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     {\flushleft\it Òîì 20, ñòð. 144--144 (2022) \hspace{65mm}{\rm\small ÓÄÊ  517.55}} %Volume, pages in Russian,
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\title[On commutative algebras]{On the Blaschke matrix product and an analogue of the Horwitz-Rubel theorem for the Blaschke matrix product}
\author{{G. Khudayberganov, J.Sh. Abdullayev}}%
\address{Gulmirza Khudayberganov
\newline\hphantom{iii} National University of Uzbekistan,
\newline\hphantom{iii} 4, University str,
\newline\hphantom{iii} 100174, Tashkent, Uzbekistan}%
\email{gkhudaiberg@mail.ru}%

\address{Jonibek Shokirovich Abdullayev
\newline\hphantom{iii} Urgench State University,
\newline\hphantom{iii} 14, Hamid Olimjon str,
\newline\hphantom{iii} 740000, Urgench, Uzbekistan}%
\email{jonibek-abdullayev@mail.ru}%

\thanks{\sc G. Khudayberganov, J.Sh. Abdullayev,
On the Blaschke matrix product and an analogue of the Horwitz-Rubel theorem for the Blaschke matrix product}
\thanks{\copyright \ 2022 G. Khudayberganov, J.Sh. Abdullayev}
\thanks{\rm The work is supported by RFFI (grant 03-01-11111)}
\thanks{\it Received  January, xx, 20xx, published  March, xx,  20xx.}%





\semrtop \vspace{1cm}
\maketitle {\small
\begin{quote}
\noindent{\sc Abstract. } We study the properties of the Blaschke product in the matrix unit circle \({{\Re }_{I}}\left( m,m \right) =\left\{ Z\in \mathbb{C}[m\times m]:Z{{Z}^{*}}<I \right\},\) and in the matrix upper half-plane  ${\Im_{\sigma }}=\left\{ Z\in \mathbb{C}\left[ m\times m \right]:ImZ>-\sigma I \right\}$. A matrix analogue of the Horwitz-Rubel theorem about the Blaschke product in the complex plane is proved.\medskip

\noindent{\bf Keywords:} Blaschke product, matrix unit circle, matrix upper half-plane, Horwitz-Rubel theorem.
 \end{quote}
}


\section{Introduction and problem statement}


The Blaschke product plays an important role in many problems of classical complex analysis. Related with the important applications of the Blaschke product in complex analysis, since the second half of the last century, interest in the study of functions of the Blaschke product type has increased and several of its analogues have been obtained (see, for example,  \cite{MikHay202}, \cite{WBla15}, \cite{Siciak05},  \cite{Peter20}, \cite{Toronto11}, \cite{GarMash},  \cite{HorRub}, \cite{AizKhu88}).

We recall the Blaschke product in $\mathbb{C}$. Let $\mathbb U=\left\{ t\in \mathbb{C}:\left| t \right|<1 \right\}$ be unit circle. The finite Blaschke product is a function of the form
$$B\left( t \right)={{e}^{i\varphi }}\prod\limits_{j=1}^{n}{\frac{t-{{t}_{j}}}{1-{{{\bar{t}}}_{j}}t}},\,\left| {{t}_{j}} \right|<1.$$
 The number  $n$ of its zeros is called the degree of the Blaschke product. The Blaschke product of degree $0$ is a constant, which module is equal to one.

The Blaschke product has the following properties (see, for example,  \cite{Garn,Kusis}):


(i) $B$ is continuous up to $\partial \mathbb U$;

(ii) $\left| B \right|=1$ to  $\partial \mathbb U$;

(iii)  $B$ has a finite number of zeros in the circle $\mathbb U$.

These three properties determine $B\left( t \right)$ up to a factor of   ${{e}^{i\varphi }}$. If a holomorphic function $f$ satisfies  (i),(ii) and (iii), and  $B$ is a finite Blaschke product with the same zeros, then   $\left| f/B \right|\le 1$ and $\left| B/f \right|\le 1$ to $\mathbb U$ by the maximum principle, and  $f/B$  is constant.

The following theorem holds (see \cite{Garn}):
\begin{theorem} Let $\left\{ {{t}_{n}} \right\}$ be a sequence of points on the circle  $\mathbb U$, such that \[\sum\limits_{n=1}^{\infty }{\left( 1-\left| {{t}_{n}} \right| \right)}<\infty \]  and $m$ is the number of ${{t}_{n}}$, which equal to 0. Then the Blaschke product
\begin{equation}\label{18}
B\left( t \right)={{t}^{m}}\prod\limits_{\left| {{t}_{n}} \right|\ne 0}{\frac{-{{t}_{n}}}{\left| {{t}_{n}} \right|}\frac{t-{{t}_{n}}}{1-{{t}_{n}}t}}
\end{equation}
converges in $\mathbb U$.
The function $B\left( t \right)$ belongs to ${{H}^{\infty }}\left( \mathbb U \right)$\footnote{The Hardy class ${{H}^{\infty }}\left( \mathbb{U} \right)$  is the set of all holomorphic and bounded  functions in $\mathbb{U}$.}, and its zeros are exactly $\left\{ {{t}_{n}} \right\}$, and each zero has multiplicity equal to the number of its occurrence in the sequence  $\left\{ {{t}_{n}} \right\}$.
Moreover, $\left| B\left( t \right) \right|\le 1$ and $$\left| B\left( {{e}^{i\theta }} \right) \right|=1$$ almost everywhere.
\end{theorem}
By definition, an infinite Blaschke product in $\mathbb U$ is a function of the form (\ref{18}).

The following Horwitz-Rubel's theorem expresses one of the most important properties of the Blaschke product (see \cite{HorRub}).

\begin{theorem} Let
\[{{A}_{1}}\left( t \right)=\prod\limits_{j=1}^{k}{\frac{t-{{a}_{j}}}{1-{{{\bar{a}}}_{j}}t}},\]
\[{{A}_{2}}\left( t \right)=\prod\limits_{j=1}^{k}{\frac{t-{{b}_{j}}}{1-{{{\bar{b}}}_{j}}t}}\]
be the Blaschke products of degree $k$, where ${{a}_{j}},{{b}_{j}}\in \mathbb{U}=\left\{ t\in \mathbb{C}:\left| t \right|<1 \right\}$. If ${{A}_{1}}\left( {{\lambda }_{j}} \right)={{A}_{2}}\left( {{\lambda }_{j}} \right)$ for $k$ distinct points ${{\lambda }_{1}},{{\lambda }_{2}},...,{{\lambda }_{k}}$ from  $\mathbb{U}$, then
$${{A}_{1}}\left( {t} \right)\equiv {{A}_{2}}\left( {t} \right).$$
\end{theorem}
Note that the specified (canonical) form for  ${{A}_{1}}\left( {{\lambda }_{j}} \right)$ and ${{A}_{2}}\left( {{\lambda }_{j}} \right)$ is essential.

Indeed, let
$\left| c \right|=1,c\ne 1$ and $$A\left( t \right)=c\prod\limits_{j=1}^{k}{\frac{t-{{a}_{j}}}{1-{{{\bar{a}}}_{j}}t}},$$ \[B\left( t \right)=\frac{t-c{{a}_{1}}}{1-{{\overline{c}}}{{\overline{a}}_{1}}t}\prod\limits_{j=1}^{k}{\frac{t-{{a}_{j}}}{1-{{{\bar{a}}}_{j}}t}},\]
where ${{a}_{j}}\ne 0$ for any $j.$ Then $A\left( {{a}_{j}} \right)=B\left( {{a}_{j}} \right)=0$ for $j=2,3,...,k$ and $$A\left( 0 \right)=B\left( 0 \right)=c\prod\limits_{j=1}^{k}{{{\left( -1 \right)}^{j}}{{a}_{j}}},$$  but $A\left( t \right) \not\equiv B\left( t \right)$. Hence, $A\not\equiv kB$ for any constant $k$.  Theorem 2 is essentially based on the possibility of representing rational functions  ${{A}_{1}}\left( t \right)$ and ${{A}_{2}}\left( t \right)$  in the form:
\begin{equation}\label{115}
{{A}_{1}}\left( t \right)=\frac{{{\alpha }_{0}}+{{\alpha }_{1}}t+...+{{\alpha }_{k-1}}{{t}^{k-1}}+{{t}^{k}}}{1+{{{\bar{\alpha }}}_{k-1}}t+...+{{{\bar{\alpha }}}_{1}}{{t}^{k-1}}+{{{\bar{\alpha }}}_{0}}{{t}^{k}}},
\end{equation}
\begin{equation}\label{116}
{{A}_{2}}\left( t \right)=\frac{{{\beta }_{0}}+{{\beta }_{1}}t+...+{{\beta }_{k-1}}{{t}^{k-1}}+{{t}^{k}}}{1+{{{\bar{\beta }}}_{k-1}}t+...+{{{\bar{\beta }}}_{1}}{{t}^{k-1}}+{{{\bar{\beta }}}_{0}}{{t}^{k}}},
\end{equation}
where the coefficients   ${{\alpha }_{0}},{{\alpha }_{1}},...,{{\alpha }_{k-1}}$ and ${{\beta }_{0}},{{\beta }_{1}},...,{{\beta }_{k-1}}$ are expressed by  ${{a}_{j}}$ and ${{b}_{j}}$, $j=1,2,...,k,$ as follows:
$${{\alpha }_{0}}={{\left( -1 \right)}^{k}}{{a}_{1}}\cdot ...\cdot {{a}_{k}},$$
$$.............................$$
$${{\alpha }_{k-2}}={{a}_{1}}{{a}_{2}}+{{a}_{1}}{{a}_{3}}+...+{{a}_{k-1}}{{a}_{k}},$$
$${{\alpha }_{k-1}}=-\left( {{a}_{1}}+....+{{a}_{k}} \right),$$
and
$${{\beta }_{0}}={{\left( -1 \right)}^{k}}{{b}_{1}}\cdot ...\cdot {{b}_{k}},$$
$$.............................$$
$${{\beta }_{k-2}}={{b}_{1}}{{b}_{2}}+{{b}_{1}}{{b}_{3}}+...+{{b}_{k-1}}{{b}_{k}},$$
$${{\beta }_{k-1}}=-\left( {{b}_{1}}+....+{{b}_{k}} \right).$$
Therefore the functions ${{A}_{1}}\left( t \right),{{A}_{2}}\left( t \right)$ are defined uniquely by $k$ different parameters
${{\alpha }_{0}},{{\alpha }_{1}},...,{{\alpha }_{k-1}}$; ${{\beta }_{0}},{{\beta }_{1}},...,{{\beta }_{k-1}}$. Further, we introduce the Blaschke product in matrix domains, and then we obtain some of its properties and give analogues of the above Theorem 2.


