\magnification=\magstep1

\def\ou{\overline{u}}

\def\RR{{\rm I\kern-1.6pt {\rm R}}}
\def\CC{{\rm \kern 3.5pt \vrule height 7.5pt
        width.7pt depth0pt \kern-3pt C \kern.5pt}}
\def\TT{{\rm T}\kern-4.8pt {\rm T} \kern2pt}

\noindent
Let $H$ be a Hilbert space, $E:={\cal L}(H)$.
We consider the mapping
$\phi:{\rm Ball}(E) \leftrightarrow
M:= \big\{ \big[{t\atop x}\big]\in E_+\times E:\ 
t^2- x^* x =1, \ t\ge 1\big\}$,
$$\phi(u) := \left[
{{(1+a^*a)(1-a^*a)^{-1}}
\atop
{2a(1-a^*a)^{-1}}} \right]
\quad{\rm with}\quad
\phi^{-1} \left[ {t\atop x}\right] =
x/(1+t) \ .$$



\noindent
\bf Remark. \rm
In the 1-dimensional commutative case $E=\CC$,
for the M\"obius transformation
$M_b(u):=(u+b)/(1-\overline{b}u)$
we have
$$\phi^\# M_b \left[ {t\atop x}\right] =
\phi\circ M_b \circ \phi^{-1} \left[ {t\atop x}\right] =
\left[{
{{1+\vert b\vert^2\over1-\vert b\vert^2} t +
{2b\over 1-\vert b\vert^2} x} \atop
{{2b\over 1-\vert b\vert^2} t +
{1+\vert b\vert^2\over1-\vert b\vert^2} x} } \right] .$$
Notice that $M_{{\rm tanh}(a)}$ is the exponential of
the vector field $[a-\overline{a}u^2]
{\partial\over\partial u}$. Hence
$$\phi^\# M_{{\rm tanh}(a)} \left[ {t\atop x}\right] =
\left[{
{ {\rm cosh}(2a) t + {\rm sinh}(2a) x }
\atop
{ {\rm sinh}(2a) t + {\rm cosh}(2a) x } 
}\right]
\qquad{\rm for} \quad
\left[ {t\atop x}\right] \in M \ .$$  
In particular
$\phi^\# M_b$ is the restriction of a 
linear mapping of $\CC^2$ to $M$.

\bigskip
We are going to study the analogues in the setting of
$E={\cal L}(H)$.

\bigskip
\noindent
\bf Notation. \rm For any 
$a\in {\cal L}(H)$ and 
$b\in{\rm Ball}({\cal L}(H))$ we write
$$A_a(u) := \big[ a-ua^*u\big]{\partial\over\partial u},
\qquad
M_b(u):= (1-bb^*)^{-1/2}(u+b)(1+b^*u)^{-1}(1-b^*b)^{1/2}
\ .$$
It is well-known that, for 
$b=[\exp(A_a)]0$, we have $M_b(u)=[\exp(A_a)]u$ \ 
$(\Vert u\Vert <1)$.

\noindent
Recall that, for fixed 
$a\in{\cal L}(H)$ and $u\in{\rm Ball}{\cal L}(H)$, 
the function $u_\tau := [\exp(\tau A_a)]u$ $(\tau\in\RR)$
is the solution of the initial value problem
${d\over d\tau} u_\tau = A_a(u_\tau)$, $u_0=u$.

