Panaite和Van Oystaeyen在文献[1]中给出了L-R Smash积和L-R Smash余积的概念, 随后在文献[2]中又给出了L-R Smash积和L-R Smash余积构成L-R Smash双积的充分条件, 并且讨论了L-R Smash双积的辫子结构.而Radford在文献[3]中讨论了普通Smash双积的积分和类群元.本文推广了文献[3]中的结论并给出了L-R Smash双积$D\natural H$的积分和类群元的一些性质.为了方便起见, 首先给出一些相关概念.

本文均采用Sweedler记号. $k$表示一个域, 代数、余代数、张量积都是域$k$上的.设$C$是余代数, 对于$c\in C$, 记$\Delta(c)=\sum c_{1}\otimes c_{2}$.在下文中我们均省略和式符号$\sum$.

定义1.1 ^[1] 设$H$是一个双代数, $(A,\rightharpoonup,\leftharpoonup)$是一个$H$-双模代数, 在张量空间$A\otimes H$上定义乘法:

$(a\otimes h)(b\otimes g)=(a\leftharpoonup g_{2})(h_{1}\rightharpoonup b) \otimes h_{2}g_{1},\forall a,b\in A,h,g\in H. $

关于此乘法和以$1_{A}\otimes 1_{H}$为单位元构成的结合代数, 称为L-R Smash积, 记作$A\natural H$.

定义1.2 ^[1] 设$H$是一个双代数, $C$是一个$H$-双余模余代数, 在张量空间$C\otimes H$上定义余乘:

$\Delta(c\otimes h)=(c_{1}^{<0>}\otimes c_{2}^{(-1)}h_{1})\otimes (c_{2}^{(0)}\otimes h_{2}c_{1}^{<1>}).$

关于此余乘和以$\varepsilon(c\otimes h)=\varepsilon_{C}(c)\varepsilon_{H}(h)$为余单位构成的余结合余代数, 称为L-R Smash余积, 记作$C\natural H$.

定义1.3 ^[2] 设$H$是一个双代数, $D$是一个$H$-双余模余代数和$H$-双模代数, 在张量空间$D\otimes H$上定义L-R Smash积和L-R Smash余积, 记作$D\natural H$.如果满足下列条件:

$ \varepsilon_{D}(1_{D})=1, \varepsilon_{D}(cd)=\varepsilon_{D}(c)\varepsilon_{D}(d),\\ \varepsilon_{D}(h\cdot d)=\varepsilon_{D}(d\cdot h)=\varepsilon_{D}(d)\varepsilon_{H}(h),\\ \rho(1_{D})=1_{D}\otimes 1_{H}, \lambda(1_{D})=1_{H}\otimes 1_{D},\\ \Delta_{D}(1_{D})=1_{D}\otimes 1_{D},\\ \rho(cd)=c^{\langle 0\rangle}d^{\langle 0\rangle}\otimes c^{\langle 1\rangle}d^{\langle 1\rangle},\\ \lambda(cd)=c^{(-1)}d^{(-1)}\otimes c^{(0)}d^{(0)},\\ \Delta_{D}(h\cdot d)=h_{1}\cdot d_{1}\otimes h_{2}\cdot d_{2},\\ \Delta_{D}(d\cdot h)=d_{1}\cdot h_{1}\otimes d_{2}\cdot h_{2},\\ \Delta_{D}(cd)=c_{1}(c_{2}^{(-1)}\cdot d_{1}^{\langle 0\rangle})\otimes (c_{2}^{(0)}\cdot d_{1}^{\langle 1\rangle})d_{2},\\ (h_{1}\cdot d)^{(-1)}h_{2}\otimes (h_{1}\cdot d)^{(0)}=h_{1}d^{(-1)}\otimes h_{2}\cdot d^{(0)},\\ (h\cdot d)^{\langle 0\rangle}\otimes (h\cdot d)^{\langle 1\rangle}=h\cdot d^{\langle 0\rangle}\otimes d^{\langle 1\rangle},\\ (d\cdot h_{2})^{\langle 0\rangle}\otimes h_{1}(d\cdot h_{2})^{\langle 1\rangle}=d^{\langle0\rangle}\cdot h_{1}\otimes d^{\langle1\rangle}h_{2},\\ (d\cdot h)^{(-1)}\otimes (d\cdot h)^{(0)}=d^{(-1)}\otimes d^{(0)}\cdot h,\\ c^{\langle0\rangle}\cdot d^{(-1)}\otimes c^{\langle1\rangle}\cdot d^{(0)}=c\otimes d.$

那么$D\natural H$构成一个双代数, 称为L-R Smash双积.

