# 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. Commented Nov 25, 2019 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. Commented Nov 25, 2019 at 20:57
• The second link deals with the crucial issue of complexification of spacetime algebra, which is part of your question. Commented Nov 25, 2019 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 $v=\gamma_1$. Commented Nov 25, 2019 at 22:10

1. The reason we usually complexify the Clifford algebra is mostly convenience: The representation theory of complex algebras is simpler in general, and if we want to restrict to real representations for some reason later on we can always do that. In particular, Dirac spinors at least exist in all dimensions, while the "real" Majorana spinors are dependent on the number of dimensions and even on the signature (depending on what, exactly, you mean by "Majorana"), see also this Q&A of mine.

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)

• Thanks a lot for the instructive response. Could you clarify a couple of things for me (again I think this is physicists abusing terminology). I'm getting a bit lost in the difference between the real and complex representation: which class does the spinor representation fall into, or is it both? And so I know i'm following, the gamma matrices present in the Dirac equation (which are complex) can form commutators that then generator the Lorentz Lie algebra? Lastly your 3rd point; the "accident" is the answer to my question haha, a happy accident I suppose. Thanks again. Commented Nov 25, 2019 at 20:49
• @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. Commented Nov 26, 2019 at 17:58
• Great answer! Just a question about point three, last sentence: Aren't spinors considered as the Minimal ideals of a Clifford algebra, so that all Clifford algebras contain spinors (or at least for $CL(n,m)$ where $n+m$ is even I think?). Commented Mar 9, 2021 at 1:00
• @R.Rankin There are many not-entirely-equivalent definitions of "spinor" in the literature, from very broad (projective but non-linear representation of $\mathfrak{so}(p,q)$) to very narrow (Dirac spinor, i.e. the irreducible representation of the Clifford algebra). I haven't heard of one involving minimal ideals. Commented Mar 9, 2021 at 15:20
• @ACuriousMind I think it's a standard definition of spinors, dating back to Marcel Riesz and his work on reoresentations. As per your other comment I've been seeking different ways of viewing the spinor bundle of a spacetime Commented Mar 9, 2021 at 19:59

The gamma matrices present in the Dirac equation generate a Clifford algebra [...]. It is argued in the gamma matrices wikipedia page that this algebra is the complexification of the spacetime algebra [...].

The Dirac algebra is normally complexified, but the gamma matrices per se don't give you the complexified algebra. The algebra generated by sums, products, and real scalar multiples of the gamma matrices is the real algebra. Even if you premultiply the gamma matrices by $$i$$, which is how they appear in the Dirac equation, you still get a real algebra (the one of opposite signature, $$-{+}{+}+$$, which is nonisomorphic). The Majorana basis makes that more obvious.

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?

No. The complex structure has to be explicitly added.

The algebra is complexified not for convenience, nor because "quantum mechanics uses complex numbers", but because the Dirac equation describes a wave with a $$U(1)$$ gauge charge, and the complex phase is really the $$U(1)$$ phase. Only with certain gauge groups do you get a complex structure. (See Why do we need complex representations in Grand Unified Theories? – though my answer only talks about the even algebra, where the details are a bit different.)

[Can one] 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 [...]?

The real Clifford algebra of any dimension and signature has an embedded copy of the Spin group (which in 3+1 dimensions is the double cover of the Lorentz group), and its Lie algebra is always the space of bivectors (which in the case of the Dirac algebra is the space of real linear combinations of pairwise products of distinct gamma matrices). The Lie bracket is the commutator, and exponentiation is $$\exp x = 1 + x + x^2/2 + \cdots$$ where the multiplication is Clifford multiplication.

The Clifford algebra is definitely an algebra of spacetime transformations. I'm not sure there is any separate concept of "spin space".

ACuriousMind wrote

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.

The last sentence isn't true: in any number of dimensions, the spinors do exist within the Clifford algebra as ideals. The easiest way to see that is that column vectors react identically under left multiplication to square matrices with only one nonzero column, or more generally square matrices whose columns are pairwise linearly dependent. Therefore, the spinors are projected out of the Clifford algebra by right multiplication by an element whose matrix representation has rank 1. An example of such an element of the Dirac algebra is $$(1+γ^5)(1+iγ^1γ^2)/4$$, which is $$\text{diag}(1,0,0,0)$$ in the chiral basis (one version of it, anyway).

The physical interpretation of these projections is the most confusing aspect of spinors to me. It's bad enough in $$3{+}1{+}U(1)$$ dimensions, and worse in $$3{+}1{+}\text{Spin}(10)$$ dimensions. The latter is interesting because of the remarkable fact that the 32-component Weyl spinor contains one complete generation of Standard Model fermions with all of the correct charges. So it seems likely that those spinors are physically relevant, but the projection to the column-vector space is quite strange and I don't understand why it would appear in nature. If you want to "worry about" something, that's what I'd recommend.

