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how is $\gamma^{\mu}$ defined in the anti commutation relation $\{\gamma_{5},\gamma^{\mu}\}$? does it make a difference if you write the index ${^\mu}$ lower? what does usually change if the index is lowered or on the contrary written above?

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$\gamma^{\mu}$ are defined from their anticommutation relation:

$$\{\gamma^{\mu},\gamma^{\nu}\}=2\eta^{\mu\nu} I_4 $$

Where I use the $(+---)$ convention for $\eta$ and $I_4$ is the 4x4 identity matrix.

There are many possible choices to express such matrices, one of them is: (Dirac basis)

$$\gamma^0= \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right] $$

$$ \gamma^1 = \left[ \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{array} \right] $$

$$ \gamma^2= \left[ \begin{array}{cccc} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & i & 0 & 0 \\ -i & 0 & 0 & 0 \end{array} \right] $$

$$ \gamma^3 = \left[ \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right] $$

then $$\gamma_{\mu}= \eta_{\mu\nu}\gamma^{\nu}$$ where $\eta_{\mu \nu}$ is the metric tensor, which means:

$\gamma_0=\gamma^0$; $\gamma_i=-\gamma^i$ for $i=1,2,3$.

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    $\begingroup$ The dirac basis is defined by what you've written. The gamma matrices are defined with their abstract properties and there are many choices of $\gamma^0$, of $\gamma^1$, etc. Not worth a downvote but I think this is important. $\endgroup$ – user12029 Apr 6 '18 at 23:19
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    $\begingroup$ Thanks for your contribution: you are completely right, I've edited my answer. $\endgroup$ – Matteo Apr 6 '18 at 23:29

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