Friday, November 29, 2024

The Spin Chronicles (Part 19): Touching circles

 For a change, this will be a very short post.

I realized the error in my thinking. We don’t need the "Klein absolute" for this. Stereographic projection is sufficient. I’m not sure why I thought that parallel lines on a plane, when mapped via inverse stereographic projection onto a sphere, would "intersect like meridians" at the poles. That was incorrect.

Bjab, as usual, was guiding me in the right direction: they don’t intersect—they are tangent. Bjab was right, and I was wrong.

 I’m humbly asking for mercy

Below is an illustration showing the inverse stereographic projection of 21 vertical lines from the (x, y) plane, all parallel to the y-axis, with x=10,9,,9,10.  As expected, they are tangent to each other at the South Pole.



Bjab was right; I was wrong. And so, I’m humbly asking for mercy. 

We will be back to the Klein absolute in the next post discussing transformations between circles.


P.S
. Here is my attempt to show how one rays from the South Pole of the sphere through a circle on this sphere generate the vertical line x=1 on the z=0 plane. View at the South Pole:



Wednesday, November 27, 2024

The Spin Chronicles (Part 18): Introducing Klein absolute


 “... there is something still lacking in this picture—the 'conformal infinity,' the 'absolutes' of projective geometry. Points where parallel lines meet...” That’s what I claimed in the previous post. It seems, however, that the idea of parallels meeting "somewhere" doesn’t sit well with the otherwise grounded tone of our series on the Clifford geometric algebra of space. Naturally, this stirred the curiosity of one particular reader (Bjab), who came armed with questions sharper than Occam's razor:

  • "What does 'meet' mean? (Different) Parallel lines don't have common points."
  • "What does 'intersect' mean?
  • "Meridians are not circles. They are semicircles."
  • "Are they really (in the common sense of the word) intersecting, or are those circles just tangent to each other?"

Ah, questions! The perpetual call to adventure for anyone foolish—or ambitious—enough to write about math. Since the cosmos obliges, so must I. Today, we’re taking a small detour (I really intend a small one) to tackle these questions head-on. And as luck would have it, this diversion ties neatly into our main subject, because it showcases the adaptability of the mathematical tools we’ve been exploring. By repurposing the same algebraic framework, we’ll reveal not just the versatility of mathematics but its elegance as well. Prepare for a deeper dive into the sublime weirdness of projective and conformal geometry!

Circles, Spheres, and Stereographic Projections

Let’s ease into this with a concept we all know and love: the humble circle. Actually, we should start with the slightly more ambitious sphere—it’s a better stand-in for the idea of "space." But hey, 3D geometry is cumbersome to scribble on a napkin or screen, so let’s stick to 2D for now. Circles are easier to draw, and, more importantly, we can rely on algebraic tools we’ve already mastered.


Circles are easier to draw


For this introductory phase, we’ll forego Clifford algebra. Don’t worry, it’ll return when we shift from static descriptions to dynamic transformations—that’s where Clifford algebra shines. After all, it’s not just about understanding objects but also how they morph, twist, and turn under various operations. These transformations form groups, and the Clifford group, as you’ll see later, is the MVP of this show. But for now, let's lay the groundwork with simpler tools.

Enter the Klein absolute. This concept will guide us into the realms of projective and conformal geometry, which are surprisingly practical—think computer vision and robotics. The journey begins with a stereographic projection on the 2D plane, using Cartesian coordinates x=(x1,x2).

Ready? Here’s the formula for (inverse) stereographic projection through the South Pole (because why not switch hemispheres for variety?):

σ(x) = (2x, (1-x·x))/(1+x·x).

Looks innocent enough, right? But let’s imagine we’re afflicted with a peculiar condition—denominatophobia. Yes, it’s a (fictional) fear of denominators. Even when they’re calm and friendly, like the one above, they still make us squirm. Now, as long as x is a real vector, there’s no reason to panic. But what if x develops a taste for the imaginary? After all, we’ve seen this happen before—constructing the Clifford algebra of space brought us to complex Minkowski spacetime! And who’s to say we won’t one day construct a Clifford algebra of a Clifford algebra? (Math is an infinite playground.)

To calm our nerves, we’ll take preventive measures. Mathematics offers a clever remedy for denominatophobia: projective space.

Projective Space: Fearless Math for Fearful Minds

We introduce a fourth coordinate, x0, alongside x1,x2,x3.
Initially, we set x0 = 1. This embeds our 2D point into 4D space:

x ⟼ ( 1, 2x/(1+x·x), (1-x·x)/(1+x·x) ).

