Characterizations of the symmetrized polydisc via another family of   domains

Sourav Pal; Samriddho Roy

arXiv:1904.03745·math.CV·October 13, 2021

Characterizations of the symmetrized polydisc via another family of domains

Sourav Pal, Samriddho Roy

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TL;DR

This paper introduces a new domain called the extended symmetrized polydisc, characterizes points within it and the original symmetrized polydisc, and explores their geometric properties, revealing differences and similarities for various dimensions.

Contribution

The paper provides new characterizations of the symmetrized polydisc and introduces the extended symmetrized polydisc, establishing their relationship and geometric properties.

Findings

01

G_n= ilde{\u00a0G}_n for n=1,2

02

G_n is a proper subset of ilde{\u00a0G}_n for n

03

ilde{\u00a0G}_n is non-convex but polynomially convex, starlike about the origin, not circled

Abstract

We find new characterizations for the points in the \textit{symmetrized polydisc} $G_{n}$ , a family of domains associated with the spectral interpolation, defined by \[ \mathbb G_n :=\left\{ \left(\sum_{1\leq i\leq n} z_i,\sum_{1\leq i<j\leq n}z_iz_j \dots, \prod_{i=1}^n z_i \right): \,|z_i|<1, i=1,\dots,n \right \}. \] We introduce a new family of domains which we call \textit{the extended symmetrized polydisc} $G_{n}$ , and define in the following way: \begin{align*} \widetilde{\mathbb G}_n := \Bigg\{ (y_1,\dots,y_{n-1}, q)\in \mathbb C^n :\; q \in \mathbb D, \; y_j = \beta_j + \bar{\beta}_{n-j} q, \; \beta_j \in \mathbb C &\text{ and }\\ |\beta_j|+ |\beta_{n-j}| < {n \choose j} &\text{ for } j=1,\dots, n-1 \Bigg\}. \end{align*} We show that $G_{n} = G_{n}$ for $n = 1, 2$ and that $G_{n} ⊊ G_{n}$ for $n\geq…

Equations320

G_{n} := {(1 \leq i \leq n \sum z_{i}, 1 \leq i < j \leq n \sum z_{i} z_{j} \dots, i = 1 \prod n z_{i}) : ∣ z_{i} ∣ < 1, i = 1, \dots, n} .

G_{n} := {(1 \leq i \leq n \sum z_{i}, 1 \leq i < j \leq n \sum z_{i} z_{j} \dots, i = 1 \prod n z_{i}) : ∣ z_{i} ∣ < 1, i = 1, \dots, n} .

\displaystyle\widetilde{\mathbb{G}}_{n}:=\Bigg{\{}(y_{1},\dots,y_{n-1},q)\in\mathbb{C}^{n}:\;q\in\mathbb{D},\;y_{j}=\beta_{j}+\bar{\beta}_{n-j}q,\;\beta_{j}\in\mathbb{C}

\displaystyle\widetilde{\mathbb{G}}_{n}:=\Bigg{\{}(y_{1},\dots,y_{n-1},q)\in\mathbb{C}^{n}:\;q\in\mathbb{D},\;y_{j}=\beta_{j}+\bar{\beta}_{n-j}q,\;\beta_{j}\in\mathbb{C}

∣ β_{j} ∣ + ∣ β_{n - j} ∣ < (j n)

G_{n} := {(1 \leq i \leq n \sum z_{i}, 1 \leq i < j \leq n \sum z_{i} z_{j} \dots, i = 1 \prod n z_{i}) : ∣ z_{i} ∣ < 1, i = 1, \dots, n} .

G_{n} := {(1 \leq i \leq n \sum z_{i}, 1 \leq i < j \leq n \sum z_{i} z_{j} \dots, i = 1 \prod n z_{i}) : ∣ z_{i} ∣ < 1, i = 1, \dots, n} .

π_{n} (z_{1}, \dots, z_{n}) = (1 \leq i \leq n \sum z_{i}, 1 \leq i < j \leq n \sum z_{i} z_{j}, \dots, i = 1 \prod n z_{i}) .

π_{n} (z_{1}, \dots, z_{n}) = (1 \leq i \leq n \sum z_{i}, 1 \leq i < j \leq n \sum z_{i} z_{j}, \dots, i = 1 \prod n z_{i}) .

Γ_{n} = \overline{G_{n}} := {(1 \leq i \leq n \sum z_{i}, 1 \leq i < j \leq n \sum z_{i} z_{j} \dots, i = 1 \prod n z_{i}) : ∣ z_{i} ∣ \leq 1, i = 1, \dots, n} .

Γ_{n} = \overline{G_{n}} := {(1 \leq i \leq n \sum z_{i}, 1 \leq i < j \leq n \sum z_{i} z_{j} \dots, i = 1 \prod n z_{i}) : ∣ z_{i} ∣ \leq 1, i = 1, \dots, n} .

G_{n} = {(s_{1}, \dots, s_{n - 1}, p) \in C^{n} : p \in D, s_{j} = β_{j} + \overset{ˉ}{β}_{n - j} p and (β_{1}, \dots, β_{n - 1}) \in G_{n - 1}} .

G_{n} = {(s_{1}, \dots, s_{n - 1}, p) \in C^{n} : p \in D, s_{j} = β_{j} + \overset{ˉ}{β}_{n - j} p and (β_{1}, \dots, β_{n - 1}) \in G_{n - 1}} .

Γ_{n} = \overline{G_{n}} = {(s_{1}, \dots, s_{n - 1}, p) \in C^{n} : p \in \overline{D}, s_{j} = β_{j} + \overset{ˉ}{β}_{n - j} p and (β_{1}, \dots, β_{n - 1}) \in Γ_{n - 1}} .

Γ_{n} = \overline{G_{n}} = {(s_{1}, \dots, s_{n - 1}, p) \in C^{n} : p \in \overline{D}, s_{j} = β_{j} + \overset{ˉ}{β}_{n - j} p and (β_{1}, \dots, β_{n - 1}) \in Γ_{n - 1}} .

\displaystyle\widetilde{\mathbb{G}}_{n}:=\Bigg{\{}(y_{1},\dots,y_{n-1},q)\in\mathbb{C}^{n}:|q|<1,\>y_{j}=\beta_{j}+\bar{\beta}_{n-j}q\>\text{ with }\;|\beta_{j}|

\displaystyle\widetilde{\mathbb{G}}_{n}:=\Bigg{\{}(y_{1},\dots,y_{n-1},q)\in\mathbb{C}^{n}:|q|<1,\>y_{j}=\beta_{j}+\bar{\beta}_{n-j}q\>\text{ with }\;|\beta_{j}|

\displaystyle 1\leq j\leq n-1\Bigg{\}}.

Φ_{j} (z, y) = ⎩ ⎨ ⎧ \frac{( j n ) q z - y _{j}}{y _{n - j} z - ( j n )} \frac{y _{j}}{( j n )} if y_{j} y_{n - j} \neq = (j n)^{2} q, if y_{j} y_{n - j} = (j n)^{2} q,

Φ_{j} (z, y) = ⎩ ⎨ ⎧ \frac{( j n ) q z - y _{j}}{y _{n - j} z - ( j n )} \frac{y _{j}}{( j n )} if y_{j} y_{n - j} \neq = (j n)^{2} q, if y_{j} y_{n - j} = (j n)^{2} q,

D_{j} (y) = z \in D sup ∣ Φ_{j} (z, y) ∣ = ∥ Φ_{j} (., y) ∥_{H^{\infty}},

D_{j} (y) = z \in D sup ∣ Φ_{j} (z, y) ∣ = ∥ Φ_{j} (., y) ∥_{H^{\infty}},

\tilde{y} = (y_{1}, \dots, y_{j - 1}, y_{n - j}, y_{j + 1}, \dots, y_{n - j - 1}, y_{j}, y_{n - j + 1}, \dots, y_{n - 1}, q) \in C^{n},

\tilde{y} = (y_{1}, \dots, y_{j - 1}, y_{n - j}, y_{j + 1}, \dots, y_{n - j - 1}, y_{j}, y_{n - j + 1}, \dots, y_{n - 1}, q) \in C^{n},

\frac{( j n ) ( y _{j} - y ˉ _{n - j} q )}{( j n ) ^{2} - ∣ y _{n - j} ∣ ^{2}} and \frac{y _{j} y _{n - j} - ( j n ) ^{2} q}{( j n ) ^{2} - ∣ y _{n - j} ∣ ^{2}}

\frac{( j n ) ( y _{j} - y ˉ _{n - j} q )}{( j n ) ^{2} - ∣ y _{n - j} ∣ ^{2}} and \frac{y _{j} y _{n - j} - ( j n ) ^{2} q}{( j n ) ^{2} - ∣ y _{n - j} ∣ ^{2}}

D_{j} (y) = ⎩ ⎨ ⎧ \frac{( j n ) ∣ y _{j} - y ˉ _{n - j} q ∣ + y _{j} y _{n - j} - ( j n ) ^{2} q}{( j n ) ^{2} - ∣ y _{n - j} ∣ ^{2}} \frac{∣ y _{j} ∣}{( j n )} \infty if ∣ y_{n - j} ∣ < (j n) and y_{j} y_{n - j} \neq = (j n)^{2} q, if y_{j} y_{n - j} = (j n)^{2} q, otherwise .

D_{j} (y) = ⎩ ⎨ ⎧ \frac{( j n ) ∣ y _{j} - y ˉ _{n - j} q ∣ + y _{j} y _{n - j} - ( j n ) ^{2} q}{( j n ) ^{2} - ∣ y _{n - j} ∣ ^{2}} \frac{∣ y _{j} ∣}{( j n )} \infty if ∣ y_{n - j} ∣ < (j n) and y_{j} y_{n - j} \neq = (j n)^{2} q, if y_{j} y_{n - j} = (j n)^{2} q, otherwise .

