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Edit File: Karpenko_MAA.tex

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\begin{document}
%% \markboth{{\footnotesize{\it \textbf{I. I. Karpenko}}}}
%% {{\footnotesize{\it \textbf{A Weyl function for a
%%         skew-symmetric operator on a Hilbert quaternion bimodule}}}}

\title{A Weyl function for a skew-symmetric operator on a Hilbert
  quaternion bimodule}

\author{I. I. Karpenko}

\address{V.~I.~Vernadsky Taurida  National University,\\
  Department of Mathematics and Information Technology,\\
  Vernadsky ave., 4, Simferopol', Crimea, Ukraine, 95007\\
  e-mail: \textit{i\_karpenko@inbox.ru}}

\begin{abstract}
  In this paper, we introduce a definition of a boundary-value
  triple and a Weyl function for a skew-symmetric operator acting on
  a Hilbert quaternion bimodule. We investigate analytic properties
  of the Weyl function, study components of its spectral
  decomposition, construct the corresponding spectral measures with
  values in the algebra of bounded $\mathbb{H}$-linear operators.
\end{abstract}

\section*{Introduction}

Quantum mechanics problems, in the quaternion setting, lead to a
study of differential operators and their proper extensions on
Hilbert quaternion modules~\cite{karpleo}. Usually, such
differential operators are skew-symmetric and, in such a case, the
study can not be reduced to the symmetric case, as it is done for
complex Hilbert spaces. A series of works of M.~M.~Malamud and
V.~A.~Derkach~\cite{karpmal,karpder} have created a certain
direction for studying proper extensions of symmetric operators on
complex Hilbert spaces. It is based on the notion of a
boundary-value triple and an abstract Weyl function. This leads to a
problem of constructing a similar machinery for studying proper
extensions of skew-symmetric operators on Hilbert quaternion
modules.

The purpose of this paper is to investigate the properties of the
Weyl function for a skew-symmetric operator, which are due to
the quaternion linearity nature of the operator.

\section{Defect submodules for skew-symmetric
  operators on $\mathbb{H}$-bimodules}

Let $\mathbb{H}$ be the division ring of quaternions,
$[q]=\{uq\overline{u}\mid u\in \mathbb{H},\ |u|=1\}$ be the
conjugacy class of the element $q$ in the multiplicative quaternion
group $\mathbb{H}^*$. Everywhere in the sequel,
$\mathbb{F}\supset\R$ is a nonreal field in $\mathbb{H}$ with a
real basis $1,f$, where $f^2=-1$. Then, by~\cite{karpsuht}, there
exists a quaternion $\varepsilon$, $ \varepsilon^2=-1$, such that
the quaternions $1,\ f,\ \varepsilon,\ f\varepsilon$ form a real
basis in $\mathbb{H}$, and any quaternion $q$ can be represented as
$q=u+v\varepsilon$, where $u,v\in \mathbb{F}$. An element of the
field $\mathbb{F}$ given by $\mathbb{F}(q):=u$ is called the
$\mathbb{F}$-part of the quaternion $q$.

Consider a Hilbert quaternion bimodule $H$ with a scalar product
$\langle \cdot,\cdot\rangle$.

\begin{definition}\label{karpdef1}
  An operator $A$ acting on a Hilbert quaternion bimodule $H$ is
  called \textit{skew-symmetric} if its domain $D(A)$ is dense in
  $H$, and for any vectors $x,y\in D(A)$,
  $$
  \langle Ax,y\rangle+\langle x,Ay\rangle=0.
  $$
\end{definition}

It is easy to show that, for a closed skew-symmetric linear operator
$A$, any quaternion $q$ with a nonzero real part is a regular type
point, that is, $||(A-R_q)x||\geq k||x||$ for all $ x\in D(A)$.
Here the set of values, $\mathfrak{Im}(A-R_{q})$, of the operator
$A-R_{q}$, where $R_{q}x:=xq$, is a closed set in $H$. In the
sequel, we always assume that the operator $A$ is closed.

As was shown in~\cite{karpenko}, for a skew-symmetric operator $A$,
the set
$\mathfrak{Im}(A-R_{q})\cap\mathfrak{Im}(A-R_{\overline{q}})$ is a
right $\mathbb{H}$-submodule that is independent of the choice of
the representative $q$ in the conjugacy class $[q]$.

Hence, we give the following definition.

\begin{definition}\label{karpdef1}
  The set $\mathfrak{N}_{q}=\Bigl(\mathfrak{Im}(A-R_{q})\cap
  \mathfrak{Im}(A-R_{\overline{q}})\Bigr)^{\bot}$, for a
  skew-symmetric operator $A$, is called \textit{a deficiency
    submodule} corresponding to the class $[q]$, where $q$ is a
  quaternion with nonzero real part.
\end{definition}

Consider the symplectic image $A^s$ of the operator $A$ with respect
to a fixed field $\mathbb{F}$. Recall that the operator $A^s$ acts
on the $\mathbb{F}$-bimodule $H^\mathbb{F}$ that is also a Hilbert
bimodule with respect to the compatible scalar product
$(x,y):=\mathbb{F}\Bigl(\langle x,y\rangle\Bigr)$. It is clear that
the operator $A^s$ is also  skew-symmetric  on
$H^\mathbb{F}$.

