\magnification=\magstep1

\def\eps{\varepsilon}

\centerline{\bf JORDAN MANIFOLDS}

\bigskip
\bigskip

\bf 1. Invariant connection on the unit ball of a complex JB*-triple
\rm
\bigskip
\noindent
In the sequel let $Z$ denote a complex JB*-triple with the triple product
$\{ .,.,.\}$ and let
$$\eqalign{
&D(a,b): x\mapsto \{ a,b,x\}, \qquad D(d):=D(a,a), \cr
&Q(a,b): x\mapsto \{ a,x,b\}, \qquad Q(a):=Q(a,a), \cr
&B(a,b):= 1 - 2D(a,b) + Q(a)Q(b), \quad B(a):=B(a,a) \cr}$$
be the usual skew-derivations, quadratic representations and Bergman operators,
respectively.
Recall that the transformation
$$T: c \mapsto  \Big[\exp\Big( \big[c - Q(z) c\big] \partial/\partial z \Big)\Big] 0$$
is a real bianalytic mapping of $Z$ onto the unit ball
${\rm Ball}(Z)$ of $Z$.
Given any point $a\in{\rm Ball}(Z)$,
it is well-known
[Sauter's Dissertation, p. 27; details: Kaup, Math. Z. 183 (1983) 503-529]
that the mapping
$$g_a := \exp \big( [T^{-1}(a) - Q(z)T^{-1}(a)] \partial/\partial z \big)$$
is a holomorphic automorphism of ${\rm Ball}(Z)$ and we have
$$g_a (x) = a + B(a)^{1/2} [ 1 + D(x,a)]^{-1}  x, \qquad \Vert x\Vert <1 .$$
\bf Remark. \rm
In [Kaup, Math. Z. 183 (1983) 503-529] one can find
$g_a (x) = a + B(a)^{1/2} [ 1 - D(x,a)]^{-1}  x$
which is obviously incorrect in the sign of the term
$D(x,a)$ as on can see on the 1-dimensional example of the
M\"obius transformation
$g_a(z)=(z+a)/(1+\overline{a}z)$.

\medskip
\noindent
\bf Definition. \rm
Given two smooth vector fields $X,Y:{\rm Ball}(Z)\to Z$, we set
$$\nabla_X Y(a) :=
{d\over dt}\Big\vert_{t=0+}
g_a^\prime(0) \big[ g_a^\prime\big( t g_a^\prime(0)^{-1} X(a) \big)\big]^{-1}
Y\big(g_a\big( t g_a^\prime(0)^{-1} X(a) \big)\big) , \qquad
\Vert a\Vert <1 .$$
\bf Theorem. \it
If $a\in {\rm Ball}(Z)$ and $g$ is any holomorphic automorphism of
${\rm Ball} (Z)$ such that
$g(0)=a$ then
$$\nabla_X Y(a) =
{d\over dt}\Big\vert_{t=0+} g^\prime(0) \big[ g^\prime\big( t
g^\prime(0)^{-1} X(a) \big)\big]^{-1} Y\big( g\big( t
g^\prime(0)^{-1} X(a) \big)\big) .$$
Furthermore we have
$$\nabla_X Y(a) = Y^\prime(a) X(a) +
2 B(a)^{1/2} D\big( B(a)^{-1/2} X(a), a \big) B(a)^{-1/2} Y(a) .$$
\bf Proof. \rm
Let $g$ be a holomorphic automorphism of ${\rm Ball}(Z)$ with
$g(0)=a$. Then we have
$$g = g \circ L \quad \hbox{for some (linear) automorphism
$L: Z\to Z$ of the product $\{ .,.,.\}$.}$$
In particular
$g^\prime(x) = g^\prime (Lx) L$ for any point $x\in{\rm Ball}(Z)$.
By writing $p_t := t g_a^\prime(0)^{-1} X(a)$ and $q_t := t g^\prime(0)^{-1} X(a)$,
we have $q_t = L^{-1} p_t$ and hence
$$\eqalign{
&{d\over dt}\Big\vert_{t=0+}
g^\prime(0) \big[ g^\prime\big( t g^\prime(0)^{-1} X(a) \big)\big]^{-1}
Y\big( g\big( t g^\prime(0)^{-1} X(a) \big)\big) =\cr
&\ = {d\over dt}\Big\vert_{t=0+}
g^\prime(0) \big[ g^\prime\big( q_t \big)\big]^{-1}
Y\big( g(q_t) \big) =
{d\over dt}\Big\vert_{t=0+}
g_a^\prime(0) L \big[ g_a^\prime\big( L L^{-1} p_t \big) L\big]^{-1}
Y\big( g_a( LL^{-1} p_t) \big) =\cr
&\ = {d\over dt}\Big\vert_{t=0+}
g_a^\prime(0) \big[ g_a^\prime\big( p_t \big)\big]^{-1}
Y\big( g_a(p_t) \big) = \nabla_X Y (a) .\cr}$$
Recall [Sauter's Dissertation, p. 27; details: Kaup, Math. Z. 183 (1983) 503-529 ]
that
$$g_a^\prime (z) = B(a)^{1/2} B(z,-a)^{-1} ,\qquad \Vert z\Vert < 1 .$$
Thus with the curve $p_t = t g_a(0)^{-1} X(a) =t B(a)^{-1/2} X(a)$ we get
$$\eqalign{
&{d\over dt}\Big\vert_{t=0+} g_a^\prime(p_t) =
B(a)^{1/2} {d\over dt}\Big\vert_{t=0+} B(p_t,-a)^{-1} = \cr
&\ = B(a)^{1/2} \Big[ - B(p_0,-a)^{-1}
{d\over dt}\Big\vert_{t=0+} B(p_t,-a) B(p_0,-a)^{-1} \Big] =\cr
&\ = - B(a)^{1/2} {d\over dt}\Big\vert_{t=0+}
B\big( t B(a)^{-1/2} X(a), -a \big) =\cr
&\ = - B(a)^{1/2} {d\over dt}\Big\vert_{t=0+} \Big[
1 - 2t D\big( B(a)^{-1/2} X(a), -a \big) +
t^2 Q\big( B(a)^{-1/2} X(a) \big) Q(a) \Big] =\cr
&\ = - 2 B(a)^{1/2} D\big( B(a)^{-1/2} X(a), a \big) .\cr}$$
It follows
$$\eqalign{
\nabla_X Y(a) &=
{d\over dt}\Big\vert_{t=0+}
g_a^\prime(0) \big[ g_a^\prime\big( p_t \big)\big]^{-1}
Y\big( g_a(p_t) \big) =\cr
&= B(a)^{1/2} \Big[ {d\over dt}\Big\vert_{t=0+}
\big[ g_a^\prime\big( p_t \big)\big]^{-1} Y\big( g_a(0)\big) +
\big[ g_a^\prime\big( 0 \big)\big]^{-1}
{d\over dt}\Big\vert_{t=0+} Y\big( g_a(p_t) \big) \Big] =\cr
&= B(a)^{1/2} \Big[ -g_a^\prime(0)^{-1}
{d\over dt}\Big\vert_{t=0+}
\big[ g_a^\prime\big( p_t \big)\big] g_a^\prime(0)^{-1} Y(a) +\cr
&\hskip2cm +
\big[ g_a^\prime\big( 0 \big)\big]^{-1} Y^\prime \big( g_a (0) \big)
g_a^\prime(0) {d\over dt}\Big\vert_{t=0+} p_t \Big] =\cr
&= B(a)^{1/2} \Big[ -B(a)^{-1/2}
\big[- 2 B(a)^{1/2} D\big( B(a)^{-1/2} X(a), a \big) \big]
B(a)^{-1/2} Y(a) +\cr
&\hskip2cm +
B(a)^{-1/2} Y^\prime (a)
B(a)^{1/2} B(a)^{-1/2} X(a) \Big] =\cr
&= 2 B(a)^{1/2} D\big( B(a)^{-1/2} X(a), a \big)
B(a)^{-1/2} Y(a) + Y^\prime(a) X(a) . \quad\hbox{Qu.e.d.}\cr}$$
\bf Notation. \rm Given a holomorphic automorphism $h$ of ${\rm Ball}(Z)$,
its {\it adjoint action} on a complete vector field $X$ on ${\rm Ball}(Z)$
will be denoted by $h^\# X$ It is well-known that
$$h^\# X = {d\over dt}\Big\vert_{t=0+} \qquad \hbox{and} \qquad
h^\# X(a) = h^\prime(a)^{-1} X\big( h(a)\big), \ \ \Vert a\Vert <1 .$$
In terms of adjoint actions, we have
$$\nabla_X Y (a) = g_a^\prime(0) {d\over dt}\Big\vert_{t=0+}
g_a^\# Y \big( \big[\exp (t g_a^\# X) \big] 0\big) .$$
Indeed, by definition,
$$\eqalign{
\nabla_X Y (a) &= {d\over dt}\Big\vert_{t=0+}
g_a^\prime(0) \big[ g_a^\prime \big( t g_a^\prime(0)^{-1} X(a) \big) \big]^{-1}
Y \big( t g_a^\prime(0)^{-1} X(a) \big) =\cr
&= g_a^\prime(0) {d\over dt}\Big\vert_{t=0+}
\big[ g_a^\prime \big( t g_a^\# X(0) \big) \big]^{-1}
Y \big( t g_a^\# X(0) \big) =\cr
&= g_a^\prime(0) {d\over dt}\Big\vert_{t=0+}
g_a^\# Y \big( \underbrace{t g_a^\# X(0)}_{\exp(t g_a^\# X) 0 + o(t)} \big) =\cr
&= g_a^\prime(0) {d\over dt}\Big\vert_{t=0+}
g_a^\# Y \big( \big[\exp (t g_a^\# X) \big] 0\big) .\cr}$$
\bf Proposition. \it $\nabla_{h^\# X} h^\# Y = h^\# \nabla_X Y$ \ for all \
$h \in{\rm Aut}\,{\rm Ball}(Z)$.
\bigskip
\noindent
\bf Proof. \rm
In terms of adjoint actions we can write
$$nabla_{h^\# X} h^\# Y (a) =
g_a^\prime(0) {d\over dt}\Big\vert_{t=0+}
\underbrace{g_a^\# h^\# Y}_{(h\circ g_a)^\# Y}
\big( \exp(t h^\# g_a^\# X) 0\big) .$$
Since $h\circ g_a$ is a holomorphic automorphism of ${\rm Ball}(Z)$
sending the origin to the point $h(a)$,
$$h \circ g_a = g_{h(a)} \circ L \qquad \hbox{for some linear operator
$L\in{\rm Aut}(Z)$}.$$
It follows
$$\nabla_{h^\# X} h^\# Y (a) =
g_a^\prime (0) {d\over dt}\Big\vert_{t=0+}
\big( g_a \circ L\big)^\# Y \big( \exp\big( t (g_{h(a)} \circ L)^\# X\big)0 \big) .$$
According to the Theorem, for any point $b\in{\rm Ball}(Z)$ we have
$$\nabla_X Y (b) = \big( g_b\circ L\big)^\prime(0)
{d\over dt}\Big\vert_{t=0+} \big( g_b\circ L\big)^\#
Y \big( \exp\big( t (g_b\circ L)^\# X\big) 0\big) ,$$
that is
$${d\over dt}\Big\vert_{t=0+} \big( g_b\circ L\big)^\# Y
\big( \exp\big( t (g_b\circ L)^\# X\big) 0\big) =
L^{-1} g_b^\prime(0)^{-1} \nabla_X Y (b) .$$
In view of this relation, we can write
$$\nabla_{h^\# X} h^\# Y (a) = g_a^\prime (0) L^{-1} g_{h(a)}^\prime (0)^{-1}
Y \big( h(a)\big) .$$
Observe that the mapping $g_{h(a)}^{-1} \circ h \circ g_a$ is the restriction
of the linear operator $L$ to the unit ball. As a consequence have
$$L = \big( g_{h(a)}^{-1} \circ h \circ g_a \big)^\prime (0) =
\big( g_{h(a)}^{-1} \big)^\prime (h(g_a(0)) \, h^\prime \big( g_a(0)\big) \,
g_a^\prime (0) = g_{h(a)}^\prime(0)^{-1} h^\prime (a) g_a^\prime(0) .$$
Hence we conclude
$$\eqalign{
\nabla_{h^\# X} h^\# Y (a) &= g_a^\prime(0) \big[ g_a^\prime(0)^{-1}
h^\prime(a)^{-1} g_{h(a)}(0^\prime \big] g_{h(a)}^\prime(0)^{-1}
\nabla_X Y \big( h(a)\big) =\cr
&= h^\prime(a)^{-1} \nabla_X Y \big( h(a)\big) = h^\# \nabla_X Y (a) .
\quad {\rm Qu.e.d.}\cr}$$

