(\varepsilon \otimes 1)(\alpha \otimes a) = \\ \sum {\beta ({b_{[0]( - 1)}}\rightharpoonup\alpha \leftharpoonup{S^{ - 1}}({b_{[1]}})) \otimes {b_{[0]}}(0)a - } \\ \beta ({b_{[0]( - 1)}}\rightharpoonup\varepsilon \leftharpoonup{S^{ - 1}}({b_{[1]}})) \times \\ ({b_{[0](0)[0]( - 1)}}\rightharpoonup\alpha \leftharpoonup{S^{ - 1}}({b_{[0](0)[1]}})) \otimes \\ {b_{[0](0)[0](0)}}a\mathop = \limits^{(5)} \sum {\beta ({b_{[0]( - 1)}}\rightharpoonup\alpha \leftharpoonup} \\ {S^{ - 1}}({b_{[1]}})) \otimes {b_{[0](0)}}a - \beta ({b_{[0]( - 1)}}\rightharpoonup\varepsilon \leftharpoonup{S^{ - 1}}({b_{[1]2}})) \times \\ ({b_{[0]( - 1)2}}\rightharpoonup\alpha \leftharpoonup{S^{ - 1}}({b_{[1]1}})) \otimes {b_{[0](0)}}a = \\ \sum {\beta ({b_{[0]( - 1)}}\rightharpoonup\alpha \leftharpoonup{S^{ - 1}}({b_{[1]}})) \otimes } {b_{[0](0)}}a - \\ \beta ({b_{[0]( - 1)}}\rightharpoonup(\varepsilon \alpha )\leftharpoonup{S^{ - 1}}({b_{[1]}})) \otimes {b_{[0](0)}}a = 0. \end{array} $

其他2个式子可类似证明.

定义3 Hk上的弱Hopf代数,AH-双余模代数.则称商代数${H^*}\bowtie A = ({H^*} \otimes A)/I$为弱广义对角交叉代数.其乘法如引理1所示.

注3 (1)若H是Hopf代数,则弱对角交叉积就是对角交叉积[3].

(2)令A=H,左右余模由余乘法给出,则弱广义对角交叉积就是通常意义下的弱量子偶D(H).

引理3 Hk上的弱Hopf代数, AH-双余模代数.则有

$ \alpha \bowtie a = (\alpha \bowtie 1)(\varepsilon \bowtie a), \;\;\;\;\;\;\;a \in A. $ (8)
$ (\varepsilon \bowtie a)(\varepsilon \bowtie b) = \varepsilon \bowtie ab, \;\;\;\;a, b \in A. $ (9)
$ (\alpha \bowtie 1)(\beta \bowtie 1) = \alpha \beta \bowtie 1, \;\;\;\;\;\;\alpha \beta \in {H^*}. $ (10)

