Rank-one holonomy and integral curvature

The first equality and integrality are exactly [F2], with the Chern-form convention and sign fixed by [F3].

Let $\gamma$ lie in one frame patch with unit frame $u$. A section is $uc$, and $\mathrm{\nabla }(uc)=u(dc+Ac)$ for $A=u^{\ast }du$; therefore a parallel section satisfies $c^{\mathrm{\prime }}=-A({\gamma }^{\mathrm{\prime }})c$. Since this is a scalar equation, its transport factor on that segment is $\exp (-{\int }_{\gamma }A)$. The compact path has a finite subdivision into local frame patches by compactness and the open-cover property; uniqueness follows directly by differentiating $c(t)\exp ({\int }_{t_0}^{t}A({\gamma }^{\mathrm{\prime }}(v))dv)$, which is constant. To identify this with Kato transport, write $P(t)=P(\gamma (t))$ and $v(t)=W(t)v_0$ with $P(0)v_0=v_0$. Intertwining gives $Pv=v$ and $P{P}^{\mathrm{\prime }}P=0$ gives $P{v}^{\mathrm{\prime }}=P[P^{\mathrm{\prime }},P]v=0$, so $v$ is parallel. Conversely, a parallel section with $Pv=v$ satisfies $v^{\mathrm{\prime }}=P^{\mathrm{\prime }}v+P{v}^{\mathrm{\prime }}=P^{\mathrm{\prime }}v=[P^{\mathrm{\prime }},P]v$, hence satisfies the same Kato equation.

At a frame change $u_{j+1}=u_j{g}_j$, equality of the represented section $u_j{c}_j=u_{j+1}{c}_{j+1}$ gives $c_{j+1}=g_j^{-1}{c}_j$. Composing the local scalar transport equations from step 1.2 with these coefficient changes gives exactly the product in the Statement. Include the final transition back to the initial frame at the closing point, so the product represents the endpoint map in the initial unit vector. Subdividing a segment splits its integral into a sum, and inserting an intermediate frame at a crossing splits a transition into two whose inverse factors multiply to the original inverse; hence refinement leaves the product unchanged. Step 1.2 identifies this parallel transport with Kato transport.

In the two-subsurface decomposition, set $A_N=u_N^{\ast }d{u}_N$ and $A_S=u_S^{\ast }d{u}_S$. From $u_S=u_{N}g$ and $g^{-1}g=1$, the product rule gives $A_S=A_N+g^{-1}dg$. For a line, $A\wedge A=0$, so the scalar curvature on each patch is $dA$. The common boundary has opposite induced orientations on the two subsurfaces. Applying [F4] gives ${\int }_{\mathrm{\Sigma }}\mathrm{tr}F={\int }_{\mathrm{\Gamma }}{A}_N+{\int }_{-\mathrm{\Gamma }}{A}_S={\int }_{\mathrm{\Gamma }}(A_N-A_S)=-{\int }_{\mathrm{\Gamma }}{g}^{-1}dg\mathrm{.}$ For one oriented component parameterized by $t\in [0,2\pi ]$, $\mathrm{\mid }g(t)\mathrm{\mid }=1$ implies $g^{-1}{g}^{\mathrm{\prime }}=ia(t)$ for a real continuous function $a$. Put $\theta (t)={\int }_0^{t}a(s)ds$; differentiating shows $g(t)e^{-i\theta (t)}$ is constant, so $g(t)=g(0)e^{i\theta (t)}$. Since $g(2\pi )=g(0)$, $\theta (2\pi )\in 2\pi \mathbb{Z}$, and hence $(2\pi i{)}^{-1}\int g^{-1}dg\in \mathbb{Z}$. Summing components shows $\mathrm{wind}(g)\in \mathbb{Z}$. Multiplying the curvature identity by $i\mathrm{/}(2\pi )$ and using step 1.1 proves the claimed Chern-number formula, with the sign fixed by the orientation of $\mathrm{\partial }{\mathrm{\Sigma }}_N$.