In projective geometry, equivalence reigns supreme. Two points ζ and ζ′
 are equivalent (ζ ∼ζ′) if there exists a scalar λ>0 such that ζ′ = λζ. These equivalence classes—essentially half-lines in R4—are denoted by [ζ]. With this setup, we can rewrite the embedding as:

[( 1, 2x/(1+x·x), (1-x·x)/(1+x·x) )] = [((1+x·x)/2, x, (1-x·x)/2 )].

Victory! The denominators are gone, leaving us free to bask in mathematical serenity. But wait—there’s more.

The Bonus: A Light Cone Emerges
The point ζ = ((1+x·x)/2, x, (1-x·x)/2 ) lies on the light cone of Minkowski space. Let’s verify this. Define:

ζ0 = (1+x·x)/2,

ζ1 = x1,

ζ2 = x2,

ζ3 = (1-x·x)/2.


Then, check the relation:

0)2-1)2 - (ζ2)2 - (ζ3)2 = 0.


Exercise 1: Prove this formula.

This light cone is the so-called Klein absolute for our 2D plane. But don’t pack up just yet—there’s more excitement in store. In the next post, circles and parallel lines will make their dramatic entrance. For now, the stage is set, the props are in place, and the mathematical actors are ready for action.

References:

[1[ Wilson, Mitchell A, "Meeting at a far meridian""Встреча на далеком меридиане" -  Movie (1977)  here.

[2] Cecil, Thomas E. "Lie Sphere Geometry", Springer 2000

P.S. A schematic drawing taken from Ref. [2] (I have a somewhat different notation):




Sunday, November 24, 2024

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

 We continue our discussion, from Part 14Part 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:


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 realtion, 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 cansider 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 pq). 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

Friday, November 22, 2024

The Spin Chronicles (Part 16): The Action surprises

 This is a continuation of Part 15, where we have discussed the action of the Clifford group G of Cl(V) on Cl(V), defined by

g: x⟼ g x τ(g).

So far we have considered one-parameter subgroups of G of the form g(t) = exp(tX), with X = -ν(X) and X even. These were X of the form X = in, where n is a real vector in V. Today we will consider the case of X odd. Thus we take X=n, where n is a unit vector in V. Since ν(n) = -n, and π(n) = -n, it follows that τ(n)=n, and therefore τ(exp(tn))=exp(tn). For x = x0+x, x- complex scalar, and x - complex vector, we need to calculate

x' = (x'0,x') = exp(tn) x exp(tn).

Again it is a question of a straightforward algebra, even somewhat easier than in the previous case, so I will give just the result. For this it is convenient to introduce the following notation. For any x in V we define x as the component of x parallel to n, and x as the complementary component perpendicular to n:

x = (x·n)n,
x = x - (x·n)n.

With this notation, for x'=exp(tn) x exp(tn), we obtain:


x'0 = cosh(2t)x0 + sinh(2t) (x·n),

x' = cosh(2t)x + sinh(2t)x0n,

x' = x.

These are the well-known formulas for the Lorentz boost from Special Relativity. But here they arise almost effortlessly, as if conjured by some hidden symmetry. Coincidence? Or perhaps this is another case of Nature’s fondness for recycling its best ideas—its cosmic version of "As above, so below" and "As below, so above."

As above, so below

Nature seems to have a habit of finding a useful tool or building block and reusing it ad nauseam, like that one song you can’t get out of your head. Consider the "eye"—an evolutionary smash hit that popped up independently across the animal kingdom. Or the cell, biology's equivalent of the Lego brick. Could Clifford algebra, Cl(V)Cl(V), be another one of Nature's favorite gadgets? It shows up everywhere: in the microcosm (spin of electrons and protons) and the macrocosm (Minkowski spacetime in Special Relativity and the tangent spaces of General Relativity). It's the ultimate multitasker—kind of like duct tape, but for the fabric of reality.

If that’s the case, maybe it’s not so far-fetched to expect quantum phenomena at a macrocosmic scale. The catch? We might just be staring them in the face without realizing it, like someone trying to find their glasses while wearing them. To uncover such phenomena, we need more than just better technology or theories—we need a fresh perspective, a new way of asking questions. After all, the universe has a funny way of hiding its best tricks in plain sight, daring us to figure out the rules of its game. There may be new phenomena on the cosmic scale as well. Think of a "condensate (infinite continuous tensor product) of Clifford algebras" etc. "More" is sometimes essentially different....

Spin Chronicles Part 27: Back to the roots

  We have to devote some space to Exercise 1 of the previous post .  Back to the roots The problems was: Prove that <ba,c> = <b,ca...