1

I am trying to find the map:

$$\psi_2(u,w,v):\mathbb{S}^2\smallsetminus\{0,0,-1\}\rightarrow\mathbb{C}$$

So this is what I have done, let $(u,v,w)$ be a point on the sphere, and let $(0,0,-1)$ be the south pole. The line between these two points can be given by:

$$\mathbf{r}(t)=(u,v,w)t+(1-t)(0,0,-1)$$ $$=(tu,tv,tw-1+t)$$ Now if I want to find the point on the plane $z=0$ I have to have: $$tw-1+t=0$$ $$\Rightarrow t=\frac{1}{1+w}$$ Which then gives: $$(x,y)=(\frac{u}{1+w},\frac{v}{1+w})$$ Which can be associated with the complex number: $$z=\frac{u}{1+w}+i\frac{v}{1+w}$$ However wikipedia says that his map should be: $$\psi(u,v,w)=\frac{u}{1+w}-i\frac{v}{1+w}$$ And I can't figure out where I went wrong. When I did the same thing for the northpole projection I got exactly the result that wikipedia got. So where am I going wrong?

Chris
  • 5,420
  • You're right. The stereographic projection of $(u, v, w)$ should be $\frac{u}{1 + w} + i \frac{v}{1 + w}$, from the south pole. The sign change of $v$ makes no sense geometrically speaking. – Theo Bendit Sep 16 '21 at 22:59
  • I’m not right though because you need the minus sign to make the diffeo between $\mathbb{CP}^1\rightarrow\mathbb{S}^2$ to work. Wikipedia says something about an orientation switch to make sure the sphere is adequately covered by the 2 projections but that doesn’t really make sense to me – Chris Sep 16 '21 at 23:40
  • Yes, Wikipedia is right about the orientation switch. Let me try to explain why. Let us say that the sphere is oriented with a normal unit vector field pointing inwards. If you do a stereographic projection from the North pole onto say the equatorial plane, then this mapping is orientation-preserving if you orient the equatorial plane with a normal unit vector field pointing in the positive z-direction. – Malkoun Sep 17 '21 at 00:42
  • On the other hand, if you do a stereographic projection from the South pole, and assuming you are using an inward normal unit vector field on the sphere again, then this mapping is orientation-preserving if you use the orientation given by the unit normal vector field on the equatorial plane pointing in the negative z-direction. So if you wish to use a chart which is compatible with the orientation, you need to use opposite orientations on the equatorial plane for the 2 stereographic projections. – Malkoun Sep 17 '21 at 00:46
  • @Christopher Oh I see. I'm too used to working with stereographic projection in the real context. – Theo Bendit Sep 17 '21 at 03:20

1 Answers1

2

You haven't gone wrong, calculation-wise. The problem, as you've identified, is that the two stereographic projection mappings from the north and south poles are not holomorphically compatible; their composition is inversion in the unit circle, $$ z \mapsto \frac{z}{|z|^{2}} = \frac{1}{\bar{z}}, $$ rather than the desired complex reciprocal map. If we take stereographic projection from the north pole as one local holomorphic chart, then the complex conjugate of stereographic projection from the south pole is holomorphically compatible, however.

Andrew D. Hwang
  • 82,053