Gauge transformation as Lie Group

Question

How can I prove that the family of Gauge transformations $$ A_\mu \mapsto A'_\mu = A_\mu +\partial_\mu \phi$$ where $\phi (x)$ is an arbitrary scalar function, is a Lie Group?

Answer 1

If you define $ g = \exp (- i \phi )$, you can rewrite this transformation as $A'_\mu = g A_\mu g^{-1} + i (\partial_\mu g ) g^{-1}$ . That is the general form that a gauge field transform for a general gauge group. In particular in your case $ g = \exp (- i \phi )$ is an element of the group U(1). Now if you do the above transformation with $g = g_1$ and then from the result you do a second transformation with $g = g_2$, what you get is exact the above gauge transformation with the element $ g = g_1 g_2$. Therefore, performing two gauge transformations is also a gauge transformation. Moreover, the transformation with $g = 1$ corresponds to the identity transformation and if you perform a transformation with an element $g$ and then another transformation with $g^{-1}$ it is equivalent to the identity transformation with $g=1$. Therefore, this set of transformations satisfy all the properties which defines a group. Alternatively, you can use your equation and perform first a transformation with a parameter $\phi_1$ and then a transformation with a parameter $\phi_2$ and you can see that it is the same as a transformation with $\phi_1 + \phi_2$. Second, you have the transformation with $ \phi = 0$ which corresponds to the identity transformation. And third, you can do a transformation with $ \phi $ and then a transformation with $ (-\phi) $ and the combination of the two transformations is equivalent to the identity transformation with $ \phi = 0 $.