\magnification=\magstep1

\parindent=0pt
\def\s{\bigskip} %sork\"oz
\baselineskip=10mm

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


\def\TT{{\bf T}}
\def\CC{{\bf C}}
\def\RR{{\bf R}}

\def\eps{\varepsilon}

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


{\bf M\"obius transformations}

Hol. aut. of the unit ball. ${\bf B}$ of a JB*-triple $({\bf E},\{\ldots\}$)

$\Phi\in{\rm Aut}({\bf B}) $ extends holomorphically to a neighborhood of $\overline{{\bf B}}$.

Canonical form [Kaup MathZ. 1983]: \quad $\Phi =M_a \circ U$

$M_a(x)= A+ {\rm Bergman}(a)^{1/2} [1 + L(x,a)]^{-1}x$, \ $U$ surj.lin ${\bf E}$-isom.


Faces: If ${\bf E}$ JBW*-triple and ${\bf F}$ is a (norm-exposed) face of $\partial{{\bf B}}$ then

$\exists\, e$ TRIP in ${\bf E}$ \quad ${\bf F} = \big\{ x\in \partial{\bf B} : \ x-e \perp e \big\} =
\big\{ M_c(e) : \ c\perp e, \ \Vert c\Vert \le 1 \big\} $.

M\"obius equivalence: \ \  $\Phi \sim \Psi$  \ \ if
$\exists\, \Theta$ M\"obius trf. with $\Psi =  \Theta \circ \Phi \circ \Theta$

{\bf Basic assumption:}

{\bf (0)} $\big[ \Phi^t : t\in\RR]$ str.cont.1prg. of M\"obius trf's

{\bf Example:} \ 1-dim. cases up to M\"obius equ.

$\Phi^t = M_{a(t)} \circ U_t$ \qquad $U_t\in{\rm Aut}({\bf E},\{\ldots\})$ \ surj.lin.isom.

$t\mapsto a(t)=\Phi^t(0)$ cont. \quad $\Rightarrow$ \quad
$t\mapsto U_t x = (2\pi i)^{-1} \int_{\vert \zeta\vert=1} \zeta^{-1} \Phi(\zeta x) \, d\zeta$ \ cont. $\forall\, x$

{\bf Candidate of infinitesimal generator:}

$\Phi^\prime := {d\over dt}\big\vert_{t=0+} \Phi^t$, \quad
${\rm dom}(\Phi^\prime) = \{ x: \exists\, v \ \ \Phi^h(x) = x+hv+o(h)\}$

{\bf Lemma.} $x\in{\rm dom}(\Phi^\prime) \iff t\mapsto \Phi^t(x) \ {\rm diff.}$

{\bf Proof.} $\Phi^h(x) = x + hv + o(h) \ \Longrightarrow \ \ \Phi^{t+h}(x) -\Phi^t(x) =
\Phi^t\big( x + hv + o(h)\big) - \Phi^t(x)=$

$= h [D_{z=x} \Phi^t(z)] v + o(h)$ \ \ In particular $x\in{\rm dom}(\Phi^\prime)
\Rightarrow x\in{\rm dom}\big( {d\over ds}\big\vert_{s=t+0}  \Phi^s \big)$.

$\Phi^{t-h} (x) - \Phi^t(x) = \Phi^{t-h}(x) - \Phi^{t-h}(x + hv +o(h)) =$

$= -[D_{z=x} \Phi^{t-h}(z)] (hv + o(h)) + o(\Vert hv + o(h)\Vert) = -h [D_{z=x} \Phi^{t-h}(z)]v + o(h)$. \ Q.~e.~d.

{\bf Corollary.} $x\in{\rm dom}(\Phi^\prime) \Longrightarrow \{ \Phi^t(x) : x\in\RR\} \subset {\rm dom}(\Phi^\prime)$.

{\bf Open problem:} $\exists ? \ [\Phi^t: t\in\RR]$ nowhere diff. in $t$?

{\bf Assumption:} ${\rm dom}(\Phi^\prime) \cap {\bf B} \ne \emptyset$ \ or (up to M\"obius equ.) \ $0\in{\rm dom}(\Phi^\prime)$, $t\mapsto a(t)$ diff.


{\bf Lemma.} $x \in {\rm dom}(\Phi^\prime) \iff t\mapsto U_t x$ diff. \qquad $ \big( U_h \in{\rm dom}(\Phi^\prime) \big)$.

{\bf Proof.} $U_tx = M_{-a(t)} \underbrace{\Phi^t}_{M_{a(t)}\circ U_t}(x)$. \qquad $(a,z)\mapsto M_a(z)$ real-anal.

$M_{c+ hv +o(h)} (u + hw + o(h)) =$

$= (c+ hv +o(h)) + B(c+ hv +o(h))^{1/2} \big( 1 + L(u+ hw +o(h),c+ hv +o(h))\big)^{-1} (u+ hw +o(h)) =$

$= M_c(u) - h (L(w,c)+L(u,v))u + h \big( 1 + L(u,c)\big)^{-1} w + o(h)$.

{\bf Corollary.} ${\rm dom}(\Phi^\prime)$ is closed to the Jordan-prod. $\{\ldots\}$

{\bf Proof.} $x,y,z \in {\rm dom}(\Phi^\prime)$ $\Rightarrow$ $t\mapsto U_t \{ xyz\} = \{(U_t x)(U_t y)(U_t z)\}$ diff.

{\bf Remark:} In particular ${\rm dom}(\Phi^\prime) =\big[$Jordan subtriple$\big] \,\cap \overline{{\bf B}}$.

\vskip-4mm
$\phantom{x}$ \hskip13mm
The orbit \ $\{ \Phi^t(0): t\in\RR\} \subset {\rm dom}(\Phi^\prime)$.

{\bf Lemma.} $x\in{\rm dom}(\Phi^\prime \Longrightarrow U_h x \in{\rm dom}(\Phi^\prime$ $(h\in\RR)$.

{\bf Proof.} $U_h x \in{\rm dom}(\Phi^\prime  \iff t \mapsto U_h U_t x$ diff.

$\Phi^{t+h} (x) = \Phi^t \circ \Phi^h (x) = M_{a(t)} \circ U_t \circ M_{a(h)} \circ U_h x =^{U \circ M_a \circ U^{-1} = M_{Ua}}$

$\phantom{x}$ \hskip5mm
$=M_{a(t)} \circ M_{U_t a(h)} \circ U_t U_h x$.

$U_t U_h x = M_{-U_t a(h)} \circ M_{-a(t)} \circ \Phi^{t+h}(x)$, \qquad
$a(h) \in {\rm dom}(\Phi^\prime) \ \Rightarrow \ t\mapsto U_t a(h)$ diff.

$t\mapsto \Phi^t$ diff., \ $t\mapsto a(t)$ diff., \ $(a,b)\mapsto M_a \circ M_b$ real-anal. ;
$\Longrightarrow t\mapsto U_t U_h x$ diff.

{\bf Notation:} \ ${\bf D}:= \overline{ {\rm dom}(\Phi^\prime)}$ \ closure in ${\bf E}$, \quad
${\bf F}:= {\rm Span}( {\bf D} )$


{\bf Proposition.} We have seen: \ ${\bf F}$ closed JB$^*$-subtriple in ${\bf E}$, \ \ ${\bf D} = \overline{{\rm Ball}({\bf F})}$,

\vskip-4mm
$\phantom{x}$ \hskip19mm
$\big\{ U_t \vert {\bf F}: t\in\RR\big\} \subset {\rm Aut}({\bf F},\{\ldots\})$, \ \
$\big\{ M_{a(t)} \vert {\bf D}: t\in\RR\big\} \subset {\rm Aut}_{{\rm hol}}({\bf D})$.

{\bf Remark.} $[\Phi^t]^{-1} = \Phi^{-t}$ $\iff$ $U_t^{-1} M_{-a(t)} = M_{a(-t)} \circ U_{-t}$ %$\iff$

\vskip-4mm
$\phantom{x}$ \hskip19mm
$\iff$ $M_{-U_t^{-1} a(t)} \circ U_t^{-1} = M_{a(-t)} \circ U_{-t}$


\vskip-4mm
$\phantom{x}$ \hskip19mm
$\iff$ $U_t^{-1} =U_{-t}$ and $-U_t^{-1} a(t) = a(-t)$.

{\bf Lemma.} ${\bf F}^{\perp \ {\rm Jordan}} = 0$.

