Pure Mathematics
Vol. 09  No. 07 ( 2019 ), Article ID: 31982 , 8 pages
10.12677/PM.2019.97102

Lie Super-Bialgebra Structures on a Super Heisenberg-Virasoro Algebra

Meijun Li

School of Mathematics and Statistics, Qingdao University, Qingdao Shandong

Received: Aug. 6th, 2019; accepted: Aug. 26th, 2019; published: Sep. 2nd, 2019

ABSTRACT

In this paper we investigate Lie super-bialgebra structures on a super Heisenberg-Virasoro algebra. We obtain sufficient and necessary conditions for this type Lie super-bialgebra structures to be triangular coboundary.

Keywords:Lie Super-Bialgebras, Yang-Baxter Equations, A Super Heisenberg-Virasoro Algebra

一类超Heisenberg-Virasoro代数的超双代数 结构

李美君

青岛大学数学与统计学院,山东 青岛

收稿日期:2019年8月6日;录用日期:2019年8月26日;发布日期:2019年9月2日

摘 要

本文主要研究了一类超Heisenberg-Virasoro代数上的超双代数结构,得到了该类超李双代数为三角余边缘的充分必要条件。

关键词 :超李双代数,Yang-Baxter方程,超Heisenberg-Virasoro代数

Copyright © 2019 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY).

http://creativecommons.org/licenses/by/4.0/

1. 引言

为了研究量子群,Drinfeld于1983年在文献 [1] [2] 中引入了李双代数。无限维李代数的双代数结构无法统一分类,文献 [3] 引入了构造三角余边缘李双代数的方法,文献 [4] 构造了Witt型和Virasoro型李双代数,文献 [5] 给出了相应分类。文献 [6] [7] [8] 研究了Schrödinger-Virasoro李代数、广义Witt型李代数和Virasoro李代数等无限维李代数的双代数结构。本文将研究超Heisenberg-Virasoro代数 L 上的超双代数

结构,它是复数域 上的无限维李超代数,以 { L n , I n , G n | n } 为一组基,且满足以下运算

[ L m , L n ] = ( m n ) L m + n , [ L m , I n ] = n I m + n , [ L m , G n ] = n G m + n , [ I n , G n ] = [ I m , I n ] = 0 , [ G m , G n ] = I m + n (1.1)

2. 预备知识

给定超向量空间 L = L 0 ¯ L 1 ¯ ,假设以下元素都是 2 -分次的,用 | x | 2 表示x的次,即 x L | x | 。引入 τ ( x y ) = ( 1 ) | x | | y | y x ξ ( x y z ) = ( 1 ) | x | ( | y | + | z | ) y z x x , y , z L

定义2.1:李超代数 ( L , φ ) 由超向量空间L和双线性映射 φ : L L L 构成,且满足

φ ( L i , L j ) L i + j Ker ( 1 τ ) Ker φ φ ( 1 φ ) ( 1 + ξ + ξ 2 ) = 0

定义2.2:李超余代数 ( L , Δ ) 由超向量空间L和线性映射 Δ : L L L 构成,且满足

Δ ( L i ) j + k = i L j L k Im Δ Im ( 1 τ ) ( 1 + ξ + ξ 2 ) ( 1 Δ ) Δ = 0

定义2.3:李超双代数 ( L , φ , Δ ) 满足: ( L , φ ) 是李超代数, ( L , Δ ) 是超余代数,且有 Δ φ ( x , y ) = x Δ y ( 1 ) | x || y | yΔx x , y L ,其中“ ”表示对角伴随作用

x ( i a i b i ) = i ( [ x , a i ] b i + ( 1 ) | x | | a i | a i [ x , b i ] ) (2.1)

U 表示L的泛包络代数,记 r = i a i b i L L ,引入的元素

r 13 = i a i 1 b i = ( τ 1 ) ( 1 r ) = ( 1 τ ) ( r 1 )

r 12 = i a i b i 1 = r 1 r 23 = i 1 a i b i = 1 r . (2.2)

其中 1 是泛包络代数 U 的单位元,则定义 c ( r ) L L L 如下

c ( r ) = [ r 12 , r 13 ] + [ r 12 , r 23 ] + [ r 13 , r 23 ] r L L .

