# Indicatrix of Invariant metrics, Maximal circularity and Scaling of domains

Huda Alzaki and Jae-Cheon Joo  Department of Mathematics, King Fahd University
of Petroleum and Minerals, 31261 Dhahran, Saudi Arabia
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We investigate how to use the indicatrix of an invariant metric to rescale a
sequence of biholomorphic maps and to ensure the convergence of the rescaled
sequence. We also use the indicatrix of the Kobayashi-Royden metric to define
the maximal circularity function which indicates how the domain is close to a
circular domain.

###### Key words and phrases:

Invariant metrics, Indicatrix, Scaling, Circularity

###### 2010 Mathematics Subject Classification:

32H02, 32M05

##  1\. Introduction

This paper consists of several observations on the celebrated work [5] by S.
Frankel which states that a convex hyperbolic domain in complex Euclidean
space admitting a discrete and cocompact subgroup of automorphisms which acts
freely is biholomorphic to a bounded symmetric domain. As a key ingredient,
Frankel proved that if $\Omega$ is a convex hyperbolic domain, then for any
sequence $\\{\varphi_{j}\\}$ of holomorphic automorphisms and any
$p\in\Omega$, the scaled sequence

(1.1)
$\tau_{j}=[d\varphi_{j}(p)]^{-1}\circ(\varphi_{j}(\cdot)-\varphi_{j}(p)),$

has a convergent subsequence. In this paper, we investigate the sequence (1.1)
about the construction, the convergence and the symmetry of limits in a
geometric point of view on invariant metrics.

Roughly speaking, by the method of scaling we mean a way of renomalizing maps
in a sequence by compositions with affine linear maps to make the new sequence
subseuentially convergent. The linear map which is composed for the scalng is
usually called the stretching factor. Although the idea of scaling is used in
a wide range of mathematics, two specific constructions are usually cited when
we deal with sequences of holomorphic automorphisms of a domain: Frankel’s
sequence (1.1) and Pinchuk’s one ([19]). In contrast with (1.1), Pinchuk’s
construction of the stretching factor is more complicated and not completely
known in general cases of domain. On the other hand, Pinchuk’s construction
has a merit that the stretching factor is completely determined by the
geometry of domain, while the stretching factor of (1.1) is
$(d\varphi_{j}(p))^{-1}$ and it depends on the map. The independence of
stretching factor from the map in Pinchuk’s constuction turned out to be quite
useful in dealing with invariants of domain, and it has been asked if or when
two constuctions are equivalent. There are some partial results about the
question ([8, 12]), however it is not answered yet in general cases. Instead,
we construct a new strecthing factor which is fully detemined by the
indicatrix of an invariant metric and prove that the scaled sequence with the
new stretching factor is indeed equivalent with (1.1) as the first result
(Theorm 2.2).

The second observation is on a condition for the convergence of (1.1). As
mentioned above, Frankel proved that the sequence is subsequentially
convergent if $\Omega$ is convex and hyperbolic by obtaining local uniform
estimates of derivatives for automorphisms up to the second order. In view of
Theorem 2.2, it is also natural to describe a geometric condition on invariant
metrics for the convergence. The condition is given by a relation between the
indicatrix of an invariant metric and the ball of the induced distance. We
call the condition the uniform domination of the domain by the given metric
(Definition 2.4). After observing the convergence of (1.1) under the uniform
domination condition, we will prove that any convex hyperbolic domain is
uniformly dominated by the Kobayashi-Royden metric in Theorem 2.8 in order to
recover the case of [5].

The last result is about a symmetry of the image of a subsequential limit of
(1.1). In [5], the author chooses a sequence of automorphisms which drags an
interior point to a regular boundary point, and proves the image of the limit
(and hence the given domain as well) admits a $1$-parameter family of
automorphisms. Let us recall that if $q$ is a regular boundary point of a
bounded symmetric domain which is not biholomorphic to the ball, then there is
a nontrivial analytic variety in the boundary which passes through $q$. On the
other hand, a boundary point $q$ of a domain is called a minimal boundary
point if the boundary of domain does not contain any nontrivial analytic
variety passing through $q$. As seen in some examples of domain of piecewise
smooth boundary ([11, 13, 14, 21]), the image of the limit of the scaled
sequence is usually more canonical in case the given sequence of automorphisms
drags an interior point to a minimal boundary point rather than to a non-
minimal boundary point. With this observation, it is natural to ask what kind
of symmetry the limit of (1.1) has, in case $\Omega$ is bounded and convex,
and $\varphi_{j}(p)$ tends to a minimal boundary point. Another question is
where the circularity of the domain comes from. Recall that the circularity is
a signature property of bounded symmetric domains in the set of bounded
homogeneous domains. We believe that these two questions are indeed same and
can be combined to the following one.

Question. Let $\Omega$ be a bounded convex domain in $\mathbb{C}^{n}$. Suppose
that there exists a sequence $\\{\varphi_{j}\\}$ of holomorphic automorphisms
and $p\in\Omega$ such that $\varphi_{j}(p)$ tends to a minimal boundary point
of $\Omega$. Then is $\Omega$ biholomorphic to a completely circular domain?

In order to study this problem in more geometric point of view, we define in
Section 3 an invariant function $c_{\Omega}$ on $\Omega$ which we call the
maximal circularity. We will see that $0<c_{\Omega}(x)\leq 1$ for any
$x\in\Omega$ and $c_{\Omega}(x)=1$ if and only if $\Omega$ is biholomorphic to
the indicatrix of the Kobayashi-Royden metric of $\Omega$ at $x$. With this,
we can ask a more general question.

Question. Let $\Omega$ be a bounded convex domain in $\mathbb{C}^{n}$. Does
$c_{\Omega}(x)$ tend to $1$ as $x$ tends to a minimal boundary point?

