# Why is the R-symmetry in $\mathcal{N}=4$ $SU(4)$ and not $U(4)$?

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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 $\mathcal{N}=4$?

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

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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 $SO(9,1) \to SO(3,1) \times SO(6)$.

This post imported from StackExchange Physics at 2014-03-07 16:30 (UCT), posted by SE-user Olof
answered Dec 7, 2013 by (210 points)
Ok, but the theory is finite so the U(1) still can't be anomalous. The N^2 U(N) currents should still be in the same multiplet as the energy momentum tensor, an anomaly in the U(1) would yield a non-zero beta function. Are you saying that there really is an anomaly that kills the U(1)? I find it more likely that the U(1) is just represented as 1 on all physical states, but I don't know why this would be

This post imported from StackExchange Physics at 2014-03-07 16:30 (UCT), posted by SE-user Dan
@Dan: I think the related anomaly should be the Konishi anomaly, but I don't remember the details.

This post imported from StackExchange Physics at 2014-03-07 16:30 (UCT), posted by SE-user Olof
Good call, I like the dimensional reduction answer

This post imported from StackExchange Physics at 2014-03-07 16:30 (UCT), posted by SE-user Dan
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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
answered Dec 7, 2013 by (955 points)
Just to clearify: the $u(1)$ generated by the diagonal matrix here is the central charge I mentioned in my answer (it's central for $m=n=2$). The $su(2,2|4)$ algebra can be extended to $u(2,2|4)$ by including a second diagonal generator with non-vanishing super trace, so of the form $\text{diagonal}(+n,-m)$.

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

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