定义2.4:1) 余边缘李超双代数 ( L , φ , Δ , r ) 是一个四元组,其中 ( L , φ , Δ ) 是李超双代数,并且有 r Im ( 1 τ ) L L ,使得 Δ = Δ r (称为r的余边缘),即对任意 x L ,有 Δ r ( x ) = ( 1 ) | r | | x | x r

2) 我们称 ( L , φ , Δ , r ) 为三角的,若满足经典的Yang-Baxter方程 (CYBE)

c ( r ) = 0 (2.3)

3) 元素 r Im ( 1 τ ) L L 称为满足修正的Yang-Baxter方程,如果

x c ( r ) = 0 x L (2.4)

文献 [1] 给出了以下两个结果,元素r满足(2.3)当且仅当满足(2.4)。设L是李超代数,且

r Im ( 1 τ ) L L ,则有 ( 1 + ξ + ξ 2 ) ( 1 Δ r ) Δ r ( x ) = x c ( r ) ,三元组 ( L , [ , ] , Δ r ) 是李双代数当且仅当r满足(2.3)。

我们可把 V = L L = V 0 ¯ V 1 ¯ 看作对角伴随作用下的 L -模。用 Der ( L , V ) = Der 0 ¯ ( L , V ) + Der 1 ¯ ( L , V ) 表示导子 D : L V 的集合,且D满足

D ( [ x , y ] ) = ( 1 ) | D | | x | x D ( y ) ( 1 ) | y | ( | D | + | x | ) y D ( x ) x , y L (2.5)

导子是偶(奇)的,若 | D | = 0 ( | D | = 1 ) 。用 Inn ( L , V ) 表示内导子 v inn 的集合,其中 v inn : x ( 1 ) | v | | x | x v 。用 H 1 ( L , V ) 表示李代数 L 系数在 L -模 V 上的一阶上同调群,则 H 1 ( L , V ) Der ( L , V ) / Inn ( L , V ) * 表示非零整数且 \ A = { n , n A } 。对任意 λ , η , ρ , ω , ν , ν ,我们可引入以下导子 D Der ( L , V )

D ( L 0 ) 0 D ( I 0 ) D ( I n ) = 2 ν ( I 0 I n I n I 0 ) n * m ,

D ( G m ) = ν ( I 0 G m G m I 0 ) + ν ( I m G 0 G 0 I m ) + ω ( I 0 I m I m I 0 ) ,

D ( L n ) = ( ( 2 n ) λ + ( n 1 ) η ) I n I 0 + ( ( n 2 ) λ + 2 n 2 η + n 2 ρ ) I 0 I n . (2.6)

D ( L 0 ) 0 表示 D ( L 0 ) 0 ( mod ( I 0 I 0 ) ) ,即 D ( L 0 ) ( I 0 I 0 )

3. 主要结果及证明过程

本文的主要结果可表述为以下两个定理。

定理3.1: H 1 ( L , V ) D

定理3.2:李双代数 ( L , [ , ] , Δ ) 是三角余边缘的当且仅当 λ = η = ρ = ω = ν = ν = 0

引理3.1:把 L 的n次张量积 L n 看作 L 对角伴随作用下的 L -模。如果对某个 r L n 和任意 x L ,使得 x r = 0 ,则 r I 0 n

证明:可运用文献 [1] [5] 相应结论的证明得到。

引理3.2:对所有 x L ,假设 v V ,使得 x v Im ( 1 τ ) ,则对某个 c ,有 v c I 0 I 0 Im ( 1 τ )

证明:可运用文献 [6] 中引理3.2的技巧得证。

定理3.1:由断言1~4得到。

断言1:如果 n * ,则 D n Inn ( L , V )

证明:首先记 y = D n ( L 0 ) n V n n * 。把 D n 作用在 [ L 0 , x j ] = j x j 上,再利用 D n ( x j ) V n + j ,有 D n ( x j ) = x j D n ( L 0 ) n = x j y x j L j ,从而 D n = y inn 为内导子。

断言2: D 0 ( L 0 ) 0 D 0 ( I 0 )

