Photon helicity bundles and exact transverse rotations

The multiplier $Wf(\mathbf{k})=f(\mathbf{k})\mathrm{/}\sqrt{2\mathrm{\mid }\mathbf{k}\mathrm{\mid }}$ is measurable on $M$, and $\mathrm{\parallel }Wf{\mathrm{\parallel }}_{L^2(d^3\mathbf{k})}^2={\int }_M\frac{\mathrm{\mid }f(\mathbf{k}){\mathrm{\mid }}^2}{2\mathrm{\mid }\mathbf{k}\mathrm{\mid }}{d}^3\mathbf{k}=\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(d\nu )}^2\mathrm{.}$ The same identity for inner products shows that $W$ is an isometry with the second-variable-linear convention of [F2].

For any $u\in {\mathcal{H}}_T^0$, set $f(\mathbf{k})=\sqrt{2\mathrm{\mid }\mathbf{k}\mathrm{\mid }}u(\mathbf{k})$. Then $\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(d\nu )}^2={\int }_{M}2\mathrm{\mid }\mathbf{k}\mathrm{\mid }\mathrm{\mid }u(\mathbf{k}){\mathrm{\mid }}^2\frac{d^3\mathbf{k}}{2\mathrm{\mid }\mathbf{k}\mathrm{\mid }}=\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(d^3\mathbf{k})}^2,$ so $f\in L^2(d\nu ;{\mathbb{C}}^3)$, $Wf=u$, and $\mathbf{k}\cdot f(\mathbf{k})=\sqrt{2\mathrm{\mid }\mathbf{k}\mathrm{\mid }}\mathbf{k}\cdot u(\mathbf{k})=0$ almost everywhere. Conversely, multiplication by the positive finite scalar $(2\mathrm{\mid }\mathbf{k}\mathrm{\mid }{)}^{-1\mathrm{/}2}$ preserves the transverse condition. Lebesgue measure and $\nu$ have the same null sets on $M$: on every annulus $1\mathrm{/}m\le \mathrm{\mid }\mathbf{k}\mathrm{\mid }\le m$ their density ratio is bounded above and below by positive constants, and these annuli cover $M$. Thus $W$ maps ${\mathcal{H}}_T^{\nu }$ onto ${\mathcal{H}}_T^0$. The isometry and surjectivity transfer completeness from the Hilbert space [F1] to ${\mathcal{H}}_T^{\nu }$, so $W$ is unitary.

Since $\mathrm{\mid }\mathbf{n}\mathrm{\mid }=1$, $P=I-\mathbf{n}{\mathbf{n}}^T$ satisfies $P^{\ast }=P=P^2$ and is the orthogonal projection onto ${\mathbf{n}}^{\perp }$; writing $z=Pz+\mathbf{n}(\mathbf{n}\cdot z)$ gives $\mathrm{\parallel }Pz{\mathrm{\parallel }}^2=\mathrm{\parallel }z{\mathrm{\parallel }}^2-\mathrm{\mid }\mathbf{n}\cdot z{\mathrm{\mid }}^2$, so $\mathrm{\parallel }P\mathrm{\parallel }\le 1$. Write $C_{\mathbf{n}}v=\mathbf{n}\times v$, whose matrix is real skew-symmetric. Expanding with ${\sum }_k{ϵ}_{ijk}{ϵ}_{\mathrm{\ell }mk}={\delta }_{i\mathrm{\ell }}{\delta }_{jm}-{\delta }_{im}{\delta }_{j\mathrm{\ell }}$ gives $(C_{\mathbf{n}}^{2}v{)}_i=n_i(\mathbf{n}\cdot v)-v_i=-[P(\mathbf{n})v{]}_i$. Antisymmetry also gives $\mathbf{n}\cdot C_{\mathbf{n}}v=0$ and $C_{\mathbf{n}}\mathbf{n}=0$, so $P{C}_{\mathbf{n}}=C_{\mathbf{n}}P=C_{\mathbf{n}}$. Consequently $h(\mathbf{n}{)}^{\ast }=h(\mathbf{n})$, $h(\mathbf{n}{)}^2=P(\mathbf{n})$, and $Ph=hP=h$.

Applying the polar-coordinate formula in [F3] to the density $(2r{)}^{-1}$ gives $d\nu =(r\mathrm{/}2)drd\mathrm{\Omega }$. Orthogonal maps preserve $d\mathrm{\Omega }$ by [F4] and fix $r$, so they preserve $\nu$.

Applying [F3] to the radial density $(2r{)}^{-1}$ gives $\nu (\{r_0\le \mathrm{\mid }\mathbf{k}\mathrm{\mid }\le r_1\})=\frac{1}{2}{\int }_{r_0}^{r_1}rdr{\int }_{S^2}d\mathrm{\Omega }=\pi (r_1^2-r_0^2)\mathrm{.}$ The annuli $\{1\mathrm{/}m\le \mathrm{\mid }\mathbf{k}\mathrm{\mid }\le m\}$ have finite measure and cover $M$, so $\nu$ is sigma-finite.