\section{Preliminaries}



Many fundamental theorems of classical complex analysis of one variable still do not have full-fledged generalizations in multidimensional complex analysis. Many formulas in use today are not complete, although each new  obtained formula is stronger and more convenient than the previous one. In the 30s and 60s of the twentieth century such scientists as E. Cartan \cite{Cart}, C.Siegel \cite{Sieg}, I.I. Pyatetsky-Shapiro \cite{Pyat} and Hua Luogeng \cite{Hua} investigated the problems of multivariate complex analysis using a matrix approach. They conducted research mainly in the classical domains\footnote{Recall that a bounded domain  $D \subset \mathbb C^{n}$ is called classical if
the group of its holomorphic automorphisms is a classical Lie group and is transitive on it.} and dealt with questions related to the theory of functions in these domains and the geometry of the domains themselves. Further, B.S. Vladimirov  \cite{VlaB84}, \cite{VlaSer85}, A.G. Sergeev  \cite{Ser88}, \cite{SerZho00}, \cite{SerZho14}, Sh. Zhou  \cite{Zho90ann}, \cite{Zho20}, G. Khenkin  \cite{Xen85}, S.G.Gindikin \cite{Gin64}, Khio-Ming  \cite{XM20ma}, G.Khudayberganov  \cite{KhKSh}, \cite{GXJL}, \cite{kuska21}, \cite{SFU21} and others continued to study the properties of holomorphic functions in matrix domains. In these scientific studies, the biholomorphic equivalence of these domains with bounded domains was widely used in constructing the theory of holomorphic extensions in unbounded domains. In the course of these investigations significant results were obtained in domains of the space  $\mathbb C^{n}$  and in matrix domains. Despite this, many unresolved problems remain in these areas, the solution of which is very important.

Great interest in complex analysis on matrix domains, in recent years, is associated with applications in mathematical physics, the theory of electrical circuits and in solving practical problems using them, including in solving problems of quantum field theory and others, interest in this direction has increased even more complex analysis. About these in the works of V.S. Vladimirov \cite{VlaB84,VlaSer85}, A.G.Sergeyeev \cite{Ser88}, \cite{SerZho00},  Sh. Chou \cite{Zho20}. Also, interest in the pipe of the future has increased due to the discovery of the Penrose transform, which allows us to reduce the study of holomorphic solutions of the most important equations of field theory to the study of the complex structure of the space of R. Penrose twistors   (see \cite{Penros15}, \cite{PenrosRind84}, \cite{PenrosRind86}, \cite{Penros68}, \cite{Penros76}), A.V. Efimova and V.P. Potapova  (see \cite{EfiPot73}), F. Barbaresco \cite{FBar13,FBar07},   E.Bedford and Y.Dadok  \cite{BedDad91}, S.A. Huggett, K.P.Tod \cite{HugTod85}.



The selection of classes of biholomorphically equivalent domains is of great importance in multidimensional complex analysis and its applications. It is fairly well known that any two simply connected domains of the same type on the complex plane map conformally onto each other (Riemann's theorem). But in the multidimensional case, the situation is completely different. For example, two simplest domains, a ball and a polydisk from the space  $\mathbb C^{n} $ do not map biholomorphically onto each other. Therefore, in multivariable complex analysis it is very important to have a stock of domains that are biholomorphically equivalent to each other. Since the set of biholomorphic automorphisms of the matrix unit circle and the matrix upper half-plane are given by linear-fractional functions of matrices, it is natural to consider Blaschke products (for these matrix domains) as a function of matrices. Such products for the matrix circle and for the matrix upper half-plane were introduced in  \cite{Khu87}.

 %\cite{Khu86}, \cite{Khu87kra}

\section{Matrix unit circle and matrix upper half-plane}

Consider the space ${{\mathbb{C}}^{{{m}^{2}}}}$, the space of  ${{m}^{2}}$  complex variables. In some questions, it is convenient to represent the point  $Z$ of this space as  $Z=({{z}_{ij}})_{i,j=1}^{m}$, i.e., as square  $\left[ m\times m \right]$-matrices. With this point representation, the space  ${{\mathbb{C}}^{{{m}^{2}}}}$ will be denoted by $\mathbb{C}[m\times m]$. Denote by ${{\mathbb{C}}^{n}}[m\times m]$ the direct product of  $n$ instances of  $[m\times m]$-matrix spaces  $\underbrace{\mathbb{C}[m\times m]\times \cdot \cdot \cdot \times \mathbb{C}[m\times m]}_{n}$.

 Let $Z=\left( {{Z}_{1}},...,{{Z}_{n}} \right)$  be a vector composed of square matrices  ${{Z}_{j}}$ of order  $m$, considered over the field of complex numbers  $\mathbb{C}$. Let us write the elements of the vector  $Z=({{Z}_{1}},...,{{Z}_{n}})$ as points  $z$ of the space ${{\mathbb{C}}^{n{{m}^{2}}}}$:
\begin{equation}\label{eq1}
z=(z_{11}^{(1)},...,z_{1m}^{(1)},...,z_{m1}^{(1)},...,z_{mm}^{(1)},...,z_{11}^{(n)},...,z_{1m}^{(n)},...,z_{m1}^{(n)},...,z_{mm}^{(n)})\in {{\mathbb{C}}^{n{{m}^{2}}}}.
\end{equation}
Hence, we can assume that $Z$  is an element of the space  ${{\mathbb{C}}^{n}}\left[ m\times m \right]$, i.e., we came to the isomorphism ${{\mathbb{C}}^{n}}\left[ m\times m \right]\cong {{\mathbb{C}}^{n{{m}^{2}}}}$.

The matrix unit circle (the classical domain of the first type according to E. Cartan's classification \cite{Cart})   is defined as the set
\[{{\Re }_{I}}\left( m,m \right) =\left\{ Z\in \mathbb{C}[m\times m]:Z{{Z}^{*}}<I \right\},\]
where ${{Z}^{*}}={\bar{Z}}'$  is  conjugate and transposed matrix to $Z$, notation $Z{{Z}^{*}}<I$  ($I={{I}_{m}}$  is the identity matrix
$\left[ m\times m \right]$) means that the Hermitian matrix $I-Z{{Z}^{*}}$ is positive definite, thus, all its eigenvalues are positive. It
is useful to note that, if  $Z\in \mathbb{C}\left[ m\times m \right]$,  then $$\det \left( I-Z{{Z}^{*}} \right)=\det \left( I-{{Z}^{*}}Z \right).$$ In addition, the conditions $I-Z{{Z}^{*}}>0$  and  $I-{{Z}^{*}}Z>0$ is equivalent. This statement is also true for rectangular matrices (see Theorem 2.1.3 in \cite{Hua}).

The boundary  ${{\Re }_{I}}\left( m,m \right)$ consists of the set
\[\partial {{\Re }_{I}}\left( m,m \right) =\left\{ Z\in \mathbb{C}[m\times m]:\det (I-Z{{Z}^{*}})=0,\,Z{{Z}^{*}}\le I \right\},\]
i.e., from the set of matrices $Z$, for which the matrix $I-Z{{Z}^{*}}$ is non-negative definite but not positive definite Hermitian matrix (its eigenvalues are non-negative and at least one of them is equal to zero).
%(its eigenvalues ??are non-negative and at least one of them is equal to zero).

