gauge groups: The B$_{µ}$ fields, and so the W ± µ and B 3 µ fields, are invariant under the action of the gauge group SU (2) . The W ± µ fields are U (1) -charged with opposite charges, whereas the B 3 µ fields behave like a U (1) -gauge potential. Let us now look at the fermion fields. Let us define ψ ′ L = U − $^{1}$ψ$_{L}$ and ψ ′ R = ψ$_{R}$ , with ψ ′ L = ( ν$_{L}$ $_{e$_{L}$}$) and ψ ′ R = e$_{R}$ . Then one obviously gets A simple computation shows that the new fermions fields have the following transforma- tion rules under the gauge groups actions: Finally, with the usual definitions cos θ$_{W}$ = g √ g $^{2}$+ g ′ 2 and sin θ$_{W}$ = g ′ √ g $^{2}$+ g ′ $_{2}$, the vector fields have the transformation rules where the electric charge e is defined as e = g sin θ$_{W}$ . 7 Z s µ = Z$_{µ}$ A s µ = A$_{µ}$ + 2 i 1 e $_{s}$− $^{1}$∂$_{µ}$ s (7) Z u µ = Z$_{µ}$ A u µ = A$_{µ}$, Z$_{µ}$ = cos θ$_{W}$ B 3 µ − sin θ$_{W}$ a$_{µ}$ A$_{µ}$ = sin θ$_{W}$ B 3 $_{µ}$+ cos θ$_{W}$ a$_{µ}$ ν s L = ν$_{L}$ e s L = s $^{2}$e$_{L}$ e s R = s $^{2}$e$_{R}$ (6) ν u L = ν$_{L}$ e u L = e$_{L}$ e u R = e$_{R}$ . D L $_{µ}$ψ$_{L}$ = U ( ∂$_{µ}$ − i g $_{2}$B$_{µ}$ + i g ′ $_{2}$a$_{µ}$ ) $_{ψ}$′ L ≡ $_{UD}$′ L $_{$_{µ}$ψ}$′L D R $_{µ}$ψ$_{R}$ = ( ∂$_{µ}$ + $_{ig}$′$_{a}$$_{µ}$$_{)}$$_{ψ}$ ′ R = D R $_{$_{µ}$ψ}$′R f ( ψ$_{L}$φψ$_{R}$ + ψ$_{R}$$_{φ}$†$_{ψ}$$_{L}$$_{) =}$ fη ( e$_{L}$e$_{R}$ + e$_{R}$e$_{L}$ ) . ( W + µ ) u = W + µ ( $_{W}$− µ ) u = $_{W}$− µ ( B 3 $_{µ}$) u = B 3 µ ( W + µ ) s = $_{s}$− $^{2}$W + µ ( $_{W}$− µ ) s = s 2 $_{W}$− µ (4) ( B 3 $_{µ}$) s = B 3 $_{µ}$+ 2 i g $_{s}$− $^{1}$∂$_{µ}$ s . (5)