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
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Hint: Use the Jacobi identity $$\sum_{\mu,\nu,\lambda~{\rm cycl.}} [D_{\mu},[ D_{\nu}, D_{\lambda}]]~=~0. $$
answered Mar 8, 2013 at 23:32
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using Jacobi identity as mentioned in one of the answers, its very easy to prove the Bianchi identity
\begin{align}\tfrac{1}{ig} [D_\mu,\,[D_\nu,\,D_\lambda]]\psi &= [D_\mu,\,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, $$D_\mu F_{\nu\lambda} + D_\nu F_{\lambda\mu} + D_\lambda F_{\mu\nu} = 0$$ which proves the desired result.
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1$\begingroup$ 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. $\endgroup$ Commented Jun 27, 2015 at 7:31
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$\begingroup$ how does the the covariant derivative act on the field strength $F_{\mu\nu}$. The context you used here just define how it acts on $\psi$, explicitly $$ \partial_{\mu} \psi^{I} + A_{\mu}{}^I_J \psi_J $$ in which $I,J$ is the group representation indices. $\endgroup$ Commented Aug 9 at 17:16
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Hint: use $$ [D_{\nu},D_{\mu}] = igF_{\nu \mu} $$