%There are many on the border
On the border lies the set
\[\Gamma_{\Re}=\left\{ Z\in \mathbb{C}[m\times m]:Z{{Z}^{*}}=I \right\},\]
which is called the skeleton of ${{\Re }_{I}}\left( m,m \right)$   (note that, $\Gamma_{\Re}$ is the Shilov boundary for ${{\Re }_{I}}\left( m,m \right)$).  It is clear that, the set  $\Gamma_{\Re}$ is the set of all unitary  $\left[ m\times m \right]$-matrices (the set of unitary matrices of order $n$ is usually denoted by  $U\left( n \right)$). It should be noted that, the set of matrices  $\left\{ Z:\det \left( I-Z{{Z}^{*}} \right)=0 \right\}$ contains a bounded component  distinguished by the condition  $Z{{Z}^{*}}\le I$, and unbounded component, for which  $Z{{Z}^{*}}\ge I$. These components intersect along the skeleton  $\Gamma_{\Re}$ (see \cite{Hua}). The following diagram shows the matrix unit circle  ${{\Re }_{I}}\left( m,m \right)$ (see Fig. 1).
\begin{center}
  \includegraphics[width=0.5\linewidth]{unitdisc.jpg}
\end{center}
%\begin{figure}[ht!]
%\centering
%\includegraphics[width=70mm]{unitdisc.jpg}

%\small{\textbf{Fig.1}}
%\end{figure}

%\begin{figure}
%\framebox[4cm]{\Huge Figure 1}
%\caption{A figure caption should be placed {\bf below} the figure.\label{fig1}}
%\end{figure}

Note that, the matrix unit circle is a bounded domain centered at ${\mathrm O}$ (${\mathrm O}$ is   zero matrix). Indeed, if we denote by  ${{z}^{j}}=\left( z_{1}^{j},z_{2}^{j},...,z_{m}^{j} \right)$   of any $j$-th row of the matrix $Z$, then the elements of the main diagonal $I-Z{{Z}^{*}}$ will be $1-{{\left| {{z}^{j}} \right|}^{2}}$, and since this matrix is positive definite, hence they are all positive. Therefore  $${{\left| Z \right|}^{2}}=\sum\limits_{j=1}^{m}{{{\left| {{z}^{j}} \right|}^{2}}}<m$$ and the domain ${{\Re }_{I}}\left( m,m \right)$ lies in the ball  $\mathbb B=\left\{ \left| Z \right|<\sqrt{m} \right\}$. Further, from \cite{Hua} it is known that, for any  \(Z\in \mathbb{C}\left[ m\times m \right]\) there exist unitary matrices  $U$ of order $m$  and $V$ of order $m$ such that
\begin{equation}
Z=U
\begin{pmatrix}
{{\lambda }_{1}} & 0 & \cdots & 0 \\
0 & {{\lambda }_{2}} & \cdots & 0 \\
\vdots  & \vdots  & \ddots & \vdots  \\
0 & 0 & \cdots & {{\lambda }_{m}}
\end{pmatrix}V
\end{equation}
for some \({{\lambda }_{1}}\ge {{\lambda }_{2}}\ge ...\ge {{\lambda }_{m}}\ge 0\). Hence it follows that
	$$\det \left( {{I}^{\left( m \right)}}-Z{\bar{Z}}' \right)=\left( 1-\lambda _{1}^{2} \right)...\left( 1-\lambda _{m}^{2} \right)=\det \left( {{I}^{\left( m \right)}}-{\bar{Z}}'Z \right).$$
Therefore, for a given \(Z\in \mathbb{C}\left[ m\times m \right]\) the relation ${{I}^{\left( m \right)}}-Z{\bar{Z}}'>0$ holds if and only if  \(1-\lambda _{s}^{2}>0\) or \({{\lambda }_{s}}<1\), \(s=1,...,m\).
On the other hand, for any  $Z\in {{\mathbb{B}}^{{{m}^{2}}}}\left( 1 \right)$, where
$${{\mathbb{B}}^{{{m}^{2}}}}\left( 1 \right)=\left\{ Z=\left( {{z}_{11}},...,{{z}_{1m}},{{z}_{21}}...,{{z}_{2m}},...,{{z}_{m1}}...,{{z}_{mm}} \right)\in {{\mathbb{C}}^{{{m}^{2}}}}:\sum\limits_{s=1}^{m}{\sum\limits_{j=1}^{m}{{{\left| {{z}_{sj}} \right|}^{2}}}}<1 \right\}$$
is the ball from space  $\mathbb C^{m^{2}}$, we have
	\[{{\left| Z \right|}^{2}}=\sum\limits_{s=1}^{m}{}\sum\limits_{j=1}^{m}{}{{\left| {{z}_{sj}} \right|}^{2}}=\sum\limits_{s=1}^{m}{}\sum\limits_{j=1}^{m}{}{{z}_{sj}}{{\bar{z}}_{sj}}=Sp\left( Z{{{\bar{Z}}}^{'}} \right)=\sum\limits_{s=1}^{m}{}\lambda _{s}^{2}<1.\]
Whence  \({{\lambda }_{s}}<1\left( s=1,...,m \right)\) and \(Z\in {{\Re }_{I}}\left( m,m \right)\). Therefore, we have the following relation
	\[{{\mathbb{B}}^{{{m}^{2}}}}\left( 1 \right)\subset {{\Re }_{I}}\left( m,m \right).\]

For $m=1$ the domain ${{\Re }_{I}}\left( m,m \right)$ coincides with the unit circle, and the $\Gamma_{\Re}$  is the unit circle in the complex plane $\mathbb{C}$.

For $m=2$  the domian $\tau $  admits the representation
\[\tau =\left\{ Z\in \mathbb{C}[2\times 2]:\,\,\,\psi (Z)<0 \right\},\]
where
\[\psi (Z)=\max \left[ {{\left| {{z}_{11}} \right|}^{2}}+{{\left| {{z}_{12}} \right|}^{2}}-1,\,\,\,{{\left| {{z}_{21}} \right|}^{2}}+{{\left| {{z}_{22}} \right|}^{2}}-1,\,\,\,{{\psi }_{0}}\left( Z \right) \right],\]
\[{{\psi }_{0}}(Z)=\det Z{{Z}^{*}}+SpZ{{Z}^{*}}-1,\]
 and $Sp\left( Z \right)$ is the trace (hole) of the matrix $Z$  (this representation can be obtained from Sylvestr's criterion for a positive definite matrix).

The matrix upper half-plane is defined as the set of matrices
\[\Im =\left\{ Z\in \mathbb{C}\left[ m\times m \right]:\operatorname{Im}Z>0 \right\},\]
where $\operatorname{Im}Z=\frac{1}{2i}\left( Z-{{Z}^{*}} \right)$. It is the usual upper half-plane when $m=1$.

The boundary $\partial \Im $  of this domain consists of matrices  $Z$, for which $\operatorname{Im}Z$ is a non-negative definite but non-positive definite Hermitian matrix (its eigenvalues are non-negative and at least one of them is equal to zero). Since the vanishing of the eigenvalues of a Hermitian matrix is expressed by a real analytic equality, then $\partial \Im $ consists of pieces of real analytic surfaces of dimension $2{{m}^{2}}-1$.

The set
\[S(\Im )=\left\{ Z:\operatorname{Im}Z=0 \right\}\,\,,\]
which lies on  $\partial \Im $, is called the skeleton of the upper half-plane   $\Im $. It consists of all Hermitian matrices. The Hermitianity condition is expressed by  ${{m}^{2}}$   independent equations, so the real dimension of  $S(\Im )$ is equal to   ${{m}^{2}}$.

