CCFU Proof 12 — Geodesic Translation Length l(A₄²) = 2·ln φ

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CCFU Proof 12 — Geodesic Translation Length l(A₄²) = 2·ln φ

05/16/2026

Given.
Let φ = (1+√5)/2, so φ² = φ + 1.
spec(A₄²) = {φ², 1/φ², 1, 1}
A₄² ∈ SO⁺(3,1)
det(A₄²) = 1
[Dependency: Proof 4]

Step 1 — Compute φ² + 1/φ².

Using φ² = φ + 1:

    φ⁴ = (φ+1)² = φ²+2φ+1 = (φ+1)+2φ+1 = 3φ+2
    φ⁴ + 1 = 3φ + 3 = 3(φ+1) = 3φ²
    φ² + 1/φ² = (φ⁴+1)/φ² = 3φ²/φ² = 3  ∎

Step 2 — Trace.

    tr(A₄²) = φ² + 1/φ² + 1 + 1 = 3 + 2 = 5  ∎

Step 3 — Translation length formula.
For a pure hyperbolic isometry in SO⁺(3,1) with
eigenvalues {e^l, e^{−l}, 1, 1}:

    cosh(l) = (tr(A₄²) − 2) / 2 = (5−2)/2 = 3/2  ∎

Step 4 — Verification.

    cosh(2·ln φ) = (φ² + 1/φ²)/2 = 3/2  ✓

Therefore l(A₄²) = 2·ln φ.  ∎

Corollary.
One step of B (the boost factor of A₄, Proof 11) translates
along its geodesic axis by ln φ.