{\bf Proof.} $y\perp^{{\rm Jordan}} x \in{\bf F}$, $x+y\in{\bf D}$ \ \  $\Longrightarrow$ \
$M_{a(t)} U_t (x+y) = \underbrace{M_{a(t)} U_t (x)}_{\Phi^t(x)} + y$ \ diff. in $t$




\bigskip
\hrule
\bigskip

{\bf Assumumption}: $e = \Phi^t(e)$ \ $\forall\, t\in\RR$ \qquad common fixed point

$\Lambda^t := {\rm D}_e \Phi^t \ 
(: z\mapsto {d\over dt}\big\vert_{t=0} \Phi^t(e+tz)$ \quad Fr\'echet derivative

$\Lambda_tz = \int_{\vert \zeta \vert =1} \zeta^{-1} \Phi^t(e+\zeta z) d\zeta$

$[\Lambda^t : t\in\RR]$ \ str.cont.1prg LIN \qquad 
${\bf Z}:={\rm dom}(\Lambda^\prime)$ dense lin. in ${\bf E}$

$\Phi=M_a U \ (=M_a \circ U)$ \qquad $t$ FIX

$w+e = \Phi( e+z ) = M_a (Uz + Ue)$

$w+e = a + B(a)^{1/2} [ 1 + L(Ue + Ue,a)]^{-1} (Uz+Ue)$

$[ 1 + L(Uz + Ue,a)] B(a)^{-1/2} (w +(e -a)) = Uz + Ue$

$\Phi(e)=e$ $\iff$ $[ 1 + L(Ue ,a)] B(a)^{-1/2} (e -a) = Ue$

$[1 + L(Uz + Ue,a)] B(a)^{-1/2} (w +(e -a)) - 
[ 1 + L(Ue ,a)] B(a)^{-1/2} (e -a) = Uz$

$[1 + L(Uz + Ue,a)] B(a)^{-1/2} w + L(Uz ,a) B(a)^{-1/2} (e -a) = Uz$

$w = B(a)^{1/2} [ 1 + L(Uz +Ue,a)]^{-1} [Uz - L(Uz,a) B(a)^{-1/2} (e-a)]$

$\Phi(z+e)-e =w = (A_z + B)^{-1}C z$

$A_z = L(Uz,a) B(a)^{-1/2}$, \ \ 
$B = [1 + L(Ue,a)] B(a)^{-1/2}$, \ \ 
$C = U + L( U \bullet, a) B(a)^{-1/2}(a-e)$


$\Lambda z = {\rm D}_e \Phi = {d\over dz}\vert_{z=0} (A_z + B)^{-1}C z = B^{-1} C z$

$\Phi^t(z+e)-e = (A_{t,z} + B_t)^{-1} C_t z$

$z\in{\bf Z}$ $\Rightarrow$ \quad $t\mapsto \Lambda^t z = B_t^{-1} C_t z$ \ diff.

$B_t^{-1} C_t z = 
B(a_t)^{1/2} [1 + L(U_t e, a_t)]^{-1} [U_t z + L(U_t z, a_t) B(a_t)^{-1/2}(a-e)]$

{\bf Question:} \ $t\mapsto B_t^{-1}C_t z$ $\Rightarrow^?$ $t\mapsto U_t z$ diff.

$a \mapsto B(a)^\alpha$, \ $u\mapsto [1 + L(u,e)]^n$ \ norm-analytic in ${\rm GL}({\bf E})$

$t\mapsto U_t z + L(U_t z, a_t) B(a_t)^{-1/2}(a-e)$ \ diff. \quad $(z\in{\bf Z})$

$L(U_t z, a_t) B(a_t)^{-1/2}(a-e) = \{ [U_t z] [a_t] [B(a_t)^{-1/2}(a-e)] \} =$

$= U_t U_{-t} \{ [U_t z] [a_t] [B(a_t)^{-1/2}(a-e)] \} =
U_t \{ z [U_{-t} a_t] [U_{-t}B(a_t)^{-1/2}(a-e)] \}$ \ \ diff. in $t$

{\bf Theorem.} If $0\in{\rm dom}(\Phi^\prime)$ and 
$\bigcap_{t\in\RR} {\rm Fix}(\Phi^t) \ne \emptyset$
then the generator $\Phi^\prime$ is of Kaup's type:

${\rm dom}(\Phi^\prime)$ subtriple in ${\bf E}$, \ \ 
$\Phi^\prime(z) = a - \{ zaz\} + iAz$ closed.

{\bf Proof.} ${\rm dom}(\Phi^\prime) = \{ x: t\mapsto U_t$ diff.$\} =
{\rm dom}(\Lambda^\prime)$ dense in ${\bf E}$, $\Lambda^\prime$ closed lin. op. 
%by Hille-Yosida.

$\Phi^t(z+e)-e = (A_{t,z} + B_t)^{-1} C_t z$

$\Psi^\prime(z+e) = 
- (A_{t,z} + B_t)^{-1} \big[ {d\over dt} (A_{t,z} + B_t) \big] (A_{t,z} + B_t)^{-1} C_t \big\vert_{t=0} +
(A_{t,z} + B_t)^{-1} {d\over dt} C_t z \big\vert_{t=0} $

$\Lambda^\prime(z) = - B_t ^{-1} \big[ {d\over dt} B_t\big] B_t^{-1} C_t \big\vert_{t=0} +
B_t^{-1}  \big[ {d\over dt} B_t\big] \big\vert_{t=0}$


Let $x_n\to x$, $\Psi^\prime(x_n)\to y$.

$z_n:=x_n-e$, \ 
 
$\ldots \ \ \ldots \ \ \ldots \ \ \ldots \ \ \ldots \ \ \ldots \ \ \ldots \ \ $
 
Let $x \in {\rm dom}(\Psi^\prime)$, \  $\Vert \Vert =1$, \ $\varphi\in{\bf E}^*$, \ 
$\langle \varphi , x \rangle = \Vert \varphi \Vert =1$

$\Phi^\prime$ is a TANGENT vector field to $\partial{bf B}$

$0 = {\rm Re} \langle \varphi \circ \overline{\kappa} , \Phi^\prime(\kappa x) \rangle$ 
\quad $\Leftarrow \vert \kappa \vert =1$

$\zeta \mapsto \langle \varphi , \Phi^\prime(\zeta x) \rangle = 
\sum_{n=0}^\infty \alpha_n \zeta^n$ \ holomorphic

${\rm Re}\big( \overline{\kappa} \sum_{n=0}^\infty \alpha_n \kappa^n\big) =0$

$\sum_{n=0}^\infty (\alpha_n \kappa^{n-1} + \overline{\alpha_n} \kappa^{1-n} ) =0$ \ 
$( \vert \kappa \vert =1 )$

$\sum_{n=-\infty}^\infty \beta_n \kappa^n =0$ \quad $\beta_n =\alpha_{n+1}$ $(n\ge 2)$, \ \  
$\beta_n =\overline{\alpha_{1-n}}$ $(n\le -2)$, \ \  

\hskip32mm
$\beta_1 =\alpha_2 + \overline{\alpha_0}$, \ \ $\beta_{-1} =\alpha_0 + \overline{\alpha_2}$, \ \ 
$\beta_0 =\alpha_1 + \overline{\alpha_1}$ 

$\alpha_n =0$ $(\vert n \vert \ge 2)$, \ \ $\alpha_1 + \overline{\alpha_1} =0$, \ \ 
$\alpha_2 = -\overline{\alpha_0}$ 

CONSIDER \ $\Omega() := \Phi^\prime(x) - \{ xbx \}$ \ INSTEAD OF $\Phi^\prime$, 
\quad $b:= \Psi^\prime(0) ={d\over dt} a(t) \big\vert_{t=0}$

This is also tangent to  $\partial{bf B}$ with $\Omega(0)=0$

$\Omega(\zeta x) = \zeta \Omega(x)$ HOMOGENITY


 










\bigskip
\hrule
\bigskip

{\bf SPIN FACTOR}

$\big( {\bf E}, \langle \cdot \vert \cdot \rangle \big)$ Hilbert space, \ \ $x\mapsto \overline{x}$ conjugation,
\quad $\langle x \vert y \rangle^{-} = \langle \overline{x} \vert \overline{y} \rangle$

$\{ xay \} = \langle x \vert a \rangle y + \langle y \vert a \rangle x -
\underbrace{\langle x \vert \overline{y} \rangle}_{\langle y \vert \overline{x} \rangle} \overline{a}$

$\big[\,$ TRIPOTENTS$\,\big] =
\Big\{ \lambda e: \ e\in{\rm Re}({\bf E}),\ \lambda\in\TT, \ \langle e\vert e\rangle =1 \Big\} \ \cup$

\vskip-4mm $\phantom{x}$ \hskip31mm
$\cup \ \Big\{ u+iv: \ u,v\in{\rm Re}({\bf E}),\ \langle u\vert u\rangle =\langle v\vert v\rangle =1/2,\
\langle u\vert v\rangle = 0\Big\}$

$U_t = \kappa_t V_t$: \ \  $V_t$ \ \ real $\langle\cdot\vert\cdot\rangle$-unitary, \
${\rm Re}({\bf E}) \to {\rm Re}({\bf E})$, $\kappa_t \in{\TT}$.




