Isometries in the symmetrized bidisc

Abstract

We provide a characterization of isometries in the sense of the Carathéodory–Reiffen metric in the symmetrized bidisc.

1 Introduction

Let $D\subset {{\mathbb {C}}}^n$ be a domain. We say that a function $F_D:D\times {{\mathbb {C}}}^n\rightarrow [0,+\infty )$ is a Finsler metric if it is an upper semi-continuous function and

$$\begin{aligned} F_D(\lambda ;t v)=|t| F_D(\lambda ;v)\quad \text { for any }\lambda \in D,v\in {{\mathbb {C}}}^n,t\in {{\mathbb {C}}}. \end{aligned}$$

Assume that $D_1\subset {{\mathbb {C}}}^{n_1}$ and $D_2\subset {{\mathbb {C}}}^{n_2}$ are domains with Finsler metrics $F_{D_1}$ and $F_{D_2}$, respectively. We say that a $C^1$ mapping $\phi :D_1\rightarrow D_2$ is an isometry if

$$\begin{aligned} F_{D_1}(\lambda ;X)=F_{D_2}(\phi (\lambda );d_{\lambda }\phi (\lambda )(X)) \quad \text { for any } \lambda \in D_1, X\in {{\mathbb {C}}}^{n_1}, \end{aligned}$$

(1)

where $d_{\lambda } \phi$ denotes the ${{\mathbb {R}}}$-differential of $\phi$ at $\lambda$. In case $n_1=1$ the equation (1) is equivalent with

$$\begin{aligned} F_{D_1}(\lambda ;1)=F_{D_2}\big (\phi (\lambda );\frac{\partial \phi }{\partial z}(\lambda )+\xi \frac{\partial \phi }{\partial {\bar{z}}}(\lambda ) \big ) \quad \text { for any } \lambda \in D_1, \xi \in {{\mathbb {T}}}, \end{aligned}$$

(2)

where ${{\mathbb {T}}}=\{t\in {{\mathbb {C}}}: |t|=1\}$ denotes the unit circle.

The paper is motivated by the question whether an isometry is holomorphic or anti-holomorphic. This problem was considered in many papers and, it seems, that in case of invariant metrics (Kobayashi metric, Carathéodory metric, etc) is very difficult (see [2, 8, 10]).

We denote by ${{\mathbb {D}}}=\{t\in {{\mathbb {C}}}: |t|<1\}$ the unit disc. For a domain $D\subset {{\mathbb {C}}}^n$ we define a biholomorphically invariant Finsler metric as follows

$$\begin{aligned} \gamma _{D}(\lambda ,X)=\sup \Bigg \{\frac{|F'(\lambda )X|}{1-|F(\lambda )|^2}: F:D\rightarrow {{\mathbb {D}}}\text { holomorphic}\Bigg \} \end{aligned}$$

for any $\lambda \in D$ and any $X\in {{\mathbb {C}}}^n$. We call $\gamma _D$ the Carathéodory-Reiffen (pseudo)metric for D (see e.g. [6], Chapter 2). According to the above definition a $C^1$-mapping $\phi :D_1\rightarrow D_2$ is a $\gamma$-isometry if

$$\begin{aligned} \gamma _{D_2}(\phi (\lambda );d_{\lambda } \phi (X))=\gamma _{D_1}(\lambda ;X) \end{aligned}$$

for any $\lambda \in D_1$ and any $X\in {{\mathbb {C}}}^{n_1}$.

Put

$$\begin{aligned} \pi :{{\mathbb {C}}}^2\ni (z,w)\rightarrow (z+w,zw)\in {{\mathbb {C}}}^2 \end{aligned}$$

and ${{\mathbb {G}}}_2=\pi ({{\mathbb {D}}}^2)$. We call ${{\mathbb {G}}}_2$ the symmetrized bidisc (see e.g. [1]). One of the main results of our paper is the following.

