\magnification=\magstep1

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

\def\CC{{\mathchoice
{\rm C\mkern-8mu\vrule height1.45ex depth-.05ex
width.05em\mkern9mu\kern-.05em}
{\rm C\mkern-8mu\vrule height1.45ex depth-.05ex
width.05em\mkern9mu\kern-.05em}
{\rm C\mkern-8mu\vrule height1.ex depth-.07ex
width.035em\mkern9mu\kern-.035em}
{\rm C\mkern-8mu\vrule height.65ex depth-.1ex
width.025em\mkern8mu\kern-.025em}}}

\def\RR{{\rm I\kern-1.6pt {\rm R}}}
\def\OO{{\rm \hbox{$\vert$}\kern-3.6pt {\rm O}}}

\def\ot{\overline{t}}
\def\ox{\overline{x}}

\def\qed{\quad\vbox{\hrule\hbox{\noindent
\vrule height10pt width 0.5pt\hskip5pt\vrule height10pt width
0.5pt}\hrule}}



\noindent
\bf 1. One-dimensional case \rm

$$\phi(a) =\left[\matrix{\phi_1(a)\cr \phi_2(a)\cr} \right]  \qquad (a\in\CC)$$

$$\phi_1(a):= {1+\vert a\vert^2 \over 1 -\vert a\vert a\vert^2} , \quad
\phi_2(a):= {2a \over 1 -\vert a\vert a\vert^2} .$$

$$\eqalign{
\phi_1(a)^2 -\big\vert \phi_2(a)\big\vert^2 &=
{1 + 2 \vert a \vert^2 + \vert a \vert^4 - 2 \vert q\vert^2
\over
\big[1 -\vert a\vert^2\big]^2} =\cr
&= {\big[1 -\vert a\vert^2\big]^2
\over
\big[1 -\vert a\vert^2\big]^2} = 1 \ .\cr}$$

$$\phi^\prime(a)u = {\partial\over\partial t}\Big\vert_{\tau=0}
\phi(a+\tau u) .$$

$$\eqalign{
\phi_1^\prime(a)u &=
{(\overline{a}u + \overline{u}a)
[1-\vert a\vert^2] - [1+\vert a\vert^2]
(\overline{a}u + \overline{u}a)
\over
\big[1 -\vert a\vert^2\big]^2} =
2 { \overline{a}u +\overline{u}a
\over \big[1 -\vert a\vert^2\big]^2} \ ,\cr \phi_2^\prime(a)u &=
{2u
[1-\vert a\vert^2] + 2a [1+\vert a\vert^2]
\over
\big[1 -\vert a\vert^2\big]^2} \ .\cr}$$

$$\big[ \phi_1^\prime(a)u \big]^2 - \big\vert \phi_2^\prime(a)u \big\vert^2 =
4 { (\overline{a}u + \overline{u}a)^2 - \vert u - a\overline{u}a \vert^2
\over
[1 -\vert a\vert^2\big]^4} \ .$$

$$\eqalign{
(\overline{a}u + \overline{u}a)^2 &- \vert u - a\overline{u}a \vert^2 = \cr
&= \underbrace{\overline{a}^2 u^2}_{(1)} +
\underbrace{2 \vert a\vert^2 \vert u\vert^2}_{(2)} +
\underbrace{\overline{u}^2 a^2}_{(3)} +
\underbrace{\vert u\vert^2}_{(4)} -
\underbrace{a \overline{u}^2 a}_{(5)} -
\underbrace{\overline{a} u^2 \overline{a}}_{(6)} -
\underbrace{\vert a\vert^4 \vert u\vert^2}_{(7)} =\cr
&= [\underbrace{(1)-(6)}_0] + [\underbrace{(3)-(5)}_0] +
\vert u\vert^2 \big( 2\vert a\vert^2 - 1- \vert a\vert^4\big) =\cr
&= - \vert u\vert^2 \big[ 1 - \vert a\vert^2 \big]^2 \ .\cr}$$

$$\big[\phi^\prime_1(a)u\big]^2 -\big\vert \phi^\prime_2(a)u\big\vert^2 =
- 4 { \vert u\vert^2
\over \big[ 1 - \vert a\vert^2 \big]^2} \ .$$

