Orbital self-adjoint generators

[(a), the complete diagonal domain and adjoint] Write $c_{lm}=⟨F,Y_{lm}⟩$. On the domain of [F2], Parseval defines $AF={\sum }_{l,m}\mathrm{\hslash }m{c}_{lm}{Y}_{lm}$ with convergence in norm, and finite sums prove symmetry. If $G\in D(A^{\ast })$ and $K=A^{\ast }G$, testing $F=Y_{lm}$ in $⟨AF,G⟩=⟨F,K⟩$ gives $⟨K,Y_{lm}⟩=\mathrm{\hslash }m⟨G,Y_{lm}⟩$. Parseval then implies $\sum {\mathrm{\hslash }}^2{m}^2\mathrm{\mid }⟨G,Y_{lm}⟩{\mathrm{\mid }}^2=\mathrm{\parallel }K{\mathrm{\parallel }}^2<\mathrm{\infty }$: $G\in D(A)$ and $K=AG$. Conversely, for such $G$, the equality holds first on finite sums and then for all $F\in D(A)$ by their coefficient truncations. Thus $A=A^{\ast }$ on exactly the full domain of [F2], and $A=L_z$. For $F\in D(A)$, both $P_{\le N}F\to F$ and $A{P}_{\le N}F\to AF$ by the two convergent coefficient square sums; hence $H$ is a graph core. All sums use the fixed countable index family; no fiber selection or uncountable basis choice occurs.

[(a),(d), adjoints and cores of the conjugates] If $Q$ is either $U(S)$ or $U(W)$, define $B=Q^{-1}AQ$ on $Q^{-1}D(A)$. For $G,K$, the adjoint identity $⟨BF,G⟩=⟨F,K⟩$ for every $F\in Q^{-1}D(A)$ is, on putting $X=QF$, exactly $⟨AX,QG⟩=⟨X,QK⟩$ for every $X\in D(A)$. By step 1.1 this holds exactly when $QG\in D(A)$ and $QK=AQG$. Thus $B^{\ast }=B$ with this full domain. Since $QH=H$, unitary transport of the graph approximations in step 1.1 gives $H$ as a core of $B$. With $L_3=A$, this proves self-adjointness, cores and the two stated conjugation/domain identities.

[(a), explicit groups and the full derivative domains] Define $E(t)F=\sum e^{-itm}{c}_{lm}{Y}_{lm}$. Parseval proves unitarity and the group law. For each $F$, a finite coefficient truncation controls the tail by $2\mathrm{\parallel }F-P_{\le N}F\mathrm{\parallel }$ and its finite head is continuous, proving strong continuity. If $F\in D(A)$, $\mathrm{\mid }(e^{-itm}-1)\mathrm{/}t\mathrm{\mid }\le \mathrm{\mid }m\mathrm{\mid }$ and convergence on finite heads with square-summable weighted tails give $(E(t)F-F)\mathrm{/}t\to -iAF\mathrm{/}\mathrm{\hslash }$. Conversely, if the quotient converges in norm to $G$, each coefficient tends to $-im{c}_{lm}$; Parseval for $G$ gives $\sum m^2\mathrm{\mid }{c}_{lm}{\mathrm{\mid }}^2=\mathrm{\parallel }G{\mathrm{\parallel }}^2<\mathrm{\infty }$ and $G=-iAF\mathrm{/}\mathrm{\hslash }$. Thus this is exactly the full strong derivative domain, not merely a core calculation. The same argument transported by $Q$ gives exactly $D(Q^{-1}AQ)$ for $Q^{-1}E(t)Q$, with derivative $-i{Q}^{-1}AQF\mathrm{/}\mathrm{\hslash }$.

[(a),(c), identification with the actual rotation groups] Pullback by $R_3(t{)}^{-1}$ replaces $\phi$ by $\phi -t$. Therefore $U_3(t)Y_{lm}=e^{-itm}{Y}_{lm}$ for every integer $m$ by [F2], including negative $m$ with its conjugation phase. Equality on the dense finite span, and boundedness, imply $U_3(t)=E(t)$ on all $L^2$. The two finite matrix identities in [F3], transported by the group law, imply $U_1(t)=U(S{)}^{-1}E(t)U(S)$ and $U_2(t)=U(W{)}^{-1}E(t)U(W)$. These are precisely the explicitly defined exponentials in the Statement. Step 2.2 gives the exact derivative domains for all three groups and their actions $-i{L}_i\mathrm{/}\mathrm{\hslash }$; it also proves uniqueness if an operator is defined by the strong derivative. In particular $L_3=L_z$ as full operators, without an AC Stone premise.

