locally conformally flat manifold - MathOverflow

locally conformally flat manifold

2011-05-13T06:42:59Z

A Riemannian manifold $(M,g)$ is locally conformally flat if it is locally conformal to $\mathbb{R}^n$ with the flat metric. I learn that Weyl tensor of a locally conformally flat manifold must vanish. I would like to ask: Is there any example of manifold $M$ such that it cannot be equipped with a metric $g$ with $(M,g)$ being locally flat? Is there any topological restriction on locally confomrally flat manifolds? Is there any classification theorem for locally conformally flat manifolds?

Answer by Dmitri for locally conformally flat manifold

2011-05-13T08:25:59Z

The simplest example is $S^n$, it is locally conformally flat with the standard metric, and is not flat for obvious reasons.

While flat manifolds are precisely quotients of $\mathbb R^n$ by discreet group of isometries, one should not expect to have a classification of conformally flat manifolds in higher dimensions. For example, already in dimension 4 it was proven by Kapovich in

M. Kapovich. Conformally ﬂat metrics on 4-manifolds. J. Diﬀerential Geom. 66 (2004), no. 2, 289–301,

that arbitrary finitely presented group can be a subgroup of a fundamental group of a conformally flat manifold.

The article of Kapovich is and from its introduction you will learn a lot on the question. 4-dimensional manifolds with LCF structure have zero signature, in dimension 3 it is known that some manfiolds don't admit conformally flat structure, first example was constructed in W. Goldman, Conformally flat manifolds with nilpotent holonomy, Transactions of AMS 278 (1983).

One more remark -- all hyperbolic manifolds (of constant negative sectional curvature) are all conformally flat. A connected sum of two conformally flat manifolds is conformally flat and so this already gives you a large collection of examples.

Answer by Robert Bryant for locally conformally flat manifold

2011-05-13T21:59:53Z

A simple obstruction is this: No compact (without boundary), simply-connected $n$-manifold that is not diffeomorphic to the $n$-sphere carries a locally conformally flat structure. The reason is that the developing map construction shows that any locally conformally flat metric $g$ on a simply-connected $M^n$ is, up to a conformal factor, the pullback of the standard metric on $S^n$ under some immersion $\phi:M^n\to S^n$. If $M$ is also compact without boundary, then $\phi$ is a covering map and, hence, a diffeomorphism.