这一节主要给出了L-R Smash双积$D\natural H$作为双代数的一些简单性质，构造了L-R Smash双积的积分和类群元.

定理2.1 设$H$是余交换的, 双代数$D\natural H$是交换的当且仅当双代数$D$和双代数$H$是交换的且对任意的$h\in H,d\in D, h\cdot d=d\cdot h$.

证 先证充分性.由已知$H$是余交换的, 可得$\Delta(h)=h_{1}\otimes h_{2}=h_{2}\otimes h_{1}$.

由已知$D$和$H$是交换的, 可得$dd^{'}=d^{'}d$, $hh^{'}=h^{'}h$.

$ (d^{'}\natural h^{'})(d\natural h)=(d^{'}\cdot h_{2})((h^{'})_{1}\cdot d)\natural (h^{'})_{2}h_{1}\\ =(d^{'}\cdot h_{1})((h^{'})_{2}\cdot d)\natural (h^{'})_{1}h_{2} =((h^{'})_{2}\cdot d)(d^{'}\cdot h_{1})\natural h_{2}(h^{'})_{1}\\ =(d\cdot (h^{'})_{2})(h_{1}\cdot d^{'})\natural h_{2}(h^{'})_{1}=(d\natural h)(d^{'}\natural h^{'}).$

所以$D\natural H$是交换的.

下证必要性.如果$D\natural H$是交换的, 则对任意$d\natural h,d^{'}\natural h^{'}\in D\natural H$, $(d^{'}\natural h^{'})(d\natural h)=(d\natural h)(d^{'}\natural h^{'}).$即

$(d^{'}\cdot h_{2})((h^{'})_{1}\cdot d)\natural (h^{'})_{2}h_{1}= (d\cdot (h^{'})_{2})(h_{1}\cdot d^{'})\natural h_{2}(h^{'})_{1}.$

(2.1)

对 (2.1) 式令$d=d^{'}=1_{D}$, 则$hh^{'}=h^{'}h$, 即$H$是交换的.

对 (2.1) 式令$h=h^{'}=1_{H}$, 则$dd^{'}=d^{'}d$, 即$D$是交换的.

对 (2.1) 式令$d^{'}=1_{D},h^{'}=1_{H}$, 则

$h_{2}\cdot d\natural h_{1}=d\cdot h_{1}\natural h_{2}.$

(2.2)

对 (2.2) 式两边同时用id $\otimes\varepsilon$作用得$h\cdot d=d\cdot h$.

定理2.2 设$H$是交换的, 双代数$D\natural H$是余交换的当且仅当双代数$D$和双代数$H$是余交换的，且对任意的$d\in D,d^{(0)}\otimes d^{(-1)}=d^{\langle 0\rangle}\otimes d^{\langle1\rangle}$.

证 先证充分性.由已知$H$是交换的, 可得$hh^{'}=h^{'}h$.

由已知$D$和$H$是余交换的, 可得$\Delta(h)=h_{1}\otimes h_{2}=h_{2}\otimes h_{1}$, $\Delta(d)=d_{1}\otimes d_{2}=d_{2}\otimes d_{1}, $

$ \Delta(d\natural h)=(d_{1}^{\langle 0\rangle}\natural d_{2}^{(-1)}h_{1})\otimes (d_{2}^{(0)}\natural h_{2}d_{1}^{\langle 1\rangle})\\ =(d_{1}^{(0)}\natural d_{2}^{\langle 1\rangle}h_{1})\otimes (d_{2}^{\langle 0\rangle}\natural h_{2}d_{1}^{(-1)})\\ =(d_{2}^{(0)}\natural d_{1}^{\langle 1\rangle}h_{2})\otimes (d_{1}^{\langle 0\rangle}\natural h_{1}d_{2}^{(-1)}) =\Delta^{cop}(d\natural h).$

所以$D\natural H$是余交换的.

