Gapped band curvature and Chern invariance

For each fixed $t$ (or for the single projector), [F1] gives the rank-one projector $Q_t={\bigwedge }^m{P}_t$ onto $\det E_{P_t}$. For any $k\in {\mathbb{T}}_{2\pi }^2$ and nonzero $v\in \mathrm{im}{Q}_t(k)$, continuity makes $U_v:=\{\mathrm{\ell }:Q_t(\mathrm{\ell })v\mathrm{\ne }0\}$ an open neighborhood on which $Q_t(\mathrm{\ell })v\mathrm{/}\mathrm{\parallel }{Q}_t(\mathrm{\ell })v\mathrm{\parallel }$ is a local frame; these neighborhoods cover the torus. Compactness gives a finite trivializing cover $U_1,\dots ,U_N$. The torus is a compact metric space with its quotient Euclidean metric, so sufficiently small open balls, followed by a finite subcover, give a finite open refinement $V_1,\dots ,V_J$ with each closure $\overline{V_j}$ contained in some $U_{i(j)}$. Put ${\rho }_j(\mathrm{\ell })=\mathrm{dist}(\mathrm{\ell },{\mathbb{T}}_{2\pi }^2\setminus V_j)$, taking ${\rho }_j=1$ if $V_j={\mathbb{T}}_{2\pi }^2$. Each ${\rho }_j$ is continuous, has cozero set $V_j$, and has support (the closure of its nonzero set) $\overline{V_j}$. Since the $V_j$ cover, ${\sum }_j{\rho }_j>0$, and the functions ${\rho }_j\mathrm{/}{\sum }_a{\rho }_a$ form a continuous partition of unity whose supports lie in the trivializing sets $U_{i(j)}$. By definition, a bundle atlas with such a subordinate partition is numerable, so [F5] applies. The torus is path-connected and the quotient square with opposite edges identified gives it a finite CW structure with one $0$-cell, two $1$-cells, and one $2$-cell. The determinant-line curvature lemma [F2] gives $\mathrm{tr}{F}_t=\mathrm{tr}(Q_t(d{Q}_t\wedge d{Q}_t)Q_t)$ pointwise, with the same orientation convention. Applying [F3] to $Q_t$ proves the displayed equality and integrality.

For a jointly $C^2$ family, the coefficient function $g(k,t)$ of the two-form $\mathrm{tr}(P_t(d{P}_t\wedge d{P}_t)P_t)$ is continuous on the compact set ${\mathbb{T}}_{2\pi }^2\times [0,1]$, so it is uniformly continuous. If $t_j\to t$, then ${\sup }_k\mathrm{\mid }g(k,t_j)-g(k,t)\mathrm{\mid }\to 0$; therefore $C(P_{t_j})\to C(P_t)$ by bounding the difference of integrals by the torus's finite area times this supremum. Thus $t\mapsto C(P_t)$ is continuous. Step 1.1 makes it integer-valued, and a continuous integer-valued function on the connected interval $[0,1]$ is constant: otherwise the intermediate-value theorem would give a noninteger value between two distinct integer values. It remains to verify the finite-range clause. The displayed finite sum makes $h(k,t)$ jointly $C^2$ on ${\mathbb{T}}_{2\pi }^2\times J$. Ordered eigenvalues of a Hermitian matrix depend continuously on its entries: the min-max formula gives $\mathrm{\mid }{\lambda }_j(A)-{\lambda }_j(B)\mathrm{\mid }\le \mathrm{\parallel }A-B\mathrm{\parallel }$, by the same Rayleigh-quotient argument as in [F6]. At any $(k_0,t_0)$, choose disjoint small positively oriented circles around the distinct eigenvalues in the selected block of $h(k_0,t_0)$, with their closed disks separated from the complementary eigenvalues. Joint continuity of $h$ and its eigenvalues keeps the selected eigenvalues inside this fixed contour union and the complementary eigenvalues outside on a neighborhood of $(k_0,t_0)$. On the contours, $(zI-h(k,t){)}^{-1}$ is jointly $C^2$, since matrix inversion is smooth on invertible matrices and the contour remains in the resolvent set. Its contour integral is therefore jointly $C^2$. For completeness, on any component circle $z=c+R{e}^{it}$ an exterior eigenvalue $\lambda$ has $q=R\mathrm{/}\mathrm{\mid }\lambda -c\mathrm{\mid }<1$. The finite geometric identity gives $(z-\lambda {)}^{-1}=-{\sum }_{k=0}^N(z-c{)}^k\mathrm{/}(\lambda -c{)}^{k+1}+r_N(z)$ with $\sup \mathrm{\mid }{r}_N\mathrm{\mid }\le q^{N+1}\mathrm{/}(\mathrm{\mid }\lambda -c\mathrm{\mid }(1-q))$. Each polynomial term has zero circle integral, since $\oint (z-c{)}^{k}dz=i{R}^{k+1}{\int }_0^{2\pi }{e}^{i(k+1)t}dt=0$, while the remainder integral has modulus at most $2\pi R\sup \mathrm{\mid }{r}_N\mathrm{\mid }\to 0$. Thus an exterior eigenvalue contributes zero. The interior formula in [F7] and this exterior calculation, applied in a finite orthonormal eigenbasis at each fixed $(k,t)$, give coefficient $1$ on each selected eigenspace and $0$ on its complement. Thus the contour integral is exactly the selected orthogonal projector. The local contour projectors agree on overlaps because they select the same ordered block, so $P(k,t)$ is jointly $C^2$ on the whole product. The general homotopy argument above now applies.

If $V$ is a global $C^2$ periodic orthonormal frame, then $A=V^{\ast }dV$ is a globally defined $C^1$ matrix-valued one-form. By [F2] and its proof, $\mathrm{tr}F=d\mathrm{tr}A$ (the trace of $A\wedge A$ vanishes). Hence $C(P)=\frac{i}{2\pi }{\int }_{{\mathbb{T}}_{2\pi }^2}d\mathrm{tr}A=0$ by [F4], since the torus has empty boundary.