证明 事实上,a, bA, α, βH*,

$ \begin{array}{l} (\alpha \bowtie 1)(\varepsilon \bowtie a) = \\ \sum {(1{'_1}\rightharpoonup\alpha \leftharpoonup{S^{- 1}}(1'{'_2}))(1{'_2}{1_{[0]( - 1)}}\rightharpoonup\varepsilon \leftharpoonup} \\ {S^{ - 1}}({1_{[1]}}){S^{ - 1}}(1'{'_2})) \otimes {1_{[0](0)}}\alpha \mathop = \limits^{(4')} \sum ( 1{'_1}\rightharpoonup\alpha \leftharpoonup\\ {S^{ - 1}}({1_{[1]2}}))(1{'_2}{1_{[0]( - 1)}}\rightharpoonup\varepsilon \leftharpoonup{S^{ - 1}}({1_{[1]1}})) \otimes \\ {1_{[0](0)}}a\mathop = \limits^{(2')(5)} \sum {({1_{( - 1)1}}\rightharpoonup\alpha \leftharpoonup} \\ {S^{ - 1}}({1_{(0)[1]2}}))({1_{( - 1)2}}\rightharpoonup\varepsilon \leftharpoonup{S^{ - 1}}({1_{(0)[1]1}})) \otimes \\ {1_{(0)[0]}}a = \sum ( {1_{[0]( - 1)}}\rightharpoonup\alpha \leftharpoonup{S^{ - 1}}({1_{[1]}})) \otimes \\ {1_{[0](0)}}a = (\varepsilon \bowtie 1)(\alpha \bowtie a) = \alpha \bowtie a.\\ (\varepsilon \bowtie a)(\varepsilon \bowtie b) = {\sum a _{[0]( - 1)}}\rightharpoonup\varepsilon \leftharpoonup{S^{ - 1}}(a_{[1]}^l) \otimes \\ {a_{[0](0)}}b\mathop = \limits^{(4)} {\sum 1 _{[0]( - 1)}}a_{( - 1)}^l\rightharpoonup\varepsilon \leftharpoonup{S^{ - 1}}({1_{[1]}}) \otimes \\ {1_{[0](0)}}{a_{(0)}}b = {\sum 1 _{[0]( - 1)}}1{'_{( - 1)}}\rightharpoonup\varepsilon \leftharpoonup{S^{ - 1}}({1_{[1]}}) \otimes \\ {1_{[0](0)}}1{'_{(0)}}ab = (\varepsilon \bowtie 1)(\varepsilon \bowtie ab) = \varepsilon \bowtie ab.\\ (\alpha \bowtie 1)(\beta \bowtie 1) = \sum {({1_{[0]( - 1)}}\rightharpoonup\alpha \leftharpoonup{S^{ - 1}}({1_{[1]}}))} \times \\ ({1_{[0](0)[0]( - 1)}}\rightharpoonup\alpha \leftharpoonup{S^{ - 1}}({1_{[0](0)[0]}})) \otimes {1_{[0](0)[0](0)}}\mathop = \limits^{(5)} \\ \sum ( {1_{[0]( - 1)1}}\rightharpoonup\alpha \leftharpoonup{S^{ - 1}}({1_{[1]2}}))({1_{[0]( - 1)2}}\rightharpoonup\alpha \leftharpoonup\\ {S^{ - 1}}({1_{[1]1}})) \otimes {1_{[0](0)}} = \sum ( {1_{[0]( - 1)}}\rightharpoonup(\alpha \beta )\leftharpoonup\\ {S^{ - 1}}({1_{[-1]}})) \otimes {1_{[0](0)}} = (\varepsilon \bowtie 1)(\alpha \beta \bowtie 1) = \alpha \beta \bowtie 1. \end{array} $

引理4 Hk上的弱Hopf代数, AH-双余模代数.若$M{ \in _{{H^*}\bowtie A}}M$,则M不仅在$a \circ m = (\varepsilon \bowtie a) \cdot m$下是左A-模,而且在$\sum {{m_{[0]}} \otimes {m_{[1]}} = } \sum ( \xi \bowtie {1_{[0](0)}}) \cdot m \otimes {S^{ - 1}}({1_{[1]}}){\xi _i}{1_{[0]( -1)}}$下是右H-余模,这里ξiH的一组基,ξiH*中相应的对偶基.

证明 由式(9),M显然是左A-模.只需证明其是右H-余模.因为对任意mMα, βH*,

$ \begin{array}{l} (i{d_M} \otimes \alpha \otimes \beta )({m_{[0][0]}} \otimes {m_{[0][1]}} \otimes {m_{[1]}}) = \\ \sum {[{\alpha _2}(1{'_{[0](0)[0]( - 1)}}\rightharpoonup{\beta _2}\leftharpoonup{S^{ - 1}}(1{'_{[0](0)[1]}}))} \bowtie \\ 1{'_{[0](0)[0](0)}}{1_{[0](0)}}] \cdot m{\beta _1}({S^{ - 1}}(1{'_{[1]2}})) \times \\ {\beta _3}({1_{[0]( - 1)}}) = \sum {[{\alpha _2}{\beta _3}\bowtie 1{'_{[0](0)}}{1_{[0](0)}}] \times } \\ m{\alpha _1}({S^{ - 1}}(1{'_{[1]}})){\alpha _3}{\beta _4}(1{'_{[0]( - 1)}}){\beta _1}({S^{ - 1}}({1_{[1]}})) \times \\ {\beta _5}({1_{[0]( - 1)}}) = \sum [\alpha {\beta _2}\bowtie {1_{[0](0)}}] \times \\ m{\beta _1}({S^{ - 1}}({1_{[1]}})){\beta _3}({1_{[0]( - 1)}}) = \\ [(\alpha \bowtie 1)(\beta \bowtie 1)] \cdot m = (\alpha \beta \bowtie 1) \cdot m.\\ (i{d_M} \otimes \alpha \otimes \beta )({m_{[0]}} \otimes {m_{[1]1}} \otimes {m_{[1]2}}) = \\ \sum {[{\alpha _2}{\beta _2}\bowtie {1_{[0](0)}}}] \cdot m{\alpha _1}({S^{ - 1}}({1_{[1]2}})){\alpha _3}({1_{[0]( - 1)1}}) \times \\ {\beta _1}({S^{ - 1}}({1_{[1]1}})){\beta _3}({1_{[0]( - 1)2}}) = \\ \sum {(({1_{[0]( - 1)}}\rightharpoonup(\alpha \beta )\leftharpoonup{S^{ - 1}}({1_{[1]}}))\bowtie {1_{[0](0)}})} \cdot m = \\ [(\varepsilon \bowtie 1)(\alpha \beta \bowtie 1)] \cdot m = (\alpha \beta \bowtie 1)m. \end{array} $

