A D=4 continuous-spin little representation has nonzero translations

The vector ${\xi }_{\rho }\mathrm{\ne }0$, so the dual orbit $S_{\rho }=\{\eta \in {\mathbb{R}}^2:\mathrm{\mid }\eta \mathrm{\mid }=\rho \}$ does not contain zero. By [F1], ${\theta }_{\rho }$ is the irreducible induction from ${\chi }_{{\xi }_{\rho }}$ and, in the induced model, translation by $a\in {\mathbb{R}}^2$ multiplies over $\eta \in S_{\rho }$ by $e^{i\eta \cdot a}$.

Choose $a=(\pi \mathrm{/}\rho )e_1$. At $\eta ={\xi }_{\rho }$ the multiplier is $e^{i{\xi }_{\rho }\cdot a}=e^{i\pi }=-1$. It remains different from $1$ on an open arc around ${\xi }_{\rho }$ by continuity. The orbit base is the circle $SO(2)$, and the full-support measure in [F2] assigns positive measure to this arc. Thus the translation operator is not the identity, proving that ${\mathbb{R}}^2$ acts nontrivially.

Choose any nonzero future null $p_{+}$ and the little-group representation ${\theta }_{\rho }$. It is irreducible by [F1], so [F3] gives an irreducible positive-energy Poincaré induction. Its little-group translations remain nontrivial by step 2.1. Hence the example lies in the continuous-spin branch with nonzero null-little-group translations; it cannot be replaced by a helicity representation, for which those translations are trivial.