36

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 $\partial E \cong S^2 \times \partial D^7$. These are classified by a map
$$f : S^2 \to BDiff_\partial(D^7)$$
to the classifying space of the group of diffeomorphisms of ...

answered Oct 21 '16 at 10:22

Oscar Randal-Williams

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16

If a section $\sigma : M \rightarrow \text{Diff}_{+}(M)$ to $\text{Diff}_{+}(M) \rightarrow M$ exists, then $M$ must be parallelizable (i.e. the tangent bundle of $M$ must be trivial). Indeed, if $\vec{e}_1,\ldots,\vec{e}_n$ is a basis for the tangent space of $M$ at $x_0$, then $\sigma(x)_{\ast}(e_1),\ldots,\sigma(x)_{\ast}(e_n)$ is a basis for the tangent ...

15

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 $e(\xi)\in H^2(M;\mathbb{Z}_{w_1(\xi)})$.
This is because the space $BO(2)$ is a generalized Eilenberg--Mac Lane space $L_{w_1}(\mathbb{Z},2)$ in the sense of
Samuel Gitler, Cohomology operations ...

15

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 are both CW complexes is necessarily a Hurewicz fibration. M. Steinberger and J. West. Covering homotopy properties of maps between CW complexes or ANRs. Proc....

15

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 un espace euclidien par des fibres compactes,
C. R. Acad. Sci. Paris 230 (1950), 2258–2260.

14

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 $S^d$ as $SO(d+1)/SO(d)$. Then all bundles in question are homogeneous vector bundles (and the isomorphisms used in the comment are compatible with the homogeneous ...

14

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 $x\in U$, and pick a local trivialization
$$
T:\pi^{-1}(U)\xrightarrow{\sim}U\times V_x.
$$
Let $f:M\to\mathbb{R}$ be a continuous function with support contained in $U$ ...

14

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 tools such as long exact sequences of homotopy groups and Serre spectral sequences of (co)homology groups become available.
It is easy to cook up examples of ...

13

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 principal G-bundles are always trivial. The only available examples of vector G-bundles are thus of the form M x G, where G is both a vector space and a Lie ...

13

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 $f : M \to S^1$ if and only if the kernel of the associated homomorphism $\pi_1(M) \to \mathbb{Z}$ is finitely generated and that homomorphism is nontrivial.

13

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 foliation of R3 and other punctured 3-manifolds by circles",
Publications Mathématiques de l'IHÉS, Tome 69 (1989), p. 215-232
http://www.numdam.org/item/...

13

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 of fields configurations, and from that function you can derive lots of quantities in the field theory.
It can happen, however, that in order to come up with, ...

12

Dylan, I am afraid that you are missing something. It is true that each element of the group gives a map of spectral sequences. But look at the fibers and my comment about isotropy groups above. If you construct the Serre spectral sequence in the usual way, there is a non-equivariantly inconsequential choice of base point in the base space. But that is no ...

12

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 transitively on spheres, so that for each such subgroup $\Gamma\subset \mathrm{Diff}(S^n)$, one can define a notion of $\Gamma$-bundles over manifolds, where a $\Gamma$-...

answered Feb 8 '15 at 12:40

Robert Bryant

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11

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:
$$M/S_n\rightarrow BS_n\stackrel{\rho_n}{\rightarrow} BO(n)$$
where $\rho_n$ is induced by the regular reprsentation $S_n\rightarrow O(n)$.
Thus what you want to compute is ...

11

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. locally path-connected spaces) which includes all cubes $[0,1]^n$ but probably not all spaces.
An important example with a simple construction is the generalized ...

11

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 theory of fibrations. Ann. Math. 78 (1963), 223-255.
The following formulations can be found in James "Topology of Stiefel manifolds" (with a couple of added ...

answered Sep 22 '17 at 19:13

Matthias Wendt

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10

Let $F$ be a general fiber of a fibration $X\to B$, where $B$ is a smooth variety. The normal bundle to $F$ in $X$ is trivial, so $K_F=K_X|_F$. Thus if $K_X$ is ample then $K_F$ is also ample and $F$ is of general type. The same argument shows that $F$ is of general type also when $K_X$ is just big.

10

As you remark, the quotient $SL_n(R)/SO_n(R)$ is contractible, and so any smooth bundle over it is smoothly trivial. So, in particular your bundle is diffeomorphic to a product.
The contractibility of $SL_n(R)/SO_n(R)$ stems from the Gram-Schmidt orthogonalization process; presumably you can use this to exhibit a specific trivialization of your bundle.

10

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 is smooth if and only if the map to the product is smooth (because it is smooth if and only if each component is smooth by the universal properties of the ...

10

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 $Diff(R^n)$-bundle has an $O(n)$-reduction.
Stewart, T. E. (1960). On groups of diffeomorphisms. Proceedings of the American Mathematical Society, 11(4), 559-563.

10

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".
In a Postikov tower for $X$, the map $p=p_n\colon P_n\to P_{n-1}$ is, by definition, a fibration with fiber equivalent to $K(A,n)$, where $A=\pi_n X$. The idea ...

9

Take the quotient $\mathbb S^2\times \mathbb S^1$ by the involution $\iota(x,y)=(-x,-y)$.

answered Aug 18 '12 at 11:01

Anton Petrunin

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9

Recall that linear, smooth, and topological $S^k$-bundles over a finite complex $X$ are classified by the sets of homotopy classes $[X, BO_{k+1}]$, $[X, B\mathrm{Diff}(S^k)]$ and $[X, B\mathrm{Homeo}(S^k)]$. There are obvious maps between the sets, and we are interested in their cokernels.
For example, if $X=S^1$, then the smooth nonlinear bundles are ...

answered Feb 19 '14 at 19:15

Igor Belegradek

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9

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 tangent bundle of $M$ has a trivial quotient of rank $1$, or equivalently a trivial subbundle of rank $1$. This is possible iff the Euler class $e(M)$ vanishes. ...

answered May 18 '15 at 5:15

Qiaochu Yuan

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9

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, two for each of the four choices of scalars, whether non orientable or orientable in each context.
Orientable : $SO(n)$, $SU(n)$, $Sp(n)$, and $G_2$ where $G_2$...

at.algebraic-topology dg.differential-geometry differential-topology characteristic-classes fibre-bundles

9

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 exponentiable. This means, in particular, that a map $B\to Homeo(F, F)$ is continuous if and only if the adjoint map $B\times F\to F$ is continuous. So under this ...

9

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 $\mathcal{S}_g(k)$ which classifies oriented surface bundles
$$\Sigma_g \to E \overset{\pi}\to B$$
equipped with a class $c \in H^2(E; \mathbb{Z})$ such that $\int_{\...

answered Jul 20 at 20:43

Oscar Randal-Williams

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8

Another way to describe Anton Petrunin's example is to start with the trivial $S^1$-bundle over $\mathbb RP^2$ and to take the fibrewise connect sum with the Hopf fibration $S^3 \to S^2$. By fibrewise connect sum I'm referring to taking the regular 2-manifold connect-sum on the base space, and then match that with fibrewise sum on the level of the bundle ...

8

Maybe the simplest example is the following. There are two fiber bundles with base and fiber both $S^2$. One is the product $S^2\times S^2$ and the other consists of two copies of the mapping cylinder of the Hopf map $S^3 \to S^2$ glued together via the identity map between their "source" ends $S^3$. The resulting space is the connected sum of two copies of $...

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