Are all fields on spacetime spinor-valued? I'm trying to understand the values that fields can take. For fermions, my understanding is that fields on spacetime take values as Dirac Spinors, which are $\mathbb{C}^4$ vectors. The vector space of Dirac Spinors is the one acted on by the matrix ring generated by the Gamma Matrices. Gamma Matrices are pairwise tensor products of Pauli Matrices, which generate the quaternions $\mathbb{H}$. Incidentally, the Clifford Algebra for spacetime $Cl_{1,3}(\mathbb{R}) \cong M(2,\mathbb{H})$, which is isomorphic to the matrix ring generated by the Gamma Matrices. $M(2,\mathbb{H})$ acts on $\mathbb{C}^4$, and therefore so do the Gamma Matrices. As a result, fields $\psi(x,t)$ for fermions on spacetime take values in the space of $\mathbb{C}^4$ spinors.
Taking a slightly more dubious route... The spacetime invariants for fermions under special relativity are given by the Minkowski Metric. The Clifford Algebra corresponding to the Minkowski Metric is $Cl_{1,3}(\mathbb{R}) \cong M(2,\mathbb{H})$, which acts on the vector space $\mathbb{C}^4$. Every relativistic field $\psi(x,t)$ on spacetime preserves the invariants of spacetime, and therefore $\psi(x,t)$ must take values in some subspace of $\mathbb{C}^4$ acted on by some sub-algebra of $Cl_{1,3}(\mathbb{R})$. For fermions, $\psi(x,t)$ takes values in the full subspace $\mathbb{C}^4$ itself.
I'm wondering, first, if that dubious route is valid. If it is valid, I'm wondering if, as a consequence, all relativistic fields on spacetime, even for bosons, take values in some subspace of $\mathbb{C}^4$ spinors.
 A: I think you are mixing up some basic facts about four-vectors and $\mathbb{C}^4$ spinors. So let us review some of them here (this is more of a long comment than an answer):

*

*$\mathbb{C}^4$ spinors transform under $\text{Spin}^\mathbb{C}(1,3)$, the complexification of the double cover of $\text{SO}(1,3)$. The latter is the transformation group of 4-vectors. See https://en.wikipedia.org/wiki/Spin_group#Complex_case


*All wavefunction which is fully in $\mathbb{C}^4$ (in the sense that all four components are complex) are not necessarily spinor valued. For example, if you come to describe with regular QM $W^+$ and $W^-$ bosons, you will have to use four complex components, which are just complex 4-vectors, and not spinors.


*The obstructions to their construction are different. While for vectors, the non-annulation of the first Stiefel-Whitney class is an obstruction, for spinors, it is the second Stiefel-Whitney class that has to vanish. See https://en.wikipedia.org/wiki/Spin_structure#SpinC_structures


*Their spins are different. Restricting ourselves to the subalgebras of $\text{Spin}^\mathbb{C}(1,3)$ and $\text{SO}(1,3)$ describing rotations, their Casimir invariants are those of $\mathfrak{spin}(3)$ in the $1/2$ representation and $\mathfrak{so}(3)$ in the $1$ representation respectively. While it is true that these two algebras happen to be isomorphic, it truly is the Casimir invariant that gives us the spin of the particles we are describing. See https://en.wikipedia.org/wiki/Representation_theory_of_SU(2)#Most_important_irreducible_representations_and_their_applications and https://en.wikipedia.org/wiki/3D_rotation_group#A_note_on_Lie_algebras


*This one is more of a physicist's fact: Their relativistic equations are not the same. On one hand, you have the Dirac equation for spin 1/2 particles, and on the other hand, you have the Proca equations for spin 1 particles. These two equations are fundamentally different since the first is to be thought of as being the "square root" of the second (more precisely the " square root" of the Klein-Gordon equation). See https://en.wikipedia.org/wiki/Dirac_equation#Mathematical_formulation


*Within the Whiteman axioms, there is the spin-statistics theorem, which tells you that spinor-valued wavefunctions do not obey the same statistics as vector-valued ones. See https://en.wikipedia.org/wiki/Spin%E2%80%93statistics_theorem#Consequences.
All these six points (which are not completely independent) indicate you that 4-vector valued wavefunctions are not spinor-valued ones. Again, this is more of a long comment than an answer, and I hope someone better than me will give you a real answer.
