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Given a compact Lie group G, acting freely on a topological manifold M. Is it true, that the orbitspace M/G is also a topological manifold and why?

flag	asked Dec 14 at 22:02 Janesis 83●2

Misha · Accepted Answer · 2012-12-15 02:01:09Z UTC

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The answer is no. Bing constructed a space $X$ (called the dogbone space) so that $X$ is not a manifold, but $X\times R$ is homeomorphic to $R^4$. In particular, $M^4=X\times S^1$ is a $4$-manifold (since its universal cover is $R^4$) and $X$ is the quotient of $M^4$ by free $S^1$-action.

answered Dec 15 at 2:01

Misha
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So why isn't $X\times R$ a manifold? ;-) – David Roberts Dec 15 at 3:35

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@David: The whole point is that $X\times R$ is a manifold ($R^4$ is a manifold last time I checked), but the quotient space $X$ is not a manifold. To find out why $X$ is not a manifold you can read Bing's paper "The cartesian product of a certain nonmanifold and a line is $E^4$", Annals of Math., 1959. – Misha Dec 15 at 3:50

I know, I just forgot the OP asked for the action of a compact group... – David Roberts Dec 15 at 9:04

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More examples of non-manifolds whose products with $\mathbb R$ are manifolds can be found on pp.33-34 of Dusan Repovs's slides at math.utk.edu/~dydak/00STS2012/Repovs7.pdf. It also has has some visualization of the dogbone space. – Igor Belegradek Dec 15 at 15:49

orbitspace of a topological manifold

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