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"?

Lie Sphere Geometry: Part 9: Spheres of negative radius

The Fourth Way – P D Ouspensky [from the oral teachings of G I Gurdjieff]

"The first step is to try not to express  negative emotions;  - fear, anger, jealousy, possessiveness, pride and so on; the second step is the study of these negative emotions themselves, making lists of them, finding their connections ….. and trying to understand that they are quite useless.

Question:  In some cases the negative emotion of fear seems useful, otherwise people would cross the road at any time without looking

You speak about instinctive fear, emotional fear is different, it is based on our imaginings of what might happen

Question: are there no negative emotions that have a use?

It sounds strange, but it is very important to understand that all negative emotions are all absolutely useless; they do not serve any useful purpose; they do not make us acquainted with new things or bring us nearer to new things; they do not give us energy; they only waste energy and create unpleasant illusions."

Well, this post is not about negative emotions. Instead, it introduces spheres of negative radius. Which concept (thanks to Anna for letting me know) has been  discussed by another Russian thinker, Pavel Florensky.

"Referring to Dante's Divine Comedy, Florensky opposes Copernicus' heliocentric system. Interprets the Michelson-Morley experience as proof of the immobility of the Earth. Declares “the notorious Foucault's experience” fundamentally unproven. Commenting on Einstein's special theory of relativity, Florensky argues that beyond the limit of the speed of light begins non-physical “that light”. This otherworld of imaginary magnitudes provides a description of the ultimate eternal reality. Based on the geocentric system, Florensky calculates the distance to this world as the distance at which a body orbiting the Earth in one day would travel at the speed of light."

This otherworld of imaginary magnitudes provides a description of the ultimate eternal reality.

We denote vectors in R⁴ by bold letters. With m in S³⊂R⁴, consider a sphere S_t(m) on S³ given by (cf. Part 4, Eq. (2))

S_t(m) = {x∈S³: x·m = cos(t)} .               (0)

For t in [0,2π), t∉{0,π},  we orient it by the normal vector
(cf. Part 4, Eq. (3))

n(x) = ( m-cos(t)x )/sin(t).               (1)

In the previous post we have derived the formulas for stereographic projection x⟼y,  S³→R³ (S³ is taken without its South Pole!), and our aim now is to identify the surface obtained by this projection from S_t(m). Let us recall the formulas for stereographic projection and its inverse (cf. Part 8, Eq. (2),(3a),(3b)). With i=1,2,3, we have:

yⁱ = xⁱ/(1+x⁰).            (2)

x⁰ = (1-y²)/(1+y²).               (3a)

xⁱ = 2yⁱ/(1+y²).                (3b)

Substituting (3a) and (3b) in (0), multiplying both sides by (1+y²), and collecting y² terms,  we obtain:

y²(m⁰+cos(t)) - 2m'·y= m⁰ - cos(t),                (4)

where

m '= m¹e₁+m²e₂+m³e₃.

There will be now two cases. The first case is when m⁰+cos(t) ≠ 0. When this happens?

Proposition 1. m⁰+cos(t)) = 0 if and only if the origin -e₀ of the stereographic projection is on the sphere S_t(m).

Proof. Using Eq. (0), with m=m⁰e₀+...+m³e₃, we see that x=-e₀ is on S_t(m) if and only if m⁰+cos(t) = 0.

⎕

Suppose now that  -e₀ is not on the sphere. We divide both sides of (4) by (m0+cos(t)):

y² - 2m'·y/(m₀+cos(t))  = (m⁰ - cos(t))/(m⁰+cos(t)) .

We now add m'²/(m⁰+cos(t))² to both sides to obtain:

(y-m'/(m⁰+cos(t)))² = (m⁰ - cos(t))/(m⁰+cos(t)) + m'²/(m⁰+cos(t))².                (5)

Since m²=1, we get m'²=1-(m⁰)², and the right hand side of (5) simplifies to

(y-m'/(m⁰+cos(t)))²= sin(t)²/(m⁰+cos(t))².                 (6)

From (6) we conclude that the image of S_t(m) is a sphere in R³ with the center at

c = m'/(m⁰+cos(t)),                (7)

and radius ρ

ρ = ε sin(t)/(m⁰+cos(t)),                 (8)

where ε = ±1.

For t>0, t<π, sin(t)>0, but what about the sign of (m⁰+cos(t))? It can be positive, but it can also be negative! This leads us to the concept of spheres with negative radius.

Oriented spheres in R³.

In the definition below the term "signed" means that ρ can have positive or negative value.

Definition 1. The oriented sphere with center p in R³ and signed radius ρ, 0≠ρ∈R, is

S_ρ(p) = {y∈R³: (y-p)²=ρ²}                (9)

with unit normal vector field

n'(y) = (p-y)/ρ.                (10)

To understand what is going on, assume first that the center of the sphere is at the origin p=0. Then n'(y) = -y/ρ. Thus the normal n'(y) is inward for ρ>0, and outward for ρ<0. If p is arbitrary, not necessarily the origin of R³, the situation is the same, just translated by p.

