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\documentclass{article}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amsthm}

\author{Vasyl Ostrovskyi}
\title{Centered one-parameter semigroups}

\newtheorem{definition}{Definition}
\newtheorem{proposition}{Proposition}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{remark}{Remark}
\theoremstyle{remark}
\newtheorem{example}{Example}

\begin{document}
\maketitle
\begin{abstract}
We introduce the class of centered one-parameter semigroups which are
continuous analogue of a single centered operator, and study their
properties. In particular, we prove the Wold decomposition for such
semigroups and give complete description of one-parameter centered
semigroups of partial isometries.
\end{abstract}

\section*{Introduction}

Let $T$ be a bounded operator and write $T_k = T^k$. Recall
\cite{mormu} that $T$ is centered if the operators $(T_kT_k^*,
T_l^*T_l)$, $k$, $l\ge1$, form a commuting family.

In this paper we introduce a continuous analogue of a centered
operator---one-parameter centered semigroup.

\begin{definition} Let $T_t$, $t\ge0$, be a strongly continuous
  semigroup of operators in a Hilbert space $H$. We say that the
  semigroup $T_t$ is centered, if the operators $(T_tT^*_t, T^*_s
  T_s)$, $t$, $s\ge0$, form a commuting family.
\end{definition}

\section{Centered one-parameter semigroups}

Let $T_t$, $t\ge0$ be a strongly continuous one-parameter semigroup. 
For every $t$
consider a polar decomposition $T_t = U_t C_t$, where $C$ is a
self-adjoint non-negative operator, $U_t$ is a partial isometry, and
$\ker U_t = \ker C_t$.

For a general semigroup $T_t$, neither $U_t$, nor $C_t$ need not be 
a semigroup. 
However, in the centered case, we have the following statement.

\begin{theorem}
Let $T_t$, $t\ge0$, be a strongly continuous semigroup, $T_t=U_tC_t$
be the polar decomposition. $T_t$ is 
centered if  and only if $U_t$, $t\ge0$, is a semigroup.
\end{theorem}

\begin{proof}
We start from the following fact.

\begin{lemma}\label{lemma1}
Any one-parameter semigroup $U_t$ of partial isometries is centered.
\end{lemma}

\begin{proof}
The statement follows from the fact that a semigroup of partial isometries 
is inverse (see, e.g., \cite{pat}). However, we give a direct proof.

Write $P_t= U_tU_t^*$, $Q_t=U_t^*U_t$, $t\ge0$. Since $U_t$ is a partial
isometry, $P_t$ and $Q_t$ are projections. We need to prove that this
family of projections is commuting.

a) First show that $P_t$ and $Q_t$, $t\ge0$ are commuting families. In
fact, we show that $P_tP_s = P_s$, $Q_tQ_s = Q_s$ for $t<s$. Then
passing to adjoints we get the needed commutation.

Let $t<s$. Then
\[
P_tP_s = P_tU_{s}U_{s}^* = P_t U_tU_{s-t}U_{s}^*.
\]
But since $P_t$ is a projection onto the image of $U_t$, we have
$P_tU_t= U_t$, and therefore, $P_tP_s= P_s$. For $Q_t$, the proof is
quite similar.

Now we show that $P_tQ_s = Q_tP_s$ for $t$, $s\ge0$. We have
\[
Q_{t+s}=U_{t+s}^*U_{t+s} = U_t^*Q_sU_t.
\]
Since $Q_{t+s}$ is a projection, $Q_{t+s} = Q_{t+s}^2$, or
\[
U_t^*Q_sU_t =  U_t^*Q_sU_tU_t^* Q_s U_t = U_t^* Q_s P_tQ_sU_t.
\]
Multiplying this equality by $U_t$ from the left, and by $U_t^*$ from
the right, we get
\[
P_tQ_sP_t = P_tQ_sP_tQ_sP_t = (P_tQ_sP_t)^2,
\]
which is possible if and only if $P_t$ and $Q_s$ commute. 
\end{proof}

