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Can the circle group $S^1$ act smoothly and freely on the Klein bottle? I'm sure there is some obvious reason why the answer is no, which eludes me right now.

We can view $K$ as the quotient of $S^1\times S^1\subset\mathbb{C}\times\mathbb{C}$ by the involution $(z_1,z_2)\to (-z_1,z_2^{-1})$. Then we get an almost-free $S^1$-action $(z,[z_1,z_2])\to [zz_1,z_2]$ with $\mathbb{Z}_2$ isotropy ($-1$ fixes the circles $[z_1,1]$ and $[z_1,-1]$).

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up vote 9 down vote accepted

No. If you had a smooth free action, the quotient would be a compact connected 1-manifold, so a circle. So the Klein bottle would be an orientable circle bundle over the circle, but there's only one and that is a torus.

So the tools I'm using are (1) when the quotient of a manifold by a free action of a compact Lie group is another manifold, (2) classification of 1-manifolds, (3) classification of circle bundles.

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Does it suffice to observe that an $SO(2)$-bundle over a circle is necessarily an orientable manifold? – Charles Rezk Jan 13 '11 at 19:35
Since the action is by diffeomorphisms I don't see a way of avoiding the argument that $Diff^+(S^1)$ has the homotopy-type of the subgroup $SO_2$. So you do have to make a structure group reduction. I suppose you could avoid this by putting a Riemann metric on the Klein bottle so that all the circle orbits had the same length -- but this is essentially the same argument. – Ryan Budney Jan 13 '11 at 19:41
The quotient map is a principal homogeneous $S^1$-space (local triviality comes from an invariant metric for instance) so the structure group is already reduced to $S^1$. – Torsten Ekedahl Jan 13 '11 at 21:56
Thanks! The sticking point for me was realising that a principal $S^1$-bundle must be orientable (since the action on the fibres is left multiplication). – Mark Grant Jan 14 '11 at 8:57

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