# Non-negative integral matrices with given spectral radius and controlled
dimension

Mehdi Yazdi

###### Abstract.

A celebrated theorem of Douglas Lind states that a positive real number is
equal to the spectral radius of some integral primitive matrix, if and only
if, it is a Perron algebraic integer. Given a Perron number $p$, we prove that
there is an integral irreducible matrix with spectral radius $p$, and with
dimension bounded above in terms of the algebraic degree, the ratio of the
first two largest Galois conjugates, and arithmetic information about the ring
of integers of its number field. This arithmetic information can be taken to
be either the discriminant or the minimal Hermite-like thickness.
Equivalently, given a Perron number $p$, there is an irreducible shift of
finite type with entropy $\log(p)$ defined as an edge shift on a graph whose
number of vertices is bounded above in terms of the aforementioned data.

##  1\. Introduction

A real algebraic integer $\lambda\geq 1$ is _Perron_ if $\lambda$ is strictly
larger than all its other Galois conjugates in absolute value. In what
follows, all matrices are considered to be square size. For a real matrix $A$,
by $A>0$ we mean that all entries of $A$ are positive. A non-negative real
matrix $A$ is _primitive_ (or _aperiodic_) if there is some natural number $k$
such that $A^{k}>0$. A non-negative real matrix is _irreducible_ if for any
two indices $i$ and $j$, there is some natural number $k=k(i,j)$ such that
$(A^{k})_{ij}>0$. The Perron–Frobenius theorem implies that

  1. I)

the spectral radius of any _integral_ primitive matrix is a Perron number; and

  2. II)

the spectral radius of any _integral_ irreducible matrix is equal to the $n$th
root of a Perron number where $n$ is the _period_ of the irreducible matrix.
Moreover, the spectral radius of an integral non-negative non-nilpotent matrix
is equal to the spectral radius of an integral irreducible one.

Lind [Lin84, Theorem 1] proved a converse to the ‘integer’ Perron–Frobenius
theorem, namely for every Perron number $\lambda$ there is an integral
primitive matrix with spectral radius equal to $\lambda$. It readily follows
that for any Perron number $\lambda$ and any natural number $n$, there is an
integral irreducible matrix with spectral radius equal to $\sqrt[n]{\lambda}$;
see [Lin84, Theorem 3].

Associated to a non-negative and non-degenerate (i.e. with no zero rows or
columns) integral matrix $A=[a_{ij}]$ with spectral radius $\lambda$ is a
_shift of finite type_ with entropy equal to $\log(\lambda)$, which is defined
as the _edge shift_ on a directed finite graph $G_{A}$ as follows. The graph
$G_{A}$ has one vertex for each row of the matrix $A$, and there are exactly
$a_{ij}$ oriented edges from the vertex $v_{i}$ to the vertex $v_{j}$. In
particular, the dimension of the matrix $A$ (i.e. its number of rows or
columns) is equal to the number of vertices of the graph $G_{A}$. The matrix
is primitive if and only if the associated shift of finite type is
topologically mixing. If the matrix is irreducible then the corresponding
shift of finite type is called _irreducible_. Irreducible shifts of finite
type are those which are topologically transitive. See e.g. [LM21].

Given a Perron number $\lambda$, the _Perron–Frobenius degree_ of $\lambda$,
$d_{PF}(\lambda)$, is the least dimension of an integral _primitive_ matrix
with spectral radius equal to $\lambda$. Clearly we have

$d_{PF}(\lambda)\geq d,$

where $d$ denotes the algebraic degree of $\lambda$ as an algebraic integer.
It is easy to see that the equality happens if $\lambda$ is quadratic (see
[Yaz21, Remark 3.1]), so we consider the case of $d\geq 3$. Lind observed that
if the _trace_ (i.e. sum of Galois conjugates) of $\lambda$ is negative, then
$d_{PF}(\lambda)$ is strictly larger than $d$; see [Lin84, page 289]. In
[Yaz21], using an idea of Lind, we gave a lower bound for $d_{PF}(\lambda)$ in
terms of the layout of the two largest (in absolute value) Galois conjugates
of $\lambda$ in the complex plane. As a corollary, it was shown that there are
examples of cubic Perron numbers with arbitrarily large Perron–Frobenius
degrees, a result previously known to Lind, McMullen, and Thurston, although
unpublished.

######  Definition 1.1.

Given a Perron number $\lambda$, define the _spectral ratio_ of $\lambda$ as
$\max_{i}\frac{|\lambda_{i}|}{\lambda}$, where $\lambda_{i}\neq\lambda$ are
the remaining Galois conjugates of $\lambda$.

######  Notation 1.2.

For a Perron number $\lambda$, let $d_{PF}^{irr}(\lambda)$ be the smallest
dimension of an integral _irreducible_ matrix with spectral radius $\lambda$.

In this paper, we give an explicit upper bound for $d_{PF}^{irr}(\lambda)$ in
terms of the algebraic degree of $\lambda$, the spectral ratio of $\lambda$,
and arithmetic information about the ring of integers
$\mathcal{O}_{\mathbb{K}}$ of the number field
$\mathbb{K}:=\mathbb{Q}(\lambda)$. This arithmetic quantity, which we call the
_minimal Hermite-like thickness_ and denote it by
$\tau_{\min}(\mathcal{O}_{\mathbb{K}})$, was previously defined by Bayer
Fluckiger [Bay06] in relation to _Minkowski’s conjecture_ ; see Definition
3.1. Intuitively, once an inner product is chosen on $\mathbb{R}^{d}$, the
Hermite-like thickness is defined as the square of the covering radius,
normalised properly, for the inclusion of the lattice
$\mathcal{O}_{\mathbb{K}}$ in $\mathbb{R}^{d}$. The minimal Hermite-like
thickness is then defined by taking the infimum of Hermite-like thickness over
an appropriate space of inner products on $\mathbb{R}^{d}$. As a corollary of
our main result, and using an inequality due to Banaszczyk and Bayer Fluckiger
(see inequality (25)), we obtain a similar bound in terms of the
_discriminant_ $D_{\mathbb{K}}$ of $\mathcal{O}_{\mathbb{K}}$ instead of
$\tau_{\min}(\mathcal{O}_{\mathbb{K}})$. See Definition 3.3.

######  Theorem 1.3.

Let $\lambda$ be a Perron number of algebraic degree $d\geq 3$ and spectral
ratio $\rho$. Set $\mathbb{K}:=\mathbb{Q}(\lambda)$. Let
$\mathcal{O}_{\mathbb{K}}$ be the ring of integers of $\mathbb{K}$, and denote
the discriminant and the minimal Hermite-like thickness of $\mathbb{K}$ by,
respectively, $D_{\mathbb{K}}$ and $\tau_{\min}(\mathcal{O}_{\mathbb{K}})$.
Then $d_{PF}^{irr}(\lambda)$ is bounded above by each of

$\Big{(}\dfrac{8d}{1-\rho}\Big{)}^{d^{2}}\tau_{\min}(\mathcal{O}_{\mathbb{K}})^{\frac{d}{2}}$

and

$\Big{(}\dfrac{8d}{1-\rho}\Big{)}^{d^{2}}\sqrt{D_{\mathbb{K}}}.$

######  Remark 1.4.

Given a natural number $n$ and an integral irreducible matrix $A$ with
spectral radius $\lambda$, one can readily construct an integral irreducible
matrix $B$ with spectral radius $\sqrt[n]{\lambda}$ such that
$\dim(B)=n\dim(A)$. Hence, Theorem 1.3 can be used to give an upper bound for
$d_{PF}^{irr}(\sqrt[n]{\lambda})$. See e.g. the proof of [Lin84, Theorem 3].

Theorem 1.3 immediately translates into the context of irreducible shifts of
finite type with a given entropy. We can derive an upper bound for the
Perron–Frobenius degree using Theorem 1.3. See also Remark 3.9 and Question
5.1. First we need to introduce a notation.

######  Notation 1.5.

Let $\mathbb{K}$ be a real number field (i.e. with at least one real place),
and $\rho\in(0,1)$. Set

$M=1+\frac{4}{1-\rho},$

and denote

$\kappa(\mathbb{K},\rho):=\max\\{d_{PF}(\alpha)\hskip 2.84526pt|\hskip
2.84526pt\alpha\in[1,M]\cap\mathbb{K}\text{ is a Perron number}\\}.$

Note that for any $M>1$, there are only finitely many Perron numbers in the
interval $[1,M]$ with degree at most $d$. Hence, $\kappa(\mathbb{K},\rho)$ can
be computed in theory.

######  Theorem 1.6.

Let $\mathbb{\lambda}$ be a Perron number of degree $d\geq 3$ and spectral
ratio $\rho$. Set $\mathbb{K}:=\mathbb{Q}(\lambda)$. Denote the bound from
Theorem 1.3 by $B(\mathbb{K},\rho)$; note that $d=[\mathbb{K}:\mathbb{Q}]$ is
uniquely determined by $\mathbb{K}$. The Perron–Frobenius degree of $\lambda$
is bounded above by

$\max\\{2^{d^{2}}B(\mathbb{K},\rho),\kappa(\mathbb{K},\rho)\\}.$

###  1.1. Previous work

In [Lin84], Lind gave a method for producing _all_ integral primitive matrices
with a given Perron number $\lambda$ as their spectral radius. Let
$B\colon\mathbb{R}^{d}\rightarrow\mathbb{R}^{d}$ be the _companion matrix_
associated to $\lambda$. In what follows, unless otherwise specified all
eigenvectors are column (right) eigenvectors; similarly for eigenspaces. Pick
an eigenvector $v$ for the eigenvalue $\lambda$, and denote the one
dimensional eigenspace of $\lambda$ by $E_{\lambda}$. Let $E$ be the positive
half-space associated to $v$, i.e. if $\pi_{1}\colon\mathbb{R}^{d}\rightarrow
E_{\lambda}$ is the projection along the complementary invariant subspace,
then

$E=\\{x\in\mathbb{R}^{d}\hskip 5.69054pt|\hskip 5.69054pt\pi_{1}(x)=rv\hskip
8.53581pt\text{for some }r>0\\}.$

######  Theorem 1.7 (Lind).

Let $\lambda$ be a Perron number of algebraic degree $d$, and $B$ and $E$ be
as above. If $A=[a_{ij}]$ is an $n$-dimensional primitive non-negative
integral matrix with spectral radius $\lambda$, then there are
$z_{i}\in\mathbb{Z}^{d}\cap E$ for $1\leq i\leq n$ such that
$Bz_{j}=\sum_{i=1}^{n}a_{ij}z_{i}$.

Conversely, if $\lambda$, $B$, and $E$ are as above, and the points
$z_{i}\in\mathbb{Z}^{d}\cap E$ and a non-negative integral matrix $A=[a_{ij}]$
satisfy $Bz_{j}=\sum_{i=1}^{n}a_{ij}z_{i}$, then every irreducible component
of $A$ has spectral radius equal to $\lambda$.

