# Limiting distributions of conjugate algebraic integers

Bryce Joseph Orloski and Naser Talebizadeh Sardari  Penn State department of
Mathematics, McAllister Building, Pollock Rd, State College, PA 16802 USA
nzt5208<EMAIL_ADDRESS>
###### Abstract.

Let $\Sigma\subset\mathbb{C}$ be a compact subset of the complex plane, and
$\mu$ be a probability distribution on $\Sigma$. We give necessary and
sufficient conditions for $\mu$ to be the weak* limit of a sequence of uniform
probability measures on a complete set of conjugate algebraic integers lying
eventually in any open set containing $\Sigma$. Given $n\geq 0$, any
probability measure $\mu$ satisfying our necessary conditions, and any open
set $D$ containing $\Sigma$, we develop and implement a polynomial time
algorithm in $n$ that returns an integral monic irreducible polynomial of
degree $n$ such that all of its roots are inside $D$ and their root
distributions converge weakly to $\mu$ as $n\to\infty$. We also prove our
theorem for $\Sigma\subset\mathbb{R}$ and open sets inside $\mathbb{R}$ that
recovers Smith’s main theorem [Smi21] as special case. Given any finite field
$\mathbb{F}_{q}$ and any integer $n$, our algorithm returns infinitely many
abelian varieties over $\mathbb{F}_{q}$ which are not isogenous to the
Jacobian of any curve over $\mathbb{F}_{q^{n}}$.

We would like to thank Professors Pagano, Sarnak and Smith for their comments
on the earlier version of this work. The authors also thank Pennsylvania State
University’s Institute for Computational and Data Sciences for allowing us to
use their ROAR servers for our numerical experiments.

##  1\. Introduction

###  1.1. Background

This paper is motivated by the following question. What compact sets $\Sigma$
in the complex plane $\mathbb{C}$ contain infinitely many sets of conjugate
algebraic integers, and how are they distributed in $\Sigma$? The ideas of
this project originated in the works of many, including Schur [Sch18], Fekete
[Fek23], Siegel [Sie45], Robinson [Rob64b], Serre [Ser19], Smyth [Smy84] and
Smith [Smi21] on algebraic integers. We introduce the basic concepts and state
our theorems which answer some open problems raised in previous works.

Let $\Sigma$ be a compact subset of the complex plane. Let

(1)
$d_{\Sigma}(n):=\max_{z_{1},\dots,z_{n}\in\Sigma}\prod_{i<j}|z_{i}-z_{j}|^{\frac{2}{n(n-1)}}.$

Fekete proved the following limit exists and called it the transfinite
diameter of $\Sigma$:

(2) $d_{\Sigma}:=\lim_{n\to\infty}d_{\Sigma}(n).$

For example, the transfinite diameter of a circle of radius $r$ is $r$. The
transfinite diameter of $\Sigma$ equals the capacity of $\Sigma$ [Ahl10,
Chapter 2]. Fekete [Fek23], generalizing some results of Schur [Sch18], proved
that if $d_{\Sigma}<1$, then there is only a finite number of irreducible
monic integral polynomials such that all of their roots lie in $\Sigma$.

Note that Fekete’s condition $d_{\Sigma}<1$ is optimal. For example, let
$\Sigma$ be the unit circle. Then primitive roots of unity give an infinite
number of irreducible integral polynomials such that their roots lie all in
$\Sigma$. The strict converse is not true. For example, the circle of radius
$r>1$ around the origin where $r$ is not an algebraic integer gives a counter-
example. However, Fekete and Szegö [FS55] proved that if $\Sigma$ is symmetric
about the real axis and $d_{\Sigma}\geq 1$, then any open set $D$ including
$\Sigma$ will contain infinitely many sets of conjugate algebraic integers.
Robinson [Rob64a] proved an analogous theorem about real point sets using the
properties of Chebyshev’s polynomials of $\Sigma$.

Recently, Smith [Smi21] studied the weak* limit of a sequence of uniform
probability measures on a set of conjugate algebraic integers inside
$\Sigma\subset\mathbb{R},$ where $d_{\Sigma}>1$ and $\Sigma$ is a countable
union of intervals. Smith used two results form the geometry of numbers
(flatness theorem and Minkowski’s second theorem) and Robinson’s method
[Rob64a] to give necessary and sufficient conditions for a probability
distribution $\mu$ on the real line to be a weak* limit of such uniform
probability measures on totally real conjugate algebraic integers.

We generalize Smith’s work to the complex plane in the next subsection. Our
method is different from Smith’s method. In particular, we do not use a
flatness theorem or the properties of square-free integral polynomials in our
paper. Both of them are crucial ingredients in Smith’s work. Instead, we prove
a new result in Theorem 1.7 which is of independent interest. Theorem 1.7
allows us to improve some results of Smith [Smi21, Proposition 3.5].

Furthermore, our results do not rely on the existence of Chebychev’s
polynomials of $\Sigma$; see subsection 2.6 and Proposition 2.16. In fact, our
method gives a new proof of Robinson’s result and Smith’s results without
using the properties of Chebychev’s polynomials; see Proposition 2.16. Our
method is based on a initial probabilistic sampling of roots with respect to
the equilibrium measure and then deforming the roots with a greedy algorithm
along a gradient vector.

Our algorithm and its implementation has no prior analogue to the best of our
knowledge. Our numerical results shows some new features of the algebraic
integers that we discuss further in subsection 1.2.2.

###  1.2. Main results

####  1.2.1. Arithmetic probability measures

Suppose that $\Sigma\subset\mathbb{C}$ is compact and symmetric about the real
axis with $d_{\Sigma}\geq 1.$ Let $\mathcal{P}_{\Sigma}$ be the space of
probability measures supported on $\Sigma$ equipped with the weak* topology.
Let

$\Sigma(\rho):=\\{z\in\mathbb{C}:|z-\sigma|<\rho\text{ for some
}\sigma\in\Sigma\\}.$

If $\Sigma\subset\mathbb{R}$, let

$\Sigma_{\mathbb{R}}(\rho):=\\{x\in\mathbb{R}:|x-\sigma|<\rho\text{ for some
}\sigma\in\Sigma\\}.$

Note that $\Sigma_{\mathbb{R}}(\rho)\subset\mathbb{R}$, and only defined for
$\Sigma\subset\mathbb{R}.$

######  Definition 1.1.

A measure $\mu\in\mathcal{P}_{\Sigma}$ is called an (real) arithmetic
probability measure if $\mu$ is the weak* limit of a sequence of distinct
uniform probability measures on a complete set of conjugate (totally real)
algebraic integers lying eventually inside ($\Sigma_{\mathbb{R}}(\varepsilon)$
for every $\varepsilon>0$) $\Sigma(\varepsilon)$ for every $\varepsilon>0$.
The set of all arithmetic probability measures is denoted (by
$\mathcal{A}_{\Sigma_{\mathbb{R}}}\subset\mathcal{P}_{\Sigma}$) by
$\mathcal{A}_{\Sigma}\subset\mathcal{P}_{\Sigma}.$

Next, we define a convex subset of $\mathcal{P}_{\Sigma}$ that includes
$\mathcal{A}_{\Sigma}.$ Let $P(x)$ be a polynomial with complex coefficients
of degree $n$. Define its associated root probability measure on the complex
plane to be

$\mu_{P}:=\frac{1}{n}\sum_{i=1}^{n}\delta_{\alpha_{i}},$

where $\alpha_{i}$ for $1\leq i\leq n$ are roots of $P$ and $\delta_{\alpha}$
is the delta probability measure at $\alpha.$ Suppose that
$\mu\in\mathcal{A}_{\Sigma}.$ Then
$\mu_{P_{n}}\stackrel{{\scriptstyle\ast}}{{\rightharpoonup}}\mu$ as
$n\to\infty$ for some sequence $\\{P_{n}\\}$ of distinct irreducible monic
polynomials with integral coefficients. Since $P_{n}$ has real coefficients,
its roots are symmetric about the real axis, so $\mu$ should also be symmetric
about the real axis. Let $Q$ be a polynomial with integral coefficients. Note
that if $P_{n}\nmid Q,$

$\int\log|Q(x)|d\mu_{P_{n}}(x)=\frac{\log|\text{Res}(Q,P_{n})|}{\deg(P_{n})}\geq
0,$

where $\text{Res}(Q,P_{n})\in\mathbb{Z}$ is the resultant of $P_{n}$ and $Q.$
By taking $n\to\infty,$ Serre [Ser19] proved that

(3) $\int\log|Q(x)|d\mu(x)\geq 0$

for every non-zero $Q(x)\in\mathbb{Z}[x].$ Let
$\mathcal{B}_{\Sigma}\subset\mathcal{P}_{\Sigma}$ be the set of all
$\mu\in\mathcal{P}_{\Sigma}$ that are symmetric about the real axis and
satisfy (3) for every non-zero $Q(x)\in\mathbb{Z}[x].$ Since $d_{\Sigma}\geq
1,$ it follows that the equilibrium measure of $\Sigma$ belongs to
$\mathcal{B}_{\Sigma}$. Note that $\mathcal{B}_{\Sigma}$ is a convex subset of
$\mathcal{P}_{\Sigma}$ and $\mathcal{A}_{\Sigma}\subset\mathcal{B}_{\Sigma}.$
Serre [AP08] proved that there exists $\Sigma\subset\mathbb{R}^{+}$ and
$\mu\in\mathcal{B}_{\Sigma}$ such that $\int_{\mathbb{R}}xd\mu<1.8984.$ Smith
[Smi21] proved that if $\Sigma\subset\mathbb{R}$ satisfies certain technical
conditions (including Serre’s construction) then
$\mathcal{A}_{\Sigma_{\mathbb{R}}}=\mathcal{B}_{\Sigma}.$ This gaves a
definite answer to the Schur–Siegel–Smyth Trace Problem. Finding
$\mu\in\mathcal{B}_{\Sigma}$ with the minimal expected value
$\int_{\mathbb{R}}xd\mu$ is a linear programming problem and finding its
optimal solution is an open problem; see [Smi21]. It is the analog of finding
the magic function in the linear programming problem formulated by Cohn-Elkies
for the sphere packing problem.

######  Theorem 1.2.

Suppose that $\Sigma\subset\mathbb{C}$ is compact and symmetric about the real
axis. If $d_{\Sigma}<1$, then
$\mathcal{A}_{\Sigma}=\mathcal{B}_{\Sigma}=\emptyset.$ Otherwise,
$d_{\Sigma}\geq 1$ and
$\mathcal{A}_{\Sigma}=\mathcal{B}_{\Sigma}\neq\emptyset.$ Moreover, if
$\Sigma\subset\mathbb{R}$ is compact then
$\mathcal{A}_{\Sigma}=\mathcal{B}_{\Sigma}=\mathcal{A}_{\Sigma_{\mathbb{R}}}.$

######  Remark 4.

Suppose that $\Sigma\subset\mathbb{R}$ is a finite union of closed intervals
with $d_{\Sigma}>1$ and $\mu\in\mathcal{B}_{\Sigma}.$ Smith’s main theorem
[Smi21, Theorem 1.6] is equivalent to [Smi21, Proposition 2.4] the existence
of a sequence of conjugate algebraic integers lying inside $\Sigma$ and
equidistributing to $\mu.$ We show that
$\mathcal{B}_{\Sigma}=\mathcal{A}_{\Sigma_{\mathbb{R}}}$ implies this
statement.

As we discussed, the circle of radius $r>1$ around the origin where $r$ is not
an algebraic integer gives a counter-example [FS55]. Knowing this, Smith
raised the question of extending his result to $\Sigma\subset\mathbb{C}.$ Let
$int_{\Sigma}$ be the interior of $\Sigma.$ If $\mu(int_{\Sigma})=1$ and
$d_{int_{\Sigma}}>1$, then Theorem 1.2 (with some extra work and using a
sequence of open sets inside $int_{\Sigma}$ and a diagonal argument) implies
the existence of a sequence of conjugate algebraic integers lying inside
$int_{\Sigma}$ and equidistributing to $\mu.$ This answers Smith’s question
when $\mu(int_{\Sigma})=1.$

Smith’s method does not directly imply Theorem 1.2. For example, his method
requires the complement of $\Sigma\subset\mathbb{C}$ to be connected. This
condition is trivially satisfied for compact $\Sigma\subset\mathbb{R},$ but
fails for some $\Sigma\subset\mathbb{C}$; e.g for the unit circle. We
introduce new ideas to prove the above theorem. This is discussed in Section
1.4.

Next, we introduce some new notation to state a quantitative version of the
Theorem 1.2 under some assumptions. Let

(5) $U_{\mu}(z):=\int\log|z-w|d\mu(w),$

which is called the logarithmic potential of $\mu$ (in other works it is with
a negative sign). A measure $\mu$ is a Hölder probability measure if there
exists $\delta>0$ and $A>0$ such that

$\mu([a,b]\times[c,d])\leq A\max(|b-a|,|c-d|)^{\delta}$

for every $a<b,c<d.$

We cite the following definition from Smith [Smi21, Definition 2.7].

######  Definition 1.3.

Fix a Hölder measure $\mu$ with support contained in the compact subset
$\Sigma\subset\mathbb{C}$. Given a complex polynomial $P$ of degree less than
or equal to $n$, define the $n$-norm of $P$ with respect to $(\mu,\Sigma)$ by

$\|P\|_{n}:=\max_{z\in\mathbb{C}}\left(e^{-nU_{\mu}(z)}|P(z)|\right).$

It follows from [ST97, Theorem 4.1] that if $\Sigma\subset\mathbb{C}$ is
compact and has empty interior with connected complement and for a sequence of
monic polynomials $P_{n}$ of degree $n$ for $n\geq 1$

$\limsup_{n}\|P_{n}\|_{n}^{1/n}\leq 1,$

then the roots of $P_{n}$ are equidistibuted to $\mu.$

######  Theorem 1.4.

Suppose $\Sigma\subset\mathbb{C}$ is compact and has empty interior with
connected complement. Suppose that $\mu\in\mathcal{B}_{\Sigma}$ is a Hölder
measure with exponent $\delta,$ and $D$ is any open set containing $\Sigma$.
For every large enough integer $n\geq 1$, there exists an irreducible
polynomial $h_{n}$ of degree n such that all roots of $h_{n}$ are inside $D$,
and

$\|h_{n}\|_{n}\leq e^{n^{1-\delta^{\prime}}}$

for some $\delta^{\prime}>0$ which only dependents on the Hölder exponent of
$\mu.$ As a result, the roots of $h_{n}$ are equidistibuted to $\mu.$
Moreover, if $\Sigma\subset\mathbb{R}$ is compact then it is possible to take
any open set $D\subset\mathbb{R}$ in the real point set topology containing
$\Sigma$ and prove the same result.

####  1.2.2. Complexity of approximating arithmetic probability measures

The next goal is to study the complexity of finding a sequence of irreducible
monic polynomials $\\{P_{n}\\}$ such that $\\{\mu_{P_{n}}\\}$ converges weakly
to a given $\mu\in\mathcal{B}_{\Sigma}.$

######  Theorem 1.5.

Let $\Sigma$ be as in Theorem 1.4. Suppose that $\mu\in\mathcal{B}_{\Sigma}$
is a Hölder probability measure and $D$ is any open set containing $\Sigma$.
There is a polynomial time algorithm in $n$ that returns an integral,
irreducible, monic polynomial $g_{n}$ of degree $n$ such that all roots of
$P_{n}$ are inside $D$, and

$\|g_{n}\|_{n}\leq e^{\frac{Cn\log(\log(n))^{3}}{\log(n)}}$

where $C>0$ depends only on $\mu$ and is independent of $n$. As a result, the
roots of $g_{n}$ are equidistibuted to $\mu.$ Moreover, if
$\Sigma\subset\mathbb{R}$ is compact then it is possible to take any open set
$D\subset\mathbb{R}$ in the real point set topology containing $\Sigma$ and
prove the same result.

We implemented a version of this algorithm for arithmetic probability measures
$\mu\in\mathcal{B}_{\Sigma}$ constructed by Serre [AP08]. Explicitly, let
$\Sigma=[a,b],$

$\mu=c\mu_{[a,b]}+(1-c)\nu_{[a,b]},$

$a=0.1715,$ $b=5.8255,$ $c=0.5004,$ $\mu_{[a,b]}$ is the equilibrium measure
on $[a,b]$, and $\nu_{[a,b]}$ is the pushforward of the equilibrium measure on
$[b^{-1},a^{-1}]$ under the map $z\to 1/z.$ Figure 1(a) shows the density of
$\mu$, and Figure 1(b) shows the number of roots of the output polynomial of
our algorithm for $n=100$.

(a) Density function of $\mu$

(b) Number of roots of $P_{100}(x)$

Figure 1. This compares the density function of $\mu$ with the histogram of
roots of our algorithm’s output with input $n=100$.

###  1.3. Applications to abelian varieties over finite fields

Recently, Shankar and Tsimerman [ST18], [ST21] conjectured that for every
$g\geq 4$, every Abelian variety defined over $\overline{\mathbb{F}_{q}}$ is
isogenous to a Jacobian of a curve defined over $\overline{\mathbb{F}_{q}}.$
By Honda Tate theory the isogeny class of simple abelian varieties over
$\mathbb{F}_{q}$ and isogenies over $\mathbb{F}_{q}$ is corresponded to
algebraic integers such that all their conjugates have norm $\sqrt{q}.$ We
apply our main theorem and Honda Tate theory to prove the following.

######  Theorem 1.6.

Given any finite field $\mathbb{F}_{q}$ and any integer $n$, our algorithm
returns infinitely many abelian varieties over $\mathbb{F}_{q}$ which are not
isogenous to the Jacobian of any curve over $\mathbb{F}_{q^{m}}$ for every
$1\leq m\leq n$.

Using the algorithm discussed in section 4 of the paper, the authors found an
explicit polynomial, with highest order terms being

$x^{290}-28x^{289}-{484}x^{288}+20784x^{287}+\cdots,$

which is irreducible with all roots inside
$[-2\sqrt{3},2\sqrt{3}]\subset\mathbb{R}$ representing an abelian variety that
is not isogenous to the Jacobian of any curve over $\mathbb{F}_{3}$ or
$\mathbb{F}_{9}$.

