1. One-dimensional case

ϕ(a) = [ \begin{matrix}ϕ_{1}(a) \\ ϕ_{2}(a) \end{matrix} ] (a ∈ C)

ϕ_{1}(a) := \frac{1 + |a|^{2}}{1 − |a|a|^{2}} , ϕ_{2}(a) := \frac{2a}{1 − |a|a|^{2}} .

\begin{aligned}ϕ_{1}(a)^{2} − |ϕ_{2}(a)|^{2} & = \frac{1 + 2|a|^{2} + |a|^{4} − 2|q|^{2}}{[1 − |a|^{2}]^{2}} = \\ & = \frac{[1 − |a|^{2}]^{2}}{[1 − |a|^{2}]^{2}} = 1 . \end{aligned}

ϕ′(a)u = \frac{∂}{∂t}|_{τ=0}ϕ(a + τu).

\begin{aligned}ϕ′_{1}(a)u & = \frac{(\overline{a}u + \overline{u}a)[1 − |a|^{2}] − [1 + |a|^{2}](\overline{a}u + \overline{u}a)}{[1 − |a|^{2}]^{2}} = 2\frac{\overline{a}u + \overline{u}a}{[1 − |a|^{2}]^{2}} , \\ ϕ′_{2}(a)u & = \frac{2u[1 − |a|^{2}] + 2a[1 + |a|^{2}]}{[1 − |a|^{2}]^{2}} . \end{aligned}

[ϕ′_{1}(a)u]^{2} − |ϕ′_{2}(a)u|^{2} = 4\frac{(\overline{a}u + \overline{u}a)^{2} − |u − a\overline{u}a|^{2}}{[1 − |a|^{2}]^{4}} .

(\overline{a}u + \overline{u}a)^{2} − |u − a\overline{u}a|^{2} =

= \underbrace{\overline{a}^{2}u^{2}}_{(1)} + \underbrace{2|a|^{2}|u|^{2}}_{(2)} + \underbrace{\overline{u}^{2}a^{2}}_{(3)} + \underbrace{|u|^{2}}_{(4)} − \underbrace{a\overline{u}^{2}a}_{(5)} − \underbrace{\overline{a}u^{2}\overline{a}}_{(6)} − \underbrace{|a|^{4}|u|^{2}}_{(7)} =

= [\underbrace{(1) − (6)}_{0}] + [\underbrace{(3) − (5)}_{0}] + |u|^{2}(2|a|^{2} − 1 − |a|^{4}) =

= −|u|^{2}[1 − |a|^{2}]^{2} .

[ϕ′_{1}(a)u]^{2} − |ϕ′_{2}(a)u|^{2} = −4\frac{|u|^{2}}{[1 − |a|^{2}]^{2}} .

Recall that the natural Riemannian metric on the unit disc B := {z ∈ C : |z| < 1}) is given by the inner products

⟨u, u⟩_{a}^{B} := |g′_{a}(0)^{−1}u|^{2} = \frac{|u|^{2}}{[1 − |a|^{2}]^{2}} (a ∈ B, u ∈ C).

Thus, in terms of the inverse Ψ := ϕ^{−1} : M ↔ B, with

t := ϕ_{1}(a), x := ϕ_{2}(a), t̃= ϕ′_{1}(a)u, x̃ = ϕ′_{2}(a), u

we have a = Ψ(t, x), u = [Ψ′(t, x)](t̃, x̃) and

t̃^{2} − |x̃|^{2} = −4⟨[Ψ′(t, x)](t̃, x̃), [Ψ′(t, x)](t̃, x̃)⟩_{Ψ(t,x)}^{B}.

2. Case of L(H) or TRO

ϕ(a) = [ \begin{matrix}ϕ_{1}(a) \\ ϕ_{2}(a) \end{matrix} ] (a ∈ L(H))

ϕ_{1}(a) := (I + a^{∗}a)(I − a^{∗}a)^{−1} , ϕ_{2}(a) := 2a(I − a^{∗}a)^{−1} .

