A3型李代數(shù)的張量積分解

魏玉麗， 王利萍， 代佳華

(北京建筑大學(xué) 理學(xué)院，北京 100044)

Kazhdan-Lusztig多項(xiàng)式的首項(xiàng)系數(shù)是Kazhdan-Lusztig理論中的核心研究對(duì)象[1]，這些系數(shù)對(duì)研究該理論起到了至關(guān)重要的作用。為了計(jì)算一些最低雙邊胞腔上的Kazhdan-Lusztig多項(xiàng)式的首項(xiàng)系數(shù)，需要知道李代數(shù)中不可約模的張量積重?cái)?shù)；而張量積中不可約模的重?cái)?shù)在李代數(shù)理論中也是一個(gè)重要的問(wèn)題。許超[2]給出了A2的不可約模的張量積分解的一個(gè)計(jì)算方法。于桂海等[3]給出了特征數(shù)大于0的代數(shù)閉域上C2型單連通半單代數(shù)群，限制支配權(quán)所對(duì)應(yīng)的不可約模的張量積分解。對(duì)于A型李代數(shù)的張量積分解，理論上有Young圖法、Klymik公式、Pieris公式。

1 預(yù)備知識(shí)

W0={e,s1,s2,s3,s1s3,s2s1,s1s2,s2s3,s3s2,s1s3s2,

s1s2s3,s1s2s1,s2s1s3,s2s3s2,s3s2s1,s1s3s2s1,

s1s3s2s3,s1s2s3s1,s2s1s3s2,s2s3s2s1,s2s1s3s2s1,

s1s3s2s1s3,s1s2s3s1s2,s2s1s3s2s1s3}

1)s0ω=ωs1,s1ω=ωs2,s2ω=ωs3,s3ω=ωs0。

2)s0ω2=ω2s2,s1ω2=ω2s3,s2ω2=ω2s0,s3ω2=ω2s1。

3)s0ω3=ω3s3,s3ω3=ω3s2,s2ω3=ω3s1,s1ω3=ω3s0。

并且有Λ=x1+x2+x3,Λ+=x1+x2+x3,Λr=α1+α2+α3。

1.2 李代數(shù)中張量積分解的相關(guān)結(jié)論

設(shè)L=sl(4,F), 為上A3型李代數(shù)。設(shè)V(λ)為首權(quán)是λ的不可約最高權(quán)L-模，V(λ)的權(quán)格為∏(λ)，且L的基本支配權(quán)為x1、x2、x3。如果μ∈Λ，定義μ在V(λ)(λ∈Λ+)內(nèi)的重?cái)?shù)為mλ(μ)=dimV(λ)μ[5]。

引理2[5]如果λ∈Λ+，則不可約L-模V=V(λ)是有限維的，且權(quán)集合∏(λ)被W0所置換，使得對(duì)于σ∈W0，有dimVμ=dimVσμ。

引理3[5](Freudenthal公式)設(shè)V=V(λ)是首權(quán)為λ(λ∈Λ+)的不可約L-模，如果μ∈Λ,則μ在V內(nèi)的重?cái)?shù)m(μ)可從如下的遞推得到：

[(λ+δ,λ+δ)-(μ+δ,μ+δ)]m(μ)=

(1)

計(jì)算m(μ)分三步：第一步確定集合D={λ,比λ低的支配權(quán)}；第二步計(jì)算和D中元素共軛的元素；第三步將前兩步的所有元素按照一定水平進(jìn)行排列。然后根據(jù)引理2～3計(jì)算出m(μ)。

引理4[5]使得V(λ)可能出現(xiàn)在V(λ′)?V(λ″)的加項(xiàng)中的λ∈Λ+,只能形如μ+λ″,μ∈∏(λ′)。當(dāng)這樣的μ+λ″都是支配權(quán)時(shí)，V(μ+λ″)出現(xiàn)在張量積內(nèi)，且重?cái)?shù)為mλ′(μ)。

