Once again I answer my own question. There is a 3 dimensional oriented vector bundle over the 4 sphere that has no completely non-zero section.Is there an example of a real vector bundle over a compact smooth manifold with all zero characteristic classes (Euler class,Stiefel-Whitney classes and Pontryagin classes) that is non-trivial?
Right. Maybe this doesn't work. Do you know the answer? Apparently there are many 3 plane bundles over S^4. Do yhey all have non-zero Pontryagin class?such an example is ex. 23.16 p. 302 of bott-tu, but could you explain why p1 is zero, since it apparently lives in in H^4 of the 4 sphere, which is Z? (The other classes apparently live in cohomology groups which are zero on the 4 sphere.)
cool! Is it easy to see that 3 plane bundles over S4 are determined by their Pontryagin class?Generalizing that example, and using again the homotopy classification of real vector bundles on spheres, as in Bott - Tu, p. 301, since SO(3) ≈ RP^3 and ∏4(RP^3) = Z/2Z, one sees there is a non trivial 3 plane bundle on S^5, for which all characteristic classes seem to lie in cohomology groups which are zero. I.e. here p1 lies again in H^4 which is now zero on S^5.
what about SO(2) bundles? Are they completely determined bt the Euler class?Generalizing that example, and using again the homotopy classification of real vector bundles on spheres, as in Bott - Tu, p. 301, since SO(3) ≈ RP^3 and ∏4(RP^3) = Z/2Z, one sees there is a non trivial 3 plane bundle on S^5, for which all characteristic classes seem to lie in cohomology groups which are zero. I.e. here p1 lies again in H^4 which is now zero on S^5.
OK.well I am almost a complete novice, and am merely reading the statements in the standard books on the subject. But if this questions means what are the circle bundles on S^5, the they are apparently all trivial (because S^1 has no higher homotopy). pages 139-140 of Steenrod's Topology of fibre bundles, the standard work for many years on the topic, lists all sphere bundles on S^5 using the same homotopy classification as in Bott -Tu page 299 (not 301).
In particular the only non trivial ones have fibers S^2, S^3 or S^4. There is one non trivial S^2 bundle, and two non trivial S^3 bundles one of which however has the same homotopy groups as the trivial bundle, as is also true for the unique non trivial S^4 bundle. Oh and several of these have sections, hence vanishing euler class I suppose.
But since the only S^1 bundle is trivial then I suppose it is indeed determined by essentially any invariant at all, or none.
I haven't had a chance to check this out.OK here's a guess, but i hardly know the meaning of the symbols. If we have a smooth SO(2) bundle on a compact orientable connected surface, then since SO(2) = U(1), we can give that bundle the structure of a smooth complex line bundle with fiber the complex numbers. These are classified by the sheaf cohomology group H^1 with coefficients in the sheaf S* of never vanishing smooth functions. Using the exponential map we get a sheaf sequence 0-->Z-->S-->S*-->0, and a corresponding cohomology sequence containing
H^1(S)-->H^1(S*)-->H^2(Z)-->H^2(S). Then since S is a "fine" sheaf (admits smooth partitions of unity) the groups with S in them are zero. Hence the group classifying smooth isomorphism classes of smooth complex line bundkles is isomorphic to H^2(Z) which on a compact connected oriented surface is Z.
The map of course is the euler number, so only the bundle with euler number 0 is trivial.
does this make sense? consullt Gunning, Lectures on Riemann surfaces, Princeton mathematical notes, 1966.
is there an intuitive way to see this?H^2 corresponds bijectively with homotopy classes [X,BU(1)], with the isomorphism given by the pullback of the generator of H^2(BU(1)). This is the Euler class of the corresponding bundle. Thus the Euler class is trivial if and only if the classifying map is nullhomotopic, in which case the bundle is trivial.
we know that the euler class of a 2 plane bundle over a manifold is zero if and only if there is a closed one form on the unit circle bundle that integrates to 1 along each fiber circle. perhaps one can directly show that if such a form exists then the bundle is trivial.is there an intuitive way to see this?