ϕ_{1}(a)^{2} − ϕ_{2}(a)^{∗}ϕ_{2}(a) =
= (I − a^{∗}a)^{−1}(I + a^{∗}a)^{2}(I − a^{∗}a)^{−1} − (I − a^{∗}a)^{−1}(4a^{∗}a)(I − a^{∗}a)^{−1} =
= (I − a^{∗}a)^{−1}[I + 2a^{∗}a + (a^{∗}a)^{2} − 4a^{∗}a](I − a^{∗}a)^{−1} =
= (I − a^{∗}a)^{−1}[I − 2a^{∗}a + (a^{∗}a)^{2}](I − a^{∗}a)^{−1} =
= (I − a^{∗}a)^{−1}[I − 2a^{∗}a + (a^{∗}a)^{2}](I − a^{∗}a)^{−1} =
= (I − a^{∗}a)^{−1}(I − a^{∗}a)^{2}(I − a^{∗}a)^{−1} =
= I .

ϕ′_{1}(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} =
= (I − a^{∗}a)^{−1}[(I − a^{∗}a)(a^{∗}x + x^{∗}a) + (I + a^{∗}a)(a^{∗}x + x^{∗}a)] =
= (I − a^{∗}a)^{−1}(a^{∗}x + x^{∗}a)(I − a^{∗}a)^{−1} .
ϕ′_{2}(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} =

= 2(I − aa^{∗})^{−1}[(I − a^{∗}a)x + a(a^{∗}x + x^{∗}a)](I − a^{∗}a)^{−1} =
= 2(I − aa^{∗})^{−1}[x + ax^{∗}a](1 − a^{∗}a)^{−1}

Thus we have

range(ϕ) = M := {(t, x) ∈ L(H) × L(H) : t = t^{∗} ≥ I, t^{2} − x^{∗}x = I} ,
ϕ : B := {a : a^{∗}a < I} ↔ M := {(t, x) : t = t^{∗} ≥ I, t^{2} − x^{∗}x = I},
Ψ := ϕ^{−1} : (t, x) ↦ x(I + t)^{−1}

Aim. Find a convenient indefinite inner product ⟨(\overline{t}_{1}, \overline{x}_{1}), (\overline{t}_{2}, \overline{x}_{2})⟩_{(t,x)} on L(H) × L(H) such that ⟨(\overline{t}, \overline{x}), (\overline{t}, \overline{x})⟩_{(t,x)}^{M} := ⟨(\overline{t}, \overline{x}), (\overline{t}, \overline{x})⟩_{(t,x)} for (t, x) ∈ M and (\overline{t}, \overline{x}) ∈ T_{(t,x)}M with the inner product

⟨(\overline{t}, \overline{x}), (\overline{t}, \overline{x})⟩_{(t,x)}^{M}:=−4⟨[Ψ′(t, x)](\overline{t}, \overline{x}), [Ψ′(t, x)](\overline{t}, \overline{x})⟩_{Ψ(t,x)}^{B} for (t, x) ∈ M, (\overline{t}, \overline{x}) ∈ T_{(t,x)}M

with the natural Riemannian inner product on the unit ball B := {a : a^{∗}a < I},

\begin{aligned}⟨u, v⟩_{a}^{B} & := [[g′_{a}(0)]^{−1}u]^{∗}[[g′_{a}(0)]^{−1}v] = \\ & = (I − a^{∗}a)^{−1/2}u^{∗}(I − aa^{∗})^{−1}v(I − a^{∗}a)^{−1/2}. \end{aligned}

Let (t, x) ∈ M and (\overline{t}, \overline{x}) ∈ T_{(t,x)}M be fixed and write

a := Ψ(t, x) = x(I + t)^{−1},
u := [Ψ′(t, x)](\overline{t}, \overline{x}) = \frac{∂}{∂τ} |_{τ=0}(x + τ\overline{x})(I + t + τ\overline{t})^{−1} =
= [\overline{x} − x(I + t)^{−1}\overline{t}](I + t)^{−1}.

Then we have t = t^{∗} = [I + x^{∗}x]^{1/2} ≥ I with a^{∗} = (I + t)^{−1}x^{∗} and

a^{∗}a = (I + t)^{−1}x^{∗}x(I + t)^{−1} = (I + t)^{−1}(t^{2} − I)(I + t)^{−1} =
= (t − I)(I + t)^{−1} ,
I − a^{∗}a = (t + I)(I + t)^{−1} − (t − I)(I + t)^{−1} =
= 2(I + t)^{−1}.