证明:把 D 0 作用在 [ L 0 , x j ] = j x j 上, j x L ,有 x j D 0 ( L 0 ) = 0 。则由引理3.1可推出 D 0 ( L 0 ) 0

类似地,通过将 D 0 作用于 [ I 0 , x ] = 0 上,有 D 0 ( I 0 ) 0

断言3:当 D 0 Der 0 ¯ ( L , V ) 时,用 D 0 u inn ( u V 0 )代替 D 0 ,我们可假设

证明:对 n D 0 ( L n ) D 0 ( I n ) D 0 ( G n ) 如下所示

D 0 ( L n ) = i ( a n , i L i L n i + b n , i L i I n i + b n , i I i L n i + c n , i I i I n i + e n , i G i G n i )

D 0 ( I n ) = i ( α n , i L i L n i + β n , i L i I n i + β n , i I i L n i + γ n , i I i I n i + f n , i G i G n i )

D 0 ( G n ) = i ( μ n , i L i G n i + μ n , i G i L n i + ν n , i I i G n i + ν n , i G i I n i )

其中,所有张量积的系数都在复数域 中,且它们的和是有限的。对于 n ,下列恒等式成立,

L 1 ( L n L n ) = ( 1 n ) L n + 1 L n + ( 1 + n ) L n L 1 n

L 1 ( I n I n ) = n I n + 1 I n + n I n I 1 n L 1 ( G n G n ) = n G n + 1 G n + n G n G 1 n

L 1 ( L n I n ) = ( 1 n ) L n + 1 I n + n L n I 1 n L 1 ( I n L n ) = n I n + 1 L n + ( 1 + n ) I n L 1 n

D 0 u inn 代替 D 0 ,其中u是 L p L p L p I p I p L p I p I p G p G p ( p )的适当的线性组合,假设对任意 i \ { 1 , 2 } j \ { 0 , 2 } k \ { 1 , 1 } m \ { 0 , 1 } ,有 a 1 , i = b 1 , j = b 1 , k = c 1 , m = e 1 , m = 0 。则 D 0 ( L 1 ) 化简为

D 0 ( L 1 ) = a 1 , 1 L 1 L 2 + a 1 , 2 L 2 L 1 + b 1 , 0 L 0 I 1 + b 1 , 2 L 2 I 1 + b 1 , 1 I 1 L 2 + b 1 , 1 I 1 L 0 + c 1 , 0 I 0 I 1 + c 1 , 1 I 1 I 0 + e 1 , 0 G 0 G 1 + e 1 , 1 G 1 G 0 .

D 0 作用在 [ L 1 , L 1 ] = 2 L 0 上,有

a 1 , i = 0 i \ { 2 , ± 1 , 0 } a 1 , 2 = a 1 , 1 + 1 3 a 1 , 0 a 1 , 1 = a 1 , 2 1 3 a 1 , 0 (3.1)

b 1 , i 1 = b 1 , i 2 = c 1 , i 3 = e 1 , i 3 = 0 i 1 \ { ± 1 , 0 } i 2 \ { 2 , 1 , 0 } i 3 \ { 1 , 0 }

b 1 , 0 = b 1 , 2 = b 1 , 1 = b 1 , 1 = 0 b 1 , 0 = 2 b 1 , 1 = 2 b 1 , 1 b 1 , 1 = 2 b 1 , 2 = 2 b 1 , 0 .

从而,进一步得到 D 0 ( L ± 1 ) 的简化式如下

D 0 ( L 1 ) = a 1 , 2 L 2 L 1 + a 1 , 0 ( L 1 L 0 + L 0 L 1 ) + a 1 , 1 L 1 L 2 + b 1 , 1 ( L 1 I 0 2 L 0 I 1 + L 1 I 2 ) + b 1 , 0 ( I 2 L 1 2 I 1 L 0 + I 0 L 1 ) + c 1 , 1 I 1 I 0 + c 1 , 0 I 0 I 1 + e 1 , 1 G 1 G 0 + e 1 , 0 G 0 G 1

D 0 ( L 1 ) = a 1 , 1 L 1 L 2 + a 1 , 2 L 2 L 1 + c 1 , 0 I 0 I 1 + c 1 , 1 I 1 I 0 + e 1 , 0 G 0 G 1 + e 1 , 1 G 1 G 0 .