Let $T_t= U_tC_t$, $t\ge0$, be a centered semigroup.
Show that $U_t$ is a semigroup. First observe that the operator
$U_tC_tU^*_t$ commutes with the family. Indeed, it is a non-negative
square root of $T_tT_t^*$. Then, since $U_s^*U_s$ is a projection onto
the co-kernel of $C_s$, we have
\begin{align*}
T_tT_s & = U_t C_t U_s C_s = U_t C_t U_s C_s (U_s^*U_s) = U_t C_t
(U_sC_s U_s^*) U_s 
\\
&= U_t (U_s C_s U_s^*) C_t U_s = U_t U_s (C_s U_s^*
C_t U_s).
\end{align*}
From the relation 
\[
C_s (U_s^* C_t U_s) = U_s^* U_s C_s U_s^* C_t U_s = U_s^* C_t U_s
C_sU_s^* U_s = (U_s^* C_tU_s) C_s
\]
we see that $C_s U_s^* C_t U_s$ is a non-negative self-adjoint
operator, its kernel coincides with $\ker U_tU_s$, and its square is
\[
C_s U_s^* C_t U_s C_s U_s^* C_t U_s = T_s^*T_t^*T_tT_s = C_{t+s}^2.
\]
Therefore, $U_{t+s} = U_tU_s$, $t,s \ge 0$.

For $t \to 0$, the operator $C_t = (T^*_t T_t)^{1/2}$ converges
strongly to the identity; from this, one can easily deduce that $U_t$
is strongly continuous.

Now let $U_t$ be a semigroup. By the Lemma above we see that it is 
centered. From $T_s T_t = T_{t+s}$ we have
\begin{equation}\label{eq:utc}
U_tC_t U_sC_s = U_{t+s}C_{t+s},
\end{equation}
and multiplying by $U_s^*U_t^*$ from the right, we get
\[
U_s^* C_t U_s C_s = C_{t+s}.
\]
Since $C_{t+s}$ is self-adjoint, taking the adjoints in the latter
equality, we conclude that the operators $U_s^* C_t U_s$ and $C_s$
commute. Then for $t>s$
\[
 C_t C_s = C_{s+(t-s)}C_s = U^*_s C_{t-s} U_s C_s C_s = C_s U^*_s
 C_{t-s} U_s C_s = C_s C_t,
\]
or $[T_t^*T_t, T_s^*T_s]=0$.

Again, multiplying \eqref{eq:utc} by $U_t^*$ from the right, and by
$U_s^*$ from the left, we get 
\begin{equation}\label{eq:ucu}
C_t U_s C_s U_s^* = U_s C_{t+s} U_s^*.
\end{equation}
Taking adjoints and observing that the right-hand side is
self-adjoint, we conclude that $C_t$ and $U_sC_s U_s^*$ commute, or
$[T_t^*T_t, T_sT_s^*]=0$.

Finally, applying \eqref{eq:ucu}, we have for $t>s$
\begin{multline*}
U_sC_s U_s^* U_tC_tU_t^*  = U_s C_s U_{t-s} C_t U_t^* =
U_sC_s(U_{t-s} C_{(t-s)+s} U_{t-s}^*) U_s^* 
\\
=U_sC_s(U_{t-s}C_{t-s}U^*_{t-s} C_s) U_s^* = U_s (C_s U_{t-s} C_{t-s}
U^*_{t-s}) C_s U^*_s 
\\
= U_s U_{t-s} C_t U_{t-s}^* C_s U^*_s 
= U_tC_tU_t^* U_sC_sU^*_s, 
\end{multline*}
or $[T_sT^*_s, T_tT^*_t]=0$. Thus, the semigroup $T_t$ is centered.
\end{proof}

\begin{remark}
If we assume that $C_t$ is also a semigroup, then we get that $C_t$ 
commutes with $U_t$ and $T_t$ is a semigroup of normal operators. 
Conversely, if $T_t$ is a semigroup of normal operators, then $T_t$ 
is centered, and both $U_t$ and $C_t$ are semigroups.
\end{remark}

\section{Semigroups of partial isometries}

Before studying general centered semigroups, we study semigroups 
of partial isometries, which are all centered. We decompose any such 
semigroup into a direct sum of up to four invariant components, and give 
a complete description of each component up to a unitary equivalence.

Notice that the problem of a description, up to a unitary equivalence, 
of a single partial isometry $U$ is very complicated (wild) problem 
\cite{halm} (see 
also \cite{ourbook} for a discussions of wild problems in 
$*$-representation theory). However, a semigroup of partial isometries 
admits a complete description up to a unitary equivalence (in this case $U^k$
is also a partial isometry for any $k\ge0$).

First, we establish a decomposition of any semigroup of partial isometries
into certain standard components (Wold decomposition).

\begin{proposition}
Let $U_t$ be a strongly continuous semigroup of partial isometries in
a Hilbert space $H$. Then $H$ can be decomposed into a direct sum of
invariant with respect to $U_t$, $U_t^*$, $t\ge0$, subspaces,
$H=H_0\oplus H_+\oplus H_- \oplus H_1$ such that:

in $H_1$ the semigroup $U_t$, $t\ge0$ is unitary, i.e., all operators
$U_t$ are unitary\textup;

in $H_+$ $U_t$ is a semigroup of completely non-unitary isometries\textup;

in $H_-$ $U_t^*$ is a semigroup of completely non-unitary
isometries\textup;

in $H_0$ $U_t$ is a semigroup of completely non-unitary partial 
isometries for which $U_t\to 0$, $U_{t}^* \to 0$, $t\to \infty$ strongly.
\end{proposition}

\begin{proof}
Indeed, let $U_t$ be a one-parameter semigroup of partial
isometries. Consider  the projections $P_t = U_tU_t^*$ and $Q_{t} =
U_t^*U_t$, $t\ge0$. According to the proof of Lemma~\ref{lemma1},
both $P_t$ and $Q_t$ are decreasing  families of projections 
as $t \to \infty$. Therefore, there exist projections 
\[
P_\infty=\lim_{t\to\infty} P_t \quad\text{and} 
\quad Q_\infty=\lim_{t\to\infty} Q_t.
\]
These projection obviously commute with $P_t$, $Q_t$, and with each other.
We claim that these projections commute with $U_t$, $U_t^*$, $t\ge0$, 
as well. Indeed, for any $t\ge0$, $s\ge0$ we have
\begin{align*}
U_tP_s &=U_tU_sU_s^* =U_tU_t^*U_tU_sU_s^* = U_tU_sU_s^*U_t^*U_t 
\\
&=P_{t+s}U_t,\quad t,s\in\mathbb{R},
\\
U_tQ_s &=U_tU_s^*U_s=U_tU_t^*U_{s-t}^*U_{s-t}U_t = 
U_{s-t}^*U_{s-t}U_tU_t^*U_t
\\
&= Q_{s-t} U_t, \quad s>t,
\end{align*}
and passing to the limit as $s \to\infty$, we conclude that $U_t$ commutes 
with $P_\infty$, $Q_\infty$. Taking adjoints, we see that $U^*_t$ commutes 
with $P_\infty$, $Q_\infty$ as well for all $t\ge0$.

Now introduce the projections $R_1=P_\infty Q_\infty$, 
$R_+ = P_\infty (I- Q_\infty)$, $R_- = (I- P_\infty) Q_\infty$,  
$R_0 = (I- P_\infty) (I-Q_\infty)$. Then $R_1 +R_++R_-+R_0=I$.
Denote the images of these projections by $H_1$, $H_+$, $H_-$, $H_0$ 
respectively. Then these spaces are invariant to the action of 
the semigroup and their direct sum is the whole space $H$. Also one sees that

---in $H_1$ the projections $Q_\infty$, $P_\infty$ are equal to the identity, 
and therefore, $P_t=I$ and $Q_t=I$, $t\ge1$ as well, and $U_t$, $t\ge1$ 
are unitary;

---in $H_+$ we have $Q_\infty=0$, $P_\infty=I$, and therefore 
$P_t=I$, $t\ge0$, $Q_t \to 0$, $t\to \infty$ strongly; thus, 
in this subspace $U_t$ is a semigroups of isometries. Assume that 
for some $s$ the isometry $U_s$ is not completely non-unitary. Then there 
exists a subspace $H'\subset H_+$ invariant with respect to $U_s$, $U_s^*$
 such that $U_s$ is unitary in $H'$. But since $(U_t^*)^nQ_tU_t^n = Q_{nt}$ 
in $H'$,  this contradicts  the condition $Q_t\to 0$, $t\to\infty$.

---in $H_-$ we have $Q_\infty=I$ and $P_\infty =0 $, 
and by similar arguments we 
conclude that in this subspace $U_t$, $t\ge0$, is a semigroup of 
completely non-unitary  co-isometries.