\bigskip
Next we proceed to the geodesic equation of the connection $\nabla$.
Recall that a smooth curve $\gamma: I\to {\rm Ball}(Z)$ on an open real
interval around $0$ is a $\nabla$-geodesic if its derivative satisfies
the equation
$\nabla_{\dot\gamma(t)} \dot\gamma(t) = 0$.
According to the Theorem, this means
$$\ddot\gamma (t) + 2 B\big( \gamma(t)\big)^{1/2}
D\big( B\big( \gamma(t)\big)^{-1/2} \dot\gamma(t) , \gamma(t) \big) B\big( \gamma(t)\big)^{-1/2}
\dot\gamma(t) = 0 .$$
First we look for particular solutions with the property $\gamma(0)=0$.
Let $0\ne v\in Z$ be any fixed vector and let $F_v$ denote the real JB*-subtriple
generated by $v$. That is
$F_v = \overline{{\rm Span}_{{\rm real}}} \big\{ D(v)^n v: \ n=0,1,\ldots\big\}$.
It is well-known that the spectrum of the
operator $D(v)\vert F_v$ is non-negative,
and, by writing
$\Omega_v := \big[ {\rm Spectrum} D(v)^{1/2} \vert F_v\vert \setminus \{ 0\}$,
the commutative Gelfand-Naimark Theorem of JB*-triples asserts that
there is a real JB*-isomorphism $T_v :{\rm Re} {\cal C}_0(\Omega_v) \to F_v$
such that $T_v ({\rm id}_{\Omega_v}) = v$ and
$T_v (\varphi\psi\chi) = \{ T_v(\varphi),T_v(\psi),T_v(\chi)\}$ for all functions
$\varphi,\psi,\chi\in{\rm Re}{\cal C}_0(\Omega_v)$.
In particular, given any solution $\varphi: I\to {\rm Re}{\cal C}_0(\Omega_v)$,
of the equation
$${d^2 \over dt^2} \varphi(t,\omega) + 2 \big[ 1 - \varphi(t,\omega)^2 \big]^{-1}
\varphi(t,\omega) \Big[ {d\over dt} \varphi(t,\omega) \Big]^2 ,\qquad \omega\in\Omega_v ,$$
the curve $\gamma(t) := T_v \varphi(t)$, $t\in I$ is a $\nabla$-geodesic.
Notice that, for any real constant $\alpha$,
the function $x(t) ={\rm artanh} (\alpha t) :=
(e^{\alpha t}-e^{-\alpha t})(e^{\alpha t}+e^{-\alpha t})^{-1}$ is the solution of the
initial value problem
$\ddot x + (1-x^2)^{-1} x \dot x^2$, $x(0)=0$, $\dot x(0)=\alpha$.
Indeed, we have $\dot x = 1-x^2$ and hence
$\ddot x = - 2 x\dot x = -2 x (1-x^2)$ and
$\ddot x + (1-x^2)^{-1} x \dot x^2 = - 2 x\dot x =
-2 x (1-x^2) + 2 (1-x^2)^{-1} x (1-x^2)^2 =0$.
Therefore we get the following.

