as before, defining the transformed fields by φ u = U $^{u}$η$^{u}$( 0 $_{1}$) , one gets (recall that the reference vector ( 0 $_{1}$) is chosen once and for all) from which we deduce that under the action of the gauge group SU (2) one has Defining in the same way φ s = U $^{s}$η$^{s}$( 0 $_{1}$) one gets U s can be uniquely written as a product of U with an element of SU (2) : Finally, under the action of the gauge group U (1) one has ̂ s ) into SU (2) in the σ$_{3}$ -direction of the group SU (2) . Using the new variables U and η in place of φ , a straightforward computation leads to The field U ∈ SU (2) then supports the (commuting) actions of SU (2) and U (1) respec- tively by left multiplication and by right multiplication in SU (2) . Here U (1) is embedded (through s ) into SU (2) in the σ$_{3}$ -direction of the group SU (2) . where φ u = $_{u}$− $^{1}$Uη   0 1   = ( $_{u}$− $^{1}$U ) η   0 1   , U u = $_{u}$− $^{1}$U, η u = η . U s =   sφ$_{2}$/η s − $^{1}$φ$_{1}$ /η − sφ$_{1}$/η s − $^{1}$φ$_{2}$ /η   , η s = √ | φ$_{1}$ | 2 + | φ$_{2}$ | 2 . ̂ s =   s 0 0 s − 1   . U s = $^{U}$̂ s, η s = η . B$_{µ}$ = $_{U}$− $^{1}$b$_{µ}$ U + 2 i g $_{U}$− $^{1}$( ∂$_{µ}$U ) . (3) D$_{µ}$φ = ( ∂$_{µ}$ − i g $_{2}$b$_{µ}$ − i g ′ 2 a$_{µ}$ ) Uη ( 0 $_{1}$) = U ( ∂$_{µ}$ − i g $_{2}$B$_{µ}$ − i g ′ 2 a$_{µ}$ ) η ( 0 $_{1}$) U s =   φ$_{2}$/η φ$_{1}$/η − φ$_{1}$/η φ$_{2}$/η     s 0 0 s − 1   = $^{U}$̂ s, with 5