Why this is the discrete bit
H² = I
No continuous path inside O(n) joins I to a reflection: det cannot go from +1 to −1 without leaving the group. Paths among reflections stay in the det = −1 component.
H u = −u
The normal flips. The mirror line (u⊥) is fixed pointwise.
block-diag(1, H) ∈ O⁺(1, n)
Time is untouched, so Λ₀₀ = 1. Orthochronous. Embeddings stay on x₀ > 0.
H₂ H₁ = R_{2ψ}
Two reflections compose to a rotation by twice the angle of the normals. Even products sit in SO⁺; odd products are the other component of O⁺.
D(σ) is a Householder only at σ = −1
σ = +1 is the identity. Every other value in (−1, 1) fails HᵀH = I.
Lie flows cannot reach a single H. so(1, 2) generators.