所以$\sum {{m_{[0][0]}} \otimes {m_{[0][1]}} \otimes {m_{[1]}} = {m_{[0]}} \otimes {m_{[1]1}} \otimes {m_{[1]2}}} $.并且$\begin{array}{l} \sum \varepsilon ({m_{[1]}}){m_{[0]}} = \sum {({1_{[0]( - 1)}}\rightharpoonup\varepsilon \rightharpoonup{S^{ - 1}}({1_{[1]}}))} \bowtie \\ {1_{[0](0)}}) \cdot m = (\varepsilon \bowtie 1) \cdot m = m. \end{array}$.

定理1 Hk上的弱Hopf代数.AH-双余模代数.则存在范畴同构$_AY{D^A}{ \cong _{{H^*}\bowtie A}}M$.

证明 若MAYDH,则M自然是左H*-模,其作用为α*m=∑α(m[1])m[0].定义

$ \begin{array}{l} (\alpha \bowtie \alpha ) \circ m = \alpha *(a \cdot m) = \\ \sum {\alpha (} {a_{[1]}}{m_{[1]}}{S^{ - 1}}({a_{[0]( - 1)}})){a_{[0](0)}}m, \\ \alpha \in {H^*}, m \in M, a \in A. \end{array} $

首先,断言上述作用是合理的.事实上,对αH*, mM, aA,

$ \begin{array}{l} ((\alpha \bowtie a)(\varepsilon \bowtie 1)) \circ m = \sum {\alpha ({a_{[0](0)[1]1}}{m_{[1]1}} \times } \\ {S^{ - 1}}({a_{[0](0)[0]( - 1)2}})) \times \\ \varepsilon ({S^{ - 1}}({a_{[1]}}){a_{[0](0)[1]2}}{m_{[1]2}}{S^{ - 1}}({a_{[0](0)[0]( - 1)1}}) \times \\ {a_{[0]( - 1)}}){a_{[0](0)[0](0)}} \cdot {m_{[0]}}\mathop = \limits^{(5)} \\ \sum {\alpha ({a_{[1]1}}{m_{[1]1}}{S^{ - 1}}({a_{[0]( - 1)2}}))\varepsilon ({S^{ - 1}}(a_{[1]2}^r){m_{[1]2}} \times } \\ {S^{ - 1}}(a_{[0]( - 1)1}^r)){a_{[0](0)}} \cdot {m_{[0]}} = \\ \sum \alpha ({a_{[1]1}}{m_{[1]1}}{S^{ - 1}}({a_{[0]( - 1)2}}))\varepsilon ({S^{ - 1}}(a_{[1]2}^r){m_{[1]2}} \times \\ {S^{ - 1}}(a_{[0]( - 1)1}^r)){a_{[0](0)}} \cdot {m_{[0]}} = \\ \sum {\alpha ({a_{[1]1}}{m_{[1]1}}{S^{ - 1}}({a_{[0]( - 1)2}}))} \times \\ \varepsilon ({a_{[1]2}}{m_{[1]2}}{S^{ - 1}}({a_{[0](1)1}})){a_{[0]0}}{m_{[0]}} \times \\ \sum \alpha ({a_{[1]}}{m_{[1]}}{S^{ - 1}}({a_{[0]( - 1)}})){a_{[0](0)}}{m_{[0]}} = (\alpha \bowtie a) \circ m.\\ ((\varepsilon \bowtie 1)(\alpha \bowtie a)) \circ m = \sum \alpha ({S^{ - 1}}({1_{[1]}}){1_{[0](0)[1]}}{a_{[1]}}{m_{[1]}} \times \\ {S^{ - 1}}({a_{[0]( - 1)}}){S^{ - 1}}({1_{[0](0)[0]( - 1)}}){1_{[0]( - 1)}}){1_{[0](0)[0](0)}} \times \\ {a_{[0](0)}}{m_{[0]}}\mathop = \limits^{(5)} \sum \alpha ({S^{ - 1}}(1_{[1]}^r){a_{[1]}}{m_{[1]}}{S^{ - 1}}({a_{[0]( - 1)}}) \times \\ {S^{ - 1}}(1_{[0]( - 1)}^r)){1_{[0](0)}}{a_{[0](0)}}{m_{[0]}}\mathop = \limits^{(2)(4)} \\ \sum {\alpha ({1_{[1]}}{a_{[1]}}{m_{[1]}}{S^{ - 1}}({a_{[0]( - 1)}}) \times } \\ {S^{ - 1}}({1_{[0]( - 1)}})){1_{[0](0)}}{a_{[0](0)}} \cdot {m_{[0]}} = \\ \sum {\alpha ({a_{[1]}}{m_{[1]}}{S^{ - 1}}({a_{[0]( - 1)}}))} {a_{[0](0)}} \cdot {m_{[0]}} = (\alpha \bowtie a) \circ m. \end{array} $