2 A3型李代數(shù)的張量積分解

2.1 ∏(λ)的計(jì)算

首先定義S(λ)(λ∈Λ+)為W0中的元素作用在λ上所得到的元素的集合，即為和λ共軛的元素的集合，同時(shí)定義H={x1,x2,x3,2x1,2x2,2x3,x1+x3,x2+x3,x1+x2}。對(duì)于λ∈H有：

S(x1)={x1,-x3,x3-x2,x2-x1}

S(x2)={x2,x1-x2+x3,x3-x1,x1-x3,

x2-x1-x3,-x2}

S(x3)={x3,-x1,x2-x3,x1-x2}

再利用編程可得：

S(2x1)={2x1,-2x3,2(x3-x2),2(x2-x1)}

S(2x2)={2x2,-2x2,2(x1-x2+x3),

2(x3-x1),2(x1-x3),2(x2-x1-x3)}

S(2x3)={2x3,-2x1,2(x2-x3),2(x1-x2)}

S(x1+x2)={2x2-x1,2x1-x2+x3,x1+x2,

x2+x3-2x1,x1-2x2+2x3,2x1-x3,

2x2-2x1-x3,2x3-x1-x2,x1-2x3,

x2-x1-2x3,x3-2x2,-x2-x3}

S(x1+x3)={x2-x1+x3,x1+x3,x1+x2-x3,

2x2-2x1-x3,2x3-x2,2x1-x2,

x2-2x1,x1+x3-2x2,x2-2x3,x3-x1-x2,

x1-x2-x3,-x1-x3}

S(x2+x3)={x2+x3,x1-x2+2x3,2x2-x3,

2x3-x1,2x1-2x2+x3,x1+x2-2x3,

2x2-x1-2x3,x3-2x1,2x1-x2-x3,

x2-2x1-x3,x1-2x2,-x1-x2}

接下來(lái)計(jì)算上述λ對(duì)應(yīng)的∏(λ)，∏(λ)中含有λ、比λ低的支配權(quán)，以及它們?cè)赪0下共軛的元素：

∏(x1)={x1,-x3,x3-x2,x2-x1}

∏(x2)={x2,x1-x2+x3,x3-x1,

x1-x3,x2-x1-x3,-x2}

∏(x3)={x3,-x1,x2-x3,x1-x2}

∏(2x1)={2x1,-2x3,2(x3-x2),2(x2-x1),x2,

x1-x2+x3,x3-x1,x1-x3,x2-x1-x3,-x2}

∏(2x2)={2x2,-2x2,2(x1-x2+x3),

2(x3-x1),2(x1-x3),2(x2-x1-x3),

x2-x1+x3,x1+x3,x1+x2-x3,

2x2-2x1-x3,2x3-x2,2x1-x2,

x2-2x1,x1+x3-2x2,x2-2x3,

x3-x1-x2,x1-x2-x3,-x1-x3,0}

∏(2x3)={2x3,-2x1,2(x2-x3),2(x1-x2),

x1-x2+x3,x3-x1,x1-x3,x2-x1-x3,-x2}

∏(x1+x3)={x2-x1+x3,x1+x3,

x1+x2-x3,2x2-2x1-x3,2x3-x2,2x1-x2,

x2-2x1,x1+x3-2x2,x2-2x3,x3-x1-x2,

x1-x2-x3,-x1-x3,0}

∏(x1+x2)={2x2-x1,2x1-x2+x3,

x1+x2,x2+x3-2x1,x1-2x2+2x3,

2x1-x3,2x2-2x1-x3,2x3-x1-x2,

x1-2x3,x2-x1-2x3,x3-2x2,

-x2-x3,x3,-x1,x2-x3,x1-x2}

∏(x2+x3)={x2+x3,x1-x2+2x3,

2x2-x3,2x3-x1,2x1-2x2+x3,x1+x2-2x3,

2x2-x1-2x3,x3-2x1,2x1-x2-x3,x2-2x1-x3,

x1-2x2,-x1-x2,x1,-x3,x3-x2,x2-x1}

2.2 A3型李代數(shù)的權(quán)重?cái)?shù)計(jì)算

命題1設(shè)V=V(λ)是首權(quán)為λ的不可約L-模，λ∈{x1,x2,x3,2x1,2x3}。如果μ∈∏(λ), 則μ在V內(nèi)的重?cái)?shù)m(μ)=1。

