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.

Lie Sphere Geometry 16: Pencils of oriented spheres

 As a schoolboy I used pencils for drawing, mainly tanks and cars. Then I learned a little bit of perspective and tried to draw Greek temples and roads converting to a point on the horizon. 

Fascinated by pencils

It was just few years ago that I have learned about the concept pencils in geometry. I remember it was a paper by a Czech mathematician Metod Saniga,  "Pencils of Conics] a Means Towards a Deeper Understanding of the Arrow of Time":

Abstract: The paper aims at giving a sufficiently complex description of the theory of pencil!generated temporal dimensions in a projective plane over reals[ The exposition starts with a succinct outline of the mathematical formalism and goes on with introducing the definitions of pencil-time and pencil-space, both at the abstract (projective) and concrete (affine) levels. The structural properties of all possible types of temporal arrows are analyzed and based on symmetry principles, the uniqueness of that mimicking best the reality is justified. A profound connection between the character of different "ordinary" arrows and the number of spatial dimensions is revealed.

The subject interested me, but the math was too difficult at that time. Now it is time to return to this subject.

Definition 16.1. Two different oriented nonpoint spheres in S³ are in oriented contact if they have a common point x ∈ S³ , they are tangent to each other at x, and if they also have the same orientation defining unit normal vector n at x.

Definition 16.2. Given a point x ∈ S³ , a pencil of oriented spheres at x is the set of all oriented spheres in pairwise oriented contact at x.

Let us now examine these concepts. For r ∈ [0, 2π) and m a unit vector in R⁴ , the equation of an oriented sphere S_r(m) is

m · x = cos(r),     x · x = 1,                 (16.1)

while the normal unit vector n at x is (for a non-point sphere)

n(x) = (m − cos(r)x) / sin(r).                 (16.2)

Suppose we have two spheres S_r(m) and S_r'(m')  in contact at some point x₀ . Let n₀ denote the common unit normal vector at the common point x₀ . Thus

n₀ = (m − cos(r)x₀)/sin(r) = (m' − cos(r')x₀)/sin(r').                 (16.3)

It follows that

m = sin(r)n₀ + cos(r)x₀ ,                 (16.4)

m'= sin(r')n₀ + cos(r')x₀ .                 (16.5)

For a nonpoint sphere r≠ 0 and r≠ π, so that | cos(r)| < 1. Therefore, given x ∈ S_r(m), the unit vectors m and x in (16.1) are linearly independent – they span a two-dimensional subspace of R⁴ . Any vector tangent to the sphere at x is orthogonal to these two vectors. So the tangent space to the sphere at x is the orthogonal complement of m and x. But from (16.2) it follows that the subspace spanned by m and x is the same as the subspace spanned by mutually orthogonal unit vectors n and x. Therefore the spheres S_r(m) and S_r'(m'), having the same x₀ and n₀ , are automatically tangent to each other at x₀ . The requirement of them being tangent to each other in Definition 16.1 is therefore redundant.

Substituting (16.4) into (16.1) we see that the spheres of the pencil defined by x₀ and n₀ are intersections of planes with the sphere S³

(sin(r)n₀ + cos(r)x₀ ) · x = cos(r),     x · x = 1.                 (16.6)

As an illustration Fig. 16.1 shows the several circles from the pencil of oriented circles for x₀ = e₂ and n₀ = e₃ . In Fig. 16.1 we have used only the values of r in (0, π].

Figure 16.1: Pencil of oriented circles for x 0 = e 2 and n 0 = e 3 , r = 2kπ/10, k = 1, . . . , 10.

This is because the spheres S_r(m) and S_{r+π mod 2π} (−m) are the same. We added the point sphere r = π, which reduces to the point x₀ , because it naturally ‘wants’ to be included.

In the next chapter we will study the representation of pencils of oriented spheres in Q - that is the space to which, as we will see, they naturally belong, and the acquire a definite geometrical and physical meaning.

Notes on Lie Sphere Geometry updated.