∥ Φ_{j} (., y) ∥_{H^{\infty}} < 1 and if y_{j} y_{n - j} = (j n)^{2} q then, in addition, ∣ y_{n - j} ∣ < (j n)

∥ Φ_{j} (., y) ∥_{H^{\infty}} < 1 and if y_{j} y_{n - j} = (j n)^{2} q then, in addition, ∣ y_{n - j} ∣ < (j n)

∥ Φ_{n - j} (., y) ∥_{H^{\infty}} < 1 and if y_{j} y_{n - j} = (j n)^{2} q then, in addition, ∣ y_{j} ∣ < (j n);

∥ Φ_{n - j} (., y) ∥_{H^{\infty}} < 1 and if y_{j} y_{n - j} = (j n)^{2} q then, in addition, ∣ y_{j} ∣ < (j n);

(j n) ∣ y_{j} - \overset{y}{ˉ}_{n - j} q ∣ + y_{j} y_{n - j} - (j n)^{2} q < (j n)^{2} - ∣ y_{n - j} ∣^{2}

(j n) ∣ y_{j} - \overset{y}{ˉ}_{n - j} q ∣ + y_{j} y_{n - j} - (j n)^{2} q < (j n)^{2} - ∣ y_{n - j} ∣^{2}

(j n) ∣ y_{n - j} - \overset{y}{ˉ}_{j} q ∣ + y_{j} y_{n - j} - (j n)^{2} q < (j n)^{2} - ∣ y_{j} ∣^{2};

(j n) ∣ y_{n - j} - \overset{y}{ˉ}_{j} q ∣ + y_{j} y_{n - j} - (j n)^{2} q < (j n)^{2} - ∣ y_{j} ∣^{2};

∣ y_{j} ∣^{2} - ∣ y_{n - j} ∣^{2} + (j n)^{2} ∣ q ∣^{2} + 2 (j n) ∣ y_{n - j} - \overset{y}{ˉ}_{j} q ∣ < (j n)^{2} and ∣ y_{n - j} ∣ < (j n)

∣ y_{j} ∣^{2} - ∣ y_{n - j} ∣^{2} + (j n)^{2} ∣ q ∣^{2} + 2 (j n) ∣ y_{n - j} - \overset{y}{ˉ}_{j} q ∣ < (j n)^{2} and ∣ y_{n - j} ∣ < (j n)

∣ y_{n - j} ∣^{2} - ∣ y_{j} ∣^{2} + (j n)^{2} ∣ q ∣^{2} + 2 (j n) ∣ y_{j} - \overset{y}{ˉ}_{n - j} q ∣ < (j n)^{2} and ∣ y_{j} ∣ < (j n);

∣ y_{n - j} ∣^{2} - ∣ y_{j} ∣^{2} + (j n)^{2} ∣ q ∣^{2} + 2 (j n) ∣ y_{j} - \overset{y}{ˉ}_{n - j} q ∣ < (j n)^{2} and ∣ y_{j} ∣ < (j n);

det B_{1} = det B_{2} = \dots = det B_{[\frac{n}{2}]} = q;

det B_{1} = det B_{2} = \dots = det B_{[\frac{n}{2}]} = q;

det B_{1} = det B_{2} = \dots = det B_{[\frac{n}{2}]} = q .

det B_{1} = det B_{2} = \dots = det B_{[\frac{n}{2}]} = q .

0 \neq = (j n) - y_{j} z - y_{n - j} w + (j n) q z w = (j n) (1 - \frac{y _{j}}{( j n )} z) (1 - \frac{y _{n - j}}{( j n )} w),

0 \neq = (j n) - y_{j} z - y_{n - j} w + (j n) q z w = (j n) (1 - \frac{y _{j}}{( j n )} z) (1 - \frac{y _{n - j}}{( j n )} w),

(j n) ∣ y_{j} - \overset{y}{ˉ}_{n - j} q ∣ < (j n)^{2} - ∣ y_{n - j} ∣^{2} \Leftrightarrow \frac{∣ y _{j} ∣}{( j n )} (j n)^{2} - ∣ y_{n - j} ∣^{2} < (j n)^{2} - ∣ y_{n - j} ∣^{2}

(j n) ∣ y_{j} - \overset{y}{ˉ}_{n - j} q ∣ < (j n)^{2} - ∣ y_{n - j} ∣^{2} \Leftrightarrow \frac{∣ y _{j} ∣}{( j n )} (j n)^{2} - ∣ y_{n - j} ∣^{2} < (j n)^{2} - ∣ y_{n - j} ∣^{2}

∣ y_{j} ∣^{2} - ∣ y_{n - j} ∣^{2} + (j n)^{2} ∣ q ∣^{2} + 2 (j n) ∣ y_{n - j} - \overset{y}{ˉ}_{j} q ∣ < (j n)^{2}

∣ y_{j} ∣^{2} - ∣ y_{n - j} ∣^{2} + (j n)^{2} ∣ q ∣^{2} + 2 (j n) ∣ y_{n - j} - \overset{y}{ˉ}_{j} q ∣ < (j n)^{2}

\Leftrightarrow

\Rightarrow

∣ y_{j} ∣ < (j n)

∣ y_{j} ∣ < (j n)

\Rightarrow

\Rightarrow

\Leftrightarrow

(j n)^{2} 1 - \frac{∣ y _{j} ∣ ^{2}}{( j n ) ^{2}} 1 - \frac{∣ y _{n - j} ∣ ^{2}}{( j n ) ^{2}} > 0.

(j n)^{2} 1 - \frac{∣ y _{j} ∣ ^{2}}{( j n ) ^{2}} 1 - \frac{∣ y _{n - j} ∣ ^{2}}{( j n ) ^{2}} > 0.

- (j n)^{2} 1 - \frac{∣ y _{j} ∣ ^{2}}{( j n ) ^{2}} 1 - \frac{∣ y _{n - j} ∣ ^{2}}{( j n ) ^{2}} < 0 and ∣ y_{j} ∣∣ y_{n - j} ∣ < (j n)^{2} .

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Characterizations of the symmetrized polydisc via another family of domains

Sourav Pal

Mathematics Department, Indian Institute of Technology Bombay, Powai, Mumbai - 400076, India.

[email protected] , [email protected]

and

Samriddho Roy

Mathematics Department, Indian Institute of Technology Bombay, Powai, Mumbai - 400076, India.

[email protected]

Abstract.

We find new characterizations for the points in the symmetrized polydisc $\mathbb{G}_{n}$ , a family of domains associated with the spectral interpolation, defined by

[TABLE]

We introduce a new family of domains which we call the extended symmetrized polydisc $\widetilde{\mathbb{G}}_{n}$ , and define in the following way:

[TABLE]

We show that $\mathbb{G}_{n}=\widetilde{\mathbb{G}}_{n}$ for $n=1,2$ and that ${\mathbb{G}}_{n}\subsetneq\widetilde{\mathbb{G}}_{n}$ for $n\geq 3$ . We first obtain a variety of characterizations for the points in $\widetilde{\mathbb{G}}_{n}$ and we apply these necessary and sufficient conditions to produce an analogous set of characterizations for the points in ${\mathbb{G}}_{n}$ . Also we obtain similar characterizations for the points in $\Gamma_{n}\setminus{\mathbb{G}}_{n}$ , where $\Gamma_{n}=\overline{{\mathbb{G}}_{n}}$ . A set of $n-1$ fractional linear transformations play central role in the entire program. We also show that for $n\geq 2$ , $\widetilde{\mathbb{G}}_{n}$ is non-convex but polynomially convex and is starlike about the origin but not circled.

Key words and phrases:

Symmetrized polydisc, extended symmetrized polydisc

2010 Mathematics Subject Classification:

32A07, 32E20, 32Q02

The named author was supported by the Seed Grant of IIT Bombay, the CPDA of Govt. of India, the INSPIRE Faculty Award (Award No. DST/INSPIRE/04/2014/001462) of DST, India and the MATRICS Award of SERB, (Award No. MTR/2019/001010) of DST, India. The second named author was supported by a Ph.D fellowship from the University Grand Commission of India.

1. Introduction

The following notations and terminologies will be used throughout the paper. We denote by $\mathbb{R}$ and $\mathbb{C}$ the sets of real numbers and complex numbers respectively. By $\mathbb{D}$ and $\mathbb{T}$ we mean the open unit disc and the unit circle respectively with centre at the origin of $\mathbb{C}$ . We follow the standard convention to denote by $\mathcal{M}_{m\times n}(\mathbb{C})$ (or $\mathbb{C}^{m\times n}$ ) the space of all $m\times n$ complex matrices and $\mathcal{M}_{n}(\mathbb{C})$ is used when $m=n$ . For a matrix $A\in\mathcal{M}_{n}(\mathbb{C})$ , $\lVert A\lVert$ is the operator norm of $A$ . Also, the spectrum and the spectral radius of a bounded operator $T$ are denoted by $\sigma(T)$ and $r(T)$ respectively.

In this article, we contribute to the understanding of the geometry of the symmetrized polydisc $\mathbb{G}_{n}$ by producing a set of necessary and sufficient conditions each of which characterizes a point in $\mathbb{G}_{n}$ , where $\mathbb{G}_{n}$ is the following family of domains:

[TABLE]

This family of domains $\{\mathbb{G}_{n}\}$ is naturally associated with the spectral Nevanlinna-Pick interpolation in the following way: a matrix $T$ belongs to the spectral unit ball $\mathcal{B}_{n}^{1}\subset\mathcal{M}_{n}(\mathbb{C})$ , i.e., $r(T)<1$ if and only if $\pi_{n}(\xi_{1},\dots,\xi_{n})\in\mathbb{G}_{n}$ (see [17]). Here $\xi_{1},\dots,\xi_{n}$ are the eigenvalues of $T$ and $\pi_{n}$ is the symmetrization map on $\mathbb{C}^{n}$ defined by

[TABLE]

For given distinct points $\eta_{1},\dots,\eta_{k}$ in $\mathbb{D}$ and matrices $T_{1},\dots,T_{k}\in\mathcal{B}_{n}^{1}$ , the spectral Nevanlinna-Pick interpolation seeks necessary and sufficient conditions under which there exists an analytic function $F:\mathbb{D}\rightarrow\mathcal{M}_{n}(\mathbb{C})$ that interpolates the data, that is, $F(\eta_{i})=T_{i}$ for $i=1,\dots,k$ . Apart from the derogatory matrices, the $n\times n$ spectral Nevanlinna-Pick problem is equivalent to a similar interpolation problem for the symmetrized polydisc $\mathbb{G}_{n}$ (see [6], Theorem 2.1). Note that a bounded domain like $\mathbb{G}_{n}$ , which has complex-dimension $n$ , is much easier to deal with than an unbounded $n^{2}$ -dimensional object like $\mathcal{B}_{n}^{1}$ . The symmetrized polydisc has attracted considerable attention in past two decades because of its rich function theory [3, 13, 19, 25, 29, 30], elegent complex geometry [4, 9, 10, 16, 18, 20, 21, 22] and associated operator theory [5, 8, 11, 12, 14, 23, 24, 28] (also see references there in). It is evident from the definition that $\mathbb{G}_{1}=\mathbb{D}$ which is convex but for $n\geq 2$ , $\mathbb{G}_{n}$ is non-convex but polynomially convex (see [5, 18]).

