The smooth transverse domain is a core for photon rotation generators

It suffices first to consider the axis $e_3$. The $z$-axis and the positive-$x$ branch cut, with the origin removed, are $\nu$-null by [F8]: on each annulus $1\mathrm{/}m\le \mathrm{\mid }\mathbf{k}\mathrm{\mid }\le m$, the mass-shell density is bounded and both sets lie in Lebesgue-null coordinate hyperplanes. To apply the Borel change-of-variables formula [F4] to an arbitrary $f\in {\mathcal{H}}_T^{\nu }$, take a finite-valued representative and apply [F11] to the Lebesgue-measurable sublevel sets of each of its six real components at rational levels. Countable Choice lets us select the corresponding Borel sets; their symmetric differences lie in a countable union of Borel null sets. Outside that union, each component equals the infimum of the rational levels whose selected Borel sets contain $\mathbf{k}$; defining each infimum to be $0$ on the union gives Borel real-valued representatives, hence a Borel vector representative $f_0=f$ almost everywhere. Since $d\nu =(2\mathrm{\mid }\mathbf{k}\mathrm{\mid }{)}^{-1}d{\lambda }_3$ has strictly positive finite density on $M$, Lebesgue-null and $\nu$-null sets there coincide. Thus $f_0$ represents the same Hilbert-space class, and [F4] applies to its coordinate pullback; the same formula shows the pullback is independent of the representative. Use the cylindrical coordinates of [F4], put $r=\sqrt{{\rho }^2+z^2}$ and ${\mathbf{n}}_0(\rho ,z)=(\rho ,0,z)\mathrm{/}r$, and define $g(\rho ,z,\varphi )=R_{e_3}(-\varphi )f_0(\rho \cos \varphi ,\rho \sin \varphi ,z)$. The rotation matrices preserve the Hermitian norm, so the measure identity and its inverse coordinate map $f_0(\mathbf{k})=R_{e_3}(\varphi )g(\rho ,z,\varphi )$ make this a unitary identification with the subspace of $L^2(\mathrm{\Omega }\times \mathbb{T},wd\rho dzd\varphi \mathrm{/}(2\pi );{\mathbb{C}}^3)$ satisfying ${\mathbf{n}}_0(\rho ,z)\cdot g(\rho ,z,\varphi )=0$. Direct substitution gives $U(R_{e_3}(t))g(\rho ,z,\varphi )=g(\rho ,z,\varphi -t)$.

Write $g_m(\rho ,z)={\int }_{\mathbb{T}}g(\rho ,z,\varphi )e^{-im\varphi }d\varphi \mathrm{/}(2\pi )$. By [F5], Parseval identifies the squared norm with ${\sum }_m\mathrm{\parallel }{g}_m{\mathrm{\parallel }}_{L^2(\mathrm{\Omega },w)}^2$, and the rotation difference quotient has Fourier coefficients $(e^{-imt}-1)g_m\mathrm{/}t$. If ${\sum }_m{m}^2\mathrm{\parallel }{g}_m{\mathrm{\parallel }}^2<\mathrm{\infty }$, then $\mathrm{\mid }(e^{-imt}-1)\mathrm{/}t\mathrm{\mid }\le \mathrm{\mid }m\mathrm{\mid }$ and dominated convergence in the square-summable coefficient space gives convergence to $(-im{g}_m{)}_m$. Conversely, if the difference quotients converge in norm, their $m$th coefficients converge to $-im{g}_m$; Fatou applied to finite partial sums and then their increasing limit gives ${\sum }_m{m}^2\mathrm{\parallel }{g}_m{\mathrm{\parallel }}^2<\mathrm{\infty }$. By [F10], this is exactly $D(J_{e_3})$; the generator acts as $J_{e_3}=-i\mathrm{\hslash }{\mathrm{\partial }}_{\varphi }$ in the rotating coordinates, and the squared graph norm is ${\sum }_{m\in \mathbb{Z}}(1+{\mathrm{\hslash }}^2{m}^2)\mathrm{\parallel }{g}_m{\mathrm{\parallel }}_{L^2(\mathrm{\Omega },w)}^2$.

Let $g$ represent any vector in $D(J_{e_3})$. For $n\ge 2$ use the explicit cutoffs ${\chi }_n$ from [F7]. They lie in $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$, satisfy $0\le {\chi }_n\le 1$, and converge pointwise to $1$. Since ${\chi }_n$ depends only on $(\rho ,z)$, it commutes with the $\varphi$-derivative. Dominated convergence [F9] therefore gives ${\chi }_{n}g\to g$ in the graph norm of [F3]. Its Fourier truncations $g^{(N)}={\sum }_{\mathrm{\mid }m\mathrm{\mid }\le N}{\chi }_n{g}_m{e}^{im\varphi }$ converge in that graph norm by the weighted Parseval sum in step 2.1.

Fix $n,N$ and put ${\eta }_n={\chi }_{2n}$; [F7] makes ${\eta }_n=1$ on the support of ${\chi }_n$ and gives compact support inside $\mathrm{\Omega }$. Each of the finitely many coefficients ${\chi }_n{g}_m$ is supported in $\mathrm{supp}{\chi }_n$ and lies in ${\mathbf{n}}_0^{\perp }$. Extend it by zero to ${\mathbb{R}}^2$. On the compact support of ${\eta }_n$, the weight in [F4] is bounded above and below by positive constants, so [F6] gives a smooth compactly supported approximation $a_m^0$ in the weighted norm. Replace it by $a_m={\eta }_{n}P({\mathbf{n}}_0)a_m^0$. Because $P({\mathbf{n}}_0){\chi }_n{g}_m={\chi }_n{g}_m$ and $\mathrm{\parallel }P\mathrm{\parallel }\le 1$, this is transverse and does not increase the approximation error. Choose the finitely many coefficient errors so that ${\sum }_{\mathrm{\mid }m\mathrm{\mid }\le N}(1+{\mathrm{\hslash }}^2{m}^2)\mathrm{\parallel }{a}_m-{\chi }_n{g}_m{\mathrm{\parallel }}^2$ is arbitrarily small. Then ${\sum }_{\mathrm{\mid }m\mathrm{\mid }\le N}{a}_m(\rho ,z)e^{im\varphi }$ is smooth, transverse, and compactly supported away from $\rho =0$; after applying $R_{e_3}(\varphi )$ and returning to $\mathbf{k}$ coordinates it is an element of $\mathcal{D}$, since both factors are $2\pi$-periodic and their derivatives match across the angular seam. Combining these finite coefficient approximations with the cutoff and Fourier truncation from step 3.1 gives graph convergence to $g$, so $\mathcal{D}$ is a core for $J_{e_3}$.

For any unit $\mathbf{a}$, choose $R\in SO(3)$ with $R{e}_3=\mathbf{a}$. The rotation representation gives $U(R)U(R_{e_3}(t))U(R{)}^{\ast }=U(R_{\mathbf{a}}(t))$; Stone uniqueness then gives $J_{\mathbf{a}}=U(R)J_{e_3}U(R{)}^{\ast }$ with the conjugated domain. Also $U(R)\mathcal{D}=\mathcal{D}$, because rotations preserve smoothness, compact support away from zero, and transversality. Applying the unitary $U(R)$ to the graph-dense approximation in step 4.1 proves that $\mathcal{D}$ is a core for $J_{\mathbf{a}}$.