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This is just an example of an important property of the GL(N) Lie Group tensor operators. It means that the tensor operator $\gamma^{\mu}$ transforms like a 4-vector under conjugation. Please see my answer to "Do the Dirac matrices form a proper four-vector?" which might have been better posted here.

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This is just an example of an important property of the GL(N) Lie Group tensor operators. It is said that the tensor operator $\gamma^{\mu}$ transforms like a 4-vector under conjugation. If you want to know the matrix elements of a tensor operator (in this case $\gamma^{\mu}$) in some representation (in this case the spin=1/2 rep) transformed to another ...

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$γ$ is an injective map from $\mathbb R^{1,3}$ into the Clifford algebra of $\mathbb R^{1,3}$, that takes each vector to itself. Since it's a linear map, you can think of it as a rank-2 tensor. Suitably interpreted, it's the identity tensor, so transforming it equivalently on both sides leaves it unchanged. That is essentially what that equation says, albeit ...

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The full symmetry group of the massless Lagrangian is actually: $$SU(2)_L\times U_L(1)\times SU(2)_R\times U_R(1)$$ which covers the additional symmetries you noticed. It can be rearranged as: $$SU(2)_V\times U_V(1)\times SU(2)_A\times U_A(1)$$ where $$U_A(1)$$ (related to your $q_1$ portion) is usually broken by the quantum anomaly, and $$SU_A(2)$$...

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The last equality on the page follows immediately from seeing that the inverse of $\gamma_{\mu} k^{\mu} + m$ is $-\frac{\gamma_{\mu} k^{\mu} - m}{k^2 + m^2}$. If we assume that all scalar products are euclidean scalar products and either the $k^{\mu}$ is numerator is considered Wick rotate, $k^{\mu} = (i k^0 ,\vec{k})$ (this would be somewhat incosistent) or ...

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