I am having some trouble locating a source for the trace identities of the gamma matrices in d dimensions. I have attempted to derive them here, where d need not be integer, please could you tell me if you think this is correct? Any additional comments on the workings would be welcomed. The defining feature that will be preserved is; $$\{\gamma_\mu, \gamma_\nu\} = 2\eta_{\mu\nu}I_d$$
The results I am interested in are;
- $tr[\gamma_\mu \gamma_\nu ] $
- $tr[\gamma_\mu\gamma_\alpha\gamma_\nu]$
- $tr[\gamma_\mu\gamma_\alpha\gamma_\nu\gamma_\beta] $
For the first one; using trace is cyclic $$ tr[\gamma_\mu\gamma_\nu] = \frac{1}{2}tr[\gamma_\mu\gamma_\nu] + \frac{1}{2}tr[\gamma_\nu\gamma_\mu] = \frac{1}{2}tr[\gamma_\mu\gamma_\nu + \gamma_\nu\gamma_\mu] = \frac{1}{2} tr[2\eta_{\mu\nu}I_d] = d\eta_{\mu\nu}$$
For the second one; start by determining the contraction of two $\gamma$ $$\{\gamma_\mu, \gamma^\mu\} = 2\gamma_\mu\gamma^\mu = 2\eta^\mu_{\mu}I_d = 2dI_d$$ so if we divide out the dimension we can insert the contraction of two gammas then draw one right through; $$tr[\gamma_\mu\gamma_\alpha\gamma_\nu] = \frac{1}{d}tr[\gamma_\delta\gamma^\delta\gamma_\mu\gamma_\alpha\gamma_\nu] = \frac{1}{d}\left(2tr[\gamma_\delta\eta^\delta_\mu\gamma_\alpha\gamma_\nu]-tr[\gamma_\delta\gamma_\mu\gamma^\delta\gamma_\alpha\gamma_\nu] \right)$$ $$= \frac{1}{d}\left(2tr[\gamma_\mu\gamma_\alpha\gamma_\nu]-2tr[\gamma_\delta\gamma_\mu\eta^\delta_\alpha\gamma_\nu]+tr[\gamma_\delta\gamma_\mu\gamma_\alpha\gamma^\delta\gamma_\nu] \right)$$ $$ = \frac{1}{d}\left(2tr[\gamma_\mu\gamma_\alpha\gamma_\nu]-2tr[\gamma_\alpha\gamma_\mu\gamma_\nu]+2tr[\gamma_\delta\gamma_\mu\gamma_\alpha\eta^\delta_\nu] -tr[\gamma_\delta\gamma_\mu\gamma_\alpha\gamma_\nu\gamma^\delta]\right)$$ then cyclic $$ = \frac{1}{d}\left(2tr[\gamma_\mu\gamma_\alpha\gamma_\nu]-2tr[\gamma_\alpha\gamma_\mu\gamma_\nu]+2tr[\gamma_\nu\gamma_\mu\gamma_\alpha] -tr[\gamma^\delta\gamma_\delta\gamma_\mu\gamma_\alpha\gamma_\nu]\right)$$ $$ = \frac{1}{d}\left(2tr[\gamma_\mu\gamma_\alpha\gamma_\nu]-2tr[\gamma_\alpha\gamma_\mu\gamma_\nu]+2tr[\gamma_\mu\gamma_\alpha\gamma_\nu] \right)-tr[\gamma_\mu\gamma_\alpha\gamma_\nu]$$ $$2tr[\gamma_\mu\gamma_\alpha\gamma_\nu] = \frac{1}{d}\left(4tr[\gamma_\mu\gamma_\alpha\gamma_\nu]-2tr[\gamma_\alpha\gamma_\mu\gamma_\nu] \right)$$ $$dtr[\gamma_\mu\gamma_\alpha\gamma_\nu] = \left(2tr[\gamma_\mu\gamma_\alpha\gamma_\nu]-tr[\gamma_\mu\gamma_\nu\gamma_\alpha] \right)$$ $$(d-3)tr[\gamma_\mu\gamma_\alpha\gamma_\nu] + tr[\gamma_\mu\gamma_\alpha\gamma_\nu+\gamma_\mu\gamma_\nu\gamma_\alpha ]=0 $$ $$(d-3)tr[\gamma_\mu\gamma_\alpha\gamma_\nu] + tr[\gamma_\mu2\eta_{\alpha\nu} ]=0 $$
now we need the trace of one $\gamma$;
$$tr[\gamma_\mu] = \frac{1}{d}tr[\gamma_\nu\gamma^\nu\gamma_\mu] = \frac{1}{d}\left( tr[\gamma_\nu2\eta^\nu_\mu]-tr[\gamma_\nu\gamma_\mu\gamma^\nu]\right)$$ Chose $\mu \neq \nu$ $$2tr[\gamma_\mu] = 0$$
so now we have; $$(d-3)tr[\gamma_\mu\gamma_\alpha\gamma_\nu] =0 $$
Finally the third one; this time simply drawing one of the existing matrices through $$tr[\gamma_\mu\gamma_\alpha\gamma_\nu\gamma_\beta] = 2\eta_{\mu\alpha}tr[\gamma_\nu\gamma_\beta] - tr[\gamma_\alpha\gamma_\mu\gamma_\nu\gamma_\beta]$$ $$tr[\gamma_\mu\gamma_\alpha\gamma_\nu\gamma_\beta] = 2\eta_{\mu\alpha}tr[\gamma_\nu\gamma_\beta] - 2\eta_{\mu\nu}tr[\gamma_\alpha\gamma_\beta] + tr[\gamma_\alpha\gamma_\nu\gamma_\mu\gamma_\beta]$$ $$tr[\gamma_\mu\gamma_\alpha\gamma_\nu\gamma_\beta] = 2\eta_{\mu\alpha}tr[\gamma_\nu\gamma_\beta] - 2\eta_{\mu\nu}tr[\gamma_\alpha\gamma_\beta] + 2\eta_{\mu\beta}tr[\gamma_\alpha\gamma_\nu]-tr[\gamma_\alpha\gamma_\nu\gamma_\beta\gamma_\mu]$$ Cycling the trace and using rules we found before; $$2tr[\gamma_\mu\gamma_\alpha\gamma_\nu\gamma_\beta] = 2d\eta_{\mu\alpha}\eta_{\nu\beta} - 2d\eta_{\mu\nu}\eta_{\alpha\beta} + 2d\eta_{\mu\beta}\eta_{\alpha\nu}$$ $$tr[\gamma_\mu\gamma_\alpha\gamma_\nu\gamma_\beta] = d\left(\eta_{\mu\alpha}\eta_{\nu\beta} - \eta_{\mu\nu}\eta_{\alpha\beta} + \eta_{\mu\beta}\eta_{\alpha\nu}\right)$$
summarising;
- $tr[\gamma_\mu] = 0$
- $tr[\gamma_\mu \gamma_\nu ] = d\eta_{\mu\nu}$
- $tr[\gamma_\mu\gamma_\alpha\gamma_\nu] = 0$
- $tr[\gamma_\mu\gamma_\alpha\gamma_\nu\gamma_\beta] =d\left(\eta_{\mu\alpha}\eta_{\nu\beta} - \eta_{\mu\nu}\eta_{\alpha\beta} + \eta_{\mu\beta}\eta_{\alpha\nu}\right) $