Take a point $q\in \mathbb{F}$ with a nonzero real part and,
following the customary notations, denote by
$\mathfrak{N}_{q}^{\circ}=
\mathfrak{Im}(A-R_{\overline{q}})^{{\bot}_\mathbb{F}}$ the
deficiency $\mathbb{F}$-submodule of the operator $A^s$. (Here
the symbol ${\bot}_\mathbb{F}$ denotes the orthogonal complement in
$H^\mathbb{F}$.) It is clear that ${\mathfrak{N}_{q}^{\circ}=
  \ker(A^*-R_{q}).}$

In the case where the skew-symmetric operator is a symplectic image
of a quaternion linear operator, we have
\begin{equation}\label{karpeq1}
  \mathfrak{N}_{q}^{\circ}=
  R_{\varepsilon}\mathfrak{N}_{\overline{q}}^{\circ}.
\end{equation}

In~\cite{karpenko} it was found that there is a relation between
deficiency submodules of the operator $A$ and its symplectic image
$A^s$. Namely, we have the following decomposition:
\begin{equation}\label{karpeq2}
  \mathfrak{N}_{q}=\mathfrak{N}_{q}^{\circ}\dotplus_\mathbb{F}
  \mathfrak{N}_{\overline{q}}^{\circ}.
\end{equation}

\begin{remark}\label{karprem1}
  In the case where $q\in \mathbb{R}$, the set
  $\mathfrak{Im}(A-R_{q})$ is a right $\mathbb{H}$-module and the
  deficiency module corresponding to $[q]=\{q\}$ is given by
  $\mathfrak{N}_{q}=\mathfrak{Im}(A-R_{q})^{\bot}$. Using
  decomposition~\eqref{karpeq2} we have in this case that
  $\mathfrak{N_{q}}=\mathfrak{N_{q}^{\circ}}$.
\end{remark}

The following directly follows from the above
properties~\cite{karpenko}.

\begin{proposition}\label{karpprop1}
  For a nonreal $q$, we have the identity
  \begin{equation}\label{karpeq3}
    \dim\left[\mathfrak{N}_{q}:\mathbb{H}\right]
    =\dim\left[\mathfrak{N}_{q}^{\circ}:\mathbb{F}\right].
  \end{equation}
\end{proposition}

Properties of deficiency submodules of symmetric operators that act
on modules over a field can naturally be transferred to deficiency
modulus of skew-symmetric operators, namely, the dimension of the
deficiency submodule $\mathfrak{N}_{q}^{\circ}$ remains constant for
all $q\in \mathbb{H},\,\,\re q>0$ and also for all $q\in
\mathbb{H},\,\,\re q<0$. Using Proposition~\ref{karpprop1}, a
similar conclusion can be made for a skew-symmetric operator acting
on a quaternion bimodule. Here, however, there is an important
exception for real numbers, namely, if $q\in \mathbb{R},\,\,p\not\in
\mathbb{R}$ and $q\re p>0$, then
$\dim\left[\mathfrak{N}_{p}:\mathbb{H}\right]=
2\dim\left[\mathfrak{N}_{q}:\mathbb{H}\right]$.

Denote $n_{+}=\dim \left[\mathfrak{N}_{q}:\mathbb{H}\right],\,\,
q>0;\,\, n_{-}=\dim\left[\mathfrak{N}_{q}:\mathbb{H}\right],\,\,
q<0$, and call the numbers $n_{\pm}$ \textit{deficiency numbers} of
the skew-symmetric operator $A$.

For a skew-symmetric operator $A$ we have von Neumann formulas,
namely, for any $q\in \mathbb{H}$, $\re q\not=0$, we have
$$
D(A^*)=D(A)\dot{+}\mathfrak{N}_{q}^{\circ}
\dot{+}\mathfrak{N}_{-\overline{q}}^{\circ}\,.
$$

Then any vector $x$ in $D(A^*)$ can be represented as
\begin{equation}\label{karpeq4}
  x=x_0+x_q+x_{-\overline{q}},
\end{equation}
where $x_0\in D(A),\, x_q\in \mathfrak{N}_{q}^{\circ},\,
x_{-\overline{q}}\in \mathfrak{N}_{-\overline{q}}^{\circ}\,.$ Here,
it should be noted that
\begin{equation}\label{karpeq5}
  A^*x=-Ax_0+x_qq-x_{-\overline{q}}\,\overline{q}\,.
\end{equation}
The proof of formulas~\eqref{karpeq4},~\eqref{karpeq5} is conducted
similarly to the proof of the corresponding formulas for a symmetric
operator on a complex Hilbert space.

\begin{remark}\label{karprem23}
  By Remark~\ref{karprem1}, for a real $q$ we have
  $$
  D(A^*)=D(A)\dot{+}\mathfrak{N}_{q}\dot{+}\mathfrak{N}_{-q},
  $$
  where $\mathfrak{N}_{q}\,\, \mathfrak{N}_{-q}$ are already
  $\mathbb{H}$-submodules of $H$.
\end{remark}

\section{Boundary-value triples of a skew-symmetric operator}

Let $A$ be a skew-symmetric operator with equal finite deficiency
indices $n_{+}=n_{-}=n$.

