Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

are there some general method in judging if a fible bundle is trivial?

At least,for vector bundles,there is a well-developed theory,that is Charicteristic Classes.the triviality of vector bundles is equivalent to the vanishment of its characteristic classes.

for principal bundles,the triviality is equivalent to the exsience of a cross section. (any good perspective on this assertion?besides,how to tell if there is a cross section)

for general fiber bundle (E,F,B,G),(here,E is total space,F the fiber,B the base space,G the structure group),we can construct its associate principal bundle (E',G,B,G),i.e.to replace the fiber F with the topological group G.there is a theorem that a fiber bundle is trivial iff its associate principal bundle is trivial.

hence the problem is reduced to find a cross section of principal bundle.

I want to know if there is some other methods that are more usable? Thank you!

share|improve this question
Are you sure that a vector bundle is trivial if all the characteristic classes vanish? What about the tangent bundle to $S^5$? The Chern, Pontrjagin, Stiefel-Whitney, and Euler class all vanish but it is not trivial. I'm not positive, but I THINK all other characteristic classes vanish. That is, if you assume the structure group is contained in $G\subseteq SO(5)$ and let $f$ be the classifying map, then I think $f^*:H^5(BG;R)\rightarrow H^5(S^5;R)$ is the 0 map for all rings $R$ and for all subgroups $G$ which the structure group reduces to. –  Jason DeVito Feb 14 '10 at 3:06
eg, $BSU(2)=HP^\infty$, so $\pi_5(BSU(2))=\pi_5S^4=Z/2$, and hence the principal bundle corresponding to $S^5\to BSU(2)$ is non-trivial, hence so is the corresponding $C^2$ vector bundle. But since $BSU(2)$ has no 5-cells $H^*(BSU(2))\to H^*(S^5)$ is trivial. Similarly you can associate a trivial vector bundle or a fiber bundle to a non-trivial principal G bundle, eg if the corresponding representation $G\to Homeo(F)$ is trivial (eg the adjoint representation for $G=S^1$.) The answer to the question is obstruction theory, I suppose. –  Paul Feb 14 '10 at 3:53
@Jason, your are right: Euler class of the tangent bundle to the odd-dimensional sphere vanishes because the sphere has zero Euler characteristic and the other characteristic classes are stable, hence they vanish as the tangent bundle to any sphere is stably trivial. On the other hand, the only parallelizable spheres are S^1, S^3, S^7. –  Igor Belegradek Feb 14 '10 at 4:32
There is an object called a connection on a fibre bundle that's sometimes handy in constructing trivializations, or at least trivializations over a ball. Depending on what situation you're in they can be useful. –  Ryan Budney Feb 14 '10 at 9:26
add comment

4 Answers

I will not really give an answer, but note that Turaev has a recent preprint on the archive (also this reply is too long for a comment!)

* Turaev, Vladimir

Abstract We study the existence problem and the enumeration problem for sections of Serre fibrations over compact orientable surfaces. When the fundamental group of the fiber is finite, a complete solution is given in terms of 2-dimensional cohomology classes associated with certain irreducible representations of this group. The proofs are based on Topological Quantum Field Theory.. Comment: 38 pages

Publication details Download http://arxiv.org/abs/0904.2692

This discusses the existence of sections for fibre bundles over surfaces. As Paul mentions above, the answer is 'obstruction theory' in general.

share|improve this answer
add comment

This also doesn't quite answer the question, but there are characteristic classes for various kinds of fiber bundles beyond simple vector bundles. The most well-developed are the so-called "Miller-Mumford-Morita" classes, which are characteristic classes of fiber bundles whose fibers are topological surfaces. For an inspired account of them, see Morita's book "Geometric of Characteristic Classes".

share|improve this answer
add comment

Here are some thoughts on the question of the posting:

  1. Suppose a principal $G$-bundle $E\to B$ for $G$ a topological group has a section $s:B\to E$. Then $(g,b)\mapsto g s(b)$ is a trivialization (recall the action of $G$ is global). The space of trivializations of a trivial bundle is not necessarily trivial, it is the gauge group $Map(B,G)$.

  2. Fiber bundles usually come equipped with a structure group $G$. (If none is specified, then it is usually assumed to be the group of all homeomorphisms or diffeomorphisms of the fiber.) Whether of not the bundle is trivial, depends on $G$. For example: the isomorphism classes of principal $G$-bundles on the sphere $S^n$ are one-to-one with $\pi_{n-1}(G)$ for connected $G$ (roughly because sphere is made of two disks, the bundle is trivial on each and we have to specify the transition maps on the intersection of the disks, which is the sphere of one dimension less). The identification is natural in $G$. For example, $\pi_1(U(n))=\mathbf{Z}$ and $\pi_1(SO(n))=\mathbf{Z}/2$ for $n\geq 3$ so there are complex rank 2 vector bundles on $S^2$ that are trivial as real vector bundles.

  3. For any topological group $G$ there is a classifying space $BG$ and the isomorphism classes of principal $G$-bundles on $X$ are one to one with the homotopy classes of maps $X\to BG$. If $G$ acts on $F$ by homeomorphisms, then the above set is one to one with the set of isomorphism classes of bundles with structure group $G$ and fiber $F$.

  4. Here is another description of the set of the isomorphism classes of principal bundles: this set is bijective with $H^1(X,\mathcal{F})$ where $\mathcal{F}$ is the ``gauge'' sheaf of groups on $X$ obtained from the presheaf $U\mapsto Map(U,G)$. This is particularly useful when $G$ is abelian, in which case we have not only $H^1$ but the higher groups as well; a particular case is the description of the isomorphism classes of line bundles on an algebraic variety $Y$ as elements of $H^1(Y,\mathcal{O}^{\*})$.

share|improve this answer
add comment

This may not be useful but you can take the vector bundle whose vector space at a point is the cohomology of the fibre at the point. Then you can ask if this is trivial.

As another possible approach; consider the principal bundle with group the group of diffeomorphisms of a fibre. Then cohomology classes for this group are characteristic classes.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.