Transverse rotations preserve helicity and generate total angular momentum

By [F1], the Axiom of Choice [A1] supplies the Countable Choice hypothesis of [F2]. Choose a measurable representative of $f$ and let $N\subset M$ be a $\nu$-null set outside which $\mathbf{k}\cdot f(\mathbf{k})=0$. Invariance of $\nu$ gives $\nu (RN)=0$. For $\mathbf{k}\mathrm{\notin }RN$, $\mathbf{k}\cdot Rf(R^{-1}\mathbf{k})=(R^{-1}\mathbf{k})\cdot f(R^{-1}\mathbf{k})=0,$ so $U(R)f$ is transverse almost everywhere. If the representative is changed on a null set $N^{\mathrm{\prime }}$, its rotated composition changes only on $R{N}^{\mathrm{\prime }}$, also null; hence $U(R)$ is well-defined on $L^2$ equivalence classes. Since $R$ preserves $d\nu$ by [F2] and preserves the Hermitian norm on ${\mathbb{C}}^3$, change of variables gives $\mathrm{\parallel }U(R)f{\mathrm{\parallel }}_{\nu }=\mathrm{\parallel }f{\mathrm{\parallel }}_{\nu }$. The same formula with $R^{-1}$ is an inverse, so $U(R)$ is unitary.

On the Lebesgue transverse space write $U_0(R)u(\mathbf{k})=Ru(R^{-1}\mathbf{k})$. Since $\mathrm{\mid }{R}^{-1}\mathbf{k}\mathrm{\mid }=\mathrm{\mid }\mathbf{k}\mathrm{\mid }$, the multiplier $W$ satisfies $W(U(R)f)(\mathbf{k})=\frac{Rf(R^{-1}\mathbf{k})}{\sqrt{2\mathrm{\mid }\mathbf{k}\mathrm{\mid }}}=R\frac{f(R^{-1}\mathbf{k})}{\sqrt{2\mathrm{\mid }{R}^{-1}\mathbf{k}\mathrm{\mid }}}=U_0(R)(Wf)(\mathbf{k})\mathrm{.}$ Thus the weighted and Lebesgue rotation actions are unitarily intertwined.

For $R\in SO(3)$, $(R\mathbf{n})\times (Rv)=R(\mathbf{n}\times v)$: pairing both sides with any real $w$ reduces the equality to invariance of the oriented scalar triple product under $R$. Hence $P(R\mathbf{n})=RP(\mathbf{n})R^{-1}$, $h(R\mathbf{n})=Rh(\mathbf{n})R^{-1}$, and $P_{\sigma }(R\mathbf{n})=R{P}_{\sigma }(\mathbf{n})R^{-1}$. Substituting $\mathbf{n}=R^{-1}\mathbf{k}\mathrm{/}\mathrm{\mid }{R}^{-1}\mathbf{k}\mathrm{\mid }$ proves that the multiplication projector $P_{\sigma }$ commutes with $U(R)$.

Put $A=[\mathbf{a}{]}_{\times }$, so $A^T=-A$. Cauchy–Schwarz gives $\mathrm{\parallel }BC{\mathrm{\parallel }}_F\le \mathrm{\parallel }B{\mathrm{\parallel }}_F\mathrm{\parallel }C{\mathrm{\parallel }}_F$, so $\mathrm{\parallel }{B}^m\mathrm{/}m!{\mathrm{\parallel }}_F\le \mathrm{\parallel }B{\mathrm{\parallel }}_F^m\mathrm{/}m!$ for every $m\ge 1$; the $m=0$ term is $I$ with $\mathrm{\parallel }I{\mathrm{\parallel }}_F=\sqrt{3}$. Hence the matrix-exponential series converges absolutely. For $B=tA$ the same estimate gives uniform convergence on bounded $t$-intervals; its Cauchy product and the binomial identity give $R_{\mathbf{a}}(s)R_{\mathbf{a}}(t)=R_{\mathbf{a}}(s+t)$, while termwise transposition gives $R_{\mathbf{a}}(t{)}^T=R_{\mathbf{a}}(-t)$. Thus $R_{\mathbf{a}}(t)$ is orthogonal. It is continuous in $t$, so its determinant is the constant $1$ (it lies in $\{+1,-1\}$ and equals $1$ at $t=0$). Hence $R_{\mathbf{a}}(t)\in SO(3)$ and $t\mapsto R_{\mathbf{a}}(t)$ is a one-parameter subgroup. Termwise differentiation gives $R_{\mathbf{a}}^{\mathrm{\prime }}(0)=A$.

