# Carathéodory extremal functions on the symmetrized bidisc

Jim Agler  Department of Mathematics, University of California at San Diego,
CA 92103, USA  ,  Zinaida A. Lykova  School of Mathematics, Statistics and
Physics, Newcastle University, Newcastle upon Tyne NE1 7RU, U.K.
<EMAIL_ADDRESS>and  N. J. Young  School of Mathematics, Statistics
and Physics, Newcastle University, Newcastle upon Tyne NE1 7RU, U.K. and
School of Mathematics, Leeds University, Leeds LS2 9JT, U.K.
<EMAIL_ADDRESS>To Rien Kaashoek in esteem and friendship

(Date: 5th May 2018)

###### Abstract.

We show how realization theory can be used to find the solutions of the
Carathéodory extremal problem on the symmetrized bidisc

$G\stackrel{{\scriptstyle\rm{def}}}{{=}}\\{(z+w,zw):|z|<1,\,|w|<1\\}.$

We show that, generically, solutions are unique up to composition with
automorphisms of the disc. We also obtain formulae for large classes of
extremal functions for the Carathéodory problems for tangents of non-generic
types.

###### Key words and phrases:

Carathéodory extremal functions, symmetrized bidisc, model formulae,
realization formulae

###### 2010 Mathematics Subject Classification:

32A07, 53C22, 54C15, 47A57, 32F45, 47A25, 30E05

Partially supported by National Science Foundation Grants DMS 1361720 and
1665260, the UK Engineering and Physical Sciences Research Council grant
EP/N03242X/1, the London Mathematical Society Grant 41730 and Newcastle
University

## Introduction

A constant thread in the research of Marinus Kaashoek over several decades has
been the power of realization theory applied to a wide variety of problems in
analysis. Among his many contributions in this area we mention his monograph
[6], written with his longstanding collaborators Israel Gohberg and Harm Bart,
which was an early and influential work in the area, and his more recent
papers [13, 10]. Realization theory uses explicit formulae for functions in
terms of operators on Hilbert space to prove function-theoretic results.

In this paper we continue along the Bart-Gohberg-Kaashoek path by using
realization theory to prove results in complex geometry. Specifically, we are
interested in the geometry of the symmetrized bidisc

$G\stackrel{{\scriptstyle\rm{def}}}{{=}}\\{(z+w,zw):|z|<1,\,|w|<1\\},$

a domain in $\mathbb{C}^{2}$ that has been much studied in the last two
decades: see [8, 9, 11, 7, 17, 18, 2], along with many other papers. We shall
use realization theory to prove detailed results about the Carathéodory
extremal problem on $G$, defined as follows (see [14, 12]).

Consider a domain (that is, a connected open set) $\Omega$ in
$\mathbb{C}^{n}$. For domains $\Omega_{1},\ \Omega_{2}$, we denote by
$\Omega_{2}(\Omega_{1})$ the set of holomorphic maps from $\Omega_{1}$ to
$\Omega_{2}$. A point in the complex tangent bundle $T\Omega$ of $\Omega$ will
be called a tangent (to $\Omega$). Thus if $\delta\stackrel{{\scriptstyle\rm
def}}{{=}}(\lambda,v)$ is a tangent to $\Omega$ then $\lambda\in\Omega$ and
$v$ is a point in the complex tangent space
$T_{\lambda}\Omega\sim\mathbb{C}^{n}$ of $\Omega$ at $\lambda$. We say that
$\delta$ is a nondegenerate  tangent if $v\neq 0$. We write $|\cdot|$ for the
Poincaré metric on $T\mathbb{D}$:

$|(z,v)|\stackrel{{\scriptstyle\rm def}}{{=}}\frac{|v|}{1-|z|^{2}}\quad\mbox{
for }z\in\mathbb{D},\,v\in\mathbb{C}.$

The Carathéodory or Carathéodory-Reiffen pseudometric [12] on $\Omega$ is the
Finsler pseudometric $|\cdot|_{\rm car}$ on $T\Omega$ defined for
$\delta=(\lambda,v)\in T\Omega$ by

$\displaystyle|\delta|_{\rm car}$ $\displaystyle\stackrel{{\scriptstyle\rm
def}}{{=}}\sup_{F\in\mathbb{D}(\Omega)}|F_{*}(\delta)|$ (0.1)
$\displaystyle=\sup_{F\in\mathbb{D}(\Omega)}\frac{|D_{v}F(\lambda)|}{1-|F(\lambda)|^{2}}.$

Here $F_{*}$ is the standard notation for the pushforward of $\delta$ by the
map $F$ to an element of $T\mathbb{D}$, given by

$\left\langle g,F_{*}(\delta)\right\rangle=\left\langle g\circ
F,\delta\right\rangle$

for any analytic function $g$ in a neighbourhood of $F(\lambda)$.

The Carathéodory extremal problem $\operatorname{{\mathrm{C}ar}}\delta$ on
$\Omega$ is to calculate $|\delta|_{\rm car}$ for a given $\delta\in T\Omega$,
and to find the corresponding extremal functions, which is to say, the
functions $F\in\mathbb{D}(\Omega)$ for which the supremum in equation
(Introduction) is attained. We shall also say that $F$ solves
$\operatorname{{\mathrm{C}ar}}\delta$ to mean that $F$ is an extremal function
for $\operatorname{{\mathrm{C}ar}}\delta$.

For a general domain $\Omega$ one cannot expect to find either $|\cdot|_{\rm
car}$ or the corresponding extremal functions explicitly. In a few cases,
however, there are more or less explicit formulae for $|\delta|_{\rm car}$. In
particular, when $\Omega=G$, $|\cdot|_{\rm car}$ is a metric on $TG$ (it is
positive for nondegenerate tangents) and the following result obtains [4,
Theorem 1.1 and Corollary 4.3]. We use the co-ordinates $(s^{1},s^{2})$ for a
point of $G$.

######  Theorem 0.1.

Let $\delta$ be a nondegenerate tangent vector in $TG$. There exists
$\omega\in\mathbb{T}$ such that the function in $\mathbb{D}(G)$ given by

(0.2) $\Phi_{\omega}(s^{1},s^{2})\stackrel{{\scriptstyle\rm
def}}{{=}}\frac{2\omega s^{2}-s^{1}}{2-\omega s^{1}}$

is extremal for the Carathéodory problem
$\operatorname{{\mathrm{C}ar}}{\delta}$ in $G$.

It follows that $|\delta|_{\rm car}$ can be obtained as the maximum modulus of
a fractional quadratic function over the unit circle [4, Corollary
4.4]111Unfortunately there is an $\omega$ missing in equation (4.7) of [4].
The derivation given there shows that the correct formula is the present one.:
if $\delta=\left((s^{1},s^{2}),v\right)\in TG$ then

$\displaystyle|\delta|_{\rm car}$
$\displaystyle=\sup_{\omega\in\mathbb{T}}|(\Phi_{\omega})_{*}(\delta)|$
$\displaystyle=\sup_{\omega\in\mathbb{T}}\left|\frac{v_{1}(1-\omega^{2}s^{2})-v_{2}\omega(2-\omega
s^{1})}{(s^{1}-\overline{s^{1}}s^{2})\omega^{2}-2(1-|s^{2}|^{2})\omega+\bar{s}-\overline{s^{2}}s^{1}}\right|.$

Hence $|\delta|_{\rm car}$ can easily be calculated numerically to any desired
accuracy. In the latter equation we use superscripts (in $s^{1},s^{2}$) and
squares (of $\omega$, $|s^{2}|$).

The question arises: what are the extremal functions for the problem
$\operatorname{{\mathrm{C}ar}}{\delta}$? By Theorem 0.1, there is an extremal
function for $\operatorname{{\mathrm{C}ar}}{\delta}$ of the form
$\Phi_{\omega}$ for some $\omega$ in $\mathbb{T}$, but are there others? It is
clear that if $F$ is an extremal function for
$\operatorname{{\mathrm{C}ar}}\delta$ then so is $m\circ F$ for any
automorphism $m$ of $\mathbb{D}$, by the invariance of the Poincaré metric on
$\mathbb{D}$. We shall say that the solution of
$\operatorname{{\mathrm{C}ar}}\delta$ is essentially unique if, for every pair
of extremal functions $F_{1},F_{2}$ for $\operatorname{{\mathrm{C}ar}}\delta$,
there exists an automorphism $m$ of $\mathbb{D}$ such that $F_{2}=m\circ
F_{1}$.

We show in Theorem 2.1 that, for any nondegenerate tangent $\delta\in TG$, if
there is a unique $\omega$ in $\mathbb{T}$ such that $\Phi_{\omega}$ solves
$\operatorname{{\mathrm{C}ar}}\delta$, then the solution of
$\operatorname{{\mathrm{C}ar}}\delta$ is essentially unique. Indeed, for any
point $\lambda\in G$, the solution of
$\operatorname{{\mathrm{C}ar}}(\lambda,v)$ is essentially unique for generic
directions $v$ (Corollary 2.7). We also derive (in Section 3) a
parametrization of all solutions of $\operatorname{{\mathrm{C}ar}}\delta$ in
the special case that $\delta$ is tangent to the ‘royal variety’
$(s^{1})^{2}=4s^{2}$ in $G$, and in Sections 4 and 5 we obtain large classes
of Carathéodory extremals for two other classes of tangents, called flat and
purely balanced tangents.