{\bf Problem.} $z\in {\bf D} \Rightarrow^{\hbox{?}} \overline{z} \in{\bf D}$ \quad $\big($i.e.
$t\mapsto U_t z$ diff. $\Rightarrow^{\hbox{?}}$ $t\mapsto U_t \overline{z}$ diff.$\big)$ \quad [YES]


{\bf Lemma.} \ $\exists\, x$ \ \ $t\mapsto U_t x, U_t\overline{x}$ diff. \ $\Longrightarrow$ \quad
$\exists\, t\mapsto\eps_t \in \{ \pm1\}$ \quad $t\mapsto\eps_t\kappa_t$ diff.


{\bf Proof.} $t\mapsto\overline{ U_t \overline{x} } = \overline{ \kappa_t V_t \overline{x} } =
\overline{\kappa_t} V_t x$ diff.

$t\mapsto \langle \kappa_t V_t x \vert \overline{\kappa_t} V_t x \rangle = \kappa_t^2$ diff.

$\forall\, h\in\RR$ \quad %$\exists\, I_h \ni t\mapsto \eps_{h,t} \in\{\pm1\}$ \quad
$\exists \, I_h$ open intv. around $h$, \qquad
${\rm Re}(\kappa_t^2 / \kappa_h^2) >0$ $(t\in I_h)$

$\ldots, J_{-2},J_{-1},J_0,J_1,J_2,\ldots$ \ \ chain of intervals \ \ $J_k \subset I_{h_k}$ $(k=0,\pm1,\ldots)$

$\exists\, k \mapsto \nu_k \in\{\pm1\}$ \qquad  $\eps_t := \nu_k {\rm sgn}(\kappa_t/\kappa_h)$ $(t\in J_k)$ well-def. and suits

{\bf Corollary.} ${\bf F} \cap {\rm conj}({\bf F}) \ne 0$ $\Rightarrow$ ${\bf F} = {\rm conj}({\bf F})$

{\bf Proof.} $0 \ne x\in {\bf F} \cap {\rm conj}({\bf F})$ $\Rightarrow$ $t\mapsto U_t x, U_t \overline{x}$ diff.
$\Rightarrow$ $t\eps_t \kappa_t, \eps_t \kappa_t^{-1}$ diff.

$z\in{\bf F}$
$\Rightarrow$ $t\mapsto \eps_t V_t z = \eps_t \kappa_t^{-1} U_t z$ diff.
$\Rightarrow$ $t\mapsto {\rm conj}\big( \eps_t \kappa_t^{-1} V_t z \big) = U_t \overline{z}$ diff.

{\bf Proposition.} ${\bf F}$ is closed under conjugation in any case.

{\bf Proof.} The only case of a JB$^*$-subtriple ${\bf H}$ such that ${\bf H} \cap {\rm conj}({\bf H}) = 0 $ is
if ${\bf H}$ is a Hilbert space spanned by a collinear grid $\{ 2^{-1/2} ( u_k + i v_k) : k\in{\cal K}\}$
where $\{  a_k, b_k: k\in{|\cal K}\}$ is $\langle \cdot\vert\cdot\rangle$-orthononormed.
Also ${\rm TRIP}({\bf H}) = \{ w + i T(w): w\in {\bf G}, \langle w\vert w\rangle =1/2\}$ with some subspace
${\bf G} \subset {\rm Re}({\bf E})$ and an isometry $T: {\rm Sphere}({\bf G}) \to {\rm Re}({\bf E})$.
The case ${\bf F} ={\bf H}$ is impossible: \ then $t\mapsto a_t = w_t+ i T(w_t)$ diff. $\Rightarrow$
$t\mapsto \overline{a_t} = w_t- i T(w_t)$ diff. $\Rightarrow$ $\{ a_t, \overline{a_t} : t\in\RR\} \subset {\bf F}$.




%{\bf Corollary.}  \ $\exists\, x$ \ \ $t\mapsto U_t x, U_t\overline{x}$ diff. \ $\Longrightarrow$ \quad ${\bf F}$ \ is closed for conjugation



{\bf Assumption without loss of gen.:} $U_t = \kappa_t V_t$, \ \ $t\mapsto \kappa_t$ diff.



{\bf Notation:} \ ${\bf F}^\perp := \{ x\in{\bf E}: \langle x \vert {\bf F}\rangle = 0\}$. \quad $(\ne {\bf F}^{\perp{\rm Jordan}}$)

{\bf Proposition.} ${\bf E} = {\bf F}$ \ (i.e. ${\bf F}^\perp =0$).

{\bf Proof.}
${\bf F} = {\rm conj}({\bf F})$ $\Rightarrow$ ${\bf F}^\perp = {\rm conj}({\bf F}^\perp)$ spin factor. \ \
${\rm dim}({\bf F}^\perp) >0$ $\Rightarrow$ $\exists\, y\in {\bf F}^\perp \quad 0\ne y = \overline{y}$

Calculate  $t\mapsto \Phi^t (y) = M_{a(t)} \circ U_t y$.

$M_a (x) = a + B(a)^{1/2} [1+L(x,a)]^{-1} x$, \quad
$B(a) = 1 - 2L(a) + Q_a^2: z\mapsto z - 2\{ a a z\} + \{ a \{ aza \} a\}$

$y\in{\bf F}^\perp, \ a\in{\bf F}$ $\Rightarrow$ \qquad $\langle y \vert f \rangle = \langle y \vert \overline{f} \rangle =0$ $(f\in{\bf F})$

$\{ fg y \} = \langle f\vert g\rangle y + \langle y\vert g\rangle f - \langle y\vert \overline{f}\rangle \overline{g} = \langle f\vert g\rangle y$, \
$\{ f y g \} = \langle f\vert y\rangle g + \langle g \vert y\rangle f - \langle g\vert \overline{f}\rangle \overline{y} = - \langle g\vert \overline{f}\rangle \overline{y}$


$x_1 + y_1 = (1 + L(y,a))^{-1} y$

$y=(1+L(y,a))(x_1+y_1) = x_1+y_1 + \{ yax_1 \} + \{ yay_1\}$

$0= x_1 - \langle y \vert \overline{y_1} \rangle \overline{a}$ \ \ (${\bf F}$-component),
\qquad
$y = y_1 + \langle x_1 \vert a \rangle y$ \ \ (${\bf F}^\perp$-component)

$\gamma = \gamma(y, y_1) := \langle y \vert \overline{y_1}\rangle = \langle y_1 \vert \overline{y}\rangle $

$x_1 = \langle y \vert \overline{y_1} \rangle \overline{a} = \gamma \overline{a}$, \qquad
$y_1 = ( 1 - \langle x_1 \vert a\rangle ) y =
(1 - \gamma \langle \overline{a} \vert a\rangle ) y$

$\gamma = \langle y_1 \vert y \rangle = (1- \gamma \langle \overline{a} \vert a \rangle ) \langle y \vert \overline{y} \rangle$,
\quad $\Longrightarrow$ \quad
$\displaystyle{
\gamma = { \langle y \vert \overline{y} \rangle \over 1 + \langle \overline{a} \vert a \rangle \langle y \vert \overline{y} \rangle}
}$

$\displaystyle{
[ 1 + L(y,a)]^{-1} y = x_1 + y_1 = \gamma \overline{a} + (1-\gamma \langle a \vert \overline{a} \rangle ) y =
{ \langle y \vert \overline{y} \rangle \overline{a} + y
\over
1 + \langle \overline{a} \vert a \rangle \langle y \vert \overline{y} \rangle }
}$

$z\perp {\bf F}$ $\Rightarrow$ \ \
$B(a)z = z - 2 \{ aa z\} + \{ a \{ a z a\} a\} = z - 2 \langle a\vert a\rangle z + \vert \langle a \vert \overline{a} \rangle \vert^2 z$

$B(a)^{1/2}z = \beta(a) z$ \ \ $\beta(a) := \sqrt{1 - 2 \langle a\vert a\rangle + \vert \langle a \vert \overline{a} \rangle \vert^2}$

$U_t y = \kappa_t V_t y$, \quad $t\mapsto \langle U_t y \vert \overline{ U_t y} \rangle = \kappa_t^2 \langle y \vert \overline{y} \rangle$ diff.

