# Gamma Matrices in Dimensional Regularization

Prove that $tr\left(\gamma_\mu\gamma_\nu\gamma_\rho\gamma_\sigma\gamma_5\right)=0$ when the spacetime dimension is not 4.

What I have tried:

We know that $\gamma_\alpha\gamma^\alpha=d\mathbb{1}$, so we can write:

$tr\left(\gamma_\mu\gamma_\nu\gamma_\rho\gamma_\sigma\gamma_5\right)=\frac{1}{d}tr\left(\gamma_\alpha\gamma^\alpha\gamma_\mu\gamma_\nu\gamma_\rho\gamma_\sigma\gamma_5\right)$

Then I thought I could commute $\gamma^\alpha$ past two gammas because if $\alpha\notin\left\{\mu,\,\nu,\,\rho,\,\sigma\right\}$, then $\left\{\gamma_\alpha,\,\gamma_\mu\right\}=0$ and somehow show that I get minus of what I started with, using the cyclicality of the trace and that $\left\{\gamma_5,\,\gamma_\mu\right\}=0$.

However, what I am not sure about is why can we always find such $\alpha$ so that $\alpha\notin\left\{\mu,\,\nu,\,\rho,\,\sigma\right\}$. I understand this is generally possible when $d\in\mathbb{R}\wedge d>4$, however, when $d\in\mathbb{C}$, this claim doesn't make any sense for me.

Can anyone provide a rigorous proof of this claim which avoids the hurdle I mentioned above?

• See also here, for an extended discussion of the issue in the context of analytic and dimensional regularisation. – Dilaton Sep 15 '14 at 8:48

As shown in chapter 47 of the book by Srednicki, using the definition of $\gamma_5$ and the relations $(\gamma^i)^2=-1$ and $(\gamma^0)^2=1$, we can show that
$\text{Tr}(\gamma_5\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=-4i\epsilon^{\mu\nu\rho\sigma}.$
• So, should we neglect this kind of term when we do dimensional regularization with $d=4-\varepsilon$? – Melquíades May 4 '14 at 21:34
• @FredericBrünner, Thanks, but I am not so happy with this solution. To show that $Tr(\gamma_5\gamma^\mu\gamma^\nu \gamma^\rho\gamma^\sigma)=-4i\varepsilon^{\mu\nu\rho\sigma}$ I use the fact that if two indices in ($\mu,\nu,\rho,\sigma$) are the same, then we pick some new index, $\lambda\notin\{\mu,\nu,\rho,\sigma\}$, and then anti-commute it to the end to get the trace is minus itself and thus zero. However, this sort of trick is exactly what I was trying to avoid when I first posed my question because when $d\neq4$ I don't feel comfortable to pick such an index. – PPR Jun 19 '14 at 17:43
• @PPR : There is a interesting discussion in this paper, chapter $7.4$ p.$213$, the precise definition of $\gamma^5$ p.$214$, and the formula $(7.22)$ p. $215$ might interest you. – Trimok Jul 9 '14 at 8:41