# what's the role of fiber bundles play in understanding the base space?

Suppose $\pi: E\to B$ be a fiber bundle, and assume $B$ is connected, I want to know:

1, What do the fiber bundles (e.g., the covering spaces) on $B$ help understand the topology, or the structure on $B$? maybe its (co-)homology groups, homotopy groups. etc.

2, Whether there are examples showing that by considering some particular fiber bundles of $B_1, B_2$, it is the easiest way to distinguish $B_1, B_2$.

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I mean, there is tons of geometry about $B$ encapsulated in the vetor bundles over $B$, for example. The phrase "K-theory" may be of particular interest to you, I suggest you google it. For a paltry example of how fiber bundles elucidate the topology of B, recall that B is contractible implies that every fiber bundle over E is trivial (assuming we're dealing with suitably nice spaces such as manifolds). I don't know enough to feel comfortable answering your question, but I think it's a great one, and I would be more than happy to put a bounty on it if it is not answered in the next few days! – Alex Youcis Apr 9 '13 at 3:59
PS, if this question is not answered in the next few days, and you would like me to take out a bounty, just @Alex Youcis me to remind me! – Alex Youcis Apr 9 '13 at 3:59
Thanks Alex, I would. – ougao Apr 9 '13 at 4:18

Understanding fiber bundles over a space gives a lot of information about that space.

The basic case of covering spaces tells you about the fundamental group. If you know the category of covering spaces over X you know the fundamental group of X. The fundamental group of X is the group of automorphism of the functor which sends a covering space to the fiber of the covering space at the base point. However, if X and Y are path-connected and have isomorphic fundamental groups, they have equivalent categories of covering spaces. Therefore the information given by covering spaces is exactly the fundamental group.

This is pretty abstract. Let me give an example. If you have a covering space $Y\to X$, pick two distinct points in the fiber over the base point of X and draw a path between these points. Its image in X is a non trivial element of the fundamental group. In practice, to find elements in the fundamental group, one needs to find covering spaces of the space which is often easier.

This description of the fundamental group in terms of covering spaces is great because it generalizes well. In algebraic geometry, it doesn't make sense to talk about curves on a variety/scheme. However, there is a good analog of covering spaces called etale maps. Using etale maps one can define what is called the etale fundamental group of a variety.

Vector bundles over a space up to isomorphisms form a monoid under the direct sum operation. If you formally add an inverse to all elements in that monoid, you get an abelian group called the 0th K-theory group $K^0(X)$. One can actually define higher K theory groups $K^n(X)$ for all $n\in\mathbb{Z}$ and it turns out that the resulting sequence of groups has formal similarities with ordinary cohomology. They send homotopy equivalence to isomorphisms, they have Mayer-Vietoris sequences. This kind of structure is called a generalized cohomology theory.

K-theory has plenty of interesting applications in homotopy theory. For instance, the best proof of the Hopf invariant one theorem uses K-theory. Another example is in the computation of stable homotopy groups of spheres. These group are very complicated but Adams was able to compute a big chunk called the image of J. This computation relies heavily on K-theory techniques.

I can't really think of a natural answer for your second question. I guess you can show that $S^1$ and $R$ are not homotopy equivalent by realizing that $S^1$ has non trivial covering spaces while $R$ does not but that's pretty boring. One example I like a lot is Milnor's proof of the existence of exotic spheres. The idea is to consider $S^3$ fiber bundles over $S^4$. Note that the seven dimensional sphere can be obtained that way (see $\mathbb{R}^8$ as $\mathbb{H}^2$ where $\mathbb{H}$ is the ring of quaternions and then consider the projection map $S^7\subset\mathbb{R}^8\to P\mathbb{H}$ the quaternionic projective space. One can show that this is a fiber bundle with fiber $S^3$). However there are other examples of such fiber bundles which are homeomorphic to the seven sphere but not diffeomorphic (Milnor construct 6 exotic spheres in that paper and it was later shown that there are 27 of them).

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@ Geoffroy, thanks for your answer, you illustrate problem 1 very well, do you have any idea on problem 2? – ougao Apr 11 '13 at 10:56
On the line of covering spaces, if a fiber bundle has discrete structure group (i.e locally constant transition functions) there is a foliation of E which is transversal to every fiber s.t every leaf L is a covering space for B. In this case there is a homomorphism $\phi: \pi_1(B,b) \rightarrow Diff^r(F)$ so even the structure of a single fiber can give information about B. – Sina Sep 18 '13 at 14:36
note that for this to be valid existence of a foliation transverse to E is also enough so discrete structure group is not necessary. – Sina Sep 18 '13 at 14:39