In Yang-Mills theory the field strength tensor $F_{\mu \nu}$ can be calculated as \begin{equation} F_{\mu\nu} \equiv \frac{i}{g} [D_\mu,D_\nu] = \partial_\mu A_\nu - \partial_\nu A_\mu -ig[A_\mu,A_\nu]. \end{equation} Where $ \ \ \ D_\mu = \partial_\mu -ig A_\mu $

Is there a physical interpretation for $[D_{\mu},D_{\nu}]$? Every book I read just gives this as the definition of the gauge tensor, because it gives the maxwell tensor in the abelian case and is Lorentz and Gauge invariant... okey, that are some good reasons but I'd like to know if there is physical motivation to use that expression. Why the strength of a gauge field should be related to the non-commutatibity of the covariant derivatives?

Background: I know QFT, some group theory and a little General Relativity. If you start with fiber bundles and stuff, go slow please.


2 Answers 2


Classically, the gauge field strength is a curvature of a connection, the same way that the Riemann tensor is. Since $F^a_{\mu\nu}$ lives in the adjoint representation of the gauge group, you can define a 4-index object very analogous to the Riemann tensor:

$$\mathcal{F}^a{}_{b\mu\nu} \equiv F^c_{\mu\nu} f_c{}^a{}_b,$$

where $f_c{}^a{}_b$ are the structure constants defined via

$$[t_c, t_b] = f_c{}^a{}_b \, t_a,$$

modulo factors of $i$ if you care to insert them. In any case, the object $\mathcal{F}^a{}_{b\mu\nu}$ contains information about parallel transport around infinitesimal loops, in the same sense that the Riemann tensor does. But the vector being translated is not a tangent vector; instead it is a vector in gauge space (or "internal" space), whose components are defined via expansion in the generators $t_a$, as in $V = V^a \, t_a$.

So, $\mathcal{F}^a{}_{b\mu\nu}$ describes the change in $V = V^a \, t_a$ as it is parallel-transported (via the covariant derivative $D_\mu \equiv \partial_\mu + A_\mu$, again modulo factors of $i$, etc.) around a small parallelogram in the directions $\partial_\mu, \partial_\nu$.

  • $\begingroup$ Why is there a need to multiply by the structure constants? You could just pull out the hidden matrix indices of the generators hidden in $A$ to the same effect, no? $\endgroup$
    – Craig
    Jun 9, 2023 at 15:22

Here are several "motivations" to consider the curvature of a gauge field. However, keep in mind that the notion of "physical" motivation is somewhat vague in non-Abelian gauge theories, and that "it coincides with electromagnetism in the Abelian case" is really a rather strong motivation already - of course we want the non-Abelian theory to reduce to electromagnetism in the Abelian case.

  1. We want a gauge invariant object to use in the action. Taking the gauge field $A$ itself or traces of it doesn't work since the additional $g\mathrm{g}^{-1}$ term under gauge transformations spoils invariance of the trace under the $gAg^{-1}$. Since the gauge field was introduced to have a covariant derivative, it seems natural to try to take the covariant derivative of it. Indeed, we find that $\mathrm{d}_A A = \mathrm{d}A + A\wedge A = F$ transforms as $F\mapsto gFg^{-1}$ under gauge transformation, so taking its trace yields a gauge-invariant object we can use to build a gauge-invariant action.

  2. It's the infinitesimal holonomy. This is the Ambrose-Singer theorem: given a connection with its notion of parallel transport, the holonomy around a closed path, often symbolically written as a path-ordered integral $\mathcal{P}\mathrm{e}^{\oint A}$, is another natural and gauge-invariant object to consider. It's physically relevant as the expectation values of Wilson and Polyakov loops are little more than the expectation value of the holonomy along these loops, and if you shrink such a loop, you find that the value of the holonomy becomes well-approximated by the value of the curvature inside the loop. Basically, this is what is "curved" about the curvature - it tells you how much the parallel transport around an infinitesimal loop based at a point deviates from the identity, i.e. "flat" space.


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