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This Wikipedia page explains that for each of the four main gamma matrices $\gamma^{\mu}$, you can find the covariant matrices $\gamma_{\mu}$ with the equation $\gamma_{\mu} = \eta_{\mu\nu}\gamma^{\mu}$. But that formula doesn't make any sense for $\gamma^5$ because $\eta_{\mu\nu}$ does not have that many indices. So what is $\gamma_5$?

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The 'five' in $\gamma_5$ is not a Lorentz index, so it doesn't make sense to lower or raise it. It can be defined in different ways, one convention is: $$\gamma_5 = \frac{i}{24}\epsilon_{\mu\nu\rho\sigma}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma} = \frac{i}{24}\epsilon^{\mu\nu\rho\sigma}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma}$$, where epsilon is the totally antisymmetric tensor.

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    $\begingroup$ Well, $\gamma_5$ may be a Lorentz index. All the matrices including $\gamma_5$ can be used to produce the generators of the Lorentz group in 5 i.e. 4+1 spacetime dimensions because the give gamma matrices still anticommute with each other, and square to $\pm 1$. Of course that we may define $\gamma_5$ and $\gamma^5$ to be either the same thing or the same thing with the opposite sign, to suit any conventions. $\endgroup$ – Luboš Motl May 31 '13 at 17:44
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$\gamma^5$ would be a 'measurer of parity', which is sensible to changes in orientation due to coordinate transformations. $\gamma^5 := i\gamma^0\gamma^1\gamma^2\gamma^3 $

if you try defining $ \gamma_5 := i\gamma_0\gamma_1\gamma_2\gamma_3 $ you probably end up with $\gamma^5=-\gamma_5$ because $\det \eta=-1$

(I haven't done the calculations, so I might be missing an extra sign or the like)

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