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I'm trying to prove a relation that is useful when studying general properties of Dirac spinors, namely, that $\left[\gamma_5,\sigma^{\mu\nu}\right]=0$ where $\sigma^{\mu\nu}\sim i\gamma^\mu\gamma^\nu$ and $\gamma^5\equiv i\gamma^0\gamma^1\gamma^2\gamma^3$. Alternatively, it is valid to write $$\gamma^5=\frac{i}{4!}\delta^{0123}_{\mu\nu\rho\sigma}\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma=-\frac{i}{4!}\epsilon^{0123}\epsilon_{\mu\nu\rho\sigma}\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma=-\frac{i}{4!}\epsilon_{\mu\nu\rho\sigma}\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma$$

for a Minkowski spacetime with $\,\delta^{\mu_1\mu_2\mu_3\mu_4}_{\nu_1\,\nu_2\,\nu_3\,\nu_4}\equiv -\,\epsilon^{\mu_1\mu_2\mu_3\mu_4}\,\epsilon_{\nu_1\nu_2\nu_3\nu_4}$ and $\epsilon^{0123}=-\epsilon_{0123}\equiv+1$.


Two properties about the delta that I will use are that

  1. If we contract $N\leq4$ of the indices in the delta (in order), that's equal to $\,N!\;\delta^{\mu_1...\mu_{4-N}}_{\nu_1...\,\nu_{4-N}}\,$.
  2. From its definition, $\,\epsilon_{\alpha_1\alpha_2\alpha_3\alpha_4}\,\delta^{\mu_1\mu_2\mu_3\mu_4}_{\nu_1\nu_2\nu_3\nu_4}=\delta^{\mu_1\mu_2\mu_3\mu_4}_{\alpha_1\alpha_2\alpha_3\alpha_4}\,\epsilon_{\nu_1\nu_2\nu_3\nu_4}\,$.

OK so I was trying to prove $\left[\gamma_5,\sigma^{\mu\nu}\right]=i \left[\gamma_5,\gamma^\mu\gamma^\nu\right]=0$ like this:

\begin{align*} \gamma_5\gamma^\mu\gamma^\nu&= -\frac{i}{4!}\,\epsilon_{\alpha_1\alpha_2\alpha_3\alpha_4}\gamma^{\alpha_1}\gamma^{\alpha_2}\gamma^{\alpha_3}\gamma^{\alpha_4}\gamma^{\mu}\gamma^{\nu}, \\[0.2cm] &= -\frac{i}{4!}\,\epsilon_{\alpha_1\alpha_2\alpha_3\alpha_4}\gamma^{\alpha_1}\gamma^{\alpha_2}\,\frac{1}{4!}\left(\delta^{\alpha_3\alpha_4\mu\nu}_{\beta_1\beta_2\beta_3\beta_4}\gamma^{\beta_1}\gamma^{\beta_2}\gamma^{\beta_3}\gamma^{\beta_4}\right), \\[0.2cm] &= -\frac{i}{4!}\,\delta^{\alpha_3\alpha_4\mu\nu}_{\alpha_1\alpha_2\alpha_3\alpha_4}\gamma^{\alpha_1}\gamma^{\alpha_2}\,\frac{1}{4!}\left(\epsilon_{\beta_1\beta_2\beta_3\beta_4}\gamma^{\beta_1}\gamma^{\beta_2}\gamma^{\beta_3}\gamma^{\beta_4}\right), \\[0.2cm] &= -\frac{i}{4!}\,\left(2\,\delta^{\,\mu\;\nu}_{\alpha_1\alpha_2}\right)\gamma^{\alpha_1}\gamma^{\alpha_2}\,\frac{1}{4!}\left(\epsilon_{\beta_1\beta_2\beta_3\beta_4}\gamma^{\beta_1}\gamma^{\beta_2}\gamma^{\beta_3}\gamma^{\beta_4}\right), \\[0.2cm] &= -\frac{i}{4!}\,(4\,\gamma^{\mu}\gamma^{\nu})\,\frac{1}{4!}\left(\epsilon_{\beta_1\beta_2\beta_3\beta_4}\gamma^{\beta_1}\gamma^{\beta_2}\gamma^{\beta_3}\gamma^{\beta_4}\right) = \frac{1}{3!}\,\gamma^{\mu}\gamma^{\nu}\gamma^5, \end{align*}

but that factor of $\frac{1}{3!}$ ruins the commutation relation. What's the mistake then?

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I think a much simpler way to prove this is to use the Leibniz rule: $$ [\gamma_5, \gamma^\mu \gamma^\nu] = [\gamma_5, \gamma^\mu] \gamma^\nu + \gamma^\mu[\gamma_5, \gamma^\nu]. $$ Then we can use the relation between commutator and anti-commutator $$ [A, B] = \{A, B\} - 2BA $$ to write the equation above in terms of anti-commutators. This is useful since we know $$ \{\gamma_5, \gamma^\mu\} = 0. $$ You will see that the result then follows immediately.

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  • $\begingroup$ Sure, the proof is easy if we use the fact that $\gamma^5$ anticommutes with $\gamma^\mu$ but, precisely, I'm seeking to improve my calculation skills with the Levi-Civitas. I do see them a lot in many formulas and it'd be useful to finally get accostumed to make calculations with them nicely. $\endgroup$ – JuanC97 Mar 1 at 2:46

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