A conical helicity packet has reduced axial spin projection

By [F7], full Choice supplies Countable Choice to the bundle definition [F1]. The set $\{\mathbf{n}\in S^2:\mathrm{\mid }\theta (\mathbf{n})-{\theta }_0\mathrm{\mid }<\delta \}$ is a nonempty open angular neighborhood of a direction with polar angle ${\theta }_0$. Choose a smaller coordinate ball whose closure lies in that neighborhood and in a trivializing patch for the helicity line from [F1]. Let $u_{\sigma }$ be its smooth unit section, choose a nonzero smooth bump $\chi$ supported in the ball by [F6], and choose a nonzero smooth radial bump $\eta$ supported in $1<r<2$. The field $\tilde{f}(\mathbf{k})=\eta (r)\chi (\mathbf{n}(\mathbf{k}))u_{\sigma }(\mathbf{n}(\mathbf{k}))$, extended by zero outside the patch, is smooth, compactly supported away from $\mathbf{k}=0$, transverse, and satisfies $P_{\sigma }\tilde{f}=\tilde{f}$. It is nonzero on an open set of positive $d\nu$ measure by [F4], so $0<\mathrm{\parallel }\tilde{f}{\mathrm{\parallel }}_{\nu }<\mathrm{\infty }$; normalizing it gives a packet of width $\delta$. Repeating for each ${\delta }_m$ with $0<{\delta }_m\le \pi$ proves the asserted existence of sequences.

Under the full-Choice hypothesis [A1], the Fock supplier [F2] gives the projected-spin expectation formula; with $\mathrm{\parallel }f{\mathrm{\parallel }}_{\nu }=1$ it yields $\mid ⟨f,S_z^{\mathrm{\prime }}f{⟩}_{\nu }-\mathrm{\hslash }\sigma \cos {\theta }_0\mid =\mathrm{\hslash }\mid {\int }_M(\cos \theta (\mathbf{k})-\cos {\theta }_0)\mathrm{\mid }f(\mathbf{k}){\mathrm{\mid }}^{2}d\nu (\mathbf{k})\mid$. On the support of $f$, [F3] bounds $\mathrm{\mid }\cos \theta -\cos {\theta }_0\mathrm{\mid }\le \mathrm{\mid }\theta -{\theta }_0\mathrm{\mid }\le \delta$. Taking the absolute value inside the integral and using normalization gives the claimed bound $\mathrm{\hslash }\delta$.

For any sequence $0<{\delta }_m\le \pi$ with ${\delta }_m\downarrow 0$, step 1.2 gives $\mathrm{\mid }⟨f_m,S_z^{\mathrm{\prime }}{f}_m{⟩}_{\nu }-\mathrm{\hslash }\sigma \cos {\theta }_0\mathrm{\mid }\le \mathrm{\hslash }{\delta }_m\to 0$, so the expectations converge to $\mathrm{\hslash }\sigma \cos {\theta }_0$.

Put $E_{{\theta }_0}=\{\mathbf{n}\in S^2:\theta (\mathbf{n})={\theta }_0\}$. If $0<{\theta }_0<\pi$, cover this latitude by two overlapping azimuth-coordinate charts. In each chart its coordinate preimage lies in a bounded rectangle $\{{\theta }_0\}\times [a,b]$, which is Lebesgue null by [F8]; the chart-area formula $\sin \theta d\theta d\varphi$ in [F4] therefore gives zero surface measure. If ${\theta }_0=0$ or $\pi$, $E_{{\theta }_0}$ is a single pole. A smooth regular sphere chart centered at that pole has a singleton preimage, null in ${\mathbb{R}}^2$ by [F8]; integrating the chart area density over that singleton again gives zero surface measure by the chart-measure identification in [F4]. Thus $E_{{\theta }_0}$ is surface-null for every stated ${\theta }_0$. For each integer $m\ge 2$, the annulus $1\mathrm{/}m\le r\le m$ has finite $\nu$-measure by [F1] and [F4], so the cone indicator restricted to it is integrable. The polar decomposition in [F4] and Fubini [F5] give $\nu (C_{{\theta }_0}\cap \{1\mathrm{/}m\le r\le m\})={\int }_{1\mathrm{/}m}^m\frac{r}{2}dr{\int }_{S^2}{\mathbf{1}}_{E_{{\theta }_0}}d\mathrm{\Omega }=0.$ These annuli cover $M$, so countable subadditivity gives $\nu (C_{{\theta }_0})=0$. If $g\in {\mathcal{H}}_T^{\nu }$ vanishes $\nu$-almost everywhere on $M\setminus C_{{\theta }_0}$, its squared norm is the sum of the integrals of $\mathrm{\mid }g{\mathrm{\mid }}^2$ over that complement and the null cone, both zero. Hence $\mathrm{\parallel }g{\mathrm{\parallel }}_{\nu }=0$ and $g$ is the zero Hilbert-space element; in particular it cannot be normalized. This is the exact null-cone conclusion in the Example.