# 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?

## 1 Answer

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