# Squeezing function corresponding to polydisk

Naveen Gupta  Department of Mathematics, University of Delhi, Delhi–110 007,
India<EMAIL_ADDRESS>and  Sanjay Kumar Pant  Department of
Mathematics, Deen Dayal Upadhyaya college, University of Delhi, Delhi–110 078,
India<EMAIL_ADDRESS>
###### Abstract.

In the present article, we define squeezing function corresponding to polydisk
and study its properties. We investigate relationship between squeezing
fuction and squeezing function corresponding to polydisk. We also give an
alternate proof for lower bound of the squeezing function of a product domain.

###### Key words and phrases:

squeezing function; extremal map; holomorphic homogeneous regular domain.

###### 2010 Mathematics Subject Classification:

32F45, 32H02

##  1\. Introduction

The work in this article is motivated by a question posed by Fornæss in his
talk [4]. Fornæss posed the question “what is analogous theory of squeezing
function when embeddings are taken into polydisk instead of unit ball?” This
question seems to rely on the fact that unit ball and unit polydisk in
$\mathbb{C}^{n}$, $n>1$ are not biholomorphic[8]. Therefore unit polydisk are
most suitable object to see what happens to the squeezing function when we
replace unit ball by it. In this paper, we consider this problem. We define
squeezing function in the case when embeddings are taken into unit polydisk —
in the definition of squeezing function — instead of unit ball and we explore
its various properties. It seems to us in the present work that the results
connected with squeezing function corresponding to polydisk do not depend in a
big way on the nature of polydisk even it not being biholomorphic to unit
ball.

The notion of squeezing function started with the work of Liu et al. [11],
[12], in 2004 and 2005, in which they studied holomorphic homogeneous
regular(HHR) manifolds. In 2009, Yeung in his paper [10], studied this notion.
He renamed the HHR property as uniform squeezing property. The formal
definition of squeezing function — motivated by the work of Liu et al. [11],
[12] and Yeung [10] — was introduced by Deng et al. [1]. In their paper [1],
the authors presented several properties of squeezing functions. Recently,
several authors have explored various geometrical and analytical aspects of
squeezing function [5],[15],[14],[16],[17], [13].

Let us denote by $B^{n}$ unit ball in $\mathbb{C}^{n}$ and let
$\Omega\subseteq\mathbb{C}^{n}$ be a bounded domain. For $z\in{\Omega},$ let

$S_{\Omega}(f,z):=\sup\\{r:B^{n}(0,r)\subseteq f(\Omega)\\},$

where $f:\Omega\to B^{n}$ is holomorphic embedding with $f(z)=0$ and
$B^{n}(0,r)$ denotes ball of radius $r$, centered at origin. Squeezing
function on $\Omega$, denoted by $S_{\Omega}$ is defined as

$S_{\Omega}(z):=\sup_{f}\\{r:B^{n}(0,r)\subseteq f(\Omega)\\},$

where supremum is taken over holomorphic embeddings $f:\Omega\to B^{n}$ with
$f(z)=0.$

The fact that squeezing function is biholomorphic invariant follows from the
definition. A bounded domain is called holomorphic homogeneous regular if its
squeezing function has a positive lower bound.

The definition of squeezing function corresponding to polydisk can be written
exactly in similar manner as the definition of squeezing function by replacing
unit ball with unit polydisk.

######  Definition 1.1.

Let us denote by $\mathbb{D}^{n}$ unit polydisk in $\mathbb{C}^{n}$ and let
$\Omega\subseteq\mathbb{C}^{n}$ be a bounded domain. For $z\in{\Omega},$ let

$T_{\Omega}(f,z):=\sup\\{r:\mathbb{D}^{n}(0,r)\subseteq f(\Omega)\\},$

where $f:\Omega\to\mathbb{D}^{n}$ is holomorphic embedding with $f(z)=0$ and
$\mathbb{D}^{n}(0,r)$ denotes polydisk of radius $r$, centered at origin.
Squeezing function corresponding to polydisk on $\Omega$, denoted by
$T_{\Omega}$ is defined as

$T_{\Omega}(z):=\sup_{f}\\{r:\mathbb{D}^{n}(0,r)\subseteq f(\Omega)\\},$

where supremum is taken over holomorphic embeddings
$f:\Omega\to\mathbb{D}^{n}$ with $f(z)=0.$

In what follows, by squeezing function we mean squeezing function
corresponding to polydisk. Relation between the squeezing function(denoted by
$T_{\Omega}$) and the squeezing function corresponding to the unit
ball(denoted by $S_{\Omega}$), for any bounded domain $\Omega$, is given by
the following lemma:

######  Lemma 1.2.

Let $\Omega\subset\mathbb{C}^{n}$ be a bounded domain. Then for every
$z\in{\Omega}$,

  1. a)

$T_{\Omega}(z)\geq\dfrac{1}{\sqrt{n}}S_{\Omega}(z);$

  2. b)

$S_{\Omega}(z)\geq\dfrac{1}{\sqrt{n}}T_{\Omega}(z).$

We say, a domain $\Omega$ is holomorphic homogeneous regular domain, if its
squeezing function has a positive lower bound. It follows from Lemma 1.2 that
if one of them has a positive lower bound then the other has it too. Therefore
holomorphic homogeneous regular domains are the same, in both the settings.

