# AdS/CFT-duality: How does the $U(1)$ decouple form the $U(N)$?

A stack of N coincident D3-branes on its world-volume describe, at the lowest order in $\alpha'$ and in absence of non-trivial background fields, a supersymmetric $U(N)$ gauge theory as explained in Becker-Becker-Schwartz section 12.3.

IN this same book they go on and argue that a $U(1)$ factor of the $U(N)$ gauge group decouples as a free theory.

More precisely, the $U(1)$ lives on the boundary and the $SU(N)$ lives in the bulk, which is why the $U(1)$ is not relevant.

• First: the boundary of what?
• Why is it not relevant if it lives on the boundary? I always learned to be cautious with boundary (effects)
• What would change to the duality, if the $U(1)$ was taken into account?
• boundary of the bulk – zzz Jun 16 '15 at 6:37
• Up to global identifications, U(N) is the same thing as SU(N) x U(1), so it is not a simple group. Even for purely field theoretical reasons, the gauge multiplets in the SU(N) and U(1) factors are living completely independently. Moreover, all the matter in SU(N) is really in adjoint and all the adjoint is neutral under the U(1). So nothing couples to the U(1) gauge field, it is decoupled! It is a bit subtle question why the AdS bulk space is really equivalent just to the SU(N) and not the whole U(N). But if it were the whole U(N), the bulk physics would have to be made of 2 indep. parts too – Luboš Motl Jun 16 '15 at 8:49
• See arxiv.org/abs/hep-th/9905111 (page 58), and references therein. I hope it helps. – Patrick El Pollo Jul 22 '16 at 1:20