## Simple question about definition of tangent bundle

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 $$p \in M$$ by $$T_p M$$. Is the definition of the tangent bundle

$$TM = \lbrace (p, T_p M)|p \in M \rbrace$$

or is it

$$TM = \lbrace (p, V)|p \in M , V \in T_p M\rbrace$$?

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.

 PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug

Recognitions:
Gold Member
Staff Emeritus
 Quote by pellman 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')

 $$TM = \lbrace (p, T_p M)|p \in M \rbrace$$
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!

 $$TM = \lbrace (p, V)|p \in M , V \in T_p M\rbrace$$?
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)

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

Recognitions:
Gold Member
Staff Emeritus

## Simple question about definition of tangent bundle

 Quote by pellman 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 $$\pi^{-1}(p)=T_p M$$
He's using the "inverse image" function, and being (very slightly) liberal with equality, since with the definition you gave, the fiber should be $\{ p \} \times T_p M$.

 Ok. That gives me enough to press on. I'm sure I will get it when I see other examples. Thanks again.