D 0 作用在 [ L 2 , L 1 ] = 3 L 1 上,结合(3.1)得

a 1 , 1 = 1 3 a 1 , 0 2 a 2 , 1 + 3 a 2 , 0 3 a 1 , 0 = 0 a 2 , 0 + 4 a 2 , 1 3 a 1 , 1 = 0 (3.2)

a 1 , 2 = 1 3 a 1 , 0 4 a 2 , 3 + a 2 , 2 3 a 1 , 2 = 0 3 a 2 , 2 + 2 a 2 , 1 + 3 a 1 , 0 = 0 . (3.3)

D 0 作用在 [ L 1 , L 2 ] = 3 L 1 上,有

a 1 , 0 = a 2 , 1 a 2 , 3 = a 2 , 0 + 4 a 2 , 3 = a 2 , 1 6 a 2 , 3 = a 2 , 2 + 4 a 2 , 3 = a 2 , i = b 2 , i 1 = b 2 , i 2 = c 2 , i 3 = e 2 , i 3 = 0

i \ { ± 1 , 0 , 2 , 3 } i 1 \ { ± 1 , 0 , 2 , 3 } i 2 \ { ± 1 , 0 , 2 , 3 } i 3 \ { 0 , 1 , 2 }

此时,由(3.2)和(3.3)得

b 2 , 0 + 3 b 2 , 1 = b 2 , 3 b 1 , 1 = b 2 , 1 3 b 2 , 1 b 1 , 1 = 3 b 2 , 2 + b 2 , 1 + 5 b 1 , 1 = 0 (3.4)

3 b 2 , 3 + b 2 , 2 = b 2 , 1 b 1 , 0 = b 2 , 1 + 3 b 2 , 0 + 5 b 1 , 0 = 3 b 2 , 3 + b 2 , 1 b 1 , 0 = 0 . (3.5)

D 0 作用在 [ L 2 , L 2 ] = 4 L 0 上,结合(3.4)和(3.5)得,则有

a 2 , 3 + a 2 , 1 = c 2 , 1 + c 2 , 1 = e 2 , 1 + e 2 , 1 = 0

b 2 , 1 = b 2 , 0 = 0 b 2 , 1 = 3 b 2 , 1 = 3 b 1 , 1 b 2 , 2 = b 2 , 2 = 2 b 1 , 1 b 2 , 1 = b 1 , 1

b 2 , 2 = b 2 , 3 = 0 b 2 , 1 = 3 b 2 , 1 = 3 b 1 , 0 b 2 , 0 = b 2 , 0 = 2 b 1 , 0 b 2 , 3 = b 1 , 0 .

u 1 = L 1 L 1 2 L 0 L 0 + L 1 L 1 u 2 = L 1 I 1 L 0 I 0 u 3 = I 1 L 1 I 0 L 0 ,用 D 0 a 2 , 3 ( u 1 ) inn b 1 , 1 ( u 2 ) inn b 1 , 0 ( u 3 ) inn 代替D,有 a 2 , 3 = b 1 , 1 = b 1 , 0 = 0 。则有

D 0 ( L ± 1 ) = c ± 1 , 0 I 0 I ± 1 + c ± 1 , ± 1 I ± 1 I 0 + e ± 1 , 0 G 0 G ± 1 + e ± 1 , ± 1 G ± 1 G 0

D 0 ( L ± 2 ) = c ± 2 , 0 I 0 I ± 2 ± c 2 , 1 I ± 1 I ± 1 + c ± 2 , ± 2 I ± 2 I 0 + e ± 2 , 0 G 0 G ± 2 ± e 2 , 1 G ± 1 G ± 1 + e ± 2 , ± 2 G ± 2 G 0 .

且系数满足以下关系式

c 1 , 1 + c 1 , 0 + c 1 , 0 + c 1 , 1 = c 2 , 2 + c 2 , 0 + c 2 , 0 + c 2 , 2 = 2 c 2 , 0 + c 2 , 1 + c 1 , 0 3 c 1 , 0 = 0

2 c 2 , 2 + c 2 , 1 + c 1 , 1 3 c 1 , 1 = 2 c 2 , 0 c 2 , 1 + c 1 , 0 3 c 1 , 0 = 2 c 2 , 2 c 2 , 1 + c 1 , 1 3 c 1 , 1 = 0 .

