# Proving identity $tr(\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\gamma^{5})=-4i\epsilon^{\mu\nu\rho\sigma}$

Im trying to proof the following identity:

$$tr(\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\gamma^{5})=-4i\epsilon^{\mu\nu\rho\sigma}$$

when $$\gamma^{\mu},\gamma^{\nu},\gamma^{\rho},\gamma^{\sigma}$$ are Dirac gamma matrices

and $$\gamma^5$$ is defined $$\gamma^5=i\gamma^0\gamma^1\gamma^2\gamma^3$$ and have the property: $$(\gamma^5)^2=I_{4x4}$$

$$\epsilon^{\mu\nu\rho\sigma}$$ is the 4 Levi-Civita tensor.

From Wikipedia: https://en.wikipedia.org/wiki/Gamma_matrices I know the following Identity:$$\gamma^5=-\frac{i}{4!}\epsilon^{\mu\nu\rho\sigma}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}$$

I tried to isolate: $$\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}$$:

$$\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}=-\frac{4!}{i}(\epsilon^{\mu\nu\rho\sigma})^{-1}\gamma^5$$

putting this relation in our original trace:

$$tr(\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\gamma^{5})=tr(-\frac{4!}{i}(\epsilon^{\mu\nu\rho\sigma})^{-1}(\gamma^5)^2)=4!i(\epsilon^{\mu\nu\rho\sigma})^{-1}tr(I_{4x4})=4\cdot4!i(\epsilon^{\mu\nu\rho\sigma})^{-1}$$

and now I stucked... how to show that $$(\epsilon^{\mu\nu\rho\sigma})^{-1}=-\frac{1}{4!}\epsilon^{\mu\nu\rho\sigma}$$ so the expression will fit?

Note that $$\operatorname{tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma^5)$$ is a totally antisymmetric function of $$4$$ spacetime indices. For example,$$\operatorname{tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma^5)+\operatorname{tr}(\gamma^\nu\gamma^\mu\gamma^\rho\gamma^\sigma\gamma^5)=2\eta^{\mu\nu}\underbrace{\operatorname{tr}(\gamma^\rho\gamma^\sigma\gamma^5)}_{0}.$$Hence $$\operatorname{tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma^5)=k\epsilon^{\mu\nu\rho\sigma}$$ with$$k:=\operatorname{tr}(\gamma^0\gamma^1\gamma^2\gamma^3\gamma^5)=\operatorname{tr}(-i(\gamma^5)^2)=\operatorname{tr}(-iI_4)=-4i.$$

You cannot divide by $$\epsilon^{\mu\nu\rho\sigma}$$, it's not a 2x2 matrix.

What you can do:

is use the anticommutation relations of gamma matrices to show that $$T^{\mu\nu\rho\sigma} = {\rm tr}\{\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma^5\}$$ is fully antisymmetric. For example, you have $${\rm tr}\{\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma^5\} = {\rm tr}\{(-\gamma^\nu\gamma^\mu + 2g^{\mu\nu})\gamma^\rho\gamma^\sigma\gamma^5\}$$ $$T^{\mu\nu\rho\sigma} = -T^{\nu\mu\rho\sigma} + 2g^{\mu\nu} {\rm tr}\{\gamma^\rho\gamma^\sigma\gamma^5\} = -T^{\nu\mu\rho\sigma}$$

(You need the fact that $${\rm tr}\{\gamma^\rho\gamma^\sigma\gamma^5\}=0$$ for that; see if you can proove it.)

Once you have the full antisymmetry, you use the fact that in 4 dimmensions, there's only one linarily independent fully antisymmetric 4-tensor, and it's $$\epsilon^{\mu\nu\rho\sigma}$$. That means that $$T^{\mu\nu\rho\sigma}=\lambda\epsilon^{\mu\nu\rho\sigma}$$ for some $$\lambda\in\mathbb C$$. Finally, Substituting $$(\mu\nu\rho\sigma)=(0123)$$ you can calculate the value of $$\lambda$$ to be $$-4i$$.

Another method utilizes the formula for $$\gamma^5$$ you've found. You have $${\rm tr}\{\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\gamma^5\} = {\rm tr}\{\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma\frac{-i}{4!}\epsilon^{\mu'\nu'\rho'\sigma'} \gamma^{\mu'}\gamma^{\nu'}\gamma^{\rho'}\gamma^{\sigma'}\} = \frac{-i}{4!}\epsilon^{\mu'\nu'\rho'\sigma'} {\rm tr}\{\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma \gamma^{\mu'}\gamma^{\nu'}\gamma^{\rho'}\gamma^{\sigma'}\}$$ Then there are long calculations using the anticommutation relations of $$\gamma$$ matrices to calculate the trace of 8 gammas and get the result.

The manipulation

$$\gamma^{5}=-\frac{i}{4!}\epsilon_{\mu\nu\sigma\tau}\gamma^{\mu}\gamma^{\nu}\gamma^{\sigma}\gamma^{\tau}\quad \Longrightarrow\quad \gamma^{\mu}\gamma^{\nu}\gamma^{\sigma}\gamma^{\tau}=4!i(\epsilon_{\mu\nu\sigma\tau})^{-1}\gamma^{5}$$ is not legal: in $$\gamma^{5}$$ the indices of $$\epsilon_{\mu\nu\sigma\tau}$$ are contracted, so you cannot simply multiply the equation by the inverse of $$\epsilon_{\mu\nu\sigma\tau}$$!

I suggest to proceed as follows:

1) Show that $$\text{Tr}\{\gamma^{\mu}\gamma^{\nu}\gamma^{\sigma}\gamma^{\tau}\gamma^{5}\}$$ is completely antisymmetric with respect to the exchange of its indices (in order to do so you first need to show that $$\text{Tr}\{\gamma^{\mu}\gamma^{\nu}\gamma^{5}\}=0$$).

2) Since $$\text{Tr}\{\gamma^{\mu}\gamma^{\nu}\gamma^{\sigma}\gamma^{\tau}\gamma^{5}\}$$ is completely antisymmetric with respect to the exchange of any of its indices, it follows that

$$\text{Tr}\{\gamma^{\mu}\gamma^{\nu}\gamma^{\sigma}\gamma^{\tau}\gamma^{5}\}=\epsilon^{\mu\nu\sigma\tau}A$$

for some $$A$$. Then

$$\epsilon_{\mu\nu\sigma\tau}\text{Tr}\{\gamma^{\mu}\gamma^{\nu}\gamma^{\sigma}\gamma^{\tau}\gamma^{5}\}=\epsilon_{\mu\nu\sigma\tau}\epsilon^{\mu\nu\sigma\tau}A=-4! A$$

so that

$$A=-\frac{1}{4!}\epsilon_{\mu\nu\sigma\tau}\text{Tr}\{\gamma^{\mu}\gamma^{\nu}\gamma^{\sigma}\gamma^{\tau}\gamma^{5}\}=-i\text{Tr}\{\gamma^{5}\gamma^{5}\}=-i\text{Tr}\{1\}=-4i$$