Theorem 3

$\begin{displaymath} \textrm{Tr}[\not{\;\!\!a}_1\cdots\not{\;\!\!a}_n] = a_1\cdo... ...t a_n \textrm{Tr}[\not{\;\!\!a}_2\cdots \not{\;\!\!a}_{n-1}] . \end{displaymath}$

(7.61)

$\begin{displaymath} \textrm{Tr}[\not{\;\!\!a}_1\not{\;\!\!a}_2\not{\;\!\!a}_3\no... ... + a_1\cdot a_4 a_2\cdot a_3 - a_1\cdot a_3 a_2\cdot a_4) . \end{displaymath}$

(7.62)

Proof: Using $\not{\;\!\!a}_1\not{\;\!\!a}_2 = - \not{\;\!\!a}_2\not{\;\!\!a}_1 + 2 a_1\cdot a_2$ ,

$\displaystyle \textrm{Tr}[\not{\;\!\!a}_1\not{\;\!\!a}_2\cdots\not{\;\!\!a}_n]$	$\textstyle =$	$\displaystyle 2a_1\cdot a_2\textrm{Tr}[\not{\;\!\!a}_3\cdots\not{\;\!\!a}_n] - \textrm{Tr}[\not{\;\!\!a}_2\not{\;\!\!a}_1\not{\;\!\!a}_3\cdots\not{\;\!\!a}_n]$
	$\textstyle =$	$\displaystyle 2a_1\cdot a_2 \textrm{Tr}[\not{\;\!\!a}_3\cdots\not{\;\!\!a}_n] - \cdots$
	$\textstyle +$	$\displaystyle 2a_1\cdot a_n \textrm{Tr}[\not{\;\!\!a}_2\cdots \not{\;\!\!a}_{n-1}] - \textrm{Tr}[\not{\;\!\!a}_2\cdots\not{\;\!\!a}_n\not{\;\!\!a}_1] .$	(7.63)

$\displaystyle b_\mu d_\nu \textrm{Tr}[\not{a}\gamma^\mu\not{c}\gamma^\nu]$	$\textstyle =$	$\displaystyle 4(a^\mu b_\mu c^\nu d_\nu + a^\nu d_\nu b_\mu c^\mu - a\cdot c b_\mu d^\mu)$
	$\textstyle =$	$\displaystyle b_\mu d_\nu 4 (a^\mu c^\nu + a^\nu c^\mu - a\cdot c g^{\mu\nu})$
$\displaystyle \Rightarrow \textrm{Tr}[\not{a}\gamma^\mu\not{c}\gamma^\nu]$	$\textstyle =$	$\displaystyle 4(a^\mu c^\nu + a^\nu b^\mu - a\cdot c g^{\mu\nu}) .$	(7.64)