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added 16 characters in body
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J.G.
J.G.
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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}\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.

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, $$D_\mu F_{\nu\lambda} + D_\nu F_{\lambda\mu} + D_\lambda F_{\mu\nu} = 0$$ which proves the desired result.

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.

fixed wrong index in last equation
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John Donne
John Donne
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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, $$D_\mu F_{\nu\lambda} + D_\nu F_{\lambda\mu} + D_\mu F_{\mu\nu} = 0$$$$D_\mu F_{\nu\lambda} + D_\nu F_{\lambda\mu} + D_\lambda F_{\mu\nu} = 0$$ which proves the desired result.

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.

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, $$D_\mu F_{\nu\lambda} + D_\nu F_{\lambda\mu} + D_\lambda F_{\mu\nu} = 0$$ which proves the desired result.

fixed latex typo
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John Donne
John Donne
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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}\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.

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.

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.

Fixed up TeX and bizarrely mis-spelled "lambda"
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DanielSank
DanielSank
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Dheeraj Shukla
Dheeraj Shukla
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