---in $H_0$ we have $Q_\infty=0$ and $P_\infty =0 $, and we have a
semigroup of  partial isometries $U_t$, $t\ge0$, which are 
completely non-unitary, and 
$U_t\to 0$, $U_{t}^* \to 0$, $t\to \infty$ strongly.
\end{proof}

The proposiotion above shows that the study of sructure of 
semigroups of partial isometries can be reduced to the study 
of their structure in each of the component listed above.

i. The structure of unitary strongly continuous one-parameter
semigroups is given by the Stone theorem, which gives the following
spectral repersentation
\[
U_t = \int _{\mathbb R} e^{it\lambda} \, dE(\lambda),
\]
where $E(\cdot)$ is a resolution of identity of some self-adjoint operator.
The latter formula gives a decomposition of any unitary one-parameter 
semigroup into irreducible one-dimensional ones, 
which have the form $U_t = e^{it\lambda}$ for some $\lambda \in \mathbb R$.

ii. The structure of one-parameter semigroups of isometries was described by
Sz.-Nagy (see \cite{nafo,nagy}). Any strongly continuous semigroup of 
completely non-unitary isometries is unitarily equivalent to 
a multiple of the following semigroup in the space $L_2(0,\infty)$
\[ 
(U_t f)(x) =\chi_{[t,\infty)}(x)\, f(x-t), \quad t\ge 0.
\]

iii. The structure of one-parameter semigroups of co-isometries can be 
obtained from (ii) by taking adjoint operators. Any such semigroup 
is unitarily equivalent to 
a multiple of the following semigroup in the space $L_2(0,\infty)$
\[ 
(U_t f)(x) = f(x+t), \quad t\ge 0.
\]

iv. We describe the structure of one-parameter semigroups of partial 
isometries in $H_0$. First, we describe irreducible semigroups, 
and then show how any semigroup in $H_0$ decomposes on irreducible ones.

\begin{theorem}
Any irreducible strongly continuous one-parameter semigroup of  partial
isometries in $H_0$ is unitarily equivalent to
the following semigroup $U_t^{(a)}$ in $L_2([0,a], dx)$ with some $a\ge0$:
\[
(U_t^{(a)}f)(x) = \chi_{[t,a]}(x)\, f(x-t).
\]
\end{theorem}

\begin{proof}
Let $U_t$, $t\ge0$, be a semigroup of partial isometries in $H=H_0$. 
Consider the subspaces $H_t = \cup_sU_s^*P_t H$, $P_t = U_tU_t^*$, in $H$.

\begin{lemma}
The subspases $H_t$ possess the following properties:

\textup{(i)} $H_t\subset H_s$, $s <t$,

\textup{(ii)} any  $H_t$ is invariant
with respect to  $U_s$, $U_s^*$ $s\ge0$.

\textup{(iii)} $\cap _t H_t = 0$, $\cup_t H_t = H$, 
\end{lemma}

\begin{proof}
Indeed, (i) follows from the fact that  $P_t$ is a decreasing family of
projections.

To show (ii), first notice that by the construction $H_t$ is
invariant with respect to $U_s^*$, $s \ge0$. For any vector of the
form $U_s^*P_tf \in H_t$ we have, if $\lambda\ge s$,
\begin{align*}
U_\lambda U_s^* P_t f &= U_{\lambda-s} P_sP_tf =
P_\lambda U_{\lambda - s} P_tf = P_\lambda
P_{t+\lambda -s} U_{\lambda - s}f
\\
& = P_{t+\lambda - s}(P_\lambda
U_{\lambda - s}f) \in H_t, 
\end{align*}
since $P_{t+\lambda - s}H \subset P_t H \subset H_t$. For $\lambda <s$
we have 
\[
U_\lambda U_s^* P_tf = P_\lambda  U_{s-\lambda}^* P_tf=
U_{s-\lambda}^* P_s P_t f = U_{s-\lambda}^* P_t(P_sf) \in H_t,
\]
therefore, $H_t$ is an invariant subspace for $U_s$, $U_s^*$,
$s\ge0$.

Now we show (iii). For $t=0$ we have $H_t=H$, thus the union is the whole 
space~$H$.