Theorem 1

Let $\phi :{{\mathbb {D}}}\rightarrow {{\mathbb {G}}}_2$ be a mapping of class $C^1$. Assume that $\phi$ is a $\gamma$-isometry. Then $\phi$ is holomorphic or anti-holomorphic.

As a corollary, we obtain the following result.

Corollary 2

Let $F:{{\mathbb {G}}}_2\rightarrow {{\mathbb {G}}}_2$ be a mapping of class $C^1$ and a $\gamma$-isometry. Then F is holomorphic or anti-holomorphic.

2 Proof of Theorem 1

For any $\lambda \in {{\mathbb {G}}}_2$, there exists an automorphism $\Phi$ of ${{\mathbb {G}}}_2$ such that $\Phi (\lambda )=(0,p)$, where $p\in [0,1)$. Define a set $\Sigma =\{(2t,t^2): t\in {{\mathbb {D}}}\}$ called the royal set. We have

$$\begin{aligned} \Sigma =\{\Phi (0): \Phi \in {{\,\textrm{Aut}\,}}({{\mathbb {G}}}_2)\}. \end{aligned}$$

Recall the following result (see [6], Chapter 7)

$$\begin{aligned} \gamma _{{{\mathbb {G}}}_2}((s,p);(X,Y))=\max \Bigg \{\frac{|(F_\eta )'_s(\lambda )X+(F_\eta )'_p(\lambda )Y|}{1-|F(\lambda )|^2}:\eta \in {{\mathbb {T}}}\Bigg \}, \end{aligned}$$

where $\lambda =(s,p)\in {{\mathbb {G}}}_2$ and $F_{\eta }(s,p)=\frac{2\eta p-s}{2-\eta s}$, $\eta \in {{\mathbb {T}}}$. Note that

$$\begin{aligned} (F_\eta )'_s(s,p)=\frac{2\eta ^2 p-2}{(2-\eta s)^2} \quad \text { and }\quad (F_\eta )'_p(s,p)=\frac{2\eta }{2-\eta s}. \end{aligned}$$

Hence,

$$\begin{aligned} \gamma _{{{\mathbb {G}}}_2}((s,p);(X,Y))=\max \Big \{\frac{|(\eta ^2 p-1)X+\eta (2-\eta s)Y|}{2\big (1-|p|^2-\Re (\eta (s-{\bar{s}}p))\big )}: \eta \in {{\mathbb {T}}}\Big \}. \end{aligned}$$

In particular, we have

$$\begin{aligned} \gamma _{{{\mathbb {G}}}_2}((0,p);(X,Y))=\frac{\max _{\omega \in {{\mathbb {T}}}}|X(1-\omega ^2p)-2Y\omega |}{2(1-p^2)}, \end{aligned}$$

where $p\in [0,1)$ and $(X,Y)\in {{\mathbb {C}}}^2$.

The main result of this section is the following.

Theorem 3

Let $(X_1,Y_1), (X_2,Y_2)\in {{\mathbb {C}}}^2$ be fixed vectors and let $p\in (0,1)$ be a fixed number. Assume that there exists a constant $C>0$ such that

$$\begin{aligned} \gamma _{{{\mathbb {G}}}_2}\big ((0,p);(X_1,Y_1)+\xi (X_2,Y_2)\big )=C\quad \text { for any }\xi \in {{\mathbb {T}}}. \end{aligned}$$

(3)

Then $(X_1,Y_1)=0$ or $(X_2,Y_2)=0$.

