Finite-rank gapped spectral cluster under a bounded perturbation

[Gap and correctly ordered Neumann inverse] By [F1], the spectral support of $H_0$ is $\sigma (H_0)$. For $z\in \mathrm{\Gamma }$, $\mathrm{\mid }z-E\mathrm{\mid }=g\mathrm{/}2$ and every other spectral point $t$ has $\mathrm{\mid }z-t\mathrm{\mid }\ge \mathrm{\mid }t-E\mathrm{\mid }-\mathrm{\mid }z-E\mathrm{\mid }\ge g\mathrm{/}2$. Thus $\mathrm{\parallel }{R}_0(z)\mathrm{\parallel }\le 2\mathrm{/}g$. The operator $H_{\lambda }$ is closed on $D$: if $x_n\to x$ and $H_{\lambda }{x}_n\to y$, then $H_0{x}_n\to y-\lambda Vx$, so closedness of $H_0$ gives $x\in D$ and $H_{\lambda }x=y$; the domain remains dense. On $D$, $z-H_{\lambda }=(I-\lambda V{R}_0(z))(z-H_0)$. Its first factor has the Neumann inverse [F2], since $\mathrm{\parallel }\lambda V{R}_0(z)\mathrm{\parallel }\le q_{\lambda }<1$. Therefore the displayed ordered series gives a two-sided inverse with range exactly $D$, and its geometric estimates give (a). Uniform convergence on the circle and each compact coupling subdisk proves norm holomorphy in $\lambda$; local inverse Neumann series give holomorphy in $z$ on the resolvent set. When $v=0$ this series is constant for all couplings.

[Idempotence from two contours] Fix $\lambda \in {\mathrm{\Omega }}_v$. Step 1.1 bounds the inverse uniformly on $\mathrm{\Gamma }$, and another inverse Neumann series gives a resolvent annulus around it. Integrals of a holomorphic operator-valued function over concentric circles in that annulus are independent of radius: differentiating ${\int }_0^{2\pi }f(E+\rho e^{it})i\rho e^{it}dt$ in $\rho$ gives the integral of the periodic $t$ derivative of $f(E+\rho e^{it})e^{it}$, hence zero. Take an outer circle for $z$ and an inner circle for $w$. The inverse equations give $R_{\lambda }(z)R_{\lambda }(w)=[R_{\lambda }(w)-R_{\lambda }(z)]\mathrm{/}(z-w)$, and also commutation of resolvents. In its double integral, the first term integrates in the outer variable to the inner projection integral because $\oint dz\mathrm{/}(z-w)=2\pi i$; the second integrates in the inner variable to zero because $\oint dw\mathrm{/}(z-w)=0$. Compact continuous integrals can be interchanged by norm Riemann sums. Annular independence identifies both circle integrals with $P_{\lambda }$, proving $P_{\lambda }^2=P_{\lambda }$. Resolvent commutation also gives $P_{\lambda }{R}_{\lambda }(z)=R_{\lambda }(z)P_{\lambda }$.

[Graph domains, commutation and holomorphy] The identity $H_{\lambda }{R}_{\lambda }(z)=z{R}_{\lambda }(z)-I$ makes the contour integrand continuous into the complete graph space of the closed $H_{\lambda }$. Integrating there proves $P_{\lambda }H\subset D$ and $H_{\lambda }{P}_{\lambda }=(2\pi i{)}^{-1}{\oint }_{\mathrm{\Gamma }}z{R}_{\lambda }(z)dz=:C_{\lambda }\in \mathcal{B}(H)$. On $D$ the other inverse equation gives $R_{\lambda }(z)H_{\lambda }=z{R}_{\lambda }(z)-I$, so integration gives $P_{\lambda }{H}_{\lambda }=C_{\lambda }$ there. Thus the range and kernel reduce the operator with its domain. The uniform coupling series makes both $P_{\lambda }$ and $C_{\lambda }$ norm holomorphic. Since $H_0{P}_{\lambda }=C_{\lambda }-\lambda V{P}_{\lambda }$, the pair $(P_{\lambda },H_0{P}_{\lambda })$ is holomorphic into $\mathcal{B}(H,H\oplus H)$ and takes values in the closed graph of $H_0$; hence $P_{\lambda }$ is holomorphic as an operator into that graph space. In particular, the finite restriction $A_{\lambda }$ is well-defined on all of $M_{\lambda }$.