We need to show that $\cap_t H_t = 0$. First notice that $P_\lambda
H_t \subset H_t$, $\lambda$, $t>0$. Indeed, this follows from the
relations $P_\lambda U_s^*P_t = U_s^* P_{s+\lambda} P_t = U_s^* P_t
P_{s+\lambda}$ for all $\lambda$, $t$, $s$.

Introduce subspaces $H_{t,\lambda} = Q_\lambda H_t$,
$H_{t,\lambda}^0 = P_{t-\lambda} H_t$, $H_{t,\lambda}^1 = H_t \ominus
H_{t,\lambda}^0 = (I-P_{t-\lambda}) H_t$, $\lambda \le t$, so that 
$H_t =H_{t,\lambda}^0\oplus H_{t,\lambda}^1 $.

First we show that $
H_{t,\lambda}^1 \subset  H_{t,\lambda}$. Indeed, 
for $\lambda <s$ we have
\[
(I-Q_\lambda)(I-P_{t-\lambda}) U_s^*P_t = 
(I-P_{t-\lambda})(I-Q_\lambda)U_s^*P_t =0
\]
since $Q_\lambda U_s^* =U_s^*$ for $\lambda <s$. For  $\lambda >s$ we have
\begin{align*}
(I-Q_\lambda)(I-P_{t-\lambda}) U_s^*P_t &=
U_s^*(I-P_{t-(\lambda-s)})(I-Q_{\lambda -s}) P_t
\\
&=  U_s^*( P_t-P_{t-(\lambda-s)} P_t)(I-Q_{\lambda -s})=0,
\end{align*}
since $P_{t-(\lambda-s)} P_t=P_t $. Then $Q_\lambda=I$ in $H_{t,\lambda}^1 $ 
and $H_{t,\lambda}^1 \subset  H_{t,\lambda}$.

Since $H_{t,\lambda} =Q_\lambda H_t$, we have 
$Q_\lambda=I$ in $H_{t,\lambda}$, and, therefore,
in $H_{t,\lambda }^1$.

The space $\cap_t H_t$ is invariant with respect to $U_\lambda$,
$U_\lambda^*$, $\lambda \ge0$. Fix any $f \in \cap _t H_t$. For  any 
$\lambda<t$ we can write a decomposition 
$f=f_{(\lambda,t)}^0 + f_{(\lambda,t)}^1$, where
$f_{(\lambda,t)}^0\in H_{t,\lambda }^0 $, 
$f_{(\lambda,t)}^1\in H_{t,\lambda }^1 $.
Since $f_{(\lambda,t)}^0=P_{t-\lambda} f \to 0$, $t\to \infty$, 
we see that $f_{(\lambda,t)}^1\to f$, $t\to \infty$. Then
\[
Q_\lambda f = Q_\lambda f_{(\lambda,t)}^0 +Q_\lambda f_{(\lambda,t)}^1 = 
Q_\lambda f_{(\lambda,t)}^0 + f_{(\lambda,t)}^1 .
\]
The latter sum converges to $f$ as $t\to \infty$, therefore, $Q_\lambda f =f$.
But $Q_\lambda\to 0$, $t\to \infty$, strongly, which implies $f=0$.
\end{proof}

The lemma above implies that for any pair $t<t'$, the 
subspace $H_t\ominus H_{t'}$ is an invariant 
subspace of the operators $U_t$, $U_t^*$, $t\ge0$. Therefore, 
in the case of irreducible one-parameter
semigroup of partial isometries there exists $a>0$ such that $H_t =
H$, $t\le a$, and $H_t=0$, $t>a$. We will show that to each $a>0$
there corresponds exactly one, up to a unitary equivalence, irreducible
one-parameter semigroup of partial isometries, and find a formula for it.

To describe irreducible one-parameter semigroups of partial isometries, 
assume that $H=H_a$ for some $a>0$ such that $H_t=0$
for all $t>a$. In fact, this implies that $P_t=0$ for all $t\ge a$, and $P_t>0$
for all $0\le t <a$. Moreover, the same holds for the family of projections 
$Q_t$, $t\ge0$. Indeed, for any $t\ge a$ we have 
\[
Q_t=Q_t^2=U_t^*U_tU_t^*U_t=U_t^*P_tU_t, \quad 
P_t = P_t^2 = U_tU_t^*U_tU_t^* = U_tQ_tU_t^*,
\]
therefore, the operators  $P_t$ and $Q_t$ are zero or non-zero simultaneously.