For $m=2$ this special case has the form
$$\Im =\left\{ Z\in \mathbb{C}\left[ 2\times 2 \right]:\left( \begin{matrix}
   \operatorname{Im}{{z}_{11}} & \frac{{{z}_{12}}-{{{\bar{z}}}_{21}}}{2i}  \\
   \frac{{{z}_{21}}-{{{\bar{z}}}_{12}}}{2i} & \operatorname{Im}{{z}_{22}}  \\
\end{matrix} \right)>0 \right\}.$$
This domain is defined by the inequalities $$\Im=\left\{\operatorname{Im}{{z}_{11}}>0,\operatorname{Im}{{z}_{11}}\operatorname{Im}{{z}_{22}}-\frac{1}{4}{{\left| {{z}_{12}}-{{{\bar{z}}}_{21}} \right|}^{2}}>0\right\},$$ and its boundary  by the equation
$$\partial \Im =\left\{\operatorname{Im}{{z}_{11}}\operatorname{Im}{{z}_{22}}=\frac{1}{4}{{\left| {{z}_{12}}-{{{\bar{z}}}_{21}} \right|}^{2}}\right\},$$ where the skeleton is the real four-dimensional plane $$S(\Im)=\left\{\operatorname{Im}{{z}_{11}}=\operatorname{Im}{{z}_{22}}=0,{{z}_{12}}={{\bar{z}}_{21}}\right\}.$$  Note that, the non-degenerate affine transformation
$$\Phi \left( Z \right)=\left( \begin{matrix}
   {{z}_{11}}+{{z}_{22}} & {{z}_{12}}+{{z}_{21}}  \\
   i\left( {{z}_{21}}-{{z}_{12}} \right) & {{z}_{11}}-{{z}_{22}}  \\
\end{matrix} \right)$$
maps $\Im $ to the domain defined by the inequalities $$\tau(2)=\left\{\operatorname{Im}{{z}_{11}}>0,{{\left( \operatorname{Im}{{z}_{11}} \right)}^{2}}>{{\left( \operatorname{Im}{{z}_{12}} \right)}^{2}}+{{\left( \operatorname{Im}{{z}_{21}} \right)}^{2}}+{{\left( \operatorname{Im}{{z}_{22}} \right)}^{2}}\right\},$$  which called the pipe of the future (see  \cite{VlaSer85}), i.e. into the tubular domain $T={{\mathbb{R}}^{4}}\left( x \right)+iC$ over the cone $$C=\left\{ y_{11}^{2}-y_{22}^{2}-y_{12}^{2}-y_{21}^{2}>0 \right\},$$  more precisely, over one cavity  ${{C}_{+}}$ of this cone for which  ${{y}_{11}}>0$ (we put ${{z}_{jk}}={{x}_{jk}}+i{{y}_{jk}}$). Then the  boundary $\partial \Im $ maps $\partial {{C}_{+}}\times {{\mathbb{R}}^{4}}\left( x \right)$, and the skeleton becomes into the real subspace ${{\mathbb{R}}^{4}}\left( x \right)$, more precisely, into the product of the vertex of the cone ${{C}_{+}}$ by it. These implementations are given in more detailed in \cite{VlaB84}, \cite{VlaSer85}.


\section{Blaschke matrix product}
In this section, using the Blaschke product for the matrix unit circle and the matrix upper half-plane, some results are obtained for one variable.

A finite Blaschke matrix product of degree $k$ in ${{\Re }_{I}}\left( m,m \right)$ we will call a function of matrices of the form 
$$B\left( Z \right)=\prod\limits_{j=1}^{k}{\left( Z-{{P}_{j}} \right){{\left( I-P_{j}^{*}Z \right)}^{-1}}},$$
where ${{P}_{j}}\in {{\Re }_{I}}\left( m,m \right)$, $j=1,2,...,k$.

\begin{theorem}
 \it The Blaschke product for the matrix unit circle has the following properties:

(i) The function $B\left( Z \right)$ is continuous up to the boundary ${{\Re }_{I}}\left( m,m \right)$;

(ii) $B\left( Z \right){{B}^{*}}\left( Z \right)=I$ on $\Gamma_{\Re}$;

(iii) $B\left( {{P}_{i}} \right)=\mathrm O,\mathrm O\in {{\Re }_{I}}\left( m,m \right),j=1,2,...,k.$
\end{theorem}
\begin{proof} i) follows from the fact that  $\det \left( I-P_{j}^{*}Z \right)\ne 0$.
 ${{P}_{j}}\in {{\Re }_{I}}\left( m,m \right)$ by condition. Since the conditions $I-Z{{Z}^{*}}>0$ and $I-{{Z}^{*}}Z>0$ are equivalent we have $P_{j}^{*}\in {{\Re }_{I}}\left( m,m \right).$
From the relation
$$I-\left( P_{j}^{*}Z \right){{\left( P_{j}^{*}Z \right)}^{*}}=I-P_{j}^{*}Z{{Z}^{*}}{{P}_{j}}=\left( I-P_{j}^{*}{{P}_{j}} \right)+P_{j}^{*}\left( I-Z{{Z}^{*}} \right){{P}_{j}}>0$$
 follows that $\left( P_{j}^{*}Z \right)\in {{\Re }_{I}}\left( m,m \right)$. Therefore, all eigenvalues of the matrix  $\left( P_{j}^{*}Z \right)\in {{\Re }_{I}}\left( m,m \right)$ lie inside the unit circle. And so  $\det \left( I-P_{j}^{*}Z \right)\ne 0$.

ii) According to \cite{Hua}, the automorphism  ${{\Re }_{I}}\left( m,m \right)$, which maps an arbitrary point $P\in {{\Re }_{I}}\left( m,m \right)$ to the origin has the form
\begin{equation}\label{110}
W=A\left( Z-P \right)\left( I-{{P}^{*}}Z \right){{B}^{-1}},
\end{equation}
where $A,B$ are $\left[ m\times m \right]$ -square matrices such that
$$\bar{A}\left( I-\bar{P}{P}' \right){A}'=I, \bar{B}\left( I-{P}'\bar{P} \right){B}'=I.$$
The inverse mapping to (\ref{110}) has the form
\begin{equation}\label{111}
Z={{\left( I-{{A}^{-1}}WB{{P}^{*}} \right)}^{-1}}\left( {{A}^{-1}}WB+P \right).
\end{equation}
Since for  $P={\mathrm O}$, it follows from (\ref{111}) that $Z={{A}^{-1}}WB,$ then the automorphism (\ref{110}) can be represented as
\begin{equation}\label{110*}
\zeta =\left( Z-P \right){{\left( I-{{P}^{*}}Z \right)}^{-1}}.
\end{equation}
Now
$$B\left( Z \right){{B}^{*}}\left( Z \right)=\prod\limits_{j=1}^{k}{\left( Z-{{P}_{j}} \right){{\left( I-P_{j}^{*}Z \right)}^{-1}}}{{\left[ \prod\limits_{j=1}^{k}{\left( Z-{{P}_{j}} \right){{\left( I-P_{j}^{*}Z \right)}^{-1}}} \right]}^{*}}=$$
$$=\left[ \left( Z-{{P}_{1}} \right){{\left( I-P_{1}^{*}Z \right)}^{-1}} \right]...\left[ \left( Z-{{P}_{k}} \right){{\left( I-P_{k}^{*}Z \right)}^{-1}} \right]{{\left[ \left( Z-{{P}_{k}} \right){{\left( I-P_{k}^{*}Z \right)}^{-1}} \right]}^{*}}...$$
$$...{{\left[ \left( Z-{{P}_{1}} \right){{\left( I-P_{1}^{*}Z \right)}^{-1}} \right]}^{*}}.$$
Due to (\ref{110*}) for $Z\in \Gamma_{\Re}$ we have
$$\left[ \left( Z-{{P}_{j}} \right){{\left( I-P_{j}^{*}Z \right)}^{-1}} \right]{{\left[ \left( Z-{{P}_{j}} \right){{\left( I-P_{j}^{*}Z \right)}^{-1}} \right]}^{*}}=I,$$
for each $j\left( j=1,2,...,k \right)$.
This implies that $B\left( Z \right){{B}^{*}}\left( Z \right)=I$ on $\Gamma_{\Re}$.

iii) Since  the function   $B(Z)$ vanishes at the points  $Z =P_{j}$ we get reality of the statement iii.

\end{proof}

\section{The Blaschke product for the matrix upper half-plane}

Let ${\Im_{\sigma }}=\left\{ Z\in \mathbb{C}\left[ m\times m \right]:\operatorname{Im}Z>-\sigma I \right\}$ be the matrix half-plane of square $\left[ m\times m \right]$ matrix in the space  $\mathbb{C}\left[ m\times m \right]$, where $\sigma >0$.


The next theorem is true.

