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homotopic maps induce isomorphic pullback bundles, and so isomorphism classes of vector bundles over X correspond to homotopy classes of maps from the grassmannian to X. But I think, in the case of X = S^1 (the 1-sphere), the isomorphism classes of 1-bundles correspond to (the generators of) π___{1}(S^1), since there is the trivial bundle, the Moebius bundle, and that's it. So my question is: am I right with this, and if yes: what needs to happen that π_n(X) = {homotopy classes of maps:grassmannian -> X}?

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up vote 8 down vote accepted

The map goes the other way: vector bundles over X correspond to homotopy classes of maps from X into a grassmannian.

Let BO(n) be the grassmannian of n-plane bundles in R-infinity. Then, if you want to know about n-dimensional real vector bundles over Sk you are led to study the homotopy classes of maps from Sk to BO(n), or in other words πk BO(n).

In particular, the fact that there are exactly two 1-bundles over S1 comes from a calculation π1 BO(1)=Z/2.

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Thank you (both) very much, now it makes sense! – knot Oct 26 '09 at 20:55

You have it backward--vector bundles on X are the same as homotopy classes of maps from X to an infinite-dimensional Grassmannian, with the correspondence being given by pulling back the tautological bundle on the Grassmannian. This equivalence holds for any paracompact X. What this means is that vector bundles are not related to homotopy groups but to cohomology--a cohomology class on a Grassmannian pulls back to one on X, given any vector bundle on X. Thus every vector bundle has certain cohomology classes on the base naturally associated to it; these are called characteristic classes and have all sorts of applications. More generally, if you look at all vector bundles on a space as a monoid under direct sum, and formally adjoin additive inverses, you get the 0th group of a generalized cohomology theory called K-theory on X. (Stable) characteristic classes can then be interpreted as natural transformations from K-theory to ordinary cohomology.

In the special case of spheres, you do get a relation to homotopy groups, in that vector bundles on S^n are homotopy classes of maps from S^n to a Grassmannian BO(k), i.e. \pi_n(BO(k)). Using the fact that O(k) is the loopspace of BO(k), this is also the same as \pi_{n-1}(O(k)). You can also see this from the fact that a vector bundle must be trivial on each hemisphere, so it is determined by the transition function between the two trivializations on the equator, which is a map from S^{n-1} to O(k). The fact that line bundles on S^1 are Z/2 is thus not about generators of \pi_1(S^1) but the fact that \pi_0(O(1))=Z/2.

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