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  Are diffeomorphisms a proper subgroup of conformal transformations?

+ 1 like - 0 dislike

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?

This post imported from StackExchange Physics at 2014-06-06 20:08 (UCT), posted by SE-user diffeomorphism
asked Jun 4, 2014 in Theoretical Physics by diffeomorphism (5 points) [ no revision ]
Does noone else want to give a +1 to @diffeomorphism for asking about clarification on the scope of the definition of diffeomorphism

This post imported from StackExchange Physics at 2014-06-06 20:08 (UCT), posted by SE-user Flint72

1 Answer

+ 3 like - 0 dislike

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 $p+q > 2$.

This post imported from StackExchange Physics at 2014-06-06 20:08 (UCT), posted by SE-user JamalS
answered Jun 4, 2014 by JamalS (895 points) [ no revision ]
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?

This post imported from StackExchange Physics at 2014-06-06 20:08 (UCT), posted by SE-user diffeomorphism

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