Imagine we live on a surface of a sphere. The sphere is really huge, on our human scale we do not feel its curvature. We think it is flat. Once our universe is flat - we use Cartesian coordinates. But those who have the ability to travel around the world find that using the Cartesian coordinates is deceptive. Thus a dictionaries have been created that allow the communication between those who travel and those who do not move beyond their villages.
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| ... on our human scale we do not feel its curvature |
One of these dictionaries is called "stereographic projection". It is a tool, a technical tool, but once we decided to play with spheres - it is necessary. So, this post is about stereographic projection, and it will cost us very little to discuss an n-dimensional unit sphere Sn in (n+1)-dimensional Euclidean space Rn. We will use the standard basis e0,e1,....,en, and coordinates x0,x1,...,xn.
The equation of Sn in Rn+1 is
(x0)2 + ... (xn)2 = 1. (0)
For the origin of the projection we choose the point with coordinates x0 = -1, x1 = ... = xn = 0, i.e the point -e0. We will project on the hyperplane x0 = 0. Let x be a point on Sn, x ≠ -e0. Thus x0 ≠ -1. Let y = y1e1+...ynen be the corresponding point on the hyperplane x0 = 0.
The line connecting -e0 to x is
y(λ)=(1-λ)(-1,0,...,0) + λ(x0,x1,...,xn) = (λ-1+λx0,λx1,....,λxn). (1)
We want the zero coordinate of y to be zero, thus 0=λ-1+λx0, thus λ = 1/(1+x0). We have no problems with the denominator since, by assumption, 1+x0>0. (Why?)
Setting y = y(λ) for this particular value of λ, from (1) we get
yi = xi/(1+x0), (i=1,...,n). (2)
Let us now calculate, using (0) and (1), the inverse transformation: given y = (0,y1,...,yn) calculate x=(x0,x1,...,xn) on Sn. From (2) we have
From which we get
x0 = (1-|y|2)/(1+|y|2). (3a)
Therefore
1+x0 = 2/(1+|y|2), and (2) gives us
xi = 2yi/(1+|y|2), (i=1,...,n). (3b).
The formulas (3a),(3b) give us the desired inverse transformation.
Let us now set n=3 and go back to our Lie spheres. Stereographic projection will map spheres on S3 into spheres in R3. Well, unless the point -e0 is not on the sphere. If it is, stereographic projection will produce an infinite plane instead of a sphere.
We have characterized spheres on S3 by (m,t) - where m is its axis and t is its arc
radius in [0,2π). A sphere in R3 , on the other hand, is parametrized by
its center and its Euclidean radius. We need to find the relation
between these different parametrizations. This we will do in the next
post.
P.S. 20-04-25 7:18
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(n+1)-dimensional Euclidean space Rn ->
ReplyDelete(n+1)-dimensional Euclidean space Rn+1
(x0)2 + ... (xn)2 = 1. ->
(x0)2 + ... (xn)2 = 1. (0)
We will project on the hyperplane x0 = 0. Let x point on Sn, x ≠ -e0. Thus x0 ≠ -1. Let ->
Bold?
Let x point ->
Let x be a point
y(λ)=λ(-1,0,...,0) + (1-λ)(x0,x1,...,xn) ->
y(λ)=(λ-1)(-1,0,...,0) + λ(x0,x1,...,xn) and following, would be more elegant.
Let us now calculate, using (1) ->
Let us now calculate, using (0)
Thank you!
DeleteThen 1-λ = 1/(1+x0). ->
DeleteShould also be deleted now.
Ouch! Thanks.
Delete"Happy Easter Sunday 2025"
ReplyDeleteArk, thank you very much!
We went slope-skiing in Kirovsk for a week and celebrated Easter with a lot of frost, sun and snow!
By the way, skiing down snow tubes, I felt like an electromagnetic wave, with its periodic fluctuations in speed and position. It's a very natural and pleasant feeling! :)
Correction: not 'fluctuations', but oscillations - well defined and perfectly harmonic! :)
Delete