\bigskip
\noindent
\bf Proposition. \it
Let $a\in{\cal L}(H)$ and $\left[ {t\atop x}\right]\in M$.
Then
$$\phi^\# A_a \left[ {t\atop x}\right] =
\left[{
{a^*x+x^*a} 
\atop
{a(1+t)+x(1+t)^{-1}x^*a}  
}\right] \ .$$
\bf Proof. \rm
Let us write
$$u:=\phi^{-1} \left[ {t\atop x}\right] =x(1+t)^{-1},\quad
\ou := A_a(u), \quad
G_\tau := \exp(\tau A_a) \ .$$
Then we have
$$\eqalign{
\phi^\# A_a \left[ {t\atop x}\right] &=
{d\over d\tau}\Big\vert_{\tau=0}
\phi \circ G_\tau \circ \phi^{-1} 
\left[ {t\atop x}\right] = 
{d\over d\tau}\Big\vert_{\tau=0}
\phi \circ G_\tau (u) =\cr
&= \phi^\prime (G_0 (u)) {d\over d\tau}\Big\vert_{\tau=0}
G_\tau (u) =
\phi^\prime (u) A_a(u) =\cr
&= \phi^\prime (u) \ou .\cr}$$ 
We calculate both $\phi^\prime(u)$ and $\ou$
in terms of $t,x$. For the first component of $\phi$,
$$\eqalign{
\phi_1(u) &= (1+u^*u)(1-u^*u)^{-1} = 
[(1-u^*u)+2u^*u] (1-u^*u)^{-1} =\cr
&= 1 + u^* \phi_2(u) \ .\cr}$$
Hence and since $\phi_1(u) =x$, 
$$\eqalign{
\phi_1^\prime(u)\ou &=
{d\over d\tau}\Big\vert_{\tau=0} 
(u+\tau \ou)^* \phi_2(u+\tau \ou) =
\ou^* \phi_2(u) + u^* \phi_2^\prime(u)\ou =\cr
&= \ou^* x + u^* \phi_2^\prime(u)\ou \ .\cr}$$ 
We can express $\phi_2^\prime(u)\ou$ in algebraic
terms of $u,\ou$ as follows:
$$\eqalign{
\phi_2^\prime(u)\ou &= 
{d\over d\tau}\Big\vert_{\tau=0} 
\phi_2(u+\tau \ou) =
{d\over d\tau}\Big\vert_{\tau=0}
2 (u+\ou)[1+(u+\ou)^*(u+\ou)]^{-1} =\cr
&=2\ou[1-u^*u]^{-1} + 
2u[1-u^*u]^{-1}(\ou^*u+u^*\ou)[1-u^*u]^{-1} \ .\cr}$$
Since 
$u=\phi^{-1}\left[{t\atop x}\right] = x(1+t)^{-1}$
and since $x^* x = t^2-1 = (t-1)(1+t)$,
here we have
$$\eqalign{
1-u^*u &= 1 - (1+t)^{-1} x^* x (1+t)^{-1} =\cr
&=1 - (1+t)^{-1} (t-1) = (1+t)^{-1} [(1+t)-(t-1)] =\cr
&= 2(1+t)^{-1} \ ,\cr
[1-u^*u]^{-1} &= {1\over 2} (1+t) \ , \cr
u[1-u^*u]^{-1} &= x(1+t)^{-1} {1\over 2} (1+t) =
{1\over 2} x \ .\cr}$$  
Hence we conclude
$$\eqalign{
\phi_2^\prime(u)\ou &=
2 \ou {1\over 2} (1+t) + 2\cdot {1\over 2} x
[\ou^* x (1+t)^{-1} + (1+t)^{-1} x^* \ou]
{1\over 2}(1+t) =\cr
&= \ou (1+t) + {1\over 2} x \ou^* x +
{1\over 2} x (1+t)^{-1} x^* \ou (1+t) \ .\cr}$$ 
We can express $\ou$ in terms of $t,x$ as
$$\ou = A_a(u) = a-ua^*u =
a - x(1+t)^{-1} a^* x(1+t)^{-1} \ .$$
Thus
$$\eqalign{
\phi_2^\prime(u)\ou &=
\overbrace{a(1+t)}^{(1)} 
\overbrace{-x(1+t)^{-1} a^* x}^{(2)} +\cr 
&\overbrace{+{1\over 2} x a^* x}^{(3)}
\overbrace{-{1\over 2} x (1+t)^{-1} x^* a (1+t)^{-1} x^*x }^{(4)} +\cr 
&\overbrace{+{1\over 2} x (1+t)^{-1} x^* a (1+t)}^{(5)} 
\overbrace{-{1\over 2} x (1+t)^{-1} x^*x (1+t)^{-1} a^*x }^{(6)}
\ .\cr}$$
The sum (2)+(3)+(6) vanishes because
$x^*x=(1+t)(t-1)$ and hence
$$\eqalign{
(2)+(3)+(6) 
&= x\left[ 
- (1+t)^{-1} + {1\over 2} - {1\over 2}(t-1)(1+t)^{-1}
\right] a^* x =\cr
&= x(1+t)^{-1} \left[
-1 + {1\over 2} + {1\over 2} t - {1\over 2} t + {1\over 2} 
\right] a^* x = 0 \ .\cr}$$
The sum (4)+(5) can also be simplified as
$$\eqalign{
(4)+(5) &=
{1\over 2} x (1+t)^{-1} x^* a \big[
-(1+t)^{-1} \!\!\!\!\!\! \overbrace{x^*x}^{(1+t)(t-1)} 
\!\!\!\!\!\! + (1+t) \big] =\cr
&= {1\over 2} x (1+t)^{-1} x^* a \left[
-(t-1) + (1+t) \right] =\cr
&= x(1+t)^{-1} x^* a \ .\cr}$$
Summing up $(1)+\cdots+(6)$, we get
$$\eqalign{
\phi_2^\prime (u)\ou &=
a (1+t) + x (1+t)^{-1} x^* a \ ,\cr
\phi_1^\prime (u)\ou &=
\ou^* x + u^* \phi_2^\prime(u) \ou =\cr
&= a^* x -(1+t)^{-1} x^* a (1+t)^{-1} +\cr
& \ \ + (1+t)^{-1} x^* \big[
a(1+t) + x(1+t)^{-1} x^*a ~big] =\cr
&= 
\overbrace{a^* x}^{(1)}
\overbrace{- (1+t)^{-1} x^* a (1+t)^{-1} x^*x}^{(2)} +\cr
&= \ \ \overbrace{+ (1+t)^{-1} x^* a (1+t)}^{(3)}
\overbrace{+ (1+t)^{-1} x^*x (1+t)^{-1} x^* a}^{(4)}
\ .\cr}$$ 
Using again the identity $x^*x =(1+t)(t-1)$, here we can write
$$\eqalign{
(2)+(3) &= (1+t)^{-1} x^*a \big[
-(1+t)^{-1} (1+t) (t-1) + (1+t) \big] =\cr
&= 2 (1+t)^{-1} x^* a \ ,\cr
(4) &= (1+t)^{-1} (1+t)(t-1) (1+t)^{-1} x^*a =
(t-1)(1+t)^{-1} x^*a =\cr
&= -\big[ (1+t) -2t \big] (1+t)^{-1} x^*a =
- x^* a + 2t (1+t)^{-1} x^* a \ .\cr}$$
Therefore
$$\eqalign{
\phi_1^\prime(u)\ou &= [(1)+(4)] + F(2)+(3)] =\cr
&= a^* x - x^* a + 2t(1+t)^{-1} x^* a + 2 (1+t)^{-1} x^* a =\cr
&= a^* x + (1+t)^{-1} \big[ -(1+t) + 2t +2 \big] x^* a =\cr
&= a^*x + x^* a \cr}$$
which completes the proof. \quad Qu.e.d. 