Our aim is to decompose the space $H$ with respect to the 
commuting family of projections $Q_t$, $t\ge0$, and describe the action of 
the partial isometries $U_t$. We have already seen that the following 
relations hold
\[
Q_sU_t = U_t Q_{s+t},\quad Q_sU_t^* = U_t^* Q_{s-t},\quad s>t, 
\quad Q_sU_t^* = U_t^*,\quad s\ge t.
\]

Applying commutative models formalism developed in \cite{ourbook} 
(see also references therein), one can obtain the following realization
\begin{align*}
H&= L_2([0,a], d\mu)\otimes \mathcal H,
\\
(Q_tf)(x) & = \chi_{[0,a-t]}(x)\, f(x),
\\
(U_tf)(x) & = \chi_{[t,a]}(x)\, u_t(x)\, \rho_+(x)\, f(x-t),
\\
(U_t^*f)(x) & = \chi_{[0,a-t]}(x)\, u_t^*(x+t)\, \rho_-(x+t)\, f(x+t).
\end{align*}
Here the Borel probability  measure $d\mu$ is such 
that $\chi_{[t,a]}(x)\,d\mu(x-t)$ and 
$\chi_{[0,a-t]}(x)\,d\mu(x+t)$ are absolutely continuous with respect to 
the measure $d\mu(x)$, 
\[
\rho_+(x) = \Bigl(\frac{\chi_{[t,a]}(x)\,d\mu(x-t)}{d\mu(x)}\Bigr)^{1/2}, \quad
\rho_-(x) =\Bigl(\frac{\chi_{[0,a-t]}(x)\,d\mu(x+t)}{d\mu(x)}\Bigr)^{1/2},
\]
$u_t(x)$ is a measurable unitary operator-valued function.
To finish the proof we show that:

i. $\mu$ is equivalent to the Lebesgue measure, so we can assume $d\mu(x)=dx$;

ii. passing to a unitarily equivalent realization one can get $u_t(x)=I$;

iii. the irreducibility implies $\dim \cal H=1$, i.e., $H=L_2([0,a],dx)$.

ii. Let $V$ be a unitary operator in $H$ commuting with the projections $Q_t$,
$t\ge0$. Then 
\end{proof} %of the theorem

\begin{theorem}
Let $S_t$, $t\ge0$, be a strongly continuous semigroup of completely
non-isometric partial isometries. Then it is unitarily equivalent to 
the direct integral
\[
S_t =\int _{a>0}^\oplus S_t^{(a)}\otimes I_a\,d\mu
\]
where $S_t$ is the following semigroup in $H_a = L_2([0,a],dx)$
\[
(S_tf)(x) = \chi_{[t,a]}(x)\, f(x-t)
\]
\end{theorem}

\begin{proof}
Indeed,
\end{proof}

\section{Wold decomposition}

The following theorem selects four classes of centered one-parameter 
semigroup, and shows that any centered semigroup can be decomposed into 
a direct sum of semigroups from these classes. This decomposition 
is similar to the Wold decomposition for isometries, which shows that 
any isometry can be decomposed into a direct sum of a unitary operator 
and pure isometry.

\begin{theorem}
Let $T_t$, $t\ge0$ be a centered semigroup in a Hilbert space $H$. The
space $H$ can be decomposed into a direct sum of invariant with
respect to $T_t$, $T_t^*$, $t\ge0$, subspaces, $H=H_0\oplus H_+ \oplus
H_-\oplus H_1$ such that

in $H_1$ the operators $T_t$ are invertible, $\ker T_t = \ker 
T_t^*=\{0\}$, $t\ge0$\textup;

in $H_+$ the semigroup is such that $\ker T_t = \{0\}$, $t\ge0$,
$\bigcup_{t\ge0} \ker T_t^* = H$\textup;

in $H_-$ the semigroup is such that
$\ker T_t^* = \{0\}$, $t\ge0$, $\bigcup_{t\ge0} \ker T_t = H$\textup;

in $H_0$ the semigroup is such that
$\bigcup_{t\ge0} \ker T_t = H$, and $\bigcup_{t\ge0} \ker T_t^* = H$.  
\end{theorem}

\begin{proof}
First we prove the statement for one-parameter centered 
semigroups of partial isometries,
and then show that  the polar decomposition is invariant with respect 
to the decomposition of $U_t$.