$t \mapsto \Phi^t(y) = M_{a(t)} \circ U_t y = a(t) + B(a(t))^{1/2} [ 1 + L(U_t y,a(t))]^{-1} U_t y =$

\vskip-2mm $\phantom{x}$ \hskip13mm
$\displaystyle{
= a(t) + \beta(a(t)) { \langle y \vert \overline{y} \rangle \overline{a(t)} + U_t y
\over
1 + \langle \overline{a(t)} \vert a(t) \rangle \langle y \vert \overline{y} \rangle }
}
$

IF ${\rm dim}({\bf F}^\perp =1$ THEN $V_t y= y$ and $T_ty = \kappa_t y$ $\Longrightarrow$ ${\rm dim}({\bf F}^\perp) =1$ impossible


CASE \ ${\rm dim}({\bf F}^\perp) >1$

We can find $y\in{\bf F}^\perp$ with $0 \ne y \perp \overline{y}$

Calculate  $t\mapsto \Phi^t (x+y) = M_{a(t)} \circ U_t (x+y)$.

$M_a (x+y) = a + B(a)^{1/2} [1+L(x+y,a)]^{-1} (x+y)$, \quad
$B(a) = 1 - 2L(a) + Q_a^2: z\mapsto z - 2\{ a a z\} + \{ a \{ aza \} a\}$

$y\in{\bf F}^\perp, \ a\in{\bf F}$ $\Rightarrow$ \qquad $\langle y \vert f \rangle = \langle y \vert \overline{f} \rangle =0$ $(f\in{\bf F})$

$\{ fg y \} = \langle f\vert g\rangle y + \langle y\vert g\rangle f - \langle y\vert \overline{f}\rangle \overline{g} = \langle f\vert g\rangle y$, \
$\{ f y g \} = \langle f\vert y\rangle g + \langle g \vert y\rangle f - \langle g\vert \overline{f}\rangle \overline{y} = - \langle g\vert \overline{f}\rangle \overline{y}$


$x_1 + y_1 = (1 + L(x+y,a))^{-1} (x+y)$

$x+y=(1+L(x+y,a))(x_1+y_1) = x_1+y_1 + \{ x a x_1 \} + \{ xa y_1\} + \{ yax_1 \} + \{ yay_1\}$


$x= x_1 + \{ xa x_1\} - \langle y \vert \overline{y_1} \rangle \overline{a}$ \ \ (${\bf F}$-component),
\qquad
$y = y_1 + \langle x \vert a \rangle y_1 + \langle x_1 \vert a \rangle y$ \ \ (${\bf F}^\perp$-component)





$\gamma_0 = \gamma_0(x_1, a) := \big( 1 - \langle x_1 \vert a \rangle \big) / ( 1 + \langle x \vert a \rangle \big)$

$y_1 = \gamma_0 y$

\DEL{

$x_1 = \langle y \vert \overline{y_1} \rangle \overline{a} = \gamma \overline{a}$, \qquad
$y_1 = ( 1 - \langle x_1 \vert a\rangle ) y =
(1 - \gamma \langle \overline{a} \vert a\rangle ) y$


\vskip3cm


$\gamma = \langle y_1 \vert y \rangle = (1- \gamma \langle \overline{a} \vert a \rangle ) \langle y \vert \overline{y} \rangle$,
\quad $\Longrightarrow$ \quad
$\displaystyle{
\gamma = { \langle y \vert \overline{y} \rangle \over 1 + \langle \overline{a} \vert a \rangle \langle y \vert \overline{y} \rangle}
}$

$\displaystyle{
[ 1 + L(y,a)]^{-1} y = x_1 + y_1 = \gamma \overline{a} + (1-\gamma \langle a \vert \overline{a} \rangle ) y =
{ \langle y \vert \overline{y} \rangle \overline{a} + y
\over
1 + \langle \overline{a} \vert a \rangle \langle y \vert \overline{y} \rangle }
}$

$z\perp {\bf F}$ $\Rightarrow$ \ \
$B(a)z = z - 2 \{ aa z\} + \{ a \{ a z a\} a\} = z - 2 \langle a\vert a\rangle z + \vert \langle a \vert \overline{a} \rangle \vert^2 z$

$B(a)^{1/2}z = \beta(a) z$ \ \ $\beta(a) := \sqrt{1 - 2 \langle a\vert a\rangle + \vert \langle a \vert \overline{a} \rangle \vert^2}$

$U_t y = \kappa_t V_t y$, \quad $t\mapsto \langle U_t y \vert \overline{ U_t y} \rangle = \kappa_t^2 \langle y \vert \overline{y} \rangle$ diff.

$t \mapsto \Phi^t(y) = M_{a(t)} \circ U_t y = a(t) + B(a(t))^{1/2} [ 1 + L(U_t y,a(t))]^{-1} U_t y =$

\vskip-2mm $\phantom{x}$ \hskip13mm
$\displaystyle{
= a(t) + \beta(a(t)) { \langle y \vert \overline{y} \rangle \overline{a(t)} + U_t y
\over
1 + \langle \overline{a(t)} \vert a(t) \rangle \langle y \vert \overline{y} \rangle }
}
$

\vskip 3cm





$y_1 = \gamma\, y$, \qquad $x_1 = \langle y \vert \overline{y_1} \rangle \overline{a} =
\overline{\gamma} \langle y \vert \overline{y} \rangle \overline{a}$

$\gamma = 1 - \langle x_1 \vert a \rangle =
1 - \overline{\gamma} \langle y \vert \overline{y} \rangle  \langle \overline{a} \vert a\rangle$



}%ENDDEL



Consider vectors $y$ with $0 \ne y \perp \overline{y}$: \quad $x = x_1 + \{ xax_1 \} - \big\langle y \big\vert \overline{\gamma_0 \strut \, y} \big\rangle \overline{a}
= x_1 + \{ x a x_1\}$

$x_1 = [1 + L(x,a)]^{-1} x$, \qquad
$\displaystyle{
y_1 = { 1 - \langle [1 + L(x,a)]^{-1} x \vert a \rangle
\over
1 + \langle  x \vert a \rangle}
= \gamma(x,a) y}$


$x_2 + y_2 = B(a)^{1/2} (x_1 +y_1)$

$M_a(x+y) = a + B(a)^{1/2} (x_1+y_1) = a + B(a)^{1/2} \big( [1 + L(x,a)]^{-1} x + \gamma(x,a) y \big] =$

\vskip-4mm $\phantom{x}$ \hskip 14mm
$= M_a(x) + \gamma(x,a) B(a)^{1/2} y$ \qquad if $y\perp \overline{y} \in{\bf F}^\perp$

$z\perp {\bf F}$ $\Rightarrow$ \ \
$B(a)z = z - 2 \{ aa z\} + \{ a \{ a z a\} a\} = z - 2 \langle a\vert a\rangle z + \vert \langle a \vert \overline{a} \rangle \vert^2 z$

$B(a)^{1/2}z = \beta(a) z$ \ \ $\beta(a) := \sqrt{1 - 2 \langle a\vert a\rangle + \vert \langle a \vert \overline{a} \rangle \vert^2}$

If $y\perp \overline{y} \in{\bf F}^\perp$ then \ $U_t y \in {\bf F}^\perp$,
$\langle U_t y \vert \overline{U_t y} \rangle = \langle \kappa_t V_t \vert \overline{\kappa_t V_t y} \rangle = \kappa_t^2 \langle y \vert \overline{y} \rangle =0$,

\vskip-4mm
$\Phi^t(x+y) = M_a(U_t x + U_t y) = M_{a(t)}(U_t x) + \beta(a(t)) \gamma(U_t x,a(t)) U_t y =$

\vskip-4mm $\phantom{x}$ \hskip 12.5mm
$= \Phi^t(x) + \beta(a(t)) \gamma(U_t x,a(t)) U_t y$

$\gamma(0,a) \equiv 0$, $t\mapsto a(t)$ diff. $\Rightarrow$

\vskip-4mm
$t \mapsto \Phi^t(y) = \underbrace{\Phi^t(0)}_{a(t)} + \beta(a(t)) y$ \ diff. \ \ whenever  \ \ $y\perp \overline{y} \in {\rm Ball}({\bf F}^\perp)$


Thus \ $0 \ne y \in {\bf F}^\perp =0$ \ contradiction if we assume \ ${\rm dim}({\bf F}^\perp) >1$





\bigskip
\hrule
\bigskip


{\bf Proof}.
${\bf F} = {\rm conj}({\bf F})$ $\Rightarrow$ ${\bf F}^\perp = {\rm conj}({\bf F}^\perp)$ spin factor. \ \
${\rm dim}({\bf F}^\perp) >1$ $\Rightarrow$ $\exists\, y\in {\bf F}^\perp \quad 0\ne y\perp \overline{y}$

Calculate the effect of $\Phi^t = M_{a(t)} \circ U_t$ on ${\bf F}^\perp$.

