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Simple question about definition of tangent bundle

by pellman
Tags: bundle, definition, simple, tangent
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pellman
#1
Nov16-08, 02:20 PM
P: 582
So I'm trying to learn about fibre bundles and I am looking at the example of a tangent bundle.

Given a differentiable manifold M. Denote the tangent space at [tex]p \in M[/tex] by [tex]T_p M[/tex]. Is the definition of the tangent bundle

[tex]TM = \lbrace (p, T_p M)|p \in M \rbrace[/tex]

or is it

[tex]TM = \lbrace (p, V)|p \in M , V \in T_p M\rbrace[/tex]?


Maybe I'm splitting hairs but there should be standard definition of one or the other, right?

I can discuss further why I think it matters but first let's just see if anyone is certain about the answer.
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Hurkyl
#2
Nov16-08, 03:03 PM
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Quote Quote by pellman View Post
Is the definition of the tangent bundle
I feel I should point out that the definition of a bundle over M is a continuous map of topological spaces with codomain M. In other words, you need to specify:

1. A topological space E, which consists of
1a. A set of points |E|
1b. A topology on |E|
2. A continuous function E --> M (often called the 'projection map', or the 'structure map')


[tex]TM = \lbrace (p, T_p M)|p \in M \rbrace[/tex]
Assuming you use the obvious projection map, this is a very boring bundle: the projection is bijective! And if you include the local triviality condition, the projection is actually a homeomorphism!

[tex]TM = \lbrace (p, V)|p \in M , V \in T_p M\rbrace[/tex]?
Assuming you use the obvious projection map and choose the appropriate topology, this is indeed a tangent bundle. (There are many tangent bundles; they're just all isomorphic)
pellman
#3
Nov16-08, 03:24 PM
P: 582
Hurkyl, you da man. Thanks for the quick response.

So bijective is bad? That's part of what I don't get. I'm following Nakahara. You can see the page I am on here http://books.google.com/books?id=cH-...um=1&ct=result

So when he says [tex]\pi^{-1}(p)=T_p M[/tex] he's being very loose with the inverse notation, right? [tex]\pi^{-1}[/tex] doesn't really exist, since [tex]\pi((p,V))=p[/tex] for every [tex]V \in T_p M[/tex]?

Hurkyl
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Nov16-08, 04:11 PM
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Simple question about definition of tangent bundle

Quote Quote by pellman View Post
So bijective is bad? That's part of what I don't get.
It would be -- roughly speaking such a bundle has only one section. If it were the tangent bundle, that would mean that there is exactly one vector field.

So when he says [tex]\pi^{-1}(p)=T_p M[/tex]
He's using the "inverse image" function, and being (very slightly) liberal with equality, since with the definition you gave, the fiber should be [itex]\{ p \} \times T_p M[/itex].
pellman
#5
Nov16-08, 05:07 PM
P: 582
Ok. That gives me enough to press on. I'm sure I will get it when I see other examples. Thanks again.


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