When we define $\mathcal{N}=4$ SYM on flat Minkowski space, the supersymmetric vacua are parametrized by scalars living in the cartan subalgebra of the gauge group. A generic point in the moduli space breaks the SU(N) to $U(1)^r$ where r is the rank of the cartan subalgebra. This is called the Coulomb branch for obvious reasons. On page 59 of this review


it is stated that when we put the theory on $\mathbb{R} \times S^3$ (so that the dual will live on global AdS instead of the Poincare patch), the Coulomb branch is lifted because the scalars are conformally coupled through a term $\int d^4x Tr(\phi^2)R$.

Why is it that the scalars are conformally coupled? It makes some sense to me that putting a CFT on a sphere should introduce some scale (radius) and could lead to some conformal anomaly (ex central charge and cylinder in string theory). But I was under the impression there is no unique way to know, given the theory in flat space, what the covariantization should be. In other words, there are lots of terms I could use to couple fields to the curvature that would vanish in the flat space limit.

In strings for instance, I could take a free scalar. That defines one CFT. I could also consider a scalar with a background charge that couples to the curvature through a term like $Q\int \phi R$. This would give me the linear dilaton CFT, a totally different CFT with different primaries etc.

So my question is why do the authors assume that the scalars are conformally coupled? Is this a general principle in QFT in curved space or is it arbitrary?

  • $\begingroup$ CFT on $R\times S^3$ is still conformal. It is similar to world sheet radial quantization. The radius $S^3$ can be rescaled out by conformal invariance. More argument can be found in hep-th/9803131. $\endgroup$ – thone Dec 8 '13 at 12:47
  • $\begingroup$ Thanks, I'll take a look. I understand that the theory is still conformal, the origin of moduli space is unaffected by the conformal coupling. I am asking specifically why the Coulomb branch is getting lifted and if this has to be the case. $\endgroup$ – Dan Dec 8 '13 at 17:53

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