Lorentz Atlas

Lie algebra

Generators of so(1, 2)

The identity component is generated by three matrices: one rotation J and two boosts K₁, K₂. Spatial Householders reach the other component of O⁺(1, n). They do not reach the two time-reversing components of O(1, n) — and should not, because embeddings live on x₀ > 0. FlorE’s continuous σ is generated by a matrix that is not in the Lie algebra. Open the Householder lab.

ε_lie ‖XᵀJ + JX‖

0

ε_tan |⟨p, Xp⟩_L|

0

ε_group of exp(tX)

0

det exp(tX)

1.0000

κ = α² + β² − ω² = -1.0000 · rotation-like. det exp(tX) = +1: Lie flows stay in SO⁺. Spatial Householders reach the other component of O⁺, not of O. π₀(O(1, n)) ≅ ℤ₂ × ℤ₂; we do not use time reversal.

so(1, 2) brackets

Three generators, three relations. Verified as matrix identities, not typeset lore.

  • [J, K₁] = K₂rotation turns a boost0
  • [J, K₂] = −K₁into another boost0
  • [K₁, K₂] = −Jtwo boosts make a rotation0

FlorE’s S is not a generator

D(e^λ) = diag(1, 1, e^λ) = exp(λ S) with S = diag(0, 0, 1). That one-parameter group exists in GL(3), not in O(1, 2).

ε_lie(S)

2.0000

ε_lie(J, Kᵢ)

0

‖SᵀJ + JS‖_F = 2. The continuous σ path is a GL(3) flow, not a Lorentz flow. Spatial parity is the nontrivial element of π₀(O⁺(1, n)) ≅ ℤ₂. The other two components of O(1, n) reverse time and send x₀ < 0.

ModelSpanConnected pieceNote
FHREℝ · Jdiag(1, SO(n)), not all of SO⁺Time axis fixed. Boosts are the missing SO⁺ directions.
LorentzKG⟨exp(so(1, n)), H⟩ = SO⁺ ⋊ ℤ₂O⁺(1, n)Boosts are exp(η K). Householder is not a Lie flow.
FlorE paperJ plus D(σ) = exp(λ S)S ∉ so(1, n)The flip interpolation is generated by diag(0,…,1), which fails ε_lie.
Hybridso(1, n) plus spatial ℤ₂O⁺(1, n) = SO⁺ ⋊ ℤ₂Orthochronous only. Time reversal would leave the upper sheet x₀ > 0.