Product space vs fiber bundle

[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...

 

[fresh_42]

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 $p(b,f)=b$.

 

[cianfa72]

$\{\,(b,F)\,|\,b\in B\,\}$

That is just a notation where the set $F$ itself appears instead of its elements, I guess

$\{\,b\,|\,b\in B\,\} \times F$

That is basically the union of sets of type $\{\,b\} \times F$ with $b \in B$

$\{\,b\,\} \times F \cong F$

that means an identification (isomorphism ?) with $F$ right ?

 

[fresh_42]

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.

 

[cianfa72]

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 ?

 

[zinq]

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.

 

[cianfa72]

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.

I'm not sure to grasp how this is related to my question in post #5

 

[zinq]

"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)).

 

[cianfa72]

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...

 

[fresh_42]

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.

 

[cianfa72]

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.

 

[fresh_42]

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 $Pr_2(a,b,c,d)=(0,b,c,d)$.

 

[cianfa72]

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 ?

 

[fresh_42]

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.

 

[cianfa72]

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

 

[mathwonk]

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.

 

[fresh_42]

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 ...

 

[cianfa72]

Sorry for the confusion: maybe my fault is due to a notation issue.

What I understand through this thread is that it should be better to use the notation A⁴ just as a short for A¹x A¹x A¹x A¹ while for a generic affine space of dimension 4 use the letter A this time dropping the superscript to denote its dimension