The above theorem of Lind gives a practical way to produce an integral
primitive matrix with a given spectral radius; see [Lin84, page 289]. However,
it does not tell us how to find such a matrix with smallest (or close to
smallest) dimension, since we are not given control over the size of the
coordinates of $z_{i}$. Nevertheless, the referee has kindly mentioned to me
that there exists a simple, but not necessarily practical, algorithm that
computes the Perron–Frobenius degree of a Perron number: Assume that $\lambda$
is given by its minimal polynomial, and denote the algebraic degree of
$\lambda$ by $d$. For every positive integer $n$, there are only finitely many
primitive $n\times n$ integral matrices with spectral radius less than or
equal to $\lambda$. One can algorithmically enumerate these. For each of them,
one can algorithmically determine whether the spectral radius is equal to
$\lambda$. So for $n=d$, we can check whether there is an integral primitive
matrix of dimension $n$ which has $\lambda$ as the spectral radius.
Recursively, if we fail at $n$, then we try at $n+1$. By Lind’s theorem we
eventually find an $n$ where we succeed; that $n$ is equal to
$d_{PF}(\lambda)$.

For matrices with non-negative integral _polynomial_ entries, the situation is
different. See the work of Boyle and Lind, which gives a uniform upper bound
(in fact a 2 by 2 matrix) in this context [BL02]. For the related topic of
inverse spectral problem for non-negative integral matrices see the works of
Boyle–Handelman [BH91] and Kim–Ormes–Roush [KOR00] and the references therein.

###  1.2. Idea of the proof

In [Thu14], Thurston gave a simpler proof of Lind’s converse to the integer
Perron–Frobenius theorem. Our proof of Theorem 1.3 follows Thurston’s
approach, while controlling the dimension of a constructed matrix.

The tensor product $\mathbb{Q}(\lambda)\otimes_{\mathbb{Q}}\mathbb{R}$ can be
identified with $\mathbb{R}^{d}$. Let $M_{\lambda}$ be the linear endomorphism
of $\mathbb{R}^{d}\cong\mathbb{Q}(\lambda)\otimes_{\mathbb{Q}}\mathbb{R}$
induced by multiplication by $\lambda$ in $\mathbb{Q}(\lambda)$. The
eigenvalues of $M_{\lambda}$ are the Galois conjugates of $\lambda$. Then
$\mathbb{R}^{d}$ decomposes into invariant subspaces of $M_{\lambda}$
corresponding to real places and pairs of conjugate complex places of
$\lambda$; see the opening paragraphs to Section 3. Fix an eigenvector for
$M_{\lambda}$ with eigenvalue $\lambda$, and consider the positive half-space
corresponding to $\lambda$.

We start with a polygonal cone with apex at the origin that lies in the
positive half-space and is invariant under $M_{\lambda}$. We then perturb the
vertices of the cone to obtain an invariant polygonal cone $\mathcal{C}$ with
_integral_ vertices; see Steps 2–4 of the proof of Theorem 1.3. It is during
this perturbation that the minimal Hermite-like thickness appears in the
picture. Since the polygonal cone has integral vertices, the semigroup $S$
generated by the set of integral points in the cone $\mathcal{C}$ under
addition of vectors is finitely generated; see Proposition 2.1. The
cardinality of a generating set for the semigroup $S$ gives an upper bound for
the dimension of an integral non-negative matrix $A$ with spectral radius
$\lambda$. Moreover, after possibly passing to an irreducible component of
$A$, an integral irreducible matrix with spectral radius $\lambda$ is
obtained; see Step 5. Finally we give an upper bound for the dimension of $A$;
see Step 6.

###  1.3. Plan of the paper

In Section 2, we present a few preliminary lemmas. The proof of Theorem 1.3 is
given in Section 3. Theorem 1.6 is proved in Section 4. In Section 5, a few
questions are posed.

###  1.4. Acknowledgement

I would like to thank Douglas Lind for sharing his intuition that some lattice
property of the ring of integers should play a role in the Perron–Frobenius
degree. Many thanks to Curtis T. McMullen and Eva Bayer Fluckiger for helpful
comments connected to Remark 3.14. I am grateful to the anonymous referee for
his/her very helpful comments, in particular for explaining how the bound for
primitive matrices in Theorem 1.6 can be obtained. During this work, the
author was supported by a Glasstone Research Fellowship in Science, a
Titchmarsh Fellowship, and a UKRI Postdoctoral Research Fellowship.

##  2\. Preliminaries

###  2.1. Lattice points

Throughout this article, by a _lattice_ $\Lambda\subset\mathbb{R}^{d}$ we mean
a lattice of full rank; i.e. a discrete subgroup of $\mathbb{R}^{d}$
isomorphic to $\mathbb{Z}^{d}$.

######  Proposition 2.1.

Let $\Lambda\subset\mathbb{R}^{d}$ be a lattice, and
$x_{1},\cdots,x_{k}\in\Lambda$. Denote by $\mathcal{C}$ the convex cone over
the points $x_{i}$ inside $\mathbb{R}^{d}$, that is

$\mathcal{C}:=\\{\alpha_{1}x_{1}+\cdots+\alpha_{k}x_{k}\hskip 5.69054pt|\hskip
5.69054pt\alpha_{i}\geq 0\hskip 8.53581pt\text{for every }i\\}.$

Let $S$ be the semigroup generated by elements of $\mathcal{C}\cap\Lambda$
under vector addition. Define the compact set $C$, and the finite set
$C_{\Lambda}\subset C$ as

$C:=\\{\alpha_{1}x_{1}+\cdots+\alpha_{k}x_{k}\hskip 5.69054pt|\hskip
5.69054pt0\leq\alpha_{i}\leq 1\hskip 8.53581pt\text{for every }i\\},\hskip
8.53581ptC_{\Lambda}:=C\cap\Lambda.$

Then $C_{\Lambda}$ is a finite generating set for the semigroup $S$.

###### Proof.

For $\alpha\in\mathbb{R}$, denote the fractional part of $\alpha$ by
$\\{\alpha\\}$, and let $\lfloor\alpha\rfloor=\alpha-\\{\alpha\\}$. Any point
$y\in S$ can be written as

$y=\alpha_{1}x_{1}+\cdots+\alpha_{k}x_{k}=\big{(}\lfloor\alpha_{1}\rfloor
x_{1}+\cdots+\lfloor\alpha_{k}\rfloor
x_{k}\big{)}+\big{(}\\{\alpha_{1}\\}x_{1}+\cdots+\\{\alpha_{k}\\}x_{k}\big{)},$

where $\alpha_{i},\lfloor\alpha_{i}\rfloor\geq 0$ for each $i$. Note that we
have $x_{i}\in C_{\Lambda}$ for each $i$, hence the first parenthesis is a sum
of elements of $C_{\Lambda}$. As $y$ and the first parenthesis are both in
$\Lambda$, the second parenthesis should represent a point in $\Lambda$ as
well. On the other hand, the coefficients of the second parenthesis are in the
interval $[0,1)$, and hence the second parenthesis lies in
$C_{\Lambda}=C\cap\Lambda$. We have written $y$ as a sum of elements of
$C_{\Lambda}$, hence $C_{\Lambda}$ is a generating set for the semigroup $S$.
Since $C$ is compact and $\Lambda$ is a lattice, $C_{\Lambda}=C\cap\Lambda$ is
a finite set. ∎

By a _Euclidean space_ of dimension $d$ we mean a $d$-dimensional vector space
$\mathbb{R}^{d}$ equipped with an inner product. A _polytope_ is the convex
hull of finitely many points in $\mathbb{R}^{d}$. Let
$\Lambda\subset\mathbb{R}^{d}$ be a lattice. If $\\{v_{1},\cdots,v_{d}\\}$ is
a basis for $\Lambda\cong\mathbb{Z}^{d}$, then a _fundamental domain_ for
$\Lambda$ is

$\\{\alpha_{1}v_{1}+\cdots+\alpha_{d}v_{d}\hskip 5.69054pt|\hskip
5.69054pt0\leq\alpha_{i}<1\text{ for every }i\\}.$

If $\mathbb{R}^{d}$ is a Euclidean space, then the _covolume_ of $\Lambda$ is
defined as the volume of any fundamental domain for $\Lambda$. A _lattice
polytope_ is a polytope whose vertices are lattice points.

######  Proposition 2.2.

Let $\mathbb{R}^{d}$ be a Euclidean space, $\Lambda\subset\mathbb{R}^{d}$ be a
lattice with covolume $\mathrm{Covol}(\Lambda)$, and $P\subset\mathbb{R}^{d}$
be a $d$-dimensional lattice polytope with volume $\mathrm{Vol}(P)$. Denote
the number of lattice points in $P$ by $|P\cap\Lambda|$. Then

$|P\cap\Lambda|\leq\frac{\mathrm{Vol}(P)}{\mathrm{Covol}(\Lambda)}\cdot(d+1)!.$

The equality happens exactly when $P$ is a $d$-simplex with
$|P\cap\Lambda|=d+1$.

###### Proof.

First consider the special case that $P$ has exactly $d+1$ vertices, and every
lattice point in $P$ is a vertex of $P$. If
$v_{0},v_{1},\cdots,v_{d}\in\mathbb{R}^{d}$ are the vertices of $P$, then the
volume of the parallelepiped formed by the vectors
$v_{1}-v_{0},\cdots,v_{d}-v_{0}$ is equal to $d!\times\mathrm{Vol}(P)$. Since
the volume of this parallelepiped is at least as large as the volume of a
fundamental domain for $\Lambda$, we have

$\displaystyle d!\times\mathrm{Vol}(P)\geq\mathrm{Covol}(\Lambda),$

implying that

$|P\cap\Lambda|=d+1\leq\frac{\mathrm{Vol}(P)}{\mathrm{Covol}(\Lambda)}\cdot(d+1)!.$

In general, decompose $P$ into $d$-simplices $\Delta_{1},\cdots,\Delta_{n}$
with disjoint interiors such that each simplex $\Delta_{i}$ contains no
lattice point except for its vertices. This can be done for example as
follows. Decompose $P$ into smaller polyhedra by coning off from one of the
vertices of $P$. Here by coning off from a point $v\in P$ we mean that for
every facet $F$ of $P$, the polyhedron which is the convex hull of $F\cup v$
is added unless its dimension is strictly smaller than that of $P$; for
example if the starting polyhedron is a polygon, then coning off from a vertex
$v$ is just decomposing the polygon into triangles via adding all the
diagonals emanating from $v$. For any of the resulting polyhedra, successively
take a lattice point inside or on the boundary, and cone off from that lattice
point. After finitely many repetitions, we arrive at the decomposition into
$\Delta_{i}$.