To complete the proof notice that $C_t$ commute with $P_t$. Therefore, 
if one construct a decomposition for a semigroup $U_t$, then 
the constructed decomposition is invariant with respect to $C_t$ 
and therefore, with respect to the whole semigroup $T_t$.
\end{proof}

\section{Commutative models}
Introduce the projections $P_t = U_t^*U_t$ and $P_{-t} = U_tU^*_t$,
$t\ge0$, and the self-adjoint operators $A_t = C_t$, $A_{-t} =
U_tC_tU^*_t$, $t\ge0$.

\begin{proposition}
The following relations hold $(s \ge0)$ for the projections $P_t$,
$t\in \mathbb{R}$
\begin{align}
P_t U_s &= U_s P_{t+s}, \quad t\ge 0,
& P_t U_s^* & = U_s^* P_{t-s}, \quad t\ge s,\notag
\\
P_t U_s &= U_s P_{t+s}, \quad t \le 0, t+s \le 0,
& P_tU_s^* & = U_s^*, \quad 0\le t\le s,\notag
\\
P_t U_s &= U_s, \quad t\le 0, t+s \ge 0,
& P_tU_s^* & = U_s^* P_{t-s}, \quad t\le 0.\label{pu}
\end{align}
and for the operators $A_t$, $t \in \mathbb{R}$:
\begin{align}
A_t U_s A_s & = U_s A_{t+s}, \quad t\ge 0,\notag
\\
A_t U_s & = U_s A_s A_{t+s}, \quad t\le 0, t+s \le 0,\notag
\\
A_t U_s A_{t+s} & = U_s A_s, \quad t\le 0, t+s \ge0.\label{aua}
\end{align} 
\end{proposition}

\begin{proof}
Just calculations.
\end{proof}

We rewrite the last relations in a slightly different form. Observe
that on $\ker A_s^\perp$ the operator $A_s$ has (possibly unbounded)
inverse $A_s^{-1}$. We define it on the whole space putting it zero on
$\ker A_s$. Then we rewrite  equalities  \eqref{aua} as
\begin{align}
A_t U_s & = U_s A_{t+s}A_s^{-1},\quad t\ge 0,\notag
\\
A_tU_s & = U_s A_{t+s} A_s, \quad t \le -s,\notag
\\
A_t U_s & = U_s A_{t+s}^{-1} A_s, \quad -s \le t \le 0. \label{au}
\end{align}
Notice that since the left-hand side is bounded, such is the right
one. Also we eliminated projections $P_s$ and $P_{t+s}$ on the kernels
of $A_s$ and $A_{t+s}$, which is possible due to \eqref{pu}.

On the space $\mathbb R^{\mathbb R}$ of all real functions introduce
the mappings 
\[
F_s  \lambda (t) = \begin{cases}\lambda(t+s)/\lambda(s),& t\ge0,\\
  \lambda(s)/\lambda(t +s), 
  & -s \le t \le 0, \\ \lambda(t +s)\lambda(s),& t \le -s.
\end{cases}
\]

\textbf{Newtext}

Let $T_t$, $t\ge0$ be a centered semigroup. First we assume that the
operators $T_t$ are invertible, i.e., have zero kernel and their image
is the whole $H$. Write the polar decomposition, $T_t
= U_tC_t$, then $C_t$ have (possibly unbounded) inverse, and $U_t$ are
unitaries.

Write $A_t = C_t$, $A_{-t} = U_t C_t^{-1} U_t^*$, $t\ge0$. Then $A_t$,
$t\in \mathbb R$ is a commuting family of positive self-adjoint
operators.

\begin{proposition}
The following relations hold for all $t\in \mathbb R$, $s\ge0$
\begin{align*}
A_t U_s & = U_s A_{t+s} A_s^{-1},
\\
A_tU_s^* & = U_s^* A_{t-s} A_{-s}^{-1}.
\end{align*}
\end{proposition}

\begin{proof}
We have the  relations
\[
U_s^*C_tU_sC_s = C_sU_s^*C_tU_s,\quad C_tU_sC_sU_s^* = U_sC_sU_s^*C_t,
\]
and taking into account that $(U_tC_tU_t^*)^{-1} = U_tC_t^{-1}U_t^*$,
we get the needed relations.
\end{proof}

