Let H be a Hilbert space, E := L(H). We consider the mapping ϕ : Ball(E) ↔ M := \{[{t \atop x}] ∈ E_{+} × E : t^{2} − x^{∗}x = 1, t ≥ 1\},

ϕ(u) := \left[{(1 + a^{∗}a)(1 − a^{∗}a)^{−1} \atop 2a(1 − a^{∗}a)^{−1}}\right] with ϕ^{−1} \left[{t \atop x}\right] = x/(1 + t) .

Remark. In the 1-dimensional commutative case E = C, for the Möbius transformation M_{b}(u) := (u + b)/(1 − \overline{b}u) we have

ϕ^{#}M_{b} \left[{t \atop x}\right] = ϕ ∘ M_{b} ∘ ϕ^{−1} \left[{t \atop x}\right] = [ {\frac{1+|b|^{2}}{1−|b|^{2}} t + \frac{2b}{1−|b|^{2}} x \atop \frac{2b}{1−|b|^{2}} t + \frac{1+|b|^{2}}{1−|b|^{2}} x}] .

Notice that M_{tanh(a)} is the exponential of the vector field [a − \overline{a}u^{2}]\frac{∂}{∂u}. Hence

ϕ^{#}M_{tanh(a)} \left[{t \atop x}\right] = \left[{cosh(2a)t + sinh(2a)x \atop sinh(2a)t + cosh(2a)x}\right] for \left[{t \atop x}\right] ∈ M .

In particular ϕ^{#}M_{b} is the restriction of a linear mapping of C^{2} to M.

We are going to study the analogues in the setting of E = L(H).

Notation. For any a ∈ L(H) and b ∈ Ball(L(H)) we write

A_{a}(u) := [a − ua^{∗}u]\frac{∂}{∂u}, 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 (∥u∥ < 1). Recall that, for fixed a ∈ L(H) and u ∈ BallL(H), the function u_{τ} := [exp(τA_{a})]u (τ ∈ IR) is the solution of the initial value problem \frac{d}{dτ} u_{τ} = A_{a}(u_{τ}), u_{0} = u.

Proposition. Let a ∈ L(H) and \left[{t \atop x}\right] ∈ M. Then

ϕ^{#}A_{a} \left[{t \atop x}\right] = \left[{a^{∗}x + x^{∗}a \atop a(1 + t) + x(1 + t)^{−1}x^{∗}a}\right] .

Proof. Let us write

u := ϕ^{−1} \left[{t \atop x}\right] = x(1 + t)^{−1}, \overline{u} := A_{a}(u), G_{τ} := exp(τA_{a}) .

Then we have

ϕ^{#}A_{a} \left[{t \atop x}\right] = \frac{d}{dτ} |_{τ=0}ϕ ∘ G_{τ} ∘ ϕ^{−1} \left[{t \atop x}\right] = \frac{d}{dτ} |_{τ=0}ϕ ∘ G_{τ} (u) =

= ϕ′(G_{0}(u))\frac{d}{dτ} |_{τ=0}G_{τ} (u) = ϕ′(u)A_{a}(u) =

= ϕ′(u)\overline{u}.

We calculate both ϕ′(u) and \overline{u} in terms of t, x. For the first component of ϕ,

\begin{aligned}ϕ_{1}(u) & = (1 + u^{∗}u)(1 − u^{∗}u)^{−1} = [(1 − u^{∗}u) + 2u^{∗}u](1 − u^{∗}u)^{−1} = \\ & = 1 + u^{∗}ϕ_{2}(u) . \end{aligned}

Hence and since ϕ_{1}(u) = x,

\begin{aligned}ϕ′_{1}(u)\overline{u} & = \frac{d}{dτ} |_{τ=0}(u + τ\overline{u})^{∗}ϕ_{2}(u + τ\overline{u}) = \overline{u}^{∗}ϕ_{2}(u) + u^{∗}ϕ′_{2}(u)\overline{u} = \\ & = \overline{u}^{∗}x + u^{∗}ϕ′_{2}(u)\overline{u} . \end{aligned}