The desired inequality follows from adding up the corresponding inequalities
for simplices $\Delta_{1},\cdots,\Delta_{n}$. Note that if $P$ is not a
$d$-simplex or $|P\cap\Lambda|>d+1$, then at least one lattice point in
$P\cap\Lambda$ is counted for more than one simplex $\Delta_{i}$, and so the
inequality is strict. ∎

###  2.2. Minkowski sum and difference

For sets $A,B\subset\mathbb{R}^{d}$, define their _Minkowski sum_ as

$A+B:=\\{a+b\hskip 5.69054pt|\hskip 5.69054pta\in A,\text{ and }b\in
B\\}\subset\mathbb{R}^{d}.$

Intuitively, $A+B$ is the union of all translates of $A$ by elements of $B$

$A+B=\bigcup_{b\in B}(A+b).$

Define the _Minkowski difference_ of $A$ and $B$ by

$A\div B:=\\{c\in\mathbb{R}^{d}\hskip 5.69054pt|\hskip 5.69054ptB+c\subseteq
A\\}.$

If $B$ is empty, $A\div B$ is, by convention, equal to $\mathbb{R}^{d}$.
Intuitively, $A\div B$ is the intersection of all translates of $A$ by the
antipodes of elements of $B$

$A\div B=\bigcap_{b\in B}(A-b).$

We have used the rather odd notation $\div$ for the Minkowski difference, in
order to distinguish it from the set

$\\{a-b\hskip 5.69054pt|\hskip 5.69054pta\in A,\text{ and }b\in
B\\}\subset\mathbb{R}^{d}.$

The Minkowski sum and difference are _not_ in general the inverse of each
other.

######  Lemma 2.3.

The following properties hold for sets $A,B,C\subset\mathbb{R}^{d}$

$\displaystyle A\subset(A+B)\div B,$ $\displaystyle A\subset B\implies A\div
C\subset B\div C,$ $\displaystyle A\subset B\implies A+C\subset B+C.$

Moreover, if $A$ and $B$ are non-empty compact, convex sets, then

$(A+B)\div B=A.$

###### Proof.

The first three properties directly follow from the definition. We sketch the
proof of the last implication, and refer the reader to e.g. [Sch13, Lemma
3.1.11, and Section 1.7] for details. Pick an inner product
$\langle\cdot,\cdot\rangle$ on $\mathbb{R}^{d}$, and denote the _support
functions_ for $A$ and $B$ by, respectively, $h_{A}$ and $h_{B}$. By
definition, for any $x\in\mathbb{R}^{d}$

$h_{A}(x):=\sup\\{\langle x,y\rangle\hskip 5.69054pt|\hskip 5.69054pty\in
A\\},$

$h_{B}$ is defined similarly. By the _separation theorem_ for convex sets, for
a non-empty compact convex set $A$ we have

$a\in A\iff\langle a,x\rangle\leq h_{A}(x)\hskip 5.69054pt\text{for all
}x\in\mathbb{R}^{d}.$

Assuming $x\in(A+B)\div B$, we would like to show that $x\in A$. By
hypothesis, $x+B\subset A+B$. Equivalently, the support function for $x+B$
does not exceed that of $A+B$ pointwise. This implies

$h_{\\{x\\}}+h_{B}\leq h_{A}+h_{B},$

using the fact that $h_{A+B}=h_{A}+h_{B}$ for non-empty compact convex sets
$A$ and $B$; see [Sch13, Theorem 1.7.5]. Cancelling $h_{B}$ from both sides
gives us the inequality $h_{\\{x\\}}\leq h_{A}$, implying that $x\in A$. ∎

##  3\. Proof of Theorem 1.3

We follow Bayer Fluckiger [Bay06] and Jarvis [Jar14] for the definitions
below. Let $\lambda$ be an algebraic integer, and
$\mathbb{K}=\mathbb{Q}(\lambda)$ be the number field obtained by adjoining
$\lambda$ to $\mathbb{Q}$. We may interpret the points of $\mathbb{K}$ as
lying in a $d$-dimensional real linear space as follows. Let
$\sigma_{1},\cdots,\sigma_{r}$ be the real embeddings of $\mathbb{K}$, and
$\sigma_{r+1},\overline{\sigma}_{r+1},\cdots,\sigma_{r+s},\overline{\sigma}_{r+s}$
be the pairwise conjugate complex embeddings of $\mathbb{K}$, where

(1) $\displaystyle r+2s=d.$

Consider the embedding

$\displaystyle\sigma\colon\mathbb{K}\longrightarrow\mathbb{R}^{r}\times\mathbb{C}^{s},$
$\displaystyle\sigma(x)=(\sigma_{1}(x),\cdots,\sigma_{r+s}(x)).$

We may identify $\mathbb{C}$ with $\mathbb{R}^{2}$, by identifying $a+bi$ with
$(a,b)$, then $\sigma$ becomes an embedding

$\sigma\colon\mathbb{K}\longrightarrow\mathbb{R}^{d}.$

The mapping $\sigma\colon\mathbb{K}\rightarrow\mathbb{R}^{d}$ identifies the
vector space $\mathbb{R}^{d}$ with the tensor product
$\mathbb{K}_{\mathbb{R}}:=\mathbb{K}\otimes_{\mathbb{Q}}\mathbb{R}$,

$\displaystyle\mathbb{K}\otimes_{\mathbb{Q}}\mathbb{R}\longleftrightarrow\mathbb{R}^{d}$
$\displaystyle x\otimes a\mapsto(\sigma x)a.$

$\mathbb{R}^{r}\times\mathbb{C}^{s}$ has a canonical involution, which is
identity on $\mathbb{R}^{r}$ and complex conjugation on $\mathbb{C}^{s}$. Let

(2)
$\displaystyle\mathfrak{B}:=\\{\alpha\in\mathbb{R}^{r}\times\mathbb{C}^{s}\hskip
5.69054pt|\hskip 5.69054pt\alpha=\bar{\alpha},\text{ and all components of
}\alpha\text{ are positive}\\}.$

Given $\alpha\in\mathfrak{B}$, define the symmetric positive definite bilinear
form $q_{\alpha}$ on $\mathbb{K}_{\mathbb{R}}$ by

$\displaystyle
q_{\alpha}\colon\mathbb{K}_{\mathbb{R}}\times\mathbb{K}_{\mathbb{R}}\longrightarrow\mathbb{R}$
$\displaystyle(x,y)\mapsto\text{Trace}(\alpha x\bar{y}).$

Here
$\text{Trace}(x_{1},\cdots,x_{r+s}):=\sum_{i=1}^{r}x_{i}+\sum_{j=r+1}^{r+s}(x_{j}+\overline{x}_{j})$
denotes the trace of
$\mathbb{K}_{\mathbb{R}}\cong\mathbb{R}^{r}\times\mathbb{C}^{s}$, and $\alpha
x\bar{y}$ denotes component wise product in
$\mathbb{R}^{r}\times\mathbb{C}^{s}$. Then $q_{\alpha}$ induces the norm
$|\cdot|_{\alpha}$ on $\mathbb{K}_{\mathbb{R}}$ given by the following
formula, where $\alpha=(\alpha_{1},\cdots,\alpha_{r+s})$ with
$\alpha_{i}\in\mathbb{R}^{>0}$

(3)
$\displaystyle|x|^{2}_{\alpha}=\sum_{i=1}^{r}\alpha_{i}\cdot|\sigma_{i}(x)|^{2}+2\sum_{j=r+1}^{r+s}\alpha_{j}\cdot|\sigma_{j}(x)|^{2}.$

Denote the ring of integers of $\mathbb{K}$ by $\mathcal{O}_{\mathbb{K}}$. Let
$(\mathcal{O}_{\mathbb{K}},q_{\alpha})$ denote the lattice
$\mathcal{O}_{\mathbb{K}}$ equipped with the inner product $q_{\alpha}$. The
_maximum_ of $(\mathcal{O}_{\mathbb{K}},q_{\alpha})$ is defined as

$\displaystyle\max(\mathcal{O}_{\mathbb{K}},q_{\alpha})=\inf\\{u\in\mathbb{R}\hskip
2.84526pt|\hskip 2.84526pt\text{for all }x\in\mathbb{K}_{\mathbb{R}},\text{
there exists }y\in\mathcal{O}_{\mathbb{K}}\text{ with }q_{\alpha}(x-y,x-y)\leq
u\\}.$

The _covering radius_ of $(\mathcal{O}_{\mathbb{K}},q_{\alpha})$ is, by
definition, the square root of $\max(\mathcal{O}_{\mathbb{K}},q_{\alpha})$.
Define the _determinant_ of $(\mathcal{O}_{\mathbb{K}},q_{\alpha})$ as the
determinant of the matrix of $q_{\alpha}$ in a $\mathbb{Z}$-basis of
$\mathcal{O}_{\mathbb{K}}$; i.e. if $\omega_{1},\cdots,\omega_{d}$ is a basis
for the abelian group $\mathcal{O}_{\mathbb{K}}\cong\mathbb{Z}^{d}$ (under
addition) then $\det(\mathcal{O}_{\mathbb{K}},q_{\alpha})$ is the determinant
of the $d\times d$ matrix $(q_{\alpha}(w_{i},w_{j}))$. With this definition,
the determinant of $(\mathcal{O}_{\mathbb{K}},q_{\alpha})$ is equal to the
_square_ of the volume of any fundamental domain for the lattice
$(\mathcal{O}_{\mathbb{K}},q_{\alpha})$. We remark that some texts define the
determinant as the volume of a fundamental domain, but we preferred to follow
Bayer Fluckiger’s convention as in [Bay06].

######  Definition 3.1.

Define the _Hermite-like thickness_
$\tau(\mathcal{O}_{\mathbb{K}},q_{\alpha})$ of
$(\mathcal{O}_{\mathbb{K}},q_{\alpha})$ as

$\displaystyle\tau(\mathcal{O}_{\mathbb{K}},q_{\alpha}):=\frac{\max(\mathcal{O}_{\mathbb{K}},q_{\alpha})}{\det(\mathcal{O}_{\mathbb{K}},q_{\alpha})^{\frac{1}{d}}}.$

Define the _minimal Hermite-like thickness_ as

$\displaystyle\tau_{\min}(\mathcal{O}_{\mathbb{K}})=\inf\\{\tau(\mathcal{O}_{\mathbb{K}},q_{\alpha})\hskip
5.69054pt|\hskip 5.69054pt\alpha\in\mathfrak{B})\\},$

where $\mathfrak{B}$ is as in (2).

######  Remark 3.2.

Although we called $\tau_{\min}(\mathcal{O}_{\mathbb{K}})$ the minimal
Hermite-like thickness, it should be noted that the minimum is taken over the
set of inner products coming from elements of $\mathfrak{B}$ and not all
possible inner products on $\mathbb{R}^{d}$.

It is clear that the concepts of maximum, covering radius, and Hermite-like
thickness can be defined more generally for a lattice in a Euclidean space;
see [Bay06].

######  Definition 3.3.

Let $\mathbb{K}$ be a number field. Assume that $\omega_{1},\cdots,\omega_{d}$
is any integral basis for $\mathcal{O}_{\mathbb{K}}$. Denote the complete list
of places of $\mathbb{K}$ by $\sigma_{1},\cdots,\sigma_{d}$. The
_discriminant_ of $\mathbb{K}$ is defined as the square of the determinant of
the $d\times d$ matrix $(\sigma_{i}(\omega_{j}))$.