The question of the essential uniqueness of solutions of
$\operatorname{{\mathrm{C}ar}}\delta$ in domains including $G$ was studied by
L. Kosiński and W. Zwonek in [15]. Their terminology and methods differ from
ours; we explain the relation of their Theorem 5.3 to our Theorem 2.1 in
Section 6. Incidentally, the authors comment that very little is known about
the set of all Carathéodory extremals for a given tangent in a domain. As far
as the domain $G$ goes, in this paper we derive a substantial amount of
information, even though we do not achieve a complete description of all
Carathéodory extremals on $G$.

The main tool we use is a model formula for analytic functions from $G$ to the
closed unit disc $\mathbb{D}^{-}$ proved in [5] and stated below as Definition
2.2 and Theorem 2.3. Model formulae and realization formulae for a class of
functions are essentially equivalent: one can pass back and forth between them
by standard methods (algebraic manipulation in one direction, lurking isometry
arguments in the other).

##  1\. Five types of tangent

There are certainly nondegenerate tangents $\delta\in TG$ for which the
solution of $\operatorname{{\mathrm{C}ar}}\delta$ is not essentially unique.
Consider, for example, $\delta$ of the form

$\delta=\left((2z,z^{2}),2c(1,z)\right)$

for some $z\in\mathbb{D}$ and nonzero complex $c$. We call such a tangent
royal: it is tangent to the ‘royal variety’

$\mathcal{R}\stackrel{{\scriptstyle\rm
def}}{{=}}\\{(2z,z^{2}):z\in\mathbb{D}\\}$

in $G$. By a simple calculation, for any $\omega\in\mathbb{T}$,

$\Phi_{\omega}(2z,z^{2})=-z,\qquad D_{v}\Phi_{\omega}(2z,z^{2})=-c,$

where $v=2c(1,z)$, so that $\Phi_{\omega}(2z,z^{2})$ and
$D_{v}\Phi_{\omega}(2z,z^{2})$ are independent of $\omega$. It follows from
Theorem 0.1 that $\Phi_{\omega}$ solves $\operatorname{{\mathrm{C}ar}}\delta$
for all $\omega\in\mathbb{T}$ and that

(1.1) $\ |\delta|_{\rm
car}=\frac{|D_{v}\Phi_{\omega}(2z,z^{2})|}{1-|\Phi_{\omega}(2z,z^{2})|^{2}}=\frac{|c|}{1-|z|^{2}}.$

Now if $\omega_{1},\ \omega_{2}$ are distinct points of $\mathbb{T}$, there is
no automorphism $m$ of $\mathbb{D}$ such that
$\Phi_{\omega_{1}}=m\circ\Phi_{\omega_{2}}$; this is a consequence of the fact
that $(2\bar{\omega},\bar{\omega}^{2})$ is the unique singularity of
$\Phi_{\omega}$ in the closure $\Gamma$ of $G$. Hence the solution of
$\operatorname{{\mathrm{C}ar}}\delta$ is not essentially unique.

Similar conclusions hold for another interesting class of tangents, which we
call flat. These are the tangents of the form

$(\lambda,v)=\left((\beta+\bar{\beta}z,z),c(\bar{\beta},1)\right)$

for some $\beta\in\mathbb{D}$ and $c\in\mathbb{C}\setminus\\{0\\}$. It is an
entertaining calculation to show that

(1.2) $|(\lambda,v)|_{\rm
car}=\frac{|D_{v}\Phi_{\omega}(\lambda)|}{1-|\Phi_{\omega}(\lambda)|^{2}}=\frac{|c|}{1-|z|^{2}}$

for all $\omega\in\mathbb{T}$. Again, the solution to
$\operatorname{{\mathrm{C}ar}}(\lambda,v)$ is far from being essentially
unique.

There are also tangents $\delta\in TG$ such that $\Phi_{\omega}$ solves
$\operatorname{{\mathrm{C}ar}}\delta$ for exactly two values of $\omega$ in
$\mathbb{T}$; we call these purely balanced tangents. They can be described
concretely as follows. For any hyperbolic automorphism $m$ of $\mathbb{D}$
(that is, one that has two fixed points $\omega_{1}$ and $\omega_{2}$ in
$\mathbb{T}$) let $h_{m}$ in $G(\mathbb{D})$ be given by

$h_{m}(z)=(z+m(z),zm(z))$

for $z\in\mathbb{D}$. A purely balanced tangent has the form

(1.3) $\delta=(h_{m}(z),ch_{m}^{\prime}(z))$

for some hyperbolic automorphism $m$ of $\mathbb{D}$, some $z\in\mathbb{D}$
and some $c\in\mathbb{C}\setminus\\{0\\}$. It is easy to see that, for
$\omega\in\mathbb{T}$, the composition $\Phi_{\omega}\circ h_{m}$ is a
rational inner function of degree at most $2$ and that the degree reduces to
$1$ precisely when $\omega$ is either $\bar{\omega}_{1}$ or
$\bar{\omega}_{2}$. Thus, for these two values of $\omega$ (and only these),
$\Phi_{\omega}\circ h_{m}$ is an automorphism of $\mathbb{D}$. It follows that
$\Phi_{\omega}$ solves $\operatorname{{\mathrm{C}ar}}\delta$ if and only if
$\omega=\bar{\omega}_{1}$ or $\bar{\omega}_{2}$.

A fourth type of tangent, which we call exceptional, is similar to the purely
balanced type, but differs in that the hyperbolic automorphism $m$ of
$\mathbb{D}$ is replaced by a parabolic automorphism, that is, an automorphism
$m$ of $\mathbb{D}$ which has a single fixed point $\omega_{1}$ in
$\mathbb{T}$, which has multiplicity $2$. The same argument as in the previous
paragraph shows that $\Phi_{\omega}$ solves the Carathéodory problem if and
only if $\omega=\bar{\omega}_{1}$.

The fifth and final type of tangent is called purely unbalanced. It consists
of the tangents $\delta=(\lambda,v)\in TG$ such that $\Phi_{\omega}$ solves
$\operatorname{{\mathrm{C}ar}}\delta$ for a unique value
$\mathrm{e}^{\mathrm{i}t_{0}}$ of $\omega$ in $\mathbb{T}$ and

(1.4)
$\left.\frac{d^{2}}{dt^{2}}\frac{|D_{v}\Phi_{\mathrm{e}^{\mathrm{i}t}}(\lambda)|}{1-|\Phi_{\mathrm{e}^{\mathrm{i}t}}(\lambda)|^{2}}\right|_{t=t_{0}}<0.$

The last inequality distinguishes purely unbalanced from exceptional tangents
– the left hand side of equation (1.4) is equal to zero for exceptional
tangents.

The five types of tangent are discussed at length in our paper [2]. We proved
[2, Theorem 3.6] a ‘pentachotomy theorem’, which states that every
nondegenerate tangent in $TG$ is of exactly one of the above five types. We
also give, for a representative tangent of each type, a cartoon showing the
unique complex geodesic in $G$ touched by the tangent [2, Appendix B].

It follows trivially from Theorem 0.1 that, for every nondegenerate tangent
$\delta\in TG$, either

  1. (1)

there exists a unique $\omega\in\mathbb{T}$ such that $\Phi_{\omega}$ solves
$\operatorname{{\mathrm{C}ar}}\delta$, or

  2. (2)

there exist at least two values of $\omega$ in $\mathbb{T}$ such that
$\Phi_{\omega}$ solves $\operatorname{{\mathrm{C}ar}}\delta$.

The above discussion shows that Case (1) obtains for purely unbalanced and
exceptional tangents, while Case (2) holds for royal, flat and purely balanced
tangents. For the purpose of this paper, the message to be drawn is that Case
(1) is generic in the following sense. Consider a point $\lambda\in G$. Each
tangent $v$ in $T_{\lambda}G$ has a ‘complex direction’ $\mathbb{C}v$, which
is a one-dimensional subspace of $\mathbb{C}^{2}$, or in other words, a point
of the projective space $\mathrm{CP}^{2}$. The directions corresponding to the
royal (if any) and flat tangents at $\lambda$ are just single points in
$\mathrm{CP}^{2}$, while, from the constructive nature of the expression (1.3)
for a purely balanced tangent, it is easy to show that there is a smooth one-
real-parameter curve of purely balanced directions (see [1, Section 1]). It
follows that the set of directions $\mathbb{C}v\in\mathrm{CP}^{2}$ for which a
unique $\Phi_{\omega}$ solves $\operatorname{{\mathrm{C}ar}}\delta$ contains a
dense open set in $\mathrm{CP}^{2}$. To summarise:

######  Proposition 1.1.

For every $\lambda\in G$ there exists a dense open set $V_{\lambda}$ in
$\mathrm{CP}^{2}$ such that whenever $\mathbb{C}v\in V_{\lambda}$, there
exists a unique $\omega\in\mathbb{T}$ such that $\Phi_{\omega}$ solves
$\operatorname{{\mathrm{C}ar}}(\lambda,v)$.

