Chirality, isospin and hypercharge

Start from γ⁵ = iγ⁰γ¹γ²γ³. It is Hermitian, squares to 1, and anticommutes with every γ^μ, so it commutes with the Lorentz generators S^(μν) = (i/4)[γ^μ, γ^ν]. Its eigenprojectors P_L = (1 − γ⁵)/2 and P_R = (1 + γ⁵)/2 are Hermitian, idempotent, mutually orthogonal and sum to one, so ψ = ψ_L + ψ_R is a Lorentz-invariant split. In the Weyl basis γ⁵ = diag(−1, −1, +1, +1) and P_L simply keeps the upper two components. Chirality is not helicity. Helicity h = σ⋅p̂/2 is an observable a boost can flip; chirality is a label a boost leaves alone. They agree only for a massless particle, and for a massive one the mismatch is of order m/E: for an electron at E = 1 GeV that is 0.511 × 10⁻³ and negligible, for a 5 MeV β-decay electron it is 0.10 and not. The charged weak current is ψ̄ γ^μ P_L ψ = ½ ψ̄ γ^μ (1 − γ⁵) ψ, and the V − A structure is nothing more than the statement that only ψ_L enters.

SU(2)_L acts on the left-handed fields through T^a = σ^a/2, the spin-½ algebra [T^a, T^b] = iε^{abc} T^c on an internal index. The first generation's left-handed fields are filed into two doublets, Q_L = (u_L, d_L) and L_L = (ν_L, e_L), each with T = ½ and T₃ = +½ for the upper member, −½ for the lower; the right-handed fields u_R, d_R and e_R are singlets with T = 0, on which every T^a is the zero matrix. The ladder operators T^± = T₁ ± iT₂ are what the charged W bosons couple to: W^+ turns d_L into u_L and e_L into ν_L, and it has no matrix element at all on a singlet. That is the whole reason the charged current is left-handed and an e_R never emits a W. Counting colour, one generation holds 6 + 2 = 8 left-handed and 3 + 3 + 1 = 7 right-handed Weyl fields, fifteen in all; a right-handed neutrino would make sixteen.

Hypercharge generates the U(1)_Y factor, normalised so that Q = T₃ + Y/2; it is defined by that relation, so Y = 2(Q − T₃) is read off the measured charges. The quark doublet gives 2(2/3 − ½) = 1/3 from u_L and 2(−1/3 + ½) = 1/3 from d_L — the same number, which is the consistency check: both members of a doublet must share one Y, and they can only do so if their charges differ by exactly one unit. The lepton doublet gives 2(0 − ½) = −1 and 2(−1 + ½) = −1. Singlets have T₃ = 0, so Y = 2Q: u_R carries 4/3, d_R carries −2/3, e_R carries −2, and the Higgs doublet (φ⁺, φ⁰) carries +1. The gauge coupling of each field then follows from D_μ = ∂_μ − ig T^a W^a_μ − ig′ (Y/2) B_μ: a singlet feels only B_μ, in proportion to its Y. The factor ½ is convention — some texts absorb it, quoting Y_Q = 1/6 — and only the products g′Y are physical.

A theory whose left- and right-handed fields carry different gauge charges is a chiral gauge theory, and its triangle diagrams break gauge invariance unless four sums over each generation's Weyl fields vanish, with every quark counted N_c = 3 times. [SU(2)]²U(1) sums Y over doublets: 3 × (1/3) + (−1) = 0. [SU(3)]²U(1) sums Y over quarks, left minus right: 2(1/3) − 4/3 + 2/3 = 0. The gravitational sum takes Y over all fields, left minus right: (6/3 − 2) − (4 − 2 − 2) = 0. The cubic sum takes Y³: (6/27 − 2) − (192/27 − 24/27 − 8) = −48/27 + 48/27 = 0. Quarks alone score +1 on the first sum and leptons alone −1, so neither sector is consistent by itself: the cancellation is generation by generation, quarks against leptons. Hold Y_L = −1 and treat Y_Q as unknown, and the doublet sum forces Y_Q = 1/N_c, so Qᵤ = ½ + 1/(2N_c). Three colours put the up quark at 2/3: the fraction is a consequence of N_c, not an accident of the quark model.

A Dirac mass term is m ψ̄ψ = m(ψ̄_L ψ_R + ψ̄_R ψ_L), because ψ̄ P_R P_L ψ vanishes. Each piece pairs a doublet with a singlet, and doublet ⊗ singlet is a doublet, not an invariant; its hypercharge is Y_R − Y_L, which for the electron is −2 + 1 = −1, not zero. So no bare mass for any quark or charged lepton is gauge-invariant before breaking, and every fermion mass must arrive through a Yukawa coupling to the Higgs: L̄_L φ e_R has hypercharge +1 + 1 − 2 = 0 and is an SU(2) singlet, and it becomes mₑ = yₑ v/√2 once φ takes its vacuum value. The neutrino has no such route. With no ν_R in the table there is no field to write L̄_L φ̃ ν_R with, and a Majorana mass for ν_L alone needs the dimension-five operator (Lφ)(Lφ)/Λ, which is not renormalisable and not in the model. Massless neutrinos are therefore a prediction of the fermion content as written, and the measured Δm² ≠ 0 is the first place that content is known to be incomplete.

The second and third generations copy the first label for label: (c_L, s_L), (t_L, b_L), (ν_μ, μ_L), (ν_τ, τ_L) are doublets with Y = 1/3 and −1, and c_R, s_R, t_R, b_R, μ_R, τ_R are singlets with Y = 4/3, −2/3 and −2. Gauge couplings cannot tell a muon from an electron; only the Yukawa couplings do, and those are inputs. That the count is three is measured, not derived: the Z lineshape at LEP gives N_ν = 2.984 ± 0.008 light neutrino species. The whole audit is five lines of NumPy. Store Y = [1/3, 1/3, 4/3, −2/3, −1, −1, −2] for u_L, d_L, u_R, d_R, ν_L, e_L, e_R, a multiplicity array n = [3, 3, 3, 3, 1, 1, 1], and a sign array s equal to +n for left-handed entries and −n for right-handed. Then (s*Y).sum() is the gravitational sum, (s*Y**3).sum() the cubic one, and n[0]*Y[0] + n[4]*Y[4] the doublet sum; all three vanish to rounding error — the cubic one prints a residue of order 10⁻¹⁵ or less rather than 0.0, because 1/3 has no exact binary form — and changing any single entry, or setting the quark multiplicity to 1, makes at least one of them fail.