\bigskip
\noindent
\bf Corollary. \it If $x$ is a normal operator and 
$\left[ {t\atop x}\right] \in M$ then
$\{ x,x^*,t\}$ commute and $x^*x =(t-1)(1+t)$ whence
$$\phi^\# A_a \left[ {t\atop x}\right] =
\left[{ 
{a^* x + x^* a}
\atop
{a^* t + t a^*}
}\right] \ .$$  

\eject

\noindent
\bf The case of the real Jordan algebra of self-adjoint operators

\bigskip
\noindent
\rm
Let henceforth only
$E={\cal A}(H) := \{ a\in{\cal L}(H) :\ a=a^* \}$.
In this case
$$\phi^\# A_a \left[ {t\atop x}\right] =
\left[{ 
{a x + x a}
\atop
{a t + t a}
}\right] = 
\left[ \matrix{ 0 & L_a +R_a \cr L_a+R_a &0 \cr} \right]
\left[ {t\atop x}\right]$$   
with the left and right multiplication operators
$L_a : {\cal L}(H) \ni z \mapsto az$ respectivaly
$R_a : {\cal L}(H) \ni z \mapsto za$.
Since $L_a$ and $R_a$ commute, 
$\exp\big( \phi^\# A_a \big)$ has a rather simple
explicite expression. Straightforward calculations
with the power series
$\sum_{n=0}^\infty n!^{-1}
\left[ { {0 \ \  L_a +R_a} \atop {L_a+R_a \ \ 0} } \right]^n
\left[ {t\atop x}\right]$,  
yield the following.