##  2\. Tangents with a unique extremal $\Phi_{\omega}$

In Section 1 we discussed extremal functions of the special form
$\Phi_{\omega},\ \omega\in\mathbb{T}$, for the Carathéodory problem in $G$.
However, there is no reason to expect that the $\Phi_{\omega}$ will be the
only extremal functions. For example, if $\delta=(\lambda,v)$ is a
nondegenerate tangent and $\Phi_{\omega_{1}},\dots,\Phi_{\omega_{k}}$ all
solve $\operatorname{{\mathrm{C}ar}}\delta$, then one can generate a large
class of other extremal functions as follows. Choose an automorphism $m_{j}$
of $\mathbb{D}$ such that $m_{j}\circ\Phi_{\omega_{j}}(\lambda)=0$ and
$D_{v}(m_{j}\circ\Phi_{\omega_{j}})(\lambda_{j})>0$ for $j=1,\dots,k$. Then
each $m_{j}\circ\Phi_{\omega_{j}}$ solves
$\operatorname{{\mathrm{C}ar}}\delta$, and so does any convex combination of
them.

Nevertheless, if there is a unique $\omega\in\mathbb{T}$ such that
$\Phi_{\omega}$ is extremal for $\operatorname{{\mathrm{C}ar}}\delta$ then the
solution of $\operatorname{{\mathrm{C}ar}}\delta$ is essentially unique.

######  Theorem 2.1.

Let $\delta$ be a nondegenerate tangent in $G$ such that $\Phi_{\omega}$
solves $\operatorname{{\mathrm{C}ar}}\delta$ for a unique value of $\omega$ in
$\mathbb{T}$. If $\psi$ solves $\operatorname{{\mathrm{C}ar}}\delta$ then
there exists an automorphism $m$ of $\mathbb{D}$ such that
$\psi=m\circ\Phi_{\omega}$.

For the proof recall the following model formula [5, Definition 2.1 and
Theorem 2.2].

######  Definition 2.2.

A $G$-_model_ for a function $\varphi$ on $G$ is a triple $(\mathcal{M},T,u)$
where $\mathcal{M}$ is a separable Hilbert space, $T$ is a contraction acting
on $\mathcal{M}$ and $u:G\to\mathcal{M}$ is an analytic map such that, for all
$s,t\in G$,

(2.1)
$1-\overline{\varphi(t)}\varphi(s)=\left\langle(1-t_{T}^{*}s_{T})u(s),u(t)\right\rangle_{\mathcal{M}}$

where, for $s\in G$,

$s_{T}\stackrel{{\scriptstyle\rm def}}{{=}}(2s^{2}T-s^{1})(2-s^{1}T)^{-1}.$

A $G$-model $(\mathcal{M},T,u)$ is unitary if $T$ is a unitary operator on
$\mathcal{M}$.

For any domain $\Omega$ we define the Schur class  $\mathcal{S}(\Omega)$ to be
the set of holomorphic maps from $\Omega$ to the closed unit disc
$\mathbb{D}^{-}$.

######  Theorem 2.3.

Let $\varphi$ be a function on $G$. The following three statements are
equivalent.

  1. (1)

$\varphi\in\mathcal{S}(G)$;

  2. (2)

$\varphi$ has a $G$-model;

  3. (3)

$\varphi$ has a unitary $G$-model $(\mathcal{M},T,u)$.

From a $G$-model of a function $\varphi\in\mathcal{S}(G)$ one may easily
proceed by means of a standard lurking isometry argument to a realization
formula

$\varphi(s)=A+Bs_{T}(1-Ds_{T})^{-1}C,\quad\mbox{ all }s\in G,$

for $\varphi$, where $ABCD$ is a contractive or unitary colligation on
$\mathbb{C}\oplus\mathcal{M}$. However, for the present purpose it is
convenient to work directly from the $G$-model.

We also require a long-established fact about $G$ [4], related to the fact
that the Carathéodory and Kobayashi metrics on $TG$ coincide.

######  Lemma 2.4.

If $\delta$ is a nondegenerate tangent to $G$ and $\varphi$ solves
$\operatorname{{\mathrm{C}ar}}\delta$ then there exists $k$ in $G(\mathbb{D})$
such that $\varphi\circ k=\mathrm{id}_{\mathbb{D}}$. Moreover, if $\psi$ is
any solution of $\operatorname{{\mathrm{C}ar}}\delta$ then $\psi\circ k$ is an
automorphism of $\mathbb{D}$.

We shall need some minor measure-theoretic technicalities.

######  Lemma 2.5.

Let $Y$ be a set and let

$A:\mathbb{T}\times Y\times Y\to\mathbb{C}$

be a map such that

  1. (1)

$A(\cdot,z,w)$ is continuous on $\mathbb{T}$ for every $z,w\in Y$;

  2. (2)

$A(\eta,\cdot,\cdot)$ is a positive kernel on $Y$ for every
$\eta\in\mathbb{T}$.

Let $\mathcal{M}$ be a separable Hilbert space, let $T$ be a unitary operator
on $\mathcal{M}$ with spectral resolution

$T=\int_{\mathbb{T}}\eta\,\mathrm{d}E(\eta)$

and let $v:Y\to\mathcal{M}$ be a mapping. Let

(2.2)
$C(z,w)=\int_{\mathbb{T}}A(\eta,z,w)\,\left\langle\mathrm{d}E(\eta)v(z),v(w)\right\rangle$

for all $z,w\in Y$. Then $C$ is a positive kernel on $Y$.

###### Proof.

Consider any finite subset $\\{z_{1},\dots,z_{N}\\}$ of $Y$. We must show that
the $N\times N$ matrix

$\begin{bmatrix}C(z_{i},z_{j})\end{bmatrix}_{i,j=1}^{N}$

is positive.

Since $A(\cdot,z_{i},z_{j})$ is continuous on $\mathbb{T}$ for each $i$ and
$j$, we may approximate the $N\times N$-matrix-valued function
$[A(\cdot,z_{i},z_{j})]$ uniformly on $\mathbb{T}$ by integrable simple
functions of the form

$[f_{ij}]=\sum_{\ell}b^{\ell}\chi_{\tau_{\ell}}$

for some $N\times N$ matrices $b^{\ell}$ and Borel sets $\tau_{\ell}$, where
$\chi$ denotes ‘characteristic function’. Moreover we may do this in such a
way that each $b^{\ell}$ is a value $[A(\eta,z_{i},z_{j})]$ for some
$\eta\in\mathbb{T}$, hence is positive. Then

(2.3)
$\begin{bmatrix}\int_{\tau}f_{ij}(\eta)\,\left\langle\mathrm{d}E(\eta)z_{i},z_{j}\right\rangle\end{bmatrix}_{i,j=1}^{N}=\sum_{\ell}b^{\ell}*\begin{bmatrix}\left\langle
E(\tau_{\ell})v_{i},v_{j}\right\rangle\end{bmatrix}_{i,j=1}^{N}$

where $*$ denotes the Schur (or Hadamard) product of matrices. Since the
matrix $\begin{bmatrix}\left\langle
E(\tau_{\ell})v_{i},v_{j}\right\rangle\end{bmatrix}$ is positive and the Schur
product of positive matrices is positive, every approximating sum of the form
(2.3) is positive, and hence the integral in equation (2.2) is a positive
matrix.

∎

######  Lemma 2.6.

For $i,j=1,2$ let $a_{ij}:\mathbb{T}\to\mathbb{C}$ be continuous and let each
$a_{ij}$ have only finitely many zeros in $\mathbb{T}$. Let $\nu_{ij}$ be a
complex-valued Borel measure on $\mathbb{T}$ such that, for every Borel set
$\tau$ in $\mathbb{T}$,

$\begin{bmatrix}\nu_{ij}(\tau)\end{bmatrix}_{i,j=1}^{2}\geq 0.$

Let $X$ be a Borel subset of $\mathbb{T}$ and suppose that

$\begin{bmatrix}a_{ij}(\eta)\end{bmatrix}_{i,j=1}^{2}\;\mbox{ is positive and
of rank }2\mbox{ for all }\eta\in X.$

Let

$C=\begin{bmatrix}c_{ij}\end{bmatrix}_{i,j=1}^{2}$

where

$c_{ij}=\int_{X}a_{ij}(\eta)\,\mathrm{d}\nu_{ij}(\eta)\qquad\mbox{ for
}i,j=1,2.$

If $\operatorname{rank}C\leq 1$ then either $c_{11}=0$ or $c_{22}=0$.

###### Proof.

By hypothesis the set

$Z\stackrel{{\scriptstyle\rm
def}}{{=}}\bigcup_{i,j=1}^{2}\\{\eta\in\mathbb{T}:a_{ij}(\eta)=0\\}$

is finite.

Exactly as in the proof of Lemma 2.5, for any Borel set $\tau$ in
$\mathbb{T}$,

(2.4)
$\begin{bmatrix}\int_{\tau}a_{ij}\,\mathrm{d}\nu_{ij}\end{bmatrix}_{ij=1}^{2}\geq
0.$

Suppose that $C$ has rank at most $1$ but $c_{11}$ and $c_{22}$ are both
nonzero. Then there exists a nonzero $2\times 1$ matrix $c=[c_{1}\,c_{2}]^{T}$
such that $C=cc^{*}$ for $i,j=1,2$ and $c_{1},\ c_{2}$ are nonzero.

