Normalisation of the $\gamma$-matrices I'm having a little difficulty with understanding the normalisation process of the $\gamma$-matrices.
In Thomson Modern Particle Physics 2013, the normalisation of the $\gamma$-matrices are quoted as:
$$
(\gamma^{\mu})^{\dagger}=\gamma^{0}\gamma^{\mu}\gamma^{0}
$$
Where $\mu=0,1,2,3$ or sometimes just $\mu=0, k$ where, obviously $k=1,2,3$. I have attempted to start this given example, but I'm not sure on the next steps. Thus far, I have:
$$
(\gamma^{0})^{\dagger}=\gamma^{0} 
$$
$$
(\gamma^{k})^{\dagger}=-\gamma^{k} 
$$
I also know that 
$$
(\gamma^{0})^{2}=I\,\,\mathrm{and}\,\,(\gamma^{k})^{2}=-I
$$
I'm just not sure how to put this together. If anyone could give a quick run through or some prods in the right direction that would be excellent.
 A: While studying the $\gamma$-matrices, I also faced the same question. Here maybe an solution about this.
First, the convention is the same:
$$(\gamma^{0})^{\dagger}=\gamma^{0}$$
$$(\gamma^{k})^{\dagger}=-\gamma^{k}$$
$$(\gamma^{0})^{2}=I\,\,\mathrm{and}\,\,(\gamma^{k})^{2}=-I$$
Additionally, here used the eqution,
$$\{\gamma^{\mu},\gamma^{\nu}\}=2g^{\mu\nu}I$$
Especially, for the $\gamma^{0}$ and $\gamma^{k}$,
$$\{\gamma^{0},\gamma^{k}\}=0$$
Then we can see,
$$(\gamma^{0} \gamma^{k})^{\dagger}=(\gamma^{k})^{\dagger} (\gamma^{0})^{\dagger}$$ and
$$\gamma^{0} \gamma^{k}=-\gamma^{k} \gamma^{0}$$ so
$$(-\gamma^{k} \gamma^{0})\dagger=(\gamma^{k})^{\dagger} (\gamma^{0})^{\dagger}$$
$$-\gamma^{0} (\gamma^{k})^{\dagger}=-\gamma^{k} \gamma^{0}$$
$$(\gamma^{k})^{\dagger}=\gamma^{0}\gamma^{k} \gamma^{0}$$
also we known that
$$(\gamma^{0})^{\dagger}=\gamma^{0}\gamma^{0} \gamma^{0}$$
we can combined these together to see
$$(\gamma^{\mu})^{\dagger}=\gamma^{0}\gamma^{\mu}\gamma^{0}$$
Noticing that here we have already selected a specific representation.
A: *

*for $(\gamma^0)^2 = I_4$, so $(\gamma^0)^\dagger = \gamma^0$;

*for $(\gamma^i)^2 = -I_4$, so $(\gamma^i)^\dagger = -\gamma^i$
Than: ${\gamma_\mu}^\dagger = \gamma^0\gamma_\mu\gamma^0$
