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The spin group ${\rm Spin}(p,q)\cong {\rm Spin}(q,p)$ is connected if $\max(p,q)\geq 2$. If we exclude multiple temporal dimensions, i.e. if we consider only Minkowskian and Euclidean signatures, then the component of ${\rm Spin}(p,q)$ that is connected to the identity is simply connected; except in the cases 2+0D and 2+1D where the fundamental group is $\...


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As one usually does, write the exponential term with a power series expansion, $$\Psi^{\dagger} \big(1 - i \beta \gamma^{5} + \mathcal{O}(\beta^2) \big) \gamma^0$$ then using the anticommutative properties $\{\gamma^5,\gamma^{\mu} \} = 0$ you can move $\gamma^{0}$ through the $\gamma^{5}$ terms, picking up a minus sign in the process. You can check the ...


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Left handed spinors $\psi_{\text{L}}$ transform as $$ \psi_{\text{L}} \mapsto \exp\left((-\mathrm{i}\boldsymbol{\theta} +\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}}{2} \right)\psi_{\text{L}}\tag{1} $$ and right handed spinors $\psi_{\text{R}}$ transform as $$ \psi_{\text{R}} \mapsto \exp\left((-\mathrm{i}\boldsymbol{\theta} -\boldsymbol{\beta})\cdot\...


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Let us evaluate the sum in the exponential: \begin{equation} -\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu}=-\frac{i}{2}(\omega_{03}S^{03}+\omega_{30}S^{30}). \end{equation} Due to $S^{\mu\nu}$ and $\omega_{\mu\nu}$ being antisymmetric, we get: \begin{equation} -\frac{i}{2}(\omega_{03}S^{03}+\omega_{30}S^{30})=-\frac{i}{2}(\omega_{03}S^{03}+(-\omega_{03})(-S^{03})) = ...


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