其次,证明它确实是左${H^*}\bowtie A$-模.若α, βH*, mM, a, bA,

$ \begin{array}{l} ((\alpha \bowtie a)(\beta \bowtie b)) \circ m\mathop = \limits^{(5)} \sum \alpha ({a_{[1]}}{b_{[1]1}}{m_{[1]1}}\\ {S^{ - 1}}({a_{[0]( - 1)3}}{b_{[0]( - 1)2}})) \times \\ \beta ({S^{ - 1}}({a_{[1]3}}){a_{[1]2}}{b_{[1]2}}{m_{[1]2}}{S^{ - 1}}({a_{[0]( - 1)2}}{b_{[0]( - 1)1}}) \times \\ {a_{[0]( - 1)1}}){a_{[0](0)}}{b_{[0](0)}}{m_{[0]}} = \\ \sum \alpha ({a_{[1]1}}{b_{[1]1}}{m_{[1]1}}{S^{ - 1}}({a_{[0]( - 1)2}}{b_{[0]( - 1)2}})) \times \\ \beta ({S^{ - 1}}(a_{[1]2}^r){b_{[1]2}}{m_{[1]2}}{S^{ - 1}}({b_{[0]( - 1)1}}){S^{ - 1}}(a_{[0]( - 1)1}^r)) \times \\ {a_{[0](0)}}{b_{[0](0)}}{m_{[0]}}\mathop = \limits^{(13){{(15)}^{[6]}}} \\ \sum \alpha ({a_{[-1]}}{1_1}{b_{[1]1}}{m_{[1]1}}{S^{ - 1}}({a_{[0]( - 1)2}}1{'_2}{b_{[0]( - 1)2}})) \times \\ \beta ({1_2}{b_{[1]2}}{m_{[1]2}}{S^{ - 1}}(1{'_1})){a_{[0](0)}}{b_{[0](0)}} \times \\ {m_{[0]}} = \sum {(\alpha \bowtie a) \circ ({b_{[0](0)}}{m_{[0]}})} \times \\ \beta ({b_{[1]}}{m_{[1]}}{S^{ - 1}}({b_{[0]( - 1)}})) = (\alpha \bowtie a) \circ ((\beta \bowtie b) \circ m) \times \\ (\varepsilon \bowtie 1) \circ m = \sum \varepsilon ({(1 \cdot m)_{[1]}}){(1 \cdot m)_{[0]}} = \\ \sum \varepsilon ({m_{[1]}}){m_{[0]}} = m. \end{array} $

最后,若M, NAYDH, f:MN既是左A-模,又是右H-余模,则断言f是左${H^*}\bowtie A-$模.因为对任意αH*, mM, aA,

$ \begin{array}{l} f((\alpha \bowtie a) \cdot m) = \\ \sum f (\alpha ({a_{[1]}}{m_{[1]}}{S^{ - 1}}({a_{[0]( - 1)}})){a_{[0](0)}} \cdot m) = \\ \sum \alpha ({a_{[1]}}{m_{[1]}}{S^{ - 1}}({a_{[0]( - 1)}})){a_{[0](0)}} \cdot f(m) = \\ (\alpha \bowtie a) \cdot f(m). \end{array} $

