\magnification=\magstep1
\parindent=0pt


\long\def\DEL#1{{}}

\newcount\SEC
\newcount\THM
\def\ADS{\advance \SEC by 1 \THM=0}
\def\ADT{\advance \THM by 1}
\def\NUM{{\the\SEC.\the\THM}}


\font\tenopen=msbm10 \font\sevenopen=msbm7 \font\fiveopen=msbm5
\newfam\openfam
\def\openo{\fam\openfam\tenopen}
\textfont\openfam=\tenopen \scriptfont\openfam=\sevenopen
\scriptscriptfont\openfam=\fiveopen

\font\tenujsym=msam10 \font\sevenujsym=msam7 \font\fiveujsym=msam5
\newfam\ujsymfam
\def\ujsym{\fam\ujsymfam\tenujsym}
\textfont\ujsymfam=\tenujsym \scriptfont\ujsymfam=\sevenujsym
\scriptscriptfont\ujsymfam=\fiveujsym

\font\Fr=eufm10 scaled \magstep1 \font\Frr=eufm7
\def\Goth#1{\hbox{\Fr #1}}

\def\RR{{\openo{R}}}
\def\CC{{\openo{C}}}
\def\QQ{{\openo{Q}}}
\def\NN{{\openo{N}}}
\def\ZZ{{\openo{Z}}}

\def\eps{\varepsilon}

\def\qed{\hskip10mm Q.e.d.}


%References
\def\ABATE{1}    %Abate
\def\BKU{2}      %Brau-Kaup-Upmeier
\def\DED{3}      %Deddens
\def\DINEEN{4}   %Dineen
\def\NAG{5}      %Engel-Nagel
\def\FRVE{6}     %Franzoni-Vesentini
\def\GOEBEL{7}   %Goebel
\def\HS{8}       %Hayden-Suffridge
\def\IST{9}      %Isidro-Stacho
\def\KAUP{10}    %Kaup1981
\def\KAUPR{11}   %Kaup Riemann mapping thm
\def\KHAS{12}    %Khaskewits -
\def\NACHBIN{13} %Nachbin
\def\RSZ{14}     %Riesz-Sz-Nagy
\def\FSZ{15}     %Sz-Nagy - Foias
\def\VESi{16}    %Vesentini Adv. Math
\def\VESii{17}   %Vesentini note
\def\VIG{18}     %Vigue


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\centerline{\bf On Vesentini's Stone type theorem for one-parameter
groups} \centerline{\bf of non-linear automorphisms in Hilbert
space}

\bigskip

\centerline{L.L. STACH\'O}

\bigskip
\bigskip
\bf Abstract. \rm

\bigskip
\bf AMS 2000 classification: \rm 47D03, 32H15, 46G20.

\ADS
\bigskip\bigskip
\noindent{\bf \the\SEC. Introduction, main results}

\bigskip
\bigskip

\hrule

\bigskip

\bf Notations \rm

\bigskip
${\bf E}$ Banach space with predual ${\bf E}_*$,

\medskip
${\bf K}$ bounded open convex subset in ${\bf E}$ with closure $\overline{{\bf K}}$,

\medskip
$\hbox{\Fr M} := \left\{ \hbox{matrices} \left[ { A \ b \atop c_* \ d} \right] \
\hbox{where} \ A\in{\cal L}({\bf E}),\ b\in{\bf E}, \ c_* \in{\bf E}_*, \ d\in\CC \right\}$, \ \
for ${\cal A}\in\hbox{\Fr M}$,
$$\Psi_{{\cal A}} (x) := (c_* x + d)^{-1} (Ax+b) \quad \hbox{with} \quad
{\rm dom}(\Psi_A)=\{ x\in{\bf E}:\ c_* x+d\ne 0\}$$

\bigskip
\hrule
\bigskip
\ADS

\bigskip
\ADT
\bf Lemma \NUM. \it Assume ${\cal K}$ is a compact topological space and let
$[f_t :\ t\in\RR_+]$ be a one-parameter semigroup of continuous maps
${\cal K}\to{\cal K}$ admitting fixed points
such that all the functions $t\mapsto f_t(x)$ are continuous. Then
also $\bigcap_{t\in\RR_+} {\rm Fix}(f_t) \ne \emptyset$.
%
\bigskip
\bf Proof. \rm
Consider any parameter $t>0$ and a point $x\in{\rm Fix}(f_t)$.
For $n=1,2,\ldots$ recursively we have
$f_{nt}(x) =f_t\big( f_{(n-1)t} (x)\big) = x$.
Thus
${\rm Fix}(f_{nt}) \supset {\rm Fix}(f_t) \ne \emptyset$ $(t\in\RR_+, \ n=1,2,\ldots)$.
From the continuity of the maps $f_t$ it follows that
${\rm Fix}(f_{1/n!})$ \ $(n=1,2,\ldots)$ is a decreasing sequence of non-empty compact sets
with non-empty intersection ${\cal X}:= \bigcap_n {\rm Fix}(f_{1/n!})$.
Since any rational number $0\ne q\in\QQ_+$ can be written in the form $q=m/n!$
for suitable integers $m,n>0$, it follows even that
$\bigcap_{q\in \QQ_+} {\rm Fix}(f_q) = {\cal X} \ne \emptyset$.
Consider any parameter $t>0$ and any point $x\in{\cal X}$.
Given any sequence $q_1,q_2,\ldots\in\QQ_+$ converging to $t$, the continuity of
the orbit $t\mapsto f_t(x)$
ensures that
$x=\lim_n f_{q_n}(x) = f_t(x)$.

\DEL{

\bigskip
%\ADT
\bf Lemma \NUM a. \it Assume $[ f_q :\ t\in\QQ_+]$ is a one-parameter semigroup of
w*-continuous maps $f_t: \overline{{\bf K}}\to\overline{{\bf K}}$.
Then $\bigcap_{q\in\QQ_+}  {\rm Fix}(f_q) \ne\emptyset$.

\bigskip
\bf Proof. \rm
By the Schauder-Tychonoff fixed point theorem, ${\rm Fix}(f_q)\ne \emptyset$ for any $q\in\QQ_+$.
Given any $q\in\QQ_+$ and $n\in\NN$ along with a fixed point $x\in{\rm Fix}(f_{q/n})$,
for $k=1,2,\ldots,n$ consecutively we have
$x=f_{q/n}\big( f_{q/n}^{k-1}(x)\big) = f_{q/n}^k(x)$ showing that
${\rm Fix}(f_{q/n}) \subset {\rm Fix}(f_q)$ for all $q\in\QQ_+$ and $n\in\NN$.
Hence
${\rm Fix}(f_{1/n!})$ \ $(n=1,2,\ldots)$ is a decreasing sequence of non-empty compact sets
with non-empty intersection ${\bf X}:= \bigcap_n {\rm Fix}(f_{1/n!})$.
Given any $0\ne q\in \QQ_+$ we can write $q=m/n!$ for suitable $m,n\in\NN$ entailing that
${\bf X} \subset {\rm Fix}(f_{1/n!}) \subset {\rm Fix}(f_{m/n!}) = {\rm Fix}(f_{q})$.


}%ENDDEL


\bigskip
\ADT
\bf Lemma \NUM. \it
Let ${\bf K}$ be a domain in a Banach space ${\bf E}$ and
let $f_t : {\bf D}_t \to {\bf E}$ $(t\in\RR_+]$ be a family of holomorphic maps
defined on open neighborhoods of $\overline{{\bf K}}$
such that the restrictions $[ f_t\vert {\bf K} : t\in\RR_+]$ form
a strongly continuous one-parameter semigroup.
Assume that for every boundary point $x\in\partial {\bf K}$ there exists
a 1-dimensional complex disc $\Delta_x$ centered in $x$
and intersecting ${\bf K}$
such that
$\Delta_x \subset \bigcap_{t\in[0,\delta_x]} {\bf D}_t$
and $\bigcup_{t\in[0,\delta_x]} f_t(\Delta_x)$ is a bounded set
for some $\delta_x>0$.
Then
$[ f_t\vert \overline{{\bf K}} : t\in\RR]$ is also a strongly continuous one-parameter semigroup.
%the functions $t\mapsto f_t(x)$ are continuous
%for every point $x\in\partial {\bf K}$.


\bigskip
\bf Proof. \rm By assumption $f_s(f_t(x)) =f_{s+t}(x)$ $(x\in {\bf K}, \ s,t\in\RR_+)$.
Since the maps $f_t\vert \overline{{\bf K}}$ are all continuous,
and it follows $f_s(f_t(x)) =f_{s+t}(x)$ $(x\in \overline{{\bf K}}, \ s,t\in\RR_+)$
that is $[f_t\vert\overline{{\bf K}}: t\in\RR_+]$ is a one-parameter semigroup of continuous maps on
the closure $\overline{{\bf K}}$.
Hence, to complete the proof, it suffices to see only that for any $x\in\partial{\bf K}$, the function
$t\mapsto f_t(x)$ is continuous on some neighborhood the origin, namely on $[0,\delta_x)$.
Fix any $x\in\partial{\bf K}$ and consider a convergent sequence $t_n\to t$ within $[0,\delta_x]$.
We show the convergence $f_{t_n}(x)\to f_t(x)$ as follows.
We can write $\Delta_x = \{ x+\zeta v: \vert \zeta\vert < 1\}$
with a suitable vector $v\in{\bf E}$.
By assumption, the functions $g_n(\zeta):= f_{t_n}(x+\zeta v)$ $(n=1,2,\ldots)$ are uniformly bounded
and holomorphic on the unit disc. Furthermore they are assumed to converge pointwise to
$g(\zeta):= f_t(x+\zeta v)$ on the non-empty open complex domain $\{ \zeta\in\CC:\ x+\zeta v \in{\bf K} \}$.
In Banach space setting,
pointwise convergence implies uniform convergence on compact sets for holomorphic maps [\NACHBIN].
In particular, we have uniform convergence for $[g_n: n=1,2,\ldots]$ on some compact disc with positive radius.
By a theorem of Vigu\'e [\VIG, \IST],
for a uniformly bounded sequence of holomorphic maps,
the uniform convergence on some subdomain entails locally uniform (and hence pointwise) convergence on the whole domain.
In particular, $g_n\to g$ pointwise and  hence $f_{t_n}(x) = g_n(0) \to g(0) =f_t(x)$.



\bigskip
\ADT
\bf Corollary \NUM. \it If ${\bf E}$ is a JB*-triple, ${\bf K}$ is its open unit ball and $[f_t: t\in\RR]$ is a strongly continuous one-parameter subgroup of ${\rm Aut}({\bf K})$
then the maps $\overline{f}_t$ obtained with graph closure from the respective $f_t$, form a
strongly continuous one-parameter group of maps $\overline{{\bf K}}\to\overline{{\bf K}}$.

\bigskip
\bf Proof. \rm
It is well-known [\KAUPR] that
we can write $f_t=M_{f_t(0)} \circ U_t$ with some invertible linear operator $U_t\in{\cal L}({\bf E})$ and a
so-called M\"obius transformation with the fractional linear form $x\mapsto f_t(0) + B_t [I+L(x,f_t(0)]^{-1}x$
where $B_t, L(x,f_t(0)) \in{\cal L}({\bf E})$ and $\Vert L(x,f_t(0))\Vert\le \Vert f_t(0)\Vert \Vert x\Vert$.
In particular each $f_t$ extends holomorphically to the ball of radius $1/\Vert f_t(0)\Vert$.
Thus the conditions required by the lemma are fulfilled since $\lim\limits_{t\to 0} f_(0) =0$ by assumption.

\DEL{
\bigskip
\ADT
\bf Lemma \NUM. \it Assume $[U^t: t\!\in\!\RR_+] \!\subset\! {\cal L}({\bf E})$ resp. $[u^t : t\!\in\!\RR_+] \!\subset\! {\bf E}^*$
are families of bounded linear operators resp. linear functionals
of a Banach space ${\bf E}$ such that the operators
$${\cal U}^t := \left[ {U^t \ 0 \atop u^t \ 1}\right] \qquad (t\in \RR_+)$$
form a strongly continuous one-parameter semigroup of ${\cal L}({\bf E}\oplus\CC)$.
Then $[U^t: t\in\RR_+]$ is a strongly continuous one-parameter semigroup of ${\cal L}({\bf E})$.
By writing $A:=dU^t/dt\vert_{t=0}$ resp. ${\cal A}:=d{\cal U}^t/dt\vert_{t=0}$ for the generators
and by setting
$$D_k :=\big\{ x\in{\rm dom}(A): [t\mapsto {\cal U}^t (x\oplus 0)]
\ \hbox{is $k$ times differentiable on $[0,\infty)$}\big\}
,$$
we have ${\rm dom}({\cal A}) \!=\! D_1\!\oplus\!\CC$.
The linear functional
$a x  \!:=\! \lim\limits_{s\downarrow 0} {1\over s} u^s x$  $(x\!\!\in\!\! D_1)$ is well-defined and
$${\cal A} = \Big[ { A \ 0 \atop a \ 0}\Big] ,  \qquad
{\cal U}^t (x\oplus \xi) = \big[U^t x\big] \oplus \big[ {\scriptstyle{\int}}_0^t \ a U^s x \ ds + \xi\big]
\quad (x\in D_2,\ \xi\in\CC).$$
\bf Proof. \rm Using the matrix form of the identity ${\cal U}^t {\cal U}^s \!=\! {\cal U}^{t+s}$, for any $x \!\in\!{\bf E}$ and $\xi\!\in\!\CC$  we get
\ADT
$$\big[ U^t U^s x \big] \oplus \big[ u^t(U^s x) + u^s x + \xi \big] = \big[ U^{t+s} x \big] \oplus \big[ u^{t+s} x + \xi \big] .
\leqno(\NUM)$$
Regarding the first components in (\NUM), we deduce the fact that $[U^t: t\in\RR_+]$ is necessarily a strongly continuous semigroup.
In particular ${\cal U}^t (0\oplus 1) = 0\oplus 1$ $(t\in\RR_+)$, whence we see that \
$0\oplus 1\in {\rm dom}({\cal A})=\{ x\oplus\xi: t\mapsto {\cal U}^t (x\oplus\xi)$ is differentiable$\}$  \ entailing also that
${\rm dom}({\cal A}) = {\rm dom}({\cal A}) + \CC (0\oplus 1) = D_1\oplus \CC$.
From (\NUM) it is also immediate that
$d{\cal U}^t(x\oplus \xi)= \big[ A U^t x\big] \oplus \big[ a U^t x\big]$ $(x\in D_1)$. In particular, if $x\in D_2$
(or only $[t\mapsto u^t x] \in {\cal C}^2[0,\infty)$), we have
$u^t x = u^0 x + \int_0^t [ d u^s x / ds] ds = 0 + \int_0^t [a U^s x] ds$ \ which completes the proof.

\bigskip
\ADT
\bf Remark \NUM. \rm
It is well-known that the linear submanifolds $D_k$ above are dense in ${\bf E}$.
Actually,
given any function $\phi\in{\cal C}^k_0(\RR)$ with compact support within $(0,\infty)$ and with $\int \phi =1$,
for any $\Goth{z} \!\in\!{\bf E} \!\oplus\! \CC$ and $h \!>\! 0$ the vectors
$\Goth{z}_h:=h^{-1} \int_0^\infty \phi(s/h) {\cal U}^s \Goth{z} \ ds$
satisfy $\Goth{z}=\lim\limits_{h\downarrow 0} \Goth{z}_h$ and the functions
$t\mapsto {\cal U}^t \Goth{z}_h =
h^{-1} \int_0^\infty  \phi\big( (s-t)/h \big) {\cal U}^{t} \Goth{z} \ ds$ belong to  ${\cal C}^k ([0,\infty),{\bf E}\oplus\CC)$.
}%ENDDEL

\bigskip
\ADT
\newcount\TriS \TriS=\SEC \newcount\TriT \TriT=\THM
\bf Lemma \NUM. \it Let ${\bf E}_1,{\bf E}_2$ be Banach spaces and suppose $[W^t: t\in\RR_+]$ is a
strongly continuous one-parameter subsemigroup of ${\cal L}({\bf E}_1)$ with the infinitesimal generator $G$.
Furthermore let $G_1\in{\cal L}({\bf E}_2,{\bf E}_1), \ G_2\in{\cal L}({\bf E}_1,{\bf E}_2)$ with
${\cal G}_1:=\left[ \matrix{G &0 \cr G_1& 0}\right]$ and ${\cal G}_2:=\left[ \matrix{G &G_2 \cr 0& 0}\right]$.
Then
$$\left[\left[ \matrix{W^t &0 \cr G_1\int_0^t W^\tau d\tau & I\cr}\right] : t\in\RR_+\right] , \quad
\left[\left[ \matrix{W^t &\int_0^t W^\tau d\tau G_2 \cr 0 &I\cr}\right] : t\in\RR_+\right] $$
are strongly continuous one-parameter subsemigroups of ${\cal L}({\bf E}_1\oplus{\bf E}_2)$ with the infinitesimal generators
%$\left[ \matrix{G &0 \cr G_1& 0}\right]$
${\cal G}_1, {\cal G}_2$, respectively.
\DEL{
{\rm (i)} $\left[\left[ \matrix{W^t &0 \cr G_1\int_0^t W^\tau d\tau & I\cr}\right] : t\in\RR_+\right]$
%$ \left[\big[ { W^t \qquad 0  \atop  G_1\int_0^t W^\tau d\tau \ I } \big] : t\in\RR_+ \ \right]$
is a
strongly continuous one-parameter subsemigroup of ${\cal L}({\bf E}_1\oplus{\bf E}_2)$ with the infinitesimal generator
$\left[ \matrix{G &0 \cr G_1& 0}\right]$;

\smallskip
{\rm (ii)} $\left[\left[ \matrix{W^t &\int_0^t W^\tau d\tau G_2 \cr 0 &I\cr}\right] : t\in\RR_+\right]$ is a
strongly continuous one-parameter subsemigroup of ${\cal L}({\bf E}_1\oplus{\bf E}_2)$ with the infinitesimal generator
$\left[ \matrix{G &G_1 \cr 0& 0}\right]$.

\smallskip
If the generator $G$ is invertible then above we have $\int_0^t W^\tau d\tau=G^{-1}[W^t-I]=[W^t-I]G^{-1}$.
}%ENDDEL

\bigskip
\bf Proof. \rm
Since ${\cal G}_1$ is a bounded perturbation of the generator ${\cal G}:= G\oplus 0$ of the semigroup $[W^t \oplus 0 : t\in\RR_+]$ with ${\rm dom}({\cal G}) = {\rm dom}(G)\oplus  E_2$,
it suffices to see only that
$${d\over dt}\Big\vert_{t=0+} \left[ \matrix{W^t &0 \cr G_1\int_0^t W^\tau d\tau & I\cr}\right] \left[ \matrix{ x \cr y} \right] =
{\cal G}_1 \left[ \matrix{ x \cr y} \right] \qquad (x\in{\rm dom}(G),\ y\in E_2)$$
which is immediate. The case of ${\cal G}_2$ is analogous.

\bigskip
\ADT
\newcount\DilS \DilS=\SEC \newcount\DilT \DilT=\THM
\bf Lemma \NUM. \it Assume $[U^t: t\in\RR_+]$ is a strongly continuous one-parameter semigroup formed by isometries of a Hilbert space ${\bf H}$.
Then there is Hilbert space $\widehat{{\bf H}}$ containing ${\bf H}$ as a subspace along with a strongly continuous one-parameter group
$[ \widehat{U}^t : t\in\RR]$ formed by isometries of $\widehat{{\bf H}}$ such that $U^t = \widehat{U}^t \big\vert {\bf H}$ $(t\ge 0)$.

\bigskip
\bf Proof. \rm
This is implicit in [\FSZ, III. Thm. 9.3].
%Below we give a short elementary proof.

\bigskip
\ADT
\newcount\DILS \DILS=\SEC \newcount\DILT \DILT=\THM
\bf Lemma \NUM. \it Assume $[U^t: t\in\RR_+]$ is a strongly continuous one-parameter semigroup of (necessarily linear) isometries of a Banach space ${\bf E}$,
${\bf E}$ is a subspace of another Banach space ${\bf F}$ and $U^{t_0}$ is
the restriction of a surjective isometry of $\widehat{{\bf E}}$ for some $t_0>0$. Then
there is subspace $\widehat{{\bf E}}$ of ${\bf F}$ along with a strongly continuous one-parameter group $[\widehat{U}^t: t\in\RR]$ of
surjective isometries of $\widehat{{\bf E}}$
such that $U^t = \widehat{U}^t \big\vert {\bf E}$ $(t\ge 0)$.

\bigskip
\bf Proof. \rm
We may assume $t_0=1$ without loss of generality and let $V$ be a surjective isometric continuation of $U^1$ to ${\bf F}$.
Notice that $V^{-1}$ a surjective ${\bf F}$-isometry with ${\bf E}\subset V^{-1}{\bf E}$ and hence, in general,
$V^n {\bf E} \supset V^{n+1}{\bf E}$ $(n\in\ZZ)$.
Observe also that we have
\ADT
$$V^n U^t x = U^s V^m x \quad \hbox{whenever $V^m x\in {\bf E}$, $n+t=m+s$ with $t,s\ge 0$ and $n,m\in\ZZ$}.
\leqno(\NUM)$$
Hence the definitions
$$\eqalign{
&\widehat{{\bf E}}_0 := \bigcup_{n=0}^\infty V^{-n}{\bf E} = \big\{ x \in{\bf F}: \exists\, n\in\ZZ \quad V^n x \in {\bf E} \big\}, \cr
&\widehat{U}^t_0 x:= \Big[ U^s V^m x: \ m\in\ZZ, \ s\ge 0, \ m+s=t, \ V^m x \in{\bf E} \Big] \qquad (x\in\widehat{{\bf E}}_0)\cr}$$
make sense. It is also immediate that each operator $\widehat{U}^t_0$ is an isometric extension of $U^t$ to the not necessarily closed subspace $\widehat{{\bf E}}_0$
and, as direct a consequence of (\NUM), they have
the one-parameter group property $\widehat{U}_0^t  \widehat{U}_0^t = \widehat{U}_0^{t+s}$ $(s,t\in\RR)$.
Thus taking the closure of $\widehat{{\bf E}}_0$ in ${\bf F}$ for $\widehat{{\bf E}}$ and setting the graph closure of $\widehat{U}_0^t$ for
$\widehat{U}^t$ suit the requirements of the lemma.










\bigskip
\hrule
\vskip2pt
\hrule
\bigskip

\ADS
\bf \the\SEC. %Strongly continuous
One-parameter semigroups of Carat\'eodory isometries of the Hilbert ball
\rm

\medskip
\noindent
Throughout this work ${\bf H}$ denotes an arbitrarily fixed complex Hilbert space
with the scalar product $\langle \cdot \vert \cdot \rangle$ and canonical norm $\Vert x\Vert :=\langle x\vert x\rangle^{1/2}$.
We also keep fixed the standard notations
${\bf B}:=\{ x\in{\bf H}: \Vert x\Vert<1\}$ ,
$a^*:=[x\mapsto \langle x\vert a\rangle]$
and $I :={\rm Id}_{{\bf H}}$
for its open unit ball, for the adjoint representation of bounded linear functionals and the unit operator.
Furthermore, given any subset or single element $X$ of ${\bf H}$, we shall write
$P_X$ for the orthogonal projection of ${\bf H}$ onto the closed (complex) subspace $\overline{{\rm Span}}(X)$ spanned by $X$.
Later on we shall regard the elements $h$ and their adjoints $h^*$ of ${\bf H}$ as column resp. row matrices.
In particular, given any vector $c\in{\bf H}$, we can write
$P_c=\big[$orthogonal projection ${\bf H}\to\CC c\big] =\Vert c\Vert^{-2} cc^*: x\mapsto \langle x\vert c\rangle c$

\medskip
$x\oplus\xi \equiv \big[ {x\atop \xi}\big]$ $(x\in{\bf H},\, \xi\in\CC)$

\medskip
${\cal L}({\bf H}\oplus\CC) \equiv \big\{ \big[ {A \ b \atop c^* \, d}\big] :\ A\in{\cal L}({\bf H}),\, b,c\in{\bf H},\, d\in\CC \big\}$

\bigskip
In general, if $A:{\bf S}_1\to{\bf H}$, $c:{\bf S}_2\to\CC$ linear maps, $b\in{\bf H}, d\in\CC$ and ${\cal S}$ is a linear submanifold in ${\bf H}\oplus\CC$,
${\cal A}:= \big[ {A \ b \atop c^* \, d}\big] \big\vert{\cal S}$ is the linear map $x\oplus\xi\mapsto (Ax+b)\oplus(c^*x+d)$ with
${\rm dom}({\cal A}):=\{ x\oplus\xi\in{\cal S}: x\in {\bf S}_1\cap{\bf S}_2\}$.

\medskip
$F({\cal A}) := \big[ x\mapsto (c^*x+d)^{-1} (Ax+b) \big]$ with ${\rm dom}\big( F({\cal A})\big) = \{ x: c^*x+d\ne 0\}$
for ${\cal A}=\big[ {A \ b \atop c^* \, d}\big]$ %\in{\cal L}({\bf H}\oplus\CC)$

\medskip
${\cal T}_x := \big[ {I \ x \atop 0 \ 1}\big]$ $(x\in{\bf H})$. Notice: ${\cal T}_x^{-1} ={\cal T}_{-x}$.

\medskip
For $c\in{\bf B}$, $\beta(c):=\sqrt{1-\Vert x\Vert^2}$ and $Q_c := P_c + \beta(c) \overline{P}_c$ with
%the complementary projection
$\overline{P}_c := I-P_c$

\medskip
${\cal M}_c := \beta(c)^{-1} \big[ {Q_c \ c \atop c^* \ 1} \big]$ lifted M\"obius transformation. Notice: ${\cal M}_{-c} ={\cal M}_c^{-1}$.

$\Theta_c := \big[x\mapsto (1+\langle x\vert c\rangle)^{-1}(Q_cx+c) \big] = F\big({\cal M}_c\big)$ M\"obius transformation, $\Theta_{-c}=\Theta_c^{-1}$.