We assume that $(X_1,Y_1)\not =0$ and $(X_2,Y_2)\not =0$. Consider polynomials

$$\begin{aligned} P_j(z)=pX_jz^2+2Y_jz-X_j,\quad z\in {{\mathbb {C}}}, j=1,2, \end{aligned}$$

and their duals,

$$\begin{aligned} Q_j(z)=z^2\overline{P_j(1/{\bar{z}})}= -{\bar{X}}_jz^2+2{\bar{Y}}_jz+p{\bar{X}}_j,\quad z\in {{\mathbb {C}}}, j=1,2. \end{aligned}$$

We have $P_j(z)Q_j(z)=z^2 |P_j(z)|^2$ for any $z\in {{\mathbb {T}}}$. Note that the equality (3) means that

$$\begin{aligned} \frac{\max _{z\in {{\mathbb {T}}}}|P_1(z)+\xi P_2(z) |}{2(1-p^2)}=C\text { for any }\xi \in {{\mathbb {T}}}. \end{aligned}$$

It is easy to see that

$$\begin{aligned} C=\frac{\max _{z\in {{\mathbb {T}}}}\big (|P_1(z)|+ |P_2(z)|\big )}{2(1-p^2)}. \end{aligned}$$

We put

$$\begin{aligned} E=\{z\in {{\mathbb {T}}}: |P_1(z)|+ |P_2(z)|=2(1-p^2)C\} \end{aligned}$$

and

$$\begin{aligned} E(\xi )=\{z\in {{\mathbb {T}}}: |P_1(z)+ \xi P_2(z)|=2(1-p^2)C\}. \end{aligned}$$

We have divided the proof of Theorem 3 into a sequence of lemmas.

Lemma 4

Assume that $z_0\in E$. Then $P_1(z_0)P_2(z_0)\not =0$.

Proof

Assume that $P_1(z_0)=0$ and that $X_1X_2\not =0$. Then

$$\begin{aligned} |P_1(z)|+|P_2(z)|\le |P_2(z_0)|\quad \text { for any }z\in {{\mathbb {T}}}. \end{aligned}$$

Put $Q(z)=z^2|P_2(z_0)|^2-P_2(z)Q_2(z)$. Then Q is a holomorphic polynomial of degree $\le 4$ such that

$$\begin{aligned} \frac{Q(z)}{z^2}\ge |P_1(z)|\big ( |P_2(z_0)|+|P_2(z)|\big )\ge 0\quad \text { for any } z\in {{\mathbb {T}}}. \end{aligned}$$

Then by the Fejér–Riesz theorem (see [4, 7], see also [1, 3]) there exists a polynomial R of degree $\le 2$ such that

$$\begin{aligned} Q(z)=z^2|R(z)|^2\quad \text { for any }z\in {{\mathbb {T}}}. \end{aligned}$$

Hence, $P_1$ has a double zero at $z_0$. We get

$$\begin{aligned} P_1(z)=pX_1(z-z_0)^2 \end{aligned}$$

and, therefore, $pX_1z_0^2=-X_1$. So, $X_1=0$. A contradiction. $\square$

Lemma 5

We have $E(\xi )\subset E$ for any $\xi \in {{\mathbb {T}}}$. Moreover, $E(\xi _1)\cap E(\xi _2)=\varnothing$ when $\xi _1\not =\xi _2$.

In particular, the set E is infinite.

Proof

It suffices to note that the equality $|a+\xi b|=|a|+|b|$ where $a,b\in {{\mathbb {C}}}{\setminus }\{0\}$ and $\xi \in {{\mathbb {T}}}$ is equivalent with $\xi =|ab|/ (b{\overline{a}})$. Now we use the previous Lemma and get that $z_0\in E$ is such that $z_0\in E(\xi )$ if and only if

$$\begin{aligned} \xi =\frac{|P_1(z_0)P_2(z_0)|}{ \overline{P_1(z_0)}P_2(z_0)}. \end{aligned}$$

$\square$

Lemma 6

Assume that the equation

$$\begin{aligned} |X_1pz^2+2Y_1z-X_1|+|X_2pz^2+2Y_2z-X_2|=1 \end{aligned}$$

(4)

has at least 9 solutions in ${{\mathbb {T}}}$. Then at least one of the following conditions hold:

• (1) $X_1=X_2=0$;

• (2) $P_j(z)=pX_j(z-\zeta _j)^2$, $j=1,2$, where $|X_1|=|X_2|=\frac{1}{2p+2}$, $\zeta _1=-\zeta _2=\frac{\epsilon i}{\sqrt{p}}$, and $\epsilon \in \{-1,1\}$.

