37 votes
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All fiber bundles over $S^2$ extendable to $\mathbb{C}P^\infty$?

No it isn't, but I had to dig quite deep to get a counterexample. Let us look at smooth $(D^7, \partial D^7)$-bundles over $S^2$, i.e $D^7 \to E \overset{\pi} \to S^2$ with an identification $\partial ...
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18 votes
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Serre fibration vs Hurewicz fibration

A short paper with references to several early counterexamples proves that (in the good category of compactly generated weak Hausdorff spaces) a Serre fibration in which the total space and base space ...
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  • 29.2k
17 votes
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Foliation of $\mathbb R^n$ by connected compact manifolds

There does not, even if you don’t require the fiber and base to be manifolds (or even connected, just that $F$ is not a single point). See Borel, Armand; Serre, Jean-Pierre, Impossibilité de fibrer ...
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  • 40.1k
16 votes
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Classification of $O(2)$-bundles in terms of characteristic classes

The $O(2)$ bundles $\xi$ over a manifold $M$ are classified by their first Stiefel-Whitney class $w_1(\xi)\in H^1(M;\mathbb{Z}/2)$ and their twisted Euler class $e(\xi)\in H^2(M;\mathbb{Z}_{w_1(\xi)})$...
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  • 33.4k
16 votes
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Is a topological fiber-bundle, whose total space admits a retraction onto a fiber, trivial?

The statement is not true. Let $\pi:V\to M$ be a vector bundle over a manifold which is non-trivial as a fiber bundle. Let $U$ be an open neighborhood of $M$ over which $V$ is trivial, fix $x\in U$, ...
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  • 8,811
16 votes

Anomaly in QFT physics v.s. determinant line bundle

The partition function should assign to each possible field configuration $\Phi$ (or field history) in your quantum field theory a number $Z(\Phi)$. That is, it should be a function on the collection ...
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15 votes

Vector bundles vs principal $G$-bundles

I believe you have answered this yourself. Principal bundles are trivial iff they admit a global smooth section. Vector bundles always admit a global smooth section (zero section). Therefore vector ...
15 votes

Is the concept of a "numerable" fiber bundle really useful or an empty generalization?

In algebraic topology, it is often more convenient to know that a map is a fibration (has the homotopy lifting property with respect to all spaces) than a fibre bundle, because then calculational ...
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  • 33.4k
14 votes
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A conjecture about parallelizable generalized spheres

It is true in general that $E_d$ is trivial. As remarked by Neil Strickland above, this boils down to showing that $TS^d\oplus\Lambda^2 TS^d$ is always a trivial bundle. To see this, represent $S^d$ ...
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  • 1,554
14 votes

Serre fibration vs Hurewicz fibration

These show up all the time in various generalized covering space theories, the reason being that you want homotopy lifting with respect to a certain class of spaces that you're interested in (e.g. ...
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14 votes

Foliation of $\mathbb R^n$ by connected compact manifolds

On the other hand, if you only mean "foliation" as in your title, and not "fibration", then there is Vogt's foliation of R^3 by circles! (But it is not C^1, only differentiable). Vogt, Elmar, "A ...
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13 votes

When can a class in $H^1(M;\mathbb{Z})$ be represented by a fiber bundle over $S^1$

If $M$ is a compact and irreducible 3-manifold, one answer is provided by a theorem of Stallings, in his 1962 paper "On fibering certain 3-manifolds": $\alpha$ is represented by a fibration $f : M \to ...
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  • 14.7k
13 votes
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Surface bundles associated to a short exact sequence of groups

(1) It is not true that these groups are precisely the fundamental groups of $S$-bundles. The correct statement is that these groups are precisely the fundamental groups of $S$-bundles over a base ...
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12 votes

Anything between vector bundles and sphere bundles?

This is more of a comment than an answer, but it's too long for a comment, so I'm putting it here. It sounds as though you are asking what sorts of groups of diffeomorphisms there are acting ...
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12 votes
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Does pullback in the category of smooth manifolds always exists?