##  2\. Basic properties

We start with the simple observation that $T_{\Omega}$, for any bounded domain
$\Omega$ is biholomorphic invariant. To see this, let
$f_{0}:\Omega_{1}\to\Omega_{2}$ be a biholomorphism and $z\in\Omega_{1}$. We
claim that $T_{\Omega_{1}}(z)=T_{\Omega_{2}}(f_{0}(z))$.

  * •

Let $f:\Omega_{1}\to\mathbb{D}^{n}$ be a holomorphic embedding with $f(z)=0$.
Take $h=f\circ g:\Omega_{2}\to\mathbb{D}^{n},$ where
$g=f_{0}^{-1}:\Omega_{2}\to\Omega_{1}$. Then $h$ is holomorphic embedding with
$h(f_{0}(z))=0$. Also note that $h(\Omega_{2})=f(\Omega_{1})$. Let $r>0$ be
such that $\mathbb{D}^{n}(0,r)\subseteq f(\Omega_{1})=h(\Omega_{2})$, which
gives us that $T_{\Omega_{2}}(f_{0}(z))\geq r$. Thus

$T_{\Omega_{2}}(f_{0}(z))\geq T_{\Omega_{1}}(z).$

  * •

On similar lines, we can prove that $T_{\Omega_{1}}(z)\geq
T_{\Omega_{2}}(f_{0}(z))$, which will give us

$T_{\Omega_{1}}(z)=T_{\Omega_{2}}(f_{0}(z)).$

###### Proof of Lemma 1.2.

Let $z\in{\Omega}$.

  1. a)

Consider a holomorphic embedding $f:\Omega\to B^{n}(0,1)$ such that $f(z)=0$.
Let $r>0$ be such that $B^{n}(0,r)\subseteq f(\Omega)\subseteq B^{n}(0,1)$.
Then by considering $f$ as a holomorphic embedding $f:\Omega\to\mathbb{D}^{n}$
and noticing that $\mathbb{D}^{n}\left(0,\dfrac{r}{\sqrt{n}}\right)\subseteq
B^{n}(0,r)$, we get part a) of the lemma.

  2. b)

Consider a holomorphic embedding $f:\Omega\to\mathbb{D}^{n}$ such that
$f(z)=0$. Let $r>0$ be such that $B^{n}(0,r)\subseteq f(\Omega)\subseteq
B^{n}(0,1)$. Let $f:\Omega\to B^{n}(0,1)$ be defined as
$g(z)=\dfrac{1}{\sqrt{n}}f(z)$. then

$B^{n}\left(0,\dfrac{r}{\sqrt{n}}\right)\subseteq\mathbb{D}^{n}\left(0,\dfrac{r}{\sqrt{n}}\right)\subseteq
g(\Omega)\subseteq B^{n}(0,1),$

which proves part b) of the lemma.

∎

Note that this lemma gives us how the two squeezing functions $T_{\Omega}$ and
$S_{\Omega}$ are related and that the holomorphic homogeneous regular domains
turns out to be the same in both the settings.

######  Remark 2.1.

The two inequalities in Lemma 1.2 can not be attained simultaneously. One can
see that it can not happen unless $T_{\Omega}(z)=0=S_{\Omega}(z)$.

The fact that for any bounded domain $\Omega$, $T_{\Omega}$ is continuous and
the existence of extremal map for $T_{\Omega}$ follows on the same lines as in
[1], with some minor changes in the arguments. For the sake of completeness,
we give the proof here. We first establish existence of extremal property for
$T_{\Omega}$. We will need the following theorem for this.

######  Result 2.2 ([1, Theorem 2.2]).

Let $\Omega\subseteq\mathbb{C}^{n}$ be a bounded domain. Let $\\{f_{i}\\}$ be
a sequence of injective holomorphic maps, $f_{i}:\Omega\to\mathbb{C}^{n}$,
with $f_{i}(x)=0$ for all $i.$ Suppose that $f_{i}\to f,$ uniformly on compact
subsets of $\Omega$, where $f:\Omega\to\mathbb{C}^{n}$. If there exists a
neighborhood $\Omega$ of $0$ such that $\Omega\subseteq f_{i}(\Omega)$ for all
$i$, then $f$ is injective.

The following theorem establishes existence of extremal maps for squeezing
function.

######  Theorem 2.3.

Let $\Omega\subseteq\mathbb{C}^{n}$ be a bounded domain. Let $z\in\Omega$,
then there exists a holomorphic embedding $f:\Omega\to\mathbb{D}^{n}$ with
$f(z)=0$ such that $\mathbb{D}^{n}(0,T_{\Omega}(z))\subseteq f(\Omega).$

###### Proof.

Let $z\in\Omega$ and $a=T(z)$. Then by definition, there exists an increasing
sequence of numbers $r_{i}$, a sequence of holomorphic embeddings
$f_{i}:\Omega\to\mathbb{D}^{n}$ with $f_{i}(z)=0$ such that

$\mathbb{D}^{n}(0,r_{i})\subseteq f_{i}(\Omega)\subseteq\mathbb{D}^{n}.$

Thus $\\{f_{i}\\}$ is locally bounded and therefore it is normal by Montel’s
theorem. Let $\\{f_{i_{k}}\\}$ be a subsequence of $f_{i}$ such that
$f_{i_{k}}\to f$; where $f:\Omega\to\mathbb{C}^{n}$ is holomorphic.

