# Uniform algebras and distinguished varieties

Sushil Gorai and Golam Mostafa Mondal  Department of Mathematics and
Statistics, Indian Institute of Science Education and Research Kolkata,
Mohanpur – 741 246<EMAIL_ADDRESS><EMAIL_ADDRESS>Department of Mathematics, Indian Institute of Science Education and Research
Pune, Pune – 411 008<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this article, we point out the connections between the distinguished
varieties introduced by Agler and McCarthy with certain uniform algebras on
bidisc studied by Samuelsson and Wold. We also prove analogues of Samuelsson-
Wold result for the domains in $\mathbb{C}^{2}$ that are the images of the
bidisc under certain proper polynomial map on $\mathbb{C}^{2}$. We also give a
description of polynomial convex hull of graph of anti-holomorphic polynomial
over the distinguished boundary of such domains. We mention the case for the
symmetrized bidisc as an example.

###### Key words and phrases:

Polynomial convexity; Uniform approximation; Wermer maximality theorem;
Symmetrized bidisc; Distinguished variety

###### 2020 Mathematics Subject Classification:

Primary: 32E30, 32E20; Secondary: 47A25

##  1\. Introduction

This article connects the theory of distinguished varieties–a well-explored
topic in operator theory, with the notions of uniform algebra generated by
holomorphic polynomials and certain pluriharmonic functions. The latter one is
also a very well-studied object in several complex variables. In particular,
we observe that the failure of uniform approximation for all continuous
functions on the distinguished boundary of certain domains in $\mathbb{C}^{2}$
by elements of holomorphic polynomials in $z_{1}$ and $z_{2},$ and some
pluriharmonic functions is the presence of certain distinguished variety in
the domain where the pluriharmonic functions become holomorphic. Before making
these precise, let us briefly mention the theory of distinguished varieties
and the theory of uniform algebras one by one.

In a seminal paper [4], Agler and McCarthy introduced the notion of
distinguished variety in the bidisc $\mathbb{D}^{2}$ as follows: A non-empty
set $V$ in $\mathbb{C}^{2}$ is said to be a distinguished variety if there
exists a polynomial $p$ in $\mathbb{C}[z,w]$ such that

$V=\\{(z,w)\in\mathbb{D}^{2}:p(z,w)=0\\}$

and such that

$\displaystyle\overline{V}\cap\partial\mathbb{D}^{2}=\overline{V}\cap\mathbb{T}^{2}.$
(1.1)

Here, $\partial\mathbb{D}^{2}$ represents the boundary of the
$\mathbb{D}^{2}$, and $\mathbb{T}^{2}$ is the distinguished boundary of
$\mathbb{D}^{2}$. A distinguished variety is an algebraic variety that exits
the bidisc through the distinguished boundary. The set $\overline{V}$ is the
closure of $V$ within $\overline{\mathbb{D}}^{2}$. We will use $\partial V$ to
denote the set described by (1.1). From a topological standpoint, $\partial V$
represents the boundary of $V$ within the zero set of $p$ instead of
encompassing the entirety of $\mathbb{C}^{2}$.

One of the fundamental results in operator theory, known as Andô’s inequality
[5], establishes that when $T_{1}$ and $T_{2}$ are commuting operators, each
with a norm not exceeding 1, the following inequality holds for any two-
variable polynomial $p$:

$\displaystyle\|p(T_{1},T_{2})\|\leq\|p\|_{\mathbb{D}^{2}}$ (1.2)

holds. Agler and McCarthy [4, Theorem 3.1] gave the following improvement of
the inequality (1.2): if $T_{1}$ and $T_{2}$ are matrices, then

$\|p(T_{1},T_{2})\|\leq\|p\|_{V},$

where $V$ is a distinguished variety that depends on $T_{1}$ and $T_{2}$.
Additionally, in [4, Theorem 1.12], the authors have shown that all
distinguished varieties in the bidisc can be expressed as

$\\{(z,w)\in\mathbb{D}^{2}:\det(\Psi(z)-wI)=0\\},$

where $\Psi$ is an analytic matrix-valued function defined on the disk, and it
is unitary on $\partial\mathbb{D}$. A similar description of distinguished
varieties in the symmetrized bidisc is given in [24].

Consider a compact subset $K$ of $\mathbb{C}^{n}$. The space of all continuous
complex-valued functions on $K$ is denoted as $\mathcal{C}(K)$, equipped with
the norm $|g|=\sup_{K}|g(z)|$. We denote the closure of the set of polynomials
in $\mathcal{C}(K)$ as $\mathcal{P}(K)$. For a collection of functions
$g_{1},\ldots,g_{N}\in\mathcal{C}(K)$, we use $[g_{1},\ldots,g_{N};K]$ to
represent the uniform algebra generated by $g_{1},\ldots,g_{N}$ on $K$. We
define the set $X={(g_{1}(z),\ldots,g_{N}(z)):z\in K}$ associated with the
uniform algebra $[g_{1},\ldots,g_{N};K]$. Using the Stone and Weierstrass
theorem, we assert that

$\displaystyle[g,\cdots,g_{N};K]=\mathcal{C}(K)$

if and only if $\mathcal{P}(X)=\mathcal{C}(X)$ and the generators
$g_{1},\ldots,g_{N}$ separate points on $K$.

If we consider $\mathcal{P}(K)$ and $\mathcal{C}(K)$ as Banach algebras, the
equality $\mathcal{P}(K)=\mathcal{C}(K)$ implies the equality of their
corresponding maximal ideal spaces. The maximal ideal space of
$\mathcal{C}(K)$ corresponds to $K$, and that of $\mathcal{P}(K)$ corresponds
to $\widehat{K}$, where $\widehat{K}$ is the polynomial convex hull of $K$
(see [13]). Here, the polynomial convex hull of $K$ is denoted as
$\widehat{K}$ and is defined as follows:

$\displaystyle\widehat{K}:=\left\\{\alpha\in\mathbb{C}^{n}:|p(\alpha)|\leq\max_{K}|p|~{}~{}\forall
p\in\mathbb{C}[z_{1},z_{2},\cdots,z_{n}]\right\\}.$

We say $K$ is polynomially convex when $\widehat{K}=K$. Thus, polynomial
convexity serves as a necessary condition for all compacts $K$ where
$\mathcal{P}(K)=\mathcal{C}(K)$ holds.

Recall that an _analytic disc_ in $\mathbb{C}^{n}$ is a holomorphic map
$\phi:\mathbb{D}\rightarrow\mathbb{C}^{n}$ which is non-constant and
continuous on $\overline{\mathbb{D}}$. Let $K\subset\mathbb{C}^{n}.$ We say an
analytic disc $\phi$ is present in $K$ if $\phi(\mathbb{D})\subset K.$ In view
of Lavrentiev’s [22] result, if $K$ be a compact subset of $\mathbb{C},$ then
$\mathscr{P}(K)=\mathcal{C}(K)$ if and only if $K=\widehat{K}$ and there does
not exist any analytic disc in $K.$ But this is far from being a sufficient
condition for polynomially convex compacts in higher dimensions. This article
discusses some results in which the presence of an analytic disc is the only
obstruction for polynomially convex compact $K$ for which
$\mathscr{P}(K)=\mathcal{C}(K).$ We now talk about the Wermer maximality
theorem. Let $\mathbb{T}^{1}$ be the unit circle in the complex plane and
$\mathcal{C}(\mathbb{T}^{1})$ be the set of all continuous complex-valued
functions on $\mathbb{T}^{1}.$ Let $\mathcal{A}$ denote the set of all
$f\in\mathcal{C}(\mathbb{T}^{1})$ which are boundary values of functions
holomorphic on $\mathbb{D}$ and continuous on $\overline{\mathbb{D}}.$ In
[23], the following question was asked:

$\textit{if }g\in\mathcal{C}(\mathbb{T}^{1})\setminus\mathcal{A},\textit{ does
the closed algebra generated by }g\textit{ and }\mathcal{A}\textit{ equal
}\mathcal{C}(\mathbb{T}^{1})?$

In [23], it is shown that if $g$ is real-valued or if $g$ satisfies a
Lipschitz condition, the algebra generated by $g$ and $\mathcal{A}$ equals
$\mathcal{C}(\mathbb{T}^{1}).$ Wermer [33] settled this question by proving
the following:

######  Result 1.1 (Wermer).

If $\mathcal{B}$ is any closed subalgebra of $\mathcal{C}(\mathbb{T}^{1})$
with $\mathcal{A}\subset\mathcal{B}\subset\mathcal{C}(\mathbb{T}^{1}).$ Then
either $\mathcal{A}=\mathcal{B}$ or $\mathcal{B}=\mathcal{C}(\mathbb{T}^{1}).$

A uniform algebra $\mathcal{U}$ defined on a compact subset $K$ is said to be
a maximal subalgebra of $\mathcal{C}(K)$ if, for any other subalgebra
$\mathcal{B}$ of $\mathcal{C}(K)$ such that
$\mathcal{U}\subset\mathcal{B}\subset\mathcal{C}(K)$, it holds that either
$\mathcal{U}=\mathcal{B}$ or $\mathcal{B}=\mathcal{C}(K)$. Result 1.1 is known
as the Wermer Maximality Theorem. A similar related result due to Wermer is
the following [34]: Let $g\in C^{1}(\overline{\mathbb{D}}).$ Assume that the
graph ${\sf Gr}_{\overline{\mathbb{D}}}(g)\subset\mathbb{C}^{2}$ of $g$ is
polynomially convex. Let $E:=\\{z\in\overline{\mathbb{D}}:\frac{\partial
g}{\partial\bar{z}}(z)=0\\}.$ Then

$\displaystyle[z,g;\overline{\mathbb{D}}]=\\{f\in
C(\overline{\mathbb{D}}):f|_{E}\in\mathcal{O}(E)\\}.$

It is natural to ask the version of these results to the higher dimensions.
The question in the higher dimension has no clear answer like the Wermer
maximality theorem. The natural object is to generalization of the second
result of Wermer, even when considering the algebra generated by polynomials
and a pluriharmonic function. For a domain $\Omega\subset\mathbb{C}^{n},$ let
$PH(\Omega)$ denote the class of all pluriharmonic function on $\Omega.$ The
works of Čirka [32], Izzo [14, 15], Samuelsson and Wold [28], and Izzo,
Samuelsson, and Wold [16] focused on the study of uniform algebras generated
by holomorphic and pluriharmonic functions in higher dimensions. In [28],
Samuelsson and Wold [28] proved the following results in the case of the
bidisc $\mathbb{D}^{2}.$

######  Result 1.2 (Samuelsson-Wold).

Let $h_{j}\in
PH(\mathbb{D}^{2})\cap\mathcal{C}^{1}(\overline{\mathbb{D}}^{2})$ for
$j=1,\cdots,N.$ Then either there exists a holomorphic disc in
$\overline{\mathbb{D}}^{2}$ where all $h_{j}$’s are holomorphic, or
$[z_{1},z_{2},h_{1},\cdots,h_{N};\overline{\mathbb{D}}^{2}]=\mathcal{C}(\overline{\mathbb{D}}^{2}).$

The following result can be thought of an analogue of the Wermer maximality
theorem in case of the bidisc.

######  Result 1.3 (Samuelsson-Wold).

Let $f_{j}\in\mathcal{C}(\mathbb{T}^{2})$ for $j=1,\cdots,N$ with $N\geq 1$,
and assume that each $f_{j}$ extends to a pluriharmonic function on
$\mathbb{D}^{2}$. Then either
$[z_{1},z_{2},f_{1},\cdots,f_{N};\mathbb{T}^{2}]=\mathcal{C}(\mathbb{T}^{2})$,
or there exists a non-trivial algebraic variety $Z\subset\mathbb{C}^{2}$ with
$\overline{V}\setminus V\subset\mathbb{T}^{2},$ and the pluriharmonic
extensions of the $f_{j}$’s are holomorphic on $Z,$ where
$V=Z\cap(\overline{\mathbb{D}^{2}}\setminus\mathbb{T}^{2}).$

######  Remark 1.4.

In Result 1.3 if not all of the functions $f_{1},\dots,f_{N}$ is holomorphic
in any analytic disc that lies in $\partial\mathbb{D}^{2}$ and
$[z_{1},z_{2},f_{1},\cdots,f_{N};\mathbb{T}^{2}]\neq\mathcal{C}(\mathbb{T}^{2})$,
then the algebraic variety that exists is a distinguished variety. As
mentioned earlier, by a result of Agler and McCarthy [4], every distinguished
variety in the bidisc is of the form
$\\{(z,w)\in\mathbb{D}^{2}:\det(\Psi(z)-wI)=0\\}$ for some matrix-valued
holomorphic function $\Psi$ on $\mathbb{D}^{2}$. Therefore, the variety that
exists in Result 1.3 is also of the above mentioned determinant form. We do
not know what connections are there with the matrix-valued funtion $\Psi$ in
[4] and the pluriharmonic functions in Result 1.3.

######  Remark 1.5.

It might occur that the variety in Result 1.3 appears in the boundary of the
bidisc. In this case, the variety is not a distinguished variety, but such
variety can also be explained from the operator theoretic point of view from a
result due to Das and Sarkar [11, Theorem 4.3]. From the proof of Result 1.3
it is clear that the form of such variety is $\\{\lambda\\}\times\mathbb{D}$
or $\mathbb{D}\times\\{\lambda\\}$ for some $\lambda\in\partial\mathbb{D}$,
which matches with the description in [11, Theorem 4.3].