I don't think you can identify Dirac spinors with the first degree (vector part) of the Clifford algebra. But a relationship does exist in 8+0 dimensions (triality) and I'm not completely sure that there isn't some vestigial 3+1 dimensional version of it. Many special things happen in $$2^n$$ and $$(2^n+1)+1$$ dimensions for $$n\in\{0,1,2,3\}$$.

The Dirac equation for non-zero mass particles mixes left and right helical modes. These are the eigenmodes of $$γ^5$$. For zero mass, they don't mix, and $$γ^5$$ is no longer involved, because the eigenmodes decouple.

$$C_{1,3}(ℝ)$$ is the algebra generated from $$\left\{γ^1,γ^2,γ^3,γ^4\right\}$$. Originally, $$γ^0$$ was designated $$γ^4$$, by the way. The complexification is $$C_{1,3}(ℂ) ≅ ℂ ⊗ C_{1,3}(ℝ)$$ is the result of tossing an extra $$i$$ into $$C_{1,3}(ℝ)$$ and making it commute with everyone. Equivalently, it's the result of tossing $$γ^5$$ into $$C_{1,3}(ℝ)$$ and making it anti-commute with the other generators, i.e. generating the Clifford algebra from $$\left\{γ^1,γ^2,γ^3,γ^4,γ^5\right\}$$ ... because $$γ^5$$ has an $$i$$ in it. That's $$C_{2,3}(ℝ)$$. It's also isomorphic to $$C_{4,1}(ℝ)$$. Let's put that in profile: $$C_{1,3}(ℂ) ≅ ℂ ⊗ C_{1,3}(ℝ) ≅ C_{4,1}(ℝ).$$

For massless particles, the mass shell is: $$\left(\frac{E}{c}\right)^2 - \left(p_x\right)^2 - \left(p_y\right)^2 - \left(p_z\right)^2 = 0.$$ For mass non-zero particles, it is: $$\left(p_x\right)^2 + \left(p_y\right)^2 + \left(p_z\right)^2 + \left(mc\right)^2 - \left(\frac{E}{c}\right)^2 = 0.$$

That's actually what it all comes down to. It really is that simple.

Split the total energy $$E$$ into the kinetic energy $$H$$ and mass-energy: $$E = mc^2 + H.$$ Then the mass shell can just as well be written as $$\left(p_x\right)^2 + \left(p_y\right)^2 + \left(p_z\right)^2 - 2mH - \frac{1}{c^2}H^2 = 0.$$ This has signature $$4+1$$ ... and it continues to have signature $$4+1$$, regardless of what you replace the $$1/c^2$$ in that equation by - even $$0$$: $$\left(p_x\right)^2 + \left(p_y\right)^2 + \left(p_z\right)^2 - 2mH = 0.$$ Consequently, $$C_{4,1}(ℝ)$$ is also the Clifford algebra for the non-relativistic Dirac equation.

This shows, more directly, just why the fifth element is required for the mass. In the non-relativistic limit its use is essential. Since the relativistic theory has the non-relativistic theory as its correspondence limit, in order for the correspondence principle to continue to be upheld, then its use must be essential in the Relativistic theory too!

I describe the geometry underlying all of this in greater depth here Seamless Morphing Between Relativistic And Non-Relativistic Dirac Equation ... and also show how you morph between the Dirac equation coupled to the Maxwell field and its non-relativistic version, the Pauli-Schrödinger equation, with the geometry.

• Your association of mass $m$ to the "fifth dimension" $\gamma_5$ can only be achieved via an axial-rotated (with rotation angle $\theta = \pi/2$) Dirac equation $i\gamma^\mu\partial_\mu\psi - i\gamma_5 m \psi = 0$. See more explanation here: physics.stackexchange.com/questions/611255/… Commented Jul 16 at 18:51

Here is a simple path of research that I myself have been using to somewhat answer your question...

You can easily extend a Clifford algebra to a space with a non-flat metric tensor. IF you assume all the elements of the Clifford objects are tensors, and thus the Clifford object itself is a scalar (from a tensor point of view), then all your equations will be generally covariant.

For example, you can do this using $$Cl_{1,3}$$ and using the electromagnetic field vector. Maxwell's equations in curved space-time then reduce to:

$$\partial F = \mu_0 J$$

where

$$F \equiv \partial A$$

$$\partial \equiv \gamma^{\mu} \frac{\partial}{\partial x^{\mu}}, A \equiv \gamma^{\mu} A_{\mu}$$

Now to get to spinors, you use tetrads to rewrite your generators as linear combinations of the original flat generators (Dirac matrices for example), and you again treat every element of some object as tensors. This then allows you to write a generally covariant version of Dirac's equation:

$$\partial \Psi = \frac{E_0}{\hbar c} \Psi \gamma^{012}$$

The trick in expanding this, is remembering to replace the derivatives of the "flat" Dirac matrices with covariant ones that use the spin-connection. This is just a trick that compensates for your use of the tetrads and changing "frame".

I'm skipping a lot of detailed steps but this DOES give you a generally covariant equation for spinors.