Inverse of the Stereographic Projection

[CMoore]

In any book on differentiable manifolds, the stereographic projection map P from the n-Sphere to the (n-1)-plane is discussed as part of an example of how one might cover a sphere with an atlas. This is usually followed by a comment such as "it is obvious" or "it can be shown" that the inverse projection P^{-1} is given by such and such. Now, it is not at all "obvious" to me algebraically how this can be accomplished. The inverse function theorem of couse can guarantee that the inverse projection exists but, being an existential theorem, provides no means for finding the inverse. How does one approach a problem such as this? Apparently, the geometry of the situation can be used to derive the inverse but I'm lousy at geometry so I'm not sure how to approach it from this angle (no pun intended) either. Thank you for any suggestions.

 

[Cexy]

I'll do it in three dimensions, it's pretty obvious how it generalizes to n dimensions. Let's have the co-ordinates on the sphere be (x,y,z) with x^2 + y^2 + z^2=1. Then we know that stereographic projection onto the xy plane takes:

(x,y,z) --> (X,Y,Z) = (x/(1-z), y/(1-z),0)

If we square X and Y and add them together, we find that

X^2 + Y^2 = (x^2 + y^2)/(1-z)^2 = (1+z)/(1-z)

We can rearrange this to give us:

z = (R^2 - 1)/(R^2 + 1)

Where I've written R^2 to mean X^2 + Y^2.

Then we know that

x = X(1-z)

And we just plus in our value for z to give us

x = 2X/(R^2 + 1)

And do a similar thing for Y, to end up with the inverse of stereographic projection, which takes

(X,Y,0) -> (x,y,z) = (2X/(R^2 + 1), 2Y/(R^2 + 1), (R^2 - 1)/(R^2 + 1))

Hope that helps... sorry for the awful notation, I couldn't be bothered to use tex!

 

[CMoore]

Thank you for your response; it was very helpful. Every time I tried the problem in the past I would end up with a jumbled up mess. I approached the it by attempting to apply the definition of the inverse mapping directly, i.e. by setting the composition of p and p^{-1} to the identity map. The key, as indicated by your solution, was the initial substituion + squaring; everything from that point was simple algebra...Thanks again!

 

[Cexy]

I have a stereographic projection question for anyone who can help!

My lecture notes have an aside about stereographic projection. Definining X to be the angle between the centre of the sphere and the point in R^2, we get that the angle at the north pole is X/2, and some simple trig gives:

r = 2 tan (X/2)

where r is the radial co-ordinate on the plane (assuming the sphere is a unit sphere). My lecturer then goes on to say (repeatedly) that

sin X = 2r/(4+r²)

Now when I work through it, there should be a 4r in the numerator, not a 2r. I'd just assume that my lecturer has made a mistake, except that he's said it repeatedly in his lecture, and in his printed notes, and no one's picked him up on it. Am I going mad!?

 

[George Jones]

You've lost me here.

If X/2 is the angle at the north pole, then I agree that r = 2 tan (X/2), and that sin X = 4r/(4+r^2), but what is X?

I'm afraid that I'm not familiar enough with this stuff to understand what "Definining X to be the angle between the centre of the sphere and the point in R^2" means. Could you be a bit more explicit?

Call the north pole N, the south pole S, the centre of the sphere C, the point on the sphere P, and the point in the plane P'. The only thing I could think of is that X is the angle SCP', but this can't be true.

Regards,
George

 

[Cexy]

You are, of course, right. Now I just have to figure out why my lecturer thinks that SCP' = 2 NCP'.

Note to self: Don't believe everything that the lecturer tells you in future!