\bigskip
In 1987 Vesentini in the pioneering works [\VESi,\VESii] established that
the mapping
$$F^\# : \big[ {\cal U}^t: t\!\in\!\RR_+ \big] \mapsto \big[ F({\cal U}^t) \vert{\bf B}: \ t\!\in\!\RR_+ \big]$$
constitutes a
one-to-one correspondence between the family $\Goth{U}$ of all
strongly continuous one-parameter semigroups of $\CC$-linear isometries of the skew metrics $\ell:=\Vert x\Vert^2-\vert \xi\vert^2$ on ${\bf H}\oplus\CC$
and the family $\Goth{F} $ of all
strongly continuous one-parameter semigroups $[\Psi^t: t\!\in\!\RR_+]$ formed by holomorphic Carat\'eodory isometries of the open unit ball ${\bf B}$.
Given $[{\cal U}^t: t\in\RR_+] \in \Goth{U}$ with the infinitesimal generator ${\cal A}$,
for the corresponding non-linear objects $\Psi^t:=F({\cal U}^t)$
%$(t\in\RR_+)$
we have
$\big\{ p\in{\bf B}: t\mapsto \Psi^t(p)$ is differentiable$\big\} =
\big\{ x\in {\bf B}:\ \big[ {x\atop 1}\big] \in{\rm dom}({\cal A})\big\}$ which is dense in the ball ${\bf B}$.
This observation gives rise to the concept of infinitesimal generator for the the non-linear semigroup
$[\Psi^t : t\in\RR_+] =[F^\# {\cal U}^t: t\in \RR_+]$ as the vector field
$$\Gamma(x) := {d\over dt} \Big\vert_{t=0+} F{\cal U}^t\Big[{x \atop 1}\Big] =
-\Big[{\cal A}\Big[{x\atop 1}\Big] \Big]_{\CC} x + \Big[{\cal A}\Big[{x\atop 1}\Big] \Big]_{{\bf H}}
$$
defined for the points $x \in {\rm dom}(\Gamma) := \big\{ x \in{\bf B}: \ \big[{x\atop 1}\big] \in{\rm dom}({\cal A})\big\}$.
%
The orbit $t\mapsto \Psi^t(0)$ is differentiable if and only if the linear generator
${\cal A}$ %:= {d\over dt}\big\vert_{t=0} {\cal W}^t$
has the form
$${\cal A}= \Big[ \matrix{iR + \nu I  &b \cr b^*  &\nu }\Big]$$
with some suitable  constant $\nu\in\CC$, a vector $r\in{\bf H}$
and a maximal symmetric, in ${\bf H}$ densely defined not necessarily bounded linear operator $R$
such that ${\rm dom}({\cal A}) = {\rm dom}(R) \oplus\CC$.
In this case the operator
$iR$ is the generator of a strongly continuous semigroup of (linear) isometries of ${\bf H}$.
Choosing any point $c\in{\rm dom}(\Gamma)$, the orbit $t\mapsto \Phi^t(0)$ with the compositions $\Phi^t:= \Theta_{-c} \circ \Psi^t \circ \Theta_c$
is differentiable. Thus any strongly continuous one-parameter semigroup $[\Psi^t : t\in\RR_+]\in\Goth{F}$
is equvialent up to a M\"obius transformation
({\it M\"obius equivalent} for short in the sequel)
to a semigroup $[\Phi^t : t\in\RR_+]\in\Goth{F}$ whose infinitesimal generator
has the form
\ADT
\newcount\VfieldS \VfieldS=\SEC
\newcount\VfieldT \VfieldT=\THM
$$\Gamma(x) = b -\big\langle x \big\vert b \big\rangle x + iRx , \quad  x\in{\rm dom}(R) \cap {\bf B}
\leqno(\NUM)$$
with some maximal symmetric operator $R$ defined densely on ${\bf H}$ and some vector $r\in {\bf H}$.
Also conversely, if $R$ is the infinitesimal generator for some strongly continuous one-parameter subsemigroup
of ${\cal L}({\bf H})$ then, for any $b\in{\bf H}$, the vector field (\the\VfieldS.\the\VfieldT) is the infinitesimal
generator of a strongly continuous one-parameter subgroup of holomorphic Carath\'eodory isometries of the ball ${\bf B}$.

\medskip
In [\BKU, \KAUPR],
the infinitesimal generators of {\it uniformly continuous one-parameter groups} of holomorphic Carath\'eodory isometries on bounded symmetric domains
are described completely even in the setting of Banach spaces:
up to a biholomorphic transformation, they are equivalent to transformation groups obtained by integration of vector fields of the form
\ $x\mapsto b - \{ x b^* x\} + iRx$ \ with some operation $\{ xy^*z \}$ in 3-variables and a bounded linear hermitian operator over the underlying space.
In particular in the case of the unit ball of a Hilbert space we have to apply the triple product
$\{ xy^* z\} = {1\over 2} \langle x\vert y\rangle z + {1\over 2} \langle z\vert y\rangle x$,
giving back the spacial case of Vesentini's formula (\NUM).
In [\KAUPR, Sec.2] even a M\"obius type explicit formula is given for the integration of the vector fields $x\mapsto r - \{ x r^* x\}$,
however the method there seems to be unsuitable in treating the general case with non-vanishing $R$.

\medskip
In this paper we develop a triangularization method resulting in an explicit formula for the vector fields (\the\VfieldS.\the\VfieldT)
in terms of fixed points $\Gamma$ and quadratures of strongly continuous one parameter semigroups of isometries of a 1-codimensional subspace of ${\bf H}$.
As a consequence we conclude that every strongly continuous one-parameter semigroup of holomorphic Carath\'odory isometries of ${\bf B}$ admits
a {\it dilation} to a strongly continuous one-parameter semigroup of surjective holomorphic Carath\'odory isometries of the unit ball of some covering Hilbert space.
Our results seem to be new even in finite dimensions.

\bigskip
\bigskip
\ADS
\bf \the\SEC. Main results
\bigskip
\rm

%On the other hand we know
Recall [\VESi] that the continuous extensions of the maps in any Abelian family of holomorphic Carath\'eodory isometries of ${\bf B}$ admit common fixed points.
Since, in general, the holomorphic Carath\'eodory isometries of a bounded balanced domain leaving the origin fixed are necessarily linear, %or antilinear,
any strongly continuous one-parameter semigroup of holomorphic Carat\'odory isomeries of ${\bf B}$ is M\"obius equivalent to a
strongly continuous one-parameter semigroup of linear isometries of ${\bf H}$.
If there is no common fixed point within the open unit ball then [\VESi] the family of common fixed points reduces to a singleton or two points in the boundary
$\partial {\bf B}=\{ x\in{\bf H}: \Vert x\Vert =1\}$.
The following explicit description up to M\"obius equivalence of these cases will be our main goal in this paper.

\bigskip
\ADT
\bf Assumptions, standard notations \NUM. \rm
Henceforth, throughout the paper

\smallskip
{(i)} \ $[\Phi^t: t\in\RR_+]$ denotes an arbitrarily fixed strongly continuous one-parameter semigroup of holomorphic Carath\'eodory isometries of ${\bf B}$
with the infinitesimal generator $(\the\VfieldS.\the\VfieldT)$;

\smallskip
{(ii)} \ $\overline{x}\in \bigcap\limits_{t\in\RR_+} \!\! {\rm Fix}(\Phi^t)$, \ \ $\Vert \overline{x}\Vert=1$; \ \
we write \ $P:={\rm P}_{\CC \overline{x}}$, $P_0:=I-P$, ${\bf H}_0 := P_0{\bf H} ={\bf H}\ominus(\CC\overline{x})$;



\bigskip
\ADT
\bf Theorem \NUM. \it We have $\overline{x} \in {\rm dom}(R)$ and the operator $i P_0 R \big\vert {\bf H}_0$
is the infinitesimal generator of some strongly continuous one-parameter semigroup $[ U_0^t: \ t\in\RR_+]$ of
${\bf H}_0$-isometries.
\break
There exist constants $\lambda,\mu\in\RR$ such that, by setting
$$S:= R-i \mu I , \quad  S_0 := P_0 S \big\vert {\bf H}_0, \quad V_0^t := e^{i\mu t} U_0^t, \quad b_0 := P_0(i Sb),$$
for every point \ $x\in {\bf B}$ \ with \ $x_0:=P_0 x$ and $P x = \xi \overline{x}$ \ we can write
\ADT
\newcount\QuadS \QuadS=\SEC
\newcount\QuadT \QuadT=\THM
$$\eqalign{
P\Phi^t(x) &= \Big[ \varphi_{\lambda,\mu}(t, x_0,\xi)^{-1} (\xi - 1) e^{-2\lambda t} +1 \Big] \overline{x} \cr
P_0\Phi^t(x) &= \varphi_{\lambda,\mu}(t, x_0,\xi)^{-1} \Big[ e^{-\lambda t} V_0^t x_0 - (\xi - 1) e^{-2\lambda t}
\big( {{\scriptstyle{\int_0^t}}} e^{\lambda \tau} V_0^\tau \, d\tau \big) b_0 \Big] \ \
\hbox{\rm where}\cr
\varphi_{\lambda,\mu}(t, x_0,\xi) &:= \Big\langle
 \big( {{\scriptstyle{\int_0^t}}} e^{-\lambda\tau} V_0^\tau \, d\tau \big) x_0 \Big\vert  b_0 \Big\rangle \!-\!
(\xi \!-\! 1) \Big\langle \big( {{\scriptstyle{\int_0^t}}} e^{-2\lambda\tau}
{{\scriptstyle{\int_0^\tau}}} e^{\lambda\sigma} V_0^\sigma \, d\sigma \, d\tau \big) b_0
\Big\vert   b_0 \Big\rangle \!+ \cr
& \hskip5mm + (\xi-1) \big( \lambda + i \mu %\langle R\overline{x} \vert \overline{x}\rangle/2
\big) e^{-2\lambda t}  + 1 . \cr}
\leqno(\NUM)$$

\ADT
\bf Corollary \NUM. \it Each strongly continuous one-parameter semigroup $[\Psi^t: t\in\RR_+]$ of
holomorphic Carath\'eodory isometries of ${\bf B}$ with exactly two joint boundary fixed points is M\"obius equivalent to some
semigroup $[\Phi^t: t\in\RR_+]$ of the form
\ADT
\newcount\TwoS \TwoS=\SEC
\newcount\TwoT \TwoT=\THM
$$\eqalign{
P\Phi^t(x) &= { (1-\lambda)(1-\xi) e^{-2\lambda t} + 1 \over 1 - (1-\xi)\lambda e^{-2\lambda t} } \overline{x} \cr
P_0\Phi^t(x) &= {e^{-\lambda t}\over 1 - (1-\xi)\lambda e^{-2\lambda t} } V_0^t x_0 \ .\cr}
\leqno(\NUM)$$
\ADT
\bf Remark \NUM. \rm
(i) The following converse is straightforward to discover from the proof later on.
Given any couple of
vectors $\overline{x},b_0\in {\bf H}$ such that $\Vert \overline{x}\Vert=1$
and $b_0\perp\overline{x}$
along with any strongly continuous one-parameter group $[V_0^t: t\in\RR_+]$ of
linear isometries of ${\bf H}_0={\bf H}\ominus(\CC\overline{x})$ and two real constants $\lambda,\mu$,
the maps $(\the\QuadS.\the\QuadT)$ form a strongly continuous one-parameter group of Carath\'eodory isometries of ${\bf B}$.

\smallskip
(ii) It seems that no analogous M\"obius equivalent simplification as $(\the\TwoS.\the\TwoT)$ is available
in the case of a unique common fixed point lying in the boundary of the unit ball.

\smallskip
(iii) In case of $\lambda\ne 0$, it is possible to express the integrated operators in $(\the\QuadS.\the\QuadT)$
in terms of the resolvents $R( \pm \lambda, iS_0)$. The case $\lambda=0$ is available hence by passing to limits for $\lambda\to 0$.
We can avoid the technical difficulties of this limiting process by means of {\it unitary dilations}.
Recall that, as direct a consequence of a far reaching result of Deddens [\DED,Main Thm.]\footnote{$^*$}{
For an elementary proof see Section 6. Appendix.},
every strongly continuous one parameter semigroup $[U^t: t\in\RR_+]$ of isometries of ${\bf H}$ admits a {\it unitary group dilation}
in the following sense: there exists a Hilbert space $\widehat{{\bf H}}$ containing ${\bf H}$ as a subspace along with a
strongly continuous one-parameter group $[{\widehat{U}}^t : t\in\RR]$ of unitary operators of $\widehat{{\bf H}}$
such that $U^t = \widehat{U}^t \big\vert {\bf H}$ $(t\in\RR_+)$.

\bigskip
Applying a unitary dilation $[ \widehat{V}_0^t : r\in\RR]$ of the isometry semigroup $[ V_0^t: t\in\RR_+]$
in (\the\QuadS.\the\QuadT), from Remark \NUM (i) we
readily obtain the following result with non-linear dilations.

\bigskip
\ADT
\bf Theorem \NUM. \it Given any strongly continuous one-parameter semigroup $[\Psi^t: t\in\RR_+]$ of holomorphic Carat\'eodory isometries of ${\bf B}$,
there is a strongly continuous one parameter group $[\widehat{\Psi}^t: t\in\RR_+]$ of surjective holomorphic Carat\'eodory isometries of the unit ball
$\widehat{{\bf B}}$ of some Hilbert space $\widehat{{\bf H}}$ containing ${\bf H}$ as a subspace such that
$\Psi^t = \widehat{\Psi}^t \big\vert {\bf B}$ $(t\in\RR_+)$.

\bigskip
\rm
Using the functional calculus of the skew self-adjoint generator $i\widehat{S}_0$
%with spectral resolution
%$\widehat{S}_0 = \int_{-\infty}^\infty \eta \ d \widehat{E}(\eta) $
for the dilation group $[\widehat{V}_0^t : t\in\RR]$ of the semigroup $[V_0^t: t\in\RR_+]$
in the setting of Theorem \NUM, we get following.

\bigskip
\ADT
\bf Corollary \NUM. \it In $(\the\QuadS.\the\QuadT)$ we can write
$$\eqalign{
\varphi_{\lambda,\mu}(t, x_0,\xi) &= \Big\langle
 x_0 \Big\vert  f_1(t,-\lambda, -\widehat{S}_0) b_0 \Big\rangle \!-\!
(\xi \!-\! 1) \Big\langle \big( f_2(t,\lambda, \widehat{S}_0) b_0
\Big\vert   b_0 \Big\rangle \!+ \cr
& \hskip5mm + (\xi-1) \big( \lambda + i \mu %\langle R\overline{x} \vert \overline{x}\rangle/2
\big) e^{-2\lambda t}  + 1 ,\cr
P_0\Phi^t(x) &= \varphi_{\lambda,\mu}(t,x_0,\xi)^{-1} \Big[ e^{-\lambda t} \exp(i t \widehat{S}_0) x_0 - (\xi - 1) e^{-2\lambda t}
f_1(t,\lambda,\widehat{S}_0) b_0 \Big] .
\cr}$$
with the bounded analytic functions $f_j(t,\lambda, \cdot) : \RR\to\CC$ $(j=1,2;\ \lambda,t\in\RR)$
$$\eqalign{
&f_1(t,\lambda,{\bf s}):= \int_0^t e^{-\lambda\tau} e^{-i {\bf s} \tau} \, d\tau=
{e^{-t(\lambda+i{\bf s})} -1 \over \lambda+i{\bf s}} = t \sum_{k=0}^\infty {(-1)^{k+1} \over (k+1)!} \big[t(\lambda+i{\bf s})\big]^{k} ,\cr
&f_2(t,\lambda,{\bf s}):= \int_0^t e^{-2\lambda t} f_1(\tau,\lambda,{\bf s}) \, d\tau =
{ e^{-2\lambda t} (\lambda-i{\bf s}) - 2\lambda e^{-t(\lambda-i{\bf s})} + i{\bf s} \over 2\lambda(\lambda^2+{\bf s}^2)(\lambda-i{\bf s})} =\cr
&\hskip14mm = \sum_{n=0}^\infty \Big[ \sum_{k=0}^n {(-2\lambda)^{n-k} \over (n-k)!} \, {(\lambda+i{\bf s})^k \over (k+1)!} \Big] {t^{m+2} \over m+2} .\cr}$$




\DEL{
\bigskip
\ADT
\bf Theorem \NUM. \it Given a strongly continuous one-parameter semigroup $[ \Psi^t: t\in\RR_+]$ of holomorphic Carath\'eodory isometries of the unit open ball ${\bf B}$
in a Hilbert space ${\bf H}$, we have the following alternatives

\medskip
\item{{\rm (i)}} either $\Psi^t= \Theta \circ U^t \circ \Theta^{-1} \vert {\bf B}$ $(t\in\RR_+)$
with some M\"obius transformation $\Theta$ and a
strongly continuous one-parameter semigroup $[U^t: t\in\RR_+]$ of linear ${\bf H}$-isometries;

\medskip
\item{{\rm (ii)}} or $\bigcap\limits_{t\in\RR_+} {\rm Fix}(\Psi^t) =\emptyset$ and there exists a unit vector $\overline{x}\in\partial{\bf B}$
along with a M\"obius transformation $\Theta$ such that
$$\Psi^t = \Theta \circ F({\cal T}{\cal V}^t{\cal T}^{-1}) \circ \Theta^{-1} \qquad (t\in\RR_+)$$
where
${\cal T}\in{\cal L}({\bf H}\oplus\CC)$ is the projective shift $h\oplus \zeta \mapsto (h+\zeta \overline{x})\oplus\zeta$ and
$[{\cal V}^t: t\in\RR_+]$ is a strongly continuous one-parameter semigroup in ${\cal L}({\bf H}\oplus\CC)$
which can be written in the matrix form
$${\cal V}^t = \left[ \matrix{
V_{11}(t) &v_{12}(t) &0 \cr
0 &1 &0 \cr
v_{31}(t)^* &\upsilon_{32}(t) &1 \cr } \right] \qquad (t\in\RR_+) $$
with respect to the orthogonal decomposition ${\bf H}\oplus\CC = [{\bf H}\ominus (\CC\overline{x})] \oplus [\CC\overline{x}] \oplus\CC$
where
$$\eqalign{
&V_{11}(t) = e^{\lambda t} V_0^t, \ \ v_{12}(t)=\int_0^t V_{11}^\tau y \, d\tau, \cr
&v_{31}(t)= -v_{12}(t)+(\lambda+i\mu) \int_0^t v_{12}(\tau) \, d\tau, \ \
\upsilon_{32}(t)= (\lambda + i\mu)t \cr}$$
with some
strongly continuous one-parameter semigroup $[V_0^t: t\in\RR_+]$ of  $[{\bf H}\ominus(\CC\overline{x})]$-isometries
and constants $\lambda,\mu\in\RR$.
}%ENDDEL

\bigskip
\bf Example \NUM. \rm
Consider the space ${\bf H}:=L^2(0,\infty)$ with the semigroup $[U^t: t\in\RR_+]$ of the isometric shifts
$U^t f:=1_{(0,t)} f([x-t]_+)$. It is well-known that its generator is the (skew symmetric) differential operator
$iR:=d/dx$ with ${\rm dom}(R)\supset C^1_0(0,\infty)$.
Given any constant $\nu\in\CC$, with the functions
$$\overline{x} := x e^{-x}, \quad b:=(\nu - iR)\overline{x} =[(\nu+1)x-1]e^{-x}$$
the vector field \
$\Gamma(f) := b \langle f \vert b\rangle f + iR f$ $\big( f\in{\rm dom}(R)\big)$ \
gives rise to a strongly continuous one-parameter semigroup $\Phi^t: t\in\RR_+]$ of Carath\'eodory isometries
of the unit ball ${\bf B}$ such that ${d\over dt} \big\vert_{t=0+} \Phi^t(f)= \Gamma(f)$ $(f\in{\rm dom}(R))$
and $\overline{x} \in \bigcap_{t\in\RR_+} {\rm Fix}(\Phi^t)$.
We can calculate the isometries $U_0^t$ %appearing implicitly in (\the\QuadT.\the\QuadS)
in the main Theorem
in terms of shifts by the aid of the following observation:

{\it The subspace ${\bf H}_1:= {\bf H}_0 \cap \big[ e^{-x} \big]^\perp$ %=\{ f\in{\bf H}: f\perp xe^{-x}, e^{-x}\}$
is $[ U^t: t\in\RR_+]$-invariant. In particular $U_0^t\big\vert{\bf H}_1=U^t\big\vert {\bf H}_1$.}

Indeed, if $f\perp xe^{-x}, e^{-x}$ then for any $t>0$ and $\alpha,\beta\in\CC$  we have
$$\eqalign{
&\big\langle U^t f \big\vert (\alpha x + \beta)e^{-x} \big\rangle = \int_0^\infty 1_{(0,t)} (x) f([x-t]_+) (\alpha x+\beta)e^{-x} dx =\cr
&= \int_0^\infty f(y) [\alpha(y+t) +\beta] e^{-(y+t)} dy =
\alpha e^{-t} \big\langle f \big\vert xe^{-x} \big\rangle + (\alpha t + \beta) e^{-t} \big\langle f \big\vert e^{-x} \big\rangle = 0
\cr} $$
Instead of using the orthogonal decomposition ${\bf H}_0={\bf H}_1 \oplus [\CC h_0]$ with $h_0:=(x-1)e^{-x} \not\in {\rm dom}(R)$,
it is convenient to pass to matrices associated with the direct sum decomposition
\def\dotplus{\buildrel.\over+}
${\bf H}_0 = {\bf H}_1 \dotplus [\CC h_1]$ e.g. with a function $h_1\in{\rm dom}(R)$ e.g. with $h_1:=(3x-2x^2)e^{-x}$.
Since
$iR h_1 = {d\over dx} h_1 = (2x^2-7x+3)e^{-x}$ and $P_0 f = f - \langle f \vert xe^{-x}\rangle \Vert xe^{-x}\Vert^{-2} xe^{-x}$ \ $(f\in{\bf H})$,
we have
$P_0 (iRh_1) = (2x^2-6x+3)e^{-x} = g \dotplus 2h_1$ with $g:=(6x^2-12x+3)e^{-x}\in{\bf H}_1$.
Thus the ${\bf H}_1 \dotplus [\CC h_1]$ matrix form of $P_0(iR)\big\vert {\bf H}_0$ is
$\big[ {[iR\vert{\bf H}_1] \atop 0} \ {g\atop 2}\big]$. Hence, by Lemma {\the\TriS.\the\TriT} we get
$$U_0^t (f \dotplus \zeta h_1) = \Big[ U^t f  + \zeta \int_0^t e^{-2\tau} U^\tau g \, d\tau \Big] \dotplus \big[ \zeta e^{2t} h_1\big] \qquad (f\in{\bf H}_1,\ \zeta\in\CC).$$


\bigskip
\bigskip
\ADS
\bf \the\SEC. Proofs
\bigskip
\rm

\bigskip
\bf Proof. \rm
The case (i) is completely treated in [\VESi]: (i) holds if and only if $\bigcap\limits_{t\in\RR_+} {\rm Fix}(\Psi^t) \ne\emptyset$
because given any joint fixed point $\overline{x}\in \bigcap\limits_{t\in\RR_+} {\rm Fix}(\Psi^t)$
$[\Theta_{\overline{x}} \circ \Psi^t \circ \Theta_{-\overline{x}}: t\in\RR_+]$
is a strongly continuous one-parameter semigroup of Carath\'eodory isometries of ${\bf B}$ leaving the origin fixed
which are known to be restrictions of linear isometries to ${\bf B}$.