We do not know yet whether or not the answers to the questions are all yes. We
will only prove in Section 3 that the answer to the last question is
affirmative in case $\Omega$ is either strongly pseudoconvex (not necessarily
convex) or a convex analytic polyhedral domain.

Acknowledgement. We would like to thank Professor Kang-Tae Kim for many
inspiring discussions about the scaling theory.

##  2\. Construction and Convergence of a scaling sequence

Let $\Omega$ be a domain in $\mathbb{C}^{n}$. By a complex Finsler metric or
simply a metric on $\Omega$, we mean an upper semicontinuous (usc) function
$F$ on $T\Omega=\Omega\times\mathbb{C}^{n}$ such that

  * (i)

$F(x;v)\geq 0$ for any $x\in\Omega$ and $v\in\mathbb{C}^{n}$ and $F(x;v)=0$
only if $v=0$, and

  * (ii)

$F(x;cv)=|c|F(x;v)$ for any $(x,v)\in\Omega\times\mathbb{C}^{n}$ and
$c\in\mathbb{C}$.

The indicatrix of $F$ at $x\in\Omega$ is the set

$I_{\Omega,F}(x):=\\{v\in\mathbb{C}^{n}:F(x;v)<1\\}.$

We say a metric on $\Omega$ is convex if the indicatrix of $F$ is a convex
domain in $\mathbb{C}^{n}$ at any point of $\Omega$.

######  Example 2.1.

Obviously, any Hermitian metric on $\Omega$ is a convex metric. In particular,
the Bergman metric or the Kähler-Einstein metric are convex metrics which are
invariant under biholomorphic maps. Another example of invariant metric on a
domain is the Kobayashi-Royden metric which is defined by

$K_{\Omega}(x;v)=\inf\\{\lambda>0:\exists
f\in\mathcal{O}(\Delta,\Omega),\,\,f(0)=x,\,\,\lambda f^{\prime}(0)=v\\},$

where $\mathcal{O}(\Delta,\Omega)$ represents the space of holomorphic
mappings from the unit disc $\Delta$ into $\Omega$. Indeed, $K_{\Omega}$ is
not necessarily a metric, since $K_{\Omega}(x;v)$ may be $0$ for a nonzero
vector $v$, however, it turns out that it is a metric for instance if $\Omega$
is bounded. We call a domain (or more generally a complex manifold) a
hyperbolic domain if $K_{\Omega}$ is a metric. It turns out that $K_{\Omega}$
is convex if $\Omega$ is convex however it is not true for general hyperbolic
domains. To make it convex, one can consider a new invariant metric
$\widehat{K}_{\Omega}$ whose indicatrix is the convex hull of the indicatrix
of $K_{\Omega}$ at each point. The metric is called the Kobayashi-Buseman
metric. See for instance [7, 16] for more details about invariant metrics.

Let $F$ be a metric on a domain $\Omega$ and $x\in\Omega$. By the upper
semicontnuity, we can find $u_{1}=u_{1}(x)\in\mathbb{C}^{n}$ with $|u_{1}|=1$
such that

$\mu_{1}(x):=F(x;u_{1})=\max\\{F(x;v):v\in\mathbb{C}^{n},\,\,|v|=1\\}.$

Inductively, if we have chosen $u_{1},\ldots,u_{\alpha}$ and
$\mu_{\beta}=F(x;u_{\beta})$ for $1\leq\beta\leq\alpha\leq n-1$, then a unit
vector $u_{\alpha+1}$ and $\mu_{\alpha+1}$ are defined by

$\mu_{\alpha+1}(x):=F(x;u_{\alpha+1})=\max\\{F(x;v):v\in
V_{\alpha},\,\,|v|=1\\},$

where $V_{\alpha}$ is the orthogonal complement of
$\\{u_{1}\ldots,u_{\alpha}\\}$ in $\mathbb{C}^{n}$. Let $L_{x}$ be the linear
transform on $\mathbb{C}^{n}$ defined by

(2.1) $L_{x}(u_{\alpha})=\mu_{\alpha}(x)e_{\alpha},\quad\alpha=1,\ldots,n,$

where $\\{e_{1},\ldots,e_{n}\\}$ is the standard Euclidean basis for
$\mathbb{C}^{n}$. We call $L_{x}$ a stretching factor of $\Omega$ at $x$
induced by the metric $F$.

For a sequence $\varphi_{j}\in\mathrm{Aut}(\Omega)$, the group of holomorphic
automorphisms of $\Omega$, S. Frankel ([5]) defined a sequence

(2.2) $\tau_{j}=[d\varphi_{j}(p)]^{-1}\circ(\varphi_{j}(\cdot)-p_{j}),$

where $p_{j}=\varphi_{j}(p)$ and $d\varphi_{j}(p)$ represents the complex
Jacobian matrix of $\varphi_{j}$ at $p$. Frankel proved that the sequence
$\\{\tau_{j}\\}$ has a convergent subsequence if $\Omega$ is hyperbolic and
convex. We prove that the linear transform $[d\varphi_{j}(p)]^{-1}$ can be
replaced by a stretching factor induced by an invariant metric.

######  Theorem 2.2.

Let $\Omega$ be a domain in $\mathbb{C}^{n}$, $p\in\Omega$ and
$\\{\varphi_{j}\\}$ be a sequence of holomorphic automorphisms. Then the
sequence (2.2) converges subsequentially if and only if

(2.3) $\sigma_{j}:=L_{j}\circ(\varphi_{j}(\cdot)-p_{j})$

converges subsequentially, where $L_{j}=L_{p_{j}}$ is a stretching factor
induced by any convex invariant metric.