For $R_1,R_2\in SO(3)$, $R_1{R}_2$ is again orthogonal with determinant $1$, and $R^{-1}=R^T$. Hence $U(R_1)U(R_2)f(\mathbf{k})=R_1{R}_{2}f(R_2^{-1}{R}_1^{-1}\mathbf{k})=U(R_1{R}_2)f(\mathbf{k}),$ and $U(I)=I$, so $U$ is a unitary representation.

First let $f\in \mathcal{D}$ be nonzero. Its compact support is a compact subset of $M$, so [F6] applied to $\mathbf{k}\mapsto \mathrm{\mid }\mathbf{k}\mathrm{\mid }$ gives $0<r_0\le \mathrm{\mid }\mathbf{k}\mathrm{\mid }\le r_1<\mathrm{\infty }$ on the support. Thus it lies in a radial annulus $K=\{r_0\le \mathrm{\mid }\mathbf{k}\mathrm{\mid }\le r_1\}$. The set $SO(3)\times K$ is closed and bounded in a finite-dimensional Euclidean space: the equations defining $SO(3)$ are closed and its matrix entries have absolute value at most $1$. Hence it is compact by [F4]. The map $(R,\mathbf{k})\mapsto Rf(R^{-1}\mathbf{k})$ is continuous there, so [F4] makes it uniformly continuous. Rotations preserve $K$ and $\nu (K)<\mathrm{\infty }$ by [F2]; therefore $R\mapsto U(R)f$ is continuous in $L^2(d\nu )$. For $f=0$ this continuity is immediate. For general $f\in {\mathcal{H}}_T^{\nu }$, choose $g\in \mathcal{D}$ with $\mathrm{\parallel }f-g{\mathrm{\parallel }}_{\nu }<\epsilon$ by [F2]. Unitarity gives $\mathrm{\parallel }U(R)f-U(R_0)f{\mathrm{\parallel }}_{\nu }\le 2\epsilon +\mathrm{\parallel }U(R)g-U(R_0)g{\mathrm{\parallel }}_{\nu },$ so continuity on $\mathcal{D}$ extends to all $f$ as $R\to R_0$.

By steps 1.1, 2.1, 2.2, and 1.4, $t\mapsto U(R_{\mathbf{a}}(t))$ is a strongly continuous one-parameter unitary group. Applying Stone's theorem [F3] gives its unique self-adjoint generator $T_{\mathbf{a}}$ with $U(R_{\mathbf{a}}(t))=e^{it{T}_{\mathbf{a}}}$. Set $J_{\mathbf{a}}=-\mathrm{\hslash }{T}_{\mathbf{a}}$; then $J_{\mathbf{a}}$ is self-adjoint and $U(R_{\mathbf{a}}(t))=e^{-it{J}_{\mathbf{a}}\mathrm{/}\mathrm{\hslash }}$.

