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How can one prove the Bianchi identity of a non-Abelian gauge theory? i.e. $$ \epsilon_{\mu \nu \lambda \sigma}(D_{\nu}F_{\lambda \sigma})^a=0 $$

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3 Answers 3

Hint: Use the Jacobi identity $$\sum_{\mu,\nu,\lambda~{\rm cycl.}} [D_{\mu},[ D_{\nu}, D_{\lambda}]]~=~0. $$

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Hint: use $$ [D_{\nu},D_{\mu}] = F_{\nu \mu} $$

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using Jacobi identity as mentioned in one of the answers, its very easy to prove the Bianchi identity

\begin{align} [D_\mu,[D_\nu,D_\lambda]]\psi &= [D_\mu,ig F_{\nu\lambda}] \psi \\ &= D_\mu (F_{\nu\lambda}\psi) - F_{\nu \lambda}D_\mu \psi \\ &= D_\mu F_{\nu\lambda}\psi + F_{nu \lambda}D_\mu\psi - F_{nu\lambda}D_\mu\psi \\ &= D_\mu F_{\nu\lambda}\psi \, . \end{align} Therefore, $$D_\mu F_{\nu\lambda} + D_\nu F_{\lambda\mu} + D_\mu F_{\mu\nu} = 0$$ which proves the desired result.

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Welcome to physics stack exchange. This site supports TeX style typesetting for mathematics. I edited your post to use this feature. Please take a look at how it was done so you can do it yourself in future posts. –  DanielSank Jun 27 at 7:31

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