\medskip
To prove (ii), let $\bigcap\limits_{t\in\RR_+} {\rm Fix}(\Psi^t)=\emptyset$.
As remarked in [\VESi], choosing any point $c\in{\bf B}$ such that the orbit $t\mapsto\Psi^t(c)$
is differentiable, the maps $\Phi^t:= \Theta_{-c} \circ \Psi^t \circ \Theta_c$ form a
strongly continuous one-parameter semigroup of Carath\'odory isometries of ${\bf B}$
with differentiable $0$-orbit.
In this case we can write
$$\Phi^t = F({\cal U}^t) \qquad (t\in\RR_+)$$
with a strongly continuous one-parameter subsemigroup $[{\cal U}^t: t\in\RR_+]$ of ${\cal L}({\bf H}\oplus\CC)$
whose generator has the matrix form
(with respect to the decomposition ${\bf H}\oplus\CC$)
$${\cal R} = \left[ \matrix{ iR & b \cr b^* & 0\cr} \right] , \quad {\rm dom}({\cal R}) = {\rm dom}(R) \oplus\CC$$
where $R$ is a densely defined closed symmetric operator on ${\bf H}$ such that $iR$ is the infinitesimal generator
of some strongly continuous one-parameter subsemigroup of isometries
in ${\cal L}({\bf H})$ and $b\in {\bf H}$.
It is also well-known [\VESi] that
the continuous extensions ${\overline{\Phi}}^t$  of the maps $\Phi^t$ to the closed unit ball $\overline{{\bf B}}$ are well-defined
and admit one or two common fixed points lying in the boundary $\partial{\bf B} =\{ x: \Vert x\Vert=1\}$.
Henceforth let
$$\overline{x}\in \partial{\bf B}\cap {\bigcap}_{t\in\RR_+} \!\!{\rm Fix}(\overline{\Phi}^t), \quad
{\bf H}_0:={\bf H}\ominus(\CC\overline{x}) , \quad
P:={\rm P}_{\CC\overline{x}}=\overline{x}\,\overline{x}^*, \quad P_0:={\rm P}_{{\bf H}_0} =I-P
.$$
Necessarily
$\overline{x} \oplus 1 =\big[ {\overline{x} \atop 1} \big]$ is a joint eigenvector of the
linear
operators ${\cal U}^t$ and for some $\nu\in\CC$,
\ADT
\newcount\BbS \BbS=\SEC
\newcount\BbT \BbT=\THM
$$\overline{x}\in{\rm dom}(R), \quad
{\cal R} [\overline{x} \oplus 1] = \nu [\overline{x} \oplus 1], \quad
b= \nu \overline{x}-iR\overline{x}, \quad \nu=\langle \overline{x} \vert b \rangle
\leqno(\NUM)$$
implying
$$\langle \overline{x} \vert R\overline{x} \rangle = 2 \, {\rm Im}(\nu), \quad
b=b_0+ \overline{\nu} \,\overline{x} \quad \hbox{where}\quad b_0:=P_0 b = -i P_0 R\overline{x} .
\leqno(\NUM^\prime)$$
Applying the projective shift ${\cal T}: x\oplus\xi\mapsto (x+\xi\overline{x})\oplus\xi$,
%and ${\cal T}^{-1}: x\oplus\xi\mapsto (x-\xi\overline{x})\oplus\xi$,
consider the operators
$${\cal V}^t := e^{-\nu t} {\cal T}^{-1} {\cal U}^t {\cal T}, \quad %(t\in\RR_+), \quad
{\cal A}:= {\cal R}-\nu (I\oplus 1), \quad{\cal B}= {\cal T}^{-1} {\cal A} {\cal T} .$$
Both ${\cal A}$ and ${\cal B}$ are bounded perturbations of the operator $(iR)\oplus 1$
with
${\rm dom}({\cal B})= {\rm dom}({\cal A}) = {\rm dom}(R)\oplus \CC$.
Here  ${\cal A}$ is the infinitesimal generator of the strongly continuous one-parameter subgroup
$[e^{-\nu t} {\cal U}^t: t\in\RR_+]$ of ${\cal L}({\bf H}\oplus\CC)$.
Since ${\cal T}$ is a bounded invertible linear operator,
$[{\cal V}^t : t\in\RR]$
is also a strongly continuous one-parameter semigroup in ${\cal L}({\bf H}\oplus\CC)$
with ${\cal B}$ as infinitesimal generator.
Straightforward ${\bf H}\oplus\CC$-matrix calculation with (\the\BbS.\the\BbT) shows that
$${\cal B} = \left[ \matrix{I &-\overline{x} \cr 0 &1\cr}\right]
\left[ \matrix{iR-\nu I &b \cr b^* &-\nu\cr}\right]
\left[ \matrix{I &\overline{x} \cr 0 &1\cr}\right] =
\left[ \matrix{B &0 \cr b^* &0\cr}\right] \quad \hbox{where} \quad
B:= i R -\nu I  - \overline{x}b^*  .$$
As being a bounded perturbation of $iR$, the operator $B$ is the
infinitesimal generator of a (unique) strongly continuous semigroup in ${\cal L}({\bf H})$
which we denote by
$[V^t: t\in\RR_+]$ in the sequel.
According to Lemma \the\TriS.\the\TriT(i) we have
\ADT
\newcount\VtS \VtS=\SEC \newcount\VtT \VtT=\THM
$${\cal V}^t = \left[ \matrix{ V^t &0 \cr b^* {{\scriptstyle{\int_0^\tau}}} \, V^\tau d\tau &1 \cr}\right]
\qquad (t\in\RR_+).
\leqno(\NUM)$$
Since
$F(e^{-\nu t} {\cal U}^t) = F({\cal U}^t)=\Phi^t$ $(t\in\RR_+)$,
in terms of the translation
$$T x := x+\overline{x} \quad (x\in{\bf H}),$$
we can interpret the semigroup
$[ {\cal V}^t : t\in\RR^+]$
as the linear representation by means of $F$ of the semigroup $[T^{-1}\circ\Phi^t\circ T: t\in\RR_+]$
formed by holomorphic isometries of the shifted ball ${\bf B}-\overline{x}$
whose continuous extensions leave the origin fixed.
Namely, due to the projective identities
$F\big( {\cal T}^{-1}{\cal V}{\cal T} \big) = T^{-1} F({\cal V}) T$
$\big({\cal V}\in{\cal L}({\bf H}\oplus\CC)\big)$,
for the points $x\in T^{-1}{\bf B} = {\bf B} -\overline{x}$ we have
$$F({\cal V}^t) x = F({\cal T}^{-1} e^{-\nu t} {\cal U}^t {\cal T})x =
%[F({e^{-\nu t} \cal U}^t)(x+\overline{x})]-\overline{x}
[T^{-1}\circ\Phi\circ T]x
=\Phi(x+\overline{x})-\overline{x} .
$$
As a consequence of (\the\VtS.\the\VtT) and since
${\cal V}^t(x\oplus 1)= \big[ V^t x\big] \oplus\big[ \int_0^t \langle V^\tau x \vert\, b\rangle d\tau + 1\big]$, we get
\ADT
\newcount\PhiS \PhiS=\SEC \newcount\PhiT \PhiT=\THM
$$\Phi^t(x) = {V^t(x-\overline{x}) \over \big\langle {{\scriptstyle{\int_0^t}}} V^\tau (x-\overline{x}) \, d\tau \big\vert b \big\rangle +1} + \overline{x}
\qquad (x\in{\bf B}, \ t\in\RR_+).
\leqno(\NUM)$$
We proceed to a quadrature formula of $V^t$
in terms of $[{\bf H}_0 \oplus\CC \overline{x}]$-matrices and isometry semigroups.
Consider the $[{\bf H}_0 \oplus\CC \overline{x}]$-matrix form
%$B=\left[ {P_0 B P_0 \ \ P_0 B \overline{x} \atop \overline{x} }\right] $
of its generator $B=iR-\nu I -\overline{x}b^*$.
%are the orthogonal projections onto the complementary subspaces $\CC\overline{x}$ and ${\bf H}_0$, %:={\bf H}\ominus(\CC\overline{x})$,
%respectively.
Since $R$ is a symmetric operator and $b=(\nu I-iR)\overline{x}$,
for any vector $z\in{\rm dom}(B)={\rm dom}(R)$
we have
$b^*z =  %\langle z \vert \nu \overline{x}- iR\overline{x} \rangle =
\langle \overline{\nu} z + iRz \vert \overline{x} \rangle$
implying also
$[\overline{x}b^*] z = \langle z \vert \nu \overline{x}- iR\overline{x} \rangle \overline{x} =
P(\overline{\nu}I + i R) z$.
Thus, by setting
$$\lambda:={\rm Re}(\nu) , \quad \mu:={\rm Im}(\nu)=\big\langle \overline{x} \big\vert R\overline{x} \big\rangle /2 , \quad S:= R-\mu I$$
we can write
$b^* =\overline{x}^* (\lambda I + iS )$, $\overline{x}b^*=P(\lambda I +i S)$,
$B= iR - \nu I -\overline{x}b^* = P_0 [iS] - \lambda (I+P)$.
Furthermore, by (\the\BbS.\the\BbT) and (\the\BbS.\the\BbT$^\prime$),
$B\overline{x} = (i S-\lambda I)\overline{x}-\langle \overline{x} \vert b \rangle \overline{x} =
-b - \nu \overline{x} = -[b_0 \oplus(2\lambda \overline{x})]$.
Thus, in terms of \ ${\bf H}_0\oplus (\CC\overline{x})\big]$-matrices,
$$B \!=\! \left[ \matrix{P_0[iS \!-\! \lambda I]\big\vert {\bf H}_0 &\!\! i P_0 S\overline{x} \cr 0 &-2\lambda\cr} \right] \!\!=\!\!
\left[ \matrix{ iS_0 \!+\!\lambda I_0 &\!-b_0 \cr 0 &0\cr} \right]
\!-\! 2\lambda \! \left[ \matrix{ I_0 &0 \cr 0 &1\cr}\right] \quad \hbox{where} \quad
\eqalign{&S_0:=P_0 S P_0\big\vert {\bf H}_0, \cr &I_0:={\rm Id}_{{\bf H}_0} .\cr}$$
Since
$\overline{x}\!\in\!{\rm dom}(S)$ and
$P_0 S P_0 \!=\! S-PS-SP+PSP \!=\! R-[\mu I + \overline{x}(S\overline{x})^*+(S\overline{x})\overline{x}^* -\langle \overline{x}\vert S\overline{x}\rangle \overline{x}\overline{x}^*]$,
the operator $P_0[iS]P_0$
is a bounded perturbation of the infinitesimal generator $iR$.
Hence its restriction $iS_0$ to ${\bf H}_0$
is also the infinitesimal generator a
strongly continuous semigroup $[V_0^t: t\in\RR_+]$ of isometries on ${\rm range}(\overline{P})={\bf H}_0$.
By Lemma \the\TriS.\the\TriT(ii) it follows
$$V^t = e^{-2\lambda t} \left[ \matrix{ e^{\lambda t} V_0^t
&{{\scriptstyle{\int_0^t}}} e^{\lambda \tau}
V_0^\tau b_0 \, d\tau \cr 0 & 1\cr}\right] \qquad (t\in\RR_+) $$
in terms of ${\bf H}_0\oplus(\CC\overline{x})$-matrices.
Therefore, for any $x=x_0+(\xi \overline{x})$ with $x_0\in{\bf H}_0$ and $\xi\in\CC$,
\ADT
\newcount\VVS \VVS=\SEC \newcount\VVT \VVT=\THM
$$V^t(x-\overline{x}) =  e^{-\lambda t} V_0^t x_0 +
e^{-2\lambda t} \big(1 - \xi \big) \Big[ {{\scriptstyle{\int_0^t}}} e^{\lambda \tau} V_0^\tau b_0 \, d\tau  - \overline{x}\Big] .
\leqno(\NUM)$$

Next we calculate the terms $b^*  V^\tau (x-\overline{x}) =
\big\langle V^t (x-\overline{x}) \big\vert \,
%\lambda \overline{x} - i S \overline{x}
b_0+\overline{\nu}\,\overline{x}\big\rangle$.
Since ${\rm range}(V_0^\tau) \!\subset\! {\bf H}_0 \!\perp\! \overline{x}$, we have
$ \big\langle V^\tau (x-\overline{x}) \big\vert \,\overline{x} \big\rangle =
-e^{-2\lambda \tau} (1-\xi)$. Also $b_0\perp \overline{x}$.
Therefore
$$\eqalign{
b^*  V^\tau (x-\overline{x}) &= \big\langle V^\tau (x-\overline{x}) \big\vert b_0 + \overline{\nu}\,\overline{x} \big\rangle =\cr
&= e^{-\lambda\tau} \big\langle V_0^\tau x_0 \big\vert b_0 \big\rangle + e^{-2\lambda\tau}(1-\xi)
\Big[ {{\scriptstyle{\int_0^\tau}}} e^{\lambda\sigma} \big\langle V_0^\sigma b_0 \big\vert b_0 \big\rangle d\sigma - \nu \Big] .\cr}$$
Substituting this into (\the\PhiS.\the\PhiT), in terms of integrals of strongly continuous isometry valued functions, we get
$$\eqalign{
P\Phi^t(x) &= \Big[ \varphi(\lambda,\xi,x_0)^{-1} (\xi - 1) e^{-2\lambda t} +1 \Big] \overline{x} \cr
P_0\Phi^t(x) &= \varphi(\lambda,\xi,x_0)^{-1} \Big[ e^{-\lambda t} V_0^t x_0 - (\xi - 1) e^{-2\lambda t}
\big( {{\scriptstyle{\int_0^t}}} e^{\lambda \tau} V_0^\tau \, d\tau \big) b_0 \Big] \ \
\hbox{where}\cr
\varphi(\lambda,\xi,x_0) &:= \Big\langle
 \big( {{\scriptstyle{\int_0^t}}} e^{-\lambda\tau} V_0^\tau \, d\tau \big) x_0 \Big\vert \ b_0 \Big\rangle -
(\xi - 1) \Big\langle \big( {{\scriptstyle{\int_0^t}}} e^{-2\lambda\tau}
{{\scriptstyle{\int_0^\tau}}} e^{\lambda\sigma} V_0^\sigma \, d\sigma \, d\tau \big) b_0
\Big\vert  \ b_0 \Big\rangle + \cr
& \hskip5mm + (\xi-1) \big( \lambda + i \mu %\langle R\overline{x} \vert \overline{x}\rangle/2
\big) e^{-2\lambda t}  + 1 . \cr}$$

\ADT
\bf Remark \NUM. \rm We can eliminate the integrals above in terms of resolvents as follows.

\DEL{
%1-st VERSION
Since $iS_0$ is the generator of the semigroup $[V_0^t : t\in\RR_0]$,
the operator $S_0$ is closed with dense domain in ${\bf H}_0$.
Since $S_0$ is also symmetric, both $\pm iS_0$ are dissipative (namely ${\rm Re}\langle \pm iS x_0 \vert x_0\rangle =0$ for $x_0\in{\rm dom}(S)$)
with the property that ${\rm range}(\pm iS_0 + \delta I_0)$ is dense in ${\bf H}$ for any $\delta \in\RR\setminus\{ 0\}$.\footnote{$^*$}{
Indeed $y_0 \perp {\rm range}(\pm iS_0 + \delta I_0)$ means $0=\langle \pm iS_0 x_0 - \delta x_0 \vert y_0\rangle$
that is  $0=\langle x_0 \vert \mp iS_0 y_0 - \delta y_0\rangle$ for $(x_0\in{\rm dom}(S))$
entailing $\mp iS_0 y_0 + \delta y_0 =0$ with $\delta \Vert y_0\Vert^2 = \pm i \langle S_0 y_0 \vert y_0\rangle \in i\RR$
which is possible only if $y=0$.}
Thus by the Lumer-Phillips theorem [\NAG, II. Thm. 3.15], also the operator $- iS_0$ generates a
strongly continuous contraction semigroup and all the values $0\ne \delta \in\RR$ belong to the resolvent set of $iS$.
In particular the operators $(iS + \delta I_0)^{-1}: {\rm range}(S) \to {\bf H}_0$ $(0\ne \delta\in\RR)$ all bounded and densely defined.
}%ENDDEL 1-ST VERSION

As being $iS_0$ the generator of the semigroup $[V_0^t : t\in\RR_0]$,
the operator $S_0$ is closed with dense domain in ${\bf H}_0$.
Since $S_0$ is also symmetric, both $\pm iS_0$ are dissipative (namely ${\rm Re}\langle \pm iS x_0 \vert x_0\rangle =0$ for $x_0\in{\rm dom}(S)$)
with the properties that both ${\rm range}(\pm iS_0 + \delta I_0)$ are dense in ${\bf H}$ for any $\delta>0$
and that the operators $(iS + \delta I_0)^{-1}: {\rm range}(S) \to {\bf H}_0$ $(0\ne \delta\in\RR)$ are all bounded and densely defined.\footnote{$^*$}{
Indeed $y_0 \perp {\rm range}(\pm iS_0 + \delta I_0)$ means $0=\langle \pm iS_0 x_0 - \delta x_0 \vert y_0\rangle$
that is  $0=\langle x_0 \vert \mp iS_0 y_0 - \delta y_0\rangle$ for $(x_0\in{\rm dom}(S))$
entailing $\mp iS_0 y_0 + \delta y_0 =0$ with $\delta \Vert y_0\Vert^2 = \pm i \langle S_0 y_0 \vert y_0\rangle \in i\RR$
which is possible only if $y=0$.
Thus by the Lumer-Phillips theorem [\NAG, II. Thm. 3.15], also the operator $- iS_0$ generates a
strongly continuous contraction (actually isometry) semigroup and all the values $0\ne \delta \in\RR$ belong to the resolvent set of $iS$.}

Given $\delta\in\RR\setminus\{ 0\}$,
by [\NAG, II. Lemma 1.3] for any $x_0\in{\rm range}(iS_0-\delta I_0)$ and $t>0$ we have
$\int_0^t e^{-\delta\tau} V_0^\tau x_0 \, d\tau = \int_0^t e^{-\delta\tau} V_0^\tau (iS_0-\delta I_0) [(iS_0-\delta I_0)^{-1} x_0] \, d\tau  =
\big( e^{-\delta t} V_0^t - I_0\big) (iS_0-\delta I_0)^{-1} x_0 $.
The boundedness of the operators $V^t$ and the resolvents $R(\delta,iS)={\rm closure}\big( (\delta I_0 - iS_0)^{-1} \big)$ establishes that
for all $t>0$ and $0\ne \lambda\in\RR$,
$$\eqalign{
&\int_0^t e^{-\lambda\tau} V_0^\tau \, d\tau = \big( I_0 - e^{-\lambda t} V_0^t \big) R(\lambda, iS_0) ,\cr
&\int_0^t e^{-2\lambda\tau} \int_0^\tau e^{\lambda\sigma} V_0^\sigma \, d\sigma \, d\tau =
\int_0^t e^{-2\lambda\tau}  \big(  I_0 - e^{\lambda \tau} V_0^\tau \big) R(-\lambda, iS_0) \, d\tau =\cr
&= {1 -e^{-2\lambda t} \over 2\lambda} R(-\lambda,iS_0) - \big( I_0 - e^{-\lambda t} V_0^t \big) R(\lambda,iS_0) R(-\lambda,iS_0) \cr} $$
with integrals of strongly continuous bounded operator valued functions. Taking (\the\PhiS.\the\PhiT) and (\the\VVS.\the\VVT) into account, hence we conclude
that for any $x=x_0+\xi\overline{x}\in{\bf B}$,
$$\eqalign{
P\Phi^t(x) &= \Big[ \varphi(\lambda,\xi,x_0)^{-1} (1-\xi) e^{-2\lambda t} +1 \Big] \overline{x} \cr
P_0\Phi^t(x) &= \varphi(\lambda,\xi,x_0)^{-1} \Big[ e^{-\lambda t} V_0^t x_0 - (1-\xi) e^{-2\lambda t} (I_0-e^{-\lambda t} V_0^t)R(-\lambda,iS_0) b_0 \Big] \cr}$$
where
$$\eqalign{
\varphi(\lambda,\xi,x_0) &:= \Big\langle
\big( I_0 - e^{-\lambda t} V_0^t \big) R(\lambda, iS_0) x_0 \Big\vert \ b_0 \Big\rangle +
(1-\xi) \Big\langle {1 -e^{-2\lambda t} \over 2\lambda} R(-\lambda,iS_0) b_0  \Big\vert \ b_0 \Big\rangle - \cr
& \  - (1-\xi) \Big\langle \big( I_0 - e^{-\lambda t} V_0^t \big) R(\lambda,iS_0) R(-\lambda,iS_0) b_0
\Big\vert  \ b_0 \Big\rangle - \cr
& \ - (1-\xi) \big( \lambda + i \mu %\langle R\overline{x} \vert \overline{x}\rangle/2
\big) e^{-2\lambda t}  + 1 . \cr}$$

\bigskip
{\bf The case of two boundary fixed points}
\bigskip
Assume $\bigcap_{t\in\RR_+} {\rm Fix}(\overline{\Phi}^t) = \{ \overline{x}_1,\overline{x}_2\} \subset \partial{bf B}$ with $\overline{x}_1\ne\overline{x}_2$.
Observe that in this case there exists $t_0$ with ${\rm Fix}(\Phi^{t_0})=\{ \overline{x}_1,\overline{x}_2\}$.
Indeed, in the contrary case,
%let $T:= \big\{ \tau\in\RR_+: {\rm Fix}(\Phi^\tau) \setminus \{ \overline{x}_1,\overline{x}_2\} \big\}$.
the whole real segment $[\overline{x}_1,\overline{x}_2]$ would be contained in ${\rm Fix}(\Phi^t)$ for every index $t$.
It is well-known that there exists a M\"obius transformation $\Theta=F({\cal M})$ and a vector $\overline{x}\in\partial{\bf B}$ such that
${\rm Fix}\big(\Theta\circ\overline{\Phi}^{t_0} \circ\Theta^{-1}\big) = \big\{ \Theta(\overline{x}_1),\Theta(\overline{x}_1) \big\} =\{ \pm\overline{x} \}$.
It follows
$\bigcap_{t\in\RR_+} {\rm Fix}\big(\Theta\circ\overline{\Phi}^t \circ\Theta^{-1}\big) =  \{ \pm\overline{x} \}$.
Since the orbits $t\mapsto \overline{\Phi}^t (\pm \overline{x})$ are differentiable (actually constants), also the orbit of $0 =-\overline{x}+\overline{x}$ is differentiable.
In particular,
$(\pm\overline{x})\oplus 1$ are joint eigenvectors of the linear representation operators ${\cal M}{\cal U}^t {\cal M}^{-1}$ with $F({\cal U}^t)=\Phi^t$.
Thus, by passing to the {\it M\"obius equivalent} semigroup
$[\Theta\circ\overline{\Phi}^t \circ\Theta^{-1}: t\in\RR_+]$ instead of  $[\overline{\Phi}^t : t\in\RR_+]$,
in this case
we may assume without loss of generality that
$$\bigcap_{t\in\RR_+} {\rm Fix}\big({\overline{\Phi}}^t\big) = \{ \pm\overline{x} \}, \quad
{\cal R}[(-\overline{x})\oplus 1] = \nu_1 [(-\overline{x})\oplus 1], \ \ {\cal R}[\overline{x}\oplus 1] = \nu_2 [\overline{x}\oplus 1]$$
for some constants $\nu_1,\nu_2\in\CC$.
%Recall that the generator ${\cal R}$ of $[{\cal U}^t: t\in\RR]$ has the form ${\cal R}=\left[ {iR \ b \atop b^* \ 0} \right]$.
Hence from (\the\BbS.\the\BbT) we see that
$b=-(\nu_1 I - iR)\overline{x}=(\nu_2 I-iR)\overline{x}$
entailing that $\overline{x}$ is an eigenvector of $R$ with eigenvalue ${1\over 2i}(\nu_1+\nu_2) =\langle R\overline{x} \vert \overline{x}\rangle \in \RR$.
%With the notation $\lambda_k := {\rm Re}(\nu_k)$, $\mu_k := {\rm Im}(\nu_k)$,
Taking also (\the\BbS.\the\BbT$^\prime$) in account we deduce
$b= 2^{-1}(\nu_2-\nu_1)\overline{x}$,
$b_0= P_0 b=0, \ \ b= -\overline{\nu_1}\overline{x} = \overline{\nu_2}\overline{x}$,
$2{\rm Im}(\nu_k) = \langle R\overline{x} \vert \overline{x} \rangle =(2i)^{-1}(\nu_1+\nu_2) =0$.
Thus, for some $\lambda \in\RR$, necessarily
$\nu_1=\lambda$, $\nu_2=-\lambda$ entailing also
$R\overline{x}=0$, $b=\lambda \overline{x}$, ${\cal R}[(\pm \overline{x})\oplus 1] = \pm\lambda [(\pm \overline{x})\oplus 1]$.
With the choice $\nu:=\nu_1$, in terms of (\the\VtS.\the\VtT) this means that $b_0=0$, $S_0=P_0 R P_0 = R$ and hence
$$\eqalign{
P\Phi^t(x) &= { (1-\lambda)(1-\xi) e^{-2\lambda t} + 1 \over 1 - (1-\xi)\lambda e^{-2\lambda t} } \overline{x} \cr
P_0\Phi^t(x) &= {e^{-\lambda t}\over 1 - (1-\xi)\lambda e^{-2\lambda t} } V_0^t x_0 \ .\cr}$$







\bigskip
\hrule
\vskip2pt
\hrule
\bigskip

\ADT
\bf Lemma \NUM. \it Assume $[ {\cal V}^t: t\in\RR_+]$ is a
strongly continuous one-parameter semigroup in ${\cal L}({\bf H}\oplus\CC)$
whose infinitesimal generator is a bounded perturbation of a skew symmetric operator and
${\cal V}^t (0 \oplus 1) = 0\oplus 1$ $(t\in\RR_+)$.
Then
there is a
strongly continuous one-parameter semigroup $[V^t: t\in\RR_+]$ in ${\cal L}({\bf H})$
along with a vector $b\in{\bf H}$ such that
$${\cal V}^t = \left[ \matrix{ V^t &0 \cr [W^t b]^* & 1\cr} \right] \quad \hbox{where} \quad
W^t b = \int_0^t V^\tau \ d\tau \qquad (t\in\RR_+).$$
In particular, the infinitesimal generator ${\cal B}$ of $[{\cal V}^t: t\in\RR_+]$
has the matrix form
$${\cal B}= \left[ {B \ 0 \atop b^* \ 0} \right] , \quad {\rm dom}({\cal B})={\rm dom}(B) \oplus \CC$$
where
$B$ is the infinitesimal generator of \ $[V^t : t\in\RR_+]$.

\bigskip
\bf Proof. \rm
Since ${\cal V}^t (0 \oplus 1) = 0\oplus 1$ $(t\in\RR_+)$,
the operators ${\cal V}^t$ have the matrix form
$${\cal V}^t = \left[ \matrix{ V^t & 0 \cr v_t^* & 1\cr} \right] , \ \ V^t\in {\cal L}({\bf H}), \ v_t \in {\bf H}\qquad (t\in\RR_+) $$
with respect to the decomposition ${\bf H}\oplus\CC$.
Considering orbits of the form $t\mapsto {\cal V}^t [x\oplus 0] = [V^t x]\oplus \langle x \vert v_t\rangle$
we see that $[V^t: t\in\RR_+]$ is a continuous one-parameter semigroup in ${\cal L}({\bf H})$.
Also $0\oplus 1\in{\rm dom}({\cal B})$ with ${\cal B}[0\oplus 1]=0$.
In particular
$x\oplus 0 = [x\oplus\xi] - \xi [0\oplus 1] \in {\rm dom}({\cal B})$
for any vector $x\oplus\xi \in{\rm dom}({\cal B})$,
showing that ${\rm dom}({\cal B}) = [X\oplus 0] + \CC [0\oplus 1]$ for some dense linear subspace of ${\bf H}$.
We have $x\in X$ if and only if the orbit
$t\mapsto {\cal V}^t [x\oplus 0] = [V^t x]\oplus \langle x \vert v_t\rangle$ is differentiable.
Hence we deduce that $x\in X$ if and only if the functions $t\mapsto V^t x$ and $t\mapsto \langle x \vert v_t\rangle$ are simultaneously differentiable.
Thus
%, by writing $B$ for the infinitesimal generator of $[V^t : t\in\RR_+]$
 $x\in X$ if and only if $x\in {\rm dom}(B)$ and $t\mapsto \langle x \vert v_t\rangle$ is differentiable.
.




\bigskip
\hrule
\bigskip

\bigskip
Henceforth
$[\Psi^t: t\in\RR_+]$ is an arbitrarily fixed strongly continuous one-parameter semigroup of Carat\'eodory isometries of ${\bf B}$.

\bigskip
\ADT
\bf Remark \NUM. \rm We know that $\bigcap\limits_{t\in\RR_+} {\rm Fix}(\overline{\Psi}^t) \ne \emptyset$.
By fixing such a common fixed point $\overline{x}\in\overline{{\bf B}}$ with $\overline{\Psi}^t(\overline{x})=\overline{x}$,
we know also that $\big[ [x\mapsto \Psi^t(x+\overline{x})-\overline{x}]: t\in\RR_+\big]$ is a strongly continuous semigroup of maps $\overline{{\bf K}}\to\overline{{\bf K}}$
with common fixed point $0$ for the shifted ball ${\bf K}:=-\overline{x}+{\bf B}$. Hence we can write
$$\Psi^t =F\big( {{\cal V}^t \big) \ \ \hbox{with} \ \ {\cal V}^t:= \cal T}_{\overline{x}} {\cal U}^t  {\cal T}_{-\overline{x}}\quad (t\in\RR_+)$$
where $[{\cal U}^t: t\in \RR_+]$ is a strongly continuous one-parameter semigroup of triangular operators of the form
described in Lemma 2.4. As suggested in [\VESi], let us consider a modification
$$\Phi^t := \Theta_{-c} \circ \Psi^t \circ \Theta_c = F\big( {\cal W}^t\big), \ \
{\cal W}^t := {\cal M}_{-c} {\cal T}_{\overline{x}} {\cal U}^t {\cal T}_{-\overline{x}} {\cal M}_c \qquad (t\in\RR_+)$$
with a M\"obius transformation $\Theta_c$ such that
the orbit $t\mapsto \Phi^t(0)=\Theta_{-c}\big(\Psi^t(c)\big)$ should be differentiable.
In terms of the generators ${\cal A}:={d\over dt}\big\vert_{t=0} {\cal U}^t$ and ${\cal R}:={d\over dt}\big\vert_{t=0} {\cal W}^t$
this means the requirement $0\oplus 1\in{\rm dom}({\cal R})$.
In any case,
\ADT
$${\cal A} = {\cal T}_{-\overline{x}} {\cal M}_c {\cal R} {\cal M}_{-c} {\cal T}_{\overline{x}} \ , \quad
{\rm dom}({\cal A}) = {\cal T}_{-\overline{x}} {\cal M}_{c} \ {\rm dom}({\cal R}) .
\leqno(\NUM)$$
In particular the suitable points $c$ with
${\cal T}_{-\overline{x}} {\cal M}_{c} (0\oplus 1) = (c-\overline{x})\oplus 1 \in {\rm dom}({\cal R}) = {\rm dom}({\cal A})$
are dense in ${\bf B}$.
%
%
\DEL{
\ADT
\bf Lemma \NUM. \it If ${\cal S}$ is a linear submanifold in ${\bf H}\oplus \CC$ and $S:= \{ x\in{\bf H} : x\oplus 0 \in{\cal S}\}$ then
we have the following alternatives: either ${\cal S}=S\oplus 0$ or ${\cal S} = [S\oplus 0] + \CC (s\oplus\sigma)$ with any vector $s\oplus\sigma\in{\cal S}$ such that $\sigma\ne 0$.