$M_a (x) = a + B(a)^{1/2} [1+L(x,a)]^{-1} x$, \quad
$B(a) = 1 - 2L(a) + Q_a^2: z\mapsto z - 2\{ a a z\} + \{ a \{ aza \} a\}$

$y\in{\bf F}^\perp$ $\Rightarrow$ \qquad $\langle y \vert f \rangle = \langle y \vert \overline{f} \rangle =0$ $(f\in{\bf F})$

$\{ fg y \} = \langle f\vert g\rangle y + \langle y\vert g\rangle f - \langle y\vert \overline{f}\rangle \overline{g} = \langle f\vert g\rangle y$, \
$\{ f y g \} = \langle f\vert y\rangle g + \langle g \vert y\rangle f - \langle g\vert \overline{f}\rangle \overline{y} = - \langle g\vert \overline{f}\rangle \overline{y}$


$x_1 + y_1 = (1 + L(x+y,a))^{-1} (x+y)$

$x+y=(1+L(x+y,a))(x_1+y_1) = x_1+y_1 + \{ xax_1 \} + \{ xay_1\} + \{ yax_1 \} + \{ yay_1\}$

$x= x_1 + \{ xa x_1 \}  - \langle y \vert \overline{y_1} \rangle \overline{a}$,
\qquad
$y = y_1 + \langle x \vert a \rangle y_1 + \langle x_1 \vert a \rangle y$

$\displaystyle{
y_1 = {1 - \langle x_1 \vert a\rangle \over 1 + \langle x \vert a\rangle} y = \gamma_0(x_1,a) y}$

Consider vectors $y$ with $y \perp \overline{y}$: \quad $x = x_1 + \{ xax_1\} - \langle y \vert \overline{\gamma_0 y} \rangle \overline{a} =
x_1 + \{ xax_1\}$

$x_1 =  [1 + L(x,a)]^{-1} x$, \quad
$\displaystyle{
y_1 = {1 - \langle [1 + L(x,a)]^{-1} x \vert a\rangle \over 1 + \langle x \vert a\rangle} y \ \ \underline{ = \gamma(x,a) y} }$
\quad $(y\perp \overline{y})$

$x_2 + y_2 = B(a)^{1/2} (x_1 +y_1)$

$M_a(x+y) = a + B(a)^{1/2} (x_1+y_1) = a + B(a)^{1/2} \big( [1 + L(x,a)]^{-1} x + \gamma(x,a) y \big] =$

\vskip-4mm $\phantom{x}$ \hskip 14mm
$= M_a(x) + \gamma(x,a) B(a)^{1/2} y$ \qquad if $y\perp \overline{y} \in{\bf F}^\perp$

$z\perp {\bf F}$ $\Rightarrow$ \ \
$B(a)z = z - 2 \{ aa z\} + \{ a \{ a z a\} a\} = z - 2 \langle a\vert a\rangle z + \vert \langle a \vert \overline{a} \rangle \vert^2 z$

$B(a)^{1/2}z = \beta(a) z$ \ \ $\beta(a) := \sqrt{1 - 2 \langle a\vert a\rangle + \vert \langle a \vert \overline{a} \rangle \vert^2}$

If $y\perp \overline{y} \in{\bf F}^\perp$ then \ $U_t y \in {\bf F}^\perp$,
$\langle U_t y \vert \overline{U_t y} \rangle = \langle \kappa_t V_t \vert \overline{\kappa_t V_t y} \rangle = \kappa_t^2 \langle y \vert \overline{y} \rangle =0$,

\vskip-4mm
$\Phi^t(x+y) = M_a(U_t x + U_t y) = M_{a(t)}(U_t x) + \beta(a(t)) \gamma(U_t x,a(t)) U_t y =$

\vskip-4mm $\phantom{x}$ \hskip 12.5mm
$= \Phi^t(x) + \beta(a(t)) \gamma(U_t x,a(t)) U_t y$

$\gamma(0,a) \equiv 0$, $t\mapsto a(t)$ diff. $\Rightarrow$

\vskip-4mm
$t \mapsto \Phi^t(y) = \underbrace{\Phi^t(0)}_{a(t)} + \beta(a(t)) y$ \ diff. \ \ whenever  \ \ $y\perp \overline{y} \in {\rm Ball}({\bf F}^\perp)$


Thus \ $0 \ne y \in {\bf F}^\perp =0$ \ contradiction if we assume \ ${\rm dim}({\bf F}^\perp) >1$



$x_1 := \Phi^t(x)$, \ $y_1 := \beta(a(t)) \gamma(U_t x,a(t)) U_t y$ \qquad $\langle y_1 \vert \overline{y_1} \rangle =0$

\vskip-4mm
$\Phi^{t+h}(x+y) =\Phi^h\big( \Phi^t(x+y) \big) = \Phi^h (x_1+y_1) = \Phi^h(x_1) + \beta(a(h)) \gamma(U_h x_1,a(h)) U_h y_1 =$

\vskip-4mm $\phantom{x}$ \hskip 12.5mm
$= \Phi^{t+h}(x) + \beta(a(h)) \gamma\big( (U_h \Phi^t(x),a(h)\big) \beta(a(t)) \gamma\big((U_t x,a(t)\big) U_h U_t y $

$\Phi^{t+h}(x+y) = \Phi^{t+h}(x) + \beta(a(t+h)) \gamma\big(U_{t+h} x,a(t+h)\big) U_{t+h} y$

$\displaystyle{
U_h U_t y = {  \beta(a(h)) \gamma\big( U_h \Phi^t(x),a(h)\big) \beta(a(t)) \gamma\big( U_t x,a(t)\big)
\over
\beta(a(t+h)) \gamma\big( U_{t+h} x,a(t+h)\big) } U_{t+h}  }$  \ \ $\big({\rm Span} \{$admissible $y\}\!=\! {{\bf F}}^\perp \big)$

$x:=0$ $\Rightarrow$ \quad $x_1=\Phi^t(x) = a(t)$, $\Phi^h(x_1)=a(t+h)$, \ $\gamma(0,a)= 1$

$U_h U_t = \lambda(h,t) U_{t+h}$, \qquad $\displaystyle{ \lambda(h,t) :=
{  \beta(a(h)) \gamma\big( U_h a(t),a(h)\big) \beta(a(t))
\over
\beta(a(t+h))  }  }$




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\DEL{
$x_1 := \Phi^t(0) = a(t)$, \ $y_1 := \beta(a(t))$ \qquad\qquad $\langle y_1 \vert \overline{y_1} \rangle =0, \quad
\gamma\big(0,a(t)\big)$

\vskip-4mm
$\Phi^{t+h}(x+y) =\Phi^h\big( \Phi^t(x+y) \big) = \Phi^h (x_1+y_1) = \Phi^h(x_1) + \beta(a(h)) \gamma(U_h x_1,a(h)) U_h y_1 =$

\vskip-4mm $\phantom{x}$ \hskip 12.5mm
$= \Phi^{t+h}(x) + \beta(a(h)) \gamma\big( (U_h \Phi^t(x),a(h)\big) \beta(a(t)) \gamma\big((U_t x,a(t)\big) U_h U_t y $

