# Spinors, Spacetime and Clifford algebra

I'm looking to understand the intrinsic connection that Clifford algebra allows one to make between spin space and spacetime. For a while now I've trying to wrap my head around how the Clifford algebra fits into this story, with the members of my department consistently telling me "not to worry about it". However, I think there's something deep to be uncovered.

The gamma matrices present in the Dirac equation generate a Clifford algebra: $$\{\gamma^{\nu}, \gamma^{\mu}\} = 2\eta^{\nu \mu} I$$. It is argued in the gamma matrices wikipedia page that this algebra is the complexification of the spacetime algebra: $$Cl_{1,3}(\mathcal{C})$$ is the complexification of $$Cl_{1,3}(\mathcal{R})$$. The answer given here (What is the role of the spacetime algebra?) would seem to suggest that this complex structure falls out naturally from the decomposition into degrees of $$Cl_{1,3}(\mathcal{R})$$. Is this the case?

Furthermore, is it the case that one can then use the gamma matrices that generate $$Cl_{1,3}(\mathcal{C})$$ to form the Lie Algebra of the Lorentz group, which to me gives the picture that these constructions in spin space can form spacetime transformations (as outlined here: Relation between the Dirac Algebra and the Lorentz group)?

Essentially (I think) the question I'm asking is does the Clifford algebra encapsulate some global space of which spacetime and the space of spinors belong - if so how then do the gamma matrices present in the Dirac equation respect and link these two spaces? Were dealing with algebras, but does the isomorphism $$SO(1,3)$$~$$SU(2)$$ x $$SU(2)$$ come into play here?

I'm not a mathematician by trade, but I think technical answers will naturally come into play here - if people could try and hold onto some physical intuition it would be much appreciated. Best wishes to all.

• See related links: physics.stackexchange.com/q/514592, physics.stackexchange.com/q/410952. – MadMax Nov 25 '19 at 17:08
• Thanks for the useful links: a couple of q's if you don't mind. I've often heard spinors are so confusing because they represent the square root of a geometry, is that what your comment regarding the flat metric being generator by the Dirac matrices encapsulates? I'm not sure what your second link is trying to say or how it relates, could you break it down please? Thankyou. – Jack Hughes Nov 25 '19 at 20:57
• The second link deals with the crucial issue of complexification of spacetime algebra, which is part of your question. – MadMax Nov 25 '19 at 22:02
• As to the meaning of "square root of a geometry", there is another interpretation: the Lorentz rotation of a spinor is one-sided as $R\psi$, as opposed to double-sided for a vector $RvR^{-1}$. Thus a $\pi$ rotation of a spinor e.g. $e^{\pi\gamma_1\gamma_2}\psi$ turns into a $2\pi$ rotation of vector $e^{\pi\gamma_1\gamma_2}ve^{-\pi\gamma_1\gamma_2} = e^{2\pi\gamma_1\gamma_2}v = v$ for say $v=\gamma_1$. – MadMax Nov 25 '19 at 22:10

2. The second degree of the Clifford algebra (complex or real doesn't matter here) is isomorphic as a Lie algebra to the Lorentz algebra (or, in the generalized version, the generalized Clifford algebra for a metric $$\eta$$ has the isometry algebra for that metric as its second degree). It is not the "gamma matrices" (= generators of the Clifford algebra, hence in particular first degree elements of it) that generate the Lorentz algebra, but their commutators $$\sigma^{ij} = [\gamma^i, \gamma^j]$$. (It may be that you are already aware of this, but this is a common point of confusion)
3. I'm not quite sure what your "does the Clifford algebra encapsulate some global space of which spacetime and the space of spinors belong" question is trying to ask, but let me point out that four dimensions - where one could identify the first degree of the Clifford algebra with both spacetime and the four-dimensional Dirac spinors - are an "accident". The Dirac spinor representation in $$d$$ dimensions is $$2^{\lfloor d/2 \rfloor}$$-dimensional, which you cannot identify with the $$d$$-dimensional first degree of the algebra in most other dimensions. Therefore, the Clifford algebra does not, in a general sense, "contain" spinors.
4. Lastly and most tangentially, there is no isomorphism $$\mathrm{SO}(1,3)\cong \mathrm{SU}(2)\times \mathrm{SU}(2)$$, regardless of how often you will read this lie in physics-oriented texts. See e.g. this answer by Qmechanic and the linked questions for details on the relation between the two groups and their algebras. The nutshell is that $$\mathrm{su}(2)\oplus\mathrm{su}(2)$$ is the compact real form of the complexification of $$\mathrm{so}(1,3)$$, hence the complex finite-dimensional representations of these algebras are equivalent, hence the projective representations of the group $$\mathrm{SO}(1,3)$$ are given equivalently by $$\mathrm{su}(2)\oplus\mathrm{su}(2)$$ representations. (For why projective representations matter, see this Q&A of mine)
• @JackHughes The Dirac representation is complex - its real version, if one exists, is the Majorana representation. And yes, the commutators of the $\gamma$-matrices are the generators of the Lorentz algebra. – ACuriousMind Nov 26 '19 at 17:58