By step 1.4, $R_{\mathbf{a}}^{\mathrm{\prime }}(0)=[\mathbf{a}{]}_{\times }$. For $f\in \mathcal{D}$, differentiating $G(t,\mathbf{k})=R_{\mathbf{a}}(t)f(R_{\mathbf{a}}(-t)\mathbf{k})$ at $t=0$ gives ${\mathrm{\partial }}_{t}G(0,\mathbf{k})=(\mathbf{a}\times f)(\mathbf{k})-(\mathbf{a}\times \mathbf{k})\cdot {\mathrm{\nabla }}_{\mathbf{k}}f(\mathbf{k})$, which equals $-\frac{i}{\mathrm{\hslash }}(L_{\mathbf{a}}+S_{\mathbf{a}}^{\mathrm{o}\mathrm{p}})f$ because $-i{S}_{\mathbf{a}}v=\mathbf{a}\times v$. If $f\mathrm{\ne }0$, use the closed radial annulus $K$ from step 2.2 containing its compact support; for $t$ near zero each $G(t,\cdot )$ is supported in $K$. The product of a compact time interval and $K$ is a closed bounded subset of ${\mathbb{R}}^4$, hence compact by [F4], and $G$ and its $t$-derivative are continuous there. Thus [F4] makes ${\mathrm{\partial }}_{t}G$ uniformly continuous. For each fixed $\mathbf{k}\in K$ and each of the six real coordinate functions $g(s)$ of $G(s,\mathbf{k})\in {\mathbb{C}}^3\cong {\mathbb{R}}^6$, set $m=(g(t)-g(0))\mathrm{/}t$ for $t\mathrm{\ne }0$. The function $g(s)-ms$ has equal endpoint values on the compact interval between $0$ and $t$. If it is nonconstant, [F6] gives an interior maximum or minimum. Difference quotients from the two sides then have opposite weak signs, so differentiability forces its derivative there to be zero. If it is constant, its derivative is zero everywhere. Thus $m=g^{\mathrm{\prime }}(c)$ for some $c$ between $0$ and $t$. Uniform continuity and $\mathrm{\mid }c\mathrm{\mid }\le \mathrm{\mid }t\mathrm{\mid }$ imply that the difference quotients of $G$ converge uniformly on $K$ to ${\mathrm{\partial }}_{t}G(0,\cdot )$. Since $\nu (K)<\mathrm{\infty }$ by [F2], this gives convergence in $L^2(d\nu )$; the quotients lie in the closed subspace ${\mathcal{H}}_T^{\nu }$ by step 1.1, so the limit is the norm derivative in that space. For $f=0$ the derivative is immediate. By the Stone derivative characterization in step 3.1, $f\in D(J_{\mathbf{a}})$ and $J_{\mathbf{a}}f=(L_{\mathbf{a}}+S_{\mathbf{a}}^{\mathrm{o}\mathrm{p}})f$.

If $f\in \mathcal{D}$, then $(\mathbf{a}\times \mathbf{k})\cdot {\mathrm{\nabla }}_{\mathbf{k}}(\mathbf{k}\cdot f)=0$; expanding this derivative and using $\mathbf{k}\cdot f=0$ gives $\mathbf{k}\cdot L_{\mathbf{a}}f=i\mathrm{\hslash }(\mathbf{k}\times f)\cdot \mathbf{a}$. The matrix definition of $S_{\mathbf{a}}$ gives $\mathbf{k}\cdot S_{\mathbf{a}}^{\mathrm{o}\mathrm{p}}f=-i\mathrm{\hslash }(\mathbf{k}\times f)\cdot \mathbf{a}$. Their sum is zero, so step 4.1 shows that $J_{\mathbf{a}}$ preserves transversality on $\mathcal{D}$.

The separate terms need not preserve transversality. For $\mathbf{a}=e_3$, choose by [F5] a smooth bump $\rho$ equal to $1$ on $B_{1\mathrm{/}8}(0)$ and supported in $B_{1\mathrm{/}4}(0)$, and translate it to $\chi (\mathbf{k})=\rho (\mathbf{k}-e_1)$. Its support misses the $e_3$ axis. Put $f(\mathbf{k})=\chi (\mathbf{k})(e_3\times \mathbf{k})$. Then $f\in \mathcal{D}$, while $(\mathbf{k}\times f)\cdot e_3=\chi (\mathbf{k})(k_1^2+k_2^2)$ is nonzero near $e_1$. By the calculation in step 5.1, both $\mathbf{k}\cdot L_{e_3}f$ and $\mathbf{k}\cdot S_{e_3}^{\mathrm{o}\mathrm{p}}f$ are nonzero there, with opposite signs. This proves that neither term separately preserves the transverse space, whereas their sum does.