R-symmetry for N=4 susy

Last year I was studying supersymmetry and since them I'm regularly thinking to one question for which I don't have answer: when one looks at the susy algebra with N generators, one sees that there is a U(N) R-symmetry. But for N=4, the group is in fact SU(4).
To explain this, one generally argues (Terning's book, McGreevy's lectures, problem set 2) using the form of the lagrangian, e.g. that the scalars are in the real antisymmetric 6 representation, and so a phase transformation is clearly not a symmetry of the lagrangian.
But I'm remember to have read somewhere a computation at the level of the algebra (with N generators) which leads to a term with (N-4) and this explains why the case N=4 is special. Unfortunately I'm not able to remember the steps of the computation or to find again the place where I have read this; it's not in usual books (Terning, Weinberg, Wess/Bagger, West, Binétruy, Freund, Müller-Kirsten/Wiedemann) nor reviews (Sohnius, Bilal, etc.). Does someone have an idea or a hint about this?
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 I'm not an expert, but it's something to do with the fact that the R-symmetry group U(4) factorizes as U(4)= SU(4)xU(1). The U(1) factor commutes with all the (superconformal)group generators and hence can be factored out, leaving the projective symmetry group PSU(2,2|4) rather than SU(2,2|4) as you would have expected.
 Do you have any reference explaining this factorization of U(1) in this case? I'm not very accustomed to supergroup.

R-symmetry for N=4 susy

I'm not sure what I said before was quite right - here is a reference (section 3).

Normally with these extended supersymmetries, the R-symmetry group is U(N), but in this case, N=4, it's SU(4)xU(1).

(I don't think what I said - that U(4)=SU(4)xU(1) is correct)
 In fact, we can always decompose as U(N) = U(1) x SU(N), where the U(1) is the usual symmetry associated to the R-charge for N=1, except for N=4 where the R-operator vanishes, and we get only SU(4). I took a look at the paper you said, and the U(1)_Z group which he speaks about is not the same as the U(1)_R, but it corresponds to the central charge(s) that one can adds (related to BPS state).

 Quote by Melsophos In fact, we can always decompose as U(N) = U(1) x SU(N), where the U(1) is the usual symmetry associated to the R-charge for N=1, except for N=4 where the R-operator vanishes, and we get only SU(4). I took a look at the paper you said, and the U(1)_Z group which he speaks about is not the same as the U(1)_R, but it corresponds to the central charge(s) that one can adds (related to BPS state).
This is briefly explained in Peter West's book Introduction to Supersymmetry and Supergravity.
 Exact, you are right, I have the book but I did not recognize since the notations differs from the usual ones. Sadly the construction is not fully explained, and I don't have access to the original paper from Hagg, Lopuszanski, Sohnius. Do you know somewhere else where this construction is detailed? In any case, thanks for showing this.

 Quote by Melsophos In fact, we can always decompose as U(N) = U(1) x SU(N), where the U(1) is the usual symmetry associated to the R-charge for N=1, except for N=4 where the R-operator vanishes, and we get only SU(4). I took a look at the paper you said, and the U(1)_Z group which he speaks about is not the same as the U(1)_R, but it corresponds to the central charge(s) that one can adds (related to BPS state).
I think you're right. As this reference (near end of page 2) points out, there are potentially 3 U(1)s involved: $U(1)_Z$ which is the central extension in the reference I gave, $U(1)_R$ (which he calls $U(1)_C$) which is the extra piece that traditionally gets removed from the R-symmetry $U(4)$ - this is the one you want, and finally $U(1)_B$ (which some people call $U(1)_Y$) which Intriligator refers to as the "bonus symmetry".

The $U(1)_R$ factor can indeed be factored out of the supergroup when $\mathcal{N}=4$ since it commutes with the generators. This is explained in the appendix here. Basically the commutator of the R-symmetry generator with the supercharge is given by $$[R^a_b,Q^c_{\alpha}]=\delta^c_bQ^a_{\alpha}-\frac{1}{4}\delta^a_bQ^c_{\alpha}$$
Contracting a and b, $\delta^a_b$ becomes $\mathcal{N}$, so the commutator vanishes when $\mathcal{N}$=4.

I assumed you'd also want the R to drop out of the anticommutator (last equation in A.2)? I'm not sure how to show that....

 Quote by sheaf I think you're right. As this reference (near end of page 2) points out, there are potentially 3 U(1)s involved: $U(1)_Z$ which is the central extension in the reference I gave, $U(1)_R$ (which he calls $U(1)_C$) which is the extra piece that traditionally gets removed from the R-symmetry $U(4)$ - this is the one you want, and finally $U(1)_B$ (which some people call $U(1)_Y$) which Intriligator refers to as the "bonus symmetry".
In another reference (I forgot which ones precisely, but it cites the one you are talking about) I read that in fact it's not very important which U(1) ones factorizes since one can always take linear combinations.

 Quote by sheaf The $U(1)_R$ factor can indeed be factored out of the supergroup when $\mathcal{N}=4$ since it commutes with the generators. This is explained in the appendix here. Basically the commutator of the R-symmetry generator with the supercharge is given by $$[R^a_b,Q^c_{\alpha}]=\delta^c_bQ^a_{\alpha}-\frac{1}{4}\delta^a_bQ^c_{\alpha}$$ Contracting a and b, $\delta^a_b$ becomes $\mathcal{N}$, so the commutator vanishes when $\mathcal{N}$=4. I assumed you'd also want the R to drop out of the anticommutator (last equation in A.2)? I'm not sure how to show that....
Your reference is really perfect, thank you! In fact, the Sohnius review "Introducing susy" has also a little section on this, but less well explained.
Sohnius explained a little what happens to this term, but I'm not sure to understand all. In short, he says the term is allowed only if the lowest helicity of the representation is different from -1.
It seems (page 2) that one can redefine the variables such that the commutator is [Q, S] ~ ... + (N-4) R, and this gives [Q, R] ~ Q.