2
$\begingroup$

Is there any way to have a scalar field that transforms non-trivially under local special conformal transformations? Just by the index structure, I can see that the possibilities are

$$\begin{align} \delta_K \phi &= \Lambda_{K}^\mu (x) \partial_\mu \phi \,, & \delta_K \phi &= x^\mu \Lambda_{K \mu} (x) \end{align}$$

where the subscript $K$ represents transformation with respect to the special conformal transformation, and $\Lambda_{K\mu} (x)$ is the parameter of special conformal transformations.

Considering the Poincare and dilatation transformations of the scalar $\phi$, I can see that both the first and the second possibilities are just orbital parts of Poincare and (inhomogeneous) dilatation transformations respectively.

Am I missing something? Is there really no way to write down a scalar that transforms non-trivially under local special conformal transformations?

$\endgroup$
9
  • $\begingroup$ Conformal transformations are those transformations which locally look like dilatations, rotations and translations, so there is no meaningful notion of "local special conformal transformation". If you think otherwise, could you explain, what a local special conformal transformation is supposed to mean? $\endgroup$ Feb 6, 2016 at 4:46
  • $\begingroup$ There are vector fields, for instance, that tranasforms under local conformal transformations, i.e. you can check the non-linear multiplet of 4D N=1 supergravity, or the Weyl multiplet of any conformal supergravity, $\delta_K b_\mu = \Lambda_{K\mu}$. Those transformations do not look like local dilatations. So, it is possible and meaningful. $\endgroup$
    – John Doe
    Feb 6, 2016 at 7:22
  • $\begingroup$ You still have not explained what do you mean by a local conformal transformation. Is this a coordinate transformation? If yes, how is it defined? If not, what is it? I am not saying that you cannot write down some formula for whatever supergravity, I am saying that the phrase "local conformal transformation" is meaningless without further clarification. $\endgroup$ Feb 6, 2016 at 8:07
  • $\begingroup$ I am also puzzled by your mentioning of conformal gravity, since I thought that conformal gravity = Weyl-invariant gravity, and has nothing to do with any sort of "local special conformal transformations". $\endgroup$ Feb 6, 2016 at 8:13
  • $\begingroup$ I assume that you know what rigid conformal transformation is. It is a coordinate transformation that solves the conformal killing equation. In the case of local conformal transformation, the parameters, both dilatations and the special conformal transformations are $x$ dependent. what do you mean what is local conformal transformation. Also, conformal gravity is, of course, invariant under the full conformal group. Weyl gravity is an example, but $-\frac1{12} \phi^2 R + \frac12 (d \phi)^2$ is another example which is not only scale invariant but invariant under the full conformal group. $\endgroup$
    – John Doe
    Feb 6, 2016 at 9:30

1 Answer 1

1
$\begingroup$

I don't completely understand the question. How a scalar transforms is completely dictated by conformal symmetry. The transformation law is

$$K_\mu \phi(x) = \big(\Delta x_\mu + x_\mu \, x_\nu \partial_\nu -x^2 \partial_\mu \big) \phi(x)$$

or if you wish

$$\delta_K \phi(x) = a^\mu K_\mu \phi(x)$$

where $a^\mu$ are the infinitesimal parameters of the special conformal transformation. The above holds for a scalar primary of scaling dimension $\Delta$. As you know, you can also have scalar descendants, which will have more complicated transformation laws. For example, if $A_\mu$ is a vector operator, then $\partial_\mu A_\mu$ is a scalar but the transformation law will have one extra term coming from the $[K_\mu,P_\nu]$ commutator.

I hope this answers your question!

$\endgroup$
7
  • $\begingroup$ First of all, i do not want to involve more fields, just a scalar. when you consider the local Poincare and dilatation transformations, these terms simply become orbital parts of translations or dilatations. For example, in a global theory, a scalar transform under the Poincare transformations as $\delta \phi = \xi^\mu \partial_\mu \phi +\lambda^{\mu\nu} x_\mu \partial_\nu \phi$. However, when you consider the LOCAL Poincare transformations, you have $\delta \phi = \xi^\mu (x) \partial_\mu \phi + \lambda^{\mu\nu} (x) x_\mu \partial_\nu \phi = \xi^{\prime \mu}(x) \partial_\mu \phi$. $\endgroup$
    – John Doe
    Feb 2, 2016 at 16:09
  • $\begingroup$ So the one that you had given, the first term can be placed into dilatation transformations by a redefinition of the dilatation parameter, and the last two terms can be placed into local translations by a redefinition of the translation parameter. $\endgroup$
    – John Doe
    Feb 2, 2016 at 16:12
  • $\begingroup$ Maybe I understand your question better now. The reason this happens is that for a primary $[K_\mu,\phi(x=0)] = 0$, by construction. Then $[K_\mu,\phi(x)]$ is generated by shifting $\phi(x)$ using $P_\mu$. This means that all the terms in $\delta_K \phi(x)$ are generated by commutators of the form $[P_\mu,K_\nu]$ and $[P_\mu,[P_\nu,K_\rho]]$; it's of the form $K_\mu \phi(x) = \ldots D + \ldots P_\mu$. (For scalars, the generator of rotations $M_{\mu \nu}$ gives zero.) $\endgroup$ Feb 2, 2016 at 16:25
  • $\begingroup$ @HansMoleman - You are implicitly assuming that $\phi$ is a primary field. $\endgroup$
    – Prahar
    Feb 4, 2016 at 5:16
  • $\begingroup$ Sure. Otherwise you will have some terms coming from $[K_\mu,P_\nu]$ terms, which are again ultimately proportional to $\ldots D^m \phi + \ldots (P_\mu) ^n \phi$. $\endgroup$ Feb 4, 2016 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.