Note that $f(z)=0$. Also, since $\mathbb{D}^{n}(0,r_{2})\subseteq
f_{2}(\Omega)$, therefore

$\mathbb{D}^{n}(0,r_{1})\subseteq\mathbb{D}^{n}(0,r_{2})\subseteq
f_{2}(\Omega).$

Similarly, it is easy to check that $\mathbb{D}^{n}(0,r_{1})\subseteq
f_{i}(\Omega)$ for all $i$, which gives us injectivity of $f$ by using Result
$\ref{injectivity}.$ Note that, since $f$ is open map, we get
$f:\Omega\to\mathbb{D}^{n}.$ Finally we need to show that
$\mathbb{D}^{n}(0,T_{\Omega}(z))\subseteq f(\Omega).$ To prove this, it is
sufficient to prove $\mathbb{D}^{n}(0,r_{j})\subseteq f(\Omega)$ for every
fixed $j$.

Since $r_{i}$ is an increasing sequence, we have
$\mathbb{D}^{n}(0,r_{j})\subseteq f_{i}(\Omega)$ for all $i>j.$ Let
$g_{i}=f_{i}^{-1}|_{\mathbb{D}^{n}(0,r_{j})}$, then $f_{i_{k}}\circ
g_{i_{k}}=\mathbbm{Id}_{\mathbb{D}^{n}(0,r_{j})}$ for $i_{k}>j$. Without loss
of generality, let us denote by $g_{i_{k}}$, a subsequence of $g_{i_{k}}$,
which exists by Montel’s theorem, converging to a function
$g:\mathbb{D}^{n}(0,r_{j})\to\mathbb{C}^{n}$, uniformly on compact subsets of
$\mathbb{D}^{n}(0,r_{j})$. We claim that $g:\mathbb{D}^{n}(0,r_{j})\to\Omega$.

Clearly, $g:\mathbb{D}^{n}(0,r_{j})\to\overline{\Omega}$. Also $g(0)=z$, this
implies that $0\in{g^{-1}(\Omega)}.$ Thus there exists a neighborhood
$\Omega_{1}$ of $0$ in $\mathbb{D}^{n}(0,r_{j})$ such that
$g(\Omega_{1})\subseteq\Omega$. Therefore $f\circ g|_{\Omega_{1}}$ is defined
and equals $Id_{\Omega_{1}}$, thus $det(J_{g}(0))\neq 0$ which further implies
that $J_{f}\neq 0$. Since $g=f^{-1}$, we get that $g$ is open. Thus
$g(\mathbb{D}^{n}(0,r_{j}))\subseteq\Omega.$

This gives us that $f\circ
g:\mathbb{D}^{n}(0,r_{j})\to\mathbb{D}^{n}(0,r_{j})$ is well defined and as
argued previously, we get that $J_{g},J_{f}\neq 0$. Hence we get that
$\mathbb{D}^{n}(0,r_{j})\subseteq f(\Omega)$ and hence the result is obtained.

∎

######  Remark 2.4.

From Theorem 2.3, it is easy to see that if $T_{\Omega}(z)=1$ for some
$z\in{\Omega}$, then $\Omega$ is biholomorphic to unit polydisk.

Let us denote by $K_{\Omega}(.,.)$, Kobayashi distance on any domain
$\Omega\subseteq\mathbb{C}^{n}$ and let us define $\sigma:[0,1)\to R^{+}$ as
$\sigma(x)=\log\dfrac{1+x}{1-x}.$ It is easy to check that $\sigma$ is one-one
with inverse given by $\sigma^{-1}(y)=\tanh\left(\dfrac{y}{2}\right)$. Now we
are ready to prove that $T_{\Omega}$ for any bounded domain $\Omega$ is
continuous.

######  Theorem 2.5.

For any bounded domain $\Omega\subseteq\mathbb{C}^{n}$, the function
$T_{\Omega}$ is continuous.

###### Proof.

Let $a\in{\Omega}$ be arbitrarily fixed. Let $\\{z^{i}\\}$ be a sequence in
$\Omega$ such that $z^{i}\to a$. We will prove that $T_{\Omega}(z^{k})\to
T_{\Omega}(a)$.

Let $\epsilon>0$ be arbitrary. For $a\in\Omega$, there exist
$f:\Omega\to\mathbb{D}^{n}$ such that $f(a)=0$ with
$\mathbb{D}^{n}(0,T_{\Omega}(a))\subseteq f(\Omega).$ Since $f$ is continuous,
there exists $N$ such that $\|f(z^{k})\|<\epsilon$ for $k>N$. For $k>N$,
consider $f^{k}:\Omega\to\mathbb{C}^{n}$ defined as

$f^{k}(z)=\dfrac{f(z)-f(z^{k})}{1+\epsilon}.$

Note that $f^{k}(\Omega)\subseteq\mathbb{D}^{n}$ and $f^{k}(z^{k})=0$. We will
first prove that
$\mathbb{D}^{n}\left(0,\dfrac{T_{\Omega}(a)-\epsilon}{1+\epsilon}\right)\subseteq
f^{k}(\Omega).$

