Is this an Accurate General description of Line Bundles?

Question

As a newbie to vector bundles, it seems like all vector bundles I have run into ( not that many, I admit) need only two charts to be trivialized; one of these charts will contain the "trouble point" (the point that prevents the top space from being a global product), and one chart that will not contain the trouble point ( I am thinking of the Möbius band over the circle here ).

It also seems like the trivializations ultimately consist of unwrapping the twist in the trouble point in the top( again, in here, in the case of the Möbius bundle, I am thinking of the lift of (1,0) in S^1 ) , unwrapping , and flattenning.

Is this an at least informally correct generalization of vector bundles ( or, at least of plane bundles, i.e., 1-bundles? )

Thanks.

Answer 1

One way to see that a given bundle is nontrivial (or requires a lot of sets in a trivialization) is to use characteristic classes. Consider the tautological bundle over $\mathbb{RP}^n$. The total Stiefel-Whitney class of this bundle is $1+\gamma$ for $\gamma \in H^1(\mathbb{RP}^n, \mathbb{Z}/2)$ nonzero (this is one of the axioms in the axiomatic approach as in Milnor-Stasheff for instance).

I claim that this implies at least $n+1$ open sets are required for a trivialization. (This implies the assertion made in elgeorges's answer.) To see this, suppose $U_1, \dots, U_m$ is an open cover on which the tautological line bundle becomes trivial. Then by naturality of the Stiefel-Whitney classes $\gamma$ becomes trivial on $H^1(U_i)$, so lies in the image of $H^1(\mathbb{RP}^n, U_i)$. But it follows that the total cup-product $\gamma \cup \dots \cup \gamma$ ($m$ times) lies in the image of $H^1(\mathbb{RP}^n, \mathbb{RP}^n) = 0$. However, since it takes a high power to kill $\gamma$ (in fact, we know the cohomology ring is $\mathbb{Z}[\gamma]/(\gamma^{n+1})$, we find that $m \geq n+1$. (This argument for the Hopf bundle, which is a fiber bundle, is explained here.)

Answer 2

Despite appearances, you should not think that a bundle is not trivial because there is a "trouble point". For example in the simplest case, the Möbius bundle (= tautological bundle) over $S^1$, all points of $S^1$ play exactly the same role: the bundle is homogeneous and there is no "trouble point". Yet this bundle is not trivial, as you already know.

It is also false that all bundles can be trivialized by using only two charts. For example the tautological bundle on $\mathbb P^{\infty} (\mathbb R) $ cannot even be trivialized by finitely many charts!

But don't let these oddities discourage you: I'm sure you' ll soon master these subtle points...

Answer 3

No. In principal one should be able to construct a bundle (even a 1-bundle) over a space that requires arbitrarily many charts to trivialize. Rather than think of an example of how to do this, I'll explain what I think you're seeing.

There is a theorem that says any vector bundle over a contractible space is trivial. You just happen to be thinking of spaces (like $S^1$ or $S^n$ for that matter) that are contractible after removing a point. Thus any vector bundle over these spaces you can trivialize with the open set of removing a point and then for the other open set you just take any small contractible neighborhood of that point.

Answer 4

I don't think you mean to ask about any "generalization" (as in your last sentence) but rather "general description" (as in your heading), but the answer is No. More complicated vector bundles over more complicated spaces than the circle will generally not admit triviailizations with just two charts. I'm not sure what "unwrapping the twist in the trouble point" might mean in higher dimensional cases, but this doesn't seem like a generally accurate description either. How would you fit, say, the tangent bundle over a torus into that framework?

Answer 5

@ elgeorges:

What I mean by a single problem point is that we only need to do a single vertical cut on the bundle to be able to "flatten it", i.e., make it be the union of a copy of two disjoint product spaces. I don't know if my statement is too informal, or maybe even wrong, but the fact that we only need two charts for the bundle would seem to reinforce this claim that there is one main problem point. I am also referring to the fact that --please correct me if I am wrong-- that all the fibers except for one are of R with one orientation, and one fiber only has orientation opposite that of all the others; it is only at the attaching parts of the fundamental domain that the bundle becomes non-trivial. If the map y-->1-y (after assignining coordinates (0,0), (0,1),(1,0),(1,1) to the sides of the square were homotopic to the map y->y , the bundle would be trivial. Is this correct?