$SO(2)_i$’s and the resulting -charges are given in Table \[table:one\]. The -charges of the complex scalars, $\phi_i$, in the three chiral multiplets are all the same and equal to $2/3$. By evaluating the usual triangle diagrams with the chiral fermions, one finds that the cubic and linear ’t Hooft anomalies for $U(1)_R^{\mathcal{N}=4}$ are given by: $$\label{thooftn4}
\begin{split}
\mathop{\rm tr} R^3_{\mathcal{N}=4} &\eql  d_{G}\,\Big [1^3-3\Big({1\over 3}\Big)^3\,\Big] = \frac{8}{9}\,d_G\;, \qquad 
\mathop{\rm tr} R_{\mathcal{N}=4}  \eql  d_{G}\,\Big [1-3\,\Big({1\over 3}\Big)\,\Big] = 0\;,
\end{split}$$ where $d_G$ is the dimension of the gauge group ($d_G=N^2-1$ for $SU(N)$).

The superconformal -symmetry current in any $\mathcal{N}=1$ SCFT lies in the same supermultiplet as the energy-momentum tensor. This can be used to show that the conformal anomaly is simply determined in terms of ’t Hooft anomalies of the -current by the following well-known formulae for the central charges [@Anselmi:1997am; @Intriligator:2003jj; @Cassani:2013dba]: $$\label{andc}
a = \frac{9}{32} \text{tr}R^3 - \frac{3}{32} \text{tr}R\;, \qquad c = \frac{9}{32} \text{tr}R^3 - \frac{5}{32} \text{tr}R\;.$$ Using and it is then straightforward to compute the central charges in $\mathcal{N}=4$ SYM $$\label{ccN=4}
a_{\mathcal{N}=4} = c_{\mathcal{N}=4} = \frac{1}{4}d_{G}\;.$$

[@ c c c c c c]{} Field & $SO(2)_1$ & $SO(2)_2$ & $SO(2)_3$ & $U(1)_R^{\cN=4}$ & $U(1)_R^{\rm LS}$\
$\lambda$ & $\phantom{-}{1\over 2}$ & 