\DEL{
\noindent
This means that

$$-4 \langle u,u \rangle_a^B =
\langle \phi^\prime(a)u, \phi^\prime(a)u \rangle_{\phi(a)}  \quad
(u\in\CC,\ a\in B:=\{ z\in \CC:\ \vert z\vert<1\})$$

\noindent
where

$$\eqalign{
&\langle u,u \rangle_a^B :=
\vert g_a^\prime(0)^{-1} u \vert^2 =
{\vert u\vert^2 \over \big[ 1-\vert a\vert^2 \big]^2} \ ,\cr
&\langle (\tau,\xi),(\tau,\xi) \rangle_{(t,x)}^M :=
\tau^2 - \vert \xi\vert^2 \qquad (\tau\in\RR,\ \xi\in\CC). \cr}$$
}%ENDDEL

\noindent
Recall that the natural Riemannian metric on the unit disc
$B:=\{ z\in \CC:\ \vert z\vert<1\})$ is given by the inner products
$$\langle u,u \rangle_a^B :=
\vert g_a^\prime(0)^{-1} u \vert^2 =
{\vert u\vert^2 \over \big[ 1-\vert a\vert^2 \big]^2}
\qquad (a\in B,\ u\in\CC).$$
Thus, in terms of the inverse $\Psi:=\phi^{-1} :M\leftrightarrow B$, with
$$t:=\phi_1(a), \ \ x:=\phi_2(a), \ \
\widetilde{t} = \phi_1^\prime(a)u, \ \
\widetilde{x} = \phi_2^\prime(a) ,u$$
we have $a=\Psi(t,x)$, $u=\big[\Psi^\prime(t,x)\big](\widetilde{t},\widetilde{x})$ and
$$\widetilde{t}^2 - \vert \widetilde{x} \vert^2 =
-4 \big\langle \big[\Psi^\prime(t,x)\big](\widetilde{t},\widetilde{x}) ,
\big[\Psi^\prime(t,x)\big](\widetilde{t},\widetilde{x}) \big\rangle_{\Psi(t,x)}^B .$$




\vfill
\eject

\noindent
\bf 2. Case of ${\cal L}(H)$ or TRO\rm

$$\phi(a) =\left[\matrix{\phi_1(a)\cr \phi_2(a)\cr} \right]  \qquad
\big(a\in{\cal L}(H)\big)$$

$$\phi_1(a) := (I+a^*a)(I-a^*a)^{-1} \ ,\quad
\phi_2(a) := 2a (I-a^*a)^{-1} \ .$$

$$\eqalign{
\phi_1(a)^2 &- \phi_2(a)^* \phi_2(a) =\cr
&= (I - a^*a)^{-1} (I + a^*a)^2 (I - a^*a)^{-1} -
(I - a^*a)^{-1} (4 a^*a) (I - a^*a)^{-1} =\cr
&= (I - a^*a)^{-1} \big[ I + 2 a^*a + (a^*a)^2 - 4 a^*a \big] (I - a^*a)^{-1} =\cr
&= (I - a^*a)^{-1} \big[ I - 2 a^*a + (a^*a)^2 \big] (I - a^*a)^{-1} =\cr
&= (I - a^*a)^{-1} \big[ I - 2 a^*a + (a^*a)^2 \big] (I - a^*a)^{-1} =\cr
&= (I - a^*a)^{-1} ( I - a^*a )^2 (I - a^*a)^{-1} =\cr
&= I \ .\cr}$$

$$\eqalign{
\phi_1^\prime(a)x &=
(a^*x + x^*a)(I-a^*a)^{-1} + (I+a^*a)(I-a^*a)^{-1} (a^*x + x^*a) (I-a^*a)^{-1} =\cr
&= (I-a^*a)^{-1} \big[ (I-a^*a) (a^*x + x^*a) +
(I+a^*a) (a^*x + x^*a) \big] =\cr
&= (I-a^*a)^{-1} (a^*x + x^*a) (I-a^*a)^{-1} \ .\cr
\phi_2^\prime(a)x &=
 2x (I-a^*a)^{-1} + 2 \underbrace{a (I-a^*a)^{-1}}_{(I-aa^*)^{-1} a}
(a^*x + x^*a) (I-a^*a)^{-1} = \cr
&= 2 (I-aa^*)^{-1}\big[ (I-a^*a)x + a(a^*x + x^*a) \big] (I-a^*a)^{-1} =\cr
&= 2 (I-aa^*)^{-1} [ x + ax^* a ] (1- a^*a)^{-1} \ \cr}$$

