Visualizing the Topology of CP^1 = S^2

[quasar987]

Like the title says, what is the easiest way to see that CP^1 is topologically just a 2-sphere? Wikipedia says that CP^1 can be identified with C u {oo} (the 1-point compactification of C) but I don't see it.

 

[arkajad]

Parametrize Hermitian idempotant 2x2 complex matrices of trace 1. Every Hermitian 2x2 matrix can be represented as a real linear combination of Pauli matrices (of trace 0) and identity (of trace 2). Write down the condition that it is an idempotent: P=P*=P^2, tr(P)=1.

 

[quasar987]

You call that an easy visualization?

 

[lavinia]

the unit 3 sphere is all complex number pairs (z,w) with |z|^2 + |w|^2 = 1.

If w is not zero then the ratio z/w is a well defined element of the complex plane. If w = 0 then map z/w to the point at infinity on the Riemann sphere. It is easy to check that this map is continuous. The set of pairs (z,w) that map to the same point on the Riemann sphere define a circle and are all of the form exp(itheta)(z,w). Thus the map of the 3 sphere onto the Riemann sphere map circles lying in planes through the origin to the same point and so maps the 3 sphere onto the complex projective line.

A good exercise is to show that if you identify all (z,w) pairs by the antipodal map to form real projective 3 space, then you actually get the tangent circle bundle of the 2 sphere.

 

[quasar987]

Thanks lavinia.

 

[adriank]

Can't you just show that the map f: C ∪ {∞} -> CP¹, defined by f(z) = (1:z) for z in C and f(∞) = (0:1), is a homeomorphism?

 

[quasar987]

This is essentially the same thing as what lavinia said.

What I hadn't realized is that CP^1 \ {[z:w] | z≠0} is a singleton.

 

[arkajad]

You call that an easy visualization?

Oh, I did not realize that $$x^2+y^2+z^2=1$$ is too difficult. Sorry.