Question on spinor brackets I have been given the Lorentz generator 
$$J_{ab}=\lambda_a \frac{\partial}{\partial \lambda^b}+a\leftrightarrow b$$
I know $\frac{\partial}{\partial \lambda^b}\lambda_a=\epsilon_{ba}$ and $A_a=A^b\epsilon_{ab}$ (? not sure about these two).
I want to show that this acting on a spinor bracket gives zero as this is supposed to be Lorentz invariant. Say I have a bracket $\langle A B \rangle = A_a B_b \epsilon^{ab}$ and now I act with $J_{ab}$ on it w.r.t $A$, i.e.
$$\begin{split}J^A_{ab}\langle A B \rangle &=(A_a \frac{\partial}{\partial A_b}+a\leftrightarrow b) A_cB_d\epsilon^{cd}\\&=A_a\epsilon_{bc}B_d\epsilon^{cd} + A_b\epsilon_{ac}B_d\epsilon^{cd}\\&=A_a\epsilon_{bc}B_{c}+A_b\epsilon_{ac}B_c\\&=?\end{split}$$
Where do I go wrong? Could someone please help me out? Why doesnt it sum to zero?
 A: Firstly, I don't think your convention for the derivative is consistent with how you lower the index on $\lambda^b$. Let us start with
$$
\frac{\partial}{\partial \lambda^b} \lambda^a = \delta_b^a
$$
If you lower the index on $\lambda^b$ you get
$$
\lambda_a = \epsilon_{ac} \lambda^c
$$
Putting this together gives
$$
\frac{\partial}{\partial \lambda^b} \lambda_a = \epsilon_{ac} \frac{\partial}{\partial \lambda^b} \lambda^c = \epsilon_{ab} = -\epsilon_{ba}
$$
which has a different sign from what you have.
Furthermore, the spinor bracket you write is Lorentz invariant if you transform both spinors at the same time. The actions of $J^A$ and $J^B$ on $\langle AB \rangle = \epsilon_{cd} A^c B^d$ is given by
\begin{align}
J_{ab}^A\, \epsilon_{cd} A^c B^d &= 
\left(A_a \frac{\partial}{\partial A^b} + A_b \frac{\partial}{\partial A^a} \right) \, \epsilon_{cd} A^c B^d \\
&= \epsilon_{cd} \left(
A_a \delta_b^c + A_b \delta_a^c
\right) B^d
= +(A_a B_b + B_b A_a) \\
%
J_{ab}^B\, \epsilon_{cd} A^c B^d &= 
\left(B_a \frac{\partial}{\partial B^b} + B_b \frac{\partial}{\partial B^a} \right) \, \epsilon_{cd} A^c B^d \\
&= \epsilon_{cd} A^c \left(
B_a \delta_b^d + B_b \delta_a^d
\right)
= -(A_a B_b + B_b A_a)
\end{align}
so if you sum them the transformation of $A$ is cancelled by the transformation of $B$.