\bf Proof. \rm Trivial.
}%ENDDEL
%
%
Recall that with some closed symmetric, in ${\bf H}$ densely defined operator $R$, with a vector $r\in{\bf H}$ and some constant $\nu\in\CC$ we have
$${\cal R} = \nu [I\oplus 1] + \left[ { iR \ r \atop r^* \ 1} \right] , \quad {\rm dom}({\cal R}) = X \oplus\CC
 \ \ \hbox{where} \ \ X:=  {\rm dom}(R) . $$
Also, with some densely defined operator $A$ and a densely defined linear functional $a$,
%${\cal A}$ has the form
$${\cal A} = \left[ {A \ 0 \atop a \ \ 0}\right]\bigg\vert Z\oplus \CC , \quad
Z :=\{ z\in{\bf H}: z\oplus 0\in{\rm dom A}\} \subset{\rm dom}(A)\cap{\rm dom}(a) .$$
Then from (\NUM) we conclude that
$Z\oplus \CC = {\rm dom}(A) = {\cal M}_{-c} {\cal T}_{\overline{x}} \ (X\oplus \CC)$ that is
$$Z= \big\{ z: \ z\oplus 0\in {\rm dom}({\cal A}) \big\} = \big\{ z:
\exists\ x\in X, \xi\in\CC \quad z\oplus 0 = {\cal M}_{-c} {\cal T}_{\overline{x}} \ (x\oplus \xi) \big\}.$$
Taking into account the straightforward relations
$$\eqalign{
&\beta(c) {\cal T}_{-\overline{x}} {\cal M}_{c} =
%\left[ {I \ -\overline{x} \atop 0 \ \ \ 1}\right] \left[ {Q_c \ c \atop c^* \ 1} \right] =
\left[ {Q_c-\overline{x}c \ \ c-\overline{x} \atop c^* \qquad 1 }\right] , \quad
\beta(c) {\cal M}_{-c} {\cal T}_{\overline{x}} =
%\left[ {Q_c \ -c \atop -c^* \ 1} \right] \left[ {I \ \overline{x} \atop 0 \ \ \ 1}\right] =
\left[ {Q_c \ \ Q_c\overline{x}-c \atop -c^* \ \ 1-c^*\overline{x} }\right] ,  \cr
& Q_c c = \big[P_c + \beta(c)(I-P_c)\big] c =c , \quad
c c^* = \big[x\mapsto \langle x\vert c\rangle c \big] = \Vert c\Vert^2 P_c, \cr
&Q_c (Q_c \!-\! cc^* ) =  \big[ P_c \!+\! \beta(c)\overline{P}_c \big]
\big[ P_c \!+\! \beta(c)\overline{P}_c \!-\! \Vert c\Vert^2 P_c \big] =\beta(c)^2 P_c \!+\! \beta(c)^2 \overline{P}_c)=
(1-\Vert c\Vert^2) I \cr}$$
we deduce that
$$\eqalign{
Z &= \big\{ (Q_c - \overline{x}c^*)x + \xi (c-\overline{x}) : \ x\in X, \xi\in\CC, \ c^*x + \xi =0 \big\} =\cr
&= \big\{ (Q_c - \overline{x}c^*)x -(c^*x) (c-\overline{x}) : \ x\in X \big\} =
\big[ Q_c \overline{x} c^* - (c-\overline{x})c^* \big] X=\cr
&= \big[ Q_c - c c^*\big] X = Q_c^{-1} X .\cr}$$


\DEL{
from (\NUM), with the vector $s\oplus\sigma := {\cal M}_{-c} {\cal T}_{\overline{x}} = (Q_c \overline{x} -c)\oplus (1-c^*x)$
we conclude that
$$\eqalign{
Z\oplus \CC &= {\rm dom}(A) = {\cal M}_{-c} {\cal T}_{\overline{x}} \ (X\oplus \CC) =
\left\{ \!\left[ {Q_c \ \ s \atop -c^* \ \sigma }\right]
\left[ {x \atop \xi} \right] \! : x\!\in\! X, \xi\!\in\!\CC \right\} \!=\! \cr
&=
\left\{ \! \left[ {Q_cx + \xi s \atop 0}\right] \!\! : {-c^*x+\sigma\xi =0, \atop x\in X} \right\}
\!+\! \CC \! \left[ { s \atop \sigma}\right] \!\!=\cr
&= \left\{ \left[ {\sigma Q_cx +(c^* x) s \atop 0}\right] \! : x\in X \right\}
+ \CC \! \left[ { s \atop \sigma}\right] .\cr}$$
Here we have
$$\eqalign{
&\sigma Q_cx +(c^* x) s = (1-c^*\overline{x}) Q_cx +(c^* x) (Q_c\overline{x} -c) =\cr
&= Q_cx -c^*\overline{x} Q_cx + Q_c\overline{x}c^* x -c^*xc =
\big[ Q_c - c^*\overline{x} Q_c + Q_c \overline{x} c^* - \Vert c\Vert^2 P_c \big] x
\cr}$$
}%ENDDEL









\bigskip
\hrule
\bigskip



\bigskip
\ADT
\bf Assumption \NUM. \it
$\big[ \Psi^t\vert{\bf B}: t\in\RR_+\big]$ is a strongly continuous one-parameter semigroup.

\bigskip
\ADT
\bf Remark \NUM. \rm
We know from [\FRVE] the following facts:

\it
${\rm (i)}$ $\Psi^t \!=\! F({\cal V}^t)$ $(t\!\in\!\RR^+)$ with suitable operators
${\cal V}^t \in {\cal L}({\bf H}\oplus\CC)$.

\rm
Actually, by writing
$c_t:= \Psi^t(0)$, $Q_t := P_t + \sqrt{1-\Vert c_t\Vert^2} \overline{P}_t$ with the orthogonal projection $P_t:{\bf H}\to\CC c_t$,
we can take
${\cal V}^t := \big[ {Q_t \ c_t \atop c_t^* \ 1} \big] \big[ {U_t \ 0 \atop 0 \ 1} \big]$
where $U_t$ is some linear isometry ${\bf H}\to{\bf H}$.
Later on we need only the following immediate consequences of this representation:

${\rm (ii)}$ $\bigcap_{t\in[0,\tau]} {\rm dom}(\Psi^t)$ is a neighborhood of $\overline{{\bf B}}$ for all $\tau \!>\! 0$;
\quad
${\rm (iii)}$  $\Psi^t(\partial{\bf B}) \subset \partial{\bf B}$ $(t\in\RR_+)$.

\bigskip
\ADT
\bf Lemma. \it  We have $\bigcap\limits_{t\in\RR_+} \! {\rm Fix} (\Psi^t) \ne \emptyset$.




\bigskip
\hrule
\bigskip

{\bf Assumptions. \it $1)$ $\Vert \overline{x} \Vert =1$; \ \ $2)$ ${\cal U}^t = \big[ {U^t \ 0 \atop u^t \ 1} \big]$ $(t\in\RR_+)$;
\ \  $3)$ $d{\cal U}^t/dt\big\vert_{t=0} = \big[ {A \ 0 \atop a \ 0}\big]$ where
$$4)\ A=\overline{P}[iS]+\lambda(I+P), \ \ ax =\big\langle (iS-\lambda I)x \big\vert \overline{x} \big\rangle, \ \
5^*)\ \overline{x} \in {\rm dom}(A)= {\rm dom}(S)$$
with $S$ closed symmetric, \ $P:=\overline{x} \overline{x}^*$ projection onto $\CC\overline{x}$, \ $\overline{P}:=I-P$.

\bigskip
\bf Then \rm
$$U^t = \exp\big( t A \big), \quad
u^t x = \int_0^t aU^s x \ ds = \int_0^t \big\langle (iS-\lambda I)x \big\vert \overline U^s x \big\rangle \ ds .$$
With the decomposition ${\bf H}= [\overline{P}{\bf H}]\oplus [\CC\overline{x}]$,
$$A=\overline{P}[iS+\lambda I] + 2\lambda P =
\left[ \matrix{ \overline{P}[iS+\lambda I] \overline{P} &\overline{P}[iS+\lambda I]\overline{x} \cr
0 &2\lambda} \right]$$
The unbounded operator
$$Y:=  \overline{P}[iS+\lambda I] \overline{P}, \quad {\rm dom}(C)={\rm dom}(S)$$
is well-defined, since (according to assumption  $5^*)$
$ \overline{P} x = x - \langle x\vert \overline{x}\rangle \overline{x} \in {\rm dom}{S}$ for every $x\in {\rm dom}(S)$.
By writing also
$$\overline{y}:=\overline{P}[iS+\lambda I] \overline{x}$$
for short, we get
$$\exp(tA)= \left[ \matrix{ \exp(tY) & w^t \cr 0 &e^{2\lambda t} \cr}\right] , \quad
d w^t/dt = Y w^t + e^{2\lambda t} \overline{y} = 2\lambda w^t + \exp(tY)\overline{y} $$
as a consequence of the equations
${d\over dt} \big[ {\exp(tY) \ w^t \atop \ \ 0 \ \ e^{2\lambda t}} \big] =
\big[ {Y \ \overline{y} \atop 0 \ 2\lambda} \big]
\big[ {\exp(tY) \ w^t \atop \ \ 0 \ \ e^{2\lambda t}} \big] =
\big[ {\exp(tY) \ w^t \atop \ \ 0 \ \ e^{2\lambda t}} \big]
\big[ {Y \ \overline{y} \atop 0 \ 2\lambda} \big]$.
Due to the skew symmetry of $iS$, each operator $\exp(tY)$ has the form $e^{\lambda t} \exp(it\overline{P}S\overline{P}) =e^{\lambda t} [$isometry] on the subspace $\overline{P}{\bf H}$.
The initial value problem
$${d w^t\over dt} = 2\lambda w^t + \exp(tY)\overline{y}, \quad w^0=0 $$
can be solved explicitly by means of the substitution $w^t = e^{2\lambda t} u^t$:
$$2\lambda e^{2\lambda t} u^t + e^{2\lambda t} {du^t\over dt} = 2\lambda e^{2\lambda t} u^t + e^{\lambda t} \exp(itS)\overline{y}, \ \
\Rightarrow \ \ u^t = \int_0^t e^{-\lambda s} \exp(is \overline{P}S\overline{P})\overline{y} \ ds $$
with the result \ \ $w^t = \int_0^t e^{\lambda (2t-s)} \exp(is \overline{P}S\overline{P})\overline{y} \ ds $, \  whence
$$\exp(tA)= e^{\lambda t}\left[ \matrix{ \exp(it\overline{P}S\overline{P}) & \int_0^t e^{t-s}\exp(is\overline{P}S\overline{P})\ ds \cr 0 & e^{\lambda t}\cr}\right] .$$


\bigskip
\hrule
\bigskip






\bigskip
\hrule
\bigskip

$$A= i \int_{-\infty}^\infty \xi \ d E(\xi)$$

$$U^t = \exp\big( t(iA+\lambda)\big) = \int_{-\infty}^\infty e^{t(i\xi+\lambda)} \ d E(\xi) $$

$$u^t z = \int_0^t a U^s z \ ds ; \qquad
av = \big\langle (iA-\lambda)v \big\vert \overline{x} \big\rangle$$

$$\eqalign{ u^t z &= \int_0^t \big\langle (iA-\lambda)U^s z \big\vert \overline{x} \big\rangle =\cr
&= \int_{\xi=-\infty}^\infty \int_{s=0}^t (i\xi - \lambda) e^{t(i\xi+\lambda)} \ ds \ d\langle E(\xi)z \vert \overline{x} \rangle =\cr
&= \int_{\xi=-\infty}^\infty { \big( e^{t(i\xi+\lambda)}\big)(i\xi-\lambda) \over i\xi+\lambda}
d \langle E(\xi)z \vert \overline{x} \rangle
\cr}$$

$$\Psi^(x) =\overline{x} +  \big[ {\cal U}^t (x-\overline{x}) \oplus 1 \big]_\CC^{-1} \big[ {\cal U}^t (x-\overline{x}) \oplus 1 \big]_{{\bf H}}$$

$${\cal U}^t (z\oplus\zeta) = \big[ U^t z \big]\oplus \big[ u^t z + \zeta \big] $$


\bigskip
\hrule
\bigskip

\ADS
\bf \the\SEC. Fractional-linear automorphisms of a circular domain in Hilbert space \rm

\bigskip
\ADT
\bf Notations \NUM. \rm
${\bf H}$ complex Hilbert space with scalar product $\langle x\vert y\rangle$ being linear in $x$;

$u_* := \big[ x\mapsto \langle x \vert u\rangle\big]$ for any vector $u\in{\bf H}$;

${\bf H}\oplus\CC =\big\{ x\oplus\xi : \ x\in{\bf H}, \xi \in \CC\big\}$ with $x\oplus \xi:=\big[ {x \atop \xi}\big]$,
$(x\oplus\xi)_{{\bf H}} := x$, $(x\oplus\xi)_{\CC} := \xi$.


${\bf K}$ bounded convex circular domain in ${\bf H}$ with center $\overline{x}$ %and $0\in \partial{\bf K}$
(that is ${\bf K}-\overline{x} = e^{it} [{\bf K}-\overline{x}]$ $t\in\RR)$.


$\Psi^t(x)$ $(t\in\RR)$ injective fractional linear transformations with $0\in{\rm Fix}(\Psi^t)$.

Without loss of generality we shall write
$$\eqalign{
&\Psi^t(x) := (1+ u_*^t x)^{-1} U^t x =
\big[ {\cal U}^t (x\oplus 1) \big]_{\CC}^{-1} \big[ {\cal U}^t (x\oplus 1) \big]_{{\bf H}} \cr
&\hbox{where} \quad U^t\in{\cal L}({\bf H}), \ u^t\in{\bf H}, \ {\cal U}^t := \left[ \matrix{ U^t & 0\cr u_*^t &1 \cr} \right] .\cr}$$
We identify the matrix ${\cal U}^t$ with the linear operator $x\oplus\xi \mapsto {\cal U}^t (x\oplus\xi)$.
\bigskip
\ADT \newcount\ASSUME \ASSUME=\THM
\bf Assumption \NUM. \it $\overline{{\bf K}} \subset {\rm dom}(\Psi^t)=\{ x\in{\bf H}: 1+ u_*^t x \ne 0\}$ and
$\big[\Psi^t : t\in\RR\big]$ is a strongly continuous one-parameter group in ${\rm Aut}({\bf K})$
with $0\in\overline{\bf K}$.

\bigskip
\ADT
\bf Remark \NUM. \rm It is immediate that Assumption \the\SEC.\the\ASSUME\ implies that
the family $\big[ {\cal U}^t: t\in\RR\big]$ is a strongly continuous one-parameter subgroup of ${\cal L}({\bf H}\oplus\CC)$,
in particular
\ADT
$${\cal U}^{t+s} ={\cal U}^t {\cal U}^s, \quad U^{t+s}=U^t U^s, \quad u_*^{t+s}= u_*^t U^s + u_*^s = [(U^s)^* u^t]_* +u_*^s \qquad (t,s\in\RR).
\leqno(\NUM)$$
\bf Proof. \rm (\NUM) is a straightforward translation of the relation $\Psi^{t+s}=\Psi^t\circ\Psi^s$.
The continuity of the functions
$t\mapsto {\cal U}^t (x\oplus\xi) = (U^t x)\oplus (u_*^t x + \xi)$ can be deduced from the assumed continuity of
the functions $t\mapsto\Psi^t(x)$ $(x\in {\bf K})$ as follows.
Given $0\ne x\in{\bf K}$ and $t_0\in\RR$ arbitrarily,
we can choose $\lambda>1$ with $\lambda x\in{\bf K}$ since ${\bf K}$ is an open set.
Since the maps $\Psi^t$ are injective (actually invertible), we can also choose a vector
$v\in{\bf H}$ with $v_*\Psi^t(x)\ne v_*\Psi^t(\lambda x)$ for $t$ in some interval $J$ around $t_0$.
Then $v_* \Psi^t(\mu x)= [\mu/(1+u_*^t x)] v_*U^tx$ that is $[\mu v_*\Psi^t(\mu x)] u_t^*x - v_*U^tx =-v_*\Psi^t(x)$
for $t\in J$ and $\mu=1,\lambda$. It follows
that $u_*^tx = [\lambda \Psi^t(x)- \Psi^t(\lambda x)]/[\lambda(\Psi^t(x)-\Psi^t(\lambda x))]$ is a continuous function of $t$
on the interval $J$. Hence we get the also continuity of $t\mapsto U^tx = (1+u_*^tx)\Psi^t(x)$ establishing the continuity of
$t\mapsto {\cal U}^t (x\oplus \xi)$ for any $\xi\in\CC$.









\bigskip
\hrule
\bigskip




\bigskip
\bigskip

\bf Acknowledgements. \rm
The author's research was supported by the project
T\'AMOP-4.1.1.C-12/1/KONV-2012-0005
\it
Impulse Lasers for Use in Materials Science and Biophotonics
\rm
of the European Union and co-financed
by the European Social Fund.


\bigskip
\bigskip

\bf References
\rm

\medskip
\item{[\ABATE]} M. Abate, \it Common fixed points of commuting holomorphic maps, \rm
Math. Ann. \bf 283 \rm (1989) 645-655.


\medskip
\item{[\DED]} J.A. Deddens, \it On extending semigroups of contractions, \rm
Trans. Amer. Math. Soc., \bf 145 \rm (1969), 233-239.

\medskip
\item{[\BKU]} H. Braun -- W. Kaup -- H. Upmeier, \it On the automorphisms of
circular and Reinhardt domains in complex Banach spaces,
\rm Manuscripta Math., \bf 25 \rm (1978), 97-133.

\medskip
\item{[\DINEEN]} S. Dineen, \it Complete holomorphic vector fields on the second dual of a Banach space,
\rm Math. Scand. \bf 59 \rm (1986) 131-142.

\medskip
\item{[\NAG]} K.-J. Engel -- R. Nagel,
One-Parameter Semigroups for Linear Evolution Equations,
Graduate Texts in Mathematics Vol. 194, Springer-Verlag, New York, 2000.

\medskip
\item{[\FRVE]} T. Franzoni -- E. Vesentini, Holomorphic Maps and Invariant Distances,
North Holland Math. Studies 40, North Holland Publ. Co., Amsterdam-New York-Oxford, 1980.

\medskip
\item{[\GOEBEL]} K. Goebel, \it Fixed points and invariant domains of holomorphic mappings of the Hilbert ball,
\rm Nonlinear Analysis, Theory, Methods and Applications, \bf 6/12 \rm (1982) 1327-1334.

\medskip
\item{[\HS]} T.L. Hayden -- T.J. Suffridge, \it Biholomorphic maps in Hilbert space have a fixed point,
\rm Pacific J. Math., \bf 38 \rm (1971) 419-422.


\medskip
\item{[\IST]} J.-M. Isidro -- L.L. Stach\'o,
{\it Holomorphic Automorphism Groups in Banach Spaces},
North Holland Math. Studies 105,
North Holland Publ. Co., Oxford--New York--Amsterdam, 1985.

\medskip
\item{[\KAUP]} W. Kaup, \it
\"Uber die Klassifikation der symmetrischen hermiteschen Mannigfaltigkeiten unendlicher Dimension I.,
\rm Math. Ann., \bf 357 \rm (1981), 463--481.


\medskip
\item{[\KAUP]} W. Kaup, \it
A Riemann mapping theorem for bounded symmetric domains in complex Banach spaces,
\rm Math. Z., \bf 83 \rm (1983), 503--529.

\medskip
\item{[\KHAS]} V. Katshkevich - S. Reich - D. Shoiket,
\it One-parameter semigroups of fractional-linear transformations, \rm
Operator Theory Advances and Applications, \bf 123 \rm (2001) 401--411.


\medskip
\item{[\NACHBIN]} L. Nachbin,
{\it Holomorphic functions, domains of holomorphy and local properties},
\rm North-Holland Pub. Co., Oxford--New York--Amsterdam, 1970.

\medskip
\item{[\RSZ]} F. Riesz -- B. Sz.-Nagy,
Vorlesungen \"uber Funktionalanalysis,
VEB Deutscher Verlag der Wissenschaften, Berlin, 1968.

\medskip
\item{[\FSZ]}
B. Sz.-Nagy -- C. Foias, {\it Harmonic Analysis of Operators on Hilbert Space}, Akadémiai Kiadó/North-Holland Publ. Co.,
Budapest--Amsterdam--London

\medskip
\item{[\VESi]} E. Vesentini,
\it Semigrioups of holomorphic isometries, \rm
Advances in Mathemetics, \bf 65 \rm (1987), 272--306.

\medskip
\item{[\VESii]} E. Vesentini,
\it Semigrioups of holomorphic isometries, \rm
in: P.M. Gauthier (ed.), Complex Potential Theory, pp. 475--548.
Kluwer Academic Pulbl., Netherlands, 1994.

\medskip
\item{[\VIG]} J.-P. Vigu\'e,
{\it Le group des automorphismes analytiques d$^{\,\prime}$un
domaine born\'e d$^{\,\prime}$un espace de Banach complexe.
Applications aux domaines born\'es symmetriques},
Ann. Sci. Ecole Norm. sup., \bf  9  \rm (1976), 203--282.


\bigskip
\bigskip
L.L. STACH\'O

Bolyai Institute

University of Szeged

Aradi V\'ertan\'uk tere 1

H-6720 SZEGED, HUNGARY

E-mail: {\tt stacho@math.u-szged.hu}

\bye


\magnification=\magstep1
\parindent=0pt


\long\def\DEL#1{{}}

\newcount\SEC
\newcount\THM
\def\ADS{\advance \SEC by 1 \THM=0}
\def\ADT{\advance \THM by 1}
\def\NUM{{\the\SEC.\the\THM}}


\font\tenopen=msbm10 \font\sevenopen=msbm7 \font\fiveopen=msbm5
\newfam\openfam
\def\openo{\fam\openfam\tenopen}
\textfont\openfam=\tenopen \scriptfont\openfam=\sevenopen
\scriptscriptfont\openfam=\fiveopen

\font\tenujsym=msam10 \font\sevenujsym=msam7 \font\fiveujsym=msam5
\newfam\ujsymfam
\def\ujsym{\fam\ujsymfam\tenujsym}
\textfont\ujsymfam=\tenujsym \scriptfont\ujsymfam=\sevenujsym
\scriptscriptfont\ujsymfam=\fiveujsym

\font\Fr=eufm10

\def\RR{{\openo{R}}}
\def\CC{{\openo{C}}}

\def\eps{\varepsilon}

\def\qed{\hskip10mm Q.e.d.}

%References
\def\HS{1}       %Hayden-Suffridge
\def\GOEBEL{2}   %Goebel
\def\FRVE{3}     %Franzoni-Vesentini
\def\ABATE{4}    %Abate
\def\DINEEN{5}   %Dineen
\def\RSZ{6}      %Riesz-Sz-Nagy
\def\BKU{7}      %Brau-Kaup-Upmeier
\def\NAG{8}      %Engel-Nagel
\def\KAUP{9}     %Kaup1981

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\centerline{\bf A Stone type theorem for non-linear automorphisms in Hilbert space}

\bigskip

\centerline{L.L. STACH\'O}

\ADS
\bigskip\bigskip
\noindent{\bf \the\SEC. Introduction, main results}

\bigskip





Throughout this work ${\bf H}$ denotes an arbitrarily fixed
infinite dimensional
complex Hilbert space
with the scalar product $\langle x \vert y \rangle$ which is linear in $x$ and conjugate linear in $y$,
giving rise to the norm $\Vert x\Vert = \langle x \vert x\rangle^{1/2}$.
We denote the open unit ball
$\{ e\in{\bf H}:\ \Vert e\Vert<1\}$ with ${\bf B}$ and
for any vector $a\in{\bf H}$ we shall write $a^*:= [x\mapsto \langle x\vert a\rangle]$
for its dual functional.

Recall that the group ${\rm Aut}({\bf B})$ of all holomorphic automorphisms of ${\bf B}$
consists the biholomorphic maps
${\bf B}\leftrightarrow{\bf B}$, and the  ${\bf H}$-unitary operators restricted to ${\bf B}$
form the isotropy subgroup of the origin of ${\rm Aut}({\bf B})$.
Stone's classical theorem on strongly continuous one-parameter groups of unitary operators
can be reformulated in terms of ${\rm Aut}({\bf B})$ as a statement that the infinitesimal
generator of a strongly continuous one-parameter subgroup of ${\rm Aut}({\bf B})$
leaving fixed the origin can be identified canonically with the restriction $iA\vert{\bf B}$
where $A$ is a possibly unbounded self-adjoint linear operator with dense domain in ${\bf H}$.
Our purpose
%in this paper
will be an analogous description for the strongly continuous
one-parameter subgroups of ${\rm Aut}({\bf B})$ formed by possibly non-linear maps,
including approximations with norm-continuous one-parameter subgroups.
%
%Actually
We establish the following main result.

\bigskip
\ADT
\bf Theorem \NUM. \it
Assume $\big[ \Psi^t:\ t\in\RR]$ is a strongly continuous one-parameter
group of holomorphic automorphisms of \ ${\bf B}$.\footnote{$^1$}{\rm
That is $\Psi^{t+s}=\Psi^t\circ\Psi^s \in{\rm Aut}({\bf B})$ for all couples $t,s\in\RR$ and the functions
$[t\mapsto\Psi^t(x)]$ are continuous $\RR\to{\bf H}$ for any fixed vector $x\in{\bf B}$.}
Then there exists a vector $\overline{x}$ with $\Vert \overline{x}\Vert \le 1$
along with a
constant $\lambda\in\RR$ and a
densely defined possibly unbounded self-adjoint operator $A:{\bf Z}\to{\bf H}$ with
dense domain such that
%the section
\ADT
$${\bf B} \cap(\overline{x} + {\bf Z}) ={\bf D} \quad \hbox{where} \quad
{\bf D}:= \big\{ x\in{\bf D}:\ t\mapsto \Psi^t(x) \ \hbox{is differentiable on} \ \RR \big\} ,
\leqno(\NUM)$$
%and we have
\vskip-6mm
\ADT
$${d\over dt}\Big\vert_{t=0} \Psi^t(x) =
-\big\langle (iA-\lambda)(x-\overline{x}) \big\vert \overline{x} \big\rangle x +
(iA+\lambda) (x-\overline{x}) \qquad (x\in{\bf D}).
\leqno(\NUM)$$
%In case $\Vert \overline{x}\Vert <1$ we have necessarily $\lambda=0$
Given any tuple $(A,\overline{x},\lambda)$ consisting of a densely defined self-adjoint operator $A:{\bf Z}\to{\bf H}$,
a vector $\overline{x}$ with $\Vert\overline{x}\Vert\le 1$ and a real number $\lambda$,
there exists (a necessarily unique) strongly continuous one-parameter group $[\Psi^t:\ t\in\RR]$
satisfying $(1.2)$ and $(1.3)$ if and only if one of the following alternatives holds: \quad
%$(1)$ $\overline{x}\!\in\!{\bf Z}$; \
%$(2)$ $\overline{x}\!\not\in\!{\bf Z},\,$
$(1)$ $\Vert \overline{x}\Vert \!=\! 1$; \
$(2)$
%$\overline{x}\!\not\in\!{\bf Z},\,$
$ \Vert \overline{x}\Vert \!<\! 1,\ \lambda \!=\! 0$.

\bigskip
\ADT
\bf Corollary \NUM. \it By setting $A_n:= \!\int_{\vert\xi\vert\!\le\! n} \xi\, dE(\xi)$,
$\overline{x}_n:=\!\int_{\vert\xi\vert\!\le\! n} dE(\xi) \overline{x}$
%$(n \!=\!1,2,\ldots)$
in terms of the spectral resolution $A=\!\int_{\xi\!\in\!\RR} \xi\, dE(\xi)$ for the self-adjoint operator in
$(1.2\!-\!3)$, the norm-continuous one-parameter subgroups $[\Psi^t_n:\ t\in\RR] \subset{\rm Aut}({\bf B})$
defined by the generator equations
$${d\over dt}\Big\vert_{t=0} \Psi_n^t(x) =
-\big\langle (iA_n-\lambda)(x-\overline{x}_n) \big\vert \overline{x} \big\rangle x +
(iA_n+\lambda) (x-\overline{x}_n) \qquad (x\in{\bf B}).
\leqno(1.3)_n $$
approximate $[\Psi^t:\ t\in\RR]$ pointwise, that is \
$\Psi_t(x)=\lim\limits_{n\to\infty} \Psi_n^t(x)$ \ $(x\in{\bf B})$.

\rm

\eject

%\bigskip
It is tempting to conjecture that all strongly continuous one-parameter groups of holomorphic automorphisms of bounded circular domains
in Banach spaces can be reconstructed similarly from a polynomial of second degree defined on some dense algebraic submanifold.
However, the techniques we use here depend heavily on Hilbert space structure.



\bigskip\bigskip
\ADS
{\bf \the\SEC. Preliminaries: linear model with joint fixed points for one-parameter groups}
% in ${\rm Aut}({\bf B})$}

\bigskip
Recall [\FRVE, Ch. VI] that the group of
${\rm Aut}({\bf B})$
all holomorphic automorphisms of ${\bf B}$ admits a matrix
representation. % in the following sense:
Namely each element $\Psi$ of ${\rm Aut}({\bf B})$ has the fractional linear form
\ADT
$$\Psi(x) = {{Ax + b}\over \langle x \vert c\rangle + d} \ , \qquad
A\in{\cal L}({\bf H}), \ \ b,c\in {\bf H}, \ \ d\in\CC
\leqno(\NUM)$$
and we have
$$\Psi_1 \!\circ\! \Psi_2(x) = \Psi_1\big(\Psi_2(x)\big) =
{Ax + b\over \langle x \vert c\rangle + d} \quad
\hbox{whenever} \quad
\left[ \matrix{A & b\cr c^* &d \cr} \right] =
\left[ \matrix{A_1 & b_1\cr c_1^* &d_1 \cr} \right]
\left[ \matrix{A_2 & b_2\cr c_2^* &d_2 \cr} \right].$$
This representation is unique up to a constant, since in (2.1) we necessarily have
$$\left[ \matrix{ A &b \cr c^* & d\cr}\right] =
d \!\left[ \matrix{ {\rm Q}_a & \!\!\!a \cr
a^* &\!\!\! 1 \cr}\right] \! \left[ \matrix{U & 0\cr 0 &1 \cr} \right]
\qquad \hbox{where} \quad a \!:=\! \Psi(0), \
U \!=(\beta_a^2 {\rm P}_a \!+\beta_a\overline{\rm  P}_a)^{-1}\Psi^\prime(0)$$
in terms of the the Fr\'echet derivative $\Psi^\prime$ and the standard notations
$${\rm P}_a \!:=\! \big[ \hbox{orthogonal projection ${\bf H}\!\to\!{\CC}a$} \big], \ \
%\overline{\rm P}_a:=1-{\rm P}_a , \ \
\beta_a \!=\!\sqrt{1 \!-\!\Vert a\Vert^2}, \ \
{\rm Q}_a:= {\rm P}_a+\beta_a (I-{\rm P}_a) .$$
We call the matrix
$$\widetilde{\Psi} :=
\left[ { {\rm Q}_a  \ a \atop a^* \ 1 }\right] \! \left[ { U \ 0 \atop 0 \ \ 1} \right] =
\left[ { {\rm Q}_a U  \ \ \ a \atop (U^*a)^* \ 1 }\right]$$
corresponding to the case with constant $d=1$ the
{\it canonical %matrix
representation} of $\Psi$.
In the sequel we shall write
$$\hbox{
${\cal H}:=\! {\bf H} \!\oplus\!\CC \!=\!
\big\{ \! \big[ {x\atop \xi}\big] \! : x\!\in\!{\bf H},\, \xi\!\in\!\CC\big\}$ }$$
and
%conveniently we
identify the matrix $M:=\big[ m_{ij} \big]_{i,j=1}^2$ where
$m_{11}\!\in\!{\cal L}({\bf H}), \, m_{12}\!\in\! {\bf H}, \, m_{2,1}\!\in\! {\bf H}^*$
and $m_{22}\!\in\!\CC$
with the linear operator $\big[ {x\atop \xi}\big] \mapsto M \big[ {x\atop \xi}\big]$ on ${\cal H}$.
Notice that,
by (2.1) we have
\ADT
$$\Psi(x) = \Big[ \widetilde{\Psi} \left[{x\atop 1}\right] \Big]_{\CC}^{-1}
\Big[ \widetilde{\Psi} \left[{x\atop 1}\right] \Big]_{{\bf H}}
%= \Pi \widetilde{\Psi}\left[{x\atop 1}\right]
\qquad (x\in{\bf B})
\leqno(\NUM)$$
where $[ \,\cdot\, ]_{\CC}$ %\big[{x\atop \xi}\big] \big]_{\CC} :=\xi$
resp. $[ \,\cdot\, ]_{{\bf H}}$ %\big[{x\atop \xi}\big] \big]_{{\bf H}} :=x$
are the standard notations for the canonical projections ${\cal H}\to\CC$ resp. ${\cal H}\to{\bf H}$.
%
It is immediate that any $\Psi \in{\rm Aut}({\bf B})$
extends holomorphically to the ball $(1-\Vert \Psi(0)\Vert)^{-1}{\bf B}$.
Hence we can define the group of all automorphisms of the closed unit ball
$\overline{{\bf B}} :=\{ x\in{\bf H}:\ \Vert x\Vert\le 1\}$
as
$${\rm Aut}(\overline{{\bf B}}) :=
\big\{ \overline{\Psi}:\ \Psi\in{\rm Aut}\,({\bf B}) \big\}  \quad \hbox{where} \quad
\overline{\Psi}:= \big[
\hbox{continuous extension of $\Psi$ to $\overline{{\bf B}} \big]$}.$$
It is also well-known [\FRVE, Ch.VI] that any mapping
$\overline{\Psi}\in{\rm Aut}(\overline{{\bf B}})$ is weakly continuous
and
preserves the Grassmann family ${\rm Aff}(\overline{{\bf B}})$ of all
complex affine
closed subspaces intersected with $\overline{{\bf B}}$.\footnote{$^1$}{If
$x \!=\!\sum_{k\!=\!1}^2 \!\lambda_k x_k$ with $\lambda_1,\lambda_2\!\!\in\!\CC$
and $\sum_{k\!=\!1}^2 \!\lambda_k \!=\!1$ then
$\overline{\Psi}(x) \!=\! \sum_{k\!=\! 1}^2 \!\alpha_k \overline{\Psi}(x_k)$ \
for some $\alpha_1,\alpha_2\!\!\in\!\CC$ with
$\sum_{k\!=\!1}^2 \!\alpha_k \!=\! 1$ \ $\big($namely
$\alpha_k\! =\! \lambda_k [1 \!+\!\langle x_k\vert U^* a\rangle]/
[1 \!+\!\langle \lambda_1 x_1 \!+\! \lambda_2 x_2 \vert U^* a\rangle]\big)$.}
By Schauder's fixed point theorem, ${\rm Fix}(\overline{\Psi})\ne\emptyset$ since
$\overline{{\bf B}}$ is weakly compact.
Moreover we have the following alternatives:

\smallskip
\hskip10mm (1) ${\rm Fix}(\overline{\Psi}) \in {\rm Aff}(\overline{{\bf B}})$,\qquad
(2) ${\rm Fix}(\overline{\Psi})$ consists of two boundary points.
%$\{ p,q \}\subset\partial {\bf B}$.
\smallskip

In case (2) from the proof of [\FRVE, Thm.VI.4.8] we see even that
$\overline{\Psi} =
\overline{\Phi}\circ \overline{\Theta}_a \circ\overline{\Phi}\,\strut^{-1}$
with a suitable automorphism $\Phi\in{\rm Aut}(\overline{{\bf B}})$ and a
{\it M\"obius shift}
\ADT
$$\Theta_a: x\mapsto {{\rm Q}_a x+a \over 1+\langle x \vert a \rangle}
\leqno(\NUM)$$
for some $0\ne a\in{\bf B}$ such that
${\rm Fix}(\overline{\Theta}_a)=\{-e,e\}$ where $e :=a/\Vert a\Vert$.

\medskip

The next result is an infinite dimensional extension for a simple special case
of a far reaching theorem of Abate [\ABATE]
established for finite dimensional uniformly convex domains.
It seems that Dineen's bidual embedding method [\DINEEN] is suitable in proving a complete
infinite dimensional analogy with uniformly convex domains in Banach spaces.
Instead, below we give a short direct proof on the basis of the special algebraic form $(2.1)$.

\ADT
\bigskip
\bf Lemma \NUM. \it  Abelian subsets of ${\rm Aut}(\overline{{\bf B}})$,
in particular one-parameter subgroups,
admit common fixed points.

\bigskip
\bf Proof. \rm
Assume $\{ \overline{\Psi}_j :\, j\!\in\!{\cal J}\} \!\subset\! {\rm Aut}(\overline{{\bf B}})$
with $\overline{\Psi}_j \!\circ\! \overline{\Psi}_k = \overline{\Psi}_k \!\circ\! \overline{\Psi}_j$
$(j,k \!\in\! {\cal J})$.
By weak continuity, for any index family ${\cal K}\subset{\cal J}$, the set
$\bigcap_{k\in{\cal K}} {\rm Fix}(\overline{\Psi}_k)$ of common fixed points
is weakly compact.
Thus, according to Riesz' intersection theorem, it suffices to see that
$\bigcap_{k\in{\cal K}} {\rm Fix}(\overline{\Psi}_k) \ne \emptyset$
for finite index families ${\cal K}$.
By proceeding to contradiction,
let ${\cal K}=\{ k_1,\ldots,k_N\}$ be a subset of ${\cal J}$ with minimal cardinality such that
$\bigcap_{k\in{\cal K}} {\rm Fix}(\overline{\Psi}_k) = \emptyset$.
Necessarily $N>1$ and ${\bf S}:=\bigcap_{n=1}^{N-1} {\rm Fix}(\overline{\Psi}_{k_n}) \ne \emptyset$
is either a weakly compact convex subset of $\overline{{\bf B}}$ or it consists of two boundary points.
Since $\overline{\Psi}_{k_N}$ commutes with all the maps $\overline{\Psi}_{k_n}$ $(n<N)$,
we have $\overline{\Psi}_{k_N}({\bf S}) \subset {\bf S}$.
Indeed, if $x\in{\bf S}$ then
$\overline{\Psi}_{k_n}\big(\overline{\Psi}_{k_N}(x)\big) =
\overline{\Psi}_{k_N}\big(\overline{\Psi}_{k_n}(x)\big) =
\overline{\Psi}_{k_N}(x) \Rightarrow \overline{\Psi}_{k_N}(x)\in{\rm Fix}(\overline{\Psi}_{k_n})$ $(n<N)$.
Hence Schauder's fixed point theorem excludes the case of ${\bf S}$ being convex.
Suppose ${\bf S}=\{p,q\}\subset\partial{\bf B}$.
Then necessarily ${\rm Fix}(\overline{\Psi}_{k_M}) =\{ p,q\}$
for some index $M<N$ and $\overline{\Psi}_{k_N}: p\leftrightarrow q$.
However, in this case we can write
$\overline{\Psi}_{k_N} =
\overline{\Phi}\circ \overline{\Theta}_a \circ\overline{\Phi}\,\strut^{-1}$
with a suitable automorphism $\overline{\Phi}\in{\rm Aut}(\overline{{\bf B}})$ and a
M\"obius shift
$\overline{\Theta}_a$
of the form $(2.3)$
where $0 \!\ne\! a\!\in\!{\bf B}$.
Then, by setting
$\overline{\Omega} \!:=\!\overline{\Phi}\,\strut^{-1}\!\circ\overline{\Psi}_{k_M}\!\circ\overline{\Phi}$
and $e \!:=\! a/\vert a\Vert$, we have
$\{ \pm e\} \!=\! {\rm Fix}(\overline{\Theta}_a)$ and $\overline{\Omega}: e\!\leftrightarrow\! -e$.
On the other hand, it is immediate that
$e=\lim_{n\to\infty}\overline{\Theta}_a^n (x)$ for every point $x\!\in\!\overline{{\bf B}}\!\setminus\!\{ e\}$.
Taking any point $f\in{\rm Fix}(\overline{\Omega})$ we get the contradiction
$e=\lim_{n\to\infty} \overline{\Theta}_a^n(f) =
\lim_{n\to\infty} \overline{\Theta}_a^n \!\!\circ \overline{\Omega}(f) =
\lim_{n\to\infty} \overline{\Omega} \!\circ \overline{\Theta}_a^n(f) = \overline{\Omega} (e) =-e$.
%\qed


\ADT
\bigskip
\bf Remark \NUM. \rm In finite dimensions, it is customary to normalize (2.1) by requiring
${\rm det} \left[ {A \ b\atop c^* d} \right] \!=\! 1$.
Thus, in case of ${\rm dim}({\bf H})=N$, in this manner one can establish a canonical identification
of ${\rm Aut}({\bf B})$
with a subgroup of the classical matrix group ${\rm SL}(N+1)$.
Though in infinite dimensions such a normalization is not available,
for one-parameter groups with common fixed point there is an alternative way as follows.


\ADT
\bigskip
\bf Definition \NUM. \rm
Let
$\big( [\Psi^t :\ t\in\RR], \overline{x} \big)$
be a couple of a
one-parameter subgroup of ${\rm Aut}({\bf B})$ with common fixed point $\overline{x}$
for the continuous extensions of its members to $\overline{{\bf B}}$:
$\overline{x}\in\overline{{\bf B}},\
\overline{\Psi}^t(\overline{x}) =\overline{x} \ \ (t\in\RR).$
In terms of the canonical representations
\DEL{
$\widetilde{\Psi^t} = \left[ {Q_t \ a_t \atop a_t^* \ 1 }\right]
\left[ { U_t \ 0 \atop 0 \ 1} \right]$
}%ENDDEL
define
$$\widehat{\Psi}^t_{\overline{x}} :=
%{1\over 1+\langle U_t \overline{x}\vert a_t\rangle}
\Big[ \widetilde{\Psi^t} \Big[ {\overline{x} \atop 1} \Big]\Big]_{\CC}^{-1}
\widetilde{\Psi^t}  =
{1\over 1+\langle U_t \overline{x}\vert a_t\rangle}
\left[ \matrix{ Q_t &a_t \cr a_t^* &1 \cr}\right]
\left[ \matrix{ U_t  &0 \cr 0 &1 \cr}\right] \qquad (t\in\RR).$$
where \ \
$a_t \!=\!\Psi_t(0)\!\in\! {\bf B}$, $U_t \!\in\!{\cal L}({\bf H})$
is a suitable unitary operator
and
$Q_t:={\rm Q}_{a_t} = P_t + \beta_t (I-P_t)$ \ with \
$P_t x :={\rm P}_{\! a_t} x \!= \Vert a_t\Vert^{- \!2}\langle x \vert a_t\rangle a_t, \
%\overline{P}_t =I \!-\! P_t, \
\beta_t \!=\! \sqrt{1 \!-\! \Vert a_t\Vert^2}$.

Later on, conveniently we shall simply write $\widehat{\Psi}^t$ instead of $\widehat{\Psi}^t_{\overline{x}}$
without danger of confusion.

\ADT
\bigskip
\bf Remark \NUM. \rm
As we have seen
$\Psi^t(x) = (1+\langle U_t x \vert a_t \rangle)^{-1} [Q_{t} U^t x + a_t]
= \big[\Theta_{a_t} \!\circ U_t \big] (x)$.

Thus, by construction we have
$$\overline{\Psi^t}(x) = \Big[ \widehat{\Psi}^t \left[{x\atop 1}\right] \Big]_{\CC}^{-1}
\Big[ \widehat{\Psi}^t \left[{x\atop 1}\right] \Big]_{{\bf H}}
%\Pi \widehat{\Psi}^t \left[{x\atop 1}\right]
\quad (x\in\overline{{\bf B}}) , \qquad
\widehat{\Psi}^t \Big[ { \overline{x} \atop 1} \Big] = \Big[ { \overline{x} \atop 1} \Big] .$$
It is worth to notice that the term $\langle U_t\overline{x}\vert a_t\rangle$ is
actually independent of $U_t$ as
\ADT
$$\langle U_t\overline{x}\vert a_t\rangle =
{\langle \overline{x}-a_t\vert a_t\rangle  \over 1 - \langle \overline{x}\vert a_t \rangle} , \qquad
\widehat{\Psi}^t  =
{ 1-\langle \overline{x}\vert a_t\rangle \over 1 -\langle a_t\vert a_t\rangle}
\left[ \matrix{ Q_t %P_t + \beta_t \overline{P}_t
&a_t \cr a_t^* &1 \cr}\right]
\left[ \matrix{ U_t  &0 \cr 0 &1 \cr}\right] .
\leqno(\NUM)$$
Proof: In general we have $P_t y = \langle y\vert a_t\rangle \langle a_t \vert a_t\rangle^{-2} a_t$
$(0\ne a_t,y\in{\bf H})$. It follows
$\langle P_t U_t \overline{x} \vert a_t\rangle = \langle U_t\overline{x}\vert a_t\rangle$
with $\langle \overline{P}_t U_t \overline{x} \vert a_t\rangle =0$ for any $t\in\RR$.
Thus multiplying the fixed point equation
$\overline{x} = \overline{\Psi}^t(\overline{x}) =
(1+\langle U_t\overline{x}\vert a_t\rangle)^{-1} (P_t+\beta_t\overline{P}_t)U_t\overline{x}$
with $\vert a_t\rangle$, we get
$(1+\langle U_t\overline{x}\vert a_t\rangle)^{-1} \langle U_t\overline{x} +a_t\vert a_t\rangle =
\langle \overline{x} \vert a_t\rangle$
whence the relations $(\NUM)$ are immediate.

\medskip
The power style indexing of $\widehat{\Psi}^t$ in $t$ is justified by the proposition below.

\ADT
\bigskip
\bf Proposition \NUM. \it Given a strongly continuous one-parameter group
$[\Psi^t:\! t\!\in\!\RR]$ in ${\rm Aut}({\bf B})$ with common fixed point $\overline{x}\in\overline{{\bf B}}$,
the family $[\widehat{\Psi}^{\, t}_{\overline{x}}:\ t\in\RR ]$
\DEL{
$$\widehat{\Psi}^t := {1\over 1+\langle U_t \overline{x}\vert a_t\rangle}
\left[ \matrix{ (P_t + \beta_t \overline{P}_t)U_t &a_t \cr [U_t^* a_t]^* &1 \cr}\right] =
{1\over 1+\langle U_t \overline{x}\vert a_t\rangle}
\left[ \matrix{ P_t + \beta_t \overline{P}_t &a_t \cr a_t^* &1 \cr}\right]
\left[ \matrix{ U_t  &0 \cr 0 &1 \cr}\right]$$
}%ENDDEL
is
a strongly continuous one-parameter group of operators in ${\cal H}$.

\bigskip
\bf Proof. \rm
Since
 $\Psi^t\!\circ\!\Psi^s \!=\! \Psi^{t+s}$ $(t,s\!\in\!\RR)$,
for the representation matrices we have
$\widehat{\Psi}^t \widehat{\Psi}^s \!=\! d_{t,s} \widehat{\Psi}^{t+s}$
with suitable constants $d_{t,s}\in\CC_*$.
The fixed point property
$\overline{\Psi}^t (\overline{x})=\overline{x}$
implies
$$\widehat{\Psi}^t \left[ {\overline{x} \atop 1}\right] =
\left[ {\overline{x} \atop 1}\right] \qquad (t\in\RR).$$
Hence necessarily $d_{t,s}\!=\!1$ $(t,s\!\in\!\RR)$, thus
$\big[ \widehat{\Psi}^t: t\!\in\!\RR\big]$ is a one-parameter matrix group.

By assumption, the function $t\mapsto a_t=\Psi^t(0)$ is norm-continuous $\RR\to{\bf B}$.
Hence we can deduce the strong continuity of the ${\bf H}$-unitary operator valued
%$\RR\to{\cal L}({\f H})$
function $t\mapsto U_t$. Namely consider any vector $x\in{\bf H}$.
To establish the norm-continuity of the function $t\mapsto U_t$,
we may assume without loss of generality that $x\in{\bf B}$.
Then, by the aid of the M\"obius shifts $(2.3)$
we can write
$$U_t x = \big[\Theta_{a_t}^{-1} \circ \Psi_t \big](x) = \Theta_{-a_t} \big( \Psi(x)\big)
\qquad (t\!\in\!\RR)
%\quad \hbox{where} \ \
%\Theta_a : z\!\mapsto\! { [{\rm P}_a \!+\! \beta_a \overline{{\rm P}}_a]z \over 1 \!+\!\langle z\vert a\rangle}
%\ \ \hbox{with} \ \ \beta_a=\sqrt{1 \!-\!\Vert a \Vert^2}
.$$
Observe that the norm continuity of $t\mapsto a_t$ implies the
continuity of $t\mapsto \langle x \vert a_t\rangle$ and $t\mapsto\beta_t \in [0,1)$
entailing the norm-continuity of
$t\mapsto P_t +\beta_t \overline{P}_t \in\overline{{\rm Ball}}\big({\cal L}({\bf H})\big)$.
Hence the required norm-continuity of
$t\mapsto U_t x = (1-\langle x\vert a_t\rangle)^{-1} [(P_t x - a_t + \beta_t \overline{P}_t x]$
is immediate.
In general, the product of two bounded strongly continuous linear operator valued functions $\RR\to{\cal L}({\bf X})$
over a normed space ${\bf X}$
is strongly continuous. Hence we conclude that the entries $(1,1),(1,2),(2,1)$ resp. $(2,2)$
of the matrices $\widehat{\Psi}^t$ are strongly continuous functions
$\RR\to{\cal L}({\bf H})$, $\RR\to{\bf H}$, $\RR\to {\bf H}^* \!\!\simeq\!{\bf H}$ resp. $\RR\to\RR$
which completes the proof.
\qed


\ADT
\bigskip
\bf Corollary \NUM. \it Given a strongly continuous one-parameter group $[{\cal T}^t: \ t\in\RR]$ in ${\cal L}({\cal H})$,
the following statements are equivalent

\hskip6mm {\rm (i)} for all $t\in\RR$, the maps $x\mapsto \big[{\cal T}^t\big[ {x\atop 1} \big]\big]_{\CC}^{-1} \big[ {\cal T}^t\big[ {x\atop 1} \big]\big]_{{\bf H}}$
belong to ${\rm Aut}({\bf B})$;

\hskip5.5mm {\rm (ii)} \vtop{
\hbox{we have ${\cal T}^t = e^{\mu t} \widehat{\Psi}^t$ $(t\in\RR)$ for some strongly continuous one-parameter subgroup}
\hbox{$[\Psi^t:\ t\in\RR]$ of ${\rm Aut}({\bf B})$
and a constant $\mu\in\CC$;}}

\hskip5mm {\rm (iii)} each operator ${\cal T}^t$ maps the cone ${\cal K}:=\big\{ \big[{x\atop\xi}\big] :\, \vert\xi\vert^2 > \Vert x\Vert^2\big\}$ onto itself;

\hskip5mm {\rm (iv)} each operator ${\cal T}^t$ maps $\partial{\cal K}:=\big\{ \big[{x\atop\xi}\big] :\, \vert\xi\vert^2 = \Vert x\Vert^2\big\}$ onto itself.

\bigskip
\bf Proof. \rm
The implication (ii)$\Rightarrow$(i) is trivial by (2.2).

\medskip
Proof of (i)$\Rightarrow$(ii): By assumption the maps
$\Psi^t(x) := \big[ {\cal T}^t\big[ {x\atop 1}\big] \big]_{\CC}^{-1} \big[ {\cal T}^t\big[ {x\atop 1}\big] \big]_{{\bf H}}$
$(t\in\RR,\, x\in{\bf B})$.
are well-defined holomorphic automorphisms of the unit ball ${\bf B}$.
By (2.2) we have ${\cal T}^t = d_t \widehat{\Psi}^t$ $(t\in\RR)$
with suitable constants $d_t\in\CC_*$.
Fixing any point $x\in{\bf B}$, the strong continuity of the group $[{\cal T}^t:\ t\in\RR]$ implies the
continuity of the function ${\cal T}^t \big[ {x\atop 1}\big]$ whence we deduce also the continuity of
$t\mapsto \Psi^t(x)$ which entails the continuity of
$t\mapsto \widehat{\Psi}^t \big[ {x\atop 1}\big] =d_t^{-1}{\cal T}^t \big[ {x\atop 1}\big]$
and hence the continuity of $t\mapsto d_t$.
%On the other hand,
By the one-parameter group property,
all the relations ${\cal T}^{t+s}={\cal T}^t {\cal T}^s$, $\widehat{\Psi}^{t+s}=\widehat{\Psi}^t \widehat{\Psi}^s$
$(t,s\in\RR)$ hold.  Therefore $d_{t+s}=d_t d_s$ $(t,s\!\in\!\RR)$ and the continuity of $t\mapsto d_t$ establishes the
existence of a constant $\mu\in\CC$ with $d_t=e^{\mu t}$ $(t\in\RR)$.

\medskip
Proof of (i)$\Leftrightarrow$(iii)$\Leftrightarrow$(iv):
Consider the projective Hilbert space ${\cal H}_* \! / \!\!\approx$
associated with ${\cal H}$
regarded as
the set of all nontrivial punctured complex rays $\CC_* \big[{x\atop\xi}\big]$
with the factor topology.\footnote{$^2$}{As usually,
${\cal H}_*:= {\cal H}\!\setminus\!\{\! 0 \!\}$
with the equivalence relation
$\big[{x\atop \xi}\big] \!\sim\! \big[{y\atop \eta}\big] :\!
\iff \CC_*\big[{y\atop \eta}\big] \!=\! \CC_*\big[{x\atop \xi}\big]$
where $\CC_*:=\CC\!\setminus\!\{\! 0\!\}$.
A subset of ${\cal H}\!/\!\approx$ is open if the union of it members (rays in ${\cal H}_*$) is open in ${\cal H}$.}
By homogeneity, any injective linear operator ${\cal T}\in{\cal L}({\cal H})$ acts holomorphically
on ${\cal H}_*\!/\!\approx$ by its factorization
${\cal T}_\approx: \CC_*\big[{x\atop\xi}\big]\mapsto \CC_* {\cal T}\big[ {x\atop \xi}\big]$.
In particular, as admitting a continuous inverse, each map ${\cal T}^t_\approx$ is a holomorphic automorphism of ${\cal H}/\approx$.
Hence the equivalences (i)$\Leftrightarrow$(iii)$\Leftrightarrow$(iv) are straightforward consequences of the facts that,
with the embedding $\Pi: x\mapsto \CC\big[{x\atop 1}\big]$ and its inverse $\pi\big( \CC\big[{x\atop \xi}\big] \big) := x/\xi$ $(\xi\ne0)$,
we have $\Pi {\bf B} := {\cal K}$, $\pi{\cal K}={\bf B}$ and
$\big[{\cal T}^t \big[{x\atop 1}\big]\big]_{\CC}^{-1} \big[{\cal T}^t \big[{x\atop 1}\big]\big]_{{\bf H}} =
\pi\circ {\cal T}^t_\approx \circ \Pi \big[{x\atop 1}\big]$ whenever $\big[ {\cal T}^t \big[{x\atop 1}\big] \big]_{\CC} \ne0$.

\bigskip
\ADT
\bf Corollary \NUM. \it Given any $\Theta \in  {\rm Aut}(\overline{{\bf B}})$,
the $\Theta$-shifted automorphisms $\Phi^t := \Theta\circ\Psi^t\circ\Theta^{-1}$
form strongly continuous one-parameter group with common fixed point $\overline{y}:=\Theta(\overline{x})$
when extended continuously to $\overline{{\bf B}}$ and \ \
$\widehat{\Phi}^t_{\overline{y}} =
e^{\mu t} \widetilde{\Theta}^{-1} \widehat{\Psi}^t_{\overline{x}} \widetilde{\Theta}$ \
$(t\!\in\!\RR)$ \ \ for some $\mu\!\in\!\CC$.



\bigskip
\bigskip

\ADS
{\bf \the\SEC. Infinitesimal generators}
\rm

\bigskip
Throughout this section, let
$\big( [\Psi^t\!\! : t \!\in\!\RR], \overline{x}\big)$
be an arbitrarily fixed couple of a strongly
continuous one-parameter group in ${\rm Aut}({\bf B})$ with a common
fixed point for the continuous extensions in $\overline{{\bf B}}$.
Recalling the Hille--Yosida theorem [\RSZ, Kap.10], Proposition 2.9
ensures that the differential \ADT
$${\cal A} :  \hbox{{\Fr h}} \mapsto
{d\over dt} \widehat{\Psi}^t \hbox{{\Fr h}}
\quad \hbox{with} \ \ {\cal D} :={\rm dom}({\cal A}) =
\big\{ \hbox{{\Fr h}}\in{\cal H} : t\mapsto \widehat{\Psi}^t \hbox{{\Fr h}} \
\hbox{is differentable on} \ \RR\big\}
\leqno(\NUM)$$
(called the {\it infinitesimal generator} of the linear model
$\big[ \widehat{\Psi}^t \!\! : t\!\in\!\RR\big]$
where $\widehat{\Psi}^t \!\equiv\! \widehat{\Psi}^t_{\overline{x}}$ for short)
is a not necessarily bounded linear map
with closed graph and
$[\widehat{\Psi}^t \!: t\!\in\!\RR]$-invariant domain being dense in ${\cal H}$.
%
Instead of the differential ${\cal A}={d\over dt}\vert_{t=0} \widehat{\Psi}^t$ of the representations
we are primarily interested in the differential
$$\Omega :={d\over dt}\Big\vert_{t=0} \Psi^t:  {\bf D} \to{\bf H}
\quad\hbox{where} \ \ {\bf D}={\rm dom}(\Omega)=\Big\{ x\in{\bf B} : \
{d\over dt}\Big\vert_{t=0} \Psi^t(x) \ \hbox{exists}\Big\} .$$
In order that we could regard the vector field $\Omega$ as a non-linear infinitesimal generator
for $[\Psi^t :\ t\in\RR]$, we should see the density of ${\bf D}$ in $\overline{{\bf B}}$.
In order to establish a non-linear Stone-type theorem, we should determine precise links to
%unbounded
self-adjoint linear operators.


\bigskip
\ADT
\bf Lemma \NUM. \it
${\bf D}$ is
$\big[ \overline{\Psi}^t: t\!\in\!\RR\big]$-invariant.
We have $\big[ {x \atop 1}\big] \!\in\! {\cal D} \!\iff\!  x\!\in\!{\bf D}$ whenever $x\!\in\!{\bf B}$.

\bigskip
\bf Proof. \rm The $\big[ \Psi^t: t\!\in\!\RR\big]$-invariance of ${\bf D}$ is clear
from the group property $\Psi^{t+s}=\Psi^t \circ \Psi^s$ $(t,s\in\!\RR)$.
Moreover even
${\rm dom}\big({d\over dt}\big\vert_{t=0} \Theta \circ \Psi^t \circ \Theta^{-1} \big) = \Theta({\bf D})$
whenever $\Theta $ is any holomorphic automorphism of ${\bf B}$.
Hence, given any point $x\in{\bf B}$, we have
$x\!\in\!{\bf D}$ if and only if $0=\Theta_{-x}(x) \!\in\!
{\rm dom}\big({d\over dt}\big\vert_{t=0} \Phi^t\big)$
with the one-parameter group of the maps $\Phi^t := \Theta_{-x} \circ \Psi^t \circ \Theta_x$
in terms of the M\"obius transformations (2.3).
That is, without loss of generality, it suffices only to see the equivalence \
$0\!\in\!{\rm dom}\big({d\over dt}\big\vert_{t=0} \Phi^t\big) \!\iff\!  \big[{0\atop 1}\big] \!\in \!{\cal D}$.
According to (2.8), by setting
$a_t:=\Psi^t(0)$ and $\hbox{\Fr a}_t:= \widehat{\Psi}^t \big[ {0\atop 1}\big]$
we have
$\hbox{\Fr a}_t=(1-\langle a_t\vert \overline{x} \rangle (1-\Vert a_t\Vert^2)^{-1} \big[ {a_t \atop 1} \big]$
$(t\in\RR)$. Hence the curves $t\mapsto a_t$ resp. $t\mapsto\hbox{\Fr a}_t$ are differentiable in the same time,
which completes the proof.

\bigskip
For later use we also introduce the notations
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                             %
%  {\cal D}_0 helyett {\bf Z} %
%  {\cal B}   \Psi^thelyett {\bf D} %
%                             %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\ADT
$${\bf Z} := \Big\{ z\in{\bf H}:\ \left[ {z \atop 0}\right] \in {\cal D} \Big\}; \qquad
Bz \!:=\!\Big[ {\cal A}\Big[{z \atop 0}\Big]\Big]_{\!{\bf H}}, \
\Lambda z \!:=\!\Big[ {\cal A}\Big[{z \atop 0}\Big]\Big]_{\!\CC} \quad
(z\in{\bf Z}) .
%\leqno(\NUM)
$$

\bigskip
\ADT
\bf Lemma \NUM. \it ${\bf Z}$ is a dense linear submanifold in ${\bf H}$ \
with \
${\bf D} = \big(\overline{x}+{\bf Z}\big) \cap {\bf B}$ and
${\cal D}= \big[ {{\bf Z} \atop 0}\big] + \CC \big[ {\overline{x} \atop 1}\big]$.
The set ${\bf D}$ is dense in ${\bf B}$
and
$${d\over dt}\Big\vert_{t=0} \overline{\Psi}^t(x) =
%-\Big\{ {\cal A} \left[ {x\atop 1}\right] \Big\}_{\CC} x +
%\Big\{ {\cal A} \left[ {x\atop 1}\right] \Big\}_{{\bf H}}
[\Lambda(\overline{x} -x)]x + B(x-\overline{x})
\qquad (x\in{\bf D}).$$

\bf Proof. \rm
By definition,
$\big[ {\overline{x} \atop 1}\big] \!\in\!{\cal D}$ with ${\cal A} \big[ {\overline{x} \atop 1}\big]\!=\! 0$
since $\widehat{\Psi}^t \!\big[ {\overline{x} \atop 1}\big] \!=\!\big[ {\overline{x} \atop 1}\big]$ $(t\!\in\!\RR)$.
Since ${\cal D}$ is closed for linear combinations, it follows that
$\big[ {{\bf Z} \atop 0}\big] + \CC \big[ {\overline{x} \atop 1}\big] = {\cal D}$
and that ${\bf Z}$ is the image of ${\cal D}$ by the bounded linear operator
$\Pi\big[ {x\atop\xi}\big] := x-\xi\overline{x}$.
Since $\Pi {\cal H}={\bf H}$ and since ${\cal D}$ is dense in ${\cal H}$,
${\bf Z}=\Pi {\cal D}$ is also dense in ${\bf H} =\Pi{\cal H}$.
From Lemma 3.2 we know that ${\bf D}={\bf B} \cap \big\{ x: \ \big[{x\atop 1}\big] \in {\cal D} \big\}$.
Hence the relation ${\bf D} = \big(\overline{x}+{\bf Z}\big) \cap {\bf B}$
along with the density of ${\bf D}$ in ${\bf B}$ is immediate.
%
Given any $x\in{\cal D}$, the relation
$\big[ {x\atop 1} \big] \in{\cal D}$  implies
${\cal A}\big[ {x\atop 1} \big] = {d\over dt}\big\vert_{t=0} \widehat{\Psi}^t \big[ {x\atop 1} \big]$.
Since
$\overline{\Psi}^t (x) =
\big\{ \widehat{\Psi}^t \big[ {x\atop 1} \big] \big\}_{\!\CC}^{\!\!-1}
\big\{ \widehat{\Psi}^t \big[ {x\atop 1} \big] \big\}_{\!{\bf H}}$
along with
$\widehat{\Psi}^0={\rm Id}$
and
${\cal A} \big[{\overline{x}\atop 1}\big] = 0$,
%{d\over dt}\big\vert_{t=0} \widehat{\Psi}^t  \big[{\overline{x}\atop 1}\big] =
%{d\over dt}\big\vert_{t=0} \big[{\overline{x}\atop 1}\big] =0 $,
we get
$$\eqalign{
{d\over dt}\Big\vert_{t=0} \!\!\!\!\overline{\Psi}^t(x) &=
-\Big[ \!\widehat{\Psi}^0 \Big[ {x\atop 1} \Big] \!\Big]_{\!\CC}^{\!\!-2} \!
\left( {d\over dt}\Big\vert_{t=0} \! \Big\{ \!\widehat{\Psi}^0 \Big[ {x\atop 1} \Big] \!\Big\}_{\!\CC} \right) \!
\Big[ \!\widehat{\Psi}^0 \Big[ {x\atop 1} \Big] \!\Big]_{\!{\bf H}} \!\!+
\Big[ \!\widehat{\Psi}^0 \Big[ {x\atop 1} \Big] \!\Big]_{\!\CC}^{\!\!-1}
{d\over dt}\Big\vert_{t=0} \!\Big\{ \!\widehat{\Psi}^0 \Big[ {x\atop 1} \Big] \!\Big\}_{\!{\bf H}} \!\!=\cr
&= - \Big[ {\cal A} \Big[ {x\atop 1} \Big] \Big]_{\CC} x +
\Big[ {\cal A} \Big[ {x\atop 1} \Big] \Big]_{\!{\bf H}} =
 - \Big[ {\cal A} \Big[ {x-\overline{x}\atop 0} \Big] \Big]_{\CC} x +
\Big[ {\cal A} \Big[ {x-\overline{x}\atop 0} \Big] \Big]_{\!{\bf H}}. \ \
%\hbox{Q.e.d.}
\cr}$$

\ADT
\bf Lemma \NUM. \it
Suppose a Hilbert space ${\bf W}$ is the orthogonal sum
of the subspaces ${\bf W}_1,{\bf W}_2$
%${\bf W_k}$ $(k\!=\!1,2)$
and ${\cal C}$ is the infinitesimal generator of a
strongly continuous one-parameter subgroup
\break
$[T^t: t\!\in\!\RR ]$ of ${\cal L}({\bf W})$.
Then,
for the cone
${\bf K} \!:=\!\big\{ w_1\!\oplus\!w_2: %{\bf w}_k\!\in\!{\bf W}_k, \,
\Vert{w}_1\Vert \!>\! \Vert{w}_2\Vert \big\}$,
we have $T^t {\bf K} \!=\!{\bf K}$ $(t\!\in\!\RR)$
if and only if ${\cal C}$ is tangent to the boundary of ${\bf K}$
that is if
\ADT
$${\rm Re}\big\langle {\cal C}({w}_1\!\oplus\!{w}_2) \big\vert {w}_1 \big\rangle =
{\rm Re}\big\langle {\cal C}({w}_1\!\oplus{w}_2) \big\vert {w}_2 \big\rangle
\quad
\big( {w}_1\!\oplus\!{w}_2 \!\in\!{\rm dom}({\cal C}) ,\
\Vert{w}_1\Vert \!=\! \Vert{w}_2\Vert\big) .
\leqno(\NUM)$$

\bf Proof. \rm
It is immediate that $T^t{\bf K\!}\!\subset\! {\bf K}$ $(t\!\in\!\RR)$
$\Rightarrow T^t\partial{\bf K}\!\subset\! \partial{\bf K}$ $(t\!\in\!\RR)$
$\Rightarrow {d\over dt}\vert_{t=0} T^t({w}_1\oplus{w}_2) \in
{\rm Tan}_{{w}_1\oplus{w}_2}({\bf K})$ for ${w}_1\oplus{w}_2\in{\rm dom}({\cal C})$
$\Rightarrow (\NUM)$.
Assume $(\NUM)$ and
let $P$ denote the canonical projection of ${\bf W}$ onto ${\bf W}_1$
and define
$T^{t,s}:=\exp(t{\cal C}+sP)$ $(s,t\in\RR)$.
By the theorem of bounded perturbations [\NAG, p.158]
the operators $T^{t,s}$ are all well-defined.
Moreover, by [\NAG, Corollary 1.7 p. 161]
(applied with $B:=sP$ and $A:={\cal C}$ there)
we have $\lim_{s\to 0} T^{t,s}{w} = T^t{w}$ $({w}\in{\rm dom}({\cal C}), t\in\RR)$.
Therefore, to establish that $T^t{\bf K}\subset {\bf K}$ $(t\in\RR)$,
it suffices to see only that \ %$(*)$
$T^{t,s}{w} \!\in\!{\bf K}$ whenever
${w}\!\in\!{\rm dom}({\cal C})\!\cap\! {\bf K}$ and $t,s\!>\! 0$.
To proceed to contradiction, let
$s,t>0$ and ${w}:={w}_1\oplus {w}_2 \in{\rm dom}({\cal C})$ with
$\Vert{w}_1\Vert > \Vert{w}_2\Vert$ but
$\Vert [T^{t,s}{w}]_1\Vert \le \Vert [T^{t,s}{w}]_2\Vert$.
The function
$\delta(\tau):=\Vert [T^{\tau,s}{w}]_1\Vert^2 - \Vert [T^{\tau,s}{w}]_2\Vert^2$
is differentiable in $\tau$ on the whole $\RR$ and
$\delta(0)> 0 \ge \delta(t)$. Thus there exist a point $t_*\in(0,t]$
such that $\delta(\tau)>0=\delta(t_*)$ $(0\le\tau<t_*)$.
Since $\delta(t_*)=0$, the vector
${w}_*:=T^{t_*,s}{w}$ belongs to $\partial{\bf K}$
and hence
${\rm Re}\langle {\cal C}{w}_* \vert [{w}_*]_1\rangle=
{\rm Re}\langle {\cal C}{w}_* \vert [{w}_*]_2\rangle$.
We get the contradiction $0\ge \delta^\prime(t_*) =
2{\rm Re}\langle ({\cal C}+sP){w}_* \vert [{w}_*]_1\rangle -
2{\rm Re}\langle ({\cal C}+sP){w}_* \vert [{w}_*]_2\rangle =
2s \Vert[{w}_*]_1\Vert^2 >0$.



\ADT
\bigskip
\bf Corollary \NUM. \it %Given any vector
%$v\in{\bf Z}$ we have
${\rm Re} \Big( -\Lambda v +
\big\langle B v \big\vert \overline{x} \big\rangle +
\big\langle B v \big\vert v \big\rangle \Big)=0$
whenever $v\in{\bf Z}$ with $\big\Vert \overline{x}+v \big\Vert =1$.


\bigskip
\bf Proof. \rm
By Corollary 2.10, we have
$\widehat{\Psi}^t {\cal K} = {\cal K}$ %$\widehat{\Psi}^t \partial{\cal K} = \partial{\cal K}$
$(t\!\in\!\RR)$
where
${\cal K}:=\big\{ \big[ {x\atop\xi}\big]:\ \vert\xi\vert >\Vert x\Vert\big\} \subset{\cal H}$.
An application of Lemma 3.4 with
${\bf W}_1:=\CC$, ${\bf W}_2:={\bf H}$,
${\bf K}:={\cal K}$, $T^t:=\widehat{\Psi}^t$, ${\cal C}:={\cal A}$
establishes that \
${\rm Re}\, \big[(\Lambda x) \overline{\xi} \big] = {\rm Re}\, \big\langle B x \big\vert x \big\rangle$
whenever $\big[ {x\atop \xi}\big]  \in{\cal D}$ and $\Vert x\Vert = \vert\xi\vert$.
We obtain the statement with the choice  $x:= v+\overline{x}$ and $\xi:=1$
if $v\in{\bf Z}$ with $\Vert v+\overline{x}\Vert=1$ because then, by Lemma 3.3,
we have $\big[{x \atop 1}\big] \in{\cal D}$.

\DEL{
\bigskip
\bf Proof. \rm
Let $v\!\in{\bf Z}$, $u := \overline{x} \!+\! v$ and assume $\Vert u\Vert \!=\! 1$.
Since
$\overline{\Psi}\big( \partial {\bf B} \big) =
\partial {\bf B}$ for any
$\Psi \in {\rm Aut}({\bf B})$ (cf. [\FRVE]), we have
$\big\Vert \overline{\Psi}^t(u) \big\Vert \!=\! 1$ \ $(t\!\in\!\RR)$.
It follows
$$0 \ = \ {d\over dt}\Big\vert_{t=0}
\Big\langle \overline{\Psi}^t(u) \,\Big\vert\, \overline{\Psi}^t(u) \Big\rangle \ = \
2 \,{\rm Re} \, \Big\langle {d\over dt}\Big\vert_{t=0} \!\!
\overline{\Psi}^t(u) \ \Big\vert \ u \Big\rangle .$$
Applying Lemma 3.3, we see that
$\big\langle [-\Lambda v] u \!+\! Bv \,\big\vert\, u \big\rangle \!=\!0$.
Taking into account that $\langle u\vert u\rangle =1$,
the statement is immediate.
%\qed
}%ENDDEL

\bigskip
\ADT
\bf Proposition \NUM. \it For some symmetric linear operator $A:{\bf Z}\to{\bf H}$ and a suitable constant $\lambda\in\RR$
which is necessarily $=0$ if $\Vert \overline{x}\Vert \ne 1$, we have $(1.3)$ as
$$B=iA+\lambda I, \qquad \Lambda z= \big\langle (iA-\lambda I)z \big\vert \overline{x} \big\rangle \quad (z\in{\bf Z}). $$

\DEL{
\it For some constant $\lambda\in\RR$
we have
$$\Lambda z = \big\langle (B \!-\! 2\lambda I) z \,\big\vert\, \overline{x} \big\rangle , \quad
{\rm Re} \big\langle \big(B \!-\!\lambda I \big)z \,\big\vert\, z \big\rangle =0 \qquad (z\in{\bf Z}) .$$
In particular, if \ $\Vert \overline{x}\Vert <1$ then necessarily $\lambda=0$ above.
}%ENDDEL


%\bigskip
\bf Proof. \rm
Consider any vector $0\ne z\in{\bf Z}$.
Let $\zeta\in \CC$ be the (unique) constant such that
$\overline{x}+\zeta z \perp z$ and define
$\varrho :=\sqrt{1-\Vert \overline{x}+\zeta z\Vert^2}$.
Actually we have
$\zeta = -\langle \overline{x} \vert z\rangle / \langle z\vert z\rangle$ and
$1\ge \Vert\overline{x} \Vert^2 = \Vert \overline{x}+\zeta z\Vert^2 + \Vert -\zeta z\Vert^2$
showing that both $\zeta$ and $\varrho$ are well-defined.
Consider the unit vectors
$$v_\varphi:= \overline{x}+\zeta z + e^{i\varphi} \varrho z \qquad (\varphi\!\in\!\RR) .$$
According to Corollary 3.6,
$${\rm Re} \Big( (\zeta \!+\! e^{i\varphi} \varrho) \big[- \Lambda z +
\big\langle Bz \,\big\vert\, \overline{x}\big\rangle \big] +
\vert \zeta \!+\! e^{i\varphi}\varrho\vert^2 \big\langle Bz\,\big\vert\, z \big\rangle \Big) =0
\qquad (\varphi\in\RR).$$
Thus the identity \ \
${\rm Re} \big( \alpha + \beta e^{i\varphi} +\gamma  e^{-i\varphi}\big)=0$
for all $\varphi\in\RR$ \ \
holds with the constants
\break
$\alpha:=
\zeta\big[ -\Lambda z + \big\langle Bz \,\big\vert\, \overline{x}\big\rangle \big] +
(\vert\zeta\vert^2 + \varrho^2) \big\langle Bz \,\big\vert\, z\big\rangle$,
$\beta:= \varrho\, \big[ -\Lambda z + \big\langle Bz \,\big\vert\, \overline{x} \big\rangle +
\overline{\zeta} \big\langle Bz \,\big\vert\, z \big\rangle\big]$
and
$ \gamma:= \varrho\zeta \big\langle Bz \,\big\vert\, z \big\rangle$.
\DEL{
$$\eqalign{
&{\rm Re} \big( \alpha + \beta e^{i\varphi} +\gamma  e^{-i\varphi}\big)=0
\quad  (\varphi\in\RR) \cr
&{\rm where} \ \ \alpha:=
\zeta\big[ -\Lambda z + \big\langle Bz \,\big\vert\, \overline{x}\big\rangle \big] +
(\vert\zeta\vert^2 + \varrho^2) \big\langle Bz \,\big\vert\, z\big\rangle,\cr
&\hskip11mm
\beta:= \varrho\, \big[ -\Lambda z + \big\langle Bz \,\big\vert\, \overline{x} \big\rangle +
\overline{\zeta} \big\langle Bz \,\big\vert\, z \big\rangle\big],
\quad \gamma:= \varrho\zeta \big\langle Bz \,\big\vert\, z \big\rangle .\cr}$$
}%ENDDEL
Since \
$2\, {\rm Re} \big( \alpha + \beta e^{i\varphi} +\gamma  e^{-i\varphi} \big) =
2\, {\rm Re} (\alpha) + (\beta+\overline{\gamma})
e^{i\varphi} + (\overline{\beta}+\gamma)e^{-i\varphi}$,
\break
we have necessarily
${\rm Re}(\alpha) = \beta+\overline{\gamma}=0$.
%The identity $\beta+\overline{\gamma}=0$ corresponds to our statement.
From the relation \ $\beta+\overline{\gamma}=0,$ \ it follows
\ADT
$$\Lambda z - \langle Bz \,\vert\, \overline{x} \rangle =
2\, \overline{\zeta}\, {\rm Re}\, \langle Bz \,\vert\, z \rangle =
-2\, \langle z \,\vert\, \overline{x} \rangle \,
{{\rm Re}\, \langle Bz \,\vert\, z \rangle \over \langle z \,\vert\, z \rangle} \ ,
\leqno(\NUM)$$
and substituting this into the relation
$0={\rm Re} (\alpha)$, we get
$$0=(\varrho^2 - \vert\zeta\vert^2) \, {\rm Re}\, \big\langle Bz \,\big\vert\, z\big\rangle =
\big(1-\Vert \overline{x}\Vert^2 \big) \, {\rm Re}\, \big\langle Bz \,\big\vert\, z\big\rangle \ .
\leqno(3.8^\prime)$$
From $(3.8)$ we see also that
$$z\mapsto{ {\rm Re} \big\langle Bz \,\big\vert\, z\big\rangle
\over  \big\langle z \,\big\vert\, z\big\rangle } =
-{1\over 2}\,
{\Lambda z - \big\langle Bz \,\big\vert\, \overline{x} \big\rangle
\over\big\langle z \,\big\vert\, \overline{x} \big\rangle }$$
is a
real valued G\^ateaux holomorphic function on the algebraically open
and in ${\bf Z}$ algebraically dense
domain $\{ z\!\in\!{\bf Z} \! :  z\not\perp \overline{x} \}$ which
is possible only if being constant on ${\bf Z}$.
\break
By writing $\lambda$ for this constant value,
%the statement is immediate
from $(3.7)$ and $(3.8^\prime)$
we conclude that
$$\Lambda z = \big\langle (B \!-\! 2\lambda I) z \,\big\vert\, \overline{x} \big\rangle , \quad
{\rm Re} \big\langle \big(B \!-\!\lambda I \big)z \,\big\vert\, z \big\rangle =0 \qquad (z\in{\bf Z}) ,$$
and, in particular, if $\Vert \overline{x}\Vert \!<\! 1$ then necessarily $\lambda \!=\! 0$ above.
By setting $A \!:=\! -i(B \!+\!\lambda I)$, hence the statement including the symmetry of $A$ is immediate.

