18
$\begingroup$

Well, when we work with a smooth manifold $M$ we can associate with each point $p\in M$ a vector space $T_p M$ of all vectors at $p$ tangent to $M$: this is the space of linear functionals obeying liebniz rule in the algebra of germs of functions at $p$.

This definition of vector is very intuitive, since it generalizes the main property of vectors in $\mathbb{R}^n$ of producing the directional derivative. Now, then we usually say: "well, we must find a way to assemble all of the tangent spaces together to have a domain and range for the derivative", then we define $TM$ as the disjoint union of all $T_p M$ and if $\pi : TM \to M$ is the projection on the first coordinate, we want to construct a vector bundle $\pi : TM \to M$.

Now, what doesn't seem clear to me is why do we want the structure of a vector bundle. The definition of a fiber bundle is meant as I understand, to make a space that locally looks like a product space, but why do we want this? Is this because $T\mathbb{R}^n = \mathbb{R}^n \times \mathbb{R}^n$ and we want to ``copy'' this behavior locally?

Also, how do we know that there's an obstruction in general to write $TM = M \times \mathbb{R}^n$? I've seem a question like this before here and there were answers based on hairy ball theorem and so on. The point is, this result needs that we first define $TM$ as it is define. If we don't know any of these theorems, how do we know that writing $TM$ that way is not possible?

Thanks very much in advance for the aid!

$\endgroup$
2
  • 2
    $\begingroup$ Well, since $M$ is locally undistinguishable from $\Bbb R^n$, the tangent bundle of $M$, whatever that may be must be locally indistinguishable from the tangent bundle of $\Bbb R^n$ which is naturally $\Bbb R^n\times\Bbb R^n$. $\endgroup$ Sep 13, 2013 at 13:39
  • 6
    $\begingroup$ In some sense you "can" write $TM=M\times \mathbb R^n$, but that is not a complete description of $TM$ -- you also need to specify how each copy of $\mathbb R^n$ represents the tangent space at each point. There's no canonical way to make this identification, and in some cases there isn't even any continuous way to do it, such as if $M$ is a Möbius strip. So it's better to leave the correspondence abstract. $\endgroup$ Sep 13, 2013 at 13:45

1 Answer 1

4
$\begingroup$

The bundle structure on the set $TM$ explains how it is a manifold in its own right, since one can construct by the bundle's local trivialization bundle charts, which map open sets of $TU\subset TM\to U\times\mathbb{R}^m$. It is especially useful for calculating objects on the (tangent) bundle, e.g. the Levi-Civita connection, in local coordinates where one needs local trivializations.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .