Timeline for Cohomology classes annihilated by pullbacks
Current License: CC BY-SA 2.5
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 9, 2010 at 12:54 | answer | added | Oscar Randal-Williams | timeline score: 3 | |
| Mar 9, 2010 at 3:46 | answer | added | Igor Belegradek | timeline score: 9 | |
| Mar 8, 2010 at 15:53 | comment | added | algori | Igor -- by the way, why don't you post this as an answer? The fact that for any fiber bundle with fiber a product of even-dimensional spheres and complex projective spaces the Leray spectral sequence collapses (your last reference) is quite impressive. | |
| Mar 8, 2010 at 15:48 | answer | added | Paul | timeline score: 1 | |
| Mar 8, 2010 at 13:25 | comment | added | Igor Belegradek | This is a correction of my earlier (deleted) comment. Playing with Serre spectral sequence (over Q) of the fiber bundle should give some insight. Suppose that B and M are connected and the bundle is orientable. Suppose that H^i(B)=0 for i<n and H^j(M)=0 for i<m where n+m=4. Then the kernel of the projection H^3(B)\to H^3(M) is the image of the transgression H^2(F)\to H^3(B), so if H^2(F)=0 the kernel is trivial. | |
| Mar 8, 2010 at 5:39 | comment | added | algori | Igor -- I agree, this argument indeed rules out homologically trivial fibrations with above-mentioned fibers. | |
| Mar 8, 2010 at 5:10 | comment | added | Igor Belegradek | algori, actually this argument might work for other fibers. Indeed, the rational type of the classifying space for fibrations with fiber F is known for certain F's, including complex projective spaces; see e.g. the googlable paper "Rational Type of Classifying Spaces for Fibrations" by Smith; discussion on bottom of page 2. | |
| Mar 8, 2010 at 4:57 | comment | added | Igor Belegradek | algori, the above argument works for any any fiber bundle whose fiber is a (rational) 2n-dimensional sphere. The key point is that BSG(2n+1) is rationally K(Q,4n) as proved in Appendix 1 in Milnor's paper "On characteristic classes of spherical fiber spaces". Thus [B, BSG(2n+1)]=H^{4n}(B, Q) whose elements vanish when restricted on any odd-dimensional cycle. So odd-dimensional characteristic classes of S^{2n}-fibrations are rationally trivial. | |
| Mar 8, 2010 at 4:56 | comment | added | algori | Igor -- yes, Smale's theorem (the smooth automorphism group of the 2-sphere can be contracted to O(3)) rules out the 2-sphere as a fiber. But still it would be interesting to see if there is a proof that the Euler class of an oriented sphere bundle vanishes, since there is a chance a similar argument would rule some other fibers such as $\mathbf{P}^2(\mathbf{C})$. | |
| Mar 8, 2010 at 4:16 | comment | added | Igor Belegradek | algori, here is an ad hoc argument for the case at hand, namely, I will show that the Euler class of any oriented S^2-bundle must be trivial over Q. If not, then the Euler class is nonzero on some 3-dimensional cycle. Restricting to the cycle gives an oriented S^2-fibration over a 3-dimensional complex X. The fibration is classified by a homotopy class in [X,BSG(3)], where SG(3) is the component of the identity of the self-maps of S^2. Now the inclusion SO(3) to SG(3) is a rational homotopy equivalence, so rationally BSG(3) is K(Q,4). Hence [X,BSG_3]=H^4(X, Q)=0 as X is 3-dimensional. | |
| Mar 8, 2010 at 3:24 | comment | added | Igor Belegradek | algori, I am pretty sure that the integral Euler class of (oriented) spherical fibrations with even dimensional fiber is torsion; whether it is in 2-torsion I do not know; I need to think a bit more as to how to prove this. | |
| Mar 8, 2010 at 3:01 | comment | added | algori | Igor -- I still don't see how you get the "minus identity" automorphism out of the mapping cylinder. | |
| Mar 8, 2010 at 2:30 | comment | added | algori | Igor -- obviously, when talking about the torsion I was referring to the integral case. Re Euler classes: if you define them via the Gysin sequence, you have to have an orientation-reversing automorphism over the identity of the base to prove that the rational Euler class of an odd-dimensional sphere bundle vanishes. This is immediate if we have a spherization of an oriented vector bundle, but how do you prove this in general? | |
| Mar 8, 2010 at 1:56 | comment | added | Igor Belegradek | @algori: you are wrong on both counts. First of all, Petya asked the question over the rationals so 2-torsion dies, second Gysin sequence over the rationals holds for any fiber bundle whose fibers are rational homology spheres. Thus the fiber cannot be rational homology sphere. I do have one clarification: for Gysin sequence to work the fiber bundle must be orientable; this holds if e.g if the base is simply-connected or if the structure group of the bundle is path-connected. | |
| Mar 8, 2010 at 1:21 | comment | added | algori | Igor -- first, this is valid only for spherizations of vector bundles, not for general bundles with fiber sphere; second, the Euler class of an odd dimensional bundle is 2-torsion but it needn't be trivial. | |
| Mar 8, 2010 at 1:05 | comment | added | Igor Belegradek | You probably know this but anyway such examples cannot be found if the fiber is a sphere (of any dimension). Indeed, the Gysin sequence implies that kernel of the map induced by the bundle projection on 3rd cohomology is a multiple of the Euler class. By dimension reasons the only nontrivial cases are those of S^2 and S^1 bundles. In the former case the Euler class is trivial. In the latter case any element in the kernel is the product of a 1-dimensional class and the 2-dimensional Euler class. | |
| Mar 7, 2010 at 23:56 | answer | added | Tim Perutz | timeline score: 1 | |
| Mar 7, 2010 at 22:22 | answer | added | Somnath Basu | timeline score: 2 | |
| Mar 7, 2010 at 21:37 | answer | added | Ben Webster♦ | timeline score: 0 | |
| Mar 7, 2010 at 20:39 | answer | added | algori | timeline score: 2 | |
| Mar 7, 2010 at 20:29 | history | asked | Petya | CC BY-SA 2.5 |