Single-particle Bloch bundles and band topology

For finitely supported $\psi$, each shift ${\psi }_n\mapsto {\psi }_{n-\delta }$ preserves the ${\mathrm{\ell }}^2$ norm, so the triangle inequality gives $\mathrm{\parallel }{H}_0\psi {\mathrm{\parallel }}_2\le ({\sum }_{\delta \in S}\mathrm{\parallel }{T}_{\delta }\mathrm{\parallel })\mathrm{\parallel }\psi {\mathrm{\parallel }}_2$. Hence $H_0$ extends uniquely to a bounded operator $H$. For finitely supported $\psi ,\varphi$, reindexing the finite sum gives $⟨H_0\psi ,\varphi ⟩={\sum }_{m,\delta }⟨T_{\delta }{\psi }_m,{\varphi }_{m+\delta }⟩={\sum }_{m,\delta }⟨{\psi }_m,T_{\delta }^{\ast }{\varphi }_{m+\delta }⟩=⟨\psi ,H_0\varphi ⟩$, because $T_{\delta }^{\ast }=T_{-\delta }$ and $S=-S$. Density and boundedness extend symmetry to all of $\mathcal{H}$, so $H=H^{\ast }$.

Under $k=2\pi x$, the functions $e^{-ik\cdot n}$ are the characters $e_{-n}(x)$. For each internal coordinate $a$, the corresponding scalar character family is complete and Parseval by [F1] and [F2]; summing the $r$ coordinate norm identities proves $\mathrm{\parallel }{\mathcal{F}}_0\psi {\mathrm{\parallel }}_{\mathcal{K}}=\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathcal{H}}$. This norm identity agrees with the conjugate-linear-first pairings fixed in the Definition. The range contains every vector-valued trigonometric polynomial, a dense subspace of $\mathcal{K}$ by [F1] coordinatewise. Hence ${\mathcal{F}}_0$ extends uniquely to an isometry $\mathcal{F}:\mathcal{H}\to \mathcal{K}$. Its range is closed because $\mathcal{H}$ is complete, and contains a dense subspace, so it is all of $\mathcal{K}$. Therefore $\mathcal{F}$ is unitary.

Direct substitution and the change of index $m=n-\delta$ give, for finitely supported $\psi$, ${\mathcal{F}}_0(H_0\psi )(k)={\sum }_{\delta \in S}{T}_{\delta }{e}^{-ik\cdot \delta }{\sum }_m{\psi }_m{e}^{-ik\cdot m}=h(k){\mathcal{F}}_0\psi (k)$. The finite sum defining $h$ is continuous and satisfies ${\sup }_k\mathrm{\parallel }h(k)\mathrm{\parallel }\le {\sum }_{\delta \in S}\mathrm{\parallel }{T}_{\delta }\mathrm{\parallel }$, so multiplication by $h$ is bounded on $L^2$. Extending the identity from the dense finitely supported sequences proves the claimed unitary equivalence. Also, $h(k{)}^{\ast }={\sum }_{\delta }{T}_{\delta }^{\ast }{e}^{ik\cdot \delta }={\sum }_{\delta }{T}_{-\delta }{e}^{-ik\cdot (-\delta )}=h(k)$, and its entries are trigonometric polynomials, hence smooth.

Write the eigenvalues of $h(k)$ in nondecreasing order as ${\lambda }_1(k)\le \cdots \le {\lambda }_r(k)$. For a Hermitian matrix $A$ with eigenbasis $e_1,\dots ,e_r$, its $j$th eigenvalue satisfies ${\lambda }_j(A)={\max }_{\dim L=r-j+1}{\min }_{\mathrm{\parallel }v\mathrm{\parallel }=1,v\in L}⟨Av,v⟩$: every such $L$ intersects $\mathrm{span}(e_1,\dots ,e_j)$ nontrivially, while $L=\mathrm{span}(e_j,\dots ,e_r)$ attains the value. Since every unit-vector Rayleigh quotient changes by at most $\mathrm{\parallel }A-B\mathrm{\parallel }$ when $A$ is replaced by $B$, this formula gives $\mathrm{\mid }{\lambda }_j(A)-{\lambda }_j(B)\mathrm{\mid }\le \mathrm{\parallel }A-B\mathrm{\parallel }$. Thus each ${\lambda }_j(k)$ is continuous.

