Is there any example of a locally conformally flat manifold that is neither a space form nor a product of space forms?
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2$\begingroup$ I don't think that products of space forms are locally conformally flat, except for products of flat space forms. $\endgroup$– Ben McKayCommented Apr 27 at 10:35
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2$\begingroup$ These aren't the only conformally flat products of space forms (the word "locally" is usually omitted). Any Fuchsian group acting on the hyperbolic $n$-space gives at infinity a conformally flat manifold diffeomorphic to $S^k\times M$ where $M$ is a hyperbolic $(n-k-1)$ manifold. By the way, one can take quotients of this by free finite group actions of hyperbolic isometries, e.g. any closed $3$-manifold modeled on $H^2\times \mathbb R$ is conformally flat. $\endgroup$– Igor BelegradekCommented Apr 27 at 12:28
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The connected sum of two locally conformally flat manifolds can be equipped with a metrics which is locally conformally flat. To see this, recall that the product metric on $\mathbb{R} \times S^{n-1}$ is (globally) conformal to the flat metric on $\mathbb{R}^n \setminus \{ 0 \}$.
Metrics constructed this way are genuinely different from space forms or products of space forms (one of sectional curvature $+1$ and one of sectional curvature $-1$; or one which is one-dimensional and one which is not). For example, in dimension four any closed spaceform has nonnegative Euler characteristic. The connected sum $(S^1 \times S^3) \# (S^1 \times S^3)$ has negative Euler characteristic, and so is not a space form. The only compact space forms with negative Euler characteristic are (quotients of) $S^2 \times \Sigma_g^2$, where $\Sigma_g^2$ is a compact surface with genus $g \geq 2$. But the fundamental group of $(S^1 \times S^3) \# (S^1 \times S^3)$ is the free product $\mathbb{Z} \ast \mathbb{Z}$, which is not the fundamental group of any $\Sigma_g^2$.
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Any open parallelizable $n$-manifold immerses into $\mathbb R^n$ (this is special case of by Hirsch-Smale h-principle), and hence is conformally flat.
There are many examples of conformally flat manifolds arising from Kleinian groups, see the survey by Kapovich, https://arxiv.org/abs/math/0701370 and references there, and an older survey Foundations of Flat Conformal Structure by Matsumoto.
A fun $4$-dimensional construction: Kapovich proved that every closed smooth spin 4-manifold appears as a connected summand of a conformally flat manifold, https://arxiv.org/abs/math/0210013. A version of this is known in dimension $3$.
There are many examples of conformally flat manifolds in dimension 3, see the above survey by Kapovich for references. Notably, it is still unknown which $\widetilde{SL}_2(\mathbb R)$ Seifert $3$-manifolds are conformally flat.
answered Apr 27 at 13:24