[Initial projection and rank throughout the disk] At zero coupling use the sign conversion in [F1]: $R_0(z)$ has multiplier $(z-t{)}^{-1}$, whose counterclockwise integral divided by $2\pi i$ is $1$ inside and $0$ outside. Thus the resolvent contour is the PVM projection on the real interior; its only unperturbed spectral point is $E$, so this is $P_0$. This evaluation follows by spectral pairings: scalar circle kernels integrate to one inside and zero outside [F3]; norm Riemann sums converge uniformly in the spectral variable, since the spectrum is at distance at least $g\mathrm{/}2$ from the circle, allowing interchange with its finite scalar measures. The circle length $\pi g$ and step 1.1 give $\mathrm{\parallel }{P}_{\lambda }-P_0\mathrm{\parallel }\le q_{\lambda }\mathrm{/}(1-q_{\lambda })$. If bounded idempotents $P,Q$ have $\mathrm{\parallel }P-Q\mathrm{\parallel }<1$, then $x\in \mathrm{ran}P$ and $Qx=0$ imply $\mathrm{\parallel }x\mathrm{\parallel }=\mathrm{\parallel }(P-Q)x\mathrm{\parallel }<\mathrm{\parallel }x\mathrm{\parallel }$ unless $x=0$; therefore $Q{\mathrm{\mid }}_{\mathrm{ran}P}$ is injective, and the reverse restriction is also injective. For $s,t\in [0,1]$, the inverse equations give $R_{s\lambda }(z)-R_{t\lambda }(z)=(s-t)\lambda R_{s\lambda }(z)V{R}_{t\lambda }(z)$, hence $\mathrm{\parallel }{P}_{s\lambda }-P_{t\lambda }\mathrm{\parallel }\le q_{\lambda }\mathrm{\mid }s-t\mathrm{\mid }\mathrm{/}(1-q_{\lambda }{)}^2$. Choose an integer $N>q_{\lambda }\mathrm{/}(1-q_{\lambda }{)}^2$ and compare the finite path $P_{j\lambda \mathrm{/}N}$, $0\le j\le N$. At each adjacent pair the two injections and finite dimension imply equal rank. Starting with rank $r$ proves rank $r$ at $\lambda$, even when the direct distance estimate is at least one. If $q_{\lambda }=0$ the path is constant.

[All spectral values of the finite restriction are inside] Put $M=M_{\lambda }$, $A=A_{\lambda }$, $P=P_{\lambda }$ and $Q=I-P$. For $z\in \mathrm{\Gamma }$, resolvent commutation and the inverse equations give $R_{\lambda }(z){\mathrm{\mid }}_M=(z-A{)}^{-1}$ on $M$. If $Au=\alpha u$ with $u\mathrm{\ne }0$, then $\alpha \mathrm{\notin }\mathrm{\Gamma }$ by those equations and $R_{\lambda }(z)u=(z-\alpha {)}^{-1}u$. The equation $Pu=u$ and its contour definition give winding number one for $\alpha$, so $\alpha \in \mathrm{\Delta }$. In a finite basis of $M$, $w-A$ is invertible exactly when its kernel is zero [F3]; thus every spectral point of $A$ is an eigenvalue and lies in $\mathrm{\Delta }$. There are at most $r$ distinct such points, since eigenvectors for distinct eigenvalues are linearly independent: in a shortest nontrivial relation among them, applying $A-\alpha I$ for one term removes that term and leaves a shorter nontrivial relation. There is at least one: otherwise $(z-A{)}^{-1}$ is holomorphic throughout a neighborhood of the closed disk, and the scalar Cauchy formula [F3] makes its contour integral zero entrywise, contradicting its integral $I_M$ and $r\ge 1$. Here the zero integral follows by applying the Cauchy formula to $(z-E)(z-A{)}^{-1}$ at $E$.