\begin{definition}\label{kdef2.1}
  A triple $\Pi=\{\mathcal{H},\Gamma_0,\Gamma_1\}$ consisting of an
  auxiliary right Hilbert $\mathbb{H}$-module $\mathcal{H}$ and
  linear mappings ${\Gamma_i: D(A^*)\to\mathcal{H},\;i=0,1}$, is
  called a boundary-value triple for a self-adjoint operator $A^*$
  if the following two conditions are satisfied:
  \begin{enumerate}
  \item $\langle A^*x,y\rangle+\langle x,A^*y\rangle=
    \langle\Gamma_{0}x,\Gamma_{1}y\rangle+\langle
    \Gamma_{1}x,\Gamma_{0}y\rangle$, where $x,y\in D(A^*)$;
  \item the mapping
    $\Gamma:=\{\Gamma_0,\Gamma_1\}:D(A^*)\to\mathcal{H}
    \oplus\mathcal{H}$, $\Gamma x:=\{\Gamma_0x,\Gamma_1x\}$ is
    surjective.
    \end{enumerate}
\end{definition}

It is not difficult to show that, for any skew-symmetric operator,
such a definition of the boundary-value triple is meaningful. For
example, the triple $\{\mathfrak{N}_{1},\Gamma_0,\Gamma_1\}$ is a
boundary-value triple for the self-adjoint operator $A^*$.

\begin{proposition}\label{kprop2.1}
  For any boundary-value triple
  $\Pi=\{\mathcal{H},\Gamma_0,\Gamma_1\}$ for a skew-symmetric
  operator $A^*$, we have
  \begin{equation}
    \ker\Gamma_0\cap\ker\Gamma_1=D(A).
  \end{equation}
\end{proposition}

\begin{proof}
  Let $x\in \ker\Gamma_0\cap \ker\Gamma_1$. Then, for any vector
  $y\in D(A^*)$ using~\ref{kdef2.1}, we have
  \begin{displaymath}
    0=\langle A^*x,y\rangle+\langle x,A^*y\rangle=
    (A^*x,y)-((A^*x)\varepsilon,y)\varepsilon+(x,A^*y)
    -(x\varepsilon,A^*y)\varepsilon.
  \end{displaymath}
  This identity is equivalent to the system
  \begin{equation}\label{keq2.1}
    \begin{cases}
      &(A^*x,y)+(x,A^*y)=0,\\
      &((A^*x)\varepsilon,y)+(x\varepsilon,A^*y)=0.
    \end{cases}
  \end{equation}
  Apply the following von Neumann formulas to the vectors $x,y$:
  $x=x_0+x_1+x_{-1}$, $y=y_0+y_1+y_{-1}$, where $x_0,y_0\in D(A)$, $
  x_1,y_1\in \mathfrak{N}_{1}$, $ x_{-1},y_{-1}\in
  \mathfrak{N}_{-1}\,.$ Here
  $A^*x=-Ax_0+x_1-x_{-1},\,A^*y=-Ay_0+y_1-y_{-1}.$

Then the first identity in system~\eqref{keq2.1} gives
  $$
  (x_1,y_1)-(x_{-1},y_{-1})=0 \qquad\forall y_1\in
  \mathfrak{N}_{1},\, \forall y_{-1}\in \mathfrak{N}_{-1}\,.
  $$
  By first setting $y_1=x_1$, $y_{-1}=0$, we obtain $x_1=0$.
  Then, setting $y_1=0$, $y_{-1}=x_{-1}$, we have $x_{-1}=0$.
  Consequently, $x=x_0\in D(A)$.

Conversely, let $x\in D(A)$. Then $\langle A^*x,y\rangle+\langle
    x,A^*y\rangle=0$ for all $ y\in D(A^*)$. In this case, we also
  have for any $y\in D(A^*)$ that
  \begin{equation}\label{keq2.2}
    {\langle\Gamma_{0}x,\Gamma_{1}y\rangle+
      \langle\Gamma_{1}x,\Gamma_{0}y\rangle=0}.
  \end{equation}

Let ${\Gamma_0x=z}$. For the pair $\{0,z\}$ there exists a vector
  $u\in D(A^*)$ such that $\Gamma_0u=0$, $\Gamma_1u=z$. Setting
  $y=u$ in~\eqref{keq2.2} we have
  \begin{equation*}
    0=\langle\Gamma_{0}x,\Gamma_{0}x\rangle.
  \end{equation*}
  Hence, it follows that $\Gamma_0x=0$. One similarly proves that
  $\Gamma_1x=0$.  Consequently, $x\in \ker\Gamma_0\cap
  \ker\Gamma_1$.
\end{proof}

\begin{proposition}\label{kprop2.2}
  We have the following identity:
  \begin{equation}
    \ker\Gamma_0+\ker\Gamma_1=D(A^*).
  \end{equation}
\end{proposition}

\begin{proof}
  Let $x\in D(A^*)$, $y=\Gamma_1x\in\mathcal{H}$.
  By~\eqref{kdef2.1}, for the pair $\{0,y\}$ there exists a vector
  $z_0\in D(A^*)$ such that $\Gamma_0z_0=0$, $\Gamma_1z_0=y$. It
  clear that $z_0\in \ker\Gamma_0$. Denote by $z_1=x-z_0$. Then
  $\Gamma_1z_1=\Gamma_1x-\Gamma_1z_0=0$ and, consequently, $z_1\in
  \ker\Gamma_1$. Thus, $x=z_0+z_1,\;z_t\in \ker\Gamma_t,\;t=0,1$.
\end{proof}

