31
votes
Accepted
If two smooth manifolds are homeomorphic, then their stable tangent bundles are vector bundle isomorphic
The result you are hoping for is in fact false.
In section 9 of Microbundles: Part I, Milnor constructs an open set $U \subset \mathbb{R}^m$. With its standard smooth structure, the (stable) tangent ...
28
votes
The quotient stack $[\mathbb{A}^n / \mathrm{GL}_n]$
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 ...
21
votes
Accepted
Is there a notion of "flat vector bundle over a topological space"?
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 $...
21
votes
Accepted
Are homology spheres stably parallelisable?
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 ...
19
votes
Generalising the Penrose Twistor Fibration
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, ...
19
votes
Is there a notion of "flat vector bundle over a topological space"?
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 ...
19
votes
Group cohomology and condensed matter
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 ...
18
votes
Accepted
Geometric interpretation of horizontal and vertical lift of vector field
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 ...
17
votes
Accepted
Splitting of tangent bundle
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,\...
16
votes
Accepted
The quotient stack $[\mathbb{A}^n / \mathrm{GL}_n]$
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 $G$-...
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 ...
Community wiki
15
votes
Accepted
Is every vector bundle over a noncompact finite-dimensional manifold a summand of a trivial bundle?
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 ...
15
votes
Accepted
Who discovered this definition of Stiefel-Whitney classes?
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 ...
14
votes
When can we cancel vector bundles from tensor products?
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 ...
14
votes
Principal bundles that can't be detected by spheres
The simplest example of a principal bundle that can't be detected by spheres:
The $S^1$-bundle over the 2-dimensional torus $T^2$ whose total space is the quotient $Heis_3(\mathbb R)/Heis_3(\mathbb Z)...
14
votes
Accepted
Infinite Grassmannian does not have the homotopy type of a finite-dimensional complex
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 $...
14
votes
Accepted
Fourth obstruction, Pontryagin and Euler class
Geometric generators for $\pi_3(SO(4))$ have been identified in §22 of Steenrod's "Topology of fibre bundles", using the identification of $S^3$ as unit quaternions. Conjugation of quaternions induces ...
14
votes
Analogy between Stiefel-Whitney and Chern classes
Here is one way I like to think of the analogy.
The maximal torus of diagonal matrices $T^{n} \subset U(n)$ gives a map $BT^n \to BU(n)$ which on integral cohomology gives an isomorphism from $H^...
13
votes
Accepted
Soft and hard part of geometry
In the context of geometry, a distinction between "soft" and "hard" was introduced by Gromov, as explained here and applied to Soft and Hard Symplectic Geometry.
In Gromov's words, 'hard' refers to ...
13
votes
Accepted
How can I endow a "locally product" CW structure on a vector bundle over a CW complex?
The authors of this book are attempting to use CW structures to justify certain cohomology isomorphisms, but this seems to be the wrong approach since some of their claims about CW structures are just ...
13
votes
Accepted
Why the Thom spectrum of $-\xi$ (or more generally of a virtual bundle) is defined as it is?
Just a quick answer to explain the original reason behind the definition and why our modern understanding of Thom spectra vindicates it.
Let $X$ be a space and $\xi$ a virtual vector bundle over $X$. ...
13
votes
Accepted
Conversion formula between "generalized" Stiefel-Whitney class of real vector bundles: O(n) and SO(n)
First I will write up what your question is asking in terms of Arun Debray's comment. I strongly suggest that when discussing questions like this, you use precise notation as in the following; I found ...
13
votes
Beauville-Laszlo for schemes
Completing the discussion under Will Sawin's answer. The question has been answered completely and affirmatively by Ben-Bassat and Temkin in their paper "Berkovich spaces and tubular descent" (Adv. ...
13
votes
Accepted
Possible mistake in Cohen notes "Immersions of manifolds and homotopy theory" (version 27 March 2022)
The statement is false in two ways. First, two immersions might not even be homotopic even though their normal bundles are both, say, trivial. Second, even if $M=\mathbb R^m$, regular homotopy classes ...
12
votes
Accepted
Is the unit tangent bundle of $S^{n}$ parallelizable?
W.Sutherland. A note on the parallelizability of sphere bundles over sphere. J. London Math. Soc. 39 (1964), 55--62.
The answer is yes.
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 ...
12
votes
Accepted
Principal bundles that can't be detected by spheres
If all maps are involved are pointed, your question is equivalent to the following: if $f : X \to BG$ is a map (where $X$ is connected) such that the induced map on $\pi_{\bullet}$ is zero, is $f$ ...
12
votes
Accepted
Vector bundles with exactly one nonzero SW-class
For $i = 2^k$ you can get an example with $X = \Bbb{RP}^{m}$ for any $m \geq 2^k$. If $L$ is the canonical line bundle on $\Bbb{RP}^{m}$ then let $E = \bigoplus_{i=1}^{2^k} L$ be the a sum of several ...
12
votes
Accepted
The relationship between flat vector bundle and flat projective bundle
Equivalently, you are asking whether every homomorphism $\pi _1(X)\rightarrow \mathrm{PGL}(n,\mathbb{C})$ lifts to a homomorphism $\pi _1(X)\rightarrow \mathrm{GL}(n,\mathbb{C})$. It is easy to find ...
12
votes
Accepted
Vector bundles over $RP^{\infty}$
You are essentially asking about the set $[B\mathbb{Z}/2,BSO(2n)]$. This bijects with the set of conjugacy classes of homomorphisms from $\mathbb{Z}/2$ to $SO(2n)$, or in other words real, oriented ...
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