Consider the domain $\Omega=\phi(\mathbb{D}^{2})$ in $\mathbb{C}^{2}$ and we
note that the distinguished boundary of $\Omega$ for the algebra
$\mathcal{A}(\Omega)$ is $\Gamma_{\Omega}=\phi(\mathbb{T}^{2}).$ We prove the
following generalization of Result 1.2 and Result 1.3 for the above domain.

######  Theorem 1.6.

Let $h_{j}\in PH(\Omega)\cap\mathcal{C}^{1}(\overline{\Omega})$ for
$j=1,\cdots,N,$ and
$\phi^{-1}(\overline{\Omega})\subset\overline{\mathbb{D}}^{2}.$ Then, either
there exists a holomorphic disc in $\overline{\Omega}$ where all $h_{j}$’s are
holomorphic, or
$[z_{1},z_{2},h_{1},\cdots,h_{N};\overline{\Omega}]=\mathcal{C}(\overline{\Omega}).$

######  Theorem 1.7.

Let $f_{j}\in\mathcal{C}(\Gamma_{\Omega})$ for $j=1,\cdots,N,N\geq 1$ and
assume that each $f_{j}$ extends to a pluriharmonic function on $\Omega.$ If
$\phi^{-1}(\Gamma_{\Omega})\subset\mathbb{T}^{2}$. If $f_{j}$ is not
holomorphic on any analytic disc present in the boundary $\partial\Omega$ for
at least one $j$, then either

$\displaystyle[z_{1},z_{2},f_{1},\cdots,f_{N};\Gamma_{\Omega}]=\mathcal{C}(\Gamma_{\Omega}),$

or there exists a distinguished variety $V$ in $\Omega$ such that the
pluriharmonic extensions of the $f_{j}$’s are holomorphic on $V.$

As a corollary we can extend Result 1.2 and Result 1.3 to the symmetrized
bidisc. Recall that the symmetrized bidisc $\mathbb{G}_{2}$ is the image of
the bidisc under the symmetrization map
$\Pi:(z_{1},z_{2})\to(z_{1}+z_{2},z_{1}z_{2})$ i.e.,

$\displaystyle\mathbb{G}_{2}=\\{(z_{1}+z_{2},z_{1}z_{2}):|z_{1}|<1,|z_{2}|<1\\}.$

Since
$\Pi^{-1}(\Pi(\overline{\mathbb{D}}^{2}))=\Pi^{-1}(\overline{\mathbb{G}}_{2})=\overline{\mathbb{D}}^{2},$
by using Result 2.1, we get that $\overline{\mathbb{G}}_{2}$ is polynomially
convex. If $f:\mathbb{G}_{2}\to\mathbb{C}$ is a holomorphic function on
$\mathbb{G}_{2},$ then $f\circ\Pi:\mathbb{D}^{2}\to\mathbb{C}$ is a symmetric
function on $\mathbb{D}^{2}.$ Therefore, if
$\mathcal{A}(\overline{\mathbb{G}}_{2})$ is the algebra of functions that are
holomorphic on $\mathbb{G}_{2}$ and continuous on $\overline{\mathbb{G}}_{2},$
then the distinguished boundary $\Gamma_{{\mathbb{G}}_{2}}$ of
$\mathbb{G}_{2}$ is the image $\Pi(\mathbb{T}^{2})$ of the torus
$\mathbb{T}^{2}$ (the distinguished boundary of $\mathbb{D}^{2}$). Since
$\mathbb{G}_{2}$ is neither convex (not even biholomorphic to any convex
domain [10]) nor smooth (not even the Lipschitz domain [8]), and hence, many
results in the theory of several complex variables does not apply to
$\mathbb{G}_{2}.$ Several authors have studied this domain over the last three
decades, and it has shown to be a domain with a highly rich complex geometry
and function theory: see, among many other articles, [31, 20, 25, 17, 12, 10,
3, 2, 1, 6, 29].

There are significant similarities and contrasts between its geometry and
function theory and those of the bidisc. Here we observe that Result 1.2 and
Result 1.3 continues to hold if the bidisc is replaced by the symmetrized
bidisc. More precisely:

######  Corollary 1.8.

Let $h_{j}\in
PH(\mathbb{G}_{2})\cap\mathcal{C}^{1}(\overline{\mathbb{G}}_{2})$ for
$j=1,\cdots,N.$ Then either there exists a holomorphic disc in
$\overline{\mathbb{G}}_{2}$ where all $h_{j}$’s are holomorphic, or

$[z_{1},z_{2},h_{1},\cdots,h_{N};\overline{\mathbb{G}}_{2}]=\mathcal{C}(\overline{\mathbb{G}}_{2}).$

######  Corollary 1.9.

Let $f_{j}\in\mathcal{C}(\Gamma_{\mathbb{G}_{2}})$ for $j=1,\cdots,N,N\geq 1$
and assume that each $f_{j}$ extends to a pluriharmonic function on
$\mathbb{G}_{2}.$ If $f_{j}$ is not holomorphic on any analytic disc present
in the boundary $\partial\mathbb{G}_{2}$ for at least one $j$, then either

$\displaystyle[z_{1},z_{2},f_{1},\cdots,f_{N};\Gamma_{\mathbb{G}_{2}}]=\mathcal{C}(\Gamma_{\mathbb{G}_{2}}),$

or there exists a distinguished variety $V$ in $\mathbb{G}_{2}$ such that the
pluriharmonic extensions of the $f_{j}$’s are holomorphic on $V.$

######  Remark 1.10.

In view of a result by Pal and Shalit [24], we see that the variety that
appears in Corollary 1.9 has the form of a zero set of a certain determinant.
However, we do not know whether a similar type of determinant form can also
given for the distinguished varieties that appear in Theorem 1.7.

##  2\. Technical Results

In this section, we provide some known results and some preliminary lemmas
that will be utilized to prove our results.

######  Result 2.1 ([30]).

If $F:\mathbb{C}^{n}\to\mathbb{C}^{n}$ is a proper holomorphic map, and if
$K\subset\mathbb{C}^{n}$ is a compact set, then the set $K$ is polynomially
convex if and only if the set $F^{-1}(K)$ is polynomially convex, and
$\mathcal{P}(K)=\mathcal{C}(K)$ if and only if
$\mathcal{P}(F^{-1}(K))=\mathcal{C}(F^{-1}(K)).$

######  Result 2.2 (Remmert Proper Mapping theorem [26, 27]).

Let $M,N$ be complex spaces, and $f:M\to N$ is a proper holomorphic map. If
$Z$ is an analytic subvariety in $M$ then $f(Z)$ is also an analytic
subvariety in $N.$ Moreover, if $Z$ is irreducible then $f(Z)$ is also
irreducible subvariety of $N.$

The following result is from the book [9, Page 29].

######  Result 2.3.

(Chirka) Let
$\Omega_{1}\subset\mathbb{C}^{p},\Omega_{2}\subset\mathbb{C}^{m},$ are open
subsets such that $\Omega=\Omega_{1}\times\Omega_{2},$ $p+m=n,$ and
$\textit{proj}_{1}:(z,w)\to z.$ Let $V$ be an analytic subset in $\Omega$ such
that $\textit{proj}_{1}:V\to\Omega_{1}$ is a proper map. Then
$\textit{proj}_{1}(V)$ is an analytic subset in $\Omega_{1}.$ Moreover, if
$\Omega=\mathbb{C}^{n},$ $\Omega_{1}=\mathbb{C}^{p},$ and $V$ is an algebraic
subset in $\mathbb{C}^{n},$ then $\textit{proj}_{1}(V)$ is also an algebraic
subset in $\mathbb{C}^{p}.$

The following lemma is well-known to experts. Since we have not found a
explicit mention of this lemma in the literature, we decided to put it here
for completeness.

######  Lemma 2.4.

Let $\Psi:\mathbb{C}^{n}\to\mathbb{C}^{n}$ be a proper polynomial map. Let $Z$
be an algebraic variety in $\mathbb{C}^{n},$ then $\Psi(Z)$ is also an
algebraic variety in $\mathbb{C}^{n}.$

###### Proof.

Consider the algebraic variety $V=\\{(\Psi(z),z):z\in Z\\}$ in
$\mathbb{C}^{n}\times\mathbb{C}^{n}$ and
$\Omega_{1}=\Omega_{2}=\mathbb{C}^{n}.$ We now show that that
$\textit{proj}_{1}:V\to\Omega_{1}$ is a proper map. Let
$K\subset\mathbb{C}^{n}$ be a compact subset of $\mathbb{C}^{n}.$ Then
$\textit{proj}_{1}^{-1}\\{K\\}=(K\times\mathbb{C}^{n})\cap V=\\{(\xi,\eta)\in
K\times\mathbb{C}^{n}:(\xi,\eta)\in V\\}=\\{(\Psi(\eta),\eta)\in
K\times\mathbb{C}^{n}:\eta\in Z\\}=$compact (since $\Psi$ is a proper map).
Therefore, $\textit{proj}_{1}:V\to\Omega_{1}$ is a proper map. Hence, by
Result 2.3, we conclude that $\textit{proj}_{1}(V)=\Psi(Z)$ is an algebraic
variety. ∎

######  Remark 2.5.

The case $\Psi=\Pi$ is available in [24, Lemma 3.1].

Let $\Psi:\mathbb{C}^{n}\to\mathbb{C}^{n}$ be a proper holomorphic polynomial
map. Let $\Omega:=\Psi(\mathbb{D}^{n})$ be a domain such that
$\Psi^{-1}(\Psi(\mathbb{D}^{n}))\subset\mathbb{D}^{n},$
$\Psi^{-1}(\Psi(\partial\mathbb{D}^{n}))\subset\partial\mathbb{D}^{n},$ and
$\Psi^{-1}(\Psi(\mathbb{T}^{n}))\subset\mathbb{T}^{n}.$ The following lemma
illustrates that every distinguished variety in $\Omega$ can be derived from a
distinguished variety in $\mathbb{D}^{n}$.

######  Lemma 2.6.

Let $Z\subset\Omega$. Then $Z$ is a distinguished variety in $\Omega$ if and
only if there is a distinguished variety $V$ in $\mathbb{D}^{n}$ such that
$\Psi(V)=Z.$

###### Proof.

Given that $\Psi$ is a proper map, it implies that $\Psi$ is onto, and
therefore, $\Psi(\Psi^{-1}(Z))=Z$. Additionally, it can be easily demonstrated
that $\Psi^{-1}(Z)$ is an algebraic variety. Let us define $V:=\Psi^{-1}(Z)$.
Now, we need to prove the following: $V\cap\partial\mathbb{D}^{n}\subset
V\cap\mathbb{T}^{n}$.

Consider an element $\alpha\in V\cap\partial\mathbb{D}^{n}.$ This implies that
$\alpha\in\Psi^{-1}(Z)\cap\partial\mathbb{D}^{n}.$ Hence, we have
$\Psi(\alpha)\in Z\cap\Psi(\partial\mathbb{D}^{n})$. Since $Z$ is a
distinguished variety, we can conclude that $\Psi(\alpha)\in
Z\cap\Psi(\partial\mathbb{T}^{n})$. Consequently, we can deduce that $\alpha$
lies in
$\Psi^{-1}(Z\cap\Psi(\mathbb{T}^{n}))=\Psi^{-1}(Z)\cap\Psi^{-1}(\Psi(\mathbb{T}^{n}))$.
By our assumption, together with this, we get that
$V\cap\partial\mathbb{D}^{n}\subset V\cap\mathbb{T}^{n}$.

Conversely, let us assume that $V$ is a subset of $\mathbb{D}^{n}$ and is a
distinguished variety. By using Lemma 2.4, we can conclude that $\Psi(V)$ is
an algebraic variety in $\Omega$. Now, we claim that $Z=\Psi(V)$ is a
distinguished variety in $\Omega$. Suppose $\alpha\in
Z\cap\Psi(\partial\mathbb{D}^{n})=\Psi(V)\cap\Psi(\partial\mathbb{D}^{n}).$ We
need to show that $\alpha$ also lies in $\Psi(\mathbb{T}^{n})$. Since
$\alpha\in Z\cap\Psi(\partial\mathbb{D}^{n})$, there exist $\eta_{1}\in V$ and
$\eta_{2}\in\partial\mathbb{D}^{n}$ such that
$\Psi(\eta_{1})=\Psi(\eta_{2})=\alpha$. Consequently, $\eta_{2}$ belongs to
$\Psi^{-1}(\Psi(\partial\mathbb{D}^{n}))$, which is a subset of
$\partial\mathbb{D}^{n}$. Thus, we have $\eta_{2}\in
V\cap\partial\mathbb{D}^{n}$, and as a result,
$\Psi(\eta_{2})\in\Psi(V\cap\partial\mathbb{D}^{n})$. This implies that
$\alpha$ lies in $\Psi(V\cap\mathbb{T}^{n})$. ∎

######  Remark 2.7.

The case $\Omega=\mathbb{G}_{2}$ is available in [24, Lemma 3.1].

######  Lemma 2.8.

Let $g:G\subset\mathbb{C}^{N}\to\mathbb{C}^{N}$ be a proper holomorphic
mapping and $q:g(G)\to\mathbb{C}$ be a continuous function. If $q\circ
g:G\to\mathbb{C}$ is holomorphic, then $q$ is holomorphic.

###### Proof.

Let us define $\Omega:=g(G).$ Since $g$ is proper holomorphic, $\Omega$ is
open. First, we assume $z\in G$ and $\det d{g}(z)\neq 0,$ where $\det dg(z)$
is the determinant of the complex Jacobian matrix of $g$ at $z.$ Then there
exists a neighborhood $V$ of $z$ and a neighborhood $W$ of $g(z)$ such that
$g^{-1}:W\to V$ is holomorphic. Therefore, $q\circ g\circ g^{-1}=q$ is
holomorphic at $g(z).$ Next, we define $X:=\\{z\in G:\det d{g}(z)=0\\}.$
Hence, $q$ is holomorphic on $\Omega\setminus g(X).$ Clearly, $X$ is an
analytic variety with $\dim_{\mathbb{C}}X\leq(N-1).$ Since $g$ is proper
holomorphic mapping, by Result 2.2, $g(X)$ is also an analytic variety in
$\Omega.$ Since $q$ is continuous on $\Omega$ and holomorphic on
$\Omega\setminus g(X),$ by Riemann’s removable singularity theorem, we can say
that $q$ is holomorphic on $\Omega.$ ∎

Let $\Psi:\mathbb{C}^{n}\to\mathbb{C}^{n}$ be a proper holomorphic map. Let
$\Omega:=\Psi(\mathbb{D}^{n})$ be a domain such that
$\Psi^{-1}(\Psi(\mathbb{D}^{n}))\subset\mathbb{D}^{n},$
$\Psi^{-1}(\Psi(\partial\mathbb{D}^{n}))\subset\partial\mathbb{D}^{n},$ and
$\Psi^{-1}(\Psi(\mathbb{T}^{n}))\subset\mathbb{T}^{n}.$ We denote the
distinguished boundary of $\Omega$ for the algebra $\mathcal{A}(\Omega)$ by
$\Gamma_{\Omega}.$ Clearly, $\Gamma_{\Omega}$ is equal to
$\Psi(\mathbb{T}^{n}).$

The following theorem might be of independent interest. We will use this in
our proofs.

######  Theorem 2.9.

Let $N\geq 1$ and $f_{1},\cdots,f_{N}\in\mathcal{C}(\Gamma_{\Omega}).$ Then
$[z_{1},\cdots,z_{n},f_{1},\cdots,f_{N};\Gamma_{\Omega}]=\mathcal{C}(\Gamma_{\Omega})$
if and only if ${\sf Gr}_{f}(\Gamma_{\Omega})$ is polynomially convex, where
$f=(f_{1},\cdots,f_{N}).$

###### Proof.

We denote $X:={\sf Gr}_{f}(\Gamma_{\Omega}).$ Since
$[z_{1},\cdots,z_{n},f_{1},\cdots,f_{N};\Gamma_{\Omega}]=\mathcal{C}(\Gamma_{\Omega})$
implies $\mathscr{P}(X)=\mathcal{C}(X),$ hence $\widehat{X}=X.$

Conversely, suppose that $\widehat{X}=X.$ We consider the proper holomorphic
map
$\Phi:\mathbb{C}_{z}^{n}\times\mathbb{C}_{w}^{N}\to\mathbb{C}_{z}^{n}\times\mathbb{C}_{w}^{N},$
define by

$\displaystyle\Phi(z,w)=(\Psi(z),w).$

Clearly,

$\displaystyle\Phi^{-1}(X)={\sf Gr}_{f\circ\Psi}(\mathbb{T}^{n})=:Y.$

Since $X$ is polynomially convex, $Y$ is also polynomially convex (by Result
2.1). Let $U$ be a neighborhood of $\mathbb{T}^{n}$ such that
$z_{1}\not=0~{}~{}\text{ on }U.$ Define
$g(z_{1},z_{2},\cdots,z_{n})=\frac{1}{z_{1}}.$ Then $g$ is holomorphic on $U.$
Also, $g$ is holomorphic on $U\times\mathbb{C}^{N}.$ Since $Y\subset
U\times\mathbb{C}^{N},$ by the Oka-Weil approximation theorem, there exists a
sequence of polynomial $P_{j}$ in $\mathbb{C}^{n}_{z}\times\mathbb{C}_{w}^{N}$
such that $P_{j}(z,w)\to g$ uniformly on $Y.$ This implies
$P_{j}(z,(f\circ\Psi)(z))\to g=\frac{1}{z_{1}}=\overline{z}_{1}$ uniformly on
$\mathbb{T}^{n}.$ Hence
$\overline{z}_{1}\in[z_{1},\cdots,z_{n},f_{1}\circ\Psi,\cdots,f_{N}\circ\Psi;\mathbb{T}^{n}].$
By the similar method we can show that
$\overline{z}_{j}\in[z_{1},\cdots,z_{n},f_{1}\circ\Psi,\cdots,f_{N}\circ\Psi;\mathbb{T}^{n}],~{}\forall_{j}\in\\{1,\cdots,n\\}.$
Hence,
$[z_{1},\cdots,z_{n},\overline{z}_{1},\cdots,\overline{z}_{n};\mathbb{T}^{n}]\subset[z_{1},\cdots,z_{n},f_{1}\circ\Psi,\cdots,f_{N}\circ\Psi;\mathbb{T}^{n}].$
Therefore,

$\displaystyle[z_{1},\cdots,z_{n},\overline{z}_{1},\cdots,\overline{z}_{n};\mathbb{T}^{n}]=\mathcal{C}(\mathbb{T}^{n})=[z_{1},\cdots,z_{n},f_{1}\circ\Psi,\cdots,f_{N}\circ\Psi;\mathbb{T}^{n}].$
(2.1)

Note that $\mathscr{P}(X)=\mathcal{C}(X)$ if and only if
$\mathscr{P}(\Phi^{-1}(X))=\mathcal{C}(\Phi^{-1}(X))$ (see Result 2.1) i.e.,
$\mathscr{P}(Y)=\mathcal{C}(Y).$ Therefore, using (2.1), we get that

$\displaystyle[z_{1},\cdots,z_{n},f_{1},\cdots,f_{N};\Gamma_{\Omega}]=\mathcal{C}(\Gamma_{\Omega}).$

∎

######  Corollary 2.10.

Let $N\geq 1,$ and
$f_{1},\cdots,f_{N}\in\mathcal{C}(\Gamma_{\mathbb{G}_{n}}).$ Then
$[z_{1},\cdots,z_{n},f_{1},\cdots,f_{N};\Gamma_{\mathbb{G}_{n}}]=\mathcal{C}(\Gamma_{\mathbb{G}_{n}})$
if and only if ${\sf Gr}_{f}(\Gamma_{\mathbb{G}_{n}})$ is polynomially convex,
where $f=(f_{1},\cdots,f_{N}).$

In [18, 19], Jimbo explored the structure of polynomial hulls concerning
graphs of antiholomorphic polynomials on the torus. For the sake of
completeness, we include Jimbo’s result from [19] here since we will use it
multiple times in this paper. Let $\mathbb{T}^{2}$ be the torus in
$\mathbb{C}^{2}$ and $P$ be an arbitrary polynomial in $\mathbb{C}^{2}.$ In
[19], Jimbo gave a description for $\widehat{{\sf
Gr}_{\overline{P}}(\mathbb{T}^{2})}.$ Let the polynomial $P(z_{1},z_{2})$ be
of degree $m$ in $z_{1}$ and of degree $n$ in $z_{2}.$ We write

$\displaystyle P(z_{1},z_{2}))=\sum_{\begin{subarray}{c}0\leq i\leq m\\\ 0\leq
j\leq n\\\ \end{subarray}}a_{ij}z_{1}^{i}z_{2}^{j}.$

Therefore, on $\mathbb{T}^{2},$ we have

$\displaystyle\overline{P(z_{1},z_{2})}$
$\displaystyle=\frac{1}{z^{m}_{1}z^{n}_{2}}\sum_{\begin{subarray}{c}0\leq
i\leq m\\\ 0\leq j\leq n\\\
\end{subarray}}\overline{a}_{ij}z_{1}^{m-i}{z_{2}}^{n-j}$
$\displaystyle=\frac{K(z_{1},z_{2})}{z^{m}_{1}z^{n}_{2}}=h(z_{1},z_{2}),\text{
where }K(z_{1},z_{2})=\sum_{\begin{subarray}{c}0\leq i\leq m\\\ 0\leq j\leq
n\\\ \end{subarray}}\overline{a}_{ij}z_{1}^{m-i}z_{2}^{n-j}.$

Hence on $\mathbb{T}^{2},$ we get that

$\displaystyle\overline{P(z_{1},z_{2})}=h(z_{1},z_{2}),\text{ where
}h(z_{1},z_{2})=\frac{K(z_{1},z_{2})}{z_{1}^{m}z_{2}^{n}}.$

We define $L:=\\{z_{1}=0,|z_{2}|\leq 1\\}\cup\\{z_{2}=0,|z_{1}|\leq 1\\}$ and

$\displaystyle
X=\left\\{(z_{1},z_{2})\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2}):\overline{P(z_{1},z_{2})}=h(z_{1},z_{2})\right\\}.$
(2.2)

We set

$\displaystyle\triangle({z}):=\begin{vmatrix}\frac{\partial
P(z)}{\partial{z_{1}}}&\frac{\partial P(z)}{\partial z_{2}}\\\\[6.45831pt]
\frac{\partial h(z)}{\partial{{z_{1}}}}&\frac{\partial
h(z)}{\partial{z_{2}}}\\\\[6.45831pt] \end{vmatrix}.$

We can write

$\displaystyle\triangle(z)=\frac{1}{z^{m+1}_{1}z^{n+1}_{2}}\prod^{l}_{j=1}q_{j}(z),$

where each $q_{j}$ is an irreducible polynomial in $\mathbb{C}^{2}.$ We define
the corresponding irreducible algebraic variety
$Z_{j}:=Z(q_{j})=\\{z\in\mathbb{C}^{2}:q_{j}(z)=0\\}.$ We assume
$\triangle({z})\not\equiv 0$ on $X.$ Therefore, each $q_{j}$ is a non-zero
holomorphic polynomial in $\mathbb{C}^{2}.$

We denote $Q_{j}=Z_{j}\cap\mathbb{T}^{2}.$

######  Result 2.11 (Jimbo).

We let $J=\\{j\in\\{1,\cdots,l\\}:\emptyset\neq
Q_{j}\neq\widehat{Q_{j}},\widehat{Q_{j}}\setminus L\subset X\\}.$

  1. (i)

If $J=\emptyset,$ then $\widehat{{\sf Gr}_{\overline{P}}(\mathbb{T}^{2})}={\sf
Gr}_{\overline{P}}(\mathbb{T}^{2}),$ and
$[z_{1},z_{2},\overline{P};\mathbb{T}^{2}]=\mathcal{C}(\mathbb{T}^{2});$

  2. (ii)

If $J\neq\emptyset,$ then

$\displaystyle\widehat{{\sf Gr}_{\overline{P}}(\mathbb{T}^{2})}={\sf
Gr}_{\overline{P}}(\mathbb{T}^{2})\cup\bigg{(}\cup_{j\in J}{\sf
Gr}_{\overline{P}}(\widehat{Q_{j}})\bigg{)}.$

##  3\. proof of Theorems 1.6 and 1.7

Note that the map $\phi:\mathbb{C}^{2}\to\mathbb{C}^{2}$ is defined as
$\phi(z)=(p_{1}(z),p_{2}(z)).$ We consider the proper holomorphic map
$\widetilde{\Psi}:\mathbb{C}^{2+N}\to\mathbb{C}^{2+N},$ defined as follows:

$\displaystyle\widetilde{\Psi}(z_{1},z_{2},w_{1},\cdots,w_{N})=\left(\phi(z_{1},z_{2}),w_{1},\cdots,w_{N}\right),$
(3.1)

where $~{}(z_{1},z_{2})\in\mathbb{C}^{2},$ and
$(w_{1},\cdots,w_{N})\in\mathbb{C}^{N}.$ Recall that
$\Omega=\phi(\mathbb{D}^{2})$ and $\Gamma_{\Omega}=\phi(\mathbb{T}^{2}).$

######  Proof of Theorem 1.6.

We claim that $\widetilde{\Psi}^{-1}({\sf Gr}_{h}(\overline{\Omega}))={\sf
Gr}_{h\circ\phi}(\mathbb{\overline{D}}^{2})$: let

$\displaystyle(\alpha,\beta)\in\widetilde{\Psi}^{-1}({\sf
Gr}_{h}(\overline{\Omega}))$
$\displaystyle\implies\widetilde{\Psi}(\alpha,\beta)\in{\sf
Gr}_{h}(\overline{\Omega})$ $\displaystyle\implies(\phi(\alpha),\beta)\in{\sf
Gr}_{h}(\overline{\Omega})$ $\displaystyle\implies\beta=h(\phi(\alpha))\text{
and }\phi(\alpha)\in\overline{\Omega}.$

Now

$\displaystyle\phi(\alpha)\in\overline{\Omega}$
$\displaystyle\implies\alpha\in\phi^{-1}(\phi(\alpha))\subset\phi^{-1}(\overline{\Omega})\subset\overline{\mathbb{D}}^{2}.$

Therefore $\widetilde{\Psi}^{-1}({\sf Gr}_{h}(\overline{\Omega}))\subset{\sf
Gr}_{h\circ\phi}(\mathbb{\overline{D}}^{2}).$
Conversely, let

$\displaystyle(p,q)\in{\sf Gr}_{h\circ\phi}(\overline{\mathbb{D}}^{2})$
$\displaystyle\implies q=(h\circ\phi)(p)\text{ and
}p\in\overline{\mathbb{D}}^{2}$ $\displaystyle\implies q=h(\phi(p))\text{ and
}\Pi(p)\in\overline{\Omega}$ $\displaystyle\implies(\phi(p),q)\in{\sf
Gr}_{h}(\overline{\Omega})$ $\displaystyle\implies\widetilde{\Psi}(p,q)\in{\sf
Gr}_{h}(\overline{\Omega})$
$\displaystyle\implies(p,q)\in\widetilde{\Psi}^{-1}\left({\sf
Gr}_{h}(\overline{\Omega})\right).$

Hence ${\sf
Gr}_{h\circ\phi}(\mathbb{\overline{D}}^{2})\subset\widetilde{\Psi}^{-1}({\sf
Gr}_{h}(\overline{\Omega})).$ Therefore, $\widetilde{\Psi}^{-1}({\sf
Gr}_{h}(\overline{\Omega}))={\sf Gr}_{h\circ\phi}(\mathbb{\overline{D}}^{2}).$
Since $\widetilde{\Psi}$ is proper holomorphic mapping and
$\widetilde{\Psi}^{-1}({\sf
Gr}_{h}(\overline{\Omega}))=G_{h\circ\phi}(\mathbb{\overline{D}}^{2}),$ by
Result 2.1, we can say that $\mathscr{P}\left({\sf
Gr}_{h}(\overline{\Omega})\right)=\mathcal{C}\left({\sf
Gr}_{h}(\overline{\Omega})\right)$ if and only if $\mathscr{P}\left({\sf
Gr}_{h\circ\phi}(\overline{\mathbb{D}}^{2})\right)=\mathcal{C}\left({\sf
Gr}_{h\circ\phi}(\overline{\mathbb{D}}^{2})\right).$ We note that $h\circ\phi$
is pluriharmonic on $\mathbb{D}^{2}$ and continuous on
$\overline{\mathbb{D}}^{2}.$ Therefore, two cases hold.

