Projector geometry, adiabatic transport and Berry curvature

Fix $x_0\in M$ and choose an orthonormal basis $e_1,\dots ,e_r$ of $\mathrm{ran}P(x_0)$. The vectors $P(x)e_j$ depend $C^2$ on $x$. Their Gram matrix is the identity at $x_0$, so its determinant remains positive on a neighborhood of $x_0$; hence the vectors remain linearly independent there. Applying the finite Gram–Schmidt formulas uses only sums, products, division by positive norms, and square roots of positive functions, so it gives a $C^2$ orthonormal frame $V(x)=(v_1(x),\dots ,v_r(x))$ for $\mathrm{ran}P(x)$ on that neighborhood. The local bundle chart is $\mathrm{\Phi }(x,z)=(x,V(x)z)$, with inverse ${\mathrm{\Phi }}^{-1}(x,v)=(x,V(x{)}^{\ast }v)$; both maps are $C^2$. On an overlap, the frame change is $z\mapsto V_{\beta }(x{)}^{\ast }{V}_{\alpha }(x)z$, a $C^2$ map into the invertible complex matrices, so the transition maps are $C^2$. These charts make $E_P$ a rank-$r$ $C^2$ complex vector bundle. For $r=0$ the bundle is the zero bundle, and for $r=d$ every $P(x)=I_d$ and the bundle is trivial.

Along $\gamma$, abbreviate $P_{\gamma }$ by $P$. Differentiating $P^2=P$ gives $P^{\mathrm{\prime }}P+P{P}^{\mathrm{\prime }}=P^{\mathrm{\prime }}$ and $P{P}^{\mathrm{\prime }}P=0$. Therefore $K=P^{\mathrm{\prime }}P-P{P}^{\mathrm{\prime }}$ is $C^1$, $K^{\ast }=-K$, and $[K,P]=P^{\mathrm{\prime }}P+P{P}^{\mathrm{\prime }}=P^{\mathrm{\prime }}$. Let $M_K={\sup }_{t\in [0,1]}\mathrm{\parallel }K(t)\mathrm{\parallel }$, set $V_0(t)=I_d$, and recursively set $V_{n+1}(t)={\int }_0^{t}K(u)V_n(u)du$. Induction gives $\mathrm{\parallel }{V}_n(t)\mathrm{\parallel }\le (M_{K}t{)}^n\mathrm{/}n!$, so $W={\sum }_{n=0}^{\mathrm{\infty }}{V}_n$ converges uniformly and satisfies $W(t)=I_d+{\int }_0^{t}K(u)W(u)du$. The fundamental theorem of calculus gives $W^{\mathrm{\prime }}=KW$; since $K\in C^1$, also $W\in C^2$. If $Z$ is the difference of two solutions, iteration of $Z(t)={\int }_0^{t}K(u)Z(u)du$ gives $\mathrm{\parallel }Z(t)\mathrm{\parallel }\le {\sup }_{[0,1]}\mathrm{\parallel }Z\mathrm{\parallel }(M_{K}t{)}^n\mathrm{/}n!$ for every $n$, whose right side tends to zero, so $Z=0$.

For a $C^1$ scalar function $f$ and a $C^1$ section $s$ with $Ps=s$, the product rule gives $\mathrm{\nabla }(fs)=Pd(fs)=P(dfs+fds)=dfs+f\mathrm{\nabla }s$, so $\mathrm{\nabla }$ satisfies the connection Leibniz rule. For two such sections $s,t$, $d(s^{\ast }t)=(Pds{)}^{\ast }t+s^{\ast }(Pdt)$ because $P=P^{\ast }$ and $Ps=s$, $Pt=t$; thus $\mathrm{\nabla }$ preserves the Hermitian fiber metric.

Differentiation yields $(W^{\ast }W{)}^{\mathrm{\prime }}=W^{\ast }(K^{\ast }+K)W=0$, hence $W^{\ast }W=I_d$; a square matrix with a left inverse is invertible, so $W$ is unitary. Also $(W^{\ast }PW{)}^{\mathrm{\prime }}=W^{\ast }(P^{\mathrm{\prime }}-[K,P])W=0$, so $W(t{)}^{\ast }P(t)W(t)=P(0)$. Multiplying by $W(t)$ on the left gives $P(t)W(t)=W(t)P(0)$, as required.

By compactness and [F2], choose finitely many $C^2$ straightening neighborhoods of $D$ with compact coordinate closures, using the local models in the Statement at points on $\mathrm{\partial }M$. At an interior boundary arc with parametrization $\gamma (s)$, choose a transverse vector $v$ and choose the signs of $s$ and $t$ so the coordinate map $(s,t)\mapsto \gamma (s)+tv$ is orientation-preserving and $D$ lies in one half-plane; its derivative is invertible, so [F8] gives a $C^2$ inverse chart. At a corner in an interior chart, extend each incident arc across its endpoint by its second-order Taylor polynomial on the missing side; the value and the first two one-sided derivatives agree at the endpoint, so this is a $C^2$ extension. Order the two arc parameters so that $(s,t)\mapsto {\gamma }_1(s)+{\gamma }_2(t)-{\gamma }_1(0)$ is orientation-preserving. Its derivative is invertible by the linear independence of the one-sided tangent vectors, so [F8] gives a $C^2$ local diffeomorphism. After shrinking, the local model is a quadrant or its reentrant complement as in the Statement. Add $M\setminus D$ to this finite open cover and use [F3] for a smooth subordinate partition of unity. On $D$, only finitely many localized forms ${\eta }_i={\rho }_i\eta$ contribute, each with compact support in one straightening chart.

