/sci/ - Science & Math » Thread #12812384

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Anonymous Tue 09 Mar 20:59:54 2021 No.12812384 View Reply Original Report

Quoted By: >>12812404 >>12812426 >>12812456 >>12812524 >>12813412

pls explain what a fiber bundle is

Anonymous

Anonymous Tue 09 Mar 2021 21:06:52 No.12812404 Report

Quoted By: >>12812417

>>12812384
It's a bundle of fibers, duh

Anonymous

Anonymous Tue 09 Mar 2021 21:10:21 No.12812417 Report

Quoted By: >>12813558

>>12812404
But I'm told the fiber bundle is the /map/. Fibers are the individual elements being mapped. I don't understand what I am working with when I try to prove something is a fiber bundle. The information I'm being given and what I'm supposed to show with it is confusing me.

Anonymous

Anonymous Tue 09 Mar 2021 21:12:42 No.12812426 Report

Quoted By: >>12812451

>>12812384
https://www.youtube.com/watch?v=2xiEEtoa-_4

Anonymous

Anonymous Tue 09 Mar 2021 21:19:27 No.12812451 Report

Quoted By:

>>12812426

.-. I think I would need to understand fiber bundles if I want to understand gauge theory. The canonical example I've been given is the cylinder over the circle. However when I look at that example, I'm not sure how I pass from one fiber to another in the base space

Anonymous

Anonymous Tue 09 Mar 2021 21:20:13 No.12812456 Report

Quoted By: >>12812472

>>12812384
Do you know what a fiber is?

Anonymous

Anonymous Tue 09 Mar 2021 21:23:35 No.12812472 Report

Quoted By:

>>12812456
I understand it to be the preimage of an element in my target space. I.e. given p \in, Y f:X -> Y. The fiber is f^-1(p).

Anonymous

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Anonymous Tue 09 Mar 2021 21:39:35 No.12812524 Report

Quoted By: >>12812536 >>12812573 >>12812738

>>12812384
the idea is that you attach the typical fibre F to every point of your space B. and you want to do it in such a way that if the point x in B varies continuously, then the fibre F_x at x also varies continuously (whatever that means).
mathematically this is implemented by organizing all fibres at distinct points into a single topological space E. the information about which part of E is a fibre of some specific point is organized in the projection E -> B: the fibre F_x at x is precisely the preimage of x.

Anonymous

Anonymous Tue 09 Mar 2021 21:41:15 No.12812536 Report

Quoted By:

>>12812524
>organized
should have been "encoded" or something like that

Anonymous

Anonymous Tue 09 Mar 2021 21:51:21 No.12812573 Report

Quoted By: >>12812610

>>12812524
You can basically follow a fiber back to its root, is that what it means?

Anonymous

Anonymous Tue 09 Mar 2021 22:00:47 No.12812610 Report

Quoted By: >>12812738

>>12812573
that's what the projection does, yes

Anonymous

Anonymous Tue 09 Mar 2021 22:41:12 No.12812738 Report

Quoted By: >>12813518

>>12812524

>>12812610
Isn't this the "trivial" projection? How does this compare to a nontrivial fiber bundle?

Anonymous

Anonymous Wed 10 Mar 2021 01:28:33 No.12813412 Report

Quoted By: >>12813415

>>12812384
Let M and Q be a topological spaces. A fiber bundle over M is a topological space P together with a surjective continuous map $\pi: P \to M$ such that for every x in M there is a neighborhood U of x in M and a homeomorphism $\Phi : \pi^{-1}(U) \to U \times Q$ , called a local trivialization of P over U, such that there exists a continuous map $\pi_1: U\times Q\to U$ such that $\pi|_{\pi^{-1}(U)} = \pi_1\circ_\phi$

Anonymous

Anonymous Wed 10 Mar 2021 01:29:35 No.12813415 Report

Quoted By:

>>12813412
last formula should be
$\pi|_{\pi^{-1}(U)} = \pi_1 \circ \Phi$
made a typo in the code

Anonymous

Anonymous Wed 10 Mar 2021 01:55:40 No.12813518 Report

Quoted By:

>>12812738
A trivial fiber bundle over $X$ is equivalent under homeomophism to a direct product space $X\times Y$ together with the projection $\pi(x,y)=x$ . For example, $\mathbb{R}^2$ over $\mathbb{R}$ . A non-trivial fiber bundle cannot be identified with a direct product. There is usually some form of "twisting" which prevents the space from being a direct product. For example, you could read on the tangent bundle over the sphere $S^2=\{ (x,y,z)\in \mathbb{R}^3 : x^2+y^2+z^2 = 1 \}$ . You could also try the integral submanifolds of $dy - z dx$ in $\mathbb{R}^3$ .

Anonymous

Anonymous Wed 10 Mar 2021 02:05:30 No.12813558 Report

Quoted By:

>>12812417
>But I'm told the fiber bundle is the /map/.
The fiber bundle is a topological space. The surjective map is part of the definition of the fiber bundle but it isn't the fiber bundle.
>Fibers are the individual elements being mapped.
Points in the topological space are being mapped to the image of the surjective map.
>The information I'm being given and what I'm supposed to show with it is confusing me.
post the problem

Anonymous

Anonymous Wed 10 Mar 2021 09:57:45 No.12814737 Report

Quoted By: >>12814766

pls explain what a differential form is
in calculus we often talk about "differentials" of the form dx in a not very precise manner - what do they mean precisely? the differential of x as a map? also what is the wedge product?

Anonymous

Anonymous Wed 10 Mar 2021 10:14:50 No.12814766 Report

Quoted By:

>>12814737
differential form is an alternating multi-linear function on tangent vectors
differential of a function is a particular example: its value on a tangent vector is the directional derivative of the function in the direction of the vector. this assignment is linear, that's definition of differentiability really.
the idea is that the value of the form on X_1,..X_k is some sort of signed "weighted" area of the parallelepiped spanned by X_1,..,X_k. integral of the differential form is sum of these.

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