西南师范大学学报(自然科学版)   2019, Vol. 44 Issue (12): 10-16.  DOI: 10.13718/j.cnki.xsxb.2019.12.003
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  • 半群$\mathscr{DOPD}$(n, r)的秩和相关秩    [PDF全文]
    李晓敏 , 罗永贵 , 赵平     
    贵州师范大学 数学科学学院, 贵阳 550025
    摘要:设自然数n≥3,$\mathscr{DOPD}$n是有限链[n]上的保序且保距部分一一奇异降序变换半群.对任意的r(0≤rn-1),记$\mathscr{DOPD}$nr)={α$\mathscr{DOPD}$n:|Im(α)|≤r}为半群$\mathscr{DOPD}$n的双边星理想.通过对秩为r的元素和星格林关系的分析,获得了半群$\mathscr{DOPD}$nr)的极小生成集和秩.确定了当0≤lr时,半群$\mathscr{DOPD}$nr)关于其星理想$\mathscr{DOPD}$nl)的相关秩.
    关键词保序    保距    保降序    部分一一奇异变换半群        相关秩    

    设[n]={1,2,…,n-1,n}(n≥3)并赋予自然数的大小序. $\mathscr{I}_n$$\mathscr{S}_n$分别表示[n]上的对称逆半群(即部分一一变换半群)和对称群,$\mathscr{SI}_n$=$\mathscr{I}_n$\$\mathscr{S}_n$是[n]上的部分一一奇异变换半群.设α$\mathscr{SI}_n$,若对任意的xy∈Dom(α),xy可推出xαyα,则称α是部分一一保序的.记$\mathscr{OI}_n$为[n]上的保序有限部分一一奇异变换半群.设α$\mathscr{OI}_n$,若对任意的xy∈Dom(α),有|-|=|x-y|,则称α是保距的.

    $ \mathscr{O}\mathscr{P}{\mathscr{D}_n} = \left\{ {\alpha \in \mathscr{O}{G_n}:|x\alpha - y\alpha | = \left| {x - y} \right|,\;\;\;\;\forall x,y \in {\rm Dom}\left( \alpha \right)} \right\} $

    则称$\mathscr{OPD}_n$为[n]上的保序且保距有限部分一一奇异变换半群.

    $ \mathscr{D}\mathscr{O}\mathscr{P}{\mathscr{D}_n} = \left\{ {\alpha \in \mathscr{O}\mathscr{P}{\mathscr{D}_n}:x\alpha \le x,\;\;\;\;\forall x \in {\rm Dom}\left( \alpha \right)} \right\} $

    则称$\mathscr{DOPD}_n$为[n]上的保序且保距有限部分一一奇异降序变换半群.

    $ \mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,r} \right) = \left\{ {\alpha \in \mathscr{D}\mathscr{O}\mathscr{P}{\mathscr{D}_n}:\left| {{\rm Im}\left( \alpha \right)} \right| \le r,0 \le r \le n - 1} \right\} $

    易见$\mathscr{DOPD}$(nr)是$\mathscr{DOPD}_n$的子半群,且对任意的α$\mathscr{DOPD}$(nr),βγ$\mathscr{DOPD}_n$,均有|Im(βαγ)|≤r,即βαγ$\mathscr{DOPD}$(nr),因而$\mathscr{DOPD}$(nr)是$\mathscr{DOPD}_n$的双边星理想.

    通常一个有限半群S的秩定义为

    $ {\rm{rank}}\left( S \right) = \min \left\{ {\left| A \right|:A \subseteq S,\langle A\rangle = S} \right\} $

    半群S及其子半群V之间的相关秩定义为

    $ r\left( {S,V} \right) = \min \left\{ {\left| A \right|:A \subseteq S,A \cap V = \mathit{\Phi },\left\langle {A \cup V} \right\rangle = S} \right\} $

    易见r(SS)=0.

    对于有限半群的秩及其相关秩的研究目前已有许多结果[1-12].文献[1]考虑了[n]上的保序有限部分一一奇异变换半群$\mathscr{OI}_n$的理想

    $ { \mathscr{K}_\mathscr{O}}\left( {n,r} \right) = \left\{ {\alpha \in \mathscr{O}{\mathscr{J}_n}:\left| {{\rm Im}\left( \alpha \right)} \right| \le r,0 \le r \le n - 1} \right\} $

    的生成集和秩,确定了半群$\mathscr{K}_{\mathscr{O}}$(nr)的秩为Cnr.文献[2]证明了半群$\mathscr{OI}_n$m偏度秩存在时一定等于n.文献[3-11]考虑了几类不同的保序且压缩的变换半群的秩和相关秩.文献[12]研究了半群$\mathscr{O}_{n}$(k)的秩.

