# Manifolds, unit 2-sphere and stereographic projection

I am always passing through this example while reading about manifolds that I don't quite get.

It is when describing the unit 2-sphere $S^2$ as an example of a manifold.

They say, initially it may be defined as the surface $x^2+y^2+z^2=1$ embedded in $R^3$. They add, it is common to use the usual spherical polar coordinates $\theta$, $\phi$ with $$x=\sin\theta \cos\phi, \qquad y=\sin\theta\sin\phi , \qquad z=\cos\theta.$$ Then, they say, that this is fine for some purposes, but it does not define a good coordinate chart at the poles $\theta=0, \pi$, since these points have no unique values of $\phi$.

Then there is the discussion of stereographic projection, a method to introduce coordinate charts to define a manifold structure. Authors say, there are two patches whose union covers the sphere, namely $M_1$, consisting of the sphere with south pole deleted, and $M_2$, which is the sphere with north pole deleted.

I don't get there pole deleted argument as well as the line in bold in the first paragraph, how does this come about?

So the function defined by $$f(\theta,\phi)=(\sin\theta\cos\phi,\sin\theta\sin\phi, \cos\theta)$$ is not a coordinate chart but rather an embedding $\mathbb{S}^2\hookrightarrow \mathbb{R}^3$. The reason it cannot be a coordinate chart is as follows:
A coordinate chart must be a homeomorphism from the open sets $\mathcal{U}_i$ of the manifold $M$ to $\mathbb{R}^n$, In this case $$f:\mathcal{U}\to \mathbb{R}^2$$ since we suppose there is a single chart $\mathcal{U}$ that covers $\mathbb{S}^2$. It must be a homeomorphism because a coordinate chart is something which "looks like the cartesian plane", and the key is that a homeomorphism such as this chart must be invertible. So the coordinate chart $g=p\circ f$ coming from stereographic projection is actually the embedding $f$ composed with a projection to the plane $p$ (it sounds like you can look up this function in your text if you want to know exactly what it looks like). But in any case, we can see why it won't work.
Consider the point $(\theta,\phi)=(0,\phi)$, which maps to $$(x,y,z)=(0,0,1)$$ under the embedding. What is the inverse function to this one? To which value of $\phi$ does the point $(0,0,1)$ map? Since $$f(0,\phi)=(0,0,1)$$ for any value of $\phi$, $f$ is not invertible so the coordinate chart $g$ cannot be either. You need two coordinate charts, each one covering one of the poles of $\mathbb{S}^2$.
• Okay, can you explain how the need for only two coordinate charts can solve this problem? Do you mean because $f(0,\phi)=(0,0,1)$ only for two values of $\phi$ did we choose two coordinate charts and not more? – Beyond-formulas May 21 '15 at 15:33
• I hope my edit(s) cleared it up. You remove one point (say, $\theta=0$) from one coordinate chart, and you remove the other ($\theta=\pi$) from the second coordinate chart. Now you have two charts which are everywhere invertible and every point ${(\theta,\phi)}$ of the sphere is included in at least one of them. – levitopher May 21 '15 at 15:37
• BTW you can choose more than 2 if you like. Say you break the range of $\theta$ $[0,\pi]$ into 10 intervals $[0,\pi/5),(\pi/10,3\pi/10)...(4\pi/5,\pi]$ (notice these sets overlap - you need that for technical reasons. They can't "just touch"). Then you would have 10 strips, each with values of $\phi$ that run 0 to $2\pi$. Each point is in at least one, and they cover $\mathbb{S}^2$. – levitopher May 21 '15 at 15:45