The symmetrized polydisc $\mathbb{G}_{n}$ and its closure $\Gamma_{n}$ are the image of polydisc $\mathbb{D}^{n}$ and the closed polydisc $\overline{\mathbb{D}^{n}}$ respectively under the symmetrization map $\pi_{n}$ . We obtain from the literature (see [17, 18]) that

[TABLE]

In [17], Costara describe the points in $\mathbb{G}_{n}$ and $\Gamma_{n}$ in the following way:

[TABLE]

Taking cue from Costara’s characterization, we introduce a new family of domains, namely the extended symmetrized polydisc $\widetilde{\mathbb{G}}_{n}$ , which we define in the following way:

[TABLE]

It is obvious that if $(s_{1},\dots,s_{n-1},p)\in\mathbb{G}_{n}$ , then $|s_{i}|<{n\choose i}$ . So, if $\left(\beta_{1},\dots,\beta_{n-1}\right)\in\mathbb{G}_{n-1}$ , then $|\beta_{j}|+|\beta_{n-j}|<{n-1\choose j}+{n-1\choose n-j}={n\choose j}$ . Therefore, it follows that $\mathbb{G}_{n}\subseteq\widetilde{\mathbb{G}}_{n}$ . It is evident that $\widetilde{\mathbb{G}_{1}}={\mathbb{G}_{1}}=\mathbb{D}$ and we shall prove that $\widetilde{\mathbb{G}_{2}}={\mathbb{G}_{2}}$ and that $\mathbb{G}_{n}\subsetneq\widetilde{\mathbb{G}_{n}}$ for $n\geq 3$ . We first characterize the points in $\widetilde{\mathbb{G}}_{n}$ and its closure $\widetilde{\Gamma}_{n}$ in several different ways. Then we use these characterizations to obtain analogous characterizations for the points in $\mathbb{G}_{n}$ and its closure $\Gamma_{n}$ . To characterize the points in $\widetilde{\mathbb{G}}_{n}$ and $\widetilde{\Gamma}_{n}$ , we introduce $n-1$ fractional linear transformations $\Phi_{1},\dots,\Phi_{n-1}$ , which play pivotal role in this paper. Also we show that $\widetilde{\Gamma}_{n}$ is neither convex nor circled but polynomially convex and starlike and hence is simply connected.

Note. This article is a part of authors’ unpublished note [26]. The techniques that are employed in the proofs of Theorems 2.2 and 2.4 are generalization in several variables of the techniques that are used in the article [2].

2. Characterizations of the points in $\widetilde{\mathbb{G}}_{n}$ and $\widetilde{\Gamma}_{n}$

For characterizing the points in $\widetilde{\mathbb{G}}_{n}$ and $\widetilde{\Gamma_{n}}$ , we introduce $(n-1)$ fractional linear transformations $\Phi_{1},\dots,\Phi_{n-1}$ in the following way.

Definition 2.1.

For $z\in\mathbb{C}$ , $y=(y_{1},\dots,y_{n-1},q)\in\mathbb{C}^{n}$ and for any $j\in\left\{1,\dots,n-1\right\}$ , let us define a meromorphic map

[TABLE]

and

[TABLE]

where ${H^{\infty}}$ denotes the Banach space of bounded complex-valued analytic functions on $\mathbb{D}$ equipped with supremum norm.

Needless to mention that for a fixed $y=(y_{1},\dots,y_{n-1},q)\in\mathbb{C}^{n}$ , the function $\Phi_{j}(.,y)$ is a Möbius transformation. It is evident from the definition that $\Phi_{j}(.,y)=\Phi_{n-j}(.,\tilde{y})$ , where

[TABLE]

which is obtained by interchanging the $i$ -th and the $j$ -th coordinates of $y$ . It can be shown by a few steps of calculations that if $|y_{n-j}|\neq{n\choose j}$ , then the image set $\Phi_{j}(\mathbb{T},y)$ is a circle whose center and radius are

[TABLE]

respectively. If $|y_{n-j}|<{n\choose j}$ , then $\Phi_{j}(.,y)$ is bounded on $\mathbb{D}$ . By the maximum modulus principle, the set $\Phi_{j}(\mathbb{D},y)$ must be lying inside the circle $\Phi_{j}(\mathbb{T},y)$ for that $j\in\{1,\dots,n-1\}$ . Now by continuity of the function $\Phi_{j}(.,y)$ , the set $\Phi_{j}(\mathbb{D},y)$ is the open disc with boundary $\Phi_{j}(\mathbb{T},y)$ . Hence for a point $y=(y_{1},\dots,y_{n-1},q)\in\mathbb{C}^{n}$ and $j\in\{1,\dots,n-1\}$ if $|y_{n-j}|<{n\choose j}$ , then the function $\Phi_{j}(.,y)$ maps $\mathbb{D}$ to the open disc with center and radius as in (2.3). Note that the point of maximum modulus of a closed disc with center $a$ and radius $r$ has modulus $|a|+r$ . Hence

[TABLE]

2.1. Characterizations for the points in $\widetilde{\mathbb{G}}_{n}$

Being armed with the functions $\Phi_{j}$ ’s and $D_{j}$ ’s, we now characterize the points in $\widetilde{\mathbb{G}}_{n}$ .

Theorem 2.2.

For a point $y=(y_{1},\dots,y_{n-1},q)\in\mathbb{C}^{n}$ , the following are equivalent:

$(1)$

$y\in\widetilde{\mathbb{G}}_{n}$ ; 2. $(2)$

${n\choose j}-y_{j}z-y_{n-j}w+{n\choose j}qzw\neq 0$ , for all $z,w\in\overline{\mathbb{D}}$ and for all $j=1,\dots,\left[\frac{n}{2}\right]$ ; 3. $(3)$

for all $j=1,\dots,\left[\frac{n}{2}\right]$ ,

[TABLE] 4. $(3^{\prime})$

for all $j=1,\dots,\left[\frac{n}{2}\right]$ ,

[TABLE] 5. $(4)$

for all $j=1,\dots,\left[\frac{n}{2}\right]$ ,

[TABLE] 6. $(4^{\prime})$

for all $j=1,\dots,\left[\frac{n}{2}\right]$ ,

[TABLE] 7. $(5)$

for all $j=1,\dots,\left[\frac{n}{2}\right]$ ,

[TABLE] 8. $(5^{\prime})$

for all $j=1,\dots,\left[\frac{n}{2}\right]$ ,

[TABLE] 9. $(6)$

$|q|<1$ * and $|y_{j}|^{2}+|y_{n-j}|^{2}-{n\choose j}^{2}|q|^{2}+2\left|y_{j}y_{n-j}-{n\choose j}^{2}q\right|<{n\choose j}^{2}$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ ;* 10. $(7)$

$\left|y_{n-j}-\bar{y}_{j}q\right|+\left|y_{j}-\bar{y}_{n-j}q\right|<{n\choose j}(1-|q|^{2})$ * for all $j=1,\dots,\left[\frac{n}{2}\right]$ ;* 11. $(8)$

there exist $\left[\frac{n}{2}\right]$ number of $2\times 2$ matrices $B_{1},\dots,B_{\left[\frac{n}{2}\right]}$ such that $\lVert B_{j}\lVert<1$ , $y_{j}={n\choose j}[B_{j}]_{11}$ , $y_{n-j}={n\choose j}[B_{j}]_{22}$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ and

[TABLE] 12. $(9)$

there exist $\left[\frac{n}{2}\right]$ number of $2\times 2$ symmetric matrices $B_{1},\dots,B_{\left[\frac{n}{2}\right]}$ such that $\lVert B_{j}\lVert<1$ , $y_{j}={n\choose j}[B_{j}]_{11}$ , $y_{n-j}={n\choose j}[B_{j}]_{22}$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ and

[TABLE]

Proof.

First we prove the equivalence of the conditions $(2)-(6)$ and $(8),(9)$ .

Case-A: We first consider the case when $y_{j}y_{n-j}={n\choose j}^{2}q$ . In this case, each of the conditions $(2)-(6)$ and $(8),(9)$ is equivalent to the pair of conditions $|y_{j}|<{n\choose j}$ and $|y_{n-j}|<{n\choose j}$ .

$(2)$

: Since we have

[TABLE]

the condition (2) is equivalent to (1). Otherwise if $|y_{j}|\geq{n\choose j}|$ , we may take $z={n\choose j}y_{j}^{-1}\in\overline{\mathbb{D}}$ or if $|y_{n-j}|\geq{n\choose j}$ , we may take $w={n\choose j}y_{n-j}^{-1}\in\overline{\mathbb{D}}$ to reach a contradiction to the equation (2). 2. $(3)$

: For the case $y_{j}y_{n-j}={n\choose j}^{2}q$ , we have $\Phi_{j}(.,y)=\dfrac{y_{j}}{{n\choose j}}$ . Thus $\lVert\Phi_{j}(.,y)\lVert<1$ if and only if $|y_{j}|<{n\choose j}$ . 3. $(3^{\prime})$

: In this case, $\Phi_{n-j}(.,y)=\dfrac{y_{n-j}}{{n\choose j}}$ . Thus $\lVert\Phi_{n-j}(.,y)\lVert<1$ if and only if $|y_{n-j}|<{n\choose j}$ . 4. $(4)$

: Since $y_{j}y_{n-j}={n\choose j}^{2}q$ , the statement $(4)$ is equivalent to

[TABLE]

which is equivalent to $|y_{n-j}|<{n\choose j}\;\text{ and }\;|y_{j}|<{n\choose j}$ . The reason of the forward implication is that, if $|y_{n-j}|>{n\choose j}$ , then the left hand side of the inequality $\eqref{1}$ becomes positive while the right hand side becomes negative, a contradiction. Also, the inequality $\eqref{1}$ is not valid for the case $|y_{n-j}|={n\choose j}$ . Hence $|y_{n-j}|<{n\choose j}$ and consequently, by $\eqref{1}$ , $|y_{j}|<{n\choose j}$ . 5. $(4^{\prime})$

: Similarly, statement $(4^{\prime})$ is equivalent to the pair of conditions $|y_{j}|<{n\choose j}$ and $|y_{n-j}|<{n\choose j}$ . 6. $(5)$

:

[TABLE]

Also by hypothesis, $|y_{n-j}|<{n\choose j}$ . Hence statement $(5)$ implies $|y_{j}|<{n\choose j}$ and $|y_{n-j}|<{n\choose j}$ , when $y_{j}y_{n-j}={n\choose j}^{2}q$ . On the other hand

[TABLE]

Hence $|y_{j}|<{n\choose j}$ and $|y_{n-j}|<{n\choose j}$ together imply statement $(5)$ for the case $y_{j}y_{n-j}={n\choose j}^{2}q$ . 7. $(5^{\prime})$

: Similarly, statement $(5^{\prime})$ is equivalent to the pair of conditions $|y_{j}|<{n\choose j}$ and $|y_{n-j}|<{n\choose j}$ . 8. $(6)$

: The fact that $|y_{j}|^{2}+|y_{n-j}|^{2}-{n\choose j}^{2}|q|^{2}+2\left|y_{j}y_{n-j}-{n\choose j}^{2}q\right|<{n\choose j}^{2}$ is equivalent to ${n\choose j}^{2}\left(1-\dfrac{|y_{j}|^{2}}{{n\choose j}^{2}}\right)+|y_{n-j}|^{2}\left(1-\dfrac{|y_{j}|^{2}}{{n\choose j}^{2}}\right)>0$ , which holds if and only if

[TABLE]

Also $|q|<1\Leftrightarrow|y_{j}||y_{n-j}|<{n\choose j}^{2}$ . Hence in this case, statement $(6)$ holds if and only if

[TABLE]

The first inequality of $\eqref{5}$ holds only if $\left(1-\dfrac{|y_{j}|^{2}}{{n\choose j}^{2}}\right)$ and $\left(1-\dfrac{|y_{n-j}|^{2}}{{n\choose j}^{2}}\right)$ have same sign (positive or negative). Then the second inequality of $\eqref{5}$ leads to the fact that both the inequalities in $\eqref{5}$ hold only if $|y_{n-j}|<{n\choose j}$ and $|y_{j}|<{n\choose j}$ . If $|y_{n-j}|<{n\choose j}$ and $|y_{j}|<{n\choose j}$ hold then clearly $\eqref{5}$ holds. Therefore, the statement $(5)$ holds if and only if $|y_{n-j}|<{n\choose j}$ and $|y_{j}|<{n\choose j}$ . 9. $(8)$

: Suppose condition $(8)$ holds. Then the matrix $B_{j}$ , as mentioned in the statement of $(8)$ , is of the form

[TABLE]

where $a_{j},b_{j}\in\mathbb{C}$ such that $\det B_{j}+a_{j}b_{j}=\dfrac{y_{j}y_{n-j}}{{n\choose j}^{2}}$ . Then $a_{j}b_{j}=\dfrac{y_{j}y_{n-j}}{{n\choose j}^{2}}-\det B_{j}=\dfrac{y_{j}y_{n-j}}{{n\choose j}^{2}}-q=0$ . Therefore, $B_{j}$ is one of the following matrices

[TABLE]

In any case $\lVert B_{j}\lVert<1$ if and only if $|y_{n-j}|<{n\choose j}$ and $|y_{j}|<{n\choose j}$ . 10. $(9)$

: Suppose condition $(9)$ holds. Then we must have

[TABLE]

Note that, $\lVert B_{j}\lVert<1$ if and only if $|y_{n-j}|<{n\choose j}$ and $|y_{j}|<{n\choose j}$ .

Case-B: Next we consider the other case when $y_{j}y_{n-j}\neq{n\choose j}^{2}q$ . First we prove the following :

[TABLE]

$(2)\Leftrightarrow(3)$ : Fix a $j\in\{1,\dots,n-1\}$ . Condition $(2)$ then implies that

[TABLE]

which is equivalent to

[TABLE]

The above mentioned condition holds only if $1-\dfrac{y_{n-j}}{{n\choose j}}w\neq 0$ for all $w\in\overline{\mathbb{D}}$ , otherwise the choice $z=0$ will lead to a contradiction. If $|y_{n-j}|<{n\choose j}$ , then $\left|\dfrac{y_{n-j}}{{n\choose j}}w\right|<1$ for all $w\in\overline{\mathbb{D}}$ and consequently $1-\dfrac{y_{n-j}}{{n\choose j}}w\neq 0$ for all $w\in\overline{\mathbb{D}}$ . If $|y_{n-j}|\geq{n\choose j}$ , then for $w=\dfrac{{n\choose j}}{y_{n-j}}\in\overline{\mathbb{D}}$ we have $1-\dfrac{y_{n-j}}{{n\choose j}}w=0$ . Hence, $1-\dfrac{y_{n-j}}{{n\choose j}}w\neq 0$ for all $w\in\overline{\mathbb{D}}$ if and only if $|y_{n-j}|<{n\choose j}$ . Therefore, condition $(1)$ holds if and only if

[TABLE]

which is equivalent to

[TABLE]

If $|y_{n-j}|<{n\choose j}$ , then the function $\Phi_{j}(.,y)$ is bounded on $\mathbb{D}$ . Then the fact that $1\notin z\Phi_{j}(\overline{\mathbb{D}},y)$ for all $z\in\overline{\mathbb{D}}$ implies that $\Phi_{j}(\overline{\mathbb{D}},y)$ does not intersect the complement of $\mathbb{D}$ . Otherwise, there exists some $\alpha\in\overline{\mathbb{D}}$ such that $|\Phi_{j}(\alpha,y)|\geq 1$ and by choosing $z=\dfrac{1}{\Phi_{j}(\alpha,y)}\in\overline{\mathbb{D}}$ we will arrive at a contradiction. Thus the conditions $|y_{n-j}|<{n\choose j}$ and $1\notin z\Phi_{j}(\overline{\mathbb{D}},y)$ for all $z\in\overline{\mathbb{D}}$ together imply that $\Phi_{j}(\overline{\mathbb{D}},y)\cap\mathbb{D}^{c}=\phi$ . On the other hand if $\Phi_{j}(\overline{\mathbb{D}},y)\cap\mathbb{D}^{c}=\phi$ , then $\Phi_{j}(.,y)$ is bounded on $\overline{\mathbb{D}}$ hence $|y_{n-j}|<{n\choose j}$ , moreover $|\Phi_{j}(\alpha,y)|<1$ for all $\alpha\in\overline{\mathbb{D}}$ and hence $|z\Phi_{j}(\alpha,y)|<1$ for all $z,\alpha\in\overline{\mathbb{D}}$ . Therefore, condition $(2)$ holds if and only if $\Phi_{j}(\overline{\mathbb{D}},y)$ does not intersect the complement of $\mathbb{D}$ , for each $j=1,\dots,\left[\frac{n}{2}\right]$ . Note that $\Phi_{j}(\overline{\mathbb{D}},y)\cap\mathbb{D}^{c}=\phi$ if and only if $\lVert\Phi_{j}(.,y)\lVert<1$ . Thus condition $(2)$ holds if and only if condition $(3)$ holds.

$(2)\Leftrightarrow(3^{\prime})$ : Fixed a $j\in\{1,\dots,n-1\}$ . Since condition $(2)$ is equivalent to

[TABLE]

by a similar argument as above, condition $(2)$ holds if and only if

[TABLE]

that is if and only if $\Phi_{n-j}(\overline{\mathbb{D}},y)$ does not intersect the complement of $\mathbb{D}$ which is same as saying $\lVert\Phi_{n-j}(.,y)\lVert<1$ .

$(3)\Leftrightarrow(4)$ : Note that

[TABLE]

Therefore $\lVert\Phi_{j}(.,y)\lVert_{H^{\infty}}<1$ if and only if

[TABLE]

Thus conditions $(3)$ and $(4)$ are equivalent.

$(3^{\prime})\Leftrightarrow(4^{\prime})$ : Similarly, this equivalence is clear from the following fact

[TABLE]

$(3)\Leftrightarrow(5)$ : Fix a $j\in\{1,\dots,n-1\}$ . Suppose condition $(3)$ holds. Since $y_{j}y_{n-j}\neq{n\choose j}^{2}q$ , the inequality $\lVert\Phi_{j}(.,y)\lVert<1$ (that is $\Phi_{j}(.,y)$ is bounded on $\mathbb{D}$ ) implies $|y_{n-j}|<{n\choose j}$ . Again if $|y_{n-j}|<{n\choose j}$ holds, then $y_{n-j}z-{n\choose j}\neq 0$ for any $z\in\overline{\mathbb{D}}$ and consequently the inequality

[TABLE]

implies $\lVert\Phi_{j}(.,y)\lVert<1$ . Then by the maximum modulus principle, condition $(2)$ holds if and only if

[TABLE]

Note that:

[TABLE]

(since $|x|<k$ if and only if $\operatorname{Re}(zx)<k$ for all $z\in\mathbb{T}$ .) Hence condition $(3)$ holds if and only if $|y_{n-j}|<{n\choose j}$ and inequality $\eqref{6}$ hold, that is, if and only if condition $(5)$ holds.

$(3^{\prime})\Leftrightarrow(5^{\prime})$ : Using the same line of argument we can show it.

Thus for the case $y_{j}y_{n-j}\neq{n\choose j}^{2}q$ conditions $(2),(3),(3^{\prime}),(4),(4^{\prime}),(5),(5^{\prime})$ are equivalent. To complete the first part of the proof we show the equivalence of $(2),(6),(8)$ and $(9)$ .

$(2)\Leftrightarrow(6)$ : write $\hat{y}=(\hat{y}_{1},...,\hat{y}_{n-1},\widehat{q})$ , where $\hat{y}_{{}_{k}}=y_{{}_{k}}$ for all $k\neq n-j$ , $\hat{y}_{n-j}={n\choose j}\bar{q}$ and $\widehat{q}=\dfrac{\bar{y}_{n-j}}{{n\choose j}}$ . That is,

[TABLE]

First note that replacing $y_{n-j}$ and $q$ by ${n\choose j}\bar{q}$ and $\dfrac{\bar{y}_{n-j}}{{n\choose j}}$ respectively, makes no difference in the inequality in condition $(5^{\prime})$ . Thus, condition $(5^{\prime})$ holds for the point $y$ ( and for the chosen $j$ ) if and only if $(5^{\prime})$ holds for the point $\hat{y}$ . Since we already proved the equivalence of $(2)$ , $(5)$ and $(5^{\prime})$ , we can conclude that the inequality in condition $(2)$ holds for the point $y$ if and only if the inequalities in $(5)$ hold for the point $\hat{y}$ . Thus, condition $(2)$ holds if and only if the following holds :

[TABLE]

which is same as saying that

[TABLE]

Hence $(2)\Leftrightarrow(6)$ . To complete the rest of the proof we need the following lemma, which can be proved by simple calculations.

Lemma 2.3.

If

[TABLE]

where $b_{j}c_{j}=\dfrac{y_{j}y_{n-j}}{{n\choose j}^{2}}-q$ . Then $\det B_{j}=q$ ,

[TABLE]

and

[TABLE]

Let us get back to the proof of Theorem 2.2.

$(6)\Rightarrow(9)$ : Suppose $(6)$ holds. Consider the matrix

[TABLE]

where $k_{j}$ is the square root(any) of $\left(\dfrac{y_{j}y_{n-j}}{{n\choose j}^{2}}-q\right)$ . Then $\det B_{j}=q$ . By Lemma 2.3, the diagonals of $1-B_{j}^{*}B_{j}$ are the following :

[TABLE]

We have proved that condition $(2)$ is equivalent to conditions $(4),(4^{\prime})$ and $(6)$ . Thus if condition $(6)$ holds, then conditions $(4)$ and $(4^{\prime})$ also hold. Then by conditions $(4)$ and $(4^{\prime})$ , we have

[TABLE]

that is, the diagonals of $1-B_{j}^{*}B_{j}$ are positive. Again by Lemma 2.3 and condition $(6)$ , we have

[TABLE]

Thus $\lVert B_{j}\lVert<1$ . Also we have $y_{j}={n\choose j}[B_{j}]_{11}$ and $y_{n-j}={n\choose j}[B_{j}]_{22}$ . Hence condition $(9)$ is satisfied.

$(9)\Rightarrow(8)$ : Obvious.

$(8)\Rightarrow(6)$ : Suppose $(8)$ holds. Then $B_{j}$ is a $2\times 2$ matrix such that $[B_{j}]_{11}=\dfrac{y_{j}}{{n\choose j}}$ , $[B_{j}]_{22}=\dfrac{y_{n-j}}{{n\choose j}}$ and $[B_{j}]_{12}[B_{j}]_{21}=\dfrac{y_{j}y_{n-j}}{{n\choose j}^{2}}-q$ . Since $\left(\left|[B_{j}]_{12}\right|-\left|[B_{j}]_{21}\right|\right)^{2}\geq 0$ , we have

[TABLE]

Since $\lVert B_{j}\lVert<1$ , by lemma 2.3 we have

[TABLE]

Therefore,

[TABLE]

The fact $\lVert B_{j}\lVert<1$ implies $|\det B_{j}|<1$ , and thus $|q|<1$ . Hence $(6)$ holds and the first part of the proof is complete.

To complete the rest of the proof we show the following : $(1)\Rightarrow(2)\Rightarrow(7)\Rightarrow(1)$ .

$(1)\Rightarrow(2)$ : Suppose $y=(y_{1},\dots,y_{n-1},q)\in\widetilde{\mathbb{G}}_{n}$ . Then $|q|<1$ , and there exists $(\beta_{1},\dots,\beta_{n-1})\in\mathbb{C}^{n-1}$ such that for each $j\in\left\{1,\dots,\left[\frac{n}{2}\right]\right\}$

[TABLE]

Hence

[TABLE]

Then, for each $j=1,\dots,[\frac{n}{2}]$ ,

[TABLE]

and

[TABLE]

Hence, for all $j=1,\dots,[\frac{n}{2}]$ ,

[TABLE]

Thus, for all $j=1,\dots,[\frac{n}{2}]$ ,

[TABLE]

Therefore, condition $(5)$ holds. Then by equivalence of conditions $(2)$ and $(5)$ , we have

[TABLE]

Hence $(2)$ holds.

$(2)\Rightarrow(7)$ : Suppose condition $(2)$ holds. Then by conditions $(5)$ and $(5^{\prime})$ ,

[TABLE]

By adding the above two inequalities we get

[TABLE]

Since $j$ is chosen arbitrarily, inequality 2.12 holds for each $j\in\left\{1,\dots,\left[\frac{n}{2}\right]\right\}$ . Hence $(2)\Rightarrow(7)$ . Further note that if inequation in condition $(2)$ holds for some $j\in\left\{1,\dots,\left[\frac{n}{2}\right]\right\}$ , then the inequality in condition $(7)$ holds for the same $j$ .

$(7)\Rightarrow(1)$ : Suppose condition $(7)$ holds. Clearly $|q|<1$ . For each $j=1,\dots,\left[\frac{n}{2}\right]$ , consider

[TABLE]

Then $\beta_{j}+\bar{\beta}_{n-j}q=y_{j}\quad\text{and}\quad\beta_{n-j}+\bar{\beta}_{j}q=y_{n-j},$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ which is same as saying that $\beta_{j}+\bar{\beta}_{n-j}q=y_{j}$ , for all $j=1,\dots,n-1$ . By condition $(7)$ of this theorem, we have

[TABLE]

Hence for the point $y=(y_{1},\dots,y_{n-1},q)\in\mathbb{C}^{n}$ , there exists $(\beta_{1},\dots,\beta_{n-1})\in\mathbb{C}^{n-1}$ such that $y_{j}=\beta_{j}+\bar{\beta}_{n-j}q$ and $|\beta_{j}|+|\beta_{n-j}|<{n\choose j}$ for all $j=1,\dots,n-1$ . Therefore $y\in\widetilde{\mathbb{G}}_{n}$ , and the proof is complete.

2.2. Characterizations of a point in $\widetilde{\Gamma}_{n}$

In this subsection, we state and prove an analog of Theorem 2.2 for $\widetilde{\Gamma}_{n}\,(=\overline{\widetilde{\mathbb{G}}_{n}})$ .

Theorem 2.4.

For a point $y=(y_{1},\dots,y_{n-1},q)\in\mathbb{C}^{n}$ , the following are equivalent:

$(1)$

$y\in\widetilde{\Gamma}_{n}$ ; 2. $(2)$

${n\choose j}-y_{j}z-y_{n-j}w+{n\choose j}qzw\neq 0$ , for all $z,w\in\mathbb{D}$ and for all $j=1,\dots,\left[\frac{n}{2}\right]$ ; 3. $(3)$

for all $j=1,\dots,\left[\frac{n}{2}\right]$ ,

[TABLE] 4. $(3^{\prime})$

for all $j=1,\dots,\left[\frac{n}{2}\right]$ ,

[TABLE] 5. $(4)$

for all $j=1,\dots,\left[\frac{n}{2}\right]$ ,

[TABLE] 6. $(4^{\prime})$

for all $j=1,\dots,\left[\frac{n}{2}\right]$

[TABLE] 7. $(5)$

for all $j=1,\dots,\left[\frac{n}{2}\right]$ ,

[TABLE] 8. $(5^{\prime})$

for all $j=1,\dots,\left[\frac{n}{2}\right]$

[TABLE] 9. $(6)$

$|q|\leq 1$ * and $|y_{j}|^{2}+|y_{n-j}|^{2}-{n\choose j}^{2}|q|^{2}+2\left|y_{j}y_{n-j}-{n\choose j}^{2}q\right|\leq{n\choose j}^{2}$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ ;* 10. $(7)$

$\left|y_{n-j}-\bar{y}_{j}q\right|+\left|y_{j}-\bar{y}_{n-j}q\right|\leq{n\choose j}(1-|q|^{2})$ * for all $j=1,\dots,\left[\dfrac{n}{2}\right]$ , and if $|q|=1$ then, in addition, $|y_{j}|\leq{n\choose j}$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ ;* 11. $(8)$

there exist $\left[\frac{n}{2}\right]$ number of $2\times 2$ matrices $B_{1},\dots,B_{\left[\frac{n}{2}\right]}$ such that $\lVert B_{j}\lVert\leq 1$ , $y_{j}={n\choose j}[B_{j}]_{11}$ , $y_{n-j}={n\choose j}[B_{j}]_{22}$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ and

[TABLE] 12. $(9)$

there exist $\left[\frac{n}{2}\right]$ number of $2\times 2$ symmetric matrices $B_{1},\dots,B_{\left[\frac{n}{2}\right]}$ such that $\lVert B_{j}\lVert\leq 1$ , $y_{j}={n\choose j}[B_{j}]_{11}$ , $y_{n-j}={n\choose j}[B_{j}]_{22}$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ and

[TABLE]

Proof.

$(1)\Leftrightarrow(2)$ : Suppose $(2)$ holds. For any $\zeta,\eta\in\overline{\mathbb{D}}$ and for any $r\in(0,1)$ we have $r\zeta,r\eta\in\mathbb{D}$ . Hence

[TABLE]

Since the above is true for any $\zeta,\eta\in\overline{\mathbb{D}}$ and any $r\in(0,1)$ , by Theorem 2.2, we have

[TABLE]

for any $r\in(0,1)$ . Therefore, $y=(y_{1},\dots,y_{n-1},q)\in\overline{\widetilde{\mathbb{G}}_{n}}=\widetilde{\Gamma}_{n}$ .

Conversely, suppose $y=(y_{1},\dots,y_{n-1},q)\in\widetilde{\Gamma}_{n}$ . Also suppose $(2)$ does not hold, that is, for some $j=1,\dots,\left[\frac{n}{2}\right]$

[TABLE]

Since $y\in\widetilde{\Gamma}_{n}$ , we have $|y_{n-j}|\leq{n\choose j}$ and hence $y_{n-j}w-{n\choose j}\neq 0$ for all $w\in\mathbb{D}$ . Therefore,

[TABLE]

and consequently $|\Phi_{j}(w,y)|>1$ for some $w\in\mathbb{D}$ . Again by Theorem 2.2 we have that, $|\Phi_{j}(w,\zeta)|<1$ whenever $w\in\mathbb{D}$ and $\zeta\in\widetilde{\mathbb{G}}_{n}$ . Since $y\in\widetilde{\Gamma}_{n}$ , we must have $|\Phi_{j}(w,y)|\leq 1$ for any $w\in\mathbb{D}$ , a contradiction. Hence $(2)$ must holds, if $y\in\widetilde{\Gamma}_{n}$ .

Now we prove the equivalence of $(2)-(6)$ and $(8),(9)$ . The proof of this part is similar to the proof of the corresponding part of Theorem 2.2. For the case $y_{y}y_{n-j}={n\choose j}^{2}q$ , each condition is equivalent to the pair of statements $|y_{j}|\leq{n\choose j}$ and $|y_{n-j}|\leq{n\choose j}$ . The explanations for each case are similar to the proof (for the corresponding case) of Theorem 2.2. In this case, we explain the reason for the parts $(4)$ and $(4^{\prime})$ as they are slightly different form the similar parts in Theorem 2.2.

: Since $y_{j}y_{n-j}={n\choose j}^{2}q$ , we have the following:

[TABLE]

Since $|y_{n-j}|>{n\choose j}$ provides a contradiction, as the left hand side of the inequality $\eqref{11}$ becomes positive while the right hand side becomes negative. Hence statement $(3)$ implies $|y_{j}|\leq{n\choose j}$ and $|y_{n-j}|\leq{n\choose j}$ . On the other hand

[TABLE]

That is, $|y_{j}|\leq{n\choose j}$ and $|y_{n-j}|\leq{n\choose j}$ together imply statement $(4)$ for the case $y_{y}y_{n-j}={n\choose j}^{2}q$ .

: Similarly, statement $(4^{\prime})$ is equivalent to the inequality

[TABLE]

which implies $|y_{j}|<{n\choose j}$ . Hence statement $(4^{\prime})$ implies $|y_{j}|\leq{n\choose j}$ and $|y_{n-j}|\leq{n\choose j}$ . Also, $|y_{j}|\leq{n\choose j}$ and $|y_{n-j}|\leq{n\choose j}$ together imply statement $(4^{\prime})$ for this case.

For the case $y_{y}y_{n-j}\neq{n\choose j}^{2}q$ , the proof is similar to the corresponding parts of Theorem 2.2. $(2)\Rightarrow(7)$ : we can have similar proof as in the corresponding part of Theorem 2.2.

$(7)\Rightarrow(10)$ : Suppose condition $(7)$ holds, then $|q|\leq 1$ . For $|q|<1$ , the part $(7)\Rightarrow(1)$ of the proof of Theorem 2.2 works here, that is, take

[TABLE]

to obtain statement $(10)$ . If $|q|=1$ , then condition $(3)$ implies $y_{j}=\bar{y}_{n-j}q$ and $y_{n-j}=\bar{y}_{j}q$ , for all $j=1,\dots,\left[\frac{n}{2}\right]$ . Hence $|y_{n-j}|=|y_{j}|\leq{n\choose j}$ . Choose $r_{1},\dots,r_{[\frac{n}{2}]}\in[0,1]$ . Consider

[TABLE]

Then for all $j=1,\dots,\left[\frac{n}{2}\right]$ we have the following

[TABLE]

Thus for either cases, there exists $(\beta_{1},\dots,\beta_{n-1})\in\mathbb{C}^{n}$ such that $y_{j}=\beta_{j}+\bar{\beta}_{n-j}q$ , $y_{n-j}=\beta_{n-j}+\bar{\beta}_{j}q$ and $|\beta_{j}|+|\beta_{n-j}|\leq{n\choose j}$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ . Hence $(10)$ holds.

$(10)\Rightarrow(2)$ : Suppose $(10)$ holds. Then $|q|\leq 1$ , and there is $(\beta_{1},\dots,\beta_{n-1})\in\mathbb{C}^{n}$ such that for any $j\in\left\{1,\dots,\left[\frac{n}{2}\right]\right\}$

[TABLE]

Hence

[TABLE]

Let $j\in\left\{1,\dots,\left[\frac{n}{2}\right]\right\}$ be arbitrary. Then

[TABLE]

and $\quad y_{n-j}-\bar{y}_{j}q=\beta_{n-j}+\bar{\beta}_{j}q-(\bar{\beta}_{j}+\beta_{n-j}\bar{q})q=\beta_{n-j}(1-|q|^{2}).$ Hence

[TABLE]

that is,

[TABLE]

Thus, condition $(5)$ holds. Since $j$ was chosen arbitrarily and condition $(2)$ is equivalent to $(5)$ , thus condition $(2)$ of this theorem holds. Consequently, the proof is complete.

Remark 2.5.

The set $\widetilde{\Gamma}_{n}$ can be described as follows :

[TABLE]

Remark 2.6.

Let $y=(y_{1},\dots,y_{n-1},q)\in\mathbb{C}^{n}$ . For any $j\in\left\{1,\dots,\left[\frac{n}{2}\right]\right\}$ , if $\tilde{y}^{j}=(\tilde{y}_{1},\dots,\tilde{y}_{n-1},\tilde{q})$ where $\tilde{y}_{k}=y_{k}$ for $k\neq j,n-j;\quad\tilde{y}_{j}=y_{n-j},\;\;\tilde{y}_{n-j}=y_{j}$ and $\tilde{q}=q$ , then $y\in\widetilde{\mathbb{G}}_{n}($ or $\widetilde{\Gamma}_{n})$ if and only if $\tilde{y}^{j}\in\widetilde{\mathbb{G}}_{n}($ or $\widetilde{\Gamma}_{n})$ .

Remark 2.7.

It was proved in [8] (see Theorem 1.1 in [8]) that the closed symmetrized bidisc $\Gamma_{2}$ has the following representation:

[TABLE]

Thus it is evident from Theorem 2.4 (and also from Remark 2.5) that $\widetilde{\Gamma_{2}}=\Gamma_{2}$ . We shall see in Section 4 that $\Gamma_{n}\subsetneq\widetilde{\Gamma}_{n}$ for all $n\geq 3$ .

3. Characterizations of the points in $\mathbb{G}_{n}$ and $\Gamma_{n}$

In this section, we find a variety of characterizations for the symmetrized polydisc $\mathbb{G}_{n}$ via the characterizations of $\widetilde{\mathbb{G}}_{n}$ as in Theorem 2.2. In [17], Costara found the following interesting characterization of a point in $\mathbb{G}_{n}$ and $\Gamma_{n}$ .

Theorem 3.1 ([17], Theorems 3.6 and 3.7).

For a point $(s_{1},\dots,s_{n-1},p)\in\mathbb{C}^{n}$ , the following are equivalent:

(1)

*The point * $(s_{1},\dots,s_{n-1},p)\in\mathbb{G}_{n}\;($ respectively $\in\Gamma_{n})$ . 2. (2)

$|p|<1\;($ *respectively $\leq 1)$ and there exists * $\left(S_{1},\dots,S_{n-1}\right)\in\mathbb{G}_{n-1}\;($ respectively $\in\Gamma_{n-1})$ such that

[TABLE]

We have mentioned before (see Remark 2.7) that $\widetilde{\Gamma}_{2}=\Gamma_{2}$ . Indeed, it is true that $\widetilde{\mathbb{G}}_{n}=\mathbb{G}_{n}$ for $n=2$ but $\mathbb{G}_{n}\subsetneqq\widetilde{\mathbb{G}}_{n}$ for $n\geq 3$ . The first part of this claim follows straight from the above result of Costara (Theorem 3.1). We show below that $\mathbb{G}_{n}\subsetneqq\widetilde{\mathbb{G}}_{n}$ for $n\geq 3$ . By Theorem 3.1, $s_{i}=S_{i}+\bar{S}_{n-i}p$ for every point $(s_{1},\dots,s_{n-1},p)\in\mathbb{G}_{n}$ , where $(S_{1},\dots,S_{n-1})\in\mathbb{G}_{n-1}$ . Needless to mention that $|S_{i}|+|S_{n-i}|<{n\choose i}$ and this implies that $(s_{1},\dots,s_{n-1},p)\in\widetilde{\mathbb{G}}_{n}$ . Now consider the point $\left(\dfrac{5}{2},\dfrac{5}{4},\dfrac{1}{2}\right)\in\mathbb{C}^{3}$ . We can write $\left(\dfrac{5}{2},\dfrac{5}{4},\dfrac{1}{2}\right)=\left(\beta_{1}+\overline{\beta}_{2}p,\beta_{2}+\overline{\beta}_{1}p,p\right)$ , where $\beta_{1}=\dfrac{5}{2},\beta_{2}=0,p=\dfrac{1}{2}$ . Also $\beta_{1},\beta_{2}$ are unique because if we write $(y_{1},y_{2},p)=\left(\dfrac{5}{2},\dfrac{5}{4},\dfrac{1}{2}\right)$ , then

[TABLE]

Since $|\beta_{1}|+|\beta_{2}|<3$ , $\left(\dfrac{5}{2},\dfrac{5}{4},\dfrac{1}{2}\right)\in\widetilde{\mathbb{G}_{3}}$ . Clearly $\left(\beta_{1},\beta_{2}\right)\notin\mathbb{G}_{2}$ , as $|\beta_{1}|>2$ . By Theorem 3.1, $\left(\dfrac{5}{2},\dfrac{5}{4},\dfrac{1}{2}\right)\notin\mathbb{G}_{3}$ . So, $\mathbb{G}_{3}\subsetneqq\widetilde{\mathbb{G}_{3}}$ . Similarly for any $n>3$ , we may choose $\left(\beta_{1},\dots,\beta_{n-1}\right)=\left(\dfrac{2n-1}{2},0,\dots,0\right)$ and $p=\dfrac{1}{2}$ to obtain a point $(y_{1},\dots,y_{n-1},p)$ , where $y_{i}=\beta_{i}+\bar{\beta}_{n-i}p,$ for $i=1,\dots,n-1$ , which is in $\widetilde{\mathbb{G}_{n}}$ but not in $\mathbb{G}_{n}$ .

3.1. Characterizations of a point in $\mathbb{G}_{n}$

Theorem 3.2.

Let $(s_{1},\dots,s_{n-1},p)\in\mathbb{C}^{n}$ with $|p|\neq 1$ . Consider the point

[TABLE]

Then $(s_{1},\dots,s_{n-1},p)\in\mathbb{G}_{n}$ if and only if $(s_{1},\dots,s_{n-1},p)\in\widetilde{\mathbb{G}}_{n}$ and $Q\in\mathbb{G}_{n-1}$ .

Proof.

Suppose $(s_{1},\dots,s_{n-1},p)\in\mathbb{G}_{n}$ . Then $|p|<1$ and $(s_{1},\dots,s_{n-1},p)\in\widetilde{\mathbb{G}}_{n}$ as $\mathbb{G}_{n}\subseteq\widetilde{\mathbb{G}}_{n}$ . By Theorem 3.1, there exists $(\beta_{1},\dots,\beta_{n-1})\in\mathbb{G}_{n-1}$ so that

[TABLE]

Then for each $j=1,\dots,n-1$ , we have $\beta_{j}=\dfrac{s_{j}-\bar{s}_{n-j}p}{1-|p|^{2}}.$ Also these $\beta_{j}^{\prime}s$ are unique. So, by Theorem 3.1,

[TABLE]

Conversely, suppose $(s_{1},\dots,s_{n-1},p)\in\widetilde{\mathbb{G}}_{n}$ and $Q\in\mathbb{G}_{n-1}$ . Then $|p|<1$ . For each $j=1,\dots,n-1$ , consider

[TABLE]

Then, we have $\beta_{j}+\bar{\beta}_{n-j}p=s_{j}$ for each $j=1,\dots,n-1$ . Since $(\beta_{1},\dots,\beta_{n-1})=Q\in\mathbb{G}_{n-1}$ , it follows from Theorem 3.1 that $(s_{1},\dots,s_{n-1},p)\in\mathbb{G}_{n}$ .

We finally arrive at the main result of this paper, which characterizes a point in $\mathbb{G}_{n}$ in several different ways.

Theorem 3.3.

Let $x=(s_{1},\dots,s_{n-1},p)\in\mathbb{C}^{n}$ and let

[TABLE]

Then the following are equivalent:

$(1)$

$x\in\mathbb{G}_{n}$ ; 2. $(2)$

$Q\in\mathbb{G}_{n-1}$ * and ${n\choose j}-s_{j}z-s_{n-j}w+{n\choose j}pzw\neq 0$ , for all $z,w\in\overline{\mathbb{D}}$ and for all $j=1,\dots,\left[\frac{n}{2}\right]$ ;* 3. $(3)$

$Q\in\mathbb{G}_{n-1}$ * and for all $j=1,\dots,\left[\frac{n}{2}\right]$ either*

[TABLE]

or

[TABLE] 4. $(4)$

$Q\in\mathbb{G}_{n-1}$ * and for all $j=1,\dots,\left[\frac{n}{2}\right]$ either*

[TABLE]

or

[TABLE] 5. $(5)$

$Q\in\mathbb{G}_{n-1}$ * and for all $j=1,\dots,\left[\frac{n}{2}\right]$ either*

[TABLE]

or

[TABLE] 6. $(6)$

$Q\in\mathbb{G}_{n-1}$ , $|p|<1$ and $|s_{j}|^{2}+|s_{n-j}|^{2}-{n\choose j}^{2}|p|^{2}+2\left|s_{j}s_{n-j}-{n\choose j}^{2}p\right|<{n\choose j}^{2}$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ ; 7. $(7)$

$Q\in\mathbb{G}_{n-1}$ * and $\left|s_{n-j}-\bar{s}_{j}p\right|+\left|s_{j}-\bar{s}_{n-j}p\right|<{n\choose j}(1-|p|^{2})$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ ;* 8. $(8)$

$Q\in\mathbb{G}_{n-1}$ * and there exist $\left[\frac{n}{2}\right]$ number of $2\times 2$ matrices $B_{1},\dots,B_{\left[\frac{n}{2}\right]}$ such that $\lVert B_{j}\lVert<1$ , $s_{j}={n\choose j}[B_{j}]_{11}$ , $s_{n-j}={n\choose j}[B_{j}]_{22}$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ and*

[TABLE] 9. $(9)$

$Q\in\mathbb{G}_{n-1}$ * and there exist $\left[\frac{n}{2}\right]$ number of $2\times 2$ symmetric matrices $B_{1},\dots,B_{\left[\frac{n}{2}\right]}$ such that $\lVert B_{j}\lVert<1$ , $s_{j}={n\choose j}[B_{j}]_{11}$ , $s_{n-j}={n\choose j}[B_{j}]_{22}$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ and*

[TABLE]

Proof.

The proof follows from Theorem 2.2 and Theorem 3.2.

