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.
    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
    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
    Jan 18 at 14:15

1 Answer 1


$D_\mu T$ 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 $D_\mu(ST)=(D_\mu S)T+SD_\mu T$.

  • $\begingroup$ For non-Abelian gauge field theory, the field strength is not gauge-invariant (rather it transforms as adjoint rep.), isn't it? $\endgroup$
    – Keyflux
    Jan 18 at 10:29

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