###  1.4. Method of proofs

In this section, we sketch proofs of our main theorems. Our main tool for
studying the distribution of the roots of polynomials is Jensen’s formula. Let
$P(z)$ be a polynomial with complex coefficients and fix $w\in\mathbb{C}.$ If
$z_{1},\dots,z_{m}$ are the roots of $P(z)$ inside the disc $|z-w|<R$ and
there is no zero on $|z-w|=R,$ then Jensen’s formula states

$\frac{1}{2\pi}\int_{0}^{2\pi}\log|P(Re^{i\theta}+w)|d\theta-\log|P(w)|=\log\frac{R^{m}}{|z_{1}-w|\dots|z_{m}-w|}.$

In particular,

(6)
$\frac{1}{2\pi}\int_{0}^{2\pi}\log|P(Re^{i\theta}+w)|d\theta-\log|P(w)|\geq
0.$

Suppose that $P$ is a monic polynomial. We have
$|P(w)|=|z_{1}-w|\cdots|z_{n}-w|$, where $n=\deg(P).$ Hence for a monic
polynomial $P$, we have

$\frac{1}{2\pi}\int_{0}^{2\pi}\frac{\log|P(Re^{i\theta}+w)|}{n}d\theta=\frac{m}{n}\log
R+\frac{1}{n}\sum_{i=m+1}^{n}\log|z_{i}-w|,$

where $|z_{i}-w|<R$ for $1\leq i\leq m$ and $|z_{i}-w|>R$ for $m+1\leq i\leq
n.$ We rewrite the above as

(7)
$\frac{1}{2\pi}\int_{0}^{2\pi}\frac{\log|P(Re^{i\theta}+w)|}{n}d\theta=\int\log(z;w,R)d\mu_{P}(z),$

where

$\log(z;w,R):=\begin{cases}\log|z-w|&\text{ if }|z-w|>R,\\\
\log{R}&\text{otherwise.}\end{cases}$

It follows from (7) that $\\{\mu_{P_{n}}\\}$ converges weakly to a given
probability measure $\mu$ if

(8) $\frac{\log|P_{n}(z)|}{n}=U_{\mu}(z)+o(1)$

for every $z\in\mathbb{C}$ such that $|z-z_{i,n}|>e^{-\sqrt{n}}$, where
$z_{i,n}$ are complex roots of $P_{n}.$ We note that
$\frac{\log|P_{n}(z)|}{n}$ is a harmonic function with singularities at roots
of $P_{n}$ and

$\frac{\log|P_{n}(z)|}{n}=\log|z|+O\left(\frac{1}{|z|}\right).$

We note that $U_{\mu}(z)$ has a similar behavior as $|z|\to\infty.$ Since
$U_{\mu}(z)$ and $\frac{\log|P_{n}(z)|}{n}$ are both harmonic functions with
similar asymptotic behavior at infinity, it follows from the mean value
theorem for harmonic functions that (8) is equivalent to

(9) $\frac{\log|P_{n}(z)|}{n}\leq U_{\mu}(z)+o(1)$

for every $z\in\mathbb{C}$ assuming $\mathbb{C}\setminus\Sigma$ is connected
and the interior of $\Sigma$ is empty. Let

(10) $K_{n}:=\left\\{p(x)\in\mathbb{R}[x]:\deg(p)\leq
n,\frac{\log|p(z)|}{n}\leq U_{\mu}(z)\text{ for every
}z\in\mathbb{C}\right\\}.$

$K_{n}$ is a symmetric, convex region inside the vector space of polynomials
with real coefficients and degree less than or equal to $n$. For proving
Theorem 1.4, we use Minkowski’s second theorem to prove the existence of
integral polynomials of degree $n$ inside $K_{n}$ up to a sub-exponential
factor in $n$. The key observation is that the necessary condition (3) implies
$K_{n}$ is well-rounded which means the successive minima of the lattice of
integral polynomials with respect to $K_{n}$ are close to each other and a
lower bound on the volume of $K$ implies the existence of integral polynomials
of degree $n$ inside $K_{n}$ up to a sub-exponential factor in $n$. The
following theorem is an important technical result that allows us to use
Minkowski’s second theorem.

######  Theorem 1.7.

Suppose that $\mu$ is a probability measure with compact support on the
complex plane. The following are equivalent:

  1. (1)

$\mu$ is invariant by complex conjugation and (3) holds for every non-zero
$Q(x)\in\mathbb{Z}[x].$

  2. (2)

$\mu$ is invariant by complex conjugation and

(11) $\int\log|Q_{m}(x_{1},\dots,x_{m})|d\mu(x_{1})\dots d\mu(x_{m})\geq 0$

for every non-zero $Q_{m}(x_{1},\dots,x_{m})\in\mathbb{Z}[x_{1},\dots,x_{m}].$

Recall that

$\Sigma(\rho):=\\{z\in\mathbb{C}:|z-\sigma|<\rho\text{ for some
}\sigma\in\Sigma\\}.$

Note that for any open set $D$ containing $\Sigma,$ there exist $\rho>0$ such
that $\Sigma(\rho)\subset D.$

For $\Sigma\subset\mathbb{C}$, we use Rouché’s theorem to show that all roots
of our irreducible polynomial are inside $\Sigma(\rho).$ This is similar to
the method of Fekete and Szegö [FS55, Theorem I].

For $\Sigma\subset\mathbb{R}$, Robinson used the Chebychev’s polynomials and
constructed an integral polynomial of degree $n$ with $n$ sign changes on
$\Sigma(\rho)\subset\mathbb{R}$ to prove roots are real and are inside
$\Sigma(\rho).$ Our results do not rely on the existence of Chebychev’s
polynomials of $\Sigma$; see subsection 2.6 and Proposition 2.16. In fact, our
method gives a new proof of Robinson’s result and Smith’s results without
using the properties of Chebychev’s polynomials; see Proposition 2.16. Our
method is based on a initial probabilistic sampling of roots with respect to
the equilibrium measure and then deforming the roots with a greedy algorithm
along a gradient vector. Proposition 2.16 allows us to improve some results of
Smith [Smi21, Proposition 4.1].

For proving Theorem 1.5, we apply a lattice algorithm due to Schnorr and find
an irreducible, integral polynomial inside
$e^{\frac{Cn\log(\log(n))^{3}}{\log(n)}}K_{n}$ in polynomial time in $n$ for
some constant $C$ dependent on $\mu$. In fact, we take the polynomial to be
monic and Eisenstein at the prime 2. Note that this differs from the strongest
bound for which we prove existence which is $e^{Cn^{1-\delta_{0}}}K_{n}$ for
some $\delta_{0}>0$. Other versions of Schnorr’s algorithm can achieve
stronger bounds at a super-polynomial run time.

##  2\. Minkowski’s theorem applied to the lattice of integral polynomials

###  2.1.  Logarithmic potentials

Recall the notation in subsections 1.2 and 1.4. In this section, we assume
that $\mu$ is supported on a compact set $\Sigma\subset\mathbb{C}$ and is a
Hölder probability measure on the complex plane with exponent $\delta>0.$
Hence for every $a<b,c<d$ and some $A>0$

$\mu([a,b]\times[c,d])\leq A\max(|b-a|,|c-d|)^{\delta}.$

Recall the definition of the logarithmic potential of $\mu$ from (5)

$U_{\mu}(z)=\int\log|z-w|d\mu(w).$

We prove a lemma on the asymptotic behaviour of logarithmic potentials.

######  Lemma 2.1.

Suppose that $\mu$ is a probability measure with compact support inside
$|z|<r$ for $z\in\mathbb{C}.$ If $|w|>2r$, we have

$U_{\mu}(w)=\log|w|+O\left(\frac{r}{|w|}\right).$

###### Proof.

We have

$\displaystyle\left|U_{\mu}(w)-\log|w|\right|\leq\int\left|\log|w-x|-\log|w|\right|d\mu(x)$
$\displaystyle=\int\left|\log|1-\frac{x}{w}|\right|d\mu(x)\ll\frac{r}{|w|}.$

∎

######  Lemma 2.2.

Suppose that $\mu$ is a Hölder probability measure on the complex plane with
exponent $\delta>0$. Then for any $z_{1},z_{2}\in\mathbb{C}$ with
$|z_{1}-z_{2}|<1/2,$ we have

$\left|U_{\mu}(z_{1})-U_{\mu}(z_{2})\right|\ll|z_{1}-z_{2}|^{\delta^{\prime}},$

where $\delta^{\prime}<\min(1/2,\delta/2)$ is any positive exponent.

###### Proof.

Suppose that $z_{1},z_{2}\in\mathbb{C}$ and $|z_{1}-z_{2}|<1/2.$ We have

$\displaystyle\left|U_{\mu}(z_{1})-U_{\mu}(z_{2})\right|$
$\displaystyle=\left|\int\log\frac{|z_{1}-w|}{|z_{2}-w|}d\mu(w)\right|.$

We define the following covering of the complex plane:

$\displaystyle U_{0}:$
$\displaystyle=\left\\{w\in\mathbb{C}:|w-z_{2}|>2\sqrt{|z_{1}-z_{2}|}\right\\},$
$\displaystyle U_{1}:$
$\displaystyle=\left\\{w\in\mathbb{C}:|z_{1}-z_{2}|<\min(|w-z_{2}|,|w-z_{1}|)\leq
2\sqrt{|z_{1}-z_{2}|}\right\\},$ $\displaystyle U_{k}:$
$\displaystyle=\left\\{w\in\mathbb{C}:|z_{1}-z_{2}|^{k}<\min(|w-z_{2}|,|w-z_{1}|)\leq|z_{1}-z_{2}|^{k-1}\right\\},$

where $k\geq 2$. Hence,

$\displaystyle\left|U_{\mu}(z_{1})-U_{\mu}(z_{2})\right|\leq\sum_{i\geq
0}\left|\int_{U_{i}}\log\frac{|z_{1}-w|}{|z_{2}-w|}d\mu(w)\right|.$

For $w\in U_{0}$, we have $\left|\frac{z_{1}-z_{2}}{z_{2}-w}\right|<1/2$. We
write the Taylor expansion of $\log$ and obtain

$\left|\int_{U_{0}}\log\frac{|z_{1}-w|}{|z_{2}-w|}d\mu(w)\right|=\left|\int_{U_{0}}\log\left|1+\frac{z_{1}-z_{2}}{z_{2}-w}\right|d\mu(w)\right|\leq\sum_{m\geq
1}\int_{U_{0}}\frac{1}{m}\left|\frac{z_{1}-z_{2}}{z_{2}-w}\right|^{m}d\mu(w)\ll|z_{1}-z_{2}|^{1/2}.$

For $w\in U_{k}$, where $k\geq 1$, we have

$\left|\int_{U_{k}}\log\frac{|z_{1}-w|}{|z_{2}-w|}d\mu(w)\right|\leq\mu(U_{k})k\big{|}\log|z_{1}-z_{2}|\big{|}\ll|z_{1}-z_{2}|^{\frac{k\delta}{2}}k\big{|}\log|z_{1}-z_{2}|\big{|},$

where we used $\mu(U_{k})\ll|z_{1}-z_{2}|^{\frac{k\delta}{2}}$, because $\mu$
is a Hölder probability measure with exponent $\delta>0$. Therefore,

$\left|U_{\mu}(z_{1})-U_{\mu}(z_{2})\right|\ll\sqrt{|z_{1}-z_{2}|}+\sum_{k\geq
1}|z_{1}-z_{2}|^{k\delta/2}k\big{|}\log|z_{1}-z_{2}|\big{|}\ll|z_{1}-z_{2}|^{1/2}+|z_{1}-z_{2}|^{\delta/2}\big{|}\log|z_{1}-z_{2}|\big{|}.$

Therefore, we have

$\left|U_{\mu}(z_{1})-U_{\mu}(z_{2})\right|\ll|z_{1}-z_{2}|^{\delta^{\prime}},$

where $\delta^{\prime}<\min(1/2,\delta/2)$ is any positive exponent. ∎

###  2.2. Minkowski’s theorem

Recall the definition of $K_{n}$ from (10):

$K_{n}=\left\\{p(x)\in\mathbb{R}[x]:\deg(p)\leq n,\text{ and
}\frac{\log|p(z)|}{n}\leq U_{\mu}(z)\text{ for every
}z\in\mathbb{C}\right\\}.$

We identify the space of polynomials with real coefficients of degree less
than or equal to $n$ with $\mathbb{R}^{n+1}$ by sending a polynomial to its
coefficients. It is easy to see that $K_{n}$ is a symmetric convex region of
$\mathbb{R}^{n+1}$.

######  Lemma 2.3.

Suppose that $\mu$ is a probability measure with compact support inside
$|z|<r$ for $z\in\mathbb{C}.$ Let $p(x)\in\lambda K_{n}$ for some $\lambda>0,$
where $\deg(p)<n$. Then

$xp(x)\in r\lambda K_{n}.$

###### Proof.

Let

$h(z):=U_{\mu}(z)-\frac{\log|zp(z)|}{n}.$

We note that $h(z)$ is a harmonic function on $|z|>r$ and outside roots of
$p.$ By Lemma 2.1, we have

$h(z)=\frac{n-\deg(p)-1}{n}\log|z|+\frac{\log|a_{p}|}{n}+O(1/|z|),$

where $a_{p}$ is the top coefficient of $p.$ By the above and since
$\deg(p)<n$ and $h(z)$ is harmonic, the minimum of $h(z)$ is obtained inside
$|z|<r$. Let $C:=\inf_{z\in\mathbb{C}}h(z)=\inf_{|z|<r}h(z).$ Since
$p(x)\in\lambda K_{n}$, we have

$|zp(z)|\leq r\lambda e^{nU_{\mu}(z)}$

for any $|z|<r.$ This implies that

$C\geq\frac{\log r\lambda}{n}.$

Therefore,

$e^{nU_{\mu}(z)-\log|zp(z)|}=e^{nh(z)}\geq e^{nC}\geq r\lambda$

for any $z\in\mathbb{C}.$ This completes the proof of our lemma. ∎

######  Lemma 2.4.

Suppose that $p(x)\in\lambda K_{n}$ for some $\lambda>0.$ We claim that

$p^{\prime}(x)\in An^{C}\lambda K_{n},$

where $p^{\prime}$ is the derivative of $p$ and $A$ and $C$ are constants that
only depend on $\mu.$

###### Proof.

Let $D(z_{0},r)$ be the disk centered at some $z_{0}\in\mathbb{C}$ and radius
$r>0.$ By Bernstein inequality [BE95, Corollary 5.1.6] for any $r>0$

$\sup_{z\in D(z_{0},r)}\left|p^{\prime}(z)\right|\leq\frac{\deg
p}{r}\sup_{z\in D(z_{0},r)}|p(z)|,$

for some constant $C$ that only depends on $\mu.$ Indeed, we have

$|p^{\prime}(z_{0})|\leq\frac{n}{r}\sup_{z\in
D(z_{0},r)}|p(z)|\leq\lambda\frac{n}{r}\sup_{z\in D(z_{0},r)}e^{nU_{\mu}(z)}.$

By Lemma 2.2, we have for any $z\in D(z_{0},r)$

$e^{nU_{\mu}(z)}=e^{n(U_{\mu}(z)-U_{\mu}(z_{0}))}e^{nU_{\mu}(z_{0})}\leq
e^{A^{\prime}nr^{\delta^{\prime}}}e^{nU_{\mu}(z_{0})}.$

for some $A^{\prime}>0$ and $\delta^{\prime}>0$ that only depends on $\mu.$ By
taking $r=n^{-C^{\prime}}$ for any $C^{\prime}>1/\delta^{\prime}$, it follows
that

$|p^{\prime}(z_{0})|\leq\lambda\frac{n}{r}e^{A^{\prime}nr^{\delta^{\prime}}}e^{nU_{\mu}(z_{0})}\leq
A\lambda n^{C^{\prime}+1}e^{nU_{\mu}(z_{0})},$

for any $z_{0}\in\mathbb{C},$ where $A=e^{A^{\prime}}$. This implies that
$p^{\prime}(x)\in An^{C}\lambda K_{n},$ where $C=C^{\prime}+1.$ ∎

Let $\Gamma_{n}$ be the lattice of integral polynomials of degree less than or
equal to $n$ in $\mathbb{R}^{n+1}$. The successive minima of $K_{n}$ on
$\Gamma_{n}$ are defined by setting the $k$-th successive minimum
$\lambda_{k}$ to be the infimum of the numbers $\lambda$ such that $\lambda
K_{n}$ contains $k+1$ linearly-independent vectors of $\Gamma_{n}$. We have
$0<\lambda_{0}\leq\lambda_{1}\leq\dots\leq\lambda_{n}<\infty$. Minkowski’s
second theorem states that

(12)
$\frac{2^{n+1}\textup{vol}(K_{n})^{-1}}{(n+1)!}\leq\lambda_{0}\lambda_{1}\dots\lambda_{n}\leq
2^{n+1}\textup{vol}(K_{n})^{-1}.$

######  Proposition 2.5.

We have

$\lambda_{m+1}\leq An^{C}\lambda_{m}$

for any $0\leq m<n$, and some constants $A$ and $C$ that only depend on $\mu.$

###### Proof.

Suppose that $p_{i}\in\Gamma_{n}$ for $0\leq i\leq m$ are the set of linearly
independent integral polynomials such that $p_{i}\in\lambda_{i}K_{n}.$ If for
some $0\leq i\leq m$, $p_{i}^{\prime}$ is linearly independent to
$\\{p_{i}:0\leq i\leq m\\}$, then by Lemma 2.4

$p^{\prime}_{i}(x)\in An^{C}\lambda_{m}K_{n}.$

Hence,

$\lambda_{m+1}\leq An^{C}\lambda_{m},$

and this implies our proposition. Otherwise,
$V_{m}:=\text{span}_{\mathbb{R}}\left<p_{0},\dots,p_{m}\right>$ is closed
under derivation. It follows that

