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

  1. May 17, 2018 #1
    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##) wich parametrize the infinitesimal ammount 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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  2. jcsd
  3. May 21, 2018 #2
    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 ##.
     
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