下证必要性.已知$D\natural H$是余交换的.则

$(d_{1}^{\langle 0\rangle}\natural d_{2}^{(-1)}h_{1})\otimes (d_{2}^{(0)}\natural h_{2}d_{1}^{\langle 1\rangle})= (d_{2}^{(0)}\natural h_{2}d_{1}^{<1>})\otimes (d_{1}^{\langle 0\rangle}\natural d_{2}^{(-1)}h_{1}).$

(2.3)

对 (2.3) 式两边用$id\otimes \varepsilon\otimes id\otimes \varepsilon$作用得$d_{1}\otimes d_{2}=d_{2}\otimes d_{1}.$即$D$是余交换的.

对 (2.3) 式两边用$\varepsilon\otimes id\otimes \varepsilon\otimes id$作用得$h_{1}\otimes h_{2}=h_{2}\otimes h_{1}.$即$H$是余交换的.

此时 (2.3) 式可以写成

$(d_{2}^{\langle 0\rangle }\natural d_{1}^{(-1)}h_{2})\otimes (d_{1}^{(0)}\natural h_{1}d_{2}^{\langle 1\rangle})= (d_{2}^{(0)}\natural h_{2}d_{1}^{\langle 1\rangle})\otimes (d_{1}^{\langle 0\rangle}\natural d_{2}^{(-1)}h_{1}).$

(2.4)

对 (2.4) 式两边同时再用$id\otimes \varepsilon\otimes id\otimes \varepsilon$作用得

$d^{\langle 0\rangle}\natural 1_{H}\otimes 1_{D}\natural hd^{\langle 1\rangle}=d^{(0)}\natural 1_{H} \otimes 1_{D}\natural hd^{(-1)}.$

(2.5)

对 (2.5) 式令$h=1_{H}$, 则得$d^{\langle 0\rangle}\otimes d^{\langle 1\rangle}=d^{(0)}\otimes d^{(-1)}$.

下面的定理给出了$D\natural H$的积分和类群元与$D,H$的积分和类群元之间的关系.

定理2.3 设$D$和$H$是双代数, $D\natural H$是L-R Smash双积, 则$g\in D\natural H$是$D\natural H$的类群元当且仅当$g=d\natural h$, 这里$d$是$D$的类群元, $h$是$H$的类群元, 并且$d$的左右余模作用是平凡的, 即

$ \lambda(d)=d^{(-1)}\otimes d^{(0)}=1_{H}\otimes d, \\ \rho(d)=d^{\langle 0\rangle}\otimes d^{\langle 1\rangle}=d\otimes 1_{H}.$

证 充分性是显然的.

下证必要性.若$g=d\natural h$是$D\natural H$的类群元, 则

$\Delta(d\natural h)=(d_{1}^{\langle 0\rangle}\natural d_{2}^{(-1)}h_{1})\otimes (d_{2}^{(0)} \natural h_{2}d_{1}^{\langle 1\rangle })=(d\natural h)\otimes(d\natural h),\\ \varepsilon(d\natural h)=\varepsilon_{D}(d)\varepsilon_{H}(h)=1.$

(2.6)

对 (2.6) 式两边同时用$\varepsilon\otimes id\otimes \varepsilon\otimes id$作用得

$\varepsilon_{D}(d)h_{1}\otimes h_{2}=\varepsilon_{D}(d)h\otimes \varepsilon_{D}(d)h.$

(2.7)

令$d=1_{D}$, 则 (2.7) 式变为$h_{1}\otimes h_{2}=h\otimes h.$即

$ \Delta(h)=h\otimes h, \\ \varepsilon(1_{D}\natural h)=1\varepsilon_{H}(h)=\varepsilon_{H}(h)=1.$

故$h$是$H$的类群元.

对 (2.6) 式两边同时再用$id\otimes \varepsilon\otimes id\otimes \varepsilon$作用得

$\varepsilon_{H}(h)d_{1}\otimes d_{2}=\varepsilon_{H}(h)d\otimes \varepsilon_{H}(h)d.$

(2.87)

令$h=1_{H}$, 则 (2.8) 式变为

$d_{1}\otimes d_{2}=d\otimes d,$

即

$\Delta(d)=d\otimes d, \\ \varepsilon(d\natural1_{H})=\varepsilon_{D}(d)1=\varepsilon_{D}(d)=1.$

故$d$是$D$的类群元.