D 0 作用在 [ L 1 , I 1 ] = I 0 上,有

α 1 , i = β 1 , i 1 = β 1 , i 2 = γ 1 , i 3 = f 1 , i 3 = α 1 , 1 + 3 α 1 , 2 = α 1 , 0 3 α 1 , 2 = α 1 , 1 + α 1 , 2 = 0

i \ { ± 1 , 0 , 2 } i 1 \ { ± 1 , 0 } i 2 \ { 0 , 1 , 2 } i 3 \ { 0 , 1 }

β 1 , 0 = 2 β 1 , 1 = 2 β 1 , 1 β 1 , 1 = 2 β 1 , 0 = 2 β 1 , 2 γ 1 , 0 + γ 1 , 1 = f 1 , 0 + f 1 , 1 = 0 .

D 0 作用在 [ L 1 , I 1 ] = I 0 上,同理有

D 0 ( I 1 ) = α 1 , 1 ( L 2 L 1 + 3 L 1 L 0 3 L 0 L 1 + L 1 L 2 ) + β 1 , 1 ( L 1 I 0 2 L 0 I 1 + L 1 I 2 ) + β 1 , 0 ( I 2 L 1 2 I 1 L 0 + I 0 L 1 ) + γ 1 , 1 ( I 1 I 0 I 0 I 1 ) + f 1 , 1 ( G 1 G 0 G 0 G 1 ) .

D 0 作用在 [ L 2 , I 1 ] = I 1 上,有

α 1 , 1 = α 1 , 2 = β 1 , 1 = β 1 , 1 = β 1 , 0 = β 1 , 2 = 0 γ 1 , 0 + γ 1 , 1 = f 1 , 0 + f 1 , 1 = 0 .

此时, D 0 ( I ± 1 ) = γ 1 , 0 ( I 0 I ± 1 I ± 1 I 0 ) + f 1 , 0 ( G 0 G ± 1 G ± 1 G 0 )

D 0 作用在 [ L 1 , G 0 ] = 0 [ L 1 , G 0 ] = 0 上,可得

D 0 ( G 0 ) = ν 0 , 0 I 0 G 0 + ν 0 , 0 G 0 I 0

D 0 ( L ± 1 ) = c ± 1 , 0 I 0 I ± 1 + c ± 1 , ± 1 I ± 1 I 0 .

D 0 作用在 [ I 1 , G 1 ] = 0 [ I 1 , G 1 ] = 0 上,分别有以下系数关系式

μ 1 1 = μ 1 , 1 = μ 1 , 2 = μ 1 , 0 = f 1 , 0 μ 1 , i 1 = μ 1 , i 2 = 0 i 1 \ { 1 , 2 } i 2 \ { 1 , 0 }

μ 1 , 2 = μ 1 , 0 = μ 1 , 1 = μ 1 , 1 = f 1 , 0 μ 1 , i 3 = μ 1 , i 4 = 0 i 3 \ { 2 , 1 } i 4 \ { 0 , 1 } .

D 0 作用在 [ L 1 , G 1 ] = G 0 上,我们可以推导出

f 1 , 0 = ν 1 , i 1 = ν 1 , i 1 = ν 1 , 1 + ν 1 , 0 ν 0 , 0 = ν 1 , 1 + ν 1 , 0 ν 0 , 0 = 0

ν 1 , i 2 = ν 1 , i 2 = ν 1 , 1 + ν 1 , 0 ν 0 , 0 = ν 1 , 1 + ν 1 , 0 ν 0 , 0 = 0 i 2 \ { 1 , 0 } .