###### Proof.

Let $F$ be a convex invariant metric on $\Omega$. Suppose that
$u_{\alpha}^{j}$ and $\mu_{\alpha}^{j}$ were chosen to construct $L_{j}$ by
(2.1) at $p_{j}=\varphi_{j}(p)$ and let

$\tilde{u}^{j}_{\alpha}=d\varphi_{j}(p)^{-1}(u^{j}_{\alpha}),\quad\alpha=1,\ldots,n.$

Since $\varphi_{j}$ preserves $F$, we see that
$F(p;\tilde{u}^{j}_{\alpha})=\mu^{j}_{\alpha}$. Therefore, the vectors
$\tilde{u}^{j}_{\alpha}/\mu^{j}_{\alpha}\in\partial I_{\Omega,F}(p),$ where

$I_{\Omega,F}(q)=\\{v\in\mathbb{C}^{n}:F(q;v)<1\\},$

the indicatrix of $F$ at $q\in\Omega$. Notice that the indicatrix is always
convex for any $q\in\Omega$. Let $A_{j}$ be the linear map on $\mathbb{C}^{n}$
defined by

$A_{j}\left(\frac{\tilde{u}^{j}_{\alpha}}{\mu^{j}_{\alpha}}\right)=e_{\alpha},\quad\alpha=1,\ldots,n.$

Then

$L_{j}=A_{j}\circ d\varphi_{j}(p)^{-1}$

by definition. Therefore, we only need to show that

(2.4) $C_{1}|v|\leq|A_{j}v|\leq C_{2}|v|,\quad\forall v\in\mathbb{C}^{n}$

for some constants $C_{1}>0$ and $C_{2}>0$ which are independent of $j$, in
order to show the equivalence of the subsequential convergence of
$\\{\sigma_{j}\\}$ and $\\{\tau_{j}\\}$.

The lower estimate is equivalent to the upper bound of $A_{j}^{-1}$. This is
obvious since

$|A_{j}^{-1}v|=\left|A_{j}^{-1}\left(\sum_{\alpha}a_{\alpha}e_{\alpha}\right)\right|=\left|\sum_{\alpha}a_{\alpha}\left(\frac{\tilde{u}^{j}_{\alpha}}{\mu^{j}_{\alpha}}\right)\right|\leq
nM|v|$

for any $v=\sum_{\alpha}a_{\alpha}e_{\alpha}\in\mathbb{C}^{n}$, where
$M=\sup\\{|w|:w\in\partial I_{\Omega,F}(p)\\}$.

This lower estimate and Cramer’s formula of inverse transform also implies
that the upper estimate in (2.4) follows from an estimate of the determinant
of $A_{j}$, that is, the upper estimate is obtained if we prove

(2.5) $|\det A_{j}|\leq C$

for some constant $C>0$ independent of $j$. The definition of $L_{j}$ implies

(2.6) $|\det L_{j}|=\prod_{\alpha}\mu^{j}_{\alpha}.$

Let $\det_{\mathbb{R}}$ be the determinant of real linear operator on
$\mathbb{R}^{2n}=\mathbb{C}^{n}$. Then

(2.7) ${\det}_{\mathbb{R}}d\varphi_{j}(p)=|\det
d\varphi_{j}(p)|^{2}=\frac{\mathrm{Vol}(I_{\Omega,F}(p_{j}))}{\mathrm{Vol}(I_{\Omega,F}(p))}=C\,\mathrm{Vol}(I_{\Omega,F}(p_{j}))$

where $\mathrm{Vol}(E)$ represents the Euclidean volume of a given region $E$
and $C^{-1}=\mathrm{Vol}(I_{\Omega,F}(p))$, since $d\varphi_{j}(p)$ maps
$I_{\Omega}(p)$ onto $I_{\Omega}(p_{j})$. Identifying the vector
$\sum_{\alpha}a_{\alpha}u^{j}_{\alpha}$ with the $n$-tuple
$(a_{1},\ldots,a_{n})$, we regard $I_{\Omega,F}(p_{j})$ as a balanced convex
set in $\mathbb{C}^{n}$. Regarding
$\mathbb{C}^{n}=\mathbb{R}^{n}+\sqrt{-1}\mathbb{R}^{n}$, let

$R_{j}=I_{\Omega,F}(p_{j})\cap\mathbb{R}^{n}=\\{(a_{1},\ldots,a_{n})\in
I_{\Omega,F}(p_{j}):a_{\alpha}\mbox{'s are real}\\}.$

Since $I_{\Omega,F}(p_{j})$ is completely circular, it is a circular region
generated by $R_{j}$ and $R_{j}$ is symmetric with respect to the origin.
Notice that $R_{j}$ is also a convex region in $\mathbb{R}^{n}$ since
$I_{\Omega,F}(p_{j})$ is convex. In order to prove the estimate (2.5), we need
the following lemma.

######  Lemma 2.3.

Let $R$ be a bounded convex domain in $\mathbb{R}^{n}$ which is symmetric with
respect to the origin. Let

$r_{1}=\min\\{|x|:x\in\partial R\\}.$

Suppose that $(r_{1},0,\ldots,0)\in\partial R$ (this means the minimum
distance to the boundary is attained at the $x_{1}$-intercept of $\partial
R$). Let

$\widetilde{R}=\\{(x_{2},\ldots,x_{n})\in\mathbb{R}^{n-1}:(0,x_{2},\ldots,x_{n})\in
R\\}.$

Suppose

$\widetilde{R}\subset\widetilde{B}:=[-r_{2},r_{2}]\times\cdots\times[-r_{n},r_{n}]$

for some $r_{2},\ldots,r_{n}>0$. Then

$R\subset B:=[-r_{1},r_{1}]\times
2\widetilde{B}=[-r_{1},r_{1}]\times[-2r_{2},2r_{2}]\times\cdots\times[-2r_{n},2r_{n}].$