Let
$z=(z_{1},\ldots,z_{n})\in{\mathbb{D}^{n}\left(0,\dfrac{T_{\Omega}(a)-\epsilon}{1+\epsilon}\right)}$,
then $|z_{i}|<\dfrac{T_{\Omega}(a)-\epsilon}{1+\epsilon}$ for all
$i=1,\ldots,n.$ Consider
$|z_{i}(1+\epsilon)+f_{i}(z^{k})|\leq|z_{i}|(1+\epsilon)+\epsilon<T_{\Omega}(a),$
for all $i=1,2,\ldots,n$ and therefore, we get
$z(1+\epsilon)+f(z^{k})\in{f(\Omega)}$. So $z\in{f^{k}(\Omega)}$, which gives
us $T_{\Omega}(z^{k})\geq\dfrac{T_{\Omega}(a)-\epsilon}{1+\epsilon}$ for all
$k>N$. Thus

(2.1) $\liminf_{k\to\infty}T_{\Omega}(z^{k})\geq T_{\Omega}(a).$

On the other hand, notice that $K_{\Omega}(z^{k},a)\to K_{\Omega}(a,a)$ as
$z^{k}\to a.$ Also for each $k$, there exists holomorphic embedding
$f^{k}:\Omega\to\mathbb{D}^{n}$ with $f^{k}(z^{k})=0$ such that
$\mathbb{D}^{n}(0,T_{\Omega}(z^{k}))\subseteq f^{k}(\Omega).$

Now,
$K_{\mathbb{D}^{n}}(f^{k}(z^{k}),f^{k}(a))=K_{\mathbb{D}^{n}}(0,f^{k}(a))\leq
K_{\Omega}(z^{k},a)\to 0$, as $k\to\infty$. Thus $f^{k}(a)\to 0$, in usual as
we know that Kobayashi metric induces standard topology[6].

Let $\epsilon>0$ be arbitrary. Since $f^{k}(a)\to 0$, therefore there exists
$N$ such that $\|f^{k}(a)\|<\epsilon$ for all $k>N$. For $k>N$, define
$f:\Omega\to\mathbb{C}^{n}$ as

$f(z)=\dfrac{f^{k}(z)-f^{k}(a)}{1+\epsilon},$

then $f(a)=0$. We claim that $f(\Omega)\subseteq\mathbb{D}^{n}.$

Let $f=(f_{1},f_{2},\ldots,f_{n})$, then it is easy to verify that
$|f_{j}(z)|<1$ for all $j=1,2,\ldots,n$, which gives us
$f(\Omega)\subseteq\mathbb{D}^{n}.$ Also, as argued in the previous case it is
easy to check that
$\mathbb{D}^{n}\left(0,\dfrac{T_{\Omega}(z^{k})-\epsilon}{1+\epsilon}\right)\subseteq(f(\Omega)).$
Therefore we get
$T_{\Omega}(a)\geq\dfrac{T_{\Omega}(z^{k})-\epsilon}{1+\epsilon}$, which
further gives us that

(2.2) $T_{\Omega}(a)\geq\limsup_{k\to\infty}T_{\Omega}(z^{k}).$

Hence the result follows from Equation 2.1 and 2.2. ∎

######  Theorem 2.6.

Let $\Omega^{\prime}\subseteq\mathbb{C}^{n}$ be a bounded domain and let
$A\subseteq\Omega^{\prime}$ be proper analytic subset. Then for
$\Omega=\Omega^{\prime}\setminus A,$

$T_{\Omega}(z)\leq\sigma^{-1}\left(K_{\Omega^{\prime}}(z,A)\right),\
z\in\Omega.$

In particular, $\Omega$ is homogeneous regular.

###### Proof.

Let $z\in{\Omega},$ and $f:\Omega\to\mathbb{D}^{n}$ be a holomorphic embedding
such that $f(z)=0.$ Since $\Omega=\Omega^{\prime}\setminus A$, therefore by
Riemann removable singularity theorem, $f$ can be extended to a holomorphic
map $\tilde{f}:\Omega^{\prime}\to\mathbb{D}^{n}$. Clearly,
$\tilde{f}(\Omega)\cap\tilde{f}(A)=\emptyset$. We know by decreasing property
of Kobayashi metric that $K_{\mathbb{D}^{n}}(\tilde{f}(z),\tilde{f}(A))\leq
K_{\Omega^{\prime}}(z,A)$, that is $K_{\mathbb{D}^{n}}(0,\tilde{f}(A))\leq
K_{\Omega^{\prime}}(z,A)$.

Let $r>0$ be such that $\mathbb{D}^{n}(0,r)\subseteq
f(\Omega)=\tilde{f}(\Omega)$. Thus,
$\mathbb{D}^{n}(0,r)\cap\tilde{f}(A)=\emptyset$. This implies that for each
$a\in A,\ K_{\mathbb{D}^{n}}(0,r)\leq K_{\mathbb{D}^{n}}(0,\tilde{f}(a))$.
Thus we get

$K_{\mathbb{D}^{n}}(0,r)\leq K_{\mathbb{D}^{n}}(0,\tilde{f}(A))\leq
K_{\Omega^{\prime}}(z,A),$

which implies that $r\leq\sigma^{-1}(K_{\Omega^{\prime}}(z,A))$. Hence we get
that $T_{\Omega}(z)\leq\sigma^{-1}\left(K_{\Omega^{\prime}}(z,A)\right).$ ∎

##  3\. Examples

In this section, we discuss squeezing function for some domains. We start with
unit polydisk and unit ball.

######  Example 3.1.

Let $\Omega_{1}=\mathbb{D}^{n}$ and $\Omega_{2}=B^{n}$.

  * •

Considering automorphisms of $\mathbb{D}^{n}$, it is easy to see that
$T_{\Omega_{1}}(z)\equiv 1$.

