D=4 massive spin one-half and spin one descent

The standard $SU(2)$ action on ${\mathbb{C}}^2$ is unitary and smooth; its tensor-square action is also unitary and smooth, and its symmetric subspace $W_2={\mathrm{Sym}}^2({\mathbb{C}}^2)$ is invariant. Hence both restricted actions ${\theta }_1$ and ${\theta }_2$ are strongly continuous unitary representations. The standard action on $W_1={\mathbb{C}}^2$ is irreducible: for every unit vector $(u,v)$, the matrix $\left(\begin{array}{cc}u& -\bar{v}\\ v& \bar{u}\end{array}\right)\in SU(2)$ sends $e_1$ to $(u,v)$, so $SU(2)$ is transitive on the unit sphere and any nonzero invariant subspace is all of $W_1$. For $W_2$, use the basis $e_1\odot e_1,e_1\odot e_2,e_2\odot e_2$. The torus element $t=\mathrm{diag}(e^{i\pi \mathrm{/}6},e^{-i\pi \mathrm{/}6})$ acts with distinct eigenvalues $e^{i\pi \mathrm{/}3},1,e^{-i\pi \mathrm{/}3}$. If a nonzero invariant subspace contains a vector with several weight terms, Lagrange interpolation polynomials in ${\theta }_2(t)$ project onto its nonzero weight components, so it contains at least one basis vector. A group-invariant subspace is invariant under the differentiated $\mathfrak{s}\mathfrak{u}(2)$ action because it is closed and contains each group orbit; complex-linear extension makes it invariant under $E_{12}$ and $E_{21}$ in ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$. These operators send $e_2\odot e_2\mapsto 2e_1\odot e_2\mapsto 2e_1\odot e_1$ and $e_1\odot e_1\mapsto 2e_1\odot e_2\mapsto 2e_2\odot e_2$ (with nonzero coefficients), and from the middle vector they reach both endpoints. Thus every nonzero invariant subspace is all of $W_2$. The torus weights are $z,z^{-1}$ for $W_1$ and $z^2,1,z^{-2}$ for $W_2$, which give the spin labels $1\mathrm{/}2$ and $1$ under the convention in the Example.

On $W_1={\mathbb{C}}^2$, $z=-I$ acts as $-I$. On the whole tensor square, it acts as $(-I)\otimes (-I)=I$, so its restriction to $W_2$ is also $I$. The kernel of $SU(2)\to SO(3)$ is exactly $\{I,z\}$ by [F3].

The momentum $p_m$ is future timelike, hence is in $C_{+}$. By [F1], each ${\theta }_{m_0}$ gives an irreducible positive-energy induction. Applying [F4] to the central kernel $\{I,z\}$, the spin-$\frac{1}{2}$ induction fails the identity condition and cannot descend; the spin-$1$ induction satisfies it and descends to the ordinary Poincaré group.