Decompose $SO(8)$ and $Spin(8)$ 8 dimensional representations to $SO(m)$ and $SO(n)$

Question

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?

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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]]

Answer 1

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.