Lie Sphere Geometry Part 15: Exercise with the pole shift

 How will the formulas of Part 14 change if North and South Pole are exchanged due to a pole shift? Immanuel Velikovsky’s hypothesis in Worlds in Collision imagines a cataclysmic pole shift triggered by planetary near-misses, such as Venus and Mars, dramatically altering Earth’s axis and climate, showcasing his bold independence of thought and vivid imagination in reinterpreting ancient myths as cosmic events. His work reflects a daring willingness to challenge conventional science, prioritizing creative synthesis over established norms.

Shifting the poles cataclysm

 In this post we follow Velikowsky's idea. Choosing the origin of the projection is our arbitrary choice. A really true geometry should not depend on such arbitrary choices. We need to look into this issue closer, and this involves straightforward calculation following the onc we have already done, but with a different formula for the stereographic projection. So, there will be nothing new in this part - just a simple exercise in applying again the techniques that we already know.

We start with repeating, almost verbatim, the derivation of the formulas for spheres

S_r(m) = {x ∈ S³: x · m = cos(r)},                 (15.1)

with normal vector field

n(x) = (m − cos(r)x)/sin(r).                 (15.2)

as in Part 9. The South Pole projection lead us, in Part 13, to the formula (13.7):

τ_s (c, ρ) = ([1 − q(c, ρ))/2 e₀ + c +(1 + q(c, ρ))/2 e₄ + ρ e₅ ],

or, using e_± = (e₄ ± e₀ )/2, to

τ_s (c, ρ) = [e₊ + q(c, ρ)e₋ + c + ρ e₅ ].                 (a)

As we will see below, the North Pole projection gives us, instead, the formula

τ'_s(c, ρ) = [(q(c, ρ) − 1)/2 e₀ + c +(q(c, ρ) + 1)/2 e₄ + ρ e₅ ],

or

τ'_s(c, ρ) = [q(c, ρ)e₊ + c + e₋ + ρ e₅ ]                 (b)

There are two distinguished points: the origin of space, and the infinity. In the first case, that of τ_s , the origin, corresponding to c = 0, ρ = 0, is represented by p_o = [e₊], while the infinity is represented by p_∞ = [e₋]. In the second case, that of τ'_s, the roles dramatically change: now p_o = [e₋], while p_∞ = [e₊ ].

The roles dramatically change: now p_o = [e₋], while p_∞ = [e₊ ].

This observation tells us that a different approach is needed. Q and Q + are homogeneous spaces. Thus any point can be chosen to represent the infinity. We know that (cf. (13.22)

(e₊, e₋ ) = − 1/2.

This suggests that if we decide to choose p_∞ = [u], then p_o = [u'] should be chosen arbitrarily, so that (u, u') ≠ 0.

This is the idea that we will follow in the future. For now, however, we need to do our boring homework. You can see it done in the complete Notes or, only Cj. 15, in the pdf displayed  below.

Lie Sphere Geometry Part 14: Q and Q+ as homogeneous spaces

 

The Erlanger Programm, delivered by Felix Klein in 1872 during his inauguration at the University of Erlangen, revolutionized mathematics by proposing that geometries should be classified by their underlying transformation groups. Klein's unifying framework emphasized that each geometry is defined by the group of transformations preserving its fundamental properties.

The Erlanger Programm, delivered by Felix Klein in 1872

We will be moving now in this direction.

14.1 The space infinity point [e∞]

In Part. 13 we have introduced, instead of e₀ and e₅ , the null vectors e₊ and e₋ . The vector e₋ is sometimes denoted e_∞ , and we will now see the reason. Let us take a sphere Sρ(c). It is represented in Q (and in Q⁺ ) by the formula:

τ_s (c, ρ) = [e₊ + q(c, ρ)e₋ + c + ρ e₄ ].                 (14.1)

We now fix the signed radius ρ, and examine the limit |c| → ∞. As far as |c| is finite and large enough so that |c| > |ρ|, q(c, ρ) is positive, we can rewrite (14.1) as (Why?)

