 #1
lpetrich
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I've done some more work on my http://homepage.mac.com/lpetrich/Science/SemisimpleLieAlgebras.zip [Broken] package, adding decompositions of representation powers. It was only recently that I discovered that they are called "plethysms".
Rep powers can be decomposed by symmetry types, a feature that can be important in their applications. For power p, the types are symmetric (p >=1), antisymmetric (p >= 2), and mixed (p >= 3), with more than one mixed type for p >= 4.
These are related to the irreps in (SU(n) fundamental rep)^{p}, and they share multiplicities. The symmetry types and irreps can be denoted with Young diagrams having p boxes.
With a lot of experimentation and trial and error, I have worked out how to do reppower decompositions, but the algorithm is hideously complicated, Though I've successfully tested it for low powers, I don't have a rigorous proof for it, and I've been unsuccessful in finding an algorithm in the professional literature. I've also had to create versions for reducible reps and algebraproduct reps, because reppower decompositions do not have any simple decompositions of those in terms of the rpd's of the component irreps of the component algebras.

As an example, consider flavor and spin symmetries of the three light quarks: SU(3) and SU(2) respectively. They can be combined into SU(6), and one can undo this combination into SU(3)*SU(2) again with
Mathematica: SubalgMultSU[ldqksp,{3,2}]; res = DoBranching[ldqksp,{1,0,0,0,0}]
Python: ldqksp = SubalgMultSU((3,2)); res = ldqksp.DoBranching((1,0,0,0,0))
Breakdown of cube of fundamental rep of SU(6):
Mathematica: MakeLieAlgebra[laqksp,{1,5}]; res = DecomposeRepPower[laqksp,{1,0,0,0,0},3]
Python: res = DecomposeRepPower((1,5), (1,0,0,0,0), 3)
Count, Irrep:
S: (1, (3, 0, 0, 0, 0))
M: (1, (1, 1, 0, 0, 0))
A: (1, (0, 0, 1, 0, 0))
Breakdown of cube of corresponding rep of SU(3)*SU(2):
Mathematica: MakeLieAlgebra[laqk, {1, 2}]; MakeLieAlgebra[lasp, {1, 1}]; res = DecomposeAlgProdRepPower[{laqk, lasp}, {{1, 0}, {1}}, 3]
Python: res = DecomposeAlgProdRepPower(((1,2),(1,1)), ((1,0),(1,)), 3)
Count, SU(3) Irrep, SU(2) Irrep
S: (1, ((3, 0), (3,))) . (1, ((1, 1), (1,)))
M: (1, ((1, 1), (3,))) . (1, ((3, 0), (1,))) . (1, ((1, 1), (1,))) . (1, ((0, 0), (1,)))
A: (1, ((1, 1), (1,))) . (1, ((0, 0), (3,)))

I'll now translate these results into typical physicist language, using rep multiplicities and spins:
Each quark: (3,1/2)
S: (10,3/2) + (8,1/2)
M: (8,3/2) + (10,1/2) + (8,1/2) + (1,1/2)
A: (8,1/2) + (1,3/2)
The symmetric one is the one that matches the lightbaryon spectrum, and not the mixed or the antisymmetric ones. This is contrary to FermiDirac statistics, and it took QCD to resolve that discrepancy. The quarks are antisymmetric in color, which gives them overall antisymmetry, thus preserving FD statistics.
Rep powers can be decomposed by symmetry types, a feature that can be important in their applications. For power p, the types are symmetric (p >=1), antisymmetric (p >= 2), and mixed (p >= 3), with more than one mixed type for p >= 4.
These are related to the irreps in (SU(n) fundamental rep)^{p}, and they share multiplicities. The symmetry types and irreps can be denoted with Young diagrams having p boxes.
With a lot of experimentation and trial and error, I have worked out how to do reppower decompositions, but the algorithm is hideously complicated, Though I've successfully tested it for low powers, I don't have a rigorous proof for it, and I've been unsuccessful in finding an algorithm in the professional literature. I've also had to create versions for reducible reps and algebraproduct reps, because reppower decompositions do not have any simple decompositions of those in terms of the rpd's of the component irreps of the component algebras.

As an example, consider flavor and spin symmetries of the three light quarks: SU(3) and SU(2) respectively. They can be combined into SU(6), and one can undo this combination into SU(3)*SU(2) again with
Mathematica: SubalgMultSU[ldqksp,{3,2}]; res = DoBranching[ldqksp,{1,0,0,0,0}]
Python: ldqksp = SubalgMultSU((3,2)); res = ldqksp.DoBranching((1,0,0,0,0))
Breakdown of cube of fundamental rep of SU(6):
Mathematica: MakeLieAlgebra[laqksp,{1,5}]; res = DecomposeRepPower[laqksp,{1,0,0,0,0},3]
Python: res = DecomposeRepPower((1,5), (1,0,0,0,0), 3)
Count, Irrep:
S: (1, (3, 0, 0, 0, 0))
M: (1, (1, 1, 0, 0, 0))
A: (1, (0, 0, 1, 0, 0))
Breakdown of cube of corresponding rep of SU(3)*SU(2):
Mathematica: MakeLieAlgebra[laqk, {1, 2}]; MakeLieAlgebra[lasp, {1, 1}]; res = DecomposeAlgProdRepPower[{laqk, lasp}, {{1, 0}, {1}}, 3]
Python: res = DecomposeAlgProdRepPower(((1,2),(1,1)), ((1,0),(1,)), 3)
Count, SU(3) Irrep, SU(2) Irrep
S: (1, ((3, 0), (3,))) . (1, ((1, 1), (1,)))
M: (1, ((1, 1), (3,))) . (1, ((3, 0), (1,))) . (1, ((1, 1), (1,))) . (1, ((0, 0), (1,)))
A: (1, ((1, 1), (1,))) . (1, ((0, 0), (3,)))

I'll now translate these results into typical physicist language, using rep multiplicities and spins:
Each quark: (3,1/2)
S: (10,3/2) + (8,1/2)
M: (8,3/2) + (10,1/2) + (8,1/2) + (1,1/2)
A: (8,1/2) + (1,3/2)
The symmetric one is the one that matches the lightbaryon spectrum, and not the mixed or the antisymmetric ones. This is contrary to FermiDirac statistics, and it took QCD to resolve that discrepancy. The quarks are antisymmetric in color, which gives them overall antisymmetry, thus preserving FD statistics.
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