Proof

From the equality $|P_1(z)|+|P_2(z)|=1$ we get $|P_2(z)|^2=(1-|P_1(z)|)^2$ and, therefore,

$$\begin{aligned} 4|P_1(z)|^2=(1+|P_1(z)|^2-|P_2(z)|^2)^2. \end{aligned}$$

We have

$$\begin{aligned} 4z^2P_1(z)Q_1(z)=(z^2+P_1(z)Q_1(z)-P_2(z)Q_2(z))^2 \end{aligned}$$

has at least 9 different solutions. Hence, it holds for any $z\in {{\mathbb {C}}}$. Using that

$$\begin{aligned} P_1(z)Q_1(z)-P_2(z)Q_2(z)=p(|X_2|^2-|X_1|^2)+\dots \end{aligned}$$

we have $|X_1|=|X_2|$. Moreover, there exists a polynomial $R_1$ such that $P_1Q_1=R_1^2$.

Let $P_1(z)=pX_1(z-\zeta _1)(z-\xi _1)$. Note that $pX_1\zeta _1\xi _1=-X_1$. Hence, $X_1=0$ or $\zeta _1\xi _1=-\frac{1}{p}$. Assume that $X_1\not =0$. Note that

$$\begin{aligned} Q_1(z)=p\overline{X_1}(1-{\bar{\zeta }}_1 z)(1-{\bar{\xi }}_1 z). \end{aligned}$$

So, $P_1Q_1=R_1^2$ if and only if $\zeta _1=\xi _1$ or ${\bar{\zeta }}_1=1/\xi _1$. Since $\zeta _1\xi _1=-\frac{1}{p}$, we get $\zeta _1=\xi _1$ and, therefore, $\zeta _1^2=-\frac{1}{p}$ and $P_1(z)=pX_1(z-\zeta _1)^2$. By similar arguments, we get $P_2(z)=pX_2(z-\zeta _2)^2$. Putting these equalities to (4) we get $\zeta _2=-\zeta _1$ and $|X_1|=|X_2|=\frac{1}{2p+2}$. $\square$

Proof of Theorem 3

For the proof, we apply Lemma 6 to the polynomials ${\tilde{P}}_j=P_j/(2C(1-p^2))$ with ${\tilde{X}}_j=X_j/(2C(1-p^2))$, ${\tilde{Y}}_j=Y_j/(2C(1-p^2))$, where $j=1,2$. Then we have ${\tilde{P}}_1(z)=p{\tilde{X}}_1(z-\zeta _0)^2$ and ${\tilde{P}}_2(z)=p{\tilde{X}}_2(z+\zeta _0)^2$, where $|\zeta _0|=\frac{1}{\sqrt{p}}$ and, $|{\tilde{X}}_1|=|{\tilde{X}}_2|=\frac{1}{2+2p}$.

There exists $\xi _0\in {{\mathbb {T}}}$ such that $\xi _0 {\tilde{X}}_2=-{\tilde{X}}_1$. Then

$$\begin{aligned} 1=\max _{z\in {{\mathbb {T}}}}| {\tilde{P}}_1(z)+\xi _0 {\tilde{P}}_2(z)|= \max _{z\in {{\mathbb {T}}}}p|{\tilde{X}}_1|\cdot |(z-\zeta _0)^2-(z+\zeta _0)^2|. \end{aligned}$$

(5)

And, therefore, we have

$$\begin{aligned} 1=4p|{\tilde{X}}_1| \cdot |\zeta _0|=\frac{2\sqrt{p}}{1+p}. \end{aligned}$$

Hence, we get $p=1$. A contradiction with the condition $p\in [0;1)$. $\square$

Lemma 7

Let $f:{{\mathbb {D}}}\rightarrow {{\mathbb {C}}}$ be a $C^1$ function such that for any $z\in {{\mathbb {D}}}$ we have $\frac{\partial f}{\partial z}(z)=0$ or $\frac{\partial f}{\partial {\bar{z}}}(z)=0$. Then f is holomorphic or anti-holomorphic in ${{\mathbb {D}}}$.

