Wedge Product of Lie Algebra Valued One-Form I've been reading about the formal structure of gauge theories and am a little confused by the notation. Could someone clarify this for me?
Suppose that $A$ is a Lie algebra valued 1-form corresponding to the gauge potential of a $U(1)$ gauge theory. Then several sources define the field strength tensor by
$$F=dA+A\wedge A$$
Surely in this case $A\wedge A$ is simply zero, since $\mathfrak{u}(1)=\mathbb{R}$ and the wedge product is an alternating construction? If this is true, why is this term included? Or am I missing something odd about Lie algebra valued 1-forms?
 A: There's something about the notation you should know before you get confused when trying to do non-abelian gauge theory. The second term in the field strength should involve a combination of the wedge product of forms and the Lie bracket: the field strength (in the case of an arbitrary gauge group $G$ with Lie algebra $\mathfrak{g}$) should actually be
$$F = dA + \tfrac{1}{2}[A \wedge A],$$
where if $\omega$ is a $\mathfrak{g}$-valued $k$-form and $\eta$ is a $\mathfrak{g}$-valued $p$-form, then
$$[\omega \wedge \eta](X_1, \dots, X_{k+p}) = \sum_{\sigma \in S_{k+p}} (-1)^{\text{sgn}(\sigma)} [\omega(X_{\sigma(1)}, \dots, X_{\sigma(k)}), \eta(X_{\sigma(k+1)}, \dots, X_{\sigma(k+p)})]$$
for any $k + p$ vector fields $X_1, \dots, X_{k+p}$. In particular, if $A$ is a $\mathfrak{g}$-valued $1$-form, then
$$[A \wedge A](X_1, X_2) = [A(X_1), A(X_2)] - [A(X_2), A(X_1)] = 2[A(X_1), A(X_2)].$$
So in components, the field strength is given by
$$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu + [A_\mu, A_\nu],$$
which is the form you'll see most frequently in the physics literature.
When the gauge group $G$ is abelian (e.g. in ${\rm U}(1)$ gauge theory), the Lie bracket on $\mathfrak{g}$ is trivial so that $[A \wedge A] \equiv 0$ and the field strength is just the exterior derivative of the gauge potential: $F = dA$.
