Spin-half curvature and a solid angle

Matrix multiplication gives $(n\cdot \sigma {)}^{\ast }=(n\cdot \sigma )$ and $(n\cdot \sigma {)}^2=I$, while $\mathrm{tr}(n\cdot \sigma )=0$. Hence $P_{+}^{\ast }=P_{+}^2=P_{+}$ and $\mathrm{tr}{P}_{+}=1$, so it is a rank-one orthogonal projection. Direct multiplication gives $(n\cdot \sigma )u_N=u_N$ and $u_N^{\ast }{u}_N=1$, so $u_N$ is a normalized local frame of its positive eigenline.

Differentiating the displayed spinor gives $u_N^{\ast }d{u}_N=i{\sin }^2(\theta \mathrm{/}2)d\varphi =i(1-\cos \theta )d\varphi \mathrm{/}2$. Taking the exterior derivative yields $d{A}_N=i\sin \theta d\theta \wedge d\varphi \mathrm{/}2$, which is the curvature in this frame by [F1]. Both the projector curvature and the spherical area form are global smooth two-forms, so their equality on the northern chart gives $F=(i\mathrm{/}2){\omega }_{S^2}$ globally.

Integrating the area form gives ${\int }_{S^2}\sin \theta d\theta d\varphi =4\pi$, hence $\frac{i}{2\pi }{\int }_{S^2}F=\frac{i}{2\pi }\frac{i}{2}(4\pi )=-1$. By [F2] this is the integral first Chern number with the stated anti-Hermitian connection convention.

A southern frame is $u_S=(e^{-i\varphi }\cos (\theta \mathrm{/}2),\sin (\theta \mathrm{/}2){)}^T=e^{-i\varphi }{u}_N$; it is regular at the south pole and together with $u_N$ covers $S^2$. Along the equator the transition is $g(\varphi )=e^{-i\varphi }$, so $(2\pi i{)}^{-1}{\int }_0^{2\pi }{g}^{-1}dg=-1$. The two-hemisphere clutching clause in [F2] therefore gives Chern number $-1$, matching the direct curvature integral in step 3.1 and fixing the orientation sign. Pull these frames back to the standard closed parameter disk filling $S$. Their inverse images form an open cover of the disk; by [F5] it has a Lebesgue number $\delta >0$. Choose $N$ with $\sqrt{2}\mathrm{/}N<\delta$, set $h=1\mathrm{/}N$, and intersect the disk with the closed axis-aligned squares $[ih,(i+1)h]\times [jh,(j+1)h]$. Retain only intersections $C_{ij}$ with nonempty interior. There are finitely many; their interiors are disjoint and their closures cover the disk, since every disk point is a limit of interior points off the grid lines and a subsequence lies in one of finitely many retained cells. Each cell has diameter at most $\sqrt{2}h<\delta$, so it lies in one frame patch. Each is a compact convex region whose boundary, after zero-length pieces are discarded, is a finite union of regular straight or circular arcs meeting at finitely many corners; the C1 Stokes item [F3] applies. On a cell $C_{ij}$, its frame connection $A_{ij}$ is $C^1$ because the filling is $C^2$, and [F3] gives ${\int }_{\mathrm{\partial }{C}_{ij}}{A}_{ij}={\int }_{C_{ij}}F$. Since one frame covers the cell, its boundary holonomy is $\exp (-{\int }_{\mathrm{\partial }{C}_{ij}}{A}_{ij})=\exp (-{\int }_{C_{ij}}F)$. Choose a unit vector at each vertex of this finite cell decomposition and write each oriented edge transport as its unit-modulus scalar in those endpoint vectors. In the product of the cell-boundary holonomies every internal edge occurs once in each direction, so its scalar and inverse cancel; only the outer boundary loop remains. Thus $\mathrm{Hol}(C)=\exp (-{\int }_{S}F)=\exp (-i\mathrm{\Omega }\mathrm{/}2)$. If a second filling is chosen, [F4] says its solid angle differs by $4\pi k$ for some integer $k$, so this exponential is independent of the filling. For a latitude circle the north-cap integral is $\mathrm{\Omega }={\int }_0^{2\pi }{\int }_0^{\theta }\sin udud\varphi =2\pi (1-\cos \theta )$.