Spin-half curvature and solid-angle filling period

Write $n=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ and use ${\sigma }_x=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),{\sigma }_y=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right),{\sigma }_z=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\mathrm{.}$ Direct multiplication of the displayed matrices gives ${\sigma }_j^2=I$ and ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=0$ for $j\mathrm{\ne }k$, while each ${\sigma }_j$ has trace zero. Hence $(n\cdot \sigma {)}^2=(n_x^2+n_y^2+n_z^2)I=I$ and $\mathrm{tr}(n\cdot \sigma )=0$, so $P_{+}$ is a rank-one orthogonal projector. On the northern chart, $u_N=(\cos (\theta \mathrm{/}2),e^{i\varphi }\sin (\theta \mathrm{/}2){)}^T$ is a unit frame and $A_N=u_N^{\ast }d{u}_N=i(1-\cos \theta )d\varphi \mathrm{/}2$, giving $d{A}_N=(i\mathrm{/}2)\omega$. On the southern chart $u_S=e^{-i\varphi }{u}_N$ is regular, and $A_S=A_N-id\varphi =-i(1+\cos \theta )d\varphi \mathrm{/}2$, so $d{A}_S=(i\mathrm{/}2)\omega$ there as well. By [F1], these local calculations identify the global curvature trace as $\mathrm{tr}{F}_{+}=(i\mathrm{/}2)\omega$.

Use orientation-preserving disk coordinates to identify each $D^2$ with $\{x\in {\mathbb{R}}^2:\mathrm{\mid }x\mathrm{\mid }\le 1\}$ oriented by $dx\wedge dy$. The explicit collar is $(q,t)\mapsto (1-t)q$, where $q=e^{i\theta }\in S^1$ and $t=1-r\ge 0$ is inward radial distance. Since $dx\wedge dy=rd\theta \wedge dt$, the induced boundary orientation at $t=0$ is increasing $\theta$ (counterclockwise). Define $h(s)=e^{-1\mathrm{/}s}$ for $s>0$ and $h(s)=0$ for $s\le 0$. For $0\le t\le ϵ\mathrm{/}2$ set ${\chi }_0(t)=h(t)\mathrm{/}(h(t)+h(ϵ\mathrm{/}2-t))$, and for $t\ge ϵ\mathrm{/}2$ set ${\chi }_0(t)=1$. Every derivative of $h$ vanishes at $0$, so ${\chi }_0$ is smooth, flat at $0$, positive for $t>0$, at most $1$, and equal to $1$ near $ϵ$; hence ${\int }_0^{ϵ}{\chi }_0(t)dt<ϵ$. Put $b(t)=h(t-ϵ\mathrm{/}4)h(3ϵ\mathrm{/}4-t)$, a nonnegative smooth function supported in $[ϵ\mathrm{/}4,3ϵ\mathrm{/}4]$ with positive integral, and define $\chi ={\chi }_0+cb$ where $c=(ϵ-{\int }_0^{ϵ}{\chi }_0)\mathrm{/}{\int }_0^{ϵ}b>0$. Then $\chi$ is positive for $t>0$, flat at $0$, equals $1$ near $ϵ$, and has integral $ϵ$. Set $\rho (t)={\int }_0^t\chi (s)ds$. Thus $\rho (0)={\rho }^{\mathrm{\prime }}(0)={\rho }^{\mathrm{\prime }\mathrm{\prime }}(0)=0$, ${\rho }^{\mathrm{\prime }}(t)>0$ for $t>0$, and $\rho (t)=t$ near $ϵ$. Replace $f_j(q,t)$ by $f_j(q,\rho (t))$ in each collar. This preserves area integral: the pullback coefficient is multiplied by ${\rho }^{\mathrm{\prime }}(t)$, and substitution $s=\rho (t)$ gives the original integral. Now form $X=D_0^2{\cup }_{S^1}(-D_1^2)$ by gluing the standard boundary circles with the identity; the first disk has orientation $dx\wedge dy$, the second has its negative, and their induced boundary orientations are opposite. Use the signed collar coordinates $s=t$ on the first disk and $s=-t$ on the second as seam charts; together with the two disk interiors they define the smooth structure on $X$. Both sides have orientation $d\theta \wedge ds$. The modified maps have the same boundary trace and all normal derivatives through order two vanish at the seam, while their tangential derivatives through order two agree there. They therefore glue to a $C^2$ map $g:X\to S^2$. The double is compact, smooth and boundaryless, and its chosen orientation gives ${\int }_X{g}^{\ast }\omega ={\int }_{D^2}{f}_0^{\ast }\omega -{\int }_{D^2}{f}_1^{\ast }\omega$.

The projector $P_{+}\circ g$ is $C^2$ and rank one. Its projected curvature is $g^{\ast }{F}_{+}$: applying [F1] to $P_{+}\circ g$, the chain rule and [F4] give $d(P_{+}\circ g)=g^{\ast }d{P}_{+}$ and preserve the ordered wedge product. Thus [F2] gives $-\frac{1}{4\pi }({\int }_{D^2}{f}_0^{\ast }\omega -{\int }_{D^2}{f}_1^{\ast }\omega )=\frac{i}{2\pi }\int \mathrm{tr}{F}_{P_{+}\circ g}\in \mathbb{Z},$ where the coefficient is $-1\mathrm{/}(4\pi )$ because $i\cdot i=-1$. Rearranging proves the stated $4\pi$ period. This is a topological ambiguity statement for the same boundary loop; no physical observation is involved.