I come up with this conclusion after reading some books and review articles on conformal field theory (CFT).

CFT is a subset of FT such that the action is invariant under conformal transformation of the fields and coordinate but leave the metric unchanged.

Is this correct?

Let me explain further and take the $\phi^4$ theory in $4$-dim as an example (just discuss classical invariance, I know that loops break the invariance). In $4$-dim consider a scalar field with conformal weight $\Delta=1$ such that \begin{align} x \to x' = \lambda x,\\ \phi'(x')=\phi'(\lambda x) = \lambda^{-1}\phi(x). \end{align} Then the action is unchanged \begin{align} S'& = \int d^4 x' \sqrt{g}\left\{\frac{1}{2}g^{\mu\nu}\partial'_{\mu}\phi'(x')\partial'_{\mu}\phi'(x')-\phi'^4(x')\right\}\\ &= \int d^4 x \lambda^4 \sqrt{g}\left\{\frac{1}{2}g^{\mu\nu}\lambda^{-4}\partial_{\mu}\phi(x)\partial_{\mu}\phi(x)-\lambda^{-4}\phi^4(x)\right\} \\ &= \int d^4 x \sqrt{g} \left\{\frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi(x)\partial_{\mu}\phi(x)-\phi^4(x)\right\}. \end{align} Note that I did not use $g'^{\mu\nu} = \lambda^2g^{\mu\nu}$, all metrics are unprimed. In this example we see that conformal invariance is realized without changing the metric. I was confused at the beginning since all textbooks and articles derive the conformal group and representations by considering the change of the metric.

If we use $g'_{\mu\nu}x'^{\mu}x'^{\nu} = g_{\mu\nu}x^{\mu}x^{\nu}$, the physical distance does not change at all and we are just choosing a new coordinate chart. My interpretation is, as what we meant by a physical scaling or transformation, we really change the distance between two points. Another reasoning is, the metric in CFTs are just background (not being integrated in the path integral) thus we do not change them. If we consider a theory including the metric as a dynamical field (we path integrate it and perhaps quantize it), the actions has to be invariant including the transformation of the metric.

Is the above correct? Please give me some comments and point out the wrong concepts if there is any. Thank you very much.

If you have time, could you please take a look at my other question.


2 Answers 2


Imposing that an action should be conformally invariant has a subtle but important difference with respect to imposing that it should be diffeomorphism invariant. Let me try to emphasize the differences between Diffeomorphisms, Conformal transformations and Weyl transformations. I will also clarify how to impose the invariance of the action, namely I'll argue whether the fields or the coordinates must be changed.


The action on the coordinates is given by a general differentiable function with differentiable inverse: $x^\mu \to {x'}^\mu = {x'}^\mu(x)$. The action on fields with spin 0, 1 and 2 is the following $$ \begin{align} \phi(x) &\to \phi'(x') = \phi(x)\,, \\ \partial_\mu\phi(x) &\to \partial'_\mu\phi'(x') = \frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu\phi(x)\,, \\ g_{\mu\nu}(x) &\to g'_{\mu\nu}(x') = \frac{\partial x^\alpha}{\partial {x'}^\mu}\frac{\partial x^\beta}{\partial {x'}^\nu}g_{\alpha\beta}(x)\,. \end{align} $$ These transformations are just changes of variables, every theory is invariant under them.