%\qed


\DEL{
\bigskip
Taking int account the obvious reversibility properties of the steps of the proof,
we conclude immediately the following technical statement.\footnote{$^3$}{Direct proof.
Given any $\lambda\in\RR$ and a vector $z\in{\bf H}$ with $\Vert \overline{x}+z \Vert=1$,
we have the following chain of equivalences:
$0= {\rm Re} \big\langle -\langle (iA-\lambda)z \vert \overline{x} \rangle (\overline{x}+z) +
(iA+\lambda I)z \big\vert \overline{x}+z \big\rangle$
iff
$0= {\rm Re} \big[ -\langle (iA-\lambda)z \vert \overline{x} \rangle +
\langle (iA+\lambda I)z \vert \overline{x}+z \rangle \big]$
iff
$0= {\rm Re} \big[ 2\lambda \langle z \vert \overline{x} \rangle +
\langle iA z \vert z \rangle + \lambda \langle z \vert z\rangle \big]$
iff
$0= {\rm Re} \big[ \lambda \big( \langle \overline{x}+z \vert \overline{x} +z \rangle
- \langle \overline{x} \vert \overline{x}\rangle
- \langle \overline{x} \vert z\rangle + \langle z \vert \overline{x} \rangle \big) +
\langle iA z \vert z \rangle \big]$
iff
$0= {\rm Re} \big[ \lambda ( 1 - \Vert \overline{x} \Vert^2 ) +
\langle iA z \vert z \rangle \big]$.
}




\bigskip
\ADT
\bf Corollary \NUM. \it
Given a possibly unbounded linear functional
$\Lambda: {\bf Z}\to\CC$ along with a possibly unbounded linear operator
$B:{\bf Z}\to {\bf H}$, we have
$\big[\Lambda(\overline{x}-x)\big]x + B(x-\overline{x}) \perp x$ for any
$x\in [\overline{x}+{\bf Z}]\cap\partial{\bf B}$
if and only if
$\Lambda z \!=\! \big\langle (B \!-\! 2\lambda I) z \,\big\vert\, \overline{x} \big\rangle , \
{\rm Re} \big\langle \big(B \!-\!\lambda I \big)z \,\big\vert\, z \big\rangle \!=\!0$ $(z\!\in\!{\bf Z})$
with some real constant $\lambda$ if $\Vert\overline{x}\Vert=1$ and
with $\lambda=0$ in case of $\Vert\overline{x}\Vert<1$.

}%ENDDEL

\bigskip
The following geometric converse of 3.7 is elementary:

\bigskip
\ADT
\bf Remark \NUM. \rm\ Given a dense linear submanifold ${\bf Z}$ in ${\bf H}$, a
symmetric linear operator $A:{\bf Z}\to{\bf H}$, a vector $\overline{x}\in\overline{{\bf B}}$,
the vector field $\Omega_\lambda(x) :=
-\langle (iA-\lambda I)(x-\overline{x}\vert\overline{x}\rangle x + (iA+\lambda I)(x-\overline{x})$
define for $\overline{x}+{\bf Z} $
is tangent to the unit sphere $\partial{\bf B}$ at the points $x\in \overline{x}+{\bf Z}$ with $\Vert x\Vert =1$
whenever either $\Vert\overline{x}\Vert=1$ and $\lambda\in\RR$ or $\Vert\overline{x}\Vert<1$ and $\lambda=0$.

\bigskip
In the sequel we proceed to the problem if the operator $A$ in Proposition 3.7 arising from the differential
${d\over dt}\big\vert_{t=0}\Psi^t$ of a strongly continuous one-parameter semigroup of ${\rm Aut}({\bf B})$
is necessarily self-adjoint
and conversely if every self-adjoint operator may appear in 3.7.




\DEL{
\bigskip
\bf Lemma. \it
The operator $B:{\bf Z}\to{\bf Z}$ is closed.

\bigskip
\bf Proof. \rm
Consider a convergent sequence
${\bf Z} \ni z_n \to z$ with limit point in ${\bf H}$
such that
$B z_n \to w$ for some $w\in{\bf H}$.
We have to show that \ $(*)$ $z\!\in\!{\bf Z}$ with $Bz \!=\! w$,
For $(*)$, it suffices to
observe that
$$\Lambda z_n = \big\langle (B-2\lambda I) z_n \big\vert \overline{x} \big\rangle
\to \big\langle w -2\lambda z \big\vert \overline{x} \big\rangle ,$$
recalling that
%, by the Hille-Yosida theorem,
the not necessarily bounded linear operator
${\cal A} \big[ { x \atop \xi} \big] =
\big[ { B(x-\xi\overline{x}) \atop \Lambda(x-\xi\overline{x})} \big]$ mapping
the linear manifold
${\cal D} := \big[ { {\bf Z} \atop 0} \big] \!\!+\! \CC \big[ { \overline{x} \atop 1} \big]$
into itself is closed.
Indeed we have
$$\Big[ { {\bf Z} \atop 0} \Big] \ni \Big[ { z_n \atop 0} \Big] \to \Big[ { z \atop 0} \Big] , \quad
{\cal A} \Big[ { z_n \atop 0} \Big] = \Big[ { B z_n \atop \Lambda z_n} \Big] \to
\Big[ { w \atop \langle w -2\lambda z \vert \overline{x} \rangle} \Big]$$
implying $(*)$ along with the relation
$\Lambda z= \langle w -2\lambda z \vert \overline{x} \rangle =
\langle (B-2\lambda I) z \vert \overline{x} \rangle$.
\qed


\ADT
\bf Definition \NUM. \rm
We shall write
$A:= -i(B+\lambda I)$. Thus henceforth
\ADT
$$Bz=(iA +\lambda I)z \, , \quad
\Lambda z = \big\langle (iA-\lambda I)z \big\vert \overline{x} \big\rangle
\qquad (z\in{\bf Z})
\leqno(\NUM)$$
where $A:{\bf Z}\to{\bf Z}$ is a densely defined %closed
symmetric linear operator
and $\lambda$ is some real number.

\bigskip
\ADT
\bf Remark \NUM. \rm
We know already that any vector field of the form
$$\overline{x}+{\bf Z} \ni \overline{x} + z \mapsto
-[\Lambda z] (\overline{x}+z)+Bz$$
is tangent to the unit sphere if and only if
$B$ and $\Lambda$ are of the form (3.10) with a symmetric operator $A$
and any real number $\lambda$ if $\Vert \overline{x}\Vert =1$ and with $\lambda=0$
in case of $\Vert\overline{x}\Vert=1$.

Our next goal is to establish that the operator $A$



In order to pass to symmetric operators. Thus for any vector
$\big[ {x\atop\xi}\big] \in{\cal D}$ we have
$${\cal A} \Big[ {x\atop\xi} \Big]= {d\over dt}\Big\vert_{t=0} \widehat{\Psi}^t \Big[ {x\atop\xi} \Big] =
\left[{ (iA+\lambda I) (x-\xi\overline{x}) \atop \langle (iA-\lambda I) (x-\xi\overline{x}) \vert \overline{x} \rangle} \right]
$$
and for any point $x \in [\overline{x}+{\bf Z}]\cap\overline{\bf B}$ we have
$${d\over dt}\Big\vert_{t=0} \Psi^t (x) =
-\big\langle (iA-\lambda I)(x-\overline{x}) \big\vert \overline{x} \big\rangle x +
(iA+\lambda I) (x-\overline{x})$$
where $A:{\bf Z}\to{\bf Z}$ is a densely defined closed symmetric linear operator
and $\lambda$ is some real number.
}%ENDDEL


\DEL{

\bigskip\bf Proposition. \it
There exists a constant $\rho\in\RR$ such that
$$\Lambda v = \big\langle (B \!+\! \rho I)v \,\big\vert\, \overline{x} \big\rangle, \quad
2\,{\rm Re}\, \big\langle Bv \big\vert v \big\rangle = \rho
\big\langle v \big\vert v \big\rangle \qquad (v \in {\bf Z}).$$
\bf Proof. \rm Consider any vector $0\ne v\in{\bf Z}$ and let
$$z:= \zeta v \quad \hbox{where} \quad
\zeta := -{ \langle \overline{x} \vert v \rangle \over \langle v
\vert v \rangle } .$$ Observe that \ $\overline{x} \!+\! z \perp z
\in{\bf Z}$. Hence we may apply the lemma with the conclusion
$\zeta\, \Lambda v = \big\langle \zeta Bv \,\big\vert\, \overline{x}
\big\rangle + 2\,{\rm Re} \big\langle \zeta Bv \,\big\vert\, \zeta
v\big\rangle$ that is $\Lambda v = \big\langle Bv \,\big\vert\,
\overline{x} \big\rangle + 2 \,\overline{\zeta}\, {\rm Re}
\big\langle Bv \,\big\vert\, v\big\rangle$ implying
$$\Lambda v = \big\langle Bv \,\big\vert\, \overline{x} \big\rangle +
2 \,{ {\rm Re} \big\langle Bv \,\big\vert\, v\big\rangle \over
\big\langle v \,\big\vert\, v\big\rangle } \, \big\langle v
\,\big\vert\, \overline{x} \big\rangle \ .$$ Hence we see that
$$v\mapsto 2 \,{ {\rm Re} \big\langle Bv \,\big\vert\, v\big\rangle
\over  \big\langle v \,\big\vert\, v\big\rangle } = {\Lambda v -
\big\langle Bv \,\big\vert\, \overline{x} \big\rangle
\over\big\langle v \,\big\vert\, \overline{x} \big\rangle }$$ is a
real valued G\^ateaux holomorphic function on the algebraically open
domain $\{ v\!\in\!{\bf Z} : \ v\not\perp \overline{x} \}$ which
is possible only if being constant. Hence the statement is
immediate. \qed

}%ENDDEL





\bigskip\bigskip
\ADS
\bf \the\SEC. The Jordan case
%$0 \in {\rm dom}\big({d\over dt}\vert_{t=0} \Psi^t \big)$
\rm

\bigskip
We continue the previous investigations
with unchanged notations but under the additional hypothesis that
\ADT
$$0 \in {\bf D} ={\rm dom}\Big( {d\over dt}\Big\vert_{t=0} \Psi^t \Big) \ .\leqno(\NUM)$$
As we know,
${\bf D} = \big\{ x\in{\bf B}:\ \big[ {x\atop 1}\big] \in {\cal D}\big\}
= {\bf B}\cap(\overline{x} + {\bf Z})$ where
${\cal D}={\rm dom}({\cal A})$ with ${\cal A}={d\over dt}\big\vert_{t=0} \widehat{\Psi}^t_{\overline{x}}$
\break
and
${\bf Z}= \big\{ x\in{\bf H}:\ \big[ {x\atop 0}\big] \in {\cal D} \big\}$
is a dense complex linear submanifold in ${\bf H}$.
Thus, as a consequence of (4.1),
for the distinguished common fixed point of the extended automorphisms
$\overline{\Psi}^t$ we have $\overline{x} \in {\bf Z} ={\rm dom}(B) ={\rm dom}(\Lambda)$.
Therefore also
$$\Big[ {0 \atop 1}\Big] \!\in\! {\cal D} \!=\! \Big[ {{\bf Z}\atop\CC} \Big], \quad
{\cal A} \Big[ {x\atop \xi}\Big] \!=\!
\Big[ {B(x-\xi \overline{x}) \atop \Lambda(x-\xi \overline{x})} \Big] \!=\!
\Big[ { B \ \ -\!B\overline{x} \atop \Lambda \ \ -\!\Lambda\overline{x} }\Big] \Big[ {x\atop \xi}\Big]\quad
(x \!\in\! {\bf Z}, \, \xi\!\in\!\CC).$$
\ADT
\bf Remark \NUM. \rm
Recall [\KAUP] that the complete holomorphic vector fields on ${\bf B}$ are the infinitesimal generators
of the uniformly continuous one-parameter subgroups of ${\rm Aut}({\bf B})$ and,
with suitable
$c\in{\bf H}$ and a bounded self-adjoint operator $C\in{\cal L}({\bf H})$
they can be written in the form $x\mapsto a - \{ x c^* x\} + iC x$
by means of the
Jordan triple product.
\ADT
$$\big\{ x y^* z \big\} := {1\over 2} \big\langle x \big\vert y\big\rangle z +
{1\over 2} \big\langle z \big\vert y\big\rangle x
\leqno(\NUM)$$

In terms of the factorization $\Psi^t=\Theta_{a_t}\circ U_t\vert {\bf B}$ we introduce the following vector resp. not necessarily
bounded symmetric linear operator:
$$b:={d\over dt}\Big\vert_{t=0} \Psi^t(0) = \lim_{t\to 0} {1\over t} a_t , \quad
R:=-i{d\over dt}\Big\vert_{t=0} U_t : x\mapsto \lim_{t\to 0} {1\over it} (U_t-I)x . $$
\DEL{
Throughout this section we assume
that
a common fixed point $\overline{x}$ of $\big[ \overline{\Psi}^t:\, t\!\in\!\RR\big]$
lyes in the boundary of the unit ball and the origin belongs to the domain
${\bf D}$ of the non-linear generator
that is we have
$$0\in {\bf D}, \quad \Vert \overline{x} \Vert =1 .
\leqno(Ab)$$

\bf Remark, \rm
Since ${\bf D}=\overline{x}+{\bf Z} \cap \overline{{\bf B}}$
and since ${\bf Z}$ is a complex linear submanifold of ${\bf H}$,
we have $0\in{\bf D}$ if and only if
$\overline{x} \in{\bf Z}$
which is equivalent to
$\big[{0\atop 1}\big]= \big[{\overline{x}\atop 1}\big] - \big[{\overline{x}\atop 0}\big] \in{\cal D}$.
In this case the terms $B\overline{x},\Lambda\overline{x}$
are also well-defined
and in their terms we can write the infinitesimal generator ${\cal A}$ in a matrix form
as
$$\eqalign{
&{\cal A} \Big[ {0\atop 1}\Big]
= {\cal A} \Big( \Big[ {\overline{x}\atop 1}\Big] - \Big[ {\overline{x}\atop 0}\Big] \Big) =
- \Big[ {B\overline{x}\atop \Lambda \overline{x}} \Big] ,\cr
&{\cal A} \Big[ {x\atop \xi}\Big] =
\Big[ {B(x-\xi \overline{x}) \atop \Lambda(x-\xi \overline{x})} \Big] =
\Big[ { B \ -\!B\overline{x} \atop \Lambda \ -\!\Lambda\overline{x} }\Big] \Big[ {x\atop \xi}\Big] .\cr}$$
Notice that in case of
$\big[{0\atop 1}\big] \not\in{\cal D}$ we have only
${\cal A}\big[{x\atop \xi}\big] =
\big[ {B \atop \Lambda} \big]\Big( [I \ -1]\big[{x\atop \xi}\big] \Big)$
while the formal product $\big[ {B \atop \Lambda} \big] [I \ -1]$ cannot be defined.
}%ENDDEL

\bigskip
\ADT
\bf Proposition \NUM. \it
Under hypothesis $(4.1)$, we have ${\bf D}={\bf B}\cap{\bf Z}$ along with ${\rm dom}(R) \supset {\bf Z}$
and the vector field
$\Omega:={d\over dt}\big\vert_{t=0}\Psi^t$ admits the Jordan form
\ADT
$$\Omega(x) = b-\{ xb^*x \} + iR x \qquad (x\in{\bf D}) .
\leqno(\NUM)$$
\bf Proof. \rm
The relation ${\bf D}={\bf B}\cap{\bf Z}$ is clear since $\overline{x}\in{\bf Z}$.
By the definition of the generator ${\cal A}$,
$${\cal A} \Big[ {0\atop 1}\Big] =
\lim_{t\to 0} {1\over t} \big( \widehat{\Psi}^t - \widehat{\Psi}^0\big) \Big[ {0\atop 1}\Big] =
\lim_{t\to 0} {1\over t} \Big\{ {1\over 1+\langle U_t \overline{x} \vert a_t \rangle}
\Big[ {a_t \atop 1} \Big] - \Big[ {0 \atop 1}\Big] \Big\} .$$
Since $\lim\limits_{t\to 0} a_t =0$
and $\Vert U_t \overline{x} \Vert\le 1$ $(t\in\RR)$,
taking $(2.8)$ into account, we see
that the limit
$$b:=\lim_{t\to 0} {1\over t} a_t = {d\over dt} \Big\vert_{t=0} \Psi^t(0)$$
is well-defined and
$$B \overline{x} = -\Big[ {\cal A} \Big[ {0\atop 1}\Big] \Big]_{{\bf H}} =-b ,
\quad
\Lambda \overline{x} = -\Big[ {\cal A} \Big[ {0\atop 1}\Big] \Big]_{\CC} =
\big\langle \overline{x} \,\big\vert\, b \big\rangle .
%\langle U_t \overline{x} \vert a_t \}rangle = t \Lambda \overline{x} + o(t) , \quad
%a_t= t b + o(t) .
$$
As a consequence we also have
$$\eqalign{
\beta_t &=\sqrt{1-\Vert a_t\Vert^2} = \sqrt{1-\Vert tb+o(t)\Vert^2} =
1 - {1\over 2} \Vert b\Vert^2 t^2 + o(t^2) ,\cr
Q_t &= P_t + \beta_t (I-P_t)= I + (1-\beta_t)P_t = I + o(t) \quad (\hbox{in operator norm}).\cr}$$
Since also $U_t z = z + o(1)$ in norm,
hence we deduce that
%$1-\beta_t = o(t)$ and therefore,
for any vector $z\in{\bf Z}$,  and $\zeta\in\CC$,
\ADT
$$\eqalign{
\widehat{\Psi}^t \Big[ {z\atop \zeta}\Big] &=
{1- \langle\overline{x}\vert a_t\rangle \over 1- \Vert a_t\Vert^2}
\left[ {Q_t U_t \ \ a_t \atop a_t^* U_t z \ \ 1 } \right] \Big[ {z\atop \zeta} \Big] =
%\big( 1 - t \langle \overline{x} \vert b\rangle +o(t) \big)
%\left[ { (I+o(t)) U_t z \atop t \langle U_t z \vert b\rangle +o(t)} \right] =\cr
%&=
\left[ { U_t z - t \langle \overline{x} \vert b\rangle z  %+ \zeta a_t
\atop t \langle U_t z \vert b\rangle + \zeta
}\right]
 + o(t) \quad \hbox{in norm}.
\cr} \leqno(\NUM)$$


\DEL{
$$\eqalign{
{1\over t} \big( \widehat{\Psi}^t - I \big) \Big[ {z\atop 0} \Big] &=
{1\over t} \left[ {(P_t + \beta_t \overline{P}_t) U_t z -
\langle U^t\overline{x} \vert a_t\rangle z
\atop
t \Lambda z + o(t) } \right] =\cr
&= \left[ {
t^{-1} (U_t-I) z + t^{-1} (1-\beta_t) \overline{P}_t U_t z -
\langle \overline{x} \vert b \rangle z
\atop
\Lambda z} \right] + o(1) .\cr}$$
}

By definition,
$\big[ {Bz  \atop \Lambda z} \big]= \lim\limits_{t\to 0}
{1\over t} \big( \widehat{\Psi}^t - I \big) \big[ {z\atop 0} \big]$.
Hence with well-defined limits we conclude that
\ADT
$$Bz =  \lim_{t\to 0} {1\over it} (U_t \!-\! I) z -
\big\langle \overline{x} \, \big\vert  b \big\rangle z
\qquad (z\in {\bf Z}) .
\leqno(\NUM)$$
The strong limit of $t^{-1}(U_t-I)\vert{\bf Z}$ is necessarily symmetric due to the fact that each $U_t$ is unitary.
Thus comparing (\NUM) with Proposition 3.7 stating that the operator $B$
has the form $iA+\lambda I$ with some $\lambda\in\RR$ and a symmetric operator $A$ with ${\rm dom}(A)={\bf Z}$,
we get
\DEL{
In particular, from Proposition 3.7
we obtain that
for any vector $z\in{\bf Z}$ we have
$i \langle Az \vert z \rangle + \lambda = \lim\limits_{t\to 0} t^{-1} \langle (U_t-I)z \vert z\rangle
- \langle \overline{x} \vert b \rangle$
whence the the symmetry of $A$ along with the properties
$0= {d\over dt}_{t=0} \Vert U_t z\Vert^2 = 2\,{\rm Re} \lim\limits_{t\to 0} t^{-1}\langle (U_t-I)z \vert z\rangle$
and $\lambda\in\RR$ implies
}%ENDDEL
\ADT
$$\lambda= -{\rm Re} \big\langle \overline{x} \,\big\vert\, b \big\rangle \, , \quad
A = \lim_{t\to 0} \hskip-4mm{\phantom{\big\vert}}^{{\rm strong}} \ \big(U^t -I \big)
- {\rm Im} \big\langle \overline{x} \,\big\vert\, b \big\rangle I \Big\vert{\bf Z}
\quad\ \hbox{in 3.7} .
\leqno(\NUM)$$
We calculate $\Omega$ by substituting
%the above values of $A$ and $\lambda$
(4.7-8)
into its form $\Omega(x)= \big[\Lambda(\overline{x}-x)\big]x + B(x-\overline{x})$
applying also the relations $B\overline{x}=-b$, $\Lambda\overline{x}=\langle x\vert b\rangle$,
$B=iA+\lambda I$, $\Lambda x=\langle (iA-\lambda I)x\vert\overline{x}\rangle = \langle (B-2\lambda I)x\vert\overline{x}\rangle$.
Namely, given any vector $x\in{\bf D}$,
taking into account the antisymmetry of the operator $iA = B \!-\!\lambda I$,
we can write
$$\eqalign{
\Omega(x) &= \big[\Lambda \overline{x}\big]x - \big[\Lambda x\big]x+ Bx - B\overline{x} =\cr
&= \langle \overline{x} \vert b \rangle x - \langle (B-2\lambda I)x \vert \overline{x}\rangle x + Bx + b = \cr
&= b + \big[ \langle \overline{x} \vert b \rangle I + B \big] x - \langle (B-\lambda I)x \vert \overline{x}\rangle x
+ \lambda \langle x \vert \overline{x}\rangle x = \cr
&= b + iR x + \langle x \vert (B - \lambda I)\overline{x} \rangle x + \lambda \langle x \vert \overline{x}\rangle x =\cr
&= b + i R x + \langle x \vert B\overline{x} \rangle x =
b + i R x - \langle x\vert b\rangle x .\cr}$$
%The proof is complete since, by Proposition 3.7 we have $\lambda=0$ in case of $\Vert \overline{x}\Vert\ne 1$.

\DEL{
\bigskip
\bf Lemma. \it We have
$(1+\langle U_t \overline{x}\vert a_t\rangle)^{-1}
\Big[  {
%P_t+\beta_t \overline{P}_t
Q_t \ a_t \atop \ \ \ a_t^* \ \ \ 1} \Big] =
\Big[ {I \ 0 \atop 0 \ 1 }\Big] +
t \Big[ \!\!\!\!
{ -\langle \overline{x} \vert b\rangle I\ b \atop \quad b^* \ -\langle \overline{x} \vert b\rangle} \Big]
+ o^{{\rm norm}}(t).$

\bigskip
\bf Proof. \rm
This is a direct consequence of the facts that
$a_t = tb+ o^{{\rm norm}}(t)$, $\beta_t=1+o(t)$,
$(1+\langle U_t \overline{x}\vert a_t\rangle)^{-1} =
(1-\langle \overline{x} \vert a_t\rangle)(1-\langle a_t \vert a_t\rangle)^{-1} =
(1-t\langle \overline{x} \vert b\rangle +o(t)) \beta^{-2} = 1-t\langle \overline{x} \vert b\rangle +o(t)$
and $P_t +\beta_t \overline{P}_t = (1-\beta_t)P_t +\beta_t (P_t+\overline{P}_t) = I +o^{{\rm norm}}(t)$
as $t\to 0$.
\qed
}%ENDDEL

\bigskip
\ADT
\bf Corollary \bf\NUM. \it %We have
${\bf Z} = {\rm dom}(R)$
%${\bf Z} = {\rm dom} \big( \lim\limits_{t\to 0} t^{-1} (U_t-I) \big)$,
that is $x\in{\bf Z}$ if and only if the limit $\lim\limits_{t\to 0} t^{-1}(U^tx-x)$ exists.

\bigskip
\bf Proof. \rm
Recall that ${\bf Z} = \big\{ x\in{\bf H} :\ {d\over dt}\big\vert_{t=0} \widehat{\Psi}^t \big[ {x\atop 0}\big]$
{\rm is well-defined}$\big\}$.
From Proposition 4.4 we know that
$iRx =\lim\limits_{t\to 0} t^{-1}(U_t-I)x$ is well-defined for every vector $x\in{\bf Z}$.
Conversely, suppose $u=\lim\limits_{t\to 0} t^{-1}(U_tx-x)$ is well-defined. Then
$U_tx= x+tu+o^{{\rm norm}}(t)$ and (4.6) establishes that
$\widehat{\Psi}^t \big[ {x\atop 0} \big] =
\big[ {x\atop 0} \big] + t \big[ {u - \langle \overline{x} \vert a_t\rangle x \atop
\langle x \vert b \rangle} \big] + o^{{\rm norm}}(t)$.
\DEL{
$$\eqalign{
\widehat{\Psi}^t \Big[ {x\atop 0} \Big] %&=
%(1+\langle U_t \overline{x} \vert a_t\rangle )
%\Big[ %{P_t +\beta_t \overline{P}_t
%{Q_t \ \ a_t \atop a_t^* \ \ \ 1 } \Big] \,
%\Big[ {U_t \ 0 \atop 0 \ \ 1} \Big] \, \Big[ {x\atop 0} \Big] =\cr
%&= \Big( \Big[ { I \ 0 \atop 0 \ 1} \Big] -
%t \Big[ { \langle \overline{x} \vert a_t\rangle \ -b \atop -b^* \ \langle \overline{x} \vert a_t\rangle} \Big]
%+ o^{{\rm norm}}(1) \Big) \,
%\Big[ {x+t u+o^{{\rm norm}}(t) \atop 0} \Big] =\cr
&= \Big[ {x\atop 0} \Big] + t \Big[ {u - \langle \overline{x} \vert a_t\rangle x \atop
\langle x \vert b \rangle} \Big] + o^{{\rm norm}}(t) . \cr}
$$
}%DEL

\bigskip
\ADT
\bf Lemma \NUM. \it We have \ \ $U_{-t} = U_t^{-1} = U_t^*$, \ $a_{-t} = -U_t^* a_t$ \ \ $(t\in\RR)$.

\bigskip
\bf Proof. \rm
Given any %parameter
$t\in\RR$, we have $\Psi^{-t} = \Psi_t^{-1}$ that is
$\Theta_{a_{-t}} U_{-t} = \big[ \Theta_{a_t} U_t \big]^{-1} =
U_t^{-1} \Theta_{a_t}^{-1} = U_t^{-1} \Theta_{-a_t} = \big[ U_t^{-1} \Theta_{-a_t} U_t\big] U_t^{-1} =
\Theta_{U_t^{-1} (-a_t)} U_t^{-1}$.
By the
%uniqueness of the
unambiguous decomposability of holomorphic automorphisms of circular domains into
M\"obius and unitary parts [\BKU], hence we deduce that
$\Theta_{a_{-t}} = \Theta_{-U_t^{-1} a_t}$ and $U_{-t} = U_t^{-1}$.
%\qed

\bigskip
\ADT
\bf Lemma \NUM. \it The operator $R$ is self-adjoint with ${\rm dom}(R)={\bf Z}$.