命題2設(shè)V=V(2x2)是首權(quán)為2x2的不可約L-模，如果μ為0權(quán)時(shí)，則m(μ)=2。如果μ不是0權(quán)且μ∈∏(2x2), 則μ在V內(nèi)的重?cái)?shù)m(μ)=1。

命題3設(shè)V=V(x1+x3)是首權(quán)為x1+x3的不可約L-模，如果μ為0權(quán)時(shí)，則m(μ)=3。如果μ不是0權(quán)且μ∈∏(x1+x3), 則μ在V內(nèi)的重?cái)?shù)m(μ)=1。

命題4設(shè)V=V(x1+x2)是首權(quán)為x1+x2的不可約L-模，如果μ∈{x3,x2-x3,x1-x2,-x1}時(shí)，則m(μ)=2。否則，m(μ)=1。

命題5設(shè)V=V(x2+x3)是首權(quán)為x2+x3的不可約L-模，如果μ∈{x1,-x2+x3,-x1+x2,-x3}時(shí)，則m(μ)=2。否則，m(μ)=1。

2.3 A3型李代數(shù)的張量積分解

定理1根據(jù)引理4，得到以下結(jié)論：

1)當(dāng)λ′∈{x1,x2,x3,2x1,2x3}時(shí)，對(duì)V(λ′)?V(λ″)進(jìn)行分解之后各項(xiàng)的重?cái)?shù)皆為1。

2)當(dāng)λ′=2x2時(shí)，V(λ″)對(duì)應(yīng)的模重?cái)?shù)為2，其余為1。

3)當(dāng)λ′=x1+x3時(shí)，V(λ″)對(duì)應(yīng)的模重?cái)?shù)為3，其余為1。

4)當(dāng)λ′=x1+x2時(shí)，V(λ″+x3),V(λ″-x1),V(λ″+x2-x3),V(λ″+x1-x2)對(duì)應(yīng)的模重?cái)?shù)為2，其余為1。

5)當(dāng)λ′=x2+x3時(shí)，V(λ″+x1),V(λ″-x3),V(λ″+x3-x2),V(λ″+x2-x1)對(duì)應(yīng)的模重?cái)?shù)為2，其余為1。

V(x1)?V(x1)=V(2x1)⊕V(x2);

V(x1)?V(x2)=V(x1+x2)⊕V(x3);

V(x1)?V(x3)=V(x1+x3)⊕V(0);

V(x1)?V(2x1)=V(3x1)⊕V(x1+x2);

V(x1)?V(2x2)=V(2x2+x1)⊕V(x2+x3);

V(x1)?V(2x3)=V(2x3+x1)⊕V(x3);

V(x1)?V(x1+x3)=

V(2x1+x3)⊕V(x1)⊕V(x2+x3);

V(x1)?V(x1+x2)=

V(2x1+x2)⊕V(2x2)⊕V(x1+x3);

V(x1)?V(x3+x2)=

V(x2)⊕V(2x3)⊕V(x1+x2+x3);

V(x1)?V(x1+x2+x3)=V(2x1+x2+x3)⊕

V(2x3+x1)⊕V(x1+x2)⊕V(2x2+x3);

V(x2)?V(x1)=V(x1+x2)⊕V(x3);

V(x2)?V(x2)=V(2x2)⊕V(x1+x3)⊕V(0);

V(x2)?V(x3)=V(x3+x2)⊕V(x1);

