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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?

diffeomorphism
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2 Answers2

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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 $p+q > 2$.

JamalS
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    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? – diffeomorphism Jun 04 '14 at 17:39
  • Wait.. this could not be right.. 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 – diffeomorphism Jun 17 '14 at 23:15
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    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. – Robin Ekman Jun 17 '14 at 23:44
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    @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? – Muphrid Jun 18 '14 at 01:34
  • @Muphrid: A coordinate transformation when viewed as an active transformation is a diffeomorphism between the same manifolds $M \to M$, so it is necessarily an isometry. So, a conformal transformation is more general than the diffeomorphism or coordinate transformations we encounter in GR. –  Jun 18 '14 at 04:51
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    @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 – Robin Ekman Jun 18 '14 at 07:31
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    if $\phi$ is an isometry. – Robin Ekman Jun 18 '14 at 07:31
  • @RobinEkman: how can you say "General relativity is not invariant under diffeomorphisms." ? Diffeos are a symmetry of the action. What do you mean? – BLS Dec 30 '15 at 06:46
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    @BLS because the usual argument is something like "the metric is then given by $g^* = \partial \phi \partial \phi , g$", i.e., the push-forward of $g$. But this is stipulating that $\phi$ is an isometry, not just a diffeomorphism. Under a general diffeomorphism $\phi$ between (pseudo-)Riemannian manifolds, there need not be a relation between the metrics. If you try to think of GR in a coordinate-independent, then what people usually mean when they say diffeomorphism invariance is coordinate-independence. But that's not unique to GR. – Robin Ekman Dec 30 '15 at 15:41
  • @RobinEkman I keep not understanding. Under change of coordinates (diffeos) $x'(x)$ the metric transforms as $g'{\mu\nu}(x') = \frac{\partial x^{\alpha}}{\partial x^{'\mu}} \frac{\partial x^{\beta}}{\partial x^{'\nu}} g{\alpha\beta}(x)$, and the EH action is invariant under this diffeo. – BLS Dec 30 '15 at 16:16
  • @BLS Change of coordinates and diffeos are NOT the same thing. Read my comment from Jun 18 '14 above. – Robin Ekman Dec 30 '15 at 16:20
  • @RobinEkman can you give an example of diffeo which is not a change of coordinate and a change of coordinate which is not a diffeo please? – BLS Jan 03 '16 at 16:54
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    @BLS Literally any diffeomorphism and literally any change of coordinates. Diffeomorphisms are maps between manifolds. Changes of coordinates are maps between open subsets of $\mathbb R^n$. They do not belong to the same classes of maps. You are not thinking in coordinate independent terms. – Robin Ekman Jan 03 '16 at 16:56
  • @RobinEkman Ok, that's right, but in physics we use to refer to change of coordinates as "diffeos", in a not very formal way. GR is indeed formulated in order to be coordinate independent. However, I think the main question here was "are conformal transformations a subset of changes of coordinates?" So the answer is yes, right? – BLS Jan 03 '16 at 17:03
  • Hi, @JamalS, I know it's a very old question, but could you explain why general relativity is not invariant under conformal transformations if it is supposed to be invariant under all diffeomorphisms? – Ayush Raj Sep 16 '22 at 18:26
  • @AyushRaj The important distinction is between active and passive transformations. General relativity is covariant with respect to coordinate transformations - if we relabel things (a passive transformation), everything should look the same. However under a diffeomorphism (an active transformation) we generate a new metric, and indeed if this new metric differs by a scalar function (making it a conformal map), it is not invariant. – JamalS Sep 19 '22 at 00:37
  • @AyushRaj It should be noted all of physics is generally covariant: it cannot be the case that a physical theory is valid in one specific coordinate system only. What distinguishes general relativity is that the background (spacetime) is a dynamical variable you solve for, whereas in other theories the background is something you set. The classical wave equation is just the Laplacian for Minkowski spacetime. – JamalS Sep 19 '22 at 00:42
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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.

Marco
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    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. – knzhou Mar 18 '20 at 23:09
  • You are right. I've edited my answer to acknowledge that, and I've clarified some points. – Marco Mar 19 '20 at 08:04