\noindent
Thus we have

$$\eqalign{
&{\rm range}(\phi) = M :=
\{ (t,x) \in {\cal L}(H)\times {\cal L}(H):
\ t=t^* \ge I, \
t^2 - x^*x =I\} \ ,\cr
&\phi : B:= \{ a:\ a^*a< I\} \leftrightarrow
M:=\{ (t,x):\ t=t^* \ge I, \ t^2 - x^*x =I\} , \cr
&\Psi := \phi^{-1} : (t,x) \mapsto x(I+t)^{-1} \cr}$$

\noindent
{\bf Aim.} Find a convenient indefinite inner product \
$\langle (\ot_1,\ox_1), (\ot_2,\ox_2) \rangle_{(t,x)}$ \
on
${\cal L}(H)\times {\cal L}(H)$
such that \quad
$\langle (\ot,\ox), (\ot,\ox) \rangle^M_{(t,x)} :=
\langle (\ot,\ox), (\ot,\ox) \rangle_{(t,x)}$
for $(t,x)\in M$ and $(\ot,\ox)\in T_{(t,x)} M$ \quad
with the inner product
$$\langle (\ot,\ox), (\ot,\ox) \rangle^M_{(t,x)} \! :=\! -4
\big\langle [\Psi^\prime(t,x)](\ot,\ox),
[\Psi^\prime(t,x)](\ot,\ox) \big\rangle_{\Psi(t,x)}^B
\ \
{\rm for} \ (t,x)\in M, \ (\ot,\ox) \in T_{(t,x)} M$$
with the natural Riemannian inner product on
the unit ball $B := \{ a: \ a^*a<I\}$,
$$\eqalign{
\langle u,v\rangle_a^B &:=
\big[[g_a^\prime(0)]^{-1} u\big]^* \big[ [g_a^\prime(0)]^{-1} v\big] =\cr
&= (I-a^*a)^{-1/2} u^* (I-aa^*)^{-1} v (I-a^*a)^{-1/2} .\cr}$$


