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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.

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    $\begingroup$ I think you're asking about one step in @DheerajShukla's answer, $D_\mu(F_{\nu\lambda}\psi)=(D_\mu F_{\nu\lambda})\psi+F_{\nu\lambda}D_\mu\psi$. Is that right? If so, do you know how $D_\mu$ is defined when it acts on quantities such as $A_\nu,\,F_{\nu\lambda},\,F_{\nu\lambda}\psi$? For example, can an answer start from a general commutator-based definition of the action on $D_\mu$? $\endgroup$
    – J.G.
    Commented Jul 10, 2021 at 18:47
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    $\begingroup$ Thank you for your comment. Yes, you're right. I just understand the covariant derivative as $D_{\mu}=\partial_{\mu}-igA_{\mu}$, but is it expressed differently when acting on $A_{\nu},\ F_{\nu\lambda},\ F_{\nu\lambda}\psi$? If so, I don't know how it is expressed. Please let me know the definition of $D_{\mu}$. $\endgroup$
    – sakata
    Commented Jul 11, 2021 at 1:25
  • $\begingroup$ I guess OP considers Peskin & Schroder (15.89) and indeed there are no description of a covariant derivative tensor product. One can define the covariant derivative $D$ so that $D(A\otimes B) = (DA)\otimes B + A \otimes (DB)$ (physicists often say this is because that "there are different legs"). $\endgroup$
    – Keyflux
    Commented Jan 18 at 14:15

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