\bigskip
\noindent
\bf Theorem. \it
Given any self-adjoint operator \ $a\in{\cal A}(H)$, the M\"obius transformation \ 
$M_{{\rm tanh}(a)}$ \ maps \ ${\rm Ball}({\cal A}(H))$ \ onto itself,
\ $\phi$ is a bijection between \ ${\rm Ball}({\cal A}(H))$ \ and \ 
$M\cap {\cal A}(H)^2 = \big\{ \big[ {t\atop x}\big]: \ 
t,x\in{\cal A}, \ t^2-x^2=1, \ t\ge 1 \big\}$ \  
and
$$\phi^\# M_{{\rm tanh}(a)} = 
\phi\circ [\exp A_a] \circ \phi^{-1} \left[ {t\atop x}\right] =
\left[{
{ {\rm cosh}(L_a+R_a) t + {\rm sinh}(L_a+R_a) x }
\atop
{ {\rm sinh}(L_a+R_a) t + {\rm cosh}(L_a+R_a) x }
}\right] \ .$$
For the terms 
${\rm cosh}(L_a+R_a) z$, ${\rm sinh}(L_a+R_a) z$
$(z=t,x)$
here we have
$$\eqalign{
&{\rm cosh}(L_a+R_a) z =
{1\over 2} [\exp (a)] z [\exp(a)] +
{1\over 2} [\exp (-a)] z [\exp(-a)] \ , \cr
&{\rm cosh}(L_a+R_a) z =
{1\over 2} [\exp (a)] z [\exp(a)] -
{1\over 2} [\exp (-a)] z [\exp(-a)] \ .\cr}$$
\bf Proof. \rm
For any $a,u\in{\cal A}(H)$, we have 
$A_a(u) = a-\{ uau \} \in{\cal A}(H)$ 
along with
$\phi_1(u)=(1+u^2)(1-u^2)^{-1}\in \in{\cal A}(H)$ and
$\phi_2(u)=2u(1-u^2)^{-1}\in \in{\cal A}(H)$.
Therefore the vector field $A_a$ is tangent to ${\cal A}(H)$ moreover
complete in ${\cal A}(H) \cap {\rm Ball}({\cal L}(H) =
{\rm Ball}({\cal A}(H)$ and
the vector field $\phi^\# A_a$ is complete in ${\cal A}(H)\cap M$.
Since $\exp(A_a) = M_{{\rm tanh}(a)}$, it follows
$$\phi^\# M_{{\rm tanh}(a)} =
\exp \phi^\# A_a = 
\exp \left.
\left[ \matrix{0 &L_a+R_a \cr L_a+R_a &0 \cr} \right] 
\right\vert M \ .$$
Here we have
\medskip
\noindent
$\displaystyle{
\exp 
\left[ \matrix{0 &L_a+R_a \cr L_a+R_a &0 \cr} \right] =
\sum_{n=0}^\infty {1\over n!} 
\left[ \matrix{0 &L_a+R_a \cr L_a+R_a &0 \cr} \right]^n =}$

\noindent
$\displaystyle{
= \sum_{k=0}^\infty {1\over (2k)!} 
\left[ \matrix{0 &L_a+R_a \cr L_a+R_a &0 \cr} \right]^{2k} +
\sum_{k=0}^\infty {1\over (2k+1)!} 
\left[ \matrix{0 &L_a+R_a \cr L_a+R_a &0 \cr} \right]^{2k+1} =}$

\noindent
$\displaystyle{
= \sum_{k=0}^\infty {1\over (2k)!} 
\left[ \matrix{(L_a+R_a)^{2k} &\!\!\!\!\!\!\!\!\!\!\!\! 0\cr 
0 &\!\!\!\!\!\!\!\!\!\!\!\!(L_a+R_a)^{2k} \cr} \right] +
\sum_{k=0}^\infty {1\over (2k+1)!} 
\left[ \matrix{0 &\!\!\!\!\!\!\!\!\!\!\!\!(L_a+R_a)^{2k+1} \cr 
(L_a+R_a)^{2k+1} &\!\!\!\!\!\!\!\!\!\!\!\!0 \cr} \right] =}$