Case I: $\mathscr{P}\left({\sf
Gr}_{h\circ\phi}(\overline{\mathbb{D}}^{2})\right)=\mathcal{C}\left({\sf
Gr}_{h\circ\phi}(\overline{\mathbb{D}}^{2})\right).$ In this case we have
$\mathscr{P}\left({\sf Gr}_{h}(\overline{\Omega})\right)=\mathcal{C}\left({\sf
Gr}_{h}(\overline{\Omega})\right).$

Case II: $\mathscr{P}\left({\sf
Gr}_{h\circ\phi}(\overline{\mathbb{D}}^{2})\right)\neq\mathcal{C}\left({\sf
Gr}_{h\circ\phi}(\overline{\mathbb{D}}^{2})\right).$ Therefore, by Result 1.2,
there exists an analytic disc
$g:\mathbb{D}\hookrightarrow{}\overline{\mathbb{D}}^{2}$ where
$(h_{j}\circ\phi)\circ g:\mathbb{D}\hookrightarrow{}\mathbb{\overline{D}}^{2}$
is holomorphic for all $j=1,\cdots,N.$ If we take $\gamma:=\phi\circ g,$ then
clearly $\gamma:\mathbb{D}\hookrightarrow{}\overline{\Omega}$ is an analytic
disc in $\overline{\Omega}$ such that $h_{j}$ is holomorphic on
$\gamma(\mathbb{D})$ (by Lemma 2.8) for all $j=1,\cdots,N.$ This proves the
theorem. ∎

###### Proof of Theorem 1.7.

Let $h_{j}$ denotes the pluriharmonic extension of $f_{j}$ to $\Omega$ and
write $h=(h_{1},\cdots,h_{N}):\overline{\Omega}\to\mathbb{C}^{N}.$ We have
$\widetilde{\Psi}$ is proper holomorphic mapping and
$\widetilde{\Psi}^{-1}({\sf Gr}_{h}(\Gamma_{\Omega}))={\sf
Gr}_{h\circ\phi}(\mathbb{T}^{2}).$ Therefore, by Result 2.1, ${\sf
Gr}_{h}(\Gamma_{\Omega})$ is polynomially convex if and only if ${\sf
Gr}_{h\circ\phi}(\mathbb{T}^{2})$ is polynomially convex. We note that
$h\circ\phi$ is pluriharmonic on $\mathbb{D}^{2}$ and continuous on
$\overline{\mathbb{D}}^{2}.$ Therefore, two cases hold.

Case I: ${\sf Gr}_{h}(\Gamma_{\Omega})$ is polynomially convex. In view of
Theorem 2.9, we have

$[z_{1},z_{2},f_{1},\cdots,f_{N};\Gamma_{\Omega}]=\mathcal{C}(\Gamma_{\Omega}).$

Case II: ${\sf Gr}_{h}(\Gamma_{\Omega})$ is not polynomially convex.
Consequently, ${\sf Gr}_{h\circ\phi}(\mathbb{T}^{2})$ is not polynomially
convex. Therefore, by Result 1.3, there exists a distinguished variety
$Z\subset\mathbb{D}^{2}$ where $(h_{j}\circ\phi)$ is holomorphic for all
$j=1,\cdots,N.$ Since $\phi$ is a proper holomorphic mapping, by Lemma 2.4, we
have $\phi(Z)$ is also an algebraic variety. Since $\phi$ is proper
holomorphic, $(h_{j}\circ\phi)$ is holomorphic on $Z,$ then $h_{j}$ is also
holomorphic on $\phi(Z)$ (by Lemma 2.8). Since $\phi$ sends distinguished
variety of $\mathbb{D}^{2}$ to distinguished variety of $\Omega$ (Lemma 2.6),
we have $\phi(Z)\cap b\Omega\subset\Gamma_{\Omega}.$ ∎

##  4\. Description of Polynomial Hull

In this section, we provide a description of the polynomial convex hull of the
graph of an anti-holomorphic polynomial over the distinguished boundary of the
domain $\Omega,$ where $\Omega$ is the image of the bidisc under certain
proper polynomial map from $\mathbb{C}^{2}$ to $\mathbb{C}^{2}.$

Let $F=(f_{1},f_{2},\cdots,f_{n}):\mathbb{C}^{n}\to\mathbb{C}^{n}$ be a proper
map. Let

$J_{f}(z)=\begin{vmatrix}\frac{\partial
f_{1}}{\partial{z_{1}}}(z)&\frac{\partial
f_{1}}{\partial{z_{2}}}(z)&\cdots&\frac{\partial
f_{1}}{\partial{z_{n}}}(z)\\\\[6.45831pt]
\vdots&\vdots&\cdots&\vdots\\\\[6.45831pt] \frac{\partial
f_{n}}{\partial{z_{1}}}(z)&\frac{\partial
f_{n}}{\partial{z_{2}}}(z)&\cdots&\frac{\partial
f_{n}}{\partial{z_{n}}}(z)\\\\[6.45831pt] \end{vmatrix}.$

The critical locus of $f$ is the complex analytic variety
$Z(J_{f})=\\{z\in\mathbb{C}^{n}:J_{f}(z)=0\\}\subset\mathbb{C}^{n}.$ The
branch locus $B(f)$ of $f$ is the image of the critical locus. Since $f$ is
proper,

$\displaystyle f:\mathbb{C}^{n}\setminus
f^{-1}(B(f))\to\mathbb{C}^{n}\setminus B(f)$

is a covering map of finite degree $d;$ $d$ is said to be the topological
degree of $f.$

######  Definition 4.1.

Two proper map $\phi,\tilde{\phi}:\mathbb{C}^{2}\to\mathbb{C}^{2}$ are said to
be equivalent if there exist $f,g\in\text{Aut}(\mathbb{C}^{2})$ such that
$\phi=f\circ\tilde{\phi}\circ g.$

Consider two holomorphic polynomials, $p_{1}$ and $p_{2}$, defined in
$\mathbb{C}^{2}$. Let $\phi(z)=(p_{1}(z),p_{2}(z))$ represent a proper
holomorphic mapping from $\mathbb{C}^{2}$ to $\mathbb{C}^{2}$, equivalent to
$\tilde{\phi}(z_{1},z_{2})=(z_{1}^{m},z_{2}^{n})$ for some natural numbers $m$
and $n$. There is a characterization due to Lamy [21] (see also Bisi and
Polizzi [7]) for $m=1$ and $n=2$ as follows: a proper polynomial map
$f:\mathbb{C}^{2}\to\mathbb{C}^{2}$ with a topological degree of 2 is
equivalent to $g(z_{1},z_{2})=(z_{1},z^{2}_{2}).$

Let $P(z_{1},z_{2})$ be any polynomial in $\mathbb{C}^{2}$. We aim to
calculate $\widehat{{\sf Gr}_{\overline{P}}(\Gamma_{\Omega})}$. It is evident
that $\widetilde{\Psi}^{-1}({\sf Gr}_{\overline{P}}(\Gamma_{\Omega}))={\sf
Gr}_{\overline{P}\circ\phi}(\mathbb{T}^{2})={\sf
Gr}_{\overline{P\circ\phi}}(\mathbb{T}^{2})$ ($\widetilde{\Psi}$ is given by
(3.1)). Consequently, ${\sf
Gr}_{\overline{P}}(\Gamma_{\Omega})=\widetilde{\Psi}\left(\operatorname{Gr}_{\overline{P\circ\phi}}(\mathbb{T}^{2})\right)$.
In this scenario, the following result holds.

######  Lemma 4.2.

$\widehat{\widetilde{\Psi}(Y)}=\widetilde{\Psi}\left(\widehat{Y}\right),$
where $Y={\sf Gr}_{\overline{P\circ\phi}}(\mathbb{{T}}^{2}).$

###### Proof.

Since $\widetilde{\Psi}$ is a proper holomorphic map, by using Result 2.1, we
have that $\widetilde{\Psi}^{-1}\left(\widehat{\widetilde{\Psi}(Y)}\right)$ is
polynomially convex. Therefore

$\displaystyle\widehat{Y}\subset\widehat{\widetilde{\Psi}^{-1}\left(\widehat{\widetilde{\Psi}(Y)}\right)}\subset\widetilde{\Psi}^{-1}\left(\widehat{\widetilde{\Psi}(Y)}\right).$

This implies,
$\widetilde{\Psi}(\widehat{Y})\subset\widehat{\widetilde{\Psi}(Y)}.$

Next, we show that
$\widetilde{\Psi}^{-1}\left(\widetilde{{\Psi}}(\widehat{Y})\right)\subset\widehat{Y}.$
To prove this, let
$(\alpha_{1},\alpha_{2},\beta)\in\widetilde{\Psi}^{-1}\left(\widetilde{{\Psi}}(\widehat{Y})\right).$
Then there exists $(\xi_{1},\xi_{2},\eta)\in\widehat{Y}$ such that
$\widetilde{\Psi}(\alpha_{1},\alpha_{2},\beta)=\Psi(\xi_{1},\xi_{2},\eta).$
This implies, $\phi(\alpha_{1},\alpha_{2})=\phi(\xi_{1},\xi_{2})$ and
$\beta=\eta.$ Since, $\phi$ is proper polynomial map and is equivalent to
$\tilde{\phi}(z_{1},z_{2})=(z^{m}_{1},z^{n}_{2}),$ there exist
$f,g\in\text{Aut}(\mathbb{C}^{2})$ such that $\phi=f\circ\tilde{\phi}\circ g.$
Then

$\displaystyle\phi(\alpha_{1},\alpha_{2})=\phi(\xi_{1},\xi_{2})$
$\displaystyle\implies$ $\displaystyle(f\circ\tilde{\phi}\circ
g)(\alpha_{1},\alpha_{2})=(f\circ\tilde{\phi}\circ g)(\xi_{1},\xi_{2})$
$\displaystyle\implies$ $\displaystyle(\tilde{\phi}\circ
g)(\alpha_{1},\alpha_{2})=(\tilde{\phi}\circ g)(\xi_{1},\xi_{2})$
$\displaystyle\implies$ $\displaystyle
g^{m}_{1}(\alpha_{1},\alpha_{2})=g^{m}_{1}(\xi_{1},\xi_{2})\text{ and
}g^{n}_{2}(\alpha_{1},\alpha_{2})=g^{n}_{2}(\xi_{1},\xi_{2}),\text{ where
}g=(g_{1},g_{2}).$ $\displaystyle\implies$
$\displaystyle(\alpha_{1},\alpha_{2})=g^{-1}\\{(\lambda^{k}_{m}g_{1}(\xi_{1},\xi_{2}),\lambda^{r}_{n}g_{2}(\xi_{1},\xi_{2})\\}=(a_{k},b_{r}),$

where $\lambda_{l}=\cos\frac{2\pi}{l}+i\sin\frac{2\pi}{l},$
$k\in\\{0,\cdots,m-1\\}$ and $r\in\\{0,\cdots,n-1\\}.$

It remains to show that $(a_{k},b_{r},\eta)\in\widehat{Y}.$ If possible,
assume that $(a_{k},b_{r},\eta)\notin\widehat{Y}$ for some
$k\in\\{0,\cdots,m-1\\},r\in\\{0,\cdots,n-1\\}.$ Then there exists a
polynomial $\chi$ in $\mathbb{C}^{2}_{z}\times\mathbb{C}_{w}$ such that

$\displaystyle|\chi(a_{k},b_{r},\eta)|>\sup_{Y}|\chi(z,w)|.$ (4.1)

Let us define $F(z_{1},z_{2}):=(\lambda^{k}_{m}z_{1},\lambda^{r}_{n}z_{2}),$
and $\tilde{F}(z_{1},z_{2},w):=((g^{-1}\circ F\circ\ g)(z),w).$ Since
$\phi^{-1}(\phi(\mathbb{T}^{2}))\subset\mathbb{T}^{2}$ (hence $(g^{-1}\circ
F\circ\ g)(z)\in\mathbb{T}^{2}$ if $z\in\mathbb{T}^{2}$), using (4.1), we get
that

