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 11: Reprojection of Spheres

 

This post is a continuation of Part 9. Working on this post gave me a real headache. All was going fine until it came to deciding the value of epsilon (it appears in (9a) and (9b) below) that takes care of the sign of the radius of the sphere in R³. I went to bed last night thinking about how to solve the contradiction I have arrived at.

It took me half the day today to figure out the solution of the problem. It was my simple algebra error, and I had to pay for it with Sisyphus  efforts.

It was my simple algebra error, 

and I had to pay for it with Sisyphus  efforts.

Let us first recall the definition from Lie Sphere Geometry: Part 9: Spheres of negative radius:

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

S_ρ(c) = {y∈R³: (y-c)² = ρ²}                (0)

with unit normal vector field

n'(y) = (c-y)/ρ.             (1)

It is an image, by the stereographic projection of some oriented sphere S_t(m) on S³.

Our aim now is to find (t,m) and to express them in terms of c and ρ. We recall that, with 0<t<π, m∈S³,  S_t(m) denotes the set

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

together with the normal vector field

n(x) = (m - cos(t)x)/sin(t).         (3)

The pair (t,m) should be such that the image of n(x(y)) has the same direction as n'(y), where x(y) is the inverse stereographic projection of y.

We start with recalling the stereographic projection formula from x = (x⁰,...,x³) in S³, with the removed "South Pole" (-1,0,0,0), to y=(y¹,y²,y³) in R³:

yⁱ  = xⁱ/(1+x⁰),      ( i=1,2,3).        (4)

It will be convenient to introduce x'=(x¹, x², x³), so that (4) can be written as a vector formula:

y = x'/(1+x⁰).                (4a)

Let us now use (4a) to write (0) as

((x'/(1+x⁰) - c)² = ρ²,

or

(x' - (1+x⁰)c)² = (1+x⁰)²ρ².         (5)

Expanding the left-hand side we obtain

x'² - 2(1+x⁰)x'·c + (1+x⁰)²c² = (1+x⁰)²ρ².        (6)

Eq. (6) does not look like (2) at all. Eq. (2) is linear in x, while (6) is a quadratic equation. However there is the following algebra magic that does the job: we have that x²=1, therefore (x⁰)²+x'²=1. Or x'²=1-(x⁰)² =(1+x⁰)(1-x⁰). Using this, and dividing both sides of (6) by (1+x⁰)≠0, we obtain

(1-x⁰) - 2x'·c + (1+x⁰)c² = (1+x⁰)ρ².        (6a)

We now collect the coefficients in terms linear in x, and move the rest to the right hand side, we also multiply both sides with (-1):

x⁰(1+ρ²-c²)+x'·(2c) = 1+c²-ρ².        (7)

Eq. (7) already has the form of Eq. (2), with m⁰=(1+ρ²-c²) and mⁱ=2cⁱ (i=1,2,3), but m should satisfy m²=1. Therefore we need to normalize. To this end we define

D =((1+ρ²-c²)²+4c²)^1/2,            (8)

and define

m=ε((1+ρ²-c²),2c)/D        (9a)

t= cos^-1(ε(1+c²-ρ²)/D),        (9b)

where ε=±1.

We have to decide now on the sign of  ε so that we have the correct direction of the normal. We know from Proposition 2 of Part 9 that for the normals to be correct we should have, in particular, the identity:

ρ = sin(t)/(m⁰+cos(t)).

Now, here, 

m⁰ = ε(1+ρ²-c²)/D,

t = cos^-1(ε(1+c²-ρ²)/D),

thus

m⁰+cos(t) = 2ε/D.

On the other hand,  we have the trigonometric identity sin(cos^-1(x))=(1-x²)^1/2. With x=ε(1+c²-ρ²)/D, we easily find that

sin(t)=(4ρ²)^1/2 /D=2|ρ|/D.

It follows that ρ=|ρ|/ε, and thus ε = sgn(ρ).

In this way we have arrived at the following Proposition:

Proposition 1. Let S_ρ(c) be a sphere in R³ of signed radius ρ∈R,  ρ≠0, and center c∈R³. The image of S_ρ(c) by the inverse stereographic projection is the oriented sphere S_t(m) in S³, with

m=ε((1+ρ²-c²),2c)/D

t = cos^-1(ε(1+c²-ρ²)/D),

where

D =((1+ρ²-c²)²+4c²)^1/2,

 

ε = sgn(ρ).

In the next post we will calculate the inverse stereographic image of an oriented plane in R³.