I've been reading this paper http://arxiv.org/abs/hep-th/0110263. In section 4, he discusses the benefits of writing the supergravity action in a superconformal way. I have a few questions regarding this:

  1. Why is (4.1) "superconformal"? I'm not sure I see how it is invariant under either conformal or SUSY transformations?

  2. Why in the second line of (4.1) is he all of a sudden talking about dilatation symmetry? Is it a general feature of superconformal actions that you pick up an extra dilatation symmetry?

  3. Assuming my previous point is correct and that the extra symmetry we pick up when going to a superconformal description is a dilatation symmetry, I still don't see what use this is? What do we gain by doing this? How is it helpful?


1 Answer 1

  1. (4.1) is NOT superconformal, it is just conformal. It is invariant under the full conformal group.

  2. Conformal = Poincare + Dilatations + Special Conformal Symmetry

(4.1) is obviously invariant under Poincare transformations. So, Toine provides you the conformal transformations of the scalar $\phi$ and the metric $g_{\mu\nu}$ in case you want to check that the action is indeed invariant under conformal symmetry transformations.

  1. The use of enhancing the number of symmetries is that it is technically much simpler to construct supergravity models when you have more symmetries. Therefore, you extend the super-Poincare symmetry to the largest possibility, which is the superconformal group, to make your life easier. Later on, you fix the extra gauge symmetries (Dilatation, Special Conformal and S-SUSY) to obtain the super-Poincare action.
  • $\begingroup$ Thanks for this answer - very helpful for conveying the main idea! Does this mean that the example in (4.1) is entirely unsupersymmetric just to convey the advantages of having extra symmetries? What would some of these advantages be: in the unsupersymmetric example of (4.1) there don't appear to be many but you said that in general SUGRA constructions things become technically easier - do you have an example of this? $\endgroup$
    – user11128
    Feb 7, 2016 at 19:42
  • $\begingroup$ Yes, 4.1 is non-supersymmetric. The advantage, as i said, is to construct models easier. You can see a recent paper arxiv.org/pdf/1512.06277.pdf . In SUGRA, the method is called the superconformal tensor calculus. a google search would give you many examples in the context of SUGRA - google.co.uk/… $\endgroup$
    – John Doe
    Feb 8, 2016 at 8:12

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