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Let $\pi: S^2-\{N\}\to \mathbb R^2$ be the stereographic projection map.

Let $\sigma:\mathbb R^2\to S^2-\{N\}$ be its inverse.

Let $p\in S^2-\{N\}$ and $x_1,x_2\in$ the tangent space of $S^2$

Would someone be nice enough to explain to me then what the following mean intuitively? And hopefully also a way to visualize them/gain some sort of physical intuition on them?

1) $(d\sigma)_{\pi(p)}$

2) $d\pi_p$

3) Why $x_1\cdot x_2=(d\pi_p(x_1),d\pi_p(x_2))_{\pi(p)}$

4) $d\pi_p\circ (d\sigma)_{\pi(p)}=$ identity

I have read the differential forms article on Wikipedia in hope to learn more, but I still don't quite get the idea. I know for example that (1) is the differential of $\sigma$ at the point $\pi(p)$ but I don't understand what that means. I hope that someone could give me a geometric picture of some kind. And if there should be such a saint out there, I would like to thank you very much (in advance).

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I will explain answer only geometrically as required:

What does $\pi $ does: $\pi $ takes (great circle- N) to line passes through origin. $\sigma $ take line passes through origin to (great circle -N) in a smooth manner.

Now for $p\in (S^2- N)$, and $v\in T_p (S^2-N)$, there is a great circle $\gamma$ which passes through $p$, $\gamma(0)= p$ and $\gamma'(0)= v$. Via map $\pi$, $\gamma$ will be map to some line $l$ passes through origin. $d\pi_p$ maps vector $v$ to speed of $l$ at $\pi(p)$.

And conversely $d\sigma_{\pi(p)}$ maps some speed vector of particular line passing through origin to the speed of the corresponding great circle at $p$.

Now in your 3rd question you are defining metric on $S^2$, This matrix is induced metric... Actually this expression just says that map $\pi$ is conformal map. Angle between two line passes through origin is same as angle between their corresponding image that is great circles.

As $\pi$ and $\sigma$ are inverse of each other hence we have $\pi \circ\sigma = Id$. Now take the derivative and use composition rule of rule of the derivative you will get your 4th one... Geometrically it says that Tangent space at $p$ is isomorphic to $\mathbb R^2$

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Thank you very much, Pradip. May I ask what do you mean by the speed of a line or of a great circle? – small potato Feb 22 '12 at 14:18
If $\gamma(t) $ is any curve either line or great circle then speed at $t$ is given by $\gamma'(t)=\frac{d}{dt}\gamma(t)|_{t=t_0}$ – zapkm Feb 22 '12 at 15:06
@smallpotato, take $\gamma(t)= 2t$ , then this is line passes through origin with constant speed $\gamma'(t)= 2$. – zapkm Feb 22 '12 at 15:44

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