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Suppose $a,b,c...=0,....,D-1$ are Lorentz indices of $SO(1,D-1)$ tangent space and consider $D$-dimensional Clifford algebra defined by the usual anticommutation relation $$\{\Gamma^a,\Gamma^b\}=2\eta^{ab} \, .$$

Let's define the fully antisymmetric product of Gamma matrices as $$ \Gamma^{a_1 a_2 ... a_n}=\Gamma^{[a_1}\Gamma^{a_2....}\Gamma^{a_n]} \qquad n=1,..D. $$

Does exist a clever analytic way to obtain a generic (anti)commutator of the form $$\{\Gamma^a,\Gamma^{bcd...}]= \,?$$

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Do the $a_i$ run from $0$ $D-1$ as in your first statement or o from 1 to $D$ as in your defintion of the antisymmetric object? It makes a difference in what you are asking. For example, in four dimensions with the $\gamma_a$, $a=1,4$ we define $\gamma_5$ as $\gamma_5 \propto \gamma_1\gamma_2\gamma_3\gamma_4$. Then $\gamma_1\gamma_2\gamma_3\gamma_4\gamma_5$ commutes with everthing and for an irreducible representation, must be a multiple of the identity. In any case he antisymmetrization is redundant as the $\gamma_i$ anticommute.

The general case works in the same way: A product of an even number of distinct gamma matrices anticommutes with any of the matrices in the product; a product of an odd number commutes with the matrices in the product.

If you have an even $D$ the product, $\gamma_1\ldots \gamma_D$, of all the gammas anticommutes with every gamma, and for an odd $D$ the product $\gamma_1\ldots \gamma_D$ commutes with all the gammas (and any sum of products of gammas) and is a multiple of the identity in the Clifford algebra. These facts just follow from that fact that any two distinct $\gamma_a$ anticommute. No fancy algebra is needed.

For products involving fewer than the whole set of gammas you need to see whether the gamma you wish to (anti)commute is in the set or not. If it is the above rules apply. If it is not, then the role of odd and even interchange.

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  • $\begingroup$ Thanks for your comment. With my notation I meant $a_1,..,a_D= 0,1,..,D-1$ but maybe is better to fix it. If I consider for example $D=11$, $\Gamma^{01..6}=\Gamma^0\Gamma^1\cdot\cdot\Gamma^6$, how can I obtain $\{\Gamma^a,\Gamma^{01..6}]$ avoiding lengthy calculations? $\endgroup$ – Marco Jul 5 at 19:27
  • $\begingroup$ I'll amend my answer. $\endgroup$ – mike stone Jul 5 at 20:57

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