How to calculate the trace of six gamma matrices multiplied to $\gamma_5$? I read from Weinberg that, the gamma matrices have the following property:
\begin{equation}
\text{Tr}\{\gamma_5 \gamma_\mu \gamma_\nu \gamma_\rho \gamma_\sigma\}=4i\epsilon_{\mu \nu \rho \sigma}
\end{equation}
This stems from the fact that
$\gamma_5 \gamma_\mu \gamma_\nu \gamma_\rho \gamma_\sigma$ is totally antisymmetric.
However, I cannot see how to extend such an argument to the case where six gamma matrices are multiplied to $\gamma_5$:
\begin{equation}
\text{Tr}\{\gamma_5 \gamma_\mu \gamma_\nu \gamma_\rho \gamma_\sigma \gamma_\psi \gamma_\xi \}=??
\end{equation}
Could anyone please help me find the right formula?
 A: @CosmasZachos cited identity $5$ here,$$\gamma^a\gamma^b\gamma^c=\eta^{ab}\gamma^c+\eta^{bc}\gamma^a-\eta^{ac}\gamma^b-i\epsilon^{gabc}\gamma_g\gamma^5.$$There are therefore sixteen terms in $\operatorname{Tr}(\gamma^5\gamma^a\gamma^b\gamma^c\gamma^d\gamma^e\gamma^f)$ in four types (two of which are very similar), which we deal with in turn. I'll take uses of $\operatorname{Tr}(XY)=\operatorname{Tr}(YX),\,(\gamma^5)^2=I_4$ to be too obvious to state.
Firstly, $\operatorname{Tr}(\gamma^5\gamma^A\gamma^B)=0$ is just identity coincidentally-also-$5$ in the next section. For the rest, use the fact $\gamma^5$ anticommutes with other gamma matrices, so$$\operatorname{Tr}(\gamma^5\gamma^A\gamma_B\gamma^5)=\operatorname{Tr}(\gamma^A\gamma_B)=\delta^A_B.$$Similarly,$$\operatorname{Tr}(\gamma^5\gamma_B\gamma^5\gamma^A)=-\operatorname{Tr}(\gamma_B\gamma^A)=-\delta^A_B.$$Finally,$$\operatorname{Tr}(\gamma^5\gamma_A\gamma^5\gamma_B\gamma^5)=-\operatorname{Tr}(\gamma^5\gamma_A\gamma_B)=0.$$We now know terms with $0$ or $2$ Levi-Civita symbols vanish, while those with $1$ survive. Hence$$\operatorname{Tr}(\gamma^5\gamma^a\gamma^b\gamma^c\gamma^d\gamma^e\gamma^f)=i(-\eta^{ab}\epsilon^{cdef}-\eta^{bc}\epsilon^{adef}+\eta^{ac}\epsilon^{bdef}-\eta^{de}\epsilon^{abcf}-\eta^{ef}\epsilon^{abcd}+\eta^{df}\epsilon^{abce}).$$
A: Very interesting discussion, and I would like to put there my version of this problem.
First, all identities only have any sense if basic conventional points are postulated. Namely, below I will use the following definition of four-dimensional Levi-Civita symbol:
$$
\varepsilon^{\alpha\beta\gamma\delta}=-\varepsilon_{\alpha\beta\gamma\delta}=
\begin{cases}
    +1,& \text{if } \alpha\beta\gamma\delta=\text{even permutation of 0123}\\
    -1,& \text{if } \alpha\beta\gamma\delta=\text{odd permutation of 0123}.
\end{cases}
$$
Next, the definition of fifth gamma-matrix is the following:
$$
\gamma^5=\gamma_5=i\gamma^0\gamma^1\gamma^2\gamma^3=-i\gamma_0\gamma_1\gamma_2\gamma_3.
$$
With such definitions
$$
\mathrm{Tr}\{{\gamma^5\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}\gamma^{\delta}}\}=\mathrm{Tr}\{{\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}\gamma^{\delta}\gamma^5}\}=-4i\varepsilon^{\alpha\beta\gamma\delta},\tag{1}\label{1}
$$
what is checked for the component $\alpha=0,\ \beta=1,\ \gamma=2,\ \delta=3$ and is carried over to the other components which are permutations of 0123.