See [Jar14, Chapters 3 and 7] for further properties of the discriminant as
well as the embedding of $\mathcal{O}_{\mathbb{K}}$ in $\mathbb{R}^{d}$.

######  Theorem 1.3.

Let $\lambda$ be a Perron number of algebraic degree $d\geq 3$ and spectral
ratio $\rho$. Set $\mathbb{K}:=\mathbb{Q}(\lambda)$. Let
$\mathcal{O}_{\mathbb{K}}$ be the ring of integers of $\mathbb{K}$, and denote
the discriminant and the minimal Hermite-like thickness of $\mathbb{K}$ by,
respectively, $D_{\mathbb{K}}$ and $\tau_{\min}(\mathcal{O}_{\mathbb{K}})$.
Then $d_{PF}^{irr}(\lambda)$ is bounded above by each of

$\Big{(}\dfrac{8d}{1-\rho}\Big{)}^{d^{2}}\tau_{\min}(\mathcal{O}_{\mathbb{K}})^{\frac{d}{2}}$

and

$\Big{(}\dfrac{8d}{1-\rho}\Big{)}^{d^{2}}\sqrt{D_{\mathbb{K}}}.$

###### Proof.

Let $\mathbb{K}_{\mathbb{R}}:=\mathbb{K}\otimes_{\mathbb{Q}}\mathbb{R}$. Let
$\sigma_{1},\cdots,\sigma_{r+s}$ be as before, and identify
$\mathbb{K}_{\mathbb{R}}$ with
$\mathbb{R}^{r}\times\mathbb{C}^{s}\cong\mathbb{R}^{d}$. A place $\sigma_{j}$
is a field homomorphism
$\sigma_{j}\colon\mathbb{Q}(\lambda)\rightarrow(\mathbb{R}\text{ or
}\mathbb{C})$ and hence $\sigma_{j}$ is completely determined by
$\sigma_{j}(\lambda)$ which is one of the Galois conjugates of $\lambda$.
Assume that $\sigma_{1}$ is the real place corresponding to $\lambda$ itself;
i.e. $\sigma_{1}(\lambda)=\lambda$. Let $M_{\lambda}$ be the linear
endomorphisms of $\mathbb{K}_{\mathbb{R}}$ induced by multiplication by
$\lambda$ in $\mathbb{K}$. The eigenvalues of $M_{\lambda}$ are the Galois
conjugates of $\lambda$. For any Galois conjugate $\lambda_{i}$, denote by
$E_{i}$ the invariant subspace for $M_{\lambda}$ with eigenvalue
$\lambda_{i}$, and let $\pi_{i}\colon\mathbb{K}_{\mathbb{R}}\rightarrow E_{i}$
be the projection along the complementary invariant subspace. Therefore,
$\pi_{i}$ is the projection onto the $i$th factor under the identification
$\mathbb{K}_{\mathbb{R}}\cong\mathbb{R}^{r}\times\mathbb{C}^{s}$.

Before going into the details, we explain the main idea when $\lambda$ is a
cubic algebraic integer that is not totally real. In order to find a non-
negative integral matrix with spectral radius $\lambda$, we follow Thurston’s
proof of Lind’s theorem. See [Thu14, pages 353–354] and [Lin84]. Let $E_{1}$
be the one dimensional invariant subspace for $M_{\lambda}$ with eigenvalue
$\lambda$, and $E_{2}$ be the two dimensional invariant subspace corresponding
to the pair of complex Galois conjugates $\\{\delta,\overline{\delta}\\}$ of
$\lambda$. The endomorphism $M_{\lambda}$ of $\mathbb{K}_{\mathbb{R}}$ leaves
$E_{1}$ and $E_{2}$ invariant, and it acts on $E_{1}\cong\mathbb{R}$ and
$E_{2}\cong\mathbb{C}$ by multiplication by, respectively, the numbers
$\lambda$ and $\delta$. Pick a large positive integer $N=N(\delta,\lambda)$
such that if $P_{\delta}$ is a regular $N$-gon inscribed in a circle of radius
$R$ around the origin in $E_{2}$, then $P_{\delta}\subset
E_{2}\cong\mathbb{C}$ is invariant under multiplication by the complex number
$\delta/\lambda$. Such an integer $N$ exists since by the Perron condition the
absolute value of $\delta/\lambda$ is strictly less than $1$. Let $v$ be an
eigenvector of $M_{\lambda}$ with eigenvalue $\lambda$, and $E_{2}^{v}$ be the
affine plane $Rv+E_{2}$. Then $E_{2}^{v}$ lies in the positive half-space $E$
corresponding to $\lambda$ and $v$.

Denote the vertices of the shifted polygon $P_{v}:=Rv+P_{\delta}\subset
E_{2}^{v}$ by $v_{1},\cdots,v_{k}$, and choose integral points
$z_{1},\cdots,z_{k}\subset\mathbb{R}^{3}$ such that the distance between
$z_{i}$ and $v_{i}$ is ‘small’. Since the cone over the points
$v_{1},\cdots,v_{k}$ lies in the positive half-space and is invariant under
$M_{\lambda}$, and the distance between $z_{i}$ and $v_{i}$ is small, it is
reasonable to expect that the cone $\mathcal{C}$ over the points $z_{i}$ also
lies in the positive half-space $E$ and is invariant under $M_{\lambda}$ for
‘large’ $R$. Let $S$ be the semigroup generated by the set of integral points
in the cone $\mathcal{C}$ under vector addition. Then $M_{\lambda}$ preserves
$S$ and induces an action $M_{\lambda}^{S}$ on $S$. Moreover, $S$ has a finite
generating set, and we can estimate an upper bound for the size $|G|$ of a
generating set $G$ using Proposition 2.1. If we write the action of
$M_{\lambda}^{S}$ on $S$ in the generating set $G$, we obtain a non-negative
integral matrix of size $|G|$ whose spectral radius is equal to $\lambda$. The
details of the proof are as follows.

Step 1: Choosing an inner product on $\mathbb{R}^{r}\times\mathbb{C}^{s}$.

Define $\mathfrak{B}$ as in (2). Pick $\alpha\in\mathfrak{B}$ and equip
$\mathbb{R}^{r}\times\mathbb{C}^{s}$ with the inner product $q_{\alpha}$. Note
that for the norm $|\cdot|_{\alpha}$ and for any $x\in\mathbb{K}_{\mathbb{R}}$

(4) $\displaystyle|x|_{\alpha}^{2}=\sum_{j=1}^{r+s}|\pi_{j}(x)|_{\alpha}^{2},$

and hence

(5) $\displaystyle|x|_{\alpha}\geq|\pi_{j}(x)|_{\alpha}\hskip
17.07164pt\text{for}\hskip 5.69054pt1\leq j\leq r+s.$

Let $\ell$ be the covering radius of $(\mathcal{O}_{\mathbb{K}},q_{\alpha})$.
Then

(6)
$\displaystyle\ell:=\max(\mathcal{O}_{\mathbb{K}},q_{\alpha})^{\frac{1}{2}}=\tau(\mathcal{O}_{\mathbb{K}},q_{\alpha})^{\frac{1}{2}}\cdot\det(\mathcal{O}_{\mathbb{K}},q_{\alpha})^{\frac{1}{2d}}.$

Step 2: Defining the polygon $P_{j}$ in the invariant subspace $E_{j}$ for
each $j>1$.

Define

(7)
$\displaystyle\rho_{j}=\frac{|\sigma_{j}(\lambda)|}{\lambda}\in\mathbb{R}\hskip
17.07164pt\text{for}\hskip 5.69054pt1<j\leq r+s.$

Hence

(8) $\displaystyle\rho=\max_{j>1}\\{\rho_{j}\\}.$

By the Perron condition, $\rho_{j}\in(0,1)$ for each $j>1$. Set

(9) $\displaystyle R_{j}=\frac{(2\sqrt{d}+4)\ell}{1-\rho}\hskip
17.07164pt\text{for}\hskip 5.69054pt1<j\leq r.$

######  Remark 3.4.

Clearly $R_{j}$ does not depend on $1<j\leq r$. However, we decided that this
notation would be more suitable if one would like to improve the bounds in the
article by substituting $\rho_{j}$ instead of $\rho$ in the definition of
$R_{j}$. Similarly, in what follows $R_{j}$ for $j>r$ will not depend on $j$.

For each real place $\sigma_{j}$ with $j>1$, define $P_{j}$ as the set of
points of distance at most $R_{j}$ from the origin in $E_{j}$; in particular
$P_{j}$ is an interval. Note, for future use, that

(10) $\displaystyle R_{j}\leq\Big{(}\frac{8\sqrt{d}}{1-\rho}\Big{)}\ell\hskip
17.07164pt\text{for}\hskip 5.69054pt1<j\leq r.$

For $j>r$, define the natural number $N_{j}\geq 3$ as the smallest positive
integer satisfying

(11) $\displaystyle N_{j}^{2}\geq\frac{2\sqrt{d}+9}{1-\rho}.$

Note, for later use, that

(12) $\displaystyle N_{j}^{2}\leq\frac{16\sqrt{d}}{1-\rho}.$

In the above, we used the fact that the smallest whole square exceeding $u\geq
16$ is less than or equal to $(\sqrt{u}+1)^{2}\leq 3u/2+1$. Set

(13) $\displaystyle R_{j}=N_{j}^{2}\ell\hskip 17.07164pt\text{for}\hskip
5.69054ptr<j\leq r+s.$

For $j>r$, define the solid polygon $P_{j}\subset E_{j}$ as a regular
$N_{j}$-gon inscribed in the circle of radius $R_{j}$ around the origin.
Define the polytope $P$ as the product of $P_{j}$ for $j>1$

$P:=\\{z\in\prod_{j>1}E_{j}\hskip 5.69054pt|\hskip 5.69054pt\pi_{j}(z)\in
P_{j}\hskip 5.69054pt\text{for every }j\\}\subset\prod_{j>1}E_{j}.$

The number of vertices of $P$ is equal to the product of the number of
vertices of $P_{j}$ for $j>1$, which is equal to
$2^{r-1}\times\prod_{j>r}N_{j}$.

Step 3: Defining the cone $\mathcal{C}$ in the positive half-space of
$\lambda$.

