46

This is more or less equivalent to Ryan's comment but with more details and a slightly different point of view.
Let $X$ be the total space of the tangent bundle, and put $Y=S^2\times\mathbb{R}^2$. If $X$ and $Y$ were homeomorphic, then their one-point compactifications would also be homeomorphic. We will show that this is impossible by considering their ...

answered Mar 27 '12 at 13:02

Neil Strickland

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41

These answers look at bit complicated so maybe there is something obviously wrong with the following argument:
Every embedded two-sphere $\Sigma \subset S^2 \times {\mathbb R}^2$ is displaceable: there is a one-parameter group (or family) of homeomorphisms $\varphi_t$ from $S^2 \times {\mathbb R}^2$ to itself such that $\varphi_T (\Sigma)$ is disjoint from $...

36

The difference is that, for a vector bundle, there is usually no natural Lie group action on the total space that acts transitively on the fibers. The fact that all of the fibers are, individually Lie groups, doesn't mean that there is a Lie group that acts on the whole space, restricting to each fiber to be a simply transitive action. The simplest example ...

26

As Igor Belegradek showed in the comments, one could find an example by finding a CW-complex $X$ and a map $X \to BO(n)$ which is not nullhomotopic, but where the restriction to every finite subcomplex is nullhomotopic. Such a map is called a phantom map. The question "is this map nullhomotopic?" has the same answer whether or not we are asking our maps to ...

26

I'll use the definition of stack as a (weak) functor from the category of schemes to that of groupoids (as opposed to the definition as a fibered category over the category of schemes).
The prestack associated to the action of $GL_n$ on $\mathbb A_n$ is, by definition, given by
$$
X \mapsto \left\{\begin{matrix}\text{Objects: maps $s:X \to \mathbb A^n$}\...

answered Oct 12 '14 at 22:20

André Henriques

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Let $U$ and $V$ be two copies of the real line and make a space $X$ by gluing them by the identity along the strictly positive half-line: $x\in U$ equals $x\in V$ for $x>0$. Now make a rank one vector bundle over this space by taking a trivial bundle over each of the lines: glue $U\times\mathbb R$ to $V\times \mathbb R$ by identifying $(x,y)\in U\times\...

answered Sep 7 '12 at 1:47

Tom Goodwillie

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As Piotr remarks, these kind of questions quickly lead to studying reflexive sheaves. I would add that one also better get acquainted with Serre's condition $S_2$. For more on this see
this and this MO answers.
Perhaps the first remark is that besides the singularities of the surface $X$ you also have to take into account the singularities of the sheaf you ...

20

Yes. Let $V$ be a real vector bundle whose base is a $d$-dimensional manifold or cell complex, and whose fibers are $r$-dimensional. Then (1) if $r>d$ then $V=W\oplus \epsilon$ where $\epsilon$ is a trivial rank one bundle, and (2) if $r>d+1$ then the rank $r-1$ bundle $W$ is determined up to isomorphism by $V$. In particular stably trivial bundles of ...

answered Mar 26 '13 at 22:00

Tom Goodwillie

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A flat vector bundle over a topological space is a bundle whose transition functions can be taken to be locally constant; equivalently, over a path-connected space, it's the same data as a principal $G$-bundle ($G = GL_n(\mathbb{R})$ or $GL_n(\mathbb{C})$ as appropriate) where $G$ is given the discrete topology. Over a reasonable space $X$ this is the same ...

answered Sep 28 '15 at 5:57

Qiaochu Yuan

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19

This may be overkill, but to elaborate on Ryan's answer in another way:
Without mentioning either boundaries or any other compactifications, we can define the intersection number of $x\in H_p$ and $y\in H_q$ for homology classes in an oriented $(p+q)$-manifold. First turn them into compactly supported cohomology classes by duality, then cup these to get ...

answered Mar 27 '12 at 13:49

Tom Goodwillie

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18

The answer is positive. Let $P$ be the principal $\mathrm{GL}_n$-bundle associated with $E$; then the space of flags is the quotient $P/B$, where $B$ is the Borel subgroup of $\mathrm{GL}_n$ consisting of upper triangular matrices. Set $Z = P/T$, where $T$ is the maximal torus consisting of diagonal matrices. A point of $Z$ is a point of $X$, plus $n$ ...