Fix $k_0$. The gap and continuity in step 1.4 let us choose disjoint small circles around the distinct eigenvalues in the selected block at $k_0$ so that, on some neighborhood of $k_0$, all selected eigenvalues stay inside their union and all complementary eigenvalues stay outside. The inverse matrix $(zI-h(k){)}^{-1}$ is $C^{\mathrm{\infty }}$ in $k$ on this fixed contour: differentiating the inverse gives ${\mathrm{\partial }}_j(zI-h{)}^{-1}=(zI-h{)}^{-1}({\mathrm{\partial }}_{j}h)(zI-h{)}^{-1}$, and repeated differentiation gives finite sums of products of smooth matrices. Therefore its contour integral is $C^{\mathrm{\infty }}$. At each $k$, [F3] diagonalizes $h(k)$ and [F4] gives $2\pi i$ for an eigenvalue inside a component circle. For an exterior eigenvalue $\lambda$, write that circle as $z=c+R{e}^{it}$ and put $q=R\mathrm{/}\mathrm{\mid }\lambda -c\mathrm{\mid }<1$. The finite geometric identity gives $(z-\lambda {)}^{-1}=-{\sum }_{j=0}^N(z-c{)}^j\mathrm{/}(\lambda -c{)}^{j+1}+r_N(z)$ with $\sup \mathrm{\mid }{r}_N\mathrm{\mid }\le q^{N+1}\mathrm{/}(\mathrm{\mid }\lambda -c\mathrm{\mid }(1-q))$. Each polynomial term integrates to zero because $\oint (z-c{)}^{j}dz=i{R}^{j+1}{\int }_0^{2\pi }{e}^{i(j+1)t}dt=0$, and the remainder integral is bounded by $2\pi R\sup \mathrm{\mid }{r}_N\mathrm{\mid }\to 0$. Thus the scalar exterior integral is zero. Summing the component circles gives coefficient $1$ on the selected eigenvalues and $0$ on their complement, proving that the matrix contour integral is exactly the selected orthogonal projector. On overlaps the local definitions project onto the same selected eigenspaces, so they agree and define the asserted global projector.

In the standard orthonormal wedge basis from [F4], the matrix entry of ${\bigwedge }^{m}T$ in row $I$, column $J$ is $\det T_{I,J}$ by [F3]. Consequently $[({\bigwedge }^{m}T{)}^{\ast }{]}_{I,J}=\overline{\det T_{J,I}}=\det (T^{\ast }{)}_{I,J}$, so $({\bigwedge }^{m}T{)}^{\ast }={\bigwedge }^m(T^{\ast })$. Applying this to $P=P^{\ast }$ shows $Q=Q^{\ast }$.

Functoriality [F2] gives $Q^2={\bigwedge }^m(P^2)={\bigwedge }^{m}P=Q$. Hence $Q$ is a Hermitian idempotent by step 1.1; for $x=Qy$ and $Qz=0$, $⟨x,z⟩=⟨Qy,z⟩=⟨y,Qz⟩=0$, so its range and kernel are orthogonal and it is the orthogonal projection onto its range.

Every decomposable wedge in the image of $Q$ is a wedge of $m$ vectors in $\mathrm{im}P$, so the full image lies in ${\bigwedge }^m\mathrm{im}P$. Conversely, for a basis $v_1,\dots ,v_m$ of the $m$-dimensional range of $P$, $Q(v_1\wedge \cdots \wedge v_m)=v_1\wedge \cdots \wedge v_m\mathrm{\ne }0$. The increasing-wedge basis theorem [F4], applied to this basis of the range, says ${\bigwedge }^m\mathrm{im}P$ is one-dimensional; thus it equals $\mathrm{im}Q$ and $Q$ has rank one.

Each matrix entry of $Q(x)$ is a polynomial minor in the entries of $P(x)$ by [F3], hence is $C^2$. Together with step 2.1 this proves that $Q$ is a $C^2$ rank-one orthogonal projector with the stated range, and its ranges form the determinant line subbundle.