    本文在文献[1-12]的基础上继续考虑保序保距且保降序部分一一奇异变换半群$\mathscr{DOPD}_n$的双边星理想$\mathscr{DOPD}$(nr)的秩和相关秩,证明了如下主要结果:

    定理1   设n≥3,0≤rn-1,则$\mathscr{J}_{r}^*$$\mathscr{DOPD}$(nr)的生成集,即$\mathscr{DOPD}$(nr)=〈$\mathscr{J}_{r}^*$〉.

    定理2  设n≥3,0≤rn-1,则

    $ {\rm{rank}}\left( {\mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,r} \right)} \right) = \left\{ {\begin{array}{*{20}{l}} 1&{r = 0}\\ {C_n^r + C_{n - 1}^r}&{1 \le r \le n - 1} \end{array}} \right. $

    定理3   设n≥3,0≤lrn-1,则

    $ r\left( {\mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,r} \right),\mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,l} \right)} \right) = \left\{ {\begin{array}{*{20}{l}} 0&{l = r}\\ {C_n^r + C_{n - 1}^r}&{0 \le l \le r} \end{array}} \right. $

    A是自然序集[n]的非空子集,符号εA表示A上的恒等变换,用Φ表示空变换.规定Φ是保距变换;Φ是部分一一保序变换.设α$\mathscr{DOPD}$(nr),用Im(α)表示α的像集,Ker(α)表示Dom(α)上的如下等价关系:

    $ \rm{Ker}\left( \alpha \right) = \{ (x,y) \in {\rm Dom}(\alpha ) \times {\rm Dom}(\alpha ):x\alpha = y\alpha \} $

    对任意的t∈Im(α),-1表示t的原像集且|-1|=1.若

    $ \left| {\rm{Im}\left( \alpha \right)} \right| = k\;\;\;\;1 \le k \le r \le n - 1 $

    则由保序性及保距性容易验证α有表示法

    $ \alpha = \left( {\begin{array}{*{20}{l}} {{a_1}}&{{a_2}}& \cdots &{{a_{i - 1}}}&{{a_i}}&{{a_{i + 1}}}& \cdots &{{a_{k - 1}}}&{{a_k}}\\ {{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}} \end{array}} \right) $

    其中

    $ {a_1} < {a_2} < \cdots < {a_{i - 1}} < {a_i} < {a_{i + 1}} < \cdots < {a_{k - 1}} < {a_k} $
    $ {b_1} < {b_2} < \cdots < {b_{i - 1}} < {b_i} < {b_{i + 1}} < \cdots < {b_{k - 1}} < {b_k} $

    对任意的jp∈{1,2,…,i-1,ii+1,…,k-1,k},有|aj-ap|=|bj-bp|,于是,令

    $ A = \left\{ {{a_1} < {a_2} < \cdots < {a_{i - 1}} < {a_i} < {a_{i + 1}} < \cdots < {a_{k - 1}} < {a_k}} \right\} $
    $ B = \left\{ {{b_1} < {b_2} < \cdots < {b_{i - 1}} < {b_i} < {b_{i + 1}} < \cdots < {b_{k - 1}} < {b_k}} \right\} $

    α= $\left( \begin{array}{l} A\\ B \end{array} \right)$,在下文的证明中用这种形式表示半群$\mathscr{DOPD}_n$中元素特点.

    为叙述方便,这里引用Green*-等价关系[13].不难验证,在半群$\mathscr{DOPD}$(nr)中,$\mathscr{L}^*$$\mathscr{R}^*$$\mathscr{J}^*$有如下刻画:对任意的αβ$\mathscr{DOPD}$(nr),有

    $ \left( {\alpha ,\beta } \right) \in {\mathscr L^ * } \Leftrightarrow {\rm{Im}}\left( \alpha \right) = {\rm{Im}}\left( \beta \right) $
    $ \left( {\alpha ,\beta } \right) \in {\mathscr R^ * } \Leftrightarrow {\rm{Ker}}\left( \alpha \right) = {\rm{Ker}}\left( \beta \right) $
    $ \left( {\alpha ,\beta } \right) \in {\mathscr{J}^ * } \Leftrightarrow \left| {{\rm{Im}}\left( \alpha \right)} \right| = \left| {{\rm{Im}}\left( \beta \right)} \right| $

    易见$\mathscr{L}^*$$\mathscr{J}^*$$\mathscr{R}^*$$\mathscr{J}^*$.记

    $ \mathscr{J}_k^ * \left\{ {\alpha \in \mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,r} \right):\left| {{\rm{Im}}\left( \alpha \right)} \right| = k} \right\}\;\;\;k = 0,1,2, \cdots ,r - 1,r $

