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We know that the $SO(8)$'s vector representation is 8 dimensional $8_v$, and $Spin(8)$ has the spinor representation $8_s$ and its conjugate representation spinor $8_c$ . Together $8_v$, $8_c$ and $8_s$ form the triality.

Naively, I expect that the decomposition $SO(8)$ (or $Spin(8)$) of the $8_c$ and $8_s$ are conjugate with each other.

However from Lie Art mathematica file, I learned that

$SO(8)$ decomposes as $SO(5)$ and $SO(3)$: $$ 8_v=(4,2), \quad 8_c=(4,2), \quad 8_s=(1,3)+(5,1), $$

  • STRANGELY, $8_v$ and $8_c$ have the similar form, are they conjugate with each other? But $8_c$ and $8_s$ do not seem to conjugate with each other. Can you illuminate why?

$SO(8)$ decomposes as $SO(6)$ and $SO(2)$: $$ 8_v=\bar{4} \text{(-1)}+4 \text{(1)}, \quad 8_c=6 \text{(0)}+1 \text{(-2)}+1 \text{(2)}, \quad 8_s=\bar{4} \text{(1)}+4 \text{(-1)}, $$

  • STRANGELY, $8_v$ and $8_s$ have the similar form, are they conjugate with each other? But $8_c$ and $8_s$ do not seem to conjugate with each other. Can you illuminate why?

My result is obtained from the Mathematica Lie Art code:

DecomposeIrrep[Irrep[D][1, 0, 0, 0], ProductAlgebra[Sp4, SU2]]

DecomposeIrrep[Irrep[D][0, 1, 0, 0], ProductAlgebra[Sp4, SU2]]

DecomposeIrrep[Irrep[D][0, 0, 1, 0], ProductAlgebra[Sp4, SU2]]

DecomposeIrrep[Irrep[D][0, 0, 0, 1], ProductAlgebra[Sp4, SU2]]

DecomposeIrrep[Irrep[D][1, 0, 0, 0], ProductAlgebra[SU4, U1]]

DecomposeIrrep[Irrep[D][0, 1, 0, 0], ProductAlgebra[SU4, U1]]

DecomposeIrrep[Irrep[D][0, 0, 1, 0], ProductAlgebra[SU4, U1]]

DecomposeIrrep[Irrep[D][0, 0, 0, 1], ProductAlgebra[SU4, U1]]

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There is an automorphism group of $\mathtt{D}_4$ that permutes these three $8$-dimensional modules. But if the subgroup you are restricting to is not normalized by this automorphism, there's no need for the modules to have similar-looking restrictions. Indeed, restricting to the $\mathtt{A}_1$ in the centre of the Dynkin diagram will produce similar decompositions.

Your $\mathtt{B}_3$ is stable under a subgroup of order $2$ of the $S_3$ of automorphisms, so two will behave the same and one (likely) will behave differently.

Notice that $\mathtt{B}_3$ is not well-defined. You have three options for the $\mathtt{B}_3$, given its images under triality.

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  • $\begingroup$ thanks voted up plus 1, you meant "an automorphism group of 𝐷4 " or "S3"? The 𝐷4 is of dihedral group order of how many elements? 4 or 8? $\endgroup$ Jul 29 '20 at 20:45
  • $\begingroup$ @anniemarieheart I mean $D_4$ as in Lie type $D_4$, i.e., $8$-dimensional orthogonal, and $B_3$ is $5$-dimensional orthogonal. The outer automorphism group of $D_4$, i.e., $\mathrm{Spin}(8)$, is isomorphic to $S_3$, but realized as genuine automorphisms you can only say that there is 'an' automorphism group, because of course there are lots of automorphisms with the same outer automorphirm image. I have edited the answer to make clear I'm not talking about $D_4$ but instead the algebraic group $D_4$, which is often denoted $\mathtt{D}_4$. $\endgroup$ Jul 29 '20 at 20:58
  • $\begingroup$ if 8v is a vector, 8s and 8c are spinors, and their triality rotats them, what makes the branch rule to have one of them differ from the other twos? I am confused what is special about one of the 8v, 8s and 8c? $\endgroup$ Jul 29 '20 at 21:31
  • $\begingroup$ SO(8) to SO(6) x SO(2) as SU4 x U1 chooses 8c special looking? $\endgroup$ Jul 29 '20 at 21:31
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    $\begingroup$ I think you don't seem to understand what tirality means. It means that there is a homomorphism from $\mathrm{Spin}(8)$ to itself, whose cube is the identity. Consequently, it moves the subgroups of the spin group around. Your group $\mathrm{Spin}(6)$ doesn't exist. There are three of them, corresponding to the three $D_3$ subdiagrams of the $D_4$ Dynkin diagram. Choosing one chooses some orientation of the Dynkin diagram, and thus fixes one of the three $\mathrm{SO}(8)$ quotients of $\mathrm{Spin}(8)$. $\endgroup$ Jul 29 '20 at 21:55

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