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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?

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    $\begingroup$ $A\wedge A$ only vanishes identically if the Lie algebra is abelian. $\endgroup$ – Robert Bryant Mar 5 '13 at 16:39
  • $\begingroup$ Ah I see - so it's just in preparation for the non-abelian general Yang-Mills theory. That makes sense - cheers! $\endgroup$ – Edward Hughes Mar 5 '13 at 17:14
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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: $F = dA$.

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    $\begingroup$ Just a comment on notation: When $G$ is a matrix Lie group, i.e,. when (as is quite common) $G$ occurs as a submanifold of $\mathrm{GL}(n,\mathbb{R})$, then the Lie algebra $\frak{g}$ of $G$ can be regarded as a subspace of the Lie algebra of $n$-by-$n$ real matrices, and, when one uses this to regard a $\frak{g}$-valued $1$-form~$A$ as a $1$-form with values in $n$-by-$n$ matrices, one has the elementary identity $[A,A] = 2 A\wedge A$, which is why one sees the curvature $2$-form sometimes written as $dA + \tfrac12[A,A]$ and other times written as $dA + A\wedge A$. Both are correct. $\endgroup$ – Robert Bryant Mar 11 '13 at 15:04

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