For any Borel set $\tau\subset X$,

$\begin{bmatrix}\int_{\tau}a_{ij}\,\mathrm{d}\nu_{ij}\end{bmatrix}\leq\begin{bmatrix}\int_{\tau}+\int_{X\setminus\tau}a_{ij}\,\mathrm{d}\nu_{ij}\end{bmatrix}=\begin{bmatrix}\int_{X}a_{ij}\,\mathrm{d}\nu_{ij}\end{bmatrix}=C=cc^{*}.$

Consequently there exists a unique $\mu(\tau)\in[0,1]$ such that

(2.5)
$\begin{bmatrix}\int_{\tau}a_{ij}\,\mathrm{d}\nu_{ij}\end{bmatrix}=\mu(\tau)C.$

It is easily seen that $\mu$ is a Borel probability measure on $X$. Note that
if $\eta\in Z$, say $a_{ij}(\eta)=0$, then on taking $\tau=\\{\eta\\}$ in
equation (2.5), we deduce that

$\mu(\\{\eta\\})c_{i}\bar{c}_{j}=0.$

Since $c_{1},c_{2}$ are nonzero, it follows that $\mu(\\{\eta\\})=0$. Hence
$\mu(Z)=0$.

Equation (2.5) states that $\mu$ is absolutely continuous with respect to
$\nu_{ij}$ on $X$ and the Radon-Nikodym derivative is given by

$c_{i}\bar{c}_{j}\frac{\mathrm{d}\mu}{\mathrm{d}\nu_{ij}}=a_{ij}$

for $i,j=1,2$. Hence, on $X\setminus Z$,

(2.6) $\mathrm{d}\nu_{ij}=\frac{c_{i}\bar{c}_{j}}{a_{ij}}\mathrm{d}\mu,\qquad
i,j=1,2.$

Pick a compact subset $K$ of $X\setminus Z$ such that $\mu(K)>0$. This is
possible, since $\mu(X\setminus Z)=1$ and Borel measures on $\mathbb{T}$ are
automatically regular. By compactness, there exists a point $\eta_{0}\in K$
such that, for every open neighbourhood $U$ of $\eta_{0}$,

$\mu(U\cap K)>0.$

Notice that, for $\eta\in\mathbb{T}\setminus Z$,

$\det\begin{bmatrix}\displaystyle\frac{c_{i}\bar{c}_{j}}{a_{ij}(\eta)}\end{bmatrix}_{i,j=1}^{2}=-\frac{|c_{1}c_{2}|^{2}\det\begin{bmatrix}a_{ij}(\eta)\end{bmatrix}}{a_{11}(\eta)a_{22}(\eta)|a_{12}(\eta)|^{2}}<0.$

Thus $[c_{i}\bar{c}_{j}a_{ij}(\eta_{0})^{-1}]$ has a negative eigenvalue.
Therefore there exists a unit vector $x\in\mathbb{C}^{2}$, an $\varepsilon>0$
and an open neighourhood $U$ of $\eta_{0}$ in $\mathbb{T}$ such that

$\left\langle\begin{bmatrix}c_{i}\bar{c}_{j}a_{ij}(\eta)^{-1}\end{bmatrix}x,x\right\rangle<-\varepsilon$

for all $\eta\in U$. We then have

$\displaystyle\left\langle\begin{bmatrix}\nu_{ij}(U\cap
K)\end{bmatrix}x,x\right\rangle$ $\displaystyle=\left\langle\int_{U\cap
K}\begin{bmatrix}c_{i}\bar{c}_{j}a_{ij}(\eta)^{-1}\end{bmatrix}\mathrm{d}\mu(\eta)x,x\right\rangle$
$\displaystyle=\int_{U\cap
K}\left\langle\begin{bmatrix}c_{i}\bar{c}_{j}a_{ij}(\eta)^{-1}\end{bmatrix}x,x\right\rangle\mathrm{d}\mu(\eta)$
$\displaystyle<-\varepsilon\mu(U\cap K)$ $\displaystyle<0.$

This contradicts the positivity of the matricial measure
$\begin{bmatrix}\nu_{ij}\end{bmatrix}$. Hence either $c_{1}=0$ or $c_{2}=0$.

∎

###### Proof of Theorem 2.1.

Let $\delta$ be a nondegenerate tangent to $G$ such that $\Phi_{\omega}$ is
the unique function from the collection
$\\{\Phi_{\eta}\\}_{\eta\in\mathbb{T}}$ that solves
$\operatorname{{\mathrm{C}ar}}\delta$. Let $\psi$ be a solution of
$\operatorname{{\mathrm{C}ar}}\delta$. We must find an automorphism $m$ of
$\mathbb{D}$ such that $\psi=m\circ\Phi_{\omega}$.

By Lemma 2.4, there exists $k$ in $G(\mathbb{D})$ such that

(2.7) $\Phi_{\omega}\circ k=\mathrm{id}_{\mathbb{D}},$

and moreover, the function

(2.8) $m\stackrel{{\scriptstyle\rm def}}{{=}}\psi\circ k$

is an automorphism of $\mathbb{D}$. Let

(2.9) $\varphi=m^{-1}\circ\psi.$

Then

(2.10) $\varphi\circ k=m^{-1}\circ\psi\circ k=m^{-1}\circ
m=\mathrm{id}_{\mathbb{D}}.$

By Theorem 2.3, there is a unitary $G$-model $(\mathcal{M},T,u)$ for
$\varphi$. By the Spectral Theorem for unitary operators, there is a spectral
measure $E(.)$ on $\mathbb{T}$ with values in $\mathcal{B}(\mathcal{M})$ such
that

$T=\int_{\mathbb{T}}\eta\;\mathrm{d}E(\eta).$

Thus, for $s\in G$,

$s_{T}=(2s^{2}T-s^{1})(2-s^{1}T)^{-1}=\int_{\mathbb{T}}\Phi_{\eta}(s)\;\mathrm{d}E(\eta).$

Therefore, for all $s,t\in G$,

$\displaystyle 1-\overline{\varphi(t)}\varphi(s)$
$\displaystyle=\left\langle(1-t_{T}^{*}s_{T})u(s),u(t)\right\rangle_{\mathcal{M}}$
(2.11)
$\displaystyle=\int_{\mathbb{T}}\left(1-\overline{\Phi_{\eta}(t)}\Phi_{\eta}(s)\right)\left\langle\mathrm{d}E(\eta)u(s),u(t)\right\rangle_{\mathcal{M}}.$

Consider $z,w\in\mathbb{D}$, put $s=k(z),\,t=k(w)$ in equation (2). Invoke
equation (2.10) and divide equation (2) through by $1-\bar{w}z$ to obtain, for
$z,w\in\mathbb{D}$,

$\displaystyle 1$
$\displaystyle=\int_{\\{\omega\\}}+\int_{\mathbb{T}\setminus\\{\omega\\}}\frac{1-\overline{\Phi_{\eta}\circ
k(w)}\Phi_{\eta}\circ k(z)}{1-\bar{w}z}\left\langle\mathrm{d}E(\eta)u\circ
k(z),u\circ k(w)\right\rangle$ (2.12) $\displaystyle=I_{1}+I_{2}$

where

$\displaystyle I_{1}(z,w)$ $\displaystyle=\left\langle E(\\{\omega\\})u\circ
k(z),u\circ k(w)\right\rangle,$ (2.13) $\displaystyle I_{2}(z,w)$
$\displaystyle=\int_{\mathbb{T}\setminus\\{\omega\\}}\frac{1-\overline{\Phi_{\eta}\circ
k(w)}\Phi_{\eta}\circ k(z)}{1-\bar{w}z}\left\langle\mathrm{d}E(\eta)u\circ
k(z),u\circ k(w)\right\rangle.$

The left hand side $1$ of equation (2.12) is a positive kernel of rank one on
$\mathbb{D}$, and $I_{1}$ is also a positive kernel. The integrand in $I_{2}$
is a positive kernel on $\mathbb{D}$ for each $\eta\in\mathbb{T}$, by Pick’s
theorem, since $\Phi_{\eta}\circ k$ is in the Schur class. Hence, by Lemma
2.5, $I_{2}$ is also a positive kernel on $\mathbb{D}$. Since $I_{1}+I_{2}$
has rank $1$, it follows that $I_{2}$ has rank at most $1$ as a kernel on
$\mathbb{D}$.

By hypothesis, $\Phi_{\eta}$ does not solve
$\operatorname{{\mathrm{C}ar}}\delta$ for any
$\eta\in\mathbb{T}\setminus\\{\omega\\}$. Therefore $\Phi_{\eta}\circ k$ is a
Blaschke product of degree $2$, and consequently, for any choice of distinct
points $z_{1},z_{2}$ in $\mathbb{D}$, the $2\times 2$ matrix

(2.14)
$\begin{bmatrix}a_{ij}(\eta)\end{bmatrix}_{i,j=1}^{2}\stackrel{{\scriptstyle\rm
def}}{{=}}\begin{bmatrix}\displaystyle\frac{1-\overline{\Phi_{\eta}\circ
k(z_{i})}\Phi_{\eta}\circ
k(z_{j})}{1-\bar{z}_{i}z_{j}}\end{bmatrix}_{i,j=1}^{2}$

is a positive matrix of rank $2$ for every
$\eta\in\mathbb{T}\setminus\\{\omega\\}$. In particular, $a_{11}(\eta)>0$ for
all $\eta\in\mathbb{T}\setminus\\{\omega\\}$.