\begin{theorem}  Transformation
\begin{equation}\label{112}
W={{\left( Z-{{A}^{*}}+2i\sigma I \right)}^{-1}}\left( Z-A \right),
\end{equation}
where $A\in {{\Im}_{\sigma }}, A-{{A}^{*}}=cI, c\in \mathbb{C}$, biholomorphically maps ${{\Im}_{\sigma }}$ to the matrix unit circle
\[{{\Re }_{I}}\left( m,m \right)=\left\{ W\in \mathbb{C}\left[ m\times m \right]:W{{W}^{*}}<I \right\}.\]
\end{theorem}
\begin{proof} First of all, let us show that the next matrix is invertable $$\left( Z-{{A}^{*}}+2i\sigma I \right).$$ Let  $\rho $ be column vector of length $m$ and
$$\left( Z-{{A}^{*}}+2i\sigma I \right)\rho =0.$$
Hence
$$\left( Z+i\sigma I \right)\rho =\left( {{A}^{*}}-i\sigma I \right)\rho ,$$
$${{\rho }^{*}}\left( {{Z}^{*}}-i\sigma I \right)={{\rho }^{*}}\left( {{A}^{*}}+i\sigma I \right).$$
Multiplying both sides of the first equality on the left by ${{\rho }^{*}}$,  the second equality on the right by $\rho $ and subtracting the second equation from the first, we get
$${{\rho }^{*}}\left( Z-{{Z}^{*}}+2i\sigma I \right)\rho =-{{\rho }^{*}}\left( A-{{A}^{*}}+2i\sigma I \right)\rho ,$$
Consequently,
$${{\rho }^{*}}\left( \operatorname{Im}Z+\sigma I \right)\rho =-{{\rho }^{*}}\left( \operatorname{Im}A+\sigma I \right)\rho.$$
Since $$\operatorname{Im}Z+i\sigma I>0,$$$$\operatorname{Im}A+i\sigma I>0,$$
 then $\rho =0.$ This means that the matrix $$\left( Z-{{A}^{*}}+2i\sigma I \right)$$ is nondegenerate, i.e. it is invertible and therefore the map (\ref{112}) is holomorphic in ${{\Im}_{\sigma }}$.
Now we need to show that (\ref{112}) maps ${{\Im}_{\sigma }}$ to ${{\Re }_{I}}\left( m,m \right)$. This follows from the following relationship:
$$I-W{{W}^{*}}=I-{{\left( Z-{{A}^{*}}+2i\sigma I \right)}^{-1}}\left( Z-A \right)\left( {{Z}^{*}}-{{A}^{*}} \right){{\left( {{Z}^{*}}-A-2i\sigma I \right)}^{-1}}=$$
$$={{\left( Z-{{A}^{*}}+2i\sigma I \right)}^{-1}}\left[ \left( Z-{{A}^{*}}+2i\sigma I \right)\left( {{Z}^{*}}-A-2i\sigma I \right)-\left( Z-A \right)\left( {{Z}^{*}}-{{A}^{*}} \right) \right]\times $$
$$\times {{\left( {{\left( Z-{{A}^{*}}+2i\sigma I \right)}^{-1}} \right)}^{*}}=$$
$$
   =G[Z{{Z}^{*}}-{{A}^{*}}{{Z}^{*}}+2i\sigma {{Z}^{*}}-ZA+{{A}^{*}}A-2i\sigma AI-2i\sigma Z+2i{{A}^{*}}\sigma I+4{{\sigma }^{2}}I- \\
 $$
$$-Z{{Z}^{*}}+A{{Z}^{*}}+Z{{A}^{*}}-A{{A}^{*}}]{{G}^{*}}=G[2i\left( \frac{A-{{A}^{*}}}{2i}+\sigma  \right){{Z}^{*}}-Z\left( \frac{A-{{A}^{*}}}{2i}+\sigma  \right)2i-$$
$$-2i\left( \frac{A-{{A}^{*}}}{2i}+\sigma  \right)2i\sigma ]{{G}^{*}}=G[2i\left( \frac{A-{{A}^{*}}}{2i}+\sigma  \right)\times \left( \frac{Z-{{Z}^{*}}}{2i}+\sigma  \right)\left( -2i \right)]{{G}^{*}}=$$
$$=4G\left( \frac{A-{{A}^{*}}}{2i}+\sigma I \right)\left( \frac{Z-{{Z}^{*}}}{2i}+\sigma I \right){{G}^{*}}=4G\left( \operatorname{Im}A+\sigma I \right)\left( \operatorname{Im}Z+\sigma I \right){{G}^{*}},$$
where  $G={{\left( Z-{{A}^{*}}+2i\sigma I \right)}^{-1}}.$

It is known that, the Hermitian matrices $B$ and $AB{{A}^{*}}$  are both positive definite if $A$ is nondegenerate. Since $A-{{A}^{*}}=cI,c\in \mathbb{C},$ then
$$B=\left( \operatorname{Im}A+\sigma I \right)\left( \operatorname{Im}Z+\sigma I \right)={{\left[ \left( \operatorname{Im}A+\sigma I \right)\left( \operatorname{Im}Z+\sigma I \right) \right]}^{*}}={{B}^{*}},$$ i.e. $B$ is Hermitian matrix. Therefore, the matrices  $I-W{{W}^{*}}$ and $B=\left( \operatorname{Im}A+\sigma I \right)\left( \operatorname{Im}Z+\sigma I \right)=\theta \left( \operatorname{Im}Z+\sigma I \right),$
are simultaneously positive definite where $\theta$ is a positive number. This says that (\ref{112}) maps  ${{\Im}_{\sigma }}$ to ${{\Re }_{I}}\left( m,m \right)$.

Now from (\ref{112}) we will find the inverse mapping
\begin{equation}\label{113}
Z=\left[ A-\left( {{A}^{*}}-2i\sigma I \right)W \right]{{\left( I-W \right)}^{-1}}.
\end{equation}
 Since for $W\in {{\Re }_{I}}\left( m,m \right)$ all its eigenvalues are less than one, then $\det \left( I-W \right)\ne 0$, i.e. the matrix  $\left( I-W \right)$ is nondegenerate. This implies that the map  (\ref{113}) is holomorphic in ${{\Re }_{I}}\left( m,m \right)$.
Due to the relation
$$I-W{{W}^{*}}=4{{\left( Z-{{A}^{*}}+2i\sigma I \right)}^{-1}}\left( \operatorname{Im}A+\sigma I  \right)\times $$
$$\times \left( \operatorname{Im}Z+\sigma I \right)({{\left( Z-{{A}^{*}}+2i\sigma I \right)}^{-1}})^{*},$$
it can be seen that for $W\in {{\Re }_{I}}\left( m,m \right)$, i.e. $I-W{{W}^{*}}>0$, the matrix $G={{\left( Z-{{A}^{*}}+2i\sigma I \right)}^{-1}},$ is non degenerate also  and hence, $\operatorname{Im}Z+\sigma I>0$, i.e. (\ref{112}) maps ${{\Re }_{I}}\left( m,m \right)$ to ${{\Im}_{\sigma }}$. The statement is proved.
\end{proof}


In a special case, when  $A=iI$ and $\sigma =0,$ the transformation (\ref{112}) is the well-known Cayley transformation. In addition, (\ref{112}) transforms the skeleton $${{\Gamma }_{\sigma }}=\left\{ Z\in \mathbb{C}\left[ m\times m \right]:\operatorname{Im}Z=-\sigma I \right\}$$ of the matrix half-plane ${{\Im}_{\sigma }}$  the set $$\Gamma_{\Re}=\left\{ W:W{{W}^{*}}=I \right\}$$ the skeleton (Shilov boundary)  ${{\Re }_{I}}\left( m,m \right)$, which consists of unitary $\left[ m\times m \right]$ matrices.

By a finite Blaschke matrix product of degree $k$ in  ${{\Im}_{\sigma }}$ we say the function of matrices (mapping $\mathbb{C}\left[ m\times m \right]\to \mathbb{C}\left[ m\times m \right]$ of the form (\ref{112}))
\begin{equation}\label{114}
{{B}_{k}}\left( Z \right)=\prod\limits_{j=1}^{k}{{{\left[ {{\left( Z-A_{j}^{*}+2i\sigma I \right)}^{-1}}\left( Z-{{A}_{j}} \right) \right]}^{{{q}_{j}}}}},
\end{equation}
where ${{A}_{j}}\in {{H}_{\sigma }},{{A}_{j}}-A_{j}^{*}={{c}_{j}}I,{{c}_{j}}\in \mathbb{C},{{q}_{j}}$ are positive integers, $j=1,2,...,k.$

 Based on Theorem 4 for the Blaschke product of the form (11), we have the following

\begin{corollary}
 The Blaschke product of the form (\ref{114}) has the following properties:


i. ${{B}_{k}}\left( Z \right)$ is continuous up to the boundary  ${{\Im}_{\sigma }}$;

ii. ${{B}_{k}(Z)}B_{j}^{*}(Z)=I$ on ${{\Gamma }_{\sigma }};$

iii. ${{B}_{k}}\left( {{A}_{j}} \right)=\mathrm O,{\mathrm O}\in {{\Re }_{I}}\left( m,m \right),j=1,2,...,k;$

iv.  Transformation   $B\left( Z \right)$
is a map of domain ${{\Im}_{\sigma }}$ to ${{\overline{\Re }}_{I}}\left( m,m \right)$.
\end{corollary}
\begin{proof} The first assertion i) follows from the invertibility of the matrices $$\left( Z-A_{j}^{*}+2i\sigma I \right)$$
for all $Z,A_{j}^{*}\in {{\Im}_{\sigma }}.$

Properties ii) is a consequence of the fact that the mapping
$$W={{\left( Z-A_{j}^{*}+2i\sigma I \right)}^{-1}}\left( Z-{{A}_{j}} \right),{{A}_{j}}\in {{\Im}_{\sigma }},$$
maps ${{\Im}_{\sigma }}$ to ${{\Re }_{I}}\left( m,m \right) $. In this case, the skeleton ${{\Im}_{\sigma }}$ maps to the skeleton  $\Gamma_ {{\Re }}$.

iii) It is clear that the function $B_{k}(Z)$ vanishes at the points  $Z =A_{j}$  , which makes the third statement true.

iv)  Note that, $Z,W\in {{\Re }_{I}}\left( m,m \right) $ implies that  $\left( ZW \right)\in {{\Re }_{I}}\left( m,m \right)$. The validity of this assertion follows from the statement above and from  the Theorem 4.
\end{proof}
\section{The Horwitz-Rubel theorem for the Blaschke matrix product}


In this subsection, we prove an analogue of the Horwitz-Rubel theorem for the Blaschke matrix product. It should be noted that in this case, Blaschke matrix products cannot be represented in the form (\ref{115}), (\ref{116}), due to the noncommutativity of matrices.

