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
Sρ(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 St(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, St(m) denotes the set
St(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 Sρ(c) be a sphere in R3 of signed radius ρ∈R, ρ≠0, and center c∈R3. The image of Sρ(c) by the inverse stereographic projection is the oriented sphere St(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.
