# (Anti)commutation relations for higher-dimensional anti-symmetrized Gamma matrices

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 $$\{\Gamma^a,\Gamma^{bcd...}]= \,?$$

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.
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.
• 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? Jul 5 '20 at 19:27