The tendimensional case is explained in some detail in appendix 4.A of volume one of GreenSchwarzWitten. Let me therefore consider the 4d case here. The calculation works essentially the same in 3d, 6d and 10d, though you can sometimes make use of Majorana and/or Weyl conditions of the spinors to simplify things.
We want to check that the expression
\begin{equation*}
f_{abc} \bigl(
\bar{\lambda}^a \gamma_{\mu} \lambda^b \, \bar{\epsilon} \, \gamma^{\mu} \lambda^c

\bar{\lambda}^a \gamma_{\mu} \lambda^b \, \bar{\lambda}^c \gamma^{\mu} \epsilon
\bigr)
\tag{1}
\end{equation*}
vanishes, where $\lambda^a$ and $\epsilon$ are fourdimensional Majorana spinors. To show this the following identities involving fourdimensional gamma matrices and arbitrary spinors will be useful
\begin{align*}
\bar{\varphi} \, \psi &= + \bar{\psi} \, \varphi , \tag{2a} \\
\bar{\varphi} \, \gamma^{\mu} \, \psi &=  \bar{\psi} \, \gamma^{\mu} \, \varphi , \tag{2b} \\
\bar{\varphi} \, \gamma^{\mu\nu} \, \psi &=  \bar{\psi} \, \gamma^{\mu\nu} \, \varphi , \tag{2c} \\
\bar{\varphi} \, \gamma^{\mu\nu\rho} \, \psi &= + \bar{\psi} \, \gamma^{\mu\nu\rho} \, \varphi , \tag{2d} \\
\bar{\varphi} \, \gamma^{0123} \, \psi &= + \bar{\psi} \, \gamma^{0123} \, \varphi , \tag{2e}
\end{align*}
where, eg., $\gamma^{\mu\nu} = \frac{1}{2} ( \gamma^\mu\gamma^\nu  \gamma^\nu\gamma^\mu)$.
Using the second identity we see that the two terms in (1) are equal.
Using the antisymmetry of $f_{abc}$ we can rewrite (1) as
\begin{equation*}
( \bar{\lambda}^a \gamma_\mu \lambda^b ) ( \bar{\epsilon} \, \gamma^\mu \lambda^c )
+
( \bar{\lambda}^c \gamma_\mu \lambda^a ) ( \bar{\epsilon} \, \gamma^\mu \lambda^b )
+
( \bar{\lambda}^b \gamma_\mu \lambda^c ) ( \bar{\epsilon} \, \gamma^\mu \lambda^a ) ,
\end{equation*}
or, if we remove the spinors,
\begin{equation*}
(\gamma^0 \gamma_\mu)_{\alpha\beta} (\gamma^0 \gamma^\mu)_{\gamma\delta}
+
(\gamma^0 \gamma_\mu)_{\delta\alpha} (\gamma^0 \gamma^\mu)_{\gamma\beta}

(\gamma^0 \gamma_\mu)_{\beta\delta} (\gamma^0 \gamma^\mu)_{\gamma\alpha} .
\end{equation*}
Finally we contract this expression with two arbitrary spinors $\psi^\alpha$ and $\varphi^\beta$ to obtain (here I have also removed an overall $\gamma^0$ and reordered the second term slightly)
\begin{equation*}
( \bar{\psi} \, \gamma_\mu \varphi ) ( \gamma^\mu )^\gamma{}_\delta
+
( \bar{\psi} \, \gamma_\mu )_\delta ( \gamma^\mu \varphi )^\gamma

( \bar{\varphi} \, \gamma_\mu )_\delta ( \gamma^\mu \psi )^\gamma .
\tag{3}
\end{equation*}
The idea is to now think of this as a $4 \times 4$ matrix. A basis of such matrices is given by the 16 matrices
\begin{equation*}
1 , \quad \gamma^\mu , \quad \gamma^{\mu\nu} , \quad \gamma^{\mu\nu\rho} , \quad \gamma^{0123} .
\tag{4}
\end{equation*}
This basis is orthogonal in the sense that if we call the sixteen matrices $\gamma^{(I)}$, with $I=1,\dots,16$, then
\begin{equation*}
\operatorname{Tr}\, \bigl( \gamma^{(I)} \gamma^{(J)} \bigr) = \operatorname{Tr} \, \bigl( \gamma^{(I)} \gamma^{(I)} \bigr) \, \delta^{IJ} .
\end{equation*}
Hence, to prove that (3) vanishes we can check that we get zero when contracting with each of the matrices in (4).
It is easy to see that the first term in (3) is zero unless we contract with a single gamma matrix, so let us first contract with $\gamma^\nu$. We then get
\begin{equation*}
( \bar{\psi} \, \gamma_\mu \varphi ) \, \operatorname{Tr}\, \bigl( \gamma^\mu \gamma^\nu \bigr)
+
\bar{\psi} \, \gamma_\mu \gamma^\nu \gamma^\mu \varphi

\bar{\varphi} \, \gamma_\mu \gamma^\nu \gamma^\mu \psi
=
4 \eta^{\mu\nu} \bar{\psi} \, \gamma_\mu \varphi  2 \, \bar{\psi} \, \gamma^\nu \varphi + 2 \, \bar{\varphi} \, \gamma^\nu \psi
=
0 ,
\end{equation*}
where I have used equation (2b) as well the second of the following useful relations for fourdimensional gamma matrices
\begin{align*}
\gamma_\mu \gamma^\mu &= 4 , \tag{5a} \\
\gamma_\mu \gamma^{\nu} \gamma^\mu &= 2\gamma^\nu , \tag{5b} \\
\gamma_\mu \gamma^{\nu\rho} \gamma^\mu &= 0 , \tag{5c} \\
\gamma_\mu \gamma^{\nu\rho\sigma} \gamma^\mu &= +2\gamma^{\nu\rho\sigma} , \tag{5d} \\
\gamma_\mu \gamma^{0123} \gamma^\mu &= 4\gamma^{0123} . \tag{5e}
\end{align*}
If we instead contract (3) with $\gamma^{\nu\rho}$ each term vanishes separately because of equation (5c). In the remaining three cases (contracting with $1$, $\gamma^{\nu\rho\sigma}$ or $\gamma^{0123}$), the first term in (3) is zero and the second two terms cancel each other because of the positive sign in (2a), (2d) and (2e).
We have now shown that all components of the $4 \times 4$ matrix in (3) vanish. Since the spinors $\psi$ and $\varphi$ were arbitrary this means that (1) is zero as well.
The above procedure of expanding in a basis of (products of) gamma matrices is equivalent to, but more transparent than, using Fierz identities.
This post imported from StackExchange Physics at 20160129 15:22 (UTC), posted by SEuser Olof