\vfill\eject


\noindent
Let \ $(t,x)\in M$ and $(\ot,\ox)\in T_{(t,x)} M$ be fixed and write
$$\eqalign{
a:= \Psi(t,x)&=x(I+t)^{-1} , \cr
u:= \big[\Psi^\prime(t,x) \big](\ot,\ox) &=
{\partial\over\partial \tau}\Big\vert_{\tau=0}
(x+\tau\ox)(I+t+\tau\ot)^{-1} =\cr
&= \big[ \ox - x (I+t)^{-1} \ot\big] (I+t)^{-1} .\cr}$$
Then we have
$t=t^* =[I+x^*x]^{1/2} \ge I$ with $a^* =(I+t)^{-1} x^*$ and
$$\eqalign{
a^*a &= (I+t)^{-1} x^* x (I+t)^{-1} = (I+t)^{-1} (t^2-I) (I+t)^{-1} =\cr
&= (t-I)(I+t)^{-1} \ ,\cr
I-a^*a
&= (t+I)(I+t)^{-1} - (t-I)(I+t)^{-1}=\cr
&= 2 (I+t)^{-1} .\cr}$$
%In particular \
%$$(I-a^*a)^{-1} ={1\over 2}(I+t) \ , \quad
%u = \big[ \ox - \underbrace{x (I+t)^{-1}}_a \ot\big] (I+t)^{-1} =
%{1\over 2} [\ox - a\ot] (I-a^*a) .$$
Therefore
$$\eqalign{
\langle u,u\rangle_a^B
&= (I-a^*a)^{-1/2} u^* (I-aa^*)^{-1} u (I-a^*a)^{-1/2} =\cr
&= {1\over 2}(I+t)^{1/2} \big[\ox^* - \ot (I+t)^{-1} x^* \big]
(I-aa^*)^{-1}
\big[ \ox - x (I+t)^{-1} \ot \big] (I+t)^{1/2} =\cr
&= {1\over 2} (I+t)^{1/2}
[\ox^* - \ot a^*] (I-aa^*)^{-1} [\ox - a\ot] (I+t)^{1/2} \ .
\cr}$$
Since \ $a^*(I-aa^*)^{-1} = (I-a^*a)^{-1} a^*$ \ and \
$(I-aa^*) (I-aa^*)^{-1} =I$, \ we get
$$\eqalign{
&a^*(I-aa^*)^{-1} = (I-a^*a)^{-1} a^* = {1\over 2} (I+t)
(I+t)^{-1} x^* = {1\over 2}  x^* \ ,\cr &(I-aa^*)^{-1} = I +
a\big[ a^* (I-aa^*)^{-1}\big] = I + {1\over 2} x (I+t)^{-1} x^* \
.\cr}$$ Hence
$$\eqalign{
&[\ox^* - \ot a^*] (I-aa^*)^{-1} [\ox - a\ot] =\cr
&= \! \ox^* \underbrace{(I-aa^*)^{-1}}_{I+x(I+t)x^*/2} \ox \!-\!
\ot \underbrace{a^* (I-aa^*)^{-1}}_{x^*/2} \ox \!-\!
\ox^* \underbrace{\big[a^* (I-a a^*)^{-1}\big]^*}_{x/2} \ot \!+\!
\ot \big[ \underbrace{a^* (I-a^*a)^{-1} }_{x^*/2}\big] \!\!\! \!\!\!
\underbrace{a}_{x(I+t)^{-1}} \!\!\!\!\!\! \ot \!=\cr
&= \ox^* \ox + {1\over 2}\Big[
\ox^* x(1+t)^{-1} x^*\ox - \ot x^* \ox - \ox^* x \ot +
\ot \big[ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\underbrace{x^* x}_{t^2-I=(t-I)(I+t)} \!\!\!\!\!\!\!\!\!\!\!\!\!\!(I+t)^{-1} \big] \ot
\Big] =\cr
&= \ox^* \ox -\ot^2 + {1\over 2}\Big[ \ox^* x(1+t)^{-1} x^* \ox -
\ot x^* \ox - \ox^* x \ot + \ot (t-I) \ot \Big] \ .\cr}$$
\bf Theorem. \it For \ $(\ot,\ox)\in T_{(t,x)} M$ \ with \ $\ot t
+ t\ot = \ox^* x + x^* \ox$ \ we conclude \rm
$$\eqalign{
&\langle (\ot,\ox),(\ot,\ox) \rangle_{(t,x)}^M =
-4 \langle u,u\rangle_a^B =\cr
&= (I+t)^{1/2} \Big\{
2\big[ \ot^2 - \ox^* \ox\big] - \Big[ \ox^* x(I+t)^{-1} x^* \ox -
\ot x^* \ox - \ox^* x \ot + \ot (t-I) \ot \Big]
\Big\} (I+t)^{1/2} \! .\cr}$$
\bf Remark. \rm In the commutative case $H=\CC \equiv {\cal L}(H)$ we have
\hfill\break
$-4 \langle (\ot,\ox),(\ot,\ox) \rangle_{(t,x)}^M =
\langle [\Psi^\prime(t,x)](\ot,\ox),
[\Psi^\prime(t,x)](\ot,\ox) \rangle_{\Psi(t,x)} =
\ot^2 - \vert \ox\vert^2$.