Let $v\in E_{1}$ be the positive (with respect to $\pi_{1}$) unit length
eigenvector for the eigenvalue $\lambda$. Here unit length is considered with
respect to the distance $|\cdot|_{\alpha}$. Let $E$ be the positive half-space
corresponding to $v$. Set

(14) $\displaystyle L=\max\\{R_{j}\hskip 5.69054pt|\hskip 5.69054pt1<j\leq
r+s\\}.$

Note, for future use, that

(15) $\displaystyle L>3\ell.$

Denote by $P_{v}$ the translated polytope $Lv+P$, and let
$\mathcal{C}_{v}\subset E$ be the cone over the polytope $P_{v}$.
Equivalently, if $\frac{1}{L}\cdot P_{j}$ is the dilation of $P_{j}$ by the
factor $\frac{1}{L}$ centered at the origin of $E_{j}$, then

(16) $\displaystyle\mathcal{C}_{v}=\\{z\in E\hskip 5.69054pt|\hskip
5.69054pt\frac{\pi_{j}(z)}{|\pi_{1}(z)|_{\alpha}}\in\frac{1}{L}\cdot
P_{j}\hskip 8.53581pt\text{for every }j>1\\}.$

Denote the vertices of $P_{v}$ by $v_{1},v_{2},\cdots,v_{k}$, where

(17) $\displaystyle k=2^{r-1}\times\prod_{j>r}N_{j}.$

Pick integral points $z_{1},z_{2},\cdots,z_{k}$ in $\mathcal{O}_{\mathbb{K}}$
such that the distance between $v_{i}$ and $z_{i}$ does not exceed the
covering radius $\ell$ of $(\mathcal{O}_{\mathbb{K}},q_{\alpha})$.

First we show that each $z_{i}$ lies in the positive half-space $E$. Identify
the subspace $E_{1}$ with $\mathbb{R}$ via

$x=rv\in E_{1}\longleftrightarrow r\in\mathbb{R}.$

It is enough to show that for each $i$ we have $\pi_{1}(z_{i})>0$. By the
triangle inequality

$\pi_{1}(z_{i})\geq\pi_{1}(v_{i})-|\pi_{1}(v_{i}-z_{i})|\geq\pi_{1}(v_{i})-|v_{i}-z_{i}|_{\alpha}\geq
L-\ell>0,$

where the second inequality holds by (5), and the last inequality is true by
(15).

Define $\mathcal{C}$ as the cone over the points $z_{1},z_{2},\cdots,z_{k}$
with apex at the origin

(18) $\displaystyle\mathcal{C}=\\{z\in E\hskip 5.69054pt|\hskip
5.69054ptz=\beta_{1}z_{1}+\cdots+\beta_{k}z_{k},\hskip 8.53581pt\beta_{i}\geq
0\hskip 8.53581pt\text{for every }i\\}.$

For each $z_{i}$, define $w_{i}$ as the intersection point of the ray through
$z_{i}$ from the origin and the affine hyperplane $Lv+E_{1}^{c}$, where
$E_{1}^{c}=\oplus_{j>1}E_{j}$ is the complementary invariant subspace to
$E_{1}$. Then

(19) $\displaystyle\mathcal{C}=\\{z\in E\hskip 5.69054pt|\hskip
5.69054ptz=\beta_{1}w_{1}+\cdots+\beta_{k}w_{k},\hskip 8.53581pt\beta_{i}\geq
0\hskip 8.53581pt\text{for every }i\\}.$

Equivalently, if $Q_{v}\subset Lv+E_{1}^{c}$ is the convex hull of
$w_{1},w_{2},\cdots,w_{k}$, then

(20) $\displaystyle\mathcal{C}=\\{z\in E\hskip 5.69054pt|\hskip
5.69054pt\frac{z}{|\pi_{1}(z)|_{\alpha}}\in\frac{1}{L}\cdot Q_{v}\\}=\\{z\in
E\hskip 5.69054pt|\hskip 5.69054pt\frac{Lz}{|\pi_{1}(z)|_{\alpha}}\in
Q_{v}\\}.$

Hence, $\mathcal{C}$ is the cone over the points $z_{1},z_{2},\cdots,z_{k}$,
or equivalently the cone over the points $w_{1},w_{2},\cdots,w_{k}$. In what
follows, we will use the two descriptions of $\mathcal{C}$ as needed.

Step 4: Showing that the cone $\mathcal{C}$ is invariant.

We need a few lemmas in order to prove that $\mathcal{C}$ is invariant.

######  Lemma 3.5.

For each $i$ we have

$\displaystyle|v_{i}|_{\alpha}\leq\sqrt{d}L$
$\displaystyle|v_{i}-w_{i}|_{\alpha}\leq(2\sqrt{d}+2)\ell.$

###### Proof.

For the first inequality, we have

$\displaystyle|v_{i}|_{\alpha}^{2}$
$\displaystyle=\sum_{j=1}^{r+s}|\pi_{j}(v_{i})|_{\alpha}^{2}\leq
L^{2}+\sum_{j>1}R_{j}^{2}\leq(r+s)L^{2}\leq dL^{2},$

where we used (4) and (14).

Assume that $w_{i}=r_{i}z_{i}$ for some positive real number $r_{i}$. Note
that

$\displaystyle|\pi_{1}(z_{i})-\pi_{1}(v_{i})|_{\alpha}=|\pi_{1}(z_{i}-v_{i})|_{\alpha}\leq|z_{i}-v_{i}|_{\alpha}\leq\ell$
$\displaystyle\implies\pi_{1}(z_{i})=\pi_{1}(v_{i})+\pi_{1}(z_{i}-v_{i})\in[L-\ell,L+\ell],$

where we used (5) in the first line above. Therefore

$\displaystyle
r_{i}=\frac{\pi_{1}(w_{i})}{\pi_{1}(z_{i})}=\frac{L}{\pi_{1}(z_{i})}\in[\frac{L}{L+\ell},\frac{L}{L-\ell}]$
$\displaystyle\implies r_{i}\leq\frac{L}{L-\ell},\hskip
8.53581pt\text{and}\hskip 8.53581pt|r_{i}-1|\leq\frac{\ell}{L-\ell}.$

Hence, by the triangle inequality

$\displaystyle|w_{i}-v_{i}|_{\alpha}=|r_{i}z_{i}-v_{i}|_{\alpha}\leq|r_{i}z_{i}-r_{i}v_{i}|_{\alpha}+|r_{i}v_{i}-v_{i}|_{\alpha}$
$\displaystyle=r_{i}|z_{i}-v_{i}|_{\alpha}+|r_{i}-1||v_{i}|_{\alpha}\leq(\frac{L}{L-\ell})\ell+(\frac{\ell}{L-\ell})\sqrt{d}L$
$\displaystyle=(\sqrt{d}+1)\frac{L\ell}{L-\ell}\leq(2\sqrt{d}+2)\ell.$

The last implication above, $L\leq 2(L-\ell)$, holds by (15). ∎

Set

(21) $\displaystyle b=(2\sqrt{d}+2)\ell.$

######  Lemma 3.6.

The following estimates hold for every $z_{i}$, where $M_{\lambda}$ is the
linear endomorphism of $\mathbb{K}_{\mathbb{R}}$ induced by multiplication by
$\lambda$ in $\mathbb{K}$.

  1. a)

$\frac{|\pi_{j}(M_{\lambda}(z_{i}))|_{\alpha}}{|\pi_{1}(M_{\lambda}(z_{i}))|_{\alpha}}\leq\rho_{j}\Big{(}\frac{R_{j}+\ell}{L-\ell}\Big{)}\hskip
17.07164pt\text{for every }j>1.$

  2. b)

$\frac{|\pi_{j}(M_{\lambda}(z_{i}))|_{\alpha}}{|\pi_{1}(M_{\lambda}(z_{i}))|_{\alpha}}\leq\frac{R_{j}-b}{L}\hskip
17.07164pt\text{for every }1<j\leq r.$

  3. c)

$\frac{|\pi_{j}(M_{\lambda}(z_{i}))|_{\alpha}}{|\pi_{1}(M_{\lambda}(z_{i}))|_{\alpha}}\leq\frac{1}{L}\bigg{(}R_{j}\big{(}1-\frac{5}{N_{j}^{2}}\big{)}-b\bigg{)}\hskip
17.07164pt\text{for every }j>r.$

###### Proof.

  1. a)

We have

$\displaystyle\frac{|\pi_{j}(M_{\lambda}(z_{i}))|_{\alpha}}{|\pi_{1}(M_{\lambda}(z_{i}))|_{\alpha}}$
$\displaystyle=\frac{|\sigma_{j}(\lambda)|}{|\sigma_{1}(\lambda)|}\cdot\frac{|\pi_{j}(z_{i})|_{\alpha}}{|\pi_{1}(z_{i})|_{\alpha}}=\rho_{j}\cdot\frac{|\pi_{j}(z_{i})|_{\alpha}}{|\pi_{1}(z_{i})|_{\alpha}}$
$\displaystyle\leq\rho_{j}\cdot\frac{|\pi_{j}(v_{i})|_{\alpha}+|\pi_{j}(z_{i}-v_{i})|_{\alpha}}{|\pi_{1}(v_{i})|_{\alpha}-|\pi_{1}(v_{i}-z_{i})|_{\alpha}}$
$\displaystyle\leq\rho_{j}\cdot\frac{|\pi_{j}(v_{i})|_{\alpha}+|z_{i}-v_{i}|_{\alpha}}{|\pi_{1}(v_{i})|_{\alpha}-|v_{i}-z_{i}|_{\alpha}}\leq\rho_{j}\Big{(}\frac{R_{j}+\ell}{L-\ell}\Big{)},$

where we have used inequality (5).

  2. b)

Assume that $\sigma_{j}$ is a real place where $1<j\leq r$, and set
$R^{\prime}_{j}=R_{j}/\ell$, $L^{\prime}=L/\ell$, and $b^{\prime}=b/\ell$.
Then

$\displaystyle\rho_{j}\Big{(}\frac{R_{j}+\ell}{L-\ell}\Big{)}\leq\frac{R_{j}-b}{L}$
$\displaystyle\iff\rho_{j}\Big{(}\frac{R^{\prime}_{j}+1}{L^{\prime}-1}\Big{)}\leq\frac{R^{\prime}_{j}-b^{\prime}}{L^{\prime}}$
$\displaystyle\iff(\rho_{j}+b^{\prime})L^{\prime}+R_{j}^{\prime}-b^{\prime}\leq(1-\rho_{j})L^{\prime}R^{\prime}_{j}.$

Hence, using $\rho_{j}\in(0,1)$, it is enough to have

$\displaystyle(1+b^{\prime})L^{\prime}+R_{j}^{\prime}\leq(1-\rho_{j})L^{\prime}R^{\prime}_{j}$
$\displaystyle\iff\frac{1+b^{\prime}}{R^{\prime}_{j}}+\frac{1}{L^{\prime}}\leq
1-\rho_{j}$
$\displaystyle\iff\frac{2\sqrt{d}+3}{R^{\prime}_{j}}+\frac{1}{L^{\prime}}\leq
1-\rho_{j}.$

On the other hand, $L^{\prime}\geq R_{j}^{\prime}$ by (14). Hence, it suffices
to have

$\displaystyle\frac{2\sqrt{d}+4}{R^{\prime}_{j}}\leq 1-\rho_{j}\iff
R^{\prime}_{j}\geq\frac{2\sqrt{d}+4}{1-\rho_{j}},$

which holds by (9).

