I am following the book by Freedman and Van-Proeyen and this question is related to exercise 6.3.
The supercurrent of a super Yang-Mills theory is given by $\mathcal{J}^{\mu} = \gamma^{\nu \rho} F^A_{\mu \nu} \gamma^{\mu} \lambda^{A}$, and we need to show $$\partial_{\mu} \mathcal{J}^{\mu} = g f_{ABC} \gamma^{\nu} \lambda^{A} \bar{\lambda}^{B} \gamma_{\nu} \lambda^C$$
is zero for certain types of spinors in certain dimensions.
In D=3,4 dimensions this is very straightforward. Essentially it is the solution of exercise 3.27 which I have done. In case other readers want to benefit let me sketch it: in the derivative of the supercurrent $f_{ABC}$ is completely anti-symmetric in all the indices. So, write the rest of the expression as a symmetric and an anti-symmetric part. The former vanishes due to contraction and the latter is shown to vanish in the following way:
we consider eq.(3.66) and we expand it
\begin{align} {(\gamma^{\mu})}_{ab}{(\gamma_{\mu})}_{cd} &= \frac{1}{2^m} \sum_{A} (-)^{r_A}(D-2 r_A) ({\Gamma}_{A})_{ad} ({\Gamma}_{A})_{cb} \\ &= \frac{1}{4} \left( 4 \cdot (1)_{ad}(1)_{cb} + 2 \cdot (\gamma_{\mu})_{ad} (\gamma^{\mu})_{cb} + 0 \cdot (\gamma_{\mu \nu})_{ad} (\gamma^{\mu \nu})_{cb} \right) \end{align} where in the above the sum is truncated at order $r \leq m$ in the notation $D=2m$ or $D=2m+1$, to account only for the linearly independent elements of the algebra. Note that in three dimensions the final term is completely absent.
From table 3.1 in the book, we see that $t_0=t_3=1$ and $t_1=t_2=-1$. Now, we can use eq.(3.63) which reads
$$(\gamma_{\mu_1 \mu_2 \cdots \mu_r})_{ab} = - t_r ~ (\gamma_{\mu_1 \mu_2 \cdots \mu_r})_{ba} $$
We symmetrize the (bcd)-indices in eq.(3.66) and observe that the term involving rank-0 Clifford elements cancels and only the rank-1 survives. We expand the RHS out, use eq.(3.63) to combine the terms that are equal, pass the RHS to the LHS and we have LHS-RHS=0 and in that way, we obtain
$$(\gamma^{\mu})_{a(b} (\gamma^{\mu})_{cd)} = 0 $$
and after contracting with three anti-commuting spinors the above symmetric equation is picking up signs and becomes
$$\gamma_{\mu} \lambda_{[A}\bar{\lambda}_B\gamma^{\mu}\lambda_{C]}=0$$
Finally, we can use the above in the derivative of the supercurrent and see that it is indeed zero; it is conserved.
Moving to six dimensions. We have -again using table 3.1- $t_0=t_3=-1$ and $t_1=t_2=1$. We can once more write explicitly eq.(3.66) as before and we have:
\begin{align} \begin{aligned} {(\gamma^{\mu})}_{ab}{(\gamma_{\mu})}_{cd} &= \frac{1}{2^m} \sum_{A} (-)^{r_A}(D-2 r_A) ({\Gamma}_{A})_{ad} ({\Gamma}_{A})_{cb} \\ &= \frac{1}{2^3} \left( 6 \cdot (1)_{ad}(1)_{cb} + 4 \cdot (\gamma_{\mu})_{ad} (\gamma^{\mu})_{cb} + 2 \cdot (\gamma_{\mu \nu})_{ad} (\gamma^{\mu \nu})_{cb} \\ + 0 \cdot (\gamma_{\mu \nu \rho})_{ad} (\gamma^{\mu \nu \rho})_{cb} \right) \end{aligned} \end{align}
Due to the values of $t_r$ if we symmetrize the indices (bcd) in eq.(3.66) the LHS vanishes. If I antisymmetrize the indices the LHS is non-zero. The pices of the RHS will give:
$$(1)_{a[d}(1)_{cb]}=0$$
using eq.(3.63) which is written above. The terms with the rank-1 and rank-2 elements are not vanishing and we can write them explicitly -doing rank-1 for illustration-
$$(\gamma_{\mu})_{a[d}(\gamma^{\mu})_{cb]} = (\gamma_{\mu})_{ad} (\gamma^{\mu})_{cb} - (\gamma_{\mu})_{ad} (\gamma^{\mu})_{bc} + \cdots \sim (\gamma_{\mu})_{a(d}(\gamma^{\mu})_{cb)}$$
having used once more eq.(3.63) and similarly, we can re-write the term involving the rank-2 piece.
And this is the point of my confusion. When I antisymmetrize the indices and expand them out using eq.(3.63) it seems to me that I can write them in the symmetrized form as done above.
If instead, I use the values of $t_r$ to argue that only the antisymmetrized parts are non-zero then I schematically I would have something like the following
$$(\gamma_{\mu})_{a[b}(\gamma^{\mu})_{cd]} \sim (\gamma_{\mu})_{a[d}(\gamma^{\mu})_{cb]} + (\gamma_{\mu \nu})_{a[d}(\gamma^{\mu \nu})_{cb]}$$
Since the LHS is already antisymmetric if I contract with three anticommuting spinors as in the case of D=3,4 dimensions previously I cannot see how to show the conservation of the supercurrent. I think that I am following the right method conceptually, see footnote 3 on page 52 of the book, unless I have misunderstood something.
I think I have managed to answer the first (above) part of the post.
For the following, I would like some explanation.
The second part of this post has to do with SYM but following the appendix of chapter 4 from the book by Green, Schwarz and Witten. They perform the analogous computation for D=10 SYM. I think I have properly understood most parts, but I have a question which is relatively simple. When they move from eq.(4.A.5) to (4.A.6) they do so by explicitly removing the spinors from the equation and adding $\gamma^0$ factors. In the $(-,+,\cdots,+)$ we have that $(\gamma^0)^2=-1$. Why are they free to remove the spinors and examine the $\gamma$-structures and also do they add the $\gamma^0$ to account for the anticommutativity of the spinors or is there something else that I am missing?
The latter method seems to be less intense and more transparent.