Let us stress that the vector field B$_{µ}$ is not obtained by a gauge transformation from the gauge field b$_{µ}$ $^{3}$. Firstly, this is obvious from the fact the U fields are some of the dynamical variables of the theory: a gauge transformation would require a fixed element in SU (2) , but, as dynamical variables, the function U is not such a fixed element. Secondly, a gauge transformation should be applied at the same time on gauge fields and to any scalar and spinor fields coupled to them. In the present situation, the scalar fields φ are not subject to a transformation accompanying (3). φ is written in terms of U but it will not be transformed any further. Finally, the following actions of the gauge groups U (1) and SU (2) on B$_{µ}$ prove that B$_{µ}$ is no more a gauge potential for SU (2) . This rules out the fact that (3) can be a gauge transformation. Indeed, for the action of s ∈ U (1) , one gets B s µ = ( U s $_{)}$− $^{1}$b s $_{µ}$U s + 2 i $_{g}$( U s $_{)}$− $^{1}$( ∂$_{µ}$ ( U $^{s}$)) = ̂ $_{s}$− 1 $_{U}$− $^{1}$b$_{µ}$ $^{U}$̂ s + 2 i g ̂ $_{s}$− 1 $_{U}$− $^{1}$[( ∂$_{µ}$U $^{)}$̂ s + U ( $^{∂$_{µ}$}$̂ s )] = $_{s}$− $^{$^{1}$B$_{µ}$}$̂ s + 2 i g ̂ $_{s}$− $^{$^{1}$∂$_{µ}$}$̂ s , whereas for the action of u ∈ SU (2) , one gets B u µ = ( U u $_{)}$− $^{1}$b u $_{µ}$U u + 2 i g ( U u $_{)}$− $^{1}$( ∂$_{µ}$ ( U $^{u}$)) = $_{U}$− $^{1}$u [ $_{u}$− $^{1}$b$_{µ}$ u + 2 i g $_{u}$− $^{1}$∂$_{µ}$ u ] $_{u}$− $^{1}$U + 2 i $_{$_{g}$U}$− $^{1}$u [( ∂$_{µ}$$_{u}$− $^{1}$) U + $_{u}$− $^{1}$( ∂$_{µ}$U )] = B$_{µ}$ . Developing B$_{µ}$ as B$_{µ}$ = B a $_{µ}$σ$_{a}$ and defining W ± µ = 1 √$_{2}$ ( B 1 µ ∓ iB 2 $_{µ}$) , such that B$_{µ}$ =   B 3 µ √$_{2}$$_{W}$ + µ √$_{2}$$_{W}$ − µ − B 3 µ   , B s µ =   B 3 µ + 2 i g s − $^{1}$∂$_{µ}$ s s − 2 √$_{2}$$_{W}$ + µ s 2 √$_{2}$$_{W}$ − µ − ( B 3 µ + 2 i g s − $^{1}$∂$_{µ}$ s )   , one gets from which we deduce the transformations of these new fields under the action of the two 6