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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?

PPR
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1 Answers1

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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

$\text{Tr}(\gamma_5\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=-4i\epsilon^{\mu\nu\rho\sigma}.$

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.

Frederic Brünner
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