Further, there is the identity
$$
\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}=\eta^{\alpha\beta}\gamma^{\gamma}-\eta^{\alpha\gamma}\gamma^{\beta}+\eta^{\beta\gamma}\gamma^{\alpha}-i\gamma^5\varepsilon^{\alpha\beta\gamma\delta}\gamma_{\delta},\tag{2}\label{2}
$$
where $\eta^{\mu\nu}$ is the Minkowski metric with the signature $\{+1, -1, -1, -1\}$ and, hence, $\gamma_0=\gamma^0$, while $\gamma_\mu=-\gamma^\mu$ with $\mu=\{1, 2, 3\}$. Such defined the identity is in agreement with the version of @J.G. because $\gamma^5$ anticommutes with any of $\gamma_\mu$ and $\varepsilon^{\delta\alpha\beta\gamma}=-\varepsilon^{\alpha\beta\gamma\delta}$.
Finally, we are ready to calculate the trace $\mathrm{Tr}\{\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}\gamma^{\rho}\gamma^{\sigma}\gamma^{\tau}\gamma^5\}$.
First, replacing $\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}$ with  the use of the identity \ref{2} one obtains
$$
\mathrm{Tr}\{\gamma^\alpha\gamma^\beta\gamma^\gamma\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}=
\eta^{\alpha\beta}\mathrm{Tr}\{\gamma^\gamma\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}-
\eta^{\alpha\gamma}\mathrm{Tr}\{\gamma^\beta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}+
\eta^{\beta\gamma}\mathrm{Tr}\{\gamma^\alpha\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}-
i\varepsilon^{\alpha\beta\gamma\delta}\mathrm{Tr}\{\gamma^5\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}.\tag{3}\label{3}
$$
First three terms are rewritten with the use of the identity \ref{1}.
As for the last term, let's consider the multiplier $\mathrm{Tr}\{\gamma^5\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}$ separately:
$$
\mathrm{Tr}\{\gamma^5\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}=\mathrm{Tr}\{\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\}
$$
because of anticommutation of $\gamma^5$ with both $\gamma_\mu$ and $\gamma^\mu$, further use the identity \ref{2} one more time:
$$
\mathrm{Tr}\{\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\}=
\eta^{\rho\sigma}\mathrm{Tr}\{\gamma_\delta\gamma^\tau\}-
\eta^{\rho\tau}\mathrm{Tr}\{\gamma_\delta\gamma^\sigma\}+
\eta^{\sigma\tau}\mathrm{Tr}\{\gamma_\delta\gamma^\rho\}-
i\varepsilon^{\rho\sigma\tau\eta}\mathrm{Tr}\{\gamma_\delta\gamma^5\gamma_\eta\}.\tag{4}\label{4}
$$
The last term vanishes (trace of product of fifth gamma-matrix with only two gamma-matrices) and every of the other terms contains the multiplier of the form
$$
\mathrm{Tr}\{\gamma_\lambda\gamma^\nu\}=\eta_{\lambda\mu}\mathrm{Tr}\{\gamma^\mu\gamma^\nu\}=4\eta_{\lambda\mu}\eta^{\mu\nu}.
$$
Using this one can rewrite \ref{4} in the form
$$
\mathrm{Tr}\{\gamma_\delta\gamma^\rho\gamma^\sigma\gamma^\tau\}=
4\eta^{\rho\sigma}\eta_{\delta\lambda}\eta^{\lambda\tau}-
4\eta^{\rho\tau}\eta_{\delta\lambda}\eta^{\lambda\sigma}+
4\eta^{\sigma\tau}\eta_{\delta\lambda}\eta^{\lambda\rho}
$$
and inserting it to \ref{3} one obtains
$$
\mathrm{Tr}\{\gamma^\alpha\gamma^\beta\gamma^\gamma\gamma^\rho\gamma^\sigma\gamma^\tau\gamma^5\}=-4i(
 \eta^{\alpha\beta}\varepsilon^{\gamma\rho\sigma\tau}
-\eta^{\alpha\gamma}\varepsilon^{\beta\rho\sigma\tau}
+\eta^{\beta\gamma}\varepsilon^{\alpha\rho\sigma\tau}
+\eta^{\rho\sigma}\varepsilon^{\alpha\beta\gamma\tau}
-\eta^{\rho\tau}\varepsilon^{\alpha\beta\gamma\sigma}
+\eta^{\sigma\tau}\varepsilon^{\alpha\beta\gamma\rho}).
$$
Note, that we have used the following transformations:
$$
\varepsilon^{\alpha\beta\gamma\delta}\eta_{\delta\lambda}\eta^{\lambda\tau}=
\varepsilon^{\alpha\beta\gamma}_{\phantom{\alpha\beta\gamma}\lambda}\eta^{\lambda\tau}=
\varepsilon^{\alpha\beta\gamma\tau}
$$
and so on.