    显然$\mathscr{J}^*_{0}$$\mathscr{J}^*_{1}$$\mathscr{J}^*_{2}$,…,$\mathscr{J}^*_{r-1}$$\mathscr{J}_{r}^*$恰好是$\mathscr{DOPD}$(nr)的r+1个$\mathscr{J}^*$-类,并且

    $ \mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,r} \right) = \left\{ {\alpha \in \mathscr{D}\mathscr{O}\mathscr{P}{\mathscr{D}_n}:\left| {{\rm{Im}}\left( \alpha \right)} \right| \le r} \right\} = \bigcup\limits_{k = 0}^r {\mathscr{J}_k^ * } $

    不难验证$\mathscr{DOPD}_n$具有如下包含关系的双边星理想链:

    $ \mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,0} \right) \subset \mathscr{D}\mathscr{O}\mathscr{P}D(n,1) \subset \mathscr{D}\mathscr{O}\mathscr{P}D(n,2) \subset \\ \cdots \subset \mathscr{D}\mathscr{O}\mathscr{P}D(n,n - 2) \subset \mathscr{D}\mathscr{O}\mathscr{P}D(n,n - 1) = \mathscr{D}\mathscr{O}\mathscr{P}{\mathscr{D}_n} $

    Xn(r)表示自然序集[n]={1,2,3,…,n-1,n}(n≥3)的所有r元子集,则Xn(r)中共有Cnr个元素,其中Cnr表示从n个元素中取出r个元素的组合数.令t=Cnr,记Xn(r)={A1A2,…,At},其中

    $ {A_i} \subseteq \left[ n \right]\;\;\;\;\left| {{A_i}} \right| = r\left( {i = 1,2, \cdots ,t - 1,t} \right) $

    定义1[4]  若对任意的A={a1 < a2 < … < ai-1 < ai < ai+1 < … < ar-1 < ar},B={b1 < b2 < … < bi-1 < bi < bi+1 < … < br-1 < br}∈Xn(r),如果对i=2,3,…,r-1,r,有ai-ai-1=bi-bi-1,则称AB同距,否则称AB不同距.

    Xn(r)按照同距概念进行分类.对任意的AXn(r),记A的同距类为[A].进一步可证:对任意的

    $ A = \left\{ {{a_1} < {a_2} < \cdots < {a_{i - 1}} < {a_i} < {a_{i + 1}} < \cdots < {a_{r - 1}} < {a_r}} \right\} \in {X_n}\left( r \right) $

    必定存在

    $ C = \left\{ {1 < {c_2} < \cdots < {c_{i - 1}} < {c_i} < {c_{i + 1}} < \cdots < {c_{r - 1}} < {c_r}} \right\} \in {X_n}\left( r \right) $

    使得CA同距,其中

    $ {c_i} = 1 + \sum\limits_{j = 2}^i {\left( {{a_j} - {a_{j - 1}}} \right)} \;\;\;\;i = 2,3, \cdots ,r - 1,r $

    本文未定义的术语及符号参见文献[14-16].

    为完成定理的证明,先给出若干引理与推论.

    引理1   对0≤k≤1,有$\mathscr{J}^*_k$$\mathscr{J}^*_{k+1}$·$\mathscr{J}^*_{k+1}$.

      设Φ是空变换,则$\mathscr{J}^*_{0}$={Φ}.令β= $\left( \begin{array}{l} 2\\ 1 \end{array} \right)$γ=$\left( \begin{array}{l} 3\\ 2 \end{array} \right)$,则βγ$\mathscr{J}_{1}$Φ=βγ,即$\mathscr{J}^*_{0}$$\mathscr{J}^*_{1}$·$\mathscr{J}^*_{1}$.

    对任意的α$\mathscr{J}^*_{1}$,不妨设α=$\left( \begin{array}{l} a\\ b \end{array} \right)$,以下分2种情形证明$\mathscr{J}^*_{1}$$\mathscr{J}^*_{2}$·$\mathscr{J}^*_{2}$.

    情形1   若a=b,由n≥3,则存在{cd}∈[n]\{a},使得

    $ A = \{ a,c\} \quad B = \{ a,d\} $

    εAεB$\mathscr{J}^*_{2}$,且α=εA·εB.

    情形2   若a>b,分两种子情形证明.

    情形2.1   若b=1,令

    $ \beta = \left( {\begin{array}{*{20}{l}} {a - 1}&a\\ {a - 1}&a \end{array}} \right)\quad \gamma = \left( {\begin{array}{*{20}{l}} a&{a + 1}\\ b&{b + 1} \end{array}} \right) $

    βγ$\mathscr{J}^*_{2}$,且α=βγ.