\begin{proposition}\label{kprop2.4}
  If $\re q\not=0$, then
  \begin{equation}\label{keq2.6}
    \ker\Gamma_t\cap\mathfrak{N}_{q}^{\circ}=\{0\},\quad   t=0,1.
  \end{equation}
  In particular, for a real nonzero $q$,
  \begin{equation}\label{keq2.6}
    \ker\Gamma_t\cap\mathfrak{N}_{q}=\{0\},\quad   t=0,1.
  \end{equation}
\end{proposition}

\begin{proof}
  Let ${x_q\in \ker\Gamma_0\cap\mathfrak{N}_{q}^{\circ}}$. In this
  case, $\langle\Gamma_{0}x_q,\Gamma_{1}x_q\rangle+
  \langle\Gamma_{1}x_q,\Gamma_{0}x_q\rangle =0 $. Then
  \begin{equation*}
    \langle A^*x_q,x_q\rangle+\langle x_q,A^*x_q\rangle=
    \langle x_qq,x_q\rangle+\langle x_q,x_qq
    \rangle=2\re q\langle x_q,x_q\rangle=0.
  \end{equation*}
  Since $\re q\not=0$, we have $x_{q}=0$.
\end{proof}

\begin{corollary}\label{kcor2.1}
  If $\re q\not=0$, we have the identity
  \begin{equation}
    \ker\Gamma_t\cap\Bigl(D(A)\dotplus
    \mathfrak{N}_{q}^{\circ}\Bigr)=D(A),\quad   t=0,1.
  \end{equation}
  In particular, for a real nonzero $q$,
  \begin{equation}
    \ker\Gamma_i\cap\Bigl(D(A)\dotplus\mathfrak{N}_{q}\Bigr)=D(A).
  \end{equation}
\end{corollary}

\begin{proof}
  The statement directly follows from
  Propositions~\ref{kprop2.1} and~\ref{kprop2.4}.
\end{proof}

\begin{theorem}\label{kteo2.1}
  For any real nonzero $q$,
  \begin{equation}
    \dim[\mathcal{H}:\mathbb{H}]=\dim[\mathfrak{N}_q:\mathbb{H}]=n.
  \end{equation}
\end{theorem}

\begin{proof}
  Let $e_1,\ldots,e_r$ be a linearly independent system in
  $\mathcal{H}$. Then, for a pair $\{e_t,0\}$ there exists a vector
  $x_t=x^{(t)}_0+x^{(t)}_{q}+x^{(t)}_{-q}\in D(A^*)$ such that
  \begin{equation}\label{keq2.13}
    \Gamma_0x_t=e_t,\;\Gamma_1x_t=0.
  \end{equation}
  Let us prove that $\{x_{q}^{(t)}\}^r_{t=1}$ is a linearly
  independent system in $\mathfrak{N}_q$. Assume that
  $\sum\limits_{t=1}^{r}x_{q}^{(t)}\alpha_t=0$. Then
  \begin{equation*}
    \sum\limits_{t=1}^{r}x_t\alpha_t=
    \sum\limits_{t=1}^{r}\Bigl(x_{0}^{(t)}+x_{-q}^{(t)}\Bigr)
    \alpha_t
    \in D(A)\dotplus\mathfrak{N}_{-q}.
  \end{equation*}
  By using identities~\eqref{keq2.13}, we get
  ${\sum\limits_{t=1}^{r}x_t\alpha_t\in\ker\Gamma_1}$.

Using Corollary~\ref{kcor2.1} we get that
  $\sum\limits_{k=1}^{r}x_t\alpha_t\in D(A)$. Hence, by
  Proposition~\ref{kprop2.1} and relations~\eqref{keq2.13},
  \begin{equation*}
    0=\Gamma_0\Bigl(\sum\limits_{t=1}^{r}x_t\alpha_t\Bigr)=
    \sum\limits_{t=1}^{r}e_t\alpha_t.
  \end{equation*}
  Consequently, $\alpha_t=0,\;t=\overline{0,r}$, and
  ${\dim[\mathcal{H}:\mathbb{H}]\leq\dim
    [\mathfrak{N}_q:\mathbb{H}]}$.

Conversely, let $x_{q}^{(1)},\ldots,x_{q}^{(r)}$ be a linearly
  independent system in $\mathfrak{N}_q$. Denote
  $e_t=\Gamma_0x_{q}^{(t)}$. Let us prove that $\{e_t\}_{t=1}^{r}$
  is a system linearly independent in $\mathcal{H}$. Assume that
  ${\sum\limits_{t=1}^re_t\alpha_t=
    \Gamma_0\Bigl(\sum\limits_{t=1}^rx_{q}^{(t)} \alpha_t\Bigr)=0}$.
  Consequently, ${\sum\limits_{t=1}^rx_{q}^{(t)}\alpha_t\in\ker
    \Gamma_0\cap\mathfrak{N}_{q}}$.  Using
  Proposition~\ref{kprop2.4} we get
  ${\sum\limits_{t=1}^rx_{q}^{(t)}\alpha_t}=0$, whence,
  $\alpha_t=0,\;t=\overline{1,r}$. Consequently,
  ${\dim[\mathfrak{N}_q:\mathbb{H}]
    \leq\dim[\mathcal{H}:\mathbb{H}]}$. Hence,
  $\dim[\mathcal{H}:\mathbb{H}]=\dim[\mathfrak{N}_q:\mathbb{H}]$.
 \end{proof}

\begin{remark}\label{krem2.1}
  Any boundary-value triple $\{\mathcal{H},\Gamma_0,\Gamma_1\}$
  defines a pair of the operators
  $A_i:=-A^*\upharpoonright\ker\Gamma_t$, $t=0,1$, that are
  skew-symmetric extensions of the operator $A$.
\end{remark}