\bigskip
\noindent \bf Theorem. \it
The $\nabla$-geodesic curves passing through the origin have the form
$$\gamma_{0,v}(t) = T_v \ {\rm artanh} \big( t\ {\rm id}_{\Omega_v}\big) ,\qquad
t\in{\bf R}, \ v\in Z$$
in terms of the Gelfand-Naimark representations $T_v$.

\bigskip
\noindent
\bf Corollary. \it
The (unique) maximal $\nabla$-geodesic $\gamma_{a,w}$
with the properties
$\gamma_{a,w}(0)=a$ and $\dot\gamma(0)=w$ has the form
$${\bf R} \ni t \mapsto g_a \Big( T_{g_a^\prime(0)^{-1} w} \
{\rm artanh}\big( t\ {\rm id}_{\Omega_{g_a^\prime(0)^{-1} w}} \big) \Big) .$$
\bf Proof. \rm
The Proposition establishes the $h^\#$-invariance of the connection $\nabla$
for holomorphic automorphisms of the unit ball.
It is a well-known consequence of this fact that
$h\circ \gamma$ is $\nabla$-geodesic whenever $\gamma$ is a $\nabla$-geodesic and
$h\in{\rm Aut}\ {\rm Ball}(Z)$.
The curve in the statement of the Corollary is thus
a $\nabla$-geodesic.
It is straightforward to see that its starting point is $a$ and its starting speed vector
is $w$.
The classical Piccard-Lindel\"of Theorem ensures
the uniqueness of the solution of the geodesic equation with given starting point
and starting speed vector. \ Qu.e.d.

