# A note on string size evolution in phantom cosmology

Soon-Tae Hong  Department of Science Education and Research Institute for
Basic Sciences,
Ewha Womans University, Seoul 120-750 Korea

###### Abstract

We analyze evolution of string size in higher-dimensional cosmology with
phantom field. Assuming that the Universe possesses the phantom field defined
in a ten-dimensional spacetime, we predict string size which is proposed to be
that of photon in nature at present. The Universe size increases as in the
standard inflationary Universe model while the photon size decreases
drastically at the early stage of the string evolution after the Big Bang.
Moreover, the photon spin in the phantom Universe is analyzed in the framework
of the stringy cosmology.

string size; higher dimensional cosmology; phantom field; photon spin

###### pacs:

14.70.Bh, 11.25.-w, 04.60.Cf, 04.50.-h, 11.10.Kk, 98.80.-k, 98.80.Es, 04.20.Jb

As it is well known, the particles in the string theory are supposed to be
extended objects witten87 ; polchinski98 . Via the string version of the
Hawking-Penrose singularity theorem hawking70 , the stringy cosmology was
proposed hong11  with a success that, with the extended string particle, one
describes precisely the Universe expansion after the Big Bang. Moreover, in
the stringy cosmology model, one has an advantage that the degrees of freedom
of the rotation and the shear are introduced naturally. In phantom cosmology
caldwell02 ; lidsey04 ; faraoni05 ; hong08 ; chimento08 ; kaeonikhom11 ; wei11
; yang12 ; astashenok12 ; novosyadlyj12 , the Universe expansion up to the
present Universe age has been studied. Motivated by this, we will investigate
the validity of the string size evolution in higher-dimensional spacetime with
phantom field. One notes that the ten-dimensional phantom cosmology has a
prominent feature which analyzes string size in itself.

The model is defined by the Lagrangian with ten-dimensional indices and
cosmological constant

${\cal L}=\sqrt{-g}\left(\frac{1}{2}R+\frac{1}{2}\alpha
g^{MN}\partial_{M}^{~{}}\sigma\partial_{N}^{~{}}\sigma-\Lambda\right),$ (1)

where $\sigma$ is the phantom field and $\alpha$ is shorthand defined as
$\alpha=8\pi G/c^{4}$. Variation of the Lagrangian with respect to the metric
yields the Einstein equations

$R_{MN}-\frac{1}{2}g_{MN}^{~{}}R+g_{MN}^{~{}}\Lambda=\alpha T_{MN}^{~{}},$ (2)

where the energy-stress tensor is given by

$T_{MN}^{~{}}=-\partial_{M}^{~{}}\sigma\partial_{N}^{~{}}\sigma+\frac{1}{2}g_{MN}^{~{}}\partial_{Q}^{~{}}\sigma\partial^{Q}\sigma.$
(3)

We take an ansatz for the metric

$g_{MN}^{~{}}={\rm diag}~{}(-1,a^{2}(t),b^{2}(t)),$ (4)

where $a(t)$ is the scale factor of the three-dimensional Universe and $b(t)$
is the scale factor of the extra six dimensions associated with the internal
torus space torus .

From (2), we find differential equations with the overdots denoting the
derivatives with respect to time,

$\displaystyle 3\frac{\ddot{a}}{a}+6\frac{\ddot{b}}{b}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\Lambda+\alpha\dot{\sigma}^{2},$
$\displaystyle\frac{\ddot{a}}{a}+2\frac{\dot{a}^{2}}{a^{2}}+6\frac{\dot{a}}{a}\cdot\frac{\dot{b}}{b}$
$\displaystyle=$ $\displaystyle\frac{1}{4}\Lambda,$
$\displaystyle\frac{\ddot{b}}{b}+5\frac{\dot{b}^{2}}{b^{2}}+3\frac{\dot{a}}{a}\cdot\frac{\dot{b}}{b}$
$\displaystyle=$ $\displaystyle\frac{1}{4}\Lambda,$ (5)

whose solutions are given by

$\displaystyle a$ $\displaystyle=$ $\displaystyle a_{0}~{}{\rm
exp}~{}\left[\frac{1}{6}\sqrt{\Lambda}ct+\frac{4n}{9}(1-e^{-\frac{3}{2}\sqrt{\Lambda}ct})\right],$
$\displaystyle b$ $\displaystyle=$ $\displaystyle b_{0}~{}{\rm
exp}~{}\left[\frac{1}{6}\sqrt{\Lambda}ct-\frac{2n}{9}(1-e^{-\frac{3}{2}\sqrt{\Lambda}ct})\right],$
(6)

where $a_{0}$ and $b_{0}$ are the initial values of $a$ and $b$ at $t=0$,
respectively, and $n$ is a model parameter to be fixed later. Up to now, we
briefly recapitulate the set-up of the phantom cosmology scheme in Ref. hong08
.

Now, we proceed to construct further novel stringy phenomenology using the
phantom cosmology. To do this, we take an ansatz that, after the Planck time
$t_{Planck}=5.39\times 10^{-44}$ second near the Big Bang, the sizes of the
Universe and the string are the same each other so that one can choose $a_{0}$
and $b_{0}$ to become the Planck length carroll04

$a_{0}=b_{0}=l_{Planck}=1.62\times 10^{-33}~{}{\rm cm}.$ (7)

We exploit the relation

$\displaystyle\Lambda$ $\displaystyle=$ $\displaystyle 12\Lambda_{*},$
$\displaystyle\Lambda_{*}$ $\displaystyle=$ $\displaystyle 2.07\times
10^{-56}~{}{\rm cm}^{-2},$ (8)

where $\Lambda_{*}$ is the four-dimensional cosmological constant. We next
define a dimensionless variable $x$ as

$\displaystyle t$ $\displaystyle=$ $\displaystyle xt_{*},$ $\displaystyle
t_{*}$ $\displaystyle=$ $\displaystyle 4.28\times 10^{17}~{}{\rm sec},$ (9)

with $t_{*}$ being the age of the Universe hartnett . Inserting $\Lambda$ in
(8) and $t$ in (9) into (6), we obtain

$\displaystyle a(x)$ $\displaystyle=$ $\displaystyle a_{0}~{}{\rm
exp}\left[1.07x+\frac{4n}{9}(1-e^{-9.61x})\right],$ (10) $\displaystyle b(x)$
$\displaystyle=$ $\displaystyle b_{0}~{}{\rm
exp}\left[1.07x-\frac{2n}{9}(1-e^{-9.61x})\right].$ (11)

Inserting the size of the our present Universe at $x=1$ davis05

$a_{*}=4.35\times 10^{28}~{}{\rm cm},$ (12)

into (10), we fix the value of $n$ to be $n=315.86$. With the above value of
$n$, we obtain

$\displaystyle a(x)$ $\displaystyle=$ $\displaystyle a_{0}~{}{\rm
exp}\left[1.07x+140.38(1-e^{-9.61x})\right],$ (13) $\displaystyle b(x)$
$\displaystyle=$ $\displaystyle b_{0}~{}{\rm
exp}\left[1.07x-70.19(1-e^{-9.61x})\right].$ (14)

The result in (14) is a formula which describes the string size evolution.

In particular, at $x=1$ in (14), we arrive at the size $b_{*}$ of the present
string:

$b_{*}=1.56\times 10^{-63}~{}{\rm cm}.$ (15)

Here, we claim that the size in (15) is that of photon in nature at present,
if we assume that our Universe possesses the phantom field, and that it is
defined in the higher-dimensional manifold with the extra dimensions as in the
standard string theory witten87 ; polchinski98 . According to the stringy
cosmology hong11 , the photon has a spin structure, due to its rotation whose
magnitude is assumed to be the same as that of the rotating Universe itself.
Here one notes that the Universe consists of celestial objects such as
galaxies, stars and planets. If we combine the results on the photon rotation
from the stringy cosmology with those of the phantom scenario, the photon has
an extremely tiny size with the spin degree of freedom. The explanation about
the photon spin phenomenology seems to be consistent with the corresponding
experiment. Here, we predict additionally the size of the photon stringy (not
point) particle in (15). Moreover, the above results on the Universe and the
string sizes indicate that, after the Universe age $t_{*}=4.28\times
10^{17}~{}{\rm sec}$ (strictly speaking after $t_{*}-t_{Planck}$), $a_{*}$
increases as in the standard inflationary Universe model, while $b_{*}$
decreases in contrast:

$\displaystyle\frac{a_{*}}{a_{0}}$ $\displaystyle=$ $\displaystyle 2.69\times
10^{61},$ $\displaystyle\frac{b_{*}}{b_{0}}$ $\displaystyle=$ $\displaystyle
9.63\times 10^{-31}.$ (16)

Figure 1: ${\rm log}_{10}[a(x)/a_{0}]$ in terms of the dimensionless time
variable $x$. Figure 2: ${\rm log}_{10}[b(x)/b_{0}]$ in terms of the
dimensionless time variable $x$.

In Fig. 1 and Fig. 2 we depict ${\log}_{10}[a(x)/a_{0}]$ and
${\log}_{10}[b(x)/b_{0}]$ in terms of the dimensionless time variable $x$
$(0\leq x\leq 1)$. Fig. 1 shows that the curve is rising up at the early stage
of the Universe evolution. Similarly, in Fig. 2 we find that the string size
is drastically falling down at the early stage of the string evolution. This
phenomenon is due to the fact that in (13) and (14) the last terms are
predominant in these curves. Moreover, these features about the Universe scale
factor seems to be consistent with those of the standard inflationary
cosmology.

In summary, we have assumed that the Universe has the phantom field in a ten-
dimensional spacetime, to predict the string size. We have claimed that the
predicted string size is the photon size in nature at present. The formulas
for the scale factors of the Universe and the string indicated that the
Universe size increases rapidly at the early stage of the evolution as in the
standard inflationary Universe model and the photon size decreases drastically
at the early stage of the evolution after the Big Bang. Moreover, the photon
spin has been investigated in the framework of the stringy cosmology with the
phantom field.

Even though the phantom field is known to be unstable carroll042 ; cline04 ;
faraoni11 , as an effective theory the phantom cosmology paradigm gives us
intriguing features as shown in this paper. Moreover, it is interesting to
note that the effective equation of state originated from the phantom field is
consistent with that of the cosmic fluid of the present superacceleration era
faraoni02 ; uzan00 ; faraoni11 ; hong08 .

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# Distance between reproducing kernel Hilbert spaces and geometry of finite
sets in the unit ball

Danny Ofek  School of Mathematical Sciences
Tel Aviv University
Tel Aviv 6997801
Israel<EMAIL_ADDRESS>,  Satish K. Pandey  Faculty of Mathematics
Technion - Israel Institute of Technology
Haifa 3200003
Israel<EMAIL_ADDRESS>http://noncommutative.space/ and
Orr Moshe Shalit  Faculty of Mathematics
Technion - Israel Institute of Technology
Haifa 3200003
Israel<EMAIL_ADDRESS>https://oshalit.net.technion.ac.il/

###### Abstract.

In this paper we study the relationships between a reproducing kernel Hilbert
space, its multiplier algebra, and the geometry of the point set on which they
live. We introduce a variant of the Banach-Mazur distance suited for measuring
the distance between reproducing kernel Hilbert spaces, that quantifies how
far two spaces are from being isometrically isomorphic as reproducing kernel
Hilbert spaces. We introduce an analogous distance for multiplier algebras,
that quantifies how far two algebras are from being completely isometrically
isomorphic. We show that, in the setting of finite dimensional quotients of
the Drury-Arveson space, two spaces are “close” to one another if and only if
their multiplier algebras are “close”, and that this happens if and only if
one of the underlying point sets is close to an image of the other under a
biholomorphic automorphism of the unit ball. These equivalences are obtained
as corollaries of quantitative estimates that we prove.

###### Key words and phrases:

Reproducing kernel Hilbert spaces, multiplier algebras, reproducing kernel
Banach-Mazur distance, multiplier Banach-Mazur distance

###### 2010 Mathematics Subject Classification:

46E22

The work of Satish K. Pandey is supported in part at the Technion by a
fellowship of the Israel Council for Higher Education. The work of O.M. Shalit
is partially supported by ISF Grant no. 195/16

##  1\. Introduction

The objective of this paper is to study the relationship between the structure
of a reproducing kernel Hilbert space (RKHS), the structure of its multiplier
algebra, and the geometry of the underlying set. We shall use freely basic
facts on RKHSs which can be found in standard introductions; see [AM02, PR16].
This continues a standing research program on the classification of complete
Pick spaces and algebras; see [APV03, ARS08, DHS15, DRS11, DRS15, Har12,
Har17, HL18, KMS13, MS17, OS20, SS16]. More generally, our study fits into a
general theme in the geometric study of RKHSs; see, e.g., [ARSW11, CD78,
McC96, Roc86] for a sample of results, or [Roc17, Roc19, Roc20, SSS18, SSS20]
for several recent examples in the commutative and the noncommutative setting.

