# Proof that $\gamma_{5}^2=1$

I need to prove that $$\gamma_{5}^2=1$$, and in order to do this I wrote:

$$\begin{equation} (\gamma_{5})^2=\gamma^{5}\gamma_{5}=\left(-\frac{i}{4!}\epsilon^{\mu\nu\rho\sigma}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma} \right)\left(-\frac{i}{4!}\epsilon_{\mu\nu\rho\sigma}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma} \right)=\frac{-1}{(4!)^2}\epsilon^{\mu\nu\rho\sigma}\epsilon_{\mu\nu\rho\sigma}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma} \gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}. \end{equation}$$

Using that $$\epsilon^{\mu\nu\rho\sigma}\epsilon_{\mu\nu\rho\sigma}=4!$$ and having already proved that $$\gamma^{\nu}\gamma_{\nu}=4I_{4}$$, $$\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma^{\mu}=4\eta_{\nu\rho}I_{4}$$ and $$\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma}\gamma^{\mu}=-2\gamma_{\sigma}\gamma_{\rho}\gamma_{\nu}$$, I arrived at:

$$\begin{equation} (\gamma_{5})^2=\frac{-1}{4!}\left(-2\gamma_{\sigma}\gamma_{\rho}\gamma_{\nu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\right)=\frac{2\cdot4^3}{4!}=\frac{16}{3}. \end{equation}$$

I have also tried the following:

$$\begin{equation} (\gamma_{5})^2=\gamma^{5}\gamma_{5}=\left(-\frac{i}{4!}\epsilon^{\mu\nu\rho\sigma}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma} \right)\left(-\frac{i}{4!}\epsilon_{\alpha\beta\lambda\theta}\gamma^{\alpha}\gamma^{\beta}\gamma^{\lambda}\gamma^{\theta} \right)=\frac{-1}{(4!)^2}\epsilon^{\mu\nu\rho\sigma}\epsilon_{\alpha\beta\lambda\theta}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma} \gamma^{\alpha}\gamma^{\beta}\gamma^{\lambda}\gamma^{\theta}=\frac{-1}{(4!)^2}\delta^{\mu\nu\rho\sigma}_{\alpha\beta\lambda\theta}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma} \gamma^{\alpha}\gamma^{\beta}\gamma^{\lambda}\gamma^{\theta}. \end{equation}$$

But I don't know how to proceed from this. What am I missing?

• Hint: In the Einstein notation one is not supposed to repeat an index more than twice. May 1 at 11:29
• Your indices are not consistent. You need to pick different names for the second product. $\alpha$, $\beta$, etc. May 1 at 11:30
• I tried to do so, but with so many indices I didn't know how to procede. May 1 at 11:31

You can do this in a much simpler way: $$(\gamma^5)^2 = (i \gamma^0 \gamma^1 \gamma^2 \gamma^3)^2 = - \gamma^0 \gamma^1 \gamma^2 \gamma^3\gamma^0 \gamma^1 \gamma^2 \gamma^3$$ Now anti-commute the $$\gamma$$-matrices next to one another and use $$\{\gamma^\mu, \gamma^\nu \}= 2 \eta^{\mu\nu}$$.

Perhaps, the simplest/fastest way:

In Dirac representation, $$\gamma ^5$$ as a product of the four gamma matrices is written as

$${\gamma ^5} = \left( {\begin{array}{*{20}{c}} 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0 \end{array}} \right).$$

First check this familiar result: $$\gamma^5=?= {\left( {{\gamma ^5}} \right)^\dagger }$$. Now, try this $$({\gamma ^5})^2=?$$

$${\left( {{\gamma ^5}} \right)^2} = -\frac{1}{{{{(4!\,)}^2}}}{\gamma ^a}{\gamma ^b}{\gamma ^c}{\gamma ^d}{\gamma _a}{\gamma _b}{\gamma _c}{\gamma _d}.$$
From $${\{\gamma ^a},{\gamma ^b}\} = 2{\eta ^{ab}}{I}$$, you have to prove $${\{\gamma ^a},{\gamma _c}\} = 2\delta _c^a {I}$$. Using this and also $${\gamma ^\mu}{\gamma_\mu}=4 I$$, you can prove your final result.