we shall now see (following a similar analysis in [@Girardello:2002pp]) they are still conserved at $N=\infty$.

In the free theory, the primary operator $J_s$ heads a short representation of the conformal group that we label $(\Delta=s+1,s)$, where $\Delta$ is the conformal dimension and $s$ the spin. The shortening condition is the conservation equation $\partial_\mu J^\mu_{\,\,\,\,\,\mu_1\dots\mu_{s-1}} = 0$. For $J_s$ to become non-conserved, there must appear on the right-hand side of this equation a non-zero operator in the representation $(s+2,s-1)$. In other words, $J_s$ must combine with another operator in this representation to form a long representation [@Girardello:2002pp], $$\begin{aligned}
 \label{confreps}
  {\mathrm{lim}}_{\epsilon \to 0}(s+1+\epsilon,s)_{\mathrm{long}} &\cong (s+1,s)_{\mathrm{short}} \oplus (s+2,s-1) {\,.}\end{aligned}$$ By acting with special conformal transformations on $\partial_\mu J^\mu_{\,\,\,\,\,\mu_1\dots\mu_{s-1}}$ one can show that in the limit in which $J_s$ is conserved, the $(s+2,s-1)$ operator in must be a primary of the conformal algebra [@Heidenreich:1980xi] (the coefficient of this operator in the equation for $d*J_s$ vanishes in this limit, but the special conformal generator acting on $d*J_s$ vanishes even faster). Now, a connected correlator of the form $\partial_\mu \left< J^\mu_{\,\,\,\,\mu_1\dots\mu_{s-1}} {\ensuremath{\mathcal{O}}}\right>$ can have a leading, $O(N^0)$ contribution in the large $N$ limit only when ${\ensuremath{\mathcal{O}}}$ is a “single-trace” operator. Therefore, at $N=\infty$, $J_s$ can 