# Classification of all equivariant structure on the Möbius line bundles

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?

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.

• 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 – Ali Taghavi Mar 16 at 12:10
• 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? – Ali Taghavi Mar 16 at 12:22
• $f_{1/2}$, 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. – Steve Costenoble Mar 16 at 14:08
• 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 $\mathbb Z/2$. – Steve Costenoble Mar 16 at 14:13
• Yes I understand thank you very much for your answer. – Ali Taghavi Mar 16 at 23:58