---

Consider the case when $C_t >0$, i.e., $\ker C_t = 0 $ for all $t$.
Write $A_t = C_t$, $t\ge0$, $A_{-t}= U_t C_tU_t^*$.

\begin{proposition}
The following relations hold
\[
A_t U_s = \begin{cases} U_s A_{t+s} A_s^{-1},& t\ge0,\\ U_s
  A_{t+s}^{-1} A_s, &-s \le t \le 0,\\ U_s A_{t+s} A_s, &t\le
  -s. \end{cases}
\]
\end{proposition}

The operators $A_t$ form a commuting family; therefore (check?!), the
following spectral decomposition holds:
\begin{align*}
H& = \int_\sigma^\oplus H_{\lambda(\cdot)} \, d\mu(\lambda(\cdot)),
\\
(A_t f(\lambda))(\cdot) & = \lambda(t) \,
f(\lambda(\cdot)),
\end{align*}
where $\sigma = \{ \lambda (\cdot) \colon \mathbb{R} \to \mathbb{R}
\mid \lambda(0) =1, \, \lambda(t) >0\}$.
In particular, if the joint spectrum of the commutiong family is
simple,
\begin{align*}
H& = L_2(\sigma, d\mu(\lambda(\cdot))),
\\
(A_t f(\lambda))(\cdot) & = \lambda(t) \, f(\lambda(\cdot)).
\end{align*}

For any $s \ge0$ introduce the mappings $F_s \colon \sigma \to
\sigma$, and  $G_s = F_s^{-1}$ as follows:
\begin{align*}
(F_s(\lambda(\cdot)))(t) & = 
  \begin{cases} \lambda(t+s)/\lambda(s), & t\ge0,\\
    \lambda(s)/\lambda(t+s), & -s\le t \le 0, \\ \lambda(t+s)\,
    \lambda(s),& t\le s,\end{cases}
\\
(G_s(\lambda(\cdot)))(t) & = 
  \begin{cases} \lambda(t-s)\,\lambda(-s), & t\ge s,\\
    \lambda(-s)/\lambda(t-s), & 0\le t \le s, \\ \lambda(t-s)/
    \lambda(-s),& t\le 0,\end{cases}
\end{align*}

According to the theorem on commutative models, for $U_s$ we have the
following representation:
\[
(U_sf(\lambda))(\cdot) = \alpha_s(\lambda(\cdot))\,
\rho_s(\lambda(\cdot)) \, f(G_s(\lambda)(\cdot)),
\]
where $\alpha_s$ is a unitary operator-valued function, $\rho$ is a
norming factor.

We consider some examples of quasi-invariant ergodic measures and the
corresponding examples of centered semigroups.

\begin{example}
Stationary points. The simples case of an ergodic quasi-invariant
measure is the measure concentrated at a stationary point. 
The only function invariant with respect to $F_s$,
$G_s$, is the identity constant. Then $C_t = I$ for all $t$, and we
have a strongly continuous unitary semigroup.
\end{example}

\begin{example}
Measures concentrated on a single orbit. Take a function
$\lambda(\cdot)\colon \mathbb{R} \to \mathbb{R}_+$, $\lambda(0)=1$,
and consider its orbit 
\[
\mathcal  O = \{\lambda_s(\cdot) , s \in \mathbb R\}, \quad
\lambda_s(t) = \begin{cases} (F_s \lambda(\cdot))(t), & s\ge 0,\\
  (G_{-s}\lambda(\cdot))(t), & s \le 0.\end{cases}
\]
The mapping $\mathbb{R} \ni s \mapsto \lambda_s(\cdot)$ establishes  a
one-to-one correspondence between points of a line and $\mathcal
O$. Take the Lebesgue measure on $\mathbb{R}$, and denote by $\mu$ its image
under this  mapping. Then $\mu$  is quasi-invariant measure on
$\sigma$ concentrated on $\mathcal O$. The inverse mapping establishes
an isomorphism between $L_2(\sigma, d\mu)$, and $L_2(\mathbb R, d
s)$. Under this isomorphism we have
\[
(A_t f) (s) = \lambda_s(t) \, f(s).
\]

\end{example}

\section{Infinitesimal generators}

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\end{document}