We can express ϕ′_{2}(u)\overline{u} in algebraic terms of u, \overline{u} as follows:

\begin{aligned}ϕ′_{2}(u)\overline{u} & = \frac{d}{dτ} |_{τ=0}ϕ_{2}(u + τ\overline{u}) = \frac{d}{dτ} |_{τ=0}2(u + \overline{u})[1 + (u + \overline{u})^{∗}(u + \overline{u})]^{−1} = \\ & = 2\overline{u}[1 − u^{∗}u]^{−1} + 2u[1 − u^{∗}u]^{−1}(\overline{u}^{∗}u + u^{∗}\overline{u})[1 − u^{∗}u]^{−1} . \end{aligned}

Since u = ϕ^{−1} \left[{t \atop x}\right] = x(1 + t)^{−1} and since x^{∗}x = t^{2} − 1 = (t − 1)(1 + t), here we have

1 − u^{∗}u = 1 − (1 + t)^{−1}x^{∗}x(1 + t)^{−1} =
= 1 − (1 + t)^{−1}(t − 1) = (1 + t)^{−1}[(1 + t) − (t − 1)] =
= 2(1 + t)^{−1} ,
[1 − u^{∗}u]^{−1} = \frac{1}{2}(1 + t) ,
u[1 − u^{∗}u]^{−1} = x(1 + t)^{−1} \frac{1}{2}(1 + t) = \frac{1}{2}x .

Hence we conclude

\begin{aligned}ϕ′_{2}(u)\overline{u} & = 2\overline{u}\frac{1}{2}(1 + t) + 2 · \frac{1}{2}x[\overline{u}^{∗}x(1 + t)^{−1} + (1 + t)^{−1}x^{∗}\overline{u}]\frac{1}{2}(1 + t) = \\ & = \overline{u}(1 + t) + \frac{1}{2}x\overline{u}^{∗}x + \frac{1}{2}x(1 + t)^{−1}x^{∗}\overline{u}(1 + t) . \end{aligned}

We can express \overline{u} in terms of t, x as

\overline{u} = A_{a}(u) = a − ua^{∗}u = a − x(1 + t)^{−1}a^{∗}x(1 + t)^{−1} .

Thus

ϕ′_{2}(u)\overline{u} = \overbrace{a(1 + t)}^{(1)} \overbrace{−x(1 + t)^{−1}a^{∗}x}^{(2)} +

\overbrace{+\frac{1}{2}xa^{∗}x}^{(3)} \overbrace{−\frac{1}{2}x(1 + t)^{−1}x^{∗}a(1 + t)^{−1}x^{∗}x}^{(4)} +

\overbrace{+\frac{1}{2}x(1 + t)^{−1}x^{∗}a(1 + t)}^{(5)} \overbrace{−\frac{1}{2}x(1 + t)^{−1}x^{∗}x(1 + t)^{−1}a^{∗}x}^{(6)} .

The sum (2)+(3)+(6) vanishes because x^{∗}x = (1 + t)(t − 1) and hence

\begin{aligned}(2) + (3) + (6) & = x [−(1 + t)^{−1} + \frac{1}{2} − \frac{1}{2}(t − 1)(1 + t)^{−1}] a^{∗}x = \\ & = x(1 + t)^{−1} [−1 + \frac{1}{2} + \frac{1}{2}t − \frac{1}{2}t + \frac{1}{2}] a^{∗}x = 0 . \end{aligned}

The sum (4)+(5) can also be simplified as

(4) + (5) = \frac{1}{2}x(1 + t)^{−1}x^{∗}a[ − (1 + t)^{−1} \overbrace{x^{∗}x}^{(1+t)(t−1)} +(1 + t)] =

= \frac{1}{2}x(1 + t)^{−1}x^{∗}a [−(t − 1) + (1 + t)] =

= x(1 + t)^{−1}x^{∗}a .

