Projected angular parts are not two independent rotation algebras

The domain $\mathcal{D}$ is invariant under each $J_i$ and $S_i^{\mathrm{\prime }}$ by their smooth differential and multiplication formulas. Write $X_i=(e_i\times \mathbf{k})\cdot {\mathrm{\nabla }}_{\mathbf{k}}$ and $C_{i}v=e_i\times v$, so $L_i=-i\mathrm{\hslash }{X}_i$ and $\mathrm{\hslash }{S}_i=i\mathrm{\hslash }{C}_i$. The product rule for the linear vector fields gives $[X_i,X_j]f=(e_j\times (e_i\times \mathbf{k})-e_i\times (e_j\times \mathbf{k}))\cdot \mathrm{\nabla }f=-{\sum }_k{ϵ}_{ijk}{X}_{k}f$. The vector triple-product identity gives $[C_i,C_j]={\sum }_k{ϵ}_{ijk}{C}_k$, while constant matrices commute with the scalar differential coefficients, so $[L_i,\mathrm{\hslash }{S}_j]=0$. Therefore $[L_i,L_j]=i\mathrm{\hslash }{\sum }_k{ϵ}_{ijk}{L}_k$ and $[\mathrm{\hslash }{S}_i,\mathrm{\hslash }{S}_j]=i\mathrm{\hslash }{\sum }_k{ϵ}_{ijk}\mathrm{\hslash }{S}_k$. Adding these identities yields $[J_i,J_j]=i\mathrm{\hslash }{\sum }_k{ϵ}_{ijk}{J}_k$ on $\mathcal{D}$.

Since $J_i$ is self-adjoint and $-S_i^{\mathrm{\prime }}$ is bounded self-adjoint, it is $J_i$-bounded with relative-bound coefficient $a=0<1$ and finite constant $b=\mathrm{\parallel }{S}_i^{\mathrm{\prime }}\mathrm{\parallel }$. Apply [F4] to $A=J_i$ and $B=-S_i^{\mathrm{\prime }}$ to conclude $L_i^{\mathrm{\prime }}=J_i-S_i^{\mathrm{\prime }}$ is self-adjoint on exactly $D(J_i)$. For every $\psi \in D(J_i)$, $\mathrm{\parallel }{L}_i^{\mathrm{\prime }}\psi \mathrm{\parallel }\le \mathrm{\parallel }{J}_i\psi \mathrm{\parallel }+\mathrm{\hslash }\mathrm{\parallel }\psi \mathrm{\parallel }$ and $\mathrm{\parallel }{J}_i\psi \mathrm{\parallel }\le \mathrm{\parallel }{L}_i^{\mathrm{\prime }}\psi \mathrm{\parallel }+\mathrm{\hslash }\mathrm{\parallel }\psi \mathrm{\parallel }$. These inequalities make the two graph norms equivalent. Hence [F3] implies that $\mathcal{D}$ is also a core for $L_i^{\mathrm{\prime }}$.

The operators $S_i^{\mathrm{\prime }}$ are multiplication by the commuting matrix fields $\mathrm{\hslash }{n}_{i}h(\mathbf{n})$. Thus on every transverse fiber $S_i^{\mathrm{\prime }}{S}_j^{\mathrm{\prime }}={\mathrm{\hslash }}^2{n}_i{n}_j{h}^2=S_j^{\mathrm{\prime }}{S}_i^{\mathrm{\prime }}$, so $[S_i^{\mathrm{\prime }},S_j^{\mathrm{\prime }}]=0$ as bounded operators, in particular on $\mathcal{D}$.

For $R\in SO(3)$, the cross-product identity $(R\mathbf{n})\times (Rv)=R(\mathbf{n}\times v)$ gives $P(R\mathbf{n})=RP(\mathbf{n})R^{-1}$ and $h(R\mathbf{n})=Rh(\mathbf{n})R^{-1}$. Also $R{S}_j{R}^{-1}={\sum }_m{R}_{mj}{S}_m$, since rotations carry $e_j$ to ${\sum }_m{R}_{mj}{e}_m$. Substituting these identities in $S_j^{\mathrm{\prime }}(\mathbf{n})=\mathrm{\hslash }P(\mathbf{n})S_{j}P(\mathbf{n})$ yields $U(R)S_j^{\mathrm{\prime }}U(R{)}^{\ast }={\sum }_m{R}_{mj}{S}_m^{\mathrm{\prime }}$. For $R=R_{e_i}(t)$, differentiating the right side at zero gives ${\sum }_m{ϵ}_{ijm}{S}_m^{\mathrm{\prime }}$. For $\psi \in \mathcal{D}$, both $\psi$ and $S_j^{\mathrm{\prime }}\psi$ lie in $D(J_i)$ by [F1]–[F2], so the product rule for the left side gives ${\frac{d}{dt}\mid }_{0}U(R_{e_i}(t))S_j^{\mathrm{\prime }}U(R_{e_i}(t){)}^{\ast }\psi =-\frac{i}{\mathrm{\hslash }}[J_i,S_j^{\mathrm{\prime }}]\psi$. Equating the two derivatives proves $[J_i,S_j^{\mathrm{\prime }}]=i\mathrm{\hslash }{\sum }_m{ϵ}_{ijm}{S}_m^{\mathrm{\prime }}$ on $\mathcal{D}$.

Using $L_i^{\mathrm{\prime }}=J_i-S_i^{\mathrm{\prime }}$ and invariance of $\mathcal{D}$, expand on $\mathcal{D}$: $[L_i^{\mathrm{\prime }},S_j^{\mathrm{\prime }}]=[J_i,S_j^{\mathrm{\prime }}]-[S_i^{\mathrm{\prime }},S_j^{\mathrm{\prime }}]=i\mathrm{\hslash }{\sum }_k{ϵ}_{ijk}{S}_k^{\mathrm{\prime }}$, and $[L_i^{\mathrm{\prime }},L_j^{\mathrm{\prime }}]=[J_i,J_j]-[J_i,S_j^{\mathrm{\prime }}]-[S_i^{\mathrm{\prime }},J_j]+[S_i^{\mathrm{\prime }},S_j^{\mathrm{\prime }}]=i\mathrm{\hslash }{\sum }_k{ϵ}_{ijk}(J_k-2S_k^{\mathrm{\prime }})=i\mathrm{\hslash }{\sum }_k{ϵ}_{ijk}(L_k^{\mathrm{\prime }}-S_k^{\mathrm{\prime }})$. Here $[S_i^{\mathrm{\prime }},J_j]=-[J_j,S_i^{\mathrm{\prime }}]=i\mathrm{\hslash }{\sum }_k{ϵ}_{ijk}{S}_k^{\mathrm{\prime }}$ because ${ϵ}_{jik}=-{ϵ}_{ijk}$. Along with step 1.3 and step 1.1, these are all stated identities. The term $-S_k^{\mathrm{\prime }}$ in the orbital commutator prevents $L_i^{\mathrm{\prime }}$ from satisfying an independent $\mathfrak{s}\mathfrak{o}(3)$ algebra; the commuting $S_i^{\mathrm{\prime }}$ also do not satisfy the nonzero $\mathfrak{s}\mathfrak{o}(3)$ brackets.