τ_s(c, ρ) = [e₊/q(c, ρ) + e₋ + c/q(c, ρ) + ρ/q(c, ρ) e₄ ].                 (14.2)

In the limit |c| → ∞ the first, third and fourth term become zero, and we get

lim_|c|→∞ τ_s(c, ρ) = [e₋ ], (14.3)

independently of the (fixed) value of ρ. We have obtained a single point of Q, which we may call the spatial infinity and denote by [e_∞ ]. This result seems to contradict the fact that in Q and in Q⁺ every point is as good as any other point, since Q and Q⁺
are homogeneous space for the action of the group SO₀(4, 2), as we will see in the next section. To understand this apparent contradiction will require some work, and we will postpone it for later.

In Exercise 1 of Part 13, I was suggesting to do the previous reasoning with a plane instead of a sphere. Anna attempted to guess the answer. I wasn't thinking carefully at that time. With the sphere we were dividing by q(c,ρ), which for |c| large enough is a positive number. But for planes we would have to divide by h, which can be positive or negative. This works perfectly for Q, but for Q⁺ it would create a strange result: e_- or -e_-, which are two different points in Q⁺. depending on the sign of h, which I do not fully understand yet.

14.2 The action of O(4, 2) on Q and Q⁺

Let G be the matrix G = diag(1, 1, 1, 1, −1, −1). Thinking of u ∈ R^4,2 as column vectors we can then write the scalar product as:

u · u'= u^TGu',                 (14.4)

where · ^T stands for the transposition.
We define the pseudo-orthogonal group O(4, 2) as the group of all real 6 × 6 matrices L satisfying

L^TGL = G.                 (14.5)

This is the isometry group of R^4,2 , and any two orthonormal basis in R^4,2 are related by a certain matrix from O(4, 2).
It will be convenient to write matrices L in a block form

L =[A, B;C, D]                , (14.6)

with A, B, C, D being, respectively, 4 × 4, 4 × 2, 2 × 4, and 2 × 2 matrices. Using this block matrix form Eq. (14.5) takes the form:

A^TA − C^TC = 1,                 (14.7)
A^TB − C^TD = 0,                 (14.8)
B^TB − D^TD = −1,                 (14.9)
where 1, 0, −1 on the right-hand side are of dimensions 4×4, 4×2, and 2×2 respectively.

We do not need the fourth condition, since it is the transpose of the second condition, and so it is not independent of the three conditions above. As an immediate consequence of these conditions we have the following Lemma:

Lemma 14.1. For every L = [ A, B; C, D] ∈ O(4, 2) we have

| det(A)| ≥ 1, | det(D)| ≥ 1.                 (14.10)

Proof. See the Notes or Ch. 14 alone below.

Definition 14.1. We denote by SO₀(4, 2) the connected component of the identity of the group O(4, 2), that is the set of all elements of O(4, 2) which can be connected to the identity element by a continuous path within the group.

It follows from the definition that SO₀(4, 2)  is a group, a proper subgroup of O(4, 2). (Why?).

As a consequence of Lemma 14.1 we obtain

Proposition 14.1. For every L = [ A, B; C, D] ∈ SO₀(4, 2) we have

det(A) ≥ 1, det(D) ≥ 1.                 (14.12)

Proof. See the Notes or Ch. 14 alone below.

Remark 14.1. The group SO(4) × SO(2) is a subgroup of SO₀(4, 2) corresponding to B = 0 and C = 0, so that

L =[A, 0;0, D],                 (14.13)

where A ∈ SO(4), D ∈ SO(2), with SO(4) and SO(2) being connected.

In line with the Erlangen program of Felix Klein (it was in 1872) geometry consists of the study of invariants under a group of transformations. We have our group–that is SO₀(4, 2). We will now look at it as a group of transformations of Q and Q⁺ , and then study constructions that are invariant under this action. Notice that the action of O(4, 2) on R^4,2 is linear, and preserves the null cone {u ∈ R^4,2 : u² = 0}, therefore it acts on Q and on Q⁺ by L[u] = [Lu].

Proposition 14.2. The actions of SO₀(4, 2) on Q and on Q⁺ are transitive.

Proof. See the Notes or Ch. 14 alone below.

Exercise 1. The action of SO(2) on S¹ is transitive. Why?

Exercise 2. The action of SO(4) on S³ is transitive. Why?

Lie Sphere Geometry Part 13: Representing oriented spheres and planes in R3 by points of the Lie quadric Q and Q+

 

The 9th of May is a Victory Day

We start with the spheres.