[Complementary inverse on its exact domain] For $w\in \mathrm{\Delta }$ define $B(w)=(2\pi i{)}^{-1}{\oint }_{\mathrm{\Gamma }}{R}_{\lambda }(z)\mathrm{/}(z-w)dz$. As in step 3.1 its range lies in $D$. The identity $(w-H_{\lambda })R_{\lambda }(z)=I-(z-w)R_{\lambda }(z)$ gives $(w-H_{\lambda })B(w)=I-P$ on $H$; its right-hand counterpart gives $B(w)(w-H_{\lambda })=I-P$ on $D$. Resolvent commutation shows $B(w)$ commutes with $P,Q$. The bounded map $B_Q(w)=QB(w)Q$ therefore maps $QH$ into $D\cap QH$ and satisfies both inverse equations for $w-H_{\lambda }$ on that exact complementary domain. No assertion that $B(w)P=0$ is needed. The idempotent splitting is $H=M\oplus QH$ and, since $PH\subset D$ and $P$ preserves $D$, its domain splitting is $D=M\oplus (D\cap QH)$. The complementary domain is dense in $QH$, since applying the bounded $Q$ to approximants from the dense $D$ preserves $D$ and approximates every vector of $QH$.

[Exactly the interior spectrum and generalized eigenspaces] If $w\in \mathrm{\Delta }\setminus \sigma (A)$, then the bounded operator $(w-A{)}^{-1}P+B_Q(w)$ maps $H$ into $D$ and is a two-sided inverse of $w-H_{\lambda }$ by the domain splitting and step 4.2. If $w\in \sigma (A)$, step 4.1 supplies a nonzero eigenvector in $M$, so $w\in \sigma (H_{\lambda })$. This proves $\sigma (H_{\lambda })\cap \mathrm{\Delta }=\sigma (A)$, with its nonempty finite character from step 4.1. For such an eigenvalue $\alpha$ and an integer $k\ge 1$, any generalized eigenvector $x\in D((H_{\lambda }-\alpha {)}^k)$ has $Qx$ in the complementary iterated domain, because $P,Q$ preserve $D$ and commute with $H_{\lambda }$. Step 4.2 makes $H_{\lambda }-\alpha$ injective on the complement; its powers are then injective on their domains, so $(H_{\lambda }-\alpha {)}^{k}Qx=0$ forces $Qx=0$. Conversely $M\subset D$ is invariant, so every generalized eigenvector of $A$ is a generalized eigenvector of the full operator. This proves (c), including the exact complementary inverse.

[Real coupling, zero branch and assembly] The bounded-perturbation adjoint identity is $D(H_{\lambda }^{\ast })=D(H_0^{\ast })=D$, $H_{\lambda }^{\ast }=H_0+\bar{\lambda }V$: subtract the everywhere-defined bounded term from the adjoint functional to identify exactly the domain of $H_0^{\ast }$. Thus $H_{\lambda }$ is self-adjoint for every real $\lambda$. For real $\lambda \in {\mathrm{\Omega }}_v$, apply [F1] to this operator, again converting $(H_{\lambda }-z{)}^{-1}$ to $R_{\lambda }(z)=-(H_{\lambda }-z{)}^{-1}$ with multiplier $(z-t{)}^{-1}$. Its spectrum misses the circle by step 1.1, so has positive distance from it by compactness of the circle and openness of the resolvent set. The same uniform scalar-kernel/Riemann-sum evaluation as in step 3.2 identifies $P_{\lambda }$ with the PVM on the real interior, hence an orthogonal spectral projection on the entire cluster. If $v=0$, the norm definition gives $V=0$, all resolvents and projections are constant for every complex coupling, and the exact PVM domain gives $H_0{P}_0=E{P}_0$; therefore the restriction is $EI$ and its spectrum is $\{E\}$. Steps 1.1–5.1 give the other assertions for arbitrary $H$ under CC, using only a finite path and finite bases; no individually ordered or holomorphic scalar branches have been introduced.