###  1.1. Background on the isomorphism problem

For $d\in\mathbb{N}\cup\\{\infty\\}$, the Drury-Arveson space $H^{2}_{d}$ is
the RKHS on the open unit ball $\mathbb{B}_{d}$ of a complex $d$-dimensional
Hilbert space, determined by the kernel $k(z,w)=\frac{1}{1-\langle
z,w\rangle}$. We write $\mathcal{M}_{d}$ for the multiplier algebra
$\operatorname{Mult}(H^{2}_{d})$ of $H^{2}_{d}$. The space $H^{2}_{d}$ can be
considered as a module over $\mathcal{M}_{d}$ and plays a prominent role in
the theory of Hilbert modules (see the survey [Sha14]). A subspace of the form
$\mathcal{H}_{X}=\overline{\operatorname{span}}\\{k_{\lambda}:x\in
X\\}=H^{2}_{d}\big{|}_{X}$, where $X\subseteq\mathbb{B}_{d}$, is called a
quotient of $H^{2}_{d}$ (the reason for this terminology is that
$\mathcal{H}_{X}$ can be naturally identified with the quotient of $H^{2}_{d}$
by the submodule of functions vanishing on $X$).

By Theorem 4.2 in [AM00], every complete Pick RKHS is isometrically isomorphic
as an RKHS to (i.e., is a rescaling of) a quotient space of the Drury-Arveson
space. Thus, the study of complete Pick spaces and their multiplier algebras
can to a large extent be carried out for quotients of $H^{2}_{d}$. It follows
from the complete Pick property that
$\mathcal{M}_{X}:=\operatorname{Mult}(\mathcal{H}_{X})=\mathcal{M}_{d}\big{|}_{X}$.
By Proposition 2.2 in [DRS15], it suffices to consider only $X$ which are
multiplier varieties, which means that $X$ is the joint zero set of an ideal
of multipliers in $\mathcal{M}_{d}$.

With the comments in the last paragraph in mind, it is natural to ask in what
way exactly does a multiplier variety $V\subseteq\mathbb{B}_{d}$ determine the
structure of $\mathcal{H}_{V}$ and of $\mathcal{M}_{V}$. By Theorems 4.4 and
5.10 in [DRS15], $\mathcal{M}_{V}$ is isometrically isomorphic to
$\mathcal{M}_{W}$ if and only if $V$ and $W$ are congruent, meaning that there
is a biholomorphic automorphism $\phi$ of $\mathbb{B}_{d}$ such that
$\phi(V)=W$, and this happens if and only if the multiplier algebras are
completely isometrically isomorphic. The methods employed (e.g., [DRS15,
Proposition 4.1]) also show that $\mathcal{H}_{V}$ is isometrically isomorphic
as a RKHS to $\mathcal{H}_{W}$ if and only $V$ and $W$ are congruent.

Next, one can ask: if $\mathcal{M}_{V}$ and $\mathcal{M}_{W}$ are merely
isomorphic, how are then $V$ and $W$ related? By [DRS15, Theorem 5.6 ] under
some reasonable regularity assumptions on $V$ and $W$, an isomorphism from
$\mathcal{M}_{V}$ onto $\mathcal{M}_{W}$ implies the existence of a
biholomorphism from $W$ onto $V$. Moreover, by [DHS15, Theorem 6.2] this
biholomorphism must be bi-Lipschitz with respect to the pseudohyperbolic
metric. Under some additional assumptions, a biholomorphism from $W$ onto $V$
gives rise to an isomorphism from $\mathcal{M}_{V}$ to $\mathcal{M}_{W}$; this
has been proved, e.g., when the varieties are homogeneous [DRS11] or finite
Riemann surfaces [KMS13].

###  1.2. What this paper is about

This paper is strongly inspired by two recent papers of Rochberg, [Roc17] and
[Roc19]. In [Roc17], a certain metric induced by RKHSs is used to show that
the Dirichlet space is not a quotient of the Drury-Arveson space $H^{2}_{d}$
for finite $d$ (it is worth noting that the same methods can be modified to
show that the Dirichlet space is not even boundedly isomorphic to a quotient
of $H^{2}_{d}$ for finite $d$ [Har]). In [Roc19], the special case of complete
Pick spaces on finitely many points is carefully investigated; the natural
metric and additional geometric constructions are studied, and shown to hold
much information on the related function spaces. In particular it is shown in
[Roc19, Theorem 7] that if $X_{1}$ and $X_{2}$ are finite sets of points in
the unit ball, then $\mathcal{H}_{X_{1}}$ and $\mathcal{H}_{X_{2}}$ are
isometrically isomorphic as RKHSs if and only if $X_{1}$ and $X_{2}$ are
congruent.

In this paper we follow Rochberg in focusing on finite sets of points
$X_{1},X_{2}$ considered as metric spaces, but here we are most interested in
the question what happens when $X_{1}$ and $X_{2}$ are not exactly congruent,
but are “close” to being so. Are then $\mathcal{H}_{X_{1}}$ and
$\mathcal{H}_{X_{2}}$ forced to be, in some sense, “close” to being
isometrically isomorphic as reproducing kernel Hilbert spaces? And conversely,
if the function spaces are “close” to being isometrically isomorphic in some
sense, are the sets then “close” to be being biholomorphic images one of the
other, or “close” to being isometric? Similarly, we are interested in
analogous questions of how the multiplier algebras, rather than the RKHSs, are
determined by the underlying sets and their geometry, and how small changes in
the structure of the underlying sets are reflected in small changes in the
structure of the algebra.

In other words, this paper studies in a quantitative manner the relationship
between the the structure of function spaces and operator algebras on sets, on
the one hand, and the geometry of the underlying sets, on the other hand.

###  1.3. Main results

To present the main results of this paper we need to briefly introduce some
more terminology. Suppose that $\rho_{\rm{ph}}$ is the pseudohyperbolic metric
on the unit ball $\mathbb{B}_{d}$. This metric induces the Hausdorff metric
$\rho_{H}$ on subsets of the ball (see Definition 3.1), which in turn gives
rise to an automorphism invariant Hausdroff distance between subsets

$\tilde{\rho}_{H}(X,Y)=\inf\left\\{\rho_{H}(X,\Phi(Y)):\Phi\in\operatorname{Aut}(\mathbb{B}_{d})\right\\};$

see Definition 3.6. We introduce a distance function $\rho_{RK}$ which is
analogous to the Banach-Mazur distance, defined as follows

$\rho_{RK}(\mathcal{H}_{1},\mathcal{H}_{2})=\log\left(\inf\left\\{\|T\|\|T^{-1}\|:T:\mathcal{H}_{1}\to\mathcal{H}_{2}\textrm{
is an RKHS isomorphism}\right\\}\right).$

This distance function quantifies how far two spaces are from being
isometrically isomorphic as reproducing kernel Hilbert spaces. Similarly, we
introduce a distance function $\rho_{M}$, defined via a similar formula, that
quantifies how far two multiplier algebras are from being completely
isometrically isomorphic as multiplier algebras (see Definitions 4.1 and 4.2).
Most of this paper is concerned with studying the reltionships between these
distance functions.

Our principal theorem, Theorem 5.4, says roughly that for two subsets
$X=\\{x_{1},\ldots,x_{n}\\}$ and $Y=\\{y_{1},\ldots,y_{n}\\}$ of the unit ball
$\mathbb{B}_{d}$ we have that $\tilde{\rho}_{H}(X,Y)$ is small if and only if
$\rho_{RK}(\mathcal{H}_{X},\mathcal{H}_{Y})$ is small, and this happens if and
only if $\rho_{M}(\mathcal{M}_{X},\mathcal{M}_{Y})$ is small.

Theorem 5.4 is a direct consequence of explcit inequalities between the
different metrics, which are the heart of this paper. Sections 4 and 5 contain
the quantitative versions of each one of the implications in the theorem; see
Propositions 4.4, 4.7, 5.1 and 5.3. These provide us with quantitative
versions of results on the isomorphism problem that were summarized in
Subsection 1.1. We hope that such quantitative results will eventually help to
resolve open problems on isomorphisms of infinite dimensional quotients of the
Drury-Arveson space.

######  Acknowledgements.

We would like to thank the referee for numerous helpful suggestions to enhance
the organization of our ideas and improve the exposition of this article.

##  2\. Preliminaries

We begin by recalling several notions from the theory of reproducing kernel
Hilbert spaces. Let $\mathcal{H}$ be a reproducing kernel Hilbert space on a
set $X$ with reproducing kernel $K$. We use $\operatorname{Mult}(\mathcal{H})$
to denote the multiplier algebra of $\mathcal{H}$. By a _kernel function_ we
mean a two-variable function $K:X\times X\to\mathbb{C}$ on a nonempty set $X$
that satisfies the following property: for every $n\in\mathbb{N}$ and for
every choice of $n$ distinct points $\\{x_{1},...,x_{n}\\}\subseteq X$, the
matrix $[K(x_{i},x_{j})]$ is positive semidefinite. We will use the notation
$K\geq 0$ to denote that the function $K$ is a kernel function. If
$\mathcal{H}$ is an RKHS on $X$ with reproducing kernel $K$, then $K$ is,
indeed, a kernel function. The element $k_{y}=K(\cdot,y)\in\mathcal{H}$ is
called _the reproducing kernel at $y\in X$_.

###  2.1. Notions of isomorphism for reproducing kernel Hilbert spaces

In the study of reproducing kernel Hilbert spaces, it is useful to have
notions for comparing two spaces and deciding whether or not and in what sense
they are alike (see [MS17, Section 2] or [AM00, Section 2.6]). Since the kinds
of questions we ask are somewhat new, it is convenient for us to introduce
some terminology, which is not completely aligned with what finds in the
literature.

For $i=1,2$, let $\mathcal{H}_{i}$ be reproducing kernel Hilbert spaces
respectively on sets $X_{i}$ with reproducing kernels
$K_{i}(x,y)=k^{i}_{y}(x)$, and let
$\mathcal{M}_{i}=\operatorname{Mult}(\mathcal{H}_{i})$ be their multiplier
algebras. The space $\mathcal{H}_{2}$ is said to be a _rescaling_ of
$\mathcal{H}_{1}$ if there is a bijection $F:X_{1}\to X_{2}$ and a nowhere-
vanishing complex valued function $x\mapsto\lambda_{x}$ defined on $X_{1}$
such that

$K_{1}(x,y)=\overline{\lambda_{x}}\lambda_{y}K_{2}(F(x),F(y))\,\,\text{ for
all }\,\,x,y\in X_{1}.$

This happens if and only if there exists a unitary
$U:\mathcal{H}_{1}\to\mathcal{H}_{2}$ that maps each one-dimensional space
$\mathbb{C}k^{1}_{x}$ onto $\mathbb{C}k^{2}_{F(x)}$, that is, there exists a
nowhere-vanishing complex valued function $x\mapsto\lambda$ defined on $X_{1}$
such that

$Uk^{1}_{x}=\lambda_{x}k^{2}_{F(x)}\,\,\,\,\text{ for all }\,\,\,\,x\in
X_{1}.$

We shall refer to the unitary $U$ as an _isometric isomorphism of reproducing
kernel Hilbert spaces_ , and say that $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$
are _isometrically isomorphic as reproducing kernel Hilbert spaces_. The point
of this definition is that spaces which are isometrically isomorphic as RKHSs
can be considered to be the same in every respect: the unitary between them
preserves the function theoretic structure, in the sense that point
evaluations are sent to point evaluations. While there is always a unitary map
from one Hilbert space onto another of the same dimension, it is much harder
for two RKHSs to be isometrically isomorphic as RKHSs (simple examples of
pairs non isometrically isomorphic RKHSs of the same dimension follow very
concretely from Proposition 5.3 below).

We are interested in understanding a coarser and more flexible structure on
RKHSs. An isomorphism of reproducing kernel Hilbert spaces from
$\mathcal{H}_{1}$ to $\mathcal{H}_{2}$ (or simply an _RKHS isomorphism_) is a
bijective bounded linear map $T:\mathcal{H}_{1}\to\mathcal{H}_{2}$ defined by

(2.1) $T(k^{1}_{x})=\lambda_{x}k^{2}_{F(x)}\,\,,\,\,x\in X_{1},$

where $x\mapsto\lambda_{x}$ is a nowhere-vanishing complex valued function and
$F:X_{1}\to X_{2}$ is a bijection. If there is an RKHS isomorphism from
$\mathcal{H}_{1}$ to $\mathcal{H}_{2}$, then clearly there is an inverse which
too is an RKHS isomorphism, and we then say that $\mathcal{H}_{1}$ and
$\mathcal{H}_{2}$ are isomorphic as RKHSs, or simply that they are isomorphic.
The point is that the linear isomorphism between the spaces preserves point
evaluations, and hence the function theoretic structure.

It is perhaps worth to warn the reader that, as far as we know, this notion of
isomorphism between RKHSs — the one described above via Equation (2.1) — is
not commonly studied, and we were led to introduce it in order to avoid
ambiguity and maintain the consistency in our definitions so that we could
keep the logical flow of the paper smooth. In this regard, our terminologies
differ from other places. For instance, in [MS17, Section 2], due to reasons
explained therein, the phrase “isomorphic RKHSs” is used differenty and it
refers to what we call isometrically isomorphic RKHSs. We also stress that the
notion of “isometric isomorphism” as defined above is precisely the same as
that of “rescaling”; we prefer this terminology because we also need to
consider a weaker notion of isomorphism, the “RKHS isomorphism”, and this
choice helps us emphasize the difference.