\bigskip
\bigskip
\bf 2. On local M\"obius transformations in real Jordan-Banach triples
\rm
\bigskip
\noindent

Throughout this section $Z$ denotes a real Jordan-Banach triple
with the norm $\Vert .\Vert$ and triple product $\{ \ldots\}$, respectively.
Thus we only assume that $Z$ is a real Banach space and
$\{\ldots\}$ is a continuous real trilinear mapping
$Z^3\to Z$ satisfying the Jordan identity
$$\{ ab \{ xyz\}\} = \{\{ abx\} yz\} - \{ a\{ bay\} z\} +
\{ xy\{ abz\}\} .$$
In particular,
the operators $D(a,b)-D(b,a):z\mapsto \{ abz\}-\{ baz\}$ are all
derivations of the triple product $\{\ldots\}$.
From the considerations in [Kaup, Math. Z. 183 (1983) 503-529] it follows that
for some $\eps>0$, the transformations
$$T_v := \exp V_v \quad \hbox{with the vector fields
$V_v:= \big[ v-\{ zvz\} \big] \partial/\partial z$}$$
with $\Vert v\Vert <\eps$ are welldefined on the ball $\eps B(Z)$ and
have the same M\"obius form
$$\eqalign{
&T_v (z) = g_{T_v(0)} (z) \quad\hbox{where}\cr
&g_a (z) := a + B(a)^{1/2} [1+D(z,a)]^{-1} z\cr}$$
as in the case of JB*-triples.
Besides the vector fields $V_v$ of polynomial degree 2, let us introduce also
the linear vector fields
$$L_\ell := \big[ \ell z \big] \partial / \partial z, \quad
\ell\in {\rm Der}(Z,\{\ldots\}) .$$
For their Poisson commutators
$\big[ f(z)\partial/\partial z,g(z)\partial/\partial z \big] :=
\big( f^\prime(z) g(z) - g^\prime(z) f(z) \big) \partial/\partial z$,
it is straightforward to check that we have
$$[L_\ell , V_v] =V_{\ell v} ,\quad
[L_\ell , L_m] = L_{[\ell,m]} ,\quad
[V_u,V_v] = L_{D(v,u)-D(u,v)} .$$
Therefore the real linear space
$$\big\{ V_v + L_\ell :\ v\in Z,\ \ell\in {\rm Der}(Z,\{\ldots\}) \big\}$$
equipped with the norm
$\Vert V_v+L_\ell \Vert := \sup\{ \Vert v-\{ zvz\} + \ell z\Vert :\
\Vert z\Vert \le 1 \}$ is a Banach-Lie algebra with the Poisson commutator.
Thus, according to the Campbell-Hausdorff formula,
for some small constant $\delta>0$ we have
$$\big[ \exp (V_u+L_\ell) \exp(V_v+L_m)\big] z =
\big[ \exp (V_{w(u,v,\ell,m)} + L_{\Delta(u,v,\ell,m)} \big] z$$
with two suitable real-analytic mappings
$$\eqalign{
&w: [\delta B(Z)]^2 \times [\delta B({\rm Der}(Z,\{\ldots\})]^2
\to Z, \cr
&\Delta: [\delta B(Z)]^2 \times [\delta B({\rm Der}(Z,\{\ldots\})]^2
\to {\rm Der}(Z,\{\ldots\}) \cr}$$
whenever
$\Vert z\Vert, \Vert u\Vert, \Vert v\Vert, \Vert\ell\Vert, \Vert m\Vert <\delta$.