Because the columns of $V$ are orthonormal, $w=v_1\wedge \cdots \wedge v_m$ has unit norm and spans the determinant line. Differentiating by the product rule gives $dw={\sum }_j{v}_1\wedge \cdots \wedge d{v}_j\wedge \cdots \wedge v_m$. The projected connection applies $P$ to each differentiated factor. Since $Pd{v}_j={\sum }_a{v}_a{A}_{aj}$, each term with $a\mathrm{\ne }j$ vanishes from a repeated wedge factor, while the $a=j$ term is $A_{jj}w$. Summing gives $Qdw=(\mathrm{tr}A)w$, so the determinant connection form is $a=\mathrm{tr}A$.

A line connection with scalar one-form $a$ has curvature $da+a\wedge a=da$, since a scalar one-form wedges with itself to zero. By [F3], $\mathrm{tr}F=\mathrm{tr}(dA)+\mathrm{tr}(A\wedge A)$. In coordinates, $\mathrm{tr}(A\wedge A)={\sum }_{i,j}{A}_{ij}\wedge A_{ji}=0$: diagonal terms vanish and each off-diagonal pair cancels because one-forms anticommute. Therefore $da=d\mathrm{tr}A=\mathrm{tr}{F}_{\mathrm{f}\mathrm{r}}=\mathrm{tr}(V^{\ast }FV)$. Finally, since $PFP=F$ and $V{V}^{\ast }=P$, cyclicity of finite-dimensional trace gives $\mathrm{tr}(V^{\ast }FV)=\mathrm{tr}(FV{V}^{\ast })=\mathrm{tr}(FP)=\mathrm{tr}F$.

If ${\phi }^{\mathrm{\prime }}=\phi +\delta \psi$ is a second representative of the same real cohomology class and $c^{\mathrm{\prime }}=c+\mathrm{\partial }b$ a second representative of the same real homology class, then ${\phi }^{\mathrm{\prime }}(c^{\mathrm{\prime }})=\phi (c)+\phi (\mathrm{\partial }b)+(\delta \psi )(c)+(\delta \psi )(\mathrm{\partial }b)=\phi (c)$: the three additional terms vanish respectively by $\delta \phi =0$, $\mathrm{\partial }c=0$, and ${\mathrm{\partial }}^2=0$. Thus the real evaluation in [F7], the real-bilinear version of the Kronecker pairing [F4], is independent of representatives. For every continuous $f:X\to Y$, the chain-level formulas $f^{\ast }\phi =\phi \circ f_{\mathrm{\#}}$ and $f_{\ast }c=f_{\mathrm{\#}}c$ give $(f^{\ast }\phi )(c)=\phi (f_{\mathrm{\#}}c)$, so $⟨f^{\ast }[\phi ],[c]{⟩}_{\mathbb{R}}=⟨[\phi ],f_{\ast }[c]{⟩}_{\mathbb{R}}$. Linearity of $\phi$ and $c$ makes the pairing real-bilinear.

In spherical coordinates $x=\sin \theta \cos \varphi$, $y=\sin \theta \sin \varphi$, $z=\cos \theta$, the unnormalized displayed two-form is $\sin \theta d\theta \wedge d\varphi$. Integrating for $0\le \theta \le \pi$ and $0\le \varphi \le 2\pi$ gives $4\pi$, so ${\omega }_{S^2}$ is a smooth closed form with integral one; its displayed ambient expression shows it extends smoothly across the coordinate poles.

Let $J$ be the natural comparison in [F3]. By [F5] and the real evaluation defined in [F7], ${\int }_{{\mathbb{T}}_{2\pi }^2}{n}^{\ast }{\omega }_{S^2}=⟨J([n^{\ast }{\omega }_{S^2}]),[{\mathbb{T}}_{2\pi }^2{]}_{\mathbb{R}}{⟩}_{\mathbb{R}}$. Naturality of $J$ and the real-chain pairing proved in step 1.1 gives ${\int }_{{\mathbb{T}}_{2\pi }^2}{n}^{\ast }{\omega }_{S^2}=⟨J([{\omega }_{S^2}]),n_{\ast }[{\mathbb{T}}_{2\pi }^2{]}_{\mathbb{R}}{⟩}_{\mathbb{R}}$.

