The title sums it pretty much. Are all diffeomorphism transformations also conformal transformations?

If the answer is that they are not, what are called the set of diffeomorphisms that are not conformal?

General Relativity is invariant under diffeomorphisms, but it certainly is not invariant under conformal transformations, if conformal transformations where a subgroup of diff, you would have a contradiction. Or I am overlooking something important?

  • $\begingroup$ Related. $\endgroup$
    – MannyC
    Commented Mar 19, 2020 at 9:50

2 Answers 2


A general diffeomorphism is not part of the conformal group. Rather, the conformal group is a subgroup of the diffeomorphism group. For a diffeomorphism to be conformal, the metric must change as,

$$g_{\mu\nu}\to \Omega^2(x)g_{\mu\nu}$$

and only then may it be deemed a conformal transformation. In addition, all conformal groups are Lie groups, i.e. with elements arbitrarily close to the identity, by applying infinitesimal transformations.

Example: Conformal Group of Riemann Sphere

The conformal group of the Riemann sphere, also known as the complex projective space, $\mathbb{C}P^1$, is called the Möbius group. A general transformation is written as,

$$f(z)= \frac{az+b}{cz+d}$$

for $a,b,c,d \in \mathbb{C}$ satisfying $ad-bc\neq 0$.

Example: Flat $\mathbb{R}^{p,q}$ Space

For flat Euclidean space, the metric is given by

$$ds^2 = dz d\bar{z}$$

where we treat $z,\bar{z}$ as independent variables, but the condition $\bar{z}=z^{\star}$ signifies we are really on the real slice of the complex plane. A conformal transformation takes the form,

$$z\to f(z)\quad \bar{z}\to\bar{f}(\bar{z})$$

which is simply a coordinate transformation, and the metric changes by,

$$dzd\bar{z}\to\left( \frac{df}{dz}\right)^{\star}\left( \frac{df}{dz}\right)dzd\bar{z}$$

as required to ensure it is conformal. We can specify an infinite number of $f(z)$, and hence an infinite number of conformal transformations. However, for general $\mathbb{R}^{p,q}$, this is not the case, and the conformal group is $SO(p+1,q+1)$, for $p+q > 2$.

  • 1
    $\begingroup$ I try to understand this intuitively. A diffeomorphism can arrange points(events) in spacetime arbitrarily. Take a diffeo close to the identity such that it can be written as a small tangent flow on each point. So what you are saying is that if the tangent flow is curly (is not a gradient of a scalar) it can't be written as a conformal close to the identity, which always have their tangent flow to be the gradient of some scalar. Is this correct? $\endgroup$ Commented Jun 4, 2014 at 17:39
  • 4
    $\begingroup$ General relativity is not invariant under diffeomorphisms. It is in general only invariant under isometries, which are conformal transformations with $\Omega^2 \equiv 1$. What people probably mean when they say that GR is diffemorphism invariant is that it is coordinate independent, but that's not a very useful statement. Any reasonable physical theory is coordinate independent. $\endgroup$ Commented Jun 17, 2014 at 23:44
  • 2
    $\begingroup$ @RobinEkman But any coordinate transformation can be viewed instead as an active transformation, a transformation promulgated by a diffeomorphism. How would you reconcile those statements? $\endgroup$
    – Muphrid
    Commented Jun 18, 2014 at 1:34
  • 1
    $\begingroup$ @Muprhid A coordinate transition function is a map between open sets $U,V \subset \mathbb R^n$. $U$ and $V$ are not naturally Lorentzian manifolds, but you can make them by pushing forward the metric from your actual Lorentzian manifold. Then the transition function is trivially an isometry. But since transition functions are generally not global, I think it is better to think of this fact as coordinate independence. If you consider global diffemorphisms $\phi : M \to N$ between Lorentzian manifolds that already have metrics, then e.g. the covariant derivative will only be conserved $\endgroup$ Commented Jun 18, 2014 at 7:31
  • 1
    $\begingroup$ if $\phi$ is an isometry. $\endgroup$ Commented Jun 18, 2014 at 7:31

I know topic this topic is 6 years old, but having just came across it, I feel the need to clarify some confusion around it, starting from the question, passing through the accepted answer and the ensuing comments. Most of this confusion sparks from two different but similar notions, that of a conformal transformatrion and that of a conformal isometry.

