# Product Space vs Fiber Bundle: Understanding the Difference

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• cianfa72
In summary, a product space can serve as an example of a trivial fiber bundle, where the base space and fiber are the same as that of the product space. However, a key difference is that a fiber bundle may not have the same structure as a product space, as it is defined by the existence of a global diffeomorphism to the product space. This means that while there may be projections on the first and second factors in a product space, they may not necessarily exist in a fiber bundle. Additionally, for any trivial fiber bundle, there exists a homeomorphism to the cartesian product space, with the fiber bundle projection being an essential part of the data.
cianfa72
TL;DR Summary
product space as example of trivial fiber bundle
Hi,

I'm not a really mathematician...I've a doubt about the difference between a trivial example of fiber bundle and the cartesian product space. Consider the product space ## B \times F ## : from sources I read it is an example of trivial fiber bundle with ##B## as base space and ##F## the fiber.

As far as I understood Fiber bundle requires fibers "attached" on base space to be actually disjoint. With that in mind should we understand (conceive) the cartesian product ## B \times F ## itself as a disjoint union where there exist for instance multiple copies of F space over B ?

hoping I was able to explain the point...

cianfa72 said:
Summary: product space as example of trivial fiber bundle

With that in mind should we understand (conceive) the cartesian product ##B \times F## itself as a disjoint union where there exist for instance multiple copies of ##F## space over ##B##?
Yes.
##B\times F = \{\,(b,f)\,|\,b\in B,f\in F\,\} = \{\,(b,F)\,|\,b\in B\,\} = \{\,b\,|\,b\in B\,\} \times F##
The fibers are ##p^{-1}(b) =\{\,b\,\} \times F \cong F## with the projection ##p(b,f)=b##.

dextercioby
fresh_42 said:
##\{\,(b,F)\,|\,b\in B\,\}##
That is just a notation where the set ##F## itself appears instead of its elements, I guess
fresh_42 said:
## \{\,b\,|\,b\in B\,\} \times F##
That is basically the union of sets of type ##\{\,b\} \times F## with ##b \in B ##
fresh_42 said:
##\{\,b\,\} \times F \cong F##
that means an identification (isomorphism ?) with ##F## right ?

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Yes, as soon as ##b## is fixed we could as well drop it. It is only a bijection. But as we talk about sets, no other structures can be expected, so a bijection is already all we can expect.

And yes, the others are just notational differences of the same set.

The crucial point is, that fiber bundles are local direct sums (in a neighborhood of ##b##), but not necessarily global.

A simple example of a non-trivial bundle is the Möbius strip. The base ##B## is here ##S^{1}## (the circle line), the fiber ##F## is a closed interval. The corresponding trivial bundle would be a cylinder from which the Möbius strip differs by twisting the fiber.

Coming back to this old thread, consider a trivial Fiber bundle. By definition we are able to find a global diffeomorphism from it to ## B \times F ## even if that's is not actually an identification though (in other words the trivial bundle and ## B \times F ## look like but are not actually the same)

For instance ## B \times F ## has a product space structure (e.g. there exist projections on the first and second factor) whereas the (trivial) fiber bundle may not have it.

Make sense ?

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I think the Möbius band (as total space) provides an excellent picture of a non-trivial fibre bundle (whose fibre is a closed interval and whose base space is the circle). Underlying this fibre bundle is the simplest non-trivial bundle, whose fibre is just two points and whose base space is the circle. The total space of this bundle is homeomorphic to a circle itself, and the bundle projection to the base space circle just wraps the total space around the base space twice, like a mapping that doubles the angle. What makes this bundle non-trivial is that if you follow the fibre once around the base space, it comes back to itself with its two points interchanged, and not by the identity mapping.

"For instance has a product space structure (e.g. there exist projections on the first and second factor) whereas the (trivial) fiber bundle may not have it."

This is not correct. For any trivial fibre bundle π : E → B, there always exists a homeomorphism

h : E → F × B

of E with the cartesian product space F × B. (The fibre bundle projection π : E → B is the most essential part of the data of a fibre bundle.)

We can in fact say more about the homeomorphism h:

If p2 : F × B → B is the projection to the second factor, then for all z ∈ E we have

π(z) = p2(h(z)).

zinq said:
If p2 : F × B → B is the projection to the second factor, then for all z ∈ E we ha π(z) = p2(h(z)).
Sure, but the point is that even for *trivial* fiber bundle, the fiber projection works as p2 : F × B → B alone.

