# Maths - Stereographic 2D Derivation

Here we look at a two dimensional euclidean space embedded in a three dimensional projective space, we are using the stereographic model to do this projection. For a more general discussion of stereographic projection see page here. In two dimensional projective space using stereographic model:

• Straight lines in euclidean space map to great circles (or semi circles) in projective space.
• Horizontal lines meet at (0,0,0) which is the point at infinity for horizontal lines.
• Vertical lines meet at (0,0,0) which is the point at infinity for vertical lines.
• infinity, in any direction, is represented by the same point (0,0,0).

From the above we can see that, as we move away from the origin, lines that cross at 90° in euclidean space do not cross at 90° in projective space. So angles away from the origin are not conserved by the projection.

## Derivation for translation between projective (stereographic) and euclidean spaces

As for the hemisphere model, since we are projecting using a straight line, there is a linear relationship between a point on the plane and its projection on the sphere:

 x y z
= λ
 u v 1

where:

• (x,y,z) = projective coordinates
• (u,v) = euclidean coordinates
• λ = expansion factor of vector: function of (x,y,z) or (u,v)

Dividing the top two rows by the bottom row gives the euclidean coodinates in terms of the projective coordinates:

 u v
=
 x/z y/z

To go the other way from the projective coordinates to the euclidean coodinates we need to use the information that the coordinates are constrained to a unit sphere centred at (0,0,1) so:

x² + y² + (z-1)² = 1

rearanging gives:

x²/z² + y²/z² + (z²-2z +1)/z² = 1/z²

u² + v² + 1 = 2z/z² = 2/z = 2/λ

λ = 2/(u² + v² + 1)

substituting this into the first vector equation gives:

 x y z
= 2/(u² + v² + 1)
 u v 1

## Alternative Coordinate System

So far I have chosen the origin of the x,y,z coordinate system as the projection point as this seems the simplest way to do it. However other sources (see Doran & Lasenby - Geometric Algebra for Physisits book on right of this page and also the Wikipedia page on stereographic projection) seem to use a coordinate system where the origin of the x,y,z coordinate system is at the centre of the sphere. So in order to relate to these other sources I will repeat the above with this coordinate system:

We will now have:

 x y z-1
= λ
 u v 1

So

 u v
=
 x/(z-1) y/(z-1)

since the origin is on the plane being projected and

x² + y² + z² = 1

dividing both sides by (z-1)² gives:

x²/(z-1)² + y²/(z-1)² + (z²-1)/(z-1)² = 0

u² + v² + (z +1)/(z-1) = 0

u² + v² + (λ+2)/λ= 0

u² + v² + 1 +2/λ= 0

λ= -2/(u² + v² + 1)

substituting this into the first vector equation gives:

 x y z-1
= -2/(u² + v² + 1)
 u v 1

adding 1 to both sides of row 3 gives:

 x y z
= -2/(u² + v² + 1)
 u v 1-(u² + v² + 1)/2

which is:

 x y z
= 1/(u² + v² + 1)
 2u 2v u² + v² - 1      Roger Penrose - The Road to Reality: Partly a 'popular science' book as it tries to minimise the number of equations (Not that I'm complaining much his book 'Spinors and space-time' went over my head in the first few pages) it still has lots of interesting results that its difficult to find elsewhere.       Spinors and Space-Time: Volume 1, Two-Spinor Calculus and Relativistic Fields (Cambridge Monographs on Mathematical Physics) by Roger Penrose and Wolfgang Rindler - This book is about the mathematics of special relativity, it very quickly goes over my head by I hope I will understand it one day.