# On the Spectrum of the Periodic Anderson-Bernoulli Model

William Wood<EMAIL_ADDRESS>Department of Mathematics at University of
California, Irvine

(August 27, 2024)

###### Abstract

We analyze the spectrum of a discrete Schrödinger operator with a potential
given by a periodic variant of the Anderson Model. In order to do so, we study
the uniform hyperbolicity of a Schrödinger cocycle generated by the
$SL(2,\mathbb{R})$ transfer matrices. In the specific case of the potential
generated by an alternating sequence of random values we show that the almost
sure spectrum consists of at most 4 intervals.

††preprint: AIP/123-QED

##  I Introduction

Discrete Schrödinger operators
$H_{{\color[rgb]{0,0,0}V}}:l^{2}(\mathbb{Z})\rightarrow l^{2}(\mathbb{Z})$

$H_{{\color[rgb]{0,0,0}V}}\phi(n)=\phi(n+1)+\phi(n-1)+V(n)\phi(n)$ (1)

are the focus of this paper. Here, $V=\\{V(n)\\}_{n\in\mathbb{Z}}$ is a
bounded, real sequence, and the spectrum of $H_{V}$ is the set

${\color[rgb]{0,0,0}\sigma(H_{V})=\\{\lambda:H_{V}-\lambda\text{ does not have
a bounded inverse}\\}.}$

The spectral type (absolutely continuous, singular continuous, pure point, or
mixed) and the topological structure of the spectrum of these operators are
objects of study, see d1 for more detail. Since (1) is a bounded, self-adjoint
operator in a Hilbert space, its spectrum $\sigma(H_{V})$ is a compact subset
of $\mathbb{R}$. If $V$ is a periodic sequence with period $m$, the spectrum
consists of, at most, $m$ intervals. If the potential is constant, the
spectrum is the set $[-2,2]+V(0)$.

There are plenty of more complicated examples, such as the Almost Mathieu
Operator. If $V_{AM}(n)=\lambda\cos(n\pi\alpha)$ where $\lambda\neq 0,$
$\alpha\notin\mathbb{Q},$ and $n\in\mathbb{Z}^{{\color[rgb]{0,0,0}+}}$, then
$\sigma(H_{AM})$ is a Cantor set. The spectrum was proven to be a Cantor set
by Avila and JitomirskayaAJ , and this problem was known as the Ten Martini
Problem. A nice survey of the subject was provided by Marx and Jitormiskaya mj
. Similarly, if the potential is given by the Fibonacci substitution sequence,
the spectrum is a Cantor set as welld1 ; d4 .

The last example is the Anderson Model, where the potential depends on random
variables. If $V(n)$ is a random potential, given by a sequence of iid random
variables, then there exists an almost-sure spectrum $\sigma_{as}$, and we
have

$\sigma_{as}(H_{V})=\sigma(\triangle)+supp\;\mu$

Here $\triangle$ is the free Laplacian, $\sigma(\triangle)=[-2,2]$,  and $\mu$
is the random variable. Additional details about this model can be found in s
.

This paper will address a model inspired by the Anderson Model. If the
sequence $\\{V(n)\\}_{n\in\mathbb{Z}}$ is given by random variables that are
not identically distributed, how can the spectrum be characterized?
Question: Suppose that $m\in\mathbb{N},$ and
$\nu_{0},\nu_{2},\cdots\;\nu_{m-1}$ are probability distributions on
$\mathbb{R}$ with finite support. Suppose that a random potential $\\{V(n)\\}$
is given by a sequence of independent random variables, such that $V(n)$ is
distributed with respect to $\nu_{n}\;(mod\;m)$. Does the spectrum
$\sigma(H_{V})$ almost surely consist of a finite number of intervals?

To the best of my knowledge, this problem is open. This paper addresses one
specific case, when $m=2$.

Let us denote by $B(p)$ a Bernoulli random variable that takes value 1 with
probability $p\in(0,1)$, and value 0 with probability $1-p$.

######  Theorem 1.

For any real $\lambda_{0},\lambda_{1},c_{0},c_{1},$ a discrete Schrödinger
operator with random potential, $V(n),$ defined by the sequence of
distributions:

$\nu(n)=\begin{cases}\lambda_{0}B(p_{0})+c_{0}&n\text{ is even}\\\
\lambda_{1}B(p_{1})+c_{1}&n\text{ is odd}\end{cases}$ (2)

the almost sure spectrum $\sigma_{as}(H_{{\color[rgb]{0,0,0}V}})$ consists of
at most 4 closed intervals

Notice that if $\lambda_{0}=0$ or $\lambda_{1}=0$, then the potential has a
constant element. In section $4E$ we provide an explicit formula for the
spectrum in terms of the parameters
$\lambda_{0},\lambda_{1},c_{0},\;\&\;c_{1}$.

After this paper was written, I became aware of the preprint d5 "Spectral
Characteristics of Schrödinger operators generated by product systems" written
by David Damanik, Jake Fillman, and Philipp Gohlke who independently proved
the same result. We agreed to post our texts to ArXiv at the same time.

##  II Preliminaries

Here, we are going to examine the uniform hyperbolicity of the cocycle
generated by transfer matrices. Specifically, we will look at the transfer
matrices of the Schrödinger operator with potential given by $(2)$. The
concept of uniform hyperbolicity is presented in subsection 2A. Uniform
hyperbolicity has similar, yet distinct, definitions when applied to cocyles
and sets of matrices. In the context of this paper, the specific cocycle
$(T,A_{E})$ generated by the transfer matrices is found to be uniformly
hyperbolic if and only if a related set of matrices is uniformly hyperbolic.

In subsection 2B, Johnson’s Theorem and transfer matrices are introduced.
Johnson’s Theoremz ; jo states that an energy $E$ belongs  to the almost sure
spectrum of the ergodic family of operators $\\{H_{V}\\}$,
$E\in\sigma_{as}(H_{V})$, if and only if the relevant cocycle $(T,A_{E})$
(generated by the transfer matrices) is not uniformly hyperbolic. For a brief
history of Johnson’s Theorem, and how it originated in a paper by Russell
Johnson in 1986, see z2 and jo . Depending on how well the potential
${\color[rgb]{0,0,0}V}(n)$ is understood, the respective cocycle can be
analyzed. In section $2C$, speculations about the broader case mentioned in
the above question will be made and shown to meet Johnson’s Theorem’s
conditions. Lemma 1 is also introduced and proven, which states the uniform
hyperbolicity of the respective cocycle is equivalent to the uniform
hyperoblicity of a related set of matrices. Last, in section $2D$, we
hypothesize about the question from section $1$.

In section 3, lemma 2, which is key to proving Theorem 1, is outlined,  The
poof of the lemma, however, is provided in section $4D$. Lemma $2$ states an
equivalence exists between the relevant set of matrices being uniformly
hyperbolic and a condition that is much easier to calculate.

In section 4, we go into proving lemma 2. The first three subsections are
dedicated to providing necessary lemmas. The second to last subsection goes
into proving lemma $2$, and the last subsection explicitly defines the
interval the complement of the spectrum.

###  II.1 Uniform Hyperbolicity

First, we will need to define what it means that a cocycle or a set of
matrices is uniformly hyperbolic aby ; y ; ca . We will begin by talking about
sets of matrices.

######  Definition 1.

A set of matrices $\\{A_{n}\\}\subset SL(2,\mathbb{R})$ is _uniformly
hyperbolic_ if there exists the constant $\lambda>1$ such that, for any
product $\prod A_{i}$ of $m$ matrices, the inequality $||\prod
A_{i}||>\lambda^{m}$ holds.

In this product, duplicates are allowed. Due to Avila, Bochi, and Yoccoz aby ,
we have the following Theorem.

######  Theorem 2.

A finite set of $2\times 2$ matrices $\\{A_{i}\\}$ is uniformly hyperbolic if
and only if there is a nontrivial open set $C\subsetneq\mathbb{P}^{1}$
(hereafter referred to as a cone) such that
$\overline{C}\subsetneq\mathbb{P}^{1}$ is a finite number of intervals and
$A_{i}(\overline{C})\subset C$ for each matrix $A_{i}$ from the set.

For this paper, the projective space, $\mathbb{P},$ will be parameterized by
$\begin{bmatrix}x\\\ y\end{bmatrix}\leftrightarrow\frac{x}{y}$. The simplest
example of a uniformly hyperbolic set of matrices is when $C$ is a principal
cone.

######  Definition 2.

If the cone $C$ consists of a single interval, it is called a _principal
cone_.

If the set $\\{A_{i}\\}$ is a set of $2\times 2$ matrices, a principal cone
forms if and only if all the unstable eigenvectors (here the unstable
eigenvector of $A_{i}$ will be expressed as  $u_{i}$) lie in a single interval
in the projective space that contains none of the stable eigenvectors (here
the stable eigenvector of $A_{i}$ will be expressed as $s_{i}$). The cone,
$C$, can be defined as that interval.

Before covering the second definition of uniform hyperbolicity, we need to
define a cocyle.

######  Definition 3.

Given a metric space $\Omega$, homeomorphism $T:\Omega\rightarrow\Omega$ and
continuous $B:\Omega\rightarrow SL(2,\mathbb{R})$, a _cocycle_ is the map

$(T,B):(\Omega,\mathbb{R}^{2})\rightarrow(\Omega,\mathbb{R}^{2})$
$(\omega,\vec{v})\mapsto(T(\omega),B(\omega)\vec{v}).$

######  Definition 4.

Denoting $(T,B)^{n}=(T^{n},B_{n})$ where

$B_{n}(\omega)=B(T^{n-1}(\omega))\cdot B(T^{n-2}(\omega))\cdots
B(T(\omega))B(\omega).$

A cocycle $(T,B)$ is _uniformly hyperbolic_ if there exist $\lambda>1$ and
$C>0$ such that $||B_{n}(\omega)||>C\lambda^{n}$ for all $n\in\mathbb{Z}^{+}$
and $\omega\in\Omega$.

