Q:

Why is the gauge potential $A_{\mu}$ in the Lie algebra of the gauge group $G$?

If we have a general gauge group whose action is  $$ \Phi(x) \rightarrow g(x)\Phi(x), $$
 with $g\in G$.
Then introducing the gauge covariant derivative $$ D_{\mu}\Phi(x) = (\partial_{\mu}+A_{\mu})\Phi(x).$$
My notes state the gauge potential $A_{\mu} \in L(G)$, $L(G)$ being the Lie Algebra of the group $G$.
What's the connection between the Lie Algebra of the group and the gauge potential?

A:

The gauge potential is an object that, when introduced in the covariant derivative, is intended to cancel the terms that spoil the linear transformation of the field under the gauge group. Every gauge transformation $g:\Sigma\to G$ (on a spacetime $\Sigma$) connected to the identity may be written as $\mathrm{e}^{\mathrm{i}\chi(x)}$ for some Lie algebra valued $\chi: \Sigma\to\mathfrak{g}$. The derivative of a transformed field is
$$ \partial_\mu(g\phi) = \partial_\mu(g)\phi + g\partial_\mu\phi = g(g^{-1}(\partial_\mu g) + \partial_\mu)\phi$$
and it is the $g^{-1}(\partial_\mu g)  = \partial_\mu\chi$ that we want to cancel here by adding the gauge field so that $D_\mu(g\phi) = gD_\mu\phi$. Since $\partial_\mu\chi$ is Lie algebra valued, so must the gauge field $A$ we add, and it has to transform as
$$ A\overset{g(x)}{\mapsto} gAg^{-1} - g^{-1}\mathrm{d} g$$
to cancel the terms we want to cancel.