If a pullback exists in the category of smooth manifold then, its underlying set of points has to be what you described simply by looking at morphism from the point. Moreover a map into the pullback ...
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  • 32.8k
11 votes
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vector bundles associated to a covering space

When you have an Euclidean vector space of type: $$\mathbb{R}^n\rightarrow M\times_{S_n}\mathbb{R}^n\rightarrow M/S_n$$ Then its classifying map $\phi:M/S_n\rightarrow BO(n)$ factors as: $$M/S_n\...
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  • 9,500
11 votes

Is a topological fiber-bundle, whose total space admits a retraction onto a fiber, trivial?

It was already pointed out that the statement is not true in the point-set sense. It is, however, true up to homotopy. This is a theorem of Dold and follows from his Partitions of unity in the ...
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11 votes

Cohomology ring of mapping torus

Not a full answer, but a very natural computational tool here (only for the additive structure) is the Leray spectral sequence $$E_2^{p,q} = H^p\big(S^1,R^qf_*\underline{\mathbf Z}\big) \Rightarrow H^{...
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10 votes
Accepted

A $\mathbb{R}^{n}$ -fiber bundle which do not admit a n-dimensional vector bundle structure

Such a fibre bundle does not exists if you suppose that it is endowed with a differentiable structure. Stewart has shown that the group of diffeomorphisms of $R^n$ retract to $O(n)$. So every $Diff(R^...
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10 votes

Classification of bundles, Postnikov towers, obstruction theory, local coefficients

I'll try to answer question 1. Unfortunately, I know of no especially convenient reference for the case of understanding general Postnikov towers; however, everything I say below is "well known". ...
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  • 25.8k
10 votes

Circle bundle with homotopically trivial fiber in the total space

Yes. By the homotopy long exact sequence $$\dots \to \pi_2(B) \to \pi_1(S^1) \to \pi_1(E) \to \dots$$ and the fact that the generator of $\pi_1(S^1)$ is homotopically trivial in $B$, we see that $\...
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  • 118k
10 votes

Cohomology ring of mapping torus

This is also an incomplete answer but from a different angle than the previous ones. Let $A=\begin{pmatrix} a&b\\c&d\end{pmatrix}$. EDIT: the answer below only deals with the orientable case $\...
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9 votes
Accepted

When can a class in $H^1(M;\mathbb{Z})$ be represented by a fiber bundle over $S^1$

Let me assume throughout this answer that $M$ is closed, oriented, and connected. Here are some necessary conditions. If you ask for a smooth fiber bundle, then a necessary condition is that the ...
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9 votes
Accepted

first Chern class of complex vector bundles and first Pontrjagin class of quaternionic vector bundles

The natural question also relates to understanding holonomy in Riemannian geometry using the idea of $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$, and Cayley numbers as scalars. There are 8 discussions, ...
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9 votes

Classification of $O(2)$-bundles in terms of characteristic classes

To complement Mark Grant's excellent answer, I'll say something more about the general case. This topic goes under the name of obstruction theory. The first observation is that a $G$-bundle on $X$ is ...
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  • 15.7k
9 votes
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Are topological fiber bundles on the same base with homeomorphic fibers isomorphic?

If $F$ is locally connected, locally compact and Hausdorff (or alternatively compact Hausdorff), then the inverse function from $Homeo(F, F)$ to itself is continuous. Moreover, such a space $F$ is ...
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9 votes
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Intersection form of surface bundle over surface

Yes, such a thing exists, but I don't know an explicit example. To see that it exists, it is clearest to me to consider the universal situation. For any $k \in \mathbb{Z}$ there is a space $\mathcal{S}...
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9 votes

Cohomology ring of mapping torus

I just want to point out that one can go quite far using algebraic topology (in this special case), although computing the cohomology ring completely may require some other ingredient. Lets call the ...
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  • 6,388
9 votes
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A question regarding isomorphism in cohomology for moduli space of stable bundles over a compact Riemann surface

Things are actually simpler. View $\Gamma _n=H^1(X,\mathbb{Z}/n)$ as the group of line bundles $L\in \operatorname{Pic}^{0}(X) $ with $L^{{\tiny \otimes }n}=\mathscr{O}_X$. The map $N_0(n,k) \times \...
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  • 34.7k
9 votes
Accepted

Does a compact Lie group action on a family of compact manifolds have diffeomorphic fixed point submanifolds?

I think the answer is yes. Since $G$ is compact, there is a $G$-invariant Riemannian metric on $M$ (by averaging any metric). The orthogonal distribution to the fiber for this metric is a $G$-...
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