Proposition 2. The image of the oriented sphere S_t(m)⊂S³, not containing -e₀, under the stereographic projection x⟼y given by (3a,3b), is the oriented sphere Sρ(c)⊂R³, where

c = m'/(m⁰+cos(t)),
ρ = sin(t)/(m⁰+cos(t)).

Proof. We have already calculated the center c of the projected sphere. It remains to be checked that the direction of the image of the normal n(x) given by (1) is the same as the direction of n'(y(x)) = (c-y(x))/ρ. From (2),(7),(8),(10) we get

n'(y(x))ⁱ =mⁱ/(ε sin(t)) - xⁱ(m0+cos(t))/(ε sin(t) (1+x⁰)).      

Now we need to find the image, let us call it  n''(x) of the normal unit vector n(x), Eq. (1),  under the stereographic projection. While coordinates transform using the formulas (2): yⁱ=yⁱ(x^μ), i=1,2,3; μ=0,1,2,3, tangent vectors transform using the matrix of partial derivatives (Why?), thus

n''(x)ⁱ = (∂yⁱ/∂x^μ) n^μ(x).

The result of this calculation is (you can use any symbolic algebra or, nowadays, probably also  online AI, software to do it for you):

n''(x)ⁱ = (1+x⁰)^-1(mⁱ/sin(t) -xⁱ(m⁰+cos(t))/(sin(t)(1+x⁰)) .        (11)

Since (1+x⁰) is positive (Why?), for n' and n'' to have the same direction we need to take ε=+1.

Notice that while stereographic projection preserves angles between tangent vectors, it does not preserve their lengths. That is why n'' is only proportional to n'. To become a unit normal it would have to be normalized. Then it should coincide with n'.

Exercise 1. Verify the last sentence.

Stereographic projection can indeed exchange the "inside" with the "outside". Here is an illustration for circles on the sphere, projected onto the plane:

Proposition 2 takes care of this problem.

The case when the origin -e₀ of the stereographic projection, the ∞ point,  is on S_t(m), i.e. when m₀+cos(t) = 0, will be discussed in the next note.

P.S. 23-04-25 19:21 I was busy writing another review

Lie Sphere Geometry Part 7: The other side of space

 The Gospel of Q is a hypothetical source document that some scholars believe was used by the authors of the Gospels of Matthew and Luke, containing sayings and teachings of Jesus. It is thought to predate the canonical Gospels and focuses primarily on Jesus' wisdom teachings without a narrative framework.

The Lie quadric Q that we have introduced in Lie Sphere Geometry Part 6: The Gospel of Q is a "canonical Q". Like the canonical Gospel, it is a mixed bag. There is some lack of coherency there. While the spheres (m,t) are considered to be oriented, the sphere S³, on which they live, is considered to be unoriented. But S³ itself can be oriented and can have inside surface and outside surface, and they can be considered as different. The spheres (m,t) can slide either on the outside or on the inside of S³. Why should we be blind to the orientation of S³ itself?

Remark. Of course once we started on our path of seeking a logical consistency, we can follow the lead even farther along this path. Why should we put S³ in R⁴ instead of on S⁴? And then S⁴ on S⁵, S⁵ on S⁶ and S⁶ on S⁷? We can then stop with S⁷, because S⁷ is "special". S³ is related to quaternions, S⁷ to octonions, and this is the end of division algebras. Or we can continue to infinity of spheres of higher and higher dimensions, until we reach the unit ball in the infinite-dimensional Hilbert space, where the orientation does not make sense any more, and  all infinite-dimensional separable Hilbert spaces are isomorphic - the natural end of our long journey.

In Part 3 we could see the animation of oriented circles gliding on the surface of the sphere. There we could see the reason for the identification (m,t) = (-m,t+π mod 2π). But now there is no such identification. Before
(m,3π/2) and (-m, π/2) represented the same oriented circle. Now they are different. (m,3π/2) is already on the inner side of the sphere, while (-m,π/2) is on the outer side. Thus we do not need any more to take the quotient by Z₂.

Let's go to the math. We modify the definition of Part 6:

Q⁺ = {[u]∈P⁺(V): (u,u) = 0},

where P⁺(V)=(V∖{0})/∼, with x∼y if and only if there exists   λ>0 such that y=λx. Topologically P⁺(V) is S⁵. (Can you see it?).

Exercise 1. Show the isomorphism of Q⁺ and S³⨉S¹.

Here is one possible speculation: if our universe indeed has two sides, the matter located on the other side would not be "visible", but still can have a gravitational influence on the matter on our side. It would modify the gravity, much like what we call "dark matter" does.