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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

http://arxiv.org/pdf/hep-th/9905111v3.pdf

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?

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  • $\begingroup$ 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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