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Given a compact Lie group G acting freely on a topological manifold M, is it true that the orbit space M/G is also a topological manifold? If so, why?

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up vote 14 down vote accepted

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.

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So why isn't $X\times R$ a manifold? ;-) –  David Roberts Dec 15 '12 at 3:35
@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 '12 at 3:50
I know, I just forgot the OP asked for the action of a compact group... –  David Roberts Dec 15 '12 at 9:04
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 '12 at 15:49

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