$\displaystyle|(\chi\circ\tilde{F})(\xi,\eta)|>\sup_{Y}|(\chi\circ\tilde{F})(z,w)|.$
(4.2)

Since $\tilde{F}\in\text{Aut}(\mathbb{C}^{3}),$ (4.2) says that
$(\xi,\eta)\notin\widehat{Y}$ and this is a contradiction. Hence
$(a_{k},b_{r},\eta)\in\widehat{Y}.$ Therefore,
$\widetilde{\Psi}^{-1}\left(\widetilde{\Psi}(\widehat{Y})\right)=\widehat{Y}.$
Since $\widetilde{\Psi}$ is proper holomorphic map, by using Result 2.1, we
can say that $\widetilde{\Psi}(\widehat{Y})$ is polynomially convex.
Therefore,
$\widehat{\widetilde{\Psi}(Y)}\subset\widetilde{\Psi}(\widehat{Y}).$ This
proves the lemma. ∎

By using Lemma 4.3, we can say that

$\displaystyle\widehat{{\sf
Gr}_{\overline{P}}({\Gamma_{\Omega})}}=\widetilde{\Psi}\left(\widehat{{\sf
Gr}_{\overline{P\circ\phi}}(\mathbb{{T}}^{2})}\right).$

Therefore, to give a description for $\widehat{{\sf
Gr}_{\overline{P}}({\Gamma_{\Omega})}},$ it is enough to compute
$\widehat{{\sf Gr}_{\overline{P\circ\phi}}(\mathbb{{T}}^{2})}.$

###  4.1.  Description of Hull on Symmetrized Bidisc

Let $P(z_{1},z_{2})$ be any polynomial in $\mathbb{C}^{2}.$ By Lemma 4.2, we
calculate $\widehat{{\sf Gr}_{\overline{P}}(\Gamma_{{\mathbb{G}}_{2}})}.$ If
we take $p_{1}(z)=z_{1}+z_{2}$ and $p_{1}(z)=z_{1}z_{2},$ then $\phi=\Pi$ and
$\widetilde{\Psi}(z,w)=(\Pi(z),w)$ is a proper map from $\mathbb{C}^{3}$ to
$\mathbb{C}^{3}.$ It is easy to show that $\Pi$ a proper polynomial map of
topological degree $2,$ and hence equivalent to $(z_{1},z^{2}_{2}).$ Clearly,
$\widetilde{\Psi}^{-1}({\sf
Gr}_{\overline{P}}({\Gamma}_{\mathbb{G}_{2}}))={\sf
Gr}_{\overline{P}\circ\Pi}(\mathbb{{T}}^{2})={\sf
Gr}_{\overline{P\circ\Pi}}(\mathbb{{T}}^{2}).$ Therefore, ${\sf
Gr}_{\overline{P}}({\Gamma}_{\mathbb{G}_{2}}))=\widetilde{\Psi}\left({\sf
Gr}_{\overline{P\circ\Pi}}(\mathbb{{T}}^{2})\right).$

By Lemma 4.2, we get that

######  Lemma 4.3.

$\widehat{\widetilde{\Psi}\left(Y\right)}=\widetilde{\Psi}\left(\widehat{Y}\right),$
where $Y={\sf Gr}_{\overline{P\circ\Pi}}(\mathbb{{T}}^{2}).$

By using Lemma 4.3, we can say that

$\displaystyle\widehat{{\sf
Gr}_{\overline{P}}({\Gamma_{\mathbb{G}_{2}})}}=\widetilde{\Psi}\left(\widehat{{\sf
Gr}_{\overline{P\circ\Pi}}(\mathbb{{T}}^{2})}\right).$

Therefore, to give a description for $\widehat{{\sf
Gr}_{\overline{P}}({\Gamma_{\mathbb{G}_{2}})}},$ it is enough to compute
$\widehat{{\sf Gr}_{\overline{P\circ\Pi}}(\mathbb{{T}}^{2})}.$

##  5\. Examples

######  Example 5.1.

Let $P(z_{1},z_{2})=z_{1}-z_{2}.$ Then
$[z_{1},z_{2},\overline{P};\Gamma_{\mathbb{G}_{2}}]\neq\mathcal{C}(\Gamma_{\mathbb{G}_{2}}).$

Explanation In view of Corollary 2.10, to demonstrate that
$[z_{1},z_{2},\overline{P};\Gamma_{\mathbb{G}_{2}}]\neq\mathcal{C}(\Gamma_{\mathbb{G}_{2}}),$
it suffices to establish that the graph of $\overline{P}$ over
$\Gamma_{\mathbb{G}_{2}}$ is not polynomially convex. To achieve this, it is
sufficient to show that the graph of $\overline{P\circ\Pi}$ over
$\Gamma_{\mathbb{G}_{2}}$ lacks polynomial convexity. Following the notation
in Result 2.11, we define
$h(z)=\frac{1}{z_{1}}+\frac{1}{z_{2}}-\frac{1}{z_{1}z_{2}}.$ Then

$\displaystyle\triangle({z})=\begin{vmatrix}\frac{\partial(P\circ\Pi)}{\partial{z_{1}}}&\frac{\partial(P\circ\Pi)}{\partial
z_{2}}\\\\[6.45831pt] \frac{\partial h}{\partial{{z_{1}}}}&\frac{\partial
h}{\partial{z_{2}}}\\\\[6.45831pt]
\end{vmatrix}=\begin{vmatrix}1-z_{2}&1-z_{1}\\\\[6.45831pt]
\frac{-1}{z^{2}_{1}}+\frac{1}{z^{2}_{1}z_{2}}&\frac{-1}{z^{2}_{2}}+\frac{1}{z^{2}_{2}z_{1}}\\\\[6.45831pt]
\end{vmatrix}$
$\displaystyle=\frac{1}{z^{2}_{1}z^{2}_{2}}(z_{1}-z_{2})(z_{1}-1)(z_{2}-1).$

We define $q_{1}:=z_{1}-1,~{}~{}q_{2}=z_{2}-1,q_{3}:=z_{1}-z_{2},$ and
$Z_{j}=\\{z\in\mathbb{C}^{2}:q_{j}(z)=0\\},j=1,2,3.$ Therefore,

$\displaystyle\Sigma$
$\displaystyle=\left\\{z\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2}):\triangle(z)=0\right\\}$
$\displaystyle=\left\\{z\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2})\right\\}\cap[\cup^{3}_{j=1}Z_{j}],$

and

$\displaystyle X$
$\displaystyle=\left\\{z\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2}):\overline{(P\circ\Pi)(z)}=h(z)\right\\}$
$\displaystyle=\left\\{z\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2}):\overline{z_{1}+z_{2}-z_{1}z_{2}}=\frac{1}{z_{1}}+\frac{1}{z_{1}}-\frac{1}{z_{1}z_{2}}\right\\}.$

Here $Q_{j}=Z_{j}\cap\mathbb{{T}}^{2}.$ Clearly,

$\displaystyle\widehat{Q_{1}}$
$\displaystyle=\\{z\in\mathbb{C}^{2}:z_{1}=1,|z_{2}|\leq 1\\}\neq Q_{1};$
$\displaystyle\widehat{Q_{2}}$
$\displaystyle=\\{z\in\mathbb{C}^{2}:z_{2}=1,|z_{1}|\leq 1\\}\neq Q_{2};$
$\displaystyle\widehat{Q_{3}}$
$\displaystyle=\\{z\in\mathbb{C}^{2}:z_{1}=z_{2},|z_{1}|\leq 1\\}\neq Q_{3}.$

It is evident that $\widehat{Q_{j}}\setminus(\mathbb{T}^{2}\cup L)\subset X$
holds true only for $j=1,2.$ On the other hand, we note that
$(\frac{1}{2},\frac{1}{2})\in\widehat{Q_{3}}\setminus(\mathbb{T}^{2}\cup L),$
yet $(\frac{1}{2},\frac{1}{2})\notin X.$ Therefore, by Result 2.11, we deduce
that:

$\displaystyle\widehat{{\sf Gr}_{\overline{P\circ\Pi}}(\mathbb{T}^{2})}={\sf
Gr}_{\overline{P\circ\Pi}}(\mathbb{T}^{2})\cup{\sf
Gr}_{\overline{P\circ\Pi}}(\widehat{Q_{1}})\cup{\sf
Gr}_{\overline{P\circ\Pi}}(\widehat{Q_{2}}).$

Hence

$\displaystyle\widehat{{\sf Gr}_{\overline{P}}(\Gamma_{\mathbb{G}_{2}})}$
$\displaystyle=\Psi\left({\sf
Gr}_{\overline{P\circ\Pi}}(\mathbb{T}^{2})\right)\cup\Psi\left({\sf
Gr}_{\overline{P\circ\Pi}}(\widehat{Q_{1})}\right)\cup\Psi\left({\sf
Gr}_{\overline{P\circ\Pi}}(\widehat{Q_{2})}\right)$ $\displaystyle={\sf
Gr}_{\overline{P}}(\Gamma_{\mathbb{G}_{2}})\cup\\{(1+z,z,w):w=\overline{P(1+z,z)},z\in\overline{\mathbb{D}}\\}$
$\displaystyle\cup\\{(1+z,z,w):w=\overline{P(1+z,z)},z\in\overline{\mathbb{D}}\\}$
$\displaystyle={\sf
Gr}_{\overline{P}}(\Gamma_{\mathbb{G}_{2}})\cup\\{(1+z,z,w):w=\overline{P(1+z,z)},z\in\overline{\mathbb{D}}\\}$
$\displaystyle={\sf
Gr}_{\overline{P}}(\Gamma_{\mathbb{G}_{2}})\cup\\{(1+z,z,1):z\in\overline{\mathbb{D}}\\}.$

######  Example 5.2.

$P(z_{1},z_{2})=z_{1}-2z_{2}.$ Then
$[z_{1},z_{2},\overline{P};\Gamma_{\mathbb{G}_{2}}]=\mathcal{C}(\Gamma_{\mathbb{G}_{2}}).$

Explanation In light of Corollary 2.10, in order to establish that
$[z_{1},z_{2},\overline{P};\Gamma_{\mathbb{G}_{2}}]=\mathcal{C}(\Gamma_{\mathbb{G}_{2}}),$
it is sufficient to demonstrate the polynomial convexity of the graph of
$\overline{P}$ over $\Gamma_{\mathbb{G}_{2}}$. To accomplish this, it is
enough to prove that the graph of $\overline{P\circ\Pi}$ over $\mathbb{T}^{2}$
is polynomially convex. Following the notation in Result 2.11, we have
$h(z)=\frac{1}{z_{1}}+\frac{1}{z_{2}}-\frac{2}{z_{1}z_{2}}.$

$\displaystyle\triangle({z})=$
$\displaystyle\begin{vmatrix}\frac{\partial(P\circ\Pi)}{\partial{z_{1}}}&\frac{\partial(P\circ\Pi)}{\partial
z_{2}}\\\\[6.45831pt] \frac{\partial h}{\partial{{z_{1}}}}&\frac{\partial
h}{\partial{z_{2}}}\\\\[6.45831pt]
\end{vmatrix}=\begin{vmatrix}1-2z_{2}&1-2z_{1}\\\\[6.45831pt]
\frac{-1}{z^{2}_{1}}+\frac{2}{z^{2}_{1}z_{2}}&\frac{-1}{z^{2}_{2}}+\frac{2}{z^{2}_{2}z_{1}}\\\\[6.45831pt]
\end{vmatrix}$ $\displaystyle=$
$\displaystyle\frac{1}{z^{2}_{1}z^{2}_{2}}(z_{1}+z_{2}-2-2z_{1}z_{2})(z_{2}-z_{1}).$

We define $q_{1}:=z_{1}+z_{2}-2-2z_{1}z_{2},~{}q_{2}:=z_{2}-z_{1}.$ and
$Z_{j}=\\{z\in\mathbb{C}^{2}:q_{j}(z)=0\\},j=1,2.$ Therefore,

$\displaystyle\Sigma$
$\displaystyle=\left\\{(z,w)\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2}):\triangle_{(z,w)}=0\right\\}$
$\displaystyle=\left\\{(z,w)\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2})\right\\}\cap[\cup^{2}_{j=1}Z_{j}],$

and

$\displaystyle X$
$\displaystyle=\left\\{z\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2}):\overline{(P\circ\Pi)(z)}=h(z)\right\\}$
$\displaystyle=\left\\{z\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2}):\overline{z_{1}+z_{2}-2z_{1}z_{2}}=\frac{1}{z_{1}}+\frac{1}{z_{2}}-\frac{2}{z_{1}z_{2}}\right\\}.$

Here $Q_{j}=Z_{j}\cap\mathbb{{T}}^{2}.$ We now claim that

$\displaystyle\widehat{Q_{1}}$
$\displaystyle=\\{z\in\mathbb{C}^{2}:z_{1}+z_{2}-2z_{1}z_{2}-2=0,|z_{1}|=1,|z_{2}|=1\\}=Q_{1}.$

Clearly,
$\widehat{Q_{1}}\subset\\{z\in\mathbb{C}^{2}:z_{1}+z_{2}-2z_{1}z_{2}-2=0,|z_{1}|\leq
1,|z_{2}|\leq 1\\}.$ Let
$(\alpha,\beta)\in\\{z\in\mathbb{C}^{2}:z_{1}+z_{2}-2z_{1}z_{2}-2=0,|z_{1}|\leq
1,|z_{2}|\leq 1\\}\setminus Q_{1}.$ First, we assume that $|\beta|<1.$ Since
$\alpha+\beta-2\alpha\beta-2=0,$ hence

$\displaystyle|2-\alpha|=|\beta||1-2\alpha|<|1-2\alpha|.$ (5.1)

Let $\alpha=u+iv.$ Then from (5.1), we get that

$\displaystyle(2-u)^{2}+v^{2}<(1-2u)^{2}+4v^{2}$ $\displaystyle\implies
4+u^{2}+v^{2}-4u<1+4(u^{2}+v^{2})-4u$ $\displaystyle\implies
u^{2}+v^{2}>1\text{ i.e., }|\alpha|>1.$

Hence, we conclude that $(\alpha,\beta)\notin\widehat{Q_{1}}.$ In the case
where $|\alpha|<1,$ we can similarly demonstrate that $|\beta|>1,$ leading to
the same conclusion, $(\alpha,\beta)\notin\widehat{Q_{1}}.$ As a result, we
establish that $Q_{1}$ is polynomially convex.

