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Is it possible to have a set of 16-dimensional matrices $\gamma_{\mu}^{a}$ such that $$\{\gamma_{\mu}^{a},\gamma_{\nu}^{b}\} = 2\delta^{ab}\eta_{\mu\nu}$$ where $\eta_{\mu\nu}$ is the Minkowski metric and $a$ and $b$ can be 1 or 2? I've been trying to construct them manually but it's tougher than I thought and I can't find any mention of this in the literature.

Edit: And also, such that $\gamma^a_{\mu}\gamma^b_{\nu}$ is not 0.

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  • $\begingroup$ ? It is not evident by inspection? a 16x16 matrix with the $\gamma_\mu^1$s in the upper left 4x4 block and zero in the lower right block, and $\gamma_\mu^2$s in the lower right and zero in the upper left does not qualify? $\gamma_\mu\oplus 0$ and $0\oplus \gamma_\mu$. Isn't this your algebra? $\endgroup$ – Cosmas Zachos Mar 9 at 16:44
  • $\begingroup$ In that case I will also have $\gamma^{a}_{\mu}\gamma^{b}_{\nu}=0$, which I do not wish. I should've added that, thanks. $\endgroup$ – user38680 Mar 9 at 16:50
  • $\begingroup$ isnt this just $C\ell(2)\otimes C\ell(1,d-1)$? if so, you know the irreps, so take as many direct sums as necessary. $\endgroup$ – AccidentalFourierTransform Mar 9 at 23:41
  • $\begingroup$ Thanks for your comment @AccidentalFourierTransform. I understand what you said and it makes sense, but I don't have enough practice in group theory to actually do what you suggested. Could you quickly sketch how one would do that? Thanks! $\endgroup$ – user38680 Mar 10 at 11:58
  • $\begingroup$ @ AccidentalFourierTransform . Alas, the $\sigma_1$ and $\sigma_2$ in different parts of the anticommutator flip the sign of the conventional Cl. If only one considered a commutator, instead... $\endgroup$ – Cosmas Zachos Mar 10 at 15:53

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