$\Phi^{t+h}(x+y) = \Phi^{t+h}(x) + \beta(a(t+h)) \gamma\big(U_{t+h} x,a(t+h)\big) U_{t+h} y$

$\displaystyle{
U_h U_t y = {  \beta(a(h)) \gamma\big( U_h \Phi^t(x),a(h)\big) \beta(a(t)) \gamma\big( U_t x,a(t)\big)
\over
\beta(a(t+h)) \gamma\big( U_{t+h} x,a(t+h)\big) } U_{t+h}  }$  \ \ $\big({\rm Span} \{$admissible $y\}\!=\! {{\bf F}}^\perp \big) $
}%ENDDEL
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%














\bigskip
\hrule
\bigskip


{\bf Fractional linear forms}

${\cal A}=\left[ \matrix{A &B \cr C &D \cr} \right] \in {\cal L}({\bf H}_1,{\bf H}_2)$, \qquad

${\cal F}({\cal A}) : X \mapsto (AX+B)(CX+D)^{-1} = \big[ {\cal A} (X\ 1)^{{\rm T}} \big]_1 \big[ {\cal A} (X\ 1)^{{\rm T}} \big]_2^{-1}$

${\cal F}({\cal A} {\cal B}) = {\cal F}({\cal A}) \circ {\cal F}({\cal B})$

$M_a = {\cal F}({\cal M}_a$, \quad ${\cal M}_a =
{\rm diag}\left( \matrix{(1-aa^*)^{-1/2} \cr (1-a^*a)^{-1/2} \cr} \right) \left[ \matrix{1 & a \cr a^* & 1 \cr} \right]$

Surj. lin. isom: $X\mapsto UXV^*$, \quad unitary $U \in {\cal L}({\bf H}_1)$, $V \in {\cal L}({\bf H}_2)$

$\Phi^t := {\cal F}({\cal A}_t)$, \qquad $\big[ \phi^t: t\in\RR \big]$ str.cont,1prg.

${\cal A}_t = {\cal M}_{a(t)} {\rm diag}( U_t,V_t)$

{\bf Attention:} $U\otimes V^* = {\cal F}\big( {\rm diag}(U,V) \big) = {\cal F}\big( \kappa {\rm diag}(U,V) \big)$ with any $\kappa\in\TT$

Adjusted str.cont.: [Stach\'o JMAA 2010, Cor. 2.6]

$\exists\, t\mapsto \kappa(t)\in\TT \quad t\mapsto \kappa(t) U_t, t\mapsto \kappa(t) V_t$ str.cont.

\bigskip
\hrule
\bigskip



{\bf Case of} ${\bf E} = {\cal L}({\bf H}_1,{\bf H}_2)$ {\bf with} $r:={\rm dim}({\bf H}_2)<\infty$

Assumptions without loss of gen. {\it up to M\"obius equ.}:

{\bf (1)} ${\cal A}_t {\cal A}_h = \lambda(t,h) {\cal A}_{t+h}$, \qquad $\lambda(t,h)\in\TT =\{ \zeta\in\CC: \vert \zeta \vert =1 \}$

{\bf (2)} ${\cal A}_t=\left[ \matrix{ A_t & B_t \cr C_t & D_t \cr} \right]$ \qquad $t\mapsto A_t,B_t,C_t,D_t$ str.cont.

{\bf (3)}$^*$ $\exists$ common fixed point (by reflexivity): \ ${\cal F}({\cal A}_t)E=E$ $(t\!\in\!\RR)$.

\vskip-5mm $\phantom{x}$ \hskip8mm
Interesting only: \ \ $\Vert E\Vert=1$, \ \ $E$ TRIP

$\lambda(t,h) = {\cal A}_{-(t+h)} {\cal A}_t {\cal A}_h$ cont. in $t,h$  \quad (prod. of unif.bded. str.cont. lin. maps) %[Engel-Nagel])

$S_t:=[ {\cal A}_t (E\ 1)^{{\rm T}} \big]_2$.

${\cal A}_t (E\ 1)^{{\rm T}} = (E\ 1)^{{\rm T}} S_t$,

$S_t S_h= \lambda(t,h) S_h S_t$


{\bf Proposition}. $[S_t : t\in\RR]$ Abelian family, \quad $\lambda(t,h)\equiv \lambda(h,t)$.


${\rm trace}\, AB={\rm trace}\, BA$.

${\rm trace}\, S_t S_h = \lambda(t,h) {\rm trace}\, S_h S_t =\lambda(t,h) {\rm trace}\, S_t S_h$.

$[\lambda(t,h)-1] {\rm trace}\, S_t S_h =0$

${\rm trace}\, S_t S_h \to {\rm trace}\, 1 ={\rm dim}({\bf H}_2)$ $(t\to 0)$.

$\exists\, \eps>0 \quad \lambda(t,h)=1 \ (\vert t\vert, \vert h\vert<\eps)$.

$S_t \smile S_h$ for $\vert t\vert, \vert h\vert<\eps$.

$u,v\in\RR$, $u/m, v/m \in (-\eps,\eps)$,

$S_u = \widetilde{\lambda} S_{u/m}^m$, \ $S_v = \widetilde{\mu} S_{v/m}^m$ \quad $\exists\, \widetilde{\lambda},\widetilde{\mu}\in \TT$,
$\Longrightarrow S_u \smile S_v$ \quad Q.e.d.

{\bf Remark:} In infinite dimensions, $AB=\lambda BA \ne 0 \not\Rightarrow A\smile B$ even if $\lambda\in\TT$.

Example: $A: e_n\mapsto e_{n+1}$ $(n=0,\pm1,\ldots)$ bilateral shift, \ $B: e_n\mapsto \lambda^n e_n$.

$S_t = \sigma_1(t) e_1\otimes e_1^* + \cdots + \sigma_r(t) e_r\otimes e_r^* +N$ \ \ Jordan normal form

$\sigma_k(t)\in\TT, \ r={\rm dim}({\bf H}_2)$, \ $N$ nilpotent, \ $N \smile \sigma_1(t) e_1\otimes e_1^* + \cdots + \sigma_r(t) e_r\otimes e_r^*$.

$S_t S_h = \lambda(t,h) S_{t+h}$

$\sigma_k(t)\sigma_k(h)=\lambda(t,h)\sigma_k(t+h)$ \quad $(k=1,\ldots,r)$,

$\lambda(t,h) = \sigma_k(t)\sigma_k(h)\sigma_k(t+h)^{-1}$.

$\kappa(t):=\sigma_1(t)^{-1}$, \quad $\widetilde{S}_t := \kappa(t) S_t$.

$\lambda(t,h) = \kappa_k(t)^{-1}\kappa_k(h)^{-1}\kappa_k(t+h)$.

$\widetilde{S}_t = e_1\otimes e_1^* + \sum_{k=2}^r \widetilde{\sigma}_k(t) e_k\otimes e_k^* + \widetilde{N}$ Jordan form, \qquad $\widetilde{\sigma}_k(t)=\kappa(t) \sigma_k(t)$.

$\widetilde{\sigma}_k(t) \widetilde{\sigma}_k(h) \widetilde{\sigma}_k(t+h)^{-1} =
\sigma_k(t)\sigma_k(t)\sigma_k(t+h)^{-1} \kappa_k(t)\kappa_k(t)\kappa_k(t+h)^{-1} =$

$\phantom{x} \hskip32mm = \sigma_k(t)\sigma_k(t)\sigma_k(t+h)^{-1} \lambda(t,h)^{-1} =1$.

$\widetilde{{\cal A}}_t := \kappa(t) {\cal A}_t$.

${\cal A}_t {\cal A}_h =\lambda(t,h) {\cal A}_{t+h} \Longrightarrow \widetilde{{\cal A}}_t \widetilde{{\cal A}}_h = \widetilde{{\cal A}}_{t+h}$.

{\bf Proposition}. $[\kappa(t)S_t : t\in\RR]$, $[\kappa(t){\cal A}_t : t\in\RR]$ str.cont.1prg.

{\bf Proof}. Proved: they are 1prg's.
%
Strong cont.: $t\mapsto \kappa(t) = \sigma_1(t)^{-1} = \langle S_t e_1 \vert e_1\rangle^{-1}$ cont.