Spheres

In Part 11 Proposition 1, using the inverse stereographic projection, to each oriented sphere S_ρ(c) in R³ we have associated an oriented sphere S_r(m) in S³. In Part 6 to each oriented sphere S_r(m) in S³ we have associated a point in the Lie quadric Q. In Part 7 we have introduced Q⁺ isomorphic to the product S³ × S¹. We will now combine these constructions. But first let me outline the rough idea that I have in mind while doing it.
Think of the sphere S_ρ(c) as a spherical wave propagated in space with the velocity of light. Then, assuming the velocity of light being a constant, its radius is related to time. Negative radius will then mean negative time. We choose the proportionality constant to be 1, which means we are choosing units in which the velocity of light is (numerically) 1. When we map our spheres to the points of Q, we will have different points of space and time represented by different points of Q or Q⁺ . Thus spacetime will be embedded in Q and Q + . That’s the rough idea. So, let us do the work now, starting with Q.

First let us recall the definitions. With R^4,2 denoting the 6-dimensional space with coordinates x⁰ , . . . , x⁵ and scalar product

x · y = x⁰y⁰+ x¹y¹+ x²y²+ x³y³ − x⁴y − x⁵y⁵,     (13.1)

We have defined Q as the quadric in the real projective space P(R^4,2 ):

Q = {[u] ∈ P(R^4,2) : u · u = 0}.         (13.2)

For a point [u] in Q the coordinates (u⁰ , . . . , u⁵ ) are called homogeneous coordinates of [u]. Given an oriented sphere S_r(m) we have associated with it the point of Q with
homogeneous coordinates (cf. Eq. (6.5)):

uⁱ= mⁱ⁻¹, (i = 1, . . . , 4), u⁴= sin(r), u⁵= cos(r).                 (13.3)

On the other hand we have the following relation (cf. Part 9 ) between signed spheres S_ρ(c) with signed radius ρ and center c ∈ R³ and oriented spheres S_r(m) not containing the origin of the stereographic projection:

c =m' / ( m⁰ + cos(r) ),                     (13.4)

ρ = sin(r) / ( m⁰ + cos(r) ),             (13.5)

where m'= (m¹ , m² , m³ ). Combining these two maps we obtain the following Proposition:

Proposition 13.1. With c = c¹e₁ + c²e₂ + c³e₃ ∈ R³ , ρ ∈ R, and

q(c, ρ) = c² − ρ²,                             (13.6)

the correspondence, denoted by τ_s between oriented spheres S_ρ(c) in R³ and points of Q
is given by:

τ_s(c, ρ) = [ ½(1 − q(c, ρ)) e₀ + c  + ρ e₄+ ½(1 + q(c, ρ)) e₅ ].                 (13.7)

Proof.  See Notes Ch. 13. But try to prove it all by yourself. It is really straightforward.

13.2 Planes

Representing oriented planes will be much simpler than spheres. We recall from Proposition 1 in Part 12 that oriented planes Π_h(n)

Π_h(n) = {y ∈ R³: y · n = h}                                 (13.11)

corresponds to oriented spheres S_r(m), so that:

r = arccos(h/√(1 + h²)),                         (13.12)

m⁰ = −h/√(1 + h²),                                 (13.13)

m' =n/√(1 + h²).                                     (13.14)

From these formulas we obtain the following:

Proposition 13.2. With n = n¹e₁ + n²e₂ + n³e₃ ∈ R³ , h ∈ R, the correspondence,
denoted by τ_p between oriented planes Π_h(n) in R³ and points of Q is given by:

τ_p (n, h) = [−he₀ + n + e₄ + he₅  ].                     (13.15)

Proof. See Notes Ch. 13. But try to prove it all by yourself. It is really straightforward.

