TITLE: On the Use of the Topology on Tangent Bundles
QUESTION [6 upvotes]: On learning about how to define smooth vector fields on a manifold $M$, I learned that one should first define a tangent bundle , $T(M)$,  as $\cup T_p(M)$ together with a topology(smooth structure). And then, a smooth vector field $X$ would be a smooth map from $M \rightarrow T(M)$ s.t. $\pi \circ X=id_M$ 
However, it is obvious that the use of the auxilary manifold $T(M)$ should not be confined to merely offering a good def. of "smooth" vector fields. Well, at least you would not be using the global topology of $T(M)$. (Since the checking of "smoothness" is done locally)
I am inclined to believe that the global topology of $T(M)$ may play a more important role in determining the properties of vectorfields on $M$. So here are my two questions: 

Can you give an example where the global topology of $T(M)$ is used to control certain properties(possibly about vec. fields) on $M$?
If the global topology on $T(M)$ is indeed important, how do we set about determining it?  I have only seen trivial examples where for $M=S^1 or \mathbb{R}^n$, $T(M)=M\times \mathbb{R}^n$. But surely, there are examples where $T(M)\neq M\times \mathbb{R}^n$. And how do we determine the topology in that case?

REPLY [8 votes]: Every manifold with trivial tangent bundle is orientable: this is immediate from the definition.
Hence every  non-orientable  manifold is an example of a manifold with non-trivial tangent bundle.
Easy examples are the Möbius band, the real projective plane $\mathbb P^2(\mathbb R)$ and the Klein bottle.
The point of view of orientability has the advantage that it is completely elementary and self-contained: no input from topology id s needed.  
(This is definitely not meant as a criticism of Paul's and Zhen's great answers which I have just upvoted: the theorems they allude to are quite interesting and should eventually be learned too)   
Edit Let me show you how easy it is to prove non-orientability!  
1) A useful remark
If $M$ is orientable and if $(U,x)$ and $(V,y)$ are two charts with connected domains $U,V$ , then the change of coordinates $\phi =y\circ x^{-1}: x(U\cap V) \to y(U\cap V)$ has a jacobian $Jac(\phi) =det (\frac {\partial y_i}{\partial x_j})$ which does not change sign on  $x(U\cap V)\subset \mathbb R^n$.
[This is remarkable because $U\cap V$ and $x(U\cap V)$  need not be connected even though $U$ and $V$ are]   
2) Application: $\mathbb P^2(\mathbb R)$ is not orientable
Consider the two charts $x :U\to \mathbb R^2:[1:v:w]\mapsto (v,w)$ and $y :V\to \mathbb R^2:[u:1:w]\mapsto (u,w)$ , where the domain $U$ (resp. $V$) is  the set of $[u:v:w]\in \mathbb P^2(\mathbb R)$ with $u\neq 0$ (resp. $v\neq 0$).
 The change of coordinates is the diffeomorphism
 $$\phi=y\circ x^{-1}: \mathbb R^*\times \mathbb R \to \mathbb R^*\times \mathbb R:(u,v)\mapsto  (\frac {1}{u},\frac {v}{u})$$
whose jacobian $(Jac (\phi))(u,v)= -\frac {1}{u^3}$ does change sign on its (disconnected!) domain $\mathbb R^*\times \mathbb R$.
Hence $\mathbb P^2(\mathbb R)$ is not orientable according to the useful remark above.