Projector geometry, adiabatic transport and Berry curvature — Examples

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 $\mathrm{\Omega }={\int }_0^{2\pi }{\int }_0^{\theta }\sin udud\varphi =2\pi (1-\cos \theta )$.

On ${\mathbb{C}}^2$, let $e_{\uparrow }=(1,0{)}^T$, $e_{\downarrow }=(0,1{)}^T$, and $H(s)=E_0(2s-1)\mathrm{diag}(1,-1)$ for $s\in [0,1]$. For $s<1\mathrm{/}2$ the ordered upper eigenvalue is $E_0(1-2s)$ and its projection is $P_{\mathrm{u}\mathrm{p}}(s)=e_{\downarrow }{e}_{\downarrow }^{\ast }$; for $s>1\mathrm{/}2$ it is $E_0(2s-1)$ and its projection is $P_{\mathrm{u}\mathrm{p}}(s)=e_{\uparrow }{e}_{\uparrow }^{\ast }$. At $s=1\mathrm{/}2$, $H(s)=0$, so the degenerate upper eigenspace is all of ${\mathbb{C}}^2$ and its spectral projection is $I_2$; the one-sided rank-one projections therefore have no continuous rank-one extension. The gap between the two eigenvalues is $2E_0\mathrm{\mid }2s-1\mathrm{\mid }$, which closes at $s=1\mathrm{/}2$.

The propagator from $s=0$ solves $i\mathrm{\hslash }{U}_T^{\mathrm{\prime }}(s)=TH(s)U_T(s)$ and is $U_T(s)=\exp \text{\hspace{0.17em}⁣}(-\frac{iT{E}_0}{\mathrm{\hslash }}(s^2-s)\mathrm{diag}(1,-1)),$ as direct differentiation verifies. In particular $U_T(1)=I_2$.

At the endpoints, $P_{\mathrm{u}\mathrm{p}}(0)=e_{\downarrow }{e}_{\downarrow }^{\ast }$ and $P_{\mathrm{u}\mathrm{p}}(1)=e_{\uparrow }{e}_{\uparrow }^{\ast }$. Therefore $\parallel (I-P_{\mathrm{u}\mathrm{p}}(1))U_T(1)P_{\mathrm{u}\mathrm{p}}(0)\parallel =\mathrm{\parallel }{e}_{\downarrow }{e}_{\downarrow }^{\ast }\mathrm{\parallel }=1$ for every $T>0$. No finite constant $C$ independent of $T$ can bound this leakage by $C\mathrm{/}T$. The ordered projection fails the continuity and uniform-gap hypotheses in [F1].

The constant projection $P_{\downarrow }=e_{\downarrow }{e}_{\downarrow }^{\ast }$ commutes with every $H(s)$ and has zero leakage for all $T$. At the crossing it is not the spectral projection onto the degenerate eigenspace, which is all of ${\mathbb{C}}^2$; it is a continuous invariant rank-one subspace, not a coherently followed gapped spectral cluster. Thus the example refutes a gap-free uniform tracking claim for the ordered branch; it does not assert that every choice of state at a crossing causes transitions.