##A_\mu^a=0## in global gauge symmetries ?

[JuanC97]

Hi, this question is related to global and local SU(n) gauge theories.
First of all, some notation: $A$ will be the gauge field of the theory (i.e: the 'vector potential' in the case of electromagnetic interactions) also known as 'connection form'.

In components: $A_\mu$ can be expanded in terms of the basis given by the generators $F_a$ of a SU(n) group using some coefficients (or components) $A_\mu^a$ so one can write $A_\mu=A_\mu^aF_a$.

As you can see in the image I uploaded (from the book "An elementary primer for Gauge theory") one can identify $A_\mu^a$ wih the partial derivatives of the parameters $\theta^a$ (in the last line you have $A_\mu=\partial_\mu\theta^a\,F_a$) which parametrize the infinitesimal amount of transformation that is made by the operator $U(dx)$ on an arbitrary vector u.

The question is: If the transformation is global, i.e: $\theta^a\neq f(x)$, then... $A_\mu^a=0$ implies that is impossible to deffine a connection?.

 

[JuanC97]

Maybe I should rephrase the statement. Before thinking a little bit about it I concluded that $A_\mu^a=0$ doesn't imply that is impossible to define a connection, it implies that it's not necessary since the partial derivative of a scalar field would be already a gauge covariant derivative.

Anyway, you could still add a connection and demand it to transform as $A_\mu' = U A_\mu U^{-1} - \frac{1}{q}(\partial_\mu U) U^{-1}$ where $q$ is the gauge coupling parameter, and... in the case of global transformations, the part of $A_\mu' = U A_\mu U^{-1}$ would be enough to ensure that the gauge covariant derivative $D_\mu\Psi = (\partial_\mu-iqA_\mu)\Psi$ would be covariant in the sense of $(D_\mu\Psi)'$ being just $U(D_\mu\Psi)$.

But... my point is that, given a global transformation, one could have $A_\mu^a\neq 0$ and, still, the gauge covariant derivative would satisfy the desired property of gauge covariance but, in other scenario, one could have $A_\mu^a=0$ and the gauge covariant derivative would be just the usual partial derivative which seems equally valid to me. The question that I have now is if the relation $A_\mu^a=\partial_\mu\theta^a$ is mandatory or if... somehow... it is realted to some kind of gauge fixing which could be done in a totally arbitrary way that doesn't imply having $A_\mu^a=0$.