Moreover, each $a_{ij}$ has only finitely many zeros in $\mathbb{T}$, as may
be seen from the fact that $a_{ij}$ is a ratio of trigonometric polynomials in
$\eta$. To be explicit, if we temporarily write $k=(k^{1},k^{2}):\mathbb{D}\to
G$, then equation (2.14) expands to $a_{ij}(\eta)=P(\eta)/Q(\eta)$ where

$\displaystyle P(\eta)$
$\displaystyle=4\left(1-\overline{k^{2}(z_{i})}k^{2}(z_{j})\right)-2\eta\left(k^{1}(z_{j})-\overline{k^{1}(z_{i})}k^{2}(z_{j})\right)$
$\displaystyle\hskip
56.9055pt-2\bar{\eta}\left(\overline{k^{1}(z_{i})}-\overline{k^{2}(z_{i})}k^{1}(z_{j})\right),$
$\displaystyle Q(\eta)$ $\displaystyle=(1-\bar{z}_{i}z_{j})(2-\eta
k^{1}(z_{i}))^{-}(2-\eta k^{1}(z_{j})).$

Let

$\nu_{ij}=\left\langle E(\cdot)u\circ k(z_{i}),u\circ k(z_{j})\right\rangle.$

Clearly $[\nu_{ij}(\tau)]\geq 0$ for every Borel subset $\tau$ of
$\mathbb{T}\setminus\\{\omega\\}$. By definition (2),

$I_{2}(z_{i},z_{j})=\int_{\mathbb{T}\setminus\\{\omega\\}}a_{ij}\,\mathrm{d}\nu_{ij}$

for $i,j=1,2$. Moreover, by equation (2.12),

$[I_{2}(z_{i},z_{j})]\leq[I_{1}(z_{i},z_{j})]+[I_{2}(z_{i},z_{j})]=\begin{bmatrix}1&1\\\
1&1\end{bmatrix}.$

It follows that

(2.15)
$\begin{bmatrix}\int_{\mathbb{T}\setminus\\{\omega\\}}a_{ij}\,\mathrm{d}\nu_{ij}\end{bmatrix}=[I_{2}(z_{i},z_{j}]=\kappa\begin{bmatrix}1&1\\\
1&1\end{bmatrix}$

for some $\kappa\in[0,1]$. We may now apply Lemma 2.6 with
$X=\mathbb{T}\setminus\\{\omega\\}$ to deduce that $\kappa=0$ and hence
$I_{2}(z_{i},z_{j})=0$. In particular,

$0=I_{2}(z_{1},z_{1})=\int_{\mathbb{T}\setminus\\{\omega\\}}a_{11}\,\mathrm{d}\nu_{11}.$

Since $a_{11}>0$ on $\mathbb{T}\setminus\\{\omega\\}$, it follows that
$\nu_{11}(\mathbb{T}\setminus\\{\omega\\})=0$, which is to say that

(2.16) $E(\mathbb{T}\setminus\\{\omega\\})u\circ k(z_{1})=0.$

Since $z_{1},z_{2}$ were chosen arbitrarily in
$\mathbb{T}\setminus\\{\omega\\}$, we have $I_{2}\equiv 0$ and therefore, by
equation (2.12),

(2.17) $1=I_{1}=\left\langle E(\\{\omega\\})u\circ k(z),u\circ
k(w)\right\rangle$

for all $z,w\in\mathbb{D}$. It follows that

$\|E(\\{\omega\\})u\circ k(z)-E(\\{\omega\\})u\circ k(w)\|^{2}=0$

for all $z,w$, and hence that there exists a unit vector $x\in\mathcal{M}$
such that

$E(\\{\omega\\})u\circ k(z)=x$

for all $z\in\mathbb{D}$.

In equation (2), choose $t=k(w)$ for some $w\in\mathbb{D}$. Since
$\Phi_{\omega}\circ k=\mathrm{id}_{\mathbb{D}}$, we have for all $s\in G$,

$\displaystyle 1-\bar{w}\varphi(s)$ $\displaystyle=1-\overline{\varphi\circ
k(w)}\varphi(s)$
$\displaystyle=\int_{\\{\omega\\}}+\int_{\mathbb{T}\setminus\\{\omega\\}}\left(1-\overline{\Phi_{\eta}\circ
k(w)}\Phi_{\eta}(s)\right)\left\langle\mathrm{d}E(\eta)u(s),u\circ
k(w)\right\rangle$ $\displaystyle=(1-\bar{w}\Phi_{\omega}(s))\left\langle
u(s),x\right\rangle+$ $\displaystyle\hskip
28.45274pt\int_{\mathbb{T}\setminus\\{\omega\\}}\left(1-\overline{\Phi_{\eta}\circ
k(w)}\Phi_{\eta}(s)\right)\left\langle\mathrm{d}E(\eta)u(s),u\circ
k(w)\right\rangle.$

In view of equation (2.16), the scalar spectral measure in the second term on
the right hand side is zero on $\mathbb{T}\setminus\\{\omega\\}$. Hence the
integral is zero, and so, for all $s\in G$ and $w\in\mathbb{D}$,

(2.18) $\displaystyle 1-\bar{w}\varphi(s)$
$\displaystyle=(1-\bar{w}\Phi_{\omega}(s))\left\langle u(s),x\right\rangle.$

Put $w=0$ to deduce that

$\left\langle u(s),x\right\rangle=1$

for all $s\in G$, then equate coefficients of $\bar{w}$ to obtain
$\varphi=\Phi_{\omega}$. Hence, by equation (2.8),

$\psi=m\circ\varphi=m\circ\Phi_{\omega}$

as required. ∎

On combining Theorem 2.1 and Proposition 1.1 we obtain the statement in the
abstract.

######  Corollary 2.7.

Let $\lambda\in G$. For a generic direction $\mathbb{C}v$ in
$\mathrm{CP}^{2}$, the solution of the Carathéodory problem
$\operatorname{{\mathrm{C}ar}}(\lambda,v)$ is essentially unique.

It will sometimes be useful in the sequel to distinguish a particular
Carathéodory extremal function from a class of functions that are equivalent
up to composition with automorphisms of $\mathbb{D}$. Consider any tangent
$\delta\in TG$ and any solution $\varphi$ of
$\operatorname{{\mathrm{C}ar}}\delta$. The functions $m\circ\varphi$, with $m$
an automorphism of $\mathbb{D}$, also solve
$\operatorname{{\mathrm{C}ar}}\delta$, and among them there is exactly one
that has the property

$m\circ\varphi(\lambda)=0\quad\mbox{ and }\quad
D_{v}(m\circ\varphi)(\lambda)>0,$

or equivalently,

(2.19) $(m\circ\varphi)_{*}(\delta)=(0,|\delta|_{\rm car}).$

We shall say that $\varphi$ is well aligned at $\delta$ if
$\varphi_{*}(\delta)=(0,|\delta|_{\rm car})$.

With this terminology the following is a re-statement of Theorem 2.1.

######  Corollary 2.8.

If $\delta$ is a nondegenerate tangent in $G$ such that $\Phi_{\omega}$ solves
$\operatorname{{\mathrm{C}ar}}\delta$ for a unique value of $\omega$ in
$\mathbb{T}$ then there is a unique well-aligned solution of
$\operatorname{{\mathrm{C}ar}}\delta$. It is expressible as
$m\circ\Phi_{\omega}$ for some automorphism $m$ of $\mathbb{D}$.

##  3\. Royal tangents

At the opposite extreme from the tangents studied in the last section are the
royal tangents to $G$. Recall that these have the form

(3.1) $\delta=\left((2z,z^{2}),2c(1,z)\right)$

for some $z\in\mathbb{D}$ and nonzero complex number $c$. As we observed in
Section 1,

$|\delta|_{\rm car}=\frac{|c|}{1-|z|^{2}}$

and all $\Phi_{\omega},\omega\in\mathbb{T}$, solve
$\operatorname{{\mathrm{C}ar}}\delta$. In this section we shall describe all
extremal functions for $\operatorname{{\mathrm{C}ar}}\delta$ for royal
tangents $\delta$, not just those of the form $\Phi_{\omega}$.

######  Theorem 3.1.

Let $\delta\in TG$ be the royal tangent

(3.2) $\delta=\left((2z,z^{2}),2c(1,z)\right)$

for some $z\in\mathbb{D}$ and $c\in\mathbb{C}\setminus\\{0\\}$. A function
$\varphi\in\mathbb{D}(G)$ solves $\operatorname{{\mathrm{C}ar}}\delta$ if and
only if there exists an automorphism $m$ of $\mathbb{D}$ and
$\Psi\in\mathcal{S}(G)$ such that, for all $s\in G$,

(3.3)
$\varphi(s)=m\left({\tfrac{1}{2}}s^{1}+\tfrac{1}{4}((s^{1})^{2}-4s^{2})\frac{\Psi(s)}{1-{\tfrac{1}{2}}s^{1}\Psi(s)}\right).$

###### Proof.

We shall lift the problem $\operatorname{{\mathrm{C}ar}}\delta$ to a
Carathéodory problem on the bidisc $\mathbb{D}^{2}$, where we can use the
results of [3] on the Nevanlinna-Pick problem on the bidisc.