More general definitions of hyperbolicity are availabley , but we do not use
them here.

###  II.2 Johnson’s Theorem and Transfer Matrices

Given a dynamical system $(\Omega,T,\mu)$ with compact metric space $\Omega$,
probability measure $\mu$, ergodic $T$, and a continuous function
$f:\Omega\rightarrow\mathbb{R}$, an ergodic potential $V$ of a Schrödinger
operator is defined by the equation

${\color[rgb]{0,0,0}V}_{\omega}(n)=f(T^{n}\omega)$ (3)

for some $\omega\in\Omega$. To provide a simple example, if $\Omega$ is a
finite set and $T$ is a bijection, then ${\color[rgb]{0,0,0}V}_{\omega}(n)$
would be a periodic potential. Any potential with period m can be defined as
such with $\Omega=\mathbb{Z}_{m}$.

In order to study operators with an ergodic potential, it would be useful to
define a transfer matrix. First, if, the continuous functions
$f:\Omega\rightarrow\mathbb{R}$ and $g:\mathbb{R}\rightarrow SL(2,\mathbb{R})$
are given, then $(T,g\circ f(\cdot))$ would be a $SL(2,\mathbb{R})-$cocycle.
Here ${\color[rgb]{0,0,0}(T,g\circ
f)}:(\Omega,{\color[rgb]{0,0,0}\mathbb{R}^{2}})\rightarrow(\Omega,{\color[rgb]{0,0,0}\mathbb{R}^{2}})$
acts by $(\omega,\vec{v})\mapsto(T\omega,g\circ f(\omega)\cdot\vec{v})$. Next,
given some Schrödinger operator $H$ with ergodic potential
${\color[rgb]{0,0,0}V_{\omega}}$, assuming there is some $E$ such that
$E\phi(n)=H_{\omega}\phi(n)$ and taking into account equation $(1)$, we have

$\begin{bmatrix}\phi(n+1)\\\
\phi(n)\end{bmatrix}=\begin{bmatrix}E-{\color[rgb]{0,0,0}V_{\omega}}(n)&-1\\\
1&0\end{bmatrix}\cdot\begin{bmatrix}\phi(n)\\\ \phi(n-1)\end{bmatrix}$

We arrive at the transfer matrix

$A_{E}(n)=g_{E}\circ{\color[rgb]{0,0,0}V_{\omega}}(n)=\begin{bmatrix}E-{\color[rgb]{0,0,0}V_{\omega}}(n)&-1\\\
1&0\end{bmatrix}$ (4)

where $g_{E}:\mathbb{R}\rightarrow SL(2,\mathbb{R})$ is given by

$g_{E}(x)=\begin{bmatrix}E-x&-1\\\ 1&0\end{bmatrix},$

and ${\color[rgb]{0,0,0}V_{\omega}}(n)$ is the potential at the $n^{th}$
place. Transfer matrices are $SL(2,\mathbb{R})$ matrices and satisfy the
property

$\begin{bmatrix}\phi(m+1)\\\ \phi(m)\end{bmatrix}=A_{E}(m)\cdot
A_{E}(m-1)\cdots A_{E}(n)\begin{bmatrix}\phi(n)\\\ \phi(n-1)\end{bmatrix}$

assuming $E\phi=H_{\omega}\phi$. Define

$\footnotesize A^{m}(n,E)=\begin{cases}A_{E}(m+n-1)\cdot A_{E}(m+n-2)\cdots
A_{E}(n)&\text{ if }m>0\\\ I&\text{ if }m=0\\\ A_{E}(n-m)^{-1}\cdots
A_{E}(n-2)^{-1}\cdot A_{E}(n-1)^{-1}&\text{ if }m<0\end{cases}$

Denote $A^{m}_{E}=A_{E}(m-1)\cdot A_{E}(m-2)\cdots A_{E}(1)\cdot A_{E}(0)$.

In order to study the spectrum of the Schrödinger operator, we will use the
notion of uniform hyperbolicity in definition 4 applied to Schrödinger
cocycles.

We have Johnson’s Theoremz ; jo :

######  Theorem 3 (Johnson’s Theorem).

If $(\Omega,T,\mu)$ is an ergodic dynamical system, $\Omega$ is a compact
metric space, $T$ is a homeomorphism, $\mu$ is a $T$-invariant measure with
the support of the measure being $\Omega$ and $f:\Omega\rightarrow\mathbb{R}$
is continuous, then the almost sure spectrum $\sigma_{as}(H)$ of the operator
(1) with an ergodic potential defined by this dynamical system is given by

$\sigma_{as}(H)=\\{E:(T,A_{E})\text{ is not uniformly hyperbolic}\\}$

###  II.3 Hyperbolicity Locus and Anderson-Bernoulli Model

The paper’s main result, Theorem 1, applies to a specific random potential
$(2)$. This opens the question about a more general class of potentials.
Specifically, how would we characterize the spectrum if the potential is
defined by a periodic sequence of more than two distributions  as defined in
the question in section $1$. The model would become more complicated and some
of the tools in this paper will not apply. In fact, I would conjecture that
with large enough period, a result similar to Theorem $1$ would not hold.

Denote a random potential as being defined by $m$ distributions like:

$\nu(n)=\begin{cases}\lambda_{0}B(p_{0})+c_{0}&n\equiv 0\mod m\\\
\lambda_{1}B(p_{1})+c_{1}&n\equiv 1\mod m\\\ \hskip 20.00003pt\vdots&\hskip
10.00002pt\vdots\\\ \lambda_{m-1}B(p_{m-1})+c_{m-1}&n\equiv m-1\mod
m\end{cases}$ (5)

Here, $\lambda_{0},\lambda_{1},\cdots,\lambda_{m-1}$ and
$c_{0},c_{1},\cdots,c_{m-1}$ are real numbers, $m$ is fixed, and the Bernoulli
distribution $B$ takes on the values $0$ or $1$ with probability $0<p_{i}<1$
$(i=0,1,\cdots,m-1)$. Denote $X=\\{0,1\\}^{\mathbb{Z}},$
$\Omega=\mathbb{Z}_{m}\times X,$ and define the cocycle

${\color[rgb]{0,0,0}(T,A_{E}):\Omega\times\mathbb{R}^{2}\rightarrow\Omega\times\mathbb{R}^{2}.}$

Assuming $\omega=(k,x)\in\Omega,$ then we can define $k\in\mathbb{Z}_{m}$ and
$x\in X$, and furthermore, we can be define $x$ as a sequence $(\cdots
x_{n-1},x_{n},x_{n+1},\cdots)$ such that $x_{i}$ is $0$ or $1$. Defining
$T_{1}:X\rightarrow X$ to be the shift operator such that
$(T_{1}x)_{n}=x_{n+1}$, we can define

$\displaystyle T:(k,x)\mapsto(k+1\;mod\;m,\;T_{1}(x))$ (6)
$\displaystyle(T,A_{E})(\omega,\vec{v})=(T(\omega),A_{E,\omega}\vec{v}).$

Let $A_{E,\omega}$ be a transfer matrix from $(4)$ corresponding to potential
$V_{\omega}(k)$ from $(3)$. We will denote the potential defined by $\nu$ from
$(5)$ as $V_{(0,x)}$ and show it is dynamically defined and corresponds to an
ergodic family of operators. We have:

$\begin{array}[]{c}{\color[rgb]{0,0,0}f((k,x))=\lambda_{k}\cdot
x_{0}+c_{k}}\\\ \\\ V_{(0,x)}(0)=f(0,x)={\color[rgb]{0,0,0}\lambda_{0}\cdot
x_{0}+c_{0}}\\\ \\\ V_{(0,x)}(n)=f(T^{n}(0,x))=(\lambda_{n\;mod\;m})\cdot
x_{n}+(c_{n\;mod\;m})\\\ \\\
A_{E,(k,x)}=\begin{bmatrix}E-{\color[rgb]{0,0,0}V_{(0,x)}(k)}&-1\\\
1&0\end{bmatrix}\\\ \\\
{\color[rgb]{0,0,0}(T,A_{E}):\big{(}(k,x),\vec{v}\big{)}}\mapsto\big{(}(k+1\;mod\;m,T_{1}x),A_{E,(k,x)}\cdot\vec{v}\big{)}\end{array}$

Notice that we only need to consider $V_{\omega}$ with $\omega=(0,x)$, as
$V_{k,x}$ would involve shifting the distributions of $\nu$ by $k$ positions.

We can use the counting, probability measure $\mu_{0}$ on $\mathbb{Z}_{m}$.
Using probabilities $\\{p_{i}\\}_{i=1}^{m}$ from $(5)$, probability measure
$\mu_{1}$ over $X$ can be defined by

$\mu_{1}(\\{x:x_{n}=1\\})=p_{n\mod m},$

and finally $\mu$ over $\Omega$ can be the product of probability measures
$\mu_{0}\times\mu_{1}$. The systems $(\Omega_{0},T_{0})$ and
$(\Omega_{1},T_{1})$ are ergodic, and due to $T_{1}$ being mixing,
$(\Omega,T)$ is ergodic. Additionally, $\Omega$ is compact and $T-$invariant.
The cocycle $(T,A_{E}),$ is thus a Schrödinger cocycle of an ergodic family of
operators. Because $(\Omega,T)$ is ergodic and satisfies the conditions of
Theorem 3 (Johnson’s Theorem), the almost sure spectrum is equal to the set

${\color[rgb]{0,0,0}\sigma_{as}(H_{V})=\\{E:(T,A_{E})\text{ is not uniformly
hyperbolic}\\}}.$

The following lemma can be used to better analysis when the respective cocycle
is uniformly hyperbolic.