反之,若$M{ \in _{{H^*}\bowtie A}}M$,由引理4知,M既是左A-模,又是右H-余模.对任意αH*, mM, aA,有

$ \begin{array}{l} \sum ( i{d_M} \otimes \alpha )({a_{[0](0)}}) \cdot {m_{[0]}} \otimes {a_{[1]}}{m_{[1]}}{S^{ - 1}}({a_{[0]( - 1)}}))\mathop = \limits^{(5)} \\ \sum ( {a_{[0]( - 1)2}}\rightharpoonup{\alpha _2}\leftharpoonup{S^{ - 1}}({a_{[1]1}})\bowtie {a_{[0](0)}}) \circ m{\alpha _1} \times \\ ({a_{[1]2}}){\alpha _3}({S^{ - 1}}({a_{[0]( - 1)1}}))\mathop = \limits^{{{(7)}^{[6]}}(4)} \\ \sum ( {S^{ - 1}}({({1_{[0]( - 1)}}a_{( - 1)}^l)^l})\rightharpoonup\alpha \leftharpoonup{S^{ - 1}}({1_{[1]}})\bowtie \\ {1_{[0](0)}}{a_{(0)}}) \circ m\mathop = \limits^{(2)} \\ \sum ( {1_{[0]( - 1)}}\rightharpoonup\alpha \leftharpoonup{S^{ - 1}}({1_{[1]}})\bowtie {1_{[0](0)}}a) \circ m = \\ [({1_{[0]( - 1)1}}\rightharpoonup\alpha \leftharpoonup{S^{ - 1}}({1_{[1]2}}))({1_{[0]( - 1)2}}\rightharpoonup\varepsilon \leftharpoonup{S^{ - 1}}({1_{[1]1}}))\bowtie \\ {1_{[0](0)}}a] \circ m = \sum ( i{d_M} \otimes \alpha )({(a \cdot m)_{[0]}} \otimes \\ {(a \cdot m)_{[1]}})\sum ( i{d_M} \otimes \alpha )({a_{[0]}} \cdot {m_{[0]}} \otimes {a_{[1]}}{m_{[1]}}) = \\ \sum ( {a_{[0][0]( - 1)}}\rightharpoonup{\alpha _2}\leftharpoonup{S^{ - 1}}({a_{[0][1]}})\bowtie \\ {a_{[0][0](0)}}{1_{[0](0)}}) \circ m{\alpha _1}({a_{[1]}}{S^{ - 1}}({1_{[1]}})){\alpha _3}({1_{[0]( - 1)}})\mathop = \limits^{(4)} \\ \sum ( {1_{[0]( - 1)}}{a_{( - 1)}}\rightharpoonup\alpha \leftharpoonup{S^{ - 1}}({1_{[1]}})\bowtie {1_{[0](0)}}{a_{(0)}}) \circ m = \\ \sum ( {1_{[0]( - 1)}}\rightharpoonup{\alpha _1}\leftharpoonup{S^{ - 1}}({1_{[1]}})\bowtie \\ {1_{[0](0)}}{a_{(0)}}) \circ m{\alpha _2}({a_{( - 1)}})\mathop = \limits^{(6)} \sum [({1_{[0]( - 1)}}\rightharpoonup{\alpha _1}\leftharpoonup\\ {S^{ - 1}}({1_{[1]}}))({1_{[0](0)[0]( - 1)}}\rightharpoonup\varepsilon \leftharpoonup{S^{ - 1}}({1_{[0](0)[0]}}))\bowtie \\ {1_{[0](0)[0](0)}}{a_{(0)}}] \circ m{\alpha _2}({a_{( - 1)}}) = \sum ( i{d_M} \otimes \alpha ) \times \\ ({({a_{(0)}} \cdot m)_{[0]}} \otimes {({a_{(0)}} \cdot m)_{[1]}}{a_{( -1)}}). \end{array} $

M, N是左${H^*}\bowtie A-$模,f:MN是左${H^*}\bowtie A-$模同态,则由引理4知,f既是左A-模,亦是右H-余模.定理1得证.

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