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 $[z\_1,-1]$).



Is there a smooth free circle action on the Klein bottle?
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. 