$V_{m}=\text{span}_{\mathbb{R}}\left<1,x,\dots,x^{m}\right>.$

This implies that there exists $0\leq l\leq m$ such that
$\deg(p_{j})\leq\deg(p_{l})=m$ for every $0\leq j\leq m.$ Let
$q(x):=xp_{l}(x).$ Since $\deg q>m,$ $q$ is linearly independent to $V_{m}.$
By Lemma 2.3,

$q(x)\in A\lambda_{m}K_{n}$

for some constant $A>0$ that only depends on $\mu.$ Hence,

$\lambda_{m+1}\leq A\lambda_{m}$

and this completes the proof of our proposition. ∎

###  2.3. Lower bound on the volume of $K_{n}$

In this subsection, we give a lower bound on the Euclidean volume of $K_{n}$
as a subset of $\mathbb{R}^{n+1}.$ Our method is to find $n+1$ linearly
independent polynomials of degree at most $n$ inside $K_{n}$ and compute the
volume of the simplex given by the convex hull of the origin and these $n+1$
points. Since $K_{n}$ is symmetric, there are $2^{n+1}$ simplexes with the
same volume and disjoint interior inside $K_{n}$ by choosing different signs
for the $n+1$ vertices. The total volume of these simplexes give a lower bound
for the volume of $K_{n}.$

####  2.3.1. Finding a simplex inside $K_{n}$

For simplicity in this section, we assume that $n$ is odd and we find a
simplex inside $K_{n}$ and estimate its volume. Recall that $\mu$ is invariant
by the complex conjugation. Let $[a,b]\times[-c,c]\subset\mathbb{C}$ be a
rectangle containing $\Sigma.$ If $\Sigma\subset\mathbb{R}$, we take
$[a,b]\subset\mathbb{R}$ containing $\Sigma$. Fix $0<M\in\mathbb{Z}$ and
$0<L\in\mathbb{R}$ where $n^{1/10}<M<n^{1/3}$, and $L<nM^{-2}.$ We partition
each side of the rectangle into $2M$ equal length sub-intervals, and obtain a
partition of the rectangle as

$[a,b]\times[-c,c]=\bigcup_{0\leq i,j<2M}B_{ij},$

where $B_{ij}:=[a_{i},a_{i+1}]\times[c_{j},c_{j+1}]$ and
$a_{i}:=a+\frac{i(b-a)}{2M}$ and $c_{j}:=-c+\frac{j(2c)}{2M}.$ Since $\mu$ is
a Hölder measure with exponent $\delta$, we have

(13) $\mu(B_{ij})\ll M^{-\delta}$

for every $i,j.$ Similarly, for $\Sigma\subset\mathbb{R}$, we write

$[a,b]=\bigcup_{0\leq i<2M}B_{i},$

where $B_{i}:=[a_{i},a_{i+1}]$ and $a_{i}:=a+\frac{i(b-a)}{2M}.$ Let

(14) $n_{ij}:=\lfloor(n+1)\mu(B_{ij})\rfloor+\varepsilon_{ij}\ll
nM^{-\delta},$

where $\varepsilon_{ij}\in\mathbb{Z}$ and $|\varepsilon_{ij}|\leq L$ are
chosen such that $\sum_{i,j}n_{ij}=n+1$ and $n_{ij}=n_{i(2M-j-1)}.$ Since we
assumed $n$ is odd, this is possible. By a covering argument, it is possible
to find $z_{ijk}\in B_{ij}$ for $0\leq i,j<2M$ and $1\leq k\leq n_{ij},$ such
that $\overline{z_{ijk}}=z_{i(2M-j-1)k}$ and for every $w\in\partial B_{ij}$
and every $1\leq k,k^{\prime}\leq n_{ij}$

(15) $\begin{split}|z_{ijk}-w|&\gg M^{-1}\\\ |z_{ijk}-z_{ijk^{\prime}}|&\gg
n^{-\frac{1}{2}}M^{-1+\frac{\delta}{2}}\gg n^{-5/6}.\end{split}$

Similarly, for $\Sigma\subset\mathbb{R}$, we define
$n_{i}:=\lfloor(n+1)\mu(B_{i})\rfloor+\varepsilon_{i}\ll nM^{-\delta}$ and
$z_{ik}\in B_{i}$ for $k\leq n_{i}$ such that $|\varepsilon_{i}|<L,$
$\sum_{i}n_{i}=n+1$ and for every $w\in\partial B_{i}$ and every $1\leq
k,k^{\prime}\leq n_{i}$

(16) $\begin{split}|z_{ik}-w|&\gg M^{-1}\\\ |z_{ik}-z_{ik^{\prime}}|&\gg
n^{-1}M^{-1+\delta}\gg n^{-4/3}.\end{split}$

For two compact subsets $B,B^{\prime}\subset\mathbb{C}$, let

$d(B,B^{\prime}):=\inf_{\begin{subarray}{c}z\in B\\\ z^{\prime}\in
B^{\prime}\end{subarray}}|z-z^{\prime}|.$

Let

$I(\mu):=\int\log|z_{1}-z_{2}|d\mu(z_{1})d\mu(z_{2})=\int U_{\mu}(z)d\mu(z).$

######  Proposition 2.6.

We have

(17) $\sum_{\begin{subarray}{c}ij,i^{\prime}j^{\prime}\\\
d(B_{ij},B_{i^{\prime}j^{\prime}})<M^{-1}\end{subarray}}\sum_{\begin{subarray}{c}k,k^{\prime}\\\
i^{\prime}j^{\prime}k^{\prime}\neq
ijk\end{subarray}}\frac{\log|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}|}{(n+1)^{2}}=O(\log(n)M^{-\delta}),$

and

(18) $\sum_{ijk}\sum_{ijk\neq
i^{\prime}j^{\prime}k^{\prime}}\frac{\log|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}|}{(n+1)^{2}}=I(\mu)+O\left(\log(n)M^{-\delta}+\frac{M^{2}\log(n)}{n}+M^{-\delta/2}\log(M)^{1/2}\right).$

Similarly, for $\Sigma\subset\mathbb{R}$, we have

(19) $\sum_{\begin{subarray}{c}i,i^{\prime}\\\
d(B_{i},B_{i^{\prime}})<M^{-1}\end{subarray}}\sum_{\begin{subarray}{c}k,k^{\prime}\\\
i^{\prime}k^{\prime}\neq
ik\end{subarray}}\frac{\log|z_{ik}-z_{i^{\prime}k^{\prime}}|}{(n+1)^{2}}=O(\log(n)M^{-\delta}),$

and

(20) $\sum_{ik}\sum_{ik\neq
i^{\prime}k^{\prime}}\frac{\log|z_{ik}-z_{i^{\prime}k^{\prime}}|}{(n+1)^{2}}=I(\mu)+O\left(\log(n)M^{-\delta}+\frac{M^{2}\log(n)}{n}+M^{-\delta/2}\log(M)^{1/2}\right).$

###### Proof.

Note that for a pair $(i,j)$, there are at most $O(1)$ pairs
$(i^{\prime},j^{\prime})$ where $d(B_{ij},B_{i^{\prime}j^{\prime}})<M^{-1}$.
By (14) and (15), we have

$\sum_{\begin{subarray}{c}ij,i^{\prime}j^{\prime}\\\
d(B_{ij},B_{i^{\prime}j^{\prime}})<M^{-1}\end{subarray}}\sum_{\begin{subarray}{c}k,k^{\prime}\\\
i^{\prime}j^{\prime}k^{\prime}\neq
ijk\end{subarray}}\frac{\log|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}|}{(n+1)^{2}}\ll\sum_{\begin{subarray}{c}ij,i^{\prime}j^{\prime}\\\
d(B_{ij},B_{i^{\prime}j^{\prime}})<M^{-1}\end{subarray}}\frac{n_{ij}n_{i^{\prime}j^{\prime}}\log(n)}{(n+1)^{2}}=O(\log(n)M^{-\delta}).$

This completes the proof of the first part of our proposition. We have

(21) $I(\mu)=\sum_{ij}\sum_{i^{\prime}j^{\prime}}\int_{z\in
B_{ij}}\int_{z^{\prime}\in
B_{i^{\prime}j^{\prime}}}\log|z-z^{\prime}|d\mu(z)d\mu(z^{\prime}).$

Similarly, we have

$\sum_{\begin{subarray}{c}ij,i^{\prime}j^{\prime}\\\
d(B_{ij},B_{i^{\prime}j^{\prime}})<M^{-1}\end{subarray}}\int_{z\in
B_{ij}}\int_{z^{\prime}\in
B_{i^{\prime}j^{\prime}}}\log|z-z^{\prime}|d\mu(z)d\mu(z^{\prime})=O(\log(n)M^{-\delta}).$

Suppose that $d(B_{ij},B_{i^{\prime}j^{\prime}})>M^{-1}.$ We have

(22)
$\frac{1}{(n+1)^{2}}\sum_{k}\sum_{k^{\prime}}\log|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}|=\int_{z\in
B_{ij}}\int_{z^{\prime}\in
B_{i^{\prime}j^{\prime}}}\log|z-z^{\prime}|d\mu(z)d\mu(z^{\prime})\\\
+O\left(\frac{M_{1}L}{n}\left(\mu(B_{ij})+\mu(B_{i^{\prime}j^{\prime}})\right)+\mu(B_{ij})\mu(B_{i^{\prime}j^{\prime}})M_{2}\right).$

where $M_{1}:=\left|\sup_{\begin{subarray}{c}z\in B_{ij}\\\ z^{\prime}\in
B_{i^{\prime}j^{\prime}}\end{subarray}}\log|z-z^{\prime}|\right|$ and
$M_{2}:=\sup_{\begin{subarray}{c}z_{1},z_{2}\in B_{ij}\\\
z_{1}^{\prime},z_{2}^{\prime}\in
B_{i^{\prime}j^{\prime}}\end{subarray}}\left|\log|z_{1}-z_{1}^{\prime}|-\log|z_{2}-z_{2}^{\prime}|\right|.$
We have

$M_{1}\ll\log(n)$

and

$M_{2}\ll\frac{M^{-1}}{d(B_{ij},B_{i^{\prime}j^{\prime}})}.$

Therefore,

$\left|\sum_{ijk}\sum_{ijk\neq
i^{\prime}j^{\prime}k^{\prime}}\frac{\log|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}|}{(n+1)^{2}}-I(\mu)\right|\ll\log(n)M^{-\delta}\\\
+\sum_{\begin{subarray}{c}ij,i^{\prime}j^{\prime}\\\
d(B_{ij},B_{i^{\prime}j^{\prime}})>M^{-1}\end{subarray}}\frac{\log(n)L}{n}\left(\mu(B_{ij})+\mu(B_{i^{\prime}j^{\prime}})\right)+\mu(B_{ij})\mu(B_{i^{\prime}j^{\prime}})\frac{M^{-1}}{d(B_{ij},B_{i^{\prime}j^{\prime}})}.$

Note that

$\sum_{\begin{subarray}{c}ij,i^{\prime}j^{\prime}\\\
d(B_{ij},B_{i^{\prime}j^{\prime}})>M^{-1}\end{subarray}}\frac{\log(n)L}{n}\left(\mu(B_{ij})+\mu(B_{i^{\prime}j^{\prime}})\right)\ll\frac{\log(n)M^{2}L}{n}.$

By Cauchy-Schwarz inequality,

$\sum_{ij,i^{\prime}j^{\prime}}\mu(B_{ij})\mu(B_{i^{\prime}j^{\prime}})\frac{M^{-1}}{d(B_{ij},B_{i^{\prime}j^{\prime}})}\leq\left(\sum_{ij,i^{\prime}j^{\prime}}\mu(B_{ij})^{2}\mu(B_{i^{\prime}j^{\prime}})\right)^{1/2}\left(\sum_{ij,i^{\prime}j^{\prime}}\frac{M^{-2}\mu(B_{i^{\prime}j^{\prime}})}{d(B_{ij},B_{i^{\prime}j^{\prime}})^{2}}\right)^{1/2},$

where the sum is over $ij,i^{\prime}j^{\prime}$ such that
$d(B_{ij},B_{i^{\prime}j^{\prime}})\geq M^{-1}.$ We have

$\sum_{ij,i^{\prime}j^{\prime}}\mu(B_{ij})^{2}\mu(B_{i^{\prime}j^{\prime}})\ll
M^{-\delta}\sum_{ij,i^{\prime}j^{\prime}}\mu(B_{ij})\mu(B_{i^{\prime}j^{\prime}})=M^{-\delta}.$

Moreover,

$\sum_{ij,i^{\prime}j^{\prime}}\frac{M^{-2}\mu(B_{i^{\prime}j^{\prime}})}{d(B_{ij},B_{i^{\prime}j^{\prime}})^{2}}\leq\max_{i^{\prime}j^{\prime}}\sum_{ij}\frac{M^{-2}}{d(B_{ij},B_{i^{\prime}j^{\prime}})^{2}}\ll\log(M).$

Therefore,

$\left|\sum_{ijk}\sum_{ijk\neq
i^{\prime}j^{\prime}k^{\prime}}\frac{\log|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}|}{(n+1)^{2}}-I(\mu)\right|\ll\log(n)M^{-\delta}+\frac{\log(n)M^{2}L}{n}+M^{-\delta/2}\log(M)^{1/2}.$

∎

Let

(23) $p_{M,n,\mu}(x):=\prod_{i,j,k}(x-z_{ijk}),$

and for $e,f<2M$ and $g\leq n_{ef}$

(24)
$\begin{split}p_{M,n,\mu}^{efg+}(x)&:=\frac{\frac{p_{M,n,\mu}(x)}{(x-z_{efg})}+\frac{p_{M,n,\mu}(x)}{(x-\overline{z_{efg}})}}{2},\\\
p_{M,n,\mu}^{efg-}(x):&=\frac{\frac{p_{M,n,\mu}(x)}{(x-z_{efg})}-\frac{p_{M,n,\mu}(x)}{(x-\overline{z_{efg}})}}{2i\textup{Im}(z_{efg})}.\end{split}$

Similarly, for $\Sigma\subset\mathbb{R}$, let

(25) $p_{M,n,\mu}(x):=\prod_{i,k}(x-z_{ik}),$

and for $e<2M$ and $g\leq n_{e}$

(26) $p_{M,n,\mu}^{eg}(x):=\frac{p_{M,n,\mu}(x)}{(x-z_{eg})}.$

######  Lemma 2.7.

We have $p_{M,n,\mu}(x)\in\mathbb{R}[x]$ and

$p_{M,n,\mu}^{efg+}(x)=\frac{p_{M,n,\mu}(x)}{(x-z_{efg})(x-\overline{z_{efg}})}(x-\textup{Re}(z_{efg}))=p_{M,n,\mu}^{e(2M-f-1)g+}(x)$

and

$p_{M,n,\mu}^{efg-}(x)=\frac{p_{M,n,\mu}(x)}{(x-z_{efg})(x-\overline{z_{efg}})}=-p_{M,n,\mu}^{e(2M-f-1)g-}(x).$

In particular, $p_{M,n,\mu}^{efg\pm}(x)\in\mathbb{R}[x],$
$\deg(p_{M,n,\mu}^{efg+}(x))=n$, and $\deg p_{M,n,\mu}^{efg-}(x)=n-1$.

###### Proof.

It follows easily from $\overline{z_{ijk}}=z_{i(2M-j-1)k}.$ ∎

######  Proposition 2.8.

Suppose that $n\geq 2\max\\{|a|,|b|,|c|\\}.$ We have

$p_{M,n,\mu}^{efg\pm}(x)\in\lambda^{efg\pm}K_{n}$

for some $\lambda^{efg\pm}>0$, where

$\frac{\log(\lambda^{efg\pm})}{n}\leq
C\left(\frac{M^{2}L\log(n)}{n}+M^{-\delta/2}\log(M)^{1/2}\right)$

for some constant $C$ that only depends on $\mu.$ In particular, by taking
$M=\lfloor n^{1/3}\rfloor$ and $L=n^{1/3-\delta/6},$ we obtain

(27) $|\lambda^{efg\pm}|\leq e^{Cn^{1-\frac{\delta}{6}}\log(n)}.$

###### Proof.

We give the proof for $p_{M,n,\mu}^{efg+}(x)$; the other case follows from a
similar argument. Note that $\deg(p_{M,n,\mu}^{efg+})=n$ and

$\frac{\log|p_{M,n,\mu}^{efg+}(z)|}{n}=\int\log|z-x|d\mu_{p_{M,n,\mu}^{efg+}}=U_{\mu_{p_{M,n,\mu}^{efg+}}}(z).$

Hence if $|z|\geq n$ and $n\geq 2\max\\{|a|,|b|,|c|\\}$, then by Lemma 2.1,

$\frac{\log|p_{M,n,\mu}^{efg+}(z)|}{n}-U_{\mu}(z)=O\left(\frac{1}{n}\right).$

So without loss of generality, we assume that $|z|<n$. For $z\in\mathbb{C},$
we have

(28)
$\frac{\log|p_{M,n,\mu}^{efg+}(z)|}{n}-U_{\mu}(z)=\sum_{i,j}\left(\sum_{k\leq
n_{ij}}\frac{\log|z-z_{ijk}|}{n}-\int_{B_{ij}}\log|z-x|d\mu(x)\right)\\\
+\frac{\log\frac{|z-\textup{Re}(z_{efg})|}{|z-z_{efg}||z-\overline{z_{efg}}|}}{n}.$

Let $d(z,B_{ij}):=\inf_{w\in B_{ij}}|z-w|.$ As noted previously, there are at
most $O(1)$ pairs $(i,j)$ such that $d(z,B_{ij})<M^{-1}$. First, we give an
upper bound on the sum in (28) over $i,j$ where $d(z,B_{ij})\leq M^{-1}$.
Suppose $z_{hrs}\in B_{ij}$ has minimal distance to $z$ among $z_{ijk}$ where
$k\leq n_{ij}$ and $d(z,B_{ij})\leq M^{-1}$. By (13), (14) and (15), we have

(29) $\sum_{i,j}\left(\sum_{k\leq
n_{ij}}\frac{\log|z-z_{ijk}|}{n}-\int_{B_{ij}}\log|z-x|d\mu(x)\right)=\frac{\log|z-z_{hrs}|}{n}+O(\log(n)M^{-\delta}),$

where the sum is over $i,j$, where $d(z,B_{ij})\leq M^{-1}.$ This is because
by (15), $|z_{ijk}-z_{ijk^{\prime}}|\gg n^{-1}.$

