Visualizing the Topology of CP^1 = S^2

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In summary, the easiest way to see that CP^1 is topologically a 2-sphere is by identifying it with the 1-point compactification of C. Every Hermitian 2x2 matrix can be parametrized as a real linear combination of Pauli matrices and identity, with the condition that it is an idempotent being P=P*=P^2 and tr(P)=1. This can be visualized by mapping the unit 3-sphere to the Riemann sphere, where circles lying in planes through the origin are mapped to the same point. Another approach is to show that the map f: C ∪ {∞} -> CP1, defined by f(z) = (1:z
  • #1
quasar987
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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.
 
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  • #2
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.
 
  • #3
You call that an easy visualization? :tongue2:
 
  • #4
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.
 
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  • #5
Thanks lavinia.
 
  • #6
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?
 
  • #7
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.
 
  • #8
quasar987 said:
You call that an easy visualization? :tongue2:

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

1. What is CP^1 and S^2?

CP^1 is the complex projective line, also known as the Riemann sphere. It is a one-dimensional complex manifold that can be visualized as a sphere with each point representing a line in the complex plane. S^2 is the two-dimensional sphere, also known as the surface of a ball. It is a two-dimensional manifold that can be visualized as a round, three-dimensional object.

2. How are CP^1 and S^2 related?

CP^1 and S^2 are topologically equivalent, meaning they have the same underlying structure. They are both compact, connected, and orientable manifolds. In fact, CP^1 can be obtained by gluing two copies of S^2 together along their boundaries.

3. What is the significance of visualizing CP^1 = S^2?

Visualizing CP^1 = S^2 can help us understand the concept of complex projective spaces and their relation to spheres. It also has applications in physics, particularly in quantum mechanics, where CP^1 is used to represent the state space of a spin-1/2 particle.

4. How can CP^1 = S^2 be visualized?

CP^1 can be visualized as a sphere with each point representing a complex line passing through the origin in the complex plane. S^2 can be visualized as a sphere with each point representing a point on the surface of a ball. By identifying the points on the boundary of CP^1 with the points on the surface of S^2, we can visualize the two manifolds as one.

5. Are there any other ways to visualize CP^1 = S^2?

Yes, there are several other ways to visualize CP^1 = S^2. For example, you can think of CP^1 as a sphere with antipodal points identified, or as the set of all lines through the origin in three-dimensional space. You can also use stereographic projection to map CP^1 to the plane, which can make it easier to visualize certain properties of the manifold.

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