Weyl invariant scalar field theory

[StatusX]

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

$$g_{\mu \nu} \rightarrow e^\Omega g_{\mu \nu}$$

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

where $\Omega$ 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 $\Omega$), 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 $\nabla^2 + \frac{1}{4} R$, but I'm pretty sure the coefficient on R is different here, and in any case, that's an operator on spinors, not scalars.

 

[Haelfix]

Hmm, good question.

That looks very stringy to me, and I've seen that before someplace. It looks very similar to terms in the Polyakov action + a worldsheet Einstein Hilbert-Dilaton like term (which has the Euler characteristic floating around, which I think is related to the geometrical reason you were looking for). Alternatively it has the form of a nonlinear sigma model written in some sort of conformal frame.

Observe however that

(Delta^2 + 1/2*(d-2)/(d-1) *R)*Phi =0
(factors of 2 might be off)

 

[Entropia]

First of all, it is important to note that Weyl invariance is a very special property that is not present in most field theories. It is a symmetry that relates to the conformal invariance of a theory, which means that the theory is invariant under scale transformations. This is a very powerful symmetry that can greatly simplify calculations and provide deeper insights into the structure of a theory.

In the case of Weyl invariant scalar field theory, the key to understanding why it is Weyl invariant lies in the structure of the action. The first term in the action is the standard kinetic term for a scalar field, which is invariant under Weyl transformations as long as the metric g_{\mu \nu} is also transformed accordingly. The second term, however, is more interesting.

The term involving the Ricci scalar R can be thought of as a coupling between the scalar field and the gravitational field. In general, this term is not Weyl invariant, but in this specific case where the coefficient is chosen to be \frac{1}{4} \left(\frac{d-2}{d-1}\right), the Weyl invariance is restored. This is because this coefficient is precisely the one needed to cancel out the linear and quadratic terms that arise from the Weyl transformation of the scalar field.

To understand why this coefficient is so special, we need to look at the structure of the Weyl transformation for the metric g_{\mu \nu}. Under this transformation, the metric transforms as g_{\mu \nu} \rightarrow e^\Omega g_{\mu \nu}. This means that the Ricci scalar R also transforms as R \rightarrow e^{-2\Omega} R. So, in order for the term involving R to be Weyl invariant, we need the exponent of e in the transformation of R to be canceled out by the exponent of e in the transformation of the scalar field \phi.

This is precisely what happens when the coefficient is chosen to be \frac{1}{4} \left(\frac{d-2}{d-1}\right). In this case, the exponent of e in the transformation of R is -2\Omega, while the exponent of e in the transformation of \phi is \frac{2-d}{4} \Omega. These two exponents cancel out, leading to a Weyl invariant theory.

In summary, the Weyl