    情形2.2   若b≥2,令

    $ \beta = \left( {\begin{array}{*{20}{c}} {a - 2}&a\\ {a - 2}&a \end{array}} \right)\quad \gamma = \left( {\begin{array}{*{20}{c}} {a - 1}&a\\ {b - 1}&b \end{array}} \right) $

    βγ$\mathscr{J}^*_{2}$,且α=βγ.

    引理2   对2≤kr-1,3≤rn-1,有$\mathscr{J}^*_k$$\mathscr{J}^*_{k+1}$·$\mathscr{J}^*_{k+1}$.

      对任意的α$\mathscr{J}^*_k$,设α的标准表示为

    $ \alpha = \left( {\begin{array}{*{20}{l}} {{a_1}}&{{a_2}}& \cdots &{{a_{i - 1}}}&{{a_i}}&{{a_{i + 1}}}& \cdots &{{a_{k - 1}}}&{{a_k}}\\ {{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}} \end{array}} \right) $

    其中

    $ {a_1} < {a_2} < \cdots < {a_{i - 1}} < {a_i} < {a_{i + 1}} < \cdots < {a_{k - 1}} < {a_k} $
    $ {b_1} < {b_2} < \cdots < {b_{i - 1}} < {b_i} < {b_{i + 1}} < \cdots < {b_{k - 1}} < {b_k} $

    对任意的jp∈{1,2,…,i-1,ii+1,…,k-1,k},有

    $ \left| {{a_j} - {a_p}} \right| = \left| {{b_j} - {b_p}} \right|\;\;\;\;{b_j} \le {a_j} $

    以下分4种情形证明存在βγ$\mathscr{J}^*_{k+1}$使得α=βγ.

    情形1   若存在j∈{2,3,…,i-1,ii+1,…,k-1,k},使得aj-aj-1≥3.令

    $ \beta = \left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}& \cdots &{{a_{i - 1}}}&{{a_i}}&{{a_{i + 1}}}& \cdots &{{a_{j - 1}}}&{{a_j} - 1}&{{a_j}}&{{a_{j + 1}}}& \cdots &{{a_{k - 1}}}&{{a_k}}\\ {{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{j - 1}}}&{{b_j} - 1}&{{b_j}}&{{b_{j + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}} \end{array}} \right) $
    $ \gamma = \left( {\begin{array}{*{20}{c}} {{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{j - 1}}}&{{b_j} - 2}&{{b_j}}&{{b_{j + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}}\\ {{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{j - 1}}}&{{b_j} - 2}&{{b_j}}&{{b_{j + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}} \end{array}} \right) $

    βγ$\mathscr{J}^*_{k+1}$,且α=βγ.

    情形2   若存在jp∈{2,3,…,i-1,ii+1,…,k-1,k},且jp,使得aj-aj-1≥2且ap-ap-1≥2,不失一般性,不妨设j < p.令

    $ \beta = \\ \left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}& \cdots &{{a_{j - 1}}}&{{a_j} - 1}&{{a_j}}& \cdots &{{a_p} - 1}&{{a_p}}& \cdots &{{a_{i - 1}}}&{{a_i}}&{{a_{i + 1}}}& \cdots &{{a_{k - 1}}}&{{a_k}}\\ {{b_1}}&{{b_2}}& \cdots &{{b_{j - 1}}}&{{b_j} - 1}&{{b_j}}& \cdots &{{b_p} - 1}&{{b_p}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}} \end{array}} \right) $
    $ \gamma = \\ \left( {\begin{array}{*{20}{c}} {{b_1}}&{{b_2}}& \cdots &{{b_{j - 1}}}&{{b_j}}& \cdots &{{b_{p - 1}}}&{{b_p} - 1}&{{b_p}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}}\\ {{b_1}}&{{b_2}}& \cdots &{{b_{j - 1}}}&{{b_j}}& \cdots &{{b_{p - 1}}}&{{b_p} - 1}&{{b_p}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}} \end{array}} \right) $

    βγ$\mathscr{J}^*_{k+1}$,且α=βγ.

    情形3   若存在j∈{2,3,…,i-1,ii+1,…,k-1,k},使得aj-aj-1=2,且对任意的

    $ p \in \left\{ {2,3, \cdots ,j - 2,j - 1} \right\} \cup \left\{ {j + 1,j + 2, \cdots ,k - 1,k} \right\} $

    ap-ap-1=1.由此可见:a1=1,必有ak < n;或ak=n,必有a1>1.否则由a1=1且ak=n可得α$\mathscr{J}^*_{n-1}$,即k=n-1,与2≤kr-1,3≤rn-1矛盾.利用保序性和保距性,类似地可得到:b1=1,必有bk < n;或bk=n,必有b1>1.

