Potential $A_\mu$ in the Lie algebra of a group $G$ : question about the "proof" I have read this topic : Why is the gauge potential $A_{\mu}$ in the Lie algebra of the gauge group $G$?
It explained why the potential is in the Lie algebra of the group. But there are things I still don't totally get in the global logic and I need your help with my "proofs" of it.
What I understood is that we want to build a derivative that will ensure us to have the following property :
$$ D_\mu (g \phi)=g D_\mu \phi $$
Where $g$ is in the group of symmetry of the theory.
Because as soon as we have this, we will be able to easily construct gauge invariant Lagrangians.
We know that :
$$ \partial_\mu(g(x) \phi(x))=[\partial_\mu g(x)] \phi(x) + g(x) [\partial_\mu \phi(x)] $$
So, in practice, what we want to do is to change our $\partial_\mu \rightarrow D_\mu $ such that we cancel the first term of the rhs above.
If I want to do this, I will then try with :
$$ D_\mu=\partial_\mu + A_\mu $$
Where $A_\mu$ is an operator that acts on $\phi$ but I don't know it is in the Lie algebra at this point.
Such that $ D_\mu (g \phi)=g D_\mu \phi $
Thus, what we need is :
$$A_\mu(g \phi)=g A_\mu \phi - (\partial_\mu g) \phi \tag{1}$$
And here start the things I don't totally get.
In practice we have to define two fields : $A$ and $A'$. But why couldn't we work with only one field $A$ that would follow $(1)$ ?
Is it because, if we try to use the same $A$ for all $\phi$ linked by the group transformation, we find that $(1)$ is a too strong condition. Then we "relax" it by saying that when $\phi$ changes, $A$ changes also. And that is why we have a transformation law for $A$ ?
Also, if I assume we have two different fields, then I would have :
$$A'_\mu(g \phi)=g A_\mu \phi - (\partial_\mu g) \phi$$
$$A'_\mu g =g A_\mu - (\partial_\mu g)$$
$$A'_\mu =g A_\mu g^{-1} - (\partial_\mu g) g^{-1}$$
And I know it is not the good sign on the second part of the rhs of last line. So there is a mistake in my logic somewhere, but I don't know where.
Remark : I know there is a close link with covariant derivative and differential geometry but I would like an answer really close to my formulation of the question. Else I think I would be lost. So if it is possible to avoid notion of differential geometry it would be very nice !
 A: We don't "define two fields".When we make a gauge transformation, a priori all dynamical fields can transform under it. $A$ is a dynamical field, so a gauge transformation goes as $\phi \mapsto \phi' = g\phi,A\mapsto A'$, where the expression for $A'$ is as of yet unknown. $\phi'$ and $A'$ are not "different operators", they are physically the same operators/fields as before, just gauge transformed. We then carry out the computation you did to determine what the transformation behaviour $A'$ of the gauge field is. As you know, it turns out $A'$ is not simply $A$, but transforms non-trivially under a gauge transformation - note that setting $A' = A$ in your equations would simply give inconsistent equations that cannot be true for any $A$.
In your computations, you aren't really wrong (except maybe for signs but I have no patience to chase these), you're just not doing the correct step to conclude where $A$ lives, which is: "Muliply by $1 = g g^{-1}$"
Your last equation is:
\begin{align} A'_\mu & = g A_\mu g^{-1} - (\partial_\mu g)g^{-1} = gA_\mu g^{-1} - g g^{-1}(\partial_\mu g) g^{-1} \\
& = g\left(A_\mu - g^{-1} (\partial_\mu g)\right)g^{-1}\end{align}
and the following are facts: 


*

*You can only add/subtract two things that live in the same vector space

*$g^{-1}\partial_\mu g$ is an element of the Lie algebra, as you can see by writing $g(x) = \exp(\chi(x))$ for a Lie-algebra valued function $\chi$.

*Conjugating, i.e. multiplying by $g$ from one side and $g^{-1}$ from the other, a Lie algebra element by a Lie group element $g$ yields a Lie algebra element (this is the adjoint rep of the Lie group upon its own algebra).
Therefore, since $g^{-1}\partial_\mu g$ is algebra-valued, so is $A_\mu$, and therefore also $A'_\mu$.
