The Spin Chronicles (Part 21): Rene Magritte and Flammarion

 

We’re on a journey. A journey to where, you ask? Honestly, I’m not entirely sure myself. Is the ultimate goal to solve all the mysteries of the universe? (No pressure, right?) Or maybe just crack open one of them, like an intellectual piñata? Or perhaps it's something subtler, something closer to René Magritte's enigmatic wisdom: “Each thing we see hides something else we want to see.”

But Magritte, ever the paradoxical philosopher, also declared: “I think we are responsible for the universe, but that doesn’t mean we decide anything.” Really, René? Nothing? Well, I’ll take my victories where I can, starting with today’s decision to write this blog. I have no clue what cosmic ripples this choice will send out (but hey, I hear butterflies start hurricanes), yet I’ll cling to the notion that I’m somehow fulfilling my obligations to the universe by clicking away at my keyboard.

I have no clue what cosmic ripples this choice will send out 

So, where were we? Ah, yes—the journey. It turns out, we’re not exactly on the straight path. We’ve taken a detour. A scenic route, if you will. After Part 17, the plan was to dive straight into options (3) and (4) for two-sided actions of the Clifford group on Cl(V):

1. $g \cdot u = g u \tau(g)$
2. $g \cdot u = g u \nu(g)$
3. $g \cdot u = \pi(g) u \tau(g)$
4. $g\cdot u=\pi (g)u\nu (g)$

Instead, I decided to veer off-road. Why? Because sometimes you’ve got to set aside the GPS of mathematical rigor and take a little detour to explore something... different. Something fun. So, let’s play with the action of the Clifford group on the null cone in Minkowski space and the induced action on the $(x, y)$-plane.

At first blush, this might sound like one of those “because I can” exercises in pure abstraction. But hold on—this isn’t some random doodle in the margins of physics. It has a gorgeous physical interpretation! The null cone in Minkowski space represents light emanating from a single spacetime point. Picture a laboratory at time $t=0$, where light bursts forth in all directions. By $t = 1$, that light forms a sphere around the lab. Think of it as the universe’s way of saying, “Let there be geometric algebra!”

On this sphere, we can project the stars of the universe—the celestial sphere—much like Flammarion’s famous engraving, which tantalizingly invites us to peek beyond the cosmic veil:

Now, let’s spice things up a bit. What happens if we apply a Lorentz transformation to our lab? A rotation is straightforward—the sky spins, and the stars gracefully whirl like a cosmic waltz. But a Lorentz boost? That’s a whole different party. Enter relativistic aberration: the stars’ positions shift dramatically, bending and dancing under the influence of spacetime’s warping lens.

We’re going to calculate this starry choreography using the Clifford group. (And if you’re getting déjà vu, it’s because I hinted at this back in my September 4, 2024 post, “Monkeys, UFOs, and Quantum Fractals: A Mind-Bending Journey through Relativity and Imagination.”)

Here’s the twist: we’ll go beyond. We’ll project the celestial sphere onto the $(x, y)$-plane of our lab. But don’t expect a straightforward mapping. The projection originates from the South Pole of the celestial sphere—a point conspicuously absent in Flammarion’s image. (That’s because it’s “beneath” the feet of our starry-eyed astronomer. Practical, but disappointing.)

This is the plan. And in the next post, we’ll dive into the algebra to unravel what happens to the projected images of stars. Will they remain stars? Will they smear into streaks of cosmic glory? Stay tuned. Who knows—this little exploration might even shake up your view of the heavens. Or at least provoke a disagreement or two.

P.S. 04-12-24 Updated the pdf file.

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?