[(a), polynomial differentiation with the correct inverse-curve sign] For $F={\mathrm{r}\mathrm{e}\mathrm{s}}_l(p)\in V_l$, set $g_i(s,\mathrm{\Omega })=\mathrm{\nabla }p(R_i(-s)\mathrm{\Omega })\cdot J_i{R}_i(-s)\mathrm{\Omega }$. The correct chain derivative is $\frac{d}{ds}p(R_i(-s)\mathrm{\Omega })=-g_i(s,\mathrm{\Omega })$, directly from $R_i(-s{)}^{\mathrm{\prime }}=-J_i{R}_i(-s)$ in [F3]. The fundamental theorem gives $(U_i(t)F-F)\mathrm{/}t=-t^{-1}{\int }_0^t{g}_i(s,\cdot )ds$. Uniform continuity on $[-\delta ,\delta ]\times S^2$ proves convergence uniformly, and hence in $L^2$, to $-g_i(0,\cdot )=-D_{i}p{\mathrm{\mid }}_{S^2}$. By step 2.2, $F\in D(L_i)$ and $-i{L}_{i}F\mathrm{/}\mathrm{\hslash }=-\mathrm{r}\mathrm{e}\mathrm{s}(D_{i}p)$, so $L_{i}F=\mathrm{r}\mathrm{e}\mathrm{s}(-i\mathrm{\hslash }{D}_{i}p)$. The curve $p\circ R_i(-t)$ lies in the finite-dimensional polynomial space $H_l^{\mathrm{p}\mathrm{o}\mathrm{l}}$ by [F1]; its coefficientwise derivative at zero also lies there, because homogeneity and the harmonic equations are finite linear coefficient constraints. This derivative is $-D_{i}p$, proving that $D_{i}p$ is harmonic homogeneous of degree $l$, including the zero polynomial. Thus $L_i{V}_l\subseteq V_l$ and $L_{i}H\subseteq H$.

[(a), finite blocks and their ODE identification] Put $A_i^{(l)}=L_i{\mathrm{\mid }}_{V_l}$. This is an everywhere-defined Hermitian matrix: symmetry follows from self-adjointness on the whole space and $V_l\subseteq D(L_i)$. The quarter turns preserve $V_l$, so $A_1^{(l)}$ and $A_2^{(l)}$ are unitarily conjugate to the diagonal $A_3^{(l)}$ with entries $\mathrm{\hslash }m$. Consequently $\mathrm{\parallel }{A}_i^{(l)}\mathrm{\parallel }\le \mathrm{\hslash }l$. Polynomial differentiation at time $t$, or the finite group law and step 4.1, gives $(U_i(t){\mathrm{\mid }}_{V_l}{)}^{\mathrm{\prime }}=(-i\mathrm{/}\mathrm{\hslash })A_i^{(l)}{U}_i(t){\mathrm{\mid }}_{V_l}$ with initial value the identity. The absolutely convergent matrix exponential power series solves this equation; uniqueness follows by differentiating its inverse times any other solution. Hence the finite polynomial action equals $\exp (-it{A}_i^{(l)}\mathrm{/}\mathrm{\hslash })$. Orthogonal block truncation then identifies its all-vector extension with the groups of step 3.1. Explicitly $D(L_i)=\{F:{\sum }_l\mathrm{\parallel }{A}_i^{(l)}{P}_{l}F{\mathrm{\parallel }}^2<\mathrm{\infty }\}$ and $L_{i}F={\sum }_l{A}_i^{(l)}{P}_{l}F$: this is the diagonal domain of step 1.1 transported by a block-preserving quarter turn. Thus no hidden narrower domain or additional time-continuity hypothesis is introduced.

[(b), commutators] The polynomial product rule gives $[x_a{\mathrm{\partial }}_b,x_c{\mathrm{\partial }}_d]=x_a{\delta }_{bc}{\mathrm{\partial }}_d-x_c{\delta }_{ad}{\mathrm{\partial }}_b$. Substituting $D_1=x_2{\mathrm{\partial }}_3-x_3{\mathrm{\partial }}_2$, $D_2=x_3{\mathrm{\partial }}_1-x_1{\mathrm{\partial }}_3$ gives $[D_1,D_2]=-D_3$, and cyclic permutation and antisymmetry give $[D_i,D_j]=-{\sum }_k{\epsilon }_{ijk}{D}_k$. All products are defined on $H$ by step 4.1. Therefore $[L_i,L_j]=(-i\mathrm{\hslash }{)}^2[D_i,D_j]={\mathrm{\hslash }}^2{\sum }_k{\epsilon }_{ijk}{D}_k=i\mathrm{\hslash }{\sum }_k{\epsilon }_{ijk}{L}_k$.

[(c), the chart action] The chain rule gives ${\mathrm{\partial }}_{\phi }p(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta )=\mathrm{\nabla }p(\mathrm{\Omega })\cdot (e_3\times \mathrm{\Omega })=D_{3}p(\mathrm{\Omega })$. Apply step 4.1 and linearity to finite harmonic sums. Together with the full equality $L_3=L_z$ of step 3.1 and the domain conjugations of step 2.1, this proves every clause (a)-(d).