  3. c)

Now assume that $\sigma_{j}$ is a complex place where $j>r$. Then

$\displaystyle\rho_{j}\Big{(}\frac{R_{j}+\ell}{L-\ell}\Big{)}\leq\frac{1}{L}\bigg{(}R_{j}\big{(}1-\frac{5}{N_{j}^{2}}\big{)}-b\bigg{)}$
$\displaystyle\iff\rho_{j}(R_{j}+\ell)\leq\Big{(}\frac{L-\ell}{L}\Big{)}\Big{(}R_{j}\big{(}1-\frac{5}{N_{j}^{2}}\big{)}-b\Big{)}.$

By (14), we have $L\geq R_{j}=N_{j}^{2}\ell$, and so

$\displaystyle\frac{L-\ell}{L}=1-\frac{\ell}{L}\geq 1-\frac{1}{N_{j}^{2}}.$

Set $b^{\prime}=b/\ell$. After substituting $R_{j}=N_{j}^{2}\ell$ and dividing
both sides by $\ell$, it is enough to have

$\displaystyle\rho_{j}(N_{j}^{2}+1)\leq\Big{(}1-\frac{1}{N_{j}^{2}}\Big{)}\Big{(}N_{j}^{2}-5-b^{\prime}\Big{)}.$

Multiplying both sides by $N_{j}^{2}$ and then expanding, the last inequality
is equivalent to

$\displaystyle N_{j}^{4}(1-\rho_{j})+(5+b^{\prime})\geq
N_{j}^{2}(b^{\prime}+6+\rho_{j}).$

So, using $\rho_{j}\in(0,1)$ and neglecting the positive term $5+b^{\prime}$,
it suffices to have

$\displaystyle N_{j}^{4}(1-\rho_{j})\geq N_{j}^{2}(b^{\prime}+7)\iff
N_{j}^{2}\geq\frac{b^{\prime}+7}{1-\rho_{j}}=\frac{2\sqrt{d}+9}{1-\rho_{j}},$

and the last inequality holds by (11).

∎

Recall that $\mathcal{C}$ is the cone over the points $w_{1},\cdots,w_{k}$
with apex at the origin, i.e. the cone over the polytope $Q_{v}$ (i.e. the
convex hull of $w_{i}$). For each $i$, the point $w_{i}$ of $Q_{v}$ is of
distance at most $b=(2\sqrt{d}+2)\ell$ from the corresponding point $v_{i}$ of
$P_{v}$. Since $b$ is ‘small’, we expect the polytope $Q_{v}$ to contain a
‘smaller version’ of $P_{v}$. More precisely, define

$\displaystyle P_{v}^{b}=\\{x\in P_{v}\hskip 5.69054pt|\hskip
5.69054pt\text{dist}(x,\partial P_{v})\geq b\\},$

where

$\displaystyle\text{dist}(X,Y):=\min\\{|x-y|_{\alpha}\hskip 5.69054pt|\hskip
5.69054ptx\in X,\hskip 5.69054pty\in Y\\}.$

######  Lemma 3.7.

We have $P_{v}^{b}\subset Q_{v}$.

###### Proof.

Let $B$ be the ball of radius $b$ around the origin in
$E_{1}^{c}=\oplus_{j>1}E_{j}$. We show that

$P_{v}\subset Q_{v}+B,$

where $Q_{v}+B$ denotes the Minkowski sum. To see this, pick an arbitrary
point $x\in P_{v}$ and write it as a convex combination

$x=\sum_{i=1}^{k}\alpha_{i}v_{i},$

with $0\leq\alpha_{i}\leq 1$ and $\sum_{i=1}^{k}\alpha_{i}=1$. Then

$x=\sum_{i=1}^{k}\alpha_{i}w_{i}+\sum_{i=1}^{k}\alpha_{i}(v_{i}-w_{i}).$

If we set $y=\sum_{i=1}^{k}\alpha_{i}w_{i}$ and
$z=\sum_{i=1}^{k}\alpha_{i}(v_{i}-w_{i})$, then $y\in Q_{v}$. Moreover, by the
triangle inequality $z\in B$, since each term $v_{i}-w_{i}$ lies in $B$. This
shows that $P_{v}\subseteq Q_{v}+B$.

If we denote the Minkowski difference by $\div$, then we have

$P_{v}\div B\subset(Q_{v}+B)\div B=Q_{v},$

where the last equality holds since $Q_{v}$ and $B$ are non-empty, compact,
and convex. See Lemma 2.3. Therefore, it suffices to show that
$P_{v}^{b}\subset P_{v}\div B$. It follows from the definition that
$P_{v}^{b}+B\subset P_{v}$. Hence, assuming that $P_{v}^{b}$ is non-empty, we
have

$P_{v}^{b}=(P_{v}^{b}+B)\div B\subset P_{v}\div B,$

where we used Lemma 2.3 twice. This completes the proof. ∎

Define

$\displaystyle P_{j}^{b}:=\\{x\in P_{j}\hskip 5.69054pt|\hskip
5.69054pt\text{dist}(x,\partial P_{j})\geq b\\},$

and set

$\displaystyle V:=Lv+\\{x\in E_{1}^{c}\hskip 5.69054pt|\hskip
5.69054pt\pi_{j}(x)\in P_{j}^{b}\hskip 8.53581pt\text{for every }j>1\\},$
$\displaystyle W:=Lv+\\{x\in E_{1}^{c}\hskip 2.84526pt|\hskip
2.84526pt|\pi_{j}(x)|_{\alpha}\leq R_{j}-b\hskip 5.69054pt\text{for }1<j\leq
r,\hskip 2.84526pt|\pi_{j}(x)|_{\alpha}\leq
R_{j}(1-\frac{5}{N_{j}^{2}})-b\hskip 5.69054pt\text{for }j>r\\}.$

######  Lemma 3.8.

We have $W\subset V\subset P_{v}^{b}\subset Q_{v}$.

###### Proof.

The inclusion $W\subset V$ follows from the following simple fact: Let $T$ be
a regular $N$-gon inscribed in a circle of radius $R$ centered at the point
$O$. Every point in $\partial T$ is of distance at least $R(1-5/N^{2})$ from
$O$.

For completeness, we give a proof of the above fact. After scaling, we may
assume that $R=1$. The minimum distance $d(O,x)$ for $x\in\partial T$ is
obtained by drawing the perpendicular from $O$ to a side of $T$. Such a
perpendicular has length $\cos(\frac{\pi}{N})$. Hence

$d(O,x)\geq\cos(\frac{\pi}{N})>1-\frac{\pi^{2}}{2N^{2}}>1-\frac{5}{N^{2}},$

where we have used the inequality $\cos(x)>1-\frac{x^{2}}{2}$ for $0<x<1$.

The inclusion $P_{v}^{b}\subset Q_{v}$ was proved in Lemma 3.7. Now we show
that $V\subset P_{v}^{b}$. Pick arbitrary points $x\in V$ and $y\in\partial
P_{v}=\partial(Lv+P)$. Since $P=\prod_{j>1}P_{j}$ is a product, there is $j>1$
such that $\pi_{j}(y)\in\partial P_{j}$. Therefore

$\text{dist}(x,y)\geq\text{dist}(\pi_{j}(x),\pi_{j}(y))\geq\text{dist}(P_{j}^{b},\partial
P_{j})\geq b.$

In the above, we used the fact that projection onto a non-empty closed convex
set in a Euclidean space is distance decreasing; see e.g. [Sch13, Theorem
1.2.1]. Hence $x\in P_{v}^{b}$, proving the inclusion $V\subset P_{v}^{b}$. ∎

Using (20), in order to show that $\mathcal{C}$ is invariant, it is enough to
prove that for every $z_{i}$

$\displaystyle
L\cdot\frac{M_{\lambda}(z_{i})}{|\pi_{1}(M_{\lambda}(z_{i}))|_{\alpha}}\in
W\subset Q_{v}.$

Equivalently, we would like to show that for every $j>1$ and every $z_{i}$ the
following inequalities hold

$\displaystyle\frac{|\pi_{j}(M_{\lambda}(z_{i}))|_{\alpha}}{|\pi_{1}(M_{\lambda}(z_{i}))|_{\alpha}}\leq\frac{R_{j}-b}{L}\hskip
8.53581pt\text{for }1<j\leq r,$
$\displaystyle\frac{|\pi_{j}(M_{\lambda}(z_{i}))|_{\alpha}}{|\pi_{1}(M_{\lambda}(z_{i}))|_{\alpha}}\leq\frac{1}{L}\bigg{(}R_{j}\big{(}1-\frac{5}{N_{j}^{2}}\big{)}-b\bigg{)}\hskip
8.53581pt\text{for }j>r.$

But the above inequalities hold by Lemma 3.6. This finishes the proof of the
invariance of $\mathcal{C}$.

Step 5: Defining a non-negative integral matrix $A$, where each irreducible
component of $A$ has spectral radius equal to $\lambda$.

Let $S$ be the semigroup generated by elements of
$\mathcal{C}\cap\mathcal{O}_{\mathbb{K}}$ under vector addition. By
Proposition 2.1, the set of integral points (i.e. elements of
$\mathcal{O}_{\mathbb{K}}$) in the compact region

(22) $\displaystyle
C:=\\{\alpha_{1}z_{1}+\alpha_{2}z_{2}+\cdots+\alpha_{k}z_{k}\hskip
2.84526pt|\hskip 2.84526pt0\leq\alpha_{i}\leq 1\hskip 8.53581pt\text{for every
}i\\}$

generate $S$. Enumerate the set of points in $C\cap\mathcal{O}_{\mathbb{K}}$
by $c_{1},\cdots,c_{n}$. Recall that $M_{\lambda}$ acts on $S$. Moreover,
$M_{\lambda}$ can be represented by the companion matrix $B$ when written in
the basis $\\{1,\lambda,\cdots,\lambda^{d-1}\\}$ of
$\mathbb{K}_{\mathbb{R}}=\mathbb{Q}(\lambda)\otimes\mathbb{R}$. Let
$A=[a_{ij}]$ be a non-negative integral matrix corresponding to the action of
$B$ on $S$ in the basis $\\{c_{1},\cdots,c_{n}\\}$. Hence

(23) $\displaystyle Bc_{j}=\sum_{i=1}^{n}a_{ij}c_{i},$

where $B\colon\mathbb{R}^{d}\rightarrow\mathbb{R}^{d}$ is the companion matrix
associated with $\lambda$. There might be various choices for $A$, when
$Bc_{j}$ can be written as a non-negative linear combination of
$c_{1},\cdots,c_{n}$ in more than one way.