###### Proof of Lemma 2.3.

We notice first that the supporting hyperplanes at $(\pm r_{1},0,\ldots,0)$
are

$x_{1}=\pm r_{1},$

since $r_{1}$ is the minimum distance from the origin to $\partial R$.
Therefore

$x=(x_{1},\ldots,x_{n})\in R\Longrightarrow|x_{1}|\leq r_{1}.$

We also observe that we may assume $\widetilde{R}=\widetilde{B}$. Otherwise,
we only need to replace $R$ by the convex hull of $\widetilde{B}\cup R$. Since
$\partial\widetilde{R}=\partial\widetilde{B}$ contains boxes $\\{\pm
r_{2}\\}\times[-r_{3},r_{3}]\times\cdots\times[-r_{n},r_{n}]$, the supporting
hyperplanes of $R$ at $(0,\pm r_{2},0,\ldots,0)\in\partial R$ also contains
those boxes. This means if

$x_{2}=\alpha_{1}x_{1}+\alpha_{3}x_{3}+\cdots+\alpha_{n}x_{n}\pm r_{2}$

are the equation of the supporting hyperplanes, then
$\alpha_{3}=\cdots=\alpha_{n}=0.$ Then from the condition that the distance
from the origin to the supporting hyperplanes are greater than equal to
$r_{1}$, we have

$\frac{r_{2}}{\sqrt{\alpha_{1}^{2}+1}}\geq r_{1}.$

Thereore,

$|\alpha_{1}|\leq\sqrt{\left(\frac{r_{2}}{r_{1}}\right)^{2}-1}.$

Since $x=(x_{1},\ldots,x_{n})$ lies between two support hyperplanes
$x_{2}=\alpha_{1}x_{1}\pm r_{2}$ and $|x_{1}|\leq r_{1}$ if $x\in R$, we
conclude that

$|x_{2}|\leq\sqrt{r_{2}^{2}-r_{1}^{2}}+r_{2}\leq 2r_{2}.$

Similarly, we can also show that $|x_{\alpha}|\leq 2r_{\alpha}$ for any
$\alpha=2,\ldots,n$ whenever $x=(x_{1},\ldots,x_{n})\in R$. This yields the
conclusion. ∎

We resume the proof of Theorem 2.2. Let

$R_{j,1}=\\{(x_{n-1},x_{n})\in\mathbb{R}^{2}:(0,\ldots,0,x_{n-1},x_{n})\in
R_{j}\\}.$

Applying Lemma 2.3, we see that

$R_{j,1}\subset
B_{j,1}:=[-1/\mu^{j}_{n-1},1/\mu^{j}_{n-1}]\times[-2/\mu^{j}_{n},2/\mu^{j}_{n}],$

since $\widetilde{R}_{j,1}=[-1/\mu^{j}_{n},1/\mu^{j}_{n}].$ Applying the lemma
to

$R_{j,2}=\\{(x_{n-2},x_{n-1},x_{n})\in\mathbb{R}^{3}:(0,\ldots,0,x_{n-2},x_{n-1},x_{n})\in
R_{j}\\},$

we see that

$R_{j,1}\subset
B_{j,2}:=[-1/\mu^{j}_{n-2},1/\mu^{j}_{n-2}]\times[-2/\mu^{j}_{n-1},2/\mu^{j}_{n-1}]\times[-2^{2}/\mu^{j}_{n},2^{2}/\mu^{j}_{n}],$

since $\widetilde{R}_{j,2}=R_{j,1}\subset B_{j,1}.$ Repeating this procedure,
we conclude that

$R_{j}\subset
B_{j}:=\prod_{\alpha=1}^{n}[-2^{\alpha-1}/\mu^{j}_{\alpha},2^{\alpha-1}/\mu^{j}_{\alpha}].$

Therefore, the indicatrix $I_{\Omega,F}(p_{j})$ is a subset of the circular
region in $\mathbb{C}^{n}$ generated by $B_{j}$ which is the polydisc

$P_{j}:=\\{z=(z_{1},\ldots,z_{n})\in\mathbb{C}^{n}:|z_{\alpha}|\leq
2^{\alpha-1}/\mu^{j}_{\alpha},\,\,\alpha=1,\ldots,n\\}.$

This implies that

$\mathrm{Vol}(I_{\Omega,F}(p_{j}))\leq\mathrm{Vol}(P_{j})\leq
C\prod_{\alpha}\left(\frac{1}{\mu^{j}_{\alpha}}\right)^{2}$

for some constant $C>0$ independent of $j$. Then (2.6) and (2.7) yields the
desired estimate (2.5). ∎

We will now investigate a condition of metric under that the Frankel sequence
(2.2) has a convergent subsequence. Let $F$ be a metric on a domain $\Omega$.
We denote by $d_{\Omega,F}$ the induced distance function, that is,

$d_{\Omega,F}(x,y)=\inf\left\\{\int_{0}^{1}F(c(t);c^{\prime}(t))\,dt:c\in
C(x,y)\right\\}$

where $C(x,y)$ denotes the space of all piecewise differentiable (real) curve
in $\Omega$ joining $x$ and $y$. For $r>0$, we denote the $d_{\Omega,F}$-ball
of radius $r$ by $B_{\Omega,F}(x;r)$, that is,

$B_{\Omega,F}(x;r)=\\{y\in\Omega:d_{\Omega,F}(x,y)<r\\}.$

######  Definition 2.4.

Let $\Omega$ be a domain in $\mathbb{C}^{n}$. We say that $\Omega$ is
uniformly dominated  by a metric $F$ on $\Omega$ if

  * •

$F$ is complete (i.e. $d_{\Omega,F}$ is Cauchy complete),

  * •

For each $r>0$, there is $\lambda(r)>0$ such that

(2.8) $B_{\Omega,F}(x;r)-x\subset\lambda(r)I_{\Omega,F}(x)$

for any $x\in\Omega$, where $B_{\Omega,F}(x;r)-x=\\{y-x:y\in
B_{\Omega,F}(x;r)\\}$, the Euclidean translation.