To take another example: consider the affine space A4 and the cartesian product A1x A3 of affine spaces of dimension 1 and 3 respectively.

I believe A4 is an instance of trivial fiber bundle and in fact does exist a global homeomorphism (actually an isomorphism)

h : A4 → A1x A3

making A4 ≅ A1x A3. Nevertheless the structure of A4 as affine space is really different from the structure of cartesian product space A1x A3 I believe...

cianfa72 said:
Nevertheless the structure of ##A^4## as affine space is really different from the structure of cartesian product space ##A^1x A^3## I believe...
How are they different? By numbering? This isn't a topological feature. I do not see any differences.

fresh_42 said:
How are they different? By numbering? This isn't a topological feature. I do not see any differences.

A1x A3 has got well-defined projections on first and second factor namely Pr1 and Pr2 whereas A4 has not.

weirdoguy
cianfa72 said:
A1x A3 has got well-defined projections on first and second factor namely Pr1 and Pr2 whereas A4 has not.
Sure it has: ##Pr_1(a,b,c,d)=(a,0,0,0)## and ##Pr_2(a,b,c,d)=(0,b,c,d)##.

weirdoguy
fresh_42 said:
Sure it has: ##Pr_1(a,b,c,d)=(a,0,0,0)## and ##Pr_2(a,b,c,d)=(0,b,c,d)##.
Thus A4 and A1x A3 are just isomorphic or are really the same space ?

cianfa72 said:
Thus A4 and A1x A3 are just isomorphic or are really the same space ?
Formally: isomorphic. Practical: identical. Just try to define ##A^4## without references to combining dimensions. Or put another way: What is 4 dimensional? You automatically get the possibility to add ##4=1+3##. This isomorphism is very, very natural.

fresh_42 said:
Formally: isomorphic. Practical: identical. Just try to define ##A^4## without references to combining dimensions. Or put another way: What is 4 dimensional? You automatically get the possibility to add ##4=1+3##. This isomorphism is very, very natural.
Sorry to be pedantic: to me A4 is not defined as A1x A1x A1x A1 but it is just an affine space of dimension 4 (with its translation vector space E4).

As for example shown here we can naturally "endow" the cartesian product A1x A3 with an affine structure yielding an affine space of dimension 4. Nevertheless it's an affine space with very specific properties that let me say a "generic" affine space does not have

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This equivalence explains my favorite math joke:

Q: what do you get when you cross an elephant with a chicken?
A: the trivial elephant bundle on a chicken.

Auto-Didact and fresh_42
mathwonk said:
This equivalence explains my favorite math joke:

Q: what do you get when you cross an elephant with a chicken?
A: the trivial elephant bundle on a chicken.
The other way round would make more sense because of the sections. However, if people like elephants ...

mathwonk

## 1. What is the difference between product space and fiber bundle?

Product space and fiber bundle are two mathematical concepts used to describe the relationships between different spaces. Product space is a Cartesian product of two or more spaces, while fiber bundle is a topological space that locally looks like a product space. In other words, product space is a global description of a space, while fiber bundle is a local description.

## 2. How are product spaces and fiber bundles related?

Product spaces and fiber bundles are related in that a fiber bundle can be seen as a generalization of a product space. A fiber bundle has an additional structure that allows for a smooth transition between different product spaces, making it a more flexible and powerful mathematical tool.

## 3. What is the significance of product spaces and fiber bundles in science?

Product spaces and fiber bundles have many applications in different areas of science, including physics, engineering, and computer science. They are used to describe and model complex systems and phenomena, such as electromagnetic fields, fluid dynamics, and data structures.

## 4. Can you give an example of a product space and a fiber bundle?

An example of a product space is the Cartesian coordinates system, where the x, y, and z axes are three different spaces that can be multiplied together to describe a point in 3D space. An example of a fiber bundle is a Möbius strip, where each point on the strip can be described by a combination of two coordinates, but the strip itself has an additional twist that makes it a non-trivial bundle.

## 5. How are product spaces and fiber bundles visualized?

Product spaces and fiber bundles can be visualized using different techniques, such as graphs, diagrams, and 3D models. For product spaces, each space is represented as a separate axis, and the combination of coordinates gives a point in the space. For fiber bundles, the additional structure is usually shown as a twisting or bending of the space, which can be represented visually using different techniques.

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