魏玉麗, 王利萍, 代佳華
(北京建筑大學 理學院,北京 100044)
Kazhdan-Lusztig多項式的首項系數是Kazhdan-Lusztig理論中的核心研究對象[1],這些系數對研究該理論起到了至關重要的作用。為了計算一些最低雙邊胞腔上的Kazhdan-Lusztig多項式的首項系數,需要知道李代數中不可約模的張量積重數;而張量積中不可約模的重數在李代數理論中也是一個重要的問題。許超[2]給出了A2的不可約模的張量積分解的一個計算方法。于桂海等[3]給出了特征數大于0的代數閉域上C2型單連通半單代數群,限制支配權所對應的不可約模的張量積分解。對于A型李代數的張量積分解,理論上有Young圖法、Klymik公式、Pieris公式。
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。
設L=sl(4,F), 為上A3型李代數。設V(λ)為首權是λ的不可約最高權L-模,V(λ)的權格為∏(λ),且L的基本支配權為x1、x2、x3。如果μ∈Λ,定義μ在V(λ)(λ∈Λ+)內的重數為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,且權集合∏(λ)被W0所置換,使得對于σ∈W0,有dimVμ=dimVσμ。
引理3[5](Freudenthal公式)設V=V(λ)是首權為λ(λ∈Λ+)的不可約L-模,如果μ∈Λ,則μ在V內的重數m(μ)可從如下的遞推得到:
[(λ+δ,λ+δ)-(μ+δ,μ+δ)]m(μ)=
(1)
計算m(μ)分三步:第一步確定集合D={λ,比λ低的支配權};第二步計算和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行排列。然后根據引理2~3計算出m(μ)。
引理4[5]使得V(λ)可能出現在V(λ′)?V(λ″)的加項中的λ∈Λ+,只能形如μ+λ″,μ∈∏(λ′)。當這樣的μ+λ″都是支配權時,V(μ+λ″)出現在張量積內,且重數為mλ′(μ)。
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,同時定義H={x1,x2,x3,2x1,2x2,2x3,x1+x3,x2+x3,x1+x2}。對于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}
接下來計算上述λ對應的∏(λ),∏(λ)中含有λ、比λ低的支配權,以及它們在W0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}
命題1設V=V(λ)是首權為λ的不可約L-模,λ∈{x1,x2,x3,2x1,2x3}。如果μ∈∏(λ), 則μ在V內的重數m(μ)=1。
命題2設V=V(2x2)是首權為2x2的不可約L-模,如果μ為0權時,則m(μ)=2。如果μ不是0權且μ∈∏(2x2), 則μ在V內的重數m(μ)=1。
命題3設V=V(x1+x3)是首權為x1+x3的不可約L-模,如果μ為0權時,則m(μ)=3。如果μ不是0權且μ∈∏(x1+x3), 則μ在V內的重數m(μ)=1。
命題4設V=V(x1+x2)是首權為x1+x2的不可約L-模,如果μ∈{x3,x2-x3,x1-x2,-x1}時,則m(μ)=2。否則,m(μ)=1。
命題5設V=V(x2+x3)是首權為x2+x3的不可約L-模,如果μ∈{x1,-x2+x3,-x1+x2,-x3}時,則m(μ)=2。否則,m(μ)=1。
定理1根據引理4,得到以下結論:
1)當λ′∈{x1,x2,x3,2x1,2x3}時,對V(λ′)?V(λ″)進行分解之后各項的重數皆為1。
2)當λ′=2x2時,V(λ″)對應的模重數為2,其余為1。
3)當λ′=x1+x3時,V(λ″)對應的模重數為3,其余為1。
4)當λ′=x1+x2時,V(λ″+x3),V(λ″-x1),V(λ″+x2-x3),V(λ″+x1-x2)對應的模重數為2,其余為1。
5)當λ′=x2+x3時,V(λ″+x1),V(λ″-x3),V(λ″+x3-x2),V(λ″+x2-x1)對應的模重數為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).
本文借助李代數中的張量積分解知識,通過計算機編程,得到了以下結論:
1)對于基本支配權λ∈H,可以求出V(λ)的權格∏(λ)。
2)對于μ∈∏(λ),根據Freudenthal公式計算出A3型李代數的權重數m(μ)。
3)對于基本支配權λ∈H,根據計算出的權格∏(λ)和權重數m(μ),通過李代數的張量積分解計算出A3型李代數的張量積分解。
可以看到:在計算A3型李代數的張量積分解時,權重數m(μ)的計算起著至關重要的作用。隨著首權λ的系數的增加,重數的計算越發(fā)復雜。