###  2.2. Notions of isomorphism for multiplier algebras

Next, we want to consider what an isomorphism of multiplier algebras should
be. First, to be called a morphism between multplier algebras, a map
$\varphi:\mathcal{M}_{1}\to\mathcal{M}_{2}$ will be required to preserve the
algebraic structure, that is, to be an algebra homomorphism. Additionally,
since multiplier algebras are algebras of functions on sets, a natural
requirement for a morphism in this category is that point evaluations in
$\mathcal{M}_{2}$ are pulled back to point evaluations in $\mathcal{M}_{1}$.
Concretely, this means that $\varphi$ is implemented by composition with a
function $G:X_{2}\to X_{1}$, i.e. $\varphi(f)=f\circ G$. Note that this isn’t
actually a separate assumption as every such map must be a homomorphism.
Conversely, there are certain conditions under which an isomorphism of
multiplier algebras can only arise as a composition with a bijection; this is
the case in the finite dimensional case considered in this paper, and also
holds more generally (see [SS16]).

Finally, the multiplier algbera $\mathcal{M}_{i}$ can identified with a
concrete operator algebra on $\mathcal{H}_{i}$ by identifying every multiplier
$f\in\mathcal{M}_{i}=\operatorname{Mult}(\mathcal{H}_{i})$ with the
multiplication operator $M_{f}:h\mapsto fh$, $h\in\mathcal{H}_{i}$. Recall
that given a map $\varphi:\mathcal{M}_{1}\to\mathcal{M}_{2}$ between operator
algebras, the completely bounded norm (or cb norm, for short) is defined to be
$\|\varphi\|_{cb}=\sup_{n}\|\varphi^{(n)}\|$, where $\varphi^{(n)}$ is the map
$\varphi^{(n)}:M_{n}(\mathcal{M}_{1})=\mathcal{M}_{1}\otimes M_{n}\to
M_{n}(\mathcal{M}_{2})=\mathcal{M}_{2}\otimes M_{n}$ between the matrices over
$\mathcal{M}_{1}$ and $\mathcal{M}_{2}$, given by
$\varphi^{(n)}=\varphi\otimes{\bf id}_{M_{n}}$ (see [Pau02]). A map $\varphi$
is said to be completely bounded if $\|\varphi\|_{cb}<\infty$ and completely
isometric if $\varphi^{(n)}$ is isometric for all $n$. In the theory of (not-
necessarily selfadjoint) operator algebras, experience has taught us that the
natural morphisms are the completely bounded homomorphisms. We say that
$\varphi$ is a complete isomorphism if it is a bijective completely bounded
isomorphism with a completely bounded inverse. In the finite dimensional case
every isomorphism is a complete isomorphism, but keeping in mind the
appropriate notion of morphisms can serve to guide us to discover the correct
theorems and their proofs.

After the above discussion on what are natural conditions that an isomorphism
of multiplier algebras should have, we define a multiplier algebra isomorphism
between multiplier algebras
$\mathcal{M}_{1}=\operatorname{Mult}(\mathcal{H}_{1})$ and
$\mathcal{M}_{2}=\operatorname{Mult}(\mathcal{H}_{2})$ to be a complete
isomorphism $\varphi:\mathcal{M}_{1}\to\mathcal{M}_{2}$ that is implemented as

$\varphi(f)=f\circ G\,\,,\,\,f\in\mathcal{M}_{1},$

where $G:X_{2}\to X_{1}$ is a bijection. If such an isomorphism exists then we
say that $\mathcal{M}_{1}$ and $\mathcal{M}_{2}$ are isomorphic as multiplier
algebras. If $\varphi$ is completely isometric then we say that
$\mathcal{M}_{1}$ and $\mathcal{M}_{2}$ are completely isometrically
isomorphic as multiplier algebras.

As a testimony to the naturality of the the above definition, we note that it
is not hard to see that if $T:\mathcal{H}_{1}\to\mathcal{H}_{2}$ is an
isomorphism of RKHS then the map $M_{f}\mapsto(T^{*})^{-1}M_{f}T^{*}$ induces
an isomorphism of multiplier algebras,
$\varphi:\mathcal{M}_{1}\to\mathcal{M}_{2}$ (see Proposition 4.4). The paper
[DHS15] contains a continuum of natural weighted Hardy spaces on the unit disc
such that their multiplier algebras are non-isomorphic. It follows therefore
that these spaces are not isomorphic as RKHSs.

Furthermore, if $T$ is an isometric isomorphism of RKHSs, then the induced
isomorphism $\varphi:\mathcal{M}_{1}\to\mathcal{M}_{2}$ is a completely
isometric isomorphism of multiplier algebras. On the other hand, a completely
isometric isomorphism of multiplier algebras need not imply the existence of
an isometric isomorphism between the spaces. Indeed, the Hardy space and the
Bergman spaces on the unit disc both have $H^{\infty}$ as their multiplier
algebra, and in fact the identity map from $H^{\infty}$ to itself is a
completely isometric isomorphism that is implemented by composition with the
identity map on the disc. However, the Hardy and Bergman spaces on the disc
are not isometrically isomorphic as RKHSs (to see this one can use the fact
that the Hardy space has the complete Pick property while the Bergman space
does not, see [AM02]).

###  2.3. Some useful facts about isomorphisms of RKHSs and multiplier
algebras on finite subsets of the ball

We record some simple observations on the spaces
$\mathcal{H}_{X}=H^{2}_{d}\big{|}_{X}$ and their multiplier algebras
$\mathcal{M}_{X}$ as defined in the introduction, where
$X\subseteq\mathbb{B}_{d}$ is finite. As vector spaces both $\mathcal{H}_{X}$
and $\mathcal{M}_{X}$ can be identified with the space of complex functions on
$X$.

Two such finite dimensional spaces $\mathcal{H}_{X}$ and $\mathcal{H}_{Y}$ are
isomomorphic as RKHSs if and only their multiplier algebras are isomorphic as
multiplier algebras, and this happens precisely when $X$ and $Y$ have the same
cardinality. Indeed, if $F:Y\to X$ is a bijection, then $k_{y}\mapsto
k_{F(y)}$ extends to an isomorphism of RKHSs. If
$T:\mathcal{H}_{X}\to\mathcal{H}_{Y}$ is an isomorphism of RKHSs, then we
noted two paragraphs ago that conjugation with the adjoint of $T$ gives rise
to an isomorphism of multiplier algebras
$\varphi:\mathcal{M}_{X}\to\mathcal{M}_{Y}$ (note that in the finite
dimensional setting all linear maps are bounded, and in fact completely
bounded). Finally, every isomorphism (in the mere algebraic sense)
$\varphi:\mathcal{M}_{1}\to\mathcal{M}_{2}$ gives rise to a bijection $F:Y\to
X$ that induces it. For completeness, we explain this below.

We let ${\mathfrak{M}}(\mathcal{M}_{X})$ denote the maximal ideal space of
$\mathcal{M}_{X}$. The maximal ideal space plays an important role in the
isomorphism problem [SS16], but in the case of general subsets $X$ in the ball
it can be quite a wild and unwieldy topological space. Luckily, in the case of
finite sets all of the subtleties disappear. For every $x\in X$ there is a
multiplicative linear evaluation functional
$\pi_{x}\in{\mathfrak{M}}(\mathcal{M}_{X})$ given by

$\pi_{x}(f)=f(x).$

It is easy to see that ${\mathfrak{M}}(\mathcal{M}_{X})=\\{\pi_{x}:x\in X\\}$,
and therefore we can identify ${\mathfrak{M}}(\mathcal{M}_{X})$ with $X$ by
identifying the evaluation functional $\pi_{x}$ at a point $x$ with the point
$x$.

Suppose that $Y\subseteq\mathbb{B}_{d}$ is a finite set, and that $F:Y\to X$
is a function. Then we can define $\varphi:\mathcal{M}_{X}\to\mathcal{M}_{Y}$
by

$\varphi(f)=f\circ F$

for all $f\in\mathcal{M}_{X}$. Clearly, $\varphi$ is a unital homomorphism,
which is an isomorphism if and only if $F$ is bijective. Conversely, if
$\varphi:\mathcal{M}_{X}\to\mathcal{M}_{Y}$ is a unital homomorphism, then we
can define a map
$\varphi^{*}:{\mathfrak{M}}(\mathcal{M}_{Y})\to{\mathfrak{M}}(\mathcal{M}_{X})$
by

$\varphi^{*}(\pi)=\pi\circ\varphi$

for all $\pi\in{\mathfrak{M}}(\mathcal{M}_{Y})$. Since every multiplicative
linear functional on $\mathcal{M}_{X}$ can be identified with a unique point
in the underlying set $X$, we obtain a map $F:Y\to X$. We leave the reader
with the straightforward task of verifying that the maps $F\mapsto\varphi$ and
$\varphi\mapsto F$ defined above are mutual inverses and in particular, if $F$
is the map determined by $\varphi^{*}$, then $\varphi(f)=f\circ F$.

###  2.4. The pseudohyperbolic metric

######  Definition 2.1.

The _pseudohyperbolic metric_ $\rho_{\rm{ph}}$ on the open unit ball
$\mathbb{B}_{d}$ is given by

$\rho_{\rm{ph}}(z,w):=\left\|\Psi_{w}(z)\right\|=\left\|\Psi_{z}(w)\right\|,\hskip
14.22636ptz,w\in\mathbb{B}_{d},$

where $\Psi_{w}$ is the elementary automorphism

$\Psi_{w}(z)=\frac{w-P_{w}z-(1-\|w\|^{2})^{1/2}P_{w}^{\perp}z}{1-\langle
z,w\rangle}$

that maps $w$ to $0$ and $0$ to $w$. Here $P_{w}$ is the orthogonal projection
onto the span of $w$ and $P_{w}^{\perp}=I-P_{w}$ (see [Rud80, Section 2.2]).
In particular, when $d=1$, we have

$\Psi_{w}(z)=\frac{w-z}{1-\overline{w}z}\,\,\text{ and
}\,\,\rho_{\rm{ph}}(z,w)=\left|\frac{w-z}{1-\overline{w}z}\right|.$

The pseudohyperbolic metric and the Euclidean metric are not equivalent, but
they are equivalent on every ball $r\mathbb{B}_{d}$ of radius strictly less
than $1$. Therefore, when working in a compact neighborhood of a fixed finite
subset of the ball it essentially makes no difference whether we work with the
Euclidean or the pseudohyperbolic metric. The pseudohyperbolic metric is
better suited to study the function spaces of interest to us, but sometimes it
is convenient to use the Euclidean metric, and we shall make use of both.

##  3\. Distances between subsets of the ball

In this section we will discuss several ways in which one can measure
distances between subsets of the ball, and we shall record a few useful facts
about them.

In a metric space $(M,\rho)$, we define, for a set $F\subseteq M$ and a point
$x\in M$, the distance from $x$ to $F$ by $\rho(x,F)=\inf\\{\rho(x,y):y\in
F\\}$.

######  Definition 3.1.

Let $X,Y$ be two sets in a metric space $(M,\rho)$. The Hausdorff distance
(induced by $\rho$) between $X$ and $Y$ is defined to be

$\rho_{H}(X,Y)=\max\left\\{\max_{x\in X}\rho(x,Y),\max_{y\in
Y}\rho(y,X)\right\\}.$

For $X=\\{x_{1},\ldots,x_{n}\\}$ and $Y=\\{y_{1},\ldots,y_{n}\\}$ we define
their symmetric distance to be

$\rho_{s}(X,Y)=\min_{\sigma\in
S_{n}}\max\\{\rho(x_{i},y_{\sigma(i)}):i=1,\ldots,n\\},$

where $S_{n}$ is the symmetric group on $\\{1,\ldots,n\\}$.

######  Lemma 3.2.

If $X=\\{x_{1},\ldots,x_{n}\\}$ and $Y=\\{y_{1},\ldots,y_{n}\\}$ are subsets
of a metric space $(M,\rho)$, then $\rho_{H}(X,Y)\leq\rho_{s}(X,Y)$.

######  Proof..

Given $x_{i}\in X$, observe that for each permutation $\sigma\in S_{n}$ we
have

$\rho(x_{i},Y)\leq\rho(x_{i},y_{\sigma(i)}),$

and hence, taking maximum over all $i$’s, we obtain

$\max_{i\in\\{1,...,n\\}}\rho(x_{i},Y)\leq\max_{i\in\\{1,...,n\\}}\rho(x_{i},y_{\sigma(i)}).$

Since the above inequality holds for every permutation in $S_{n}$, we get

(3.1) $\max_{i}\rho(x_{i},Y)\leq\min_{\sigma\in
S_{n}}\max_{i}\rho(x_{i},y_{\sigma(i)})=\rho_{s}(X,Y).$

Similarly, $\max_{j}\rho(y_{j},X)\leq\rho_{s}(Y,X)=\rho_{s}(X,Y)$, thus

$\rho_{H}(X,Y)=\max\left\\{\max_{x\in X}\rho(x,Y),\max_{y\in
F}\rho(y,X)\right\\}\leq\rho_{s}(X,Y).$

∎

######  Proposition 3.3.

Let $X=\\{x_{1},\ldots,x_{n}\\}$ and $Y=\\{y_{1},\ldots,y_{n}\\}$ be subsets
of a metric space $(M,\rho)$ and set $\delta:=\min\\{\rho(x_{i},x_{j}):i\neq
j\\}.$ If $\rho_{H}(X,Y)<\delta/2$, then $\rho_{s}(X,Y)=\rho_{H}(X,Y)$.

