Bianchi identity of gauge theory How to prove Bianchi identity?
\begin{align*}
\varepsilon^{\mu\nu\rho\sigma}D_{\nu}F_{\rho\sigma}=0
\end{align*}
using Jacobi identity;
\begin{align*}
\epsilon^{\mu\nu\rho\sigma}[D_{\mu},[D_{\rho},D_{\sigma}]]=0
\end{align*}
where covarient derivative is given as
\begin{align*}
D_{\mu}=\partial_{\mu}-igA_{\mu}
\end{align*}
I know the same question was asked on this site before; Bianchi identity of a non-Abelian gauge theory?
In this answer, he used the fact that covariant derivation satisfies the Leibniz rule. So, I would like to know why this fact holds.
 A: $D_\mu T$ is defined for a tensor $T$ in any of several equivalent ways, e.g. by these two axioms:

*

*The $\partial_\nu T$ coefficient in $D_\mu T$ is $\delta_\mu^\nu$, while higher-order derivatives are absent;

*$D_\mu T$ gauge-transforms like $T$.

Since $F_{\nu\lambda}$ is carefully defined so as to be gauge-invariant, $D_\mu F_{\nu\lambda}=\partial_\mu F_{\nu\lambda}$. This is a counterexample to $D_\mu T=\partial_\mu T-igA_\mu T$, which is really only true in a special case, namely where $T$ has $\psi$'s transformation rule, e.g. $T=\psi$ or $T=F_{\nu\lambda}\psi$. So$$\begin{align}D_\mu(F_{\nu\lambda}\psi)&=\partial_\mu(F_{\nu\lambda}\psi)-igA_\mu F_{\nu\lambda}\psi\\&=(\partial_\mu F_{\nu\lambda})\psi+F_{\nu\lambda}\partial_\mu\psi-F_{\nu\lambda}igA_\mu\psi\\&=(\partial_\mu F_{\nu\lambda})\psi+F_{\nu\lambda}(\partial_\mu\psi-igA_\mu\psi)\\&=(D_\mu F_{\nu\lambda})\psi+F_{\nu\lambda}D_\mu\psi.\end{align}$$In fact, you can show tensors $S,\,T$ satisfy $D_\mu(ST)=(D_\mu S)T+SD_\mu T$.