\bigskip
\hrule
\bigskip


{\bf Assumption without loss of gen.:}

{\bf (1)$^*$} $\big[ {\cal A}_t: t\in \RR \big]$, $\big[ S_t: t\in \RR \big]$ str.cont.1prg.
\ \ $S_t := \big[ {\cal A}_t (E\ 1)^{{\rm T}}\big]_2^{-1} =(C_t E + D_t)^{-1}$

$\displaystyle{
{\cal A}^\prime:= {d\over dt}\Big\vert_{t=0} {\cal A}_t =
\left\{ \left[ \matrix{ x \cr y \cr} \right] \mapsto
{d\over dt}\Big\vert_{t=0} \left[ \matrix{ A_t & B_t \cr C_t & D_t \cr}\right]
\left[ \matrix{ x \cr y \cr} \right] \right\} }$

${\bf D} := {\rm dom}({\cal A}^\prime)$ dense lin. submanifold in ${\cal L}({\bf H}_1,{\bf H}_2)$

Observation: $t\mapsto \Phi^t(X)$ diff. \ whenever \
$\left[{Xy\atop y}\right] \in{\bf D}$ \ $\forall\, y\in{\bf H}_2$.

Proof: \ $X\in{\cal L}({\bf H}_1,{\bf H}_2)$ $\Longrightarrow$ \ \
since ${\rm dim}({\bf H}_2) <\infty$,

$t\mapsto{\cal A}_t \left[{X\atop 1}\right]$ diff. \ $\iff$ \
$t\mapsto{\cal A}_t \left[{X\atop 1}\right]y =
{\cal A}_t \left[{Xy\atop y}\right]$ diff. $\forall\, y\in{\bf H}_2$.


${\cal A}^\prime$ is of ${\bf H}_1\oplus{\bf H}_2$-split matrix form if
${\bf D} = {\bf D}_1 \oplus {\bf D}_2$

$r:={\rm dim}({\bf H}_2) <\infty$, $\{ e_1,\ldots,e_r\}$ lin.indep. basis in ${\bf H}_2$

${\cal A}^\prime$ is ${\bf H}_1\oplus{\bf H}_2$-split $\iff$ $0\oplus e_k =\left[ {0 \atop e_k} \right] \in {\bf D}$ $(k=1,\ldots,r)$.

{\bf Lemma}. $\exists\, F \in{\cal L}({\bf H}_1,{\bf H}_2)$ \quad $\Vert F\Vert<1$, \ \
${\cal M}_F {\cal A}^\prime {\cal M}_F^{-1}$ is ${\bf H}_1\oplus{\bf H}_2$-split.

{\bf Proof.} $E=\sum\limits_{k=1}^r g_k \otimes e_k^*$, \quad
$\{ g_1,\ldots,g_r\}$, $\{ e_1,\ldots,e_r\}$ ortn. \ ($E$ TRIP fixed point)

$g_k\oplus e_k = \left[{Ee_k \atop e_k}\right] \in {\bf D}$,
%\quad $(k=1,\ldots,r)$,
\qquad ${\bf D} =  {\bf D}_1^0\oplus 0 + \sum\limits_{k=1}^r \CC g_k\oplus e_k$, \quad
${\bf D}_1^0 := \{ x: x\oplus 0 \in {\bf D}\}$

$\exists ? \ \ F= \sum\limits_{k=1}^r f_j\otimes e_j^*$, \ \
$\exists ? \ \ h_k \in{\bf D}$ \qquad
${\cal M}_F (h_k\oplus e_k) =  0\oplus e_k$

${\cal M}_F := {\rm diag}\left[ \matrix{ (1-FF^*)^{-1/2} \cr (1-F^*F)^{-1/2} \cr}\right]
\left[ \matrix{ 1 & F \cr F^* & 1\cr} \right]$, \ \ ${\cal M}_F^{-1} = {\cal M}_{-F}$

${\rm dom}\big( {\cal M}_F {\cal A}^\prime {\cal M}_F^{-1} \big) = {\cal M}_F {\rm dom}({\cal A}^\prime) =
\left[ \matrix{ 1 & F \cr F^* & 1\cr} \right] {\bf D}$ \quad if \ $\Vert F\Vert <1$


$\left[ \matrix{ 1 & F \cr F^* & 1\cr} \right] h_k \oplus e_k =
\left[ \matrix{ 1 & F \cr F^* & 1\cr} \right] \left[ \matrix{ h_k \cr e_k \cr} \right] =
\left[ \matrix{ h_k + Fe_k \cr F^* h_k + e_k \cr} \right]$

Choice for $f_k,h_k$: \ \ $\Vert F \Vert<1$, \ $F^* h_k + e_k = 0$, \
$h_k = d_k + g_k$ with $d_k\in{\bf D}_1^0$

$H :=\sum\limits_{k=1}^r d_k \oplus e_k^*$, \ \
$H=E+\Delta$ with $\Delta := \sum\limits_{k=1}^r d_k \oplus e_k^*$, \


$(F^*H + 1) e_k = F^*h_k + e_k$, \quad
$\left[ \matrix{ 1 & F \cr F^* & 1\cr} \right] \left[ \matrix{ H \cr 1 \cr} \right] =
\left[ \matrix{ E + F \cr F^*H + 1 \cr} \right]$


Requirement: $F^*H= -1 (= -{\rm id}_{{\bf H}_2})$, \quad $\Vert F\Vert <1$.

Choice for $F$: \ $F:= H \Theta$, \ $\Theta\in{\cal L}({\bf H}_2)$.

$F^*H = \Theta^* H^*H$, \quad $\Theta=\Theta^* = -(H^*H)^{-1}$

%$G$ dense in ${\cal L}({\bf H}_1,{\bf H}_2)$ \quad
%(density of $\{ G(g_1,\ldots,g_r): g_1\in{\bf D}_1,\ldots,g_r\in{\bf D}_r\}$).

%$G^*G$ is dense in ${\cal L}_+({\bf H}_2)$.

$F^*F = \Theta^* H^* H \Theta = (H^* H)^{-1} H^* H (H^* H)^{-1} = (H^* H)^{-1}=$

$\phantom{x}$ \hskip10mm
$= \big[ (E+\Delta)^* (E+\Delta)\big]^{-1}$

 $\Rightarrow$ the requirement can be fulfilled.

{\bf Special choice for $\Delta$} to \ $M_F(E)$ TRIP

$\{ d_1,\ldots, d_r\} = \lambda\cdot$[ORTN] = $\lambda \Delta_0$ \ with \quad $\langle d_k \vert e_\ell\rangle =0$,

Possible since ${\bf D}_0^1$ is of {\it finite codim.}

$H^* H = (E+\Delta)^* (E+\Delta) = 1 + \vert\lambda\vert^2$ in ${\cal L}({\bf H}_2)$

$E, \Delta_0$ \ collinear TRIPs, \quad $F\in \RR E + \RR \Delta_0$, $\Rightarrow$
$M_F(E)$ TRIP.

\bigskip
\hrule
\bigskip


{\bf By means of M\"obius equivalence, we may assume:}

%$\left[ \matrix{A_t & B_t \cr C_t & D_t \cr} \right] \left[ \matrix{ E \cr 1 \cr} \right] =
%\left[ \matrix{ E S_t \cr S_t \cr} \right] = \left[ \matrix{ E \cr 1 \cr} \right] S_t$,
%\quad
{\bf (4)}
$\displaystyle{ {\cal A}^\prime:=
{d\over dt}\Big\vert_{t=0} \left[ \matrix{A_t & B_t \cr C_t & D_t \cr} \right] = \left[ \matrix{A & B \cr C & D \cr} \right] }$
\ \ ${\bf H}_1\oplus{\bf H}_2$-split,

$\phantom{x}$ \hskip5mm
$A: [{\bf H}_1{\rm -dense}] \,\to {\bf H}_1$, \
$C\in{\cal L}({\bf H}_1,{\bf H}_2)$, \ $D\in{\cal L}({\bf H}_2,{\bf H}_2)$.

${\cal A}_t= {\cal M}_{a(t)} {\rm diag}(U_t, V_t) =
{\rm diag}\left( \matrix{ [1-a(t)a(t)^*]^{-1/2} \cr [1-a(t)^*a(t)]^{-1/2} \cr} \right)
\left[ \matrix{ 1 & a(t) \cr a(t)^* & 1 \cr} \right] {\rm diag}(U_t, V_t)$.