Next, we give an upper bound on the above sum over $i,j$, where
$d(z,B_{ij})\geq M^{-1}$. Suppose that $d(z,B_{ij})\geq M^{-1},$ we have

$\left|\sum_{k\leq
n_{ij}}\frac{\log|z-z_{ijk}|}{n}-\int_{B_{ij}}\log|z-x|d\mu(x)\right|\leq\frac{M_{1}L}{n}+\mu(B_{ij})M_{2}.$

where $M_{1}:=\left|\sup_{z_{1}\in B_{ij}}\log|z-z_{1}|\right|$ and
$M_{2}:=\sup_{z_{1},z_{2}\in B_{ij}}\left|\log|z-z_{2}|-\log|z-z_{1}|\right|.$
Because we know $d(z,B_{ij})\geq M^{-1},$ we have

$M_{1}\ll\max(|\log(z)|,\log(M))\ll\log(n).$

Suppose that $z_{1},z_{2}\in B_{ij}.$ Let $r_{1}:=|z-z_{1}|$ and
$r_{2}:=|z-z_{2}|$ and assume that $r_{1}\leq r_{2}.$ By the mean value
theorem, we have

$M_{2}=\frac{\left|r_{1}-r_{2}\right|}{r_{3}}$

for some $r_{3},$ where $r_{1}\leq r_{3}\leq r_{2}.$ We have
$|r_{1}-r_{2}|\leq|z_{1}-z_{2}|\leq M^{-1}$ and

$r_{3}\gg d(z,B_{ij}).$

Hence,

$M_{2}\ll\frac{M^{-1}}{d(z,B_{ij})}.$

Therefore,

(30) $\sum_{i,j}\left|\sum_{k\leq
n_{ij}}\frac{\log|z-z_{ijk}|}{n}-\int_{B_{ij}}\log|z-x|d\mu(x)\right|\ll\frac{M^{2}L\log(n)}{n}+\sum_{i,j}\mu(B_{ij})\frac{M^{-1}}{d(z,B_{ij})},$

where the sum is over $i,j,$ where $d(z,B_{ij})\geq M^{-1}$. By the Cauchy-
Schwarz inequality,

$\sum_{i,j}\mu(B_{ij})\frac{M^{-1}}{d(z,B_{ij})}\leq\left(\sum_{i,j}\mu(B_{ij})^{2}\right)^{1/2}\left(\sum_{i,j}\frac{M^{-2}}{d(z,B_{ij})^{2}}\right)^{1/2}\ll
M^{-\delta/2}\log(M)^{1/2},$

where we used $\sum_{i,j}\mu(B_{ij})^{2}\ll
M^{-\delta}\sum_{i,j}\mu(B_{ij})=M^{-\delta}$ and
$\sum_{i,j}\frac{M^{-2}}{d(z,B_{ij})^{2}}\ll\log(M)$ where the sum is over
$i,j,$ with $d(z,B_{ij})\geq M^{-1}.$ Therefore by (30) and the above
inequality, we have

(31) $\sum_{i,j}\left|\sum_{k\leq
n_{ij}}\frac{\log|z-z_{ijk}|}{n}-\int_{B_{ij}}\log|z-x|d\mu(x)\right|\ll\frac{M^{2}L\log(n)}{n}+M^{-\delta/2}\log(M)^{1/2},$

where the sum is over $i,j,$ where $d(z,B_{ij})\geq M^{-1}.$ Finally, by (28),
(29) and (31), we obtain

(32)
$\frac{\log|p_{M,n,\mu}^{efg+}(z)|}{n}-U_{\mu}(z)=\frac{\log\frac{|z-z_{hrs}||z-\textup{Re}(z_{efg})|}{|z-z_{efg}||z-\overline{z_{efg}}|}}{n}\\\
+O\left(\log(n)M^{-\delta}+\frac{M^{2}L\log(n)}{n}+M^{-\delta/2}\log(M)^{1/2}\right).$

By the definition of $z_{hrs},$
$\frac{|z-z_{hrs}||z-\textup{Re}(z_{efg})|}{|z-z_{efg}||z-\overline{z_{efg}}|}\leq
1$ and by letting $M=\lfloor n^{1/3}\rfloor$ and $L=n^{1/3-\delta/6}$, we
obtain

$\frac{\log|p_{M,n,\mu}^{efg+}(z)|}{n}\leq
U_{\mu}(z)+O(n^{-\frac{\delta}{6}}\log(n)).$

∎

####  2.3.2. Lower bound on the volume of the simplex

Let

$A:=\left[p_{M,n,\mu}^{ijk\pm}\right],$

where $0\leq i<2M,0\leq j<M,k\leq n_{ij}$ be the square matrix of size
$(n+1)\times(n+1)$ with coefficients of $p_{M,n,\mu}^{efg\pm}$ as its column
vectors.

######  Proposition 2.9.

We have

$|\det(A)|=2^{-\frac{n+1}{2}}\prod_{ijk}\frac{1}{|\textup{Im}(z_{ijk})|^{1/2}}\prod_{ijk<i^{\prime}j^{\prime}k^{\prime}}|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}|,$

where $0\leq i<2M,0\leq j<2M,k\leq n_{ij}$

###### Proof.

Let

$B:=\left[\frac{p_{M,n,\mu}(x)}{(x-z_{ijk})}\right],$

where $0\leq i<2M,0\leq j\leq M-1,k\leq n_{ij}$ be the square matrix of size
$(n+1)\times(n+1)$ with coefficients of $\frac{p_{M,n,\mu}(x)}{(x-z_{ijk})}$
as its column vectors. By (24) for fixed indices $efg$, we have

$\left[p_{M,n,\mu}^{efg+},p_{M,n,\mu}^{efg-}\right]=\left[\frac{p_{M,n,\mu}(x)}{(x-z_{efg})},\frac{p_{M,n,\mu}(x)}{(x-\overline{z_{efg}})}\right]\begin{bmatrix}1/2&\frac{1}{2i\textup{Im}(z_{efg})}\\\
1/2&-\frac{1}{2i\textup{Im}(z_{efg})}\end{bmatrix}.$

Hence,

(33)
$|\det(A)|=|\det(B)|2^{-\frac{n+1}{2}}\prod_{ijk}\frac{1}{|\textup{Im}(z_{ijk})|^{1/2}}$

where $0\leq i<2M,0\leq j<2M,k\leq n_{ij}.$ Let

$V:=[z_{ijk}^{e}]$

be the Vandermonde matrix of size $(n+1)\times(n+1),$ where $0\leq e<n+1.$ It
is well-known that

$|\det
V|=\prod_{ijk}\prod_{ijk<i^{\prime}j^{\prime}k^{\prime}}|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}|.$

We have

$VB=\textup{diag}\left[\prod_{ijk\neq efg}(z_{efg}-z_{ijk})\right].$

Hence,

$|\det(B)|=\prod_{ijk}\prod_{ijk<i^{\prime}j^{\prime}k^{\prime}}|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}|.$

Therefore, by (33)

$|\det(A)|=2^{-\frac{n+1}{2}}\prod_{ijk}\frac{1}{|\textup{Im}(z_{ijk})|^{1/2}}\prod_{ijk<i^{\prime}j^{\prime}k^{\prime}}|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}|.$

∎

######  Proposition 2.10.

We have

$\textup{vol}(K_{n})\gg n^{-cn^{2-\frac{\delta}{6}}}e^{\frac{n^{2}}{2}I(\mu)}$

where $c$ is a constant dependents only on $\mu$.

###### Proof.

By Proposition 2.8, $e^{Cn^{1-\frac{\delta}{6}}\log(n)}K_{n}$ includes the
simplex with vertices $\left\\{0,p_{M,n,\mu}^{ijk\pm}\right\\}$ with volume
$\frac{|\det(A)|}{(n+1)!}$, where $0\leq i<2M,0\leq j\leq M-1,k\leq n_{ij}.$
Since $K_{n}$ is symmetric, there are $2^{n+1}$ simplexes with the same volume
and disjoint interior inside $e^{Cn^{1-\frac{\delta}{6}}\log(n)}K_{n}$ by
choosing different signs for the $n+1$ vertices $\left\\{0,\pm
p_{M,n,\mu}^{ijk\pm}\right\\}$. Hence,

$\textup{vol}(K_{n})\geq\frac{2^{n+1}}{(n+1)!}e^{-Cn^{2-\frac{\delta}{6}}\log(n)}|\det(A)|.$

By Proposition (2.9),

$\textup{vol}(K_{n})\geq\frac{2^{\frac{n+1}{2}}}{(n+1)!}e^{-Cn^{2-\frac{\delta}{6}}\log(n)}\prod_{ijk}\frac{1}{|\textup{Im}(z_{ijk})|^{1/2}}\prod_{ijk<i^{\prime}j^{\prime}k^{\prime}}|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}|.$

We have that $|\textup{Im}(z_{ijk})|\leq c$ where
$\text{supp}(\mu)\in[a,b]\times[-c,c]$ with $c\geq 1$. Therefore

$\prod_{ijk}\frac{1}{|\textup{Im}(z_{ijk})|^{1/2}}\prod_{ijk<i^{\prime}j^{\prime}k^{\prime}}|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}|\geq
c^{-n/2}\prod_{ijk<i^{\prime}j^{\prime}k^{\prime}}|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}|.$

From (18) using $M=\lfloor n^{1/3}\rfloor$, we have
$\prod_{ijk<i^{\prime}j^{\prime}k^{\prime}}|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}|=e^{\frac{n^{2}}{2}I(\mu)+O(n^{2-\frac{\delta}{6}}\log(n))}$
and so for some (potentially negative) constant $C^{\prime}$, this is at least
$e^{\frac{n^{2}}{2}I(\mu)+C^{\prime}n^{2-\frac{\delta}{6}}\log(n)}$. This
gives

$\textup{vol}(K_{n})\gg\frac{2^{\frac{n+1}{2}}}{(n+1)!c^{n/2}}e^{\frac{n^{2}}{2}I(\mu)+(C^{\prime}-C)n^{2-\frac{\delta}{6}}\log(n)}\gg\frac{n^{(C^{\prime}-C)n^{2-\frac{\delta}{6}}}e^{\frac{n^{2}}{2}I(\mu)}}{\left(n+1\right)^{n+1}c^{n}}\gg
n^{\tilde{C}n^{2-\frac{\delta}{6}}}e^{\frac{n^{2}}{2}I(\mu)}$

since $\frac{(n+1)^{n+1}}{n^{n}}\sim(n+1)e$ for some constant $\tilde{C}$. ∎

######  Proposition 2.11.

Suppose that $\mu\in\mathcal{B}_{\Sigma}.$ There exists an integral polynomial
$p_{0}$ with $\deg(p_{0})\leq n$ such that

$\frac{\log|p_{0}(x)|}{n}\leq
U_{\mu}(x)-\frac{I(\mu)}{2}+C\log(n)n^{-\delta/6},$

where $C$ is a constant independent of $n.$ Moreover,

$I(\mu)\geq 0.$

###### Proof.

By Proposition (2.10) there are constants $c,c^{\prime}$ such that for all
$n>0$, $\textup{vol}(K_{n})\geq
c^{\prime}n^{cn^{2-\frac{\delta}{6}}}e^{\frac{n^{2}}{2}I(\mu)}$. By
Minkowski’s second theorem (12), there exists
$p_{0}\in\mathbb{Z}[x]\cap\lambda_{0}K_{n}$ of degree at most $n$, where

(34) $\lambda_{0}\leq 2\left(\textup{vol}(K_{n})\right)^{-1/(n+1)}\ll
n^{cn^{1-\frac{\delta}{6}}}e^{-\frac{n}{2}I(\mu)}.$

By definition of $K_{n},$

$\frac{\log|p_{0}(x)|}{n}\leq U_{\mu}(x)+\frac{\log(\lambda_{0})}{n}$

By using (34), we obtain

$\frac{\log|p_{0}(x)|}{n}\leq
U_{\mu}(x)-\frac{I(\mu)}{2}+c\log(n)n^{-\frac{\delta}{6}}.$

By taking the expected value of the above inequality, we have

$0\leq\int\frac{\log|p_{0}(x)|}{n}d\mu(x)\leq\frac{I(\mu)}{2}+c\log(n)n^{-\frac{\delta}{6}},$

where we used the assumption $\mu\in\mathcal{B}_{\Sigma}$ for the first
inequality. By letting $n\to\infty$, we have

$I(\mu)\geq 0.$

This completes the proof of our proposition. ∎

###  2.4. Proof of Theorem 1.7

In this section, we prove Theorem 1.7. Suppose that $\mu$ is a probability
measure with compact support on the complex plane and that for every non-zero
$Q(x)\in\mathbb{Z}[x]$,

$\int\log|Q(z)|d\mu(z)\geq 0.$

First, we reduce the proof of Theorem 1.7 to the case of Hölder probability
measure by the following proposition.

######  Proposition 2.12.

Suppose that $\Sigma\subset\mathbb{C}$ is compact and
$\mu\in\mathcal{B}_{\Sigma}.$ For any $\rho>0,$ there exists a sequence of
Hölder probability measure $\mu_{n}\in\mathcal{B}_{\Sigma(\rho)}$ such that
$\lim_{n\to\infty}\mu_{n}=\mu$.

###### Proof.

Let $\lambda_{r}$ be the uniform probability measure on the circle of radius
$r$ centered at the origin. Let $\mu*\lambda_{r}$ be the convolution of $\mu$
with $\lambda_{r}.$ We show that $U_{\mu*\lambda_{r}}(z)\geq U_{\mu}(z)$ for
every $z\in\mathbb{C}$ and every $r\geq 0.$ In fact, we have

$U_{\mu*\lambda_{r}}(z)-U_{\mu}(z)=\int\left(\int_{0}^{1}\log|z-x+re^{2\pi
i\theta}|-\log|z-x|d\theta\right)d\mu(x)\geq 0,$

where we used Jensen’s formula (6) to show the inner integral is positive.
Therefore, $\mu*\lambda_{r}\in\mathcal{B}_{\Sigma(\rho)}$ for any $r\leq\rho.$
Define

$\mu_{n}:=\mu*\lambda_{\rho/n}.$

It is clear from the definition that $\lim_{n\to\infty}\mu_{n}=\mu$ and
$\mu_{n}$ is a Hölder measure with exponent at least 1. This completes the
poof of our proposition. ∎

###### Proof of Theorem 1.7.

Part (1) is the special case of part (2) for $m=1.$ So, it is enough to show
that part (1) implies part (2). Suppose that $\mu$ satisfies the conditions of
part part (1), which is equivalent to $\mu\in\mathcal{B}_{\Sigma}$ by
definition.

Next, we reduce the theorem to the case where $\mu$ is Hölder. By Proposition
2.12, there exists a sequence of Hölder probability measure
$\mu_{n}\in\mathcal{B}_{\Sigma(\rho)}$ such that
$\lim_{n\to\infty}\mu_{n}=\mu$. Suppose that Theorem 1.7 holds for Hölder
probability measures. Then

$\int\log|Q(z_{1},\dots,z_{m})|d\mu_{n}(z_{1})\dots d\mu_{n}(z_{m})\geq 0,$

for every $n.$ Note that $\log|Q(z_{1},\dots,z_{m})|$ is an upper
semicontinuous function. Hence, by monotone convergence theorem, we have

$\int\log|Q(z_{1},\dots,z_{m})|d\mu(z_{1})\dots
d\mu(z_{m})\geq\limsup_{n\to\infty}\int\log|Q(z_{1},\dots,z_{m})|d\mu_{n}(z_{1})\dots
d\mu_{n}(z_{m})\geq 0.$

So without loss of generality we assume that $\mu\in\mathcal{B}_{\Sigma}$ is
Hölder. The proof is by induction on $m,$ and part (1) is the base of our
induction hypothesis. We assume that part (2) holds for $m$ and we show that
it holds for $m+1.$ Suppose that
$Q(x_{1},\dots,x_{m+1})\in\mathbb{Z}[x_{1},\dots,x_{m+1}].$ Without loss of
generality, we assume that $Q$ is irreducible in
$\mathbb{Z}[x_{1},\dots,x_{m+1}]$ and does not belong to
$\mathbb{Z}[x_{m+1}].$ Let

$U_{Q}(z_{m+1}):=\int\log|Q(z_{1},\dots,z_{m},z_{m+1})|d\mu(z_{1})\dots
d\mu(z_{m}).$

It is enough to show that

(35) $\int U_{Q}(z)d\mu(z)\geq 0.$

We fix $z_{i}\in\mathbb{C}$ for $1\leq i\leq m$ and consider $Q$ as a
polynomial in $x_{m+1}$ variable and assume that it has degree $d$ in
$x_{m+1}.$ Then

$Q(z_{1},\dots,z_{m},x_{m+1})=\sum_{i=0}^{d}a_{i}(z_{1}\dots,z_{m})x_{m+1}^{i}=a_{d}(z_{1},\dots,z_{m})\prod_{i=1}^{d}(x_{m+1}-\xi_{i}(z_{1},\dots,z_{m}))$

where $\xi_{i}(z_{1},\dots,z_{m})\in\mathbb{C}$ are complex roots of $Q$. We
define the measure $\mu_{Q}$ on $\mathbb{C}$ as follows

(36) $\mu_{Q}:=\int\sum_{i}\delta_{\xi_{i}(z_{1},\dots,z_{m})}d\mu(z_{1})\dots
d\mu(z_{m})$

where $\delta_{a}$ is the delta probability measure at point $a\in\mathbb{C}.$
We have

$U_{Q}(z)=a_{Q}+\int\log|z-x|d\mu_{Q}(x)$

where

(37) $a_{Q}:=\int\log|a_{d}(z_{1},\dots,z_{m})|d\mu(z_{1})\dots d\mu(z_{m}).$

Note that by induction hypothesis, we have

$a_{Q}\geq 0.$

By Proposition 2.11, there exists an integral polynomial $P_{0}(x)$ with
$\deg(P_{0})\leq n$ such that

$\frac{\log|P_{0}(x)|}{n}\leq\int\log|z-x|d\mu(z)+Cn^{-\delta}.$

We take the expected value of the above inequality with respect to $d\mu_{Q}$
and obtain

$a_{Q}+\int\frac{\log|P_{0}(x)|}{n}d\mu_{Q}(x)\leq
a_{Q}+\int\log|z-x|d\mu(z)d\mu_{Q}(x)+\frac{Cd}{n^{\delta}}=\int
U_{Q}(z)d\mu(z)+\frac{Cd}{n^{\delta}}.$

It is enough to show that

(38) $na_{Q}+\int\log|P_{0}(x)|d\mu_{Q}(x)\geq 0,$

since (35) follows by letting $n\to\infty.$ In fact, we prove that for every
integral polynomial $P(x),$

(39) $\deg(P)a_{Q}+\int\log|P(x)|d\mu_{Q}(x)\geq 0.$

This implies (38), since $\deg(P_{n})\leq n$ and $a_{Q}\geq 0.$ For proving
(39), we consider $P$ as an integral polynomial in $x_{m+1}$. Define

$R(Q,P)(x_{1},\dots,x_{m}):=Res(Q(x_{1},\dots,x_{m+1}),P(x_{m+1}))\in\mathbb{Z}[x_{1},\dots,x_{m}]$

to be the resultant of $Q$ and $P$ as polynomials in $x_{m+1}$ variable. By
our assumption, $Q$ is irreducible and does not divide $P(x_{m+1}).$ Hence,
$Res(Q(x_{1},\dots,x_{m+1}),P(x_{m+1}))\neq 0.$ By our induction hypothesis,

$\int\log R(Q,P)(z_{1},\dots,z_{m})d\mu(z_{1})\dots d\mu(z_{m})\geq 0.$

We note that

$\log
R(Q,P)(z_{1},\dots,z_{m})=\deg(P)\log|a_{d}(z_{1},\dots,z_{d})|+\sum_{i=1}^{d}\log|P(\xi_{i}(z_{1},\dots,z_{m}))|$

Hence,

$\int\deg(P)\log|a_{d}(z_{1},\dots,z_{d})|+\sum_{i=1}^{d}\log|P(\xi_{i}(z_{1},\dots,z_{m}))|d\mu(z_{1})\dots
d\mu(z_{m})\geq 0.$

The above implies (39) by definitions of $\mu_{Q}$ and $a_{Q}$ in (36) and
(37). This complete our proof and implies

$\int\log|Q(z_{1},\dots,z_{m},z_{m+1})|d\mu(z_{1})\dots
d\mu(z_{m})d\mu(z_{m+1})\geq 0.$

∎

###  2.5. Lower bound on the product of Minkowski’s minima factors

First, we give a lower bound on the product of the Minkowski minima factors by
using Theorem 1.7.

