Finding the Metric in Stereographic Projection

[mnb96]

Hello,
if we consider the stereographic projection $\mathcal{S}^2\rightarrow \mathbb{R}^2$ given in the form:

$$(X,Y) = \left( \frac{x}{1-z} , \frac{y}{1-z} \right)$$

how can I find the metric in X,Y coordinates?

-- Should I first express the projection in spherical coordinates, then find the inverse projection $\mathbb{R}^2\rightarrow \mathcal{S}^2$?

 

[tiny-tim]

hello mnb96!

start with spherical coordinates, and use a simplification for sin/(1 - cos)

 

[Bacle]

I think they may be referring to the Riemannian metric. If this is so, then
the R-metric is a 2-form, so you just need to use the stereo projection
to pullback a 2-form.

 

[mnb96]

oK, thanks to you both.
I tried the following:

I parametrized the S^2 sphere with coordinates $(\theta,\phi)$, then I wrote down the transformation from the S^2 sphere (with spherical coordinates) to the XY plane (with polar coordinates):

$$R = \frac{\sin\theta}{1-\cos\theta}$$

$$\alpha = \phi$$

Then I simply used the polar->cartesian transformation to obtain:

$$X(\theta,\phi) = \frac{\sin\theta \cos\phi}{1-\cos\theta}$$

$$Y(\theta,\phi) = \frac{\sin\theta \sin\phi}{1-\cos\theta}$$

At this point I was ready to compute the metric as:

$$ds^2 = dX^2 + dY^2 = \frac{1}{(1-\cos\theta)^2} \left( d\theta^2 + \sin^2 \theta \cdot d\phi^2 \right)$$

Was this correct?
@tiny-tim: I don't know what you meant by simplification for sin(x)/(1-cos(x))

 

[tiny-tim]

sinx/(1 - cosx) = 2sin(x/2)cos(x/2) / 2sin²(x/2) = cot(x/2)

 

[mnb96]

sinx/(1 - cosx) = 2sin(x/2)cos(x/2)/2sin²(x/2) = cot(x/2)

Ah, you meant that!
Thanks!

-- However, the end of the story was that we were forced to use spherical coordinates because we needed to parametrize a 2-dimensional surface with two parameters, and not three-parameters x,y,z plus a quadratic constraint.

 

[tiny-tim]

(have a theta: θ and a phi: φ )

i think you're missing the obvious …

using coordinates θ and φ, you already have X = cot(θ/2)cosφ, Y = cot(θ/2)sinφ …

so change one of the coordinates again! ​

 

[mnb96]

(have a theta: θ and a phi: φ )

i think you're missing the obvious …

using coordinates θ and φ, you already have X = cot(θ/2)cosφ, Y = cot(θ/2)sinφ …

so change one of the coordinates again! ​

Uhm...ah yes! maybe I got what you were trying to tell me :)
Darn, you were just suggesting me to use the geometrical definition of 'cotangent'. This way everything becomes almost trivial.

Actually now that I think about it, the cotangent operator itself does a stereographic projection of the 1-sphere onto the real line. Stupid me.

I think I missed it because I tried to directly work on the stereographic projection for the n-sphere.

 

[tiny-tim]

he he!