Furthermore, consider
$\widehat{Q_{2}}=\\{z\in\mathbb{C}^{2}:z_{1}=z_{2},|z_{1}|\leq 1\\}\neq
Q_{2}.$ Notably,
$(\frac{1}{2},\frac{1}{2})\in\widehat{Q_{2}}\setminus(\mathbb{T}^{2}\cup L),$
while $(\frac{1}{2},\frac{1}{2})\notin X.$ Hence, by Result 2.11, we can
deduce that:

$\displaystyle\widehat{{\sf Gr}_{\overline{P\circ\Pi}}(\mathbb{T}^{2})}={\sf
Gr}_{\overline{P\circ\Pi}}(\mathbb{T}^{2}).$

This implies:

$\displaystyle\widehat{{\sf
Gr}_{\overline{P}}(\Gamma_{\mathbb{G}{2}})}=\Psi\left({\sf
Gr}_{\overline{P\circ\Pi}}(\mathbb{T}^{2})\right)={\sf
Gr}_{\overline{P}}(\Gamma_{\mathbb{G}_{2}}).$

######  Example 5.3.

Let
$p_{1}(z_{1},z_{2})=2z_{1}+z^{2}_{2},~{}~{}p_{2}(z_{1},z_{2})=z_{1}-z^{2}_{2},~{}P(z_{1},z_{2})=z_{1}-z_{2}$
and $\phi(z_{1},z_{2})=(p_{1}(z_{1},z_{2}),p_{2}(z_{1},z_{2})).$ Therefore
$\Omega=\phi(\mathbb{D}^{2}).$ Then
$[z_{1},z_{2},\overline{P};\Gamma_{\Omega}]=\mathcal{C}(\Gamma_{\Omega}).$

Explanation According to Theorem 2.9, it follows that
$[z_{1},z_{2},\overline{P};\Gamma_{\Omega}]=\mathcal{C}(\Gamma_{\Omega})$ if,
and only if, ${\sf Gr}_{\overline{P}}(\Gamma_{\Omega})$ exhibits polynomial
convexity. Furthermore, the polynomial convexity of ${\sf
Gr}_{\overline{P}}(\Gamma_{\Omega})$ is equivalent to the polynomial convexity
of ${\sf Gr}_{\overline{P\circ\phi}}(\mathbb{T}^{2})$.

Here
$\overline{P\circ\phi}=\overline{z_{1}+2z^{2}_{2}}=\frac{1}{z_{1}}+\frac{2}{z^{2}_{2}}=:h(z)$
on $\mathbb{T}^{2}.$

$\displaystyle\triangle({z})$
$\displaystyle=\begin{vmatrix}\frac{\partial(P\circ\phi)}{\partial{z_{1}}}&\frac{\partial(P\circ\phi)}{\partial
z_{2}}\\\\[6.45831pt] \frac{\partial h}{\partial{{z_{1}}}}&\frac{\partial
h}{\partial{z_{2}}}\\\\[6.45831pt]
\end{vmatrix}=\begin{vmatrix}1&4z_{2}\\\\[6.45831pt]
\frac{-1}{z^{2}_{1}}&\frac{-4}{z^{3}_{2}}\\\\[6.45831pt]
\end{vmatrix}=\frac{1}{z^{2}_{1}z^{3}_{2}}(z_{1}+z^{2}_{2})(z^{2}_{2}-z_{1}).$

We define $q_{1}:=z_{1}+z^{2}_{2},~{}q_{2}:=z^{2}_{2}-z_{1},$ and
$Z_{j}=\\{z\in\mathbb{C}^{2}:q_{j}(z)=0\\},j=1,2.$ Therefore,

$\displaystyle\Sigma$
$\displaystyle=\left\\{(z,w)\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2}):\triangle_{(z,w)}=0\right\\}$
$\displaystyle=\left\\{(z,w)\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2})\right\\}\cap[\cup^{2}_{j=1}Z_{j}],$

and

$\displaystyle X$
$\displaystyle=\left\\{z\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2}):\overline{(P\circ\phi)(z)}=h(z)\right\\}$
$\displaystyle=\left\\{z\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2}):\overline{z_{1}+2z^{2}_{2}}=\frac{1}{z_{1}}+\frac{2}{z^{2}_{2}}\right\\}.$

Here $Q_{j}=Z_{j}\cap\mathbb{{T}}^{2}.$ Clearly,

$\displaystyle\widehat{Q_{1}}$
$\displaystyle=\\{z\in\mathbb{C}^{2}:z_{1}+z^{2}_{2}=0,|z_{1}|\leq
1,|z_{2}|\leq 1\\}\neq Q_{1},~{}\text{ and}$ $\displaystyle\widehat{Q_{2}}$
$\displaystyle=\\{z\in\mathbb{C}^{2}:z^{2}_{2}-z_{1}=0,|z_{1}|\leq
1,|z_{2}|\leq 1\\}\neq Q_{2}.$

It is easy to see that $\widehat{Q_{j}}\setminus(\mathbb{T}^{2}\cup
L)\nsubseteq X$ for $j=1,2.$ Therefore, by a Result 2.11, we get that

$\displaystyle\widehat{{\sf Gr}_{\overline{P\circ\phi}}(\mathbb{T}^{2})}={\sf
Gr}_{\overline{P\circ\phi}}(\mathbb{T}^{2}).$

Hence

$\displaystyle\widehat{{\sf Gr}_{\overline{P}}(\Gamma_{\Omega})}$
$\displaystyle=\Psi\left({\sf
Gr}_{\overline{P\circ\phi}}(\mathbb{T}^{2})\right)={\sf
Gr}_{\overline{P}}(\Gamma_{\Omega}).$

######  Example 5.4.

Let
$p_{1}(z_{1},z_{2})=z_{1}+z_{2},~{}~{}p_{2}(z_{1},z_{2})=z^{2}_{1}+z^{2}_{2},~{}P(z_{1},z_{2})=z^{2}_{1}+z_{2}$
and $\phi(z_{1},z_{2})=(p_{1}(z_{1},z_{2}),p_{2}(z_{1},z_{2})).$ Therefore
$\Omega=\phi(\mathbb{D}^{2}).$ Then
$[z_{1},z_{2},\overline{P};\Gamma_{\Omega}]\neq\mathcal{C}(\Gamma_{\Omega}).$

Explanation Based on Theorem 2.9, we can assert that
$[z_{1},z_{2},\overline{P};\Gamma_{\Omega}]\neq\mathcal{C}(\Gamma_{\Omega})$
if, and only if, ${\sf Gr}_{\overline{P}}(\Gamma_{\Omega})$ lacks polynomial
convexity. Furthermore, ${\sf Gr}_{\overline{P}}(\Gamma_{\Omega})$ possesses
polynomial convexity if, and only if, ${\sf
Gr}_{\overline{P\circ\phi}}(\mathbb{T}^{2})$ is polynomially convex.
Therefore, it is enough to show that ${\sf
Gr}_{\overline{P\circ\phi}}(\mathbb{T}^{2})$ is not polynomailly convex.

Here $P\circ\phi=2(z^{2}_{1}+z_{1}z_{2}+z^{2}_{2}).$ Hence,

$\displaystyle\overline{P\circ\phi}=\overline{2(z^{2}_{1}+z_{1}z_{2}+z^{2}_{2})}=2\left(\frac{1}{z^{2}_{1}}+\frac{1}{z^{2}_{2}}+\frac{1}{z_{1}z_{2}}\right)=:h(z)\text{
on }\mathbb{T}^{2}.$

$\displaystyle\triangle({z})$
$\displaystyle=\begin{vmatrix}\frac{\partial(P\circ\phi)}{\partial{z_{1}}}&\frac{\partial(P\circ\phi)}{\partial
z_{2}}\\\\[6.45831pt] \frac{\partial h}{\partial{{z_{1}}}}&\frac{\partial
h}{\partial{z_{2}}}\\\\[6.45831pt]
\end{vmatrix}=\begin{vmatrix}2(2z_{1}+z_{2})&2(2z_{2}+z_{1})\\\\[6.45831pt]
2(\frac{-2}{z^{3}_{1}}-\frac{-1}{z^{2}_{1}z_{2}})&2(\frac{-2}{z^{3}_{2}}-\frac{-1}{z_{1}z^{2}_{2}})\\\\[6.45831pt]
\end{vmatrix}$
$\displaystyle=\frac{8\alpha^{-1}}{z^{3}_{1}z^{3}_{2}}(z_{1}+z_{2})(z_{2}-z_{1})(z_{1}-\alpha
z_{2})(z_{2}-\alpha z_{1}),\text{ where }\alpha=e^{\frac{2\pi i}{3}}.$

We define $q_{1}:=z_{1}+z_{2},~{}q_{2}:=z_{2}-z_{1},~{},q_{3}=z_{1}-\alpha
z_{2},~{}q_{4}=z_{2}-\alpha z_{1},$ and
$Z_{j}=\\{z\in\mathbb{C}^{2}:q_{j}(z)=0\\},j=1,2,3,4.$ Therefore,

$\displaystyle\Sigma$
$\displaystyle=\left\\{z\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2}):\triangle(z)=0\right\\}$
$\displaystyle=\left\\{z\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2})\right\\}\cap[\cup^{3}_{j=1}Z_{j}],$

and

$\displaystyle X$
$\displaystyle=\left\\{z\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2}):\overline{(P\circ\phi)(z)}=h(z)\right\\}$
$\displaystyle=\left\\{z\in\overline{\mathbb{D}}^{2}\setminus(L\cup\mathbb{T}^{2}):\overline{2(z^{2}_{1}+z_{1}z_{2}+z^{2}_{2})}=2\left(\frac{1}{z^{2}_{1}}+\frac{1}{z^{2}_{2}}+\frac{1}{z_{1}z_{2}}\right)\right\\}.$

Here $Q_{j}=Z_{j}\cap\mathbb{{T}}^{2}.$ Clearly,

$\displaystyle\widehat{Q_{1}}$
$\displaystyle=\\{z\in\mathbb{C}^{2}:z_{1}+z_{2}=0,|z_{1}|\leq 1,|z_{2}|\leq
1\\}\neq Q_{1};$ $\displaystyle\widehat{Q_{2}}$
$\displaystyle=\\{z\in\mathbb{C}^{2}:z_{2}-z_{1}=0,|z_{1}|\leq 1,|z_{2}|\leq
1\\}\neq Q_{2};$ $\displaystyle\widehat{Q_{3}}$
$\displaystyle=\\{z\in\mathbb{C}^{2}:z_{1}-\alpha z_{2}=0,|z_{1}|\leq
1,|z_{2}|\leq 1\\}\neq Q_{3};$ $\displaystyle\widehat{Q_{4}}$
$\displaystyle=\\{z\in\mathbb{C}^{2}:z_{2}-\alpha z_{1}=0,|z_{1}|\leq
1,|z_{2}|\leq 1\\}\neq Q_{4}.$

Again $\widehat{Q_{j}}\setminus(\mathbb{T}^{2}\cup L)\nsubseteq X$ for
$j=1,2,$ and $\widehat{Q_{j}}\setminus(\mathbb{T}^{2}\cup L)\subset X$ for
$j=3,4.$ Therefore, by Result 2.11, we get that

$\displaystyle\widehat{{\sf Gr}_{\overline{P\circ\phi}}(\mathbb{T}^{2})}={\sf
Gr}_{\overline{P\circ\phi}}(\mathbb{T}^{2})\cup{\sf
Gr}_{\overline{P\circ\phi}}(\widehat{Q_{3}})\cup{\sf
Gr}_{\overline{P\circ\phi}}(\widehat{Q_{4}}).$

Hence

$\displaystyle\widehat{{\sf Gr}_{\overline{P}}(\Gamma_{\Omega})}$
$\displaystyle=\Psi\left({\sf
Gr}_{\overline{P\circ\phi}}(\mathbb{T}^{2})\right)={\sf
Gr}_{\overline{P}}(\Gamma_{\Omega})\cup\Psi\left({\sf
Gr}_{\overline{P\circ\phi}}(\widehat{Q_{3}})\right)\cup\Psi\left({\sf
Gr}_{\overline{P\circ\phi}}(\widehat{Q_{4}})\right).$

Acknowledgements. We would like to express our sincere gratitude to Professor
Franc Forstnerič for pointing out Result 2.3 in [32] and showing us the proof
of Lemma 2.4. The first named author was partially supported by a Matrics
Research Grant (MTR/2017/000974) of SERB, Dept. of Science and Technology,
Govt. of India, for the beginning of this work and is supported by a Core
Research Grant (CRG/2022/003560) of SERB, Dept. of Science and Technology,
Govt. of India, for the later part of the work. The second named author’s work
received partial support from an INSPIRE Fellowship (IF 160487) provided by
the Dept. of Science and Technology, Govt. of India, during the early stage of
this work. Presently, this research is supported by a research grant from SERB
(Grant No. CRG/2021/005884), Dept. of Science and Technology, Govt. of India.