3.2. Characterizations of a point in $\Gamma_{n}$

Recall from the literature (see [18]) that the distinguished boundary of $\Gamma_{n}$ is the smallest closed subset $b\Gamma_{n}$ of $\subset\Gamma_{n}$ such that every complex-valued function on $\Gamma_{n}$ , that is analytic in the interior $\mathbb{G}_{n}$ and continuous on $\Gamma_{n}$ , attains its maximum modulus on $b\Gamma_{n}$ . Evidently, $b\Gamma_{n}$ is a subset of the topological boundary $\partial\Gamma_{n}$ of $\Gamma_{n}$ . In [15], the points in the distinguished boundary $b\Gamma_{n}$ of $\Gamma_{n}$ are characterized in the following way:

Theorem 3.4 ([15], Theorem 2.4).

For $(s_{1},\dots,s_{n-1},p)\in\mathbb{C}^{n}$ the following are equivalent:

(1)

$(s_{1},\dots,s_{n-1},p)\in b\Gamma_{n}$ * ;* 2. (2)

$(s_{1},\dots,s_{n-1},p)\in\Gamma_{n}$ * and $|p|=1$ ;* 3. (3)

$|p|=1$ , $s_{j}=\bar{s}_{n-j}p$ and $\left(\dfrac{n-1}{n}s_{1},\dfrac{n-2}{n}s_{2},\dots,\dfrac{1}{n}s_{n-1}\right)\in\Gamma_{n-1}$ ; 4. (4)

$|p|=1$ * and there exists $(\beta_{1},\dots,\beta_{n-1})\in b\Gamma_{n-1}$ such that*

$s_{j}=\beta_{j}+\bar{\beta}_{n-j}p$ * , for $j=1,\dots,n-1$ .*

Theorem 3.4 provides several independent descriptions of the points in $b\Gamma_{n}$ . Thus, in order to characterize the points in $\Gamma_{n}$ it suffices if we find characterizations for the points in $\Gamma_{n}\setminus b\Gamma_{n}$ .

Theorem 3.5.

Let $(s_{1},\dots,s_{n-1},p)\in\mathbb{C}^{n}$ . Then the following are equivalent:

(1)

$(s_{1},\dots,s_{n-1},p)\in\Gamma_{n}\setminus b\Gamma_{n}$ ; 2. (2)

$(s_{1},\dots,s_{n-1},p)\in\Gamma_{n}$ * and*

[TABLE]

Proof.

Suppose $(s_{1},\dots,s_{n-1},p)\in\Gamma_{n}\setminus b\Gamma_{n}$ . Then $(s_{1},\dots,s_{n-1},p)\in\widetilde{\Gamma}_{n}$ and $|p|\leq 1$ . By Theorem 3.4, we have $|p|<1$ . By Theorem 3.1, there exists $(\beta_{1},\dots,\beta_{n-1})\in\Gamma_{n-1}$ such that

[TABLE]

Since $|p|<1$ , we have that

[TABLE]

Thus, $\left(\dfrac{s_{1}-\bar{s}_{n-1}p}{1-|p|^{2}},\dfrac{s_{2}-\bar{s}_{n-2}p}{1-|p|^{2}},\dots,\dfrac{s_{n-1}-\bar{s}_{1}p}{1-|p|^{2}}\right)\in\Gamma_{n-1}$ . Since $(s_{1},\dots,s_{n-1},p)\notin b\Gamma_{n}$ , by Theorem 3.4, $\left(\dfrac{s_{1}-\bar{s}_{n-1}p}{1-|p|^{2}},\dfrac{s_{2}-\bar{s}_{n-2}p}{1-|p|^{2}},\dots,\dfrac{s_{n-1}-\bar{s}_{1}p}{1-|p|^{2}}\right)\in\Gamma_{n-1}\setminus b\Gamma_{n-1}$ . Thus $(1)\Rightarrow(2)$ .

Conversely, suppose (2) holds. Then $|p|\leq 1$ . First suppose $|p|<1$ and consider $\beta_{j}=\dfrac{s_{j}-\bar{s}_{n-j}p}{1-|p|^{2}}.$ Then, we have $\beta_{j}+\bar{\beta}_{n-j}p=s_{j}$ for each $j=1,\dots,n-1$ . By hypothesis $(\beta_{1},\dots,\beta_{n-1})\in\Gamma_{n-1}$ . Hence by Theorem 3.7 in [17], we have $(s_{1},\dots,s_{n-1},p)\in\Gamma_{n}$ . Now suppose $|p|=1$ . Since $\left(\dfrac{s_{1}-\bar{s}_{n-1}p}{1-|p|^{2}},\dfrac{s_{2}-\bar{s}_{n-2}p}{1-|p|^{2}},\dots,\dfrac{s_{n-1}-\bar{s}_{1}p}{1-|p|^{2}}\right)\notin b\Gamma_{n-1}$ , by Theorem 3.4 we can conclude that $(s_{1},\dots,s_{n-1},p)\notin b\Gamma_{n}$ . Hence $(1)$ holds.

The following theorem, which is another main result of this article, is a consequence of Theorems 2.4 and 3.5 and it provides several independent characterizations of a point in $\Gamma_{n}\setminus b\Gamma_{n}$ .

Theorem 3.6.

Let $x=(s_{1},\dots,s_{n-1},p)\in\mathbb{C}^{n}$ and let

[TABLE]

Then the following are equivalent:

$(1)$

$x\in\Gamma_{n}\setminus b\Gamma_{n}$ ; 2. $(2)$

$Q\in\mathbb{G}_{n-1}$ * and ${n\choose j}-s_{j}z-s_{n-j}w+{n\choose j}pzw\neq 0$ , for all $z,w\in\mathbb{D}$ and $1\leq j\leq\left[\frac{n}{2}\right]$ ;* 3. $(3)$

$Q\in\mathbb{G}_{n-1}$ * and for all $j=1,\dots,\left[\frac{n}{2}\right]$ either*

[TABLE]

or

[TABLE] 4. $(4)$

$Q\in\mathbb{G}_{n-1}$ * and for all $j=1,\dots,\left[\frac{n}{2}\right]$ either*

[TABLE]

or

[TABLE] 5. $(5)$

$Q\in\mathbb{G}_{n-1}$ * and for all $j=1,\dots,\left[\frac{n}{2}\right]$ either*

[TABLE]

or

[TABLE] 6. $(6)$

$Q\in\mathbb{G}_{n-1}$ , $|p|<1$ and $|s_{j}|^{2}+|s_{n-j}|^{2}-{n\choose j}^{2}|p|^{2}+2\left|s_{j}s_{n-j}-{n\choose j}^{2}p\right|\leq{n\choose j}^{2}$ for $j=1,\dots,\left[\frac{n}{2}\right]$ ; 7. $(7)$

$Q\in\mathbb{G}_{n-1}$ * and $\left|s_{n-j}-\bar{s}_{j}p\right|+\left|s_{j}-\bar{s}_{n-j}p\right|\leq{n\choose j}(1-|p|^{2})$ for $j=1,\dots,\left[\frac{n}{2}\right]$ ;* 8. $(8)$

$Q\in\mathbb{G}_{n-1}$ * and there exist $\left[\frac{n}{2}\right]$ number of $2\times 2$ matrices $B_{1},\dots,B_{\left[\frac{n}{2}\right]}$ such that $\lVert B_{j}\lVert\leq 1$ , $s_{j}={n\choose j}[B_{j}]_{11}$ , $s_{n-j}={n\choose j}[B_{j}]_{22}$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ and $\det B_{1}=\det B_{2}=\det B_{\left[\frac{n}{2}\right]}=p$ ;* 9. $(9)$

$Q\in\mathbb{G}_{n-1}$ * and there exist $\left[\frac{n}{2}\right]$ number of $2\times 2$ symmetric matrices $B_{1},\dots,B_{\left[\frac{n}{2}\right]}$ such that $\lVert B_{j}\lVert\leq 1$ , $s_{j}={n\choose j}[B_{j}]_{11}$ , $s_{n-j}={n\choose j}[B_{j}]_{22}$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ and $\det B_{1}=\det B_{2}=\det B_{\left[\frac{n}{2}\right]}=p$ .*

Proof.

The proof is similar to the proof of Theorem 3.3.

4. Some more geometric properties of $\widetilde{\mathbb{G}}_{n}$ and $\widetilde{\Gamma}_{n}$

We further study the geometry of the domain $\widetilde{\mathbb{G}}_{n}$ and its closure $\widetilde{\Gamma}_{n}$ and obtain a few important features. We prove that $\widetilde{\Gamma}_{n}$ is not convex but is polynomially convex. Since the closed symmetrized bidisc $\Gamma_{2}$ is not convex (see Section 2 of [5] and [18]), it follows that $\widetilde{\Gamma_{2}}$ is not convex because $\Gamma_{2}=\widetilde{\Gamma_{2}}$ (see Remark 2.7). Below we show that $\widetilde{\Gamma}_{n}$ is not convex for any $n\geq 2$ . Also, we show that both $\widetilde{\mathbb{G}}_{n}$ and $\widetilde{\Gamma}_{n}$ are starlike about the origin but not circled. It follows from here that $\widetilde{\mathbb{G}}_{n}$ is simply connected. For the convenience of a reader, we begin with a few definitions from the literature.

Definition 4.1.

A compact set $K\subset\mathbb{C}^{n}$ is called polynomially convex if for any $y\in\mathbb{C}^{n}\setminus K$ , there exists a polynomial in $n$ variables, say $P$ , such that

[TABLE]

Definition 4.2.

A set $S\subset\mathbb{C}^{n}$ is said to be starlike if there exists a $z_{0}\in S$ such that for all $z\in S$ the line segment joining $z$ and $z_{0}$ , lies in $S$ .

Definition 4.3.

A set $S\subset\mathbb{C}^{n}$ is said to be circular if $z\in S$ implies that $(e^{\iota\theta}z_{1},e^{\iota\theta}z_{2},...,e^{\iota\theta}z_{n})\in S$ for all $0\leq\theta<2\pi$ .

Consider the points $a=(n,0,\dots,0,ni,i)\in\mathbb{C}^{n}$ and $b=(-ni,0\dots,0,n,-i)\in\mathbb{C}^{n}$ . Then $a,b\in\widetilde{\Gamma}_{n}$ , because, both the points $a$ and $b$ satisfy condition $(7)$ of Theorem 2.4. The mid point of the line joining $a$ and $b$ is

[TABLE]

The point $c\notin\widetilde{\Gamma}_{n}$ . This is because the inequality in condition $(7)$ of Theorem 2.4, corresponding to the point $c$ and $j=1$ , does not hold. Indeed, after substituting the values in that particular inequality we get the following

[TABLE]

Hence, the line segment joining $a$ and $b$ does not entirely lie within $\widetilde{\Gamma}_{n}$ and consequently $\widetilde{\Gamma}_{n}$ is not convex.

We now show that $\widetilde{\Gamma}_{n}$ is polynomially convex. The techniques that we use to prove this are similar to that in [2], where the authors established that the tetrablock is polynomially convex.

