Why is the R-symmetry in $\mathcal{N}=4$ $SU(4)$ and not $U(4)$? In $\mathcal{N}=4$ SYM we have 4 supercharges. Naively, I would have thought that the R-symmetry would be $U(4)$. I know that in theories with less SUSY the $U(1)$ can be anomalous. But $\mathcal{N}=4$ is a SCFT, and the R-charge appears in the SUSY algebra on the r.h.s. of the anticommutators $\{S,Q\}$ and cannot be anomalous without breaking SCFT. So why do we lose the $U(1)$ in $\mathcal{N}=4$?
 A: Only the $su(4)$ generators appear on the right hand side of the $u(2,2|4)$ commutation relations, so superconformal invariance does not prevent an anomaly in the $u(1)$ reducing the symmetry to $su(2,2|4)$. In $\mathcal{N}=4$ SYM the central charge is furthermore zero, so the actual symmetry is $psu(2,2|4)$.
The breaking of the generator with non-zero supertrace is related to the Konishi anomaly. 
However, an easier way to see that the R-symmetry is $SU(4)$ is to construct $\mathcal{N}=4$ SYM from 10D $\mathcal{N}=1$ SYM by compactification on $T^6$. The R-symmetry then comes from the breaking $SO(9,1) \to SO(3,1) \times SO(6)$.
A: 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 $u(2,2|4)$