    b1>1时,则b1-1≥1.令

    $ \beta = \left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}& \cdots &{{a_{i - 1}}}&{{a_i}}&{{a_{i + 1}}}& \cdots &{{a_{j - 1}}}&{{a_j} - 1}&{{a_j}}&{{a_{j + 1}}}& \cdots &{{a_{k - 1}}}&{{a_k}}\\ {{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{j - 1}}}&{{b_j} - 1}&{{b_j}}&{{b_{j + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}} \end{array}} \right) $
    $ \gamma = \left( {\begin{array}{*{20}{c}} {{b_1} - 1}&{{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{j - 1}}}&{{b_j}}&{{b_{j + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}}\\ {{b_1} - 1}&{{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{j - 1}}}&{{b_j}}&{{b_{j + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}} \end{array}} \right) $

    βγ$\mathscr{J}^*_{k+1}$,且α=βγ.

    bk < n时,则bk+1≤n.令

    $ \beta = \left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}& \cdots &{{a_{i - 1}}}&{{a_i}}&{{a_{i + 1}}}& \cdots &{{a_{j - 1}}}&{{a_j} - 1}&{{a_j}}&{{a_{j + 1}}}& \cdots &{{a_{k - 1}}}&{{a_k}}\\ {{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{j - 1}}}&{{b_j} - 1}&{{b_j}}&{{b_{j + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}} \end{array}} \right) $
    $ \gamma = \left( {\begin{array}{*{20}{c}} {{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{j - 1}}}&{{b_j}}&{{b_{j + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}}&{{b_k} + 1}\\ {{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{j - 1}}}&{{b_j}}&{{b_{j + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}}&{{b_k} + 1} \end{array}} \right) $

    βγ$\mathscr{J}^*_{k+1}$,且α=βγ.

    情形4   对任意的j∈{2,3,…,i-1,ii+1,…,k-1,k}使得aj-aj-1=1,利用保序性和保距性可知:对任意的j∈{2,3,…,i-1,ii+1,…,k-1,k},bj-bj-1=1.由2≤kr-1,3≤rn-1可知kn-2,即k+2≤n.

    如果a1≠1,分以下3种子情形证明:

    情形4.1   如果b1=1,则bk < n-1.令

    $ \beta = \left( {\begin{array}{*{20}{c}} {{a_1} - 1}&{{a_1}}&{{a_2}}& \cdots &{{a_{i - 1}}}&{{a_i}}&{{a_{i + 1}}}& \cdots &{{a_{k - 1}}}&{{a_k}}\\ 1&2&3& \cdots &i&{i + 1}&{i + 2}& \cdots &k&{k + 1} \end{array}} \right) $
    $ \gamma = \left( {\begin{array}{*{20}{c}} 2&3&4& \cdots &i&{i + 1}&{i + 2}& \cdots &{k + 1}&{k + 2}\\ {{b_1}}&{{b_2}}&{{b_3}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_k}}&{{b_k} + 1} \end{array}} \right) $

    βγ$\mathscr{J}^*_{k+1}$,且α=βγ.

    情形4.2   如果2=b1a1,则

    $ 1 \le {b_1} - 1 \le {a_1} - 1 $

    $ \beta = \left( {\begin{array}{*{20}{c}} {{a_1} - 1}&{{a_1}}&{{a_2}}& \cdots &{{a_{i - 1}}}&{{a_i}}&{{a_{i + 1}}}& \cdots &{{a_{k - 1}}}&{{a_k}}\\ {{b_1} - 1}&{{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}} \end{array}} \right) $
    $ \gamma = \left( {\begin{array}{*{20}{c}} {{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}}&n\\ {{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}}&n \end{array}} \right) $

    βγ$\mathscr{J}^*_{k+1}$,且α=βγ.

    情形4.3   如果3≤b1a1n,则

    $ 1 \le {b_1} - 2 < {b_1} - 1 \le {a_1} - 1 $

    $ \beta = \left( {\begin{array}{*{20}{c}} {{a_1} - 1}&{{a_1}}&{{a_2}}& \cdots &{{a_{i - 1}}}&{{a_i}}&{{a_{i + 1}}}& \cdots &{{a_{k - 1}}}&{{a_k}}\\ {{b_1} - 1}&{{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}} \end{array}} \right) $
    $ \gamma = \left( {\begin{array}{*{20}{c}} {{b_1} - 2}&{{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}}\\ {{b_1} - 2}&{{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}} \end{array}} \right) $

    βγ$\mathscr{J}^*_{k+1}$,且α=βγ.

    如果akn,分以下2种子情形证明:

    情形4.4   如果1=b1a1,令

    $ \beta = \left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}& \cdots &{{a_{i - 1}}}&{{a_i}}&{{a_{i + 1}}}& \cdots &{{a_{k - 1}}}&{{a_k}}&{{a_k} + 1}\\ {{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}}&{{b_k} + 1} \end{array}} \right) $
    $ \gamma = \left( {\begin{array}{*{20}{l}} {{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}}&{{b_k} + 2}\\ {{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}}&{{b_k} + 2} \end{array}} \right) $

    βγ$\mathscr{J}^*_{k+1}$,且α=βγ.

