This is somewhat of a continuation of my previous question.
I had stated there that a conformal spinor ($V$) of $SO(n,2)$ can be created by taking a direct sum of two $SO(n-1,1)$ spinors $Q$ and $S$ as,
$V = (Q, C\bar{S})$
- Here $C$ is the "charge conjugation operator" and it went unanswered as to what is the definition of that in general.
Given a set of gamma matrices $\gamma_\mu$ ( $0\leq \mu \leq n-1$) whose pair wise commutators give a representation of the $so(n-1,1)$ one can define a set of $\Gamma_a$ ($-1\leq a \leq n (or\text{ }n+1\text{ if n is even)}$) whose pair-wise commutators again represent the Lie algebra $so(n,2)$. It is such that if $S_{ab}$ are a basis for $so(n,2)$ then it will be true that $[S_{ab},V_\alpha] = (\frac{i}{4}[\Gamma_a,\Gamma_b])_\alpha ^\beta V_\beta$
If one is having an $\cal{N}$ extended supersymmetry then the claim is that,
$Q$ and $S$ will have two indices, a spinor $SO(n)$ index $i$ and a vector $SO(\cal{N})$ index $\mu$ as $S^i _\mu$ and $Q^i _ \mu$
Further it will hold that $S_i ^\mu = (Q^i_\mu)^\dagger$
- I would like to know why the above two claims are true.
Especially since $Q$ and $S$ are $SO(n-1,1)$ spinors then why should they have $SO(n)$ spinor indices? (...the "argument" that I have seen seems to go like - it is so since $SO(n)\times SO(2)$ is the maximal compact subgroup of $SO(n,2)$ whose spinor $V$ is and hence the $SO(n)$ index on $S$ and $C$..)
I would like to know of a more precise argument for this and how are these $i$ and $\mu$ indices raised and lowered and about why or how the hermiticity relation follows.