######  Proof..

Since $\rho_{H}(X,Y)<\delta/2$, it follows that $\max_{x\in
X}\rho(x,Y)<\delta/2$ which in turn implies that for every $x\in X$,
$\rho(x,Y)<\delta/2$. This amounts to: for every $i\in\\{1,...,n\\}$, there
exists a $j\in\\{1,...,n\\}$ such that $y_{j}\in B(x_{i};\delta/2)$. But since
$X$ and $Y$ are finite sets with same cardinality, we infer that $y_{j}$ is
the only element of $Y$ that is contained in $B(x_{i};\delta/2)$. Let us
rename and denote this $y_{j}$ by $y_{i}$, so that for every
$i\in\\{1,...,n\\}$, we have a unique $y_{i}\in B(x_{i};\delta/2)$. Then
$\rho(x_{i},Y)=\rho(x_{i},y_{i})$.

Similarly, for every $y_{j}\in Y$, there is one and only one element of $X$
that is contained in $B(y_{j};\delta/2)$, and since, from the arguments in the
above paragraph, there is already $x_{j}$ in $X$ with
$\rho(y_{j},x_{j})<\delta/2$, it follows that
$\rho(y_{j},X)=\rho(y_{j},x_{j})$. Consequently, we get

$\rho_{H}(X,Y)=\max\left\\{\rho(x_{i},y_{i}):i\in\\{1,...,n\\}\right\\}.$

But this is same as $\min_{\sigma\in
S_{n}}\max\\{\rho(x_{i},y_{\sigma(i)}):i=1,\ldots,n\\}$; for if we choose any
permutation other than the identity permutation, which leads to measuring the
distance of $x_{i}$ from $y_{i}$, it would lean to measuring the distance of
one of the the elements of $X$, say $x_{i}$, to a point in $Y$ that is not in
$B(x_{i};\delta/2)$ thereby forcing the value of
$\max_{i}\\{(x_{i},y_{\sigma(i)})\\}$ to be strictly greater than $\delta/2$.
This proves that $\rho_{H}(X,Y)=\rho_{s}(X,Y)$. ∎

######  Corollary 3.4.

If $\rho_{s}(X,Y)<\delta/2$, then $\rho_{s}(X,Y)=\rho_{H}(X,Y)$.

The following example shows that in Proposition 3.3, the constant $\delta/2$
cannot be replaced by a larger constant.

######  Example 3.5.

Let $t,\delta>0$ and assume also that $\delta<t/10<1/10$. Consider two subsets
$X=\\{0,\delta,t+\delta/2\\}$ and $Y=\\{\delta/2,t,t+\delta\\}$ in
$\mathbb{R}$ endowed with the usual metric. Since all the points are on a
straight line, it is easy to see that $\rho_{s}(X,Y)=t-\delta$, while
$\rho_{H}(X,Y)=\delta/2$.

This example can be modified to show that no constant $c$ strictly larger than
$\delta/2$ can work in Proposition 3.3 even when we restrict attention to the
pseudohyperbolic metric on the disc. Indeed, from Definition 2.1 it follows
$\rho_{\rm{ph}}(z,w)/|z-w|\to 1$ uniformly as $z,w\to 0$. Then for
sufficiently small $t$ we will have $\rho_{H}(X,Y)\approx\delta/2<c$ while
$\rho_{s}(X,Y)\approx t-\delta$.

Recall that $\operatorname{Aut}(\mathbb{B}_{d})$ denotes the group of
biholomorphic automorphisms (or simply, automorphisms) of the unit ball
$\mathbb{B}_{d}$, and that two subsets of the unit ball are said to be
congruent if one is the image of the other under an automorphism.
Automorphisms preserve the pseudohyperbolic metric [Rud80, Theorem 8.1.4],
hence congruent subsets of the ball are isometric as metric spaces. Moreover,
the RKHSs induced on two finite subsets of the ball by the Drury-Arveson space
are isometrically isomorphic if and only if they are congruent [Roc19, Theorem
7]. For this reason, we would like a measure of distance between subsets of
the ball that is blind to automorphisms.

######  Definition 3.6.

Let $\rho$ be a metric on $\mathbb{B}_{d}$, and let $X$ and $Y$ be two subsets
of $\mathbb{B}_{d}$. The automorphism invariant Hausdorff distance between $X$
and $Y$ is defined to be

$\tilde{\rho}_{H}(X,Y)=\inf\left\\{\rho_{H}(X,\Phi(Y)):\Phi\in\operatorname{Aut}(\mathbb{B}_{d})\right\\}.$

If $X=\\{x_{1},\ldots,x_{n}\\}$ and $Y=\\{y_{1},\ldots,y_{n}\\}$ are finite
sets of the same cardinality, then we define their automorphism invariant
symmetric distance to be

$\tilde{\rho}_{s}(X,Y)=\inf\left\\{\rho_{s}(X,\Phi(Y)):\Phi\in\operatorname{Aut}(\mathbb{B}_{d})\right\\}.$

Because $\operatorname{Aut}(\mathbb{B}_{d})$ is a group this expression is
symmetric in $X$ and $Y$.

######  Lemma 3.7.

Let $\rho$ be a metric on $\mathbb{B}_{d}$. If $X$ and $Y$ are finite subsets
of $\mathbb{B}_{d}$ with the same cardinality, then

$\tilde{\rho}_{H}(X,Y)\leq\tilde{\rho}_{s}(X,Y).$

######  Proof..

Notice that for any $\Psi\in\operatorname{Aut}(\mathbb{B}_{d})$, we have by
Lemma 3.2

$\tilde{\rho}_{H}(X,Y)\leq\rho_{H}(X,\Psi(Y))\leq\rho_{s}(X,\Psi(Y)).$

Taking infimum over all automorphisms in $\mathbb{B}_{d}$ we get the result. ∎

######  Proposition 3.8.

Let $X=\\{x_{1},\ldots,x_{n}\\}$ and $Y=\\{y_{1},\ldots,y_{n}\\}$ be subsets
of the metric space $(\mathbb{B}_{d},\rho)$, and set
$\delta:=\min\\{\rho(x_{i},x_{j}):i\neq j\\}.$ If
$\tilde{\rho}_{H}(X,Y)<\delta/2$, then
$\tilde{\rho}_{s}(X,Y)=\tilde{\rho}_{H}(X,Y)$.

######  Proof..

By Lemma 3.7, it suffices to show that
$\tilde{\rho}_{s}(X,Y)\leq\tilde{\rho}_{H}(X,Y)$. Let us write
$D=\tilde{\rho}_{H}(X,Y)$, and let $\varepsilon>0$ be small enough so that
$D+\varepsilon<\delta/2$. By definition of $D$, we can find an automorphism
$\Psi_{\varepsilon}\in\operatorname{Aut}(\mathbb{B}_{d})$ such that

$\rho_{H}(X,\Psi_{\varepsilon}(Y))<D+\varepsilon<\delta/2,$

which, by Proposition 3.3, yields
$\rho_{s}(X,\Psi_{\varepsilon}(Y))=\rho_{H}(X,\Psi_{\varepsilon}(Y))$. Thus,

$\tilde{\rho}_{s}(X,Y)=\inf\left\\{\rho_{s}(X,\Phi(Y)):\Phi\in\operatorname{Aut}(\mathbb{B}_{d})\right\\}<D+\varepsilon.$

Since this is true for all $\varepsilon$ as above we conclude that
$\tilde{\rho}_{s}(X,Y)\leq\tilde{\rho}_{H}(X,Y)$. ∎

######  Corollary 3.9.

Let $\rho$ be a metric on $\mathbb{B}_{d}.$ Let $X$ and $Y$ be finite subsets
of $\mathbb{B}_{d}$ with the same cardinality and let
$\delta=\min\\{\rho(x_{i},x_{j}):i\neq j\\}$. If
$\tilde{\rho}_{s}(X,Y)<\delta/2$, then
$\tilde{\rho}_{s}(X,Y)=\tilde{\rho}_{H}(X,Y)$.

Thus, once sets are close enough, $\tilde{\rho}_{s}$ and $\tilde{\rho}_{H}$
are the same. We choose to continue working $\tilde{\rho}_{s}$ because it is
more convenient, keeping in mind the equivalence.

We close this section with an additional suggestion for measuring distances
between subsets of the ball which might be relevant to our endeavours. A
function $F=(F_{1},\ldots,F_{m})\in\mathbb{B}_{d}\to\mathbb{C}^{m}$ will be
called a vector valued multiplier if its component functions are all
multipliers, i.e., if $F_{1},\ldots,F_{m}\in\mathcal{M}_{d}$. We then define
the multipier norm of a vector valued multiplier to be $\|F\|=\|M_{F}\|$,
where we think of $M_{F}$ as the “row multiplication operator”
$M_{F}=(M_{F_{1}},\ldots,M_{F_{m}}):H^{2}_{d}\oplus\cdots\oplus H^{2}_{d}\to
H^{2}_{d}$ that acts as

$M_{F}(h_{1}\oplus\cdots\oplus h_{m})=\sum_{i=1}^{m}F_{i}h_{i}.$

If $X,Y$ are both finite subsets of the unit ball
$\mathbb{B}_{d}\subseteq\mathbb{C}^{d}$ with $n$ points each, one can always
find vector valued multipliers (in fact, vector valued polynomials)
$F,G:\mathbb{B}_{d}\to\mathbb{C}^{d}$ such that $F\big{|}_{X}$ is a bijection
onto $Y$ and $G\big{|}_{Y}$ is its inverse. This allows us to define the
following measure of difference.

######  Definition 3.10.

Let $X=\\{x_{1},\ldots,x_{n}\\}$ and $Y=\\{y_{1},\ldots,y_{n}\\}$ be two
subsets of the unit ball $\mathbb{B}_{d}$. The multiplier discrepancy between
$X$ and $Y$ is defined to be

$\delta_{mult}(X,Y)=\inf\left\\{\max\\{\|G\|,\|F\|\\}:F,G\textrm{ are
multipliers },G\big{|}_{Y}=\left(F\big{|}_{X}\right)^{-1}\right\\}.$

One can show that $\delta_{mult}(X,Y)\geq 1$ whenever $n>1$ (if $X$ and $Y$
are singleton sets they can be mapped from one to the other using a constant
function which may even be zero). If $X$ and $Y$ are congruent then
$\delta_{mult}(X,Y)=1$, and a weak compactness argument, together with the
methods of [DRS15, Section 4] shows the converse.

######  Remark 3.11.

One might think that the following could be a good way to measure distance
between sets in our context:

$\tilde{\delta}_{mult}(X,Y)\stackrel{{\scriptstyle?}}{{=}}\inf\left\\{\|G\|\|F\|:F,G\textrm{
are multipliers },G\big{|}_{Y}=\left(F\big{|}_{X}\right)^{-1}\right\\}.$

However, the above definition is not promising. For example, consider
$X=\\{0,r\\}$. Then $\tilde{\delta}_{mult}(X,cX)=1$ for every $c>0$, while
$\tilde{\rho}_{H}(X,{cX})$ grows with diminishing $c$.

##  4\. Distances for reproducing kernel Hilbert spaces and multiplier
algebras

In this section we define notions of distance on the sets of all RKHSs and all
multiplier algebras, and we describe a quantitative relation between the two
notions in the case of complete Pick spaces.

######  Definition 4.1.

For $i=1,2$, let $\mathcal{H}_{i}$ be a reproducing kernel Hilbert space on a
set $X_{i}$ with kernel $K_{i}$. Define

$\delta_{RK}(\mathcal{H}_{1},\mathcal{H}_{2})=\inf\left\\{\|T\|\|T^{-1}\|:T:\mathcal{H}_{1}\to\mathcal{H}_{2}\textrm{
is an RKHS isomorphism}\right\\}.$

The reproducing kernel Banach-Mazur distance between these spaces is defined
to be

$\rho_{RK}(\mathcal{H}_{1},\mathcal{H}_{2})=\log\left(\delta_{RK}(\mathcal{H}_{1},\mathcal{H}_{2})\right).$

The reproducing kernel Banach-Mazur distance is always defined when
$\mathcal{H}_{1},\mathcal{H}_{2}$ are finite dimensional and of the same
dimension, and in this case one can show that
$\rho_{RK}(\mathcal{H}_{1},\mathcal{H}_{2})=0$ if and only if
$\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ are isometrically isomorphic as RKHSs.
If there is no RKHS isomorphism $T:\mathcal{H}_{1}\to\mathcal{H}_{2}$ then one
can say that $\rho_{RH}(\mathcal{H}_{1},\mathcal{H}_{2})=\infty$, but we will
not deal with this case in this paper.

If $\varphi:\mathcal{M}_{1}\to\mathcal{M}_{2}$ is a homomorphism between
multiplier algebras, then we let $\|\varphi\|_{cb}$ denote the completely
bounded norm of $\varphi$. This makes sense, because multiplier algebras are
operator algebras (see Section 2.2).