Let
$${{B}_{1}}\left( Z \right)=\prod\limits_{j=1}^{k}{\left[ \left( Z-{{P}_{j}} \right){{\left( I-P_{j}^{*}Z \right)}^{-1}} \right]},$$
$${{B}_{2}}\left( Z \right)=\prod\limits_{j=1}^{k}{\left[ \left( Z-{{Q}_{j}} \right){{\left( I-Q_{j}^{*}Z \right)}^{-1}} \right]},$$
where ${{P}_{j}},{{Q}_{j}}\in {{\Re }_{I}}\left( m,m \right),j=1,2,...,k,$ Blaschke matrix products of degree $k$ for the matrix unit disc.

The following theorem is true.


\begin{theorem} Let be ${{P}_{j}},{{Q}_{j}}$ be diagonal matrices. If   ${{B}_{1}}\left( {{\nu }_{j}} \right)={{B}_{2}}\left( {{\nu }_{j}} \right)$ for diagonal matrices ${{\nu }_{j}}$ from ${{\Re }_{I}}\left( m,m \right)$, such that $\det \left( {{\nu }_{l}}-{{\nu }_{j}} \right)\ne 0,l\ne j,\,l,j=1,...,2,$ then
$${{B}_{1}}\left( Z \right)\equiv {{B}_{2}}\left( Z \right).$$
\end{theorem}


To prove Theorem 5, we first prove the following.

\begin{lemma} If ${{P}_{j}},{{Q}_{j}}$ are diagonal matrices, then for ${{B}_{1}}$ and ${{B}_{2}}$ there exists at least one matrix $U\in \Gamma_{\Re}$, such that
$${{B}_{1}}\left( U \right)={{B}_{2}}\left( U \right).$$
\end{lemma}

\begin{proof} The equality ${{B}_{1}}\left( Z \right)={{B}_{2}}\left( Z \right)$  is equivalent to the equality
\begin{equation}\label{*}
\prod\limits_{j=1}^{k}{\left[ \left( Z-{{P}_{j}} \right){{\left( I-P_{j}^{*}Z \right)}^{-1}} \right]}=\prod\limits_{j=1}^{k}{\left[ \left( Z-{{Q}_{j}} \right){{\left( I-Q_{j}^{*}Z \right)}^{-1}} \right]}.
\end{equation}
For $Z{{Z}^{*}}=I$, i.e. $Z\in \Gamma_{\Re}$, the matrices  $Z-{{P}_{j}}$ and $Z-{{Q}_{j}}$ are nondegenerate, i.e.
$\det \left( Z-{{P}_{j}} \right)\ne 0$, $\det \left( Z-{{Q}_{j}} \right)\ne 0,j=1,2,...,k,$ so from (\ref{*}) we get
\begin{equation}\label{117}
\prod\limits_{j=1}^{k}{\left[ \left( Z-{{P}_{j}} \right){{\left( I-P_{j}^{*}Z \right)}^{-1}} \right]}{{\left[ \prod\limits_{j=1}^{k}{\left[ \left( Z-{{Q}_{j}} \right){{\left( I-Q_{j}^{*}Z \right)}^{-1}} \right]} \right]}^{-1}}=I.
\end{equation}
Consider a matrix function of a scalar variable $t\in \mathbb{C}$ of the form
$$\varphi \left( t \right)=\prod\limits_{j=1}^{k}{\left[ \left( tI-{{P}_{j}} \right){{\left( I-P_{j}^{*}t \right)}^{-1}} \right]}{{\left[ \prod\limits_{j=1}^{k}{\left[ \left( tI-{{Q}_{j}} \right){{\left( I-Q_{j}^{*}t \right)}^{-1}} \right]} \right]}^{-1}}.$$
Note that, the elements of $\varphi \left( t \right)$ are holomorphic in the disc  $\left\{ \left| t \right|<1 \right\}=\mathbb U$ and continuous functions on $\overline{\mathbb U}$.

If $t\overline{t}=1$ then, due to the diagonality of  ${{P}_{j}},{{Q}_{j}}$, the function $\varphi \left( t \right)$ can be presented in the form
$$\varphi \left( t \right)=\prod\limits_{j=1}^{k}{\left[ \left( tI-{{P}_{j}} \right){{\left( tI-{{Q}_{j}} \right)}^{-1}} \right]}{{\left[ \overline{\left( tI-{{P}_{j}} \right){{\left( tI-{{Q}_{j}} \right)}^{-1}}} \right]}^{-1}}.$$
Then the elements of  $\varphi \left( t \right)$ are functions of the form
$${{\varphi }_{ss}}\left( t \right)={{\prod\limits_{j=1}^{k}{\left[ \left( tI-{{P}_{j}} \right){{\left( tI-{{Q}_{j}} \right)}^{-1}} \right]}}_{ss}}{{\left[ \overline{\left( tI-{{P}_{j}} \right){{\left( tI-{{Q}_{j}} \right)}^{-1}}} \right]}_{ss}}^{-1},$$ $s=1,2,...,m.$
Assuming that  $${{P}_{j}}=\left( p_{j}^{ss} \right),{{Q}_{j}}=\left( q_{j}^{ss} \right),i=1,2,...,m,j=1,2,...,k,$$ we can rewrite the function ${{\varphi }_{ss}}\left( t \right)$ in the form
$${{\varphi }_{ss}}\left( t \right)=\prod\limits_{j=1}^{k}{\left[ \frac{t-p_{j}^{ss}}{t-q_{j}^{ss}}/{{\overline{\left( \frac{t-p_{j}^{ss}}{t-q_{j}^{ss}} \right)}}^{-1}} \right]}=\frac{{{g}_{ss}}\left( t \right)}{\overline{{{g}_{ss}}\left( t \right)}},$$
where $${{g}_{ss}}\left( t \right)=\prod\limits_{j=1}^{k}{\frac{t-p_{j}^{ss}}{t-q_{j}^{ss}}}.$$
If we represent ${{g}_{ss}}\left( t \right)$  in exponential form, i.e.
$${{g}_{ss}}\left( t \right)=\left| {{g}_{ss}}\left( t \right) \right|{{e}^{i\arg {{g}_{ss}}\left( t \right)}},$$
then
 ${{\varphi }_{ss}}\left( t \right)={{e}^{2i\arg {{g}_{ss}}\left( t \right)}}$
(here we assume that $\arg {{g}_{ss}}\left( t \right)$ exists). We are interested in the values  ${{t}_{s}},\left| {{t}_{s}} \right|=1,$ such that  ${{\varphi }_{ss}}\left( {{t}_{s}} \right)=1$ for each fixed $s$, i.e.
$$\cos \left( 2\arg {{g}_{ss}}\left( {{t}_{s}} \right) \right)+i\sin \left( 2\arg {{g}_{ss}}\left( {{t}_{s}} \right) \right)=1.$$
It follows $$\cos \left( 2\arg {{g}_{ss}}\left( {{t}_{s}} \right) \right)=1,$$
$$\sin \left( 2\arg {{g}_{ss}}\left( {{t}_{s}} \right) \right)=0.$$
Hence, $\arg {{g}_{ss}}\left( {{t}_{i}} \right)=\pi n$ for some $n$. Thus, \[{{\varphi }_{ss}}\left( {{e}^{i{{\psi }_{s}}}} \right)=1,\] where \[{{e}^{i{{\psi }_{s}}}}={{t}_{s}},0\le \psi_{s} \le 2\pi.\]
Let us now show that for every fixed $s,$ $s=1,2,...,m,$  there exists  $\arg {{g}_{ss}}\left( t \right)$.  Denote  $$\delta =\underset{\begin{smallmatrix}
 t\in \left\{ p_{j}^{ss},q_{j}^{ss} \right\} \\
 \zeta \in \partial \mathbb U
\end{smallmatrix}}
{\mathop{\min }}\,\left| t-\zeta  \right|>0.$$ Then the functions  ${{g}_{ss}}\left( t \right)$ are holomorphic and do not vanish in \[{{\mathbb U}_{\delta }}=\left\{ t\in \mathbb{C}:\left| t \right|>1-\delta  \right\},\] since ${{P}_{j}},{{Q}_{j}}\in {{\Re }_{I}}\left( m,m \right)$, thats why $p_{j}^{ss},q_{j}^{ss}\in \mathbb U=\left\{ \left| t \right|<1 \right\}.$
Consider the analytic function
$$\ln {{g}_{ss}}\left( t \right)={{h}_{ss}}\left( t \right),$$
which admits a holomorphic branch in ${{\mathbb U}_{\delta }}$, corresponding to the function ${{g}_{ss}}\left( t \right)$ on $\infty $. Since  ${{g}_{ss}}\left( \infty  \right)=1$, then we choose a branch with the condition ${{h}_{ss}}\left( \infty  \right)=0$.
Let  $T(t)= \frac{1}{t}.$ Then
$$T\left( {{\mathbb U}_{\delta }} \right)=\left\{ t\in \mathbb{C}:\left| t \right|<\frac{1}{1-\delta } \right\}\supset \mathbb U, T\left( \partial \mathbb U \right)=\partial \mathbb U$$ and $$T\left( \infty  \right)=0.$$
It's clear that
$${{h}_{ss}}\left( \frac{1}{t} \right)\in \mathcal{O}\left( \left\{ t\in \mathbb{C}:\left| t \right|<\frac{1}{1-\delta } \right\} \right).$$
Therefore, by virtue of the mean value theorem for harmonic functions (we apply this theorem in $\mathbb U$ for harmonic function $\operatorname{Im}{{h}_{ss}}\left( \frac{1}{t} \right)$)  we get
$$\operatorname{Im}{{h}_{ss}}\left( \infty  \right)=\frac{1}{2\pi }\int\limits_{0}^{2\pi }{\operatorname{Im}{{h}_{ss}}\left( {{e}^{i\psi }} \right)d\psi }.$$
From here
$$0=\frac{1}{2\pi }\int\limits_{0}^{2\pi }{\operatorname{Im}{h}_{ss}\left( {{e}^{i\psi }} \right)d\psi }.$$
Therefore, there exists a value ${{\psi}_{s}},\, 0\le{{\psi}_{s}}\le2\pi$, such that $\operatorname{Im}{{h}_{ss}}\left( {{e}^{i{{\psi }_{s}}}} \right)=0.$ Since
\[{{h}_{ss}}\left( t \right)=\ln {{g}_{ss}}\left( t \right)=\ln \left| {{g}_{ss}}\left( t \right) \right|+i\arg {{g}_{ss}}\left( t \right)=\operatorname{Re}{{h}_{ss}}\left( t \right)+i\cdot \operatorname{Im}{{h}_{ss}}\left( t \right),\]
then
\[{{h}_{ss}}\left( {{e}^{i{{\psi }_{s}}}} \right)=\operatorname{Re}{{h}_{ss}}\left( {{e}^{i{{\psi }_{s}}}} \right)+i\cdot \operatorname{Im}{{h}_{ss}}\left( {{e}^{i{{\psi }_{s}}}} \right)=\ln \left| {{g}_{ss}}\left( {{e}^{i{{\psi }_{s}}}} \right) \right|=\ln {{g}_{ss}}\left( {{e}^{i{{\psi }_{s}}}} \right)=0.\]
Thus ${{\varphi }_{ss}}\left( {{e}^{i{{\psi }_{s}}}} \right)=1,s=1,2,...,m.$
Now, using the Cauchy formula for functions of matrices (see e.g. \cite{Lankas82}) for all $Z\in {{\Re }_{I}}\left( m,m \right)$,  we have \begin{equation}\label{118}
\varphi \left( Z \right)=\frac{1}{2\pi i}\int\limits_{\partial \mathbb U}{\varphi \left( t \right){{\left( tI-Z \right)}^{-1}}dt}.
\end{equation}
The left side of (\ref{118}) is the same as the left side of (\ref{117}) and it is continuous on  $\Gamma_{\Re}$.
At the points ${{t}_{s}}={{e}^{i{{\psi }_{s}}}},0\le {{\psi }_{s}}\le 2\pi ,$ the function ${{\varphi }_{ss}}\left( t \right)$ takes the value 1, hence at the points
\[U=\left( \begin{matrix}
   {{e}^{i{{\psi }_{1}}}} & 0 & ... & 0  \\
   0 & {{e}^{i{{\psi }_{2}}}} & ... & 0  \\
   ... & ... & ... & ...  \\
   0 & 0 & ... & {{e}^{i{{\psi }_{m}}}}  \\
\end{matrix} \right)\in \Gamma_{\Re}\]
the value of the matrix function $\varphi(z)$ equals to $I$, i.e.,
 $$\varphi \left( Z \right)=\prod\limits_{j=1}^{k}{\left[ \left( Z-{{P}_{j}} \right){{\left( I-P_{j}^{*}Z \right)}^{-1}} \right]}{{\left[ \prod\limits_{j=1}^{k}{\left[ \left( Z-{{Q}_{j}} \right){{\left( I-Q_{j}^{*}Z \right)}^{-1}} \right]} \right]}^{-1}}=I.$$
This means that  $${{B}_{1}}\left( U \right)={{B}_{2}}\left( U \right).$$ The lemma is proved.
\end{proof}