\bigskip
\noindent
\bf Proposition. \it There exists $\delta^\prime >0$ such that
$$\Vert v\Vert,\Vert\ell\Vert < \delta^\prime \ {\rm and} \
[\exp (V_v+L_\ell)]0 =0 \quad \Longrightarrow \ v=0 .$$
\bf Proof. \rm
We can choose $\delta^\prime$ to be so small that
the mapping $u\mapsto [\exp(V_u)]0$ be injective on the ball
$\delta^\prime B(Z)$ and
the terms
$$[\exp (t L_m)] [\exp(V_u+L_m)] [\exp(-tL_m)]z, \quad
\vert t\vert, \Vert z\Vert, \Vert u\Vert, \Vert m\Vert <\delta^\prime$$
be all welldefined. Consider any couple $v,\ell$ with
$\Vert v\Vert,\Vert\ell\Vert < \delta^\prime$ and
$[\exp (V_v+L_\ell)]0 =0$.
Then, for all $-\delta^\prime < t<\delta^\prime$,
$$\eqalign{
0 &= [\exp (t L_\ell)] [\exp(V_v+L_\ell)] [\exp(-tL_\ell)]0 =\cr
&= \Big[ \exp\Big( \big[ \exp (t L_\ell^\#) \big] (V_v+L_\ell) \Big)\Big] 0 =\cr
&= \Big[ \exp \big( V_{\exp t\ell)v} + L_\ell \big)\Big]0 .\cr}$$
%Here we have
%$V_{\exp t\ell)v} + L_\ell = (V_v + L_\ell) +t V_{\ell v} + o(t)$ and hence
%$$0 = \Big[ \exp \big( V_{\exp t\ell)v} + L_\ell \big)\Big]0 =
%\big[ \exp(V_v+L_\ell + t V_{\ell v})\big]0 + o(t) .$$
Notice that the points
$$x_{s,t} := \Big[ \exp \big( s [ V_{(\exp t\ell)v} + L_\ell] \big) \Big]0, \quad
\vert s\vert <1+\eps^\prime,\ \vert t\vert <\delta^\prime$$
are welldefined for some $\eps^\prime>0$
and they satisfy the differential equation
$${\partial\over \partial s} x_{s,t}  = (\exp t\ell)v -
\{ x_{s,t} [(\exp t\ell)v] x_{s,t} \} + \ell x_{s,t} .$$
Moreover, since $\big[ \exp \big( V_{\exp t\ell)v} + L_\ell \big)\big]0 =0$
for $\vert t\vert< \delta^\prime$, it follows
$$\eqalign{
0 &= {\partial^2 \over \partial s \partial t}\Big\vert_{s=1,\ t=0} x_{s,t} =
{\partial^2 \over \partial t \partial s}\Big\vert_{s=1,\ t=0} x_{s,t} =\cr
&= \ell v - 2 \Big\{
\Big( {\partial\over\partial t}\Big\vert_{t=0,\ s=1} x_{s,t} \Big) v x_{s,t} \Big\} +
\ell {\partial\over\partial t}\Big\vert_{t=0,\ s=1} x_{s,t} .\cr}$$
However, here we have
${\partial\over\partial t}\Big\vert_{t=0,\ s=1} x_{s,t} =0$ which implies
that necessarily $\ell v =0$.
It follows
$[ L_\ell, V_v] = V_{\ell v} =0$ and therefore
$$0 = [\exp (V_v+L_\ell)]0 = [\exp V_v][\exp L_\ell]0 =
[\exp V_v]0 .$$
The choice of $\delta^\prime$ with $\Vert v\Vert< \delta^\prime$ ensures that
$v=0$. Qu.e.d.

\bigskip
\noindent
\bf Corollary. \it There exists $0<\delta^{\prime\prime} (<\delta^\prime)$
such that given any $v\in Z$ and $\ell \in{\rm Der}()Z.\{\ldots\})$ with
$\Vert v\Vert, \Vert \ell \Vert <\delta^{\prime\prime}$,
for some $m\in{\rm Der}(Z,\{\ldots\})$ we have
$$\big[ \exp (V_v + L_\ell)\big] z =
g_{a(v,\ell)}\big( [ \exp m ] z \big),
\quad \Vert z\Vert <\delta^{\prime\prime}$$
where $a(v,\ell) := \big[ \exp (V_v+L_\ell)\big]0$.
In particular, for sufficiently small vectors $p,q\in Z$,
$$g_p \circ g_q = g_{g_p(q)} \circ [\exp m_{p,q}]$$
with a real-analytic mapping $(p,q)\mapsto m_{p,q} \in{\rm Der}Z,\{\ldots\})$.

\bigskip
\bigskip
\bf 3. Jordan manifolds
\bigskip
\noindent
\rm
Throughout this section $Z$ dentotes a Jordan-Banach triple
with the norm $\Vert .\Vert$ and triple product $\{\ldots\}$, respectively.
By a
{\it M\"obius transformation}
on $Z$ we mean a bianalytic mapping $\Phi:U\to Z$
defined in a neighborhood $U$ of the origin in $Z$
such that for some $\ell\in{\rm Der}(Z)$
we have
$\Phi(z)= g_a (\exp\ell z)$, $z\in U$
with the customary notation
$g_a(z) := a + B(a)^{1/2} [1-D(z,a)]^{-1}$.

\bigskip
\noindent
\bf Definition. \rm
By a {\it Jordan manifold} modeled with $Z$ we mean
a a real-analytic manifold $M$ with an atlas
${\cal A} =\{ X_p:\ p\in M\}$ such that
for each point $p\in M$ we have
$X_p:U_p\to M$ with $X_p(0) =p$ where $U_p$ is an open
%connected
neighborhood
of the origin in $Z$
and the
chart transition maps $X_p^{-1}\circ X_q$ are M\"obius transformations
whenever $q\in X_p(U_p)$.