## References

  * [1] J. Agler, Z. Lykova, and N. Young.  Geodesics, retracts, and the norm-preserving extension property in the symmetrized bidisc.  Mem. Amer. Math. Soc., 258(1242):vii+108, 2019.
  * [2] J. Agler, Z. Lykova, and N. J. Young.  A geometric characterization of the symmetrized bidisc.  J. Math. Anal. Appl., 473(2):1377–1413, 2019.
  * [3] J. Agler and N. J. Young.  The hyperbolic geometry of the symmetrized bidisc.  J. Geom. Anal., 14(3):375–403, 2004.
  * [4] Jim Agler and John E. McCarthy.  Distinguished varieties.  Acta Math., 194(2):133–153, 2005.
  * [5] T. Andô.  On a pair of commutative contractions.  Acta Sci. Math. (Szeged), 24:88–90, 1963.
  * [6] T. Bhattacharyya, S. Pal, and S. S. Roy.  Dilations of $\Gamma$-contractions by solving operator equations.  Adv. Math., 230(2):577–606, 2012.
  * [7] C. Bisi and F. Polizzi.  On proper polynomial maps of $\mathbb{C}^{2}$.  J. Geom. Anal., 20(1):72–89, 2010.
  * [8] D. Chakrabarti and S. Gorai.  Function theory and holomorphic maps on symmetric products of planar domains.  J. Geom. Anal., 25(4):2196–2225, 2015.
  * [9] E. M. Chirka.  Complex analytic sets, volume 46 of Mathematics and its Applications (Soviet Series).  Kluwer Academic Publishers Group, Dordrecht, 1989.  Translated from the Russian by R. A. M. Hoksbergen.
  * [10] C. Costara.  The symmetrized bidisc and Lempert’s theorem.  Bull. London Math. Soc., 36(5):656–662, 2004.
  * [11] B. Krishna Das and Jaydeb Sarkar.  Ando dilations, von Neumann inequality, and distinguished varieties.  J. Funct. Anal., 272(5):2114–2131, 2017.
  * [12] A. Edigarian and W. Zwonek.  Geometry of the symmetrized polydisc.  Arch. Math. (Basel), 84(4):364–374, 2005.
  * [13] T. W. Gamelin.  Uniform algebras.  Prentice-Hall, Inc., Englewood Cliffs, N. J., 1969.
  * [14] A. J. Izzo.  Uniform algebras generated by holomorphic and pluriharmonic functions.  Trans. Amer. Math. Soc., 339(2):835–847, 1993.
  * [15] A. J. Izzo.  Uniform algebras generated by holomorphic and pluriharmonic functions on strictly pseudoconvex domains.  Pacific J. Math., 171(2):429–436, 1995.
  * [16] A. J. Izzo, Samuelsson K. H., and E. F. Wold.  Presence or absence of analytic structure in maximal ideal spaces.  Math. Ann., 366(1-2):459–478, 2016.
  * [17] M. Jarnicki and P. Pflug.  On automorphisms of the symmetrized bidisc.  Arch. Math. (Basel), 83(3):264–266, 2004.
  * [18] T. Jimbo.  Polynomial hulls of graphs of antiholomorphic functions.  volume 57, pages 157–163. 2003.  Japanese Association of Mathematical Sciences 2001 Annual Meeting (Tennoji).
  * [19] T. Jimbo.  Polynomial hulls of graphs on the torus in $C^{2}$.  Sci. Math. Jpn., 62(3):335–342, 2005.
  * [20] L. Kosiński and W. Zwonek.  Nevanlinna-Pick problem and uniqueness of left inverses in convex domains, symmetrized bidisc and tetrablock.  J. Geom. Anal., 26(3):1863–1890, 2016.
  * [21] S. Lamy.  Sur la structure du groupe d’automorphismes de certaines surfaces affines.  Publ. Mat., 49(1):3–20, 2005.
  * [22] M. A. Lavrentiev.  Sur les fonctions d’une variable complexe, représentables par des séries de polynomes.  1936\.
  * [23] Z. L. Leĭbenzon.  On the ring of continuous functions on a circle.  Uspehi Matem. Nauk (N.S.), 7(4(50)):163–164, 1952.
  * [24] S Pal and O. M. Shalit.  Spectral sets and distinguished varieties in the symmetrized bidisc.  J. Funct. Anal., 266(9):5779–5800, 2014.
  * [25] P. Pflug and W. Zwonek.  Description of all complex geodesics in the symmetrized bidisc.  Bull. London Math. Soc., 37(4):575–584, 2005.
  * [26] R. Remmert.  Projektionen analytischer Mengen.  Math. Ann., 130:410–441, 1956.
  * [27] R. Remmert.  Holomorphe und meromorphe Abbildungen komplexer Räume.  Math. Ann., 133:328–370, 1957.
  * [28] H. Samuelsson and E. F. Wold.  Uniform algebras and approximation on manifolds.  Invent. Math., 188(3):505–523, 2012.
  * [29] J. Sarkar.  Operator theory on symmetrized bidisc.  Indiana Univ. Math. J., 64(3):847–873, 2015.
  * [30] E. L. Stout.  Polynomial convexity, volume 261 of Progress in Mathematics.  Birkhäuser Boston, Inc., Boston, MA, 2007.
  * [31] M. Trybuła.  Invariant metrics on the symmetrized bidisc.  Complex Var. Elliptic Equ., 60(4):559–565, 2015.
  * [32] E. M. Čirka.  Approximation by holomorphic functions on smooth manifolds in ${\bf C}^{n}$.  Mat. Sb. (N.S.), 78 (120):101–123, 1969.
  * [33] J. Wermer.  On algebras of continuous functions.  Proc. Amer. Math. Soc., 4:866–869, 1953.
  * [34] J. Wermer.  Polynomially convex disks.  Math. Ann., 158:6–10, 1965.


A note on simple zeros related to Dedekind zeta functions

Wei Zhang

Abstract We give a conditional lower bound on the number of non-trivial simple
zeros for the Dedekind zeta function $\zeta_{K}(s)$, where $K$ is a quadratic
number field. The conditional result is given by assuming a Lindelöf on
average (in the $L^{6}$ sense) for both $\zeta(s)$ and $L(s,\chi)$, which can
be seen as a stronger version of Conrey-Gonek-Ghosh’s [1] conditional result.
This improves upon the work of Wu and Zhao [8], who had a similar result.

Keywords: Simple zeros, zeta function.

2000 Mathematics Subject Classification:11M06, 11M41.

##  1\. Introduction

A folklore conjecture in analytic number theory is that for any number field
$K$, almost all zeros of $\zeta_{K}(s)$ in the critical strip are simple. This
conjecture is attractive because of how hard it is to prove to the best of my
knowledge of the existing literature, we still do not know if $\zeta_{K}(s)$
has infinitely many simple zeros if $[K:Q]\geq 3$ and it was not known if even
the simplest case of $\zeta(s)$ (Riemann zeta function) has infinitely many
simple zeros in the critical strip until it was observed independently by
Heath-Brown [3] and Selberg [6] that Levinson’s method [5] yields a positive
proportion of simple zeros on the critical line.

In this note, we are interested in the non-real zeros of $\zeta_{K}(s)$ in the
critical strip

$\mathcal{R}=\\{s=\sigma+it:0<\sigma<1,0<t<T\\}.$

Let $K$ be a certain quadratic extension of the $\mathbb{Q}$ and let
$\zeta_{K}(s)$ be the corresponding Dedekind zeta function. Then we have

$\zeta_{K}(s)=\sum_{\mathfrak{a}}\frac{1}{(\mathfrak{Na})^{s}}=\sum_{n=1}^{\infty}\frac{a_{K}(n)}{n^{s}},\
\ \Re(s)>1,$

where $\mathfrak{a}$ suns over the non-zero integral ideals of $K,$
$\mathfrak{Na}$ is the norm of $\mathfrak{a},$ and
$a_{K}(n)=\\#\\{\mathfrak{a}:\mathfrak{Na}=n\\}$ is the ideal counting
function of $K.$ It is well known that

$N_{K}(T)=\\#\\{\rho_{K}:\zeta_{K}(\rho_{K})=0,\rho_{K}\in\mathcal{R}\\}\sim\pi^{-1}T\log
T.$

Moreover, it is also generally believed that all the non-real zeros of
$\zeta_{K}(s)$ in $\mathcal{R}$ are also simple zeros. However, one even
cannot confirm that whether there are infinitely many non-real simple zeros in
$\mathcal{R}$ before 1986. In 1986, Conrey, Ghosh and Gonek [1] introduced a
new method to deal with the simple zeros for such type problems. For
sufficiently large $T,$ Conrey, Ghosh and Gonek [1] proved that

$N^{\prime}_{K}(T)\gg T^{\frac{6}{11}},$

where

$N^{\prime}_{K}(T)=\\#\\{\rho_{K}:\zeta_{K}(\rho_{K})=0,\zeta^{\prime}_{K}(\rho_{K})\neq
0,\rho_{K}\in\mathcal{R}\\}.$

This implies that there are infinitely many simple zeros for the
$\zeta_{K}(s)$ associated to the quadratic number fields. The exponent in this
result depends upon progress towards the Lindelöf hypothesis. In fact, a
better result is also given in [1]. Precisely, for any $\varepsilon>0,$
sufficiently large $T$ and

$\theta=\max\left\\{\frac{1}{1+6c},\frac{\sqrt{1+16c+16c^{2}}-1-4c}{4c}\right\\},$

it is proved in [1] that

$N^{\prime}_{K}(T)\gg T^{\theta-\varepsilon},$

where $c$ is determined by the subconvexity
$\zeta(1/2+it)\ll(|t|+1)^{c+\varepsilon}.$ One can find that the Lindelöf
hypothesis implies that

$N^{\prime}_{K}(T)\gg T^{1-\varepsilon},$

which means that one can get almost to a positive proportion. It is worth
emphasizing that the zeros what they find are also simple zeros of $\zeta(s),$
where $\zeta(s)$ is the well known Riemann zeta function. In another paper
[2], the authors also proved that the Riemann Hypothesis implies that more
than $1/54$ of the non-real zeros of $\zeta_{K}(s)$ are simple.

Very recently, in [8], some results in the above are improved by showing
(unconditionally) that

$N^{\prime}_{K}(T)\gg T^{\frac{6}{7}-\varepsilon}.$

In [8], some conditional results were also given. To state these conditional
results precisely, we need introduce two conjectures.

Conjecture 1. Let $N(\sigma,T)$ denote the number of zeros of $\zeta(s)$ for
$\beta\geq\sigma$ and $|\gamma|\leq T.$ Then for any $\varepsilon>0$ and
$1/2\leq\sigma\leq 1,$ we have

$N(\sigma,T)\ll T^{A(\sigma)(1-\sigma)+\varepsilon}.$

Conjecture 2. Let $q\geq 1$ be a fixed integer, and let $\chi$ be a real
primitive character to the modulus $q$. For $\bar{s}=1/2+it,$ any
$\varepsilon>0,$ any real number $k>4$ and sufficiently $T,$ we have

$\int_{1}^{T}\left|L(\bar{s},\chi)\right|^{k}\
\mathrm{d}t\ll_{q}T^{1+\varepsilon},$

where $L(s,\chi)=\sum_{n=1}^{\infty}\chi(n)n^{-s}$ $(\Re s>1)$ is the
Dirichlet $L$-function.

In [8], Wu and Zhao also proved that

  * •

If Conjecture 1 is true with $A(\sigma)=2$, then

$N^{\prime}_{K}(T)\gg T^{1-\varepsilon}.$

  * •

If Conjecture 2 is true with $k=\bf{8}$, then

$N^{\prime}_{K}(T)\gg T^{1-\varepsilon}.$

Surprisingly, we find that $k=\bf{6}$ for Conjecture 2 is sufficient enough
for this purpose. We also remark that in this method, only with a real number
$k\in(4,6),$ one cannot get the desired conclusion. Now we intend to give a
concrete reason for this. We assume that Conjecture 2 is true with certain $k$
and assume that

$N(\sigma,T)\ll T^{(2l+2)(1-\sigma)/(l+2-2\sigma)+\varepsilon},$

which is from the result of [9]. Then in the process of this paper, we can
obtain that

$S(T,1)-S(T,a)\ll T^{1-B(\sigma,k,l)(\sigma-1/2)+(1-2/k)\varepsilon},$

where

$B(\sigma,k,l)=2l(1-2/k)/(l+2-2\sigma)-1.$

Then we have

$B(\sigma,k,l)=\frac{2l(1-2/k)-l-2+2\sigma}{l+2-2\sigma}.$

We need $2l(1-2/k)-l-2+2\sigma>0.$ Hence for $1/2+\varepsilon<\sigma<1,$ we
have

$\frac{k}{k-4}<l+\varepsilon.$

When we choose $k={\bf 6}$ in Conjecture 1, Conjecture 1 just implies the
density result we needed of $l=3,\theta=1,\alpha=1$ in [9] (see also in [4]).