######  Definition 4.2.

Given two RKHSs $\mathcal{H}_{i}$ ($i=1,2$) with multiplier algebras
$\mathcal{M}_{i}=\operatorname{Mult}(\mathcal{H}_{i})$ we define

$\delta_{M}(\mathcal{M}_{1},\mathcal{M}_{2})=\inf\left\\{\|\varphi\|_{cb}\|\varphi^{-1}\|_{cb}:\varphi:\mathcal{M}_{1}\to\mathcal{M}_{2}\textrm{
a multiplier algebra isomorphism}\right\\}.$

The multiplier Banach-Mazur distance between $\mathcal{M}_{1}$ and
$\mathcal{M}_{2}$ is defined to be

$\rho_{M}(\mathcal{M}_{1},\mathcal{M}_{2})=\log\left(\delta_{M}(\mathcal{M}_{1},\mathcal{M}_{2})\right).$

Again, this distance is always defined when $\mathcal{H}_{1}$ and
$\mathcal{H}_{2}$ are of the same finite dimension. In the case of finite
dimensional quotients of $H^{2}_{d}$ that we are interested in, every
isomorphism is a multiplier algebra isomorphism (see Section 2.3), so we have
in this case

$\delta_{M}(\mathcal{M}_{1},\mathcal{M}_{2})=\inf\left\\{\|\varphi\|_{cb}\|\varphi^{-1}\|_{cb}:\varphi:\mathcal{M}_{1}\to\mathcal{M}_{2}\textrm{
is an algebra isomorphism}\right\\}.$

######  Remark 4.3.

Some readers might wonder why we work with the completely bounded norm of the
map $\varphi$ and $\varphi^{-1}$ rather than the operator norm of these maps,
and whether we can obtain similar results if we work with the operator norm
instead. The completely bounded norm is a natural choice for operator
algebras, as we briefly discussed in Section 2.2; for a reader not familiar
with this idea, we recommend [Pau02]. We do not know whether our results hold
with the norm instead of the completely bounded norm.

The goal of this section is to study the connection between $\rho_{RK}$ and
$\rho_{M}$. We start with the easy direction, which holds true for arbitrary
RKHSs.

######  Proposition 4.4.

For every pair of RKHSs $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$, it holds that

$\rho_{M}(\mathcal{M}_{1},\mathcal{M}_{2})\leq\rho_{RK}(\mathcal{H}_{1},\mathcal{H}_{2})^{2}.$

######  Proof..

Let $T:\mathcal{H}_{1}\to\mathcal{H}_{2}$ be an RKHS isomorphism given by
(2.1), that is, $T(k^{1}_{x})=\lambda_{x}k^{2}_{F(x)}$ for every $x\in X_{1}$.
Given $f\in\mathcal{M}_{1}$, we have the multiplication operator $M_{f}$ on
$\mathcal{H}_{1}$. Consider the bounded operator
$(T^{*})^{-1}M_{f}T^{*}:\mathcal{H}_{2}\to\mathcal{H}_{2}$. We claim that it
is a multiplication operator on $\mathcal{H}_{2}$, i.e., there exists a
function $h\in\mathcal{M}_{2}$ such that $(T^{*})^{-1}M_{f}T^{*}=M_{h}$. To
this end, observe that for every $x\in X_{1}$, we have

$TM_{f}^{*}(k_{x}^{1})=\lambda_{x}\overline{f(x)}k^{2}_{F(x)}=\lambda_{x}M^{*}_{f\circ
F^{-1}}(k^{2}_{F(x)})=M^{*}_{f\circ
F^{-1}}(\lambda_{x}k^{2}_{F(x)})=M^{*}_{f\circ F^{-1}}T(k^{1}_{x}),$

which yields $TM_{f}^{*}T^{-1}=M^{*}_{f\circ F^{-1}}$, and our claim is
proved. Thus $h=f\circ F^{-1}$ is a multiplier, which leads us to define
$\varphi:\mathcal{M}_{1}\to\mathcal{M}_{2}$ via

$\varphi(f)=f\circ F^{-1}.$

It is then easy to see that $\|\varphi\|_{cb}\leq\|T\|\|T^{-1}\|$. The
conclusion
$\rho_{M}(\mathcal{M}_{1},\mathcal{M}_{2})\leq\rho_{RK}(\mathcal{H}_{1},\mathcal{H}_{2})^{2}$
follows readily, since $\varphi^{-1}$ is also given by composition and is
implemented by conjugation with $T$. ∎

The following two lemmas will be used to prove a converse to Proposition 4.4.

######  Lemma 4.5.

Let $v_{1},\ldots,v_{n}$ be distinct elements in the unit ball
$\mathbb{B}_{d}\subseteq\mathbb{C}^{d}$. For all $\varepsilon>0$ and
$r\in(0,1)$, there exists $N\in\mathbb{N}$ such that the following matrix
inequality holds:

$\varepsilon\left[\sum_{k=0}^{N}\langle
v_{i},v_{j}\rangle^{k}\right]\geq\left[\sum_{k=N+1}^{\infty}\langle
u_{i},u_{j}\rangle^{k}\right]$

for every $n$-tuple $u_{1},\ldots,u_{n}\in r\mathbb{B}_{d}$.

######  Proof..

Fix $\varepsilon>0$ and let $A=\left[\langle
v_{i},v_{j}\rangle\right]_{i,j=1}^{n}$ be the $n\times n$ Grammian matrix of
vectors $\\{v_{1},...,v_{n}\\}$. Similarly let $u_{1},\ldots,u_{n}\in
r\mathbb{B}_{d}$ and put $B=\left[\langle
u_{i},u_{j}\rangle\right]_{i,j=1}^{n}$.

If $T$ is a linear operator on $\mathbb{C}^{n}$, then we let
$\operatorname{Im}(T)=\\{Tv:v\in\mathbb{C}^{n}\\}$ denote the image of $T$.
For an arbitrarily chosen natural number $M$ we define

$\mathcal{G}_{M}:=\operatorname{Im}\left(\sum_{k=0}^{M}A^{\circ k}\right),$

where $A^{\circ k}$ is the Schur product of $A$ with itself $k$ times. These
form an increasing sequence
$\mathcal{G}_{1}\subseteq\mathcal{G}_{2}\subseteq...\subseteq\mathbb{C}^{n}$
of subspaces of $\mathbb{C}^{n}$, and so they stabilize after a certain point.
Thus, there exists a natural number $M_{0}$ such that

$\mathcal{G}_{1}\subseteq\mathcal{G}_{2}\subseteq...\subseteq\mathcal{G}_{M_{0}}=\mathcal{G}_{M_{0}+1}=\mathcal{G}_{M_{0}+2}=...\subseteq\mathbb{C}^{n},$

or, in other words,

$\mathcal{G}_{M}=\mathcal{G}_{M_{0}}\subseteq\mathbb{C}^{n}\,\,\text{ for all
}\,\,M\geq M_{0}.$

Let $K$ be the Szego kernel in the ball with kernel function
$k_{w}(z)=(1-\langle z,w\rangle)^{-1}$. Since the kernel functions
$k_{v_{1}},\ldots,k_{v_{n}}$ are linearly independent, the matrix

$\left[K(v_{i},v_{j})\right]=\left[\frac{1}{1-\langle
v_{i},v_{j}\rangle}\right]=\sum_{k=0}^{\infty}A^{\circ k}$

is invertible, so we have that $\mathcal{G}_{M}=\mathbb{C}^{n}$ for all $M\geq
M_{0}$.

From $\max\\{\|u_{i}\|^{2}:i=1,\ldots,n\\}\leq r^{2}<1$ we have the elementary
estimate $\|B^{\circ k}\|\leq nr^{2k}$ for the norm of $B^{\circ k}$, which
yields the operator inequality $B^{\circ k}\leq nr^{2k}\cdot I$ for all $k$.
Consequently,

$\sum_{k=M+1}^{\infty}B^{\circ k}\leq\frac{nr^{2(M+1)}}{1-r^{2}}I\,\,\text{
for all }\,\,M\in\mathbb{N}.$

On the other hand, since $\sum_{k=0}^{M_{0}}A^{\circ k}$ is strictly positive
definite, there exists $c>0$ such that

$\sum_{k=0}^{M}A^{\circ k}\geq cI\,\,\text{ for all }\,\,M\geq M_{0}.$

Now, we choose a natural number $N\geq M_{0}$ which satisfies

$\frac{nr^{2(N+1)}}{1-r^{2}}\leq c\varepsilon.$

This yields

$\varepsilon\sum_{k=0}^{N}A^{\circ k}\geq\sum_{k=N+1}^{\infty}B^{\circ k},$

which is the desired inequality. ∎

######  Lemma 4.6.

Let $v_{1},\ldots,v_{n}$ be distinct elements in the unit ball
$\mathbb{B}_{d}\subseteq\mathbb{C}^{d}$. For all $\varepsilon>0$, there exists
$N\in\mathbb{N}$ such that the following matrix inequality holds:

$\left[\sum_{k=0}^{N}\langle
v_{i},v_{j}\rangle^{k}\right]\geq\frac{1}{1+\varepsilon}\left[\sum_{k=0}^{\infty}\langle
v_{i},v_{j}\rangle^{k}\right]=\frac{1}{1+\varepsilon}\left[\frac{1}{1-\langle
v_{i},v_{j}\rangle}\right].$

######  Proof..

Fix $\varepsilon>0$ and let $A=\left[\langle
v_{i},v_{j}\rangle\right]_{i,j=1}^{n}$ be the $n\times n$ Grammian matrix as
above. Then we need to find an $N\in\mathbb{N}$ such that the following
inequality holds:

$\sum_{k=0}^{N}A^{\circ
k}\geq\frac{1}{1+\varepsilon}\sum_{k=0}^{\infty}A^{\circ k}.$

Subtracting $\frac{1}{1+\varepsilon}\sum_{k=0}^{N}A^{\circ k}$ from both
sides, we seek an $N$ such that the following equivalent inequality holds:

$\varepsilon\sum_{k=0}^{N}A^{\circ k}\geq\sum_{k=N+1}^{\infty}A^{\circ k}.$

Invoking Lemma 4.5 for $u_{i}=v_{i}$ ($i=1,\ldots,n$), the proof is complete.
∎

Now we are ready to prove that closeness of the multiplier algebras implies
closeness of the function spaces.

######  Proposition 4.7.

Let $X_{1}$ be a finite subset of $\mathbb{B}_{d}$, and let
$\mathcal{H}_{1}=\mathcal{H}_{X_{1}}$ and
$\mathcal{M}_{1}=\operatorname{Mult}\mathcal{H}_{1}$. Put

$r:=\max\\{\rho_{\rm ph}(x,y):x,y\in X_{1}\\}$

and fix some $R\in(0,1)$. For every $\varepsilon>0$ there exists
$N_{0}\in\mathbb{N}$, which depends only on $\varepsilon$, on $R$ and on
$X_{1}$, such that the following holds: for every
$X_{2}\subseteq\mathbb{B}_{d}$, if
$\delta_{M}(\mathcal{M}_{1},\mathcal{M}_{2})<r^{-1}R$, then

$\delta_{RK}(\mathcal{H}_{1},\mathcal{H}_{2})\leq(1+\varepsilon)\delta_{M}(\mathcal{M}_{1},\mathcal{M}_{2})^{N_{0}},$

where $\mathcal{H}_{2}=\mathcal{H}_{X_{2}}$ and
$\mathcal{M}_{2}=\operatorname{Mult}\mathcal{H}_{2}$.

######  Proof..

Let us assume that $G:X_{2}\to X_{1}$ is a bijection that induces via
composition an isomorphism $\varphi:\mathcal{M}_{1}\to\mathcal{M}_{2}$ such
that

$\|\varphi\|_{cb}\|\varphi^{-1}\|_{cb}<r^{-1}R.$

Since the norm of an isomorphism is at least $1$, we can consider constants
$C_{1},C_{2}$ such that

(4.1) $\|\varphi\|_{cb}\leq C_{1},\|\varphi^{-1}\|_{cb}\leq C_{2},\,\,\textrm{
and }\,\,C_{1},C_{2}<r^{-1}R.$

We will find natural numbers $M,N$ (which do not depend on $G$) such that the
map $T:\mathcal{H}_{2}\to\mathcal{H}_{1}$ given by $T:k^{2}_{x}\mapsto
k^{1}_{G(x)}$ satisfies
$\|T\|\|T^{-1}\|\leq(1+\varepsilon)C_{1}^{N}C_{2}^{M}$. This will give the
asserted estimate with $N_{0}=\max\\{M,N\\}$.

We can assume, without any loss of generality, that $0\in X_{1}\cap X_{2}$ and
that $G(0)=0$ — if this is not the case, then we apply an automorphism of the
ball to the points, which induces a reproducing kernel Hilbert space
isomorphism between the function spaces (see [DRS15, Section 4] for details).
Note that, under this assumption, we have

$\max\\{\|x\|:x\in X_{1}\\}=r,$

since $\rho_{\rm ph}(x,0)=\|x\|$ and the pseudohyperbolic metric is invariant
under automorphisms of the unit ball.