Therefore

⟨u, u⟩_{a}^{B} = (I − a^{∗}a)^{−1/2}u^{∗}(I − aa^{∗})^{−1}u(I − a^{∗}a)^{−1/2} =

= \frac{1}{2}(I + t)^{1/2}[\overline{x}^{∗} − \overline{t}(I + t)^{−1}x^{∗}](I − aa^{∗})^{−1}[\overline{x} − x(I + t)^{−1}\overline{t}](I + t)^{1/2} =

= \frac{1}{2}(I + t)^{1/2}[\overline{x}^{∗} − \overline{t}a^{∗}](I − aa^{∗})^{−1}[\overline{x} − a\overline{t}](I + t)^{1/2} .

Since a^{∗}(I − aa^{∗})^{−1} = (I − a^{∗}a)^{−1}a^{∗} and (I − aa^{∗})(I − aa^{∗})^{−1} = I, we get

\begin{matrix}a^{∗}(I − aa^{∗})^{−1} = (I − a^{∗}a)^{−1}a^{∗} = \frac{1}{2}(I + t)(I + t)^{−1}x^{∗} = \frac{1}{2}x^{∗} , \\ (I − aa^{∗})^{−1} = I + a[a^{∗}(I − aa^{∗})^{−1}] = I + \frac{1}{2}x(I + t)^{−1}x^{∗} . \end{matrix}

Hence [\overline{x}^{∗} − \overline{t}a^{∗}](I − aa^{∗})^{−1}[\overline{x} − a\overline{t}] =

=\overline{x}^{∗} \underbrace{(I − aa^{∗})^{−1}}_{I+x(I+t)x^{∗}/2} \overline{x}−\overline{t} \underbrace{a^{∗}(I − aa^{∗})^{−1}}_{x^{∗}/2} \overline{x}−\overline{x}^{∗} \underbrace{[a^{∗}(I − aa^{∗})^{−1}]^{∗}}_{x/2} \overline{t}+\overline{t}[ \underbrace{a^{∗}(I − a^{∗}a)^{−1}}_{x^{∗}/2} ] \underbrace{a}_{x(I+t)^{−1}} \overline{t}=

= \overline{x}^{∗}\overline{x} + \frac{1}{2}[\overline{x}^{∗}x(1 + t)^{−1}x^{∗}\overline{x} − \overline{t}x^{∗}\overline{x} − \overline{x}^{∗}x\overline{t} + \overline{t}[ \underbrace{x^{∗}x}_{t^{2}−I=(t−I)(I+t)} (I + t)^{−1}]\overline{t}] =

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

Theorem. For (\overline{t}, \overline{x}) ∈ T_{(t,x)}M with \overline{t}t + t\overline{t} = \overline{x}^{∗}x + x^{∗}\overline{x} we conclude

\begin{matrix}⟨(\overline{t}, \overline{x}), (\overline{t}, \overline{x})⟩_{(t,x)}^{M} = −4⟨u, u⟩_{a}^{B} = \\ = (I + t)^{1/2}\{2[\overline{t}^{2} − \overline{x}^{∗}\overline{x}] − [\overline{x}^{∗}x(I + t)^{−1}x^{∗}\overline{x} − \overline{t}x^{∗}\overline{x} − \overline{x}^{∗}x\overline{t} + \overline{t}(t − I)\overline{t}]\}(I + t)^{1/2}. \end{matrix}

Remark. In the commutative case H = C ≡ L(H) we have −4⟨(\overline{t}, \overline{x}), (\overline{t}, \overline{x})⟩_{(t,x)}^{M} = ⟨[Ψ′(t, x)](\overline{t}, \overline{x}), [Ψ′(t, x)](\overline{t}, \overline{x})⟩_{Ψ(t,x)} = \overline{t}^{2} − |\overline{x}|^{2}.

3a. Spectral integral for D(φ(a), ψ(a)) in L(H)

Let a ∈ L(H) be fixed and let

Ω := Sp(a^{∗}a) \ {0} = Sp(aa^{∗}) \ {0} , α := Id_{Ω} : Ω ∋ ω ↦ ω .

It s well-known that, there is a positive Borel measure μ on Ω and there are surjective partial isometries Q_{1}, Q_{2} : H → L^{2}(Ω, μ) such that, by writing M_{φ} for the multiplication operator

M_{φ}(f) := φf (f ∈ L^{2}(Ω, μ) ,

we have

a = Q_{1}^{∗}M_{α}Q_{2} .