    情形4.5   如果1≠b1n-ka1,令

    $ \beta = \\ \left( {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}& \cdots &{{a_{i - 1}}}&{{a_i}}&{{a_{i + 1}}}& \cdots &{{a_{k - 1}}}&{{a_k}}&{{a_k} + 1}\\ {n - k}&{n - k + 1}& \cdots &{n - k - 2 - i}&{n - k - 1 - i}&{n - k - i}& \cdots &{n - 2}&{n - 1}&n \end{array}} \right) $
    $ \gamma = \\ \left( {\begin{array}{*{20}{c}} {n - k - 1}&{n - k}&{n - k + 1}& \cdots &{n - k - 2 - i}&{n - k - 1 - i}&{n - k - i}& \cdots &{n - 2}&{n - 1}\\ {{b_1} - 1}&{{b_1}}&{{b_2}}& \cdots &{{b_{i - 1}}}&{{b_i}}&{{b_{i + 1}}}& \cdots &{{b_{k - 1}}}&{{b_k}} \end{array}} \right) $

    βγ$\mathscr{J}^*_{k+1}$,且α=βγ.

    定理1的证明   由引理1和引理2可知,任意的α$\mathscr{DOPD}$(nr)可以表达成$\mathscr{DOPD}$(nr)的顶端$\mathscr{J}^*$-类$\mathscr{J}_{r}^*$中秩为r的若干元素的乘积或α$\mathscr{J}_{r}^*$.换句话说,$\mathscr{J}_{r}^*$$\mathscr{DOPD}$(nr)的生成集,即

    $ \mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,r} \right) = \left\langle {\mathscr{J}_r^*} \right\rangle $

    引理3   设自然数n≥3,则rank($\mathscr{DOPD}$(n,0))=1,rank($\mathscr{DOPD}$(n,1))=Cn1+Cn-11=2n-1.

      由引理1的证明过程易知

    $ \mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,0} \right) = \mathscr{J}_0^ * = \left\{ \mathit{\Phi } \right\} $

    显然有

    $ {\rm{rank}}\left( {\mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,0} \right)} \right) = 1 $

    首先,容易验证

    $ \begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {{\alpha _1} = \left( {\begin{array}{*{20}{l}} 2\\ 1 \end{array}} \right)}&{{\alpha _2} = \left( {\begin{array}{*{20}{l}} 3\\ 2 \end{array}} \right)}&{{\alpha _3} = \left( {\begin{array}{*{20}{l}} 4\\ 3 \end{array}} \right)} \end{array}}\\ \cdots \\ {\begin{array}{*{20}{c}} {{\alpha _{i - 1}} = \left( {\begin{array}{*{20}{c}} i\\ {i - 1} \end{array}} \right)}&{{\alpha _i} = \left( {\begin{array}{*{20}{c}} {i + 1}\\ i \end{array}} \right)}&{{\alpha _{i + 1}} = \left( {\begin{array}{*{20}{l}} {i + 2}\\ {i + 1} \end{array}} \right)} \end{array}}\\ \cdots \\ {\begin{array}{*{20}{c}} {{\alpha _{n - 2}} = \left( {\begin{array}{*{20}{l}} {n - 1}\\ {n - 2} \end{array}} \right)}&{{\alpha _{n - 1}} = \left( {\begin{array}{*{20}{c}} n\\ {n - 1} \end{array}} \right)}&{{\alpha _n} = \left( {\begin{array}{*{20}{l}} 1\\ 1 \end{array}} \right)} \end{array}}\\ {\begin{array}{*{20}{c}} {{\alpha _{n + 1}} = \left( {\begin{array}{*{20}{l}} 2\\ 2 \end{array}} \right)}& \cdots &{{\alpha _{2n - 2}} = \left( {\begin{array}{*{20}{l}} {n - 1}\\ {n - 1} \end{array}} \right)}&{{\alpha _{2n - 1}} = \left( {\begin{array}{*{20}{l}} n\\ n \end{array}} \right) \in \mathscr{J}_1^ * } \end{array}} \end{array} $

    $ M = \left\{ {{\alpha _1},{\alpha _2}, \cdots ,{\alpha _{i - 1}},{\alpha _i},{\alpha _{i + 1}}, \cdots ,{\alpha _{n - 1}},{\alpha _n},{\alpha _{n + 1}}, \cdots ,{\alpha _{2n - 2}},{\alpha _{2n - 1}}} \right\} $

    $ \left| M \right| = C_n^1 + C_{n - 1}^1 = 2n - 1 $

    其次,对任意的α$\mathscr{J}^*_{1}$,必存在ij∈{1,2,3,…,n-1,n},使得α=$\left( \begin{array}{l} i\\ j \end{array} \right)$,则当i=j时,有α=αn+i-1;当i>j时,有α=αi-1αi-2αj+2αj+1αj.

