# 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.

• The transformations of a Dirac spinor belong to $\text{Spin}^\mathbb{C}(1,3)$, the double cover of the Lorentz group, while the transformations of (vector) bosons belong to the Lorentz group. This means that (vector) bosons and Dirac fermions are different objects and so are sections of different bundles. Dec 9, 2022 at 10:15
• @JeanbaptisteRoux Do you mean the Lorentz Group $O(1,3)$ or the proper Lorentz Group $SO(1,3)$? Both preserve Minkowski spacetime invariants, so both should be a sub-algebras of $Cl_{1,3}(\mathbb{R})$, and so they should act on $\mathbb{C}^4$ spinors. Even if they're part of different bundles, the Lorentz group shares the same parent algebra, and so it seems vector bosons should be a subgroup of the same parent spinors. I'll need to split my response over two comments. Dec 9, 2022 at 13:42
• (Part 2 of my comment) I don't actually understand $Spin^\mathbb{C}(1,3)$ well enough to see how it's a sub-algebra of $Cl_{1,3}(\mathbb{R})$. Maybe something like $Spin^C(1,3) \subseteq SO(1,3) \times_{\mathbb{Z}_2} S^1 \subseteq O(1,3)$? Is this supposed to be a double cover of the Lorentz Group or the Proper Lorentz Group? I know logically this group should be equivalent to the one generated by the Gamma Matrices, though I don't understand the group well enough to say how. Dec 9, 2022 at 13:42
• First of all, and as per the Wikipedia entry on the spin group, $\text{Spin}(1,3)$ is the double cover of $\text{SO}(1,3)$. Secondly, the Clifford algebra spanned by the gamma matrices is $\text{C}\ell_{1,3}(\mathbb{R})_\mathbb{C}$, which indeed acts on $\mathbb{C}^4$ spinors. What do you mean by "the Lorentz group shares the same parent algebra, and so it seems vector bosons should be a subgroup of the same parent spinors?" Dec 9, 2022 at 14:33
• "I believe the double cover of SO(1,3) is just the Lorentz Group O(1,3)" Double cover of SO(1,3) is something else, Spin(1,3). And incidentally, double cover of O(1,3) is a less known group, Pin(1,3) Dec 10, 2022 at 22:13

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.

• Thank you. There's a lot here for me to unpack. I think I understand part of my confusion. Spinors are vectors associated with a particular group within $Cl_{1,3}(\mathbb{R})$, but there are other groups within $Cl_{1,3}(\mathbb{R})$. It might be true that the vectors assigned to $\psi(x,t)$ are vector subspaces of $\mathbb{C}^4$ (unclear), but they won't all be spinors. I want to unpack a bit more of your response before accepting it, though it seems correct to me at the moment. Dec 10, 2022 at 9:12