Curvature and the obstruction to a global polarization basis

In spherical coordinates, ${\mathrm{\partial }}_{\theta }{e}_{\theta }=-n$, ${\mathrm{\partial }}_{\theta }{e}_{\varphi }=0$, ${\mathrm{\partial }}_{\varphi }{e}_{\theta }=\cos \theta e_{\varphi }$, and ${\mathrm{\partial }}_{\varphi }{e}_{\varphi }=-\cos \theta e_{\theta }-\sin \theta n$. Using $e_{\theta }\cdot e_{\theta }=e_{\varphi }\cdot e_{\varphi }=1$, $e_{\theta }\cdot e_{\varphi }=0$, and $e_{\theta }\cdot n=e_{\varphi }\cdot n=0$, direct expansion gives $v_{\sigma }^{†}{\mathrm{\partial }}_{\theta }{v}_{\sigma }=0$ and $v_{\sigma }^{†}{\mathrm{\partial }}_{\varphi }{v}_{\sigma }=-i\sigma \cos \theta$. Thus $A_{\sigma }=i{v}_{\sigma }^{†}d{v}_{\sigma }=\sigma \cos \theta d\varphi$.

Differentiating the preceding local form yields $d{A}_{\sigma }=-\sigma \sin \theta d\theta \wedge d\varphi$. The globally defined form ${\omega }_{S^2,n}(X,Y)=n\cdot (X\times Y)$ is smooth because its coefficients are polynomial in the embedded coordinates and tangent vectors; since ${\mathrm{\partial }}_{\theta }n=e_{\theta }$ and ${\mathrm{\partial }}_{\varphi }n=\sin \theta e_{\varphi }$, it satisfies ${\omega }_{S^2}=\sin \theta d\theta \wedge d\varphi$. Therefore $d{A}_{\sigma }=-\sigma {\omega }_{S^2}$ away from the poles. On any overlap neighborhood where two unit frames differ by a smooth real phase $\chi$, [F1] changes $A$ by $-d\chi$, so [F4] gives $d{A}^{\mathrm{\prime }}=dA$; the local curvatures glue smoothly, and equality with the smooth form $-\sigma {\omega }_{S^2}$ extends by continuity to the poles.

To compute the flux with [F3], take the boundary $O=\{\mathrm{\mid }{x}_1\mathrm{\mid }+\mathrm{\mid }{x}_2\mathrm{\mid }+\mathrm{\mid }{x}_3\mathrm{\mid }=1\}$ of the unit octahedron. Fix one order of its six vertices. For each of its eight triangular faces, let ${\tau }_f:{\mathrm{\Delta }}^2\to f$ be the affine simplex using that global vertex order, and compose it with the radial projection $r(x)=x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel }$ to $S^2$. The denominator is nonzero on $O$, so each face map extends smoothly to a neighborhood of its simplex. Give each map coefficient $+1$ or $-1$ so its orientation agrees with the outward sphere orientation. A shared edge then has the same ordered affine edge map in the two adjacent faces and opposite boundary coefficients, so the finite sum $z$ of the eight maps satisfies $\mathrm{\partial }z=0$. Each spherical octant has ${\omega }_{S^2}=\sin \theta d\theta \wedge d\varphi$ and area ${\int }_0^{\pi \mathrm{/}2}\sin \theta d\theta {\int }_0^{\pi \mathrm{/}2}d\varphi =\pi \mathrm{/}2$; the eight octants cover the sphere and their edges contribute no two-dimensional integral. Hence ${\int }_z{\omega }_{S^2}=8(\pi \mathrm{/}2)=4\pi$, and step 2.1 gives ${\int }_{S^2}{F}_{\sigma }:={\int }_z{F}_{\sigma }=-4\pi \sigma$. Dividing by $2\pi$ gives $-2\sigma$.

Suppose a global smooth unit frame $v$ existed. Then $A=i{v}^{†}dv$ would be a global smooth one-form and its curvature would be $F_{\sigma }=dA$. Applying [F2] to the finite cycle $z$ from step 3.1 gives ${\int }_z{F}_{\sigma }={\int }_{z}dA={\int }_{\mathrm{\partial }z}A=0$, contradicting ${\int }_z{F}_{\sigma }=-4\pi \sigma \mathrm{\ne }0$ by the nonzero values of $\sigma$ in [F1] and step 3.1. Therefore no global smooth unit frame exists.