则β,γ∈J2且α=βγ.
引理2对2≤k≤r-1, 2≤r≤n-1, 有Jk⊆Jk+1·Jk+1.
证明: 对任意的α∈Jk, 设α的标准表示为
这里每个Ai(i=1,2,…,k-1,k)都是凸集, 并且A1A2>…>Ak-1>Ak,a1由于2≤k≤r-1≤n-2, 因此必存在i∈{1,2,…,k-1,k}, 使得Ai≥2. 若α是保序的, 则记x=minAi; 若α是反序的, 则记x=maxAi. 下面分3种情形证明存在β,γ∈Jk+1, 使得α=βγ.
情形1) 若a1≠1, 令
则β,γ∈Jk+1且α=βγ.
情形2) 若ak≠n, 令
则β,γ∈Jk+1且α=βγ.
情形3) 若a1=1且ak=n, 结合2≤k≤n-2知, 存在j∈{2,3,…,k-1,k}, 使得aj-aj-1>1.
① 如果i则β,γ∈Jk+1且α=βγ.
② 如果i=j, 令
则β,γ∈Jk+1且α=βγ.
③ 如果i>j, 令
则β,γ∈Jk+1且α=βγ.
2.1 定理1的证明
由引理1和引理2可知, 对任意的α∈LD(n,r)都可以表示为LD(n,r)的顶端J-类Jr中秩为r的元素的乘积或α∈Jr. 即Jr是LD(n,r)的生成集,LD(n,r)=〈Jr〉.
引理3设α,β∈LD(n,r), 若(α,β),(α,αβ)∈J, 则(αβ,β)∈L, (α,αβ)∈R.
证明: 设α,β∈LD(n,r), 若(α,β),(α,αβ)∈J, 则Imα=Imβ=Im(αβ). 再由Im(αβ)⊆Imβ, Kerα⊆Ker(αβ)与Xn的有限性知, Im(αβ)=Imβ, Kerα=Ker(αβ), 即(αβ,β)∈L, (α,αβ)∈R.
注意到当r=1时, J1中共有n个L-类和1个R-类, 且每个H=R∩L仅有一个保序的元素, 因此, 有:
推论2设自然数n≥3, 则rank(LD(n,1))=n.
2) 这m个幂等元都是保序变换.
其次, 对任意的α∈Jr, 分两种情形验证α∈〈M〉, 即Jr⊆〈M〉.
1) 若存在i,j∈{1,2,…,m-1,m}, 使得Kerα=Kerαi, Imα=Imαj.
① 若α是保序的, 则当iα=αiαi+1…αm-1αmα1α2α3…αi-1αi…αm-1αm;
当i=j=m时, 有α=αmα1α2α3…αm-1αm; 当i=jj时, 有
α=αiαi+1…αm-1αmα1α2α3…αi-1αiαi+1…αm-1αmα1α2…αj-1αj.
② 若α是反序的, 则当iα=αiαi+1…αj-1αj…αm-1αmα1α2…αi-1αiαi+1…αj-1αj;
当iα=αiαi+1…αm-1αmα1α2…αi-1αi;
当i>j时, 有α=αiαi+1…αm-1αmα1α2…αj-1αj.
① 若α是保序的, 则当j=1时, 有β=βi; 当2≤j≤m时, 有
α=αjαj+1…αm-1αmα1α2…αj-1αjαj+1…αm-1αmβi.
② 若α是反序的, 则当1≤j≤m时, 有α=αjαj+1…αm-1αmβi.
2.2 定理2的证明
2.3 定理3的证明
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