Supercurrent conservation for super-Yang-Mills in D=3,4,6,10 dimensions 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. 
 A: I think I have figured it out. I will sketch and mention the basic relations and considerations with the hope that they will be useful to other users and also if you see a mistake you can point it out. 
All equations below are references to the book.
Let us contract eq.(3.65) with three spinors; namely $\bar{\lambda}^{\gamma} \psi_{\delta} \chi_{\beta}$ and obtain the following relation. 
\begin{align}
\bar{\lambda} \psi \chi_{\alpha} = - \frac{1}{2^m} \sum_{A} \bar{\lambda} \Gamma_{A} \chi \Gamma^{A} \psi_{\alpha}
\end{align}
or, one can contract with the result of exercise 3.26 to include the rank-1 Clifford elements in the LHS of the equation.
An equivalent approach would be to write a Fierz rearrangement for the spinor bilinear of the supercurrent $\lambda^{A} \bar{\lambda}^{B}$ by following the spirit of the derivation in eq.(3.70) which is exercise 3.28 on page 53. If I am not mistaken this should be written as 
\begin{align}
\lambda^{B} \bar{\lambda}^C = - \frac{1}{2^m} \sum_{A} \bar{\lambda}^C \Gamma_{A} \lambda^{B} \Gamma^{A}
\end{align}
And now we need to take a step back. Recall that in the supecurrent $f_{ABC}$ is completely antisymmetric in all indices so we need to consider only terms that are antisymmetric. 
Looking at the dimensionality of spacetime only does not work. One has to take into account the nature of the spinors. This takes us back to table 3.2 on page 59 where we observe that for symplectic-Weyl spinors  in six dimensions spinor bilinears need to have the insertion of a rank-3 Clifford element and no other choice is giving an antisymmetric result (of course the same argument holds mod 4 due to the properties of the $t_r$, but we want to consider only the linearly independent elements of the algebra). 
So, the only thing that will consistently contribute is a bilinear of the form 
\begin{align}
\lambda^A \gamma^{\mu \nu \rho} \bar{\lambda}^B
\end{align}
But in the supercurrent the above term is found in between a contraction and the schematic that we are interested in is: 
\begin{align}
\gamma_{\omega} \gamma^{\mu \nu \rho} \gamma^{\omega} 
\end{align}
which vanishes by virtue of eq.(3.24) on page 43. 
So, to wrap it up: if the above understanding is correct -and it seems so- the previous mistake was that I was looking at table 3.1 to deduce the values of the various $t_r$ while neglecting that the only description -so far- for symplectic Weyl spinors is given on Table 3.2. 
The above is related to the first part of my question. If anyone could give a definite answer for the GSW trick which is the second part please feel free to do so, as I have not found another explanation as the one I originally wrote down. 