For the first part of the estimate, we find $N$ such that the norm of the map
$T:k^{2}_{x}\mapsto k^{1}_{G(x)}$ can be bounded as follows:

$\|T\|\leq(1+\varepsilon)^{1/2}C_{1}^{N}.$

By assumption, for every matrix valued multiplier $f$ in the unit ball of
$\mathcal{M}_{1}\otimes M_{m,n}(\mathbb{C})$, the composition $f\circ G$ is a
multiplier in $\mathcal{M}_{2}$ of norm at most $C_{1}$. This holds, in
particular, for $f$ the identity map, which is a row contraction. Thus,
$\hat{G}=f\circ G$, with $f$ being the identity map, is a vector valued
multiplier of norm at most $C_{1}$. Let us write $G$ for $\hat{G}$ and $C$ for
$C_{1}$. Consider the following kernel function on $X_{2}$

$K_{G/C}(x,y):=K_{1}(G(x)/C,G(y)/C)=\frac{1}{1-C^{-2}\langle
G(x),G(y)\rangle}.$

Our immediate goal is to show that $K_{2}\geq K_{G/C}$ (our proof of this fact
is inspired by ideas from [Har17, Section 3]). Since $G/C$ is a multiplier in
$\mathcal{M}_{2}$ of norm at most one, it follows that

$K_{2}/K_{G/C}=K_{2}(1-(G/C)^{*}(G/C))\geq 0,$

that is, $K_{2}/K_{G/C}$ is a kernel function. By the assumption made above
$G/C(0)=0$ and consequently, $K_{G/C}$ is normalized at $0$ (meaning that
$K_{G/C}(x,0)=1$ for all $x$), which in turn implies that $K_{2}/K_{G/C}$ is
normalized at $0$. The Schur complement of the row and column corresponding to
$0$ in $K_{2}/K_{G/C}$ is $K_{2}/K_{G/C}-1$, and so it follows that the
function $K_{2}/K_{G/C}-1$ is a kernel function. Since

$K_{2}-K_{G/C}=K_{G/C}\left(K_{2}/K_{G/C}-1\right),$

we infer, by the Schur product theorem, that

$K_{2}\geq K_{G/C},$

as required. Thus we have the inequality of $n\times n$ matrices

(4.2) $\left[\sum_{k}\langle
x_{i},x_{j}\rangle^{k}\right]\geq\left[\sum_{k}C^{-2k}\langle
G(x_{i}),G(x_{j})\rangle^{k}\right],$

where $x_{1},\ldots,x_{n}$ are the points in $X_{2}$, and
$G(x_{1}),\ldots,G(x_{n})$ are the points in $X_{1}$.

To finish off the estimate for $\|T\|$ we use Lemma 4.6. Fixing
$\varepsilon>0$, the lemma applied to the set
$X_{1}=\\{v_{i}=G(x_{i}):i=1,\ldots,n\\}$ gives us an $N\in\mathbb{N}$ such
that

$C^{-2N}\sum_{k=0}^{N}\langle
G(x_{i}),G(x_{j})\rangle^{k}\geq\frac{C^{-2N}}{1+\varepsilon}\sum_{k=0}^{\infty}\langle
G(x_{i}),G(x_{j})\rangle^{k}.$

On the other hand,

$\sum_{k=0}^{\infty}C^{-2k}\langle
G(x_{i}),G(x_{j})\rangle^{k}\geq\sum_{k=0}^{N}C^{-2k}\langle
G(x_{i}),G(x_{j})\rangle^{k}\geq C^{-2N}\sum_{k=0}^{N}\langle
G(x_{i}),G(x_{j})\rangle^{k},$

so by combining with (4.2) we conclude that the matrix inequality

$\left[K_{1}(G(x_{i}),G(x_{j}))\right]\leq(1+\varepsilon)C^{2N}\left[K_{2}(x_{i},x_{j})\right]$

holds. This means that the map $T:k^{2}_{x}\mapsto k^{1}_{G(x)}$ extends to a
linear map of norm at most $(1+\varepsilon)^{1/2}C_{1}^{N}$.

We next bound the norm of the map $T^{-1}:k^{1}_{G(x)}\mapsto k^{2}_{x}$. We
will show that there is an integer $M$ such that
$\|T^{-1}\|\leq(1+\varepsilon)^{1/2}C_{2}^{M}$. Recall that we are assuming
that composition with $G^{-1}$ gives rise to an isomorphism of cb norm less
than or equal to $C_{2}$. Let us write $C$ for $C_{2}$, and using similar
notation as before, we find that $K_{1}\geq K_{{G^{-1}}/C}$, an equality which
can be rewritten as

$\left[K_{1}(G(x_{i}),G(x_{j}))\right]=\left[\sum_{k}\langle
G(x_{i}),G(x_{j})\rangle^{k}\right]\geq\left[\sum_{k}C^{-2k}\langle
x_{i},x_{j}\rangle^{k}\right].$

As above, we find that $G^{-1}$ must be a multiplier of norm at most $C$.
Since $G^{-1}$ maps $0$ to $0$, the Schwarz lemma implies that $\|x\|\leq rC$
for all $x\in X_{2}$. Invoking the assumption $C=C_{2}<r^{-1}R$, we find that
$\|x\|\leq R<1$ for all $x\in X_{2}$.

Invoking Lemma 4.5 we find $M$ such that the matrix inequality
$\sum_{k=M+1}^{\infty}\langle x_{i},x_{j}\rangle^{k}\leq\varepsilon
K_{1}(G(x_{i}),G(x_{j}))$ holds. Note that an $M$ can be found that works for
all the sets of points satisfying $\max\\{\|x_{i}\|^{2}:i=1,\ldots,n\\}\leq
R^{2}$. But then we find

$(1+\varepsilon)K_{1}(G(x_{i}),G(x_{j}))\geq C^{-2M}\sum_{k=0}^{M}\langle
x_{i},x_{j}\rangle+\sum_{k=M+1}^{\infty}\langle x_{i},x_{j}\rangle^{k}\geq
C^{-2M}K_{2}(x_{i},x_{j}).$

We conclude that the matrix inequality

$\left[K_{2}(x_{i},x_{j})\right]\leq(1+\varepsilon)C^{2M}\left[K_{1}(G(x_{i}),G(x_{j}))\right]$

holds, and this shows that $\|T^{-1}\|\leq(1+\varepsilon)^{1/2}C_{2}^{M}$.
Combining the estimates for $\|T\|$ and $\|T^{-1}\|$, and recalling that the
argument works for any pair of constants as in (4.1), the proof is complete. ∎

######  Corollary 4.8.

If $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ are complete Pick spaces that live
on sets $X_{1},X_{2}$ with $n$ points each, then for every $\varepsilon>0$
there exists $N\in\mathbb{N}$, which depends only on $\varepsilon$ and on
$\mathcal{H}_{1}$, such that

$\delta_{RK}(\mathcal{H}_{1},\mathcal{H}_{2})\leq(1+\varepsilon)\delta_{M}(\mathcal{M}_{1},\mathcal{M}_{2})^{N},$

whenever $\delta_{M}(\mathcal{M}_{1},\mathcal{M}_{2})$ is sufficiently close
to $1$.

######  Proof..

If $\mathcal{H}_{i}$ and $\mathcal{H}^{\prime}_{i}$ are isometrically
isomorphic as RKHSs for $i=1,2$, then

$\delta_{RK}(\mathcal{H}_{1},\mathcal{H}_{2})=\delta_{RK}(\mathcal{H}^{\prime}_{1},\mathcal{H}^{\prime}_{2})$

and also

$\delta_{M}(\operatorname{Mult}(\mathcal{H}_{1}),\operatorname{Mult}(\mathcal{H}_{2}))=\delta_{M}(\operatorname{Mult}(\mathcal{H}^{\prime}_{1}),\operatorname{Mult}(\mathcal{H}^{\prime}_{2})),$

and so the result follows from 4.7 and the universality of the Drury-Arveson
space [AM00]. ∎

######  Remark 4.9.

We probably cannot expect to replace the bound in the above corollary with a
simple formula that holds uniformly for all $\mathcal{H}_{1}$, as in
Proposition 4.4. It is worth noting the example at the end of Section 3 in
[Har17], where it is shown that two complete Pick spaces defined on the same
(infinite) subvariety of the unit ball can have completely boundedly
isomorphic (via the identity map) multiplier algebras, while being different
as reproducing kernel Hilbert spaces. The point is that the maxim complete
Pick spaces are determined by their multiplier algebras should be understood
with care.

##  5\. Structure of spaces and algebras versus geometry of point sets

In this section we prove the main results of the paper, which tie the geometry
of a set $X$ to the structure of the reproducing kernel Hilbert space
$\mathcal{H}_{X}$ and its multiplier algebra $\mathcal{M}_{X}$ in a
quantitative way.

######  Proposition 5.1.

Let $X=\\{x_{1},\ldots,x_{n}\\}$ and $Y=\\{y_{1},\ldots,y_{n}\\}$ be two
subsets of the unit ball $\mathbb{B}_{d}$. Define $r=\max\\{\|z\|:z\in X\cup
Y\\}$ and $A=[K_{1}(x_{i},x_{j})]$ and $B=[K_{2}(y_{i},y_{j})]$. Then

$\delta_{RK}(\mathcal{H}_{X},\mathcal{H}_{Y})\leq\left(1+\frac{4nr(1-r^{2})^{-2}}{\min\\{\lambda_{\min}(A),\lambda_{\min}(B)\\}}\rho_{s}(X,Y)\right)^{2},$

where $\lambda_{\min}(\cdot)$ denotes the minimal eigenvalue of a matrix, and
$\rho$ denotes the Euclidean distance in the ball.

######  Proof..

Given that $X,Y\subseteq\mathbb{B}_{d}$, let us write
$\mathcal{H}_{1}=H^{2}_{d}\big{|}_{X}$ and
$\mathcal{H}_{2}=H^{2}_{d}\big{|}_{Y}$ so that

$K_{i}(z,w)=\frac{1}{1-\langle z,w\rangle}\,\,\text{ for }\,\,i=1,2.$

Let us further assume that $\rho_{s}(X,Y)=\varepsilon.$ Then there exists a
bijection $\sigma$ in $S_{n}$ such that
$\max\\{\rho(x_{i},y_{\sigma(i)}):i=1,...,n\\}=\varepsilon$ and we may as well
assume that $\sigma$ is the identity. Consider the canonical bijection $F:X\to
Y$ obtained from $\sigma$, the bijection given by $x_{i}\mapsto
y_{\sigma(i)}=y_{i}$, and use it to define a linear map from $\mathcal{H}_{1}$
to $\mathcal{H}_{2}$ by

$T(k^{1}_{x_{i}})=k^{2}_{F(x_{i})}=k^{2}_{y_{i}}.$

Clearly, $T$ is bijective and bounded, thus $T$ is an RKHS isomorphism. We are
interested in computing a bound on the norm of $T$. Let $\|A\|_{2}$ denote the
Frobenius norm of a matrix $A$, that is
$\|A\|_{2}=\sqrt{\sum_{i,j}|a_{ij}|^{2}}=\operatorname{Tr}(A^{*}A)^{1/2}$. We
put $A=[K_{1}(x_{i},x_{j})]$ and $B=[K_{2}(y_{i},y_{j})]$, observe that
$B-A\leq\lambda_{\max}(B-A)I$ and $\lambda_{\min}(A)I\leq A$. This implies
that

$B-A\leq\left(\frac{\lambda_{\max}(B-A)}{\lambda_{\min}(A)}\right)A.$

Consequently,

(5.1)
$B=A+(B-A)\leq\left(1+\frac{\lambda_{\max}(B-A)}{\lambda_{\min}(A)}\right)A.$

Let

(5.2) $\mu=1+\frac{\lambda_{\max}(B-A)}{\lambda_{\min}(A)}.$

Then, using (5.1) and (5.2), for every
$\alpha_{1},...,\alpha_{n}\in\mathbb{C}$, we have the following computation.

$\displaystyle\left\|T\left(\sum_{j=1}^{n}\alpha_{j}k^{1}_{x_{j}}\right)\right\|^{2}$
$\displaystyle=\left\|\sum_{j=1}^{n}\alpha_{j}k^{2}_{y_{j}}\right\|^{2}$
$\displaystyle=\sum_{i,j=1}^{n}\overline{\alpha}_{i}\alpha_{j}K_{2}(y_{i},y_{j})$
$\displaystyle\leq\mu\sum_{i,j=1}^{n}\overline{\alpha}_{i}\alpha_{j}K_{1}(x_{i},x_{j})$
$\displaystyle=\mu\left\|\sum_{j=1}^{n}\alpha_{j}k^{1}_{x_{j}}\right\|^{2}.$

It follows that $\|T\|\leq\mu$. To estimate $\mu$, we will use the inequality

$\displaystyle\left|K_{1}(x_{i},x_{j})-K_{2}(y_{i},y_{j})\right|$
$\displaystyle=\left|\frac{1}{1-\langle x_{i},x_{j}\rangle}-\frac{1}{1-\langle
y_{i},y_{j}\rangle}\right|$
$\displaystyle\leq\frac{1}{(1-r^{2})^{2}}(2r\varepsilon+\varepsilon^{2})$
$\displaystyle\leq\frac{4r\varepsilon}{(1-r^{2})^{2}},$

where the first inequality is an elementary estimate, and the second
inequality follows from $\varepsilon=\max_{k=i,j}\|x_{k}-y_{k}\|\leq 2r$. Then

$\displaystyle\mu$ $\displaystyle\leq
1+\frac{\lambda_{\max}(B-A)}{\lambda_{\min}(A)}$ $\displaystyle\leq
1+\frac{\|B-A\|_{2}}{\lambda_{\min}(A)}$ $\displaystyle\leq
1+\frac{\sqrt{\sum_{i,j=1}^{n}(K_{2}(y_{i},y_{j})-K_{1}(x_{i},x_{j}))^{2}}}{\lambda_{\min}(A)}$
$\displaystyle\leq 1+\frac{4nr(1-r^{2})^{-2}}{\lambda_{\min}(A)}\varepsilon.$

By arguing in the reverse direction, we obtain the asserted estimate. ∎

Proposition 5.1 shows how closeness of the sets imply closeness of the
corresponding reproducing kernel Hilbert spaces. Our next task is to prove the
converse. For this, we require a technical lemma regarding the Procrustes
problem which is of independent interest. We will denote the group of $d\times
d$ unitary matrices by $\mathcal{U}(d)$.