\vfill
\eject

\noindent
\bf 3a. Spectral integral for $D\big( \varphi(a),\psi(a)\big)$ in ${\cal L}(H)$
\rm

\bigskip
\noindent
Let \ $a\in {\cal L}(H)$ \ be fixed and let
$$\Omega := {\rm Sp} (a^*a)\setminus\{ 0\} =
{\rm Sp}(aa^*)\setminus\{ 0\} \ , \quad
\alpha := {\rm Id}_\Omega :\Omega \ni\omega \mapsto \omega \ .$$
It s well-known that,
there is a positive Borel measure \ $\mu$ on $\Omega$ \
and there are surjective partial isometries \
$Q_1,Q_2: H\to L^2(\Omega,\mu)$ \ such that,
by writing $M_\varphi$ for the multiplication operator
$$M_\varphi (f) := \varphi f \quad (f\in L^2(\Omega,\mu) \ ,$$
we have
$$a = Q_1^* M_\alpha Q_2 \ .$$
For any couple of Borelian sets $S,T\subset \Omega$, define the
projection $P_{S\times T} : {\cal L}(H) \to {\cal L}(H)$ by
$$P_{S\times T} X :=
Q_1^* M_{1_S} Q_1 X Q_2^* M_{1_T} Q_2 \qquad
(X\in {\cal L}(H))$$
where $1_Z$ stands for the indicator function of the set $Z$ (with
$1_Z(\omega)=1$ for $\omega\in Z$ and
$1_Z(\omega)= 0$ for $\omega\in \Omega\setminus Z$).
Furthermore, let \
$P_{\{0\}\times\{ 0\}} X := (I-Q_1^* Q_1)X(I-Q_2^* Q_2)X$ \ and
$$P_{S\times\{ 0\}} X := Q_1^* M_{1_S} Q_1 X (I-Q_1^* Q_1)
 \ , \quad
P_{\{ 0\}\times T}X := (I-Q_1^* Q_1)X Q_2^* M_{1_T} Q_2 \ .$$
%
%\bigskip
%\noindent
\bf Proposition. \it Let \
$\varphi,\psi:\Omega\cup\{ 0\}\to\CC$ \ be a couple of continuous functions
such that \ $\varphi(0)\!=\! \psi(0)\!=\! 0$. \ Then
$$D\big( \varphi(a),\psi(a)\big) =
\int_{(\lambda_1,\lambda_2)\in[\Omega\cup\{ 0\}]^2} {1\over 2}
\big[ \varphi(\lambda_1) \overline{\psi(\lambda_1)} +
\varphi(\lambda_2) \overline{\psi(\lambda_2)} \big] \,
P\big( d(\lambda_1,\lambda_2) \big)$$
with integration in Cauchy sense.

\bigskip
\noindent
\bf Remark. \rm
Given a bounded finitely additive vector valued measure
$V: {\cal A} \to B$ where ${\cal A}$ is an algebra of subetes of a set ${\cal S}$
and $B$ is any Banach space, a function $f:{\cal S}\to \CC$
is said to be $V$-integrable in Cauchy sense if
it can be uniformly approximated by a sequence of $V$-simple functions
($V$-measurable functions with finite range).
For a $V$-simple function $g:{\cal S}\to\CC$,
using any representation of the form
$g:= \sum_{k=1}^N \lambda_k 1_{A_k}$, the vector
$\int g(\omega)\, V(d\omega) := \sum_{k=1}^N \lambda_k V(A_k)$
is unambigously defined.
If $f$ is the uniform limit of $V$-simple functions then
the definition
$\int g(\omega)\, V(d\omega) := \lim_{n\to\infty}
\int g_n(\omega)\, V(d\omega)$ is unambigous with any
uniformly convergent sequence $g_n\to g$ of $V$-simple functions.
In the Proposition we take \
${\cal S}:=[\Omega\cup\{ 0\}]^2$, \
${\cal A}:= \big\{$finite disjoint unions from
$\big\{ S\times T:\ S,T$ Borel$\subset\Omega\cup\{ 0\}\big\}$.
%with \
%$P\big( \bigcup_k^\bullet S_k\times T_k \big)X :=
%\sum_k Q_1 M_{1_{S_k}} X Q_1^* X Q_2 M_{1_{T_k}} Q_2^*$.


