The lepton sector and family number

γ⁵ = iγ⁰γ¹γ²γ³ is Hermitian and squares to the identity, so its eigenvalues are ±1 and P_L = (1 − γ⁵)/2, P_R = (1 + γ⁵)/2 are orthogonal projectors onto its two eigenspaces: P_L² = P_L, P_L P_R = 0, P_L + P_R = 1. Chirality is that eigenvalue, and it is Lorentz-invariant. Helicity is a different operator, h = Σ⋅p̂, and for a massive particle it is not invariant: travel fast enough and you overtake the particle, so p̂ reverses while the spin does not. Now write a Dirac mass term in chiral blocks: m ψ̄ψ = m(ψ̄_L ψ_R + ψ̄_R ψ_L), which is purely off-diagonal. A mass is precisely the operator that turns one chirality into the other, so chirality is a good quantum number only in the massless limit. A chirality eigenstate of energy E therefore carries a wrong-helicity component of amplitude of order m/2E — about 0.5% for the 50 MeV electrons in muon decay, and utterly negligible for reactor neutrinos. The two words agree only when m ≪ E.

The gauge group acts on the projected fields, not the full ones. Three left-chiral doublets Lᵢ = P_L(νᵢ, ℓᵢ)ᵀ carry weak isospin ½ and hypercharge −½; the right-chiral charged leptons ℓᵢR are singlets with Y = −1; and the minimal model contains no ν_R. Three consequences follow at once. The charged current is (g/√2) W⁺_μ ν̄ᵢ γ^μ P_L ℓᵢ, pure V−A by construction with g = 0.653 for every family, and matching it onto a contact interaction below M_W gives G_F/√2 = g²/8M_W², so G_F = 1.166 × 10⁻⁵ GeV⁻². The neutral current couples to T³ and Q, both diagonal in the same basis that diagonalises the charged-lepton masses, so the Z changes no flavour at tree level. And with no ν_R there is no partner for a renormalisable Dirac mass m ν̄_L ν_R — neutrinos are massless in the minimal Lagrangian for want of a field, not by choice. How many such doublets exist? The Z's invisible width answers it: LEP found N_ν = 2.984 ± 0.008, revised to 2.996 ± 0.007 once the Bhabha luminosity was recomputed. Three, with no fourth below M_Z/2.

Far below M_W the charged current collapses to a four-fermion contact operator of strength G_F, and dimensions finish the job. [Γ] = GeV while [G_F²] = GeV⁻⁴, so the only mass in the problem must enter to the fifth power: Γ(ℓ → ℓ′ν ν̄) = G_F² m_ℓ⁵/192π³, the 192π³ coming out of the three-body phase-space integral. Put the muon in. m_μ = 0.105658 GeV gives m_μ⁵ = 1.3168 × 10⁻⁵ GeV⁵; with G_F² = 1.3604 × 10⁻¹⁰ GeV⁻⁴ and 192π³ = 5953.2, Γ = 3.009 × 10⁻¹⁹ GeV, so τ = ħ/Γ = 2.187 μs. The measured lifetime is 2.19698 μs, 0.44% longer, and the exact phase-space factor f(mₑ²/m_μ²) = 0.99981 together with the O(α) QED correction of −0.42% closes the gap to 0.002%. That agreement is partly a definition — G_F is fixed by this very equation — and the physics is that the same G_F then reproduces the tau. The electron is the one case the formula cannot touch: it is the lightest charged particle, so charge conservation alone leaves it nothing to decay into, and its width is exactly zero rather than merely small.

SU(2) gauge invariance forces gₑ = g_μ = g_τ, because a non-abelian group allows one coupling for every doublet, but a prediction still has to be tested, and experiment checks it family against family. The cleanest test lives inside the τ, where one parent feeds both light families and everything hadronic cancels from the ratio. Only phase space differs: with f(x) = 1 − 8x + 8x³ − x⁴ − 12x² ln x and x = (m_μ/m_τ)² = 3.536 × 10⁻³, f = 0.97256, while for the electron f = 1 to six figures, so the prediction is Γ(τ→μνν̄)/Γ(τ→eνν̄) = 0.9726. The measured ratio is 0.9762 ± 0.0028, and since widths go as g⁴ that fixes g_μ/gₑ = 1.0019 ± 0.0014. A second test comes from the pion and probes the Lorentz structure of the current rather than its strength: Γ(π→eν)/Γ(π→μν) = (mₑ/m_μ)²[(m_π² − mₑ²)/(m_π² − m_μ²)]² = 1.283 × 10⁻⁴, four orders of suppression that exist only because a γ^μ current, V−A included, has to produce a wrong-helicity charged lepton. The measurement, 1.2327 × 10⁻⁴, agrees once radiative corrections are applied — whereas a pseudoscalar coupling, which pays no helicity penalty, would give about 5.5, the phase-space ratio alone.

Write the most general renormalisable, gauge-invariant Lagrangian you can with the Standard Model's lepton content. The kinetic terms are flavour-blind, so on their own they carry a global U(3)_L × U(3)ₑ symmetry — 18 real rotation parameters. The only other term is the Yukawa (Yₑ)ᵢⱼ L̄ᵢ Φ ℓ_Rj, and a general complex 3×3 matrix also carries 18 real parameters. Take its singular-value decomposition, Yₑ = V_L Σ V_R† with Σ = diag(yₑ, y_μ, y_τ) real and non-negative, then rotate L → V_L†L and ℓ_R → V_R†ℓ_R. Both unitaries vanish from the Lagrangian without leaving a trace anywhere else, because nothing else in it distinguishes the families. What remains is three masses, mᵢ = yᵢ v/√2 with v = 246 GeV, and the diagonal phase rotations that Σ cannot fix: U(1)ₑ × U(1)_μ × U(1)_τ. Those three phases are Lₑ, L_μ and L_τ. Nobody imposed them — they are the leftovers of a counting exercise. The quark sector makes the contrast sharp: there are two Yukawa matrices, no single rotation of the shared quark doublet diagonalises both, and their relative rotation survives as the CKM matrix.

Allow a single non-renormalisable term and the accident is over. The unique dimension-five operator you can build from Standard Model fields is Weinberg's (LΦ)(LΦ)/Λ, which after electroweak breaking hands the neutrinos Majorana masses m_ν ≈ v²/Λ; taking m_ν ≈ 0.05 eV puts Λ ≈ 1.2 × 10¹⁵ GeV. Now the lepton sector has two matrices, only their relative rotation is physical, and that rotation is the PMNS matrix — oscillation is the measurement that family number is broken. The charged sector is a separate question. μ → eγ proceeds through a loop with the neutrino masses inside, and PMNS unitarity cancels everything in the loop function that does not depend on mᵢ, leaving Δm²/M_W² ≈ 4 × 10⁻²⁵ per term: B(μ→eγ) ≈ 3 × 10⁻⁵⁵. The bounds are 1.5 × 10⁻¹³ from MEG II, below 1.0 × 10⁻¹² for μ→eee, and below 7 × 10⁻¹³ for μN→eN conversion, with Μ2e and COMET aiming near 10⁻¹⁷. Forty-two orders separate prediction from bound, so these are not tests of neutrino mass. They are background-free searches for TeV-scale flavour structure, and one clean event would settle it.