C2 projector curvature has integral Chern number

If $\mathrm{\Sigma }=\mathrm{\varnothing }$, the integral is the empty sum $0$ by [F11], and $[\mathrm{\Sigma }]=0$ by [F6], so its pairing with $c_1(E_P)$ is $0$ and the conclusion holds. Henceforth assume $\mathrm{\Sigma }\mathrm{\ne }\mathrm{\varnothing }$. Choose a finite coordinate-chart cover of the compact surface $\mathrm{\Sigma }$ and a smooth partition $\{{\varphi }_j{\}}_{j=1}^N$ subordinate to it by [F3]. In each chart, every matrix entry of ${\varphi }_{j}P$ is a $C^2$ function supported compactly inside the coordinate domain; extend it by zero to ${\mathbb{R}}^2$. Start with $\rho (y)=c\exp (-1\mathrm{/}(1-\mathrm{\mid }y{\mathrm{\mid }}^2))$ for $\mathrm{\mid }y\mathrm{\mid }<1$ and $\rho (y)=0$ otherwise, where $c>0$ normalizes its integral to one. The extension is smooth because every derivative at the boundary is a finite sum of terms bounded by $t^{-m}{e}^{-1\mathrm{/}t}$ as $t\downarrow 0$, and these tend to zero. Put ${\rho }_{ϵ}(y)={ϵ}^{-2}\rho (y\mathrm{/}ϵ)$ and convolve each real and imaginary entry with it. For every multi-index $\alpha$ with $\mathrm{\mid }\alpha \mathrm{\mid }\le 2$, ${\mathrm{\partial }}^{\alpha }({\rho }_{ϵ}\ast f)={\rho }_{ϵ}\ast ({\mathrm{\partial }}^{\alpha }f)$ and $\mathrm{\parallel }{\rho }_{ϵ}\ast ({\mathrm{\partial }}^{\alpha }f)-{\mathrm{\partial }}^{\alpha }f{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\sup }_{\mathrm{\mid }y\mathrm{\mid }\le ϵ}\mathrm{\parallel }{\mathrm{\partial }}^{\alpha }f(\cdot -y)-{\mathrm{\partial }}^{\alpha }f{\mathrm{\parallel }}_{\mathrm{\infty }}⟶0,$ because each compactly supported derivative is uniformly continuous. For sufficiently small $ϵ$, the convolved supports remain inside their charts. Summing the finitely many pulled-back approximants gives a smooth matrix field $Q_{ϵ}\to P$ in $C^2$; replacing it by $(Q_{ϵ}+Q_{ϵ}^{\ast })\mathrm{/}2$ keeps it smooth Hermitian and preserves this convergence.

Fix $ϵ$ so $\delta :={\sup }_{\mathrm{\Sigma }}\mathrm{\parallel }{Q}_{ϵ}-P\mathrm{\parallel }<1\mathrm{/}4$. For each $x\in \mathrm{\Sigma }$, define $R_{ϵ}(x)=\frac{1}{2\pi i}{\oint }_{\mathrm{\mid }z-1\mathrm{\mid }=1\mathrm{/}2}(zI-Q_{ϵ}(x){)}^{-1}dz\mathrm{.}$ On this contour $\mathrm{\parallel }(zI-P(x){)}^{-1}\mathrm{\parallel }=2$, so the Neumann series shows $zI-Q_{ϵ}(x)$ is invertible and its inverse has norm at most $4$. The finite-dimensional spectral theorem [F4] diagonalizes $Q_{ϵ}(x)$. For an eigenvalue $\lambda$ inside the circle, the Cauchy integral formula [F4] gives ${\oint }_{\mathrm{\Gamma }}(z-\lambda {)}^{-1}dz=2\pi i$. For an exterior eigenvalue put $c=1$, $R=1\mathrm{/}2$, and $q=R\mathrm{/}\mathrm{\mid }\lambda -c\mathrm{\mid }<1$. On $\mathrm{\Gamma }$, the geometric expansion $\frac{1}{z-\lambda }=-{\sum }_{k=0}^N\frac{(z-c{)}^k}{(\lambda -c{)}^{k+1}}+r_N(z),{\sup }_{\mathrm{\Gamma }}\mathrm{\mid }{r}_N\mathrm{\mid }\le \frac{q^{N+1}}{\mathrm{\mid }\lambda -c\mathrm{\mid }(1-q)}$ has a remainder whose contour integral tends to zero, since the circle has length $2\pi R$. Each polynomial term integrates to zero, as the parametrization $z=c+R{e}^{it}$ gives ${\oint }_{\mathrm{\Gamma }}(z-c{)}^{k}dz=i{R}^{k+1}{\int }_0^{2\pi }{e}^{i(k+1)t}dt=0$. Thus the exterior eigenvalue contributes zero. Applying these two scalar calculations to the diagonalization shows that $R_{ϵ}(x)$ is the orthogonal spectral projection onto eigenvalues inside the circle. The same contour is valid for $Q_t=P+t(Q_{ϵ}-P)$ for every $t\in [0,1]$, so these projections vary continuously with $t$; their integer-valued ranks are constant and equal the rank one of $P$. The resolvent identity gives $\mathrm{\parallel }{R}_{ϵ}-P{\mathrm{\parallel }}_{\mathrm{\infty }}\le \frac{1}{2\pi }\pi (4)\delta (2)=4\delta <1.$ The contour integral also shows $R_{ϵ}$ is smooth in $x$.