18

If you let $T=S^1 \times D^2$ be the solid torus and pick an embedding $i: T \to \mathrm{int}(T)$ which multiplies by 2 in $\pi_1$, the direct limit $X = \varinjlim(T \xrightarrow{i} T \xrightarrow{i} \dots)$ is a smooth 3-dimensional manifold (non-compact, but admitting a proper embedding into $\mathbb{R}^4$). Its homotopy type is $K(\mathbb{Z}[\frac12],1)$...

18

According to Kamber-Tondeur (1967), a principal $G$-bundle over a space $X$
is flat if it is induced from the universal covering bundle of $X$ by a homomorphism $\pi_1X\to G$. In the differentiable case this is equivalent to the existence of a connection with curvature zero [15, Lemma 1].
(...)
A vector bundle is called flat, if its associated ...

answered Sep 28 '15 at 2:44

Francois Ziegler

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17

Yes, it is true that (over an algebraically closed field) the only positive-dimensional smooth projective variety on which every algebraic vector bundle splits as a sum of line bundles is $\mathbb P^1$.
More generally Ballico has proved that if every algebraic vector bundle splits as a sum of line bundles on a reduced but maybe reducible positive-...

answered Mar 24 '13 at 19:42

Georges Elencwajg

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17

This is definitely false in the algebraic or holomorphic setting, even in dimension 1. There is a well-known example (see this post) of a rank 2 vector bundle $E$ on $\mathbb{C}\times \mathbb{P}^1$ such that $E_{|\{t\}\times \mathbb{P}^1 }=\mathcal{O}_{\mathbb{P}^1}^2$ for $t\neq 0$, but
$E_{|\{0\}\times \mathbb{P}^1}=\mathcal{O}_{\mathbb{P}^1}(-1)\oplus \...

17

Yes, there is such a twistor fibration over each $S^{2n}$, and the resulting manifold is a complex manifold endowed with a holomorphic $n$-plane field transverse to the fibers of the mapping. Namely, one writes $S^{2n} = \mathrm{SO}(2n{+}1)/\mathrm{SO}(2n)$ and then, using the inclusion $\mathrm{U}(n)\subset\mathrm{SO}(2n)$, one has the coset fibration
$$
...

17

Yes, they have stably trivial tangent bundles. A remark to this effect can be found on page 70 of
M. Kervaire "Smooth Homology Spheres and their Fundamental Groups"
but it is a little terse. It is essentially the argument as for homotopy spheres, due to Kervaire--Milnor, with a little elaboration. Let me try to explain it.
Let $M^n$ be a $\mathbb{Z}$-...

answered Jun 16 '16 at 20:31

Oscar Randal-Williams

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17

The geometric interpretation for $1$-cocyles.
Recall the following construction due to Bisson and Joyal.
Let $p:P\rightarrow B$ be a covering space over the connected manifold $B$. Suppose that the fibres of $p$ are finite. For every topological space $X$, the polynomial functor $p(X)=\{ (u,b),b\in B, u:p^{-1}(b)\rightarrow X\}$ $p(X)$ is a total space of ...

16

The answer is yes. The third homology group distinguishes these spaces, namely $H_3(C_2(E_k))=\mathbb{Z}_k$ (up to extension).
This shows that configuration spaces of homotopic open manifolds are easier to tell apart than configuration spaces of homotopic closed manifolds. The latter have the same
additive homology, and one needs finer invariants, like ...

16

The category of maps from a test object $T$ to a quotient stack $[X/G]$ has the following general form. Objects are pairs $(P, f)$, where $P$ is a $G$-torsor over $T$, and $f: P \to X$ is a $G$-equivariant map. Morphisms $(P,f) \to (P',f')$ are torsor isomorphisms $g: P \to P'$ satisfying $f = f' g$. Here, $X$ is the vector representation $\mathbf{O}^n$, ...

