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ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)
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Gap closings and Chern regions in the two-band lattice model

Example

Assume Axiom of Choice and Countable Choice. For M∈R, consider the two-orbital finite-range Bloch Hamiltonian

hM(k)=sin⁡k1 σ1+sin⁡k2 σ2+(M+cos⁡k1+cos⁡k2) σ3,

on the oriented torus T2π2. Its spectral gap closes exactly when M∈{−2,0,2}. For each fixed M outside this set, the lower and upper band Chern numbers in the convention C(P)=i2π∫tr⁡(P(dP∧dP)P) are

Parameter rangedeg⁡(nM)C(P−)C(P+)
M<−2000
−2<M<0+1+1−1
0<M<2−1−1+1
M>2000

Each Chern number is constant on its parameter interval. At the three closing values a separated band projector, and therefore its Chern number, is not defined.

Facts & Assumptions

Given: Axiom of Choice, Countable Choice, the real parameter M, the stated Bloch Hamiltonian, the torus orientation, and the Chern convention in the example.

[A1]

Axiom of Choice and Countable Choice are the principles in The Axiom of Choice and The Axiom of Countable Choice (ACω).

[F1]

Two-band Chern number and degree formula gives C(P−)=deg⁡(nM) and C(P+)=−deg⁡(nM) for nM=dM/∣dM∣ when dM is nowhere zero.

[F2]

Regular-value formula for degree computes the degree of a smooth map between closed oriented surfaces by the signed number of preimages of any regular value.

Verification

Given: [A1], [F1], [F2], the Bloch matrix, and the standard orientations on the torus and sphere.

1.1givenalgebra

Write dM=(sin⁡k1,sin⁡k2,M+cos⁡k1+cos⁡k2). The band energies are ±∣dM∣, so the gap is 2∣dM∣ and closes exactly if all three components vanish. The first two vanish only when each kj is 0 or π modulo 2π; the third component at (0,0),(π,0),(0,π),(π,π) is respectively M+2,M,M,M−2. Thus a zero exists exactly at M=−2,0,2. For every other fixed M, compactness of the torus and continuity give min⁡k∣dM(k)∣>0, so the two bands have a uniform gap. The sine and cosine terms are finite Fourier sums with nearest-neighbor support, hence this is a finite-range Hermitian Bloch matrix.

2.1F2step 1.1givenalgebra

Fix M outside the closing set and let N=(0,0,1)∈S2. The preimages of N under nM=dM/∣dM∣ are exactly the four candidate points above for which the third component is positive. Near such a point, use the oriented coordinates (n1,n2) at N. The derivative matrix is diagonal with entries cos⁡k1/d3 and cos⁡k2/d3, where d3=M+cos⁡k1+cos⁡k2>0; its determinant is nonzero. Therefore N is a regular value, and the local orientation signs at (0,0),(π,0),(0,π),(π,π), whenever present, are respectively +1,−1,−1,+1. By [F2], degree is the sum of the signs for those points with d3>0.

3.1A1F1step 2.1givenalgebra∎

If M<−2, no candidate has d3>0 and the sum is 0. If −2<M<0, only (0,0) contributes, giving +1. If 0<M<2, (0,0),(π,0),(0,π) contribute +1−1−1=−1. If M>2, all four contribute +1−1−1+1=0. Substitution into [F1] gives every table entry. For any two values in one of the four intervals, the closed segment between them remains in that interval; the joint function ∣dM(k)∣ is positive on the compact torus times that segment, so it has a positive minimum and supplies a gap-preserving deformation.

Depends on

Dependency tree · 2 levels

Sources