Simple question about definition of tangent bundle

1. Nov 16, 2008

pellman

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.

2. Nov 16, 2008

Hurkyl

Staff Emeritus
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')

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!

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)

3. Nov 16, 2008

pellman

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...&hl=en&sa=X&oi=book_result&resnum=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 $$V \in T_p M$$?

4. Nov 16, 2008

Hurkyl

Staff Emeritus
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.

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

5. Nov 16, 2008

pellman

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