V(x2)?V(2x1)=V(2x1+x2)⊕V(x1+x3);

V(x2)?V(2x2)=V(3x2)⊕V(x1+x2+x3)⊕V(x2);

V(x2)?V(2x3)=V(2x3+x2)⊕V(x1+x3);

V(x2)?V(x1+x3)=V(x1+x2+x3)⊕V(2x1)⊕

V(2x3)⊕V(x2);

V(x2)?V(x1+x2)=V(x1+2x2)⊕V(x2+x3)⊕

V(2x1+x3)⊕V(x1);

V(x2)?V(x3+x2)=V(2x2+x3)⊕V(2x3+x1)⊕

V(x1+x2)⊕V(x3);

V(2x3)?V(2x2)=V(2x3+2x2)⊕V(2x1)⊕

V(3x2)⊕V(x1+x2+x3)⊕V(x2);

V(x3)?V(x1)=V(x1+x3)⊕V(0);

V(x3)?V(x2)=V(x3+x2)⊕V(x1);

V(x3)?V(x3)=V(2x3)⊕V(x2);

V(x3)?V(2x1)=V(2x1+x3)⊕V(x1);

V(x3)?V(2x2)=V(2x2+x3)⊕V(x1+x2);

V(x3)?V(2x3)=V(3x3)⊕V(x2+x3);

V(x3)?V(x1+x3)=V(x1+2x3)⊕

V(x1+x2)⊕V(x3);

V(x3)?V(x1+x2)=V(x1+x2+x3)⊕

V(x2)⊕V(2x1);

V(x3)?V(x3+x2)=V(x2+2x3)⊕

V(2x2)⊕V(x1+x3);

V(x3)?V(x1+x3+x2)=V(x1+x2+2x3)⊕

V(x2+x3)⊕V(2x1+x3);

V(2x1)?V(2x1)=V(4x1)⊕V(2x2)⊕

V(2x1+x2)⊕V(x1+x3);

V(2x1)?V(2x2)=V(2x1+2x2)⊕V(2x3)⊕

V(3x2)⊕V(x1+x2+x3)⊕V(x2);

V(2x1)?V(2x3)=V(2x1+2x3)⊕V(0)⊕

V(x2+2x3)⊕V(x1+x3);

V(2x1)?V(4x1)=V(6x1)⊕V(2x1+2x2)⊕

V(x2+4x1)⊕V(x3+3x1);

V(2x1)?V(4x2)=V(4x2+2x1)⊕V(2x2+2x3)⊕

V(5x2)⊕V(x1+3x2+x3)⊕V(3x2);

V(2x1)?V(4x3)=V(4x3+2x1)⊕V(2x3)⊕

V(x2+4x3)⊕V(x1+3x3);

V(2x3)?V(2x1)=V(2x1+2x3)⊕V(0)⊕

V(2x1+x2)⊕V(x1+x3);

V(2x3)?V(2x3)=V(4x3)⊕V(2x2)⊕

V(x2+2x3)⊕V(x1+x3);

V(2x3)?V(4x1)=V(4x1+2x3)⊕V(2x1)⊕

V(4x1+x2)⊕V(3x1+x3);

V(2x1)?V(2x1+2x3)=

V(4x1+2x3)⊕V(2x1)⊕V(2x2+2x3)⊕

V(2x1+x2+2x3)⊕V(x1+3x3)⊕V(3x1+x3)⊕

V(x1+x2+x3);

V(2x1)?V(2x1+2x2)=V(4x1+2x2)⊕

V(2x1+2x3)⊕V(4x2)⊕

V(2x1+3x2)⊕V(3x1+x2+x3)⊕

V(x1+2x2+x3)⊕V(2x1+x2);

V(2x3)?V(4x2)=V(4x2+2x3)⊕V(2x1+2x2)⊕

V(5x2)⊕V(x1+3x2+x3)⊕V(3x2);