By [F1], $n_{\ast }[{\mathbb{T}}_{2\pi }^2{]}_{\mathbb{Z}}=\mathrm{deg}(n)[S^2{]}_{\mathbb{Z}}$. The coefficient map $\iota :C_{\bullet }(X;\mathbb{Z})\to C_{\bullet }(X;\mathbb{R})$ sends each finite sum ${\sum }_j{a}_j{\sigma }_j$ to ${\sum }_j{a}_j{\sigma }_j$ with the same coefficients viewed in $\mathbb{R}$. It commutes with boundaries because [F6] defines the singular boundary as an integer linear combination of faces and $\iota$ is additive; it commutes with $n_{\mathrm{\#}}$ because [F6] postcomposes each simplex without changing its coefficient. Hence it induces a coefficient-change map on homology and carries each integral fundamental class to its real fundamental class: it sends the positive local orientation generator $1$ to $1$, so uniqueness of the class with those local restrictions in [F2] identifies the image. Applying $\iota$ to the degree equation gives $n_{\ast }[{\mathbb{T}}_{2\pi }^2{]}_{\mathbb{R}}=\mathrm{deg}(n)[S^2{]}_{\mathbb{R}}$. By the real bilinearity in [F7] and steps 2.1, ${\int }_{{\mathbb{T}}_{2\pi }^2}{n}^{\ast }{\omega }_{S^2}=\mathrm{deg}(n)⟨J([{\omega }_{S^2}]),[S^2{]}_{\mathbb{R}}{⟩}_{\mathbb{R}}=\mathrm{deg}(n){\int }_{S^2}{\omega }_{S^2}=\mathrm{deg}(n)$ by [F5] and step 1.2.

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.

Direct multiplication of the displayed Pauli matrices gives ${\sigma }_a{\sigma }_b={\delta }_{ab}I+i{\sum }_c{ϵ}_{abc}{\sigma }_c$, $\mathrm{tr}{\sigma }_a=0$, and $\mathrm{tr}({\sigma }_a{\sigma }_b)=2{\delta }_{ab}$. Thus $(n\cdot \sigma {)}^2=\mathrm{\mid }n{\mathrm{\mid }}^{2}I=I$, and $h=d_{0}I+\mathrm{\mid }d\mathrm{\mid }(n\cdot \sigma )$ has eigenvalues $d_0\pm \mathrm{\mid }d\mathrm{\mid }$ with the stated orthogonal projectors. Since $\mathrm{\mid }d\mathrm{\mid }$ is continuous and positive on compact ${\mathbb{T}}^2$, its minimum is positive, so the two bands are uniformly separated; $n$ and $P_s$ are smooth.

Differentiating $P_s=(I+sn\cdot \sigma )\mathrm{/}2$ gives ${\mathrm{\partial }}_j{P}_s=(s\mathrm{/}2)({\mathrm{\partial }}_{j}n)\cdot \sigma$. The Pauli product identity yields $[a\cdot \sigma ,b\cdot \sigma ]=2i(a\times b)\cdot \sigma$, hence $[{\mathrm{\partial }}_1{P}_s,{\mathrm{\partial }}_2{P}_s]=\frac{i}{2}({\mathrm{\partial }}_{1}n\times {\mathrm{\partial }}_{2}n)\cdot \sigma$. Using $\mathrm{tr}(P_s{\sigma }_a)=s{n}_a$, which follows from the trace identities in step 1.1, we obtain $\mathrm{tr}(P_s[{\mathrm{\partial }}_1{P}_s,{\mathrm{\partial }}_2{P}_s])=\frac{is}{2}n\cdot ({\mathrm{\partial }}_{1}n\times {\mathrm{\partial }}_{2}n)$. In dimension two this is the coefficient of $\mathrm{tr}(P_s(d{P}_s\wedge d{P}_s)P_s)$: cyclicity and $P_s^2=P_s$ give $\mathrm{tr}(P_{s}X{P}_s)=\mathrm{tr}(P_{s}X)$. Multiplying by $i\mathrm{/}(2\pi )$ therefore gives $C(P_s)=-s(4\pi {)}^{-1}\int n\cdot ({\mathrm{\partial }}_{1}n\times {\mathrm{\partial }}_{2}n)d{k}_{1}d{k}_2$.

The normalized area form in [F3] pulls back to $(4\pi {)}^{-1}n\cdot ({\mathrm{\partial }}_{1}n\times {\mathrm{\partial }}_{2}n)d{k}_1\wedge d{k}_2$, by the defining outward area form on the unit sphere. Hence its integral is $\mathrm{deg}(n)$ by [F3]. Substitution into step 2.1 gives $C(P_s)=-s\mathrm{deg}(n)$, which is the two displayed upper- and lower-band formulas.