$A_t = [1-a(t)a(t)^*]^{-1/2} U_t$, \quad $B_t=[1-a(t)a(t)^*]^{-1/2} a(t) V_t$,

$C_t=[1-a(t)a(t)^*]^{-1/2} a(t)^* U_t$, \quad $D_t=[1-a(t)a(t)^*]^{-1/2} V_t$


${\rm dom}({\cal A}^\prime) = {\bf D}_1 \oplus {\bf H}_2$, \qquad ${\bf D}_1= {\rm dom}(A)= {\rm dom} \big( {d\over dt}\big\vert_{t=0} U_t \big)$.

$t\mapsto a(t) = B_t D_t^{-1}$ is differentiable, \quad
$a(t) = t b + o(t)$ \ at $t=0$

${\cal A}^\prime = \left[ \matrix{ A^\prime & B^\prime \cr C^\prime & D^\prime \cr} \right] =
\left[ \matrix{ U^\prime & b \cr b^* & V^\prime \cr} \right]$, \quad
$U^\prime := {d\over dt}\big\vert_{t=0} U_t$, \  $V^\prime := {d\over dt}\big\vert_{t=0} V_t$.

\bigskip
\hrule
\bigskip


${\cal A}_t \left[ \matrix{ E \cr 1 \cr} \right] =
\left[ \matrix{A_t & B_t \cr C_t & D_t \cr} \right] \left[ \matrix{ E \cr 1 \cr} \right] =
\left[ \matrix{ E S_t \cr S_t \cr} \right] = \left[ \matrix{ E \cr 1 \cr} \right] S_t$

$[ S_t: t\in\RR]$ str.cont.1prg, \quad $S^\prime :=  {d\over dt}\big\vert_{t=0} S_t =
 {\rm gen}[ S_t: t\in\RR]$

$y\in{\bf H}_2$ \ $\Rightarrow$ \ $t\mapsto {\cal A}_t \left[ {Ey \atop y} \right] =
\left[ {E \atop 1} \right] S_t y$ \ diff., \quad
$\left[ {Ey \atop y} \right] \in {\rm dom}({\cal A}^\prime)$, \ \ $Ey \in{\bf D}_1$.

{\bf Projective translation:} \
${\cal T}:= \left[ \matrix{1 & E \cr  & 1 \cr} \right]$, \quad ${\cal T}^{-1}:= \left[ \matrix{1 & -E \cr  & 1 \cr} \right]$

${\cal B}_t := {\cal T}^{-1} {\cal A}_t {\cal T}$,
\quad ${\cal B}^\prime := {\cal T}^{-1} {\cal A} {\cal T}$

${\cal A}^\prime ={\rm gen} \big[ {\cal A}_t: t\in\RR\big]$, \quad ${\cal B}^\prime =
{\rm gen} \big[  {\cal B}_t : t\in\RR\big]$, \quad
${\rm dom}({\cal B}^\prime) = {\cal T}^{-1} ({\bf D}_1\oplus{\bf H}_2)$.

${\rm dom}({\cal B}^\prime) =
\big\{ [d \!-\! Ey] \oplus y  : d\in{\bf D}_1,\ y\in{\bf H}_2) \big\}
= {\bf D}_1 \oplus{\bf H}_2 (= {\rm dom}({\cal A}^\prime))$.


${\cal T}^{-1} \left[ \matrix{A_t & B_t \cr C_t & D_t \cr} \right] {\cal T} =
\left[ \matrix{1 & -E \cr  & 1 \cr} \right] \left[ \matrix{A_t & A_t E + B_t \cr C_t & C_t E + D_t \cr} \right] =
\left[ \matrix{1 & -E \cr  & 1 \cr} \right] \left[ \matrix{A_t & E S_t \cr C_t & S_t \cr} \right] =$

$\phantom{x} \hskip25mm = \left[ \matrix{A_t - E C_t & {\bf 0} \cr C_t & S_t \cr} \right]$.


${\cal B}^\prime  = {\cal T}^{-1} {\cal A}^\prime {\cal T} =
\left[ \matrix{A^\prime -EC^\prime & 0 \cr C^\prime & S^\prime \cr} \right] =
\left[ \matrix{U^\prime -Eb^* & 0 \cr b^* & b^*E + V^\prime \cr} \right]$

$W_t := \big[ {\cal B}_t \big]_{11}$ str.cont.1prg. \quad
$W^\prime ={\rm gen} [W_t: t\!\in\!\RR] = A^\prime -EC^\prime = U^\prime -Eb^*$

$S_t := \big[ {\cal B}_t \big]_{22}$ str.cont.1prg. \quad
$S^\prime = {\rm gen} [S_t: t\!\in\!\RR] = C^\prime E + D^\prime = b^*E + V^\prime$


Triangular lemma [Stach\'o JMAA 2016, Lemma 3.8] $\Rightarrow$

${\cal B}^\prime = {\rm gen}\bigg[ \underbrace{\left[ \matrix{W_t & 0 \cr \int_0^t  S_{t-h} C^\prime W_h \, dh & S_t\cr} \right] }_{{\cal B}_t} : t\in\RR \bigg]$

$\Psi^t = {\cal F} \left( {\cal B}_t \right) : \ X \mapsto
W_t X \big[ \int_0^t S_{t-h} C^\prime W_h X \, dh + S_t \big]^{-1}$,

${\cal A}^\prime = {\cal T} {\cal B}^\prime {\cal T}^{-1} = {\rm gen}\big[ {\cal A}_t :  t\in\RR\big]$, \quad $T:= {\cal F}({\cal T}): X \mapsto X+E$

$\Phi^t = {\cal F}\big({\cal A}_t \big) = {\cal F}\big( {\cal T} {\cal B}_t {\cal T}^{-1} \big) = T \circ \Psi_t \circ T^{-1}$

{\bf Closed integrated form:} \quad For  all $X\in{\rm Ball}\big({\cal L}({\bf H}_1,{\bf H}_2)\big)$,

\vskip-10mm
$$\Phi^t(X) = %E + \Psi^t (X \!-\! E) =
E + W_t (X \!-\! E) \bigg[ \int_0^t S_{t-h} \underbrace{C^\prime}_{b^*} W_h (X \!-\! E) \, dh + S_t \bigg]^{-1} .$$

\vskip-5mm
{\bf Vector fields}

$\Phi^t (X) \in {\cal L}({\bf H}_1,{\bf H}_2)$ \quad $\big[ {\bf H}_1 \to {\bf H}_2$ operators$\big]$

$t\mapsto \Phi^t (X)$ \ diff. \ \ $\iff$ \ \ $t\mapsto \Phi^t (X)y$ \ diff. $\forall\, y$
\quad ($\Leftarrow$ ${\rm dim}({\bf H}_2) < \infty$.)

If \ ${\rm ran}(X) \subset {\bf D}_1 \big( = {\rm dom}([{\cal A}^\prime]_{11}) \big)$ \ then

$t\mapsto \Phi^t (X)y = [A_t X + B_t] [C_t X + D_t]^{-1} y$ \quad diff. $\forall\, y$

$\Phi^\prime := {d \over dt} \big\vert_{t=0} \Phi^t$, \qquad
${\rm dom}(\Phi^\prime)= \{ X: {\rm ran}(X) \subset {\bf D}_1\}$

{\bf Kaup type formula} up to M\"obius equ.:

$\Phi^\prime(X)y =  {d \over dt} \big\vert_{t=0} [A_t X + B_t] [C_t X + D_t]^{-1}y =
[A^\prime X + B^\prime]y - X[C^\prime X + D^\prime]y  =$

$\phantom{x}$ \hskip20mm
$= \big[ b - X b^* X + U^\prime X - X V^\prime  \big]y$

\bigskip
\hrule
\bigskip

$\Phi^t(E)=E$ \ \ $(t\in \RR) \quad \Rightarrow \quad \Phi^\prime(E)=0$

$0 = b - E b^* E + U^\prime E - E V^\prime$

$\Phi^\prime(E)=0 \iff E^* \Phi^\prime(E)=0 \iff V^\prime = E^* U^\prime E + E^*b - b^*E$ \quad since \ $E^*E=1$
%\Rightarrow^{E^*E=1}\Rightarrow V^\prime = E^* U^\prime E + E^*b - b^*E \Rightarrow

${\cal B}^\prime  =
\left[ \matrix{U^\prime -Eb^* & 0 \cr b^* & b^*E + V^\prime \cr} \right] =
\left[ \matrix{U^\prime -Eb^* & 0 \cr b^* & E^* U^\prime E + E^*b \cr} \right]$













\bye