######  Lemma 1.

Consider an ergodic family of Schrödinger operators with random potentials
expressible as a possible outcome of the distribution $(5)$. The
$SL(2,\mathbb{R})-$cocycle $(T,A_{E})$ is uniformly hyperbolic if and only if
the set of distinct matrices

$\\{M:M=A^{m}_{E,(0,x)}\text{ for some }(0,x)\in{\color[rgb]{0,0,0}\Omega}\\}$

is uniformly hyperbolic for any $m\in\mathbb{Z}^{+}$.

Remark: There are at most 2 distinct elements in $\\{A_{E,(0,x)}:\text{ for
some }(0,x)\in{\color[rgb]{0,0,0}\Omega}\\}$. These elements are expressible
as

$\begin{bmatrix}E-\lambda_{0}x_{0}-c_{0}&-1\\\ 1&0\end{bmatrix}$

for $x_{0}\in\\{0,1\\}$. Similarly,
$\\{{\color[rgb]{0,0,0}A^{2}_{E,(0,x)}:(0,x)\in\Omega}\\}$ has at most 4
distinct elements, expressible as

$\begin{bmatrix}E-\lambda_{0}x_{0}-c_{0}&-1\\\
1&0\end{bmatrix}\cdot\begin{bmatrix}E-\lambda_{1}x_{1}-c_{1}&-1\\\
1&0\end{bmatrix}$

for $x_{0},x_{1}\in\\{0,1\\}$. The set
$\\{A^{m}_{E,(0,x)}:(0,x)\in{\color[rgb]{0,0,0}\Omega}\\}$ has at most $2^{m}$
matrices.

###### Proof.

$(\leftarrow)$ If the set of matrices is uniformly hyperbolic, then there
exists $\lambda>1$ such that for any ordered set $I$ of the elements in
$\\{M:M=A^{m}_{E,(0,x)}\text{ for some }(0,x)\in\Omega\\}$ where $|I|=k$,
$\big{|}\big{|}\prod_{I}M_{i}\big{|}\big{|}>\lambda^{k}$. Here $I$ can have
repeated elements. For any $(0,x)\in\Omega$ and any $k$, there exists $|I|=k$,
such that $\prod_{I}M_{i}=A^{km}_{E,(0,x)}.$ Therefore

$\Big{|}\Big{|}A^{km}_{E,(0,x)}\Big{|}\Big{|}=\Big{|}\Big{|}{\color[rgb]{0,0,0}\prod_{I}M_{i}}\Big{|}\Big{|}>\lambda^{k}=\sqrt[m]{\lambda}^{mk}.$

It is necessary, however, to bound
$\Big{|}\Big{|}A^{j}_{E,(0,x)}\Big{|}\Big{|}$ for any $j$. Next, a bound for
$\Big{|}\Big{|}A^{j}_{E,(0,x)}\Big{|}\Big{|}$ will be provided for $0<j<m.$
Given that for fixed $E,(0,x),$ and $j\in(0,m)$, we have

$\inf_{||\vec{v}||=1}\Big{|}\Big{|}A^{j}_{E,(0,x)}(\vec{v})\Big{|}\Big{|}=\Big{|}\Big{|}\big{(}A^{j}_{E,(0,x)}\big{)}^{-1}\Big{|}\Big{|}^{-1}\in(0,1)$

Keeping $E$ fixed, there are at most $\sum_{0<j<m}2^{j}$ matrices expressible
as $A^{j}_{E,(0,x)}$. So, there exists $C$ such that

$\min_{\begin{subarray}{c}0<j<m\\\
(0,x)\in\Omega\end{subarray}}\inf_{||\vec{v}||=1}\Big{|}\Big{|}A^{j}_{E,(0,x)}(\vec{v})\Big{|}\Big{|}\Big{/}\sqrt[m]{\lambda}^{j}=C>0.$

For $j\in(0,m),$ this gives the inequality

$\inf_{||\vec{v}||=1}\Big{|}\Big{|}A^{j}_{E,(0,x)}(\vec{v})\Big{|}\Big{|}\geq
C\sqrt[m]{\lambda}^{j},$

and this brings the conclusion that for all $x\in X$:

$\begin{array}[]{c}\Big{|}\Big{|}A^{mk+j}_{E,(0,x)}\Big{|}\Big{|}=\sup_{||\vec{v}||=1}\Big{|}\Big{|}A^{j}_{E,(mk\;mod\;m,x)}\cdot
A^{mk}_{E,(0,x)}\vec{v}\Big{|}\Big{|}\geq\\\ \\\
\inf_{||\vec{v}||=1}\Big{|}\Big{|}A^{j}_{E,(0,x)}(\vec{v})\Big{|}\Big{|}\cdot\Big{|}\Big{|}A^{mk}_{E,(0,x)}\Big{|}\Big{|}>C\sqrt[m]{\lambda}^{mk+j}.\end{array}$

By definition 4, the cocycle is uniformly hyperbolic.

$(\rightarrow)$ Assume the cocycle $(T,A)$ is uniformly hyperbolic, then there
exists a $\lambda>1$ and a $C>0$ such that for any $\omega\in\Omega$

$\Big{|}\Big{|}A^{k}_{E,\omega}\Big{|}\Big{|}>C\lambda^{k}$

For any ordered set $I$ (with $|I|=k$) consisting of the matrices from the set
$\\{M_{i}\\}$, there exists ${\color[rgb]{0,0,0}(0,x)}\in\Omega$ such that
$A_{E,{\color[rgb]{0,0,0}(0,x)}}^{mk}=\prod_{I}M_{i}$. Therefore,

$\begin{array}[]{c}C\lambda^{mk}<\big{|}\big{|}\prod_{I}M_{i}\big{|}\big{|}\Rightarrow
C\big{(}\lambda^{m}\big{)}^{k}<\big{|}\big{|}\prod_{I}M_{i}\big{|}\big{|}\end{array}.$

For a set of matrices to be uniformly hyperbolic, we technically need to
express such an inequality without a constant $C$. To address this, we note
that for any $\prod_{i}M_{i},$ there exists $(0,x)\in\Omega$ such that for any
$n\in\mathbb{Z}^{+}$, $(A_{E,(0,x)}^{nmk})=(\prod_{I}M_{i})^{n}$, and we have

$C\lambda^{nmk}<\Big{|}\Big{|}A^{nmk}_{E,(0,x)}\Big{|}\Big{|}=\Big{|}\Big{|}\Big{(}\prod_{i}M_{i}\Big{)}^{n}\Big{|}\Big{|}\leq\Big{|}\Big{|}\prod_{i}M_{i}\Big{|}\Big{|}^{n}.$

Taking the $n^{th}$ root,

$C^{1/n}(\lambda^{m})^{k}<\Big{|}\Big{|}\prod_{i}M_{i}\Big{|}\Big{|},$

and allowing $n$ to be arbitrarily, we get large,

$(\lambda^{m})^{k}\leq\Big{|}\Big{|}\prod_{i}M_{i}\Big{|}\Big{|}\Rightarrow(\lambda^{m/2})^{k}<\Big{|}\Big{|}\prod_{i}M_{i}\Big{|}\Big{|}.$

Therefore, the cocycle being uniformly hyperbolic from definition $4$ implies
the set of matrices is uniformly hyperbolic from definition $1$.

∎

###  II.4 Conjecture

By Johnson’s Theorem and lemma $3$, the almost-sure spectrum $\sigma(H_{V})$
consisting of an infinite number of intervals is equivalent to the set of
matrices being uniformly hyperbolic over an infinite number of intervals over
$E.$ Theorem $1$ states that if the potential of an ergodic family is given by
$(5)$ and $m=2$, the almost sure spectrum is a finite number of intervals. For
larger $m$ this theorem does not apply. Using the paper by Avila, Bochi, and
Yoccoz aby , we can consider a geometric approach to this question. There
exists open region $\mathcal{H}\subset SL(2,\mathbb{R})$ defined by

${\color[rgb]{0,0,0}\scriptsize\mathcal{H}=\\{(A_{1},\cdots,A_{m})\in
SL(2,\mathbb{R})^{m}:{\color[rgb]{0,0,0}\\{A_{i}\\}_{i=1}^{m}}\text{ is
uniformly hyperbolic}\\}.}$

We can parameterize a path in $SL(2,\mathbb{R})^{2^{m}}$ by $E$ and identify
the intervals where the path intersects $\mathcal{H}$. If the set of matrices
defined in lemma 1 parametized by $E$ define a path in
$SL(2,\mathbb{R})^{2^{m}}$ and the spectrum consists of infinitely many
intervals, then a path must intersect the boundary of $\mathcal{H}$ infinitely
often.

######  Theorem 4 (Avila-Bochi-Yoccoz aby , Theorem 4.1).

Define $\mathcal{H}\subset SL(2,\mathbb{R})^{N}$ to be a region of $N$-tuples
of matrices such that the matrices as a set is uniform hyperbolic, and let
$(A_{1},\cdots,A_{N})$ belong to the boundary of $\mathcal{H}$. Then one of
the following possibilities hold:

  1. 1.

There exists product $\prod_{I}A_{i}$ which is parabolic for some ordered set
$I$

  2. 2.

There exists product $\prod_{I}A_{i}$ which is the identity for some ordered
set $I$

  3. 3.