######  Proposition 2.5.

Let $\Omega$ be a domain and $p\in\Omega$ fixed. Assume that $\Omega$ is
uniformly dominated by an invariant metric $F$. Then

$\mathcal{S}=\\{\tau_{\varphi}:\Omega\rightarrow\mathbb{C}^{n}:\varphi\in\mathrm{Aut}(\Omega)\\}$

is a normal family, where
$\tau_{\varphi}=[d\varphi(p)]^{-1}\circ(\varphi(\cdot)-\varphi(p))$,

###### Proof.

For any $r>0$ and $\varphi\in\mathrm{Aut}(\Omega)$,

$\displaystyle\tau_{\varphi}(B_{\Omega,F}(p;r))$ $\displaystyle=$
$\displaystyle[d\varphi(p)]^{-1}(B_{\Omega,F}(\varphi(p);r)-\varphi(p))$
$\displaystyle\subset$
$\displaystyle\lambda(r)[d\varphi(p)]^{-1}(I_{\Omega,F}(\varphi(p)))$
$\displaystyle=$ $\displaystyle\lambda(r)I_{\Omega,F}(p)$

by (2.8), since $F$ is invariant. Therefore, the family $\mathcal{S}$ is
uniformly bounded on $B_{\Omega,F}(p;r)$. Since $F$ is complete, this implies
that $\mathcal{S}$ is uniformly bounded on each compact subset of $\Omega$.
The Montel theorem yields the conclusion. ∎

######  Example 2.6.

Let $\mathcal{H}=\\{z\in\mathbb{C}:\mathrm{Im}\,z>0\\}$ the upper half plane.
Let $F$ be the Poincaré metric on $\mathcal{H}$ which coincides with the
Kobayashi-Royden metric $K_{\mathcal{H}}$. The explicit formulas of the metric
and the distance are as follows.

(2.9) $F(z;v)=\frac{|v|}{2\mathrm{Im}\,z},\quad
d_{\mathcal{H},F}(z_{1},z_{2})=2\tanh^{-1}\left|\frac{z_{2}-z_{1}}{z_{2}-\bar{z}_{1}}\right|.$

We will show (2.8) for a suitable $\lambda$ at every $x=a+\sqrt{-1}b.$ By the
translation invariance of $F$, we assume without loss of generality
$x=\sqrt{-1}b$ for $b>0$. Therefore, the indicatrix $I_{\mathcal{H},F}(x)$ is
the disc $\Delta(2b)=\\{\zeta\in\mathbb{C}:|\zeta|<2b\\}$ and the distance
ball $B_{\mathcal{H},F}(x;r)$ is the disc centered at $\sqrt{-1}b(r)$ of
radius $\sqrt{b(r)^{2}-b^{2}}$ where

$b(r)=\left(\frac{1+\tanh^{2}(r/2)}{1-\tanh^{2}(r/2)}\right)b.$

Therefore, if we define $\lambda_{\mathcal{H}}(r)$ by

(2.10)
$\lambda_{\mathcal{H}}(r)=(2b)^{-1}\left(b(r)-b+\sqrt{b(r)^{2}-b^{2}}\right)=\frac{\tanh(r/2)}{1-\tanh(r/2)},$

then, we see that
$B_{\mathcal{H},F}(x;r)-x\subset\lambda_{\mathcal{H}}(r)I_{\mathcal{H},F}(x)$
for any $x=\sqrt{-1}b\in\mathcal{H}$. Since $b$ is arbitrary, this implies
that $\mathcal{H}$ is uniformly dominated by the Poincaré metric.

######  Remark 2.7.

It may be necessary to point out that the uniform domination condition of a
metric is not holomorphic invariant. More precisely, even if $\Omega$ is
uniformly dominated by a metric $F$ and
$\varphi:\widehat{\Omega}\rightarrow\Omega$ is a biholomorphic map,
$\widehat{\Omega}$ is not necessarily uniformly dominated by $\varphi^{*}F$.
The reason is that the definition of the uniform domination depends on a
Euclidean translation of sets, which is not preserved by biholomorphic maps in
general. Therefore, the uniform domination condition is just invariant under
affine transforms. This is not an unexpected result, since the convergence of
Frankel’s scaling sequence is not a holomorphic invariant property, either. It
is also an affine invariant one.

It would be an interesting question to ask about conditions of invariant
metrics and domains which imply the uniform domination. The following theorem
is an answer to this question in case of the Kobayshi-Royden metric. Together
with Proposition 2.5, it provides a geometric proof of the convergence of
Frankel’s scaling sequence on convex domains.

In order to simplify symbols, we denote the indicatrix and the distance ball
by $I_{\Omega}(x)$ and $B_{\Omega}(x;r)$, respectively, if the metric is the
Kobayashi-Royden metric.

