1
vote
0answers
24 views

Induced measurable subbundle

Let $G_k(\mathbb{R}^m)=\{ W: W$ is subspace of $\mathbb{R}^m, \dim W=k \}$ measurable and suppose that the application $$ \displaystyle{\begin{array}{rccl} h:&Z&\longrightarrow& ...
0
votes
0answers
26 views

express skew-commutative product of Whitney sum of vector bundles in tensor products

Let $\xi$ and $\eta$ be vector bundles over a paracompact space $B$ and $\xi\oplus\eta$ be their Whitney sum. Can we write $\Lambda(\xi\oplus\eta)\cong \Lambda(\xi)\otimes \Lambda(\eta)$ as (graded) ...
3
votes
0answers
91 views

measurable function induces a measurable bundle

Greetings I am preparing a work on bundles and I found this statement Let $V$ a topological vector space with $\dim V=n$ and $(E,\pi,M)$ a vector bundle continuous over $M$ (compact space). If ...
0
votes
1answer
19 views

generation of sub bundle

Let $M$ differentiable manifold with $\dim M=n$. If $(TM,\pi,M)$ be the fiber bundle tangent. Consider the family $E=\lbrace E_x\rbrace _{x\in M}$ such that $E_x \subset T_xM$ and $\dim E_x=k$ for ...
7
votes
3answers
133 views

let $\xi$ be an arbitrary vector bundle. Is $\xi\otimes\xi$ always orientable?

Let $\xi=(E,p,B)$ be a line bundle (not nec. orientable). Then the tensor product $\xi\otimes \xi$ is orientable. I obtain this by choosing $b\in U\cap V$, $U,V$ open in $B$ such that ...
3
votes
2answers
109 views

tensor product of two vector bundles

Let $\xi$ and $\eta$ be two vector bundles over the same base space $B$. Then $\xi\otimes \eta$ is orientable if and only if $\xi$ and $\eta$ are both orientable. How to prove this true or not true? ...
0
votes
0answers
39 views

finite-dimensional continuous vector bundle

Let $M$ a compact metric space and $\pi: F \rightarrow M$ a finite-dimensional continuos vector bundle over $M$, endowed with a continuous Riemannian metric. I was wondering if it will be true that: ...
-4
votes
1answer
291 views

Another vector bundles over of a Riemannian manifold

How is that any other vector bundle over a Riemannian manifold can be turn into a Riemannian manifold as well? I think that the question is of interest due to the presence in math.SE of several ...
0
votes
0answers
40 views

Seifert manifolds

Seifert fiber space is a PFB. The theorem states that every principal fiber bundle (PFB) admits a connection form, so how can we define the connection 1-form on it? Or how can I find a book or article ...
3
votes
0answers
47 views

how to obtain a generalized Morse function out of a fiber bundle?

Let $M\to E\to B$ be a smooth fiber bundle. In "Parametrized Morse Theory and Its Applications,(Proceedings of the ICM, 1990)", K. Igusa says that if dim $B$$<$dim $M$, then, there exists a smooth ...
0
votes
0answers
56 views

Sections on the Tautological Line bundle $E(\gamma_n)$..

I have a question about the tautological line bundle over $\mathbb R\mathbb P^n$. Recall, this bundle is that whose total space is $$E(\gamma_n):=\{([x], v)\in\mathbb R\mathbb P^n\times \mathbb ...
13
votes
1answer
274 views

Can we generalize the regular value theorem even beyond the Ehresmann's theorem?

The formulation is complicated, but the answer may be some clever usage of the partition of unity, because locally the answer is given by the regular value theorem and the whole problem is to glue it ...
7
votes
2answers
119 views

$\mathbb{S}^2$ as a fibre bundle

I know, by the Hopf fibration, that $\mathbb{S}^3$ is an $\mathbb{S}^1$-fibre bundle over $\mathbb{S}^2$. Can $\mathbb{S}^2$ be an $\mathbb{S}^1$-fibre bundle over some manifold $M$?
7
votes
0answers
304 views

When is a fibration a fiber bundle?

In this question I am using Wiki's definitions for fibration and fiber bundle. I want to be general in asking my question, but I am mostly interested in smooth compact manifolds and smooth fibrations ...
2
votes
1answer
51 views

Is a manifold $M$ which is a bundle over “perpendicular” submanifolds $A$ and $B$ always a Cartesian product?

Suppose $M$ is manifold with submanifolds $A$ and $B$ whose intersection is a single point $x$. Suppose further that $M$ can be written as a fiber bundle over $A$ with $B$ being the fiber at $x$, and ...
3
votes
1answer
180 views

Trivialisation of the normal bundle of $S^1$

I'd like to show that the normal bundle of $S^1$ is trivial. That is, I want to find a homeomorphism $\varphi : NS^1 \to S^1 \times \mathbb R$ such that $\varphi \mid_{N_s S^1}$ is linear for every $s ...
2
votes
2answers
69 views

Is it true that $R^2 = E(2)/U(1)$?

(Just so we're clear: that the Lie group of planar translations $R^2$ is isomorphic to a quotient of the 2D Euclidean Lie group $E(2)$ and the circle group $U(1)$.) I am trying to prove that $R^2 = ...
5
votes
0answers
120 views

Alternate pullback bundle construction

If $\pi : F \to N$ is a fiber bundle, and $\phi : M \to N$ (here, $M$ and $N$ are manifolds), then the standard way to define the pullback bundle of $F$ by $\phi$ is $$\phi^* F := \{(m,f) \in M ...