Higher rank $\gamma$-matrix question I read that the higher rank $\gamma$ matrices can be written as alternate commutators and anti-commutators. For example, the rank 3 gamma matrix can be written as
$$\gamma^{123} = \frac{1}{2}\{\gamma^{1}, \gamma^{23} \}, \tag{1}$$
where 
$$\gamma^{23} = \frac{1}{2}[\gamma^{2},\gamma^{3}]. \tag{2}$$
Now if we put (2) into (1) we get four terms and an overall factor of 1/2. Despite that, if we take the permutations of 1,2,3 we get 6 elements, namely the symmetric
$123, 312, 231$ and the anti-symmetric $132, 321, 213$. Thus we have 6 elements and we should have an overall factor of $1/3! = 1/6$.
My question is: is there some mistake in the definition of (1)?
P.S. Note that $\gamma^{1 \ldots d} = \gamma^{[1}\gamma^2 \ldots \gamma^{d]}$
 A: The definition (pay attention to not confuse generic tensor with tensor components) of the antisymmetric gamma tensor is:
$$\gamma^{\mu_1 \mu_2 \dots \mu_r}=\gamma^{[\mu_1 \mu_2 \dots \mu_r]}$$ 
For the highest rank you have $r=D$, so you have to use all the possible indices. For example, in components, you have the identity:
$$\gamma^{1 2 3}=\frac{1}{3!}  (\gamma^{1 2 3}-\gamma^{132}-\gamma^{21 3}+\gamma^{231}-\gamma^{321}+\gamma^{312})$$ 
that is true, using the anticommutativity of the gamma matrices and the fact that in components $\gamma^{1 2 3}=\gamma^{1 }\gamma^{2}\gamma^{ 3}$. Using this reasoning you can show that your expression (1) is true. (you have two factors 1/2, so a total 1/4 and four terms)
More explicitly: 
$$\gamma^{123} = \frac{1}{2}\{\gamma^{1}, \gamma^{23} \}= \frac{1}{2}\left(\gamma^{1} \gamma^{23}+ \gamma^{23}\gamma^{1} \right)=\frac{1}{4}\left(\gamma^{1} \gamma^{2}\gamma^{3}-\gamma^{1} \gamma^{3}\gamma^{2}+\gamma^{2} \gamma^{3}\gamma^{1}-\gamma^{3} \gamma^{2}\gamma^{1} \right)=\gamma^{1} \gamma^{2}\gamma^{3}=\gamma^{123}$$
A: Maybe I am missing something, but in your eq. (2) there are two equal terms on the right-hand side, and the factor is 1/2 to take that into account. In your eq.(1) there are also two equal terms on the right-hand side, and the factor is 1/2 to take that into account. There is no summation over all permutations in your formulas. 
