Exact Bloch transform and regularity of gapped band projectors

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.