对 (2.6) 式令$h=1_{H}$, 则

$\Delta(d\natural 1_{H})=(d_{1}^{\langle 0\rangle}\natural d_{2}^{(-1)}1_{H})\otimes (d_{2}^{(0)} \natural1_{H}d_{1}^{\langle 1\rangle})=(d\natural 1_{H})\otimes(d\natural 1_{H}).$

(2.9)

对 (2.9) 式两边同时用$\varepsilon\otimes id \otimes id \otimes\varepsilon $作用得

$d^{(-1)}\otimes d^{(0)}=1_{H}\otimes d,$

即$d$的左-$H$余模作用是平凡的.

对 (2.9) 式两边同时用$id\otimes \varepsilon \otimes \varepsilon \otimes id$作用得

$d^{\langle 0\rangle }\otimes d^{\langle 1\rangle}=d\otimes 1_{H},$

即$d$的右-$H$余模作用是平凡的.

定理2.4 (1) 设$D$和$H$是双代数, $D\natural H$是L-R Smash双积.如果$x_{D}\natural x_{H}$是$D\natural H$的右积分, 则$x_{D}$是$D$的右积分, $x_{H}$是$H$的右积分.

(2) 设$D\natural H$是一个有限维Hopf代数, 如果$D\natural H$是半单的, 则存在左积分$x_{D}\in D,x_{H}\in H$, 使得$\varepsilon(x_{D})=\varepsilon(x_{H})=1$.

证 (1) 若$x_{D}\natural x_{H}$是$D\natural H$的右积分, 则对任意的$d\natural h$下式成立:

$(x_{D}\natural x_{H})(d\natural h)=x_{D}\natural x_{H}\varepsilon(d)\varepsilon(h),$

即

$(x_{D}\cdot h_{2})((x_{H})_{1}\cdot d)\natural (x_{H})_{2}h_{1}=x_{D}\natural x_{H}\varepsilon(d)\varepsilon(h).$

(2.10)

对 (2.10) 式两边同时用$\varepsilon\otimes id$作用得

$\varepsilon(x_{D})\varepsilon( d)1_{D}\natural x_{H}h= \varepsilon(x_{D})\varepsilon( d)1_{D}\natural \varepsilon(h)x_{H}.$

故$x_{H}h=\varepsilon(h)x_{H}$.即$x_{H}$是$H$的右积分.

令$h=1_{H}$, 则

$(x_{D}\natural x_{H})(d\natural 1_{H})=x_{D}\natural x_{H}\varepsilon(d)\varepsilon(1_{H}),$

即

$x_{D}((x_{H})_{1}\cdot d)\natural (x_{H})_{2}=x_{D}\natural x_{H}\varepsilon(d).$

(2.11)

对 (2.11) 式两边同时用$id\otimes \varepsilon$作用得

$x_{D}(x_{H}\cdot d)=x_{D}\varepsilon(x_{H}\cdot d).$

故$x_{D}$是$D$的右积分.

(2) 设$D\natural H$是半单的, 则存在左积分$d\natural h\in D\natural H$, 使得$\varepsilon_{D}(d)\varepsilon_{H}(h)=1$.由于$d\natural h$是$D\natural H$的左积分, 对任意的$b\natural 1_{H}$, 下式成立:

$(b\natural 1_{H})(d\natural h)=\varepsilon(b\natural 1_{H})(d\natural h),$

即

$b\cdot h_{2}d\natural h_{1}=\varepsilon(b)d\natural h.$

(2.12)

对 (2.12) 式两边同时用$id\otimes \varepsilon$作用得

$(b\cdot h)d=\varepsilon(b)\varepsilon(h)d=\varepsilon(b\cdot h)d.$

故$d$是$D$的左积分.

令$x_{D}=\varepsilon(h)d$, 显然$x_{D}$是$D$的左积分, 且$\varepsilon(x_{D})=1$.

同理, 对任意的$1_{D}\natural g$, 下式成立:

$(1_{D}\natural g)(d\natural h)=\varepsilon(1_{D}\natural g)(d\natural h),$

即

$g_{1}\cdot d\natural g_{2}h=\varepsilon(g)(d\natural h).$

(2.13)

对 (2.13) 式两边同时用$ \varepsilon\otimes id$作用得

$\varepsilon(d)\natural gh=\varepsilon(d)\natural \varepsilon(g)h.$

故$h$是$H$的左积分.

令$x_{H}=\varepsilon(d)h$, 显然$x_{H}$是$H$的左积分, 且$\varepsilon(x_{H})=1$.