\section{The Weyl function for a skew-symmetric operator}

Let $A$ be a closed skew-symmetric operator,
$\Pi=\{\mathcal{H},\Gamma_0,\Gamma_1\}$ a boundary-value triple for
the operator $A^*$. For an arbitrary quaternion $q\in\pi(A),\,\,\re
q\not=0$, where $\pi(A)$ is the regularity field for the operator
$A$, consider an operator $M(q)$ defined by
\begin{equation}\label{keq3.1}
  M(q)\Gamma_0x_{q}=
  \Gamma_1x_{q},\quad x_{q}\in\mathfrak{N}^{\circ}_{q}.
\end{equation}
If $q$ is a nonreal quaternion, it naturally generates the field
$\mathbb{F}=\mathbb{R}\langle1,q\rangle$, and
$\mathfrak{N}^{\circ}_{q}$ is a $\mathbb{F}$-submodule. It is clear
that identity~\eqref{keq3.1} in this case gives rise to an
$\mathbb{F}$-linear operator. For a real quaternion $q$, we already
have deficiency $\mathbb{H}$-submodules and, consequently, the
operator $M(q)$ is linear over the division ring $\mathbb{H}$.

Hence, relation~\eqref{keq3.1} defines a mapping
$M(\cdot):\pi(A)\to[\mathcal{H}^F]$, for nonreal $q$, and a mapping
$M(\cdot):\pi(A)\to[\mathcal{H}]$ for a real $q$.

\begin{definition}\label{karpdef3.1}
  For a closed skew-symmetric operator $A$, the operator-valued
  function $M(\cdot)$ is called a \textit{Weyl function}
  corresponding to the boundary-value triple $\Pi$.
\end{definition}

Let us study properties of the operator-valued function $M$. To this
end, fix a nonreal field $\mathbb{F}\supset\R$ in $\mathbb{H}$, with
a real basis formed by quaternions $1,f$, where $f^2=-1$, and
introduce the following notation:
$\pi_{\mathbb{F}}(A):=\pi(A)\cap\mathbb{F}$. Consider the operator
$B:=R_fA^s$, which is clearly a symmetric operator on the
$\mathbb{F}$-module $\mathcal{H}^F$, and
$\Pi=\{\mathcal{H},R_f\Gamma_0,\Gamma_1\}$ is a boundary-value
triple for the operator $B^*$. Denote the deficiency submodules of
the operator $B$ by $\mathfrak{N}_{q}^{\circ}(B)$ Since, for any
$q\in \pi_{\mathbb{F}}(A)$, $(B^*-R_{q})x=-R_f(A^*-R_{fq})x$, we
have
\begin{equation}\label{keq3.2}
  \mathfrak{N}_{q}^{\circ}(B)=\mathfrak{N}_{fq}^{\circ}.
\end{equation}

This relation allows to find a relation between the Weyl
functions for the operators $A$ and $B$ in the field $\mathbb{F}$.

Indeed, by~\cite{karpmal}, the Weyl function for the operator $B$
is defined as follows:
\begin{equation*}
  M_B(q)R_f\Gamma_0x_{q}=\Gamma_1x_{q},
  \quad x_{q}\in\mathfrak{N}^{\circ}_{q}(B).
\end{equation*}
Using identities~\eqref{keq3.2}, the latter relation means that
\begin{equation}\label{keq3.3}
  M(fq)=M_B(q)R_f,\,\,M(q)=M_B(-qf)R_f.
\end{equation}

As is known~\cite{karpmal}, the Weyl function for a symmetric
operator $B$ is a strong Nevanlinna function for which $0\in
\rho(\im(M_B(i)))$. Consequently, the function $M$ has the following
properties.
\begin{enumerate}
\item The function $M$ is analytic on the domain
  $\mathbb{F}^{+}=\{q\in \mathbb{F}\,|\,\re q>0\}$ and its values
  are operators accretive on $\mathcal{H^{\mathbb{F}}}$, i.e., $\re
  M(q)\geq0,\,\,q\in \mathbb{F}^{+}$;
\item $0\in \rho(M(-1))$;
\item $M(-\overline{q})=-M(q)^*$.
\end{enumerate}

Using the definition of an analytic function of a quaternion
variable~\cite{karpsuht}, property~(1) can be reformulated as
follows.

The function $M$ is analytic on the domain $\mathbb{H}^{+}=\{q\in
\mathbb{H}\,|\,\re q>0\}$ and its values, on this domain, are
operators accretive on $\mathcal{H^{\mathbb{F}}}$, that is, $\re
M(q)\geq0,\,\,q\in \mathbb{H}^{+}$.

