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

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1. May 17, 2018 at 11:34 PM

### 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 $A_\mu=\partial_\mu\theta^a\,F_a$) 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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