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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?

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3 Answers 3

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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$$k:=\operatorname{tr}(\gamma^0\gamma^1\gamma^2\gamma^3\gamma^5)=\operatorname{tr}(-i(\gamma^5)^2)=\operatorname{tr}(-iI_4)=-4i.$$

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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.

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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

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

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