\noindent
$\displaystyle{ 
= \left[ \matrix{{\rm cosh}(L_a+R_a) &\!\!\!\!\!\!\! 0\cr 
0 &\!\!\!\!\!\!\!{\rm cosh}(L_a+R_a) \cr} \right] +
\left[ \matrix{0 &\!\!\!\!\!\!\!{\rm sinh}(L_a+R_a) \cr 
{\rm sinh}(L_a+R_a)^{2k+1} &\!\!\!\!\!\!\!0 \cr} \right] .}$

\medskip
\noindent
In general, the operations $L_x$ and $R_x$ commute 
($L_x(R_y u) =x(uy)=(xu)y =L_x(R_yu$). 
Therefore
$$\eqalign{
{\rm cosh}(L_a+R_a) &= {1\over 2} \exp(L_a+R_a) + {1\over 2} \exp(-L_a-R_a) =\cr
&= {1\over 2} \exp(L_a) \exp(R_a) + {1\over 2} \exp(-L_a) \exp(-R_a) =\cr
&= {1\over 2} L_{\exp(a)} R_{\exp(a)} +
{1\over 2} L_{\exp(a)} R_{\exp(a)}: \ z \mapsto 
{1\over 2} e^a z e^a + {1\over 2} e^{-a} z e^{-a} .\cr}$$
Similarly 
${\rm sinh}(L_a+R_a) :\ z \mapsto 
{1\over 2} e^a z e^a - {1\over 2} e^{-a} z e^{-a}$
which completes the proof. \quad Qu.e.d.

\vfill
\eject

\noindent
\bf Case of JB*-triples

\bigskip
\rm

\noindent
Throughout this section let $E$ be a JB*-triple with the triple product
$\{ uvw \}$ and structure derivations $D(a,b):u\mapsto \{ abu\}$.
We define the mapping
$\phi: {\rm Ball}(E) \to {\cal L}(E) \times E$ as
$$\phi(u) := 
\left[{ 
{\phi_1(u)}
\atop
{\phi_2(u)}
}\right]
\qquad\hbox{where} \qquad
\eqalign{
&\phi_2(u) := 2 \big[ 1 - D(u,u) \big]^{-1} u \ ,\cr
&\phi_1(u) := 1 + D\big( \phi_1(u) , u \big) \ .
\cr}$$
We shall also write $M := {\rm range}(\phi) =
\big\{ \phi(u):\  u \in {\rm Ball}(E) \big\}$.

\bigskip
\noindent
\bf Lemma. \it On $M$ we have \quad 
$\phi^{-1}
\left[{ t\atop x }\right] = (1+t)^{-1} x \ .$

\bigskip
\noindent
\bf Proof. \rm
Let $u\in{\rm Ball}(E)$. We have to see that \ 
$u = (1+t)^{-1} x$ \ that is \ 
$(1+t)u=x$ \ with
$$x := 2 \big[ 1 - D(u,u)\big]^{-1} u \ , \qquad
t:= 1 + D(x,u) \ .$$
Recall that, using the customary notation \ 
$u^{2k+1} := D(u,u)^k u$ \ $(k=0,1,\ldots)$, \ we have the identity \
$\{ u^{2k+1} u^{2\ell+1} u^{2m+1}\} = u^{2(k+\ell+m)+1}$. 
Therefore
$$\eqalign{
&x = 2 \big[ 1 - D(u,u)\big]^{-1} u = 2 \sum_{k=0}^\infty D(u,u)^k = 
2 \sum_{k=0}^\infty u^{2k+1} \ ,\cr
&(1+t)u =
\big[ 2 +D(x,u) \big] u =
\Big[ 2 + 2 \sum_{k=0}^\infty D\big( u^{2k+1}, u \big) \Big] u =
2u + 2 \sum_{k=0}^\infty D\big( u^{2k+1},u \big) u \Big] =\cr
&\hskip12mm = 2u^1 + 2 \sum_{k=0}^\infty u^{2k+3} =
2 \sum_{k=0}^\infty u^{2k+1} = x \ . \quad {\rm Qu.e.d.}\cr}$$