Moreover, one can also give conditional results for this problem by assuming
Conjecture 2 ($k\in(4,6)$) with assumed results that are weaker than the
density hypothesis but stronger than

$N(\sigma,T)\ll T^{A(\sigma)(1-\sigma)+\varepsilon},$

where $A(\sigma)=8/(5-2\sigma).$

######  Theorem 1.1.

If Conjecture 2 is true with $k=\bf{6}$, then

$N^{\prime}_{K}(T)\gg T^{1-\varepsilon}.$

######  Remark 1.

It is worth remarking that if the $L^{6}$-Lindelöf on average could be
replaced by an $L^{4}$-Lindelöf on average, then the theorem would become
unconditional. Since we have known how to prove $L^{4}$-Lindelöf-on-average
for Dirichlet $L$-functions by the classical work of Ingham on the fourth
moment of $\zeta(s)$ (see Titchmarsh [7] or [4]).

##  2\. Preliminaries

For proof of Theorem 1.1, the method in this paper relies heavily on the
relation

$\zeta_{K}(s)=\zeta(s)L(s,\chi),$

where $\chi$ is a real primitive character to the modulus $q=|D|$, where $D$
is the discriminant of $K.$ It is well known that

$\zeta(1-s)=2(2\pi)^{-s}\Gamma(s)\cos(\pi s/2)\zeta(s),$

and

$\displaystyle L(1-s,\chi)$
$\displaystyle=\frac{q^{s}}{\sum_{l=1}^{q}\chi(l)e(l/q)}(2\pi)^{-s}\Gamma(s)(e^{\pi
is/2}+\chi(-1)e^{-\pi is/2})L(s,\chi),$

and

$\zeta^{\prime}_{K}(s)=\zeta^{\prime}(s)L(s,\chi)+\zeta(s)L^{\prime}(s,\chi).$

Let $\rho=\beta+i\gamma$ denote the non-trivial zeros of $\zeta(s)$. Then it
is implied that (for instance, see (8) in [1] and (2.6)-(2.8) in [8])

$\displaystyle|L(1-\rho,\chi)|\ll T^{\beta-1/2}|L(\rho,\chi)|$ (2.1)

and

$\displaystyle|\zeta^{\prime}(1-\rho)|\ll
T^{\beta-1/2}|\zeta^{\prime}(\rho)|.$ (2.2)

For $1/2\leq a\leq 1,$ let

$S(T,a)=\sum_{\rho\in\mathcal{R},1-a\leq\beta\leq
a}\zeta^{\prime}(\rho)L(1-\rho,\chi).$

Moreover, let

$N^{\prime}_{K}(T,a)=\\#\\{\rho_{K}:\rho_{K}=\beta_{K}+i\gamma_{K}\in\mathcal{R},\rho_{K}\
\textup{is\ a\ simple\ zero\ of}\ \zeta_{K}\ \textup{and}\
1/2\leq\beta_{K}\leq a\\}.$

Then we can introduce the following lemmas.

######  Lemma 2.1.

For some $1/2\leq a\leq 5/8,$ suppose that $|S(T,a)|\gg T.$ Then

$N^{\prime}_{K}(T,a)\gg T^{\frac{2a}{4a-1}-\varepsilon}.$

###### Proof.

See Lemma 8 in [8]. ∎

######  Lemma 2.2.

With the notations above, we have

$S(T,1)=L(1,\chi)\frac{T}{4\pi}\log^{2}T+O(T\log T).$

###### Proof.

This is (18) in [1]. ∎

######  Lemma 2.3.

For $k\geq 1$ a fixed integer let $1/2\leq\sigma_{k}^{*}<1$ denote the infimum
of all numbers $\sigma_{k}^{*}$ for which

$\int_{1}^{T}|\zeta(\sigma_{k}^{*}+it)|^{2k}dt\ll T^{1+\varepsilon}$

holds for any $\varepsilon>0.$ Then the asymptotic formula

$\int_{1}^{T}|\zeta(\sigma+it)|^{2k}dt=T\sum_{n=1}^{\infty}d_{k}(n)^{2}n^{-2\sigma}+R(k,\sigma,T)$

holds for $\sigma_{k}^{*}<\sigma<1$ with

$R(k,\sigma,T)\ll T^{(2-\sigma-\sigma_{k}^{*})/(2-2\sigma_{k}^{*})}.$

###### Proof.

See Lemma 8.4 in [4]. ∎

######  Lemma 2.4.

If Conjecture 2 is true with $k={\bf 6},$ then for $1/2\leq\sigma\leq 1,$ we
have

$\displaystyle N(\sigma,T)\ll T^{\frac{8-8\sigma}{5-2\sigma}+\varepsilon},$
(2.3) $\displaystyle\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\
\beta\geq\sigma\end{subarray}}|\zeta^{\prime}(\rho)|^{6}\ll
T^{1+\varepsilon},$ (2.4)

and

$\displaystyle\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\
\beta\geq\sigma\end{subarray}}|L(\rho,\chi)|^{6}\ll T^{1+\varepsilon}.$ (2.5)

###### Proof.

(2.3) can be obtained by Theorem 1 in [9] for the case $l=3,$ $\alpha=1,$ and
$\theta=1.$ (see also in [4]). Now we begin to prove (2.4). By Cauchy’s
theorem,

$\zeta^{\prime}(\rho)=\frac{1}{\Delta}\int_{\Delta}^{2\Delta}\frac{1}{2\pi
i}\int_{|s-\rho|=\delta}\frac{\zeta(s)}{(s-\rho)^{2}}dsd\delta,$

where $\Delta=(\log T)^{-1}.$ Hence if $1/2\leq\beta<1,$ then using Hölder’s
inequality we have

$|\zeta^{\prime}(\rho)|^{6}\ll
T^{\varepsilon}\int_{1/2-2\Delta}^{1+2\Delta}\int_{1}^{T+1}|\zeta(\sigma+it)|^{6}dtd\sigma.$

Since the number of $\rho$ in a square of side length 1 is $\ll(\log T),$ then
by Lemma 2.3 we have

$\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\ \beta\geq
1/2\end{subarray}}|\zeta^{\prime}(\rho)|^{6}\ll
T^{\varepsilon}\int_{1/2-2\Delta}^{1+2\Delta}\int_{1}^{T+1}|\zeta(\sigma+it)|^{6}dtd\delta\ll
T^{1+\varepsilon}.$

(2.5) can be proved by adapting the argument of [1, 8] and Lemma 2.3 in
assuming Conjecture 2 with $k=6.$ ∎

##  3\. Proof of Theorem 1.1

Then we will prove the following lemma.

######  Lemma 3.1.

For any sufficiently small $\varepsilon>0$ and $a\geq
1/2+10\sqrt{\varepsilon},$ if Conjecture 2 is true with $k={\bf 6},$ we have

$S(T,a)\asymp T\log^{2}T.$

###### Proof.

Recall that

$S(T,1)-S(T,a)\ll\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\ \beta\geq
a\end{subarray}}|\zeta^{\prime}(\rho)L(1-\rho,\chi)|+\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\
\beta\geq a\end{subarray}}|\zeta^{\prime}(1-\rho)L(\rho,\chi)|.$

Then by (2.1) and (2.2), one has

$\displaystyle S(T,1)-S(T,a)\ll\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\
\beta\geq
a\end{subarray}}T^{\beta-\frac{1}{2}}|\zeta^{\prime}(\rho)L(\rho,\chi)|.$
(3.1)

By using the Riemann-Stieltjes integral (or the partial summation), one can
obtain that

$\displaystyle\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\ \beta\geq
a\end{subarray}}T^{\beta-\frac{1}{2}}|\zeta^{\prime}(\rho)L(\rho,\chi)|$
$\displaystyle=\int_{a}^{1}T^{\sigma-1/2}dF(\sigma,\mathcal{R}),$

where

$F(\sigma,\mathcal{R}):=\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\
\beta\geq\sigma\end{subarray}}|\zeta^{\prime}(\rho)L(\rho,\chi)|.$

Then applying integration by parts, we have

$\displaystyle\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\ \beta\geq
a\end{subarray}}T^{\beta-\frac{1}{2}}|\zeta^{\prime}(\rho)L(\rho,\chi)|$
$\displaystyle\ll
T^{1-\frac{1}{2}}F(1,\mathcal{R})-T^{a-\frac{1}{2}}F(a,\mathcal{R})$
$\displaystyle-\int_{a}^{1}T^{\sigma-1/2}F(\sigma,\mathcal{R})(\log
T)d\sigma.$

This gives that

$\displaystyle\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\ \beta\geq
a\end{subarray}}T^{\beta-\frac{1}{2}}|\zeta^{\prime}(\rho)L(\rho,\chi)|$
$\displaystyle\ll\max_{a\leq\sigma\leq
1}T^{\sigma-\frac{1}{2}}\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\
\beta\geq\sigma\end{subarray}}|\zeta^{\prime}(\rho)L(\rho,\chi)|(\log T)$
$\displaystyle\ll\max_{a\leq\sigma\leq
1}T^{\sigma-\frac{1}{2}+\varepsilon}\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\
\beta\geq\sigma\end{subarray}}|\zeta^{\prime}(\rho)L(\rho,\chi)|.$

By Hölder’s inequalities, we have

$\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\
\beta\geq\sigma\end{subarray}}|\zeta^{\prime}(\rho)L(\rho,\chi)|\leq
N(\sigma,T)^{\frac{2}{3}}\left(\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\
\beta\geq\sigma\end{subarray}}|\zeta^{\prime}(\rho)|^{6}\right)^{\frac{1}{6}}\left(\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\
\beta\geq\sigma\end{subarray}}|L(\rho,\chi)|^{6}\right)^{\frac{1}{6}}.$

Recall Lemma 2.4 and the 6th moment hypothesis, one can get

$\displaystyle\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\
\beta\geq\sigma\end{subarray}}|\zeta^{\prime}(\rho)L(\rho,\chi)|\ll
T^{\frac{8-8\sigma}{5-2\sigma}\times\frac{2}{3}+\frac{1}{3}+\varepsilon}.$

Then

$\displaystyle\sum_{\begin{subarray}{c}\rho\in\mathcal{R}\\\
\beta\geq\sigma\end{subarray}}|\zeta^{\prime}(\rho)L(\rho,\chi)|\ll
T^{1+\frac{3-6\sigma}{5-2\sigma}\times\frac{2}{3}+\varepsilon}\ll
T^{1-\frac{4}{5-2\sigma}\times(\sigma-\frac{1}{2})+\varepsilon}.$

Hence by (3.1), we have

$\displaystyle S(T,1)-S(T,a)$ $\displaystyle\ll\max_{a\leq\sigma\leq
1}T^{\sigma-1/2}T^{1-\frac{4}{5-2\sigma}\times(\sigma-\frac{1}{2})+\varepsilon}$
$\displaystyle\ll\max_{a\leq\sigma\leq
1}T^{1-\frac{4}{5-2\sigma}\times(\sigma-\frac{1}{2})+\frac{5-2\sigma}{5-2\sigma}\times(\sigma-\frac{1}{2})+\varepsilon}.$

This gives that

$S(T,1)-S(T,a)\ll\max_{a\leq\sigma\leq
1}T^{1-\frac{2}{5-2\sigma}\times(\sigma-\frac{1}{2})^{2}+\varepsilon}.$

For suitable $\varepsilon^{\prime}=10\sqrt{\varepsilon},$ $a\geq
1/2+\varepsilon^{\prime}$ and $a\leq\sigma\leq 1,$ we can get that

$\displaystyle S(T,1)-S(T,a)$ $\displaystyle\ll
T^{1-\frac{25}{1-5\sqrt{\varepsilon}}+\varepsilon}$ $\displaystyle\ll
T^{1-24\varepsilon}$ $\displaystyle\ll T^{1-\varepsilon}.$

Recall (see Lemma 2.2) that

$S(T,1)=L(1,\chi)\frac{T}{4\pi}\log^{2}T+O(T\log T).$

Hence for $a\geq 1/2+\varepsilon^{\prime},$ we can obtain that

$S(T,a)\asymp T\log^{2}T.$

This completes the proof of Lemma 3.1. ∎

Now our theorem can be deduced by Lemma 2.1, Lemma 3.1 and the choice of
$a=1/2+\varepsilon^{\prime}.$

$\mathbf{Acknowledgements}$ I am deeply grateful to the referee(s) for
carefully reading the manuscript and making useful suggestions.

## References

  * [1] J.B. Conrey, A. Ghosh and S.M. Gonek, Simple zeros of the zeta function of a quadratic number field. I. Invent. Math. 86, 563-576 (1986)
  * [2] J.B. Conrey, A. Ghosh and S.M. Gonek, Simple zeros of the zeta function of a quadratic number field. II, In Analytic Number Theory and Diophantine Problems (Stillwater, OK, 1984) (Progress in Mathematics 70) Birkh auser (Boston, MA, 1987) 87-114.
  * [3] D.R. Heath-Brown, Simple zeros of the Riemann zeta-function on the critical line. Bull. Lond. Math. Soc. 11 (1979), 17- 18\.
  * [4] A. Ivić, The Riemann Zeta-function. John Wiley $\&$ Sons, New York, 1985.
  * [5] N. Levinson, More than one third of zeros of Riemann’s zeta-function are on $\sigma=1/2$. Advances in Math., 13 (1974), 383-436.
  * [6] A. Selberg, On the zeros of Riemann’s zeta-function. Skr. Nor. Vidensk.-Akad. Oslo, I 10 (1942), 1-59.
  * [7] E.C. Titchmarsh, The Theory of the Riemann Zeta-function. revised by D. R. HeathBrown, 2nd edn., Clarendon Press (Oxford, 1986).
  * [8] X.S. Wu and L.L. Zhao, On simple zeros of the Dedekind zeta function of a quadratic number field. Mathematika 65, 851-861 (2019)
  * [9] Y.B. Ye and D.Y. Zhang, Zero density for automorphic $L$ -functions. J. Number Theory 133, 3877-3901 (2013)