且有 D 0 ( I ± 1 ) = γ 1 , 0 ( I 0 I ± 1 I ± 1 I 0 )

D 0 作用在 [ L 2 , G 1 ] = G 1 [ L 2 , G 1 ] = G 1 上,可得 e 2 , 0 = e 2 , 1 = e 2 , 2 = 0

从而有 D 0 ( L ± 2 ) = c ± 2 , 0 I 0 I ± 2 ± c 2 , 1 I ± 1 I ± 1 + c ± 2 , ± 2 I ± 2 I 0

D 0 作用在 [ G 1 , G 1 ] = I 0 上,有

D 0 ( G 0 ) = ν 0 , 0 ( I 0 G 0 G 0 I 0 )

D 0 ( G ± 1 ) = ν 1 , 0 ( I 0 G ± 1 G ± 1 I 0 ) + ν 1 , 1 ( I ± 1 G 0 G 0 I ± 1 )

且满足 ν 0 , 0 = ν 1 , 0 + ν 1 , 1 γ 1 , 0 = 2 ν 1 , 0

根据以下记法: c 1 , 1 = λ c 2 , 0 = η c 2 , 1 = κ c 2 , 2 = ρ ν 1 , 0 = ν ν 1 , 1 = ν ,利用数学归纳法,并考虑(1.1)中所有的李括号可得

D 0 ( L 0 ) 0 D 0 ( I 0 ) D 0 ( I n ) = 2 ν ( I 0 I n I n I 0 )

D 0 ( L n ) = ( ( 2 n ) λ + ( n 1 ) η ) I n I 0 + ( ( n 2 ) λ + 2 n 2 η + n 2 ρ ) I 0 I n

D 0 ( G m ) = ν ( I 0 G m G m I 0 ) + ν ( I m G 0 G 0 I m ) n * m .

断言4:当 D 0 Der 1 ¯ ( L , V ) 时,用 D 0 u inn ( u V 0 )代替 D 0 ,我们可假设 D 0 ( L ) = D ( L )

证明:对 n D 0 ( L n ) D 0 ( I n ) D 0 ( G n ) 如下所示

D 0 ( L n ) = i ( a n , i L i G n i + a n , i G i L n i + b n , i I i G n i + b n , i G i I n i )

D 0 ( I n ) = i ( α n , i L i G n i + α n , i G i L n i + β n , i I i G n i + β n , i G i I n i )

D 0 ( G n ) = i ( μ n , i L i L n i + ν n , i L i I n i + ν n , i I i L n i + ω n , i I i I n i + λ n , i G i G n i )

其中,所有张量积的系数都在复数域 中,且它们的和是有限的。对于 n ,下列恒等式成立,

L 1 ( L n G n ) = ( 1 n ) L n + 1 G n + n L n G 1 n L 1 ( I n G n ) = n I n + 1 G n + n I n G 1 n

L 1 ( G n L n ) = n G n + 1 L n + ( n + 1 ) G n L 1 n L 1 ( G n I n ) = n G n + 1 I n + n G n I 1 n .

D 0 u inn 代替 D 0 ,其中u是 L p G p G p L p I p G p G p I p 的适当的线性组合,,假设对于任意 i \ { 0 , 2 } j \ { 1 , 1 } k \ { 0 , 1 } ,有 a 1 , i = a 1 , i = b 1 , k = b 1 , k = 0 。则 D 0 ( L 1 ) 可写成

D 0 ( L 1 ) = a 1 , 0 L 0 G 1 + a 1 , 2 L 2 G 1 + a 1 , 1 G 1 L 2 + a 1 , 1 G 1 L 0 + b 1 , 0 I 0 G 1 + b 1 , 1 I 1 G 0 + b 1 , 0 G 0 I 1 + b 1 , 1 G 1 I 0 .

D 0 作用在 [ L 1 , L 1 ] = 2 L 0 上,记 u 1 = L 0 G 0 + L 1 G 1 u 2 = G 1 L 1 L 0 G 0 ,用 D 0 a 1 , 1 ( u 1 ) inn a 1 , 0 ( u 2 ) inn 代替D,可得

D 0 ( L ± 1 ) = b ± 1 , 0 I 0 G ± 1 + b ± 1 , ± 1 I ± 1 G 0 + b ± 1 , 0 G 0 I ± 1 + b ± 1 , ± 1 G ± 1 I 0 .

