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In non-abelian gauge theory, such as P & S's chapter 15, eq. (15.89), we also have Bianchi identity.

Start with $$\epsilon^{\mu\nu\lambda\sigma}[D_\nu,[D_\lambda,D_\sigma]]=0$$ and use $[D_\mu,D_\nu]=-ig\vec{F_{\mu \nu}}$, where $\vec{F_{\mu \nu}}=F_{\mu \nu}^at^a$, we obtain $$\epsilon^{\mu\nu\lambda\sigma}(D_\nu \vec{F_{\lambda \sigma}})=0$$

I think this formula is general, which includes abelian case. However, I find contradiction in Abelian case.

In Abelian case, $[D_\mu,D_\nu]=ieF_{\mu \nu}$, where $F_{\mu \nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$, and $D_\nu=\partial_\nu+ieA_\nu$, so we have $$\begin{aligned}&\epsilon^{\mu\nu\lambda\sigma}(D_\nu F_{\lambda \sigma}) \\ &=\epsilon^{\mu\nu\lambda\sigma}(\partial_\nu F_{\lambda \sigma}+ieA_\nu F_{\lambda \sigma} )=0 \end{aligned}$$ However, we already know that $\epsilon^{\mu\nu\lambda\sigma}\partial_\nu F_{\lambda \sigma}=0$ from $\partial_\nu F_{\lambda \sigma}+\partial_\lambda F_{\sigma \nu}+\partial_\sigma F_{\nu \lambda}=0$. Which means we must have $$\epsilon^{\mu\nu\lambda\sigma}A_\nu F_{\lambda \sigma}=2\epsilon^{\mu\nu\lambda\sigma}A_\nu\partial_\lambda A_\sigma=0 $$

But this seems not equal to zero.

Qmechanic
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Daren
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  • Which reference puts an arrow over $F_{\mu\nu}$? – Qmechanic Feb 16 '23 at 11:45
  • @Qmechanic Since according to P & S's book eq.(15.32), $[D_\mu,D_\nu]=-igF_{\mu \nu}^a t^a$. So I thought there should be an arrow to contain the Lie algebra. Would my thought right? There are also Lie algebras inside $D_\mu$. – Daren Feb 16 '23 at 12:22
  • Aha, an arrow to indicate a vector. An arrow is a bit unusual notation in this context. – Qmechanic Feb 16 '23 at 12:29
  • Thank you, so I think the correct bianchi identity for non-abelian should be $\epsilon^{\mu \nu \lambda \sigma}[D_\nu,F_{\lambda \sigma}^a t^a]=0$, right? – Daren Feb 16 '23 at 12:37
  • $\uparrow$ Yes. – Qmechanic Feb 16 '23 at 12:45

1 Answers1

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OP's question seems spurred by an ambiguity in the notation for differential operators, cf. e.g. this related Phys.SE post.

There is a Lie-bracket/commutator in the 2nd Bianchi identity: $$ \epsilon^{\mu\nu\lambda\sigma}[D_{\nu},F_{\lambda\sigma}]=0. $$

In the abelian case the Lie-bracket is zero, so we can replace the covariant derivative with a partial derivative: $$ \epsilon^{\mu\nu\lambda\sigma}[\partial_{\nu},F_{\lambda\sigma}]=0. \qquad\text{(Abelian case)}$$

Qmechanic
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