What is the dimension of the $A_\mu$ field?

Question

Just as the title say: what is the dimension of the $A_\mu$ field?
(If that's of any importance, I'm interested in the dimension of $A$ to figure out the dimension of the coupling constant $g$)
I have conflicting sources, some saying that it is always $1$ because $A$ is a connection and the integral $\int A = \int A_\mu dx^\mu$ is always dimensionless or similar arguments (like this answer), while some saying that it is $\frac{d}{2}-1$ because the Yang-Mills term $F^2$ has dimensions $d$ (like this other answer). The arguments seems both convincing. Which one is it?

Answer 1

There is a slight difference between the mass dimension and the scaling dimension. As you sketched in your self-answer, the mass dimension can vary depending on the conventions of the Lagrangian. What is important, however, is the scaling dimension, and that is robust.

A foolproof way to read off the scaling dimension of your gauge field is through the propagator. In momentum space, for a gauge field $A_\mu$, you have $$ \Delta_{\mu\nu}(k):= \left<A_\mu(k)A_\nu(-k)\right> = \frac{\eta_{\mu\nu}}{\left\Vert\,k\,\right\Vert^{2}}+\text{gauge dependent terms},$$ where I suppress possible colour indices. Therefore, in position space, in $d$ dimensions, you get $$ \Delta_{\mu\nu}(x,y):= \left<A_\mu(x)A_\nu(y)\right> = \frac{\eta_{\mu\nu}}{\left\Vert\,x-y \,\right\Vert^{d-2}}+\text{gauge dependent terms}.$$ Hence, the scaling dimension of $A_\mu$ is $$ \big[A_\mu\big] = \frac{d-2}{2}.$$ Note that it coincides with the mass dimension, in the conventions where $L=-\frac{1}{4}\text{tr}\ F_{\mu\nu} F^{\mu\nu}$.

Answer 2

I think I found my answer and the reason why the two sources were conflicting: it depends if one defines the Yang-Mills Lagrangian to be $$-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}\qquad\text{or}\qquad-\frac{1}{4g^2}F^{\mu\nu}F_{\mu\nu}$$ In the first case, $\left[F\right]=m^\frac{d}{2}\Rightarrow\left[A\right]=m^{\frac{d}{2}-1}$ and $\left[g\right]=m^\frac{4-d}{2}$;
while in the second case $\left[F\right]=m^2\Rightarrow\left[A\right]=m^1$ but one still has $\left[g\right]=m^\frac{4-d}{2}$.