D=4 helicity and both continuous-spin central parities

The circles $SO(2)=\mathbb{R}\mathrm{/}(2\pi \mathbb{Z})$ and $\mathrm{S}\mathrm{p}\mathrm{i}\mathrm{n}(2)=\mathbb{R}\mathrm{/}(4\pi \mathbb{Z})$ are abelian. In an irreducible unitary representation of either circle, each operator commutes with the entire representation, so [F3] makes every operator scalar; the representation is therefore a one-dimensional continuous character. Pull it back to $\mathbb{R}$. By [F4] it is $e^{isr}$ for a unique real $s$. It descends to $SO(2)$ exactly when $e^{2\pi is}=1$, i.e. $s\in \mathbb{Z}$, and to $\mathrm{S}\mathrm{p}\mathrm{i}\mathrm{n}(2)$ exactly when $e^{4\pi is}=1$, i.e. $s\in \frac{1}{2}\mathbb{Z}$. Thus the zero-translation little-group labels in the Example are exhaustive.

For $\rho >0$, [F2] gives a single induced little-group representation for $E$, because its point stabilizer in $SO(2)$ is trivial. For the double cover the stabilizer is $\{\pm 1\}$, whose two one-dimensional characters have values $ϵ=+1$ and $ϵ=-1$ on $z$. Thus the two double-cover little representations are inequivalent and exhaust this orbit's unitary dual.

In $\mathrm{S}\mathrm{p}\mathrm{i}\mathrm{n}(2)=\mathbb{R}\mathrm{/}(4\pi \mathbb{Z})$ the element $z=[2\pi ]$ has order two. The character ${\theta }_s$ sends it to $e^{2\pi is}$, which is $+1$ for integer $s$ and $-1$ for noninteger half-integer $s$. Applying [F6] to the full Poincaré induction gives precisely the stated helicity descent outcomes.

Each little representation in step 1.2 is irreducible by [F2]. In a double-cover branch it is induced from $H={\mathbb{R}}^2⋊\{1,z\}$ with $\sigma ={\chi }_{\xi }\otimes ϵ$. The element $z$ is central in $\tilde{E}$ and belongs to $H$, so it fixes every quotient point; its Radon--Nikodym factor is $1$ almost everywhere. By [F7], $({\theta }_{\rho ,ϵ}(z)F)(x)=F(z^{-1}x)=F(x{z}^{-1})=\sigma (z^{-1}{)}^{-1}F(x)=ϵF(x)$, so $z$ acts on this intermediate little-group representation as $ϵI$ on its Hilbert completion. Now its induction to the full Poincare group is irreducible and positive-energy by [F5], and [F6] transfers this computed central value to that induction. Thus the $+1$ branch factors through the base group and the $-1$ branch fails the necessary identity condition. The base group has no cover kernel. This verifies both continuous-spin central parities and their descent.

The zero translation orbit ($\rho =0$) and the nonzero spheres ($\rho >0$) are disjoint and exhaust the dual by [F2]. Steps 1.1–2.2 therefore give the stated helicity and continuous-spin branches, including the distinct base and double-cover central actions.