There exists products $\prod_{I}A_{i}$, $\prod_{J}A_{j}$, and $\prod_{K}A_{k}$
such that $(\prod_{I}A_{i})\cdot u(\prod_{J}A_{j})=s(\prod_{K}A_{k})$ for some
ordered sets $I$, $J$, $K$.  Here $u(\cdot)$ is the respective unstable
eigenvector and $s(\cdot)$ is the respective stable eigenvector.

Further research in this field can be seen in ca . The paper by Avila, Bocci,
and Yoccoz aby outlines an example in section 4.7 such that a path interests
the boundary of $\mathcal{H}$ infinitely often. It would seem possible that
for a large enough period, the spectrum may consist of an infinite number of
intervals.

##  III Key Lemma

Theorem 1 states that if a family of ergodic Schrödinger operators has a
potential given by $(2)$, then the almost sure spectrum consists of a finite
number of intervals. Applying lemma 1 with $m=2$ , we simply need to find a
way to calculate the set $\\{E:\\{A_{E,(0,x)}^{2}\\}\text{ is uniformly
hyperbolic}\\}$. We can calculate the set using the following lemma.

######  Lemma 2.

The set of matrices

$\bigg{\\{}AC,AD,BC,BD:\begin{array}[]{c}A,B,C,D\text{ are transfer matrices
and}\\\ A_{11}\geq B_{11}\text{ and }C_{11}\geq D_{11}\end{array}\bigg{\\}}$

is uniformly hyperbolic if and only if the matrices $AC,AD,BC,\;\&\;BD$ are
individually hyperbolic.

If the set is uniformly hyperbolic, all the matrices are hyperbolic. In the
regions where all the matrices are all hyperbolic, they turn out to form a
principal cone. We will identify the principal cones in the projective space
in section $4$, proving lemma $2$.

##  IV Calculations

In this section, we will provide a few technical lemmas necessary to prove
lemma $2$. We will begin with defining the appropriate transfer matrices and
their products. Then we will define the intervals over which these products
are hyperbolic. After that, over these intervals the corresponding set of
matrices will be shown to have a principal cone, proving the set is uniformly
hyperbolic.

###  IV.1 Transfer Matrices

With potential ${\color[rgb]{0,0,0}V}$ given by the equation

$\nu(n)=\begin{cases}\lambda_{0}B(p_{0})+c_{0}&n\text{ is even}\\\
\lambda_{1}B(p_{1})+c_{1}&n\text{ is odd}\end{cases}.$

It is enough to consider the case where,
$\lambda_{0},\lambda_{1},\;c_{1}{\color[rgb]{0,0,0}\geq}0,$ and $c_{0}=0$. For
any $n$, $\nu(n)$ and $\nu(n+1)$ can take on at most 2 values, so we can
define the constants

${\color[rgb]{0,0,0}\begin{array}[]{c}c_{0}=\min\limits_{n}\\{\nu(2n)\\}\\\
\\\ \lambda_{0}=\max\limits_{n}\\{\nu(2n)\\}-\min\limits_{n}\\{\nu(2n)\\}\geq
0\\\ \\\ c_{1}=\min\limits_{n}\\{\nu(2n+1)\\}\\\ \\\
\lambda_{1}=\max\limits_{n}\\{2n+1\\}-\min\limits_{n}\\{2n+1\\}\geq
0\end{array}}.$

We can also assume that $c_{0}\leq c_{1}$ without loss of generality. If a
constant is added to the potential, the spectrum of the operator is shifted by
that value, so we can shift the potential by $-c_{0}$ without altering the
spectral type or topolgical structure. This redefines the constants $c_{0}=0$
and $c_{1}$ as a nonnegative number. The four possible transfer matrices can
now be written as

$\begin{array}[]{c}A=\begin{bmatrix}E&-1\\\ 1&0\end{bmatrix}\hskip
10.00002ptB=\begin{bmatrix}E-\lambda_{0}&-1\\\ 1&0\end{bmatrix}\\\ \\\
C=\begin{bmatrix}E-c_{1}&-1\\\ 1&0\end{bmatrix}\hskip
10.00002ptD=\begin{bmatrix}E-\lambda_{1}-c_{1}&-1\\\
1&0\end{bmatrix}\end{array}$

and these will be the matrices $A,B,C,$ and $D$ in lemma $2$. For the context
of this paper it is useful to use this expression. Notice that
$A^{2}_{E,(0,x)}$ to be one of the 4 matrices $AC,AD,BC,$ or $BD$, which is
the set

$\scriptsize{\color[rgb]{0,0,0}\begin{array}[]{c}\\{AC,AD,BC,BD\\}=\Bigg{\\{}\begin{bmatrix}E&-1\\\
1&0\end{bmatrix}\cdot\begin{bmatrix}E-c_{1}&-1\\\
1&0\end{bmatrix},\begin{bmatrix}E&-1\\\
1&0\end{bmatrix}\cdot\begin{bmatrix}E-\lambda_{1}-c_{1}&-1\\\
1&0\end{bmatrix},\\\ \\\ \begin{bmatrix}E-\lambda_{0}&-1\\\
1&0\end{bmatrix}\cdot\begin{bmatrix}E-c_{1}&-1\\\
1&0\end{bmatrix},\begin{bmatrix}E-\lambda_{0}&-1\\\
1&0\end{bmatrix}\cdot\begin{bmatrix}E-\lambda_{1}-c_{1}&-1\\\
1&0\end{bmatrix}\Bigg{\\}}\\\ \\\ =\Bigg{\\{}\begin{bmatrix}E(E-c_{1})-1&-E\\\
(E-c_{1})&-1\end{bmatrix},\begin{bmatrix}E(E-\lambda_{1}-c_{1})-1&-E\\\
(E-\lambda_{1}-c_{1})&-1\end{bmatrix}\\\ \\\
\begin{bmatrix}(E-\lambda_{0})(E-c_{1})-1&-(E-\lambda_{0})\\\
(E-c_{1})&-1\end{bmatrix},\begin{bmatrix}(E-\lambda_{0})(E-\lambda_{1}-c_{1})-1&-(E-\lambda_{0})\\\
(E-\lambda_{1}-c_{1})&-1\end{bmatrix}\Bigg{\\}}.\end{array}}$

In the order listed, these matrices will be denoted as
$A_{1},A_{2},A_{3},\;\&\;A_{4}$ such that

$\begin{array}[]{cc}A_{1}=AC&A_{2}=AD\\\ \\\ A_{3}=BC&A_{4}=BD.\end{array}$
(7)

To prove lemma $2$, we need to show that if each of these four matrices is
hyperbolic, then the set of matrices $\\{A_{1},A_{2},A_{3},A_{4}\\}$ is
uniformly hyperbolic. Let us recall that an $SL(2,\mathbb{R})$ matrix is
hyperbolic if and only if an absolute value of its trace is greater than $2$.
In order to do that, we will identify the intervals over $E$ where the
absolute value of the traces of all four matrices are greater than 2.

For each of these four matrices the eigenvectors can be easily calculated. We
will benefit from the following definition and lemma first.

######  Definition 5.

Denoted $r\in\mathbb{R}$, the _signed spectral radius_ of a matrix $A\in
SL(2,\mathbb{R})$ satisfies

$\begin{array}[]{c}sign(r)=sign(tr(A))\\\ \\\ |r|=\text{spectral radius of
}A\end{array}.$

In terms of $\\{A_{i}\\}$, the respective signed spectral radius will be
written as $r_{i}$.

######  Lemma 3.

Let $r$ be a signed spectral radius of $A{\color[rgb]{0,0,0}\in
SL(2,\mathbb{R})}$. Then the following hold:

  1. 1.

If $A$ is hyperbolic, $r$ is the unstable eigenvalue, and $r^{-1}$ is the
stable eigenvalue.

  2. 2.

$r\neq 0$ as it is a root of $\lambda^{2}-tr(A_{i})\lambda+1=0$.

  3. 3.

$r>0$ if and only if the eigenvalues are positive.

  4. 4.

$r<0$ if and only if the eigenvalues are negative.

###### Proof.

By definition, $r$ is a root of $\lambda^{2}-tr(A_{i})\lambda+1=0$. Therefore
$r$ is real if and only if $A$ has strictly real roots. Because $r$ is the
eigenvalue of $A$ with a larger absolute value, and so the other eigenvalue of
$A$ must be the inverse of $r$ and so

$|r^{-1}|<1<|r|$

The trace is positive if and only if the roots of the polynomial are positive.
The roots are the eigenvectors $r$ and $r^{-1}$.

∎

We can now, more easily, explicity express the unstable $(u_{i})$ and stable
$(s_{i})$ eigenvectors of $A_{i}$.

######  Lemma 4.

For any hyperbolic $A{\color[rgb]{0,0,0}\in SL(2,\mathbb{R})}$, denote by $r$
its signed spectral radius. Then $r$ is a zero of its minimal polynomial, and
either the vectors

$\begin{bmatrix}-A_{22}+r\\\ A_{21}\end{bmatrix}\hskip 10.00002pt\&\hskip
10.00002pt\begin{bmatrix}-A_{22}+r^{-1}\\\ A_{21}\end{bmatrix}$

are eigenvectors $u_{A}$ and $s_{A}$ or one of them is the zero vector.

For $A_{1},\;A_{2},\;A_{3},\;\&\;A_{4}$, assuming $A_{i}$ is hyperbolic, the
stable and unstable eigenvectors are given by:

$u_{i}=\frac{1+r_{i}}{(A_{i})_{21}}\hskip 10.00002pt\&\hskip
10.00002pts_{i}=\frac{1+r_{i}^{-1}}{(A_{i})_{21}}$

in the projective space parameterized by $\begin{bmatrix}x\\\
y\end{bmatrix}\leftrightarrow\frac{x}{y}$.

