# Commutator of Gauge Covariant derivatives

What is the physical meaning of $$[D_{\mu}, D_{\nu}] ~\propto~ F_{\mu, \nu},$$ where $D_{\mu}$ is the gauge covariant derivative and $F_{\mu,\nu}$ is the field strength?

Is it just a definition? Does the commutator tell us something about the coupling between the $A_{\mu}$ fields? Is a Lie Algebra involved (could $[,]$ be a Lie bracket)?

• – Phoenix87 May 7 '15 at 11:38

Commutators are Lie brackets, in this case on the algebra of differential operators. Asking whether $[\dot{},\dot{}]$ is a Lie bracket doesn't really make sense (since, for matrix groups, the Lie bracket is the matrix commutator, anyway).
The field strength is not "just a definition", it is the natural curvature associated to the principal connection that is the gauge field $A_\mu$. There are two ways in which this curvature arises geometrically:
1. By applying the exterior covariant derivative $\mathrm{d}_A = \mathrm{d} + A \wedge$ twice to a $p$-form $\omega$: $$\mathrm{d}_A\mathrm{d}_A\omega = F \wedge \omega$$ As such, you can think of $F$ as being the obstruction to $\mathrm{d}^2 = 0$, which would hold in "flat" space with $F=0$.
2. As the holonomy around an infinitesimal rectangular loop $\gamma$ around an area element $\sigma_{\mu\nu}\mathrm{d}x^\mu \wedge \mathrm{d}x^\nu$ as $$h(\gamma) = 1 + \sigma^{\mu\nu}F_{\mu\nu} + \mathrm{O}(\sigma^2)$$ where you should physically think of the holonomy as a generalized relative phase a particle travelling along $\gamma$ will acquire (e.g the phase acquired in the Aharonov-Bohm effect is nothing but a $\mathrm{U}(1)$ holonomy).
And, indeed, as a term in the Lagrangian, $\mathrm{Tr}(F\wedge\star F) = F^a_{\mu\nu}F^{a\mu\nu}$ contains the self-interaction of the gauge field, and the $[A,A]$-term in $F = [D,D]$ produces cubic and quartic interactions of $A$ with itself if the theory is non-Abelian (for an Abelian theory, the commutator $[A,A]$ vanishes, and $F^a_{\mu\nu}F^{a\mu\nu}$ contains only kinetic terms like $\partial_\mu A_\nu \partial^\mu A^\nu$).