# Question About Elementary Fiber Bundle Example

• A
• Ghost Repeater
In summary: U1 and U2...into two parts: one where t_{12}(θ) = t_{12}(\theta) and one where t_{12}(\theta) = -t_{12}(θ). In summary, the two choices for B in Nakahara's example are a cylinder (no twist) or a Mobius strip (with a twist).
Ghost Repeater
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:

"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."
Then he says (and this is the bit where he loses me):

'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)"
My questions are:

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?

I think he actually means B = (-π,0). There are two intersections of U1 and U2. Think of it like a strip of paper and you are gluing the two ends together. There are two possibilities. In Case I you glue the two ends without a twist, in which case you have a cylinder. This is a trivial bundle, and it can be covered with a single chart. In Case II you glue the two ends together with a twist, reversing the orientation of the two ends. This is a Mobius strip. In this case there is no universal covering of the strip. You need two different overlapping charts to cover the whole bundle. I've tried to illustrate with the figure below.

#### Attachments

• Mobius.png
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phyzguy said:
I think he actually means B = (-π,0). There are two intersections of U1 and U2. Think of it like a strip of paper and you are gluing the two ends together. There are two possibilities. In Case I you glue the two ends without a twist, in which case you have a cylinder. This is a trivial bundle, and it can be covered with a single chart. In Case II you glue the two ends together with a twist, reversing the orientation of the two ends. This is a Mobius strip. In this case there is no universal covering of the strip. You need two different overlapping charts to cover the whole bundle. I've tried to illustrate with the figure below.

View attachment 221290

However, I am still a bit perplexed by these objects A and B. Mathematically, what are they? And exactly what is the difference between B and A that allows t to be mapped to its inverse, whereas this is not allowed on A?

Is A one intersection of U1 and U2, while B is the other? So they are sets? And so you're saying that on set A we map t to t, but once we 'reach' set B we have the option of mapping t to t OR mapping t to -t? And if we choose the first option we get a 'cylinder' (no twist), while if we choose the second, we get a 'mobius band' (twist)?

Ghost Repeater said:
Is A one intersection of U1 and U2, while B is the other? So they are sets? And so you're saying that on set A we map t to t, but once we 'reach' set B we have the option of mapping t to t OR mapping t to -t? And if we choose the first option we get a 'cylinder' (no twist), while if we choose the second, we get a 'mobius band' (twist)?

Yes, that is how I understand it. So in the Mobius strip case, in region A the transition function is $t_{12}(\theta):t \rightarrow t$, while in region B it is $t_{12}(\theta):t \rightarrow -t$

I just want to make sure I understand how the various parts of the 'machinery' work together. We have two coordinate patches, U1 and U2, on the base manifold. On each of these we have a local trivialization. In any region where U1 and U2 intersect, we have a transition function, which takes us from the local trivialization on U1 to the one on U2 (or the other way round).

Then we can partition the region of overlap between U1 and U2 into two regions, A and B. On A we define a transition function which maps fiber element t to itself, and then on B we can define either a) a transition function which maps fiber element t to itself or b) a transition function which maps fiber element t to -t. If we define the former on B, we get a trivial 'cylinder' bundle, whereas if we define the latter, we get a non-trivial 'Mobius band' bundle.

phyzguy said:
Yes, that is how I understand it. So in the Mobius strip case, in region A the transition function is $t_{12}(\theta):t \rightarrow t$, while in region B it is $t_{12}(\theta):t \rightarrow -t$
Isn't it the other way round? For the Mobius band you twistto glue t with -t , and, if no twisting happens, you get a cylinder? Or did I misread?

"There are two intersections of U1 and U2."

Actually, there is a continuum (an uncountable infinity) of intersections. On the other hand, the intersection set (there is just one of these) has two connected components.

## 1. What is an elementary fiber bundle?

An elementary fiber bundle is a mathematical construct that describes the relationship between a base space and a fiber space. It consists of a base space, a fiber space, and a projection map that maps each point in the base space to a fiber in the fiber space.

## 2. Can you provide an example of an elementary fiber bundle?

One example of an elementary fiber bundle is the Möbius strip, which consists of a rectangular base space and a fiber space that is a single line segment. The projection map twists the line segment as it is mapped onto the base space, creating a one-sided surface.

## 3. What is the difference between a fiber bundle and an elementary fiber bundle?

A fiber bundle is a generalization of an elementary fiber bundle and can have more complex fibers and base spaces. In an elementary fiber bundle, both the base space and fiber space are topologically equivalent to a single point, while in a fiber bundle, they can be more complex topological spaces.

## 4. How are elementary fiber bundles used in mathematics?

Elementary fiber bundles are used in mathematics to study and understand the relationship between spaces and how they can be transformed. They are also used in physics to model physical phenomena, such as electromagnetic fields and particle interactions.

## 5. What are some real-life applications of elementary fiber bundles?

Elementary fiber bundles have various applications in physics, including quantum mechanics, general relativity, and gauge theories. They are also used in computer graphics to create visual effects and simulations. In addition, they have applications in engineering, such as in the design of complex structures and materials.

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