Proof

Put $h=\frac{\partial f}{\partial z}$. Then h is a continuous function on ${{\mathbb {D}}}$ and on a set $\{z\in {{\mathbb {D}}}: h(z)\not =0\}$ we have $\frac{\partial f}{\partial {\bar{z}}}\equiv 0$. Hence, h is holomorphic on ${{\mathbb {D}}}{\setminus } h^{-1}(0)$. By Radó’s theorem, h is holomorphic in ${{\mathbb {D}}}$. If $\{z\in {{\mathbb {D}}}: h(z)=0\}$ is a discrete, locally finite set, then f is holomorphic in ${{\mathbb {D}}}$. For otherwise it is equal to ${{\mathbb {D}}}$ and, therefore, f is an anti-holomorphic function. $\square$

Proof of Theorem 1

It suffices to show that for any $(X_1,Y_1), (X_2,Y_2)\in {{\mathbb {C}}}^2$, any $(s,p)\in {{\mathbb {G}}}_2$, and any constant $C>0$ such that

$$\begin{aligned} \gamma _{{{\mathbb {G}}}_2}\big ((s,p);(X_1,Y_1)+\xi (X_2,Y_2)\big )=C\quad \text { for any }\xi \in {{\mathbb {T}}}, \end{aligned}$$

(6)

we have $(X_1,Y_1)=0$ or $(X_2,Y_2)=0$. There exists an automorphism $\Phi$ of ${{\mathbb {G}}}_2$ such that $\Phi (s,p)=(0,{\tilde{p}})$, where ${\tilde{p}}\ge 0$. Then

$$\begin{aligned}{} & {} \gamma _{{{\mathbb {G}}}_2}\big ((s,p);(X_1,Y_1)+\xi (X_2,Y_2)\big )=\nonumber \\{} & {} \quad \gamma _{{{\mathbb {G}}}_2}\big ((0,{\tilde{p}});\Phi '(s,p)(X_1,Y_1)+\xi \Phi '(s,p)(X_2,Y_2)\big ). \end{aligned}$$

(7)

Since $\det \Phi '\not =0$, we get $(X_1,Y_1)=0$ or $(X_2,Y_2)=0$. $\square$

3 Carathéodory isometries

For a domain $D\subset {{\mathbb {C}}}^n$, we define another biholomorphically invariant function. For any $\lambda _1,\lambda _2\in D$ we put

$$\begin{aligned} c_{D}(\lambda _1,\lambda _2)=\sup \{\rho (F(\lambda _1),F(\lambda _2)): F:D\rightarrow {{\mathbb {D}}}\text { holomorphic}\} \end{aligned}$$

We say that a mapping $\Phi :D_1\rightarrow D_2$ between domains $D_1,D_2$ is a Carathéodory isometry if

$$\begin{aligned} c_{D_1}(\lambda _1,\lambda _2)=c_{D_2}(\Phi (\lambda _1),\Phi (\lambda _2))\quad \text { for any }\lambda _1,\lambda _2\in D_1. \end{aligned}$$

We say that $\Phi$ is local c-isometry if for any $\lambda _0\in D_1$ there exists a neighborhood $U_0\subset D_1$ of $\lambda _0$ such that

$$\begin{aligned} c_{D_1}(\lambda _1,\lambda _2)=c_{D_2}(\Phi (\lambda _1),\Phi (\lambda _2))\quad \text { for any }\lambda _1,\lambda _2\in U_0. \end{aligned}$$

Note that any c-isometry is a local c-isometry and any local c-isometry is a $\gamma$-isometry (see e.g. [10]). In particular, we have

Theorem 8

Let $\phi :{{\mathbb {D}}}\rightarrow {{\mathbb {G}}}_2$ be a local c-isometry of class $C^1$. Then $\phi$ is holomorphic or anti-holomorphic.

