I am trying to prove that the supercurrent $J^\mu = \gamma^{\nu \rho} F^A_{\nu \rho} \gamma^\mu \lambda^A $ is conserved in ${\cal N}=1$ SUSY Yang-Mills theory ( basically trying to reproduce equation 6.11 in the textbook "Supergravity" by Freedman and Proeyen ). Here $A$ is index in the adjoint representation of gauge group $G$. $F$ is the field strength tensor for the non abelian gauge field $A_\mu(x)$ . $\lambda^A(x)$ is the gaugino field (superpartner of $A_\mu(x)$) and thus carries an adjoint index $A$.
The equations of motion of the theory are: $$ D^\mu F^A_{\mu \nu} = -\frac{1}{2} g f_{BC}^{\ \ \ \ A} \bar{\lambda}^B \gamma_\nu \lambda^C \tag{A}$$ $$ \gamma^\mu D_\mu \lambda^A = 0 \tag{B}$$
The Bianchi identity is: $$ D_\mu F^A_{\nu \rho} + D_\nu F^A_{\rho \mu } + D_\rho F^A_{\mu \nu} = 0 \tag{C} $$
Here is how I proceed. Firstly the partial derivative distributes over field F and field $\lambda$ giving: $$ \partial_\mu J^\mu = \partial_\mu F^A_{\nu \rho} \gamma^{\nu \rho }\gamma^\mu \lambda^A + F^A_{\nu \rho} \gamma^{\nu \rho }\gamma^\mu \partial_\mu \lambda^A .\tag{D}$$
Since $F$ and $\lambda$ transform in adjoint representation, they have similar action of covariant derivative $D$ on them, namely:
$$ D^\mu \lambda^A = \partial_\mu \lambda^A + g f_{BC}^{\ \ \ \ A} A_{\mu}^B \lambda^C \tag{E.1}$$ $$ D^\mu F_{\nu \rho}^A = \partial_\mu F_{\nu \rho}^A + g f_{BC}^{\ \ \ \ A} A_{\mu}^B F_{\nu \rho}^C. \tag{E.2}$$
With these definitions, we can write equation (D) as follows (the terms proportional to $g$ will simply cancel when we use antisymmetry):
$$ \partial_\mu J^\mu = D_\mu F^A_{\nu \rho} \gamma^{\nu \rho }\gamma^\mu \lambda^A + F^A_{\nu \rho} \gamma^{\nu \rho }\gamma^\mu D_\mu \lambda^A. \tag{F}$$
I am writing equation (F) to match it with what's given in the textbook equation 6.11, and also so that then we can use equations A,B,C for simplification. Now the rightmost term in equation (F) vanishes due to the equation of motion (B). And then I use Bianchi identity (C) on the remaining term to write:
$$ \partial_\mu J^\mu = - D_\nu F^A_{\rho \mu } \gamma^{\nu \rho} \gamma^\mu \lambda^A - D_\rho F^A_{\mu \nu } \gamma^{\nu \rho} \gamma^\mu \lambda^A . \tag{G}$$
Then using the antisymmetry of indices, I bring both of the terms to same index structure and write:
$$ \partial_\mu J^\mu = - 2 \ D_\rho F^A_{\mu \nu } \gamma^{\nu \rho} \gamma^\mu \lambda^A. \tag{I}$$
This is not quite what we have in the textbook. In the textbook, they get: $$ \partial_\mu J^\mu = - 2 \ D^\mu F^A_{\mu \nu } \gamma^\nu \lambda^A. \tag{6.11}$$
This is where I am struggling. Can someone help me out?
FYI, to complete the story, equation 6.11 can then be simplified by using equation of motion (A) and fierz identities. And it gives zero thus proving that the supercurrent in indeed conserved. By Haag-Lopuszanski-Sohnius theorem then it follows that this theory must have supersymmetry.
Please feel free to comment in general also to impart more knowledge or to correct if I said something incorrectly.