Lind’s argument [Lin84, pages 288–289] shows that every irreducible component
of $A$ has spectral radius $\lambda$. We follow the proof given in [LM21,
Lemma 11.1.10] briefly for the reader’s convenience. Replace $A$ by one of its
irreducible components; this amounts to taking a minimal subset of
$\\{c_{1},\cdots,c_{n}\\}$ for which (23) holds. Let $\mu$ be the spectral
radius of $A$. Let $e_{i}$ be the $i$th unit vector in $\mathbb{R}^{n}$, and
define the linear map $P\colon\mathbb{R}^{n}\rightarrow\mathbb{R}^{d}$ by
$P(e_{i})=c_{i}$. Then by (23) we have $PA=BP$. Since $A$ is non-negative and
irreducible, by the Perron–Frobenius theorem $A$ has a positive eigenvector
$v_{\mu}$ corresponding to $\mu$. Since $Pv_{\mu}$ is a positive linear
combination of the $c_{j}$, and each $c_{j}$ satisfies $\pi_{1}(c_{j})>0$, we
necessarily have $\pi_{1}(Pv_{\mu})>0$ and hence $Pv_{\mu}\neq 0$. Moreover,

$B(Pv_{\mu})=P(Av_{\mu})=\mu(Pv_{\mu}).$

Hence, $Pv_{\mu}$ is an eigenvector for $B$ with eigenvalue $\mu$ . But the
only eigenvectors of $B$ with positive $E_{1}$ component lie in $E_{1}$. Hence
$Pv_{\mu}$ is a multiple of $v$ (the eigenvector with eigenvalue $\lambda$)
and

$\displaystyle\mu(Pv_{\mu})$ $\displaystyle=B(Pv_{\mu})=\lambda(Pv_{\mu})$
$\displaystyle\implies\mu=\lambda.$

######  Remark 3.9.

In [Thu14, page 354], Thurston gave a new proof of Lind’s converse to the
integer Perron–Frobenius theorem. Thurston assumed that the matrix constructed
via his method can be taken to be primitive as well. If this is the case, then
Theorem 1.3 readily upgrades to give an upper bound for the Perron–Frobenius
degree of a Perron number.

Step 6: Bounding the dimension of the matrix $A$.

By Proposition 2.2, the number of integral points in $C$ is at most

$\displaystyle\frac{\textsc{Vol}(C)}{\text{Covol}(\mathcal{O}_{\mathbb{K}})}\cdot(d+1)!=\frac{\textsc{Vol}(C)}{\text{det}(\mathcal{O}_{\mathbb{K}},q_{\alpha})^{\frac{1}{2}}}\cdot(d+1)!.$

We give an upper bound for $\textsc{Vol}(C)$ via a series of lemmas.

######  Lemma 3.10.

For every $j>1$ and every $w_{i}$, we have

$|\pi_{j}(w_{i})|_{\alpha}<2R_{j}.$

###### Proof.

By the triangle inequality

$\displaystyle|\pi_{j}(w_{i})|_{\alpha}$
$\displaystyle\leq|\pi_{j}(v_{i})|_{\alpha}+|\pi_{j}(w_{i}-v_{i})|_{\alpha}$
$\displaystyle\leq|\pi_{j}(v_{i})|_{\alpha}+|w_{i}-v_{i}|_{\alpha}$
$\displaystyle\leq R_{j}+(2\sqrt{d}+2)\ell<2R_{j}$ $\displaystyle\iff
R_{j}>(2\sqrt{d}+2)\ell.$

In the first line above (5) is used. The second line follows from Lemma 3.5,
and the last inequality holds by (9), (11), and (13). ∎

######  Lemma 3.11.

Let $M:=2k\sqrt{d}$, where $k$ is the number of vertices of $P_{v}$. Then

$C\subset\\{rx\in\mathbb{R}^{d}\hskip 5.69054pt|\hskip 5.69054ptx\in
Q_{v},\hskip 2.84526pt\text{and}\hskip 5.69054pt0\leq r\leq M\\}.$

###### Proof.

Since $Q_{v}$ is the convex hull of $w_{i}$, the ray from the origin and
passing thorough an arbitrary point $p$ of $C$ intersects $Q_{v}$ at some
point $w$. Therefore, if we define $M_{1}$ as

$M_{1}=\frac{\max_{p\in C}|p|_{\alpha}}{\min_{w\in Q_{v}}|w|_{\alpha}},$

then

$C\subset\\{rx\in\mathbb{R}^{d}\hskip 5.69054pt|\hskip 5.69054ptx\in
Q_{v},\hskip 2.84526pt\text{and}\hskip 5.69054pt0\leq r\leq M_{1}\\}.$

It is enough to show that $M_{1}\leq M$. For every point $w$ in $Q_{v}$

$|w|_{\alpha}\geq|\pi_{1}(w)|_{\alpha}=L.$

On the other hand if $p$ is a point in $C$, then by the triangle inequality we
have

$|p|_{\alpha}\leq k\cdot\max_{i}|z_{i}|_{\alpha}\leq
k\cdot\max_{i}(|v_{i}|_{\alpha}+\ell)\leq k(\sqrt{d}L+\ell)<2k\sqrt{d}L,$

where we used Lemma 3.5 for the implication $|v_{i}|_{\alpha}\leq\sqrt{d}L$.
Combining the previous inequalities gives

$M_{1}=\frac{\max_{p\in C}|p|_{\alpha}}{\min_{w\in
Q_{v}}|w|_{\alpha}}\leq\frac{2k\sqrt{d}L}{L}=2k\sqrt{d}=M.$

∎

######  Lemma 3.12.

We have

$\textsc{Vol}(Q_{v})\leq
2^{2d-2}\times\prod_{j>1}^{r}R_{j}\times\prod_{j>r}R_{j}^{2},$

where $\textsc{Vol}(Q_{v})$ is the $(d-1)$-dimensional volume of $Q_{v}$.

###### Proof.

We have

$\displaystyle\textsc{Vol}(Q_{v})\leq\prod_{j>1}^{r}\textsc{Length}(\pi_{j}(Q_{v}))\times\prod_{j>r}\textsc{Area}(\pi_{j}(Q_{v})),$

where Vol, Length and Area are with respect to the inner product $q_{\alpha}$.
For $j>1$, define $\textbf{R}_{j}$ as

$\textbf{R}_{j}=\max_{w\in Q_{v}}|\pi_{j}(w)|_{\alpha}.$

Therefore

$\textsc{Vol}(Q_{v})\leq\prod_{j>1}^{r}(2\textbf{R}_{j})\times\prod_{j>r}(\pi\textbf{R}_{j}^{2}).$

By Lemma 3.10 and the triangle inequality we have $\textbf{R}_{j}<2R_{j}$. The
lemma now follows from the inequalities $\textbf{R}_{j}<2R_{j}$ and $\pi<4$,
and the equality $r+2s=d$. ∎

######  Lemma 3.13.

We have

$\textsc{Vol}(C)<2^{d^{2}+6d-1}\times
d^{\frac{ds}{4}+\frac{3d}{2}-1}\times\frac{\ell^{d}}{(1-\rho)^{d+\frac{ds}{2}}},$

where $C$ is defined as in (22).

###### Proof.

Note that

(24) $\displaystyle M=2k\sqrt{d}=2^{r}\times\prod_{j>r}N_{j}\times\sqrt{d}\leq
2^{r}\times\Big{(}\frac{16\sqrt{d}}{1-\rho}\Big{)}^{\frac{s}{2}}\times\sqrt{d}<2^{d}\times\frac{d^{\frac{s}{4}+1}}{(1-\rho)^{\frac{s}{2}}},$

where we used (17), (12), and the identity $r+2s=d$. By Lemma 3.11

$\displaystyle\textsc{Vol}(C)$
$\displaystyle\leq\textsc{Vol}(\\{rx\in\mathbb{R}^{d}\hskip 2.84526pt|\hskip
2.84526ptx\in Q_{v},\hskip 2.84526pt\text{and}\hskip 5.69054pt0\leq r\leq
M\\}).$

The upper bound above is the volume of a cone over $M\cdot Q_{v}$, where
$M\cdot Q_{v}$ denotes a dilation of $Q_{v}$ with the factor of $M$ from the
point $0\in\mathbb{R}^{d}$. The height of this cone is $ML$, since the height
of $Q_{v}$ measured perpendicularly from its apex along the $E_{1}$-axis is
$L$. Hence, the volume of the cone is equal to

$\frac{(ML)\textsc{Vol}(M\cdot
Q_{v})}{d}=\frac{M^{d}L}{d}\textsc{Vol}(Q_{v}).$

Therefore, we have

$\displaystyle\textsc{Vol}(C)\leq\frac{M^{d}L}{d}\times\textsc{Vol}(Q_{v})$
$\displaystyle\leq\Big{(}2^{d}\times\frac{d^{\frac{s}{4}+1}}{(1-\rho)^{\frac{s}{2}}}\Big{)}^{d}\times\frac{L}{d}\times
2^{2d-2}\times\prod_{j>1}^{r}R_{j}\times\prod_{j>r}R_{j}^{2},$

where we used Lemma 3.12. Moreover, using (10), (12), and (13) we have

$\displaystyle\prod_{j>1}^{r}R_{j}\times\prod_{j>r}R_{j}^{2}$
$\displaystyle\leq\prod_{j>1}^{r}\Big{(}\frac{8\sqrt{d}\ell}{1-\rho}\Big{)}\times\prod_{j>r}\Big{(}\frac{16\sqrt{d}\ell}{1-\rho}\Big{)}^{2}$
$\displaystyle=2^{3r+8s-3}\times
d^{\frac{d-1}{2}}\times\frac{\ell^{d-1}}{(1-\rho)^{d-1}}.$

Combining with the previous upper bound for $\textsc{Vol}(C)$, and using
$L<\frac{16\sqrt{d}\ell}{1-\rho}$, we have

$\displaystyle\textsc{Vol}(C)\leq 2^{d^{2}+2d+3r+8s-1}\times
d^{\frac{ds}{4}+\frac{3d}{2}-1}\times\frac{\ell^{d}}{(1-\rho)^{d+\frac{ds}{2}}}\leq
2^{d^{2}+6d-1}\times
d^{\frac{ds}{4}+\frac{3d}{2}-1}\times\frac{\ell^{d}}{(1-\rho)^{d+\frac{ds}{2}}},$

where we used the relation $r+2s=d$.

∎

Therefore, the number of integral points in $C$ is at most

$\displaystyle\frac{\textsc{Vol}(C)}{\text{det}(\mathcal{O}_{\mathbb{K}},q_{\alpha})^{\frac{1}{2}}}\cdot(d+1)!$
$\displaystyle\leq 2^{d^{2}+6d}\times
d^{\frac{ds}{4}+\frac{5d}{2}-1}\times\frac{\ell^{d}}{\text{det}(\mathcal{O}_{\mathbb{K}},q_{\alpha})^{\frac{1}{2}}}\times\frac{1}{(1-\rho)^{d+\frac{ds}{2}}}$
$\displaystyle=2^{d^{2}+6d}\times
d^{\frac{ds}{4}+\frac{5d}{2}-1}\times\tau(\mathcal{O}_{\mathbb{K}},q_{\alpha})^{\frac{d}{2}}\times\frac{1}{(1-\rho)^{d+\frac{ds}{2}}}.$

In the first line above, we used the inequality

$(d+1)!=(d+1)d!<(2d)d^{d-1}=2d^{d}.$

The second line above uses the definition of $\ell$, and Definition 3.1. By
taking infimum over all $\alpha\in\mathfrak{B}$, we obtain the upper bound

$\displaystyle 2^{d^{2}+6d}\times
d^{\frac{ds}{4}+\frac{5d}{2}-1}\times\frac{\tau_{\min}(\mathcal{O}_{\mathbb{K}})^{\frac{d}{2}}}{(1-\rho)^{d+\frac{ds}{2}}},$

for the number of integral points in $C$ and hence for the dimension of the
matrix $A$.