######  Lemma 5.2.

Let $A,B\in M_{d\times n}(\mathbb{C})$ such that
$||A^{*}A-B^{*}B||_{2}<\varepsilon.$ Then we have:

$\min_{W\in\mathcal{U}(d)}\|A-WB\|_{2}^{2}\leq
d(2\|A\|_{2}\varepsilon^{1/2}+\varepsilon).$

######  Proof..

For $1\leq i\leq d$, let $\sigma_{i}(M)$ denote the $i$th singular value of a
matrix $M$. By Paragraph 7.4.5 in [HJ13] (or Example 7.4.8 of [HJ85])

$\min_{W\in\mathcal{U}(d)}\|A-WB\|_{2}^{2}=\sum_{i=1}^{d}\sigma_{i}^{2}(A)+\sigma_{i}^{2}(B)-2\sigma_{i}(AB^{*}).$

We will now show that the right hand side is less than or equal than
$d(2\|A\|_{2}\varepsilon^{1/2}+\varepsilon)$, and that will complete the
proof.

Notice that $\sigma_{i}(A^{*}A)=\sigma_{i}(AA^{*})=\sigma_{i}^{2}(A)$ and
similarly:

$|\sigma_{i}^{2}(AB^{*})-\sigma_{i}^{2}(AA^{*})|=|\sigma_{i}(AB^{*}BA^{*})-\sigma_{i}(AA^{*}AA^{*})|.$

Using the inequality $|x-y|\leq\sqrt{|x^{2}-y^{2}|}$, we get:

$|\sigma_{i}(AB^{*})-\sigma_{i}^{2}(A)|\leq{|\sigma_{i}(AB^{*}BA^{*})-\sigma_{i}(AA^{*}AA^{*})|}^{1/2}.$

Recall the fact that for any $X,Y\in M_{d\times n}(\mathbb{C})$ we have
$|\sigma_{i}(X)-\sigma_{i}(Y)|\leq\|X-Y\|_{2}$ (see, e.g., [HJ13, Corollary
7.3.5]). We use this fact and the sub-multiplicativity of the Frobenius norm
to estimate for all $1\leq i\leq d$:

$\displaystyle\sigma_{i}^{2}(A)+\sigma_{i}^{2}(B)-2\sigma_{i}(AB^{*})$
$\displaystyle\leq
2|\sigma_{i}^{2}(A)-\sigma_{i}(AB^{*})|+|\sigma_{i}^{2}(A)-\sigma_{i}^{2}(B)|$
$\displaystyle\leq
2{|\sigma_{i}(AB^{*}BA^{*})-\sigma_{i}(AA^{*}AA^{*})|}^{1/2}+\|A^{*}A-B^{*}B\|_{2}$
$\displaystyle\leq 2{\|AB^{*}BA^{*}-AA^{*}AA^{*}\|_{2}}^{1/2}+\varepsilon$
$\displaystyle\leq 2\|A\|_{2}\|B^{*}B-A^{*}A\|_{2}^{1/2}+\varepsilon$
$\displaystyle\leq 2\|A\|_{2}\varepsilon^{1/2}+\varepsilon$

as required. ∎

######  Proposition 5.3.

Let $X=\\{x_{1},\dots,x_{n}\\}$ be a finite subset of $\mathbb{B}_{d}$. Put
$r:=\max_{i,j}\\{\rho_{\rm ph}(x_{i},x_{j})\\}$. There exists a constant $C>0$
which depends only on $n,d$ and $r$, such that for any
$Y=\\{y_{1},\dots,y_{n}\\}$ with
$\delta_{RK}(\mathcal{H}_{X},\mathcal{H}_{Y})<2$:

$\tilde{\rho}_{s}(X,Y)\leq
C(\delta_{RK}(\mathcal{H}_{X},\mathcal{H}_{Y})-1)^{1/4},$

where $\tilde{\rho}_{s}$ denotes the automorphism invariant symmetric distance
induced by the pseudohyperbolic metric $\rho_{\rm{ph}}$.

######  Proof..

Let $X,Y\subset\mathbb{B}_{d}$ be as in the statement of the proposition. Let
$2>\alpha>\delta_{RK}(\mathcal{H}_{X},\mathcal{H}_{Y})$. Then there exists an
RKHS isomorphism $T:\mathcal{H}_{X}\rightarrow\mathcal{H}_{Y}$ that satisfies:

(5.3) $\|T\|\|T^{-1}\|<\alpha.$

We will prove this implies:

$\tilde{\rho}_{s}(X,Y)\leq\widetilde{C}(\alpha-1)^{1/4}.$

Where $\widetilde{C}>0$ is a constant which depends only on $r,n$ and $d$. By
taking the infimum of the right hand side of the inequality over all
$2>\alpha>\delta_{RK}(\mathcal{H}_{X},\mathcal{H}_{Y})$ we obtain the result.

Assume $T$ is given by the formula:

$T(k_{x_{i}})=\lambda_{i}k_{y_{i}}.$

For some $\lambda\in\mathbb{C}^{n}$. Because all the quantities that appear in
the conditions of the proposition (as well as the norm of $T$ and its inverse)
are invariant under automorphisms of the ball, we may assume $x_{1}=y_{1}=0$
without loss of generality (see [DRS15, Section 4]). It now suffices to prove
the inequality for the symmetric distance induced by the Euclidean metric,
since the two metrics are equivalent inside the closed ball of radius $r<1$
centered at the origin.

After rescaling $T$ by a complex number of unit modulus, we may assume that
$\lambda_{1}\geq 0$. After rescaling $T$ by a positive number, we may assume
as well that $\|T\|=\|T^{-1}\|\leq\sqrt{{\alpha}}$. Under these assumptions,
(5.3) implies for any $x\in\mathcal{H}_{X}$ with $\|x\|=1$:

(5.4) $\alpha^{-1}\leq\|T^{-1}\|^{-2}\leq\|Tx\|^{2}\leq\|T\|^{2}\leq\alpha.$

Because $\alpha-1$ is greater then $1-\alpha^{-1}$, we have
$|\langle(I-T^{*}T)x,x\rangle|=|\|x\|^{2}-\|Tx\|^{2}|\leq\alpha-1$.
Define $\varepsilon=\alpha-1$ to get

(5.5) $\|I-T^{*}T\|\leq\varepsilon.$

Notice that for any $j$ we have $\rho(0,x_{j})=\|x_{j}\|$, therefore:

(5.6) $\max_{i}\\{\|x_{i}\|\\}\leq r.$

We now bound the difference between the kernel functions. Denote
$k(x_{i},x_{j})$ and $k(y_{i},y_{j})$ by $k_{ij}^{x}$ and $k_{ij}^{y}$
respectively. Furthermore, denote
$\overline{\lambda_{i}}{\lambda_{j}}=\lambda_{ij}$. For any $1\leq i,j\leq n$
we estimate:

$\displaystyle|k_{ij}^{x}-\lambda_{ij}k_{ij}^{y}|$
$\displaystyle=|\langle(I-T^{*}T)k_{x_{j}},k_{x_{i}}\rangle|$
$\displaystyle\leq\|I-T^{*}T\|\|k_{x_{i}}\|\|k_{x_{j}}\|$
$\displaystyle\leq\frac{\varepsilon}{(1-\|x_{i}\|^{2})^{1/2}(1-\|x_{j}\|^{2})^{1/2}}$

and so, after putting $\delta=\frac{\varepsilon}{1-r^{2}}$,

(5.7) $|k_{ij}^{x}-\lambda_{ij}k_{ij}^{y}|\leq\delta.$

Dividing both sides of (5.7) by $|k^{y}_{ij}k^{x}_{ij}|$ which is bounded from
below by $\frac{1}{4}$ we get:

$\displaystyle 4\delta$ $\displaystyle\geq|1-\langle
y_{i},y_{j}\rangle-\lambda_{ij}(1-\langle x_{i},x_{j}\rangle)|$
$\displaystyle=|\langle x_{i},x_{j}\rangle-\langle
y_{i},y_{j}\rangle-(1-\lambda_{ij})(\langle x_{i},x_{j}\rangle-1)|$
$\displaystyle\geq|\langle x_{i},x_{j}\rangle-\langle
y_{i},y_{j}\rangle|-|1-\lambda_{ij}|(|\langle x_{i},x_{j}\rangle|+1).$

This implies that

(5.8) $|\langle x_{i},x_{j}\rangle-\langle y_{i},y_{j}\rangle|\leq
4\delta+2|1-\lambda_{ij}|.$

Our next goal is to bound $|1-\lambda_{ij}|$. Recalling the assumptions
$x_{1}=y_{1}=0$, $\lambda_{1}\geq 0$, and plugging $i=1$ into (5.7) we get

$|1-\lambda_{1}\lambda_{j}|\leq\delta,$

and in particular, for $j=1$,

$|1-\lambda_{1}|\leq|1-\lambda_{1}^{2}|\leq\delta.$

Thus, using $|\lambda_{j}|\leq\|T\|\|k_{x_{j}}\|/\|k_{y_{j}}\|\leq
2\sqrt{\alpha}/(1-r^{2})^{1/2}$,

$\displaystyle|\lambda_{j}-1|$
$\displaystyle\leq|\lambda_{j}-\lambda_{1}\lambda_{j}|+|\lambda_{1}\lambda_{j}-1|$
$\displaystyle\leq|\lambda_{j}|\delta+\delta\leq
2\sqrt{\alpha}/(1-r^{2})^{1/2}\delta+\delta$ $\displaystyle\leq
3\sqrt{\alpha}/(1-r^{2})^{1/2}\delta.$

Now we can finally estimate
$|1-\lambda_{ij}|=|1-\overline{\lambda_{i}}{\lambda_{j}}|$:

$\displaystyle|1-\lambda_{ij}|$
$\displaystyle\leq|\overline{\lambda_{i}}{\lambda_{j}}-\lambda_{1}\lambda_{j}|+|\lambda_{1}\lambda_{j}-1|$
$\displaystyle\leq|\lambda_{j}||{\lambda_{i}}-\lambda_{1}|+\delta$
$\displaystyle\leq 12\alpha/(1-r^{2})\delta+\delta\leq
13\alpha/(1-r^{2})\delta.$

By (5.8), we have for all $i$ and $j$:

$|\langle x_{i},x_{j}\rangle-\langle y_{i},y_{j}\rangle|\leq
4\delta+26\alpha/(1-r^{2})\delta\leq\frac{30\varepsilon\alpha}{(1-r^{2})^{2}}.$

Let $A$ and $B$ be the $d\times n$ matrices with columns given by the
coordinate vectors of $x_{1},\dots,x_{n}$ and $y_{1},\dots,y_{n}$
respectively. This yields

$\|A^{*}A-B^{*}B\|_{2}\leq C\varepsilon$

for $C=\frac{30\alpha n}{(1-r^{2})^{2}}$. Because
$\|A\|_{2}=\sqrt{\sum\|x_{i}\|_{2}^{2}}<\sqrt{n}$, by Lemma 5.2 there exists a
unitary matrix $W\in\mathcal{U}(d)$ for which:

$\|A-WB\|_{2}\leq[d(2n^{1/2}(C\varepsilon)^{1/2}+C\varepsilon)]^{1/2}\leq
2\sqrt{Cdn^{1/2}}\varepsilon^{1/4}.$

The last inequality holds because $\alpha<2$ and therefore $\varepsilon<1$,
and $C\geq 1$. Letting $\widetilde{C}=2\sqrt{Cdn^{1/2}}$ and recalling that
$\varepsilon=\alpha-1$, we get

$\tilde{\rho}_{s}(X,Y)=\tilde{\rho}_{s}(X,W(Y))\leq\widetilde{C}(\alpha-1)^{1/4},$

as we wanted to show. This finishes the proof. ∎

The following theorem summarizes the main results of this paper in qualitative
form.