  * •

Note that, using Theorem 2.3, and the fact that $B^{n}(0,1)$ and
$\mathbb{D}^{n}\,,n>1$ are not biholomorphic, we get that
$T_{\Omega_{2}}(z)<1$, Also by Lemma 1.2
$T_{\Omega_{2}}(z)\geq\frac{1}{\sqrt{n}}S_{\Omega_{2}}(z)$. Since we know that
$S_{\Omega_{2}}(z)\equiv 1$, we get $\frac{1}{\sqrt{n}}\leq
T_{\Omega_{2}}(z)\leq 1$. Also by [2, Proposition 2], we get that
$T_{\Omega_{2}}(z)\leq\frac{1}{\sqrt{n}}$. Thus,
$T_{\Omega_{2}}(z)\equiv\frac{1}{\sqrt{n}}.$

######  Example 3.2.

Take $\Omega^{\prime}=B^{n}(0,1)$, $A=(0,0,\ldots,0)$ and
$\Omega=\Omega^{\prime}\setminus A$. Using Theorem 2.6, we get that
$T_{\Omega}(z)\leq\|z\|.$

Let $z\in{\Omega}$. Consider a holomorphic embedding
$f:\Omega\to\mathbb{D}^{n}$ with $f(z)=0$. Then $f$ can be extended to a
holomorphic embedding $F:B^{n}\to\mathbb{D}^{n}$.

Let $r>0$ be such that $\mathbb{D}^{n}(0,r)\subseteq
f(\Omega)\subseteq\mathbb{D}^{n}$, and therefore $\mathbb{D}^{n}(0,r)\subseteq
F(B^{n}(0,1))\subseteq\mathbb{D}^{n}$, therefore by [2, Proposition 2] , we
get that

(3.1) $T_{\Omega}(z)\leq\dfrac{1}{\sqrt{n}}\ \mbox{for all}\ z\in{\Omega}.$

Let $z\in{\Omega}$ be such that $\|z\|<\frac{1}{\sqrt{n}}$. Consider an
automorphism $\phi_{z}$ of $B^{n}$, which maps $z$ to $0$ and $0$ to $z$. Thus
$f=\phi_{z}|_{\Omega}:\Omega\to\mathbb{D}^{n}$ is a holomorphic embedding with
$f(z)=0$. We claim that $\mathbb{D}^{n}(0,\|z\|)\subseteq
f(\Omega)=B^{n}\setminus\\{z\\}.$ To see this, let
$w=(w_{1},w_{2},\ldots,w_{n})\in{\mathbb{D}^{n}(0,\|z\|)}.$ Then clearly
$z\notin{\mathbb{D}^{n}(0,\|z\|)}$. Also,
$\|w\|^{2}=\sum_{i=1}^{n}|w_{i}|^{2}<n\|z\|^{2}<1$. Thus we get our claim. So
we get that $T_{\Omega}(z)\geq\|z\|$. We also have $T_{\Omega}(z)\leq\|z\|$ by
equation 3.1 and therefore $T_{\Omega}(z)=\|z\|$ on
$B^{n}\left(0,\frac{1}{\sqrt{n}}\right)\setminus\\{0\\}$ and hence on
$\overline{B^{n}\left(0,\frac{1}{\sqrt{n}}\right)}$ by continuity of
$T_{\Omega}$.

Also, for $\|z\|>\frac{1}{\sqrt{n}}$, by Lemma 1.2 and Equation 3.1, we have
that $\frac{\|z\|}{\sqrt{n}}\leq T_{\Omega}(z)\leq\frac{1}{\sqrt{n}}.$ Here we
have used that $S_{\Omega}(z)=\|z\|,$ by [1, Corollary 7.3].

##  4\. Classical symmetric domains

We now study squeezing function for classical symmetric bounded domain. A
classical domain is one of the following four types:

$R_{1}(r,s)=\\{Z=(z_{jk}):I-Z\overline{Z}^{\prime}>0,\mbox{where }\ Z\
\mbox{is an}\ r\times s\ \mbox{matrix}\\}\ (r\leq s),$

$R_{II}(p)=\\{Z=(z_{jk}):I-Z\overline{Z}^{\prime}>0,\mbox{where }\ Z\ \mbox{is
a symmetric matrix of order}\ p\\},$

$R_{III}(q)=\\{Z=(z_{jk}):I-Z\overline{Z}^{\prime}>0,\mbox{where }\ Z\
\mbox{is a skew symmetric matrix of order}\ q\\},$

$R_{IV}(n)=\\{z=(z_{1},z_{2},\ldots,z_{n})\in{\mathbb{C}^{n}}:1+|zz^{\prime}|^{2}-2zz^{\prime}>0,\
1-|zz^{\prime}|>0\\}$.