Let $\pi:\mathbb{D}^{2}\to G$ be the ‘symmetrization map’,

$\pi(\lambda^{1},\lambda^{2})=(\lambda^{1}+\lambda^{2},\lambda^{1}\lambda^{2})$

and let $k:\mathbb{D}\to\mathbb{D}^{2}$ be given by $k(\zeta)=(\zeta,\zeta)$
for $\zeta\in\mathbb{D}$.

Consider the royal tangent $\delta$ of equation (3.2) and let

$\delta_{zc}=\left((z,z),(c,c)\right)\in T\mathbb{D}^{2}.$

Observe that

$\pi^{\prime}(\lambda)=\begin{bmatrix}1&1\\\
\lambda^{2}&\lambda^{1}\end{bmatrix},$

and so

(3.4)
$\pi_{*}(\delta_{zc})=\left(\pi(z,z),\pi^{\prime}(z,z)(c,c)\right)=\left((2z,z^{2}),2c(1,z)\right)=\delta,$

while

$k_{*}((z,c))=(k(z),k^{\prime}(z)c)=\left((z,z),(c,c)\right)=\delta_{zc}.$

Consider any $\varphi\in\mathbb{D}(G)$. Figure 1 illustrates the situation.

Figure 1.

It is known that every Carathéodory problem on the bidisc is solved by one of
the two co-ordinate functions $F_{j}(\lambda)=\lambda^{j}$ for $j=1$ or $2$
(for a proof see, for example, [2, Theorem 2.3]). Thus

$\displaystyle|\delta_{zc}|_{\rm car}^{\mathbb{D}^{2}}$
$\displaystyle=\max_{j=1,2}\frac{|D_{(c,c)}F_{j}(z,z)|}{1-|F_{j}(z,z)|^{2}}$
$\displaystyle=\frac{|c|}{1-|z|^{2}}$ $\displaystyle=|\delta|_{\rm car}.$

Here of course the superscript $\mathbb{D}^{2}$ indicates the Carathéodory
extremal problem on the bidisc.

Hence, for $\varphi\in\mathbb{D}(G)$,

$\displaystyle\varphi\circ\pi\mbox{ solves
}\operatorname{{\mathrm{C}ar}}\delta_{zc}$
$\displaystyle\iff|(\varphi\circ\pi)_{*}(\delta_{zc})|=\frac{|c|}{1-|z|^{2}}$
$\displaystyle\iff|\varphi_{*}\circ\pi_{*}(\delta_{zc})|=\frac{|c|}{1-|z|^{2}}\quad\mbox{
by the chain rule}$
$\displaystyle\iff|\varphi_{*}(\delta)|=\frac{|c|}{1-|z|^{2}}\quad\hskip
34.14322pt\mbox{ by equation \eqref{dedezc}}$ (3.5)
$\displaystyle\iff\varphi\mbox{ solves }\operatorname{{\mathrm{C}ar}}\delta.$

Next observe that a function $\psi\in\mathbb{D}(\mathbb{D}^{2})$ solves
$\operatorname{{\mathrm{C}ar}}\delta_{zc}$ if and only if $\psi\circ k$ is an
automorphism of $\mathbb{D}$. For if $\psi\circ k$ is an automorphism of
$\mathbb{D}$ then it satisfies

$|(z,c)|=|(\psi\circ k)_{*}(z,c)|=|\psi_{*}\circ
k_{*}(z,c)|=|\psi_{*}(\delta_{zc})|,$

which is to say that $\psi$ solves $\operatorname{{\mathrm{C}ar}}\delta_{zc}$.
Conversely, if $\psi$ solves $\operatorname{{\mathrm{C}ar}}\delta_{zc}$ then
$\psi\circ k$ is an analytic self-map of $\mathbb{D}$ that preserves the
Poincaré metric of a nondegenerate tangent to $\mathbb{D}$, and is therefore
(by the Schwarz-Pick lemma) an automorphism of $\mathbb{D}$. On combining this
observation with equivalence (3) we deduce that

$\displaystyle\varphi\mbox{ solves }\operatorname{{\mathrm{C}ar}}\delta\iff$
$\displaystyle\mbox{ there exists an automorphism }m\mbox{ of }\mathbb{D}$
(3.6) $\displaystyle\mbox{ such that }m^{-1}\circ\varphi\circ\pi\circ
k=\mathrm{id}_{\mathbb{D}}.$

For a function $f\in\mathbb{D}(\mathbb{D}^{2})$, it is easy to see that
$f\circ k=\mathrm{id}_{\mathbb{D}}$ if and only if $f$ solves the Nevanlinna-
Pick problem

(3.7) $(0,0)\mapsto
0,\qquad({\tfrac{1}{2}},{\tfrac{1}{2}})\mapsto{\tfrac{1}{2}}.$

See [3, Subsection 11.5] for the Nevanlinna-Pick problem in the bidisc. Hence

$\displaystyle\varphi\mbox{ solves }\operatorname{{\mathrm{C}ar}}\delta$
$\displaystyle\iff\mbox{ there exists an automorphism }m\mbox{ of
}\mathbb{D}\mbox{ such that }$ (3.8) $\displaystyle
m^{-1}\circ\varphi\circ\pi\mbox{ solves the Nevanlinna-Pick problem
\eqref{NPD2}.}$

In [3, Subsection 11.6] Agler and McCarthy use realization theory to show the
following.

A function $f\in\mathcal{S}(\mathbb{D}^{2})$ satisfies the interpolation
conditions

(3.9) $f(0,0)=0,\qquad f({\tfrac{1}{2}},{\tfrac{1}{2}})={\tfrac{1}{2}}$

if and only if there exist $t\in[0,1]$ and $\Theta$ in the Schur class of the
bidisc such that, for all $\lambda\in\mathbb{D}^{2}$,

(3.10)
$f(\lambda)=t\lambda^{1}+(1-t)\lambda^{2}+t(1-t)(\lambda^{1}-\lambda^{2})^{2}\frac{\Theta(\lambda)}{1-[(1-t)\lambda^{1}+t\lambda^{2}]\Theta(\lambda)}.$

Inspection of the formula (3.10) reveals that $f$ is symmetric if and only if
$t={\tfrac{1}{2}}$ and $\Theta$ is symmetric. Hence the symmetric functions in
$\mathcal{S}(\mathbb{D}^{2})$ that satisfy the conditions (3.9) are those
given by

(3.11) $\displaystyle f(\lambda)$
$\displaystyle={\tfrac{1}{2}}\lambda^{1}+{\tfrac{1}{2}}\lambda^{2}+\tfrac{1}{4}(\lambda^{1}-\lambda^{2})^{2}\frac{\Theta(\lambda)}{1-{\tfrac{1}{2}}(\lambda^{1}+\lambda^{2})\Theta(\lambda)}$

for some symmetric $\Theta\in\mathcal{S}(\mathbb{D}^{2})$. Such a $\Theta$
induces a unique function $\Psi\in\mathcal{S}(G)$ such that
$\Theta=\Psi\circ\pi$, and we may write the symmetric solutions $f$ of the
problem (3.9) in the form $f=\tilde{f}\circ\pi$ where, for all
$s=(s^{1},s^{2})$ in $G$,

(3.12)
$\tilde{f}(s)={\tfrac{1}{2}}s^{1}+\tfrac{1}{4}((s^{1})^{2}-4s^{2})\frac{\Psi(s)}{1-{\tfrac{1}{2}}s^{1}\Psi(s)}.$

Let $\varphi$ solve $\operatorname{{\mathrm{C}ar}}\delta$. By the equivalence
(3), there exists an automorphism $m$ of $\mathbb{D}$ such that
$m^{-1}\circ\varphi\circ\pi$ solves the Nevanlinna-Pick problem (3.7). Clearly
$m^{-1}\circ\varphi\circ\pi$ is symmetric. Hence there exists
$\Psi\in\mathcal{S}(G)$ such that, for all $\lambda\in\mathbb{D}^{2}$,

(3.13)
$m^{-1}\circ\varphi(s)={\tfrac{1}{2}}s^{1}+\tfrac{1}{4}((s^{1})^{2}-4s^{2})\frac{\Psi(s)}{1-{\tfrac{1}{2}}s^{1}\Psi(s)}.$

Thus $\varphi$ is indeed given by the formula (3.3).

Conversely, suppose that for some automorphism $m$ of $\mathbb{D}$ and
$\Psi\in\mathcal{S}(G)$, a function $\varphi$ is defined by equation (3.3).
Let $f=m^{-1}\circ\varphi\circ\pi$ Then $f$ is given by the formula (3.11),
where $\Theta=\Psi\circ\pi$. Hence $f$ is a symmetric function that satisfies
the interpolation conditions (3.9). By the equivalence (3), $\varphi$ solves
$\operatorname{{\mathrm{C}ar}}\delta$. ∎

##  4\. Flat tangents

In this section we shall give a description of a large class of Carathéodory
extremals for a flat tangent. Recall that a flat tangent has the form

(4.1) $\delta=\left((\beta+\bar{\beta}z,z),c(\bar{\beta},1)\right)$

for some $z\in\mathbb{D}$ and $c\neq 0$, where $\beta\in\mathbb{D}$. Such a
tangent touches the ‘flat geodesic’

$F_{\beta}\stackrel{{\scriptstyle\rm
def}}{{=}}\\{(\beta+\bar{\beta}w,w):w\in\mathbb{D}\\}.$

The description depends on a remarkable property of sets of the form
$\mathcal{R}\cup F_{\beta},\ \beta\in\mathbb{D}$: they have the norm-
preserving extension property in $G$ [2, Theorem 10.1]. That is, if $g$ is any
bounded analytic function on the variety $\mathcal{R}\cup F_{\beta}$, then
there exists an analytic function $\tilde{g}$ on $G$ such that
$g=\tilde{g}|\mathcal{R}\cup F_{\beta}$ and the supremum norms of $g$ and
$\tilde{g}$ coincide. Indeed, the proof of [2, Theorem 10.1] gives an explicit
formula for one such $\tilde{g}$ in terms of a Herglotz-type integral. Let us
call the norm-preserving extension $\tilde{g}$ of $g$ constructed in [2,
Chapter 10] the special extension of $g$ to $G$.

