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

• I
• JuanC97
In summary, the conversation discusses global and local SU(n) gauge theories, specifically focusing on the gauge field and its components. The question is raised about the relationship between the gauge field and the partial derivatives of the parameters used to define the infinitesimal transformation. It is argued that for global transformations, the gauge covariant derivative can be satisfied without requiring the gauge field components to be zero. The question remains about whether the relation between the gauge field components and the partial derivatives is necessary or if it is related to a gauge fixing process.

#### 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?.

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• primer.png
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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 ##.