Note that for a bounded homogeneous domain $R$, its squeezing function is
constant, using biholomorphic invariance of squeezing functions. Let us denote
this constant by $t(R).$

######  Theorem 4.1.

$\frac{1}{\sqrt{n}\sqrt{r}}\leq
t\left(R_{I}(r,s)\right)\leq\frac{1}{\sqrt{r}},\ n=rs,$
$\frac{1}{\sqrt{n}\sqrt{p}}\leq
t\left(R_{II}(p)\right)\leq\frac{1}{\sqrt{p}},\ n=\frac{p(p+1)}{2},$
$\frac{1}{\sqrt{n}\sqrt{\left[\frac{q}{2}\right]}}\leq
t\left(R_{III}(q)\right)\leq\frac{1}{\sqrt{\left[\frac{q}{2}\right]}},\
n=\frac{q(q-1))}{2},$ $\frac{1}{\sqrt{n}\sqrt{2}}\leq
t\left(R_{IV}(n)\right)\leq\frac{1}{\sqrt{2}}.$

######  Theorem 4.2.

If $R_{1},R_{2},\ldots,R_{k}$ are classical symmetric domains, and
$D=R_{1}\times R_{2}\times\ldots\times R_{k}$ then

$s(D)\frac{1}{\sqrt{n}}\leq t(D)\leq s(D),$

where $n=n_{1}+n_{2}+\ldots+n_{k}$, $n_{i}$ is complex dimension of $R_{i}\
i=1,2,\ldots,k$ and $s(D)$ is, squeezing function(which is constant since
domain is homogeneous) of $D$ — corresponding to unit ball — given by

$s(D)=\left[s(R_{1})^{-2}+s(R_{2})^{-2}+\ldots+s(R_{k})^{-2}\right]^{-\frac{1}{2}}.$

We will need the following key lemma, whose proof is based on the method of
Kubota [3].

######  Lemma 4.3.

Let $\Omega$ be a bounded homogeneous domain in $\mathbb{C}^{n}$ satisfying
the conditions:

  1. (1)

$\\{z=(z_{1},\ldots,z_{n}):|z_{\alpha_{j}}|<1,\ \mbox{for}\ j=1,\ldots,m\
\mbox{and}\ z_{\alpha}=0\ \mbox{for the other}\
\alpha^{\prime}s\\}\subseteq\Omega$, where $1\leq\alpha_{1}\leq\ldots
a_{m}\leq n;$

  2. (2)

for each $(1\leq j\leq m)$,

$\\{z=(z_{1},\ldots,z_{n}):|z_{\alpha_{j}}|=1\ \mbox{and}\ z_{\alpha}=0\
\mbox{for the other}\ \alpha^{\prime}s\\}\subseteq\partial\Omega.$

If $f:\Omega\to\mathbb{D}^{n}$ is holomorphic embedding and if
$\mathbb{D}^{n}(0,s)\subseteq f(\Omega),$ then $s\leq\frac{1}{\sqrt{m}}$.

###### Proof.

We have that $f=(f_{1},f_{2},\ldots,f_{n}),$ where each
$f_{i}:\Omega\to\mathbb{D},$ where $\mathbb{D}$ is unit ball in $\mathbb{C}.$
Since $\mathbb{D}^{n}(0,s)\subseteq
f(\Omega)=(f_{1}(\Omega),f_{2}(\Omega),\ldots,f_{n}(\Omega))$, therefore
$B(0,s)\subseteq f_{i}(\Omega)$, for $1\leq i\leq n$. Thus for $i,$ $1\leq
i\leq n$, we have that

(4.1) $s\leq\liminf_{r\uparrow 1}|f_{i}(0,\ldots,re^{i\theta_{j}},\ldots,0)|.$

Define a function $g:\mathbb{D}^{m}\to\mathbb{D}^{n}$ by
$g=(g_{1},g_{2},\ldots,g_{n}),$ where

$g_{i}(\zeta_{1},\zeta_{2},\ldots,\zeta_{m})=f_{i}(0,\ldots,\zeta_{1},\zeta_{2},\ldots,\zeta_{m},\ldots,0);$

where $\zeta_{j}$ is in $\alpha_{j}$th position on right hand side expression.
Let us write taylor series expansion of $g_{i}$ as $g_{i}(\zeta)=\sum
a_{v}^{(\alpha)}\zeta^{v}$, where $v$ is multi index and $a_{0}^{(\alpha)}=0.$

By condition (1) and using Equation 4.1 for all components of $g$, we get

(4.2)
$ns^{2}\leq\frac{1}{2\pi}\int_{0}^{2\pi}\|g(0,\ldots,e^{i\theta_{j}},\ldots,0)\|^{2},$

where $g(0,\ldots,e^{i\theta_{j}},\ldots,0)$ is defined as $\lim_{r\to
1}g(0,\ldots,re^{i\theta_{j}},\ldots,0)$ and $e^{\theta_{j}}$ lies in the
$\alpha_{j}$th position. Notice that the expression on the right side of the
inequaltiy 4.2 equals $\sum_{\alpha=1}^{n}\sum|a_{0\ldots v_{j}\ldots
0}^{(\alpha)}|^{2}$. Thus summing it over all positions
$\alpha_{1},\alpha_{2},\ldots,\alpha_{m},$ we get

$\begin{split}mns^{2}&\leq\sum_{\alpha=1}^{n}\sum|a_{0\ldots v_{1}\ldots
0}^{(\alpha)}|^{2}+\sum_{\alpha=1}^{n}\sum|a_{0\ldots v_{2}\ldots
0}^{(\alpha)}|^{2}+\ldots+\sum_{\alpha=1}^{n}\sum|a_{0\ldots v_{m}\ldots
0}^{(\alpha)}|^{2}\\\ &\leq\sum_{\alpha=1}^{n}\sum|a_{v_{1}v_{2}\ldots
v_{m}}^{(\alpha)}|^{2}\\\ &\leq\int_{\mathbb{T}^{m}}\|g\|^{2}dh\\\ &\leq
n.\end{split}$

Here we have used that $\|g\|^{2}\leq n$ a.e. on $\mathbb{T}^{m}$. Thus we get
that $ms^{2}\leq 1$, hence we get our result. ∎

Now the proof of Theorem 4.1 and Theorem 4.2 — using the method of Kubota [3]
— follows from Lemma 1.2 and Lemma 4.3.