It is also clear that property~(3) allows to talk about analyticity
of the function $M$ on the domain $\mathbb{H}^{-}=\{q\in
\mathbb{H}\,|\,\re q<0\}$, and $\re M(q)\leq0,\,\,q\in
\mathbb{H}^{-}$.

Let us study the way the values of the function $M$ depend on the
choice of the field. Consider the fields
$\mathbb{F}_t=\mathbb{R}\langle 1,f_t\rangle,\,\,t=1,2$. The
quaternions $f_1$ and $f_2$ belong to the same conjugacy class and,
consequently, satisfy the relation $f_2=uf_1\overline{u},\,\,|u|=1$.
Then we also have that $\mathbb{F}_2=u\mathbb{F}_1\overline{u}$. Let
$q\in \mathbb{F}_1, \,\,p=uq\overline{u}\in \mathbb{F}_2$.

By the definition of the operator-valued function $M$, we have that
if $g=M(q)h$ then $h=\Gamma_0x_{q},\,\,g=\Gamma_1x_{q},\,\,x_{q}\in
\mathfrak{N}_{q}^{\circ}$. In this case, using the identity
$\mathfrak{N}_{p}^{\circ}=R_{\overline{u}}\mathfrak{N}_{q}^{\circ}$
we have $R_{\overline{u}}h=\Gamma_0(x_{q}
{\overline{u}})=\Gamma_0x_{p},\,\, R_{\overline{u}}g=\Gamma_1(x_{q}
{\overline{u}})=\Gamma_1x_{p},\,\,x_{p}\in
\mathfrak{N}_{p}^{\circ}$. Hence,
$M(p)R_{\overline{u}}h=R_{\overline{u}}g=R_{\overline{u}}M(q)h$.
This permits to formulate the following proposition addressing the
above problem.

\begin{proposition}\label{kprop3.1}
  Let $q\in \mathbb{F}_1,\,\,p=uq\overline{u}\in \mathbb{F}_2$. Then
  $M(p)=R_{\overline{u}}M(q)R_{u}$.
\end{proposition}

Hence, values the function $M$ takes on every field $\mathbb{F}$ are
determined up to a unitary equivalence that is defined by operators
of the form $R_u,|u|=1$.

Since the quaternions $q$ and $\overline{q}$ belong to the same
conjugacy class, and $\overline{q}=\varepsilon q(-\varepsilon)$, a
similar reasoning leads to the relation
\begin{equation}\label{keq3.4}
  M(\overline{q})R_{\varepsilon}=R_{\varepsilon}M(q).
\end{equation}

These results permit to consider the problem of integral
representation of the Weyl function for a skew-symmetric operator.
Let us first use the spectral decomposition of the Weyl function for
a symmetric operator on a Hilbert space, see e.g.~\cite{karpmal}. To
this end, let us fix a nonreal field $\mathbb{F}\supset\R$ in
$\mathbb{H}$ and consider the symmetric operator $B:=R_fA^s$ on the
$\mathbb{F}$-module $\mathcal{H}^F$. Since the field $\mathbb{F}$
and the complex field are isomorphic, we can assert that there is an
operator-valued measure
$\Sigma_{\mathbb{F}}:\mathfrak{B}(\mathbb{R})\rightarrow
[\mathcal{H}^{\mathbb{F}}]$, non-orthogonal and unbounded in
general, and an operator $C_0\in
[\mathcal{H}^{\mathbb{F}}],\,\,C_0=C_0^*$, such that
\begin{equation}\label{keq3.5}
  M_B(z)=C_0+\int_{\mathbb{R}}\Bigl(\frac{1}{t-z}-\frac{t}{1+t^2}\Bigr)
  \Sigma_{\mathbb{F}}(dt),\,\,z\in\mathbb{F}^{+}\cup\mathbb{F}^{-}.
\end{equation}
Here the function $M_B$ defines, in a unique way, an unbounded
spectral measure $\Sigma_{\mathbb{F}}$ via the Stieltjes inversion
formula
\begin{equation}\label{keq3.6}
  \Sigma_{\mathbb{F}}\bigl((a,b)\bigr)=
  s-\lim_{\delta\to+0}s-\lim_{\eta\to+0}\:\frac{1}{\pi }
  \!\!\!\int_{[a+\delta,\,b-\delta]}\!\!\!\!\!\!
  \im M_B(x+\eta f)\,dx.
\end{equation}
Relation~\eqref{keq3.3}, in this case, allows to obtain the
following representation for the function $M$:
\begin{equation}\label{keq3.7}
  M(q)=C_0R_f+\int_{\mathbb{R}}
  \Bigl(R_{(t+qf)^{-1}}-R_{t(1+t^2)^{-1}}\Bigr)
  R_f\Sigma_{\mathbb{F}}(dt),\qquad
  q\in\mathbb{F}^{+}\cup\mathbb{F}^{-}.
\end{equation}

