# Labelling of states using $R$-symmetry for ${\cal N=4}$ SUSY

In Modern Supersymmetry by John Terning, page 13, the states of the massless supermultiplets of $N=4$ SUSY are labelled by the helicity and representation of R-symmetry under which they transform. However, for $N$ other than 4, only the helicity and degeneracy are used to label the states. Is there a reason why $N=4$ is singled out in this regard, or is it just a notational convention?

Further, for massive states, he uses spin and representation of R-symmetry to label the states for all $N$, so why was a similar treatment by replacing spin with helicity not done for the massless supermultiplets?

$\mathcal{N}$ in the $\mathcal{N}$ supersymmetry denotes the number of the supercharge spinors $Q_\alpha^a$. The $R$-symmetry comes from the $U(\mathcal{N}$) rotations in terms of $a$ index.

Now, when we construct the massless supermultiplet (usually choosing the frame of reference $p^\mu=E(1,0,0,1)$) we note (from the commutation relations) that one of each supercharge components $Q_1^a$ annihilates any state from that supermultiplet whereas the other component $Q_2^a$ is a helicity raising operator. Likewise $Q_{\dot{2}}^a$ is a helicity lowering operator. If $|-h_0\rangle$ is a state with some minimal helicity $-h_0$ then the states with helicity $-h_0+\frac{m}{2}$ can be constructed by application helicity raising operator $m$ times, \begin{equation} Q_2^{a_1}\ldots Q_2^{a_m}|-h_0\rangle \end{equation} with complete antisymmetry in $\{a_k\}$ indices thanks to $\{Q_\alpha^a,Q_\beta^b\}=0$. The degeneracy is a dimension of this state subspace i.e. the number of linearly independent states with the same helicity.

Now think what is the nature of $Q_2^{a_1}\ldots Q_2^{a_m}$ from the point of view of $R$-symmetry (more precisely its $SU(\mathcal{N})$ subgroup as $U(1)$ factor will simply give those states an irrelevant phase). This is a component of some antisymmetric tensor with $m$ indices. The result of the rotation will be a linear superposition of the components of the same tensor i.e. of other states with helicity $-h_0+\frac{m}{2}$. I.e. if you take all states with the same helicity from one supermultiplet they form a multiplet of $R$-symmetry. The aforementioned degeneracy is exactly the dimension of the corresponding representation.

The case $\mathcal{N}=4$ is special only in the following sense. If you start with $-h_0=-1$ the maximum helicity you can reach is $h_{max}=-1+\frac{\mathcal{N}}{2}$. For $\mathcal{N}=1,2,3$ those are $h_{max}=-\frac{1}{2},0,+\frac{1}{2}$ correspondingly. That means that this supermultiplet is not $\mathcal{CPT}$-invariant and you have to supplement it with extra states constructed by application of $\mathcal{CPT}$ operator on the original states. In case of $\mathcal{N}=2,3$ that obviously doubles the degeneracy of states $h=0$ (and for $\mathcal{N}=3$ of $h=\pm 1/2$). However for $\mathcal{N}=4$ you can reach $h_{max}=+1$ and get $\mathcal{CPT}$-invariant multiplet automatically.

Concerning Terning, it should be stressed that he notes very briefly the connection between degeneracy and the $R$-symmetry representation dimensions. E.g. he writes

Note that the degeneracy also tells us the dimension of representation of the $SU(2)_R$ subgroup of the $U(2)_R$ symmetry that each state belongs to.

But frankly from my quick look at this book I find it somewhat sloppy and skipping some points that may be important for a student just starting to learn susy (he doesn't even write full commutation relations of susy!) It looks more like a quick reference guide than a true textbook.