In an oriented coordinate plane, a compactly supported $C^1$ form $adx+bdy$ satisfies ${\int }_{{\mathbb{R}}^2}({\mathrm{\partial }}_{x}b-{\mathrm{\partial }}_{y}a)=0$, ${\int }_{y\ge 0}({\mathrm{\partial }}_{x}b-{\mathrm{\partial }}_{y}a)={\int }_{\mathbb{R}}a(x,0)dx$, and ${\int }_{x,y\ge 0}({\mathrm{\partial }}_{x}b-{\mathrm{\partial }}_{y}a)={\int }_0^{\mathrm{\infty }}a(x,0)dx-{\int }_0^{\mathrm{\infty }}b(0,y)dy$. For complex coefficients, apply the calculation to real and imaginary parts. Put the support in a rectangle; [F9] makes the coefficients and their first derivatives Riemann integrable, and [F10] makes them uniformly continuous. Double Riemann sums and iterated one-dimensional Riemann sums have the same limit, since their difference is bounded by the rectangle area times the modulus of continuity at the mesh diameter. Then [F7] evaluates the one-dimensional integrals of ${\mathrm{\partial }}_{x}b$ and ${\mathrm{\partial }}_{y}a$ at their endpoints, and the compact support removes the artificial outer edges. The half-plane and quadrant boundary signs are exactly the outward-normal-first convention [F6]. On the reentrant model, subtract the missing open quadrant from the plane; the two axis integrals have the opposite orientation, so the same identity holds.

A $C^2$ straightening map preserves the exterior derivative of a $C^1$ one-form by the chain rule: the terms involving its second derivatives cancel by symmetry. The compact-support change-of-variables formula [F4] carries each local area identity to $D$, while [F6] carries the axis orientations to the oriented boundary arcs. Summing over the partition from step 1.1 gives ${\int }_{D}d\eta ={\int }_{\mathrm{\partial }D}\eta$. The empty region and empty boundary give zero on both sides.

Differentiate $P^2=P$ in a tangent direction $X$ to obtain $(dP)(X)P+P(dP)(X)=(dP)(X)$. Multiplying on both sides by $P$ gives $P(dP)(X)P=0$. For a $C^2$ section $s$ with $Ps=s$, expand $F_{\mathrm{\nabla }}(X,Y)s$ by the curvature definition in the Statement. The second-derivative terms cancel by $X(ds(Y))-Y(ds(X))-ds([X,Y])=0$, leaving $P(dP(X)ds(Y)-dP(Y)ds(X))$. Substituting $ds(X)=dP(X)s+Pds(X)$ and using $P(dP)P=0$ leaves $P(dP(X)dP(Y)-dP(Y)dP(X))Ps$. This is $P(dP\wedge dP)Ps$, proving the projector formula.

In a local frame write a section as $s=Va$. Then $\mathrm{\nabla }(Va)=Pd(Va)=V(V^{\ast }dVa+da)=V(da+Aa)$, where $A=V^{\ast }dV$. Differentiating $V^{\ast }V=I_r$ gives $(dV{)}^{\ast }V+V^{\ast }dV=0$, so $A^{\ast }=-A$. Applying $d+A$ twice gives $dA+A\wedge A$; expressing the already computed curvature in this frame gives $F_{\mathrm{f}\mathrm{r}}=V^{\ast }FV=dA+A\wedge A$.

For $V^{\mathrm{\prime }}=Vg$, direct differentiation gives $A^{\mathrm{\prime }}=V^{\mathrm{\prime }\ast }d{V}^{\mathrm{\prime }}=g^{-1}Ag+g^{-1}dg$, since $g^{\ast }=g^{-1}$. The frame-coordinate covariant derivative changes from $d+A$ to $g^{-1}(d+A)g$ on coefficient sections. Squaring this identity gives $d{A}^{\mathrm{\prime }}+A^{\mathrm{\prime }}\wedge A^{\mathrm{\prime }}=g^{-1}(dA+A\wedge A)g$, which is the claimed curvature covariance.

For a path the definitions give $W^{\mathrm{\prime }}=[P^{\mathrm{\prime }},P]W$; [F2] proves that its unique solution is unitary and satisfies $P(t)W(t)=W(t)P(0)$. This is projector parallel transport and uses no Schrödinger evolution.

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 $P$.

Divide the evolution equation by $\mathrm{\hslash }$ and set $\tau =T\mathrm{/}\mathrm{\hslash }>0$. Then $i{\mathrm{\partial }}_s{U}_T=\tau H(s)U_T$, exactly the scale convention of [F1], with the same $H$, $P$, $K$, and $X$. The supplier's hypotheses hold: $H$ is $C^2$, the cluster is coherently followed, and its complementary spectral distance is uniformly at least $g>0$. Therefore its estimate gives $\mathrm{\parallel }(I-P(s))U_T(s)P(0)\mathrm{\parallel }\le C\mathrm{/}\tau =\mathrm{\hslash }C\mathrm{/}T$.

Suppose $\mathrm{rank}P=1$ and $HP=EP$. By [F2], $W^{\mathrm{\prime }}=KW$, $P(s)W(s)=W(s)P(0)$, and $W$ is unitary. Set $V(s)=W(s)\exp (-i\tau {\int }_0^{s}E(u)du)$. For $v\in \mathrm{ran}P(0)$, intertwining gives $W(s)v\in \mathrm{ran}P(s)$, so $H(s)W(s)v=E(s)W(s)v$. Differentiating therefore gives $V^{\mathrm{\prime }}(s)v=(-i\tau H(s)+K(s))V(s)v$, with $V(0)v=v$. The defined $U_A$ satisfies the same matrix initial-value equation $U_A^{\mathrm{\prime }}=(-i\tau H+K)U_A$; uniqueness of finite-dimensional linear ODE solutions yields $U_A(s)v=V(s)v$ for every such $v$, which is the stated factorization.