- #1
Ghost Repeater
- 32
- 5
My question regards the classic example of trivial vs non-trivial fiber bundles, the 'cylinder' and the 'Mobius band.' I'm using Nakahara's 'Geometry, Topology, and Physics', specifically Example 9.1.
Here's the text:
i) The construction of the problem confuses me a little, because how can two open sets A and B make an intersection? Why did Nakahara take the intersection of U1 and U2 and split it up this way?
ii) How do we know that we have these two choices on B? It seems like it's supposed to be self-evident, but it isn't to me. Why do we have these two choices on B but not on A?
Here's the text:
Then he says (and this is the bit where he loses me):"Let E be a fibre bundle E → S1 with a typical fibre F = [-1, 1]. Let U1 = (0, 2π) and U2 = (-π,π) be an open covering of S1 and let A = (0,π) and B = (-π,π) be the intersection U1 ∩ U2. The local trivializations φ1 and φ2 are given by
φ1-1(u) = (θ, t), φ2-1(u) = (θ, t)
for θ ∈ A and t ∈ F. The transition function t12(θ), θ ∈ A, is the identity map t12(θ): t → t."
My questions are:'We have two choices on B:
I) φ1-1(u) = (θ, t), φ2-1(u)=(θ, t)
II) φ1-1(u)=(θ, t), φ2-1(u) = (θ, -t)"
i) The construction of the problem confuses me a little, because how can two open sets A and B make an intersection? Why did Nakahara take the intersection of U1 and U2 and split it up this way?
ii) How do we know that we have these two choices on B? It seems like it's supposed to be self-evident, but it isn't to me. Why do we have these two choices on B but not on A?