Recently in [5], authors have found lower bound for squeezing function
corresponding to unit ball of the product domain. We give its alternate proof
here.

######  Theorem 4.4.

Let $\Omega_{i}\subseteq\mathbb{C}^{n_{i}},\ i=1,2,\ldots,k$ be bounded
domains and
$\Omega=\Omega_{1}\times\Omega_{2}\times\ldots\Omega_{k}\subseteq\mathbb{C}^{n},\
n=n_{1}+n_{1}+\ldots+n_{k},$. Then for any
$a=(a_{1},a_{2},\ldots,a_{k})\in\Omega,\ a_{i}\in{\Omega_{i}}\ i=1,2,\ldots,k$
, we have

$S_{\Omega}(a)\geq\left[(S_{\Omega_{1}}(a_{1}))^{-2}+(S_{\Omega_{2}}(a_{2}))^{-2}+\ldots+(S_{\Omega_{k}}(a_{k}))^{-2}\right]^{-\frac{1}{2}}.$

We will need the following easy to prove lemma to prove this theorem.

######  Lemma 4.5.

If $B^{n_{i}}(0,1)\subseteq A_{i}\subseteq
B^{n_{i}}(0,r_{i})\subseteq\mathbb{C}^{n_{i}}\ i=1,2,\ldots,k$, then

$B^{n}(0,1)\subseteq A_{1}\times A_{2}\times\ldots A_{k}\subseteq
B^{n}\left(0,\sqrt{r_{1}^{2}+r_{2}^{2}+\ldots+r_{k}^{2}}\right),$

where $n=n_{1}+n_{2}+\ldots+n_{k}.$

###### Proof of theorem 4.4.

Let $f_{i}:\Omega_{i}\to B^{n_{i}}(0,1)$ be holomorphic embedding with
$f_{i}(a_{i})=0$ such that $B^{n_{i}}(0,S_{\Omega_{i}}(a_{i}))\subseteq
f_{i}(\Omega_{i})\subseteq B^{n_{i}}(0,1)$; $i=1,2,\ldots k$.

Thus $B^{n_{i}}(0,1)\subseteq r_{i}f_{i}(\Omega_{i})\subseteq
B^{n_{i}}(0,r_{i}),$ where $r_{i}=(S_{\Omega_{i}}(a_{i}))^{-1}$ and therefore
by Lemma 4.5, we get that

(4.3) $B^{n}(0,1)\subseteq r_{1}f_{1}(\Omega_{1})\times
r_{2}f_{2}(\Omega_{2})\times\ldots r_{k}f_{k}(\Omega_{k})\subseteq
B^{n}\left(0,\sqrt{r_{1}^{2}+r_{2}^{2}+\ldots+r_{k}^{2}}\right).$

Now consider $f:\Omega\to B^{n}(0,1)$ defined as

$f(z_{1},z_{2},\ldots,z_{k})=\left(\frac{r_{1}f_{1}(z_{1})}{\sqrt{r_{1}^{2}+r_{2}^{2}+\ldots+r_{k}^{2}}},\frac{r_{2}f_{2}(z_{2})}{\sqrt{r_{1}^{2}+r_{2}^{2}+\ldots+r_{k}^{2}}},\ldots,\frac{r_{k}f_{k}(z_{k})}{\sqrt{r_{1}^{2}+r_{2}^{2}+\ldots+r_{k}^{2}}}\right).$

Notice that $f$ is holomorphic embedding with $f(a)=0$, and hence our result
follows from equation 4.3. ∎

In [3], Kubota proved that equality in Theorem 4.4 is obtained, when the
product of classical symmetric domains is considered. That is

(4.4)
$S_{\Omega}(a)=\left[(S_{\Omega_{1}}(a_{1}))^{-2}+(S_{\Omega_{2}}(a_{2}))^{-2}+\ldots+(S_{\Omega_{k}}(a_{k}))^{-2}\right]^{-\frac{1}{2}}.$

We give here the following example to show that such an equality does not hold
in case of squeezing function.

######  Example 4.6.

Let us consider $\Omega_{1}=B^{3}(0,1)\subseteq\mathbb{C}^{3}$ and
$\Omega_{2}=B^{2}(0,1)\subseteq\mathbb{C}^{2}.$

Let $\Omega=\Omega_{1}\times\Omega_{2}\subseteq\mathbb{C}^{5}$ and
$a=(a_{1},a_{2})\in\Omega,\ a_{i}\in{\Omega_{i}},\ i=1,2.$ Then by Example
3.2, we get $T_{\Omega_{1}}(a_{1})=\frac{1}{\sqrt{3}}$ and
$T_{\Omega_{2}}(a_{2})=\frac{1}{\sqrt{2}}$. Thus
$\left[(T_{\Omega_{1}}(a_{1}))^{-2}+(T_{\Omega_{2}}(a_{2}))^{-2}\right]^{-\frac{1}{2}}=\frac{1}{\sqrt{5}}.$
Also by Theorem 4.7, we have that
$T_{\Omega}(a)\geq\min\left({T_{\Omega_{1}}(a_{1}),T_{\Omega_{2}}(a_{2})}\right)$,
therefore we get

$T_{\Omega}(a)\geq\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{5}}.$

Thus we see that Equation 4.4 does not hold.