\medskip
\noindent
Let $\left[{t \atop x}\right] \in M$ be arbitrarily given and write
$$u := \phi^{-1} \left[{t \atop x}\right] ,\qquad
\ou := A_a(u) = a - \{ uau \} \ .$$
Our aim is to express the vector
$\phi^\# A_a \left[{t \atop x}\right]$ 
in terms of $x,t$.
We know already that
$$\eqalign{
&\phi_2(v) = 2\big[ 1 - D(v,v) \big]^{-1} v , \qquad
\phi_1(v) = 1 + D\big( \phi_2(v), v \big) 
\qquad \big( v\in{\rm Ball}(E) \big) ,\cr
&x = \phi_2(u) =2 \big[ 1 - D(u,u)\big]^{-1} u \ , \qquad
t= \phi_1(u) = 1 + D(x,u) \ .\cr}$$
Hence
$$\eqalign{
\phi_1^\prime(u)\ou &= {d\over d\tau} \Big\vert_{\tau=0}
D\big( \phi_2 (u+\tau \ou), u+\tau \ou\big) =
D\big( \phi_2^\prime (u) \ou , u\big) +
D\big( \phi_2 (u), \ou\big) =\cr
&= D\big( \phi_2^\prime (u) \ou , u\big) +
D\big( x,\ou \big) \ .\cr}$$
The algebraic term for $\phi_2^\prime(u)\ou$ is
$$\eqalign{
\phi_2(u)\ou &= {d\over d\tau} \Big\vert_{\tau=0}
2 \big[ 1 - D\big( u+\tau\ou , u+\tau\ou \big) \big]^{-1} ( u+\tau\ou ) =\cr
&= 2 \big[ 1 - D\big( u , u \big) \big]^{-1} 
\big[ D\big( \ou, u\big) + D\big( u,\ou \big) \big] 
\big[ 1 - D\big( u , u \big) \big]^{-1} u +
2 \big[ 1 - D\big( u , u \big) \big]^{-1} \ou=\cr
&= \big[ 1 - D\big( u , u \big) \big]^{-1} 
\big\{ \big[ D\big( \ou, u\big) +  D\big( u,\ou \big) \big] x + 
2 \ou \big\} \ .\cr}$$
Since $D\big( \ou, u\big) x = \{ \ou u x\} = \{ x u \ou\} =
D\big( x,u \big) \ou$ and $t=1+D\big( x,u\big)$, it follows
$$\phi_2^\prime(u)\ou = 
\big[ 1 - D\big( u , u \big) \big]^{-1} \big\{ (1+t) \ou + D(u,\ou) x \big\} \ .$$
Observe that, by writing
$${\rm Der}_*(E,\{\ldots\}) :=
\big\{ \delta\in{\cal L}(E):\ \delta\{ abc\} =
\{[\delta a]bb\} - \{ a[\delta b]c +\{ ab[\delta c]\} \ (ab,c\in E)\big\}$$
for the {\it Jordan derivations} of $(E,\{\ldots\})$,
we have
$$D(x,u)=D(u,x) \in {\rm Der}_*(E,\{\ldots\}) \ .$$
Indeed, \ 
$D(x,u) = \sum_{k=0}^\infty D( u^{2k+1} u, u) = 
\sum_{k=0}^\infty D( u, u^{2k+1} u) = D(u,x)$ \ and, \ by the Jordan identity, \ \ 
$D(x,u)\{ abc\} = \{ [D(x,u)a]bc\} - \{ a[D(u,x)b]c\} +
\{ ab[D(x,u)c]\} = \{ [D(x,u)a]bc\} - \{ a[D(x,u)b]c\} +
\{ ab[D(x,u)c]\}$. Therefore, since $u=(1+t)^{-1}x$,
$$\eqalign{
(1+t) \ou &= (1+t)a - (1+t)\{ uau\} =\cr
&= (1+t)a - 2\{ [(1+t)u]au \} + \{ u[(1+t)a]u\} =\cr
&= (1+t)a + \{ u[(1+t)a]u\} - 2\{ xau\} \ ,\cr
\phi_2(u)\ou &= \big[ 1 - D\big( u , u \big) \big]^{-1} 
\big\{ (1+t) \ou + D(u,\ou) x \big\} =\cr
&= \big[ 1 - D\big( u , u \big) \big]^{-1} 
\big\{ (1+t)a + \{ u[(1+t)a]u\} - 2\{ xau\} + D(u,\ou)x \big\} 
\ .\cr}$$

\bye