######  Proposition 2.13.

We have

$\prod_{i=0}^{m-1}\lambda_{i}\geq\frac{1}{m!}e^{(-nm+\frac{(m-1)m}{2})I_{\mu}}.$

###### Proof.

By definition of $\\{\lambda_{i}:0\leq i\leq m-1\\}$, there exists linearly
independent integral polynomials $\\{P_{0},\dots,P_{m-1}\\}$, where

$|P_{i}(z)|\leq\lambda_{i}e^{nU_{\mu}(z)}.$

for every $z\in\mathbb{C}.$ Let $\\{z_{1},\dots,z_{m}\\}\subset\mathbb{C}.$
Therefore,

(40) $\det[P_{i}(z_{j})]\leq
m!\prod_{i=0}^{m-1}\lambda_{i}e^{nU_{\mu}(z_{i})}.$

On the other hand, we have

$\det[P_{i}(x_{j})]=Q_{m}(x_{1},\dots,x_{m})\prod_{i<j}(x_{i}-x_{j})$

for some $Q_{m}\in\mathbb{Z}[x_{1},\dots,x_{m}].$ By taking the average of the
above, we obtain

$\int\log|\det[P_{i}(z_{j})]|d\mu(z_{1})\dots
d\mu(z_{m})=\frac{(m-1)m}{2}I_{\mu}+\int\log|Q_{m}(z_{1},\dots,z_{m})|d\mu(z_{1})\dots
d\mu(z_{m}),$

where

$I(\mu):=\int\log|z_{1}-z_{2}|d\mu(z_{1})d\mu(z_{2}).$

By second part of Theorem (1.7), we have

$\int\log|Q_{m}(z_{1},\dots,z_{m})|d\mu(z_{1})\dots d\mu(z_{m})\geq 0.$

Hence,

$\int\log|\det[P_{i}(z_{j})]|d\mu(z_{1})\dots
d\mu(z_{m})\geq\frac{(m+1)m}{2}I_{\mu}.$

By (40), we obtain

$mnI_{\mu}+\sum_{i}\log|\lambda_{i}|+\log(m!)\geq\int\log|\det[P_{i}(z_{j})]|d\mu(z_{1})\dots
d\mu(z_{m})\geq\frac{(m+1)m}{2}I_{\mu}$

Therefore,

$\prod_{i}\lambda_{i}\geq\frac{1}{m!}e^{(-nm+\frac{(m-1)m}{2})I_{\mu}}.$

∎

######  Proposition 2.14.

We have

$\lambda_{n}\leq n^{cn^{1-\frac{\delta}{12}}}.$

###### Proof.

Let $m=n-n^{\delta_{1}}$ where $0<\delta_{1}<1$ is specified later. By
Proposition 2.13,

$\prod_{i=0}^{m}\lambda_{i}\geq\frac{1}{m!}e^{-nm+\frac{(m-1)m}{2}I_{\mu}}.$

By Minkowski’s second theorem and Proposition 2.10,

$\lambda_{0}\lambda_{1}\dots\lambda_{n}\leq 2^{n+1}\textup{vol}(K_{n})^{-1}\ll
2^{n+1}n^{cn^{2-\frac{\delta}{6}}}e^{-\frac{n^{2}}{2}I(\mu)}.$

By combining the above inequalities, we have

$\prod_{i=m+1}^{n}\lambda_{i}\ll
n^{cn^{2-\frac{\delta}{6}}}e^{-\frac{(n-m)^{2}}{2}I(\mu)}.$

By Proposition 2.11, $I(\mu)\geq 0,$ and we have

$\prod_{i=m+1}^{n}\lambda_{i}\ll n^{cn^{2-\frac{\delta}{6}}}.$

Hence, by taking $\delta_{1}=1-\delta/12,$ we obtain

$\lambda_{m+1}\leq n^{cn^{2-\frac{\delta}{6}-\delta_{1}}}\leq
n^{cn^{1-\frac{\delta}{12}}}.$

By Proposition 2.5, we have

$\lambda_{n}\leq\lambda_{m}(An)^{C(n-m)}\leq n^{cn^{1-\frac{\delta}{12}}}.$

This completes the proof of our proposition. Note that we may change constant
$c$ to a larger constant in the above lines of argument. ∎

###  2.6. Equilibrium measure

In this subsection, we apply the results of our previous section to the
equilibrium measure and prove Proposition 2.15. Our proof is based on Fekete
and Szegö [FS55, Theorem D] which is also used by Robinson [Rob64a] and Smith
[Smi21, Proposition 4.1] to prove a similar result. All these results are
essential in controlling the roots to remain inside $\Sigma(\rho).$

Suppose that $\Sigma$ is compact and its complement is open and connected. Let
$D$ be any open set containing $\Sigma$. Let

$\Sigma(\rho):=\\{z\in\mathbb{C}:|z-\sigma|<\rho\text{ for some
}\sigma\in\Sigma\\},$

where $\Sigma(\rho)\subset D.$ Note that for any open set $D$ containing
$\Sigma,$ there exist $\rho>0$ such that $\Sigma(\rho)\subset D.$

Let $\mu_{eq}$ be the equilibrium measure of $\Sigma.$ Fix any $\rho>0$ and
$z_{0}\in\Sigma.$ It is well known that

(41) $U_{\mu_{eq}}(z)\geq\log(d_{\Sigma})+\delta_{\rho,\Sigma}$

for any $z\notin\Sigma(\rho),$ where $\delta_{\rho,\Sigma}>0$ is a constant
that only depends on $\Sigma$ and $\rho.$ Let $p_{M,n,\mu_{eq}}(x)$ be the
polynomial constructed in (23) associated to the equilibrium measure. By (32)
and (41) there exists some $R>d_{\Sigma}$ and $M_{0},n_{0}\in\mathbb{Z}$ such
that

$p_{M_{0},n_{0},\mu_{eq}}(z)\geq R^{n_{0}}$

for any $z\notin\Sigma(\rho).$ Fix $z_{0}\in\Sigma(\rho)$ and
$p_{M_{0},n_{0},\mu_{eq}}(z),$ and define

(42) $T_{m}(x):=(x-z_{0})^{r}p_{M_{0},n_{0},\mu_{eq}}(x)^{q},$

where $m=n_{0}q+r$ is the Euclidean division of $m$ by $n_{0}$. Let $m\geq h$
be two positive integers. It follows that

(43) $T_{m}(z)\geq CR^{m-h}T_{h}(z)$

for any $z\notin\Sigma(\rho),$ where $C$ only depends on fixed parameters
$\rho$ and $n_{0}.$

######  Proposition 2.15.

Given compact subset $\Sigma\subset\mathbb{C}$ with connected complement, a
monic polynomial with real coefficients $p(x)$ and $\rho>0$, there exists
fixed integers $l_{0}$ and $k_{0}$ independent of $p(x)$ and a monic
polynomial $Q_{m}(x)$ with $\deg(Q_{m})=m$ for every $m=kl_{0},$ where
$k>k_{0}$ such that

$|Q_{m}(z)|\geq\kappa^{m}$

for every $z\notin\Sigma(\rho)$ where $\kappa>d_{\Sigma}.$ Moreover, the
coefficient of $x^{i}$ in $p(x)Q_{m}(x)$ is an even integer for every
$\deg(p)\leq i\leq\deg(p)+m-1$, and all complex roots of $Q_{m}(x)$ are inside
$\Sigma(\rho).$

###### Proof.

Recall $T_{m}(x)$ in (42) for $m\in\mathbb{Z},$ and define

$q_{k,l}(x):=\left(T_{k}(x)+a_{1}T_{k-1}(x)+\dots+a_{k}T_{0}(x)\right)^{l}$

where $|a_{i}|\leq\frac{1}{l}$ are chosen recursively for $1\leq i\leq k$ such
that $q_{k,l}(x)p(x)$ has even coefficients for $x^{\deg(p)+\deg(q_{k,l})-i},$
where $1\leq i\leq k.$ Note that this is possible since the coefficient of
$x^{\deg(p)+\deg(q_{k,l})-j}$ for $j\leq l$ in $q_{k,l}(x)p(x)$ is given by

$la_{j}+F_{j}(a_{1},\dots,a_{j-1})$

where $F_{j}(a_{1},\dots,a_{j-1})$ is a polynomial in terms of $a_{i}$ for
$i<j.$ By (43), we have

$\displaystyle|q_{k,l}(z)|$
$\displaystyle=\left|\left(T_{k}(z)+a_{1}T_{k-1}(z)+\dots+a_{k}T_{0}(z)\right)^{l}\right|$
$\displaystyle\geq\left|T_{k}(z)\left(1-\frac{C}{l}\sum_{j\geq
1}\frac{1}{R^{j}}\right)\right|^{l}\geq|T_{k}(z)|^{l}\left(1-\frac{C}{l(R-1)}\right)^{l}.$

We fix an integer $l_{0}>\frac{C}{R-1}$, and let

$C^{\prime}:=\left(1-\frac{C}{l_{0}(R-1)}\right)^{l_{0}}>0.$

Hence,

(44) $|q_{k,l_{0}}(z)|\geq C^{\prime}T_{k}(z)^{l_{0}}\geq
C^{\prime\prime}|p_{M_{0},n_{0},\mu_{eq}}(z)|^{\left\lfloor\frac{k}{n_{0}}\right\rfloor
l_{0}}$

where $C^{\prime\prime}$ is a fixed constant depending on our fixed
parameters. Finally, let $m:=l_{0}k$ and

$Q_{m}(x):=q_{k,l_{0}}(x)-\sum_{i=0}^{m-k}b_{i}T_{i}(x),$

where $|b_{i}|\leq 1$ are chosen recursively for $0\leq i\leq m-k$ such that
$Q_{m}(x)p(x)$ has even coefficients for $x^{j},$ where $\deg(p)\leq
j\leq\deg(p)+m-k.$ By (44), (43), and (42), we have

$\displaystyle|Q_{m}(z)|\geq|q_{k,l_{0}}(z)|-\sum_{i=0}^{m-k}|T_{i}(z)|\geq|p_{M_{0},n_{0},\mu_{eq}}(z)|^{\frac{kl_{0}}{n_{0}}}\left(C^{\prime\prime}-\frac{C_{1}}{R^{k-1}(R-1)}\right).$

Our proposition follows by taking $k>k_{0}$ where $k_{0}$ is a fixed constant.
∎

Let $\Sigma\subset\mathbb{R}$ be a finite union of closed intervals. For
applications to abelian varieties and the trace problem, we use the following
proposition. The statement of this proposition is motivated by [Smi21,
Proposition 4.1] which is based on Robinson [Rob64a]. All these results in
Robinson’s [Rob64a] Smith’s[Smi21], and our work are essential in controlling
the roots to remain inside $\Sigma_{\mathbb{R}}(\rho).$ Smith and Robinson
used the properties of the Chebyshev’s polynomial to prove their result and
their method are very similar to the proof of Proposition 2.15. Our proof is
conceptually different.

######  Proposition 2.16.

Suppose that $\Sigma\subset\mathbb{R}$ is a finite union of closed intervals
with $d_{\Sigma}\geq 1$. Given a monic polynomial with real coefficients
$p(x)$ of degree $n$, $\rho>0$, and every $m\leq n,$ there exists a monic
polynomial $Q_{m}(x)$ with $\deg(Q_{m})=m$ such that

  1. (1)

The degree $i$ coefficient of the product $pQ_{m}$ is an even integer for
$n-m\leq i\leq n-1$.

  2. (2)

Take $X$ to be the set of roots of $Q_{m}$. Then $X$ is a subset of
$\Sigma_{\mathbb{R}}(\rho)\subset\mathbb{R}$, and we have

$\frac{\log|Q_{m}(z)|}{m}=\frac{\min_{\alpha\in
X}\log|z-\alpha|}{m}+U_{\mu_{eq}}(z)+O(m^{-\frac{\delta}{6}}\log(m))$

for all $z\in\mathbb{C}$ and some $\delta>0,$ where $\mu_{eq}$ is the
equilibrium measure for $\Sigma_{\mathbb{R}}(\rho/10)\subset\mathbb{R}.$

  3. (3)

Given any root $\alpha$ of $Q_{m}$, and given any $\alpha^{\prime}$ that is
either a root of $p$ or a boundary point of $\Sigma_{\mathbb{R}}(\rho)$, we
have $n^{-3}\leq|\alpha-\alpha^{\prime}|.$

###### Proof.

Let $P_{M,m,\mu_{eq}}$ be the polynomial constructed in (25) associated to the
equilibrium measure for $\Sigma_{\mathbb{R}}(\rho/10)$ with roots
$x_{1},\dots,x_{m},$ such that $M=m^{1/3},$ and by a covering argument we pick
$x_{1},\dots,x_{m}$ such that for every $x_{i}$ we have
$n^{-3}\leq|x_{i}-\alpha^{\prime}|$ for any $\alpha^{\prime}$ that is either a
root of $p$, any other $x_{j}\neq x_{i}$ or a boundary point of
$\Sigma_{\mathbb{R}}(\rho).$ By (32) and (41), this polynomial satisfies all
the conditions above except the first one. Our method is to deform the roots
of $P_{M,m,\mu_{eq}}$ with a greedy algorithm so that it satisfies all the
properties of Proposition 2.16. We start with the degree $n-1$ coefficient of
$pP_{M,m,\mu_{eq}}.$ We pick $\lceil 10/\rho\rceil$ of them say
$x_{1},\dots,x_{10/\rho}$ and deform each of them less than $\rho/10$ so that
the degree $n-1$ coefficient of $pQ_{m}$ becomes even and the roots remain
inside $\Sigma_{\mathbb{R}}(\rho).$ We show that this is possible. Let
$Q_{m,0}(x):=P_{M,m,\mu_{eq}}(x)=x^{m}+\sum_{i=1}^{m}a_{i}x^{m-i}.$ The
coefficient of $x^{\deg(p)+m-j}$ for $j\leq l$ in $Q_{m,0}(x)p(x)$ is given by

$a_{j}+F_{j}(a_{1},\dots,a_{j-1})$

where $F_{j}(a_{1},\dots,a_{j-1})$ is a polynomial in terms of $a_{i}$ for
$i<j.$ We define $x_{i}(t):=x_{i}+t\frac{\rho}{10}$ for $i\leq 10/\rho$ and
call the deformed polynomial $Q_{m,0,t}.$ By the intermediate value theorem
there exists $|t|\leq 1$ such that the degree $n-1$ coefficient of
$pQ_{m,0,t}$ becomes an even number and roots remains inside
$\Sigma_{\mathbb{R}}(\rho).$ Suppose that after deformation, condition (3) is
violated for $x_{1},\dots,x_{k}.$ By (16), it is possible to pick $k$ other
roots of $Q_{m,0}(x)$ say $y_{1},\dots,y_{k}$ such that there are no other
roots close to $y_{i}$ within distance $n^{-4/3}.$ We updated $x_{1}$ and
$y_{1}$ by $x_{1}+t$ and $y_{1}-t.$ By a covering argument, there exists $t\ll
n^{-2}$ such that $x_{1}+t$ is away from other roots with distance at least
$n^{-3}.$ Similarly, update $x_{i},y_{i}$ so that the third condition is
satisfied. Denote the updated polynomial by $Q_{m,1}.$ Note that $Q_{m,1}$
satisfies the second and third conditions above, and the degree $n-1$
coefficient of $pQ_{m,1}$ is even.

