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 $\gamma^{1 \ldots d} = \gamma^{[1}\gamma^2 \ldots \gamma^{d]}$

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    $\begingroup$ Are you talking about p.41 of Supergravity? (Freedman-Van Proeyen) $\endgroup$ – Rexcirus Dec 20 '14 at 13:36
  • $\begingroup$ yes! this is exactly what I am talking about $\endgroup$ – Marion Dec 20 '14 at 13:37

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:

$$\gamma^{123} = \frac{1}{2}\{\gamma^{1}, \gamma^{23} \}= \frac{1}{2}\left(\gamma^{1} \gamma^{23}+ \gamma^{23}\gamma^{1} \right)=\frac{1}{4}\left(\gamma^{1} \gamma^{2}\gamma^{3}-\gamma^{1} \gamma^{3}\gamma^{2}+\gamma^{2} \gamma^{3}\gamma^{1}-\gamma^{3} \gamma^{2}\gamma^{1} \right)=\gamma^{1} \gamma^{2}\gamma^{3}=\gamma^{123}$$

  • $\begingroup$ Sure, I agree with what you re saying. But if you expand (1) you end up with 4 terms. Two that have $\gamma^1$ in the beginning and two that have it in the end. I am still missing the two terms that have it in the middle. $\endgroup$ – Marion Dec 20 '14 at 14:25
  • $\begingroup$ Ok, I see what I was making wrong. Thanks a lot @Rexcirus. How stupid of mine! $\endgroup$ – Marion Dec 20 '14 at 14:32
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    $\begingroup$ Don't worry! It can happen to anyone! XD $\endgroup$ – Rexcirus Dec 20 '14 at 14:35

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.

  • $\begingroup$ Which are the equal terms? $\gamma^2 \gamma^3 \neq \gamma^3 \gamma^2 $ $\endgroup$ – Marion Dec 20 '14 at 14:09
  • $\begingroup$ @Marion: $\gamma^2 \gamma^3 = -\gamma^3 \gamma^2$ $\endgroup$ – akhmeteli Dec 20 '14 at 14:22
  • $\begingroup$ Sure, I know this. $\endgroup$ – Marion Dec 20 '14 at 14:29
  • $\begingroup$ @Marion: Then what seems to be the problem? The anticommutator in (2) includes these two equal terms. $\endgroup$ – akhmeteli Dec 20 '14 at 14:42
  • $\begingroup$ I would forget to use this fact and would try to fully expand the whole 3-rank tensor. It is really simple though indeed. $\endgroup$ – Marion Dec 20 '14 at 14:56

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