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Is there a classification of all equivariant structures of the M?bius line bundle $\ell\to S^1$?.

For example the antipodal action of $\mathbb{Z}/2\mathbb{Z}$ on $S^1$ cannot be lifted to the total space $\ell$ to get an equivariant structure. But what about the general case?

In particular, let $\phi_{\theta}$ be the irrational rotation of the circle by $\theta$. Can the action of $\mathbb{Z}$ on $S^1$ given by $n.x=\phi_{\theta}^n(x)$ be lifted to an action on the total space of the M?bius bundle to give us an equivariant bundle?

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Most of my work is with actions of compact Lie groups, so forgive me if I make any errors below related to other groups.

I'm not sure how general a classification you're looking for, but I'm going to assume a framework in which I can say something: Consider a group $G$ acting on the base space $S^1$ by rotations, so the action is given by a group homomorphism $G\to S^1$ and the usual action of $S^1$ on itself. Can that action be extended to an action on the M?bius bundle $\mathscr l$ by bundle maps?

Consider the space of unit vectors in $\mathscr l$. This space is a copy of $S^1$ and the projection to the base space is the double cover $\delta\colon S^1\to S^1$, $\delta(e^{i\theta}) = e^{2i\theta}$. We can also think of $\delta$ as the principle $\mathbb Z/2\mathbb Z$-bundle associated to $\mathscr l$.

Proposition If $G$ acts on the base space via $f\colon G\to S^1$, then the action extends to $\mathscr l$ if and only if $f$ lifts to a homomorphism $\bar f\colon G\to S^1$ such that $\delta\bar f = f$.

Proof: For clarity, write $B$ for the base space of $\mathscr l$ and $E$ for the set of unit vectors, so that each is a copy of $S^1$ and the projection $E\to B$ is the double cover $\delta$. If $G$ acts on $\mathscr l$, with its action on $B$ being by rotations, then it clearly acts on $E$ by rotations as well, which will be given by a homomorphism $\bar f\colon G\to E$ such that $\delta\bar f = f$. Conversely, given such an $\bar f$, it defines an action of $G$ on $E$ covering the action on $B$, and this extends to an action of $G$ on $\mathscr l$ by bundle maps. QED

So then you would want to determine what homomorphisms $f$ lift up the double cover. I don't know a general answer, but I suspect that, if you restrict to monomorphisms $f$, then the answer is that this is the case iff $G$ contains no element of order 2. This is certainly the case for finite subgroups of $S^1$, but there may be topological issues for infinite subgroups. (Or I could be missing something blindingly obvious, always a possibility.)

On the other hand, we can answer your last question: The homomorphism $f_\theta\colon \mathbb Z \to S^1$ given by $f_\theta(n) = e^{2\pi in\theta}$, with $\theta$ irrational, does lift to $\bar f_\theta$, given by $\bar f_\theta(n) = e^{\pi in\theta}$. In fact, there's no need to assume that $\theta$ is irrational, this works for any $\theta$, it's just that $f$ won't be a monomorphism unless $\theta$ is irrational, so in that case we can't think of this as lifting the action of a subgroup.

Note that I'm playing a bit fast and loose with topologies here. The proposition above can be interpreted as dealing with continuous $f$ and a continuous lifting/action. But thinking about topologies, you have to remember that, when $\theta$ is irrational, $f_\theta$ is a monomorphism but not a homeomorphism onto its image, so you have to be careful thinking of this as a lifting of the action of a subgroup.

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  • $\begingroup$ Thank you But I thinkt some thing is missing in your answer. You wrote "in fact, there's no need to assume that θ is irrational". The antipodal action is a rotation, by $180$ degree, but this action can never introduce an equivariant structure $\endgroup$ – Ali Taghavi Mar 16 at 12:10
  • $\begingroup$ because when a group G act on X freely and $E\to X$ is G-equivariant then $E/G \to X/G$ is a vector bundle furthermore $q^*(E/G)=E$ where $q$ is the quotion map $q:X\to X/G$. Regarding the antpodal action of $\mathbb{Z}/2\mathbb{Z}$ on $S^1$, the quotion map behaves like $z\mapsto z^2$. But it is easy to prove that the pull back of every line bundle on $S^1$, under the square map, is trivial bundle, a contradiction, since Mobious is non trivial.Am I missing some thing?does your argument really work? $\endgroup$ – Ali Taghavi Mar 16 at 12:22
  • $\begingroup$ $f_{1/2}$, which would be rotation by 180 degrees, is not a free action of $\mathbb Z$. It would be a free action of $\mathbb Z/2$, but that action won't lift. $\endgroup$ – Steve Costenoble Mar 16 at 14:08
  • $\begingroup$ Put another way: The antipodal action of $\mathbb Z/2$ on the base space $B$ "lifts" to an action of $\mathbb Z/4$ on $E$, with $\mathbb Z/4$ acting on $B$ via the surjection $\mathbb Z/4\to \mathbb Z/2$. The actions of $\mathbb Z$ come from its surjections onto $\mathbb Z/4$ and $\mathbb Z/2$. $\endgroup$ – Steve Costenoble Mar 16 at 14:13
  • $\begingroup$ Yes I understand thank you very much for your answer. $\endgroup$ – Ali Taghavi Mar 16 at 23:58

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