The Spin Chronicles (Part 20):Group action on the sphere and the spin idea

 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. 

Projective geometry is about light rays

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.

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: $L(u)v=uv$
• From the right: $R(u)v = vu$
  

Now, here’s the juicy part: when you let a Clifford group element $g$ act on something from the left, $g \cdot x$, it drags you out of the real Minkowski space. (Thanks a lot, $g$.) To reel things back in, you need to hit it from the right with something like $\tau(g)$. This leads to a beautiful commuting relationship:

$$L(u)R(w)=R(w)L(u)\mathrm{.}$$

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?

The Spin Chronicles (Part 17): When The Field appear

 We continue our discussion, from Part 14, Part 15, and Part 16, of actions of the Clifford group G on the Clifford geometric algebra of space Cl(V). In Part 14 we have listed four different actions:

How Leonardo da Vinci pictured the Last Supper

1). g·u = g u τ(g),

2). g·u = g u ν(g),

3). g·u = π(g) u τ(g),

4). g·u  = π(g) u ν(g).

We have already seen that 1) leads to Lorentz transformations of coordinates in a complex Minkowski space of Special Relativity. Today we will discuss the second action. Let us recall that G consists of all invertible elements of Cl(V). It is defined by the condition gν(g) = ν(g)g =1.  Action 2) can thus be written as

g·u = g u g^-1,

and so it consists of automorphisms of the algebra. In particular g·i = i. The scalar part of u does not transform at all under this action, only the vector part transforms! What can be the relation, if any, to special relativity in this case? For action 1) the scalar part behaved as the time component of a Minkowski space event. Now the scalar part does not transform at all. The mystery will be solved immediately after, anticipating the result, I introduce the notation for this discussion. The vector part consists of three real and three imaginary numbers. What can it be? A physicist guesses immediately: write u as u = E+iB,

and see what happens!

Writing g = exp(tX), we find that X+ν(X)=0. Again we will consider two cases: X odd and X even. Let us start with X even, thus π(X) = X. As in the case 1) we set X = in, where n is a real unit vector. In this case, for g(t) = exp(itn) we have ν(g(t))=g(-t)=τ(g(t)). The action is the same as in the case 1) - E and B transform as vectors (or covectors?) under 3D rotations in the plane perpendicular to n. The same formulas as for x in the case 1).

Now we consider the case of X odd. Thus we set X=n. Then exp(tn) is rather easily calculated:

exp(tn) = cosh(t)1 + sinh(t)n.

Exercise 1. Verify the last formula.

Then calculating

(E'+iB') ≡ exp(tn)(E+iB)exp(-tn) = (cosh(t)1 + sinh(t)n) (E+iB) (cosh(t)1 - sinh(t)n)

is a matter of a straightforward algebra using the bi-quaternion multiplication formula

(p₀,p)(q₀,q) = (p₀q₀ + p_·q,  p₀q + q₀p ₊ i p⨯q). We obtain

E' = ....

B' = ...

I will not give the results now. I will not take  the pleasure from the reader to find it himself/herself. To find and then compare with what can be found on internet.

Anyway, it looks like Cl(V) contains not only spacetime, but also "the field strength". If so, then a field would be a function on Cl(V) (treated as complexified space-time) with values in Cl(V) (treated this time as a receptacle for field strengths). What kind of a function? V.V. Kassandrov suggested the answer: it should be "analytic" function, "analytic" in a special way. Then analyticity would automatically lead to something similar to Maxwell equations. But this is still in the future - I am only learning this stuff.  So far we are dealing with kinematics, no dynamics yet. Yet there is something else still lacking in this picture - the "conformal infinity", the "absolutes" of projective geometry. Points where parallel lines meet.

Morrice Kline, Projective Geometry, Scientific American, 192, p. 80-86, 1966