Theorem 4.4.

$\widetilde{\Gamma}_{n}$ * is polynomially convex.*

Proof.

Consider a point $y=(y_{1},\dots,y_{n-1},q)\in\mathbb{C}^{n}\setminus\widetilde{\Gamma}_{n}$ . To show $\widetilde{\Gamma}_{n}$ is polynomially convex, it is enough to find a polynomial $f$ in $n$ variables, such that $|f|\leq 1$ on $\widetilde{\Gamma}_{n}$ and $|f(y)|>1\,$ . Let $|y_{j}|>{n\choose j}$ for some $j=1,\dots,\left[\frac{n}{2}\right]$ . Then $f\left((x_{1},\dots,x_{n})\right)=\dfrac{x_{j}}{{n\choose j}}$ has the required property, since for any $x\in\widetilde{\Gamma}_{n}$ , $|x_{j}|\leq{n\choose j}$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ . If $|q|>1$ then take $f\left((x_{1},\dots,x_{n})\right)=x_{n}\,$ . Thus assume $y$ to be such that $|y_{j}|\leq{n\choose j}$ for all $j=1,\dots,n-1$ ; and $|q|<1$ . Since $y\notin\widetilde{\Gamma}_{n}$ , by condition $(3)$ of Theorem 2.2 there exist some $j\in\left\{1,\dots,\left[\frac{n}{2}\right]\right\}$ such that $\lVert\Phi_{j}(.,y)\lVert>1$ . That is, there exist $z\in\mathbb{D}$ and some $j\in\left\{1,\dots,\left[\frac{n}{2}\right]\right\}$ such that $\left|\Phi_{j}(z,y)\right|>1$ . Fix that $j$ and $z$ . Note the following cases :

: If $y_{j}y_{n-j}={n\choose j}^{2}q$ , then $|\Phi_{j}(z,y)|=\dfrac{|y_{j}|}{{n\choose j}}$ . Hence $|y_{j}|>{n\choose j}$ and then the function $f\left((x_{1},\dots,x_{n})\right)=\dfrac{x_{j}}{{n\choose j}}$ is the required function.

: If $y_{j}y_{n-j}\neq{n\choose j}^{2}q$ , then $\Phi_{j}(z,y)=\dfrac{{n\choose j}qz-y_{j}}{y_{n-j}z-{n\choose j}}$ . For $m\in\mathbb{N}$ , define a polynomial $f_{m}$ in $n$ variables as follows

[TABLE]

Let

[TABLE]

Then $\widetilde{\Gamma}_{n}\subset\mathcal{W}$ and also $y\in\mathcal{W}$ . For $x\in\mathcal{W}$ , we have $\left|\dfrac{x_{n-j}z}{{n\choose j}}\right|\leq|z|<1$ and hence $\Phi_{j}(z,x)$ can be written in the following form

[TABLE]

For $x\in\mathcal{W}$ , we also have

[TABLE]

Then for any $x\in\mathcal{W}$ and for any $m\in\mathbb{N}$ , we have

[TABLE]

Let $0<\epsilon<\dfrac{1}{3}\left(|\Phi_{j}(z,y)|-1\right)$ . Note that $z\in\mathbb{D}$ implies $|z|^{m+1}\rightarrow 0$ as $m\rightarrow\infty$ . Hence there exists $k\in\mathbb{N}$ , large enough, such that $\dfrac{2|z|^{k+1}}{1-|z|}<\epsilon$ . Since

[TABLE]

we have

[TABLE]

Again since $\widetilde{\Gamma}_{n}\subset\mathcal{W}$ and $|\Phi_{j}(z,x)|\leq 1$ for all $x\in\widetilde{\Gamma}_{n}$ , we have

[TABLE]

Note that, $|\Phi_{j}(z,y)|>1+3\epsilon$ . Since $y\in\mathcal{W}$ , we have $|\Phi_{j}(z,y)-f_{k}(y)|<\epsilon$ , that is

[TABLE]

Therefore,

[TABLE]

Now take $f=(1+\epsilon)^{-1}f_{k}$ . Then $|f(x)|\leq 1$ for all $x\in\widetilde{\Gamma}_{n}$ , and

[TABLE]

Hence $f$ is the required polynomial.

Theorem 4.5.

$\widetilde{\mathbb{G}}_{n}$ * and $\widetilde{\Gamma}_{n}$ are both starlike about $(0,\dots,0)$ but not circular. Hence $\widetilde{\mathbb{G}}_{n}$ is simply connected.*

Proof.

Let $y=(y_{1},\dots,y_{n-1},q)\in\widetilde{\Gamma}_{n}$ and let $0\leq r<1$ . To prove that $\widetilde{\mathbb{G}}_{n}$ and $\widetilde{\Gamma}_{n}$ are starlike about $(0,0,0)$ , it is enough to show that $ry\in\widetilde{\mathbb{G}}_{n}$ for all such $y$ and $r$ . First note that, for $r>0$ , $z\in\mathbb{C}$ and $j=1,\dots,n-1$ we have

[TABLE]

Since $y\in\widetilde{\Gamma}_{n}$ , by condition $(3)$ of Theorem 2.4, we have $\lVert\Phi_{j}(.,y)\lVert\leq 1$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ . Hence, for each $j=1,\dots,\left[\frac{n}{2}\right]$

[TABLE]

Again for any $z\in\overline{\mathbb{D}}$ and for all $j=1,\dots,\left[\frac{n}{2}\right]$ , we have

[TABLE]

This is because $r|zy_{n-j}|^{2}<|zy_{n-j}|^{2}\leq|y_{n-j}|^{2}\leq{n\choose j}^{2}$ (as $y\in\widetilde{\Gamma}_{n}$ ). Hence from equation $\eqref{starlike}$ , we have

[TABLE]

for any $r\in[0,1)$ , $z\in\overline{\mathbb{D}}$ and for all $j=1,\dots,\left[\frac{n}{2}\right]$ . Therefore,

[TABLE]

whenever $y\in\widetilde{\Gamma}_{n}$ , $z\in\overline{\mathbb{D}}$ and $0\leq r<1$ . Thus, $\lVert\Phi_{j}(.,ry)\lVert<1$ for all $j=1,\dots,\left[\frac{n}{2}\right]$ , whenever $y\in\widetilde{\Gamma}_{n}$ and $0\leq r<1$ . Therefore, by Theorem 2.2, $ry\in\widetilde{\mathbb{G}}_{n}$ whenever $y\in\widetilde{\Gamma}_{n}$ and $0\leq r<1$ . Hence $\widetilde{\mathbb{G}}_{n}$ and $\widetilde{\Gamma}_{n}$ are starlike about $(0,\dots,0)$ .

Next we show that $\widetilde{\Gamma}_{n}$ is not a circular, which implies that $\widetilde{\mathbb{G}}_{n}$ is not a circular domain either. Let $y=(y_{1},\dots,y_{n-1},q)\in\mathbb{C}^{n}$ be such that $y_{j}={n\choose j}$ for $j=1,\dots,\left[\frac{n}{2}\right]$ . and $q=1$ , that is,

[TABLE]

To prove $\widetilde{\Gamma}_{n}$ is not a circular, we show that $y\in\widetilde{\Gamma}_{n}$ but $iy\notin\widetilde{\Gamma}_{n}$ . Note that, for the above mentioned point $y$ , we have $y_{j}y_{n-j}={n\choose j}^{2}q$ , and

[TABLE]

and

[TABLE]

Therefore, by condition $(4)$ of Theorem 2.2, $y\in\widetilde{\Gamma}_{n}$ . Now consider the point

[TABLE]

Then for each $j=1,\dots,\left[\frac{n}{2}\right]$ , we have ${n\choose j}^{2}-|\tilde{y}_{n-j}|^{2}=0$ . Again for each $j=1,\dots,\left[\frac{n}{2}\right]$ ,

[TABLE]

Thus $\tilde{y}$ does not satisfy condition $(4)$ of Theorem 2.2 and hence $\tilde{y}=iy\notin\widetilde{\Gamma}_{n}$ . Thus $\widetilde{\Gamma}_{n}$ is not circular. Thus it follows from here that $\widetilde{\mathbb{G}}_{n}$ is simply connected.

Acknowledgement. We are thankful to the referee for making fruitful suggestions which help in improving the exposition of the article.

Bibliography30

The reference list from the paper itself. Each links out to its DOI / PubMed record.

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Characterizations of the symmetrized polydisc via another family of domains

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1. Introduction

2. Characterizations of the points in G~n\widetilde{\mathbb{G}}_{n}Gn​ and Γ~n\widetilde{\Gamma}_{n}Γn​

Definition 2.1**.**

2.1. Characterizations for the points in G~n\widetilde{\mathbb{G}}_{n}Gn​

Theorem 2.2**.**

Proof.

Lemma 2.3**.**

2.2. Characterizations of a point in Γ~n\widetilde{\Gamma}_{n}Γn​

Theorem 2.4**.**

Proof.

Remark 2.5**.**

Remark 2.6**.**

Remark 2.7**.**

3. Characterizations of the points in Gn\mathbb{G}_{n}Gn​ and Γn\Gamma_{n}Γn​

Theorem 3.1** ([17], Theorems 3.6 and 3.7).**

3.1. Characterizations of a point in Gn\mathbb{G}_{n}Gn​

Theorem 3.2**.**

Proof.

Theorem 3.3**.**

Proof.

3.2. Characterizations of a point in Γn\Gamma_{n}Γn​

Theorem 3.4** ([15], Theorem 2.4).**

Theorem 3.5**.**

Proof.

Theorem 3.6**.**

Proof.

4. Some more geometric properties of G~n\widetilde{\mathbb{G}}_{n}Gn​ and Γ~n\widetilde{\Gamma}_{n}Γn​

Definition 4.1**.**

Definition 4.2**.**

Definition 4.3**.**

Theorem 4.4**.**

Proof.

Theorem 4.5**.**

Proof.

2. Characterizations of the points in $\widetilde{\mathbb{G}}_{n}$ and $\widetilde{\Gamma}_{n}$

Definition 2.1.

2.1. Characterizations for the points in $\widetilde{\mathbb{G}}_{n}$

Theorem 2.2.

Lemma 2.3.

2.2. Characterizations of a point in $\widetilde{\Gamma}_{n}$

Theorem 2.4.

Remark 2.5.

Remark 2.6.

Remark 2.7.

3. Characterizations of the points in $\mathbb{G}_{n}$ and $\Gamma_{n}$

Theorem 3.1 ([17], Theorems 3.6 and 3.7).

3.1. Characterizations of a point in $\mathbb{G}_{n}$

Theorem 3.2.

Theorem 3.3.

3.2. Characterizations of a point in $\Gamma_{n}$

Theorem 3.4 ([15], Theorem 2.4).

Theorem 3.5.

Theorem 3.6.

4. Some more geometric properties of $\widetilde{\mathbb{G}}_{n}$ and $\widetilde{\Gamma}_{n}$

Definition 4.1.

Definition 4.2.

Definition 4.3.

Theorem 4.4.

Theorem 4.5.