13.3 Vectors e₊ and e₋

The terms in Eq. (13.7) can be collected in a different way:

τ_s (c, ρ) = [(e₅ + e₀)/2 + q(c, ρ)(e₅ − e₀)/2 + c + ρ e₄ ].                 (13.19)

This suggest introducing, instead of e₀ and e₅ another basis vectors

e₊ = ( e₅ + e₀)/2,     e₋ = (e₅ − e₀)/2.                 (13.20)

Vectors e₊ and e₋ are null vectors (Why?):

(e₊ , e₊ ) = (e₋ , e₋ ) = 0,                     (13.21)

and satisfy (Why?):

(e₊ , e₋ ) = −1/2.                         (13.22)

Eqs. (13.19) and (13.15) read now:

τ_s (c, ρ) = [e₊ + q(c, ρ)e₋ + c + ρ e₄ ],                 (13.23)

τ_p (n, h) = [0 · e₊ + 2he₋ + n + e₄ ].                     (13.24)

13.4 What about Q⁺ ?

We have obtained the formulas in Propositions 13.1 and 13.2 using the intermediary tool of the stereographic projection. But once we have them, we can use them simple as definitions of embeddings. As such they make a perfect sense also in Q⁺ if we interpret the equivalence classes [·] within Q⁺ . The same concerns (13.23) and (13.24), both being well defined in Q⁺ .

Exercise 1. Take a plane, any n, and move to infinity. Which particular element of Q would correspond to such a "plane at infinity"?

Sunday Special: Notes on Lie Sphere Geometry

Today it is short. Just to inform my readers, that I have just finished collecting my posts on Lie sphere geometry published so far into a single pdf document. It was a marathon.

Notes on Lie Sphere Geometry

It is fresh, I will be improving it, fixing possible errors. At the same time we will be digging deeper into the Lie quadric Q.

The pdf is available here.

Spheres as Objects - May 1 Special

 We humans are poor creatures. We are bound by our genetics, imprisoned within three spatial dimensions, residing on a beautiful but otherwise insignificant planet in the vast Cosmos—a planet periodically bombarded by comets that wipe out most of its life.

And yet, we strive to understand as much of the world—outside and inside us—as it is possible for us to grasp. What we perceive with our senses at any given moment is just one side of a multifaceted reality. Plato's allegory of the cave reveals only a small part of the whole truth.

Since today is May 1st, it is more than appropriate to quote from Mario Bunge’s 1950 paper, “The Inexhaustible Electron” (Science & Society, Vol. 14, No. 2, Spring 1950, pp. 115–121):

“The electron is as inexhaustible as the atom; nature is infinite, but it exists infinitely.”
—Lenin, Materialism and Empirio-Criticism (1908)

Lenin wrote this during the crisis of modern physics, at a time when the classical theory of the electron had been given a consistent form, mainly through the work of H. A. Lorentz (1853–1928). The central problem of this theory—which was then expected to explain all the properties of matter—was the structure of the electron.

What is the nature of the electron’s mass: mechanical, electromagnetic, or both? What is the nature of the field within the electron? What is its radius? These were some of the questions that intrigued physicists forty years earlier.

The electron is as inexhaustible

Physicist and philosopher Mario Bunge wrote those words 75 years ago. Today, the situation is not much better. In 1995, J. Keller and Z. Oziewicz organized an international conference, Theory of the Electron, in Mexico City. A group photo shows 37 participants, and the conference proceedings span 500 pages. In 2002, J. Keller published a 275-page monograph, The Theory of the Electron (Kluwer, 2002), summarizing just one such theory.

Why would the electron be considered inexhaustible? Clearly, it is more than just a point. But if it is not a point, then what is it? What “shape” does it have?

I think a better approximation would be to imagine the electron as a sphere. But what lies inside this sphere? Perhaps it contains a window. A window to where? To other dimensions—what else could it be?

And how many of these other dimensions are there? Well, that depends on how you choose to look at them. Which face of the multidimensional universe do you want to observe? One face might have a finite number of dimensions; another might well be infinite-dimensional.

This should not be surprising. We need to get used to it. One face of a quantum object may appear as a “particle,” another as a “wave.” We can see only one face at a time. Sometimes it will be sharp, sometimes it will be blurry.

Anyway, we are discussing spheres—in fact, we are discussing their geometry. We treat oriented spheres as “objects.” We are learning how to navigate in the space of these objects.

Object Oriented Programming

Lie sphere geometry reminds me of "object-oriented programming," which caught my attention in the past. We will return to the math in the next post.