C^\infty(T^*{\mathcal{M}})\otimes C^\infty(D{\mathcal{M}})\otimes C^\infty(D^*{\mathcal{M}})$ by setting its components $\gamma^{\,\,\,A}_{\alpha\,\,\,\,\,B}$ in the frame $e^\alpha_a\otimes E_A\otimes E^B$ equal to the matrix elements $(\gamma_\alpha)^{A}_{\,\,\,\,B}$. For example, it can be shown that $$\label{split gamma 0}
\gamma_0=\delta^{AB}E_A\otimes E^+_B \, .$$

The Dirac equation {#dirac background 2}
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The metric $g$ determines a connection $\Gamma$ in the usual way via the covariant derivative operator $\nabla: C^\infty(T{\mathcal{M}})\mapsto C^\infty(T^*{\mathcal{M}}\otimes T{\mathcal{M}})$. One may equally define a connection $\sigma$ and covariant derivative on the spinor and cospinor bundle, which we also denote by $\nabla$, $$\nabla: \left\{  \begin{array}{c} C^\infty(D{\mathcal{M}}) \\ C^\infty(D^*{\mathcal{M}})
\end{array}\right.  \mapsto \left\{  \begin{array}{c}
C^\infty(T^*{\mathcal{M}}\otimes D{\mathcal{M}}) \\ C^\infty(T^*{\mathcal{M}}\otimes D^*{\mathcal{M}}) \end{array}\right. \, .$$ Given a local section $E$, $f\in C^\infty(D{\mathcal{M}})$ may be decomposed $f=f^AE_A$, then $\nabla_af$ has components $$\nabla_\alpha f^A = \partial_\alpha f^A + \sigma^A_{\alpha B}f^B$$ in the frame $e^\alpha_a\otimes E_A$ where the connection $\sigma$ has elements given by $\sigma^A_{\alpha B} =
-\frac{1}{4}\Gamma^\beta_{\alpha\delta}\gamma_{\beta\,\,\,\,C}^{\,\,A}\gamma^{\delta
C}_{\,\,\,\,\,\, B}$.

We are now in a position to define the equation of motion the spinors will satisfy, i.e. the Dirac equation. The Dirac operator $${\nabla\hspace{-3mm}{/}}: \left\{  \begin{array}{l} C^\infty(D{\mathcal{M}}) \\ C^\infty(D^*{\mathcal{M}}) \end{array}
\right. \mapsto
\left\{ \begin{array}{l} C^\infty (D{\mathcal{M}}) \\ C^\infty(D^*{\mathcal{M}})
\end{array} \right.$$ maps (co)spinor fields into (co)spinor fields by $$\begin{aligned}
{\nabla\hspace{-3mm}{/}}f &=& ({\nabla\hspace{-3mm}{/}}f)^A E_A = \eta^{\alpha\beta}\gamma^{\,\,\,
A}_{\alpha\,\,\,\, B}(\nabla_\beta f^B) 