######  Theorem 5.4.

Fix $n\geq 2$. Let $X=\\{x_{1},\ldots,x_{n}\\}\subset\mathbb{B}_{d}$ and let
$Y^{(k)}=\\{y^{(k)}_{1},\ldots,y^{(k)}_{n}\\}$ be a sequence of subsets of
$\mathbb{B}_{d}$. Let $\tilde{\rho}_{s}$ be the automorphism invariant
symmetric distance induced by the pseudohyperbolic metric $\rho_{\rm{ph}}$.
Put $\mathcal{H}=H^{2}_{d}\big{|}_{X}$,
$\mathcal{M}=\operatorname{Mult}(\mathcal{H})$,
$\mathcal{H}_{k}=H^{2}_{d}\big{|}_{Y^{(k)}}$ and
$\mathcal{M}_{k}=\operatorname{Mult}(\mathcal{H}_{k})$. Then, the following
statements are equivalent.

  1. (1)

$\tilde{\rho}_{s}(X,Y^{(k)})\xrightarrow{k\to\infty}0$.

  2. (2)

$\rho_{RK}(\mathcal{H},\mathcal{H}_{k})\xrightarrow{k\to\infty}0$.

  3. (3)

$\rho_{M}(\mathcal{M},\mathcal{M}_{k})\xrightarrow{k\to\infty}0$.

  4. (4)

$\inf\left\\{\|\varphi\|_{cb}\|\varphi^{-1}\|_{cb}:\varphi:\mathcal{M}\to\mathcal{M}_{k}\textrm{
is an algebra isomorphism}\right\\}\xrightarrow{k\to\infty}1$.

Moreover, if $(3)$ holds, that is, if
$\rho_{M}(\mathcal{M},\mathcal{M}_{k})\xrightarrow{k\to\infty}0$, then we have
$\delta_{mult}(X,Y^{(k)})\xrightarrow{k\to\infty}1$.

######  Proof..

$(1)\implies(2)$ follows from Proposition 5.1, noting that after transforming
$X$ so that one of its points is at the origin, the assumptions of that
proposition hold uniformly for $X$ and all $Y^{(k)}$. $(2)\implies(1)$ follows
from Proposition 5.3. $(2)\implies(3)$ follows from Proposition 4.4 and
$(3)\implies(2)$ from 4.7. The equivalence $(3)\Longleftrightarrow(4)$ holds
because the expression in $(4)$ is equal to
$\delta_{M}(\mathcal{M},\mathcal{M}_{k})$ by the remarks following Definition
4.2 (see also the discussion in Section of [SS16]). The final assertion
follows from Proposition 5.5 below. ∎

######  Proposition 5.5.

Let $X=\\{x_{1},\ldots,x_{n}\\}$ and $Y=\\{y_{1},\ldots,y_{n}\\}$ be two
subsets of the unit ball $\mathbb{B}_{d}$. Then

$\delta_{mult}(X,Y)\leq\delta_{M}(\mathcal{H}_{1},\mathcal{H}_{2}).$

######  Proof..

Let $\varphi:\mathcal{M}_{1}\to\mathcal{M}_{2}$ be an isomorphism. Let
$z=(z_{1},\ldots,z_{d})$ denote the coordinate functions on $X_{1}$, which
generate the algebra $\mathcal{M}_{1}=\operatorname{Mult}\mathcal{H}_{1}$,
where $\mathcal{H}_{1}=H^{2}_{d}\big{|}_{X_{1}}$. Consider
$F=(F_{1},\ldots,F_{d})=\varphi(z)=(\varphi(z_{1}),\ldots,\varphi(z_{d}))$. As
$z$ is a row contraction, we find that $\|F\|\leq\|\varphi\|_{cb}$. Since $F$
is the map corresponding to the adjoint
$\varphi^{*}:{\mathfrak{M}}(\mathcal{M}_{2})\to{\mathfrak{M}}(\mathcal{M}_{1})$
associated to $\varphi$ as in Section 2, we have that $F:Y\to X$. In the same
way we obtain a map $G:X\to Y$ associated with $\varphi^{-1}$ satisfying
$\|G\|\leq\|\varphi^{-1}\|_{cb}$, and one readily shows that $G=F^{-1}$. Since
an isomorphism is unit preserving, we have $\|\varphi\|,\|\varphi^{-1}\|\geq
1$, and so

$\max\\{\|F\|,\|G\|\\}\leq\|\varphi\|_{cb}\|\varphi^{-1}\|_{cb}.$

Taking the infimum over all isomorphisms
$\varphi:\mathcal{M}_{1}\to\mathcal{M}_{2}$ (which are all induced by
composition with multipliers) we obtain
$\delta_{mult}(X,Y)\leq\delta_{M}(\mathcal{H}_{1},\mathcal{H}_{2})$. ∎

The proposition establishes the implication

$\rho_{M}(\mathcal{M},\mathcal{M}_{k})\xrightarrow{k\to\infty}0\Longrightarrow\delta_{mult}(X,Y^{(k)})\xrightarrow{k\to\infty}1$

from Theorem 5.4. Example 6.3 below shows that the converse does not hold.

##  6\. Examples

We close this paper by considering several examples highlighting what goes
wrong when some of our assumptions are dropped.

The first two simple examples show that when we drop the assumption that the
two varieties have the same finite cardinality, a small Hausdorff distance
between varieties does not imply any kind of relationship between the RKHSs or
the multiplier algebras.

######  Example 6.1.

Let $V=\\{0\\}\subset\mathbb{D}$ and $W=\\{0,\varepsilon\\}\subset\mathbb{D}$.
One readily sees that $\rho_{H}(V,W)=\varepsilon$. On the other hand,
$\mathcal{H}_{V}$ and $\mathcal{M}_{V}$ are one-dimensional, whereas
$\mathcal{H}_{W}$ and $\mathcal{M}_{W}$ are two-dimensional, and no
isomorphism exists between the spaces corresponding to the different
varieties.

The following is perhaps a more interesting example, in which the varieties
have the same cardinality.

######  Example 6.2.

We define two sequences in the unit disc: $x_{n}=1-2^{-n}$ and
$y_{n}=1-(1-\varepsilon^{n})2^{-n}$, where $\varepsilon\in(0,1)$. Clearly both
sequences are Blaschke sequences, $\sum|1-x_{n}|,\sum|1-y_{n}|<\infty$, and so
they are equal to the zero sets of bounded analytic functions on the disc. We
can therefore define two multiplier varieties $V=\\{x_{n}\\}_{n=1}^{\infty}$
and $W=\\{x_{n}\\}_{n=1}^{\infty}\cup\\{y_{n}\\}_{n=1}^{\infty}$. Clearly $V$
and $W$ have the same cardinality. If $\varepsilon$ is sufficiently small,
then the Hausdorff distance between $V$ and $W$ with respect to the either the
hyperbolic metric or the Euclidean metric can be seen to be

$\rho_{H}(V,W)=\sup_{n}\left|\frac{x_{n}-y_{n}}{1-x_{n}y_{n}}\right|\leq\varepsilon.$

On the other hand, the condition of Hayman and Newman for a sequence of
positive numbers to be interpolating (see p. 204 in [Hof62]) shows that $V$ is
an interpolating sequence while $W$ is not, so $\mathcal{M}_{V}$ and
$\mathcal{M}_{W}$ cannot be isomorphic as algebras [DRS15, Theorem 6.3]. This
means also that the RKHSs cannot be related by an isomorphism of RKHSs.

In Theorem 5.4 the set $X$ and the spaces $\mathcal{H},\mathcal{M}$ are held
fixed. This is required in the proofs, because the bounds we prove relating
the different distance functions depend on one of the sets. Our next example
shows that this is really necessary.

######  Example 6.3.

In this example we show that the map $X\mapsto\mathcal{H}_{X}$ is not
uniformly continuous on the space of two-point subsets in $\mathbb{D}$ with
diameter less than $1/2$.

We consdier sets of the form $X=\\{0,x\\}$ and $Y=\\{0,y\\}$. Then
$\tilde{\rho}_{s}(X,Y)\leq 2\max\\{|x|,|y|\\}$. We will show that
$\tilde{\rho}_{s}(X,Y)$ can be arbitrarily small, while
$\rho_{RK}(\mathcal{H}_{X},\mathcal{H}_{Y})\geq\log\alpha$ where
$\alpha:=1+10^{-2}$. Suppose, with the intent of obtaining a contradiction,
that this is not the case. Then there exists some $\delta>0$, such that
$|x|,|y|<\delta$ implies that
$\delta_{RK}(\mathcal{H}_{X},\mathcal{H}_{Y})<\alpha$. Fixing $|x|<\delta$, we
find that for all $|y|<\delta$ there are
$\lambda_{1}(y),\lambda_{2}(y)\in\mathbb{C}$ such that

$\begin{pmatrix}|\lambda_{1}(y)|^{2}&\lambda_{1}(y)\overline{\lambda_{2}(y)}\\\
\lambda_{2}(y)\overline{\lambda_{1}(y)}&\frac{|\lambda_{2}(y)|^{2}}{1-|x|^{2}}\end{pmatrix}\leq\alpha\begin{pmatrix}1&1\\\
1&\frac{1}{1-|y|^{2}}\end{pmatrix}.$

As in the proof of Proposition 5.3, one can show that $\lambda_{1}(y)$ and
$\lambda_{2}(y)$ stay in a neighborhood of $1$ (this is where we use that
$\alpha$ is close to $1$), and therefore for every sequence of $y$’s tending
to $0$, there is a subsequence such that the left hand side in the above
inequality converges to a strictly positive matrix. On the other hand, the
right hand side converges to a singular matrix as $y\to 0$, so in the limit we
obtain an impossible inequality.

After the first version of this paper appeared on the arxiv, Michael Hartz
shared with us an example that shows that the implication

$\delta_{mult}(X,Y^{(k)})\xrightarrow{k\to\infty}1\Longrightarrow\rho_{M}(\mathcal{M}_{X},\mathcal{M}_{Y^{(k)}})\xrightarrow{k\to\infty}0$

does not hold, in general [Har]. This led us to the following example and
comments, and to several other improvements in the paper. We are grateful to
Dr. Hartz for contacting us and permitting us to include his ideas in this
paper.

######  Example 6.4.

Fix $r\in(0,1/6)$ and let $X=\\{0,r\\}$. Let $y_{k}=1-2^{-k}$ and put
$Y^{(k)}=\\{y_{k},y_{k+1}\\}$. We first show that
$\lim_{k\to\infty}\delta_{mult}(X,Y^{(k)})=1$.

As $\rho_{\rm{ph}}\left(k/2^{k},k/2^{k+1}\right)\xrightarrow{k\to\infty}0<r$,
Pick’s interpolation theorem for two points means that for all large enough
$k$ there exists a function $h_{k}\in H^{\infty}$ that satisfies
$h_{k}(0)=k/2^{k}$, $h_{k}(r)=k/2^{k+1}$, and $\|h_{k}\|_{\infty}\leq 1$.
Letting $f_{k}=1-k^{-1}h_{k}$, we have $f_{k}(0)=y_{k}$, $f_{k}(r)=y_{k+1}$
and $\|f_{k}\|_{\infty}\xrightarrow{k\to\infty}1$. On the other hand since
$\rho_{\rm{ph}}(y_{k},y_{k+1})\xrightarrow{k\to\infty}1/3>r$, Pick’s theorem
implies that there is $g_{k}\in H^{\infty}$ such that $g_{k}(y_{k})=0$,
$g_{k}(y_{k+1})=r$, and $\|g_{k}\|_{\infty}\leq 1$. It follows that
$\limsup_{k}\delta_{mult}(X,Y^{(k)})\leq 1$. But for two-point sets we always
have the inequality $\delta_{mult}(X,Y^{(k)})\geq 1$, whence

$\lim_{k\to\infty}\delta_{mult}(X,Y^{(k)})=1.$

Since
$\rho_{\rm{ph}}(y_{k},y_{k+1})\xrightarrow{k\to\infty}1/3>r=\rho_{\rm{ph}}(0,r)$,
the convergnece $\tilde{\rho}_{s}(X,Y^{(k)})\xrightarrow{k\to\infty}0$ cannot
hold. Theorem 5.4 implies that
$\rho_{M}(\mathcal{M}_{X},\mathcal{M}_{Y^{(k)}})\xrightarrow{k\to\infty}1$
also cannot hold.

Alternatively, we can see that
$\rho_{M}(\mathcal{M}_{X},\mathcal{M}_{Y^{(k)}})\xrightarrow{k\to\infty}1$
doesn’t hold also as follows: suppose that
$\varphi:\mathcal{M}_{X}\to\mathcal{M}_{Y^{(k)}}$ is an isomorphism that is
induced by a bijection $G:Y^{(k)}\to X$. Then by the proof of Proposition 6.2
in [DHS15],

$\|\varphi^{-1}\|\geq\frac{\rho_{\rm{ph}}(y_{k},y_{k+1})}{2\rho_{\rm{ph}}(G(y_{k}),G(y_{k+1}))}=\frac{\rho_{\rm{ph}}(y_{k},y_{k+1})}{2\rho_{\rm{ph}}(0,r)}\xrightarrow{k\to\infty}\frac{1/3}{2r}.$

This shows that
$\liminf_{k}\rho_{M}(\mathcal{M}_{X},\mathcal{M}_{Y^{(k)}})\geq\frac{1}{6r}>1$,
proving what we claimed.

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