Weyl transformations

Weyl transformations do not act on the coordinates but simply rescale the fields $$ \begin{align} \phi(x) &\to \tilde{\phi}(x) = \Omega^{-\Delta}(x)\phi(x)\,, \\ \partial_\mu\phi(x) &\to \partial_\mu\tilde{\phi}(x) = \partial_\mu(\Omega^{-\Delta}(x)\phi(x))\,, \\ g_{\mu\nu}(x) &\to \tilde{g}_{\mu\nu}(x) = \Omega^{2}(x)g_{\mu\nu}(x)\,. \end{align} $$

Conformal transformations

Conformal transformations are coordinate transformations that preserve the metric up to a scale factor, that is $x^\mu \to {x'}^\mu$ and $$ \frac{\partial x^\alpha}{\partial {x'}^\mu}\frac{\partial x^\beta}{\partial {x'}^\nu} g_{\alpha\beta}(x)= \Omega^{-2}(x)g_{\mu\nu}(x)\,.\tag{1} $$ In OP's example ${x'}^\mu = \lambda x^\mu$ and $\Omega(x) = \lambda$. Now, this is the key point: we don't want to impose that the theory is invariant under this change of variables, this is always true because it is a particular case of diffeomorphism invariance. We want instead to compare theories with the same metric. That is, we always imagine a conformal transformation as a diffeomorphism satisfying (1) followed by a Weyl rescaling that cancels the $\Omega$ factor. Therefore the transformation laws are $$ \begin{align} \phi(x) &\to \tilde{\phi}'(x') = \Omega^{-\Delta}(x)\phi(x)\,, \\ \partial_\mu\phi(x) &\to \partial'_\mu\tilde{\phi}'(x') = \frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu(\Omega^{-\Delta}(x)\phi(x))\,, \\ g_{\mu\nu}(x) &\to \tilde{g}'_{\mu\nu}(x') = g_{\mu\nu}(x)\,. \end{align} $$ With the notation "tilde prime" I denote the composition of the two last transformations.

Invariance of the action

Consider a generic Lagrangian that depends on the field $\phi$, its first derivative and the metric tensor. The action then reads $$ S = \int d^4x \, \mathcal{L}[g_{\mu\nu}(x),\phi(x),\partial_\mu\phi(x)]\,. $$ The transformation acts only on the fields and leaves $x$ unchanged since it is a dummy variable (integrated over). Proving diffeomorphism invariance amounts to requiring $$ \begin{align} S &= \int d^4x\, \mathcal{L}\left[g'_{\mu\nu}(x),\phi'(x),\partial'_\mu\phi'(x)\right] = \\ &=\int d^4x'\, \mathcal{L}\left[g'_{\mu\nu}(x'),\phi'(x'),\partial'_\mu\phi'(x')\right] = \\ &=\int d^4x\,|\Omega(x)|^4 \mathcal{L}\left[\Omega^{-2}(x)g_{\mu\nu}(x),\phi(x),\frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu\phi(x)\right]\,, \end{align} $$ in the second step I just renamed $x$ to $x'$, it's a trivial operation. In the last step I used the transformation properties of the measure and of the fields (See Di Francesco - Mathieu - Sénéchal for example).

A Weyl rescaling requires simply $$ S = \int d^4x\, \mathcal{L}\left[\Omega^2(x)g_{\mu\nu}(x),\Omega^{-\Delta}(x)\phi(x),\partial_\mu(\Omega^{-\Delta}(x)\phi(x))\right]\,. $$ Finally, the requirement for conformal invariance can be obtained by composing the two transformations above $$ S = \int d^4x\, |\Omega(x)|^4\mathcal{L}\left[g_{\mu\nu}(x),\Omega^{-\Delta}(x)\phi(x),\frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu(\Omega^{-\Delta}(x)\phi(x))\right]\,. $$ This is consistent with what OP says. Notice that since diffeomorphism invariance is always true, just proving Weyl invariance is enough. The converse is not obvious and it is object of current research. We could say that by imposing conformal invariance we are imposing Weyl invariance only by those $\Omega$'s that make $g_{\mu\nu}$ and $\Omega^2 g_{\mu\nu}$ diffeomorphic.

