I'm a bit confused about the gauge transformation properties of non-abelian gauge fields, and I just wanted some clarification. I keep seeing the statement that "gauge fields transform in the adjoint representation", but I have my doubts.

If we have a theory with a gauge symmetry corresponding to some simple, compact Lie group $G$, then we define the gauge covariant derivative $D_\mu$ as:


Where $T^a\in\mathfrak{g}$ form a basis of the Lie algebra $\mathfrak{g}$ of $G$. This definition doesn't assume any representation of $T^a$, since this is determined by the representation of the field on which $D_\mu$ acts. I.e if we had a field $\psi$ which transforms in some representation $\Pi$ of a simple, compact Lie group $G$, $\psi\mapsto\Pi(g)\psi$, then we would have:

$$D_\mu\psi=\big(\partial_\mu-igA^a_\mu \pi(T^a)\big)\psi$$

Where $\pi(T^a)$ is the corresponding representation of $\mathfrak{g}$ that induces the representation $\Pi(g)$ after exponentiating. In this case, we require that the gauge covariant derivative has the same gauge transformation properties as $\psi$, namely $D_\mu\psi\mapsto\Pi(g)D_\mu\psi$ for some $g\in G$. This means that we must have:


Question 1) I know that objects transforming in the adjoint representation transform as $x\mapsto gxg^{-1}$. This is obviously very similar to this expression, but I don't think they are the same. Is it therefore correct to say in this case that $D_\mu$ transforms in the adjoint representation, or rather that it transforms "adjointly" to $\psi$?

From the expression $D_\mu\mapsto \Pi(g)D_\mu \Pi^{-1}(g)$ we find:

$$A^a_\mu\pi(T^a)\mapsto \Pi(g)\Big(A^a_\mu\pi(T^a)+\frac{i}{g}\partial_\mu\Big)\Pi^{-1}(g)$$

If we consider $\Pi(g)=\exp\Big(i\alpha^a(x)\pi(T^a)\Big)$, then for an infinitesimal transformation we can expand to first order in $\alpha$ to find:

$$A^a_\mu\mapsto f^{abc}A^b_\mu\alpha^c+\frac{1}{g}\partial_\mu\alpha^a$$

The first term in this expression is reminscent of the adjoint rep of the Lie algebra, so Question 2) is this what people refer to when they say that the gauge field transforms in the adjoint?

Sorry if there are any errors or glaring misunderstandings, I'm just trying to get my head around the terminology (and possibly the maths, who knows).


You may find the following illuminating: S. Weinberg, "The Quantum Theory of Fields", Vol. II, page 4.

Going from (15.1.10) to (15.1.11) makes evident why the gauge field must transform in the adjoint representation (15.1.6).

| cite | improve this answer | |
  • $\begingroup$ Fantastic - thanks. I’ll give this a read and see if it clears things up. Was anything I said in my question incorrect? I think half the battle is getting terminology correct so I know what I’m saying/reading to be honest! $\endgroup$ – arcturus7 Jan 24 '19 at 16:24

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.