Summing up (1) + ⋯ + (6), we get

ϕ′_{2}(u)\overline{u} = a(1 + t) + x(1 + t)^{−1}x^{∗}a ,
ϕ′_{1}(u)\overline{u} = \overline{u}^{∗}x + u^{∗}ϕ′_{2}(u)\overline{u} =

= a^{∗}x − (1 + t)^{−1}x^{∗}a(1 + t)^{−1}+

+ (1 + t)^{−1}x^{∗}[a(1 + t) + x(1 + t)^{−1}x^{∗}a big] =

= \overbrace{a^{∗}x}^{(1)} \overbrace{−(1 + t)^{−1}x^{∗}a(1 + t)^{−1}x^{∗}x}^{(2)} +

= \overbrace{+(1 + t)^{−1}x^{∗}a(1 + t)}^{(3)} \overbrace{+(1 + t)^{−1}x^{∗}x(1 + t)^{−1}x^{∗}a}^{(4)} .

Using again the identity x^{∗}x = (1 + t)(t − 1), here we can write

(2) + (3) = (1 + t)^{−1}x^{∗}a[ − (1 + t)^{−1}(1 + t)(t − 1) + (1 + t)] =
= 2(1 + t)^{−1}x^{∗}a ,
(4) = (1 + t)^{−1}(1 + t)(t − 1)(1 + t)^{−1}x^{∗}a = (t − 1)(1 + t)^{−1}x^{∗}a =
= −[(1 + t) − 2t](1 + t)^{−1}x^{∗}a = −x^{∗}a + 2t(1 + t)^{−1}x^{∗}a .

Therefore

ϕ′_{1}(u)\overline{u} = [(1) + (4)] + F (2) + (3)] =
= a^{∗}x − x^{∗}a + 2t(1 + t)^{−1}x^{∗}a + 2(1 + t)^{−1}x^{∗}a =
= a^{∗}x + (1 + t)^{−1}[ − (1 + t) + 2t + 2]x^{∗}a =
= a^{∗}x + x^{∗}a

which completes the proof. Qu.e.d. Corollary. If x is a normal operator and \left[{t \atop x}\right] ∈ M then \{x, x^{∗}, t\} commute and x^{∗}x = (t − 1)(1 + t) whence

ϕ^{#}A_{a} \left[{t \atop x}\right] = \left[{a^{∗}x + x^{∗}a \atop a^{∗}t + ta^{∗}}\right] .

The case of the real Jordan algebra of self-adjoint operators

Let henceforth only E = A(H) := {a ∈ L(H) : a = a^{∗}}. In this case

ϕ^{#}A_{a} \left[{t \atop x}\right] = \left[{ax + xa \atop at + ta}\right] = [ \begin{matrix}0 & L_{a} + R_{a} \\ L_{a} + R_{a} & 0 \end{matrix} ] \left[{t \atop x}\right]

with the left and right multiplication operators L_{a} : L(H) ∋ z ↦ az respectivaly R_{a} : L(H) ∋ z ↦ za. Since L_{a} and R_{a} commute, exp (ϕ^{#}A_{a}) has a rather simple explicite

expression. Straightforward calculations with the power series ∑_{n=0}^{∞} n!^{−1} \left[{0 L_{a}+R_{a} \atop L_{a}+R_{a} 0}\right]^{n} \left[{t \atop x}\right], yield the following.

Theorem. Given any self-adjoint operator a ∈ A(H), the Möbius transformation M_{tanh(a)} maps Ball(A(H)) onto itself, ϕ is a bijection between Ball(A(H)) and M ∩ A(H)^{2} = \{[{t \atop x}] : t, x ∈ A, t^{2} − x^{2} = 1, t ≥ 1\} and

ϕ^{#}M_{tanh(a)} = ϕ ∘ [exp A_{a}] ∘ ϕ^{−1} \left[{t \atop x}\right] = \left[{cosh(L_{a} + R_{a})t + sinh(L_{a} + R_{a})x \atop sinh(L_{a} + R_{a})t + cosh(L_{a} + R_{a})x}\right] .

For the terms cosh(L_{a} + R_{a})z, sinh(L_{a} + R_{a})z (z = t, x) here we have

cosh(L_{a} + R_{a})z = \frac{1}{2}[exp(a)]z[exp(a)] + \frac{1}{2}[exp(−a)]z[exp(−a)] ,
cosh(L_{a} + R_{a})z = \frac{1}{2}[exp(a)]z[exp(a)] − \frac{1}{2}[exp(−a)]z[exp(−a)] .

