TITLE: Is there a smooth free circle action on the Klein bottle?
QUESTION [6 upvotes]: 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 $[z_1,-1]$).

REPLY [11 votes]: 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.