###### Proof.

The matrix

$A=\begin{bmatrix}A_{11}&A_{12}\\\ A_{21}&A_{22}\end{bmatrix}$

is $SL(2,\mathbb{R})$. This means that the two eigenvalues are $r$ and
$r^{-1}$ if it is hyperbolic. Because $det(A-rI)=0$, the rows of $A-r$ are
multiples of each other. This is also true for $A-r^{-1}I.$ For all $E$ where
$A_{i}$ is hyperbolic, the second row of $A_{i}-r_{i}$ (or $A_{i}-r_{i}^{-1}$)
is not $\begin{bmatrix}0&0\end{bmatrix}$. If

$\begin{bmatrix}A_{21}&A_{22}-r_{i}\end{bmatrix}\neq\begin{bmatrix}0&0\end{bmatrix},$

then

$\begin{bmatrix}A_{11}-r&A_{12}\\\
A_{21}&A_{22}-r\end{bmatrix}\cdot\vec{u}=\begin{bmatrix}0\\\ 0\end{bmatrix}$

if and only if

$\begin{bmatrix}A_{21}&A_{22}-r\end{bmatrix}\cdot\vec{u}=\begin{bmatrix}0\end{bmatrix}.$

This gives

$\vec{u}\propto\begin{bmatrix}r-A_{22}\\\ A_{21}\end{bmatrix}$

assuming that $A$ is hyperbolic (here $\vec{u}\propto\vec{v}$ means there is
some scalar $\lambda$ such that $\vec{u}=\lambda\vec{v}$). The same holds true
for $r^{-1}$, with

$\vec{s}\propto\begin{bmatrix}r^{-1}-A_{22}\\\ A_{21}\end{bmatrix}$

proving the lemma.∎

Because

${\color[rgb]{0,0,0}A_{i}=\begin{bmatrix}(A_{i})_{11}&E\\\
(A_{i})_{21}&-1\end{bmatrix}}$

We can explicitly define $u_{i}$ and $s_{i}$, which we do in table $1$ below.
For the rest of the paper, we will use the notation $r_{i}$ for the spectral
radius of $A_{i},$ $u_{i}\in\mathbb{P}$ for the unstable eigenvector of
$A_{i}$, and $s_{i}\in\mathbb{P}$ for the stable eigenvector of $A_{i}$.
Defined explicitly in the table below are the values $r_{i},$ $u_{i},$ and
$s_{i}$.

Eigenvectors
---
|  $r$ |  Expression
$u_{1}$ |  $r_{1}=\frac{2-E(E-c_{1})+sgn(2-E(E-c_{1}))\sqrt{(2-E(E-c_{1}))^{2}-4}}{2}$ |  $\frac{1+r_{1}}{E-c_{1}}$
$s_{1}$ |  $r_{1}=\frac{2-E(E-c_{1})+sgn(2-E(E-c_{1}))\sqrt{(2-E(E-c_{1}))^{2}-4}}{2}$ |  $\frac{1+r_{1}^{-1}}{E-c_{1}}$
$u_{2}$ |  $r_{2}=\frac{2-E(E-c_{1}-\lambda_{1})+sgn(2-E(E-c_{1}-\lambda_{1}))\sqrt{(2-E(E-c_{1}-\lambda_{1}))^{2}-4}}{2}$ |  $\frac{1+r_{2}}{E-c_{1}-\lambda_{1}}$
$s_{2}$ |  $r_{2}=\frac{2-E(E-c_{1}-\lambda_{1})+sgn(2-E(E-c_{1}-\lambda_{1}))\sqrt{(2-E(E-c_{1}-\lambda_{1}))^{2}-4}}{2}$ |  $\frac{1+r_{2}^{-1}}{E-c_{1}-\lambda_{1}}$
$u_{3}$ |  $r_{3}=\frac{2-(E-\lambda_{0})(E-c_{1})+sgn(2-(E-\lambda_{0})(E-c_{1}))\sqrt{(2-(E-\lambda_{0})(E-c_{1}))^{2}-4}}{2}$ |  $\frac{1+r_{3}}{E-c_{1}}$
$s_{3}$ |  $r_{3}=\frac{2-(E-\lambda_{0})(E-c_{1})+sgn(2-(E-\lambda_{0})(E-c_{1}))\sqrt{(2-(E-\lambda_{0})(E-c_{1}))^{2}-4}}{2}$ |  $\frac{1+r_{3}^{-1}}{E-c_{1}}$
$u_{4}$ |  $r_{4}=\frac{2-(E-\lambda_{0})(E-c_{1}-\lambda_{1})+sgn(2-(E-\lambda_{0})(E-c_{1}-\lambda_{1}))\sqrt{(2-(E-\lambda_{0})(E-c_{1}-\lambda_{1}))^{2}-4}}{2}$ |  $\frac{1+r_{4}}{E-c_{1}-\lambda_{1}}$
$s_{4}$ |  $r_{4}=\frac{2-(E-\lambda_{0})(E-c_{1}-\lambda_{1})+sgn(2-(E-\lambda_{0})(E-c_{1}-\lambda_{1}))\sqrt{(2-(E-\lambda_{0})(E-c_{1}-\lambda_{1}))^{2}-4}}{2}$ |  $\frac{1+r_{4}^{-1}}{E-c_{1}-\lambda_{1}}$

Table 1: For every eigenvector, there is a corresponding $r$ value it depends
on.

###  IV.2 Intervals Over $E$

We will now find the intervals in $E$ where all four matrices are hyperbolic.

######  Proposition 1.

Over $E$, there are at most 5 intervals over which every matrix of the set
$\\{A_{i}\\}$ is hyperbolic.

The set $\\{E:A_{i}\text{ is hyperbolic for all }i\\}$ is the intersection of
open intervals defined by

$\bigcap_{i=1}^{4}\\{E:|tr(A_{i})|>2\\},$ (8)

and this lemma can be proven by explicitly calculating the intervals over $E$
where the statement $|tr(A_{i})|>2$ is true for all $i$. These intervals
depend on the 3 possible orderings of the elements of the set
$\\{\lambda_{0},c_{1},c_{1}+\lambda_{1}\\}.$ These orderings are listed below,
and within the list are sublists, defining the intervals that make up the set
$\\{E:A_{i}\text{ is hyperbolic for all }i\\}.$ Here, we are assuming
$c_{0}=0,c_{1}\geq 0,\lambda_{0}\geq 0,\lambda_{1}\geq 0$, as mentioned in
section $4A$.

  1. 1.

If we assume $\lambda_{0}\leq c_{1}\leq c_{1}+\lambda_{1},$ then the list of
intervals making up $\\{E:A_{i}\text{ is hyperbolic for all }i\\}$ are below.

  1. a)

$E<0$ and $E(E-c_{1})>4$

  2. b)

$0<E<\lambda_{0}$ and $(E-\lambda_{0})(E-c_{1})>4$

  3. c)

$\lambda_{0}<E<c_{1}$

  4. d)

$c_{1}<E<c_{1}+\lambda_{1}$ and $(E-\lambda_{0})(E-c_{1})>4$

  5. e)

$E>c_{1}+\lambda_{1}$ and $(E-\lambda_{0})(E-\lambda_{1}-c_{1})>4$

  1. 2.

If $c_{1}<\lambda_{0}<c_{1}+\lambda_{1},$ then $\\{E:A_{i}\text{ is hyperbolic
for all }i\\}$ is the union of intervals listed below.

    1. a)

$E<0$ and $E(E-c_{1})>4$

    2. b)

$0<E<c_{1}$ and $(E-\lambda_{0})(E-c_{1})>4$

    3. c)

$c_{1}<E<\lambda_{0}$ and $E(E-c_{1})>4$ and
$(E-\lambda_{0})(E-\lambda_{1}-c_{1})>4$

    4. d)

$\lambda_{0}<E<c_{1}+\lambda_{1}$ and $(E-\lambda_{0})(E-c_{1})>4$

    5. e)

$E>c_{1}+\lambda_{1}$ and $(E-\lambda_{0})(E-\lambda_{1}-c_{1})>4$

  1. 3.

If $c_{1}\leq c_{1}+\lambda_{1}\leq\lambda_{0},$ the the list of intervals
over which the matrices are hyperbolic are listed below.

    1. a)

$E<0$ and $E(E-c_{1})>4$

    2. b)

$0<E<c_{1}$ and $(E-\lambda_{0})(E-c_{1})>4$

    3. c)

$c_{1}<E<c_{1}+\lambda_{1}$ and $(E-\lambda_{0})(E-c_{1})>4$ and
$E(E-c_{1})>4$

    4. d)

If $c_{1}+\lambda_{1}<E<\lambda_{0}$ and $E(E-\lambda_{1}-c_{1})>4$

    5. e)

If $E>\lambda_{0}$ and $(E-\lambda_{0})(E-\lambda_{1}-c_{1})>4$

It is worth noting that these arent the explicit intervals but just defining
conditions. Depending on the constants, some of these conditions are not met
for any $E$. In the case of the trivial example, where
$c_{1}=\lambda_{0}=\lambda_{1}=0$ (looking at either ordering 1 or ordering
3), only conditions $a$ and $e$ are possible. The explicit intervals are
defined in section $4E$.

###  IV.3 Eigenvector Combinatorics

Before we can prove lemma $2$, we need some technical lemmas which can be used
to locate the eigenvectors in $\mathbb{P}^{1}$ in relation to each other.

######  Lemma 5.

Given distinct transfer matrices $A,B,C$, if the matrices $AB$ and $AC$ are
hyperbolic, then they cannot share an eigenvector. Similarly, $BA$ and $CA$
cannot share an eigenvector if they are hyperbolic.

