content_image/349974.jpg content_image/349976.jpg content_image/349979.jpg content_image/349978.jpg
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 

whereas for the action of u ∈ SU (2) , one gets 

<img src='content_image/349974.jpg'>

<img src='content_image/349976.jpg'>

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 

one gets 

<img src='content_image/349979.jpg'>

<img src='content_image/349978.jpg'>

from which we deduce the transformations of these new fields under the action of the two 