V(2x1)?V(2x3+2x2)=V(2x1+2x2+2x3)⊕

V(2x2)⊕V(4x3)⊕V(2x3+3x2)⊕V(x1+x2+3x3)⊕

V(x1+2x2+x3)⊕V(2x3+x2);

V(2x1)?V(2x1+2x2+2x3)=V(4x1+2x2+2x3)⊕

V(2x1+2x2)⊕V(2x1+4x3)⊕V(2x3+4x2)⊕

V(2x1+3x2+2x3)⊕V(3x1+x2+3x3)⊕

V(x1+2x2+3x3)⊕V(3x1+2x2+x3)⊕

V(x1+3x2+x3)⊕V(2x1+x2+2x3);

V(2x3)?V(4x3)=V(6x3)⊕V(2x2+2x3)⊕

V(x2+4x3)⊕V(x1+3x3);

V(2x3)?V(2x1+2x3)=V(2x1+4x3)⊕V(2x1+2x2)⊕

V(2x3)⊕V(2x1+x2+2x3)⊕V(x1+3x3)⊕

V(3x1+x3)⊕V(x1+x2+x3);

V(2x3)?V(2x1+2x2)=V(2x1+2x2+2x3)⊕

V(4x1)⊕V(2x2)⊕V(2x1+3x2)⊕V(3x1+x2+x3)⊕

V(x1+2x2+x3)⊕V(2x1+x2);

V(2x3)?V(2x3+2x2)=V(2x2+4x3)⊕V(4x2)⊕

V(2x1+2x3)⊕V(3x2+2x3)⊕V(x1+x2+3x3)⊕

V(x1+2x2+x3)⊕V(x2+2x3);

V(2x2)?V(2x1)=V(2x1+2x2)⊕V(2x3)⊕

V(x1+x2+x3)⊕V(3x1+x3)⊕

V(x2)⊕2V(2x1);

V(2x3)?V(2x1+2x3+2x2)=V(2x1+2x2+4x3)⊕

V(2x1+4x2)⊕V(4x1+2x3)⊕

V(2x2+2x3)⊕V(2x1+3x2+2x3)⊕

V(3x1+x2+3x3)⊕V(x1+2x2+3x3)⊕

V(3x1+2x2+x3)⊕V(x1+3x2+x3)⊕

V(2x1+x2+2x3);

V(2x2)?V(2x2)=V(4x2)⊕V(2x1+2x3)⊕

V(0)⊕V(x1+2x2+x3)⊕V(x2+2x3)⊕

V(x2+2x1)⊕V(x1+x3)⊕2V(2x2);

V(2x2)?V(2x3)=V(2x3+2x2)⊕V(2x1)⊕

V(x1+3x3)⊕V(x1+x2+x3)⊕

V(x2)⊕2V(2x3);

V(2x2)?V(4x1)=V(4x1+2x2)⊕

V(2x1+2x3)⊕V(3x1+x2+x3)⊕

V(5x1+x3)⊕V(2x1+x2)⊕2V(4x1);

V(x2)?V(x1+x2+x3)=V(x1+2x2+x3)⊕

V(2x1+2x3)⊕V(x2+2x3)⊕

V(x2+2x1)⊕V(2x2)⊕V(x1+x3);

V(2x2)?V(4x3)=V(4x3+2x2)⊕

V(2x1+2x3)⊕V(x1+5x3)⊕

V(x1+x2+3x3)⊕V(2x3+x2)⊕2V(4x3);

V(2x2)?V(4x2)=V(6x2)⊕V(2x1+2x2+2x3)⊕

V(2x2)⊕V(x1+4x2+x3)⊕V(3x2+2x3)⊕

V(3x2+2x1)⊕V(x1+2x2+x3)⊕2V(4x2);

