Here is a follow up of on strange facts of $Spin(8)$ group Decompose $SO(8)$ and $Spin(8)$ 8 dimensional representations to $SO(m)$ and $SO(n)$
We know that the $Spin(8)$'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.
Let us do $Spin(8)$ decomposes as $Spin(5)$ and $Spin(3)$: $$ 8_v=(4,2), \quad 8_c=(4,2), \quad 8_s=(1,3)+(5,1). \tag{a} $$
$Spin(8)$ decomposes as $Spin(6)$ and $Spin(2)$: $$ 8_v=\bar{4} \text{(-1)}+4 \text{(1)}, \quad 8_c=6 \text{(0)}+1 \text{(-2)}+1 \text{(2)}, \quad 8_s=\bar{4} \text{(1)}+4 \text{(-1)}, \tag{b} $$
I seem to know $$8_v=(4,2)$$ in (a) is related to $$8_v=\bar{4} \text{(-1)}+4 \text{(1)}$$ in (b), simply that 4 and $\bar{4}$ in $Spin(6)$ become the same as 4 in $Spin(5)$.
STRANGELY, $$8_c=(4,2)$$ in (a) seems NOT related to $$8_c=6 \text{(0)}+1 \text{(-2)}+1 \text{(2)}$$ in (b). Can you illuminate why? or how can they be related? (By restricting $Spin(6)$ to $Spin(5)$, etc?)
STRANGELY, $$8_s=(1,3)+(5,1)$$ in (a) seems NOT related to $$8_s=\bar{4} \text{(1)}+4 \text{(-1)}$$ in (b). Can you illuminate why? or how can they be related? (By restricting $Spin(6)$ to $Spin(5)$, etc?)
My main point is that it seems to be natural to have $8_c=(4,2)$ in (a) related to $8_s=\bar{4} \text{(1)}+4 \text{(-1)}$ in (b). So $8_c$ from $Spin(5)\times Spin(3)$ mutates to $8_s$ from $Spin(6)\times Spin(2)$? Do we know a precise math formulation?
My main point is that it seems to be natural to have $8_s=(1,3)+(5,1)$ in (a) related to $8_c=6 \text{(0)}+1 \text{(-2)}+1 \text{(2)}$ in (b). So $8_s$ from $Spin(5)\times Spin(3)$ mutates to $8_c$ from $Spin(6)\times Spin(2)$? Do we know a precise math formulation?