###### Proof.

  * •

To prove the first statement, given matrices $AB$ and $AC$, assume they share
an eigenvector $\vec{v}$. Then

$AB\vec{v}=\xi_{1}AC\vec{v}=\xi_{2}\vec{v}$

for constants $\xi_{1}$ and $\xi_{2}$. Therefore

$B\vec{v}=\xi_{1}C\vec{v}\Rightarrow(B-\xi_{1}C)\vec{v}=0$

$det(B-\xi C)=(1-\xi_{1})^{2}$. For this to be zero, $\xi_{1}=1.$ This makes
$(B-C)\vec{v}=\begin{bmatrix}B_{11}-C_{11}&0\\\
0&0\end{bmatrix}\cdot\vec{v}=0$ in which case
$\vec{v}\propto\begin{bmatrix}0\\\ 1\end{bmatrix}$. For this to be an
eigenvalue of $AB,$ we get the equation

$AB\cdot\begin{bmatrix}0\\\ 1\end{bmatrix}=A\cdot\begin{bmatrix}-1\\\
0\end{bmatrix}=\xi_{2}\begin{bmatrix}0\\\ 1\end{bmatrix},$

which gives the equations $A=\begin{bmatrix}0&-1\\\ 1&0\end{bmatrix}$ and
$\xi_{2}=-1$. This makes $AB$ not hyperbolic.

  * •

To prove the second statement, assume matrices $BA$ and $CA$ which share an
eigenvector $\vec{v}$.

$BA\vec{v}=\xi_{1}CA\vec{v}=\xi_{2}\vec{v}$

Therefore,

$B(A\vec{v})=\xi_{1}C(A\vec{v})\Rightarrow(B-\xi_{1}C)(A\vec{v})=0.$

This gives the equation $A\vec{v}\propto\begin{bmatrix}0\\\ 1\end{bmatrix}$,
which implies the equation $\xi_{2}\vec{v}=B\left(A\vec{v}\right)\propto
B\begin{bmatrix}0\\\ 1\end{bmatrix}\propto\begin{bmatrix}-1\\\
0\end{bmatrix}$. Given the statement $A\vec{v}\propto\begin{bmatrix}0\\\
1\end{bmatrix},$ then the equations $A=\begin{bmatrix}0&-1\\\
1&0\end{bmatrix},$ and $\xi_{2}=\xi_{1}=1$ are true, making the product
parabolic and proves the lemma.

∎

Recognizing that the matrices $\\{A_{i}\\}$ are products of transfer matrices,
we can use this lemma to deduce some properties about the unstable and stable
matrices.

######  Corollary 1.

Assuming all the matrices in $\\{A_{i}\\}$ are hyperbolic and using $(7)$, if
$A_{i}$ and $A_{j}$ share an eigenvector (with $i<j$), then
$(i,j)\in\\{(1,4),(2,3)\\}$.

Over the intervals where the matrices are hyperbolic, the functions $u_{i}(E)$
and $s_{i}(E)$ in the projective space are well-defined and continuous over
intervals of $E$. Below we have the unstable eigenvector (red) and stable
eigenvector (blue) of matrix $A_{1}$ in $\mathbb{P}^{1}$ as a function of $E$
to help illustrate what the graph can look like.

Figure 1: Eigenvectors of $A_{1}$. Here $c_{0}=0$

Next we have graphed the eigenvectors of matrices
$A_{1},\;A_{2},\;A_{3}\;\&\;A_{4}$.  The matrices $A_{2},A_{3},\;\&\;A_{4}$
can be seen as perturbed versions of $A_{1}.$ If any of the graphs intersect
they are eigenvectors of $A_{2}$ and $A_{3}$, or they are eigenvectors of
$A_{1}$ and $A_{4}$.

Figure 2: Eigenvectors of $A_{1}$ to $A_{4}$.
$\lambda_{0}=1,\;\lambda_{1}=1.4,\;c_{0}=0,$ and $c_{1}=2$

Only a few of the eigenvector graphs can intersect, while others bound each
other over intervals of $E$. This brings us to the lemma below.

######  Lemma 6.

  * •

Define open interval $J$ over $E$ such that $A_{1}(E)$ and
$A_{2}(E,\lambda_{1})$ are hyperbolic, and there exists $\lambda_{1}^{*}>0$
such that $A_{2}(E,\lambda_{1})$ is hyperbolic for all
$\lambda_{1}\in[0,\lambda_{1}^{*}]$. Over $J$, the eigenvector $u_{2}$ (resp
$s_{2}$) is greater than $u_{1}$ (resp $s_{1}$) if and only if $\partial
u_{2}/\partial\lambda_{1}|_{\lambda_{1}=0}\geq 0$ (resp $\partial
s_{2}/\partial\lambda_{1}|_{\lambda_{1}=0}\geq 0$).

  * •

Define open interval $J$ over $E$ such that $A_{3}(E)$ and
$A_{4}(E,\lambda_{1})$ are hyperbolic, and there exists $\lambda_{1}^{*}>0$
such that $A_{4}(E,\lambda_{1})$ is hyperbolic for all
$\lambda_{1}\in[0,\lambda_{1}^{*}]$. Over $J$ the eigenvector $u_{4}$ (resp
$s_{4}$) is greater than $u_{3}$ (resp $s_{3}$) if and only if $\partial
u_{4}/\partial\lambda_{1}|_{\lambda_{1}=0}\geq 0$ (resp $\partial
s_{4}/\partial\lambda_{1}|_{\lambda_{1}=0}\geq 0$).

  * •

Define open interval $J$ over $E$ such that $A_{1}(E)$ and
$A_{3}(E,\lambda_{0})$ are hyperbolic, and there exists $\lambda_{0}^{*}>0$
such that $A_{3}(E,\lambda_{0})$ is hyperbolic for all
$\lambda_{0}\in[0,\lambda_{0}^{*}]$. Over $J$ the eigenvector $u_{3}$ (resp
$s_{3}$) is greater than $u_{1}$ (resp $s_{1}$) if and only if $\partial
u_{3}/\partial\lambda_{0}|_{\lambda_{0}=0}\geq 0$ (resp $\partial
s_{3}/\partial\lambda_{0}|_{\lambda_{0}=0}\geq 0$).

  * •

Define open interval $J$ over $E$ such that $A_{2}(E)$ and
$A_{4}(E,\lambda_{0})$ are hyperbolic, and there exists $\lambda_{0}^{*}>0$
such that $A_{4}(E,\lambda_{0})$ is hyperbolic for all
$\lambda_{0}\in[0,\lambda_{0}^{*}]$. Over $J$ the eigenvector $u_{4}$ (resp
$s_{4}$) is greater than $u_{2}$ (resp $s_{2}$) if and only if $\partial
u_{4}/\partial\lambda_{0}|_{\lambda_{0}=0}\geq 0$ (resp $\partial
s_{4}/\partial\lambda_{0}|_{\lambda_{0}=0}\geq 0$).

###### Proof.

  * •

If $\lambda_{1}=0,$ then $A_{1}=A_{2}.$ Over any interval $J$ where both
$A_{1}$ and $A_{2}$ are hyperbolic, both $\partial
u_{2}/\partial\lambda_{1}|_{\lambda_{1}=0}$ and $\partial
s_{2}/\partial\lambda_{1}|_{\lambda_{1}=0}$ are well-defined and not constant
over $J$. Because of corollary 1, $sign(u_{2}-u_{1})$ and $sign(s_{2}-s_{1})$
are well-defined and fixed for all $\lambda_{1}\in(0,\lambda_{1}^{*}]$ over
the interval $J$.

  * •

If $\lambda_{1}=0,$ then $A_{3}=A_{4}.$ Over any interval $J$ where both
$A_{3}$ and $A_{4}$ are hyperbolic, both $\partial
u_{4}/\partial\lambda_{1}|_{\lambda_{1}=0}$ and $\partial
s_{4}/\partial\lambda_{1}|_{\lambda_{1}=0}$ are well-defined and not constant
over $J$. Because of corollary 1, $sign(u_{4}-u_{3})$ and $sign(s_{4}-s_{3})$
are well-defined and fixed for all $\lambda_{1}\in(0,\lambda_{1}^{*})$ over
$J$.

  * •

If $\lambda_{0}=0,$ then $A_{1}=A_{3}.$ Over any interval $J$ where both
$A_{1}$ and $A_{3}$ are hyperbolic, both $\partial
u_{3}/\partial\lambda_{0}|_{\lambda_{0}=0}$ and $\partial
s_{3}/\partial\lambda_{0}|_{\lambda_{0}=0}$ are well-defined and not constant
over $J$. Because of corollary 1, $sign(u_{3}-u_{1})$ and $sign(s_{3}-s_{1})$
are well-defined and fixed for all $\lambda_{0}\in(0,\lambda_{0}^{*})$ over
$J$.

  * •

If $\lambda_{0}=0,$ then $A_{4}=A_{2}.$ Over any interval where both $A_{2}$
and $A_{4}$ are hyperbolic, both $\partial
u_{4}/\partial\lambda_{0}|_{\lambda_{0}=0}$ and $\partial
s_{4}/\partial\lambda_{0}|_{\lambda_{0}=0}$ are well-defined and not constant
over $J$. Because of corollary 1, $sign(u_{4}-u_{2})$ and $sign(s_{4}-s_{2})$
are well-defined and fixed for all $\lambda_{0}\in(0,\lambda_{0}^{*})$ and
over the interval.

From table $1$ in section $4A$ we have explicit expressions of $u_{i}$, and
the partial derivatives are explicitly defined in the table below.