\bigskip
\noindent
\bf Proof. \rm
By definition,
$\varphi(a):= Q_1^* M_\varphi Q_2$ and
$\psi(a):= Q_1^* M_\psi Q_2$.
Let $\varphi_n$ be the simple function assuming the value
$\varphi(k/n)$ on the set
$I_{k,n} :=[(k-1)/n,k/n)\cap \Omega$ (for $n=1,2,\ldots$ with
$k=1,\ldots,n\cdot{\rm entier}(\Vert a\Vert)+1)$.
We define $\psi_n(a)$ analogously.
Then
$$D\big(\varphi_n(a),\psi_n(a)\big) %\, X
=
\sum_{k,\ell} {1\over 2}
\big[ \varphi(k/n) \overline{\psi(k/n)} +
\varphi(\ell/n) \overline{\psi(\ell/n)} \big]\,
P \big( I_{n,k}\times I_{n,\ell} \big)% X
\ .$$
For $n\to\infty$ we have
$\varphi_n\to\varphi$, $\psi_n\to\psi$ uniformly,
implying
$\varphi_n(a) \to \varphi(a)$, \ $\psi_n(a)\to\psi(a)$ \
and
$D\big( \varphi_n(a), \psi_n(a) \big) \to D\big( \varphi(a), \psi(a) \big)$
in ${\cal L}\big( {\cal L}(H)\big)$.
Hence the statement is immediate.
\qed


\vfill\eject

\noindent
\bf 3b. Spectral integral for $D\big( \varphi(a),\psi(a)\big)$ in ${\cal H}_3(\OO)$

\bigskip
\noindent
\rm
Given $a\in {\cal H}_3(\OO)$, there is an
orthogonal frame (in Jordan sense)
$G_0:=\{ g_1,g_2,g_3\}$ and there are scalars
$\lambda_1\ge\lambda_2\ge\lambda_3\ge0$
such that
$$a=\sum_{\ell=1}^3 \lambda_\ell g_\ell , \quad
D \big( \varphi(a),\psi(a)\big) =
\sum_{\ell=1}^3
\varphi(\lambda_\ell)\overline{\psi(\lambda_\ell)}
D \big( g_\ell,g_\ell \big) \ .$$
The frame $G_0$ can be extended to a covering grid
$$G=\big\{ g_j :\ j=1,\ldots 27\big\}$$
of linearly independent tripotents such that
$$D\big( g_j,g_j\big) g_k = \langle \gamma_j,\gamma_k\rangle g_k \quad (j,k=1,\ldots,27)$$
where \ $\big\{ \gamma_j:\ j=1,\ldots 27\big\} \subset \RR^27$ is the 1-part of the
3-graded root system $E_6$.
Actually, after a suitable change of indices
we may assume that
$$\eqalign{
G=G_0 \cup G_{\{ 1,2 \}} \cup G_{\{ 2,3 \}} \cup G_{\{ 1,3 \}}
 \ \ \hbox{with} \ \
&G_{\{ 1,2 \}} := \{ g_j:\ 4\le j\le 11\}, \cr
&G_{\{ 2,3 \}} := \{ g_j:\ 12\le j\le 19\}, \cr
&G_{\{ 1,3 \}} := \{ g_j:\ 20\le j\le 27\} \cr}$$
and. for \ $\ell =1,2,3$, \ we have
$$\eqalign{
&\big\{ g\in G:\ D\big( g_\ell,g_\ell \big) g = g \big\} = \{ g_\ell\} \ ,\cr
&\hbox{$
\big\{ g\in G:\ D\big( g_\ell,g_\ell \big) g = (1/2) g \big\} =
\bigcup_{m \in\{ 1,2,3\}\setminus\{\ell\}} G_{\{ \ell,m\}} \ ,$}\cr
&\big\{ g\in G:\ D\big( g_\ell,g_\ell \big) g =0  \big\} =
\big[ G_0 \setminus \{ g_\ell\} \big] \cup G_{\{1,2,3\}\setminus\{\ell\}} \ .\cr}$$
It follows
$$D\big( \varphi(a),\psi(a)\big) g_j =
\sum_{\ell=1}^3
\varphi(\lambda_\ell)\overline{\psi(\lambda_\ell)}
\big\langle \gamma_\ell, \gamma_j\rangle g_j \qquad
(j=1,\ldots,27).$$
Or alternatively,
with the projections
$$\eqalign{
&P_{\{\ell\}} : {\cal H}_3(\OO) \to \CC g_\ell \hskip22mm (1\le \ell \le 3),\cr
&P_{\{ \ell,m\}} : {\cal H}_3(\OO) \to \sum_{g \in G_{\{ \ell,m\}}} \CC g
\qquad (1\le \ell< m\le 3) \cr}$$
we can write
$$D\big( \varphi(a),\psi(a)\big) =
\sum_{1\le \ell \le m\le 3}
\Big[ {1\over 2} \varphi(\lambda_\ell)\overline{\psi(\lambda_\ell)}
+ {1\over 2} \varphi(\lambda_m)\overline{\psi(\lambda_m)}\Big]
P_{\{\ell,m\}} \ .$$