V(2x2)?V(2x1+2x3)=V(2x1+2x2+2x3)⊕

V(4x3)⊕V(4x1)⊕V(2x2)⊕V(x1+x2+3x3)⊕

V(3x1+3x3)⊕V(3x1+x2+x3)⊕

V(x1+2x2+x3)⊕V(x2+2x3)⊕

V(2x1+x2)⊕V(x1+x3)⊕2V(2x1+2x3);

V(2x2)?V(2x1+2x2)=V(2x1+4x2)⊕

V(4x1+2x3)⊕V(2x2+2x3)⊕

V(2x1)⊕V(x1+3x2+x3)⊕V(3x1+2x2+x3)⊕

V(2x1+x2+2x3)⊕V(4x1+x2)⊕V(3x2)⊕

V(3x1+x3)⊕V(x1+x2+x3)⊕2V(2x1+2x2);

V(2x2)?V(2x3+2x2)=V(4x2+2x3)⊕

V(4x3+2x1)⊕V(2x1+2x2)⊕

V(2x3)⊕V(x1+2x2+3x3)⊕V(x1+3x2+x3)⊕

V(x2+4x3)⊕V(2x1+x2+2x3)⊕V(x1+3x3)⊕

V(3x2)⊕V(x1+x2+x3)⊕2V(2x3+2x2);

V(x1+x2)?V(2x1)=V(x1+2x2)⊕

V(3x1+x2)⊕V(x2+x3)⊕2V(2x1+x3)⊕2V(x1);

V(x1+x2)?V(2x2)=V(2x1+x2+x3)⊕

V(x1+3x2)⊕V(x1+2x3)⊕

V(x3)⊕2V(2x2+x3)⊕2V(x1+x2);

V(2x2)?V(2x1+2x2+2x3)=

V(2x1+4x2+2x3)⊕V(4x1+4x3)⊕V(2x2+4x3)⊕

V(4x1+2x2)⊕V(4x2)⊕V(2x1+2x3)⊕

V(x1+3x2+3x3)⊕V(3x1+2x2+3x3)⊕

V(3x1+3x2+x3)⊕V(x1+4x2+x3)⊕

V(2x1+x2+4x3)⊕V(4x1+x2+2x3)⊕

V(3x2+2x3)⊕V(3x1+3x3)⊕V(2x1+3x2)⊕

V(x1+x2+3x3)⊕V(3x1+x2+x3)⊕

V(x1+2x2+x3)⊕2V(2x1+2x2+2x3);

V(x1+x2)?V(2x3)=V(x1+x2+2x3)⊕

V(2x1+x3)⊕V(x1)⊕2V(3x3)⊕2V(x3+x2);

V(x1+x2)?V(4x1)=V(3x1+2x2)⊕

V(5x1+x2)⊕V(2x1+x2+x3)⊕

2V(4x1+x3)⊕2V(3x1);

V(x1+x2)?V(4x2)=V(2x1+3x2+x3)⊕

V(x1+5x2)⊕V(x1+2x2+2x3)⊕

V(2x2+x3)⊕2V(4x2+x3)⊕2V(x1+3x2);

V(x1+x2)?V(4x3)=V(4x3+x1+x2)⊕

V(2x1+3x3)⊕V(x1+2x3)⊕

2V(5x3)⊕2V(x2+3x3);

V(x1+x2)?V(2x1+2x3)=

V(x1+2x2+2x3)⊕V(3x1+x2+2x3)⊕

V(x2+3x3)⊕V(4x1+x3)⊕V(2x2+x3)⊕

V(3x1)⊕V(x1+x2)⊕2V(2x1+3x3)⊕

2V(2x1+x2+x3)⊕2V(x1+2x3);

V(x1+x2)?V(2x1+2x2)=

V(x1+4x2)⊕V(4x1+x2+x3)⊕V(3x1+3x2)⊕

V(3x2+x3)⊕V(3x1+2x3)⊕V(x1+x2+2x3)⊕

V(2x1+x3)⊕2V(2x1+2x2+x3)⊕2V(3x1+x2)⊕

2V(x1+2x2);

