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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?

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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. –  José Figueroa-O'Farrill May 13 '11 at 9:32
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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.

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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.

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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. –  Renato G Bettiol Sep 5 '13 at 22:15
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@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. –  Robert Bryant Sep 6 '13 at 8:11
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