\section{Proof of Theorem 5}

 Let $$R\left( Z \right)={{B}_{1}}\left( Z \right)\cdot B_{2}^{-1}\left( Z \right).$$ Then
$$R\left( Z \right)=\prod\limits_{j=1}^{k}{\left[ \left( Z-{{P}_{j}} \right){{\left( I-P_{j}^{*}Z \right)}^{-1}} \right]}{{\left[ \prod\limits_{j=1}^{k}{\left[ \left( Z-{{Q}_{j}} \right){{\left( I-Q_{j}^{*}Z \right)}^{-1}} \right]} \right]}^{-1}}.$$
Since ${{B}_{1}}\left( Z \right)$ and ${{B}_{2}}\left( Z \right)$ are automorphisms of ${{\Re }_{I}}\left( m,m \right)$, then $$R\left( Z \right){{R}^{*}}\left( Z \right)=I$$ on $\Gamma_{\Re}$.  Hence $$R\left( Z \right){{R}^{*}}\left( {{Z}^{{{*}^{-1}}}} \right)=I$$ on $\Gamma_{\Re}$. By the condition of the theorem we have  $R\left( {{\nu }_{j}} \right)=I$. Hence, $R\left( \nu _{j}^{{{*}^{-1}}} \right)=I$ for all $j=1,2,...,k$. Consider the matrix function  $$R\left( t \right)=\left( {{R}_{ss}}\left( t \right) \right)_{1}^{m},\,  t\in \mathbb{C}.$$ Put
\[{{\nu }_{j}}=\left( \begin{matrix}
   {{\left( {{\nu }_{j}} \right)}_{11}} & 0 & ... & 0  \\
   0 & {{\left( {{\nu }_{j}} \right)}_{22}} & ... & 0  \\
   ... & ... & ... & ...  \\
   0 & 0 & ... & {{\left( {{\nu }_{j}} \right)}_{mm}}  \\
\end{matrix} \right).\]
Then ${{R}_{ss}}\left( {{\left( {{\nu }_{j}} \right)}_{ss}} \right)=1$ and ${{R}_{ss}}\left( \frac{1}{{{\left( \overline{{{\nu }_{j}}} \right)}_{ss}}} \right)=1$ for all $s=1,..., m, j=1,..., k.$

By virtue of Lemma 1, we have $R\left( U \right)=I$ for some
\[U=\left( \begin{matrix}
   {{e}^{i{{\psi }_{1}}}} & 0 & ... & 0  \\
   0 & {{e}^{i{{\psi }_{2}}}} & ... & 0  \\
   ... & ... & ... & ...  \\
   0 & 0 & ... & {{e}^{i{{\psi }_{m}}}}  \\
\end{matrix} \right)\in \Gamma_{\Re}\]
i.e. ${{R}_{ss}}\left( {{e}^{i{{\psi }_{s}}}} \right)=1, s=1,..., m.$
Therefore, for every fixed $s$, the rational function ${{R}_{ss}}\left( t \right)$  of degree $2k$ takes the value 1 at $2k+1$ points ${{e}^{i{{\psi }_{s}}}};{{\left( {{\nu }_{1}} \right)}_{ss}},...,{{\left( {{\nu }_{k}} \right)}_{ss}};\frac{1}{{{\left( {{\overline{\nu }}_{1}} \right)}_{ss}}},...,\frac{1}{{{\left( {{\overline{\nu }}_{k}} \right)}_{ss}}}.$
Since ${{R}_{ss}}\left( t \right)\equiv 1$, we have $${{B}_{1}}\left( Z \right)\equiv {{B}_{2}}\left( Z \right).$$  The Theorem 5 is proved.

\section{Some problems of multidimensional complex analysis in matrix domains}

Recently, scientists have achieved many significant results in the classical fields, and at the same time, a number of open problems have been formulated. The importance of studying classical domains is that they are not reducible. These domains are, "model" domains of the multidimensional space ${{\mathbb{C}}^{n}}$ in some meaning. Moreover, scientific work are  developing in matrix balls, associated with classical domains in ${{\mathbb{C}}^{n}}\left[ m\times m \right]$. In this section, we present such areas and some unsolved issues of multivariable complex analysis on them.