######  Theorem 2.8.

Let $\Omega$ be a hyperbolic convex domain in $\mathbb{C}^{n}$. Then $\Omega$
is uniformly dominated by the Kobayashi-Royden metric $K_{\Omega}$. Indeed,

$B_{\Omega}(x;r)-x\subset 2\lambda_{\mathcal{H}}(r)I_{\Omega}(x)$

for each $r>0$ and $x\in\Omega$, where $\lambda_{\mathcal{H}}$ is the function
defined as (2.10)

###### Proof.

Let $x\in\Omega$ be a point and $v\in\mathbb{C}^{n}$ be arbitrary nonzero
vector. We denote by $l_{v}$ the complex line defined by

$l_{v}=\\{\zeta v:\zeta\in\mathbb{C}\\}$

and $l_{v}+x=\\{x+\zeta v:\zeta\in\mathbb{C}\\}$ by $l_{x,v}$. Let
$\Omega_{x,v}=\Omega\cap l_{x,v}$. Since $\Omega$ is convex and hyperbolic,
$\Omega_{x,v}$ is also a convex hyperbolic planar domain. Let $p=p(x,v)$ be a
point on $\partial\Omega_{x,v}$ such that

$d:=|x-p|=\mathrm{dist}(x,\partial\Omega_{x,v})$

where $\mathrm{dist}$ represents the Euclidean distance. Then since
$\Omega_{x,v}$ contains the disc centered at $x$ of radius $d$,

(2.11) $K_{\Omega}(x;v)\leq
K_{\Omega_{x,v}}(x;v)\leq\frac{|v|}{d}=2K_{\mathcal{H}}(\sqrt{-1}d;v).$

By a unitary change which map $(x-p)/|x-p|$ to $e_{1}$ and the translation
$z\rightarrow z-p$, we may first assume that

  * •

$p$ is the origin,

  * •

$l_{x,v}$ is the $z_{1}$-plane,

  * •

$\Omega_{x,v}$ lies on the upper half plane in $z_{1}$-plane and the real axis
is the supporting line of $\Omega_{x,v}$,

  * •

$x=\sqrt{-1}d$.

Applying a linear transform which leaves $z_{1}$-plane invariant, we can also
assume that

  * •

the supporting hyperplane of $\Omega$ at $p=O$ is
$\Pi=\\{\mathrm{Im}\,z_{1}=0\\}$ and as a result
$\Omega\subset{\mathcal{H}}\times\mathbb{C}^{n-1}$.

It can be achieved by the linear map

$e_{1}\rightarrow e_{1},\,\,v_{2}\rightarrow e_{2},\ldots,v_{n}\rightarrow
e_{n},$

where $v_{2},\ldots,v_{n}$ are linearly independent complex vectors which are
tangent to $\Pi\cap\sqrt{-1}\Pi$. Recall that the Kobayashi distance decreases
under holomorphic map, and if $p_{1},p_{2}\in X$ and $q_{1},q_{2}\in Y$ for
two complex manifolds $X$ and $Y$, then

$d_{X\times
Y}((p_{1},q_{1}),(p_{2},q_{2}))=\max\\{d_{X}(p_{1},p_{2}),d_{Y}(q_{1},q_{2})\\}.$

Therefore, if $y\in\Omega_{x,v}\subset\mathcal{H}\times\\{O\\}$, then

(2.12) $d_{\Omega}(x,y)\geq
d_{\mathcal{H}\times\mathbb{C}^{n-1}}(x,y)=d_{\mathcal{H}}(x,y),$

by identiying $\mathcal{H}\times\\{O\\}$ with $\mathcal{H}$. Combining (2.11)
and (2.12), we conclude that

$(B_{\Omega}(x;r)-x)\cap l_{v}\subset 2\lambda_{\mathcal{H}}(I_{\Omega}(x)\cap
l_{v}).$

The conclusion follows since $x\in\Omega$ and $v\in\mathbb{C}^{n}$ are
arbitrary. ∎

##  3\. Maximal circularity function

Let $\Omega$ be a bounded domain in $\mathbb{C}^{n}$. For $x\in\Omega$, let
$\mathcal{B}_{\Omega}(x)$ be the class of bounded holomorphic embeddings
$\varphi:\Omega\rightarrow\mathbb{C}^{n}$ such that $\varphi(x)=0$. Let

$\displaystyle c_{\Omega}(x)$ $\displaystyle:=$
$\displaystyle\sup\left\\{\frac{r}{R}:\exists\varphi\in\mathcal{B}_{\Omega}(x),R>0,r>0\mbox{
such that }\right.$ $\displaystyle\hskip
56.9055pt\left.rI_{\varphi(\Omega)}(0)\subset\varphi(\Omega)\subset
RI_{\varphi(\Omega)}(0)\right\\},$

where $I_{\Omega}(x)$ represents the Kobayashi indicatrix of $\Omega$ at
$x\in\Omega$. Obiously, $0<c_{\Omega}(x)\leq 1$, and if $\Omega$ is a bounded
complete circular domain, then $\Omega$ is same with its Kobayashi indicatrix
at the origin ([1]) and hence $c_{\Omega}(0)=1$. With ths reason, we call
$c_{\Omega}$ the maximal circularity function or simply the maximal
circularity of $\Omega$. The first observation is that the converse is also
true.

######  Proposition 3.1.

Let $\Omega$ be a bounded domain in $\mathbb{C}^{n}$. Suppose that
$c_{\Omega}(x)=1$ at a point $x\in\Omega$. Then $\Omega$ is biholomorphic to a
complete circular domain.

###### Proof.

Suppose $\varphi_{j}\in\mathcal{B}_{\Omega}(x)$ such that

(3.2) $r_{j}I_{j}\subset\varphi_{j}(\Omega)\subset R_{j}I_{j}$

for some $R_{j}\geq r_{j}>0$ and $r_{j}/R_{j}\rightarrow 1$ as
$j\rightarrow\infty$, where

$I_{j}:=I_{\varphi_{j}(\Omega)}(0)=d\varphi_{j}(x)(I_{\Omega}(x)).$

Let

$\psi_{j}=[d\varphi_{j}(x)]^{-1}\circ\varphi_{j}.$

Then $\psi_{j}\in\mathcal{B}_{\Omega}(x)$, and acting $[d\varphi_{j}(x)]^{-1}$
to (3.2), we have