Partial Derivatives
---
|  $\partial/\partial\lambda_{0}$ |  $\partial/\partial\lambda_{1}$
$u_{1}$ |  0 |  0
$s_{1}$ |  0 |  0
$u_{2}$ |  0 |  $\frac{1+(E-c_{1}-\lambda_{1})(\partial r)/(\partial\lambda_{1})}{(E-c_{1}-\lambda_{1})^{2}}$
$s_{2}$ |  0 |  $\frac{1-r^{-2}(E-c_{1}-\lambda_{1})(\partial r)/(\partial\lambda_{1})}{(E-c_{1}-\lambda_{1})^{2}}$
$u_{3}$ |  $\frac{(\partial r)/(\partial\lambda_{0})}{E-c_{1}}$ |  0
$s_{3}$ |  $-\frac{(\partial r)/(\partial\lambda_{0})}{r^{2}(E-c_{1})}$ |  0
$u_{4}$ |  $\frac{(\partial r)/(\partial\lambda_{0})}{E-c_{1}-\lambda_{1}}$ |  $\frac{1+(E-c_{1}-\lambda_{1})(\partial r)/(\partial\lambda_{1})}{(E-c_{1}-\lambda_{1})^{2}}$
$s_{4}$ |  $-\frac{(\partial r)/(\partial\lambda_{0})}{r^{2}(E-c_{1}-\lambda_{1})}$ |  $\frac{1-r^{-2}(E-c_{1}-\lambda_{1})(\partial r)/(\partial\lambda_{1})}{(E-c_{1}-\lambda_{1})^{2}}$
Table 2: In $\mathbb{P}$, the partial derivatives of the eigenvectors are
defined here.

∎

###  IV.4 Existence of Principal Cone

Here, we prove that over the intervals defined in section $4B$, the
eigenvectors of the matrices $\\{A_{i}\\}$ form a principal cone, which will
prove lemma $2$. The proof relies heavily on lemmas $3$, $4$, $5$, and $6$ as
well as table $1$ from the end of $4A$. We will also need the derivatives of
the unstable and stable matrices with respect to $\lambda_{0}$ and
$\lambda_{1}$ from table $5$ at the end of the proof of lemma $6$. Similar to
subsection $4B$, the 3 orderings of the set
$\\{\lambda_{0},c_{1},c_{1}\lambda_{1}\\}$ are all addressed separately.

###### Proof.

Same as in section $4B$, we have a list of the different orderings of
$\\{c_{1},\lambda_{0},c+1+\lambda_{1}\\}$, and there are sublists stating the
different intervals over which the matrices are all hyperbolic. Here, they are
listed as scenarios, and an ordering of the eigenvectors are presented and
then proven to be true.

  1. 1.

If we assume $\lambda_{0}\leq c_{1}$

    1. (Scenario a)

To show $u_{4}<u_{2},u_{3}<u_{1}<s_{1}<s_{2},s_{3}<s_{4}<0:$

Over this region, the trace of all matrices are greater than $2$. Therefore,
we can conclude that $r_{i}>1$ for all $i$, and, using lemma $3$ and table
$1$, all the eigenvectors are negative and $u_{i}<s_{i}$. Over this interval,
we have the inequalities $\partial/\partial\lambda_{i}(u_{j})\leq
0\leq\partial/\partial\lambda_{i}(s_{j})$ for all possible $i,j$. From lemma
$6$, we get the order the eigenvectors, and hence a principal cone can be
defined.

    1. (Scenario b)

To show $u_{4}<u_{3}<s_{3}<s_{4},s_{1},s_{2}<0<u_{2}<u_{1}:$

The matrices $A_{1}$ & $A_{2}$ have negative traces, and $A_{3}$ & $A_{4}$
have positive traces. This means $r_{3},r_{4}>0>r_{1},r_{2}$ from lemma $3$.
If $r_{i}<0,$ then $u_{i}$ and $s_{i}$ will have opposite signs which can be
seen from lemma $4$ and, more explicitly, table $1$ in subsection $4A$. Over
this region we have the inequality, $u_{4}<u_{3}<s_{3}<s_{4}<0,$ because
$\partial/\partial\lambda_{1}(u_{4})\leq
0\leq\partial/\partial\lambda_{1}(s_{4})$ and lemma $6$. Due to corollary $1$,
$s_{1}$ is bounded by $s_{3}$. It can be easily calculated that
$0>s_{1}>s_{3}$, because $\lim_{E\rightarrow 0^{+}}s_{1}\nearrow 0,$ while
$s_{3}(0)<0$ and continuous. On top of that, $s_{1}<s_{2}$ because of lemma
$6$ and the inequality $\partial/\partial\lambda_{1}(s_{2})\geq 0$ holds over
this interval. In the projective space, $\infty=-\infty$ is an element, and so
the principal cone can be defined as $(0,\frac{u_{3}+s_{3}}{2})$, which
contains the element $\infty$.

    1. (Scenario c)

To show $s_{1},s_{2},s_{3},s_{4}<0<u_{1},u_{2},u_{3},u_{4}:$

Over this entire interval, the traces of all four matrices are negative,
making the eigenvectors different signs. All four matrices have positive
unstable eignevectors and negative stable eigenvectors. The principal cone can
be defined as $(0,\infty).$

    1. (Scenario d)

To show $s_{2},s_{4}<0<s_{1}<s_{3}<u_{3}<u_{1},u_{2},u_{4}:$

The trace of $A_{1}$ and $A_{3}$ are positive, and the trace of $A_{2}$ and
$A_{4}$ are negative. This means that $r_{1},r_{3}>1$ and $r_{2},r_{4}<-1.$
Because of lemma $3$ and $4$, we get the inequalities
$s_{4},s_{2}<0<s_{1},s_{3}.$ Similarly, from lemma $3$, $4$, and table $1$, we
get the inequalities $0<s_{3}<u_{3}$ and $0<s_{1}<u_{1}$. To show $u_{3}$
serves as a lower bound for the unstable eigenvectors, the following two
inequalities will suffice. We can calculate
$\partial/\partial\lambda_{1}(u_{j})\leq 0$, so $u_{3}<u_{1}$ and
$u_{4}<u_{2}$ by lemma $6$. By corollary 1, $u_{3}$ bounds $u_{4}$. As $E$
approaches the upper bound of this interval defining scenario d,
$u_{4}\nearrow\infty$ and $u_{3}$ does not. Therefore $u_{3}$ bounds $u_{4}$
from below. This makes $u_{3}$ a lower bound of the unstable eigenvectors.
Because $\partial/\partial\lambda_{1}(s_{3})>0,$ by lemma $6$, we get the
inequality $s_{3}>s_{1}$. This is enough to give the inequalities for this
scenario.

    1. (Scenario e)

To show $0<s_{1}<s_{2},s_{3}<s_{4}<u_{4}<u_{2},u_{3}<u_{1}:$

Over this interval, assuming $(E-\lambda_{0})(E-\lambda_{1}-c_{1})>4$ which is
necessary,
$\partial/\partial\lambda_{i}(u_{j})<0<\partial/\partial\lambda_{i}(s_{j})$
for all possible $i,j$ and $s_{4}<u_{4}$. All possible eigenvectors are
positive, and by lemma $6$ the partial derivatives point to $s_{i}<s_{4}$ for
all $i\neq 4$. Similarly, $u_{i}>u_{4}>s_{4}$ for all $i\neq 4.$

  2. 2.

If we assume $c_{1}\leq\lambda_{0}<c_{1}+\lambda_{1}$

    1. (Scenario a)

To show $u_{4}<u_{2},u_{3}<u_{1}<s_{1}<s_{2},s_{3}<s_{4}<0:$

Here the proof is the same as scenario a if $\lambda_{0}\leq c_{1}$.

    1. (Scenario b)

To show $u_{4}<u_{3}<s_{3}<s_{4},s_{2},s_{1}<0<u_{1}<u_{2}:$

Here the proof is the same as scenario b if $\lambda_{0}\leq c_{1}$ .

    1. (Scenario c)

Show $u_{3}<u_{4}<s_{4}<s_{2}<0<s_{3}<s_{1}<u_{1}<u_{2}:$

Over this interval, $tr(A_{2}),\;tr(A_{3})<-2$ and $tr(A_{4}),\;tr(A_{1})>2$,
so we can determine the signs of the eigenvectors. We get the inequalities
$s_{2},s_{4},u_{3},u_{4}<0<s_{3},s_{1},u_{1},u_{2}$. The matrices $A_{3}$ and
$A_{4}$ are hyperbolic for

${\color[rgb]{0,0,0}E\in\bigg{(}c_{1},\frac{\lambda_{0}-\lambda_{1}-c_{1}-\sqrt{(\lambda_{0}-\lambda_{1}-c_{1})^{2}-4\lambda_{0}(\lambda_{1}+c_{1})+16}}{2}\bigg{)}}$

and $\lim_{E\rightarrow c_{1}^{+}}u_{3}\searrow-\infty.$ The eigenvector
$u_{4}$ satisfies the inequality $-\infty<u_{4}|_{E=c_{1}}<0$, however, so we
can deduce the inequalities $u_{3}<u_{4}<s_{4}<0$ using lemma $4$, corollary
$1$, and $r_{4}>0$. Similarly, $A_{3}$ and $A_{1}$ are hyperbolic over

${\color[rgb]{0,0,0}E\in\Bigg{(}\frac{c_{1}+\sqrt{c_{1}^{2}+16}}{2},\lambda_{0}\Bigg{)}}$

as $\lim_{E\rightarrow\lambda_{0}^{+}}s_{3}\searrow 0$, but
$s_{1}|_{E=\lambda_{0}}>0$ and $s_{1}$ is continuous. Using lemma $4$ and
$r_{1}>0$, this is sufficient to show that $0<s_{3}<s_{1}<u_{1}$. Additionally
$A_{2}$ and $A_{4}$ are hyperbolic over

${\color[rgb]{0,0,0}E\in\bigg{(}0,\frac{\lambda_{0}-\lambda_{1}-c_{1}-\sqrt{(\lambda_{0}-\lambda_{1}-c_{1})^{2}-4\lambda_{0}(\lambda_{1}+c_{1})+16}}{2}\bigg{)}}.$

Using the limit $\lim_{E\rightarrow 0}s_{2}\nearrow 0$, the eigenvector
satisfying $s_{4}|_{E=0}<0,$ and continuity of $s_{4}$, corollary $1$ gives us
$s_{4}<s_{2}<0.$ Lastly, $A_{1}$ and $A_{2}$ are hyperbolic over

$E\in\Bigg{(}\frac{c_{1}+\sqrt{c_{1}^{2}+16}}{2},c_{1}+\lambda_{1}\Bigg{)}$

Here, $u_{2}>0$ by lemma 3, $\lim_{E\rightarrow
c_{1}+\lambda_{1}^{+}}u_{2}\nearrow\infty$, and
$u_{1}|_{E=c_{1}+\lambda_{1}}<\infty$, so by corollary 1 we have the
inequality $0<u_{1}<u_{2}$. This completes the inequalities needed to define
the principal cone.