The restriction $R_{ϵ}(x){\mathrm{\mid }}_{\mathrm{ran}P(x)}$ is injective: if $R_{ϵ}v=0$ and $Pv=v$, then $\mathrm{\parallel }v\mathrm{\parallel }=\mathrm{\parallel }(P-R_{ϵ})v\mathrm{\parallel }\le \mathrm{\parallel }P-R_{ϵ}\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$, which forces $v=0$. Both ranges are one-dimensional, so these restrictions form a continuous complex line-bundle isomorphism $E_P\cong E_{R_{ϵ}}$ and [F6] gives $c_1(E_P)=c_1(E_{R_{ϵ}})$. The compact surface has finitely many components by [F6]; apply the disconnected line-class convention of [F5] on each component and sum the pairings using the fundamental-class direct sum. In a local unit frame $u$ for $E_{R_{ϵ}}$, its connection form is $A=u^{\ast }du$ and its scalar curvature is $dA$. The rank-one curvature matrix is $F_{R_{ϵ}}=u(dA)u^{\ast }$, hence $\mathrm{tr}{F}_{R_{ϵ}}=dA={\mathrm{\Omega }}_{{\mathrm{\nabla }}_{ϵ}}$. The smooth integral here is the chart-partition integral [F10], agreeing with the continuous convention [F11]; its coordinate independence uses the common-refinement change-of-variables argument [F8]. Applying [F5] to the smooth Hermitian line bundle $E_{R_{ϵ}}$ and its projected Hermitian connection, then evaluating on $[\mathrm{\Sigma }]$ using [F6] and [F7], gives $\frac{i}{2\pi }{\int }_{\mathrm{\Sigma }}\mathrm{tr}{F}_{R_{ϵ}}=⟨c_1(E_{R_{ϵ}}),[\mathrm{\Sigma }]⟩=⟨c_1(E_P),[\mathrm{\Sigma }]⟩\in \mathbb{Z}\mathrm{.}$

The contour formula for $R_{ϵ}$ and its first two coordinate derivatives show $R_{ϵ}\to P$ in $C^2$: differentiating $(zI-Q_{ϵ}{)}^{-1}$ once gives $(zI-Q_{ϵ}{)}^{-1}({\mathrm{\partial }}_j{Q}_{ϵ})(zI-Q_{ϵ}{)}^{-1}$, and differentiating twice gives a finite sum of products of uniformly bounded resolvents with first and second derivatives of $Q_{ϵ}$. Thus in each chart of the fixed finite partition, $R_{ϵ}$ and its first derivatives converge uniformly to $P$ and its first derivatives. On the compact supports of that partition, let ${\delta }_0$ and ${\delta }_1$ be the largest of the zeroth- and first-coordinate-derivative errors, and choose $M\ge 1$ bounding all first derivatives of $P$ and $R_{ϵ}$ for sufficiently small $ϵ$. For the coordinate coefficient $F_{12}(P)=P({\mathrm{\partial }}_{1}P{\mathrm{\partial }}_{2}P-{\mathrm{\partial }}_{2}P{\mathrm{\partial }}_{1}P)P$, submultiplicativity and $\mathrm{\parallel }P\mathrm{\parallel }=\mathrm{\parallel }{R}_{ϵ}\mathrm{\parallel }=1$ give $\mathrm{\parallel }{F}_{12}(R_{ϵ})-F_{12}(P)\mathrm{\parallel }\le 4M^2{\delta }_0+4M{\delta }_1$; the trace difference is at most $d$ times this bound. Hence the continuous coefficients in the fixed finite chart-partition formula [F11] converge uniformly. Put each compact chart support in the interior of a closed nondegenerate rectangle $Q_j$. By [F12], both real and imaginary parts of each continuous zero-extended coefficient and its difference are Riemann integrable there. For a real continuous $h$ on $Q_j$, put $c={\sup }_{Q_j}\mathrm{\mid }h\mathrm{\mid }$. Every Darboux sum of the constant functions $\pm c$ is $\pm c\mathrm{vol}(Q_j)$, since the cell volumes sum to the rectangle volume. Monotonicity in [F9] applied to $-c\le h\le c$ therefore gives $\mathrm{\mid }{\int }_{Q_j}h\mathrm{\mid }\le c\mathrm{vol}(Q_j)$. Applying this separately to the real and imaginary parts of a complex coefficient $g$ and using $\mathrm{\mid }\mathrm{Re}g\mathrm{\mid },\mathrm{\mid }\mathrm{Im}g\mathrm{\mid }\le \mathrm{\mid }g\mathrm{\mid }$ gives $\mathrm{\mid }{\int }_{Q_j}g\mathrm{\mid }\le 2\mathrm{vol}(Q_j){\sup }_{Q_j}\mathrm{\mid }g\mathrm{\mid }$. Finite linearity then gives $\mid {\int }_{\mathrm{\Sigma }}\mathrm{tr}{F}_{R_{ϵ}}-{\int }_{\mathrm{\Sigma }}\mathrm{tr}{F}_P\mid \le 2{\sum }_j\mathrm{vol}(Q_j){\sup }_{Q_j}\mathrm{\mid }{\tilde{f}}_j^{ϵ}-{\tilde{f}}_j\mathrm{\mid }⟶0$. The continuous-integral convention [F11] and this estimate show that the limit is the integral in the Statement. Step 2.1 gives the same integer for every sufficiently small $ϵ$, so taking the limit proves the stated equality for $P$.