# Are diffeomorphisms a proper subgroup of conformal transformations?

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?

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 $p+q > 2$.

• 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 4 '14 at 17:39
• 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
• @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 1:34
• @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 7:31
• if $\phi$ is an isometry. – Robin Ekman Jun 18 '14 at 7:31