- #1

StatusX

Homework Helper

- 2,571

- 2

I'm not sure if this is the right place for this question, so feel free to move it. Anyway, my question is, is there any good reason why the following field theory should be Weyl invariant in an arbitrary dimension d>1:

[tex] S = \int d^d x \sqrt{g} \left( g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi + \frac{1}{4} \left(\frac{d-2}{d-1}\right) R \phi^2 \right) [/itex]

Here [itex]\phi[/itex] is a scalar field and [itex]R[/itex] is the Ricci scalar associated to the metric [itex]g_{\mu \nu}[/itex]. By Weyl invariant I mean invariant under the scalings:

[tex] g_{\mu \nu} \rightarrow e^\Omega g_{\mu \nu} [/tex]

[tex] \phi \rightarrow e^{\frac{2-d}{4} \Omega} \phi [/itex]

where [itex]\Omega[/itex] is an arbitrary scalar function. If one calculates the change in the action under this transformation, one finds two separate types of terms have to cancel (one linear and one quadratic in [itex]\Omega[/itex]), and the coefficients seem to miracuously work out so that this happens. I'm guessing there's some simple geometric reason why this has to be so, but I can't think of it.

It might be related to the fact that the dirac operator commutes with conformal rescalings, and squares to something like [itex] \nabla^2 + \frac{1}{4} R[/itex], but I'm pretty sure the coefficient on R is different here, and in any case, that's an operator on spinors, not scalars.

[tex] S = \int d^d x \sqrt{g} \left( g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi + \frac{1}{4} \left(\frac{d-2}{d-1}\right) R \phi^2 \right) [/itex]

Here [itex]\phi[/itex] is a scalar field and [itex]R[/itex] is the Ricci scalar associated to the metric [itex]g_{\mu \nu}[/itex]. By Weyl invariant I mean invariant under the scalings:

[tex] g_{\mu \nu} \rightarrow e^\Omega g_{\mu \nu} [/tex]

[tex] \phi \rightarrow e^{\frac{2-d}{4} \Omega} \phi [/itex]

where [itex]\Omega[/itex] is an arbitrary scalar function. If one calculates the change in the action under this transformation, one finds two separate types of terms have to cancel (one linear and one quadratic in [itex]\Omega[/itex]), and the coefficients seem to miracuously work out so that this happens. I'm guessing there's some simple geometric reason why this has to be so, but I can't think of it.

It might be related to the fact that the dirac operator commutes with conformal rescalings, and squares to something like [itex] \nabla^2 + \frac{1}{4} R[/itex], but I'm pretty sure the coefficient on R is different here, and in any case, that's an operator on spinors, not scalars.

Last edited: