# Vectyor bundles with all zero characteristic classes

Gold Member
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?

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Gold Member
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?
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.

mathwonk
Homework Helper
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.)

mathwonk
Homework Helper
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.

Gold Member
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.)
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?

Gold Member
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.
cool! Is it easy to see that 3 plane bundles over S4 are determined by their Pontryagin class?

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Gold Member
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?

mathwonk
Homework Helper
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.

Gold Member
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.
OK.

I know that there is a theorem that says that the Euler characteristic of a manifold is zero if and only if the tangent bundle has a non-zero section. For orientable surfaces these are SO(2) bundles and one section means two sections so the bundle is trivial.

But the only orientable surface with zero Euler characteristic is the torus.

But a surface may have some other SO(2) bundle that has zero Euler class but maybe no section. I have a feeling though that this doesn't happen. Will do some reading today.

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mathwonk
Homework Helper
you raise an interesting point about "classification" of vector bundles. I.e. although the isomorphism classes of vector bundles of rank 2 over a surface S is bijectively equivalent with the set of homotopy classes of maps from S to the grassmannian of 2 planes in high diml euclidean space, so what? How does one calculate that homotopy set?

well let me make a guess. assume an orientable 2 plane bundle on a smooth orientable surface S can be given the structure of a holomorphic line bundle on a riemann surface structure for S. Then one knows I believe that these bundles are classified by their 1st chern clas, i.,e. euler number of zeroes of a section, up to smooth equivalence, hence none exist that are non trivial and yet have sections.

why do i say this? I think it follows from sheaf cohomology and the vanishing of cohomology of "fine" sheaves... mumble mumble.

mathwonk
Homework Helper
mumble mumble, 1: reduce structure group from SO(2) to SU(1) giving a complex structure of a complex line bundle, then 2: show all such are smoothly classified by euler number, then 3: find a complex holomorphic bundle in each equivalence class,.... see if you can make some sense of this.... or are we already done after step 2?

mathwonk
Homework Helper
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.

mathwonk
Homework Helper
a more succinct version of Gunning's book, revised in 2008, is free at http://www.math.princeton.edu/~gunning/book.html [Broken]

the relevant statement is equation 1.14 page 10 there.

the discussion in the earlier notes is much more detailed culminating on pages 99-100.

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Gold Member
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.
I haven't had a chance to check this out.

I tried to directly calculate the transition functions from the assumption that the Euler class is a coboundary using the formulas in Bott and Tu. For a manifold with only two coordinate charts, i.e. a sphere one can show by simple calculation that the transition function must be constant and from that triviality is immediate. For three charts and more I have no idea.

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.

About the SO(3) bundle over S^4, you can show that isomorphism classes of oriented 3-plane bundles over S^4 are isomorphic to Z_2. I believe (but am not completely sure) that this corresponds to the parity of the second chern class of the associated SU(2) bundle.

Gold Member
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.
is there an intuitive way to see this?

mathwonk
Homework Helper
the brown representability theorem implies that cohomology corresponds to homotopy classes of maps to a space with certain homotopy groups. so this could follow from knowing the homotopy groups of BU(1). I.e. that it looks enough like an Eilenberg Maclane "K pi" space.

Indeed the following notes give exactly this example and explain why BU(1) is a K(Z,2) space, i.e. has homotopy precisely in dimension 2, and nowhere else, hence qualifies as a classifying space for the cohomology functor H^2.

http://www.math.sunysb.edu/~mbw/notes/orals/Classifying%20Spaces%20and%20Representability%20Theorems.pdf [Broken]

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