Let us use the following natural notations in~\eqref{keq3.5}:
$D_0:=C_0R_f$, $ tf\rightarrow \lambda$, $\textbf{f}:=\{tf\,|\,t\in
\mathbb{R}\}$, $
\sigma_{\mathbb{F}}(d\lambda):=\Sigma_{\mathbb{F}}(-fd\lambda)$.
Then the corresponding integral representation over the field
$\mathbb{F}$ will become
\begin{equation}\label{keq3.8}
  M(q)=D_0+\int_{\textbf{f}}\Bigl(R_{(q-\lambda)^{-1}}-
  R_{\lambda(1-\lambda^2)^{-1}}\Bigr)
  \sigma_{\mathbb{F}}(d\lambda),\qquad
  q\in\mathbb{F}^{+}\cup\mathbb{F}^{-}.
\end{equation}
Here $D_0\in [\mathcal{H}^{\mathbb{F}}]$, $D_0^*=-D_0$, and
$\sigma_{\mathbb{F}}:\mathfrak{B}(\textbf{f})\rightarrow
[\mathcal{H}^{\mathbb{F}}]$ is an operator-valued measure. Applying
the identity $\im M_B(q)=-\re M(qf)=\re M(\overline{q}f)$ in the
Stieltjes inversion formula for this operator-valued measure, we
have
\begin{equation}\label{keq3.9}
  \sigma_{\mathbb{F}}\bigl((af,bf)\bigr)=
  s-\lim_{\delta\to+0}s-\lim_{\eta\to+0}\:\frac{1}{\pi }
  \!\!\!\int_{[(a+\delta),\,(b-\delta)]}\!\!\!\!\!\!
  \re M(\eta+ xf)\,dx.
\end{equation}

In what follows, we will need special properties of the spectral
measure $\sigma_{\mathbb{F}}$ that come from specifics of
$\mathbb{H}$-modules.

\begin{proposition}\label{kprop3.2}
  If $\sigma_{\mathbb{F}}$ is an operator-valued measure over
  $\mathbb{F}\supset\R$ for a skew-symmetric operator $A\in L[H]$,
  then for any set $\alpha\in\mathfrak{B}(\mathbf{f})$ we have
  \begin{equation}\label{keq3.10}
    R_{\varepsilon}\sigma_{\mathbb{F}}(\alpha)=
    \sigma_{\mathbb{F}}(-\alpha)R_{\varepsilon}.
  \end{equation}
\end{proposition}

\begin{proof}
  Indeed, using relations~\eqref{keq3.4}, we have
  $R_{\varepsilon}\re M(\eta+ xf)=\re M(\eta- xf)R_{\varepsilon}$,
  hence, since $R_{\varepsilon}$ is continuous, we have
  \begin{align*}
    &R_{\varepsilon}\sigma_{\mathbb{F}}(af,bf)=
    s-\lim_{\delta\to+0}s-\lim_{\eta\to+0}\:\frac{1}{\pi}
    \!\!\!\int_{[a+\delta,\,b-\delta]}\!\!\!\!\!\!
    \re M(\eta+ xf)\,dx=\\
    &=s-\lim_{\delta\to+0}s-\lim_{\eta\to+0}\:\frac{1}{\pi }
    \!\!\!\int_{[a+\delta,\,b-\delta]}\!\!\!\!\!\!
    \re M(\eta- xf)\,dxR_{\varepsilon}.
  \end{align*}
  Replacing $y=-x$ in the last expression we get
  \begin{align*}
    &R_{\varepsilon}\sigma_{\mathbb{F}}(af,bf)=s-\lim_{\delta\to+0}s-\lim_{\eta\to+0}\:
    \frac{1}{\pi}    \!\!\!\int_{[-b+\delta,\,-a-\delta]}\!\!\!\!\!\!
    \re M(\eta+y)\,dxR_{\varepsilon}=\\
    &=\sigma_{\mathbb{F}}(-bf,-af)R_{\varepsilon}
  \end{align*}
  for any real $a$, $b$, $a<b$. Since $R_{\varepsilon}$ is countably
  additive and continuous, the latter implies~\eqref{keq3.7} for any
  Borel subset $\alpha$ of the ``axis'' $\mathbf{f}$.
\end{proof}

\begin{corollary}\label{karpcor3.1}
  The operator $D_0$ in representation~\eqref{keq3.8} is
  $\mathbb{H}$-linear.
\end{corollary}

\begin{proof}
  Indeed, by Proposition~\ref{kprop3.2},
  \begin{align*}
    &-R_{\varepsilon}M(\overline{q})R_{\varepsilon}=
    -R_{\varepsilon}D_0R_{\varepsilon}-\int_{\textbf{f}}\Bigl(
    R_{\varepsilon} (R_{(\overline{q}-\lambda)^{-1}}-
    R_{\lambda(1-\lambda^2)^{-1}}\Bigr)
    \sigma_{\mathbb{F}}(d\lambda)R_{\varepsilon}=\\
    &-R_{\varepsilon}D_0R_{\varepsilon}-\int_{\textbf{f}}\Bigl
    (R_{(q+\lambda)^{-1}}+ R_{\lambda(1-\lambda^2)^{-1}}\Bigr)
    R_{\varepsilon}\sigma_{\mathbb{F}}(d\lambda)R_{\varepsilon}=\\
    &-R_{\varepsilon}D_0R_{\varepsilon}-\int_{\textbf{f}}\Bigl
    (R_{(q+\lambda)^{-1}}+ R_{\lambda(1-\lambda^2)^{-1}}\Bigr)
    \sigma_{\mathbb{F}}(-d\lambda)R_{\varepsilon}^2=\\
    &-R_{\varepsilon}D_0R_{\varepsilon}+\int_{\textbf{f}}\Bigl
    (R_{(q+\lambda)^{-1}}+ R_{\lambda(1-\lambda^2)^{-1}}\Bigr)
    \sigma_{\mathbb{F}}(-d\lambda).
  \end{align*}
  Making the change of variable $\lambda\rightarrow
  \overline{\lambda}=-\lambda$ in the integrand, we have
  \begin{align*}
    &-R_{\varepsilon}M(\overline{q})R_{\varepsilon}=
    -R_{\varepsilon}D_0R_{\varepsilon}+\int_{\textbf{f}}\Bigl
    (R_{(q-\lambda)^{-1}}- R_{\lambda(1-\lambda^2)^{-1}}\Bigr)
    \sigma_{\mathbb{F}}(d\lambda).
  \end{align*}
  Then, by~\eqref{keq3.4}, $D_0=-R_{\varepsilon}D_0R_{\varepsilon}$
  or $D_0R_{\varepsilon}=R_{\varepsilon}D_0$, whence it follows that
  the operator $D_0$ is homogeneous with respect to an arbitrary
  quaternion.
\end{proof}