Bayer Fluckiger showed in [Bay06, Propositon 4.2], as a corollary of the work
of Banaszczyk [Ban93, Theorem 2.2], that for any $\alpha\in\mathfrak{B}$

(25)
$\displaystyle\tau(\mathcal{O}_{\mathbb{K}},q_{\alpha})\leq\frac{d}{4}\cdot
D_{\mathbb{K}}^{\frac{1}{d}}.$

This gives the upper bound

$2^{d^{2}+5d}\times
d^{\frac{ds}{4}+3d-1}\times\frac{\sqrt{D_{\mathbb{K}}}}{(1-\rho)^{d+\frac{ds}{2}}}$

for the dimension of $A$. Both parts of the theorem now follow from the crude
estimates

$2^{d^{2}+6d}\leq 8^{d^{2}},\hskip 17.07164pt\frac{ds}{4}+3d-1<d^{2},\hskip
17.07164ptd+\frac{ds}{2}<d^{2}.$

∎

The bound in Theorem 1.3 is perhaps enormous compared to the Perron–Frobenius
degree, so we have not tried to make the constants optimal. The point is to
have an explicit bound in terms of data that we believe are relevant to the
Perron–Frobenius degree; see Question 5.2.

######  Remark 3.14.

Denote the _covering conjecture_ for dimension $d$ by $C_{d}$.

######  Conjecture 3.15 (Covering conjecture).

The covering radius of any well-rounded unimodular lattice $L$ in
$\mathbb{R}^{d}$ (with the standard norm $|\cdot|$) satisfies

$\sup_{x\in\mathbb{R}^{d}}\inf_{y\in L}|x-y|\leq\frac{\sqrt{d}}{2}.$

Equality happens if and only if $L=g\cdot\mathbb{Z}^{d}$ for some
$g\in\mathrm{SO}_{d}(\mathbb{R})$.

See McMullen [McM05] for the definition of _well-rounded lattice_ , and the
application of the covering conjecture to _Minkowski’s conjecture_. The
covering conjecture is proved for $d\leq 10$; see [KR20, KR16] and the
references therein. Moreover, it is known to be false for $d\geq 30$; see
[RSW17].

In [Bay06, page 313], Bayer Fluckiger pointed out that McMullen’s results
[McM05] together with the covering conjecture $C_{d}$ imply that for any
totally real $\lambda$ of degree $d$ we have
$\tau_{\min}(\mathcal{O}_{\mathbb{K}})\leq\frac{d}{4}$. Hence, for any $d\leq
10$ and for any totally real Perron number $\lambda$ of degree $d$

$d_{PF}^{irr}(\lambda)\leq\Big{(}\frac{8d}{1-\rho}\Big{)}^{d^{2}}.$

It would be interesting to know which number fields satisfy an inequality
similar to that of $\tau_{\min}(\mathcal{O}_{\mathbb{K}})\leq\frac{d}{4}$ with
upper bound only depending on the degree $d$. As pointed out kindly by
McMullen to me, it is easy to see that such a bound does not exist for
imaginary quadratic fields $\mathbb{Q}(\sqrt{n})$ for $n<0$. On the other
hand, Bayer Fluckiger showed that the inequality
$\tau_{\min}(\mathcal{O}_{\mathbb{K}})\leq\frac{d}{4}$ holds for fields of the
form $\mathbb{K}=\mathbb{Q}(\zeta_{p^{r}})$ or
$\mathbb{K}=\mathbb{Q}(\zeta_{p^{r}}+\zeta^{-1}_{p^{r}})$, where $p$ is an odd
prime number, $r$ is a natural number, and $\zeta_{p^{r}}$ is a primitive
$p^{r}$th root of unity; see respectively [Bay06, page 319, line 4] and
[Bay06, Lemma 8.5].

######  Example 3.16 (Pisot numbers).

A real algebraic integer $\alpha>1$ is called _Pisot_ if all other Galois
conjugates of $\alpha$ lie in the unit circle $\\{z\in\mathbb{C}\hskip
5.69054pt|\hskip 5.69054pt|z|<1\\}$. Note a Pisot number is always Perron. We
collect a few facts about Pisot numbers.

  1. (1)

A number field is called _real_ if it has at least one real place. It is clear
that every number field containing a Pisot number with the same degree should
be real. Pisot [Pis38] proved that every real number field $\mathbb{K}$ of
degree $d$ contains a Pisot number of degree $d$. Moreover, the set of Pisot
numbers of degree $d$ in $\mathbb{K}$ is closed under multiplication; see
[Mey72, Corollary in page 33].

  2. (2)

The smallest Pisot number is the largest root $p$ of $x^{3}-x-1$ and is known
as the _plastic constant_. This was identified as the smallest known Pisot
number by Salem [Sal44], and Siegel proved it to be the smallest possible
Pisot number [Sie44]. In particular, if $\lambda$ is a Pisot number with
spectral ratio $\rho$, then $\rho<p^{-1}$ implying that

$\frac{1}{1-\rho}<\frac{1}{1-p^{-1}}.$

Let $\mathbb{K}$ be a real number field of degree $d\geq 3$. Let $\lambda$ be
any Pisot number in $\mathbb{K}$ of degree $d$, and note that
$\mathbb{Q}(\lambda)=\mathbb{K}$. By Theorem 1.3, $d_{PF}^{irr}(\lambda)$ is
bounded above by

$\Big{(}\dfrac{8d}{1-p^{-1}}\Big{)}^{d^{2}}\sqrt{D_{\mathbb{K}}}.$

Note this upper bound only depends on $\mathbb{K}$ and not on the Pisot number
$\lambda\in\mathbb{K}$. Every number field $\mathbb{K}$ has only finitely many
subfields $\mathbb{K}^{\prime}\subset\mathbb{K}$. Hence there is a similar
upper bound depending only on $\mathbb{K}$ for an arbitrary Pisot number
$\lambda\in\mathbb{K}$ (with any degree).

##  4\. Primitive matrices

In this section, we use Theorem 1.3 to derive an upper bound for the
Perron–Frobenius degree of a Perron number $\lambda$. A useful observation is
that if there exists an irreducible matrix $A$ with spectral radius
$\lambda-1$, then $\lambda$ is the spectral radius of the _primitive_ matrix
$I+A$. This is because $I+A$ is irreducible and has positive trace; hence it
is primitive.

######  Theorem 1.6.

Let $\mathbb{\lambda}$ be a Perron number of degree $d\geq 3$ and spectral
ratio $\rho$. Set $\mathbb{K}:=\mathbb{Q}(\lambda)$. Denote the bound from
Theorem 1.3 by $B(\mathbb{K},\rho)$, and let $\kappa(\mathbb{K},\rho)$ be as
in Notation 1.5. The Perron–Frobenius degree of $\lambda$ is bounded above by

$\max\\{2^{d^{2}}B(\mathbb{K},\rho),\kappa(\mathbb{K},\rho)\\}.$

###### Proof.

We may assume that $\lambda\geq M=1+\frac{4}{1-\rho}$. First we show that
$\lambda-1$ is Perron. The Galois conjugates of $\lambda-1$ are equal to
$\lambda_{i}-1$. We have

$\displaystyle\frac{|\lambda_{i}-1|}{\lambda-1}$
$\displaystyle\leq\frac{|\lambda_{i}|+1}{\lambda-1}\leq\frac{\rho\lambda+1}{\lambda-1}=\rho+\frac{\rho+1}{\lambda-1}\leq$
$\displaystyle\rho+\frac{2}{\lambda-1}\leq\rho+\frac{1-\rho}{2}<1,$

where we used the assumptions $\rho<1$ and $\lambda\geq 1+\frac{4}{1-\rho}$.
Denote the spectral ratio for $\lambda-1$ by $\mu$. Note that

(26) $\displaystyle 1-\mu\geq\frac{1-\rho}{2}.$

In fact, we just showed that for every $i$

$\displaystyle\frac{|\lambda_{i}-1|}{\lambda-1}\leq\rho+\frac{1-\rho}{2},$

hence

$\displaystyle 1-\frac{|\lambda_{i}-1|}{\lambda-1}\geq
1-(\rho+\frac{1-\rho}{2})=\frac{1-\rho}{2}.$

Inequality (26) in particular implies that

(27) $\displaystyle B(\mathbb{K},\mu)\leq 2^{d^{2}}B(\mathbb{K},\rho).$

Now $\lambda-1$ is a Perron number of spectral ratio $\mu$ and it generates
the same number field $\mathbb{K}$ as $\lambda$. By Theorem 1.3, there is an
integral irreducible matrix $A$ with spectral radius $\lambda-1$ and dimension
at most $B(\mathbb{K},\mu)$. Then $I+A$ is a primitive matrix of the same
dimension and with spectral radius $\lambda$.

∎

##  5\. Questions

In Theorem 1.3, for technical reasons, we constructed an integral
_irreducible_ matrix with spectral radius equal to a given Perron number,
although it would have been more natural to construct an integral _primitive_
matrix instead. This motivates the following:

######  Question 5.1.

  1. (1)

Are there upper bounds for $d_{PF}(\lambda)$ in terms of
$d_{PF}^{irr}(\lambda)$?

  2. (2)

Does $d_{PF}(\lambda)=d_{PF}^{irr}(\lambda)$ always hold?

The lower bound for the Perron–Frobenius degree in [Yaz21] is in terms of the
two largest Galois conjugates in the complex plane. Therefore, the following
is a natural question.

######  Question 5.2.

Let $d$ and $\rho$ denote, respectively, the algebraic degree and the spectral
ratio. Is there an upper bound $B=B(d,\rho)$ for $d_{PF}^{irr}(\lambda)$
(respectively $d_{PF}(\lambda)$) where $\lambda$ is an arbitrary Perron
number?

We expect the above question to have a negative answer, meaning that in
Theorem 1.3 the arithmetic information on $\mathcal{O}_{\mathbb{K}}$ cannot be
overlooked. A unit algebraic integer $\lambda$ is _bi-Perron_ if all other
Galois conjugates of $\lambda$ lie in the annulus $\\{c\in\mathbb{C}\hskip
2.84526pt|\hskip 2.84526pt\lambda^{-1}<|z|<\lambda\\}$ except possibly for
$\pm\lambda^{-1}$. See [McM14].

######  Question 5.3.

How can the bound in Theorem 1.3 be improved for special classes of algebraic
integers such as totally real Perron numbers, Pisot numbers, Salem numbers, or
bi-Perron numbers?

In particular, we ask the following about Pisot numbers.

######  Question 5.4.

Let $d$ denote the algebraic degree. Is there an upper bound $B=B(d)$ for
$d_{PF}^{irr}(\lambda)$ (respectively $d_{PF}(\lambda)$) where $\lambda$ is an
arbitrary Pisot number?

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