    由此可见,$\mathscr{J}^*_{1}$⊆〈M〉.结合定理1知$\mathscr{DOPD}$(n,1)=〈M〉.注意到|M|=Cn1+Cn-11=2n-1.

    引理4   设n≥3,2≤rn-1,则在$\mathscr{J}_{r}^*$中存在基数为Cnr+Cn-1r的集合M,使得$\mathscr{J}_{r}^*$⊆〈M〉.

      首先,构造$\mathscr{J}_{r}^*$中基数为Cnr+Cn-1r的集合M.

    对任意的AXn(r)={A1A2,…,At}(t=Cnr+Cn-1r),不妨设

    $ \left[ A \right] = \left\{ {A = {B_1},{B_2}, \cdots ,{B_{m - 1}},{B_m}} \right\} $

    其中

    $ m < t = C_n^r + C_{n - 1}^r $
    $ \min {B_1} < \min {B_1} < \cdots < \min {B_{m - 1}} < \min {B_m} $

    m=1,只有α1=$\left( \begin{array}{l} A\\ A \end{array} \right)$=εA$\mathscr{J}_{r}^*$.

    若2 < mt=Cnr+Cn-1r,容易验证

    $ \begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {{\alpha _1} = \left( {\begin{array}{*{20}{l}} {{B_2}}\\ {{B_1}} \end{array}} \right)}&{{\alpha _2} = \left( {\begin{array}{*{20}{l}} {{B_3}}\\ {{B_2}} \end{array}} \right)}&{{\alpha _3} = \left( {\begin{array}{*{20}{l}} {{B_4}}\\ {{B_3}} \end{array}} \right)} \end{array}}\\ \cdots \\ {{\alpha _{i - 1}} = \left( {\begin{array}{*{20}{c}} {{B_i}}\\ {{B_{i - 1}}} \end{array}} \right)\quad {\alpha _i} = \left( {\begin{array}{*{20}{c}} {{B_{i + 1}}}\\ {{B_i}} \end{array}} \right)\quad {\alpha _{i + 1}} = \left( {\begin{array}{*{20}{c}} {{B_{i + 2}}}\\ {{B_{i + 1}}} \end{array}} \right)}\\ \cdots \\ {{\alpha _{m - 1}} = \left( {\begin{array}{*{20}{c}} {{B_m}}\\ {{B_{m - 1}}} \end{array}} \right)\quad {\alpha _m} = \left( {\begin{array}{*{20}{c}} {{B_1}}\\ {{B_1}} \end{array}} \right)\quad {\alpha _{m + 1}} = \left( {\begin{array}{*{20}{c}} {{B_2}}\\ {{B_2}} \end{array}} \right)}\\ \cdots \\ {\begin{array}{*{20}{c}} {{\alpha _{2m - 2}} = \left( {\begin{array}{*{20}{l}} {{B_{m - 1}}}\\ {{B_{m - 1}}} \end{array}} \right)}&{{\alpha _{2m - 1}} = \left( {\begin{array}{*{20}{l}} {{B_m}}\\ {{B_m}} \end{array}} \right) \in \mathscr{J}_r^ * } \end{array}} \end{array} $

    对其余的保降序同距类也用类似的方式进行构造,可以得到集合

    $ M = \left\{ {{\alpha _1},{\alpha _2}, \cdots ,{\alpha _{m - 1}},{\alpha _m},{\alpha _{m + 1}}, \cdots ,{\alpha _{2m - 1}},{\alpha _2}m,{\alpha _{2m + 1}}, \cdots ,{\alpha _{t - 1}},{\alpha _t}} \right\} $

    t1t2分别表示当|[A]|=1和|[A]|≥2时生成元的个数,若1∈AnA,则|[A]|=1,若1∈AnA,则|[A]|≥2,则

    $ {t_1} = C_{n - 2}^{r - 2}\;\;\;\;{t_2} = 2\left( {C_n^r - C_{n - 2}^{r - 2}} \right) - C_{n - 2}^{r - 1} $

    $ \begin{array}{l} t = {t_1} + {t_2} = C_{n - 2}^{r - 2} + 2\left( {C_n^r - C_{n - 2}^{r - 2}} \right) - C_{n - 2}^{r - 1} = C_{n - 2}^{r - 2} + 2C_n^r - 2C_{n - 2}^{r - 2} - C_{n - 2}^{r - 1} = \\ \;\;\;\;\;2C_n^r - \left( {C_{n - 2}^{r - 2} + C_{n - 2}^{r - 1}} \right) = 2C_n^r - C_{n - 1}^{r - 1} = \\ \;\;\;\;\;C_n^r + \left( {C_n^r - C_{n - 1}^{r - 1}} \right) = C_n^r + C_{n - 1}^r \end{array} $

    其次,对任意的α$\mathscr{J}_{r}^*$,验证α∈〈M〉,即$\mathscr{J}_{r}^*$⊆〈M〉.

