# Matter content of $\mathcal{N}=4$ SYM

I've always heard it said that the matter content of $\mathcal{N} = 4$ Super-Yang-Mills theory is uniquely fixed by supersymmetry, but I have never quite understood why this must be so.

Could anyone provide a simple watered-down explanation for why this must be the case?

Existence of supersymmetry in a theory implies that all fields must occur in supermultiplets. Let us now try to figure out what the field content of an ${\cal N} = 4$ SYM could possibly be.

Now, suppose the lowest helicity in a supermultiplet is $h$. For massless representations, the highest helicity in the multiplet is then $h + \frac{1}{2}{\cal N} = h+2$. Now, note that we are interested in a field theory, i.e. with no gravity. We must therefore have no fields with helicities $\pm\frac{3}{2}$ and $\pm2$. Thus, $h \geq -1$ and $h+2\leq+1 \implies h\leq-1$. The two inequalities imply that only one choice of $h$ is allowed, namely $h=-1$.

Thus, there is a unique ${\cal N}=4$ field theory, namely one with lowest helicity $h=-1$ in the supermultiplet. Now, let us count the number of states here. Out of the $4{\cal N} = 16$ supercharges, 8 identically vanish on massless representations. Out of the remaining 8, 4 are raising and 4 are lowering operators. Let us use the 4 raising operators on the $h=-1$ state to create the rest of the multiplet.

$1$ state with $h=-1$

${4\choose 1}=4$ states with $h=-1+\frac{1}{2} = -\frac{1}{2}$.

${4\choose 2}=6$ states with $h=-1 + 2\frac{1}{2}=0$.

${4\choose 3}=4$ states with $h=-1 + 3\frac{1}{2}=+\frac{1}{2}$.

${4\choose 4}=1$ states with $h=-1 + 4\frac{1}{2}=+1$.

The $h=\pm1$ states together forms the gauge boson, the 4 $h=\pm\frac{1}{2}$ states form the 4 Weyl spinors and the 6 $h=0$ states form 6 scalars.

More generally of course, in an ${\cal N}=4$ theory, we may have as many copies of the above field content as we want, say $n$.

Thus, the most general field content of a ${\cal N}=4$ theory is $n$ gauge bosons, $4n$ Weyl spinors and $6n$ scalars. The $n$ gauge bosons may together transform in the adjoint of any gauge group such that $\dim G = n$. Since the gauge boson must transform in the adjoint and the spinors and scalars are in the same multiplet, the Weyl spinors and scalars must also transform in the adjoint. The only freedom in the choice of ${\cal N}=4$ theories then, is the gauge group $G$ (In $SU(N)$ theories $n=N^2-1$).