$\varphi=\widetilde{\varphi}\circ\iota$.

It is worth pointing out that spin structures do not exist in general for an arbitrary manifold; their existence is determined by the *second Stiefel-Whitney class*. In essence, the requirement that the second Stiefel-Whitney class vanishes ensures consistency between the (transition functions of the) fibre group of the tangent bundle and the (lift to the transition functions of the) spin group. It is known that there exist spin structures over orientable manifolds if and only if the second Stiefel-Whitney class vanishes and that every four-dimensional globally hyperbolic manifold admits a spin structure. We now assume that an arbitrary spin structure has been chosen and is fixed for the remainder of this discussion.

We may now define spinor fields on a curved manifold by saying they are sections of the associated ${\mathrm{Spin }}_0(1,3)$ bundle $${D{\mathcal{M}}}= {S{\mathcal{M}}}\ltimes_{{\mathrm{Spin }}_0(1,3)}{\mathbb{C}}^4 \,  .$$ The fibre of ${D{\mathcal{M}}}$ at $x\in{\mathcal{M}}$ is the equivalence class $[T,z]_x$ where $T\in S_x{\mathcal{M}}$ and $z\in{\mathbb{C}}^4$ is a *column* vector and the equivalence relation is: $[\widetilde{T},\widetilde{z}]_x=[T,z]_x$ if and only if $\widetilde{T}=R^{-1}_ST$ and $\widetilde{z}=Sz$ for some $S\in {\mathrm{Spin }}_0(1,3)$. The bundle $D{\mathcal{M}}$ has fibre ${\mathbb{C}}^4$ at every point and left action given by $L_S
[T,z]_x = [T,Sz]_x$. The dual bundle $$D^*{\mathcal{M}}=S{\mathcal{M}}\ltimes_{{\mathrm{Spin }}_0({\mathcal{M}},g)}{{\mathbb{C}}}^4 