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?

  • 1
    $\begingroup$ 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. $\endgroup$ – user10001 Nov 17 '19 at 14:32
  • $\begingroup$ @user10001 Thanks a lot for your comment. I recalled this point. And no problem now... $\endgroup$ – Wein Eld Nov 17 '19 at 14:44
  • $\begingroup$ see answer here: physics.stackexchange.com/q/483197 $\endgroup$ – 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.


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