By a {\it uniform Jordan manifold} modelled with $Z$ we mean
a a real-analytic manifold $M$ with an atlas
${\cal A} =\{ X_p:\ p\in M\}$ such that
$X_p: \eps B(Z) \to M$ and $X_p(0)=p$ for all $p\in M$
with some common $\eps>0$ and the chart transition maps
$X^{-1}_p\circ X_q$ are M\"obius transformations
whenever $p\in M$ and $q \in X_p\big( \delta B(Z)\big)$
with a common constant $\eps>\delta>0$.

\bigskip
\noindent
\bf Example. \rm
If $Z$ is a JB*-triple then its unit ball
$M:=B(Z)$ is a uniform Jordan manifold with the charts
$X_p := g_p\vert B(Z)$, $p\in M$. In this case
we may choose $\eps=\delta=1$.

\bigskip
\noindent
\bf Example. \rm
Let $Z$ be any Jordan-Banach triple.
Then we can find a constant $\varrho>0$
such that
the sections of the mapping
$[\varrho B(Z)]^2 \ni(p,z)\mapsto
g_p(z)$
are real-bianalytic for any fixed $p$ and $z$, respectively.
Then the ball
$M:= \varrho B(Z)$ with the topology from $Z$
and with the charts
$X_{p} (z) := g_p(z)$ defined for
$z\in U_p := \{ u\in Z:\ \Vert u\Vert, \Vert g_p(u) \Vert < \varrho\}$
is a Jordan manifold which is not uniform in general.

\bigskip
\noindent
\bf Lemma. \it
Assume $U,V$ are domains in a Banach space $W$ and let
$T:U\leftrightarrow V$ be a smooth diffeomorphism between them.
Then given any couple $X,Y:U\to W$ of smooth vector fields on $U$,
for their transforms
$\widetilde{X} := T^\# X$ respectively
$\widetilde{Y} := T^\# Y$ on $V$ we have\footnote{$^*$}{\rm
For notation: $^\prime$ means Fr\'echet derivative. In particular,
$T^\prime (u) w \!=\! {d\over dt}\big\vert_{t=0} T( u \!+\! t w)$,
$T^{\prime\prime}(u) w_1 w_2 = {\partial^2 \over \partial t_1 \partial t_2}
\big\vert_{t_1=t_2=0} T(u+t_1 w_1+t_2 w_2)$ \ and \
$T^\# X (v) = T^\prime \big( T^{-1}(v)\big) X\big( T^{-1}(v)\big)$.}
$$\widetilde{Y}^\prime \widetilde{X} (v) =
T^{\prime\prime} \big(T^{-1}(v)\big) X\big( T^{-1}(v) \big) Y\big( T^{-1}(v) \big)
+ T^\prime \big(T^{-1}(v)\big) Y^\prime \big( T^{-1}(v) \big)
X\big( T^{-1}(v) \big) ,\ \ v\!\in\! V .$$
\bf Proof. \rm
By definition,
$\widetilde{Y}^\prime \widetilde{X} (v) =
{d\over dt}\big\vert_{t=0} \widetilde{Y} \big( v + t \widetilde{X}(v)\big)$.
Therefore
$$\eqalign{
\widetilde{Y}^\prime \widetilde{X} (v) &=
{d\over dt}\Big\vert_{t=0}
T^\prime \Big( T^{-1} \big( v + t\widetilde{X}(v) \big)\Big)
Y \Big( T^{-1} \big( v + \widetilde{X}(v) \big)\Big) = \cr
&= \Big[ {d\over dt}\Big\vert_{t=0}
T^\prime \Big( T^{-1} \big( v + t\widetilde{X}(v) \big)\Big) \Big]
Y \big( T^{-1} (v)\big) + \cr
& \ \ + T^\prime \big( T^{-1}(v)\big) {d\over dt}\Big\vert_{t=0}
Y \big( v + t \widetilde{X}(v)\big) =\cr
&= T^{\prime\prime} \Big( T^{-1} (v) \big)
\Big[ {d\over dt}\Big\vert_{t=0}
T^{-1} \big( v + t\widetilde{X}(v) \big) \Big] Y\big( T^{-1}(v)\big) +\cr
& \ \ + T^\prime \Big( T^{-1} (v) \big) Y^\prime\big( T^{-1}(v)\big)
{d\over dt}\Big\vert_{t=0} T^{-1} \big( v + t\widetilde{X}(v) \big) =\cr
&= T^{\prime\prime} \Big( T^{-1} (v) \big)
\Big[ T^\prime \big( T^{-1} (v)\big)^{-1} \widetilde{X}(v) \Big]
Y\big( T^{-1}(v)\big) +\cr
& \ \ + T^\prime \Big( T^{-1} (v) \big) Y^\prime \big( T^{-1}(v)\big)
\big[ T^\prime \big( T^{-1}(v)\big)\big]^{-1}
\widetilde{X}\big( T^{-1}(v) \big) =\cr
&= T^{\prime\prime} \Big( T^{-1} (v) \big)
X\big( T^{-1}(v) \big) Y\big( T^{-1}(v) \big) + \cr
& \ \ + T^\prime \Big( T^{-1} (v) \big)
Y^\prime \big( T^{-1}(v) \big) X\big( T^{-1}(v) \big) . \ \ \hbox{Qu,e.d.}\cr}$$