We construct $Q_{m,l}$ for $l\geq 1$ recursively as follows. Suppose that
$Q_{m,l}$ satisfies the second and third conditions above above, and the
degree $i$ coefficient of the product $pQ_{m,l}$ is an even integer for
$n-l\leq i\leq n-1$, we want to make the coefficient of $n-l-1$ even as well
without changing the higher degree coefficients by deforming the roots of
$Q_{m,l}$ as we did in the previous paragraph.

Suppose that $l\leq\log\log(n).$ We pick $n^{\varepsilon}$ distinct $l$-tuples
$\boldsymbol{x}:=[x_{1},\dots,x_{l}]$ among roots inside
$\Sigma_{\mathbb{R}}(\rho/10)$ for some $\varepsilon\geq 0$ such that
$|x_{i}-x_{j}|\gg 1/l$ . Fix $\boldsymbol{x}=[x_{1},\dots,x_{l}]$, and let
$s_{k}:=\sum_{i=1}^{l}x_{i}^{k},$ and $e_{k}:=\sum_{1\leq
i_{1}<\dots<i_{k}\leq l}x_{i_{1}}\dots x_{i_{k}},$ where $e_{0}=1$ and
$s_{0}=l.$ Let $\nabla s_{k}=k[x_{i}^{k-1}]$ be the gradient of $s_{k}.$ Let

$v_{l}:=\left[\frac{1}{\prod(x_{1}-x_{i})},\frac{1}{\prod(x_{2}-x_{i})},\dots,\frac{1}{\prod(x_{l}-x_{i})}\right].$

It follows that $v_{l}$ is orthogonal to $\nabla s_{k}$ for $1\leq k\leq l-1$
and $\left<v_{l},\nabla s_{l}\right>=l.$ We note that

$e_{k}=\frac{1}{k}\sum_{j=1}^{k}(-1)^{j-1}e_{k-j}s_{j}.$

This implies that $\left<v_{l},\nabla e_{k}\right>=0$ for every $0\leq k\leq
l-1$ and $\left<v_{l},\nabla e_{l}\right>=1.$ We deform $\boldsymbol{x}$ with
the following ODE $\frac{d\boldsymbol{x}(t)}{dt}=v_{l}(t)$ and the initial
condition $\boldsymbol{x}(0)=\boldsymbol{x}.$ We note that each coordinate of
$v_{l}$ is less than $l^{l}\leq\frac{n^{\varepsilon}\rho}{10}$. Therefore, by
the intermediate value theorem, it is possible to deform $n^{\varepsilon}$
distinct $l$-tuples $[x_{1},\dots,x_{l}]$ along the direction of $v_{l}$ at
most $\rho/10$ amount and make the coefficient of $n-l-1$ also even. Suppose
that after deformation, condition (3) is violated for $x_{1},\dots,x_{k},$
where $k\leq n^{\varepsilon}$. By (16), it is possible to pick $kl\leq
n^{\varepsilon}$ other roots of $Q_{m,0}(x)$ say $y_{i,j}$ for $1\leq i\leq k$
and $1\leq j\leq l$ such that there are no other roots close to $y_{i,j}$
within distance $n^{-4/3}$, $|y_{i,k}-y_{i,k^{\prime}}|\gg 1/l$ for $k\neq
k^{\prime},$ and $|x_{i}-y_{i,j}|\gg 1/l$. We deform $l+1$ tuple
$\boldsymbol{x}_{i}:=\left[x_{i},y_{i,1},\dots,y_{i,l}\right]$ with the
following ODE $\frac{d\boldsymbol{x}_{i}(t)}{dt}=v_{l+1,i}(t)$ and the initial
condition $\boldsymbol{x}_{i}(0)=\boldsymbol{x}_{i}$ where

$v_{i,l+1}(t):=\left[\frac{1}{\prod_{j}(x_{i}-y_{1,j})},\frac{1}{(y_{i,1}-x_{i})\prod_{j}(y_{i,1}-y_{i,j})},\dots,\frac{1}{(y_{i,l}-x_{i})\prod_{j}(y_{i,l}-y_{i,j})}\right].$

By a covering argument, there exists $t\ll n^{-2+\varepsilon}$ such that
$x_{1}(t)$ is away from other roots with distance at least $n^{-3}$. Denote
the updated polynomial by $Q_{m,l+1}.$ Note that $Q_{m,l+1}$ satisfies second
and third conditions above, and the degree $i$ coefficient of the product
$pQ_{m,l+1}$ is an even integer for $n-l-1\leq i\leq n-1$.

Suppose that $n^{\varepsilon}\geq l\geq\log\log(n)$ for some $\varepsilon>0.$
We pick $x_{1},\dots,x_{l}$ according to the equilibrium measure among the
roots inside $\Sigma(\rho/10)$. As before, let
$\boldsymbol{x}=[x_{1},\dots,x_{l}]$ and

$v_{l}:=\left[\frac{1}{\prod(x_{1}-x_{i})},\frac{1}{\prod(x_{2}-x_{i})},\dots,\frac{1}{\prod(x_{l}-x_{i})}\right].$

By (32) and (41), the size of $v_{l}$ is exponentially small and we have

$|v_{l}|\leq e^{-ld_{\Sigma}}.$

We deform $\boldsymbol{x}$ with the following ODE
$\frac{d\boldsymbol{x}(t)}{dt}=v_{l}(t)$ and the initial condition
$\boldsymbol{x}(0)=\boldsymbol{x}.$ By the intermediate value theorem there
exists $|t|\leq 1$ so that the degree $n-l-1$ coefficient of $pQ_{m,l+1}$ is
even. It is clear that the perturbation remains inside $\Sigma(\rho).$ As
before, we can insure the third condition is satisfied by picking $y_{i,j}$
according to the equilibrium measure and deform the roots so that they are
away from each other with distance at least $n^{-3}.$

Finally, suppose that $m\geq l\geq n^{\varepsilon}.$ We pick
$x_{1},\dots,x_{l}$ according to the equilibrium measure among the roots
inside $\Sigma_{\mathbb{R}}(\rho/10)$. As before, let
$\boldsymbol{x}=[x_{1},\dots,x_{l}]$ and

$v_{l}:=\left[\frac{1}{\prod(x_{1}-x_{i})},\frac{1}{\prod(x_{2}-x_{i})},\dots,\frac{1}{\prod(x_{l}-x_{i})}\right].$

By (32) and (41), the size of $v_{l}$ is smaller than any negative power of
$n$

$|v_{l}|\leq e^{-ld_{\Sigma}}\leq e^{-n^{\varepsilon}}\ll n^{-A}$

for any $A>0.$ We deform $\boldsymbol{x}$ with the following ODE
$\frac{d\boldsymbol{x}(t)}{dt}=v_{l}(t)$ and the initial condition
$\boldsymbol{x}(0)=\boldsymbol{x}.$ By mean value theorem there exits $|t|\leq
1$ so that the degree $n-l-1$ coefficient of $pQ_{m,1}$ is even. This time the
perturbation does not change conditions (2) and (3), since the perturbation is
smaller than $n^{-A}$ for any $A.$ This completes the proof of our
proposition. ∎

##  3\. Proofs of Theorem 1.4 and Theorem 1.2

###  3.1. Proof of Theorem 1.4

In this subsection, we assume that $\Sigma\subset\mathbb{C}$ is compact and
has empty interior with connected complement. We also assume that
$\mu\in\mathcal{B}_{\Sigma}$ is a Hölder measure.

###### Proof of Theorem 1.4.

Suppose that $\mu\in\mathcal{B}_{\Sigma}$ is a Hölder measure with exponent
$\delta,$ and $D$ is any open set containing $\Sigma$. Fix $\rho>0$ such that
$\Sigma(\rho)\subset D.$ We apply Proposition 2.15 to $p_{M,n-m,\mu}(x)$
defined in (23) and $m:=n^{1-\delta_{0}}$ for some $\delta_{0}>0$ that we
specify later and obtain $Q_{m}(x)$ such that

$p_{M,n-m,\mu}(x)Q_{m}(x)=x^{n}+\sum_{i=0}^{n-1}a_{i}x^{i}$

where $a_{i}$ are even for $n-m\leq i\leq n-1.$ Let

$r(x):=\frac{1}{2}+\sum_{i=0}^{n-1-m}\frac{a_{i}}{4}x^{i}.$

We write $r(x)$ in terms of linearly independent integral polynomials
$\\{P_{0},\dots,P_{n-1-m}\\}$ obtained from Minkowski’s successive minima of
$K_{n-1-m}$, and obtain

$r(x)=\sum_{i=0}^{n-1-m}\alpha_{i}P_{i}(x).$

Let

$w(x):=\sum_{i=0}^{n-1-m}\lfloor\alpha_{i}\rfloor P_{i}(x),$

and

$h_{n}(x):=x^{n}+\sum_{i=n-m}^{n-1}a_{i}x^{i}+4w(x)-2.$

By the Eisenstein criteria at the prime 2, $w(x)$ is irreducible. Moreover,

$h_{n}(x)=p_{M,n-m,\mu}Q_{m}(x)+\sum_{i=0}^{n-1-m}\beta_{i}P_{i}(x),$

where $|\beta_{i}|=4|\alpha_{i}-\lfloor\alpha_{i}\rfloor|<4.$ Note that

$|P_{i}(z)|\leq\lambda_{i}e^{(n-m)U_{\mu}(z)}$

for every $z\in\mathbb{C},$ and by Proposition 2.14,

$\lambda_{i}\leq n^{cn^{1-\frac{\delta}{12}}}.$

Moreover, by (32) and Proposition 2.15

$\left|p_{M,n-m,\mu}(z)Q_{m}(z)\right|\geq
e^{(n-m)U_{\mu}(z)}\kappa^{m}n^{-Cn^{1-\frac{\delta}{6}}}.$

for every $z\notin\Sigma(\rho)$, where $M=\lfloor n^{1/3}\rfloor.$ By taking
$\frac{\delta}{12}>\delta_{0}>0$ and $m=n^{1-\delta_{0}}$, it follows that

$\left|p_{M,n-m,\mu}Q_{m}(z)\right|>\left|\sum_{i=0}^{n-m-1}\beta_{i}P_{i}(z)\right|$

for large enough $n$ and every $z\notin\Sigma(\rho).$ By Rouché’s theorem,
$h_{n}(x)$ and $p_{M,n,\mu}Q_{m}(z)$ has the same number of roots inside
$\Sigma(\rho).$ Since all roots of $p_{M,n-m,\mu}Q_{m}(x)$ are inside
$\Sigma(\rho),$ all roots of $h_{n}(x)$ are also inside $\Sigma(\rho).$ By
Lemma 2.1, we have

$\frac{\log|Q_{m}(z)|}{m}\leq\max(\log|z|,0)+C_{\Sigma},$

where $C_{\Sigma}$ is a constant that only depends on $\Sigma.$ Since
$m=n^{1-\delta_{0}},$ we have

$\frac{\log|h_{n}(z)|}{n}\leq U_{\mu}(z)+O(n^{-\delta_{0}}).$

This completes the proof of Theorem 1.4. For $\Sigma\subset\mathbb{R}$, our
argument is similar. We apply Proposition 2.16 instead of Proposition 2.15 and
$p_{M,n-m,\mu}(x)Q_{m}(x)$ which has $n$ distinct roots which are $n^{-3}$
apart. We use the intermediate value theorem and the fact that we have $n$
sign changes on $\Sigma(\rho)$ to prove all roots are real and inside
$\Sigma(\rho)$ instead of the Rouché’s theorem. ∎

###  3.2. Proof of Theorem 1.2

In this section, we show that Theorem 1.4 implies Theorem 1.2. To prove
Theorem (1.4), we assumed that $\mu$ is a Hölder measure, the support of $\mu$
has empty interior, and that its complement is connected. Here we show that
these conditions are unnecessary for proving Theorem 1.2. By Proposition 2.12,
for any $\rho>0,$ there exists a sequence of Hölder probability measure
$\mu_{n}\in\mathcal{B}_{\Sigma(\rho)}$ such that
$\lim_{n\to\infty}\mu_{n}=\mu$. So by a diagonal argument the the proof is
reduced to the case where $\mu$ is Hölder.

Recall $\Sigma\subset[a,b]\times[-c,c]$ as in section (2.3) and
$[a,b]\times[-c,c]=\bigcup_{0\leq i,j<2M}B_{ij}$, where we define
$B_{ij}=[a_{i},a_{i+1}]\times[c_{j},c_{j+1}]$ and $a_{i}=a+\frac{i(b-a)}{2M}$
and $c_{j}=-c+\frac{j(2c)}{2M}$. Furthermore, $z_{ijk}\in B_{ij}$ for each
$1\leq k\leq n_{ij}=\lfloor(n+1)\mu(B_{ij})\rfloor+\epsilon_{ij}$ are chosen
with distinct imaginary parts. We now define a measure which imitates $\mu$ by
being uniformly distributed on a union of intervals around the $z_{ijk}$. Let
$\tilde{\mu}$ be the measure supported on
$\bigcup_{ijk}\\{z:|\textup{Re}(z-z_{ijk})|<\frac{1}{n},\textup{Im}(z)=\textup{Im}(z_{ijk})\\}$
where $\tilde{\mu}$ restricted to each interval is the one-dimensional
Lebesgue measure on that interval. Now let $\tilde{\nu}$ be the equilibrium
measure on the support of $\tilde{\mu}$ and $\tilde{\nu}_{\epsilon}$ be the
normalized pushforward of $\tilde{\nu}$ under the scaling map
$x\mapsto(1+\epsilon)x$. Finally, define
$\mu_{\epsilon,\epsilon^{\prime}}=(1-\epsilon^{\prime})\tilde{\mu}+\epsilon^{\prime}\tilde{\nu}_{\epsilon}$
for $\epsilon,\epsilon^{\prime}\in(0,1)$.

######  Theorem 3.1.

For all $Q\in\mathbb{Z}[x]$ and $0<\epsilon,\epsilon^{\prime}<\frac{1}{2}$,
$\int\log|Q(x)|d\mu_{\epsilon,\epsilon^{\prime}}(x)\geq 0$ for large enough
$n$.

###### Proof.

First, we relate $U_{\tilde{\mu}}$ with $U_{\mu}$. We compute

$U_{\tilde{\mu}}(x)=\int\log|x-z|d\tilde{\mu}=\sum_{ijk}\int_{\frac{-1}{2n}}^{\frac{1}{2n}}\log|x-y-
z_{ijk}|dy.$

Let $z_{hrs}$ be the closest root of $P_{M,n,\mu}$ to $x$. Then $\forall
ijk\neq hrs$, $|x-z_{ijk}|\gg n^{-5/6}$ by (15) choosing $M\sim n^{1/3}$. So
for $ijk\neq hrs$, we have that

$\int_{\frac{-1}{2n}}^{\frac{1}{2n}}\log|x-y-
z_{ijk}|dy-\frac{1}{n}\log|x-z_{ijk}|=\int_{\frac{-1}{2n}}^{\frac{1}{2n}}\log\left|1-\frac{y}{x-z_{ijk}}\right|dy\leq\int_{\frac{-1}{2n}}^{\frac{1}{2n}}\frac{|y|dy}{|x-z_{ijk}|}\ll
n^{-7/6}$

Thus if $|x-z_{hrs}|\geq 2\cdot\text{diam}(\text{supp}(\mu))$, we have by
Lemma 2.1 that

$U_{\tilde{\mu}}(x)=U_{\mu}(x)+O(n^{-1/6}),$

and otherwise,

$U_{\tilde{\mu}}(x)=\int_{\frac{-1}{2n}}^{\frac{1}{2n}}\log|x-y-
z_{hrs}|dy+\frac{1}{n}\sum_{ijk\neq hrs}\log|x-z_{ijk}|+O(n^{-1/6})$

In the latter case, $\int_{\frac{-1}{2n}}^{\frac{1}{2n}}\log|x-y-
z_{hrs}|dy\ll\max(n^{-1},\int_{\frac{-1}{2n}}^{\frac{1}{2n}}\log|y|dy)\ll
n^{-1}\log(n)$. Therefore, we have that
$U_{\tilde{\mu}}(x)=U_{\mu_{p_{M,n,\mu}}}(x)-\frac{1}{n}\log|x-z_{hrs}|+O(n^{-1/6})$.
From our proof of Proposition 2.8,

$\frac{\log|p_{M,n,\mu}^{efg+}(x)|}{n}-U_{\mu}(x)=\frac{\log\frac{|x-z_{hrs}||x-\textup{Re}(z_{efg})|}{|x-z_{efg}||x-\overline{z_{efg}}|}}{n}\\\
+O\left(\log(n)M^{-\delta}+\frac{M^{2}\log(n)}{n}+M^{-\delta/2}\log(M)^{1/2}\right).$

Taking $M\sim n^{1/3}$, we have now shown that

$U_{\tilde{\mu}}(x)=U_{\mu}(x)+O\left(n^{-1/6}+n^{\frac{-\delta}{6}}\log(n)^{\frac{1}{2}}\right)$

Therefore, in either case, we have
$U_{\tilde{\mu}(x)}=U_{\mu}(x)+O(n^{-1/6}+n^{-\delta/6}\log(n)^{1/2})$.