\bigskip
\bf Proof. \rm
In view of (4.6),
and since $U_t x = x+it Rx +o^{{\rm norm}}(t)$ for any
$x\in {\bf Z}={\rm dom}(R)={\rm dom}({d\over dt}\big\vert_{t=0}U_t)$,
we conclude that
$$\eqalign{
{d\over dt}\Big\vert_{t=0} \widehat{\Psi}^t \Big[ {x\atop \xi}\Big]
%&= {d\over dt}\Big\vert_{t=0} (1+\langle U_t \overline{x}\vert a_t\rangle)^{-1}
%\Big[  { %P_t+\beta_t \overline{P}_t
%Q_t \ \ a_t \atop a_t^* \ \ 1} \Big]
%\Big[ {U_t \ 0 \atop 0 \ \ 1} \Big] \Big[ { x\atop \xi}\Big] =\cr
%&= {d\over dt}\Big\vert_{t=0} (1+\langle U_t \overline{x}\vert a_t\rangle)^{-1}
%\Big[  { %P_t+\beta_t \overline{P}_t
%Q_t \ \ a_t \atop a_t^* \ \ \ 1} \Big] \Big[ { x\atop \xi}\Big] +
%{d\over dt}\Big\vert_{t=0} \Big[ {U_t \ 0 \atop 0 \ \ 1} \Big] \Big[ { x\atop \xi}\Big] =\cr
&= \Big[\hskip-3mm { iR \!-\! \langle \overline{x} \vert b\rangle I \ \ \ \ b \atop \qquad\ \ b^*  \quad \ \ -\!\langle \overline{x} \vert b\rangle} \Big]
\Big[ { x\atop \xi}\Big] \cr}
\quad \hbox{for any} \ \ \Big[ {x\atop \xi}\Big] \in{\cal D} = \Big[ {{\bf Z}\atop \CC}\Big].$$
The linear operator in ${\cal L}({\cal H})$ with matrix
$\big[ \!\!\!\! {-\!
\langle \overline{x} \vert b\rangle I \ \ b \atop \quad b^* \quad -\langle \overline{x} \vert b\rangle} \big] $
is bounded. Since ${\cal A} ={d\over dt}\big\vert_{t=0} \widehat{\Psi}^t$ with domain ${\cal D}$
is the generator of a strongly continuous semigroup in ${\cal L}({\cal H})$,
by the theorem of bounded perturbations [\NAG], also
the operator with matrix
$\big[ {iR \ 0 \atop \ 0 \ 0}\big]$ with domain ${\cal D}$ is the
generator of a
strongly continuous one-parameter subgroup of ${\cal L}({\cal H})$
entailing that $iR$ is the
generator of a strongly continuous group $\big[ V_t: t\in\RR\big]$ in ${\cal L}({\bf H})$.
Since $U_{-t}=U_t^{-1}=U_t^*$, the arguments on sun adjoint semigroups in [\NAG, p. 69]
show that $\lim\limits_{t\to 0} t^{-1}(U_t^* -I) =-iR$ is the generator of the sun adjoint group
$\big[ V_t^*: t\in\RR\big]= \big[ V_{-t}: t\in\RR\big]$
and we have $-iR=(iR)^*$ which completes the proof.
%\qed




\bigskip
\ADT
\bf Theorem \NUM. \it Any vector field of the form $(4.5)$
where $R$ is a not necessarily bounded self-adjoint operator with dense domain ${\bf Z} \subset {\bf H}$,
is the infinitesimal generator defined on ${\bf D}:={\bf Z} \cap{\bf B}$
of a pointwise continuous one-parameter group
$[\Phi^t:\ t\in\RR]$ of holomorphic automorphisms of ${\bf B}$.\footnote{$^3$}{\rm
That is, for all $x\in {\bf D}:={\bf Z}\cap{\bf B}$,
the functions $t\mapsto \Phi^t(x)$ range in ${\bf Z}$, they are differentiable and satisfy
the identity ${d\over dt} \Phi^t(x)={\cal V}\big(\Phi^t(x)\big)$ $(t\in\RR)$.}

\bigskip
\bf Proof. \rm
It suffices to see that
there is a strongly (i.e. pointwise) continuous one-parameter group
$\big[ {\cal V}^t:\ t\in\RR\big]$ of
bounded linear operators of the space ${\cal H}$
%${\bf H}\oplus \CC$
such that
$${d\over dt}\Big\vert_{t=0} {\cal V}^t \Big[ {x \atop \xi} \Big] =
\Big[ {iR \ b \atop b^* \ \ 0}\Big] \Big[ {x \atop \xi} \Big] \quad (x\in{\bf Z},\ \xi\in\CC),
\qquad {\cal V}^t {\cal K} \subset {\cal K} := \Big\{ \Big[ {x\atop \xi}\Big] : \ \Vert x\Vert^2 \ge \vert \xi\vert^2 \Big\}. $$
Namely, in this case the maps
$$\Phi^t (x) := \Big[ {\cal V}^t \Big[ {x\atop 1} \Big] \Big]_{\CC}^{-1} \Big[ {\cal V}^t \Big[ {x\atop 1} \Big] \Big]_{{\bf H}}
\qquad (t\in\RR,\ x\in{\bf D})$$
suit the requirements of the theorem since
$x\in{\bf D} \Rightarrow \big[ {x\atop 1}\big] \in{\cal K}
\Rightarrow {\cal V}^t \big[ {x\atop 1}\big] \Rightarrow \Phi^t(x)\in{\bf D}$ and
$x\!\in\!{\bf D} \!\Rightarrow\! {d\over dt}\big\vert_{t=0} \! \Phi^t(x) \!=\!
- \!\big[ {\cal V}^0 \big[ {x\atop 1} \big] \big]_{\!\CC}^{\!-2}
{d\over dt}\big\vert_{t=0} \big[ {\cal V}^t \big[ {x\atop 1} \big] \big]_{\!\CC}
\big[ {\cal V}^0 \big[ {x\atop 1} \big] \big]_{\!{\bf H}} +
\big\{ {\cal V}^0 \big[ {x\atop 1} \big] \big\}_{\!\CC}^{\!-1}
{d\over dt}\big\vert_{t=0} \big[ {\cal V}^t \big[ {x\atop 1} \big] \big]_{\!{\bf H}}$
$=
- \big[ \big[ {iR \ b \atop b^* \ 0}\big] \big[ {x\atop 1}\big] \big]_{\CC} x \!+\!
\big[ \big[ {iR \ b \atop b^* \ 0}\big] \big[ {x\atop 1} \big] \big]_{\!{\bf H}} =
-\langle x\vert b\rangle x + iRx +b = \Omega(x)$.
%
Notice that, by Corollary 2.10, a strongly continuous one parameter group of linear operator
leaves the cone ${\cal K}$ invariant if all its members map the boundary
$\partial {\cal K} \!\!=\! \big\{ \!\big[ {x\atop \xi} \big] \!\! : \Vert x\Vert \!\!=\!\!\vert \xi\vert\big\} \!\!=\!
\big\{ \!\big[ {x\atop e^{i\tau}\Vert x\Vert} \big] \!\! :  x\!\!\in\!\!{\bf H},\, \tau\!\!\in\!\!\RR \big\}$
\break
into itself.
\DEL{
\footnote{$^*$}{In general, if a Hilbert space ${\bf W}$ is the orthogonal
%direct
sum
of the subspaces ${\bf W_k}$ $(k\!=\!1,2)$ and $C$ is the infinitesimal generator of a
strongly continuous one-parameter group $[T^t: t\!\in\!\RR ]$ in ${\cal L}({\bf W})$
then we have $T^t {\bf C} \!=\!{\bf C}$ $(t\!\in\!\RR)$ for the cone
${\bf C} \!:=\!\big\{ {\bf w}_1\!\oplus\!{\bf w}_2: {\bf w}_k\!\in\!{\bf W}_k, \,
\Vert{\bf w}_1\Vert \!>\! \Vert{\bf w}_2\Vert \big\}$
if and only if $C$ is tangent to the boundary of ${\bf C}$
that is if
\hfill\break
$(*)$ ${\rm Re}\big\langle\! C({\bf w}_1\!\oplus\!{\bf w}_2),\!{\bf w}_1 \!\big\rangle \!=\!
{\rm Re}\big\langle\! C({\bf w}_1\!\oplus{\bf w}_2),\!{\bf w}_2 \!\big\rangle$ whenever
${\bf w}_1\!\oplus\!{\bf w}_2 \!\in\!{\rm dom}(C)$ with
$\Vert{\bf w}_1\Vert \!=\! \Vert{\bf w}_2\Vert$.
\hfill\break
Proof. It is immediate that $T^t{\bf C\!}\!\subset\! {\bf C}$ $(t\!\in\!\RR)$
$\Rightarrow T^t\partial{\bf C}\!\subset\! \partial{\bf C}$ $(t\!\in\!\RR)$
$\Rightarrow {d\over dt}\vert_{t=0} T^t({\bf w}_1\oplus{\bf w}_2) \in
{\rm Tan}_{{\bf w}_1\oplus{\bf w}_2}({\bf C})$ for ${\bf w}_1\oplus{\bf w}_2\in{\rm dom}(C)$
$\Rightarrow (*)$.
Assume $(*)$ and
let $P$ denote the canonical projection of ${\bf W}$ onto ${\bf W}_1$
and define
$T^{t,s}:=\exp(tC+sP)$ $(s,t\in\RR)$.
By the theorem of bounded perturbations [\NAG, p.158]
the operators $T^{t,s}$ are all well-defined.
Moreover, by [\NAG, Corollary 1.7 p. 161]
(applied with $B:=sP$ and $A:=C$ there)
we have $\lim_{s\to 0} T^{t,s}{\bf w} = T^t{\bf w}$ $({\bf w}\in{\rm dom}(C), t\in\RR)$.
Therefore, to establish that $T^t{\bf C}\subset {\bf C}$ $(t\in\RR)$,
it suffices to see only that \ $(**)$
$T^{t,s}{\bf w} \!\in\!{\bf C}$ whenever
${\bf  w}\!\in\!{\rm dom}(C)\!\cap\! {\bf C}$ and $t,s\!>\! 0$.
To proceed to contradiction, let
$s,t>0$ and ${\bf w}:={\bf w}_1\oplus {\bf w}_2 \in{\rm dom}(C)$ with
$\Vert{\bf w}_1\Vert > \Vert{\bf w}_2\Vert$ but
$\Vert [T^{t,s}{\bf w}]_1\Vert \le \Vert [T^{t,s}{\bf w}]_2\Vert$.
The function
$\delta(\tau):=\Vert [T^{\tau,s}{\bf w}]_1\Vert^2 - \Vert [T^{\tau,s}{\bf w}]_2\Vert^2$
is differentiable in $\tau$ on the whole $\RR$ and
$\delta(0)> 0 \ge \delta(t)$. Thus there exist a point $t_*\in(0,t]$
such that $\delta(\tau)>0=\delta(t_*)$ $(0\le\tau<t_*)$.
Since $\delta(t_*)=0$,
${\bf w}_*:=T^{t_*,s}{\bf w} \in\partial{\bf C}$
and hence
${\rm Re}\langle C{\bf w}_*,[{\bf w}_*]_1\rangle=
{\rm Re}\langle C{\bf w}_*,[{\bf w}_*]_2\rangle$.
We get the contradiction $0\ge \delta^\prime(t_*) =
2{\rm Re}\langle (C+sP){\bf w}_*,[{\bf w}_*]_1\rangle -
2{\rm Re}\langle (C+sP){\bf w}_*,[{\bf w}_*]_2\rangle =
2s \Vert[{\bf w}_*]_1\Vert^2 >0$.
}
}%ENBDDEL
Therefore it suffices to check that
there is a (necessarily unique) strongly continuous one-parameter group
in ${\cal L}({\cal H})$ with domain ${\bf Z}\oplus\CC=\big[ {{\bf Z}\atop\CC}\big]$
%${\bf Z}\oplus\CC$
such that
$${d\over dt} {\cal V}^t \Big[ {x \atop \xi} \Big] =
\Big[ {iR \ b \atop b^* \ \ 0}\Big] {\cal V}^t \Big[ {x \atop \xi} \Big] ,
\quad \left\Vert \Big[ {\cal V}^t \Big[ { x\atop \Vert x\Vert } \Big] \Big]_{{\bf H}} \right\Vert=
\left\vert \Big[ {\cal V}^t \Big[ { x\atop \Vert x\Vert } \Big] \Big]_{\CC} \right\vert
\qquad (x\in{\bf Z},\ t\in\RR) . $$
By Stone's theorem, the ${\cal H}$-unitary operators
${\cal W}^t \big[{x\atop\xi}\big] := \big[{\exp(itR)x \atop \xi}\big]$
form a strongly continuous one-parameter group whose infinitesimal generator
is defined on ${\rm dom}(R) \oplus\CC ={\bf Z}\oplus\CC$ with the diagonal matrix
$\big[ {iR \ 0 \atop 0 \ \ 0}\big]$.
Since the matrix $\big[ {0 \ \ b \atop b^* \ 0} \big]$ represents a bounded linear operator
in ${\cal H}$,
by the theorem of bounded perturbations [\NAG],
there is a strongly continuous one-parameter group
$\big[ {\cal V}^t: \, t\in\RR\big]$
whose generator is defined on
${\bf Z}\oplus\CC$ with the matrix $\big[ {iR \ b \atop b^* \ 0}\big]$.
In particular
${d\over dt} {\cal V}^t \big[ {x\atop \xi}\big] =
\big[ {iR \ b \atop b^* \ 0}\big] {\cal V}^t \big[ {x\atop \xi}\big]$ $(t\in\RR,\ x\in{\bf Z})$.
To complete the proof, we show that necessarily
$${d\over dt} \left[ \Big\Vert\Big[ {\cal V}^t \Big[ { x\atop \Vert x\Vert } \Big] \Big]_{{\bf H}} \Big\Vert^2 -
\Big\vert \Big[ {\cal V}^t \Big[ { x\atop \Vert x\Vert } \Big] \Big]_{\CC} \Big\vert^2 \right] =0
\qquad (t\in\RR,\ x\in{\bf Z}). $$
Consider any vector
$x\in{\bf Z}$ and write $\big[ {x_t \atop \xi_t} \big] :=  {\cal V}^t \big[ {x\atop \Vert x\Vert }\big]$
for all parameters $t\in\RR$.
Then
$$\eqalign{
{d\over dt} \Big[ \Vert x_t \Vert^2 - \vert \xi_t\vert^2 \Big] &=
2\,{\rm Re} \Big[ \big\langle dx_t/dt \big\vert x_t \big\rangle - \big( d\xi_t/dt) \overline{\xi_t} \Big] =\cr
&= 2\,{\rm Re} \left\{ \Big\langle \Big[ \Big[ {iR \ b \atop b^* \ \ 0}\Big]\Big[ {x_t\atop\xi_t}\Big] \Big]_{\!{\bf H}}
\Big\vert x_t \Big\rangle -
\Big[ \Big[ {iR \ b \atop b^* \ \ 0}\Big]\Big[ {x_t\atop\xi_t}\Big] \Big]_{\!\CC}
\big( d\xi_t/dt) \overline{\xi_t} \right\} =\cr
&= 2\,{\rm Re} \left[ \big\langle iRx_t + \xi_t b \big\vert x_t \big\rangle -
\big\langle x_t \big\vert b\big\rangle \overline{\xi_t} \right] =\cr
&= -2\,{\rm Im} \big\langle Rx_t \big\vert x_t \big\rangle +
2\,{\rm Im}\big( \big\langle \xi b\big\vert x_t\big\rangle - \big\langle x_t \big\vert \xi b\big\rangle \big)
=0 . %\hbox{\qed}
\cr}$$




\bigskip
\bigskip

\ADS
\bf \the\SEC. Proof of Theorem 1.1 and Corollary 1.4

\bigskip
\bf Proof of Theorem 1.1. \rm
Consider any pointwise continuous one-parameter subgroup $\big[ \Psi^t:\ t\in\RR\big]$ of ${\rm Aut}\big({\bf B}\big)$
and let us fix any common fixed point $\overline{x}\in\overline{{\bf B}}$
of the continuous extensions
of the maps $\Psi^t$ $(t\in\RR)$ (guaranteed by Lemma 2.4).
From Proposition 3.7, we know already that (1.2-3) hold
for some dense linear complex submanifold ${\bf Z}$ of the underlying Hilbert space
with a symmetric linear operator $A$ with ${\rm dom}(A)={\bf Z}$.
We have to see that $A$ is even self-adjoint in any case and, conversely
every self-adjoint operator with domain ${\bf Z}$ may appear in (1.3)
with any constant $\lambda\in\RR$ if $\overline{x}\in\partial{\bf B}$ or $\lambda=0$ if $\overline{x}\in{\bf B}$.
In order to establish a link to the Jordan case,
fix any point
$c \!\in\! {\bf D}\!=\!{\rm dom}\big({d\Psi^t\over dt}\big\vert_{t=0}\big)$
and let
%introduce the automorphisms
$$\Phi^t := \Theta_{-c} \Psi^t \Theta_c \quad (t\in\RR), \quad \overline{y}:=\Theta_c(\overline{x})$$
by means of the M\"obius transformations (2.3).
%
Clearly $\big[ \overline{\Phi}^t:\ t\in\RR\big]$ is a
strongly continuous one-parameter subgroup of
${\rm Aut}\big({\bf B}\big)$
such that
\ADT
$$0=\Theta_{-c} (c) \in \Theta_{-c} \left({\rm dom}\Big( {\partial \Psi^t \over dt}\Big\vert_{t=0}\Big) \right) =
{\rm dom}\Big( {\partial \Phi^t \over dt}\Big\vert_{t=0}\Big) , \quad
\overline{y} \in
%=\Theta_c(\overline{x} \in \Theta_c\Big( \bigcap_{t\in\RR} {\rm Fix}\big( \overline{\Psi}^t\big) \Big) =
\bigcap_{t\in\RR} {\rm Fix}\big( \overline{\Phi}^t\big) .
\leqno(\NUM)$$
Thus we can apply the results of Section 4 in particular Lemma 4.11 to $[\Phi^t:\ t\in\RR]$ to conclude
that there is a dense complex linear submanifold ${\bf Y} \subset {\bf H}$ along with a self-adjoint operator
$R$ with ${\rm dom}(R)={\bf Y}$ and a vector $b\in{\bf H}$ such that
$$\widehat{\Phi}^t_{\overline{y}} = e^{-\langle \overline{x} \vert b\rangle t}
\exp \Big( t \Big[ { iR \ \ b \atop b^* \ \ 0}\Big] \Big) \qquad (t\in\RR). $$
Hence Corollary 2.11 establishes the existence of a constant $\nu\in\CC$ with
$$\widehat{\Psi}^t_{\overline{x}} = e^{\nu t} \beta^{-2} \widetilde{\Theta_c} \widehat{\Phi}^t_{\overline{y}} \widetilde{\Theta_{-c}}
\qquad (t\in\RR)$$
due to the identity
$\widetilde{\Theta_c} \widetilde{\Theta_{-c}} = \beta^2 {\cal I} = (1-\Vert c\Vert^2) \big[ {I \ 0\atop 0 \ 1}\big]$
for the canonical representations
$$\widetilde{\Theta_{\pm c}} = \Big[ {Q \ \ \pm c \atop \pm c^* \ \ 1 }\Big]
\quad \hbox{where} \ \ Q:=\beta I + (1-\beta) P , \ \beta:=\sqrt{1-\Vert c\Vert^2}, \ P:= {\rm P}_c .$$
By passing to infinitesimal generators, with $\mu:=\nu-\langle \overline{x} \vert b\rangle$, we get
\ADT
$${d\widehat{\Psi}^t_{\overline{x}}\over dt}\Big\vert_{t=0} \Big[{ x\atop \xi}\Big] = \mu \Big[ {x\atop \xi}\Big] +
\beta^{-2} \widetilde{\Theta_c} \Big[ {iR \ \ b\atop b^* \ 0}\Big] \widetilde{\Theta_{-c}} \Big[ {x\atop \xi}\Big] \quad
\hbox{for} \ \ \Big[{ x\atop \xi}\Big]\in{\rm dom}\Big( {d\widehat{\Psi}^t_{\overline{x}}\over dt}\Big\vert_{t=0}\Big).
\leqno(\NUM)$$
From Lemma 3.3 and (5.1) we see that
$$\Big[ {{\bf Z} \atop 0}\Big] + \CC\Big[ {\overline{x} \atop \xi}\Big] =
{\rm dom}\Big( {d\widehat{\Psi}^t_{\overline{x}}\over dt}\Big\vert_{t=0}\Big) =
\widetilde{\Theta_c} {\rm dom}\Big( {d\widehat{\Phi}^t_{\overline{y}}\over dt}\Big\vert_{t=0}\Big) =
\widetilde{\Theta_c} \Big[ {{\bf Y} \atop \CC}\Big] .$$
Thus, given any vector $z\in{\bf H}$, we have $z\in{\bf Z}$ if and only if
$\big[ {z \atop 0}\big] = \widetilde{\Theta_c} \big[ {y\atop \eta}\big]$ for some $y\in{\bf Y}$ and $\eta\in\CC$
that is if $z=Qy-\langle y\vert c\rangle c =\big [Q-\Vert c\Vert^2  P] y  $ for some $y\in{\bf Y}$.
It follows
\ADT
$${\bf Z} = \big[ Q - \Vert c\Vert^2 P\big] {\bf Y} = Q^{-1} {\bf Y}
\leqno(\NUM)$$
because the operators $P,Q$ commute, we have $\beta^2=1-\Vert c\Vert^2>0$ and
$Q \big[ Q - \Vert c\Vert^2 P\big] = \big[ \beta I + (1-\beta) P] \big[ \beta I + (\beta^2 -\beta)P \big] =
\beta^2 I$.
We are now ready to establish the self-adjointness of the operator $A$ in (1.3).
By (5.2) we have
$$(iA + \lambda I) z = \left[ {d\widehat{\Psi}^t_{\overline{x}} \over dt} \Big\vert_{t=0} \Big[ {z\atop 0}\Big] \right]_{{\bf H}} =
\mu z + \beta \big[ iQRQ z  - Qbc^*z + cb^*Q z \big] \quad (z\in{\bf Z}) .$$
That is, with the bounded self-adjoint operator $S:=i\beta [  Qbc^* - cb^*Q z] = [iQbc^*]+[iQbc^*]^*$ we have
$A =  \beta Q R Q + S + i(\lambda-\mu) I$. We know the symmetry of $A$ already entailing the relations $\mu=\lambda$
with $A=\beta QRQ +S$. Here the operator $QRQ$ self-adjoint with ${\rm dom}(QRQ)=Q^{-1}{\rm dom}(R)=Q^{-1}{\bf Y} ={\bf Z} = {\rm dom}(A)$
since $R$ is a self-adjoint operator with ${\rm dom}(R)={\bf Y}$ while $Q$ is an invertible bounded self-adjoint operator,
Therefore, as being the bounded self-adjoint perturbation, the operator $A$ is necessarily self-adjoint.

To see the converse, we need only to check the reversibility of some of our previous arguments.
Assuming $A$ to be self-adjoint in (1.2-3), it is the theorem of bounded perturbations [\NAG]
ensures  that the operator
${\cal A}\big[{z+\xi \overline{x}\atop\xi}\big] :=
\big[ {(iA+\lambda I)z \atop \langle (iA-\lambda I) z\vert \overline{x} \rangle (z+\xi \overline{x}) } \big] $
$(z\in{\bf Z},\ \xi\in\CC)$
is the infinitesimal generator of a strongly continuous one-parameter subgroup $[ {\cal U}^t : t\in\RR]$ of ${\cal L}({\bf H}$ with graph
being tangent to the boundary of cone ${\cal K}$ in Corollary 2.10(iii) and
we have ${\cal K} = {\cal U}^t {\cal K}$ $(t\in\RR)$. Hence
the holomorphic maps $\Psi^t(x) := \big[ {\cal U}^t \big[ {x\atop 1}\big] \big]_{\CC}^{-1} \big[ {\cal U}^t \big[ {x\atop 1}\big] \big]_{{\bf H}}$
are well-defined on the unit ball ${\bf B}$ leaving it invariant, and form a strongly continuous one-parameter subgroup in ${\rm Aut}({\bf B})$.

\bigskip
\bf Proof of Corollary 1.4. \rm


\bigskip
\bigskip
\hrule
\bigskip





\vfill
\eject

\bf References
\rm

\medskip
\item{[\HS]} T.L. Hayden -- T.J. Suffridge, \it Biholomorphic maps in Hilbert space have a fixed point,
\rm Pacific J. Math., \bf 38 \rm (1971) 419-422.

\medskip
\item{[\GOEBEL]} K. Goebel, \it Fixed points and invariant domains of holomorphic mappings of the Hilbert ball,
\rm Nonlinear Analysis, Theory, Methods and Applications, \bf 6/12 \rm (1982) 1327-1334.


\medskip
\item{[\FRVE]} T. Franzoni -- E. Vesentini, Holomorphic Maps and Invariant Distances,
North Holland Math. Studies 40, North Holland Publ. Co., Amsterdam-New York-Oxford, 1980.

\medskip
\item{[\ABATE]} M. Abate, \it Common fixed points of commuting holomorphic maps, \rm
Math. Ann. \bf 283 \rm (1989) 645-655.

\medskip
\item{[\DINEEN]} S. Dineen, \it Complete holomorphic vector fields on the second dual of a Banach space,
\rm Math. Scand. \bf 59 \rm (1986) 131-142.

\medskip
\item{[\RSZ]} F. Riesz -- B. Sz.-Nagy,
Vorlesungen \"uber Funktionalanalysis,
VEB Deutscher Verlag der Wissenschaften, Berlin, 1968.

\medskip
\item{[\BKU]} H. Braun -- W. Kaup -- H. Upmeier, \it On the automorphisms of
circular and Reinhardt domains in complex Banach spaces,
\rm Manuscripta Math., \bf 25 \rm (1978), 97-133.

\medskip
\item{[\NAG]} K.-J. Engel -- R. Nagel,
One-Parameter Semigroups for Linear Evolution Equations,
Graduate Texts in Mathematics Vol. 194, Springer-Verlag, New York, 2000.

\medskip
\item{[\KAUP]} W. Kaup, \it
\"Uber die Klassifikation der symmetrischen hermiteschen Mannigfaltigkeiten unendlicher Dimension I.,
\rm Math. Ann., \bf 357 \rm (1981), 463--481.


\bye
