Covariant gamma matrices Covariant gamma matrices are defined by
$$\gamma_{\mu}=\eta_{\mu\nu}\gamma^{\nu}=\{\gamma^{0},-\gamma^{1},-\gamma^{2},-\gamma^{3}\}.$$

The gamma matrix $\gamma^{5}$ is defined by
$$\gamma^{5}\equiv i\gamma^{0}\gamma^{1}\gamma^{2}\gamma^{3}.$$

Is the covariant matrix $\gamma_{5}$ then defined by
$$\gamma_{5} = i\gamma_{0}(-\gamma_{1})(-\gamma_{2})(-\gamma_{3})?$$
 A: Indeed geometric interpretation of $\gamma_5$  is related to the volume form
$$
V=\frac 1 {4!} \epsilon_{\mu\nu\alpha\beta} dx^\mu \wedge dx^\nu \wedge dx^\alpha \wedge dx^\beta = \frac 1 {4!} \sqrt{-g} \varepsilon_{\mu\nu\alpha\beta} dx^\mu \wedge dx^\nu \wedge dx^\alpha \wedge dx^\beta = \sqrt{-g} dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3 =\sqrt{-g}  d^4x
$$
Your $\gamma^5$ can be written as
$$
\gamma^5 := \frac i {4!} \epsilon_{\mu\nu\alpha\beta} \;\gamma^\mu  \gamma^\nu \gamma^\alpha \gamma^\beta \;,
$$
which can be shown equivalents to
\begin{eqnarray}
\gamma^5 &=& i\;\sqrt{-\eta}\;\varepsilon_{0123}\; \gamma^0 \gamma^1\gamma^2 \gamma^3\;,\\
&=& i \gamma^0 \gamma^1\gamma^2 \gamma^3\;.
\end{eqnarray}
So, the most natural way to define $\gamma_5$ must be
$$
\gamma_5 := \frac i {4!} \epsilon^{\mu\nu\alpha\beta} \;\gamma_\mu  \gamma_\nu \gamma_\alpha \gamma_\beta \;,
$$
Consequently, we have
\begin{eqnarray}
\gamma_5 &=& i \Big( \frac{-1}{\sqrt{-\eta}}  \Big) \varepsilon^{0123} \;\gamma_0  \gamma_1 \gamma_2 \gamma_3 \;,\\
&=& -i \;\gamma_0  \gamma_1 \gamma_2 \gamma_3 \;,\\
&=& -i\; \gamma^0(-\gamma^1)(-\gamma^2)(-\gamma^3)\;,\\
&=&   i \gamma^0 \gamma^1\gamma^2 \gamma^3\;\\
&=& \gamma^5
\end{eqnarray}
So the position of 5 does not matter. 
A: *

*by the definition of the $\epsilon$ symbol:
$-\frac{i}{4!} \epsilon_{\mu\nu\rho\sigma} = -\frac{i}{4!}(\gamma^0\gamma^1\gamma^2\gamma^3 - \gamma^0\gamma^1\gamma^3\gamma^2 + ... + \gamma^3\gamma^2\gamma^1\gamma^0) = i\gamma^0\gamma^1\gamma^2\gamma^3 = \gamma_5$
