# Proving identity $\mathrm{Tr}[\gamma^{\mu}\gamma^{\nu}] = 4 \eta^{\mu\nu}$

In the lecture notes accompanying a course I'm following, it is stated that $$\DeclareMathOperator{\Tr}{Tr} \Tr\left[\gamma^{\mu}\gamma^{\nu}\right] = 4 \eta^{\mu\nu}$$

Yet when I try to prove this, I find something different as follows: $$\begin{eqnarray} \Tr\left[\gamma^{\mu}\gamma^{\nu}\right] & = & \gamma^\mu\gamma_\mu \\ & = &\eta_{\nu\mu}\gamma^\nu\gamma^\mu \\ & = &\frac12\left(\eta_{\nu\mu}+\eta_{\mu\nu}\right)\gamma^\nu\gamma^\mu \\ &=&\frac12\left(\eta_{\nu\mu}\gamma^\nu\gamma^\mu+\eta_{\mu\nu}\gamma^\nu\gamma^\mu\right) \\ &=&\frac12\eta_{\mu\nu}\left(\gamma^\nu\gamma^\mu+\gamma^\mu\gamma^\nu\right) \\ &=&\eta_{\mu\nu}\eta^{\mu\nu}I_4 = 4I_4 \end{eqnarray}$$ Where I used the anticommutator relation $$\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}I_4$$. With $$I_4$$ the identity matrix.

I assume that I made a mistake somewhere, or are these statements equivalent?

• The mistake is in the first step: $\Tr\left[\gamma^{\mu}\gamma^{\nu}\right] = \gamma^\mu\gamma_\mu$ – Thomas Fritsch Jan 26 '19 at 11:30
• Oh dear, I see.. I'll give it another go! Thank you – Simon Jan 26 '19 at 11:37

The key ingredients are the cyclicity of the trace and the anti-commutation relations of the $$\gamma$$ matrices. It is $$\begin{eqnarray} \Tr\left[\gamma^{\mu}\gamma^{\nu}\right] & = & \Tr\left[2\eta^{\mu\nu}I_4-\gamma^{\nu}\gamma^{\mu}\right] \\ & = &8\eta^{\mu\nu}-\Tr\left[\gamma^{\mu}\gamma^{\nu}\right], \end{eqnarray}$$ where in the first line we used the anti-commutation relation $$\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}I_4$$ and in the second line the cyclicity of the trace $$\Tr\left[\gamma^{\mu}\gamma^{\nu}\right]=\Tr\left[\gamma^{\nu}\gamma^{\mu}\right]$$. The sought for expression then follows by rearranging the terms.