It is a simple calculation to show that $\mathcal{R}$ and $F_{\beta}$ have a
single point in common.

By equation (1.2), for $\delta$ in equation (4.1)

$|\delta|_{\rm car}=\frac{|c|}{1-|z|^{2}}.$

######  Theorem 4.1.

Let $\delta$ be the flat tangent

(4.2) $\delta=\left((\beta+\bar{\beta}z,z),c(\bar{\beta},1)\right)$

to $G$, where $\beta\in\mathbb{D}$ and $c\in\mathbb{C}\setminus\\{0\\}$. Let
$\zeta,\eta$ be the points in $\mathbb{D}$ such that

$(2\zeta,\zeta^{2})=(\beta+\bar{\beta}\eta,\eta)\in\mathcal{R}\cap F_{\beta}$

and let $m$ be the unique automorphism of $\mathbb{D}$ such that

$m_{*}((z,c))=(0,|\delta|_{\rm car}).$

For every function $h\in\mathcal{S}(\mathbb{D})$ such that $h(\zeta)=m(\eta)$
the special extension $\tilde{g}$ to $G$ of the function

(4.3) $g:\mathcal{R}\cup F_{\beta}\to\mathbb{D},\quad(2w,w^{2})\mapsto
h(w),\quad(\beta+\bar{\beta}w,w)\mapsto m(w)$

for $w\in\mathbb{D}$ is a well-aligned Carathéodory extremal function for
$\delta$.

###### Proof.

First observe that there is indeed a unique automorphism $m$ of $\mathbb{D}$
such that $m_{*}((z,c))=(0,|\delta|_{\rm car})$, by the Schwarz-Pick Lemma.
Let

$k(w)=(\beta+\bar{\beta}w,w)\quad\mbox{ for }w\in\mathbb{D},$

so that $F_{\beta}=k(\mathbb{D})$ and $k_{*}((z,c))=\delta$. By the definition
(4.3) of $g$, $g\circ k=m$.

Consider any function $h\in\mathcal{S}(\mathbb{D})$ such that
$h(\zeta)=m(\eta)$. By [2, Lemma 10.5], the function $g$ defined by equations
(4.3) is analytic on $\mathcal{R}\cup F_{\beta}$.

We claim that the special extension $\tilde{g}$ of $g$ to $G$ is a well-
aligned Carathéodory extremal function for $\delta$. By [2, Theorem 10.1],
$\tilde{g}\in\mathbb{D}(G)$. Moreover

$\displaystyle(\tilde{g})_{*}(\delta)$ $\displaystyle=(\tilde{g})_{*}\circ
k_{*}((z,c))$ $\displaystyle=(\tilde{g}\circ k)_{*}((z,c))$
$\displaystyle=(g\circ k)_{*}((z,c))$ $\displaystyle=m_{*}((z,c))$
$\displaystyle=(0,|\delta|_{\rm car})$

as required. Thus the Poincaré metric of $(\tilde{g})_{*}(\delta)$ on
$T\mathbb{D}$ is

$|(\tilde{g})_{*}(\delta)|=|(0,|\delta|_{\rm car})|=|\delta|_{\rm car}.$

Therefore $(\tilde{g})_{*}$ is a well aligned Carathéodory extremal function
for $\delta$. ∎

Clearly the map $g\mapsto\tilde{g}$ is injective, and so this procedure yields
a large class of Carathéodory extremals for $\delta$, parametrized by the
Schur class.

######  Remark 4.2.

In the converse direction, if $\varphi$ is any well-aligned Carathéodory
extremal for $\delta$, then $\varphi$ is a norm-preserving extension of its
restriction to $\mathcal{R}\cup F_{\beta}$, which is a function of the type
(4.3). Thus the class of all well-aligned Carathéodory extremal functions for
$\delta$ is given by the set of norm-preserving analytic extensions to $G$ of
$g$ in equation (4.3), as $h$ ranges over functions in the Schur class taking
the value $m(\eta)$ at $\zeta$. Typically there will be many such extensions
of $g$, as can be seen from the proof of [2, Theorem 10.1]. An extension is
obtained as the Cayley transform of a function defined by a Herglotz-type
integral with respect to a probability measure $\mu$ on $\mathbb{T}^{2}$. In
the proof of [2, Lemma 10.8], $\mu$ is chosen to be the product of two
measures $\mu_{\mathcal{R}}$ and $\mu_{\mathcal{F}}$ on $\mathbb{T}$;
examination of the proof shows that one can equally well choose any measure
$\mu$ on $\mathbb{T}^{2}$ such that

$\mu(A\times\mathbb{T})=\mu_{\mathcal{R}}(A),\quad\mu(\mathbb{T}\times
A)=\mu_{\mathcal{F}}(A)\quad\mbox{ for all Borel sets }A\mbox{ in
}\mathbb{T}.$

Thus each choice of $h\in\mathcal{S}(\mathbb{D})$ satisfying
$h(\zeta)=m(\eta)$ can be expected to give rise to many well-aligned
Carathéodory extremals for $\delta$.

##  5\. Purely balanced tangents

In this section we find a large class of Carathéodory extremals for purely
balanced tangents in $G$ by exploiting an embedding of $G$ into the bidisc.

######  Lemma 5.1.

Let

$\Phi=(\Phi_{\omega_{1}},\Phi_{\omega_{2}}):G\to\mathbb{D}^{2}$

where $\omega_{1},\omega_{2}$ are distinct points in $\mathbb{T}$. Then $\Phi$
is an injective map from $G$ to $\mathbb{D}^{2}$.

###### Proof.

Suppose $\Phi$ is not injective. Then there exist distinct points
$(s^{1},s^{2})$, $(t^{1},t^{2})\in G$ such that
$\Phi_{\omega_{j}}(s^{1},s^{2})=\Phi_{\omega_{j}}(t^{1},t^{2})$ for $j=1,2$.
On expanding and simplifying this relation we deduce that

$s^{1}-t^{1}-2\omega_{j}(s^{2}-t^{2})-\omega_{j}^{2}(s^{1}t^{2}-t^{1}s^{2})=0.$

A little manipulation demonstrates that both $(s^{1},s^{2})$ and
$(t^{1},t^{2})$ lie on the complex line

$\ell\stackrel{{\scriptstyle\rm
def}}{{=}}\\{(s^{1},s^{2})\in\mathbb{C}^{2}:(\omega_{1}+\omega_{2})s^{1}-2\omega_{1}\omega_{2}s^{2}=2\\}.$

However, $\ell$ does not meet $G$. For suppose that $(s^{1},s^{2})\in\ell\cap
G$. Then there exists $\beta\in\mathbb{D}$ such that

$\displaystyle s^{1}$ $\displaystyle=\beta+\bar{\beta}s^{2},$ $\displaystyle
2\omega_{1}\omega_{2}s^{2}$
$\displaystyle=(\omega_{1}+\omega_{2})s^{1}-2=(\omega_{1}+\omega_{2})(\beta+\bar{\beta}s^{2})-2.$

On solving the last equation for $s^{2}$ we find that

$s^{2}=-\bar{\omega}_{1}\bar{\omega}_{2}\frac{2-(\omega_{1}+\omega_{2})\beta}{2-(\bar{\omega}_{1}+\bar{\omega}_{2})\bar{\beta}},$

whence $|s^{2}|=1$, contrary to the hypothesis that $(s^{1},s^{2})\in G$.
Hence $\Phi$ is injective on $G$. ∎

######  Remark 5.2.

$\Phi$ has an analytic extension to the set
$\Gamma\setminus\\{(2\bar{\omega}_{1},\bar{\omega}_{1}^{2}),(2\bar{\omega}_{2},\bar{\omega}_{2}^{2})\\}$,
where $\Gamma$ is the closure of $G$ in $\mathbb{C}^{2}$. However this
extension is not injective: it takes the constant value
$(-\bar{\omega}_{2},-\bar{\omega}_{1})$ on a curve lying in $\partial G$.