V(x1+x2)?V(2x3+2x2)=V(2x1+x2+3x3)⊕

V(x1+3x2+2x3)⊕V(x1+4x3)⊕

V(2x1+2x2+x3)⊕V(x1+2x2)⊕V(3x3)⊕

V(x2+x3)⊕2V(2x2+3x3)⊕

2V(3x2+x3)⊕2V(x1+x2+2x3);

V(x2+x3)?V(2x1)=

V(2x1+x2+x3)⊕V(x1+2x3)⊕

V(x3)⊕2V(3x1)⊕2V(x1+x2);

V(x1+x3)?V(2x3)=3V(2x3)⊕V(x1+3x3)⊕

V(x1+x2+x3)⊕V(x2);

V(x2+x3)?V(2x2)=V(3x2+x3)⊕

V(x1+x2+2x3)⊕V(2x1+x3)⊕

V(x1)⊕2V(x1+2x2)⊕2V(x2+x3);

V(x2+x3)?V(2x3)=V(3x3+x2)⊕V(2x2+x3)⊕

V(x1+x2)⊕2V(x1+2x3)⊕2V(x3);

V(x1+x2)?V(2x1+2x2+2x3)=

V(x1+4x2+2x3)⊕V(4x1+x2+3x3)⊕

V(3x1+3x2+2x3)⊕V(3x3+3x2)⊕

V(3x1+4x3)⊕V(4x1+2x2+x3)⊕

V(4x2+x3)⊕V(x1+x2+4x3)⊕

V(3x1+2x2)⊕V(x1+3x2)⊕V(2x1+3x3)⊕

V(2x1+x2+x3)⊕2V(2x1+2x2+3x3)⊕

2V(2x1+3x2+x3)⊕2V(3x1+x2+2x3)⊕

2V(x1+2x2+2x3);

V(x2+x3)?V(4x1)=

V(4x1+x2+x3)⊕V(3x1+2x3)⊕

V(2x1+x3)⊕2V(5x1)⊕2V(3x1+x2);

V(x2+x3)?V(4x2)=V(5x2+x3)⊕

V(x1+3x2+2x3)⊕V(2x1+2x2+x3)⊕

V(x1+2x2)⊕2V(x1+4x2)⊕2V(3x2+x3);

V(x2+x3)?V(4x3)=V(5x3+x2)⊕

V(2x2+3x3)⊕2V(3x3)⊕

V(x1+x2+2x3)⊕2V(x1+4x3);

V(x2+x3)?V(2x1+2x3)=

V(2x1+x2+3x3)⊕V(2x1+2x2+x3)⊕

V(x1+4x3)⊕V(3x1+x2)⊕V(x1+2x2)⊕

V(3x3)⊕V(x2+x3)⊕2V(3x1+2x3)⊕

2V(2x1+x3)⊕2V(x1+x2+2x3);

V(x2+x3)?V(2x1+2x2)=

V(2x1+3x2+x3)⊕V(3x1+x2+2x3)⊕

V(x1+2x2+2x3)⊕V(4x1+x3)⊕

V(2x2+x3)⊕V(3x1)⊕V(x1+x2)⊕

2V(3x1+2x2)⊕2V(2x1+x2+x3)⊕2V(x1+3x2);

V(x2+x3)?V(2x3+2x2)=

V(3x2+3x3)⊕V(x1+x2+4x3)⊕

V(4x2+x3)⊕V(2x1+3x3)⊕

V(x1+3x2)⊕V(2x1+x2+x3)⊕

V(x1+2x3)⊕2V(x1+2x2+2x3)⊕

2V(2x2+x3)⊕2V(x2+3x3);

