# Mismatch between conformal generators and conformal transformations as changes of variables

Introduction

It is known that under changes of coordinates different fields transform according to their tensorial nature (scalar, vector, etc.) like$$^{[1]}$$

$$$$\phi(x)\rightarrow\phi'(x')=\phi(x)$$$$

$$$$V_{\mu}(x)\rightarrow V_{\mu}'(x')=\frac{\partial x^\nu}{\partial x'^{\mu}}V_\nu(x)$$$$

$$$$g_{\mu \nu}(x)\rightarrow g_{\mu \nu}'(x')=\frac{\partial x^\rho}{\partial x'^{\mu}}\frac{\partial x^\sigma}{\partial x'^{\nu}}g_{\rho \sigma}(x)$$$$

and so on. We can write an action for a massive free scalar field that is invariant under any change of coordinates

$$$$S=\int d^Dx \sqrt{|g|(x)}\Big\{g^{\mu \nu}(x)\partial_\mu\phi(x)\partial_\nu\phi(x) -m^2\phi^2(x)\Big\}.$$$$

If we apply a coordinate change should transform the metric $$g$$ and the field $$\phi$$. And since the variables are dummy indeces we can change them as well to get

$$$$S'=\int d^Dx' \sqrt{|g'|(x')}\Big\{g'^{\mu \nu}(x')\partial'_\mu\phi'(x')\partial'_\nu\phi'(x') -m^2\phi'^2(x')\Big\}.$$$$

The differential $$d^D x$$ and the jacobian $$\sqrt{|g|(x)}$$ transform in opposite ways so they cancel out. In the same way, the change on the metric cancels out with the change in the derivatives. Lastly, the scalar field itself doesn't change so we end up with the same action. This is the statement that this action is invariant under changes of coordinates.

However, scale transformations

$$$$x^\mu \rightarrow x'^\mu =\lambda x^\mu$$$$

are a specific type of coordinate transformation and thus this action is also invariant under those. It is commonly known that theories with a mass term break the scale invariance, so the actual meaning of scale invariance in this context is not the same. According to an answer here Simple conceptual question conformal field theory what we actually mean by scale transformation is a change of coordinates $$x'^\mu =\lambda x^\mu$$ followed by a Weyl transformation that rescales everything

$$$$\phi(x)\rightarrow \tilde{\phi}(x)=\lambda^{-\Delta}\phi(x)$$$$

$$$$g_{\mu \nu}(x)\rightarrow \tilde{g}_{\mu \nu}(x)=\lambda^{2}g_{\rho \sigma}(x)$$$$

Where $$\lambda^2$$ is set to cancel the factor that the coordinate transformation induced on the metric. Now the transformed action is

\begin{align} \tilde{S}'= & \int d^Dx' \sqrt{|\tilde{g}'|(x')}\Big\{\tilde{g}'^{\mu \nu}(x')\partial'_\mu\tilde{\phi}'(x')\partial'_\nu\tilde{\phi}'(x') -m^2\tilde{\phi}'^2(x')\Big\} \\ = & \int d^Dx \sqrt{|g|(x)}\Big\{g^{\mu \nu}(x)\partial_\mu\phi(x)\partial_\nu\phi(x) -\lambda^{-2\Delta}m^2\phi^2(x)\Big\}. \end{align}

The kinetic term is scale invariant if $$\Delta=\frac{D-2}{2}$$ when we add the Weyl rescaling but now the mass term breaks the symmetry. This action is only scale invariant when the mass is $$m=0$$.

Conflict with the Conformal Group

On the previous calculation the factor $$\lambda^{-\Delta}$$ in the transformation of the field happens because of the Weyl Transformation, not because of the conformal transformation (which is just a change of coordinates and thus $$\phi'(x')=\phi(x)$$). However, in many books we see that the generators of the conformal group are$$^{[2]}$$

\begin{align} P_{\mu} =&-i\partial_{\mu} \\ D =&-i\big(x^{\mu}\partial_{\mu}+\Delta\big)\\ L_{\mu\nu} =&S_{\mu\nu}+i\big(x_{\mu}\partial_{\nu}-x_{\nu}\partial_{\mu}\big)\\ K_{\mu} =&i\Big[x^{2}\partial_{\mu}-2x_{\mu}x^{\nu}\partial_{\nu}-\Delta x_{\mu}-iS_{\mu\nu}x^{\nu}\Big]. \end{align}

Scale transformations ($$D$$) and Lorentz rotations ($$L_{\mu\nu}$$) can affect the fields even when we transform both the fields and the coordinates as

$$$$\phi_\mu(x)\rightarrow \phi_\mu'(x')=e^{i\theta^{ab}\big(S_{ab}\big)_{\mu\nu}}\phi_\nu (x)$$$$

for rotations and

$$$$\phi_\mu(x)\rightarrow \phi_\mu'(x')=\lambda^{-\Delta}\phi_\mu (x)$$$$

for scale transformations. Now, although it is clear that the rotation is compatible with the definition of change of coordinates that we did for vectorial objects, the scale transformations make no sense. How is this last transformation compatible with the fact that conformal transformations are just a change of coordinates? The calculations done in the introduction of the post assumed that conformal transformations (as coordinate transformations) don't change fields, while the Weyl transformation is neccesary to produce the scale transformation for $$\phi$$. But in this last result of the conformal group, the change of coordinates changes the field on its own.

References:

[2] Section 4.2.1. in Francesco's CFT

• I think I already answered this elsewhere, but it's because every CFT source ever is just incredibly sloppy. The sociological reason is that almost zero time is spent on classical CFT, 99% of work on the subject is on quantum CFTs, which don't even have an action half the time. So there's no particular reason or incentive to use correct logic in the classical case when you're gonna jettison it after 5 pages and never see it again. Yes, it sucks. I wasted like two whole afternoons just trying to find the correct definition of a conformal transformation. – knzhou Apr 4 at 23:12
• And yes, you would think that people that work on one of the most mathematical abstruse subsubfields of physics would at the very least correctly and clearly define the central object of their study, but, well, we don't live in such a world. – knzhou Apr 4 at 23:16
• Hi! Thanks for the quick answer. I agree with all that you said. The thing I don't understand is that the transformation $\phi'(x')=\lambda^{-\Delta}\phi(x)$ seems to be due to a Weyl rescaling in the post by Mane.Andrea but then CFT books derive the generators of the conformal group thinking only in terms of a change of variables and they seem to get the same transformation rule, which is the opposite to what a change of variables should do. – P. C. Spaniel Apr 4 at 23:17
• And mane.andrea's answer in the post you linked is completely correct. A conformal transformation in CFT is not just a special diffeomorphism. If it were, then conformal symmetry would not have any power at all, because it would be implied by diffeomorphism symmetry, which every reasonable theory has. A real conformal transformation is a special diffeomorphism composed with a Weyl transformation that cancels its effect on the metric. It is the Weyl transformation that allows the $\lambda^{-\Delta}$ term. – knzhou Apr 4 at 23:17
• The books get the right answer because the algebra of real, actual conformal transformations matches the algebra of conformal changes of coordinates. (The extra scale factors just "come along for the ride".) Then they just look for general representations of this algebra. – knzhou Apr 4 at 23:18