Let $M$ be a manifold of dimension $n>3$ and $g$ a Riemannian metric on $M$ which is conformally equivalent to a flat one. Are there formulas (different from $R(u,v)w=\nabla_u\nabla_vw-\nabla_v\nabla_uw-\nabla_{[u,v]}w$) that allow to compute easily the curvature tensor of a conformally flat manifold such as $(M,g)$?
[Math] Curvature tensor of a conformally flat manifold
differential-geometryriemannian-geometry
Related Solutions
This is an old question but it deserves a correct answer. As it turns out, the open 2-dimensional disk admits continuum of conformally inequivalent pseudo Riemannian metrics, see
Uncountably many $C^0$ conformally distinct Lorentz surfaces and a finiteness theorem, by Robert W. Smyth, Proc. Amer. Math. Soc. 124 (1996), 1559-1566.
Edit. Furthermore, there are Lorentz metrics on the open 2-disk which do not embed conformally in the Lorentzian plane, see p. 117 of
T. Weinstein, An Introduction to Lorentz Surfaces, de Gruyter, 1996.
Last thing: In his answer Luboš Motl confused the Riemannian and pseudo Riemannian cases.
To the aside, "flat" and "locally flat" are often used interchangeably to refer to your definition, but there is also a stronger condition of "global flatness" which requires all holonomy groups to be trivial (or, equivalently, the existence of a parallel global frame). Throughout, I'll talk only about the local case, so every statement need only hold locally.
Note that a metric $g$ has vanishing curvature $R_g=0$ if and only if it is locally the pullback of a Euclidean metric in the manner you describe. Either of this could serve as the definition of flatness, but I'll use only the former condition for simplicity.
By definition, a metric $g$ is conformally flat if there exists a conformally related metric $\tilde{g}=e^{2f}g$ with $R_{\tilde g}=0$. This does not mean that $R_g=0$, since conformally related metrics do not have the same Riemann curvature. The Ricci and scalar curvature will also differ in general (see here), but the Weyl tensor will be the same.
Using the example of the stereographic projection $\sigma:S^n\setminus\{p\}\to\mathbb{R}^n$, we have the round metric $\mathring{g}$ and the pullback of the Euclidean metric $\widetilde{g}=\sigma^*\delta$ on $S^n\setminus\{p\}$, which are conformally related. The pullback metric has vanishing curvature $R_{\widetilde{g}}=0$, but the round metric has curvature $\mathring{R}_{abcd}=\mathring{g}_{ac}\mathring{g}_{bd}-\mathring{g}_{ad}\mathring{g}_{bc}$
Best Answer
A good place to start is the wikipedia article which describes how various curvatures transform under conformal change a Riemannian metric. Now, apply the formula for the transformation of the full curvature tensor to your situation, taking into account that initially you have a flat metric.