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?

  • 5
    $\begingroup$ $A\wedge A$ 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

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: $F = dA$.

  • 10
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.