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This is the followed up question of LieArt --- 3 different 8 dimensional Irreducible representation of SO(8) and their decompositions,

Since $$ \mathrm{SO}(8) \supset \mathrm{SU}(2) \times \mathrm{SU}(2) \times \mathrm{SU}(2) \times \mathrm{SU}(2), $$ I am trying to figure how to do the decompositions of irreducible representation of $SO(8)$ into four components of $\mathrm{SU}(2) \times \mathrm{SU}(2) \times \mathrm{SU}(2) \times \mathrm{SU}(2)$ (and their irreducible-representation),

Naively, the code should be,

DecomposeIrrep[Irrep[SO8][8], ProductAlgebra[SU2, SU2, SU2, SU2]]

This outputs the warning message: enter image description here

But it does not work. There are indeed 3 different 8 dimensional Irreducuble representations of $\mathrm{SO}(8)$, LieArt --- 3 different 8 dimensional Irreducible representation of SO(8) and their decompositions, but I also do not see how to call them out individually, at least not from

Irrep[SO8][IrrepPrime[8, 0]]

Irrep[SO8][IrrepPrime[8, 1]]

Irrep[SO8][IrrepPrime[8, 2]]

My question is what is the correct way to write the code for

 DecomposeIrrep[Irrep[SO8][8], ProductAlgebra[SU2, SU2, SU2, SU2]]?
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  • $\begingroup$ Did you try and increase the value of $MaxDynkinDigit (ass suggested by the error message)? $\endgroup$ – AccidentalFourierTransform Oct 1 '18 at 1:31
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When in doubt, use Dynkin labels instead of (the more ambiguous and less precise) dimension notation:

In any case, the particular branching rule has not been implemented (yet?):

DecomposeIrrep[Irrep[D][1, 0, 0, 0], ProductAlgebra[SU2, SU2, SU2, SU2]]
DecomposeIrrep[Irrep[D][0, 0, 1, 0], ProductAlgebra[SU2, SU2, SU2, SU2]]
DecomposeIrrep[Irrep[D][0, 0, 0, 1], ProductAlgebra[SU2, SU2, SU2, SU2]]

which yields the error message

Either $D_4$ does not have $A_1\times A_1\times A_1\times A_1$ as a subalgebra or the branching rule is implemented with a different ordering of $A_1\times A_1\times A_1\times A_1$ or not at all. Try changing the order in $A_1\times A_1\times A_1\times A_1$.

Oh well. Maybe you can get in contact with the author(s) of the package and ask them directly about this branching rule, and whether they intend to implement it. Or do it yourself: have a look at the code of the package to understand how other branching rules where implemented, and include this one yourself.


That being said, I believe the correct branching rule is \begin{equation} \begin{aligned} \boldsymbol{8_v}&\mapsto(\boldsymbol 2,\boldsymbol 2,\boldsymbol 1,\boldsymbol 1)\oplus(\boldsymbol 1,\boldsymbol 1,\boldsymbol 2,\boldsymbol 2)\\ \boldsymbol{8_c}&\mapsto(\boldsymbol 2,\boldsymbol 1,\boldsymbol 1,\boldsymbol 2)\oplus(\boldsymbol 1,\boldsymbol 2,\boldsymbol 2,\boldsymbol 1)\\ \boldsymbol{8_s}&\mapsto(\boldsymbol 2,\boldsymbol 1,\boldsymbol 2,\boldsymbol 1)\oplus(\boldsymbol 1,\boldsymbol 2,\boldsymbol 1,\boldsymbol 2) \end{aligned} \end{equation}

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