Lift of finite group action on circle to double cover

Question

Suppose we have a finite group G acting on the circle. The circle has a double cover, corresponding to the subgroup $2\mathbb{Z}$ of the fundamental group $\mathbb{Z}$ of the circle. We can lift the group action to the double cover, obtaining an an action of some central extension $\tilde{G}$ of G by $\mathbb{Z}_2$. My question is: when is $\tilde{G}$ a direct product $\tilde{G}\simeq G\times\mathbb{Z}_2$?

Answer 1

For each $g \in G$ you have given me a diffeomorphism $\rho_g: S^1 \to S^1$. Now choose a lift $\overline \rho_g: S^1 \to S^1$ so that $\rho_g(z) = [\overline \rho_g(z^2)]$, that is, $\overline \rho_g$ is a double cover of $\rho_g$.

This gives you a 2-cocycle in $H^2(G; \Bbb Z/2)$, by $f(g,h) = \overline \rho_g \overline \rho_h (\overline \rho_{gh})^{-1}$. By the assumption that the $\overline \rho_g$ double-cover $\rho_g$, and the fact that $\rho_g$ is a group action, this expression reduces (in the quotient) to the identity. It follows that this expression is either the identity or the antipodal map, which we can think of as being an element of $\Bbb Z/2$. It is straightforward to check that this is indeed a cocycle.

Your central extension is trivial iff this group cohomology class is zero.

In many circumstances you can guarantee that this group cohomology class is zero simply because that cohomology group is zero. This is true automatically for any group of odd order (for which an extension with fiber $\Bbb Z/2$ automatically splits as a product, central or no). Sometimes you may find that this group is zero by some explicit calculation

Assuming $G$ is finite, the universal coefficient theorem gives a homological condition that is necessary and sufficeient for $H^2(G;\Bbb Z/2)$ to be zero is as follows.

You require both i) $G^{\text{ab}}$ has no 2-torsion. ii) The Schur multiplier, which may be calculated explicitly from a presentation due to work of Hopf, is odd order.

If $G$ has finitely generated homology groups, (but is not necessarily finite), then the first condition is unchanged but the second condition is replaced by: "The Schur multiplier is finite of odd order"; finiteness is no longer automatic.

For general $G$ you must write the conditions less explicitly: $\text{Ext}(G^{\text{ab}}, \Bbb Z/2) = 0$ and $\text{Hom}(H_2(G;\Bbb Z), \Bbb Z/2) = 0$.