######  Theorem 5.3.

Let $\delta=(\lambda,v)$ be a purely balanced tangent to $G$ and let
$\Phi_{\omega}$ solve $\operatorname{{\mathrm{C}ar}}\delta$ for the two
distinct points $\omega_{1},\omega_{2}\in\mathbb{T}$. Let $m_{j}$ be the
automorphism of $\mathbb{D}$ such that $m_{j}\circ\Phi_{\omega_{j}}$ is well
aligned at $\delta$ for $j=1,2$ and let

(5.1)
$\Phi=(\Phi^{1},\Phi^{2})=(m_{1}\circ\Phi_{\omega_{1}},m_{2}\circ\Phi_{\omega_{2}}):G\to\mathbb{D}^{2}.$

For every $t\in[0,1]$ and every function $\Theta$ in the Schur class of the
bidisc the function

$\displaystyle F$ $\displaystyle=t\Phi^{1}+(1-t)\Phi^{2}+$ (5.2)
$\displaystyle\hskip
28.45274ptt(1-t)(\Phi^{1}-\Phi^{2})^{2}\frac{\Theta\circ\Phi}{1-[(1-t)\Phi^{1}+t\Phi^{2}]\Theta\circ\Phi}$

is a well-aligned Carathéodory extremal function for $\delta$.

###### Proof.

By Lemma 5.1, $\Phi$ maps $G$ injectively into $\mathbb{D}^{2}$. By choice of
$m_{j}$,

$(m_{j}\circ\Phi_{\omega_{j}})_{*}(\delta)=(0,|\delta|_{\rm car}).$

Hence

$\Phi_{*}(\delta)=\left((0,0),|\delta|_{\rm car}(1,1)\right),$

which is tangent to the diagonal $\\{(w,w):w\in\mathbb{D}\\}$ of the bidisc.
Since the diagonal is a complex geodesic in $\mathbb{D}^{2}$, we have

$|\Phi_{*}(\delta)|_{\rm car}=(0,|\delta|_{\rm car}).$

As in Section 3, we appeal to [3, Subsection 11.6] to assert that, for every
$t\in[0,1]$ and every function $\Theta$ in the Schur class of the bidisc, the
function $f\in\mathbb{C}(\mathbb{D}^{2})$ given by

(5.3)
$f(\lambda)=t\lambda^{1}+(1-t)\lambda^{2}+t(1-t)^{2}\frac{\Theta(\lambda)}{1-[(1-t)\lambda^{1}+t\lambda^{2}]\Theta(\lambda)}$

solves $\operatorname{{\mathrm{C}ar}}(\Phi_{*}(\delta))$. For every such $f$
the function $F\stackrel{{\scriptstyle\rm def}}{{=}}f\circ\Phi:G\to\mathbb{D}$
satisfies

$F_{*}(\delta)=(f\circ\Phi)_{*}(\delta)=f_{*}(\Phi_{*}(\delta))=(0,|\delta|_{\rm
car}).$

Thus $F$ is a well-aligned Carathéodory extremal for $\delta$. On writing out
$F$ using equation (5.3) we obtain equation (5.3). ∎

######  Remark 5.4.

The range of $\Phi$ is a subset of $\mathbb{D}^{2}$ containing $(0,0)$ and is
necessarily nonconvex, by virtue of a result of Costara [8] to the effect that
$G$ is not isomorphic to any convex domain. $\Phi(G)$ is open in
$\mathbb{D}^{2}$, since the Jacobian determinant of
$(\Phi_{\omega_{1}},\Phi_{\omega_{2}})$ at $(s^{1},s^{2})$ is

$\frac{4(\omega_{1}-\omega_{2})(1-\omega_{1}\omega_{2}s^{2})}{(2-\omega_{1}s^{1})^{2}(2-\omega_{2}s^{1})^{2}}$

which has no zero in $G$. Carathéodory extremals $F$ given by equation (5.3)
have the property that the map $F\circ\Phi^{-1}$ on $\Phi(G)$ extends
analytically to a map in $\mathbb{D}(\mathbb{D}^{2})$. There may be other
Carathéodory extremals $\varphi$ for $\delta$ for which
$\varphi\circ\Phi^{-1}$ does not so extend. Accordingly we do not claim that
the Carathéodory extremals described in Theorem 5.3 constitute all extremals
for a purely balanced tangent.

##  6\. Relation to a result of L. Kosiński and W. Zwonek

Our main result in Section 2, on the essential uniqueness of solutions of
$\operatorname{{\mathrm{C}ar}}\delta$ for purely unbalanced and exceptional
tangents, can be deduced from [15, Theorem 5.3] and some known facts about the
geometry of $G$. However, the terminology and methods of Kosiński and Zwonek
are quite different from ours, and we feel it is worth explaining their
statement in our terminology.

Kosiński and Zwonek speak of left inverses of complex geodesics where we speak
of Carathéodory extremal functions for nondegenerate tangents. These are
essentially equivalent notions. By a complex geodesic in $G$ they mean a
holomorphic map from $\mathbb{D}$ to $G$ which has a holomorphic left inverse.
Two complex geodesics $h$ and $k$ are equivalent if there is an automorphism
$m$ of $\mathbb{D}$ such that $h=k\circ m$, or, what is the same, if
$h(\mathbb{D})=k(\mathbb{D})$. It is known (for example [4, Theorem A.10])
that, for every nondegenerate tangent $\delta$ to $G$, there is a unique
complex geodesic $k$ of $G$ up to equivalence such that $\delta$ is tangent to
$k(\mathbb{D})$. A function $\varphi\in\mathbb{D}(G)$ solves
$\operatorname{{\mathrm{C}ar}}\delta$ if and only if $\varphi\circ k$ is an
automorphism of $\mathbb{D}$. Hence, for any complex geodesic $k$ and any
nondegenerate tangent $\delta$ to $k(\mathbb{D})$, to say that $k$ has a
unique left inverse up to equivalence is the same as to say that
$\operatorname{{\mathrm{C}ar}}\delta$ has an essentially unique solution.

Kosiński and Zwonek also use a different classification of types of complex
geodesics (or equivalently tangent vectors) in $G$, taken from [16]. There it
is shown that every complex geodesic $k$ in $G$, up to composition with
automorphisms of $\mathbb{D}$ on the right and of $G$ on the left, is of one
of the following types.

  1. (1)

$k(z)=(B(\sqrt{z})+B(-\sqrt{z}),B(\sqrt{z})B(-\sqrt{z})$

where $B$ is a non-constant Blaschke product of degree $1$ or $2$ satisfying
$B(0)=0$;

  2. (2)

$k(z)=(z+m(z),zm(z))$

where $m$ is an automorphism of $\mathbb{D}$ having no fixed point in
$\mathbb{D}$.

These types correspond to our terminology from [2] (or from Section 1) in the
following way. Recall that an automorphism of $\mathbb{D}$ is either the
identity, elliptic, parabolic or hyperbolic, meaning that the set
$\\{z\in\mathbb{D}^{-}:m(z)=z\\}$ consists of either all of $\mathbb{D}^{-}$,
a single point of $\mathbb{D}$, a single point of $\mathbb{T}$ or two points
in $\mathbb{T}$.

  1. (1a)

If $B$ has degree $1$, so that $B(z)=cz$ for some $c\in\mathbb{T}$ then, up to
equivalence, $k(z)=(0,-c^{2}z)$. These we call the flat geodesics. The general
tangents to flat geodesics are the flat tangents described in Section 1, that
is $\delta=\left((\beta+\bar{\beta}z,z),c(\bar{\beta},1)\right)$ for some
$\beta\in\mathbb{D},\,z\in\mathbb{D}$ and nonzero $c\in\mathbb{C}$.

  2. (1b)

If $B(z)=cz^{2}$ for some $c\in\mathbb{T}$ then $k(z)=(2cz,c^{2}z^{2})$. Thus
$k(\mathbb{D})$ is the royal variety $\mathcal{R}$, and the tangents to
$k(\mathbb{D})$ are the royal tangents.

  3. (1c)

If $B$ has degree $2$ but is not of the form (1b), say
$B(z)=cz(z-\alpha)/(1-\bar{\alpha}z)$ where $c\in\mathbb{T}$ and
$\alpha\in\mathbb{D}\setminus\\{0\\}$, then

$k(z)=\frac{\left(2c(1-|\alpha|^{2})z,c^{2}z(z-\alpha^{2})\right)}{1-\bar{\alpha}^{2}z}.$

Here $k(\mathbb{D})$ is not $\mathcal{R}$ but it meets $\mathcal{R}$ (at the
point $(0,0)$). It follows that $k(\mathbb{D})$ is a purely unbalanced
geodesic and the tangents to $k(\mathbb{D})$ are the purely unbalanced
tangents.

  4. (2a)

If $m$ is a hyperbolic automorphism of $\mathbb{D}$ then $k(\mathbb{D})$ is a
purely balanced geodesic and its tangents are purely balanced tangents.

  5. (2b)

If $m$ is a parabolic automorphism of $\mathbb{D}$ then $k(\mathbb{D})$ is an
exceptional geodesic, and its tangents are exceptional tangents.

With this description, Theorem 5.3 of [15] can be paraphrased as stating that
a complex geodesic $k$ of $G$ has a unique left inverse (up to equivalence) if
and only if $k$ is of one of the forms (1c) or (2b). These are precisely the
purely unbalanced and exceptional cases in our terminology, that is, the cases
of tangents $\delta$ for which there is a unique $\omega\in\mathbb{T}$ such
that $\Phi_{\omega}$ solves $\operatorname{{\mathrm{C}ar}}\delta$, in
agreement with our Theorem 2.1.

The authors prove their theorem with the aid of a result of Agler and McCarthy
on the uniqueness of solutions of $3$-point Nevanlinna-Pick problems on the
bidisc [3, Theorem 12.13]. They also use the same example from Subsection 11.6
of [3] which we use for different purposes in Sections 3 and 5.

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