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:


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?

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    $\begingroup$ See also here, for an extended discussion of the issue in the context of analytic and dimensional regularisation. $\endgroup$ – 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


We can now use a fundamental property of the Levi-Civita symbol, namely the fact that the number of its indices has to agree with the number of spacetime dimensions. If this is not the case, it vanishes. Hence, this proves the initial assertion.

  • $\begingroup$ So, should we neglect this kind of term when we do dimensional regularization with $d=4-\varepsilon$? $\endgroup$ – Melquíades May 4 '14 at 21:34
  • $\begingroup$ Even though it is formally zero, it might make sense to keep it in some cases. I am, however, not experienced enough in that area to make a general statement. $\endgroup$ – Frederic Brünner May 4 '14 at 21:36
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    $\begingroup$ @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. $\endgroup$ – PPR Jun 19 '14 at 17:43
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    $\begingroup$ @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. $\endgroup$ – Trimok Jul 9 '14 at 8:41

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