Now for any $Q(x)=\prod_{i=1}^{m}(x-\alpha_{i})\in\mathbb{Z}[x]$, we have that

$\int\log|Q(x)|d\mu_{\epsilon,\epsilon^{\prime}}=(1-\epsilon^{\prime})\int\log|Q(x)|d\tilde{\mu}+\epsilon^{\prime}\int\log|Q(x)|d\tilde{\nu}_{\epsilon}$

Using our calculation above,

$\int\log|Q(x)|d\tilde{\mu}=\sum_{i=1}^{m}U_{\tilde{\mu}}(\alpha_{i})=\int\log|Q(x)|d\mu+mO(n^{-1/6}+n^{\frac{-\delta}{6}}\log(n)^{\frac{1}{2}})$

Since $I_{\tilde{\nu}}\geq I_{\mu}\geq 0$, we have that

(45)
$I(\tilde{\nu}_{\epsilon})=\int\int\log|(1+\epsilon)(x-y)|d\tilde{\nu}d\tilde{\nu}=I(\tilde{\nu})+\log|1+\epsilon|>I(\tilde{\nu})+\epsilon-\epsilon^{2}.$

We now bound $I(\tilde{\nu})$. Since $\tilde{\nu}$ is the equilibrium measure
on the support of $\tilde{\mu}$, $I(\tilde{\mu})\leq I(\tilde{\nu})$. To
compute $I(\tilde{\mu})$, we first see that
$\int_{\frac{-1}{2n}}^{\frac{1}{2n}}\int_{\frac{-1}{2n}}^{\frac{1}{2n}}\log|x-y|dxdy\ll
n^{-2}\log(n)$. Therefore

$I(\tilde{\mu})=\sum_{ijk\neq
i^{\prime}j^{\prime}k^{\prime}}\int_{\frac{-1}{2n}}^{\frac{1}{2n}}\int_{\frac{-1}{2n}}^{\frac{1}{2n}}\log|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}+x-y|dxdy+O(n^{-1}\log(n)).$

By the same argument before, this is $\frac{1}{n^{2}}\sum_{ijk\neq
i^{\prime}j^{\prime}k^{\prime}}\log|z_{ijk}-z_{i^{\prime}j^{\prime}k^{\prime}}|+O(n^{-1/6})$.
Applying (18) with $M\sim n^{-1/3}$,
$I(\tilde{\mu})=O(n^{-1/6}+n^{\frac{-\delta}{6}}\log(n))$. Therefore for large
enough $n$, $|I(\tilde{\mu})|\leq\frac{\epsilon}{4}$. Using (45), we have
$I(\tilde{\nu}_{\epsilon})>\frac{\epsilon}{4}$ since $\epsilon<1/2$. Thus
since

$\int\log|Q(x)|d\mu_{\epsilon,\epsilon^{\prime}}=(1-\epsilon^{\prime})\int\log|Q(x)|d\mu+m\left(\epsilon^{\prime}I(\tilde{\nu}_{\epsilon})+(1-\epsilon^{\prime})O(n^{-1/6}+n^{\frac{-\delta}{6}}\log(n)^{\frac{1}{2}})\right)$

by equation (1.4) of [ST97], this shows that
$\int\log|Q(x)|d\mu_{\epsilon,\epsilon^{\prime}}>0$ for sufficiently large
$n$. So for each $0<\epsilon,\epsilon^{\prime}<\frac{1}{2}$, there is a large
enough $n$ for which $\int\log|Q(x)|d\mu_{\epsilon,\epsilon^{\prime}}>0$ as
desired. ∎

###### Proof of Theorem 1.2.

Suppose that $d_{\Sigma}<1.$ Note that

$\log d_{\Sigma}=I(\mu_{eq})=\sup_{\mu\in\mathcal{P}_{\Sigma}}I(\mu),$

where $\mu_{eq}$ is the equilibrium measure of $\Sigma.$ Hence,

$I(\mu)<0$

for every $\mu\in\mathcal{P}_{\Sigma}.$ By Theorem 1.7 for polynomial
$Q(x,y)=x-y$,

$\mathcal{B}_{\Sigma}=\emptyset.$

Since $\mathcal{A}_{\Sigma}\subset\mathcal{B}_{\Sigma},$ this completes the
proof of Theorem 1.2 if $d_{\Sigma}<1.$

Otherwise, $d_{\Sigma}\geq 1$ and $\mu_{eq}\in\mathcal{B}_{\Sigma}.$ This
implies that $\mathcal{B}_{\Sigma}\neq\emptyset.$ Denote by
$\tilde{\mu}_{\epsilon,\epsilon^{\prime}}$ the measure described in Theorem
3.1 for the minimal value of $n$ giving the theorem. This gives us that
$\tilde{\mu}_{\frac{1}{m},\frac{1}{m}}\stackrel{{\scriptstyle\ast}}{{\rightharpoonup}}\mu$
as $m\to\infty$. Further, for each $m$, there is a sequence $P_{m,n}$ of
polynomials with
$\mu_{P_{m,n}}\stackrel{{\scriptstyle\ast}}{{\rightharpoonup}}\tilde{\mu}_{\frac{1}{m},\frac{1}{m}}$
as $n\to\infty$. Taking a subsequence $P_{m,n_{m}}$, we have
$\mu_{P_{m,n_{m}}}\stackrel{{\scriptstyle\ast}}{{\rightharpoonup}}\mu$ as
$m\to\infty$. Therefore, we have shown that the assumptions that the support
of $\mu$ must have non-empty interior and have its complement be connected are
unnecessary, and Theorem 1.2 follows from Theorem 1.4. ∎

##  4\. The Algorithm

###  4.1. Lattice Algorithms

The algorithm we develop requires access to short vectors in a lattice. As
seen in the work by Micciancio [Mic01], finding a vector whose length is
within a constant multiple of the shortest vector in a lattice is NP-hard to
compute. This is known as the shortest vector problem or SVP. It is still
desirable in general to have access to short vectors in a lattice. An
algorithm known as the LLL-algorithm, developed by Lenstra, Lenstra, and
Lovász, finds a basis of vectors in a lattice in polynomial time [LLL82] where
the shortest vector of the output has length within an exponential factor of
the shortest vector of the lattice. In the implementation of this algorithm,
we use the built-in method in PARI/GP named qflll. When the qflll method is
given a matrix $A$ whose columns generate a lattice, it returns a
transformation matrix $T$ so that $AT$ is an “LLL-reduced” basis of the
lattice. While the LLL-algorithm is quite reasonable in practice, the
asymptotic bounds are not strong enough to prove a sub-exponential factor.

In 1986, Schnorr [Sch87] devised a parametrized family of algorithms attacking
the SVP problem which output reduced bases somewhere between the stringent
requirements of Korkine-Zolotarev reduction and the looser requirements of LLL
reduction based on the input parameter. Careful parameter selection achieves
slightly sub-exponential constant factors in polynomial time. Schnorr proved
that if the algorithm produces a basis $b_{1},\dots,b_{n}$, then $||b_{1}||$
is within a subexponential factor of a shortest vector in the lattice. We
prove that in addition, if $\lambda_{1},\dots,\lambda_{n}$ are so that
$\lambda_{i}$ is the least value for which there are $i$ linearly independent
vectors of norm at most $\lambda_{i}$, then for each $i=1,\dots,n$,
$||b_{i}||$ is within a sub-exponential factor of $\lambda_{i}$. We now
introduce some notation from Schnorr’s work and recall some of his results.

Let $b_{1},\dots,b_{n}$ be a basis. Then let $b_{i}^{*}$ be the projection of
$b_{i}$ onto the orthogonal complement of
$\text{span}\\{b_{1},\dots,b_{i-1}\\}$ for $i=1,\dots,n$. We can get
$b_{1}=b_{1}^{*},b_{2}^{*},\dots,$ and $b_{n}^{*}$ by Gram-Schmidt process
without normalizing. Now define $\mu_{i,j}:=\frac{\langle
b_{i},b_{j}^{*}\rangle}{||b_{j}^{*}||^{2}}$.

######  Definition 4.1.

We say $b_{1},\dots,b_{n}$ is size-reduced if $\mu_{i,j}\leq 1/2$ for all
$i>j$.

######  Definition 4.2.

Let $\Lambda_{i}$ be the lattice spanned by the projections of
$b_{i},b_{i+1},\dots,b_{n}$ onto the orthogonal complement of
$\text{span}\\{b_{1},\dots,b_{i-1}\\}$. Then $b_{1},\dots,b_{n}$ is Korkine-
Zolotarev reduced if $||b_{i}^{*}||$ is minimal among non-zero vectors in
$\Lambda_{i}$.

######  Definition 4.3.

Let $k\mid n$. We say that $b_{1},\dots,b_{n}$ is semi $k$-reduced if it is
size-reduced and it satisfies both

(46) $||b_{ik}^{*}||^{2}\leq 2||b_{ik+1}^{*}||^{2},$

and the components of $b_{ik+j}$ for $j=1,\dots,k$ orthogonal to
$b_{1},\dots,b_{ik-1}$ are Korkine-Zolotarev reduced for each
$i=0,\dots,\frac{n}{k}-1$. Call this property (KZ).

Lastly, the constant $\alpha_{k}$ is defined as
$\max\frac{||b_{1}||^{2}}{||b_{k}^{*}||^{2}}$ where the max is over all
Korkine-Zolotarev reduced bases on lattices of rank $k$. Corollary 2.5 in
Schnorr’s paper shows that $\alpha_{k}\leq k^{1+\ln k}$.

######  Theorem 4.4.

Let $b_{1},\dots,b_{n}$ be a semi $k$-reduced basis and $\lambda_{j}$ be the
least real number so there are $j$ linearly independent vectors in the lattice
spanned by $b_{1},\dots,b_{n}$ of norm at most $\lambda_{j}$. Then
$||b_{j}||\leq k^{\frac{n}{k}\ln k+O(\frac{n}{k}+\ln k)}\lambda_{j}$.

The essence is a combination of the proofs of Theorems 2.3 and 3.4 in [Sch87].

###### Proof.

Fix $1\leq j\leq n$. Suppose that $v_{1},\dots,v_{j}$ are linearly independent
vectors in the lattice with $||v_{i}||=\lambda_{i}$ for $i=1,\dots,j$. For
each $i$, write $v_{i}=\sum_{s=1}^{n}c_{s,i}b_{s}$. Let $s_{i}$ be the largest
index for which $c_{s_{i},i}\neq 0$. There is some $i=1,\dots,j$ so that
$s_{i}\geq j$ by linear independence. If $c_{s,i}^{*}$ is so that
$v_{i}=\sum_{s=1}^{n}c_{s,i}^{*}b_{s}^{*}$, then
$c_{s_{i},i}=c_{s_{i},i}^{*}\in\mathbb{Z}$. This implies that
$\lambda_{j}=||v_{j}||\geq||v_{i}||\geq||c_{s_{i},i}^{*}b_{s_{i}}^{*}||\geq||b_{s_{i}}^{*}||$.
Now note that for each $0<s<t\leq k$ and $m$ so that $(m+1)k\leq n$, then
$||b_{mk+s}^{*}||\leq\alpha_{k}||b_{mk+t}^{*}||$ as stated in the proof of
Theorem 2.3 of [Sch87] and call this inequality $(\ast)$.

Using the fact that $b_{1},\dots,b_{n}$ is size reduced,

$||b_{j}||=\left|\left|b_{j}^{*}+\sum_{s=1}^{j}\mu_{j,s}b_{s}^{*}\right|\right|\leq\sum_{s=1}^{j}||b_{s}^{*}||.$

To bound $||b_{s}^{*}||$ for $1\leq s\leq j$, suppose $s=m_{1}k+t_{1}$ and
$s_{i}=m_{2}k+t_{2}$ for $0\leq t_{2},t_{2}<k$. Then

$\displaystyle||b_{s}^{*}||$
$\displaystyle\leq\alpha_{k}||b_{(m_{1}+1)k}^{*}||$ $\displaystyle\text{by
property (KZ) and }(\ast)$ $\displaystyle\leq
2\alpha_{k}||b_{(m_{1}+1)k+1}^{*}||$ $\displaystyle\text{by }(6)$
$\displaystyle\leq(2\alpha_{k})^{m_{2}-m_{1}}||b_{m_{2}k+1}||$ by induction
$\displaystyle\leq(2\alpha_{k})^{m_{2}-m_{1}}\alpha_{k}||b_{s_{i}}^{*}||$
$\displaystyle\text{by property (KZ) and }(\ast)\text{ since }j\leq s_{i}$
$\displaystyle\leq(2\alpha_{k})^{\frac{n}{k}-\lfloor\frac{s}{k}\rfloor}\alpha_{k}\lambda_{j}$
$\displaystyle\because||b_{s_{i}}^{*}||\leq\lambda_{j}$

Since $\alpha_{k}\leq k^{1+\ln k}$, we have $||b_{s}^{*}||\leq
k^{(\frac{n}{k}+1)(\ln k+2)-\lfloor\frac{s}{k}\rfloor\ln k}$. Therefore,

$\displaystyle||b_{j}||$ $\displaystyle\leq k^{\frac{n}{k}\ln
k+O(\frac{n}{k}+\ln k)}\sum_{s=0}^{\infty}k^{-\lfloor\frac{s}{k}\rfloor\ln
k}\lambda_{j}$ $\displaystyle\leq k^{\frac{n}{k}\ln k+O(\frac{n}{k}+\ln
k)}k\sum_{s=0}^{\infty}k^{-s\ln k}\lambda_{j}$ $\displaystyle\text{since every
group of }k\text{ is equal}$ $\displaystyle\leq k^{\frac{n}{k}\ln
k+O(\frac{n}{k}+\ln k)}\left(1-k^{-\ln k}\right)^{-1}\lambda_{j}$
$\displaystyle\leq k^{\frac{n}{k}\ln k+O(\frac{n}{k}+\ln k)}\lambda_{j}$
$\displaystyle\text{since }(1-k^{-\ln k})^{-1}\leq 3\text{ for }k\geq 2.$

This completes the proof. ∎

According to Schnorr’s paper, this algorithm takes $O(n^{4}\log
B+k^{\frac{k}{2}+o(k)}n^{2}\log B)$ arithmetical steps on $O(n\log B)$
integers where $B$ is the Euclidean length of the longest input vector. In our
case, $B$ is a constant depending on $\mu$. Taking
$k=\frac{\log(n)}{\log(\log(n))}$, this is $O(n^{4}+n^{2}\log(n)^{\frac{3\log
n}{2\log\log n}})$ which is contained in $O(n^{4})$ and thus this choice of
$k$ makes the algorithm polynomial time in $n$. In this case, the factor in
Theorem 4.4 is at most $e^{\frac{2n\log(\log(n))^{3}}{\log(n)}}$ for large
enough $n$; in particular, it is sub-exponential.

###  4.2. Geometry of numbers

There are two different sub-methods of our algorithm regarding geometry. The
first finds a short basis of a given lattice. The second finds an integral
polynomial near a real polynomial where distance is measured with respect to a
given basis. In these two algorithms, $P$ is an integral, square-free, monic
polynomial with roots $\alpha_{1},\dots,\alpha_{n}$.

####  4.2.1. Finding Short Lattice Bases

To find short lattice bases, we implement a version of Proposition 2.11. This
theorem shows the existence of a short basis of polynomials in a given convex
space. In particular, let $K$ be the set of real polynomials of degree at most
$n$ with $\log|P(x)|\leq nU_{\mu}(x)$. There are linearly-independent,
integral polynomials $P_{0},\dots,P_{n}$ so that for each $0\leq k\leq n$, if
$P_{k}\in\lambda K$ where $\lambda\in\mathbb{R}$ is minimal, then there is no
integral polynomial in $\rho K$ for any $\rho<\lambda$ linearly-independent
from $P_{0},\dots,P_{k-1}$. We showed that the largest of these $\lambda$’s is
sub-exponential in $n$ in Proposition 2.14. Thus $\\{P_{0},\dots,P_{n}\\}$
forms a basis of lattice points in a subexponential multiple of $K$. By
Theorem 4.4, we can compute explicit integral polynomials which are within a
sub-exponential factor of $K$ in polynomial time in $n$. This is formalized in
Corollary 4.5.

######  Corollary 4.5.

If $\mu\in\mathcal{B}_{\Sigma}$ is a Hölder probability measure, then we can
compute a basis $Q_{0},\dots,Q_{n}$ of integral polynomials of degree at most
$n$ in polynomial time where each $Q_{i}\in
e^{Cn\frac{\log(\log(n))^{3}}{\log(n)}}K_{n}$ for some fixed constant $C$
depending only on $\Sigma$ and $\mu$. More specifically, this runs in
$O(n^{4})$ time and assuming $n$ is sufficiently large, we can take $C=3$.

###### Proof.

First compute the basis $p_{M,n,\mu}^{efg\pm}$ of polynomials of degree less
than $n$ as defined in (24) choosing $M=\lfloor n^{1/3}\rfloor$. Call this
basis $\mathcal{S}$. We want to find an integral basis which is short in the
basis $\mathcal{S}$ by Proposition 2.8. Now write
$\mathcal{E}:=\\{1,x,\dots,x^{n-1}\\}$ in the basis $\mathcal{S}$ and apply
Schnorr’s algorithm to get a semi $k$-reduced basis choosing $k=\frac{\log
n}{\log\log n}$. We showed at the end of Section 4.1 that this runs in
$O(n^{4})$ time and that the output polynomials $Q_{0},\dots,Q_{n}$ are so
that $Q_{i}\in e^{2n\cdot\frac{\log(\log(n))^{3}}{\log
n}}n^{cn^{1-\frac{\delta}{12}}}K_{n}$ by Proposition 2.14 for each $i$ and
fixed $c$ for large enough $n$. It follows that $Q_{i}\in
e^{3n\cdot\frac{\log(\log(n))^{3}}{\log n}}K_{n}$ for large enough $n$. ∎

One way of computing the integral polynomials is as follows. Suppose $T$ is a
matrix so that $AT$ is a reduced form (as in Schnorr’s algorithm) of $A$ where
$A$ is the matrix of $\mathcal{E}$ written in base $\mathcal{S}$. Then $AT$
has linearly independent columns representing such polynomials in base
$\mathcal{S}$. So $T$ has columns representing those polynomials in base
$\mathcal{E}$ as desired. This gives the desired $n+1$ linearly independent
integer polynomials. Further note that that doing the change of basis via
matrix inversion is numerically unstable for large matrices. Fortunately, we
can compute the change of basis matrix explicitly. As is typically done, the
time bounds given in this paper is in number of arithmetic operations.
Numerical stability is not studied in this paper.

####  4.2.2. Finding Close Integer Polynomials

The following discussion allows us to algorithmically produce polynomials like
those whose existence is proved in Theorem 3.2 of [Smi21]. The intention of
this method is to find an integral polynomial close to a given real polynomial
in polynomial time where distance is measured with respect to a given basis.
In our case, we use the basis produced by Corollary 4.5.

