The Spin Chronicles (Part 15): The Action starts

Understanding the meaning of the Universe—and, by extension, our purpose in this dizzying 3D playground—is impossible without first grasping the concept of spinors. Yes, spinors. Everything spins. Planets pirouette around stars, galaxies swirl like cosmic pinwheels, and deep inside you, photons and electrons are spinning their little hearts out. Without spin, there would quite literally be "no nothing." But what exactly is this mysterious "spin"? And what, pray tell, is a "spinor"?

To get the most precise answers, we must summon the sharpest tool in humanity’s intellectual shed: mathematics. Forget poetry or philosophy for now (though they’re lovely companions); math is the only game in town when it comes to rigor. The groundwork for this particular game was laid by none other than the French mathematical collective known as Nicolas Bourbaki. This was not an  ordinary group of thinkers—they were the masterminds who struck the match, poured the fuel, and launched the metaphorical rocket of modern mathematical formalism.

The Bourbakists

I, too, am a humble traveler on this rocket of discovery, squinting over the shoulders of these giants to glimpse what lies ahead. 

The Bourbaki

And let me share with you one of the gems they left behind, a definition from Bourbaki’s magnum opus, ÉLÉMENTS DE MATHÉMATIQUE, ALGÈBRE CHAPITRE IX:

DÉFINITION 2: On appelle groupe de Clifford de Q (resp. groupe de Clifford spécial de Q), le groupe multiplicatif des éléments inversibles de C(Q) (resp. C+(Q)) tels que sEs⁻¹ = E.
Dans ce no nous noterons G et G⁺ le groupe de Clifford et le groupe de Clifford spécial de Q.

If you’re already scratching your head, don’t worry—it’s not your fault. Mathematics is as much a language of symbols as it is a puzzle of patience. But let’s break it down: Their E is our V, their Q is the Euclidean quadratic form of V, and their C(Q) corresponds to our Cl(V). Meanwhile, their C⁺(Q) is a subalgebra of Cl(V), comprising those nice even elements that remain unchanged by π.

Here’s the twist: where Bourbaki insists on the condition  sEs⁻¹ = E. we’re a bit more liberal. We drop that requirement, making our Clifford group G bigger and, dare I say, more fun. Why stop at “sensible” constraints when you can explore the wild possibilities of mathematical freedom?

And here’s where it gets spicy: this expansion of the Clifford group isn’t just a technical curiosity—it unlocks surprises. It’s a reminder that while mathematicians love their neat, self-contained universes, Nature herself isn’t so tidy. She operates with her own version of simplicity, one that often leaves mathematicians muttering into their coffee mugs.

In short, if you think Bourbaki nailed it, wait until you see what happens when we take the leash off. Welcome to the uncharted territory where Clifford algebras meet Nature’s playful chaos.

Note: The idea of enlarging the Clifford group is not new (is there anything really-truly new under the Sun?). C.f. for instance William Baylis, "Lecture 4: Applications of Clifford Algebras in Physics, 4.3 Paravectors and relativity", in "Lectures on Clifford (geometric) algebras and applications" [edited by] Rafal Ablamowicz and Garret Sobczyk], Springer 2004, and references therein.

This post is a continuation from The Spin Chronicles Part 14, where we have introduced our "Clifford group G" defined as the set of elements u = (p₀,p)  in Cl(V) for which (p₀)²-p² = 1. Another form of the definition of G is

G = {g ∈ Cl(V): g ν(g) = 1}.

Exercise 1: use the definition of the multiplication in Cl(V) written in a bi-quaternion form to verify that these two definitions are indeed equivalent

In this post we will take a closer look at the action of G on Cl(V) defined by

g: u ⟼ guτ(g).

To this end we will analyze the action of one-parameter subgroups g(t) = exp(t X). In order for g(t) to be in G, we must have X + ν(X) = 0.

Exercise 2. Verify the last statement.

First we take the case of X even. Thus we have two conditions on X:

a) ν(X) = -X,

b) π(X) = X.

A general solution of these two conditions is X = in, where n is a real vector.

Exercise 3. Verify the last statement. Without loss of generality we can choose n to be a unit vector. Remember that n is a vector in Cl(V), therefore we have nn=1. Using this property calculation of exp(itn) is easy. We get:

exp(itn) = cos(t)1 + i sin(t) n.

Exercise 4. Use the power series expansion of exp(itn) to derive the above formula.

We can easily verify that we have

τ(g(t)) = g(-t).

Indeed, τ(exp(itn)) = exp(τ(itn)) = exp(-iτ(tn)) = g(-t), where we used the fact that τ(i) = -i, and that t and n are real, t - scalar, and n - vector.

Taking now u=(x₀,x), where (x₀,x) are complex, and setting it is a matter of simple algebra using the multiplication formula for bi-quaternions (c.f. Exercise 1, Part 11) to get for u(t) = g(t) u g(-t), and for n=(0,0,1),  the formula

x₀(t) = x₀,

x₁(t) = cos(2t)x₁ + sin(2t) x₂,

x₂(t) = -sin(2t)x₁ +cos(2t) x₂

x₃(t) = x₃.

It follows that this action of g(t) on Cl(V) is nothing else but a simple rotation by the angle 2t of the vector part. It rotates the same way both real and imaginary components of the vector. The scalar component x₀ does not change under these rotations.

In the next post we will consider the case of odd X.

P.S. For a very nice exposition of the standard mathematical approach to spinor groups see "Clifford Algebras, Clifford Groups, and a Generalization of the Quaternions" by Jean H. Gallier available on Academia.edu. Really nicely written paper, but in the whole of 27 pages you will not meet the concept of "spinor". This proves that the concept of a spinor is more difficult to write about than the concept of a "spinor group". Or, perhaps, that there are too many of definitions of spinors, and it is not clear which one will win in the future. And yes, I did not forget, we are getting there, slowly crawling ....