15
$\begingroup$

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?

$\endgroup$
  • 5
    $\begingroup$ Small nitpick: the Weyl tensor is the obstruction to local conformal flatness only in dimension $> 3$. In dimension 3, the obstruction is measured by the Cotton-York tensor and in dimension 2 all surfaces are locally conformally flat due to the existence of isothermal coordinates. $\endgroup$ – José Figueroa-O'Farrill May 13 '11 at 9:32
18
$\begingroup$

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 and the uniformization problem for $3$-manifolds, 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.

$\endgroup$
18
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Does it follow from this that any locally conformally flat metric $g$ on a simply-connected closed manifold $M$ is "globally" conformally equivalent to a round sphere? (This statement, where "locally" and "globally" are a bit fuzzy, can be found in en.wikipedia.org/wiki/Conformally_flat_manifold, and I basically wanted to check that the conclusion is not only that the metrics are locally conformally equivalent, but actually conformally equivalent). In particular, if $g$ is a metric on $S^n$ not conformal to the round metric, then it is not locally conformally flat. $\endgroup$ – Renato G. Bettiol Sep 5 '13 at 22:15
  • 5
    $\begingroup$ @Renato: Yes, that is the correct conclusion. A simply-connected, closed (= compact without boundary) conformal $n$-manifold that is (locally) conformally flat is globally conformally equivalent to $S^n$. However, the above argument only applies when $n>2$. When $n=2$, the statement is true, but it doesn't follow from any developing map argument; when $n=2$, one has to appeal to the uniformization theorem. $\endgroup$ – Robert Bryant Sep 6 '13 at 8:11
1
$\begingroup$

Another topological obstruction by Kuiper,1950 is the following:

Universal cover of a compact, LCF space with an infinite Abelian fundamental group must be $\mathbb R^n$ or $\mathbb R \times \mathbb S^{n−1}$.

Using this, for example you can show that $\mathbb S^2\times \mathbb T^2$ does not admit any LCF metric. See the following for references and introduction to the subject:

Kalafat, M. - Locally conformally flat and self-dual structures on simple 4-manifolds. Proceedings of the Gökova Geometry-Topology Conference 2012, 111–122, Int. Press, Somerville, MA, 2013.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.