######  Theorem 4.6.

Given a Hölder probability measure $\mu\in\mathcal{B}_{\Sigma}$, there is a
polynomial time algorithm which takes in a real polynomial $Q(z)$ of degree
less than $n$ and produces an integer polynomial $H$ of degree less than $n$
such that $\frac{\log(Q-H)(x)}{n}\leq
U_{\mu}(x)+Cn\cdot\frac{\log(\log(n))^{3}}{\log(n)}$. The run-time is
$O(n^{4})$ where $C$ is dependent only on $\mu$ and $\Sigma$ and assuming $n$
is sufficiently large, we can choose $C=4$.

###### Proof.

Let $Q_{0},\dots,Q_{n-1}$ be the output of our algorithm described in
Corollary 4.5. Call this basis $\mathcal{Q}$. Suppose the $j$-th coordinate of
$[Q]_{\mathcal{Q}}$ is $c_{j}$. Take $\tilde{c}_{j}$ to be the nearest integer
(rounding up if $c_{j}\in\frac{1}{2}\mathbb{Z}$), and let $H$ be the
polynomial so that $[H]_{\mathcal{Q}}=\begin{pmatrix}\tilde{c}_{0}\\\
\vdots\\\ \tilde{c}_{n-1}\end{pmatrix}$. Then $H$ is an integer polynomial
since $[H]_{\mathcal{Q}}$ is an integer linear combination of the integral
polynomials in the basis $\mathcal{Q}$. We know
$Q-H=\sum_{j=1}^{n}(c_{j}-\tilde{c}_{j})Q_{j}$. Thus
$Q-H\leq\frac{n}{2}e^{3n\cdot\frac{\log(\log(n))^{3}}{\log(n)}}e^{nU_{\mu}}$
and so $Q-H\leq e^{\frac{4n\log(\log(n))^{3}}{\log(n)}}e^{nU_{\mu}}$ for large
enough $n$ as desired. For an analysis of the run-time, we note that Schnorr’s
algorithm is the bottleneck. It runs in $O(n^{4})$ time by Corollary 4.5. ∎

###  4.3. Main Algorithm

Here we prove Theorem 1.5. We have already covered the main concepts. By
Corollary 4.5, we can get a basis of integral polynomials in a sub-exponential
multiple of $K_{n}$. If we now apply Theorem 4.6 to
$\frac{1}{4}(p_{M,n,\mu}-x^{n})+\frac{1}{2}$ to get $H(x)$, we could output
$P_{n}(x):=x^{n}+4H(x)-2$. This gives us an Eisenstein polynomial in a sub-
exponential multiple of $K_{n}$. If we simply want a polynomial $P$ with small
$n$-norm, this would be sufficient; however, we have to introduce some extra
steps to ensure $\text{supp}(\mu_{P_{n}})\subset D$ for a given open
$D\supset\Sigma$. The proof will imitate the proof of Theorem 1.4.

###### Proof of Theorem 1.5.

Suppose that $\mu\in\mathcal{B}_{\Sigma}$ is a Hölder measure with exponent
$\delta$, let $D$ be any open set containing $\Sigma$, and choose $\rho>0$ so
that $\Sigma(\rho)\subset D$. Apply Proposition 2.15 to $p_{M,n-m,\mu}$ with
$m$ chosen later to get the polynomial $Q_{m}(x)$. Note that the proof of
Proposition 2.15 explicitly computes $Q_{m}(x)$ in polynomial time. As in the
proof of Theorem 1.4, let $p_{M,n-m,\mu}(x)Q_{m}(x)$ be given by
$x^{n}+\sum_{i=0}^{n-1}a_{i}x^{i}$. Apply Theorem 4.6 to

$h(x):=\frac{1}{4}\left(p_{M,n-m,\mu}(x)Q_{m}(x)-x^{n}-\sum_{i=n-m}^{n-1}a_{i}x^{i}\right)+\frac{1}{2}$

to get an integer polynomial $H(x)$. Finally,

$p(x):=x^{n}+\sum_{i=n-m}^{n-1}a_{i}x^{i}+4H(x)-2$

is an Eisenstein polynomial at $2$. We see that

(47) $|p_{M,n-m,\mu}(x)Q_{m}(x)-p(x)|=|4\left(h(x)-H(x)\right)-2|\leq
e^{Cn\cdot\frac{\log(\log(n))^{3}}{\log(n)}}e^{(n-m)U_{\mu}(x)}$

for some constant $C$ depending only on $\mu$ and $\Sigma$ by Theorem 4.6. As
in the proof of Theorem 1.4, we have

$|p_{M,n-m,\mu}(x)Q_{m}(x)|\geq\kappa^{m}n^{-cn^{1-\frac{\delta}{6}}}e^{(n-m)U_{\mu}(x)}$

for every $x\not\in\Sigma(\rho)$ where $M=\lfloor n^{1/3}\rfloor$. For large
enough $n$, we can choose $m<n$ so that
$m=\lceil\frac{2Cn}{\log(\kappa)}\cdot\frac{\log(\log(n))^{3}}{\log(n)}\rceil$
where we choose $C=4$ as in Theorem 4.6. Then for sufficiently large $n$,

$|p_{M,n-m,\mu}(x)Q_{m}(x)|\geq|p_{M,n-m,\mu}(x)Q_{m}(x)-p(x)|$

for all $x\not\in\Sigma(\rho)$. By Rouché’s theorem, since
$p_{M,n-m,\mu}(x)Q_{m}(x)$ has all its roots in $\Sigma(\rho)$, so does
$p(x)$. Furthermore by (47),

$||p||_{n}\leq
e^{Cn\cdot\frac{\log(\log(n))^{3}}{\log(n)}}+||p_{M,n-m,\mu}(x)Q_{m}(x)||_{n}\leq
e^{\tilde{C}n\cdot\frac{\log(\log(n))^{3}}{\log(n)}}$

for some constant $\tilde{C}$ dependent only on $\Sigma$ and $\mu$. ∎

##  5\. Applications

###  5.1. Numerical Data

We implemented the algorithm described above with some simplifications.111All
examples in section 5 can be found at https://github.com/Bryce-
Orloski/Limiting-Distributions-of-Conjugate-Algebraic-Integers-Applications.
Firstly, we used the LLL-algorithm implemented in PARI/GP named qflll.
Secondly, instead of choosing $z_{ijk}$ as indicated in the paper, we use
reasonable sampling methods as described in each numerical example given. We
also do not apply the extra step of forcing the roots in a $\rho$-neighborhood
of the measure’s support. So these algorithmic outputs do not employ the full
power of the algorithm, but as we see, it still outputs strong results
quickly. All of the examples given in this subsection were computed using
Pennsylvania State University’s ROAR servers and gave the output in at most
two minutes.

The first example we discuss is pictured in Figure 1. The measure depicted is
one in a family which was constructed by Serre [AP08]. In particular, it is
the probability distribution $\mu$ on $\Sigma=[a,b]$ with
$\mu=c\mu_{[a,b]}+(1-c)\nu_{[a,b]}$ where $a=0.1715,$ $b=5.8255,$ $c=0.5004,$
$\mu_{[a,b]}$ is the equilibrium measure on $[a,b]$, and $\nu_{[a,b]}$ is the
pushforward of the equilibrium measure on $[b^{-1},a^{-1}]$ under the map
$z\to 1/z$. We compute the sample points of $\mu$ using by taking the $k^{th}$
sample point to be the inverse distribution of $\mu$ at $k/n$. The algorithm
ran on the ROAR servers for roughly $3$ seconds when asked to compute a degree
100 polynomial (with the sample points pre-computed). The output polynomial
has all real roots with their histogram being displayed in Figure 1(b). The
endpoints and interval width in Figure 1(b) are approximate.

(a) Plot of $|z^{2}-1|=1$

(b) Plotted roots or degree 200 polynomial

(c) Mapping of output roots via $z\mapsto z^{2}-1$

Figure 2. These demonstrate the effectiveness of the algorithm on a
lemniscate.

The second example is pictured in Figure 2. Figure 2(a) depicts the support of
the pull-back of the uniform distribution on the unit circle via
$|z^{2}-1|=1$. We sampled this support via this distribution and ran the
algorithm on 200 samples. It finished in under two minutes and a plot of its
roots are given in Figure 2(b). We want this to sample the pull-back of the
uniform distribution on the circle. To see that this is well sampled, Figure
2(c) plots the image of these roots under the map $z\mapsto z^{2}-1$ and we
see that it roughly approximates the uniform distribution on the unit circle.

Our last example is depicted in Figure 3. Here we sampled the annulus
$1\leq|z|\leq 2$ with 100 complex numbers by sampling 50 using the Monte Carlo
method and taking their complex conjugates. This is shown in Figure 3(a).
Using these samples in our algorithm, it returned a degree 100 monic,
integral, irreducible polynomial in under two minutes whose roots are plotted
in Figure 3(b).

(a) Plot of 50 sampled points from the annulus $1\leq|z|\leq 2$ and their
conjugates

(b) Plotted roots of 100 degree polynomial from algorithm given the samples as
input

Figure 3. The left shows 50 points sampled uniformly from $1\leq|z|\leq 2$ and
right plots the roots of the output of the algorithm which tried to
approximate the distribution of the left plot.

Notice how we cannot provably show the desired convergence with the LLL-
algorithm, and even with Schnorr’s algorithm, the convergence we prove is very
slow. However, here we are applying the LLL-algorithm and in each of Figures
1, 2(c), and 3, we see that the plots of the roots of the polynomial produced
by the algorithm are very close to the plots of the corresponding samples.

###  5.2. Applications to Abelian varieties

Recently, Shankar and Tsimerman [ST18], [ST21] conjectured that for every
$g\geq 4$, every Abelian variety defined over $\overline{\mathbb{F}_{q}}$ is
isogenous to a Jacobian of a curve defined over $\overline{\mathbb{F}_{q}}.$
Given a finite field $\mathbb{F}_{q}$ and any arbitrary large extension
$\mathbb{F}_{q^{n}}$, we prove that there are infinitely many abelian
varieties over $\mathbb{F}_{q}$ which are not isogenous to the Jacobian of any
curve over $\mathbb{F}_{q^{n}}$. We use Honda Tate theory and construct an
arithmetic measure with some conditions on its moments. First, we introduce
our moment conditions which are motivated by the work of Tsfasman and Vladut
[TV97].

Let $X$ be a finite curve defined over $\mathbb{F}_{q}$. Let
$N_{r}(X):=\\#X(\mathbb{F}_{q^{r}})$ be the number of $\mathbb{F}_{q^{r}}$ of
$X.$ By Weil’s formula, we have

(48) $0\leq N_{r}(X)=q^{r}+1-2q^{r/2}\sum_{j=1}^{g}\cos(r2\pi\theta_{j}),$

where $\sqrt{q}e^{2\pi i\theta_{i}}$ are Frobenius eigenvalues. We have

$N_{r}(X)=\sum_{d|r}dM_{d}(X),$

where $M_{d}(X)$ is the number of points with degree $d.$ By Mobius inversion
formula,

$M_{r}(X)=\frac{1}{r}\sum_{d|r}\mu(d)N_{\frac{r}{d}}(X).$

In particular, we have

$\sum_{d|r}\mu(d)N_{\frac{r}{d}}(X)\geq 0$

for every $r\geq 0.$

Let $S_{\sqrt{\alpha}}:=\left\\{\sqrt{\alpha}e^{2\pi
i\theta}:\theta\in[0,1)\right\\}$ be the circle of radius $\sqrt{\alpha}$
centered at the origin, and defined the following probability measure on
$S_{\sqrt{\alpha}}$

$d\mu_{q}:=\left(1+\sum_{k=1}^{\infty}\frac{c}{k^{2}}\cos(k2\pi\theta)\right)d\theta.$

where $c=\frac{6}{\pi^{2}10}.$ Note that

(49) $0\leq 0.9d\theta\leq d\mu_{q}\leq 1.1d\theta.$

######  Proposition 5.1.

Suppose that $\alpha>1.65$ then $U_{\mu_{\alpha}}(z)\geq 0$ for any
$z\in\mathbb{C}.$ As a result, $d\mu_{\alpha}$ is an arithmetic measure.

###### Proof.

It is enough to show that $U_{\mu_{\alpha}}(z)\geq 0$ for any
$z\in\mathbb{C}.$ Suppose that $|z|\geq 2\sqrt{\alpha}$, then

$U_{\mu_{\alpha}}(z)=\int_{S_{\sqrt{\alpha}}}\log|z-\sqrt{\alpha}e^{2\pi
i\theta}|d\mu_{\alpha}\geq\log\sqrt{\alpha}\geq 0.$

Otherwise, suppose that $|z|<2\sqrt{\alpha}.$ By (49), we have

$\begin{split}U_{\mu_{\alpha}}(z)&=\int_{S_{\sqrt{\alpha}}}\log|z-\sqrt{\alpha}e^{2\pi
i\theta}|d\mu_{\alpha}\\\ &=\int_{|z-\sqrt{\alpha}e^{2\pi
i\theta}|<1}\log|z-\sqrt{\alpha}e^{2\pi
i\theta}|d\mu_{\alpha}+\int_{|z-\sqrt{\alpha}e^{2\pi
i\theta}|>1}\log|z-\sqrt{\alpha}e^{2\pi i\theta}|d\mu_{\alpha}\\\ &\geq
1.1\int_{|z-\sqrt{\alpha}e^{2\pi i\theta}|<1}\log|z-\sqrt{\alpha}e^{2\pi
i\theta}|d\theta+0.9\int_{|z-\sqrt{\alpha}e^{2\pi
i\theta}|>1}\log|z-\sqrt{\alpha}e^{2\pi i\theta}|d\theta\\\
&=1.1\int_{S_{\sqrt{\alpha}}}\log|z-\sqrt{\alpha}e^{2\pi
i\theta}|d\theta-0.2\int_{|z-\sqrt{\alpha}e^{2\pi
i\theta}|>1}\log|z-\sqrt{\alpha}e^{2\pi i\theta}|d\theta\\\ &\geq
1.1\log(\sqrt{\alpha})-0.2\log(3\sqrt{\alpha})>0.\end{split}$

This completes the proof of our propositions. ∎

######  Corollary 5.2.

There exists a sequence $\\{A_{g}\\}$ of abelian varities over
$\mathbb{F}_{q}$ such that $\dim(A_{g})=g$ and their Frobenious eigenvalues
equidistributes with $\mu_{q}.$

###### Proof.

Let $\alpha_{n}:=\sqrt{q}-\frac{1}{10n},$
$\Sigma_{n}:=[-2\alpha_{n},2\alpha_{n}]$ and
$h_{n}(z):=z+\frac{\alpha_{n}}{z}.$ Note that $h_{n}$ is a conformal map with
fixed point at infinity and derivative 1 infinity that sends
$\mathbb{C}\backslash S_{\alpha_{n}}$ to the $\mathbb{C}\backslash\Sigma_{n}$.
Let $h_{n}d\mu_{\alpha_{n}}$ be the push-forward of $\mu_{\alpha_{n}}$ by
$h_{n}.$ By conformal in-variance of the potential function and Proposition
5.1, $h_{n}d\mu_{\alpha_{n}}$ has a positive potential function and hence
$h_{n}d\mu_{\alpha_{n}}$ is arithmetic. It follows from Theorem 1.2 that there
exists a sequence of irreducible polynomial $\\{p_{m}\\}$ with real roots
contained in $\Sigma_{n}(\frac{1}{100n})\subset[2\sqrt{q},-2\sqrt{q}]$ with
root distribution converging to $h_{n}d\mu_{\alpha_{n}}.$ Let

$f_{m}(x):=x^{\deg{p_{m}}}p_{m}(x+\frac{q}{x}).$

Note that $f_{m}(x)$ has all its roots on $S_{\sqrt{q}}.$ Now by a diagonal
argument and letting $n\to\infty$ and taking $m$ large enough it follows that
there exists a sequecen of $\\{A_{g}\\}$ of abelian varities over
$\mathbb{F}_{q}$ such that $\dim(A_{g})=g$ and their Frobenious eigenvalues
equidistributes with $\mu_{q}.$

∎

######  Theorem 5.3.

Let $\\{A_{g}\\}$ be any family of abelian varieties over $\mathbb{F}_{q}$
such that $\dim(A_{g})=g$ and their Frobenius eigenvalues distribution
converges to $\mu_{q}.$ Given any integer $r\geq 0$, there exists $N$ such
that if $g\geq N$ then $A_{g}$ is not isogenous to the Jacobian of any curve
over $\mathbb{F}_{q^{r}}$.

###### Proof.

Suppose the contrary that there exists a sub-sequence $\\{A_{g_{i}}\\}$ of
abelian varieties over $\mathbb{F}_{q}$ such that their Frobenius eigenvalues
equidistribute with $\mu_{q}$ and also they are isogenous to Jacobian of
curves $\\{X_{g_{i}}\\}$ over $\mathbb{F}_{q^{r}}$ for some $r\geq 0.$ By
(48), it follows that

$0\leq\frac{N_{r}(X_{g_{i}})}{g_{i}}=\frac{q^{r}+1-2q^{r/2}\sum_{j=1}^{g_{i}}\cos(r2\pi\theta_{j})}{g_{i}}.$

By taking the limit of the above as $g_{i}\to\infty$, we obtain

$0\leq-2q^{r/2}\lim_{g_{i}\to\infty}\frac{\sum_{j=1}^{g_{i}}\cos(r2\pi\theta_{j})}{g_{i}}=-2q^{r/2}\int\cos(r2\pi\theta)d\mu_{q}=-q^{r/2}\frac{c}{r^{2}}<0,$

which is a contradiction. This completes the proof of our theorem. ∎

As was mentioned in the introduction, we constructed a polynomial using the
algorithm described in section 4 of the paper. Like the other results in
section 5, this was implemented with the LLL-algorithm and without using the
construction that forces the roots to lie inside a desired support. The
highest order terms were

$x^{290}-28x^{289}-{484}x^{288}+20784x^{287}+\cdots.$

The largest coefficient of this polynomial has 105 digits. The roots all lie
inside $[-2\sqrt{3},2\sqrt{3}]\subset\mathbb{R}$ and so by Honda-Tate theory,
it represents an abelian variety over $\mathbb{F}_{3}$. Computing $N_{2}$ of
this variety, we get $-2$. Thus it is not the Jacobian of any curve over
$\mathbb{F}_{9}$.

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