Massive spinless shell reduction is a single KKS orbit

In a rest frame $P=(mc,0)$, $W=0$ is equivalent to $J^{ij}=0$, since its three spatial components are an invertible multiple of the three spatial angular-momentum components. The remaining $J^{0i}$ are produced explicitly by $x^0=0$, $x^i=-J^{0i}\mathrm{/}(mc)$, since $J=x\wedge P$ then has exactly those components and $J^{ij}=0$. Lorentz transforming this representative gives a position representative for every future timelike $P$ with $W=0$. Equivalently for a transverse representative $x\cdot P=0$, $x^a=J^{ab}{P}_b\mathrm{/}{P}^2$, as substitution of $J=x\wedge P$ gives $J^{ab}{P}_b=x^a{P}^2-(x\cdot P)P^a$. This characterization is Lorentz covariant.

H8's quotient point $(a,p)$ maps to $P^0=E\mathrm{/}c$, $P^i=p_i$, $J^{0i}=-a^i{P}^0$, and $J^{ij}=a^i{p}_j-a^j{p}_i$. The inverse is $p_i=P^i$, $a^i=-J^{0i}\mathrm{/}{P}^0$, smooth since $P^0>0$. These mutually inverse smooth coordinates make the image an embedded six-dimensional graph and show the quotient map is a diffeomorphism onto the charge locus of step 1.1.

A proper orthochronous boost sends each future massive $P$ to rest. By step 1.1 its $W=0$ charge then has only $J^{0i}$ components, and a translation removes those components. Thus every pair in the locus is in the orbit of $(mc,0;0)$; conversely $P^2$ and $W=0$ are invariant under the coadjoint action. The stabilizer at rest consists of spatial rotations and translations parallel to $P$, dimension $3+1=4$, so the orbit dimension is $10-4=6$. The six smooth orbit coordinates $p_i=P^i$, $a^i=-J^{0i}\mathrm{/}{P}^0$ from step 2.1 have injective differential: if their differentials vanish, the graph formulas in step 2.1 make every component of $(P,J)$ have zero differential, and the coadjoint-orbit inclusion is an immersion. Since both manifolds have dimension six, the inverse function theorem makes this coordinate map a local diffeomorphism. The bijection in step 2.1 is therefore a global diffeomorphism for the canonical orbit smooth structure.

Let $\rho :{\mathrm{\Sigma }}_m^{+}\to B_m$ be the H8 quotient and $\iota :{\mathrm{\Sigma }}_m^{+}\hookrightarrow T^{\ast }{\mathbb{R}}^4$ the inclusion. H8 states ${\rho }^{\ast }{\omega }_{\mathrm{r}\mathrm{e}\mathrm{d}}={\iota }^{\ast }\omega$. The cotangent action of $g=(a,\mathrm{\Lambda })$ preserves this future shell: $P^{\mathrm{\prime }}=\mathrm{\Lambda }P$ preserves its norm and future cone, and it maps the full characteristic line $(x+\alpha P,p)$ to $(\mathrm{\Lambda }x+a+\alpha \mathrm{\Lambda }P,{\mathrm{\Lambda }}^{-T}p)$, which is exactly a characteristic line with momentum $P^{\mathrm{\prime }}$. The inverse group element gives the converse. Therefore the quotient action is well-defined. It is smooth because the global section $\sigma (a^i,p_i)$ given by $x^0=0$, $x^i=a^i$ is smooth and $g\cdot b=\rho (g\cdot \sigma (b))$ is its explicit expression; fibre preservation gives the group law independently of the chosen section. The cotangent action is symplectic by [L1], so pulling back along the surjective submersion $\rho$ gives ${\rho }^{\ast }(g^{\ast }{\omega }_{\mathrm{r}\mathrm{e}\mathrm{d}})=g^{\ast }{\iota }^{\ast }\omega ={\iota }^{\ast }\omega ={\rho }^{\ast }{\omega }_{\mathrm{r}\mathrm{e}\mathrm{d}}$, hence the quotient action preserves ${\omega }_{\mathrm{r}\mathrm{e}\mathrm{d}}$. Differentiating $\rho (\exp (-s\xi )\cdot z)=\exp (-s\xi )\cdot \rho (z)$ at $s=0$ gives $d\rho ({\xi }_{{\mathrm{\Sigma }}_m^{+}})={\xi }_{B_m}$ with the library's same fundamental-field convention. The moment components from [L1] are constant on each characteristic fibre since $P$ is fixed and $(x+\alpha P)\wedge P=x\wedge P$. Thus for any tangent $v$ on $B_m$ choose a lift $\tilde{v}$ and obtain $d{\bar{\mu }}^{\xi }(v)=d{\mu }^{\xi }(\tilde{v})=-{\omega }_{\mathrm{r}\mathrm{e}\mathrm{d}}({\xi }_{B_m},v)$. On the orbit [F2] gives the same moment equation. Its fundamental fields span each tangent space, so these equations determine both two-forms pointwise and show the diffeomorphism in step 3.1 pulls KKS back to ${\omega }_{\mathrm{r}\mathrm{e}\mathrm{d}}=d{a}^i\wedge d{p}_i$.