Central kernel characters control descent of Poincaré inductions

A covering homomorphism has discrete kernel by [F2]. For fixed $\gamma \in \ker c$ the map $k\mapsto k\gamma k^{-1}$ is continuous from connected $K$ into that discrete kernel; it is therefore constant, equal to $\gamma$. Thus $\ker c$, and hence $\mathrm{\Gamma }$, is central in $K$. It acts trivially on translations because $c(\gamma )=1$, so $(0,\gamma )$ is central in $G$. Since $\gamma$ maps to the identity Lorentz transformation, it fixes $p$ and lies in $K_p$.

In the induced model from [F4]–[F5], centrality gives $g_{\gamma }^{-1}x=x{g}_{\gamma }^{-1}$ for $g_{\gamma }=(0,\gamma )$. Since $\gamma \in K_p$, this action fixes every point of $G\mathrm{/}{H}_p$, so it preserves the measure and its Radon–Nikodym derivative is $1$ almost everywhere. The density is continuous by [F5] and the measure has full support by [F7]; if the density differed from $1$ at a point, continuity would give a nonempty open set where it differed, which has positive measure, a contradiction. Thus the cocycle factor is $1$ everywhere. Covariance then yields $(U(g_{\gamma })F)(x)=F(x{g}_{\gamma }^{-1})=({\chi }_p\otimes \theta )(g_{\gamma }^{-1}{)}^{-1}F(x)=\theta (\gamma )F(x)\mathrm{.}$ Hence $U(g_{\gamma })$ is the constant fibre operator $\theta (\gamma )$. Since $\gamma$ is central in $K_p$, $\theta (\gamma )$ commutes with the irreducible representation $\theta$; [F6] says it is scalar. AC is used here by the induction-action and Schur suppliers.

A representation factors through $G\mathrm{/}\{(0,\gamma ):\gamma \in \mathrm{\Gamma }\}\cong \bar{G}$ exactly when every element of this kernel acts as the identity: necessity follows by applying a factored representation to the identity coset, and sufficiency follows by defining the quotient action on cosets, which is well-defined precisely when the kernel acts trivially. By step 2.1 this is equivalent to $\theta (\gamma )=I$ for every $\gamma \in \mathrm{\Gamma }$.

For $D\ge 4$, the only nontrivial connected-cover kernel is the order-two deck group $\{1,z\}$ by [F3]. Since $z^2=1$, the scalar $\theta (z)$ from step 2.1 satisfies $\theta (z{)}^2=I$, hence is $+I$ or $-I$. Applying step 3.1 to $\mathrm{\Gamma }=\{1,z\}$ proves the stated descent dichotomy.

In $D=3$, [F3] gives $z$ as rotation coordinate $2\pi$ in the massive universal-cover stabilizer $\mathbb{R}$ and as generator $(0,1)$ in the null stabilizer $\mathbb{R}\times \mathbb{Z}$. The massive character therefore has ${\theta }_s(z^N)=e^{2\pi iNs}$, which equals $1$ exactly when $Ns\in \mathbb{Z}$. The null character has ${\theta }_{t,\alpha }(z^N)=e^{2\pi iN\alpha }$, which equals $1$ exactly when $N\alpha \in \mathbb{Z}$; the parameter $t$ does not enter because $z$ has zero coordinate in the real null-translation factor. Step 3.1 gives both iff statements.

At zero momentum, $K_p=K$ and the inducing representation is an arbitrary irreducible $\sigma$ of $K_{\mathrm{\infty }}$. Applying step 3.1 with $\mathrm{\Gamma }=⟨z^N⟩$ gives descent exactly when $\sigma (z^N)=I$. In $D=2$ the Lorentz group has no nontrivial connected cover kernel, so there is no additional descent condition. These cases exhaust the stated consequences of the general criterion.