V(x2+x3)?V(2x1+2x2+2x3)=

V(2x1+3x2+3x3)⊕V(3x1+x2+4x3)⊕

V(2x1+4x2+x3)⊕V(x1+2x2+4x3)⊕

V(4x1+3x3)⊕V(3x1+3x2)⊕

V(4x1+x2+x3)⊕V(3x2+x3)⊕

V(3x1+2x3)⊕V(x1+x2+2x3)⊕

2V(3x1+2x2+2x3)⊕2V(2x1+2x2+x3)⊕

2V(2x1+x2+3x3)⊕2V(x1+3x2+2x3);

V(x1+x3)?V(2x1)=

3V(2x1)⊕V(x1+x2+x3)⊕V(3x1+x3)⊕V(x2);

V(x1+x3)?V(2x2)=3V(2x2)⊕V(x1+2x2+x3)⊕

V(2x3+x2)⊕V(2x1+x2)⊕V(x1+x3);

V(x1+x3)?V(4x1)=3V(4x1)⊕

V(3x1+x2+x3)⊕V(5x1+x3)⊕V(2x1+x2);

V(x1+x3)?V(4x2)=3V(4x2)⊕V(x1+4x2+x3)⊕

V(3x2+2x3)⊕V(2x1+3x2)⊕V(x1+2x2+x3);

V(x1+x3)?V(4x3)=3V(4x3)⊕

V(x1+5x3)⊕V(x1+x2+3x3)⊕V(x2+2x3);

V(x1+x3)?V(2x1+2x3)=3V(2x1+2x3)⊕

V(x1+x2+3x3)⊕V(3x1+3x3)⊕V(3x1+x2+x3)⊕

V(x1+2x2+x3)⊕V(x2+2x3)⊕

V(x2+2x1)⊕V(x1+x3);

V(x1+x3)?V(2x1+2x2)=

3V(2x1+2x2)⊕V(x1+3x2+x3)⊕

V(3x1+2x2+x3)⊕V(2x1+x2+2x3)⊕

V(4x1+x2)⊕V(3x2)⊕

V(3x1+x3)⊕V(x1+x2+x3);

V(x1+x3)?V(2x3+2x2)=

3V(2x3+2x2)⊕V(x1+2x2+3x3)⊕

V(x1+3x2+x3)⊕V(x2+4x3)⊕

V(2x1+x2+2x3)⊕V(x1+3x3)⊕

V(3x2)⊕V(x1+x2+x3);

V(x1+x3)?V(2x1+2x3+2x2)=

3V(2x1+2x2+2x3)⊕V(x1+3x2+3x3)⊕

V(3x1+2x2+3x3)⊕V(3x1+3x2+x3)⊕

V(x1+4x2+x3)⊕V(2x1+x2+4x3)⊕

V(4x1+x2+2x3)⊕V(3x2+2x3)⊕

V(3x1+3x3)⊕V(2x1+3x2)⊕V(x1+x2+3x3)⊕

V(3x1+x2+x3)⊕V(x1+2x2+x3).

3 結(jié)論

本文借助李代數(shù)中的張量積分解知識(shí)，通過(guò)計(jì)算機(jī)編程，得到了以下結(jié)論：

1)對(duì)于基本支配權(quán)λ∈H,可以求出V(λ)的權(quán)格∏(λ)。

2)對(duì)于μ∈∏(λ)，根據(jù)Freudenthal公式計(jì)算出A3型李代數(shù)的權(quán)重?cái)?shù)m(μ)。

3)對(duì)于基本支配權(quán)λ∈H,根據(jù)計(jì)算出的權(quán)格∏(λ)和權(quán)重?cái)?shù)m(μ)，通過(guò)李代數(shù)的張量積分解計(jì)算出A3型李代數(shù)的張量積分解。

可以看到：在計(jì)算A3型李代數(shù)的張量積分解時(shí)，權(quán)重?cái)?shù)m(μ)的計(jì)算起著至關(guān)重要的作用。隨著首權(quán)λ的系數(shù)的增加，重?cái)?shù)的計(jì)算越發(fā)復(fù)雜。