  • 2
    $\begingroup$ This answer looks great to me, but if it really is this way, then why do books seem to go out of their way to hide it? Does, e.g. di Francesco ever say anything explicit about this? $\endgroup$
    – knzhou
    Commented Dec 22, 2018 at 16:11
  • $\begingroup$ Sorry but I am confused about something. The transformation of the metric tensor under a general diffeomorphism comes up in GR by imposing that the line element $ds^2$ doesn't change under the change of coordinates. But this doesn't allow for instance a rescaling, where distances grow. So from this it seems that conformal transformations are not a subset of general diffeomorphisms, while in the definition you give of conformal transformations they are clearly a subset. $\endgroup$
    – MBolin
    Commented Mar 28, 2019 at 18:26
  • $\begingroup$ Moreover, if you take your definition of a conformal transformation and consider a simple rotation (conformal transformation as far as I know), it is clear that you don't have $\frac{\partial x^{\alpha}}{\partial x'^{\mu}} \frac{\partial x^{\beta}}{\partial x'^{\nu}} g_{\alpha \beta}(x) = \Omega^{-2}(x) g_{\mu \nu}(x)$. $\endgroup$
    – MBolin
    Commented Mar 28, 2019 at 19:03
  • 1
    $\begingroup$ I just wanted to comment that I think that under diffeomorphism invariance you do want to make sure you change the variables of integration. Even though they correspond to dummy variables, you need to be careful because in general diffeomorphisms change the domain of integration. This is an essential step in deriving, for example, Noether's theorem. $\endgroup$ Commented Feb 28, 2020 at 16:09
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    $\begingroup$ @Sigh_at_psi The free scalar field is not Weyl invariant, even though it will be scale invariant with $\Delta = (D-2)/2$. You can however make it fully Weyl invariant if add a term of the form $\xi R \phi^2$ to the Lagrangian ($R$ being the Ricci scalar). If you tune the coupling constant $\xi$ correctly, this will take care of the cross terms involving derivatives of $\Omega$ that you mentioned, due to the change in $R$ after a Weyl transformation. $\endgroup$ Commented Nov 16, 2021 at 22:52

I think in your case it is better to see conformal transformation this way:

  1. You don't change coordinates
  2. Field with conformal dimension $\Delta$ transforms as $\widetilde \Phi = \lambda^{- \Delta} \Phi$
  3. Field $\phi$ has $\Delta=1$ and metrics $g_{\mu\nu}$ has $\Delta = 2$.

So parts of the action change just as you wrote, but for different reasons. Volume form change beacause $\sqrt{\det \widetilde g} = \sqrt{\det \widetilde g_{\mu\nu}}= \lambda^{4} \sqrt {\det g}$. Derivative $\partial_\mu$ doesn't change, however term $g^{\mu\nu}$ changes as $\widetilde g^{\mu\nu}=\lambda^{-2}g^{\mu\nu}$. And the $\phi$ scales the same way.

This gives the same result as you wrote in your question, but here it is more clear that we do actual physical transformation, not just change of coordinates.

In your case, you make conformal transformation as I just described it along with coordinate transformation $x \rightarrow \lambda x$. All together the fields transform as you wrote $dx^\mu \rightarrow \lambda dx^\mu, g_{\mu\nu}\rightarrow g_{\mu\nu}$ e.t.c., except that metric actually changes $\widetilde{ ds}^2 = \lambda^2 g_{\mu\nu} dx^\mu dx^\nu=\lambda^2 ds^2$.

  • $\begingroup$ Great! This is another side of my question. It seems that we can think of it like either changing the metric or coordinate but never both. If I regard it as a change of metric and the field contents, then how is it differ from Weyl symmetry? Thank you very much. $\endgroup$
    – user260822
    Commented Dec 29, 2015 at 9:51
  • $\begingroup$ Weyl transformation is a local change of metric and here we've been discussing the global one. $\endgroup$ Commented Dec 29, 2015 at 17:05
  • $\begingroup$ Conformal can be also local. Do you mean that they are the same? difference between conformal and weyl $\endgroup$
    – user260822
    Commented Dec 30, 2015 at 6:29

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