\begin{proposition}\label{kprop3.3}
  Let $\mathbb{F}_j$, $j=1,2$, be arbitrary fields in $\mathbb{H}$,
  and $\mathbb{F}_2=u\mathbb{F}_1\overline{u},\,\,|u|=1$. Then the
  spectral measures corresponding to the Weyl function for a
  skew-symmetric operator $A$ satisfies the relation
  \begin{equation}\label{keq3.11}
    \sigma_{\mathbb{F}_2}(\alpha)=R_{\overline{u}}\,\sigma_{\mathbb{F}_1}(\alpha)R_u,
    \,\,\alpha\in \mathfrak{B}(\mathbf{f}).
  \end{equation}
\end{proposition}
%
\begin{proof}
  The proof immediately follows from Proposition~\ref{kprop3.1} and
  the Stieltjes inversion formula~\eqref{keq3.9}.
\end{proof}

Let now $\mathbb{F}$ be a fixed field, $\sigma_F$ an unbounded
spectral measure for the Weyl function $M$.

\begin{theorem}\label{karpteo3.1}
  Let $\mathbf{f}_+=\{\tau f\mid \tau\geqslant 0\}$. Then the
  operator-valued function
  $\mathfrak{B}(\mathbf{f}_+)\ni\alpha\mapsto
  \sigma(\alpha)=\sigma_{\mathbb{F}}(\alpha)+
  \sigma_{\mathbb{F}}(-\alpha)$ satisfies the following:
  \begin{enumerate}
  \item $\sigma(\alpha)\in L[\mathcal{H}]\
    (\alpha\in\mathfrak{B}(\mathbf{f}_+));$
  \item $\forall\alpha\in\mathfrak{B}(\mathbf{f}_+)$, $E(\alpha)$ is
    a nonnegative self-adjoint operator on $\mathcal{H}$;
  \item the function $\sigma$ is countably additive
    \textnormal{(}with respect to the strong operator
    topology\textnormal{)};
  \item the function $\sigma$ does not depend on the choice of the
    field $\mathbb{F}$.
\end{enumerate}
\end{theorem}

\begin{proof}
  \textit{Property (1)}. Proposition~\ref{kprop3.2} gives at once
  that
  \begin{equation*}
    \sigma(\alpha)R_{\varepsilon}=\sigma_{\mathbb{F}}(\alpha)R_{\varepsilon}+
    \sigma_{\mathbb{F}}(-\alpha)R_{\varepsilon}=
    R_{\varepsilon}\sigma_{\mathbb{F}}(-\alpha)+R_{\varepsilon}\sigma_{\mathbb{F}}(\alpha)
    =R_{\varepsilon}\sigma(\alpha).
  \end{equation*}
  Whence it follows that the mapping $\sigma(\alpha)$ is homogeneous
  for any quaternion.

\textit{Properties (2) and (3)} follow form the corresponding
  properties of the operator-valued measure $\sigma_{\mathbb{F}}$.

Let us now prove property~(4). Indeed, if
  $\mathbb{F}'=u\mathbb{F}\overline{u}$, then
  \begin{align*}
    &\sigma'(\alpha)=\sigma_{\mathbb{F}'}(\alpha)+\sigma_{\mathbb{F}'}(-\alpha)=
    R_{\overline{u}}\Bigl(\sigma_{\mathbb{F}}(\alpha)+\sigma_{\mathbb{F}}(-\alpha)\Bigr)R_U=
    R_{\overline{u}}\sigma(\alpha)R_U=\sigma(\alpha) ,
  \end{align*}
  since the operator $\sigma(\alpha)$ is linear.
\end{proof}

Hence, the operator-valued function
$\sigma\colon\mathfrak{B}(\mathbf{f}_+)\mapsto L[\mathcal{H}]$ is an
operator-valued measure. It makes sense to call this function a
\textit{spectral-valued measure} for the Weyl function $M$ of a
skew-symmetric operator $A$. The support of this spectral measure
completely determines classes of the spectrum of the self-adjoint
extension $A_0$ described in Remark~\ref{karprem23}.

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\bibitem{karpsuht} I. I. Karpenko, A. I.  Sukhtaev, D. L.
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\end{document}