It is not an answer, but maybe some information which could be useful :

In an other post, it has been also noticed that the commutators of the $R$-symmetry generators with supercharge generators are:

$[R^a_b,Q^c_{\alpha}]=\delta^c_bQ^a_{\alpha}-\frac{1}{4}\delta^a_bQ^c_{\alpha}$

So, taking the trace (on $a,b$), with $\mathcal N=4$, gives a null commutator, so the $U(1)$ symmetry may be factored out.

In "Pierre Ramond, Group Theory, A Physicist's Survey, Cambridge, p $218$", it is stated that the superalgebra $su(n|m)$, may be wiritten in terms of hermitian matrices of the form :

$$\begin {pmatrix}SU(n) &( \textbf n, \bar{\textbf m})\\( \bar {\textbf n}, \textbf m)&SU(m)\end {pmatrix} + \begin {pmatrix} n&0\\0&m\end {pmatrix}$$
where the diagonal matrix generates the $U(1)$. (so the even elements form the Lie algebra $SU(n) \times SU(m) \times U(1)$)

It is stated that : "when $n=m$, the supertrace of the diagonal matrix vanishes, and the $U(1)$ decouples from the algebra".

This is also true for non-compact versions of the superalgebra, for instance $su(2,2|4)$ (here $n=m=4$)

[EDIT]

Correction : Following Olof's comment below, the $U(1)$ here is the (zero )central charge, so the symmetry group is really $psu(2,2|4)$ (see Olof's answer). It is not the $U(1)$ necessary to extend $su(2,2|4)$ to $u(2,2|4)$

This post imported from StackExchange Physics at 2014-03-07 16:30 (UCT), posted by SE-user Trimok