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

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]]

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.
• @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 $\mathtt{D}_4$. Jul 29 '20 at 20:58
• 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 $\mathrm{Spin}(8)$. Jul 29 '20 at 21:55