We give the following theorem which gives us lower bound for squeezing
function of product domain.

######  Theorem 4.7.

Let $\Omega_{i}\subseteq\mathbb{C}^{n_{i}},\ i=1,2,\ldots,k$ be bounded
domains and
$\Omega=\Omega_{1}\times\Omega_{2}\times\ldots\Omega_{k}\subseteq\mathbb{C}^{n},\
n=n_{1}+n_{1}+\ldots+n_{k},$. Then for any
$a=(a_{1},a_{2},\ldots,a_{k})\in\Omega,\ a_{i}\in{\Omega_{i}}\
i=1,2,\ldots,k$, we have

$T_{\Omega}(a)\geq\min_{1\leq i\leq n}T_{\Omega_{i}}(a_{i}).$

In particular, product of holomorphic homogeneous regular domains is
holomorphic homogeneous regular.

We will need an observation to prove the above theorem, which we mention below
as a lemma.

######  Lemma 4.8.

If
$\mathbb{D}^{n_{i}}(0,1)\subseteq\Omega_{i}\subseteq\mathbb{D}^{n_{i}}(0,r_{i}),$,
where $\Omega_{i}$ are domains in $\mathbb{C}^{n_{i}},\ i=1,2,\ldots,k$ then

$\mathbb{D}^{n}(0,1)\subseteq\Omega_{1}\times\Omega_{2}\times\ldots\times\Omega_{k}\subseteq\mathbb{D}^{n}(0,r),$

where $n=n_{1}+n_{2}+\ldots+n_{k}$ and $r=\max_{1\leq i\leq
n}T_{\Omega_{i}}(a_{i}).$

###### Proof.

It is obvious that
$\mathbb{D}^{n}(0,1)\subseteq\Omega_{1}\times\Omega_{2}\times\ldots\times\Omega_{k}$.

Also for
$z=(z_{1},z_{2},\ldots,z_{k})\in{\Omega_{1}\times\Omega_{2}\times\ldots\times\Omega_{k}},$
we have $z_{i}\in{\mathbb{D}^{n_{i}}(0,r_{i})}.$ Therefore it is clear that
$z\in{\mathbb{D}^{n}(0,r)},$ where $r=\max{1\leq i\leq k}\ r_{i}$.

∎

###### Proof of theorem 4.7.

By theorem 2.3, for each $a_{i}\in{\Omega_{n_{i}}}$, there is a holomorphic
embedding $f_{i}:\Omega_{i}\to\mathbb{D}^{n_{i}}(0,1)$, with $f_{i}(a_{i})=0$
such that

$\mathbb{D}^{n}(0,T_{\Omega_{i}}(a_{i}))\subseteq
f_{i}(\Omega_{i})\subseteq\mathbb{D}^{n_{i}}(0,1),\ i=1,2,\ldots,k.$

Thus we have

$\mathbb{D}^{n_{i}}(0,1)\subseteq
r_{i}f_{i}(\Omega_{i})\subseteq\mathbb{D}^{n_{i}}(0,r_{i}),\ i=1,2,\ldots,k,$

where $r_{i}=\left(T_{\Omega_{i}}(a_{i})\right)^{-1}$. Therefore by Lemma 4.8,
we get that

(4.5) $\mathbb{D}^{n}(0,1)\subseteq r_{1}f_{1}(\Omega_{1})\times
r_{2}f_{2}(\Omega_{2})\times\ldots\times
r_{k}f_{k}(\Omega_{k})\subseteq\mathbb{D}^{n}(0,r),$

where $r=\max_{1\leq i\leq k}r_{i}$.

Consider $f:\Omega\to\mathbb{D}^{n}(0,1)$ defined as

$f(z)=\left(\frac{r_{1}}{r}f_{1}(z_{1}),\frac{r_{2}}{r}f_{2}(z_{2}),\ldots,\frac{r_{k}}{r}f_{k}(z_{k})\right).$

Now our result follows from Equation 4.5.

∎

######  Remark 4.9.

For $\Omega_{1}=B^{n}(0,1)$ and $\Omega_{2}=\mathbb{D}^{n}(0,1),$ we have
$T_{\Omega_{1}}\equiv\frac{1}{\sqrt{n}}$, $T_{\Omega_{2}}\equiv 1$,
$S_{\Omega_{1}}\equiv 1$ and $S_{\Omega_{2}}\equiv\frac{1}{\sqrt{n}}$.
Therefore $T_{\Omega_{1}}=\frac{1}{\sqrt{n}}S_{\Omega_{1}}$ and
$S_{\Omega_{2}}=\frac{1}{\sqrt{n}}T_{\Omega_{2}}.$ Also for classical domain
$R_{1}(r,s),$ if we take $r=1,$ then $R_{1}(1,s)=B^{s}(0,1)$, and therefore
$T_{R_{1}(1,s)}\equiv\frac{1}{\sqrt{s}}\equiv\frac{1}{\sqrt{s}}S_{R_{1}(1,s)}$.
In Example 3.2, we noticed that such an equality does not hold for the
punctured unit ball. We strongly feel that for a bounded homogeneous domain
$\Omega$, one of the two inequalities in Lemma 1.2 holds.

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