I have some questions about the construction of $\mathcal{N}=2$ supermultiplets for chiral matter. I know that the supermultiplet should not include spin one states since they are always in the adjoint representation. So my first question is: why are spin-one modes always in adjoint representation?

To avoid confusion, I will denote the four supersymmetry generators as $Q^A_\alpha$, where $\alpha$ is the spinor index and $A=1,2$.

Firstly, 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 an $SU(2)$ doublet.

Secondly, BPS states. In the presence of a 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 Alvarez-Gaume's 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 Alvarez-Gaume was saying the two states are exactly the same. So what's missing in my thinking? Thanks a lot.

  • $\begingroup$ This is a very brief answer to your first question but hopefully it provides a starting point. Gauge bosons (spin 1 particles) are always in the adjoint representation of the gauge group by construction. The QFT justification is in Peskin and Schroeder, the math justification is that gauge bosons describe a connection on the underlying manifold. $\endgroup$
    – Heterotic
    Commented Feb 5, 2014 at 22:32

1 Answer 1


Spin one fields are in the adjoint representation because of the specific transformation behaviour of such modes under gauge symmetries. This is a symmetry requirement on the lagrangian.

Regarding your second question: the two multiplets are different. In one case we have a central charge and in the other we do not. However, the issue is essentially one of terminology. The references I am most familiar with (see for example "Modern Supersymmetry" by Terning) identify the short multiplet with the BPS state, whereas the "massless multiplet or "short multiplet" you refer to is called the hypermultiplet.


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