The difference of N=2 short multiplets and BPS states
I have some questions about the construction of $\mathcal{N}=2$ supermultiplet for chiral matter. I knew from somewhere that the supermultiplet should not include spin one states since they are always in adjoint representation. So my first question is: why are spin-one modes always in adjoint representation?
To avoid confusion, I will denote the four $SUSY$ generators as $Q^A_\alpha$, where $\alpha$ is the spinor index and $A$ equals to $1$ or $2$.
First, massless multiplet or short multiplet. we know $Q^A_2$ generates states with zero norm. Then we can just focus on $Q^A_1$. The supermultiplet can be constructed in this way: $$|\Omega_{-\frac{1}{2}}>\ Q^{\dagger 1}\dot{1} |\Omega{-\frac{1}{2}}>, Q^{\dagger 2}\dot{1} |\Omega{-\frac{1}{2}}>\ Q^{\dagger 2}\dot{1} Q^{\dagger 1}\dot{1} |\Omega_{-\frac{1}{2}}>$$ The two spin-zero states in the second line form a $SU(2)$ doublet.
Second, BPS states. 656
In the presence of central charge $Z$, one can write the four generators in terms of $A_\alpha$ and $B_\alpha$ after linear transformation with ${ B_\alpha, B^\dagger_\beta}=\delta_{\alpha\beta}(M-\sqrt{2}Z)$ and ${A_\alpha, A^\dagger_\beta}=\delta_{\alpha\beta}(M+\sqrt{2}Z)$. When $M=\sqrt{2}Z$, one can get ${ B_\alpha, B^\dagger_\beta}=0$. Thus, the multiplet will be: $$|\Omega_0>\ A^\dagger_\dot{1}|\Omega_0>, A^\dagger_\dot{2}|\Omega_0>\ A^\dagger_\dot{1} A^\dagger_\dot{2}|\Omega_0>$$ Again there are only four states, the same as the short multiplet.
So do the two fermion states in BPS states form some kind of doublet as in the short multiplet? it's not so obvious since the relation between $A^\dagger_\dot{1}$ and $A^\dagger_\dot{2}$ are different from that of $Q^1_1$ and $Q^2_1$.
From the above discussion, is it legitimate to conclude that BPS states are different from the short multiplet in $\mathcal{N}=2$? for instance, in the short multiplet, the two scalars form a doublet while I don't see that in BPS states. But in <PERSON> review paper hep-th/9701069 section 2.9, he mentioned that the BPS state belongs to a $\mathcal{N}=2$ short multiplet and the two scalar modes form a doublet while the two fermions are singlet. It's like <PERSON> was saying the two states are exactly the same. So what's missing in my thinking? Thanks a lot. 656