For any couple of Borelian sets S, T ⊂ Ω, define the projection P_{S×T} : L(H) → L(H) by

P_{S×T} X := Q_{1}^{∗}M_{1_{S}}Q_{1}XQ_{2}^{∗}M_{1_{T}} Q_{2} (X ∈ L(H))

where 1_{Z} stands for the indicator function of the set Z (with 1_{Z}(ω) = 1 for ω ∈ Z and 1_{Z}(ω) = 0 for ω ∈ Ω \ Z). Furthermore, let P_{{0}×{0}}X := (I − Q_{1}^{∗}Q_{1})X(I − Q_{2}^{∗}Q_{2})X and

P_{S×{0}}X := Q_{1}^{∗}M_{1_{S}}Q_{1}X(I − Q_{1}^{∗}Q_{1}) , P_{{0}×T} X := (I − Q_{1}^{∗}Q_{1})XQ_{2}^{∗}M_{1_{T}} Q_{2} .

Proposition. Let φ, ψ : Ω ∪ {0} → C be a couple of continuous functions such that φ(0)=ψ(0)=0. Then

D(φ(a), ψ(a)) = ∫_{(λ_{1},λ_{2})∈[Ω∪\{0\}]^{2}} \frac{1}{2}[φ(λ_{1})\overline{ψ(λ_{1})} + φ(λ_{2})\overline{ψ(λ_{2})}] P (d(λ_{1}, λ_{2}))

with integration in Cauchy sense.

Remark. Given a bounded finitely additive vector valued measure V : A → B where A is an algebra of subetes of a set S and B is any Banach space, a function f : S → C 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 : S → C, using any representation of the form g := ∑_{k=1}^{N} λ_{k}1_{A_{k}}, the vector ∫ g(ω) V (dω) := ∑_{k=1}^{N} λ_{k}V (A_{k}) is unambigously defined. If f is the uniform limit of V - simple functions then the definition ∫ g(ω) V (dω) := lim_{n→∞} ∫ g_{n}(ω) V (dω) is unambigous with any uniformly convergent sequence g_{n} → g of V -simple functions. In the Proposition we take S := [Ω∪{0}]^{2}, A := {finite disjoint unions from {S ×T : S, T Borel⊂ Ω∪{0}}.

Proof. By definition, φ(a) := Q_{1}^{∗}M_{φ}Q_{2} and ψ(a) := Q_{1}^{∗}M_{ψ}Q_{2}. Let φ_{n} be the simple function assuming the value φ(k/n) on the set I_{k,n} := [(k − 1)/n, k/n) ∩ Ω (for n = 1, 2, … with k = 1, …, n · entier(∥a∥) + 1). We define ψ_{n}(a) analogously. Then

D(φ_{n}(a), ψ_{n}(a)) = ∑_{k,ℓ} \frac{1}{2}[φ(k/n)\overline{ψ(k/n)} + φ(ℓ/n)\overline{ψ(ℓ/n)}] P (I_{n,k} × I_{n,ℓ}) .

For n → ∞ we have φ_{n} → φ, ψ_{n} → ψ uniformly, implying φ_{n}(a) → φ(a), ψ_{n}(a) → ψ(a) and D(φ_{n}(a), ψ_{n}(a)) → D(φ(a), ψ(a)) in L(L(H)). Hence the statement is immediate.

3b. Spectral integral for D(φ(a), ψ(a)) in H_{3}(|O)

Given a ∈ H_{3}(|O), there is an orthogonal frame (in Jordan sense) G_{0} := {g_{1}, g_{2}, g_{3}} and there are scalars λ_{1} ≥ λ_{2} ≥ λ_{3} ≥ 0 such that

a = ∑_{ℓ=1}^{3} λ_{ℓ}g_{ℓ}, D(φ(a), ψ(a)) = ∑_{ℓ=1}^{3} φ(λ_{ℓ})\overline{ψ(λ_{ℓ})}D(g_{ℓ}, g_{ℓ}) .