$r_{j}I_{\Omega}(x)\subset\psi_{j}(\Omega)\subset R_{j}I_{\Omega}(x).$

Notice that $R_{j}<R$ for some $R>0$. Otherwise, we may assume
$R_{j}\rightarrow\infty$ as $j\rightarrow\infty$. Then
$r_{j}\rightarrow\infty$ as well. Let $\epsilon>0$ be chosen small enough that
$B_{\epsilon}\subset I_{\Omega}(x)$, where $B_{\epsilon}$ is the Euclidean
$\epsilon$-ball centered at the origin. Let $v\in\mathbb{C}^{n}$ be a nonzero
vector. Then since $d\psi_{j}(x)=Id$ and $\Omega$ is hyperbolic,

$0<K_{\Omega}(x;v)=K_{\psi_{j}(\Omega)}(0;v)\leq
K_{r_{j}B_{\epsilon}}(0;v)=\frac{|v|}{r_{j}\epsilon}\rightarrow 0$

as $j\rightarrow\infty$. This is a contradiction. Choosing a subsequence, we
may assume that $\psi_{j}$ is convergent by Montel’s theorem and
$r_{j}\rightarrow R_{0}$, $R_{j}\rightarrow R_{0}$ for some $R_{0}\geq 0$.
Since $d\psi_{j}(x)=Id$, the limit $\psi$ of the sequence $\psi_{j}$ is also a
1-1 holomorphic mapping and $\psi(\Omega)=R_{0}I_{\Omega}(x)$. Therefore,
$R_{0}>0$ and $\psi:\Omega\rightarrow R_{0}I_{\Omega}(x)$ is a biholomorphic
mapping. ∎

Notice that the defintion of $c_{\Omega}$ is similar to that of the squeezing
function defined in [2] (cf. [3, 9, 15]) which is motivated by [17, 18, 20].
The defintion of the squeezing function can be generalized to the case of more
arbitrary circular models as follows (see also [6]). Let $D$ be a bounded
complete circular domain in $\mathbb{C}^{n}$. Then the squeezing function
$s^{D}_{\Omega}(x)$ of $\Omega$ at $x\in\Omega$ modelled on $D$ is defined by

$\displaystyle s_{\Omega}^{D}(x)$ $\displaystyle:=$
$\displaystyle\sup\left\\{\frac{r}{R}:\exists\varphi\in\mathcal{B}_{\Omega}(x),R>0,r>0\mbox{
such that }\right.$ $\displaystyle\hskip
56.9055pt\left.rD\subset\varphi(\Omega)\subset RD\right\\}.$

Thanks to the similarity of definitions (3) and (3), the following natural
comparison between $c_{\Omega}$ and $s^{D}_{\Omega}$ can be verified.

######  Proposition 3.2.

Let $\Omega$ be a bounded domain in $\mathbb{C}^{n}$ and $x\in\Omega$. Then
for any bounded complete circular domain $D\subset\mathbb{C}^{n}$,

$c_{\Omega}(x)\geq\left(s^{D}_{\Omega}(x)\right)^{2}.$

###### Proof.

Let $\varphi\in\mathcal{B}_{\Omega}(x)$ such that $\varphi(\Omega)\subset D.$
and $r>0$ be chosen that $rD\subset\varphi(\Omega).$ Since $D$ and $rD$ are
completely circular, Barth’s theorem ([1]) implies that

$I_{D}(0)=D,\quad I_{rD}(0)=rD.$

Then we have

$rD=I_{rD}(0)\subset I_{\varphi(\Omega)}(0)\subset I_{D}(0)=D$

from the decreasing property of the Kobayashi metric. Therefore,

$rI_{\varphi(\Omega)}(0)\subset rD\subset\varphi(\Omega)\subset
D\subset\frac{1}{r}I_{\varphi(\Omega)}(0).$

This implies $c_{\Omega}(x)\geq r^{2}.$ Taking the supremum for $r$, we get
the conclusion. ∎

If $\Omega$ is strongly pseudoconvex, then it turned out that the squeezing
function $s_{\Omega}(x)=s^{B^{n}}_{\Omega}(x)$ modelled on the unit ball tends
to $1$ as $x$ goes to a boundary point ([4, 15]). Therefore, we have
immediately the following theorem as a corollary.

######  Theorem 3.3.

If $\Omega$ is a bounded domain in $\mathbb{C}^{n}$ with $C^{2}$-smooth
strongly pseudoconvex boundary, then $c_{\Omega}(x)\rightarrow 1$ as
$x\rightarrow p\in\partial\Omega$.

Another example of domains is polyhedral domains. Recall that a bounded domain
$\Omega$ in $\mathbb{C}^{n}$ is called a polyhedral domain if there are smooth
real-valued functions $\rho_{1},\ldots,\rho_{N}$ defined on a neighborhood of
$\overline{\Omega}$ such that

  * •

$\Omega=\\{z\in\mathbb{C}^{n}:\rho_{1}(z)<0,\ldots,\rho_{N}(z)<0\\}$,

  * •

on $\partial_{\alpha}\Omega:=\\{z\in\partial\Omega:\rho_{\alpha}(z)=0\\}$ for
$\alpha=1,\ldots,N$, $\nabla\rho_{\alpha}$ never vanishes,

  * •

each $\partial_{\alpha}\Omega$ is Levi flat.

$\Omega$ is said to be normal if it satisfies in addition

  * •

on each
$\partial_{\alpha_{1}\cdots\alpha_{k}}\Omega:=\\{z\in\partial\Omega:\rho_{\alpha_{1}}(z)=\cdots=\rho_{\alpha_{k}}(z)=0\\}$,
$\nabla\rho_{\alpha_{1}}$,…, $\nabla\rho_{\alpha_{k}}$ are linearly
independent over $\mathbb{C}$.

Imitating the idea of [10, 11] about the scaling of convex normal polyhedral
domains, we can prove the following.