    1. (Scenario d)

To show $s_{2},s_{4}<0<s_{4}<s_{1},s_{3}<u_{3}<u_{1},u_{2},u_{4}:$

Here the proof is the same as scenario d if $\lambda_{0}\leq c_{1}$ .

    1. (Scenario e)

To show $s_{1}<s_{2},s_{3}<s_{4}<u_{4}<u_{2},u_{3}<u_{1}:$

Here the proof is the same as scanrio e if $\lambda_{0}\leq c_{1}.$

  3. 3.

If we assume $c_{1}+\lambda_{1}<\lambda_{0}$

    1. (Scenario a)

To show $u_{4}<u_{2},u_{3}<u_{1}<s_{1}<s_{2},s_{3}<s_{4}<0:$

Here the proof is the same as scenario a if $\lambda_{0}\leq c_{1}.$

    1. (Scenario b)

To show $u_{4}<u_{3}<s_{3}<s_{4},s_{2},s_{1}<0<u_{1}<u_{2}:$

Here the proof is the same as scenario b if $\lambda_{0}\leq c_{1}$ .

    1. (Scenario c)

Show $u_{3}<u_{4}<s_{4}<s_{2}<0<s_{3}<s_{1}<u_{1}<u_{2}:$

Here the proof is the same if $c_{1}\leq\lambda_{0}<c_{1}+\lambda_{1}$ in
scenario c.

    1. (Scenario d)

To show $u_{4}<u_{3}<0<s_{1},s_{3},s_{4}<s_{2}<u_{2}<u_{1}:$

The spectral radii satisfy the inequality $r_{3},r_{4}<0<r_{1},r_{2}$, So we
can conclude that $u_{4},u_{3}<0$, but all the other eigenvectors are greater
than $0.$ By lemma 4, we can also conclude that $s_{2}<u_{2}$. By lemma $6$
and $0<\partial/\partial\lambda_{1}(s_{2})$, we can deduce that $s_{1}<s_{2}$
by lemma 4. Given that $\partial/\partial\lambda_{1}(s_{4})>0$, so
$s_{3}<s_{4}$ by lemma $6$. Given the limit
$\lim_{E\rightarrow\lambda_{1}+c_{1}^{+}}s_{4}\searrow 0$, as well as the
inequality $s_{2}|_{E=\lambda_{1}+c_{1}}>0,$ with $s_{2}$ being continuous, we
have the inequality $0<s_{4}<s_{2}$ by corollary $1$. Last, we need the
inequality $u_{2}<u_{1}$, which is true by
$\partial/\partial\lambda_{1}(u_{2})<0$ by lemma $6$.

    1. (Scenario e)

To show $s_{1}<s_{2},s_{3}<s_{4}<u_{4}<u_{2},u_{3}<u_{1}:$

Here the proof is the same as scenario e if $\lambda_{0}<c_{1}.$

∎

###  IV.5 The Explicit Gaps in the Spectrum

We can conclude that for all of the intervals listed as scenarios $a$ to $e$,
the 4 matrices $A_{1},A_{2},A_{3},\&A_{4}$ are hyperbolic and the set is
uniformly hyperbolic. The spectrum therefore consists of at most 4 intervals,
which are the complement of the intervals in $E$ over which the set of
matrices is uniformly hyperbolic. Below the intervals are explicitly
calculated.

######  Proposition 2.

The complement of the spectrum of $H_{V}$ from theorem 1 can be explicitly
calculated, depending on the ordering of the set
$\\{\lambda_{0},c_{1},c_{1}+\lambda_{1}\\},$ where $\lambda_{i},c_{1}\geq 0$

  1. 1.

If we assume $\lambda_{0}\leq c_{1}\leq c_{1}+\lambda_{1}$, then

${\color[rgb]{0,0,0}\mathbb{R}\backslash\sigma(H_{V})}=\bigg{(}-\infty,\frac{1}{2}\Big{(}c_{1}-\sqrt{c_{1}^{2}+16}\big{)}\Bigg{)}\cup\bigg{(}0,\frac{1}{2}\Big{(}\lambda_{0}+c_{1}-\sqrt{(\lambda_{0}+c_{1})^{2}-4(c_{1}\cdot\lambda_{0}-4)}\Big{)}\bigg{)}\cup(\lambda_{0},c_{1})\cup$
$\bigg{(}\frac{1}{2}\Big{(}\lambda_{0}+c_{1}+\sqrt{(\lambda_{0}+c_{1})^{2}-4(c_{1}\lambda_{0}-4)}\Big{)},c_{1}+\lambda_{1}\bigg{)}\cup\Big{(}\frac{1}{2}\Big{(}\lambda_{0}+\lambda_{1}+c_{1}+\sqrt{(\lambda_{0}+\lambda_{1}+c_{1})^{2}-4\lambda_{0}\cdot(\lambda_{1}+c_{1})+16},\infty\bigg{)}$

  2. 2.

If we assume $c_{1}<\lambda_{0}<c_{1}+\lambda_{1}$, then then gaps in the
spectrum consist of the intervals

$\mathbb{R}\backslash\sigma(H_{V})=\bigg{(}-\infty,\frac{1}{2}\Big{(}c_{1}-\sqrt{c_{1}^{2}+16}\Big{)}\bigg{)}\cup\bigg{(}0,\frac{1}{2}\Big{(}\lambda_{0}+c_{1}-\sqrt{(\lambda_{0}+c_{1})^{2}-4(c_{1}\lambda_{0}-4)}\Big{)}\bigg{)}\cup$
$\bigg{(}\frac{1}{2}\Big{(}c_{1}+\sqrt{c_{1}^{2}+16}\Big{)},\frac{1}{2}\Big{(}\lambda_{0}+\lambda_{1}+c_{1}-\sqrt{(\lambda_{0}+\lambda_{1}+c_{1})^{2}-4\lambda_{0}(\lambda_{1}+c_{1})+16}\Big{)}\bigg{)}\cup$
$\bigg{(}\frac{1}{2}\Big{(}\lambda_{0}+c_{1}+\sqrt{(\lambda_{0}+c_{1})^{2}-4(c_{1}\lambda_{0}-4)}\Big{)},c_{1}+\lambda_{1}\bigg{)}\cup\bigg{(}\frac{1}{2}\Big{(}\lambda_{0}+\lambda_{1}+c_{1}+\sqrt{(\lambda_{0}+\lambda_{1}+c_{1})^{2}-4(\lambda_{0}(\lambda_{1}+c_{1})-4)}),\infty\bigg{)}$

  3. 3.

If we assume $c_{1}\leq c_{1}+\lambda_{1}\leq\lambda_{0}$, then the gaps in
the spectrum consist of the intervals

$\mathbb{R}\backslash\sigma(H_{V})=\bigg{(}-\infty,\frac{1}{2}\Big{(}c_{1}-\sqrt{c_{1}^{2}+16}\Big{)}\bigg{)}\cup\bigg{(}0,\frac{1}{2}\Big{(}\lambda_{0}+c_{1}-\sqrt{(\lambda_{0}+c_{1})^{2}-4(c_{1}\lambda_{0}-4)}\Big{)}\bigg{)}\cup$
$\bigg{(}\frac{1}{2}\Big{(}c_{1}+\sqrt{c_{1}^{2}+16}\Big{)},\frac{1}{2}\Big{(}\lambda_{0}+\lambda_{1}+c_{1}-\sqrt{(\lambda_{0}+\lambda_{1}+c_{1})^{2}-4(\lambda_{0}(\lambda_{1}+c_{1})-4)}\Big{)}\bigg{)}\cup$
$\bigg{(}\frac{1}{2}\Big{(}c_{1}+\lambda_{1}+\sqrt{(c_{1}+\lambda_{1})^{2}+16}\Big{)},\lambda_{0}\Bigg{)}\cup\bigg{(}\frac{1}{2}\Big{(}\lambda_{0}+\lambda_{1}+c_{1}+\sqrt{(\lambda_{0}+\lambda_{1}+c_{1})^{2}-4(\lambda_{0}(\lambda_{1}+c_{1})-4)}\Big{)},\infty\bigg{)}$

Note that in all scenarios, some of the intervals may be the empty set if the
lower bound is greater than or equal to the upper bound. We can conclude that
$H_{\nu}-E$ has a bounded inverse if and only if $E$ is not in one of these,
and the spectrum consists of the complement of these 5 intervals.

This proposition follows from lemma 1 by direct computation.

##  V Acknowledgements

I would like to acknowledge Anton Gorodetski for helping edit and put together
this paper. The author was supported in part by NSF grant DMS–1855541 (PI
A.Gorodetski).

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