Lie Sphere Geometry Part 17: Pencils in Q

 We continue with pencils  form Part. 16. 

Pencils and all that

Given two mutually orthogonal unit vectors x₀ and n₀ in R⁴ we have constructed a pencil of oriented spheres {S_r (m(r)) : 0 < r ≤ π} in R³ , where

m(r) = sin(r)n₀ + cos(r)x₀ .                 (17.1)

In Part. 6. Eq. (6.5) with each pair (m, r), m ∈ S³ , 0 ≤ r < 2π we associated a point in Q:

(m, r) → [m + sin(r)e₄ + cos(r)e₅ ].                 (17.2)

With x₀ and n₀ as above let

u_x0,n0(r) = sin(r)n₀ + cos(r)x₀ + sin(r)e₄ + cos(r)e₅ .                     (17.3)

Then, combining (17.1) and (17.2) we obtain a family p_x0,n0(r), 0 ≤ r < 2π, of points of Q:

p_x0,n0(r) = [u_x0,n0(r)].                     (17.4)

It makes sense to extend this definition to the whole interval [0, 2π), and this is what we
do.

Definition 17.1. Given any two points p = [u], p'= [u'] in Q, we write p ⊥ p' if the scalar product (u, u') of u and u' in R^4,2 is zero. In that case we say that p and p' are orthogonal. Notice that the definition makes sense since the condition (u, u') = 0 does not depend on the choice of representatives of the equivalent classes. We also notice that for any point p in Q we have p ⊥ p. (Why?).

Proposition 17.1. For any two points p = p_x0,n0(r), p'= p_x0,n0(r') we have p ⊥ p'.

Proof. Vectors x , n₀ , e₄ , e₅ are mutually orthogonal with (x₀ , x₀ ) = (n₀ , n₀ ) = 1 and (e4, e4 ) = (e5 , e5 ) = −1. The result follows by calculating the scalar product of u_x0,n0(r)
and u_x0,n0(r'). We obtain sin(r) sin(r') + cos(r) cos(r') − sin(r) sin(r') − cos(r) cos(r') =0. QED

Let

u₁ = u_x0,n0(0) = x₀ + e₅ ,                     (17.5)

u₂ = u_x0,n0(π/2) = n₀ + e4 .                 (17.6)

Then (u₁)²= (u₂)²= 0, and (u₁ , u₂ ) = 0. Vectors u₁ and u₂ therefore span a totally isotropic plane in R^4,2 , where by ”totally isotropic” we mean that any two vectors of this plane are orthogonal.
In projective geometry a projective line is, by definition, the image (by equivalence
relation u ∼ u' if and only if u = λu) of a two-dimensional plane. In fact any point on the projective line spanned by x₀ and n₀ corresponds to an element of the pencil. This follows from the following proposition.

Proposition 17.2. For every point p of the isotropic line {[αu₁ + βu₂ ] : α, β ∈ R, α² +β² > 0} there exists r ∈ [0, 2π) such that p = p_x0,n0(r).

Proof. The condition α² +β² > 0 excludes the case of α = β = 0. It is necessary since in the definition of any projective space P(V ) we exclude the origin of V. Using the definitions of u₁ and u₂ we have

αu₁ + βu₂ = αx₀ + βn₀ + αe₄ + βe₅                         (17.7)

=√(α² + β²)(α'x₀ + β'n₀ + α'e₄ + β'e₅ ),                 (17.8)

where α' = α/√(α² + β²), β'= β/√(α² + β²), and α'²+β'² =1.

Thus there exists r ∈[0, 2π) such that α'= sin(r), β'= cos(r). But [αu₁ + βu₂ ] = [α'u₁ + β'u₂ ] .QED

The construction of pencils of oriented spheres may look like an otherwise useless
exercise. It is like trampoline workout. Certainly it can

• Improve Heart Health and Cardiovascular System,
• Improve Joint Health,
• Develop Balance and Coordination,
• Help With Weight Loss,
• Improve Mental Health.

But how long can we or should we do it?

But how long can we or should we do it? The answer is: all our lives! Studying pencils of spheres we study properties of light, and light is the foundation of being. We did not see any light yet, but that is because we are just entering the theater. We need to have a wider view, and for this we need, in particular, relate our pencils to more general concepts, first of geometry and algebra, and then of physics. In the next chapter we will relate our construction to linear algebra concepts: Quadratic spaces and Witt’s theorem.

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:

(ζ₀)²- (ζ₁)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):

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