######  Theorem 3.4.

Let $\Omega=\\{z\in\mathbb{C}^{n}:\rho_{1}(z)<0,\ldots,\rho_{N}(z)<0\\}$ be a
convex normal polyhedral domain in $\mathbb{C}^{n}$. Let
$q\in\partial_{\alpha_{1}\cdots\alpha_{n}}\Omega$ for some distinct indices
$\alpha_{1},\ldots,\alpha_{n}.$ Then $c_{\Omega}(x)\rightarrow 1$ as
$x\rightarrow q$.

###### Proof.

Reordering indices, we may assume that
$\\{\alpha_{1},\ldots,\alpha_{n}\\}=\\{1,\ldots,n\\}$ without loss of
generality. We can also assume that $|\nabla\rho_{\alpha}|\equiv 1$ on
$\partial_{j}\Omega$ for each $\alpha=1,\ldots,N$. By Proposition 3.2, we only
need to prove that

$\sigma(x):=s^{\Delta^{n}}_{\Omega}(x)\rightarrow 1\quad\mbox{as $x\rightarrow
q$},$

where $\Delta^{n}$ represents the unit polydisc in $\mathbb{C}^{n}$. Notice
that $\partial_{\alpha}\Omega$ is foliated by analytic varieties since it is
Levi flat, and each of them is indeed a part of affine hyperplane because
$\Omega$ is convex. Therefore, there are $n$ hyperplanes $H_{1},\ldots,H_{n}$
such that

  * •

$H_{1}\cap\cdots\cap H_{n}=\\{q\\}$,

  * •

$V_{\alpha}:=H_{\alpha}\cap\partial\Omega$ is the maximal subvariety in
$\partial_{j}\Omega$ containing $q$.

Taking a linear affine change of coordinates, we assume that

  * •

$q=O=(0,\ldots,0)$ the origin,

  * •

$H_{\alpha}=\\{z_{\alpha}=0\\}$,

  * •

$\\{\mathrm{Re}\,z_{\alpha}=0\\}$ is a supporting (real) hyperplanes of
$\Omega$ for $\alpha=1,\ldots,n$,

  * •

$\Omega\subset\\{\mathrm{Re}\,z_{1}<0,\ldots\mathrm{Re}\,z_{n}<0\\}=\mathcal{H}^{n}$,

where $\mathcal{H}=\\{\zeta\in\mathbb{C}:\mathrm{Re}\,\zeta<0\\}$, the left
half plane in $\mathbb{C}$. Let us suppose that $x\in\Omega$ is sufficiently
close to $O$. Then among all affine hyperplanes which are tangent to
$\partial_{\alpha}\Omega$, there is a unique $h_{\alpha}=h_{\alpha}(x)$ which
is closest to $x$, $\alpha=1,\ldots,n$. Let $p=p(x)\in\partial_{12\cdots
n}\Omega$ be the intersection point of those hyperplanes, that is,

$h_{1}\cap\cdots\cap h_{n}=\\{p\\}.$

Let

$v_{\alpha}=v_{\alpha}(x)=\nabla\rho_{\alpha}(p),$

and let $A_{x}$ be the affine transform defined by

  * •

$A_{x}(p)=0$,

  * •

$dA_{x}(v_{\alpha})=e_{\alpha}$ for $\alpha=1,\ldots,n$.

It is obvious that $A_{x}\rightarrow Id$ locally uniformly on $\mathbb{C}^{n}$
as $x\rightarrow O$, since $p\rightarrow O$ and $v_{\alpha}\rightarrow
e_{\alpha}$ as $x\rightarrow O$ by the construction. Notice also that
$A_{x}(\Omega)\subset\mathcal{H}^{n}$ and $A_{x}(x)=(x_{1},\ldots,x_{n})$
where $x_{\alpha}<0$ and $x_{\alpha}\rightarrow 0$ as $x\rightarrow O$ for all
$\alpha=1,\ldots,n$. Let $\Phi:\mathcal{H}^{n}\rightarrow\Delta^{n}$ be the
Cayley transform such that $\Phi(O)=(1,\ldots,1)$ and
$\Phi(-1,\ldots,-1)=O=(0,\ldots,0)$. Then
$\Phi(A_{x}(x))=(1-r_{1},\ldots,1-r_{n})$ for some $r_{\alpha}>0$ which tends
to $0$ as $x\rightarrow O$. The normality condition implies that
$\partial_{1\cdots n}\Omega$ is a smooth real $n$-dimensional manifold and for
each $\alpha=1,\ldots,n$, $\pi_{\alpha}(\Phi(A_{x}(\Omega)))$ has a smooth
boundary near $1$, where $\pi_{\alpha}$ denotes the natural projection
$\mathbb{C}^{n}\ni(z_{1},\ldots,z_{n})\rightarrow z_{\alpha}\in\mathbb{C}$.
Therefore, there is $\epsilon>0$ such that
$\Delta_{\epsilon}^{n}\subset\Phi(A_{x}(\Omega))$ for all $x\in\Omega$ which
is sufficiently close to $O$, where $\Delta_{\epsilon}$ represents the disc in
$\mathbb{C}$ centered at $1-\epsilon$ of radius $\epsilon$. Let
$\Psi_{x}\in\mathrm{Aut}(\Delta^{n})$ such that

$\Psi_{x}(1,\ldots,1)=(1,\ldots,1)\quad\mbox{and}\quad\Psi_{x}(1-r_{1},\ldots,1-r_{n})=(0,\ldots,0)$

. For each $0<r<1$, then

$r\Delta^{n}\subset\Psi_{x}(\Delta_{\epsilon}^{n})\subset\Psi_{x}\circ\Phi\circ
A_{x}(\Omega)$

for any $x\in\Omega$ which is suficiently close to $O$. Therefore,

$\liminf_{\Omega\ni x\rightarrow O}s^{\Delta^{n}}_{\Omega}(x)\geq r$

for each $0<r<1$, and this yields the conclusion. ∎

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