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.
Ark, I am deeply impressed by the illustration and sympathize very much to the Sisyphus labour you have to perform... But wouldn't it be easier to do all the calculation for the case c=0? The coordinates of center should not affect the orientation aspect, right?
ReplyDeleteBut that would be relaying on intuitive properties that are true, but not yet mathematically proven. Soon we will discuss the symmetry group, and we will come in particular to translations, and how they act on R^3 and on S^3 being a part of the conformal group SO(4,2).
Delete"The coordinates of center should not affect the orientation aspect, right?"
That is true, but it needs a mathematical framework and a proof.
And it is also good to know that stereographic projection while it maps spheres into spheres, it does not map centers into centers.
Ok, Ark, thank you for the comprehensive answer. Apparently I was using too loosely the phrase from Part 9: "If p is arbitrary, not necessarily the origin of R3, the situation is the same, just translated by p".
DeleteIndeed, this was in the context of "To understand what is going on...", to get an intuitive feeling, not in the course of a rigorous mathematical reasoning.
DeleteSometimes our intuitive feelings are very hard to prove (Fermat's Last Theorem, Riemann Hypothesis), or impossible to prove (Axiom of Choice, equivalent to Zorn's Lemma).
DeleteAnd sometimes our intuitions proved to be wrong. A classic example is the belief in Euclidean geometry as the only true geometry. For centuries, mathematicians assumed Euclid's fifth postulate (the parallel postulate) was self-evidently true and universally applicable.
DeleteA less well-known example is the belief that the regular tetrahedron tiles space. Aristotle, in the 4th century BCE, stated that regular tetrahedra could perfectly fill space without gaps, a claim that was accepted and repeated in commentaries for nearly 1,800 years.
DeleteSo interesting, didn't know about tetrahedron. I have a little book "Intuition and mathematics". Will have a look at it, probably find some else nontrivial examples.
DeleteI believe that intuition can be trained, but it requires even more effort than training brain and body.
However their ->
ReplyDeleteHowever there
m=ε((1+ρ2-c2),c)/D (9a) ->
m=ε((1+ρ2-c2),2c)/D (9a) (and below)
In thi way ->
In this way
be s sphere ->
be a sphere
Thanks!
DeleteConstraints of c and ρ ?
DeleteOK. Added to Proposition 1. Thanks.
Delete