Given $f\in {\mathcal{H}}_T^{\nu }$, put $u=Wf\in {\mathcal{H}}_T^0$ by [F1]. Write $u=(u_1,u_2,u_3)$. For each of the three components, [F2] and [A1] give a compactly supported smooth $v_j$ with $\mathrm{\parallel }{u}_j-v_j{\mathrm{\parallel }}_{L^2(d^3\mathbf{k})}<\epsilon \mathrm{/}(3\sqrt{3})$. The three component choices are permitted by Countable Choice, and $v=(v_1,v_2,v_3)$ then satisfies $\mathrm{\parallel }u-v{\mathrm{\parallel }}_{L^2(d^3\mathbf{k})}<\epsilon \mathrm{/}3$.

For each $m$, [F5] supplies a smooth cutoff ${\rho }_m$ equal to $1$ on ${\bar{B}}_{1\mathrm{/}m}(0)$ and supported in $B_{2\mathrm{/}m}(0)$; set ${\chi }_m=1-{\rho }_m$. By [F3], ${\lambda }_3(B_{2\mathrm{/}m}(0))=4\pi {\int }_0^{2\mathrm{/}m}{r}^{2}dr=(4\pi \mathrm{/}3)(2\mathrm{/}m{)}^3$. Since $v$ is bounded, $\mathrm{\parallel }{\chi }_{m}v-v{\mathrm{\parallel }}_{L^2(d^3\mathbf{k})}\le \mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}\sqrt{{\lambda }_3(B_{2\mathrm{/}m}(0))}⟶0.$ Choose $m$ large enough that this norm is below $2\epsilon \mathrm{/}3$. The field ${\chi }_{m}v$ is smooth, compactly supported away from $0$, and can therefore be projected with the smooth matrix $P(\mathbf{k}\mathrm{/}\mathrm{\mid }\mathbf{k}\mathrm{\mid })$.

The operators $P_{\sigma }$ are self-adjoint and $P_{\sigma }^2=\frac{P+2\sigma h+h^2}{4}=P_{\sigma },P_{+}{P}_{-}=0,P_{+}+P_{-}=P\mathrm{.}$ Moreover $h{P}_{\sigma }=\sigma P_{\sigma }$. The trace of $P$ is $2$ and the trace of $h=i{C}_{\mathbf{n}}$ is $0$, so $\mathrm{tr}{P}_{\sigma }=1$; a finite-dimensional orthogonal projection has rank equal to its trace, hence each range is a complex line.

The matrices $P_{\sigma }(\mathbf{n})$ depend smoothly on $\mathbf{n}$. For any ${\mathbf{n}}_0$, choose $v\in {\mathbb{C}}^3$ with $P_{\sigma }({\mathbf{n}}_0)v\mathrm{\ne }0$. Continuity makes $P_{\sigma }(\mathbf{n})v$ nonzero on a neighborhood of ${\mathbf{n}}_0$, and normalizing this smooth vector gives a smooth unit local section. Thus the ranges form the stated smooth line subbundles.

Set $g_m(\mathbf{k})=P(\mathbf{k}\mathrm{/}\mathrm{\mid }\mathbf{k}\mathrm{\mid }){\chi }_m(\mathbf{k})v(\mathbf{k})$. It is smooth, transverse, and compactly supported away from $0$. Since $Pu=u$ for the transverse field $u$ and $\mathrm{\parallel }P(\mathbf{k}\mathrm{/}\mathrm{\mid }\mathbf{k}\mathrm{\mid })\mathrm{\parallel }\le 1$ by step 1.1, $\mathrm{\parallel }{g}_m-u{\mathrm{\parallel }}_{L^2(d^3\mathbf{k})}\le \mathrm{\parallel }{\chi }_{m}v-v{\mathrm{\parallel }}_{L^2(d^3\mathbf{k})}+\mathrm{\parallel }v-u{\mathrm{\parallel }}_{L^2(d^3\mathbf{k})}\mathrm{.}$ The choices in steps 1.4 and 1.5 make the right side less than $\epsilon$. The inverse image $W^{-1}{g}_m=\sqrt{2\mathrm{\mid }\mathbf{k}\mathrm{\mid }}{g}_m$ is smooth and compactly supported away from $0$, hence belongs to $\mathcal{D}$. As $W$ is unitary by [F1], $\mathcal{D}$ is dense in ${\mathcal{H}}_T^{\nu }$.

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.

