In Part 18 we have met the Klein absolute, which, for the case of the 2D (x,y) plane, happens to be the null cone (ζ0)2- (ζ1)2 - (ζ2)2 - (ζ3)2 = 0 in the 4D Minkowski space. As it is usual in projective geometry we remove the origin ζ=0 from this null cone - we do not need it.
Let us recall that we have added two extra dimensions x0 and x3. We added x3 for the stereographic projection, and then we added x0 to get rid of denominators and, instead of doing the stereographic projection we employed the projective space.
Our null cone ζ2 = 0 is invariant under the action ζ⟼g ζ τ(g) of the Clifford group G of the Clifford algebra Cl(V). The action is linear. We already know that g
ζ τ(g) = Λ(g)ζ, where Λ(g) is a Lorentz transformation. We have
calculated explicitly these transformations for typical one-parameter
subgroups of G and obtain either rotations or Lorentz boosts. Now, since
the action is linear, it defines the action on equivalence classes of
the equivalence relation "∼" defining the projection (see Part 18). If
ζ=λη, λ>0, then Λ(g)ζ = λΛ(g)η. Therefore g acts on the equivalence classes of "∼". Let us take a closer look at the structure of the set of these equivalence classes. What do we get? We rewrite ζ2 = 0 as
(ζ1)2
+ (ζ2)2
+ (ζ3)2 = (ζ0)2.
Since we have removed the origin from the null cone, ζ0 must be non zero.
Exercise 1. Why?
Thus either ζ0 > 0, or ζ0 < 0. We choose λ = 1/|ζ0| and obtain a unique point in the equivalence class of
ζ with the zero coordinate equal +1 or -1. Let us concentrate on the case +1. We obtain the equation
(ζ1)2
+ (ζ2)2
+ (ζ3)2 =1 for this unique point. It represents a
point on the two-dimensional sphere. It stereographically projects onto a
point in (x,y) plane except for the South Pole of the sphere with ζ3 = -1.
Let us return to our embedding formula from Part 18.
After some thinking, and prompted by Bjab's observation I have changed the embedding there to get rid of unnecessary minuses. It became
ζ0
= (1+x·x)/2,
ζ1
= x1,
ζ2 = x2,
ζ3 = (1-x·x)/2.
Now ζ0 + ζ3 = 1. If we have any point ζ' on the cone with ζ'0 + ζ'3 > 0, we can always rescale it by a unique positive λ to get ζ0 + ζ3 = 1, and then read the coordinates from ζ (the rescaled ζ') the two coordinates on the plane. The line with ζ0 + ζ3 = 0 form an equivalence class defining one point on the sphere - its South Pole, the Infinite Point on the plane.
Exercise 2. If ζ0 + ζ3
= 0 on the null cone, then, necessarily ζ1 = ζ2 = 0. Why?
Now we can return to the action of the Clifford group on Minkowski space, its null cone, the sphere, and the plane. We have a distinguished point on the sphere - its South Pole. Thus we can extract a special subgroup of transformations, namely those that do not move that distinguished point. These transformations will transform the (x,y) plane into itself - they will be easy to interpret. But there will be also other transformations that move the infinity point into some other point. Inverse of such a transformation will move a point on the plane into Infinity Point on the sphere, producing a singularity on the plane. We will analyze all this in the next post.
Where Have All the Spinors Gone?
We’ve got "spinors" boldly declared in the title of this series, yet somewhere along the way, they’ve managed to slip out of sight. This is unacceptable! Spinors should be front and center, the main act, not some backstage crew hiding in the shadows. So, just to keep them from becoming the forgotten middle child of mathematical objects, let’s bring them back into focus.
Recently, I received a seven-page paper from V.V. Varlamov. The topic? Clifford algebras and spinors, inspired by Rozenfeld's work on non-Euclidean geometry. I opened it eagerly, confident that by page seven, I’d emerge enlightened, ready to declare, "At last, I understand spinors!"
I did not.
The math in this paper was so advanced, it left me feeling like a freshman who wandered into a graduate-level seminar by mistake. My head began to spin—not in the cool quantum way, but in the “I need aspirin” way. The formulas weren’t just over my head; they were in orbit. It became clear that to keep my sanity intact, I needed to ditch the "follow the paper" approach and start thinking in my way.
So, here’s the mental road trip I embarked on:
We’ve got our trusty geometric Clifford algebra of space. It’s an elegant, eight-dimensional creature, cozy with bi-quaternions and complexified Minkowski space. Lovely. What’s more, this algebra isn’t just lounging around—it acts. And it acts on itself, no less. How?
- From the left:
- From the right:
Now, here’s the juicy part: when you let a Clifford group element act on something from the left, , it drags you out of the real Minkowski space. (Thanks a lot, .) To reel things back in, you need to hit it from the right with something like . This leads to a beautiful commuting relationship:
This is clean, elegant math. But what does it have to do with the quantum physics of spin? Here’s where my mind started making some leaps—and possibly doing a few backflips.
Left, Right, and Quantum Duality
In quantum theory, there’s always this duality: the "observer" versus the "system under observation." For spins, we’ve got the laboratory frame with its neat axes, along with a physicist who’s busy assigning complex spin vectors in Hilbert space. And then we have the spin itself—precessing, pirouetting, doing its quantum dance.
In my mental model, I associate the left action with the quantum system itself and the right action with the observer. Or maybe it’s the other way around? I’ll admit, this part’s still a work in progress. The details need ironing out, like a wrinkly shirt you’re not sure is clean or dirty, but you’re wearing it anyway.
Enter the Hairy Ball Theorem
Somehow, all these thoughts led me to the Hairy Ball Theorem. Yes, that theorem from topology—the one that proves you can’t comb a hairy ball flat without creating a cowlick. If you’re wondering what this has to do with spinors, quantum dualities, or Clifford algebra, congratulations—you’re just as confused as I am.
But don’t worry, this will all (hopefully) become clearer in future posts. For now, I’ll leave you with this cliffhanger: Can spinors help us avoid cowlicks in quantum mechanics?