The frame G_{0} can be extended to a covering grid

G = {g_{j} : j = 1, … 27}

of linearly independent tripotents such that

D(g_{j}, g_{j})g_{k} = ⟨γ_{j}, γ_{k}⟩g_{k} (j, k = 1, …, 27)

where {γ_{j} : j = 1, … 27} ⊂ IR^{2}7 is the 1-part of the 3-graded root system E_{6}. Actually, after a suitable change of indices we may assume that

G = G_{0} ∪ G_{{1,2}} ∪ G_{{2,3}} ∪ G_{{1,3}} with G_{{1,2}} := {g_{j} : 4 ≤ j ≤ 11},
G_{{2,3}} := {g_{j} : 12 ≤ j ≤ 19},
G_{{1,3}} := {g_{j} : 20 ≤ j ≤ 27}

and. for ℓ = 1, 2, 3, we have

{g ∈ G : D(g_{ℓ}, g_{ℓ})g = g} = {g_{ℓ}} ,
{g ∈ G : D(g_{ℓ}, g_{ℓ})g = (1/2)g} = ⋃_{m∈{1,2,3}\{ℓ}} G_{{ℓ,m}} ,
{g ∈ G : D(g_{ℓ}, g_{ℓ})g = 0} = [G_{0} \ {g_{ℓ}}] ∪ G_{{1,2,3}\{ℓ}} .

It follows

D(φ(a), ψ(a))g_{j} = ∑_{ℓ=1}^{3} φ(λ_{ℓ})\overline{ψ(λ_{ℓ})}⟨γ_{ℓ}, γ_{j}⟩g_{j} (j = 1, …, 27).

Or alternatively, with the projections

\begin{matrix}P_{\{ℓ\}} : H_{3}(|O) → Cg_{ℓ}  (1 ≤ ℓ ≤ 3), \\ P_{\{ℓ,m\}} : H_{3}(|O) → ∑_{g∈G_{\{ℓ,m\}}} Cg  (1 ≤ ℓ < m ≤ 3) \end{matrix}

we can write

D(φ(a), ψ(a)) = ∑_{1≤ℓ≤m≤3} [\frac{1}{2}φ(λ_{ℓ})\overline{ψ(λ_{ℓ})} + \frac{1}{2}φ(λ_{m})\overline{ψ(λ_{m})}]P_{\{ℓ,m\}} .

4. Jordan case

(E, {…}) JB^{∗}-triple, D(a, b)x := {abx}, Q(a, b)x := {axb},

B(a, b)x {:= I − 2D(a, b) + Q(a, a)Q(b, b) = \atop = [I − D(a, b)]^{2} ,}

M_{a}(x) := a + B(a, a)^{1/2}[I + D(x, a)]^{−1}x, M_{−a} = [M_{a}]^{−1},

M′_{a}(x)v = ∂/∂t|_{t=0}M_{a}(x + tv) = ; = B(a, a)^{1/2}[I + D(x, a)]^{−2}v = ; = B(a, a)^{1/2}B(x, −a)^{−1}v.

Remark. In the commutative case E = C, \{xyz\} = x\overline{y}z we have B(a, b)z = [1 − 2a\overline{b} + a^{2}\overline{b}^{2}]z and B(a, a)z = [1 − |a|^{2}]^{2}z. Also M_{a}(z) = (z + a)/(a − \overline{a}z). The identity ϕ_{1}(a)^{2} − |ϕ_{2}(a)|^{2} = 1 can be written as ϕ_{1}(a)^{2} − D(ϕ_{2}(a), ϕ_{2}(a)) = I. Here ϕ_{2}(a) = 2a[1 − |a|^{2}]^{−1} = 2[I − D(a, a)]^{−1}a = 2[M′_{a}(0)]^{−1}a = 2[(M_{a})^{−1}]′(M_{a}(0))a. Thus ϕ_{2}(a) = 2M′_{−a}(a)a

ϕ(a) = [ \begin{matrix}ϕ_{1}(a) \\ ϕ_{2}(a) \end{matrix} ] (a ∈ E)

range(ϕ) = \{ \left[{t \atop x}\right] ∈ L(E) × E : t^{2} − D(x, x) = I, Sp(t) > 0\}

\begin{aligned}ϕ_{2}(a) & := 2[I − D(a, a)]^{−1}a = 2M′_{−a}(a)a , \\ ϕ_{1}(a) & := [I + D(ϕ_{1}(a), ϕ_{1}(a))]^{1/2} =^{?} B(ϕ_{2}(a), −ϕ_{2}(a))^{1/4} . \end{aligned}

Conjecture. A natural candidate for a trailer space (as subspace of L(E) × E) would be the Jordan supertriple of (E, {…}).