# A basic question about how to apply the gauge covariant derivative in Yang-Mills theory

I am sorry if this question is too stupid...

We know that Yang-Mills equation (without source) can be written as $$D^\mu F_{\mu\nu}=0,\tag{1}$$ where $$D^{\mu}=\partial^\mu-ig A^{\mu}$$ and $$A^\mu=A^{\mu a} T^a,\\ F_{\mu\nu}=F_{\mu\nu}^a T^a.$$ Here $$T^a$$ are the generators of the gauge group and satisfy $$[T^a,T^b]=if^{abc}T^c. \tag{2}$$

So far, everything is fine. But usually, we also say that Eq. (1) can be written as (see, e.g., wikipedia)

$$\partial^\mu F_{\mu\nu}^a+g f^{abc}A^{\mu b}F^{c}_{\mu\nu}=0.$$ I was wondering how to derive this equation from Eq. (1). From Eq. (1), we have $$(\partial^\mu-igA^{\mu a}T^a)(F^b_{\mu\nu}T^b)=0.$$ So we have $$-igA^{\mu a}T^a F^b_{\mu\nu}T^b{\stackrel{?}{=}}g f^{abc} A^{\mu a}F^b_{\mu\nu} T^c.$$ Apparently, relation (2) has been used. But how can we use this relation? How can I see that the "a" and "b" are antisymmetric in $$A^{\mu a} F^b_{\mu\nu}$$ such that we can take $$T^a T^b\rightarrow T^{[a}T^{b]}=[T^a,T^b]/2$$? Also, what about the factor $$1/2$$ here?

• In the operator $D^{\mu}=\partial^\mu-ig A^{\mu}$, $A^{\mu}$ will act on $F^{\mu\nu}$ via commutator operation rather than simple matrix multiplication. – user10001 Nov 17 '19 at 14:32
• @user10001 Thanks a lot for your comment. I recalled this point. And no problem now... – Wein Eld Nov 17 '19 at 14:44
• see answer here: physics.stackexchange.com/q/483197 – MadMax Nov 18 '19 at 18:06

1. If an object, say, a field $$\phi\in V$$ lives in a Lie algebra representation $$\rho: \mathfrak{g}\to {\rm End}(V)$$, then it is implicitly understood that the gauge covariant derivative $$D_{\mu}~=~\partial_{\mu}-ig~A_{\mu}\tag{1}$$ should be interpreted as $$D_{\mu}\phi~=~\partial_{\mu}\phi-ig~\rho(A_{\mu})\phi.\tag{2}$$ If $$\Phi\in {\rm End}(V)$$ is (isomorphic to) a matrix-valued field, then the gauge covariant derivative is $$D_{\mu}\Phi~=~\partial_{\mu}\Phi-ig~[\rho(A_{\mu}),\Phi]_C,\tag{3}$$ where $$[\cdot,\cdot]_C$$ denotes the commutator.
2. In particular, since the field strength $$F_{\nu\lambda}\in\mathfrak{g}$$ is Lie algebra valued, the gauge covariant derivative is \begin{align}D_{\mu}F_{\nu\lambda} ~=~&\partial_{\mu}F_{\nu\lambda}-ig~{\rm ad}(A_{\mu})F_{\nu\lambda}\cr ~=~&\partial_{\mu}F_{\nu\lambda}-ig~[A_{\mu},F_{\nu\lambda}],\end{align}\tag{4} where $${\rm ad}:\mathfrak{g}\to {\rm End}(\mathfrak{g})$$ denotes the adjoint representation.
In a Lie-algebra representation $$\rho: \mathfrak{g}\to {\rm End}(V)$$, this becomes \begin{align} D_{\mu}\rho(F_{\nu\lambda}) ~=~&\partial_{\mu}\rho(F_{\nu\lambda})-ig~\rho([A_{\mu},F_{\nu\lambda}])\cr ~=~&\partial_{\mu}\rho(F_{\nu\lambda})-ig~[\rho(A_{\mu}),\rho(F_{\nu\lambda})]_C,\end{align}\tag{5} where $$[\cdot,\cdot]$$ denotes the Lie bracket. Be aware that the representation map $$\rho$$ is often implicitly implied.