4 Applications

In [5] the authors gave a proof of the description of automorphisms of the symmetrized bidisc. One of the main step in the proof is to show that for any automorphism $\Phi :{{\mathbb {G}}}_2\rightarrow {{\mathbb {G}}}_2$ we have $\Phi (\Sigma )\subset \Sigma$. We show this property by using our approach.

Corollary 9

Let $F:{{\mathbb {G}}}_2\rightarrow {{\mathbb {G}}}_2$ be a holomorphic mapping such that F is a $\gamma$-isometry at 0. Then $F(0)\in \Sigma$. In particular, if F is a $\gamma$-isometry on $\Sigma$, then $F(\Sigma )\subset \Sigma$.

Proof

Assume that $F(0)=(0,p)$, where $p\in (0,1)$. Then

$$\begin{aligned} \gamma _{{{\mathbb {G}}}_2}(0;(X,Y))=\gamma _{{{\mathbb {G}}}_2}((0,p); a_1X+b_1Y, a_2X+b_2Y)\quad \text { for any }X,Y\in {{\mathbb {C}}}, \end{aligned}$$

where $a_1=\frac{\partial F_1}{\partial s}(0)$, $b_1=\frac{\partial F_1}{\partial p}(0)$, $a_2=\frac{\partial F_2}{\partial s}(0)$, $b_2=\frac{\partial F_2}{\partial p}(0)$. Take pairs $(X,Y)=(1,\xi )$, where $\xi \in {{\mathbb {T}}}$. Recall that

$$\begin{aligned} \gamma _{{{\mathbb {G}}}_2}(0;(X,Y))=\frac{|X|}{2}+|Y|. \end{aligned}$$

In this way we get a contradiction. $\square$

Now we can prove a Vigué type result (see e.g. [9]).

Corollary 10

Let $F:{{\mathbb {G}}}_2\rightarrow {{\mathbb {G}}}_2$ be a holomorphic mapping such that F is a $\gamma$-isometry at 0. Then F is an automorphism of ${{\mathbb {G}}}_2$.

Proof

By the above Corollary, we may assume that $F(0)=0$. Then

$$\begin{aligned} \frac{|X|}{2}+|Y|=\frac{|a_1X+b_1Y|}{2}+ |a_2X+b_2Y|\quad \text { for any }X,Y\in {{\mathbb {C}}}, \end{aligned}$$

where $a_1=\frac{\partial F_1}{\partial s}(0)$, $b_1=\frac{\partial F_1}{\partial p}(0)$, $a_2=\frac{\partial F_2}{\partial s}(0)$, $b_2=\frac{\partial F_2}{\partial p}(0)$.

By taking $X=0$ and later $Y=0$ we get $\frac{|b_1|}{2}+|b_2|=1$ and $\frac{|a_1|}{2}+|a_2|=\frac{1}{2}$. So,

$$\begin{aligned}{} & {} \frac{|X|}{2}+|Y|=\frac{|a_1X+b_1Y|}{2}+ |a_2X+b_2Y|\le \nonumber \\{} & {} \quad \frac{|a_1X|+|b_1Y|}{2}+ |a_2X|+|b_2Y|\le \frac{|X|}{2}+|Y|. \end{aligned}$$

(8)

We get $a_1b_1=0$ and $a_2b_2=0$. Therefore, $a_1=0$, $|a_2|=\frac{1}{2}$, $b_2=0$, $|b_1|=2$ or $b_1=0$, $|b_2|=1$, $a_2=0$, $|a_1|=1$. We have $|\det F'(0)|=1$. From the Cartan theorem we get that F is an automorphism. $\square$

Remark 11

The author thanks the anonymous referee for her/his helpful comments that improved the presentation of the results.