Thanks, I did gather this much from the WP page. I'm only considering the case where the function ##\varphi## is defined for te whole manifold, not just locally. Could you perhaps give me a counterexample either in three or more dimensions in which a non-constant curvature Riemannian manifold is conformally flat? Perhaps there is some obvious counterexample, checking the infinite possible non-constant curvatures looks like a daunting task.Ben Niehoff said:Conformally flat manifolds always have a metric tensor of the form
$$ds^2 = e^{2\varphi} (\delta_{ab} \, dx^a \, dx^b)$$
So from that you can learn all the necessary facts about the curvature. It should be easy for you to compute all the curvatures. It shouldn't be hard for you to show the following:
1. In 2d, all manifolds are (locally) conformally flat.
2. In 3d, a manifold is (locally) conformally flat if and only if the Cotton tensor vanishes.
3. In 4d and higher, a manifold is (locally) conformally flat if and only if the Weyl tensor vanishes.
(In 3d, the Weyl tensor vanishes identically).
Oops, my bad, you are right. But at least in the negative constant curvature case of hyperbolic space, it is not only locally conformally flat but also globally, then? (since there is no topologic obstruction, i.e. it is homeomorphic to ##\mathbb{R}^n##)Ben Niehoff said:If you want it to be globally conformally flat, then your constant-curvature assumption doesn't work in the first place. A sphere has constant curvature, but is not conformally flat.
Ok. Thanks. I take it you mean that the curvature remains nonpositive?In general, any manifold with positive curvature will probably be closed (there are theorems that involve more specific criteria, I think), so they can't be conformally flat. However, negative-curvature manifolds can generally avoid being closed.
So, I recommend doing the following: Take the ansatz I wrote for a conformally-flat metric, and work out its curvatures in terms of the function ##\varphi##. The Ricci scalar will be related to the Laplacian of ##\varphi##, for example. Next, just choose ##\varphi## to be any smooth function such that the curvature remains negative. You will get a smooth manifold which is probably homeomorphic to ##\mathbb{R}^n## (you may need to check for any non-trivial topology, but I think generically you won't have it). And then you'll have a non-constant-curvature manifold which is conformally flat.
RockyMarciano said:Oops, my bad, you are right. But at least in the negative constant curvature case of hyperbolic space, it is not only locally conformally flat but also globally, then? (since there is no topologic obstruction, i.e. it is homeomorphic to ##\mathbb{R}^n##)
Ok. Thanks. I take it you mean that the curvature remains nonpositive?
Conformal flatness is a property of a Riemannian manifold, which means that the metric tensor on the manifold can be locally transformed into a constant multiple of the Euclidean metric tensor through a conformal transformation.
Conformal flatness is a stronger condition than intrinsic flatness, as it requires the metric tensor to be transformed into a constant multiple of the Euclidean metric tensor through a conformal transformation, while intrinsic flatness only requires the metric tensor to be locally isometric to the Euclidean metric tensor.
Conformal flatness has several important consequences, such as simplifying the calculation of geodesics, making it easier to define and compute curvature, and allowing for the use of conformal invariants in studying the manifold.
Yes, a Riemannian manifold can be conformally flat and still have non-zero curvature. Conformal flatness only guarantees that the metric tensor can be transformed into a constant multiple of the Euclidean metric tensor, but does not necessarily mean that the curvature is zero.
Conformal flatness is closely related to conformal symmetry, as both involve conformal transformations. However, conformal flatness only requires the metric tensor to be transformed into a constant multiple of the Euclidean metric tensor, while conformal symmetry requires the metric tensor to be transformed into itself.