Proof. For any a, u ∈ A(H), we have A_{a}(u) = a − {uau} ∈ A(H) along with ϕ_{1}(u) = (1 + u^{2})(1 − u^{2})^{−1} ∈∈ A(H) and ϕ_{2}(u) = 2u(1 − u^{2})^{−1} ∈∈ A(H). Therefore the vector field A_{a} is tangent to A(H) moreover complete in A(H) ∩ Ball(L(H) = Ball(A(H) and the vector field ϕ^{#}A_{a} is complete in A(H) ∩ M. Since exp(A_{a}) = M_{tanh(a)}, it follows

ϕ^{#}M_{tanh(a)} = exp ϕ^{#}A_{a} = exp [ \begin{matrix}0 & L_{a} + R_{a} \\ L_{a} + R_{a} & 0 \end{matrix} ]| M .

Here we have

exp [ \begin{matrix}0 & L_{a} + R_{a} \\ L_{a} + R_{a} & 0 \end{matrix} ] = ∑_{n=0}^{∞} \frac{1}{n!} [ \begin{matrix}0 & L_{a} + R_{a} \\ L_{a} + R_{a} & 0 \end{matrix} ]^{n} =

= ∑_{k=0}^{∞} \frac{1}{(2k)!} [ \begin{matrix}0 & L_{a} + R_{a} \\ L_{a} + R_{a} & 0 \end{matrix} ]^{2k} + ∑_{k=0}^{∞} \frac{1}{(2k + 1)!} [ \begin{matrix}0 & L_{a} + R_{a} \\ L_{a} + R_{a} & 0 \end{matrix} ]^{2k+1} =

= ∑_{k=0}^{∞} \frac{1}{(2k)!} [ \begin{matrix}(L_{a} + R_{a})^{2k} & 0 \\ 0 & (L_{a} + R_{a})^{2k} \end{matrix} ] + ∑_{k=0}^{∞} \frac{1}{(2k + 1)!} [ \begin{matrix}0 & (L_{a} + R_{a})^{2k+1} \\ (L_{a} + R_{a})^{2k+1} & 0 \end{matrix} ] =

= [ \begin{matrix}cosh(L_{a} + R_{a}) & 0 \\ 0 & cosh(L_{a} + R_{a}) \end{matrix} ] + [ \begin{matrix}0 & sinh(L_{a} + R_{a}) \\ sinh(L_{a} + R_{a})^{2k+1} & 0 \end{matrix} ] .

In general, the operations L_{x} and R_{x} commute (L_{x}(R_{y}u) = x(uy) = (xu)y = L_{x}(R_{y}u). Therefore

cosh(L_{a} + R_{a}) = \frac{1}{2} exp(L_{a} + R_{a}) + \frac{1}{2} exp(−L_{a} − R_{a}) =

= \frac{1}{2} exp(L_{a}) exp(R_{a}) + \frac{1}{2} exp(−L_{a}) exp(−R_{a}) =

= \frac{1}{2}L_{exp(a)}R_{exp(a)} + \frac{1}{2}L_{exp(a)}R_{exp(a)} : z ↦ \frac{1}{2}e^{a}ze^{a} + \frac{1}{2}e^{−a}ze^{−a}.

Similarly sinh(L_{a} + R_{a}) : z ↦ \frac{1}{2}e^{a}ze^{a} − \frac{1}{2}e^{−a}ze^{−a} which completes the proof. Qu.e.d.

Case of JB*-triples Throughout this section let E be a JB*-triple with the triple product {uvw} and structure derivations D(a, b) : u ↦ {abu}. We define the mapping ϕ : Ball(E) → L(E) × E as

ϕ(u) := \left[{ϕ_{1}(u) \atop ϕ_{2}(u)}\right]  where  \begin{matrix}ϕ_{2}(u) := 2[1 − D(u, u)]^{−1}u , \\ ϕ_{1}(u) := 1 + D(ϕ_{1}(u), u) . \end{matrix}

We shall also write M := range(ϕ) = {ϕ(u) : u ∈ Ball(E)}. Lemma. On M we have ϕ^{−1} \left[{t \atop x}\right] = (1 + t)^{−1}x . Proof. Let u ∈ Ball(E). We have to see that u = (1 + t)^{−1}x that is (1 + t)u = x with

x := 2[1 − D(u, u)]^{−1}u , t := 1 + D(x, u) .