In the sequel let $\big( M,\{ X_p:\ p\in M\}\big)$ be a Jordan-manifold
with charts $X_p: Z \supset U_p\to M$ satisfying $X_p(0)=p$, $p\in M$ and
with M\"obius transition maps $X_p^{-1}\circ X_q$ for couples of point
lying sufficiently close together.

\def\X{{\bf X}}
\def\Y{{\bf Y}}
\def\R{{\bf R}}

\bigskip
\noindent
\bf Proposition,
\it
Let $p\in M$ be an arbitrarily given point,
let $\X,\Y$ be two smooth vector fields on $M$
and define $\R := \nabla_{\X} \Y$.
Then, for any vector $w\in U_p$,
the image $R:=[X_p^{-1}]^\# \R$ of the vector field $\R$
by means of the local coordinate $X_p$ can be espressed
in terms of the image vector fields
$X:=[X_p^{-1}]^\# \X$ and $Y:=[X_p^{-1}]^\# \Y$
as
$$R(w) = T_w^\prime(w)^{-1} T_w^{\prime\prime}(w) X(w) Y(w) +
Y^\prime (w) X(w)
\quad \hbox{\rm where} \ \ T_w := X_{X_p(w)}^{-1} \circ X_p .$$
\bf Proof. \rm
Let $w\in U_p$ be arbitrarily fixed and write $q:=X_p(w)$.
Then, with the mapping $T:= T_w$
we have
$$T(w) = X_q^{-1}\circ X_p(w) = X_q^{-1} (q) =0 .$$
Also in terms of $T$ we can write
$$\eqalign{
[X_q^{-1}]^\# \X &= [X_q^{-1}]^\# X_p^\# [X_p^{-1}]^\# \X =\cr
&= [X_q^{-1}\circ X_p]^\# [X_p^{-1}]^\# \X =\cr
&= T^\# X \cr}$$
and similarly
$[X_q^{-1}]^\# \Y = T^\# Y$ and
$[X_q^{-1}]^\# \R = T^\# R$.
Hence
$$\eqalign{
[X_q^{-1}]^\# \R (0) &= \big[ T^\# \big( [X_p^{-1}]^\# \R \big) \big] (0) =\cr
&= T^\prime \big( T^{-1}(0)\big) \big[ [ X_p^{-1}]^\# \R \big]
\big( T^{-1}(0)\big) =\cr
&= T^\prime(w) \Big[ \big[ [X_p^{-1}]^\# \R \big] (w) \Big] =\cr
&= T^\prime(w) R(w) .\cr}$$
By axiomatic assumption we have \quad
$[X_q^{-1}]^\# \R (0) =
\Big[\big[ [X_q^{-1}]^\# \Y\big]^\prime (0)\Big]
\Big[\big[ [X_q^{-1}]^\# \X\big] (0)\big]$.
That is
$$\eqalign{
R(w) &= [X_p^{-1}]^\# \R (w) =\cr
&= T^\prime(w)^{-1}
\Big[ \big[ [X_q^{-1}]^\# \Y\big]^\prime (0) \Big]
\Big[ \big[ [X_q^{-1}]^\# \X\big] (0) \Big] =\cr
&= T^\prime(w)^{-1}
\big[ (T^\# Y)^\prime (0)\big]
\big[ T^\# X (0) \big] =\cr
&= T^\prime(w)^{-1} T^{\prime\prime}(w) X(w) Y(w) +
Y^\prime(w) X(w) \cr}$$
in view of the lemma and the fact that $w=T^{-1}(0)$. \ \
Qu.e.d.








\bye