E. Cartan gave a classification of all bounded symmetric domains (see \cite{Cart,Hua}):
$${{\Re }_{I}}\left( m,k \right)=\left\{ Z\in \mathbb{C}\left[ m\times k \right]:{{I}^{\left( m \right)}}-Z{{\overline{Z}}^{\prime }}>0 \right\},$$
$${{\Re }_{II}}\left( m \right)=\left\{ Z\in \mathbb{C}\left[ m\times m \right]:{{I}^{\left( m \right)}}-Z\overline{Z}>0,\,\forall {Z}'=Z \right\},$$
$${{\Re }_{III}}\left( m \right)=\left\{ Z\in \mathbb{C}\left[ m\times m \right]:{{I}^{\left( m \right)}}+Z\overline{Z}>0,\,\forall {Z}'=-Z \right\},$$
\[{{\Re }_{IV}}\left( n \right)=\left\{ \,\,z\in {{\mathbb{C}}^{n}}\,:\,{{\left| \left\langle z, z \right\rangle  \right|}^{2}}-2{{\left| z \right|}^{2}}+1>0,\,\,\,\left| \left\langle z,z \right\rangle  \right|<1 \right\},\]
where ${{I}^{\left( m \right)}}$ is the identity matrix of order  $m$, ${{\overline{Z}}^{\prime }}$ is the complex conjugate matrix of the transposed matrix  ${Z}'$\footnote{Then condition $H>0$ for a Hermitian matrix $H$ means, as usual, that  $H$ is positive definite.}. In the theory of functions of single complex variable, is often studied the theory of functions in the unit circle, since in general, all symmetric domains are equivalent to the unit circle  $\mathbb U=\left\{ z\in \mathbb{C}:\left| z \right|<1 \right\}$, hence the of symmetric classical domains given above play an important role in the theory of functions of several complex variables. All these domains are homogeneous, symmetric, convex, complete, circular domains centered at ${\mathrm O}$ (${\mathrm O}$ is zero matrix) and biholomorphically non-equivalent, so the complex analysis for  each of is then constructed differently.

In the space   ${{\mathbb{C}}^{n}}\left[ m\times m \right]$ we define the matrix "scalar" \,product as follows:
\[\left\langle Z,W \right\rangle ={{Z}_{1}}W_{1}^{*}+\cdots +{{Z}_{n}}W_{n}^{*},\]
where $W_{j}^{*}$ is the  conjugated and transpose matrix of the matrix  ${{W}_{j}}$. The matrix balls  $\mathbb{B}_{m,n}^{(1)}$, $\mathbb{B}_{m,n}^{(2)}$  and $\mathbb{B}_{m,n}^{(3)}$  of the first, second and third types, respectively, have the form (see  \cite{Ser88}, \cite{KhKSh}):
$$\mathbb{B}_{m,n}^{(1)}=\left\{ ({{Z}_{1}},...,{{Z}_{n}})=Z\in
{{\mathbb{C}}^{n}}\left[ m\times m \right]:I-\left\langle Z,Z
\right\rangle >0  \right\},$$
$$\mathbb{B}_{m,n}^{(2)}=\left\{ Z\in
{{\mathbb{C}}^{n}}\left[ m\times m \right]:I-\left\langle Z,Z
\right\rangle >0,\,\,\,\,\,\forall {{{{Z}'}}_{\nu}}={{Z}_{\nu
}},\,\,\nu =1,...,n \right\},$$
and
$$\mathbb{B}_{m,n}^{(3)}=\left\{ (Z\in
{{\mathbb{C}}^{n}}\left[ m\times m \right]:I-\left\langle Z,Z
\right\rangle >0,\, \right.\forall \left. {{{{Z}'}}_{\nu}}=-{{Z}_{\nu }},\,\,\,\,\nu =1,...,n \right\}.$$


\ \ Skeletons (Shilov boundaries) of matrix balls   $\mathbb{B}_{m,n}^{(k)}$, denoted by  $\mathbb{X}_{m,n}^{(k)},\,\,\,\,k=1,2,3$, i.e.,
$$\mathbb{X}_{m,n}^{(1)}=\left\{ Z\in {{\mathbb{C}}^{n}}\left[ m\times m
\right]:\left\langle Z,Z \right\rangle =I \right\},$$
$$\mathbb{X}_{m,n}^{(2)}=\left\{ Z\in {{\mathbb{C}}^{n}}\left[ m\times m
\right]:\left\langle Z,Z \right\rangle
=I,\,\,\,\,\,{{{{Z}'}}_{\nu}}={{Z}_{\nu }},\,\,\nu =1,2,...,n
\right\},$$ $$\mathbb{X}_{m,n}^{(3)}=\left\{ Z\in {{\mathbb{C}}^{n}}\left[
m\times m \right]:\left\langle Z,Z \right\rangle =I,\,
\right.\left. {{{{Z}'}}_{\nu}}=-{{Z}_{\nu }},\,\,\,\,\nu =1,2,...,n
\right\}.$$
 \ \ Note that, $\mathbb{B}_{1,1}^{(1)}$, $\mathbb{B}_{1,1}^{(2)}$  and    $\mathbb{B}_{2,1}^{(3)}$ are unit circles, and   $\mathbb{X}_{1,1}^{(1)}$, $\mathbb{X}_{1,1}^{(2)}$, and  $\mathbb{X}_{2,1}^{(3)}$ are unit circles in the complex plane $\mathbb{C}$.

\ \ If $n=1, m>1$, then   $\mathbb{B}_{m,1}^{(k)},\,\,\,\,k=1,2,3$ are classical domains of the first, second and third types (according to the classification of E. Cartan, and the skeletons   $\mathbb{X}_{m,1}^{(1)}$, $\mathbb{X}_{m,1}^{(2)}$, and   $\mathbb{X}_{m,1}^{(3)}$  are unitary, symmetric unitary, and skew-symmetric unitary matrices, respectively.

The first type of matrix ball was considered by A. G. Sergeev \cite{Ser88}, G. Khudaiberganov \cite{KhKSh}, Kosbergenov \cite{Kos99}. Integral formulas for matrix balls of the second and third types were obtained in  \cite{KhRUS}, \cite{KhURZM}. In paper \cite{UJvo19} the volumes of matrix balls of the second and third types are calculated. The total volumes of these domains are necessary to find holomorphic kernels of integral formulas for these domains (Bergmann, Cauchy--Szeg\"{o}, Poisson kernels, etc.). Note that the matrix balls of the first, second, and third types are biholomorphically equivalent to the matrix Siegel domain of the second type in the space ${{\mathbb{C}}^{n}}\left[ m\times m \right]$ (see, for example,  \cite{KhKSh}). Are the matrix balls $\mathbb{B}_{m,n}^{(1)}$, $\mathbb{B}_{m,n}^{(2)}$ and $\mathbb{B} _{m,n}^{(3)}$ classical domains? This question still remains open. However, in particular, they coincide with the classical domains, and this consideration is reflected in this article. In general, although the matrix domains have not been sufficiently studied, they still remain one of the most relevant areas in which leading scientists conduct their research.

Naturally the question arises, how the Blaschke product will look like in the above classical domains ${{\Re }_{I}}\left( m,k \right),{{\Re }_{II}}\left( m \right),{{\Re }_{II}}\left( m \right),{{\Re }_{II}}\left( n \right)$ and in matrix balls $\mathbb{B} _{m,n}^{(1)}$, $\mathbb{B}_{m,n}^{(2)}$, $\mathbb{B}_{m,n}^{(3 )}$, and what do the properties of the Blaschke product look like in the areas mentioned in the previous sections of this article? We present some unsolved problems related to classical domains and matrix balls $\mathbb{B}_{m,n}^{(1)}$, $\mathbb{B}_{m,n}^{(2) }$, $\mathbb{B}_{m,n}^{(3)}$ associated with classical domains:

\textbf{1.}\emph{ Obtaining a criterion for holomorphic continuation to the matrix ball $\mathbb{B}_{m,n}^{(2)}$ and $\mathbb{B}_{m,n}^{ (3)}$ of functions defined on a piece of its skeleton (Shilov's boundary).}

\textbf{2.} \emph{Define the Laplace-Hua Lo-Keng operator for the matrix balls $\mathbb{B}_{m,n}^{(2)}$ and $\mathbb{B}_{m, n}^{(3)}$.}

\textbf{3.} \emph{Introduce the concepts of $A-$ harmonic functions for matrix balls $\mathbb{B}_{m,n}^{(2)}$ and $\mathbb{B}_{m, n}^{(3)}$.}

\textbf{4.} \emph{Solve the Dirichlet problem for the matrix balls $\mathbb{B}_{m,n}^{(2)}$ and $\mathbb{B}_{m,n}^ {(3)}$.}


\textbf{5.} \emph{Obtain  of the expansion analogues of the Laurent–Hua Lo-Keng series in matrix balls $\mathbb{B}_{m,n}^{(2)}$ and $\mathbb{B}_{ m,n}^{(3)}$.}

\textbf{6.} \emph{Describe the convergence domains of Laurent series with respect to matrix balls $\mathbb{B}_{m,n}^{(1)}$, $\mathbb{B}_{m,n}^ {(2)}$ and $\mathbb{B}_{m,n}^{(3)}$.}


\textbf{7.} \emph{Find an orthonormal basis for $\mathbb{B}_{m,n}^{(2)}$ and $\mathbb{B}_{m,n}^{(3) }$.}






%In this article, we discuss some aspects of   commutative algebras.

%\begin{theorem}
%Text of the theorem
%\end{theorem}

%\begin{proof}
%Text of the proof.
%\end{proof}
%\subsection{Introduction and problem statement}
\bigskip
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\end{document}