Recall that, using the customary notation u^{2k+1} := D(u, u)^{k}u (k = 0, 1, …), we have the identity {u^{2k+1}u^{2ℓ+1}u^{2m+1}} = u^{2(k+ℓ+m)+1}. Therefore

x = 2[1 − D(u, u)]^{−1}u = 2 ∑_{k=0}^{∞} D(u, u)^{k} = 2 ∑_{k=0}^{∞} u^{2k+1} ,

(1 + t)u = [2 + D(x, u)]u = [2 + 2 ∑_{k=0}^{∞} D(u^{2k+1}, u)]u = 2u + 2 ∑_{k=0}^{∞} D(u^{2k+1}, u)u] =

= 2u^{1} + 2 ∑_{k=0}^{∞} u^{2k+3} = 2 ∑_{k=0}^{∞} u^{2k+1} = x . Qu.e.d.

Let \left[{t \atop x}\right] ∈ M be arbitrarily given and write

u := ϕ^{−1} \left[{t \atop x}\right] , \overline{u} := A_{a}(u) = a − \{uau\} .

Our aim is to express the vector ϕ^{#}A_{a} \left[{t \atop x}\right] in terms of x, t. We know already that

\begin{matrix}ϕ_{2}(v) = 2[1 − D(v, v)]^{−1}v,  ϕ_{1}(v) = 1 + D(ϕ_{2}(v), v)  (v ∈ Ball(E)), \\ x = ϕ_{2}(u) = 2[1 − D(u, u)]^{−1}u ,  t = ϕ_{1}(u) = 1 + D(x, u) . \end{matrix}

Hence

\begin{aligned}ϕ′_{1}(u)\overline{u} & = \frac{d}{dτ} |_{τ=0}D(ϕ_{2}(u + τ\overline{u}), u + τ\overline{u}) = D(ϕ′_{2}(u)\overline{u}, u) + D(ϕ_{2}(u), \overline{u}) = \\ & = D(ϕ′_{2}(u)\overline{u}, u) + D(x, \overline{u}) . \end{aligned}

The algebraic term for ϕ′_{2}(u)\overline{u} is

ϕ_{2}(u)\overline{u} = \frac{d}{dτ} |_{τ=0}2[1 − D(u + τ\overline{u}, u + τ\overline{u})]^{−1}(u + τ\overline{u}) =

= 2[1 − D(u, u)]^{−1}[D(\overline{u}, u) + D(u, \overline{u})][1 − D(u, u)]^{−1}u + 2[1 − D(u, u)]^{−1}\overline{u} =

= [1 − D(u, u)]^{−1}\{[D(\overline{u}, u) + D(u, \overline{u})]x + 2\overline{u}\} .

Since D(\overline{u}, u)x = \{\overline{u}ux\} = \{xu\overline{u}\} = D(x, u)\overline{u} and t = 1 + D(x, u), it follows

ϕ′_{2}(u)\overline{u} = [1 − D(u, u)]^{−1}\{(1 + t)\overline{u} + D(u, \overline{u})x\} .

Observe that, by writing

Der_{∗}(E, {…}) := {δ ∈ L(E) : δ{abc} = {[δa]bb} − {a[δb]c + {ab[δc]} (ab, c ∈ E)}

for the Jordan derivations of (E, {…}), we have

D(x, u) = D(u, x) ∈ Der_{∗}(E, {…}) .

Indeed, D(x, u) = ∑_{k=0}^{∞} D(u^{2k+1}u, u) = ∑_{k=0}^{∞} 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,

(1 + t)\overline{u} = (1 + t)a − (1 + t)\{uau\} =
= (1 + t)a − 2{[(1 + t)u]au} + {u[(1 + t)a]u} =
= (1 + t)a + {u[(1 + t)a]u} − 2{xau} ,
ϕ_{2}(u)\overline{u} = [1 − D(u, u)]^{−1}\{(1 + t)\overline{u} + D(u, \overline{u})x\} =
= [1 − D(u, u)]^{−1}\{(1 + t)a + \{u[(1 + t)a]u\} − 2\{xau\} + D(u, \overline{u})x\} .