D 0 依次作用在 [ L 2 , L 1 ] = 3 L 1 [ L 1 , L 2 ] = 3 L 1 [ L 2 , L 2 ] = 4 L 0 上,

我们可以推导出 D 0 ( L ± 2 ) 如下

D 0 ( L ± 2 ) = b ± 2 , 0 I 0 G ± 2 ± b 2 , 1 I ± 1 G ± 1 + b ± 2 , ± 2 I ± 2 G 0 + b ± 2 , 0 G 0 I ± 2 ± b 2 , 1 G ± 1 I ± 1 + b ± 2 , ± 2 G ± 2 I 0 .

且系数满足以下关系

2 b 2 , 0 + b 2 , 1 + b 1 , 0 3 b 1 , 0 = 2 b 2 , 2 + b 2 , 1 + b 1 , 1 3 b 1 , 1 = 0 (3.6)

2 b 2 , 0 + b 2 , 1 + b 1 , 1 3 b 1 , 0 = 2 b 2 , 2 + b 2 , 1 + b 1 , 1 3 b 1 , 1 = 2 b 2 , 0 b 2 , 1 + b 1 , 0 3 b 1 , 0 = 0 (3.7)

2 b 2 , 2 b 2 , 1 + b 1 , 1 3 b 1 , 1 = 2 b 2 , 0 b 2 , 1 + b 1 , 0 3 b 1 , 0 = 2 b 2 , 2 b 2 , 1 + b 1 , 1 3 b 1 , 1 = 0 .(3.8)

D 0 依次作用在 [ L 1 , I 1 ] = I 0 [ L 1 , I 1 ] = I 0 [ L 2 , I 1 ] = I 1 上,我们可将 D 0 ( I ± 1 ) 化简为

D 0 ( I ± 1 ) = β 1 , 0 ( I 0 G ± 1 I ± 1 G 0 ) + β 1 , 0 ( G 0 I ± 1 G ± 1 I 0 ) .

D 0 作用在 [ L ± 1 , G 0 ] = 0 [ I 1 , G 0 ] = 0 上,可得

D 0 ( G 0 ) = ω 0 , 0 I 0 I 0 + λ 0 , 0 G 0 G 0 .

D 0 作用在 [ L ± 1 , G 1 ] = ± G 0 [ L 2 , G 1 ] = G 1 上,再结合(3.6)~(3.8),可以推导出

D 0 ( L ± 1 ) = D 0 ( L ± 2 ) = D 0 ( G 0 ) = 0

D 0 ( G ± 1 ) = ω 1 , 0 ( I 0 I ± 1 I ± 1 I 0 ) + λ 1 , 0 ( G 0 G ± 1 G ± 1 G 0 ) .

D 0 作用在 [ I 1 , G 1 ] = 0 上,有 D 0 ( I ± 1 ) = 0

D 0 作用在 [ G 1 , G 1 ] = I 0 上,有 D 0 ( G ± 1 ) = ω 1 , 0 ( I 0 I ± 1 I ± 1 I 0 )

根据以下记法: ω 1 , 0 = ω ,利用数学归纳法,并考虑(1.1)所有李括号可推导出

D 0 ( G 0 ) = D 0 ( L m ) = D 0 ( I m ) = 0 D 0 ( G n ) = ω ( I 0 I n I n I 0 ) m n * .

则定理3.1最终由归纳法 D = D 0 可证得。

( L , [ , ] , Δ ) L 上的超李双代数结构。由定理3.1得,在(2.6)中提到对某个 r L L ,有 Δ = Δ r 当且仅当 λ = η = ρ = ω = ν = ν = 0 。结合 Im Δ Im ( 1 τ ) 和引理3.2,可以推导出对某个 c ,有 r c I 0 I 0 Im ( 1 τ ) ,则引理3.1得出 c ( r ) I 0 I 0 。因此, ( L , [ , ] , Δ ) 是一个三角余边缘的超李双代数当且仅当 λ = η = ρ = ω = ν = ν = 0 。故定理3.2得证。

文章引用

李美君. 一类超Heisenberg-Virasoro代数的超双代数结构
Lie Super-Bialgebra Structures on a Super Heisenberg-Virasoro Algebra[J]. 理论数学, 2019, 09(07): 783-790. https://doi.org/10.12677/PM.2019.97102

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