15

The answer (to both questions (a) and (b)) is YES (assuming $B$ is a smooth manifold). A proof can be found on Walschap's book "Metric Structures in Differential geometry", p. 77, Lemma 7.1.
For the OP's convenience, here's a sketch of the proof. Choose an open cover of $B$ such that your vector bundle is trivial over each element. From general results in ...

15

I find the following viewpoint helpful to translate between the different incarnation of a connection.
To every vector bundle $\pi: E \to M$ (in your case $E = TM$) we have an associated exact sequence of vector bundles (sometimes called the Atiyah sequence, at least in the principal bundle case):
$$ 0 \to V E \to TE \to \pi^* TM \to 0 $$
Here $VE$ denotes ...

14

You shouldn't expect vector bundles to form an abelian category because they are projective module objects over local ring objects in a category of sheaves over a space. In other words, there is really no more reason to expect this of vector bundles than you would of a category of finitely generated projective modules over a ring.
It might help to add that ...

answered Sep 13 '12 at 16:44

14

A spin structure on a real vector spaceÂ V equipped with a real quadratic formÂ Î¼
is an invertible bimodule (i.e., a Morita equivalence)
from Cl(V,Î¼) to Cl(Rdim(V),Î½).
Here Î½ is the direct sum of dim(V) copies of the canonical
quadratic form onÂ R.
A spinc structure on a complex vector spaceÂ V equipped with a complex quadratic formÂ Î¼
is an invertible bimodule
...

answered Feb 24 '13 at 13:59

Dmitri Pavlov

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14

No, because the Wu formulae express $\mathrm{Sq}^j(w_i)$ in terms of $w_k$'s, so if the $x_i$ you choose don't satisfy this formula, they cannot possibly arise as Stiefel--Whitney classes.

answered Apr 22 '14 at 8:38

Oscar Randal-Williams

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14

The proposition you refer to holds for any space homotopy equivalent to a finite dimensional CW complex. Here are the main points.
The property that any vector bundle $E$ over a space has a complementary bundle $E^\prime$ is preserved by homotopy equivalences.
Any finite dimensional CW complex is homotopy equivalent to a smooth manifold. (A classical result ...

answered Oct 29 '15 at 14:11

Igor Belegradek

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14

We have $H^*(BO(k); \mathbb{Z}_2) \cong \mathbb{Z}_2[w_1, \dots, w_k]$ where $\deg w_i = i$. In particular, $H^n(BO(k); \mathbb{Z}_2) \neq 0$ for every $n$ as $w_1^n$ is a non-zero element. Therefore $BO(k)$ cannot be homotopy equivalent to a finite-dimensional CW complex.
That is, the degrees of the usual choice of generators of the $\mathbb{Z}_2$ ...

answered Aug 7 '16 at 13:35

Michael Albanese

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14

For the relation $Sq(U) = \Phi(w)$, where $Sq$ is the total Steenrod squaring operation, $U$ is the Thom class, $\Phi$ is the Thom isomorphism and $w$ is the total Stiefel-Whitney class, I would cite Rene Thom's 1951 thesis, published as "Espaces fibres en spheres et carres de Steenrod" in Ann. Sci. Ecole Norm. Sup. (3) 69 (1952). In the introduction he ...

14

To expand on my comment above, suppose that $TS^{2k}\cong\xi\oplus\eta$ for some non-trivial vector bundles $\xi$ and $\eta$ over $S^{2k}$ of dimensions $m$ and $\ell$, respectively. Hence $0<m,\ell<2k$. Since $S^{2k}$ is simply-connected, both $\xi$ and $\eta$ are oriented, hence possess Euler classes
$$
e(\xi)\in H^m(S^{2k};\mathbb{Z}),\qquad e(\eta)\...

13

According to this preprint, over a connected proper algebraic variety $X$ there is a universal reductive group $G$ such that isomorphism classes of vector bundles of rank $n$ are in bijection with isomorphism classes of $n$-dimensional representations of $G$. Furthermore the component group is $\pi_1(X)$.
So the cancellation property you seek holds for any ...

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