Let me clarify (I'm following Wald's General Relativity book, appendix D on conformal transformations): In the context of General Relativity (to which the OP is referring) a conformal transformation of parameter $\Omega:M\to \mathbb{R}$ ( $\Omega(p)>0\forall p$ ) (by other authors called a Weyl transformation) of a spacetime $(M,g)$ gives another spacetime $(M,g^\prime)$ where the metric $g^\prime$ is related to $g$ by $$g^\prime=\Omega^2 g$$ and theories whose predictions are invariant under this operation are said to be conformally invariant, or to posses Weyl symmetry. According to this definition, a conformal transformation has nothing to do with the group of diffeomorphisms $\phi:M\to M$, in fact whatever atlas (read set of coordinates systems) we use to cover $M$, this is not changed by the conformal transformation (please recall that a manifold $M$ is essentially defined to be topological space with an atlas covering it, while the metric of a spacetime is an additional structure imposed a manifold: changing the metric through a conformal transformation don't do anything to the manifold).

So if the OP is referring to a conformal/Weyl transformation, the answare to its question is: the group of conformal transformations of the metric and the group of diffeomorphisms are two different groups, and neither of the two is a subgroup of the other.

In particular notice that a given tensor $T$ transform under a conformal transformation of the metric only through its functional dependence on the metric: if $T=t(g)$ in the spacetime $(M,g)$, then in the spacetime $(M,g^\prime)$ we have $$T^\prime = t(g^\prime) = t(\Omega^2 g)$$ and the actual relation between $T$ and $T^\prime$ depends on the actual functional form of $t(g)$, rather than on the tensor structure of $T$ (as it would be the case if a conformal transformation was a particular diffeomorphism).

Different is the story of conformal isometries: a conformal isometry is indeed a special kind of diffeomorphism, one for which the the pullback of the metric at a point $p$ is given by $$\phi_\star g_{\phi(p)} = \Omega^2(p) g_p $$Being a conformal isometry a diffeomorphisms, it transforms all tensors according to their tensor structure. Relating to the OP question, the group of conformal isometries is a subgroup of the group of diffeomorphisms of M.

Regarding the last point in the OP question: General Relativity is indeed not "invariant under conformal transformations", but actually it is not even "invariant under diffeomorphisms" and neither it is "invariant under any isometries". Rather, its equations are tensorial in nature (they are in the form $T=0$ for some tensor $T$), therefor their validity is preserved under the action of all diffeomorphisms, including all isometries and all conformal isometries: $$\phi_\star T=0 \iff T=0$$ Now, according to the principle of general covariance, and if one disregard fermions, the metric $g$ is the only geometrical quantity that can enter the equations of general relativity, but this does not means that each of these equations is invariant under isometries: this is not true even of the Einstein field equation $$G=8\pi T$$ because the Einstein tensor $G$ and the stress energy tensor $T$ are rank-(0,2) tensors.

The last point I want to cover, is the confusion that emerged in the comments regarding the relation between diffeomorphisms and changes of coordinates. First, a diffeomorphisms $\phi:M\to M$ maps points of $M$ to other points of $M$ (active view point), and is defined to be an homeomorphism preserving the smooth structure of $M$. As such it sends charts and atlases of $M$ to other charts and atlases of $M$ in a smooth and invertible way (if $\varphi : U\subset M \to \mathbb{R}^k$ is a chart, $\phi|_U \circ \varphi$ in another one, etc), thus inducing a change of coordinates (passive view). On the other direction, a smooth change of coordinates is precisely one that sends an atlas covering $M$ to another one in a smooth and invertible manner, thus inducing a smooth and invertible map between points of $M$. So actually diffeomorphisms and coordinate changes are in 1-to-1 correspondence, provided that coordinate changes "do not forget to cover some points of $M$", that is they map atlases to atlases, not merely charts to charts. If a given coordinate change instead "forgets some points", this change of coordinate will only induce a local diffeomorphism.

  • 2
    $\begingroup$ The accepted answer is not wrong. The problem is that different textbooks mean completely different things when they say "conformal transformation". What Wald calls a conformal transformation is called a Weyl transformation in other books. $\endgroup$
    – knzhou
    Commented Mar 18, 2020 at 23:09
  • $\begingroup$ You are right. I've edited my answer to acknowledge that, and I've clarified some points. $\endgroup$
    – Marco
    Commented Mar 19, 2020 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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