The vectors ${\mathbf{e}}_{\theta },{\mathbf{e}}_{\phi }$ are real unit vectors orthogonal to each other and to $\mathbf{n}$, and $\mathbf{n}\times {\mathbf{e}}_{\theta }={\mathbf{e}}_{\phi }$, $\mathbf{n}\times {\mathbf{e}}_{\phi }=-{\mathbf{e}}_{\theta }$. Hence $e_{\sigma }$ has unit norm and satisfies $h(\mathbf{n})e_{\sigma }=\sigma e_{\sigma }$, so it is a local frame of ${\mathcal{L}}_{\sigma ,\mathbf{n}}$. Direct differentiation gives ${\mathbf{e}}_{\theta }\cdot d{\mathbf{e}}_{\phi }=-\cos \theta d\phi$ and ${\mathbf{e}}_{\phi }\cdot d{\mathbf{e}}_{\theta }=\cos \theta d\phi$; the diagonal terms vanish. Therefore $e_{\sigma }^{†}d{e}_{\sigma }=\frac{i\sigma }{2}({\mathbf{e}}_{\theta }\cdot d{\mathbf{e}}_{\phi }-{\mathbf{e}}_{\phi }\cdot d{\mathbf{e}}_{\theta })=-i\sigma \cos \theta d\phi \mathrm{.}$ Multiplying by $-i$ gives $A_{\sigma }=-\sigma \cos \theta d\phi$, and differentiating gives $F_{\sigma }=\sigma \sin \theta d\theta \wedge d\phi$.

In Cartesian coordinates, the north and south gauges for $\sigma =+1$ are $e_{+}^N=\frac{1}{\sqrt{2}}\left(\begin{array}{c}(1+n_3)\mathrm{/}2-(n_1+i{n}_2{)}^2\mathrm{/}[2(1+n_3)]\\ i(1+n_3)\mathrm{/}2+i(n_1+i{n}_2{)}^2\mathrm{/}[2(1+n_3)]\\ -(n_1+i{n}_2)\end{array}\right),e_{+}^S=\frac{1}{\sqrt{2}}\left(\begin{array}{c}(n_3-1)\mathrm{/}2+(n_1-i{n}_2{)}^2\mathrm{/}[2(1-n_3)]\\ i(1-n_3)\mathrm{/}2+i(n_1-i{n}_2{)}^2\mathrm{/}[2(1-n_3)]\\ -(n_1-i{n}_2)\end{array}\right)\mathrm{.}$ Their denominators are nonzero on the north and south charts respectively, so they are smooth there. Since $e_{-}^N=\overline{e_{+}^N}$ and $e_{-}^S=\overline{e_{+}^S}$, the gauges for $\sigma =-1$ are smooth as well. Each has unit norm because it differs from $e_{\sigma }$ by a phase. On the overlap, $e_{\sigma }^N=e^{2i\sigma \phi }{e}_{\sigma }^S$, and the transition map $e^{i\phi }\mapsto e^{2i\sigma \phi }$ has degree $2\sigma$.

Since $F_{\sigma }=\sigma d\mathrm{\Omega }$, [F2] gives ${\int }_{S^2}{F}_{\sigma }=4\pi \sigma$, hence the normalized Berry flux is $2\sigma$. The skew-Hermitian curvature is ${\mathrm{\Omega }}_{\sigma }=d{\omega }_{\sigma }=i{F}_{\sigma }$ because ${\omega }_{\sigma }=i{A}_{\sigma }$. Under the stated Chern-Weil convention, $c_1({\mathcal{L}}_{\sigma })[S^2]=\frac{i}{2\pi }{\int }_{S^2}i{F}_{\sigma }=-\frac{1}{2\pi }{\int }_{S^2}{F}_{\sigma }=-2\sigma \mathrm{.}$

Suppose a global smooth unit section $s$ of ${\mathcal{L}}_{\sigma }$ existed. Then $A_s=-i{s}^{†}ds$ would be a globally defined smooth real one-form with compact support because $S^2$ is compact by [F4]. On each local unit frame, writing $s=g{e}_{\sigma }$ with $\mathrm{\mid }g\mathrm{\mid }=1$ gives $A_s=A_{\sigma }-i{g}^{-1}dg$; the correction is closed because $d(g^{-1}dg)=-g^{-2}dg\wedge dg=0$, so $d{A}_s=F_{\sigma }$. The sphere is oriented and boundaryless by the Statement and [F4], so [F3] yields ${\int }_{S^2}{F}_{\sigma }={\int }_{S^2}d{A}_s={\int }_{\mathrm{\partial }{S}^2}{A}_s=0$. This contradicts ${\int }_{S^2}{F}_{\sigma }=4\pi \sigma \mathrm{\ne }0$ from step 2.1. Therefore no global smooth unit helicity eigenvector exists.