Lie Sphere Geometry Part 12: Preimages of planes

Oriented planes in R³ are images by stereographic projection of those oriented spheres in S³ that contain the point -e₀. In this post we use the same method as we did it in Part 11 when looking for preimages of oriented spheres. At the end we will have to decide again about the value of  ε to obtain the correct orientation. While for spheres it required some work with somewhat complicated algebraic expression, here it will be much easier - a merry and fast slide downhill.

Easy epsilon

With n a unit vector in R³, and h a real number, consider the oriented plane Π_h(n) in R³:

Π_h(n) = {y∈R³: y·n = h}.   (1)

We expect it to be an image, by the stereographic projection, of a certain oriented sphere

S_t(m) =  {x∈S³: x·m = cos(t)},     0<t<π ,     (2)

that contains the point -e₀=(-1,0,0,0) - the origin of the stereographic projection. Our aim is to find an explicit expression of (t,m) through (n,h). Let us stress that the formulas (1) and (2) should be understood having in mind the fact Π_h(n) and S_t(m)are more than just "sets". They include the orientations, and orientations are contained in the pairs (n,h) and (m,t). Pairs (n,h) and (-n,-h) (resp. (m,t) and (-m,π-t)) define the same sets, but correspond to opposite "orientations".

We will proceed the same way as we have done it with spheres in Part 11. This time it will be even simpler, because the equation (1) defining Π_h(n) is linear. Thus, with x=(x0,...,x3)∈R⁴, ||x||=1, we substitute the stereographic projection formula

yⁱ  =  xⁱ/(1+x⁰),         (i=1,2,3)        (3)

into (1) to obtain

x'·n = (1+x⁰)h = h+x⁰h,                (4)

where x'=(x¹,x²,x³).

Then we rewrite (4) as

- x⁰h + x'·n = h,  

which suggests the candidate for m, namely (-h,n). But m should have the norm 1, while (-h,n) has the norm squared h²+n²=1+h². We must  also remember that (n,-h) and (-n,h) lead to the same sphere as a set, so our general solution is

m = (-εh, εn )/√(1+h²),

t = arccos(εh/√(1+h²)),       0<t<π

where ε=±1, and we have to decide the sign taking into account the orientations. To this end we follow the same path as we have taken in Part 11 with the spheres. From Part 10, Proposition 1 therein, we know that to have the correct orientation we should have h=cot(t).  Now, cos(arccos( εh/√(1+h²) ) )=εh/√(1+h²), while sin( arccos(εh/√(1+h²)) )=(1-(εh/√(1+h²))² )^1/2= 1/√(1+h²). Thus cot(t) = εh, therefore ε=+1. 

Et voilà: ε=+1!

This way we have arrived at the following Proposition:

Proposition 1. With n a unit vector in R³, and h a real number, the pre-image of the oriented plane Π_h(n)with respect to the stereographic projection is the oriented sphere S_{arccos(h/√(1+h²))}( (-h, n )/√(1+h²) ).

Exercise 1. Verify that m⁰+cos(t)=0 as it should be. (Why?)

In the next post we will return to the 6D universe and we will place our 3D world of  spheres and planes in there.

Lie Sphere Geometry Part 10: Planes

 Spheres of infinite radius, those that cross the infinity point,  are just planes. Spheres are mysterious, planes are mysterious too, perhaps even more so.

Even more so...

This post is a straightforward continuation of Lie Sphere Geometry: Part 9: Spheres of negative radius. The stereographic projection of the oriented sphere

S_t(m) = {x∈S³: x·m = cos(t)},     0<t<π ,      (0)

leads to the following equation (Eq. (4) of Part 9) for its image in R³:

y²(m⁰+cos(t)) - 2m'·y= m⁰ - cos(t),                (1)

where

m '= m¹e₁+m²e₂+m³e₃.                (2)

We have already considered the case of m⁰+cos(t) ≠ 0. We consider now the case of m⁰+cos(t) = 0, i.e. m⁰ = -cos(t). Eq (1) then becomes

m'·y = cos(t).                 (3)     

Definition 1. With n∈S²⊂R³ and h∈R, the oriented plane Π_h(n) with unit normal n and signed height h, is a pair (n,h) and  associated with it the set [Π_h(n)] defined as:

[Π_h(n)] = {y∈R³: y·n = h}.                (4)

Remark. To see the meaning of h, suppose we orient the Cartesian axes of R³ so that n is the unit vector in the z-direction. Then h is the value of the z-coordinate at which the plane intersects the z-axis. Thus the name "signed height". Notice that while Π_h(n)  and Π_-h(-n) define the same set, these are two different oriented planes, as they have opposite normals.

Proposition 1. The image of the oriented sphere S_t(m)⊂S³ , containing -e₀, under the stereographic projection x⟼y  is the oriented Π_h(n') , where n' = m'/sin(t), h = cot(t).

Proof.  Comparing Eq. (3) with (4) we get instantly that n' has to be ε=±1 times the normalized vector m'. Now, (m⁰)²+m'²=1, and m⁰ = -cos(t). It follows that m'² = 1- cos²(t) = sin(t)². Since 0<t<π, we deduce that ||m'|| = sin(t).

Thus n'=ε m'/sin(t), h = ε cot(t). In Part 9 we have calculated (Eq. (11)) the image n'' of n:

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

which now reduces to

n''(x)ⁱ = (1+x⁰)^-1(mⁱ/sin(t)).

Since 1+x⁰ > 0, we deduce that ε = 1.

⎕

In the next post we will deal with the inverse stereographic projection of oriented spheres and planes.

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