\vfill
\eject

\bigskip
\noindent
\bf 4. Jordan case \rm

\bigskip
\noindent
$(E,\{\ldots\})$ \ JB$^*$-triple,

$D(a,b)x := \{ abx\}$, \quad $Q(a,b)x := \{ axb\}$,

$B(a,b)x$ \vtop{
\hbox{$:= I - 2D(a,b) + Q(a,a)Q(b,b) =$}
\hbox{$= [I-D(a,b)]^2 \ ,$}}

$M_a(x) := a + B(a,a)^{1/2}[I+D(x,a)]^{-1}x,$ \quad $M_{-a} = [M_a]^{-1}$,

$M_a^\prime(x)v$ \vtop{
\hbox{$= \partial/\partial t\big\vert_{t=0} M_a(x+tv) =$}
\hbox{$= B(a,a)^{1/2} [I+D(x,a)]^{-2} v =$}
\hbox{$= B(a,a)^{1/2} B(x,-a)^{-1} v$.}}


\bigskip
\noindent
\bf Remark. \rm
In the commutative case \ $E =\CC$, $\{ xyz\} = x\overline{y}z$ \ we have
\hfill\break
$B(a,b)z = [1 - 2 a \overline{b} + a^2 \overline{b}^2 ]z$ \ and \
$B(a,a)z = [1-\vert a\vert^2]^2 z$. \
Also \
$M_a(z)=(z+a)/(a-\overline{a}z)$.
\hfill\break
The identity \ $\phi_1(a)^2-\vert\phi_2(a)\vert^2 =1$ \
can be written as \
$\phi_1(a)^2 -D\big( \phi_2(a),\phi_2(a) \big) =I$.
\hfill\break
Here \
$\phi_2(a) =2a[1-\vert a\vert^2]^{-1} =
2[I-D(a,a)]^{-1}a =
2 [M_a^\prime(0)]^{-1}a
= 2 [(M_a)^{-1}]^\prime(M_a(0))a$.
\hfill\break
Thus \
$\phi_2(a) = 2 M_{-a}^\prime(a)a$


$$\phi(a) =\left[\matrix{\phi_1(a)\cr \phi_2(a)\cr} \right]  \qquad
\big(a\in E\big)$$


$${\rm range}(\phi) = \Big\{ \left[ {t\atop x}\right] \in
{\cal L}(E)\times E :\ t^2 - D(x,x) = I, \ {\rm Sp}(t) >0 \Big\} \ $$

$$\eqalign{
\phi_2(a) &:= 2[I-D(a,a)]^{-1} a = 2 M_{-a}^\prime(a)a\ ,\cr
\phi_1(a) &:= [I+D(\phi_1(a),\phi_1(a))]^{1/2} \ \buildrel{?}\over{=}
\ B\big( \phi_2(a),-\phi_2(a) \big)^{1/4} \ .\cr}$$
\bf Conjecture. \rm A natural candidate for a trailer space
(as subspace of ${\cal L}(E)\times E$) would be the
\it Jordan supertriple \rm of $(E,\{\ldots\})$.









\bye