    对任意的α$\mathscr{J}_{r}^*$,必存在AXn(r),使得Im(α),Dom(α)∈[A].不失一般性,可设α=$\left( \begin{array}{l} B_i\\ B_j \end{array} \right)$,其中,BiBj∈[A]={A=B1B2,…,Bm-1Bm}且ij∈{1,2,…,m-1,m}.

    若|[A]|=1,则α=εA=εIm(α).

    若|[A]|≥2,则当i=j时,有α=αm+i-1;当i>j时,有α=αi-1αi-2αj+1αj.

    为叙述方便,这里引用符号αij=$\left( \begin{array}{l} B_i\\ B_j \end{array} \right)$.

    引理5   设自然数n≥3,则M是半群$\mathscr{DOPD}$(nr)唯一的极小生成集.

      对任意的αstαmnMstmn∈{1,2,…,t-1,t},当t=m时,有αst·αmn=αsn;当tm时,有αst·αmn$\mathscr{DOPD}$(nr-1).

    对任意的αstαmnMstmn∈{1,2,…,t-1,t},当t=m时,有αst·αmn=αsnM;当tm时,有αst·αmn$\mathscr{DOPD}$(nr-1).

    对任意的αstA1αmnA2A1∉[A2],有αst·αmn$\mathscr{DOPD}$(nr-1).

    定理2的证明   由引理3与引理4可知,任意的α$\mathscr{J}_{r}^*$可以表达为M中若干元素的乘积或αM,即$\mathscr{J}_{r}^*$⊆〈M〉.再由定理1知,M$\mathscr{DOPD}$(nr)的生成集,即$\mathscr{DOPD}$(nr)=〈M〉,其中M的定义见引理4与引理5的证明过程.注意到|M|=Cnr+Cn-1r,进一步有

    $ {\rm{rank}}\left( {\mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,r} \right)} \right) \le C_n^r + C_{n - 1}^r $

    因此,结合引理5,有rank($\mathscr{DOPD}$(nr))=Cnr+Cn-1r.

    定理3的证明   当l=r时,显然有r($\mathscr{DOPD}$(nr),$\mathscr{DOPD}$(nl))=0.

    当0≤l < r时,由定理1与定理2的证明过程可知

    $ \begin{array}{*{20}{c}} {\mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,r} \right) = \left\langle M \right\rangle }&{M \cap \mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,l} \right) = \left\{ \mathit{\Phi } \right\}}&{\left| M \right| = C_n^r + C_{n - 1}^r} \end{array} $

    再由相关秩的定义,可知

    $ r\left( {\mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,r} \right),\mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,l} \right)} \right) = C_n^r + C_{n - 1}^r $

    $ r\left( {\mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,r} \right),\mathscr{D}\mathscr{O}\mathscr{P}\mathscr{D}\left( {n,l} \right)} \right) = \left\{ {\begin{array}{*{20}{c}} 0&{l = r}\\ {C_n^r + C_{n - 1}^r}&{0 \le l < r} \end{array}} \right. $
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    On Rank and Relative Rank of Semigroup $\mathscr{DOPD}$(n, r)
    LI Xiao-min , LUO Yong-gui , ZHAO Ping     
    School of Mathematics Science, Guizhou Normal University, Guiyang 550025, China
    Abstract: Let $\mathscr{DOPD}$n be the semigroup of all order-preserving and distance-preserving partial one-to-one singular order-decreasing transformations on a finite-chain[n] (n ≥ 3), and let $\mathscr{DOPD}$(n, r)=$\mathscr{DOPD}$(n, r)={α$\mathscr{DOPD}$n:|Im(α)| ≤ r} be the two-sided star ideal of the semigroup $\mathscr{DOPD}$n for an arbitrary integer r such that 0 ≤ rn-1. By analyzing the elements of rank r and star Green's relations, the minimal generating set and rank of the semigroup $\mathscr{DOPD}$(n, r) are obtained, respectively. Furthermore, the relative rank of the semigroup $\mathscr{DOPD}$(n, r) with respect to itself each star ideal $\mathscr{DOPD}$(nl) is determined for 0 ≤ lr.
    Key words: order-preserving    distance-preserving    order-decreasing    partial one-to-one singular transformation semigroup    rank    relative rank    
    X